082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

ENGINEERING
MATHEMATICS
A Foundation for Electronic, Electrical,
Communications and Systems Engineers
FIFTH EDITION
Anthony Croft • Robert Davison
Martin Hargreaves • James Flint

EngineeringMathematics

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FifthEdition
EngineeringMathematics
AFoundationforElectronic,Electrical,
CommunicationsandSystemsEngineers
AnthonyCroft
LoughboroughUniversity
RobertDavison
MartinHargreaves
CharteredPhysicist
JamesFlint
LoughboroughUniversity
Harlow, England • London • New York • Boston • San Francisco • Toronto • Sydney
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Third edition publishedunder the PrenticeHall imprint 2001 (print)
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ISBN: 978-1-292-14665-2 (print)
978-1-292-14667-6 (PDF)
978-1-292-14666-9 (ePub)
BritishLibraryCataloguing-in-PublicationData
A catalogue recordfor the print edition is available from the British Library
LibraryofCongressCataloging-in-PublicationData
Names: Croft, Tony,1957– author.
Title: Engineering mathematics:afoundation for electronic,electrical,
communicationsand systemsengineers/AnthonyCroft, Loughborough
University,Robert Davison,De Montfort University,Martin Hargreaves,
De Montfort University,James Flint,LoughboroughUniversity.
Description:Fifth edition.| Harlow, England ; NewYork : Pearson,2017. ‖
Revised edition of: Engineeringmathematics : a foundation for electronic,
electrical,communications,and systemsengineers/AnthonyCroft, Robert
Davison,Martin Hargreaves.3rd editon.2001.| Includesindex.
Identifiers:LCCN2017011081| ISBN 9781292146652 (Print) | ISBN 9781292146676
(PDF) | ISBN 9781292146669 (ePub)
Subjects: LCSH: Engineeringmathematics.| Electrical
engineering–Mathematics.| Electronics–Mathematics.
Classification:LCC TA330 .C76 2017 | DDC 510–dc23
LC record available at https://lccn.loc.gov/2017011081
A catalog record for the print edition is available from the Library of Congress
10987654321
21 20 19 18 17
Print edition typesetin 10/12 Times Roman by iEnerziger Aptara ® , Ltd.
Printedin Slovakia by Neografia
NOTE THATANYPAGECROSS REFERENCES REFER TO THE PRINTEDITION

ToKate,TomandHarvey -- A.C.
ToKathy-- R.D.
Tomyfatherandmother --M.H.
ToSuzanne,AlexandraandDominic --J.F.

Contents
Preface
Acknowledgements
xvii
xix
Chapter1 Reviewofalgebraictechniques 1
1.1 Introduction 1
1.2 Laws of indices 2
1.3 Number bases 11
1.4 Polynomial equations 20
1.5 Algebraicfractions 26
1.6 Solution of inequalities 33
1.7 Partial fractions 39
1.8 Summation notation 46
Review exercises 1 50
Chapter2 Engineeringfunctions 54
2.1 Introduction 54
2.2 Numbers and intervals 55
2.3 Basic concepts of functions 56
2.4 Review of some common engineering functions and techniques 70
Review exercises 2 113
Chapter3 Thetrigonometricfunctions 115
3.1 Introduction 115
3.2 Degreesand radians 116
3.3 The trigonometric ratios 116
3.4 The sine, cosine and tangent functions 120
3.5 The sincxfunction 123
3.6 Trigonometric identities 125
3.7 Modellingwavesusing sint and cost 131
3.8 Trigonometric equations 144
Review exercises 3 150

viii Contents
Chapter4 Coordinatesystems 154
4.1 Introduction 154
4.2 Cartesiancoordinate system – two dimensions 154
4.3 Cartesiancoordinate system – three dimensions 157
4.4 Polar coordinates 159
4.5 Some simple polar curves 163
4.6 Cylindricalpolar coordinates 166
4.7 Spherical polar coordinates 170
Review exercises 4 173
Chapter5 Discretemathematics 175
5.1 Introduction 175
5.2 Set theory 175
5.3 Logic 183
5.4 Boolean algebra 185
Review exercises 5 197
Chapter6 Sequencesandseries 200
6.1 Introduction 200
6.2 Sequences 201
6.3 Series 209
6.4 The binomial theorem 214
6.5 Power series 218
6.6 Sequences arising from the iterativesolution
of non-linear equations 219
Review exercises 6 222
Chapter7 Vectors 224
7.1 Introduction 224
7.2 Vectors and scalars: basic concepts 224
7.3 Cartesiancomponents 232
7.4 Scalar fields and vector fields 240
7.5 The scalar product 241
7.6 The vector product 246
7.7 Vectors ofndimensions 253
Review exercises 7 255
Chapter8 Matrixalgebra 257
8.1 Introduction 257
8.2 Basic definitions 258

Contents ix
8.3 Addition,subtraction and multiplication 259
8.4 Using matrices in the translationand rotation of vectors 267
8.5 Some special matrices 271
8.6 The inverse of a 2 × 2 matrix 274
8.7 Determinants 278
8.8 The inverse of a 3 × 3 matrix 281
8.9 Applicationto the solution of simultaneous equations 283
8.10 Gaussian elimination 286
8.11 Eigenvaluesand eigenvectors 294
8.12 Analysis of electricalnetworks 307
8.13 Iterative techniquesfor the solution of simultaneous equations 312
8.14 Computer solutions of matrix problems 319
Review exercises 8 321
Chapter9 Complexnumbers 324
9.1 Introduction 324
9.2 Complex numbers 325
9.3 Operations with complex numbers 328
9.4 Graphical representationof complex numbers 332
9.5 Polar form of a complex number 333
9.6 Vectors and complex numbers 336
9.7 The exponential form of a complex number 337
9.8 Phasors 340
9.9 De Moivre’s theorem 344
9.10 Loci and regions of the complex plane 351
Review exercises 9 354
Chapter10 Differentiation 356
10.1 Introduction 356
10.2 Graphical approach to differentiation 357
10.3 Limits and continuity 358
10.4 Rate of change at a specific point 362
10.5 Rate of change at a general point 364
10.6 Existence of derivatives 370
10.7 Common derivatives 372
10.8 Differentiationas a linear operator 375
Review exercises 10 385
Chapter11 Techniquesofdifferentiation 386
11.1 Introduction 386

x Contents
11.2 Rules of differentiation 386
11.3 Parametric, implicit and logarithmicdifferentiation 393
11.4 Higher derivatives 400
Review exercises 11 404
Chapter12 Applicationsofdifferentiation 406
12.1 Introduction 406
12.2 Maximum points and minimum points 406
12.3 Points of inflexion 415
12.4 The Newton–Raphson method for solving equations 418
12.5 Differentiationof vectors 423
Review exercises 12 427
Chapter13 Integration 428
13.1 Introduction 428
13.2 Elementary integration 429
13.3 Definite and indefinite integrals 442
Review exercises 13 453
Chapter14 Techniquesofintegration 457
14.1 Introduction 457
14.2 Integration by parts 457
14.3 Integration by substitution 463
14.4 Integration using partialfractions 466
Review exercises 14 468
Chapter15 Applicationsofintegration 471
15.1 Introduction 471
15.2 Average value of a function 471
15.3 Root mean square value of a function 475
Review exercises 15 479
Chapter16 Furthertopicsinintegration 480
16.1 Introduction 480
16.2 Orthogonal functions 480
16.3 Improper integrals 483
16.4 Integral propertiesof the delta function 489
16.5 Integration of piecewise continuous functions 491
16.6 Integration of vectors 493
Review exercises 16 494

Contents xi
Chapter17 Numericalintegration 496
17.1 Introduction 496
17.2 Trapezium rule 496
17.3 Simpson’s rule 500
Review exercises 17 505
Chapter18 Taylorpolynomials,TaylorseriesandMaclaurinseries 507
18.1 Introduction 507
18.2 Linearizationusing first-order Taylor polynomials 508
18.3 Second-order Taylor polynomials 513
18.4 Taylor polynomials of the nth order 517
18.5 Taylor’s formula and the remainder term 521
18.6 Taylor and Maclaurinseries 524
Review exercises 18 532
Chapter19 OrdinarydifferentialequationsI 534
19.1 Introduction 534
19.2 Basic definitions 535
19.3 First-order equations: simple equationsand separation
of variables 540
19.4 First-order linear equations:use of an integratingfactor 547
19.5 Second-order linear equations 558
19.6 Constant coefficient equations 560
19.7 Series solution of differentialequations 584
19.8 Bessel’s equationand Bessel functions 587
Review exercises 19 601
Chapter20 OrdinarydifferentialequationsII 603
20.1 Introduction 603
20.2 Analogue simulation 603
20.3 Higher order equations 606
20.4 State-space models 609
20.5 Numerical methods 615
20.6 Euler’s method 616
20.7 Improved Euler method 620
20.8 Runge–Kuttamethod of order 4 623
Review exercises 20 626
Chapter21 TheLaplacetransform 627
21.1 Introduction 627
21.2 Definitionof the Laplace transform 628

xii Contents
21.3 Laplace transforms of some common functions 629
21.4 Properties of the Laplace transform 631
21.5 Laplace transform of derivativesand integrals 635
21.6 Inverse Laplacetransforms 638
21.7 Using partialfractions to find the inverse Laplacetransform 641
21.8 Finding the inverse Laplace transform using complex numbers 643
21.9 The convolution theorem 647
21.10Solving linear constant coefficient differential
equations using the Laplace transform 649
21.11Transfer functions 659
21.12Poles, zeros and the s plane 668
21.13Laplace transforms of some special functions 675
Review exercises 21 678
Chapter22 Differenceequationsandtheztransform 681
22.1 Introduction 681
22.2 Basic definitions 682
22.3 Rewriting difference equations 686
22.4 Block diagramrepresentationof difference equations 688
22.5 Design of a discrete-time controller 693
22.6 Numerical solution of difference equations 695
22.7 Definitionof the z transform 698
22.8 Sampling a continuous signal 702
22.9 The relationshipbetweenthe z transform and the
Laplace transform 704
22.10Properties of the z transform 709
22.11Inversion of z transforms 715
22.12The z transform and difference equations 718
Review exercises 22 720
Chapter23 Fourierseries 722
23.1 Introduction 722
23.2 Periodic waveforms 723
23.3 Odd and even functions 726
23.4 Orthogonality relationsand other useful identities 732
23.5 Fourier series 733
23.6 Half-range series 745
23.7 Parseval’s theorem 748
23.8 Complex notation 749
23.9 Frequency response of a linear system 751
Review exercises 23 755

Contents xiii
Chapter24 TheFouriertransform 757
24.1 Introduction 757
24.2 The Fourier transform – definitions 758
24.3 Some propertiesof the Fourier transform 761
24.4 Spectra 766
24.5 The t−ω dualityprinciple 768
24.6 Fourier transforms of some special functions 770
24.7 The relationship betweenthe Fourier transform
and the Laplace transform 772
24.8 Convolution and correlation 774
24.9 The discrete Fourier transform 783
24.10Derivationof the d.f.t. 787
24.11Using the d.f.t.to estimate a Fourier transform 790
24.12Matrixrepresentationof the d.f.t. 792
24.13Some propertiesof the d.f.t. 793
24.14The discrete cosine transform 795
24.15Discrete convolution and correlation 801
Review exercises 24 821
Chapter25 Functionsofseveralvariables 823
25.1 Introduction 823
25.2 Functions of more than one variable 823
25.3 Partial derivatives 825
25.4 Higher order derivatives 829
25.5 Partial differentialequations 832
25.6 Taylor polynomials and Taylor series in two variables 835
25.7 Maximum and minimum points of a function of two variables 841
Review exercises 25 846
Chapter26 Vectorcalculus 849
26.1 Introduction 849
26.2 Partial differentiationof vectors 849
26.3 The gradient of a scalar field 851
26.4 The divergence of a vector field 856
26.5 The curl of a vector field 859
26.6 Combining the operators grad, div and curl 861
26.7 Vector calculus and electromagnetism 864
Review exercises 26 865

xiv Contents
Chapter27 Lineintegralsandmultipleintegrals 867
27.1 Introduction 867
27.2 Line integrals 867
27.3 Evaluation of line integralsin two dimensions 871
27.4 Evaluation of line integralsin three dimensions 873
27.5 Conservative fields and potentialfunctions 875
27.6 Double and triple integrals 880
27.7 Some simple volume and surface integrals 889
27.8 The divergence theorem and Stokes’ theorem 895
27.9 Maxwell’s equationsin integral form 899
Review exercises 27 901
Chapter28 Probability 903
28.1 Introduction 903
28.2 Introducing probability 904
28.3 Mutuallyexclusive events: the additionlaw of probability 909
28.4 Complementary events 913
28.5 Concepts from communication theory 915
28.6 Conditional probability:the multiplicationlaw 919
28.7 Independent events 925
Review exercises 28 930
Chapter29 Statisticsandprobabilitydistributions 933
29.1 Introduction 933
29.2 Random variables 934
29.3 Probability distributions–discrete variable 935
29.4 Probability density functions – continuous variable 936
29.5 Mean value 938
29.6 Standard deviation 941
29.7 Expected value of a random variable 943
29.8 Standard deviationof a random variable 946
29.9 Permutations and combinations 948
29.10The binomial distribution 953
29.11The Poisson distribution 957
29.12The uniform distribution 961
29.13The exponentialdistribution 962
29.14The normal distribution 963
29.15Reliabilityengineering 970
Review exercises 29 977

Contents xv
AppendixI Representing a continuous function and a sequence
as a sum of weighted impulses 979
AppendixII The Greek alphabet 981
AppendixIII SI units and prefixes 982
( ) n
AppendixIV The binomial expansion of
982
n−N
n
Index 983

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Preface
Audience
This book has been written to serve the mathematical needs of students engaged in a
firstcourseinengineeringatdegreelevel.Itisprimarilyaimedatstudentsofelectronic,
electrical,communicationsandsystemsengineering.Systemsengineeringtypicallyencompasses
areas such as manufacturing, control and production engineering. The textbook
will also be useful for engineers who wish to engage in self-study and continuing
education.
Motivation
Engineers are called upon to analyse a variety of engineering systems, which can be
anything from a few electronic components connected together through to a complete
factory.Theanalysisofthesesystemsbenefitsfromtheintelligentapplicationofmathematics.Indeed,
many cannot beanalysed withouttheuseofmathematics.Mathematics
is the language of engineering. It is essential to understand how mathematics works in
order to master the complex relationships present in modern engineering systems and
products.
Aims
Therearetwomainaimsofthebook.Firstly,wewishtoprovideanaccessible,readable
introductiontoengineeringmathematicsatdegreelevel.Thesecondaimistoencourage
the integration of engineering and mathematics.
Content
The first three chapters include a review of some important functions and techniques
thatthereadermayhavemetinpreviouscourses.Thismaterialensuresthatthebookis
self-contained and provides a convenient reference.
Traditional topics in algebra, trigonometry and calculus have been covered. Also included
are chapters on set theory, sequences and series, Boolean algebra, logic, difference
equations and theztransform. The importance of signal processing techniques is
reflectedbyathoroughtreatmentofintegraltransformmethods.ThustheLaplace,zand
Fourier transformshave been given extensive coverage.
Inthelightoffeedbackfromreaders,newtopicsandnewexampleshavebeenadded
in the fifth edition. Recognizing that motivation comes from seeing the applicability
of mathematics we have focused mainly on the enhancement of the range of applied
examples. These include topics on the discrete cosine transform, image processing, applicationsinmusictechnology,communicationsengineeringandfrequencymodulation.

xviii Preface
Style
The style of the book is to develop and illustrate mathematical concepts through examples.
We have tried throughout to adopt an informal approach and to describe mathematical
processes using everyday language. Mathematical ideas are often developed
by examples rather than by using abstract proof, which has been kept to a minimum.
This reflects the authors’ experience that engineering students learn better from practical
examples, rather than from formal abstract development. We have included many
engineering examples and have tried to make them as free-standing as possible to keep
thenecessaryengineeringprerequisitestoaminimum.Theengineeringexamples,which
havebeencarefullyselectedtoberelevant,informativeandmodern,rangefromshortillustrativeexamplesthroughtocompletesectionswhichcanberegardedascasestudies.
A further benefit is the development of the link between mathematics and the physical
world. An appreciation of this link is essential if engineers are to take fulladvantage of
engineeringmathematics.Theengineeringexamplesmakethebookmorecolourfuland,
moreimportantly,theyhelpdeveloptheabilitytoseeanengineeringproblemandtranslateitintoamathematicalformsothatasolutioncanbeobtained.Thisisoneofthemost
difficult skills that an engineer needs to acquire. The ability to manipulate mathematical
equations is by itself insufficient. It is sometimes necessary to derive the equations
corresponding to an engineering problem. Interpretation of mathematical solutions in
terms of the physical variables is also essential. Engineers cannot afford to get lost in
mathematical symbolism.
Format
Importantresultsarehighlightedforeasyreference.Exercisesandsolutionsareprovided
at the end of most sections; it is essential to attempt these as the only way to develop
competence and understanding is through practice. A further set of review exercises is
providedattheendofeachchapter.Inadditionsomesectionsincludeexercisesthatare
intended to be carried out on a computer using a technical computing language such as
MATLAB ® ,GNUOctave,MathematicaorPython ® .TheMATLAB ® commandsyntax
is supported in several software packages as well as MATLAB ® itself and will be used
throughout the book.
Supplements
A comprehensive Solutions Manual is obtainable free of charge to lecturers using this
textbook. Itisalsoavailable fordownload via the web atwww.pearsoned.co.uk/croft.
Finallywehopeyouwillcometoshareourenthusiasmforengineeringmathematics
and enjoy the book.
AnthonyCroft
RobertDavison
MartinHargreaves
JamesFlint
March2017

Acknowledgements
Wearegratefultothefollowingforpermissiontoreproducecopyrightmaterial:
Tables
Table29.7fromBiometrikaTablesforStatisticians,Vol.1,NewYork:Holt,Rinehart&
Winston(Hays,W.L. and Winkler,R.L. 1970) Table 1,© Cambridge UniversityPress.
Text
General Displayed Text on page xviii from https://www.mathworks.com/products/
matlab.html,MATLAB ® isaregisteredtrademarkofTheMathWorks,Inc.;GeneralDisplayed
Text xviii from Mathematica, https://www.wolfram.com/mathematica/, © Wolfram;
General Displayed Text xviii from https://www.python.org/, Python ® and the
Python logos are trademarks or registered trademarks of the Python Software Foundation,
used by Pearson Education Ltd with permission from the Foundation; General
Displayed Text on page 291 from http://www.blu-raydisc.com/en/, Blu-ray Disc is
a trademark owned by Blu-ray Disc Association (BDA); General Displayed Text on
page 291 from http://wimaxforum.org/home, WiMAX ® is a registered trademarks of
the WiMAX Forum. This work is produced by Pearson Education and is not endorsed
by any trademark owner referenced inthispublication.

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1 Reviewofalgebraic
techniques
Contents 1.1 Introduction 1
1.2 Lawsofindices 2
1.3 Numberbases 11
1.4 Polynomialequations 20
1.5 Algebraicfractions 26
1.6 Solutionofinequalities 33
1.7 Partialfractions 39
1.8 Summationnotation 46
Reviewexercises1 50
1.1 INTRODUCTION
This chapter introduces some algebraic techniques which are commonly used in engineering
mathematics. For some readers this may be revision. Section 1.2 examines the
laws of indices. These laws are used throughout engineering mathematics. Section 1.3
looks at number bases. Section 1.4 looks at methods of solving polynomial equations.
Section 1.5 examines algebraic fractions, while Section 1.6 examines the solution of
inequalities. Section 1.7 looks at partial fractions. The chapter closes with a study of
summation notation.
Computers are used extensively in all engineering disciplines to perform calculations.
Some of the examples provided in this book make use of the technical computing
language MATLAB ® , which is commonly used in both an academic and industrial
setting.
Because MATLAB ® and many other similar languages are designed to compute not
just with single numbers but with entire sequences of numbers at the same time, data
is entered in the form of arrays. These are multi-dimensional objects. Two particular
typesofarrayarevectorsandmatriceswhicharestudiedindetailinChapters7and8.

2 Chapter 1 Review of algebraic techniques
Apart from being able to perform basic mathematical operations with vectors and
matrices, MATLAB ® has, in addition, a vast range of built-in computational functions
whicharestraightforwardtousebutneverthelessareverypowerful.Manyofthesehighlevelfunctionsareaccessiblebypassingdatatothemintheformofvectorsandmatrices.
A small number of these special functions are used and explained in this text. However,
to get the most out of a technical computing language it is necessary to develop
a good understanding of what the software can do and to make regular reference to the
manual.
1.2 LAWSOFINDICES
Considertheproduct6 ×6×6×6×6.Thismaybewrittenmorecompactlyas6 5 .We
call5theindexorpower. Thebaseis6.Similarly,y ×y×y×ymaybewrittenasy 4 .
Here the base isyand the index is4.
Example1.1 Write the following using index notation:
(a) (−2)(−2)(−2) (b) 4.4.4.5.5 (c)
Solution (a) (−2)(−2)(−2) may bewritten as (−2) 3 .
(b) 4.4.4.5.5 may be writtenas 4 3 5 2 .
yyy
xxxx
(d) aa(−a)(−a)
bb(−b)
(c)
yyy y3
may bewritten as
xxxx x 4.
(d) Note that (−a)(−a) = aa since the product of two negative quantities is positive.
Soaa(−a)(−a) =aaaa =a 4 . Alsobb(−b) = −bbb = −b 3 . Hence
aa(−a)(−a)
bb(−b)
= a4
−b = 3 −a4 b 3
Example1.2 Evaluate
(a) 7 3 (b) (−3) 3 (c) 2 3 (−3) 4
Solution (a) 7 3 = 7.7.7 = 343
(b) (−3) 3 = (−3)(−3)(−3) = −27
(c) 2 3 (−3) 4 = 8(81) = 648
Most scientific calculators have an x y button to enable easy calculation of expressions
ofasimilarform tothoseinExample 1.2.
1.2.1 Multiplyingexpressionsinvolvingindices
Consider the product (6 2 )(6 3 ). We maywrite thisas
(6 2 )(6 3 ) = (6.6)(6.6.6) = 6 5

1.2 Laws of indices 3
So
6 2 6 3 =6 5
This illustratesthefirst lawof indiceswhich is
a m a n =a m+n
When expressions with the same base aremultiplied, the indices areadded.
Example1.3 Simplifyeach of the following expressions:
(a) 3 9 3 10 (b) 4 3 4 4 4 6 (c) x 3 x 6 (d) y 4 y 2 y 3
Solution (a) 3 9 3 10 = 3 9+10 = 3 19
(b) 4 3 4 4 4 6 = 4 3+4+6 = 4 13
(c) x 3 x 6 =x 3+6 =x 9
(d) y 4 y 2 y 3 =y 4+2+3 =y 9
Engineeringapplication1.1
Powerdissipationinaresistor
The resistor is one of the three fundamental electronic components. The other two
are the capacitor and the inductor, which we will meet later. The role of the resistor
is to reduce the current flow within the branch of a circuit for a given voltage. As
current flows through the resistor, electrical energy is converted into heat. Because
the energy is lost from the circuit and is effectively wasted, it is termed dissipated
energy. The rate of energy dissipationisknown as the power,P, and isgiven by
P=I 2 R (1.1)
whereI isthecurrentflowingthroughtheresistorandRistheresistancevalue.Note
that the current is raised to the power 2. Note that power, P, is measured in watts;
current,I, ismeasured inamps;and resistance,R, is measured inohms.
Thereisanalternativeformulaforpowerdissipationinaresistorthatusesthevoltage,V,
across the resistor. To obtain this alternative formula we need to use Ohm’s
law,whichstatesthatthevoltageacrossaresistor,V,andthecurrentpassingthrough
it, are relatedby the formula
V=IR (1.2)
From Equation (1.2) wesee that
I = V R
(1.3)
Combining Equations (1.1) and (1.3) gives
( ) V 2
P = R = V ·V V2
·R =
R R R R
➔

4 Chapter 1 Review of algebraic techniques
Note that in this formula forP, the voltage is raised to the power 2. Note an important
consequence of this formula is that doubling the voltage, while keeping the
resistance fixed, results in the power dissipation increasing by a factor of 4, that is
2 2 .Alsotreblingthevoltage,forafixedvalueofresistance,resultsinthepowerdissipation
increasing by a factor of 9,thatis3 2 .
Similar considerations can be applied to Equation 1.1. For a fixed value of resistance,
doubling the current results in the power dissipation increasing by a factor of
4,andtreblingthecurrentresultsinthepowerdissipationincreasingbyafactorof9.
Consider the product 3(3 3 ). Now
3(3 3 ) = 3(3.3.3) = 3 4
Also, using the first law of indices we see that 3 1 3 3 = 3 4 . This suggests that 3 is the
same as 3 1 . This illustratesthe general rule:
a=a 1
Raising a number tothe power 1 leaves the number unchanged.
Example1.4 Simplify (a) 5 6 5 (b) x 3 xx 2
Solution (a) 5 6 5 = 5 6+1 = 5 7 (b) x 3 xx 2 =x 3+1+2 =x 6
1.2.2 Dividingexpressionsinvolvingindices
Consider the expression 45
4 3:
4 5
4 = 4.4.4.4.4
3 4.4.4
= 4.4 by cancelling 4s
= 4 2
This serves toillustratethesecond lawof indiceswhich is
a m
a n =am−n
When expressions with the same baseare divided, the indices aresubtracted.
Example1.5 Simplify
(a) 59
5 7 (b) (−2)16
(−2) 13
(c) x9
x 5
(d) y6
y
Solution (a)
(b)
5 9
5 = 7 59−7 = 5 2
(−2) 16
(−2) = 13 (−2)16−13 = (−2) 3

1.2 Laws of indices 5
(c)
(d)
x 9
x 5 =x9−5 =x 4
y 6
y =y6−1 =y 5
Consider the expression 23
23. Usingthe second law of indices we may write
2 3
2 = 3 23−3 = 2 0
But, clearly, 23
2 3 = 1,and so 20 = 1.This illustratesthe general rule:
a 0 =1
Any expression raised tothe power 0 is1.
1.2.3 Negativeindices
Consider the expression 43
45. We can write thisas
4 3
4 = 4.4.4
5 4.4.4.4.4 = 1
4.4 = 1 4 2
Alternatively, using the second law of indices we have
4 3
4 5 = 43−5 = 4 −2
So wesee that
4 −2 = 1 4 2
Thus we are able to interpret negative indices. The sign of an index changes when the
expression is inverted. Ingeneral wecan state
a −m = 1
a m a m = 1
a −m
Example1.6 Evaluate the following:
(a) 3 −2 2
(b) (c) 3 −1 (d) (−3) −2 (e) 6−3
4 −3 6 −2
Solution (a) 3 −2 = 1 3 = 1 2 9
2
(b)
4 = −3 2(43 ) = 2(64) = 128
(c) 3 −1 = 1 3 = 1 1 3
(d) (−3) −2 = 1
(−3) = 1 2 9
6 −3
(e)
6 = −2 6−3−(−2) = 6 −1 = 1 6 = 1 1 6

6 Chapter 1 Review of algebraic techniques
Example1.7 Write the following expressions using only positive indices:
(a) x −4 (b) 3x −4 (c) x−2
y −2
(d) 3x −2 y −3
Solution (a) x −4 = 1 x 4
(b) 3x −4 = 3 x 4
(c)
x −2
y −2 =x−2 y 2 = y2
x 2
(d) 3x −2 y −3 = 3
x 2 y 3
Engineeringapplication1.2
Powerdensityofasignaltransmittedbyaradioantenna
Aradioantennaisadevicethatisusedtoconvertelectricalenergyintoelectromagnetic
radiation,whichis thentransmitted todistant points.
An idealtheoreticalpointsourceradioantenna whichradiates the same power in
alldirectionsistermedanisotropicantenna.Whenittransmitsaradiowave,thewave
spreads out equally in all directions, providing there are no obstacles to block the
expansionofthewave.Thepowergeneratedbytheantennaisuniformlydistributed
on the surfaceofanexpanding sphereofarea,A, given by
A = 4πr 2
whereristhe distance fromthe generatingantenna tothe wave front.
The power density,S, provides an indication of how much of the signal can potentially
be received by another antenna placed at a distance r. The actual power
received depends on the effective area or aperture of the antenna, which is usually
expressed inunitsofm 2 .
Electromagneticfieldexposurelimitsforhumansaresometimesspecifiedinterms
of a power density. The closer a person is to the transmitter, the higher the power
densitywill be. So a safe distanceneeds tobedetermined.
Thepowerdensityistheratioofthepowertransmitted,P t
,totheareaoverwhich
itisspread
S =
power transmitted
area
= P t
4πr 2 = P t
4π r−2 W m −2
Note that r in this equation has a negative index. This type of relationship is
knownasaninversesquarelawandisfoundcommonlyinscienceandengineering.
Note that if the distance,r, is doubled, then the area,A, increases by a factor of
4 (i.e. 2 2 ). If the distance is trebled, the area increases by a factor of 9 (i.e. 3 2 ) and
so on. This means that as the distance from the antenna doubles, the power density,
S, decreases to a quarter of its previous value; if the distance trebles then the power
density isonly a ninth ofits previous value.

1.2 Laws of indices 7
1.2.4 Multipleindices
Consider the expression (4 3 ) 2 . This may be writtenas
(4 3 ) 2 =4 3 .4 3 =4 3+3 =4 6
This illustratesthethirdlaw ofindiceswhich is
(a m ) n =a mn
Note that the indicesmandnhave been multiplied.
Example1.8 Writethe following expressions usingasingleindex:
(a) (3 2 ) 4 (b) (7 −2 ) 3 (c) (x 2 ) −3 (d) (x −2 ) −3
Solution (a) (3 2 ) 4 = 3 2×4 = 3 8
(b) (7 −2 ) 3 = 7 −2×3 = 7 −6
(c) (x 2 ) −3 =x 2×(−3) =x −6
(d) (x −2 ) −3 =x −2×−3 =x 6
Consider the expression (2 4 5 2 ) 3 . We see that
(2 4 5 2 ) 3 = (2 4 5 2 )(2 4 5 2 )(2 4 5 2 )
=2 4 2 4 2 4 5 2 5 2 5 2
= 2 12 5 6
This illustratesageneralization of the thirdlaw of indices which is
(a m b n ) k =a mk b nk
Example1.9 Remove the brackets from
(a) (2x 2 ) 3 (b) (−3y 4 ) 2 (c) (x −2 y) 3
Solution (a) (2x 2 ) 3 = (2 1 x 2 ) 3 = 2 3 x 6 = 8x 6
(b) (−3y 4 ) 2 = (−3) 2 y 8 = 9y 8
(c) (x −2 y) 3 =x −6 y 3
Engineeringapplication1.3
Radarscattering
It has already been shown in Engineering application 1.2 that the power density of
an isotropic transmitter of radio waves is
S = P t
4π r−2 W m −2
➔

8 Chapter 1 Review of algebraic techniques
It is possible to use radio waves to detect distant objects. The technique involves
transmittingaradiosignal,whichisthenreflectedbackwhenitstrikesatarget.This
weakreflectedsignalisthenpickedupbyareceivingantenna,thusallowinganumber
ofpropertiesofthetargettobededuced,suchasitsangularpositionanddistancefrom
the transmitter. This system is known as radar, which was originally an acronym
standing for RAdioDetectionAndRanging.
When the wave hits the target it produces a quantity of reflected power. The
power depends upon the object’s radar cross-section (RCS), normally denoted by
theGreeklower caselettersigma, σ,andhaving unitsofm 2 .Thepower reflected at
the object,P r
, isgiven by
P r
=Sσ= P t σ
4π r−2 W
SomemilitaryaircraftusespecialtechniquestominimizetheRCSinordertoreduce
the amount of power they reflect and hence minimize the chance of being detected.
If the reflected power at the target is assumed to spread spherically, when it
returns tothe transmitter position itwill have the power density,S r
, given by
S r
=
power reflected attarget
area
= P r
4π r−2 W m −2
Substituting forthe reflected power,P r
, gives
power reflected attarget
S r
=
area
= P t σ W m −2
(4π) 2r−4
=
(
Pt σ
4π r−2 )
4π
r −2 =
P t σ ( ) r
−2 2
4π×4π
Notethattheproductofthetwor −2 termshasbeencalculatedusingthethirdlawof
indices.
Thisexampleillustratesoneofthemainchallengeswithradardesignwhichisthat
thepowerdensityreturnedbyadistantobjectisverymuchsmallerthanthetransmitted
power, even for targets with a large RCS. For theoretical isotropic antennas, the
received power density depends upon the factorr −4 . This factor diminishes rapidly
forlarge values ofr, thatis, as the object being detected gets further away.
In practice, the transmit antennas used are not isotropic but directive and often
scantheareaofinterest.Theyalsomakeuseofreceiveantennaswithalargeeffective
area which can produce a viable signalfrom the smallreflected power densities.
1.2.5 Fractionalindices
Thethirdlawofindicesstatesthat (a m ) n =a mn .Ifwetakea = 2,m = 1 2 andn=2we
obtain
(2 1/2 ) 2 = 2 1 = 2
So when 2 1/2 is squared, the result is 2. Thus, 2 1/2 is a square root of 2. Each positive
number has two square roots and so
2 1/2 = √ 2 = ±1.4142 ...

1.2 Laws of indices 9
Similarly
(2 1/3 ) 3 = 2 1 = 2
so that2 1/3 isacube rootof 2:
√
2 1/3 = 3 2=1.2599...
Ingeneral 2 1/n isannthrootof 2.The general lawstates
x 1/n isannthrootofx
Example1.10 Write the following usingasingle positive index:
(a) (3 −2 ) 1/4 (b) x 2/3 x 5/3 (c) yy −2/5 (d) √ k 3
Solution (a) (3 −2 ) 1/4 = 3 −2×1 4 = 3 −1/2 = 1
3 1/2
(b) x 2/3 x 5/3 =x 2/3+5/3 =x 7/3
(c) yy −2/5 =y 1 y −2/5 =y 1−2/5 =y 3/5
(d) √ k 3 = (k 3 ) 1/2 =k 3×1 2 =k
3/2
Example1.11 Evaluate
(a) 8 1/3 (b) 8 2/3 (c) 8 −1/3 (d) 8 −2/3 (e) 8 4/3
Solution We notethat 8 maybewritten as2 3 .
(a) 8 1/3 = (2 3 ) 1/3 = 2 1 = 2
(b) 8 2/3 = (8 1/3 ) 2 = 2 2 = 4
(c) 8 −1/3 = 1
8 1/3 = 1 2
(d) 8 −2/3 = 1
8 2/3 = 1 4
(e) 8 4/3 = (8 1/3 ) 4 = 2 4 = 16
Engineeringapplication1.4
Skindepthinaradialconductor
When an alternating current signal travels along a conductor, such as a copper wire,
most of the current is found near the surface of the conductor. Nearer to the centre
of the conductor, the current diminishes. The depth of penetration of the signal,
termed the skin depth, into the conductor depends on the frequency of the signal.
Skin depth, illustrated in Figure 1.1, is defined as the depth at which the current
➔

10 Chapter 1 Review of algebraic techniques
densityhasdecayedtoapproximately37%ofthatattheedge.Skindepthisimportant
because it affects the resistance of wires and other conductors: the smaller the skin
depth, the higher the effective resistance and the greater the loss due toheating.
At low frequencies, such as those found in the domestic mains supply, the skin
depthissolargethatoftenitcan beneglected; however, invery large-diameter conductors
and smaller conductors at microwave frequencies it becomes important and
has tobe taken into account.
The skin depth, δ, isgiven by
( ) 2 1/2
δ =
ωµσ
where µisamaterialconstantknownasthepermeabilityoftheconductor, ω isthe
angular frequency of the signal and σ isthe conductivity of the conductor.
d
Figure1.1
Cross-sectionofaradialconductor
illustratingaskindepth δ.
EXERCISES1.2
1 Evaluate
(a) 2 3 (b) 3 2 5 13
(c)
5 12
19 −11
(d)
19 −13 (e) (2 1/4 ) 8 (f) (−4) −2
(g) 4 −1/2 (h) (9 1/3 ) 3/2 √ √
(i) 32 2
√
(j) 0.01 (k) 81 3/4
2 Useascientificcalculator to evaluate
(a) 10 1.2 (b) 6 −0.7 (c) 6 2.5
(d) (3 −1 4 2 ) 0.8
3 Expresseachofthe followingexpressions usinga
single positiveindex:
(a) x 4 x 7 (b) x 2 (−x)
(c)
x 2
x
(d)
x −2
x −1
(e) (x −2 ) 4 (f) (x −2.5 x −3.5 ) 2
4 Simplify asmuch aspossible
(a)
(c)
x 1/2
x 1/3 (b) (16x 4 ) 0.25
( ) 1/3
27 2xy 2
y 3 (d)
(2xy) 2
(e) √ a 2 b 6 c 4 (f) (64t 3 ) 2/3

1.3 Number bases 11
Solutions
1 (a) 8 (b) 9 (c) 5 (d) 361
1 1
(e) 4 (f) (g) (h) 3
16 2
(i) 8 (j) 0.1 (k) 27
2 (a) 15.8489 (b) 0.2853
(c) 88.1816 (d) 3.8159
3 (a) x 11 (b) −x 3 (c) x
1 1 1
(d) (e)
x x 8 (f)
x 12
4 (a) x 1/6 (b) 2x (c)
(d)
1
2x
3
y
(e) ab 3 c 2 (f) 16t 2
1.3 NUMBERBASES
The decimal system of numbers in common use is based on the 10 digits 0, 1, 2, 3, 4,
5, 6, 7, 8 and 9. However, other number systems have important applications in computerscienceandelectronicengineering.Inthissectionweremindthereaderofwhatis
meant by a number in the decimal system, and show how we can use powers or indices
with bases of 2 and 16 to represent numbers in the binary and hexadecimal systems
respectively.Wefollowthisbyanexplanationofanalternativebinaryrepresentationof
a number known as binarycoded decimal.
1.3.1 Thedecimalsystem
Thenumbersthatwecommonlyuseineverydaylifearebasedon10.Forexample,253
can bewrittenas
253=200+50+3
= 2(100) + 5(10) + 3(1)
= 2(10 2 ) + 5(10 1 ) + 3(10 0 )
In this form it is clear why we refer to this as a ‘base 10’ number. When we use 10 as a
basewesaywearewritinginthedecimalsystem.Notethatinthedecimalsystemthere
are10digits:0,1,2,3,4,5,6,7,8,and9.Youmayrecallthephrase‘hundreds,tensand
units’ and as we have seen these are simply powers of 10. To avoid possible confusion
with numbers using other bases, we denote numbers in base 10 with a small subscript,
forexample, 5192 10
:
5192 10
= 5000 +100 +90 +2
= 5(1000) + 1(100) + 9(10) + 2(1)
= 5(10 3 ) + 1(10 2 ) + 9(10 1 ) + 2(10 0 )
Notethat,inthepreviousline,aswemovefromrighttoleft,thepowersof10increase.
1.3.2 Thebinarysystem
Abinarysystemusesthenumber2foritsbase.Abinarysystemhasonlytwodigits,0
and 1, and these are called binary digits or simply bits. Binary numbers are based on
powersof2.Inacomputer,binarynumbersareusuallystoredingroupsof8bitswhich
wecall abyte.

12 Chapter 1 Review of algebraic techniques
Convertingfrombinarytodecimal
Considerthebinarynumber110101 2
.Asthebaseis2thismeansthataswemovefrom
right toleft the position of each digitrepresents an increasing power of2asfollows:
110101 2
= 1(2 5 ) +1(2 4 ) +0(2 3 ) +1(2 2 ) +0(2 1 ) +1(2 0 )
= 1(32) +1(16) +0(8) +1(4) +0(2) +1(1)
=32+16+4+1
= 53 10
Hence 110101 2
and 53 10
areequivalent.
Example1.12 Convert the following todecimal: (a) 1111 2
(b)101010 2
Solution (a) 1111 2
= 1(2 3 ) +1(2 2 ) +1(2 1 ) +1(2 0 )
= 1(8) +1(4) +1(2) +1(1)
=8+4+2+1
= 15 10
(b) 101010 2
= 1(2 5 ) +0(2 4 ) +1(2 3 ) +0(2 2 ) +1(2 1 ) +0(2 0 )
=1(32)+0+1(8)+0+1(2)+0
=32+8+2
= 42 10
Convertingdecimaltobinary
Wenowlookatsomeexamplesofconvertingnumbersinbase10tonumbersinbase2,
that is from decimal to binary. We make use of Table 1.1, which shows various powers
of 2,when converting from decimal tobinary. Table 1.1 may be extended as necessary.
Table1.1
Powers of2.
2 0 1 2 4 16 2 8 256
2 1 2 2 5 32 2 9 512
2 2 4 2 6 64 2 10 1024
2 3 8 2 7 128 2 11 2048
Example1.13 Convert 83 10
toabinary number.
Solution We need to express 83 10
as the sum of a set of numbers, each of which is a power of 2.
From Table 1.1 we see that 64 is the highest number in the table that does not exceed
the given number of 83. We write
83=64+19
Wenowfocusonthe19.FromTable1.1,16isthehighestnumberthatdoesnotexceed
19. So wewrite
19=16+3

1.3 Number bases 13
giving
83=64+16+3
We now focus on the 3 and again usingTable 1.1 we may write
83=64+16+2+1
=2 6 +2 4 +2 1 +2 0
= 1(2 6 ) +0(2 5 ) +1(2 4 ) +0(2 3 ) +0(2 2 ) +1(2 1 ) +1(2 0 )
= 1010011 2
Example1.14 Express 200 10
as a binary number.
Solution From Table 1.1 we note that 128 is the highest number that does not exceed 200 so we
write
200 = 128 +72
UsingTable 1.1 repeatedly we may write
200 = 128 +72
=128+64+8
=2 7 +2 6 +2 3
= 1(2 7 ) +1(2 6 ) +0(2 5 ) +0(2 4 ) +1(2 3 ) +0(2 2 ) +0(2 1 ) +0(2 0 )
= 11001000 2
Anotherwaytoconvertdecimalnumberstobinarynumbersistodivideby2repeatedly
and note the remainder. We rework the previous two examples using thismethod.
Example1.15 Convert the following decimal numbers tobinary: (a)83 (b) 200
Solution (a) We divide by 2 repeatedly and note the remainder.
Remainder
83÷2=41r1 1
41÷2=20r1 1
20÷2=10r0 0
10÷2= 5r0 0
5÷2= 2r1 1
2÷2= 1r0 0
1÷2= 0r1 1
To obtain the binary number we write out the remainder, working from the bottom
one tothe topone. This gives
as before.
83 10
= 1010011 2

14 Chapter 1 Review of algebraic techniques
(b) We repeat the process by repeatedly dividing 200 by 2 and noting the remainder.
Remainder
200÷2=100r0 0
100÷2= 50r0 0
50÷2= 25r0 0
25÷2= 12r1 1
12÷2= 6r0 0
6÷2= 3r0 0
3÷2= 1r1 1
1÷2= 0r1 1
Readingtheremaindercolumnfromthebottomtothetopgivestherequiredbinary
number:
200 10
= 11001000 2
1.3.3 Hexadecimalsystem
We now consider the number system which uses 16 as a base. This system is termed
hexadecimal (or simply hex). There are 16 digits in the hexadecimal system: 0, 1, 2,
3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. Notice that conventional decimal digits are
insufficient to represent hexadecimal numbers and so additional ‘digits’, A, B, C, D, E,
andF,areincluded.Table1.2showstheequivalencebetweendecimalandhexadecimal
digits. Hexadecimal numbers are based on powers of16.
Table1.2
Hexadecimal numbers.
Decimal Hexadecimal Decimal Hexadecimal
0 0 8 8
1 1 9 9
2 2 10 A
3 3 11 B
4 4 12 C
5 5 13 D
6 6 14 E
7 7 15 F
Convertingfromhexadecimaltodecimal
The following example illustrates how to convert from hexadecimal to decimal. We
use the fact that as we move from right to left, the position of each digit represents an
increasing power of 16.

1.3 Number bases 15
Example1.16 Convert the following hexadecimal numbers todecimal numbers: (a)93A (b)F9B3
Solution (a) Noting thathexadecimal numbers use base16 we have
93A 16
= 9(16 2 ) +3(16 1 ) +A(16 0 )
= 9(256) +3(16) +10(1)
= 2362 10
(b) F9B3 16
= F(16 3 ) +9(16 2 ) +B(16 1 ) +3(16 0 )
= 15(4096) +9(256) +11(16) +3(1)
= 63923 10
Convertingfromdecimaltohexadecimal
Table 1.3 provides powers of 16 which help in the conversion from decimal to hexadecimal.
Table1.3
16 0 1
16 1 16
16 2 256
16 3 4096
16 4 65536
The following example illustrateshow toconvert from decimal tohexadecimal.
Example1.17 Convert 14397 toahexadecimal number.
Solution We need to express 14397 as the sum of multiples of powers of 16. From Table 1.3 we
see that the highest number that does not exceed 14397 is 4096. We express 14397 as
amultipleof4096withanappropriateremainder. Dividing14397by4096weobtain3
with a remainder of 2109. So we may write
14397 = 3(4096) +2109
Wenowfocuson2109andapplythesameprocessasabove.FromTable1.3thehighest
number thatdoes not exceed 2109 is 256:
2109 = 8(256) +61
Finally, 61 = 3(16) +13. So wehave
14397 = 3(4096) +8(256) +3(16) +13
= 3(16 3 ) +8(16 2 ) +3(16 1 ) +13(16 0 )
FromTable 1.2 wesee that13 10
isDinhexadecimal, sowe have
14397 10
= 383D 16
As with base 2 we can convert decimal numbers by repeated division and noting the
remainder. The previous example isreworked toillustratethis.

16 Chapter 1 Review of algebraic techniques
Example1.18 Convert 14397 tohexadecimal.
Solution We divide repeatedly by 16, noting the remainder.
Remainder
14397÷16=899r13 13
899÷16= 56r3 3
56÷16= 3r8 8
3÷16= 0r3 3
Recall that 13 inhexadecimal isD.Reading up the Remainder column wehave
as before.
14397 10
= 383D 16
Electronic engineers need to be familiar with the decimal, binary and hexadecimal systemsandbeabletoconvertbetweenthem.Theequivalentrepresentationsofthedecimal
numbers 0--15 are provided inTable 1.4.
Table1.4
Decimal Binary Hex Decimal Binary Hex
0 0000 0 8 1000 8
1 0001 1 9 1001 9
2 0010 2 10 1010 A
3 0011 3 11 1011 B
4 0100 4 12 1100 C
5 0101 5 13 1101 D
6 0110 6 14 1110 E
7 0111 7 15 1111 F
Convertingfrombinarytohexadecimal
There is a straightforward way of converting a binary number into a hexadecimal number.Thedigitsofthebinarynumberaregroupedintofours,orquartets,(fromtherighthand
side) and each quartetisconverted toits hex equivalent using Table 1.4.
Example1.19 Convert 1101011100111 2
into hexadecimal.
Solution Working from the right, the binary number is grouped into fours, with additional zeros
being added as necessary tothe final grouping.
0001 1010 1110 0111

1.3 Number bases 17
Table1.4isusedtoexpresseachgroupoffourasitshexequivalent.Forexample,0111 =
7 16
, and continuing inthis waywe obtain
1AE7
Thus 110101110 0111 2
= 1AE7 16
.
1.3.4 Binarycodeddecimal
We have seen in Section 1.3.2 that decimal numbers can be expressed in an equivalent
binary form where the position of each binary digit, moving from the right to the left,
represents an increasing power of 2. There is an alternative way of expressing numbers
using the binary digits 1 and 0 that is often used in electronic engineering because for
some applications it is more straightforward to build the necessary hardware. This systemiscalledbinary
coded decimal(b.c.d.).
First of all, recall how the decimal digits 0,1,2,...,9 are expressed in their usual
binary form. Note that the largest decimal digit 9 is 1001 in binary, and so we need
at most four digits to store the binary representations of 0,1,...,9. Expressing each
decimal digitas a four-digitbinary number we obtain Table 1.5.
Table1.5
Decimal digits and their four-digit
binaryrepresentations.
0 0000 5 0101
1 0001 6 0110
2 0010 7 0111
3 0011 8 1000
4 0100 9 1001
A four-digit binary number is referred to as a nibble. To express a multi-digit decimal
number, such as 347, in b.c.d. each decimal digit in turn is converted into its binary
representation as shown. Note thatanibble isused foreach decimal digit.
3 4 7
↓ ↓ ↓
0011 0100 0111
Recall from Section 1.3.2 that a byte is a group of 8 bits (binary digits). Computers
usually store numbers in 8-bit bytes so there are two common ways of encoding b.c.d.
The first is to use a whole byte for each nibble, with the first 4 bits always set to 0. So,
forexample, 347 10
can be storedas
00000011 00000100 00000111
Alternatively, eachbytecanbeusedtostoretwonibbles,inwhichcase347 10
wouldbe
storedas
00000011 01000111
Ruleshavebeendevelopedforperformingcalculationsinb.c.d.butthesearebeyondthe
scope of thisbook.

18 Chapter 1 Review of algebraic techniques
Engineeringapplication1.5
Seven-segmentdisplays
The number displays found on music systems, video and other electronic equipment
commonly employ one or more seven-segment indicators. A single sevensegment
indicator is shown in Figure 1.2(a). The individual segments are typically
illuminatedwithalight-emittingdiode(LED)orsimilaropticaldeviceandareeither
onoroff.ThesegmentsareilluminatedaccordingtothetableshowninFigure1.2(b),
where 1 indicates thatthe segment is turned on and 0 indicates thatitisturned off.
b.c.d.
number a b c d e f g
0000 1
1 1 1 1 1 0
0001
0 1 1 0 0 0 0
0010
1 1 0 1 1 0 1
a
0011
0100
1 1 1 1 0 0 1
0 1 1 0 0 1 1
f
g
b
0101
0110
1 0 1 1 0 1 1
1 0 1 1 1 1 1
e
d
c
0111
1000
1001
1 1 1 0 0 0 0
1 1 1 1 1 1 1
1 1 1 1 0 1 1
(a)
(b)
Figure1.2
(a) Seven-segment LED display. (b)Seven-segment coding.
The numbers in the microprocessor system driving the display are typically
stored in binary format, known as, binary coded decimal (b.c.d.). As an example
we consider displaying binary number 11101010 2
as a decimal number on sevensegment
displays. This represents the decimal number 234, which requires three
seven-segment displays.
Themicroprocessorfirstdividestheinputnumberby100andinthiscaseobtains
the result 2 with a remainder of 34. This can be done directly on the binary number
itself via a series of operations within the assembly language of the microprocessor
without first converting to a decimal number. The result 2 = 0010 2
is then decoded
usingFigure1.2(b),givingthebitpattern1101101whichispassedtothe‘hundreds’
display.
Theremainderof34isthendividedby10giving3withafinalremainderof4.The
number 3 = 0011 2
and so this can be outputted to the ‘tens’ display as the pattern
1111001. Finally, 4 = 0100 2
, which ispassed tothe display as the pattern 0110011.

1.3 Number bases 19
The display shows
a
a
a
f
g
b
f
g
b
f
g
b
e
d
c
e
d
c
e
d
c
Notice that prior to decoding for display, by successive division by 100 and 10
the number has been converted into separate b.c.d. digits. Integrated circuits are
available which convert b.c.d. directly into the bit patterns for display. Hence the
output bit pattern of the microprocessor may be chosen to be b.c.d. In this case it
has the advantage that fewer pins are required on the microprocessor to operate the
display.
EXERCISES1.3
1 Convert the following decimal numbersto binary
numbers:(a) 19 (b)36 (c) 100 (d)796
(e)5000
2 Convert the following binarynumbers to decimal
numbers:(a) 111 (b)10101 (c) 111001
(d)1110001 (e) 11111111
3 What isthe highestdecimal number that can be
written in binary formusingamaximum of(a)2
binarydigits (b)3binarydigits (c)4binary digits
(d)5binary digits? Canyou spotapattern? (e)Write
aformulaforthe highest decimal number that can be
written usingN binary digits.
4 Writethe decimal number 0.5 in binary.
5 Convert the following hexadecimal numbersto
decimal numbers:(a)91 (b)6C (c)A1B (d)F9D4
(e)ABCD
6 Convert the following decimal numbersto
hexadecimal numbers:(a)160 (b)396 (c)5010
(d)25000 (e)1000000
7 Calculate the highestdecimal number that can be
represented byahexadecimal number with (a)1digit
(b)2digits (c)3digits (d)4digits (e)N digits
8 Expressthedecimalnumber375asbothapurebinary
number and anumber in b.c.d.
9 Convert (a)1111111 2 (b)101010111 2 into
hexadecimal.
Solutions
1 (a) 19 10 =10011 2 (b) 100100 (c) 1100100
(d) 1100011100 (e) 1001110001000
2 (a)111 2 = 7 (b)21 (c) 57 (d)113 (e)255
3 (a)3 (b)7 (c) 15 (d)31 (e) 2 N −1
4 Thebinarysystem is basedonpowers of2. The
examples in the text can be extended to the case of
negative powers of2justasin the decimal system
numbersafterthe decimal place represent negative
powers of10.So, forexample, the binarynumber
11.101 2 isconverted to decimal asfollows:
11.101 2 =1×2 1 + 1×2 0 + 1×2 −1
+0×2 −2 +1×2 −3
=2+1+ 1 2 + 1 8
= 3 5 8

20 Chapter 1 Review of algebraic techniques
Inthe same way the binary equivalent ofthe decimal
number 0.5is 0.1.
5 (a) 91 16 =145 10 (b)6C =108 (c)2587 (d)63956
(e) 43981
6 (a) 160 10 = A0 (b)18C (c) 1392 (d)61A8
(e) F4240
7 (a)15 (b)255 (c)4095 (d)65535 (e)16 N −1
8 (a)101110111 2 (b)001101110101 bcd
9 (a)7F (b)157
1.4 POLYNOMIALEQUATIONS
Apolynomial equationhas the form
P(x) =a n
x n +a n−1
x n−1 +a n−2
x n−2 + ··· +a 2
x 2 +a 1
x +a 0
=0 (1.4)
where n is a positive whole number, a n
, a n−1
,...,a 0
are constants and x is a
variable. The constants a n
, a n−1
,...,a 2
, a 1
, a 0
are called the coefficients of the
polynomial.
The roots of an equation are those values ofxwhich satisfyP(x) = 0. So ifx =x 1
is a
rootthenP(x 1
) = 0.
Examples of polynomial equations are
7x 2 +4x−1=0 (1.5)
2x−3=0 (1.6)
x 3 −20=0 (1.7)
Thedegreeofanequationisthevalueofthehighestpower.Equation(1.5)hasdegree2,
Equation (1.6) has degree 1 and Equation (1.7) has degree 3. A polynomial equation of
degreenhasnroots.
Therearesomespecialnamesforpolynomialequationsoflowdegree(seeTable1.6).
Table1.6
Equation Degree Name
ax+b=0 1 Linear
ax 2 +bx +c = 0 2 Quadratic
ax 3 +bx 2 +cx+d=0 3 Cubic
ax 4 +bx 3 +cx 2 +dx +e =0 4 Quartic
1.4.1 Quadraticequations
Wenowfocusattentiononquadraticequations.Thestandardformofaquadraticequation
isax 2 +bx +c = 0.We look atthreemethods of solvingquadratic equations:
(1) factorization,
(2) use of a formula,
(3) completing the square.
Example 1.20 illustratessolution by factorization.

1.4 Polynomial equations 21
Example1.20 Solve
6x 2 +11x−10=0
Solution The left-hand side(l.h.s.)isfactorized:
(3x −2)(2x +5) =0
So either
3x−2=0 or 2x+5=0
Hence
x = 2 3 ,−5 2
When roots cannotbefound byfactorizationwecan make use ofaformula.
Theformulaforfindingthe roots ofax 2 +bx +c = 0 is
x = −b ± √ b 2 −4ac
2a
Example1.21 Usethe quadraticformulatosolve
3x 2 −x−6=0
Solution Comparing3x 2 −x−6 withax 2 +bx +cwesee thata = 3,b= −1 andc = −6.So
x = −(−1) ± √ (−1) 2 −4(3)(−6)
2(3)
= 1 ± √ 73
6
= −1.2573,1.5907
Engineeringapplication1.6
Currentusedbyanelectricvehicle
Personal transport systems that make use of electrical power are becoming increasinglycommon.Oneofthefactorsbehindthischangeisthattheirusecanreduceroadsidepollutioninanurbanenvironment.Electricalvehicleshavealsobecomethebase
forself-drivingcars when combinedwith electricalcontroland navigation systems.
The motor in an operational electric vehicle has to do work to overcome wind,
inertia, friction, road resistance and in order to climb inclines. The energy supply
in the form of electrical power comes from the on-board battery pack. Due to its
internal construction the battery pack has a total internal resistance,R, which serves
toreduce the power available tothe motor.
➔

22 Chapter 1 Review of algebraic techniques
Asimplified circuitdiagram of a vehicle isshown inFigure 1.3.
Internal
resistance
R
I
Terminals
V
Motor and gearbox
Drive
wheels
Battery pack
Figure1.3
Electric vehicle wiringdiagram.
The total power delivered by the battery pack is
power = voltage ×current =VI
This is shared between loss due to the internal resistance and the power, P, to the
motor. The power loss due to the internal resistance isI 2 R (see Engineering application
1.1).So the equation for the power inthe circuitis
VI=I 2 R+P
This can be rewritten into the formof a quadratic equation
RI 2 −VI+P=0
whichcan besolved tocalculate the currentinthe wireforaparticular power delivered
to the motor. It is important to know the current in order to specify the size of
the fuses,the motor controller and the wirediameters used inthe vehicle.
Consider the case where the power output is 2 kW. If the circuit parameters are
V = 150 volts, R = 1.6 ,we have
1.6I 2 −150I +2000 = 0
The solutions tothe quadratic equation are
I = −b ± √ b 2 −4ac
2a
= 77.7 A, 16.1 A
= −(−150) ± √ (−150) 2 − (4 ×1.6 ×2000)
2×1.6
Therelevantsolutiondependsontheelectricalcharacteristicsofthemotorusedinthe
circuit. In practice, the larger of the two currents would correspond to a substantial
loss inthe internal resistance and would beavoided by the correctchoice of motor.
The technical computing language MATLAB ® has the function roots which
finds the solutions of a polynomial equation. In this example we would type
roots ([1.6 -150 2000]) at the command line to obtain the results calculated
above.

1.4 Polynomial equations 23
We now introduce the method of completing the square. The idea behind completing
the square is to absorb both thex 2 and thexterm into a single squared term. Note that
thisispossible since
and so
x 2 +2kx+k 2 =(x+k) 2
x 2 +2kx=(x+k) 2 −k 2
and finally
x 2 +2kx+A=(x+k) 2 +A−k 2
Thex 2 and thexterms are both contained in the (x +k) 2 term. The coefficient ofxon
the l.h.s.
(
is 2k. The squared
)
term on the right-hand side (r.h.s.) has the form (x +k) 2 ,
coefficient ofx 2
thatis x + . The following example illustratesthe idea.
2
Example1.22 Solve the following quadraticequationsbycompletingthe square:
(a) x 2 +8x+2=0
(b) 2x 2 −4x+1=0
Solution (a) Bycomparingx 2 +8x +2withx 2 +2kx +Aweseek = 4.Thusthesquaredterm
mustbe (x +4) 2 .Now
and so
Therefore
(x+4) 2 =x 2 +8x+16
x 2 +8x=(x+4) 2 −16
x 2 +8x+2=(x+4) 2 −16+2
=(x+4) 2 −14
At this stage we have completed the square. Finally, solvingx 2 + 8x + 2 = 0 we
have
x 2 +8x+2=0
(x+4) 2 −14=0
(x+4) 2 =14
x+4=± √ 14
x=−4± √ 14 = −7.7417,−0.2583
(b) 2x 2 −4x +1 = 0maybeexpressedasx 2 −2x +0.5 = 0.Comparingx 2 −2x +0.5
with x 2 + 2kx +A we see that k = −1. Thus the required squared term must be
(x −1) 2 .Now
and so
(x−1) 2 =x 2 −2x+1
x 2 −2x=(x−1) 2 −1

24 Chapter 1 Review of algebraic techniques
and
x 2 −2x+0.5=(x−1) 2 −1+0.5
=(x−1) 2 −0.5
Finally, solvingx 2 −2x +0.5 = 0 we have
(x−1) 2 −0.5=0
(x −1) 2 = 0.5
x−1=± √ 0.5
x=1± √ 0.5 = 0.2929,1.7071
1.4.2 Polynomialequationsofhigherdegree
Example1.23 Verifythatx = 1 andx = 2 areroots of
P(x)=x 4 −2x 3 −x+2=0
Solution P(x) =x 4 −2x 3 −x +2
P(1)=1−2−1+2=0
P(2)=2 4 −2(2 3 )−2+2=16−16−2+2=0
SinceP(1) = 0 andP(2) = 0, thenx = 1 andx = 2 are roots of the given polynomial
equation and are sometimes referred to as real roots. Further knowledge is required to
find the two remainingroots,whichareknown as complex roots. This topiciscovered
inChapter9.
Example1.24 Solve the equation
P(x)=x 3 +2x 2 −37x+52=0
Solution AsseeninExample1.21aformulacanbeusedtosolvequadraticequations.Forhigher
degree polynomial equations such simple formulae do not always exist. However, if
one of the roots can be found by inspection we can proceed as follows. By inspection
P(4) = 4 3 +2(4) 2 −37(4) +52 = 0sothatx = 4isaroot.Hencex−4isafactorof
P(x). ThereforeP(x) can bewritten as
P(x)=x 3 +2x 2 −37x+52=(x−4)(x 2 +αx+β)
where α and β mustnow befound.Expanding the r.h.s.gives
Hence
P(x)=x 3 +αx 2 +βx−4x 2 −4αx−4β
x 3 +2x 2 −37x+52=x 3 +(α−4)x 2 +(β−4α)x−4β
Bycomparingconstanttermson the l.h.s. and r.h.s.wesee that
52 = −4β

so that
β=−13
Bycomparing coefficients ofx 2 we see that
2=α−4
Therefore,
α = 6
1.4 Polynomial equations 25
Hence,P(x) = (x −4)(x 2 +6x −13).Thequadraticequationx 2 +6x −13 = 0canbe
solved using the formula
x = −6 ± √ 36 −4(−13)
2
= −6 ± √ 88
2
= 1.690,−7.690
We conclude thatP(x) = 0 has roots atx = 4,x = 1.690 andx = −7.690.
EXERCISES1.4
1 Calculatethe rootsofthe following linearequations:
(a) 4x−12=0
(b) 5t+20=0
(c) t+10=2t
y
(d)
2 −1=3
(e) 0.5t−6=0
(f) 2x+3=5x−6
3x
(g)
2 −17=0
x
(h)
2 + x 3 = 1
(i) 2x−1= x 2 +2
(j) 2(y+1)=6
(k) 3(2y −1) =2(y +2)
3
(l)
2 (t+3)=2 (4t −1)
3
2 Solvethe followingquadraticequations by
factorization:
(a) t 2 −5t+6=0
(b) x 2 +x−12=0
(c) t 2 =10t −25
(d) x 2 +4x−21=0
(e) x 2 −9x+18=0
(f) x 2 =1
(g) y 2 −10y+9=0
(h) 2z 2 −z−1=0
(i) 2x 2 +3x−2=0
(j) 3t 2 +4t+1=0
(k) 4y 2 +12y+5=0
(l) 4r 2 −9r+2=0
(m)6d 2 −d−2=0
(n) 6x 2 −13x+2=0
3 Complete the square forthe followingquadratic
equations andhence findtheir roots:
(a) x 2 +2x−8=0
(b) x 2 −6x−5=0
(c) x 2 +4x−6=0
(d) x 2 −14x−10=0
(e) x 2 +5x−49=0
4 Solve the following quadraticequationsusingthe
quadraticformula:
(a) x 2 +x−1=0
(b) t 2 −3t−2=0

26 Chapter 1 Review of algebraic techniques
(c) h 2 +5h+1=0
(d) 0.5x 2 +3x−2=0
(e) 2k 2 −k−3=0
(f) −y 2 +3y+1=0
(g) 3r 2 =7r+2
(h) x 2 −70=0
(i) 4s 2 −2=s
(j) 2t 2 +5t+2=0
(k) 3x 2 = 50
5 Calculate the rootsofthe followingpolynomial
equations:
(a) x 3 −6x 2 +11x−6=0givenx=1isaroot
(b) t 3 −2t 2 −5t+6=0givent=3isaroot
(c) v 3 −v 2 −30v+72=0givenv=4isaroot
(d) 2y 3 +3y 2 −11y +3 = 0 giveny = 1.5 is aroot
(e) 2x 3 +3x 2 −7x−5=0givenx=− 5 2 isaroot.
6 Checkthat the given values are roots ofthe following
polynomial equations:
(a) x 2 +x−2=0
(b) 2t 3 −3t 2 −3t+2=0
(c) y 3 +y 2 +y+1=0
x=−2,1
(d) v 4 +4v 3 +6v 2 +3v=0
t=−1,0.5
y=−1
v=−1,0
Solutions
1 (a) 3 (b) −4 (c) 10 (d) 8
(e) 12 (f) 3 (g)
(i) 2 (j) 2 (k)
34
3
7
4
(h)
(l)
2 (a) 2,3 (b) −4,3 (c) 5
6
5
31
7
(d) −7,3 (e) 3,6 (f) −1,1
(g) 1,9 (h) −0.5,1 (i) −2,0.5
(j) −1, − 1 3
(m) − 1 2 , 2 3
(k) −2.5, −0.5 (l) 0.25,2
(n)
1
6 ,2
3 (a) (x+1) 2 −9=0,x=−4,2
(b) (x −3) 2 −14 = 0,x = −0.7417,6.7417
(c) (x +2) 2 −10 = 0,x = −5.1623,1.1623
(d) (x −7) 2 −59 = 0,x = −0.6811,14.6811
(e)
(
x + 5 ) 2 221 − = 0,x = −9.9330,4.9330
2 4
4 (a) −1.6180,0.6180
(b) −0.5616,3.5616
(c) −4.7913,−0.2087
(d) −6.6056,0.6056
(e) −1,1.5
(f) −0.3028,3.3028
(g) −0.2573,2.5907
(h) −8.3666,8.3666
(i) −0.5931,0.8431
(j) −2, −0.5
(k) −4.0825,4.0825
5 (a) 1,2, 3 (b) −2,1,3
(c) −6,3,4
(e) −2.5, −0.6180,1.6180
(d) −3.3028,0.3028,1.5
1.5 ALGEBRAICFRACTIONS
An algebraicfraction has the form
algebraicfraction = numerator polynomial expression
=
denominator polynomial expression
For example,
3t+1
t 2 +t+4 , x 3
x 2 +1
are all algebraic fractions.
and
y 2 +1
y 2 +2y+3

1.5.1 Properandimproperfractions
1.5 Algebraic fractions 27
Whenpresentedwithafraction,wecannotethedegreeofthenumerator,sayn,andthe
degree of the denominator, sayd.
A fraction is proper ifd > n, that is the degree of the denominator is greater than
the degree of the numerator. Ifd nthen the fraction isimproper.
Example1.25 Classify the following fractions as either proper or improper. In each case, state the
degree of both numerator and denominator.
x 2 +9x−6
(a)
3x 3 +x 2 +100
(b) t3 +t 2 +9t−6
t 5 +9
(v +1)(v −6)
(c)
v 2 +3v+6
(z +2) 3
(d)
5z 2 +10z +16
Solution (a) The degree of the numerator,n, is 2. The degree of the denominator,d, is 3. Since
d >nthe fraction isproper.
(b) Heren = 3 andd = 5.The fraction isproper sinced >n.
(c) Heren = 2 andd = 2,sod =nand the fraction isimproper.
(d) Heren = 3 andd = 2,sod <nand the fraction isimproper.
1.5.2 Equivalentfractions
Consider the numerical fractions 1 2 and 2 . These fractions have the same value. Similarly,
2 3 , 6 20
4
and
9 30 all have the same value. The algebraic fractions x y , 2x xt
and
2y yt all
have the same value. Fractions with the same value arecalledequivalent fractions.
The value of a fraction remains unchanged if both numerator and denominator are
multiplied or divided by the same quantity. This fact can be used to write a fraction in
many equivalent forms.Consider forexample the fractions
(a) 2 x
(b)
2(x +1)
x(x +1)
(c) 2xt
x 2 t
These are all equivalent fractions. Fraction (b) can be obtained by multiplying both numerator
and denominator of fraction (a) by (x +1), so they are equivalent. Fraction (a)
canbeobtainedbydividingnumeratoranddenominatoroffraction(c)byxt andsothey
arealsoequivalent.
Example1.26 Show that
x +1
x +7
areequivalent.
and
x 2 +4x+3
x 2 +10x+21

28 Chapter 1 Review of algebraic techniques
Solution We factorize the numerator and denominator of the second fraction:
x 2 +4x+3 (x +1)(x +3)
=
x 2 +10x+21 (x +7)(x +3)
Dividingbothnumeratoranddenominatorby (x +3)resultsin x +1
x +7 .Sothetwogiven
fractions are equivalent.
Dividingbothnumeratoranddenominatorbyx+3isoftenreferredtoas‘cancelling
x+3’.
1.5.3 Expressingafractioninitssimplestform
Consider the numerical fraction 6 . To simplify this we factorize both numerator and
10
denominator and then cancel any common factors.Thus
6
10 = 2 ×3
2 ×5 = 3 5
Thefractions 6
10 and 3 5 haveidenticalvaluesbut 3 5 isinasimplerformthan 6
10 .Itisimportanttostressthatonlyfactorswhicharecommontobothnumeratoranddenominator
can becancelled.
Example1.27 Simplify
(a)
(b)
6x
18x 2
12x 3 y 2
4x 2 yz
Solution (a) Notethat18canbefactorizedto6×3andso6isafactorcommontobothnumerator
and denominator. Alsox 2 isx×x and soxis also a common factor. Cancelling the
common factors,6andx, produces
6x
18x 2 =
6x
(6)(3)(x)(x) = 1 3x
(b) The common factors are4,x 2 andy. Cancelling these factors gives
12x 3 y 2
4x 2 yz = 3xy
z
Example1.28 Simplify (a)
4
6x+4
(b) 6t3 +3t 2 +6t
3t 2 +3t
Solution (a) Factorizingbothnumeratoranddenominatorandcancellingcommonfactorsyields
4
6x+4 = (2)(2)
2(3x +2) = 2
3x+2

1.5 Algebraic fractions 29
(b) Factorizing and cancelling common factors yields
6t 3 +3t 2 +6t
3t 2 +3t
= 3t(2t2 +t+2)
3t(t +1)
= 2t2 +t+2
t +1
Note thatthe common factor, 3t, has been cancelled.
4t+8
Example1.29 Simplify(a) (b) 2y2 −y−1
t 2 +3t+2 y 2 −2y+1
Solution The numerator and denominator arefactorized and common factors are cancelled.
4t+8
(a)
t 2 +3t+2 = 4(t +2)
(t +2)(t +1) = 4
t +1
(b)
The common factor,t +2,has been cancelled.
2y 2 −y−1 (2y +1)(y −1)
= = 2y+1
y 2 −2y+1 (y −1) 2 y −1
The common factor,y−1,has been cancelled.
1.5.4 Multiplicationanddivisionofalgebraicfractions
Tomultiplytwoalgebraicfractionstogether,wemultiplytheirnumeratorstogether,and
multiplytheirdenominatorstogether, thatis
a
b × c d = a ×c
b ×d = ac
bd
Division isperformed by inverting the second fraction and then multiplying, thatis
a
b ÷ c d = a b × d c = ad
bc
Before multiplying or dividing fractions it is advisable to express each fraction in its
simplest form.
Example1.30 Simplify
x 2 +5x+6
2x−2
× x2 −x
x 2 +3x+2
Solution Factorizing numerators and denominators produces
x 2 +5x+6
2x−2
× x2 −x (x +2)(x +3) x(x −1)
= ×
x 2 +3x+2 2(x −1) (x +1)(x +2)
=
(x +2)(x +3)x(x −1)
2(x −1)(x +1)(x +2)
Commonfactors (x +2)and (x −1)canbecancelledfromnumeratoranddenominator
togive
(x +2)(x +3)x(x −1) (x +3)x
=
2(x −1)(x +1)(x +2) 2(x +1)

30 Chapter 1 Review of algebraic techniques
Hence
x 2 +5x+6
2x−2
× x2 −x x(x +3)
=
x 2 +3x+2 2(x +1)
Example1.31 Simplify
x 2 +8x+7
x 2 −6x
÷ x +7
x 3 +x 2
Solution The second fraction isinverted togive
x 2 +8x+7
x 2 −6x
× x3 +x 2
x +7
Factorizing numerators and denominators yields
(x +1)(x +7)
x(x −6)
× x2 (x+1)
(x+7)
= (x +1)(x +7)x2 (x +1)
x(x −6)(x +7)
Common factors ofxand (x +7) arecancelled leaving
(x +1)x(x +1)
x −6
which maybe written as
x(x +1) 2
x −6
1.5.5 Additionandsubtractionofalgebraicfractions
Themethodofaddingandsubtractingalgebraicfractionsisidenticaltothatfornumerical
fractions.
Each fraction is written in its simplest form. The denominators of the fractions are
then examined and the lowest common denominator (l.c.d.) is found. This is the simplest
expression that has the given denominators as factors. All fractions are then written
in an equivalent form with the l.c.d. as denominator. Finally the numerators are
added/subtractedand placedover the l.c.d. Consider the following examples.
Example1.32 Expressasasinglefraction
2
x +1 + 4
x +2
Solution Bothfractionsarealreadyintheirsimplestform.Thel.c.d.ofthedenominators, (x+1)
and (x +2), is found. This is (x +1)(x +2). Note that this is the simplest expression
that has bothx+1 andx+2 as factors.
Each fraction iswritten inan equivalent form with the l.c.d. as denominator. So
2
x +1 iswritten as 2(x +2)
(x +1)(x +2)

1.5 Algebraic fractions 31
and
4
x +2 iswritten as 4(x +1)
(x +1)(x +2)
Finally the numerators areadded. Hence we have
2
x +1 + 4
x +2 = 2(x +2)
(x +1)(x +2) + 4(x +1)
(x +1)(x +2)
= 2(x+2)+4(x+1)
(x +1)(x +2)
6x+8
=
(x +1)(x +2)
= 6x+8
x 2 +3x+2
Example1.33 Expressasasingle fraction
x 2 +3x+2
x 2 −1
− 2
2x+6
Solution Each fraction iswritten initssimplest form:
x 2 +3x+2
x 2 −1
=
(x +1)(x +2)
(x −1)(x +1) = x +2
x −1
2
2x+6 = 2
2(x +3) = 1
x +3
Thel.c.d.is (x −1)(x +3).Eachfractioniswritteninanequivalent formwithl.c.d.as
denominator:
So
x +2 (x +2)(x +3)
=
x −1 (x −1)(x +3) , 1
x +3 =
x 2 +3x+2
x 2 −1
x −1
(x −1)(x +3)
− 2
2x+6 = x +2
x −1 − 1
x +3
(x +2)(x +3)
=
(x −1)(x +3) − (x−1)
(x −1)(x +3)
= (x+2)(x+3)−(x−1)
(x −1)(x +3)
= x2 +5x+6−x+1
(x −1)(x +3)
= x2 +4x+7
(x −1)(x +3)

32 Chapter 1 Review of algebraic techniques
Engineeringapplication1.7
Resistorsinparallel
When carrying out circuit analysis it is often helpful to reduce the complexity of a
circuit by calculating an equivalent single resistance for several resistors connected
togetherinparallel.Thissimplifiedversionoftheoriginalcircuitthenbecomesmuch
easier to understand. Figure 1.4 shows the simplest case of two resistors connected
together inparallel.
R 1
R 2
Figure1.4
Two resistorsin parallel.
The equivalent resistance,R E
, ofthis simple network isfound from the formula
1
R E
= 1 R 1
+ 1 R 2
Bycombining the fractions on the r.h.s.we see
1
R E
= R 2 +R 1
R 1
R 2
and hence
R E
= R 1 R 2
R 1
+R 2
Consider the case whenR 1
andR 2
are equal and have valueR. The equivalent resistance
then becomes
R E
=
RR
R +R = R2
2R = R 2
So
R E
= R 2 = 0.5R
Thereforetheeffectofputtingtwoequalresistorsinparallelistoproduceanoverall
equivalent resistance which ishalf thatof a singleresistor.
EXERCISES1.5
1 Classifyeachfraction aseitherproperorimproper.
(a)
(d)
x +2
x 2 +2
x 2 +2
x +2
(b)
(e)
2
x +2
x 2 +2
x 2 +1
(c)
(f)
2 +x
2
x 2 +1
x 2 +2
2 Classifyeachofthe following algebraic fractions as
properorimproper.
3t+1 10v 2 +4v−6 6−4t+t 3
(a)
t 2 (b)
−1 3v 2 (c)
+v−1 6t 2 +1
(d)
9t+1
t +1
(e)
100f 2 +1
f 3 −1
(f)
(x +1)(x +2)
(x +3) 3

1.6 Solution of inequalities 33
(g)
(h)
(j)
(y +1)(y +2)(y +3)
(y +4) 3
(z +1) 10
(2z +1) 10 (i)
3k 2 +2k−1
k 3 +k 2 −4k+1
(q +1) 10
(q 2 +1) 6
3 Expresseachfraction in itssimplestform.
(a)
y 3 +2y
2y−y 2
(c) t2 +7t+12
t 2 +5t+4
x 2 +2x+1
(e)
x 2 −2x+1
4 Simplify the following:
(a)
(b)
x +1
x +3 × x +3
x +2
4
x 2 −1 × x +1
6
(b)
(d)
5x 2 +5
10x −10
x 2 −1
x 3 −2x 2 +x
(c)
(d)
(e)
x 2 +3x
x 3 +2x 2 × x2 +4x+4
4x
4xt +4t
xt 2 −t 2 × 4x2 −4
8x+8
x 2 +2x−15
x 2 +4x−5 × x2 +3x−4
x 2 −4x+3
5 Express asasingle fraction
(a)
(b)
(c)
(d)
(e)
3
x +6 + 2
x +1
4
x +2 − 2
(x +2) 2
2x+1
x 2 +x+1 + 4
x −1
x 2 +3x−18
x 2 +7x+6 − 2x2 +7x−4
x 2 +9x+20
3(x +1) 2(x −1)
x 2 +
+4x+4 x 2 −4
Solutions
1 (a) proper (b) proper (c) improper
(d) improper (e) improper (f) improper
2 (a) proper (b) improper (c) improper
(d) improper (e) proper
(f) proper
(g) improper (h) improper (i) proper
(j) proper
y 2 +2
3 (a)
2 −y
x +1
(d)
x(x −1)
(b)
(e)
x 2 +1
(c) t +3
2x−2 t +1
x 2 +2x+1
x 2 −2x+1
4 (a)
(c)
5 (a)
(c)
(d)
x +1
x +2
(x +2)(x +3)
4x 2
5x+15
(x +1)(x +6)
(b)
(d)
6x 2 +3x+3
(x−1)(x 2 +x+1)
−x 2 +x−14
(x +1)(x +5)
(b)
(e)
2
3(x −1)
2(x +1)
t
4x+6
(x +2) 2
(e)
5x 2 −x−10
(x+2) 2 (x−2)
x +4
x −1
1.6 SOLUTIONOFINEQUALITIES
Aninequality isany expression involving one ofthe symbols >, , <, .
a >bmeansaisgreaterthanb
a <bmeansaislessthanb
a bmeansaisgreaterthanorequal tob
a bmeansaislessthanorequal tob
Just as with an equation, when we add or subtract the same quantity to both sides of an
inequalitythe inequalitystill remains.Mathematically wehave

34 Chapter 1 Review of algebraic techniques
Ifa>bthen
a +k >b +k
a −k >b −k
addingkto both sides
subtractingkfromboth sides
We can make similarstatements fora b,a <banda b.
When multiplying or dividing both sides of an inequality extra care must be taken.
Suppose we wish to multiply or divide an inequality by a quantityk. Ifkis positive the
inequality remains the same;ifkisnegative then the inequality isreversed.
Ifa>bthen
ka>kb
a
k > b k
⎫
⎬
⎭
ifkispositive
ka<kb
a
k < b k
⎫
⎬
⎭
ifkisnegative
Notethatwhenkisnegativetheinequalitychangesfrom >to <.Similarstatementscan
bemadefora b,a <banda b.Whenaskedtosolveaninequalityweneedtostate
all the values of the variable forwhich the inequality istrue.
Example1.34 Solve the following inequalities:
(a) 3t+1>t+7
(b) 2−3z6+z
Solution (a) 3t +1>t +7
2t +1>7
2t >6
t >3
subtractingt fromboth sides
subtracting 1 fromboth sides
dividing both sides by2
Hence all values oft greaterthan3satisfy the inequality.
(b) 2−3z 6+z
−3z 4 +z
−4z 4
z −1
subtracting 2 fromboth sides
subtractingzfrom bothsides
dividing both sides by −4,rememberingtoreverse
the inequality
Hence all values ofzgreaterthanorequal to −1 satisfy the inequality.
We often have inequalities of the form α β > 0, α < 0, αβ > 0and αβ < 0tosolve.It
β
isuseful tonote thatif
α
β >0theneitherα>0andβ>0orα<0andβ<0
α
β <0theneitherα>0andβ<0orα<0andβ>0
αβ>0theneitherα>0andβ>0orα<0andβ<0
αβ<0theneitherα>0andβ<0orα<0andβ>0
The following examples illustratethis.

1.6 Solution of inequalities 35
Example1.35 Solve the following inequalities:
(a)
x +1 2t+3
>0 (b)
2x−6 t +2 1
x +1
Solution (a) Consider the fraction . For the fraction to be positive requires either of the
following:
2x−6
(i) x+1>0and2x−6>0.
(ii) x+1<0and2x−6<0.
We consider both cases.
Case(i)
x+1>0andsox>−1.
2x−6>0andsox>3.
Both of these inequalities are true only whenx > 3. Hence the fraction is positive
whenx > 3.
Case(ii)
x+1<0andsox<−1.
2x−6<0andsox<3.
(b)
Bothoftheseinequalitiesaretrueonlywhenx < −1.Hencethefractionispositive
whenx < −1.
x +1
Insummary,
2x−6 >0whenx>3orx<−1.
2t+3
t +2 1
2t+3
t +2 −10
t +1
t +2 0
Wenowconsiderthefraction t +1
t +2 .Forthefractiontobenegativeorzerorequires
either ofthe following:
(i) t+10andt+2>0.
(ii)t+1 0andt+2<0.
We consider each case inturn.
Case(i) t+10andsot −1.
t+2>0andsot>−2.
Hencetheinequalityistruewhent isgreaterthan −2andlessthanorequalto −1.
We write thisas −2 <t −1.
Case(ii) t+10andsot −1.
t+2<0andsot<−2.

36 Chapter 1 Review of algebraic techniques
Itisimpossibletosatisfybotht −1andt < −2andsothiscaseyieldsnovalues
oft.
Insummary, 2t+3
t +2
1when−2<t −1.
Example1.36 Solve the following inequalities:
(a)x 2 >4 (b)x 2 <4
Solution (a) x 2 >4
x 2 −4>0
(x−2)(x+2)>0
For the product (x −2)(x +2) tobepositive requires either
(i) x−2>0andx+2>0
or
(ii) x−2<0andx+2<0.
We examine eachcase inturn.
Case(i)
Case(ii)
x−2>0andsox>2.
x+2>0andsox>−2.
Bothofthesearetrueonly whenx > 2.
x−2<0andsox<2.
x+2<0andsox<−2.
Bothofthesearetrueonly whenx < −2.
Insummary,x 2 > 4 whenx > 2 orx < −2.
(b) x 2 <4
x 2 −4<0
(x−2)(x+2)<0
For the product (x −2)(x +2) tobenegative requires either
(i) x−2>0andx+2<0
or
(ii) x−2<0andx+2>0.
We examine eachcase inturn.
Case(i)
x−2>0andsox>2.
x+2<0andsox<−2.
No values ofxarepossible.

1.6 Solution of inequalities 37
Case(ii)
x−2<0andsox<2.
x+2>0andsox>−2.
Here we havex < 2 andx > −2. This is usually written as −2 < x < 2. Thus all
values ofxbetween −2 and 2 will ensure thatx 2 < 4.
Insummary,x 2 < 4 when −2 <x<2.
The previous example illustratesageneral rule.
Ifx 2 >kthenx> √ korx<− √ k.
Ifx 2 <kthen − √ k<x< √ k.
Example1.37 Solve the following inequalities:
(a)x 2 +x−6>0
(b)x 2 +8x+1<0
Solution (a) x 2 +x−6> 0
(x−2)(x+3)>0
For the product (x −2)(x +3) tobepositive requires either
(i) x−2>0andx+3>0
or
(ii) x−2<0andx+3<0.
Case(i)
x−2>0andsox>2.
x+3>0andsox>−3.
Bothofthese inequalitiesare satisfied only whenx > 2.
Case(ii)
x−2<0andsox<2.
x+3<0andsox<−3.
Bothofthese inequalitiesare satisfied only whenx < −3.
Insummary,x 2 +x−6 > 0 when eitherx > 2 orx < −3.
(b) The quadratic expression x 2 + 8x + 1 does not factorize and so the technique of
completingthe squareisused.
Hence
x 2 +8x+1=(x+4) 2 −15
(x+4) 2 −15<0
(x+4) 2 <15

38 Chapter 1 Review of algebraic techniques
Using the resultafter Example 1.36 wemay write
− √ 15<x+4< √ 15
− √ 15−4<x< √ 15−4
−7.873 <x<−0.127
EXERCISES1.6
1 Solve the following inequalities:
y
(a) 2x > 6 (b)
4 > 0.6
(c) 3t<12 (d) z+14
(e) 3v−24 (f) 6−k −1
6−2v
(g) <1
3
(h)m 2 2
(i)x 2 <9 (j) v 2 +110
(k) x 2 +10<6 (l) 2k 2 −31
(m) 10−2v 2 6 (n) 5+4k 2 >21
(o) (v −2) 2 25 (p) (3t +1) 2 >16
2 Solve the following inequalities:
(a) x 2 −6x+8>0
(b) x 2 +6x+80
(c) 2t 2 +3t−2<0
(d) y 2 −2y−240
(e) h 2 +6h+91
(f) r 2 +6r+70
(g) x 2 +4x−6<0
(h) 4t 2 +4t+912
(i)
x +4
x −5 >1 2t−3
(j)
t +6 6
(k)
3v+12
6−2v 0 (l) x 2
x +1 > 0
x
(m)
x 2 <0
+1
3y+1
(n)
y −2 2
(o) k 3 >0 (p)x 3 >8
t 2 +6t+9
(q) < 0
t +5
(r) (x +1)(x −2)(x +3) >0
Solutions
1 (a) x>3 (b)y>2.4 (c)t<4
(d) z3 (e) v2 (f)k7
(g) v> 3 √ (h)m 2orm− √ 2
2
(i) −3<x<3
(j) −3v3
(k) nosolution
(l) √ k 2ork− √ √ 2
(m)v 2orv− √ 2
(n) k>2ork<−2
(o) −3v7
(p) t>1ort<− 5 3
2 (a) x>4orx<2
(b) −4x −2
(c) −2<t< 1 2
(d) y6ory −4
(e) −4h −2
(f) √ r 2−3orr− √ 2 −3
(g) − √ 10−2<x< √ 10−2
(h) − 3 2 t 1 2
(i) x>5
(j) t − 39 4 ort>−6
(k) −4v<3
(l) x>−1withx≠0
(m)x<0 (n) −5y<2
(o) k>0 (p) x>2
(q) t<−5
(r) x>2or−3<x<−1

1.7 Partial fractions 39
1.7 PARTIALFRACTIONS
Givenasetoffractions,wecanaddthemtogethertoformasinglefraction.Forexample,
inExample 1.32 wesaw
2
x +1 + 4
x +2 = 2(x+2)+4(x+1)
(x +1)(x +2)
= 6x+8
x 2 +3x+2
Alternatively,ifwearegivenasinglefraction,wecanbreakitdownintothesumofeasier
fractions. These simple fractions, which when added together form the given fraction,
6x+8
arecalledpartialfractions.Thepartialfractionsof
x 2 +3x+2 are 2
x +1 and 4
x +2 .
Whenexpressingagivenfractionasasumofpartialfractionsitisimportanttoclassifythefractionasproperorimproper.Thedenominatoristhenfactorizedintoaproduct
of factors which can be linear and/or quadratic. Linear factors are those of the form
x
ax + b, for example 2x − 1, + 6. Repeated linear factors are those of the form
2
(ax +b) 2 , (ax +b) 3 andsoon,forexample (3x −2) 2 and (2x +1) 3 arerepeatedlinear
factors.Quadraticfactorsarethoseoftheformax 2 +bx+c,forexample2x 2 −6x+1.
1.7.1 Linearfactors
We can calculate the partial fractions of proper fractions whose denominator can be
factorized into linear factors.The following steps areused:
(1) Factorize the denominator.
(2) Eachfactorofthedenominatorproducesapartialfraction.Afactorax+bproduces
a partialfraction of the form A whereAisan unknown constant.
ax+b
(3) Evaluate the unknown constants of the partial fractions. This is done by evaluation
using a specific value ofxor by equating coefficients.
A linear factor ax +bin the denominator produces a partial fraction of the form
A
ax+b .
Example1.38 Express
6x+8
x 2 +3x+2
as itspartialfractions.
Solution The denominator isfactorized as
x 2 +3x+2=(x+1)(x+2)

40 Chapter 1 Review of algebraic techniques
Thelinearfactor,x +1,producesapartialfractionoftheform A
x +1 .Thelinearfactor,
x + 2, produces a partial fraction of the form B . A and B are unknown constants
x +2
whose values have tobefound. So we have
6x+8
x 2 +3x+2 = 6x+8
(x +1)(x +2) = A
x +1 + B
x +2
Multiplying both sides of Equation (1.8)by (x +1) and (x +2) weobtain
(1.8)
6x+8=A(x+2)+B(x+1) (1.9)
We now evaluateAandB. There are two techniques which enable us to do this: evaluation
using a specific value of xand equating coefficients. Each isillustratedinturn.
Evaluationusingaspecificvalueofx
We examine Equation (1.9). We will substitute a specific value ofxinto this equation.
Although any value can be substituted for x we will choose a value which simplifies
the equation as much as possible. We note that substitutingx = −2 will simplify the
r.h.s.oftheequationsincethetermA(x +2)willthenbezero.Similarly,substitutingin
x = −1 will simplify the r.h.s. because the termB(x +1) will then be zero. Sox = −1
andx = −2 are two convenient values to substitute into Equation (1.9). We substitute
each inturn.
Evaluating Equation (1.9) withx = −1 gives
−6+8=A(−1+2)
2 =A
Evaluating Equation (1.9) withx = −2 gives
−4 =B(−1)
B = 4
SubstitutingA = 2,B = 4into Equation (1.8) yields
6x+8
x 2 +3x+2 = 2
x +1 + 4
x +2
2
Thus the required partialfractions are
x +1 and 4
x +2 .
The constantsAandBcould have been found by equating coefficients.
Equatingcoefficients
Equation (1.9) may be writtenas
6x+8=(A+B)x+2A+B
Equating the coefficients ofxon both sides gives
6=A+B
Equating the constant termson both sides gives
8=2A+B

1.7 Partial fractions 41
Thus we have two simultaneous equations in A and B, which may be solved to give
A = 2 andB = 4 as before.
1.7.2 Repeatedlinearfactor
We now examine proper fractions whose denominators factorize into linear factors,
where one or more of the linear factors isrepeated.
Arepeated linear factor, (ax +b) 2 , produces two partialfractions of the form
A
ax+b +
B
(ax +b) 2
Arepeated linear factor, (ax +b) 2 , leadstopartialfractions
A
ax+b +
B
(ax +b) 2
Example1.39 Express
2x+5
x 2 +2x+1
as partialfractions.
Solution Thedenominatorisfactorizedtogive (x+1) 2 .Herewehaveacaseofarepeatedfactor.
A
This repeated factor generates partialfractions
x +1 + B
(x +1) 2.
Thus
2x+5
x 2 +2x+1 = 2x+5
(x +1) 2 = A
x +1 +
Multiplying by (x +1) 2 gives
2x+5=A(x+1)+B=Ax+A+B
B
(x +1) 2
Equating coefficients ofxgivesA = 2.Evaluation withx = −1 givesB = 3.So
2x+5
x 2 +2x+1 = 2
x +1 + 3
(x +1) 2
Example1.40 Express
14x 2 +13x
(4x 2 +4x +1)(x −1)
as partialfractions.
Solution Thedenominatorisfactorizedto (2x+1) 2 (x−1).Therepeatedfactor, (2x+1) 2 ,produces
partialfractions of the form
A
2x+1 + B
(2x +1) 2

42 Chapter 1 Review of algebraic techniques
The factor, (x −1), produces a partialfraction ofthe form C
x −1 . So
14x 2 +13x
(4x 2 +4x +1)(x −1) = 14x 2 +13x
(2x +1) 2 (x −1) = A
2x+1 + B
(2x +1) + C
2 x −1
Multiplying both sides by (2x +1) 2 (x −1) gives
14x 2 +13x =A(2x +1)(x −1) +B(x −1) +C(2x +1) 2 (1.10)
The unknown constantsA,B andC can now befound.
Evaluating Equation (1.10) withx = 1 gives
27 =C(3) 2
from which
C = 3
Evaluating Equation (1.10) withx = −0.5 gives
14(−0.5) 2 +13(−0.5) =B(−0.5 −1)
from which
B = 2
Finally, comparing the coefficients ofx 2 on both sides of Equation (1.10) wehave
14=2A+4C
Since wealready haveC = 3 then
A = 1
Hence we see that
14x 2 +13x
(4x 2 +4x +1)(x −1) = 1
2x+1 + 2
(2x +1) 2 + 3
x −1
1.7.3 Quadraticfactors
We now look at proper fractions whose denominator contains a quadratic factor, that is
a factorofthe formax 2 +bx +c.
Aquadraticfactor,ax 2 +bx +c, producesapartialfraction ofthe form
Ax+B
ax 2 +bx+c
Example1.41 Notingthatx 3 +2x 2 −11x −52 = (x −4)(x 2 +6x +13), express
3x 2 +11x +14
x 3 +2x 2 −11x−52
as partialfractions.

1.7 Partial fractions 43
Solution Thedenominatorhasalreadybeenfactorized.Thelinearfactor,x−4,producesapartial
fraction of the form A
x −4 .
The quadratic factor,x 2 +6x +13, will not factorize further into two linear factors.
Thus this factor generates a partialfraction of the form
Bx+C
x 2 +6x+13 . Hence
3x 2 +11x +14
(x −4)(x 2 +6x +13) = A
x −4 + Bx+C
x 2 +6x+13
Multiplying by (x −4) and (x 2 +6x +13) produces
3x 2 +11x +14 =A(x 2 +6x +13) + (Bx +C)(x −4) (1.11)
The constantsA,BandC can now be found.
Puttingx = 4 into Equation (1.11) gives
106 =A(53)
A = 2
Equating the coefficients ofx 2 gives
3=A+B
B = 1
Equating the constant termon both sides gives
Hence
14 =A(13) −4C
C = 3
3x 2 +11x +14
x 3 +2x 2 −11x−52 = 2
x −4 + x +3
x 2 +6x+13
1.7.4 Improperfractions
The techniques of calculating partial fractions in Sections 1.7.1 to 1.7.3 have all been
applied to proper fractions. We now look at the calculation of partial fractions of improper
fractions. The techniques described in Sections 1.7.1 to 1.7.3 are all applicable
to improper fractions. However, when calculating the partial fractions of an improper
fraction, an extra term needs to be included. The extra term is a polynomial of degree
n − d, where n is the degree of the numerator and d is the degree of the denominator.
A polynomial of degree 0 is a constant, a polynomial of degree 1 has the form
Ax +B, a polynomial of degree 2 has the form Ax 2 +Bx +C, and so on. For example,ifthenumeratorhasdegree3andthedenominatorhasdegree2,thepartialfractions
will include a polynomial of degree n −d = 3 − 2 = 1, that is a term of the form
Ax +B. If the numerator and denominator are of the same degree, the fraction is improper.
The partial fractions will include a polynomial of degree n −d = 0, that is a
constantterm.

44 Chapter 1 Review of algebraic techniques
Letthedegreeofthenumeratorbenandthedegreeofthedenominatorbed.Ifn d
thenthefractionisimproper.Improperfractionshavepartialfractionsinadditionto
thosegeneratedbythefactorsofthedenominator.Theseadditionalpartialfractions
take the form ofapolynomial of degreen−d.
Example1.42 Express as partialfractions
4x 3 +10x+4
2x 2 +x
Solution Thedegreeofthenumeratoris3,thatisn = 3.Thedegreeofthedenominatoris2,that
isd = 2.Thus,the fraction isimproper.
Nown −d = 1andthisisameasureoftheextenttowhichthefractionisimproper.
Thepartialfractionswillincludeapolynomialofdegree1,thatisAx +B,inadditionto
the partialfractions generated by the factors of the denominator.
The denominator factorizes tox(2x +1). These factors generate partial fractions of
the form C x + D
2x+1 . Hence
4x 3 +10x+4
2x 2 +x
= 4x3 +10x+4
x(2x +1)
Multiplying byxand 2x +1 yields
=Ax+B+ C x +
D
2x+1
4x 3 +10x +4 = (Ax +B)x(2x +1) +C(2x +1) +Dx (1.12)
The constantsA,B,C andDcan now beevaluated.
Puttingx = 0 into Equation (1.12) gives
4 =C
Puttingx = −0.5 into Equation (1.12) gives
−1.5 = − D 2
D = 3
Equating coefficients ofx 3 gives
4=2A
A = 2
Equating coefficients ofxgives
Hence
10=B+2C+D
B=−1
4x 3 +10x+4
2x 2 +x
=2x−1+ 4 x + 3
2x+1

1.7 Partial fractions 45
EXERCISES1.7
1 Calculate the partial fractionsofthe following
fractions:
6x+14 7−2x
(a)
x 2 (b)
+4x+3 x 2 −x−2
3x+6 8 −x
(c)
2x 2 (d)
+3x 6x 2 −x−1
(e)
13x 2 +11x +2
(x +1)(2x +1)(3x +1)
2 Calculatethe partialfractionsofthe following
fractions:
2x+7 4x−5
(a)
x 2 (b)
+6x+9 x 2 −2x+1
(c)
(d)
3x 2 +8x+6
(x 2 +2x +1)(x +2)
3x 2 −3x−2
(x 2 −1)(x −1)
(e)
3x 2 +7x+6
x 3 +2x 2
3 Expressthe following aspartialfractions:
(a)
x 2 +x+2
(x 2 +1)(x +1)
(b)
(c)
(d)
(e)
5x 2 +11x+5
(2x +3)(x 2 +5x +5)
4x 2 +5
(x 2 +1)(x 2 +2)
18x 2 +7x +44
(2x −3)(2x 2 +5x +7)
2x
(x 2 −x+1)(x 2 +x+1)
4 Express the following fractions aspartialfractions:
(a)
(c)
(d)
(e)
x 2 +7x+13
x +4
x 2 +8x+2
x 2 +6x+1
x 3 −2x 2 +3x−3
x 2 −2x+1
2x 3 +2x 2 −2x−1
x 2 +x
(b)
12x−4
2x−1
Solutions
1 (a)
(c)
(e)
2 (a)
(c)
(d)
(e)
3 (a)
2
x +3 + 4
x +1
2
x − 1
2x+3
(b)
(d)
2
x +1 + 1
2x+1 − 1
3x+1
2
x +3 + 1
(x +3) 2 (b)
1
x +1 + 1
(x +1) 2 + 2
x +2
2
x −1 − 1
(x −1) 2 + 1
x +1
2
x + 3 x 2 + 1
x +2
1
x +1 + 1
x 2 +1
1
x −2 − 3
x +1
3
2x−1 − 5
3x+1
4
x −1 − 1
(x −1) 2
(b)
(c)
(d)
(e)
2x
x 2 +5x+5 + 1
2x+3
1
x 2 +1 + 3
x 2 +2
5
2x−3 + 4x−3
2x 2 +5x+7
1
x 2 −x+1 − 1
x 2 +x+1
4 (a) x+3+ 1
x +4
(c) 1+ 2x+1
x 2 +6x+1
(d) x+ 2
x −1 − 1
(x −1) 2
(e) 2x − 1 x − 1
x +1
(b) 6+ 2
2x−1

46 Chapter 1 Review of algebraic techniques
TechnicalComputingExercises1.7
1 Useatechnical computing language suchas
MATLAB ® to verifythe solutionsto the problemsin
Exercises1.7. InMATLAB ® ,the functionresidue
calculates the partialfraction expansion. Forexample,
exercise1(a)wouldbesolvedbytypingthefollowing:
b = [6 14];
a=[143];
[r,p,k] = residue(b, a)
Notice how the coefficients ofthe numerator are input
in the formb = [6 14];thisisknown asarow
vector. The concept ofavector willbe discussed in
later chapters. Fornow itisadequate to treat thisasa
horizontal list ofnumbers which are passed to
MATLAB ® in a specific order.
Similarly, the coefficients ofthe denominator are
input bya = [1 4 3].
Each vectoris arranged with the coefficient ofthe
highestpower ofxfirst.
Theresult is:
r =
p =
4.0000
2.0000
-1
-3
k =
[]
Examining the solution we notethat the output for
bothrand pisarranged asavertical list.Thisway of
representing the outputis known asacolumn vector.
We note that the numbersreturnedin columnvector p
have anegative sign.Thisis because the result
calculated containsthepoles ofthe partial fraction
expansion. These are values ofthe variable which
make the denominator ofthe fractionzero. The
significance ofthiswillbecome clear later in the text
but fornow itisadequate to note the difference in
sign from what might have been expected.
1.8 SUMMATIONNOTATION
Inengineeringweoftenwanttomeasurethevalueofavariable,suchascurrent,voltage
or pressure.
Suppose we make three measurements of a variablex. We can label these measurementsx
1
,x 2
andx 3
. Inthis context, the numbers 1,2,3arecalledsubscripts.
Inmathematics,theGreeklettersigma,written ∑ ,standsfora‘sum’.Forexample,
the sumx 1
+x 2
+x 3
is written
3∑
k=1
x k
Notethatthesubscriptkrangesfrom1to3.Askrangesfrom1to3,x k
becomesx 1
then
x 2
and thenx 3
and the sigmasign tells us toadd up these quantities.
Ingeneral,
N∑
x k
=x 1
+x 2
+···+x N
k=1
This notation is often used to express some of the fundamental equations of electrical
circuitanalysis. Sometimes ‘Summation Notation’ isknown as ‘Sigma Notation’.

1.8 Summation notation 47
Engineeringapplication1.8
Kirchhoff’scurrentlaw
Kirchhoff’scurrentlaw,oftenabbreviatedtoKCL,providesoneofthefundamental
equationsforanalysingelectricalcircuits.Thelawstatesthatthesumofthecurrents
flowing out of any junction, or node, in a circuit must equal the sum of the currents
flowing into it.
This principle is intuitive as it has a direct analogy with fluid flow in connected
water pipes. Currents flowing into a junction are considered positive; those flowing
out of a junction are negative. It is then valid to say that the sum of the currents
at a junction is zero. If there are N currents at the junction, denoted I 1
,I 2
,...,I N
,
then
I 1
+I 2
+I 3
+···+I N−1
+I N
=0
This can be expressed using the summation notation as
N∑
I k
=0
k=1
HereI k
means ‘the current,I, in branchk’. The first equation can be produced from
the summation notation by first substitutingk = 1, thenk = 2, right up tok = N.
Theexpressionbelowthesummationsymboltellsyouwheretostartandthevariable
to be substituted, and the number above the summation symbol indicates where to
stopcounting.Summationnotationisaverycompactandprecisewayofexpressing
KCL forany number ofcurrents atanode.
Consider the node shown inFigure 1.5.
Branch 1 Branch 3
1 A
2 A
3 A 2 A Figure1.5
Acircuit node with four separate branches.The currents are
Branch 2 Branch 4 given in amperes(oramps,A).
It can be seen that the total current flowing into the node is 1 + 3 = 4 amps. The
current flowing outof the node is2 +2 = 4 amps. Clearly,
Total current flowing into node =total currentflowing outof node
Alternatively, using the summation formof KCL wehave
4∑
I k
=I 1
+I 2
+I 3
+I 4
=0=1+3−2−2
k=1
Notethatforcurrentsflowingoutofthenodeanegativesignisusedandforcurrents
flowing into the node a positive sign is used. This is equivalent to considering the
currents separately as inward and outward flowing currents and equating the two.
Suppose for a moment that we did not know the current in branch 4 and,
furthermore, itwas notlabelled with anarrow toshow the direction of current flow.
➔

48 Chapter 1 Review of algebraic techniques
This situation is likely to occur in a circuit problem in electronics. There are
two options for labelling the current flow direction, and these are summarized in
Figure 1.6.
1 A
2 A
1 A
2 A
3 A I 4 3 A I 4 Figure1.6
Two different ways ofdefining the
Branch 4
Branch 4 currentdirection in Branch4.
1+3−2−I 4
=0 1+3−2+I 4
=0
I 4
=2 I 4
=−2
Note that the two solutions are both correct butI 4
= −2 has a negative sign, which
simply indicates that the current flows in the opposite direction to the arrow drawn
ontheright-handdiagram.Itdoesnotmatterwhichwayroundthearrowismarked,
as long as weobserve the sign.
Engineeringapplication1.9
Kirchhoff’svoltagelaw
Kirchhoff’s voltage law, often abbreviated to KVL, provides another of the fundamental
equations for analysing electrical circuits. The law states that the sum of the
voltages around a closed loop equals zero. It is often written down in the form of a
summation, as follows:
N∑
V k
=0
k=1
For the circuit shown in Figure 1.7 there are three possible loops to which we
could apply KVL.
V d
V a
V c
V b
V e
Figure1.7
A simple circuitto illustrate Kirchhoff’svoltage law.
In this example an ideal voltage source and resistors are used, although any components
could be substituted as KVL applies universally. Note that we ‘walk around’

1.8 Summation notation 49
thecircuitwhenwritingdowntheequations.Ifthearrowisinthedirectionoftravel
then it is given a positive sign; if it opposes the direction of travel it is given a
negative sign.
The equations are
V b
V a
V d
V d
V c
V e
V c
V a
V b
V e
V a
−V b
−V c
=0 V a
−V b
−V d
−V e
=0 V c
−V d
−V e
=0
If the equations are solved and the voltage has a negative sign it indicates that the
polarityisoppositetothedirectionofthevoltagearrowdrawnonthediagram.KVL
and KCL are the fundamental circuit laws that allow networks of electronic components
to be mathematically analysed. Although they are simple in concept they are
very powerful techniques.
EXERCISES1.8
1 Writeoutfullywhatismeantbyeachofthefollowing
expressions:
(a) ∑ 4
k=1 x k
(c) ∑ 7
k=1 x k
(b) ∑ 4
i=1 x i
(d) ∑ 3
k=1 x 2 k
(e) ∑ 4
j=1 (x j −2) 3 (f) ∑ 3
n=0 (2n +1) 2
4 Determine the current,I,at eachofthe following
circuitnodes:
(a)
3 A
I
1 A
(b)
3 A
I
1 A
2 Writeoutfully
(a) ∑ 4
k=1 (−1) k k (b) ∑ 5
k=1 (−1) k+1 k 2
3 Writethe followingsums more concisely byusing
sigma notation:
(c)
3 A
I
1 A
(d)
I 1
I 2
I 3
I
(a) 1 3 +2 3 +3 3 +···+10 3
1
(b)
1 − 1 2 + 1 3 − 1 4 +···− 1 12
(c) 1+ 1 3 + 1 5 + 1 7

50 Chapter 1 Review of algebraic techniques
5 FindV a in each ofthe followingcircuits:
(a)
6 FindV a andV b usingKVL.
1 V
1 V
2 V
3 V
V a
1 V
V a
V b
(b)
−1 V
2 V
V a
Solutions
1 (a) x 1 +x 2 +x 3 +x 4 (b)x 1 +x 2 +x 3 +x 4
(c) x 1 +x 2 +x 3 +x 4 +x 5 +x 6 +x 7
(d) x 2 1 +x2 2 +x2 3
(e) (x 1 −2) 3 + (x 2 −2) 3 + (x 3 −2) 3 + (x 4 −2) 3
(f) 1+3 2 +5 2 +7 2
2 (a) −1+2−3+4
(b) 1−4+9−16+25
∑
3 (a) 10
k=1 k 3
(b)
(c)
∑ 12 (−1) k+1
k=1 k
∑ 3n=0 1
2n+1 or ∑ 4 1
n=1 2n−1
4 Allsolved usingKCL
(a) 3−1−I=0 · · · I = 2
(b) 3+1−I=0 · · · I = 4
(c) 3+1+I=0 · · · I=−4
(d) I 1 +I 2 +I 3 −I=0, · · · I=I 1 +I 2 +I 3
orI= ∑ 3
k=1 I k
5 Both solved usingKVL
(a) 2−1−V a =0 · · ·Va =1
(b) 2+(−1) +V a =0 · · ·Va =−1
6 3−1−V a =0 · · ·Va =2
3−1−1−V b =0 · ·Vb =1
orV a −1−V b =0 · · · by substitution forV a ,
V b =1
REVIEWEXERCISES1
1 Simplify eachofthe followingasfar aspossible: 2 Simplify asfar aspossible:
(a) 7 6 7 4 6 3
(b)
6 −2 (c) (3 4 ) −2 (a) x 7 x −3 (b) (x 2 ) 4 (c)
√
(d) 3 4 6 2 (e) (3 2/3 4 1/3 ) 6 (f) 10−3
10 −4 (d) (y −2 ) −1 (e) y 1/3 yy 2
(√ ) −1 x
x

Review exercises 1 51
3 Removethe brackets and simplify:
( ) −1
(a) (2x 2 y) 3 (b) (6a 2 b 3√ c) 2 (c)
y −2
2
(d) (x 2 y −1 ) 0.5
( )
3 −1 x −2 −2
(e)
y −3
4 Expressthe following astheir partialfractions:
(a)
(c)
(e)
3x+11
(x −3)(x +7)
6x 2 −2
2x 2 −x
4x−11
2x 2 +15x+7
(b)
(d)
−3−x
x 2 −x
2x 2 −x−7
(x +1)(x −1)(x +2)
5 Convert the following intoasingle fraction:
(a)
(b)
(c)
1
x + 3
x +2 + 6
8x+4
1
s + 2 s 2 + 3s+4
8s+6
6
s + 10
s 2 − s +1
(s +2)(s +3) + s −1
(s +4)(s +3)
6 Expressthe following aspartialfractions:
(a)
(c)
(e)
(f)
(g)
(i)
(k)
(m)
(o)
(q)
(s)
5x
(x +1)(2x −3)
y +3
y 2 +3y+2
2z 2 +15z +30
(z +2)(z +3)(z +6)
(b)
(d)
24x 2 +33x +11
(2x +1)(3x +2)(4x +3)
3x+2
x 2 +5x+6
1
t 2 +3t+2
s +3
(s +1) 2 (h) 2k2 +k+1
k 3 −k
x 3
x 2 +1
s 2
s 2 +1
6d 2 +15d+8
(d +1) 2 (d +2)
−y−1
(y 2 +1)(y −1)
t 2 +t−2
(t −2) 2 (t +1)
x 3 +4x 2 +7x+5
x 2 +3x+2
(j)
(l)
t +5
(t +3) 2
8x−15
4x 2 −12x+9
(n) 2x2 +x+3
x 2 +2x+1
(p)
s 2 −8s−5
(s 2 +s+1)(s−4)
(r) 2s3 +3s 2 −s−4
s 2 +s−1
7 Solve the following quadraticequationsusingthe
quadraticformula:
(a) x 2 +10x+2=0
(b) y 2 −6y−3=0
(c) 2t 2 +2t−9=0
(d) 3z 2 −9z−1=0
(e) 5v 2 +v−6=0
8 Solve the quadratic equationsin Question7by
completing the square.
9 Solve
x 3 −4x 2 −25x+28=0
givenx = 7 is aroot.
10 Solve the following inequalities:
(a) 6t−14
(c) 1−2v<v+4
(e) (x−2) 2 36
x −3
(g)
x +1 0
x
(i)
2 3 x
(b) −63r6
(d) 2 x −2
3
(f)x 2 −2x−3<0
(h)x2 −8x+50
(j) x2 −2x−3
x −5
11 Express eachfraction in its simplestform.
3ab 2
(a)
12ab
3x 2 +3x
(d)
6x 2 −3x
(b) 6x2 y 2 z
3xy 3 (c)
z
xyz −2x 2 y 2 z
(e)
x 2 y 2 −2x 3 y 3
12 Express eachfraction in its simplestform.
x 2 +2x−15
(a)
x 2 −2x−3
2x 2 +7x−4
(c)
2x 2 −3x+1
(e)
x 3 −2x 2 +x−2
x 3 +x 2 +x+1
> 0
9t+6
12−3t
y 2 +4y−12
(b)
y 2 +13y+42
3x 2 t +3xt −3t
(d)
4x 2 z+4xz−4z
13 Express asasingle fraction in itssimplestform.
x +1
(a)
x +6 × x +6
x +2
3x−6
(b)
xy+2y × xy+3y
4x−8
(c)
(d)
(e)
x 2 −1
4
÷ x −1
6
x 2 −9x
x +1 ÷ x −9
x 3 +x 2
x 2 −5x−6
x 2 +x−42 ÷ x2 −1
x 2 +6x−7

52 Chapter 1 Review of algebraic techniques
14 Expressasasingle fraction:
(a)
(b)
5
x +6 + 3
x +1
3x
2x−1 − 4
x +5
(c)
x +1
x 2 −5x−6 + 5x
x +3
(d) x+1+ 2
x −3
(e) 2x−3+ 1
x +1 −
x
x 2 +1
Solutions
1 (a) 7 10 (b) 6 5 (c) 3 −8
(d) 3 2 6 (e) 3 4 4 2 (f) 10
2 (a) x 4 (b) x 8 (c) √ x
(d) y 2 (e) y 10/3
3 (a) 8x 6 y 3 (b) 36a 4 b 6 c (c) 2y 2
(d) xy −0.5 9x 4
(e)
y 6
2
4 (a)
x −3 + 1
x +7
3
(b)
x − 4
x −1
(c) 3+ 2 x − 1
2x−1
2
(d)
x +1 − 1
x −1 + 1
x +2
3
(e)
x +7 − 2
2x+1
5 (a)
(b)
(c)
6 (a)
(b)
(c)
(d)
(e)
(f)
19x 2 +22x +4
2x(x +2)(2x +1)
3s 3 +12s 2 +22s +12
2s 2 (4s +3)
2(3s 4 +30s 3 +120s 2 +202s +120)
s 2 (s +2)(s +3)(s +4)
1
x +1 + 3
2x−3
7
x +3 − 4
x +2
2
y +1 − 1
y +2
1
t +1 − 1
t +2
2
z +2 − 1
z +3 + 1
z +6
1
2x+1 + 3
3x+2 − 2
4x+3
1
(g)
s +1 + 2
(s +1) 2
1
(h)
k +1 + 2
k −1 − 1 k
(i) x−
x
x 2 +1
1
(j)
t +3 + 2
(t +3) 2
(k) 1− 1
s 2 +1
4
(l)
2x−3 − 3
(2x −3) 2
4
(m)
d +1 − 1
(d +1) 2 + 2
d +2
(n) 2− 3
x +1 + 4
(x +1) 2
(o)
y
y 2 +1 − 1
y −1
2s+1
(p)
s 2 +s+1 − 1
s −4
11
(q)
9(t −2) + 4
3(t −2) 2 − 2
9(t +1)
3
(r) 2s+1−
s 2 +s−1
(s) x+1+ 1
x +1 + 1
x +2
7 (a) −9.7958,−0.2042
(b) −0.4641,6.4641
(c) −2.6794,1.6794
(d) −0.1073,3.1073
(e) −1.2,1
8 (a) (x+5) 2 −23=0
(b) (y−3) 2 −12=0
[ (
(c) 2 t + 1 ) ] 2
− 19 = 0
2 4

Review exercises 1 53
[ (
(d) 3 z − 3 ) ] 2
− 31 = 0
2 12
[ (
(e) 5 v + 1 ) ] 2
− 121 = 0
10 100
9 x=−4,1,7
Solutionssame asforQuestion 7
10 (a)t 5 (b) −2r2
6
(c) v>−1 (d)x8
(e) x −4orx8
(f) −1<x<3
(g) x<−1orx3
(h) 4− √ 11x4+ √ 11
(i) √ 0<x 6,x − √ 6
(j) x>5or−1<x<3
b 2x
11 (a) (b) (c)
4 y
(d)
x +1
2x−1
(e)
z
xy
3t+2
4 −t
x +5
12 (a)
x +1
3t
(d)
4z
13 (a)
(c)
14 (a)
(b)
(c)
(d)
(e)
x +1
x +2
3(x +1)
2
y −2
(b)
y +7
x −2
(e)
x +1
(b)
3(x +3)
4(x +2)
8x+23
(x +1)(x +6)
3x 2 +7x+4
(2x −1)(x +5)
5x 2 −29x+3
(x −6)(x +3)
x 2 −2x−1
x −3
2x 4 −x 3 −x 2 −2x−2
(x +1)(x 2 +1)
(c)
x +4
x −1
(d) x 3 (e) 1

2 Engineeringfunctions
Contents 2.1 Introduction 54
2.2 Numbersandintervals 55
2.3 Basicconceptsoffunctions 56
2.4 Reviewofsomecommonengineeringfunctionsandtechniques 70
Reviewexercises2 113
2.1 INTRODUCTION
The study of functions is central to engineering mathematics. Functions can be used to
describe the way quantities change: for example, the variation in the voltage across an
electroniccomponentwithtime,thevariationinpositionofanelectricmotorwithtime
and the variation inthe strengthof a signalwith both position and time.
Inthischapterweintroduceseveralconceptsassociatedwithfunctionsbeforegoing
ontocatalogue anumber ofengineering functions inSection2.4.Muchofthematerial
ofSection2.4willalreadybefamiliartothereaderandsothissectionshouldbetreated
asareferencesectiontobedippedintowhenevernecessary.Anumberofmathematical
methods are also included in Section 2.4, most of which will be familiar but they have
been collected together inorder tomake the book complete.
When trying to understand a mathematical function it is always useful to sketch a
graphinordertoobtainanideaofitsbehaviour.Thereaderisencouragedtosketchsuch
graphs whenever a new function is met. Graphics calculators are now readily available
andtheymakethistaskrelativelyeasy.Ifyoupossesssuchacalculatorthenitwouldbe
useful to make use of it whenever a new function is introduced. Software packages are
alsoavailabletoallowsuchplotstobecarriedoutonacomputer.Thesecanbeusefulfor
plotting more complicated functions and ones that depend on more than one variable.
We examine functions of more than one variable inChapter 25.
Throughoutthebookwemakeuseofthetermmathematicalmodel.Whendoingso
wemeananidealizationofanengineeringsystemoraphysicalsituationsothatitcanbe
describedbymathematicalequations.Toreflectanengineeringsystemveryaccurately,
a sophisticated model, consisting of many interrelated equations, may be needed. Although
accurate, such a model may be cumbersome to use. Accuracy can be sacrificed

2.2 Numbers and intervals 55
in order to achieve a simple, easy-to-use model. A judgement is made as to when the
right blend of accuracy and conciseness is achieved. For example, the most common
mathematicalmodelforaresistorusesOhm’slawwhichstatesthatthevoltageacrossa
resistor equals the current through the resistor multiplied by the resistance value of the
resistor,thatisV =IR.However,thismodelisbasedonanumberofsimplifications.It
ignores any variation in current density across the cross-section of the resistor and assumes
a singlecurrent value isacceptable. Italsoignores the factthatifalargeenough
voltageisplacedacrosstheresistorthentheresistorwillbreakdown.Inmostcasesitis
worthaccepting these simplifications inorder toobtain a concise model.
Havingobtainedamathematicalmodel,itisthenusedtopredicttheeffectofchanging
elements or conditions within the actual system. Using the model to examine these
effects isoften cheaper, safer and more convenient than usingthe actual system.
2.2 NUMBERSANDINTERVALS
Numberscanbegroupedintovariousclasses,orsets.Theintegersarethesetofnumbers
{...,−3,−2,−1,0,1,2,3,...}
denotedby Z.Thenaturalnumbersare {0,1,2,3,...}andthissetisdenotedby N.The
positive integers, denoted by N + , are given by {1,2,3,...}. Note that some numbers
occur inmore than one set,thatisthe setsoverlap.
A rational number has the form p/q, where p andqare integers withq ≠ 0. For
example,5/2,7/118, −1/9and3/1areallrationalnumbers.Thesetofrationalnumbers
isdenotedby Q.Whenrationalnumbersareexpressedasadecimalfractiontheyeither
terminate orrecur infinitely.
}
5
can be expressed as2.5 These decimal fractions terminate,
2
1
can be expressed as0.125 thatisthey are of finitelength.
8
}
1
can be expressed as0.111111... These areinfinitely
9
1
can be expressed as 0.090909... recurring decimal fractions.
11
A number which cannot be expressed in the form p/q is called irrational. When
writtenasadecimalfraction,anirrationalnumberisinfiniteinlengthandnon-recurring.
The numbers π and √ 2 areboth irrational.
It is useful to introduce the factorial notation. We write 3! to represent the product
3 ×2×1. The expression 3! is read as ‘factorial 3’. Similarly 4! is a shorthand way of
writing4 ×3×2×1.Ingeneral, forany positive integer,n, we can write
n! =n(n −1)(n −2)(n −3)...(3)(2)(1)
Itisusefultorepresentnumbersbypointsontherealline.Figure2.1illustratessome
rational and irrational numbers marked on the real line. Numbers which can be represented
by points on the real line are known as real numbers. The set of real numbers
is denoted by R. This set comprises all the rational and all the irrational numbers. In
Chapter9weshallmeetcomplexnumberswhichcannotberepresentedaspointsonthe
–2 –– 3
5
2
–1 0 1 √ 2 2 2
– 3 p 4
Figure2.1
Both rationalandirrational numbers are representedonthe real line.

56 Chapter 2 Engineering functions
–6 –4 –1 0 2 3 4
Figure2.2
The intervals (−6,−4),[−1,2], (3,4]depicted onthe real line.
real line. The real line extends indefinitely to the left and to the right so that any real
number can be represented.
Sometimesweareinterestedinonlyasmallsection,orinterval,oftherealline.We
write [1, 3] to denote all the real numbers between 1 and 3 inclusive, that is 1 and 3 are
includedintheinterval.Thustheinterval[1,3]consistsofallrealnumbersx,suchthat
1 x3. The square brackets, [ ], are used to denote that the end-points are included
in the interval and such an interval is said to be closed. The interval (1, 3) consists of
all real numbers x, such that 1 < x < 3. In this case the end-points are not included
and the interval is said to be open. Brackets, (), denote open intervals. An interval may
be open at one end and closed at the other. For example, (1, 3] is open at the left and
closed at the right. It consists of all real numbersx, such that 1 < x 3, and is known
as a semi-open interval. Open and closed intervals can be represented on the real line.
A closed end-point is denoted by •; an open end-point is denoted by ◦. The intervals
(−6,−4), [−1,2] and (3, 4] are illustratedinFigure 2.2.
An upper bound of a set of numbers is any number which is greater than or equal
toeverynumberinthegivenset.So,forexample,7isanupperboundfortheset[3,6].
Clearly, 7 isgreater than every number inthe interval[3, 6].
Alowerboundofasetofnumbersisanynumberwhichislessthanorequaltoevery
number inthe given set.For example, 3 isalower bound forthe set (3.7,5).
Note that upper and lower bounds are not unique. Both 3 and 10 are upper bounds
for (1, 2). Both −1 and −3 are lower bounds for[0, 6].
TechnicalcomputinglanguagessuchasMATLAB ® usuallyhavefunctionsthatautomaticallygenerateasetofnumberswithinaparticularinterval.InMATLAB
® wecould
generate a set of time values,t, by typing:
t= 0:0.1:1
Thisgeneratesasetofrealnumbersfromtheinterval[0,1]storedinarowvectort,each
individual number being separated by an increment of 0.1. The values oft generated
are:
0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000
0.8000 0.9000 1.0000
2.3 BASICCONCEPTSOFFUNCTIONS
Loosely speaking, we can think of a function as a rule which, when given an input,
producesasingleoutput.Ifmorethanoneoutputisproduced,theruleisnotafunction.
Considerthefunctiongivenbytherule:‘doubletheinput’.If3istheinputthen6isthe
output. Ifxisthe input then 2x isthe output, as shown inFigure 2.3.
If the doubling function has the symbol f wewrite
f:x→2x
or more compactly,
f(x)=2x

2.3 Basic concepts of functions 57
f
f
double
double
3 6 x
2x
the input
the input
Figure2.3
Thefunction: ‘double the input’.
Thelastformisoftenwrittensimplyas f = 2x.If f (x)isafunctionofx,thenthevalue
of the function whenx = 3,forexample, iswrittenas f (x = 3) or simplyas f (3).
Example2.1 Given f (x) = 2x +1 find
(a) f(3)
(c) f(−1)
(e) f(2α)
(g) f(t+1)
(b) f(0)
(d) f(α)
(f) f(t)
Solution (a) f(3) = 2(3) +1 = 7
(b) f(0)=2(0)+1=1
(c) f(−1) =2(−1) +1 = −1
(d) f(α)isthevalueof f(x)whenxhasavalueof α,hence f(α) = 2α +1
(e) f(2α)=2(2α)+1=4α+1
(f) f(t)=2t+1
(g) f(t+1)=2(t+1)+1=2t+3
Observe from Example 2.1 that it is the rule that is important and not the letter being
used. Both f (t) = 2t + 1 and f (x) = 2x + 1 instruct us to double the input and then
add 1.
2.3.1 Argumentofafunction
Theinputtoafunctionisoftencalledtheargument.InExample2.1(d)theargumentis
α, whileinExample 2.1(e) the argument is2α.
Example2.2 Given f (x) = x , write down
5
(a) f(5x)
(b) f(−x)
(c) f(x+2) (d) f(x 2 )
Solution (a) f(5x) = 5x
5 =x (b) f(−x)=−x 5
(c) f(x+2)= x +2
5
Example2.3 Giveny(t) =t 2 +t, write down
( t
(a) y(t +2) (b) y
2)
(d) f(x 2 )= x2
5

58 Chapter 2 Engineering functions
Solution (a) y(t +2) = (t +2) 2 + (t +2) =t 2 +5t+6
( ( ) t t 2 ( t
(b) y = + =
2)
2 2)
t2 4 + t 2
2.3.2 Graphofafunction
A function may be represented in graphical form. The function f (x) = 2x is shown in
Figure2.4.Notethatthefunctionvaluesareplottedverticallyandthexvalueshorizontally.Thehorizontalaxisisthencalledthexaxis.Theverticalaxisiscommonlyreferred
toastheyaxis, sothat weoften write
y=f(x)=2x
f (x)
2
–2
1 x
Sincexandycan have a number of possible values, they are called variables:xis the
independent variable and y is the dependent variable. Knowing a value of the independent
variable,x, allows us to calculate the corresponding value of the dependent
variable,y. To show this dependence we often writey(x). The set of values thatxis allowedtotakeiscalledthedomainofthefunction.Adomainisoftenanintervalonthe
x axis. For example, if
–4
Figure2.4
Thefunction: f (x) = 2x.
f(x)=3x+1 −5x10 (2.1)
thedomainofthefunction, f,istheclosedinterval[−5,10].Ifthedomainofafunction
isnot explicitly given itis taken tobe the largestset possible. For example,
g(x)=x 2 −4 (2.2)
hasadomainof (−∞,∞)sincegisdefinedforeveryvalueofxandthedomainhasnot
been given otherwise. The set of values that the function takes on is called the range.
The range of f (x) in Equation (2.1) is [−14,31]; the range ofg(x) in Equation (2.2) is
[−4,∞).
Wenowconsiderplottingthefunction f (t) =t 2 for0 t < 100inatechnicalcomputing
language. Firstwegenerate a number set as shown inSection 2.2.
t= 0:1:100
Then we produce a graph of the function by using the MATLAB ® plot command to
give the following
plot(t, t.^2)
Example2.4 Consider the function, f,given by the rule:‘squarethe input’. This can bewritten as
f(x)=x 2
Theruleandthegraphof f areshowninFigure2.5.Thedomainof f is (−∞,∞)and
the range is[0,∞).

2.3 Basic concepts of functions 59
square
x x
the input
2
f
f(x)
9
4
–2 0 3 x
Figure2.5
Thefunction: ‘square the input’.
Manyvariablesofinteresttoengineers,forexamplevoltage,resistanceandcurrent,can
be related by means of functions. We try to choose an appropriate letter for a particular
variable; so, forexample,t is usedfor time andPforpower.
Engineeringapplication2.1
Functiontomodelthepowerdissipationinaresistor
Recall from Engineering application 1.1 that the power, P, dissipated by a resistor
depends on the current,I, flowing through the resistance,R. The relationship is
given by
P=I 2 R
The power dissipated in the resistor depends on the square of the current passing
throughit.InthiscaseI istheindependent variableandPisthedependent variable,
assumingRremainsconstant.Thefunctionisgivenbytherule:‘squaretheinputand
multiply by the constantR’, and the input to the function isI. The output from the
function isP. This isillustratedinFigure 2.6,for the casesR = 4 andR = 2.
I
square the
input and
multiply
by R
Figure2.6
The function:P =I 2 R.
P
P
16
8
4
2
R = 4
R = 2
–1 2 I
This model for a resistor only approximates the behaviour of the device. In practice,changesinthetemperatureoftheresistorleadtoslightchangesintheresistance
value. If the current through the resistor is excessively high then the resistor overheatsandispermanentlydamaged.Itnolongerhasthecorrectresistancevalue.The
amount of power that a resistor can handle depends on the materials that have been
usedinitsconstruction.Agoodcircuitdesignerwouldcalculatetheamountofpower
tobedissipatedandthenallowasuitablesafetymargintoensurethattheresistorcannot
beoverloaded.

60 Chapter 2 Engineering functions
2.3.3 One-to-many
Some rules relating input to output are not functions. Consider the rule: ‘take plus or
minus the square rootof the input’, that is
x→± √ x
Now, for example, if 4 is the input, the output is ± √ 4 which can be 2 or −2. Thus a
single input has produced more than one output. The rule is said to be one-to-many,
meaning that one input has produced many outputs. Rules with this property are not
functions. For a ruletobe a function there mustbe a single output forany given input.
Bydefiningarulemorespecifically,itmaybecomeafunction.Forexample,consider
therule:‘takethepositivesquarerootoftheinput’.Thisruleisafunctionbecausethere
is a single output for a given input. Note that the domain of this function is [0,∞) and
the range isalso[0,∞).
2.3.4 Many-to-oneandone-to-onefunctions
Consideragainthefunction f (x) =x 2 giveninExample2.4.Theinputs2and −2both
producethesameoutput,4,andthefunctionissaidtobemany-to-one.Thismeansthat
many inputs produce the same output. A many-to-one function can be recognized from
itsgraph.Ifahorizontallineintersectsthegraphinmorethanoneplace,thefunctionis
many-to-one. Figure 2.7 illustrates a many-to-one function,g(x). The inputsx 1
,x 2
,x 3
andx 4
all produce the same output.
Afunctionisone-to-oneifdifferentinputsalwaysproducedifferentoutputs.Ahorizontallinewillintersectthegraphofaone-to-onefunctioninonlyoneplace.Figure2.8
illustratesaone-to-one function,h(x).
Bothone-to-onefunctionsandmany-to-onefunctionsaresupportedintechnicalcomputinglanguages.Forexample,inMATLAB
® thefunction f (x) =x 2 canbedefinedby
using the command:
f = @(x) x^2;
Itis now possible totype:
f(3)
or
f(-3)
g(x)
h(x)
x 1 x 2 x 3 x 4 x
Figure2.7
Theinputsx 1 ,x 2 ,x 3 andx 4 allproduce the same
output,thereforeg(x) isamany-to-one function.
Figure2.8
Eachinput producesadifferent outputandso
h(x)is aone-to-onefunction.
x

2.3 Basic concepts of functions 61
both of which have the same result:
ans=9
affirming that f (x) =x 2 isamany-to-one function.
Noticethatthevariableusedbythefunctionisdefinedinbracketsafterthe@sign.This
indicatesthattheinputtothefunctionisxandthecommandcreatesafunctionhandle,
f. Giving the function a handle enables ittobe used elsewhere inthe program.
More complicated functions are usually created in a separate file and saved on the
computer’s internal storage devices. They can be easily reused to create sophisticated
programs. In MATLAB ® these files are saved with the file name extension .m and are
often termedm-files.
2.3.5 Parametricdefinitionofafunction
Functions are often expressed in the formy(x). For every value ofxthe corresponding
valueofycanbefoundandthepointwithcoordinates (x,y)canthenbeplotted.Sometimes
it is useful to expressxandycoordinates in terms of a third variable known as a
parameter.Commonlyweuset or θ todenoteaparameter.Thusthecoordinates (x,y)
of the points on a curve can be expressed inthe form
x=f(t)
y=g(t)
For example, given the parametric equations
x =t 2 y=2t 0t5
we can calculatexandyfor various values of the parametert. Plotting the points (x,y)
produces partof a curve known as a parabola.
2.3.6 Compositionoffunctions
Considerthefunctiony(x) = 2x 2 .Wecanthinkofy(x)asbeingcomposedoftwofunctions.Onefunctionisdescribedbytherule:‘squaretheinput’,whiletheotherfunction
isdescribed by the rule:‘double the input’. This isshown inFigure 2.9.
y(x)
8
x
g
square
the input
Figure2.9
Thefunction:y(x) =h(g(x)).
h
x 2 double
2x
the input
2
2
–2 1 x
Mathematically, ifh(x) = 2x andg(x) =x 2 then
y(x) = 2x 2 = 2(g(x)) =h(g(x))
The form h(g(x)) is known as a composition of the functions h and g. Note that the
compositionh(g(x)) isdifferent fromg(h(x)) as Example 2.5 illustrates.

62 Chapter 2 Engineering functions
Example2.5 If f (t) = 2t +3 andg(t) = t +1 writeexpressions for the compositions
2
(a) f(g(t))
(b) g(f(t))
( ) t +1
Solution (a) f(g(t)) = f
2
The ruledescribing the function f is: ‘double the inputand then add 3’. Hence,
So
( ) ( ) t +1 t +1
f = 2 +3=t+4
2 2
f(g(t))=t+4
(b) g(f(t)) =g(2t +3)
The ruleforgis: ‘add 1 tothe inputand then divide everything by 2’. So,
Hence
g(2t +3) = 2t+3+1
2
g(f(t))=t+2
Clearly f (g(t)) ≠g(f(t)).
=t+2
2.3.7 Inverseofafunction
Considerafunction f (x).Itcanbethoughtofasacceptinganinputx,andproducingan
output f (x). Suppose now that this output becomes the input to the functiong(x), and
the output fromg(x) isx, thatis
g(f(x))=x
We can think ofg(x) as undoing the work of f (x). Figure 2.10 illustrates this situation.
Theng(x) is the inverse of f, and is written as f −1 (x). Since f −1 (x) undoes the work
of f(x)wehave
f (f −1 (x)) = f −1 (f(x))=x
x
f
f (x)
g
g( f (x))=x
Figure2.10
The functiongisthe inverse of f.

2.3 Basic concepts of functions 63
Example2.6 If f (x) = 5x verifythat the inverse of f isgiven by f −1 (x) = x 5 .
Solution The function f receives an input of x, and produces an output of 5x. Hence when the
inverse function, f −1 , receives an input of 5x, itproduces anoutput ofx, thatis
f −1 (5x) =x
We introduce a new variable,z, given by
so
z=5x
Then
x = z 5
f −1 (z)=x= z 5
Writing f −1 withxas the argument gives
f −1 (x) = x 5
Example2.7 If f (x) = 2x +1,find f −1 (x).
Solution The function f receives an input of x and produces an output of 2x + 1. So when the
inverse function, f −1 , receives aninputof2x +1 itproducesanoutput ofx, thatis
f −1 (2x+1)=x
We introduce a new variable,z, defined by
z=2x+1
Rearranginggives
So
x = z −1
2
f −1 (z)=x= z −1
2
Writing f −1 withxas the argument gives
f −1 (x) = x −1
2

64 Chapter 2 Engineering functions
Example2.8 Giveng(x) = x −1 find the inverse ofg.
2
Solution We knowg(x) = x −1 ( ) x −1
2 , and sog−1 =x.Lety= x −1 so that
2 2
But,
and so
g −1 (y) =x
x=2y+1
g −1 (y)=2y+1
Using the same independent variable as forthe functiong, we obtain
g −1 (x)=2x+1
WenotethattheinversesofthefunctionsinExamples2.7and2.8arethemselvesfunctions.
They are called inverse functions. The inverse of f (x) = 2x + 1 is f −1 (x) =
x −1 x −1
, and the inverse ofg(x) = isg −1 (x) = 2x +1. This illustrates the important
point that if f (x) andg(x) are two functions and f (x) is the inverse ofg(x), then
2 2
g(x) is the inverse of f (x). It is important to point out that not all functions possess an
inverse function. Consider f (x) =x 2 , for −∞ <x<∞.
The function, f, is given by the rule: ‘square the input’. Since both a positive and
negative value of x will yield the output x 2 , the inverse rule is given by: ‘take plus or
minus the square root of the input’. As discussed earlier, this is a one-to-many rule and
soisnotafunction.Clearlynotallfunctionshaveaninversefunction.Infact,onlyoneto-onefunctionshaveaninversefunction.Supposewerestrictthedomainof
f (x) =x 2
suchthatx 0.Then f isaone-to-onefunctionandsothereisaninversefunction.The
inverse function is f −1 (x) given by
Clearly,
f −1 (x) = + √ x
f −1 (f(x))=f −1 (x 2 )=x
wherexisthepositivesquarerootofx 2 .Restrictingthedomainofamany-to-onefunction
so that a one-to-one function results is a common technique of ensuring an inverse
function can be found.
2.3.8 Continuousandpiecewisecontinuousfunctions
Wenowintroduceinaninformalwaytheconceptofcontinuousandpiecewisecontinuousfunctions.AmorerigoroustreatmentfollowsinChapter10afterwehavediscussed
limits.Figure2.11showsagraphof f (x) = 1 x .Notethatthereisabreak,ordiscontinuity,
inthe graph atx = 0.The function f (x) = 1 issaidtobe discontinuousatx = 0.
x

2.3 Basic concepts of functions 65
f (x)
f(t)
3
2
x
1
Figure2.11
Thefunction f (x) = 1 has a discontinuity at
x
x=0.
1 3 t
Figure2.12
Thefunction f (t)isapiecewise continuous
functionwith adiscontinuity att = 1.
If the graph ofafunction, f (x), containsabreak, then f (x) isdiscontinuous.
Afunctionwhosegraph has no breaksisacontinuousfunction.
Sometimes a function is defined by different rules on different intervals of the domain.
For example,consider
g(t)
6
f(t)=
{
2 0t<1
t 1t3
Thedomainis[0,3]buttheruleon[0,1)isdifferenttothaton[1,3].Thegraphof f (t)
is shown in Figure 2.12. Recall the convention of using • to denote that the end-point
isincludedand ◦todenotetheend-pointisexcluded.Notethat f (t)hasadiscontinuity
att = 1. Each component, or piece, of the graph is continuous and f (t) is said to be
piecewisecontinuous.
2
Apiecewisecontinuousfunctionhasafinitenumberofdiscontinuitiesinanygiven
interval.
1 3 t
Figure2.13
The functiong(t)is a
continuous function
on (0,3).
Notallfunctionsdefineddifferentlyondifferentintervalsarediscontinuous.Forexample,
{
2 0<t<1
g(t) =
2t 1t<3
isacontinuousfunction onthe interval (0,3), asshown inFigure 2.13.
2.3.9 Periodicfunctions
Aperiodicfunctionisafunctionwhichhasadefinitepatternwhichisrepeatedatregular
intervals. More formally wesay a function, f (t),isperiodicif
f(t)=f(t+T)
forall values oft. Theconstant,T, isknown asthe period ofthe function.

66 Chapter 2 Engineering functions
Example2.9 Figure2.14illustratesaperiodicwaveform.Itisoftenreferredtoasatriangularwaveform
because of its shape. The form of the function is repeated every two seconds, that
is
f(t)=f(t+2)
and so the function is periodic. The period is 2 seconds, that isT = 2. Note that this
function iscontinuous.
f(t)
–2 –1 0 1 2 3 4 t
Figure2.14
The triangularwaveform is aperiodic function.
Engineeringapplication2.2
Saw-toothwaveform
Figure 2.15 illustrates asaw-tooth voltage waveform. It is called a saw-tooth waveform
because its shape is similar to that of the teeth on a saw. It has many uses in
electronicengineering.Oneusewouldbetoprovideasignaltosweepabeamofelectrons
across a cathode ray tube in a uniform way and then quickly move the beam
backtothestartagain.Thistechniqueisusedinananalogueoscilloscopeandforms
a signalfor the time base.
The form of the function isrepeated every threeseconds, that is
v(t) = v(t +3)
v(t)
1
–3 0 3
6 9 t
Figure2.15
Thesaw-tooth waveform isaperiodicfunction.

2.3 Basic concepts of functions 67
Technical computing languages often have a range of built-in functions for producing
waveforms. Sometimes specialist functions are provided in a separate softwarepackage.InMATLAB
® ,thesesoftwarepackagesareknownastoolboxes.The
signalprocessingtoolboxhasafunctionforgeneratingsaw-toothwaves.Thiscanbe
accessed by typing, forexample:
t=(-2*pi:0.1:2*pi);
plot(t, sawtooth(t));
This will plot two periods of a saw-tooth wave. The first line generates a set of time
values −2π ≤t < 2πinavectorformwithaspacingof0.1betweeneachpoint.The
secondlineplotst againsttheresultofpassingthevectort tothesawtoothfunction.
The sawtooth function always produces a wave with a period of 2π. It highlights
the need to read the manual pages carefully before using a function to understand
how itwill behave.
Engineeringapplication2.3
Squarewaveform
Periodic functions may be piecewise continuous. Consider the functiong(t) defined
by
{
1 0t<1
g(t) = period = 2
0 1t<2
Thefunctiong(t)isperiodicwithperiod2.Agraphofg(t)isshowninFigure2.16.
This function is commonly referred to as a square waveform by engineers. In Figure
2.16 the open and closed end-points have been shown for mathematical correctness.Note,however,thatengineerstendtoomitthesewhensketchingfunctionswith
discontinuities and usually they use a vertical line to show the discontinuity. This
reflectsthefactthatnopracticalwaveformcaneverchangeitslevelinstantaneously:
evenveryfastrisingwaveformsstillhaveafiniterisetime.Thefunctionhasdiscontinuities
att = ...,−3,−2,−1,0,1,2,3,4,5,....
g(t)
1
–5 –4 –3 –2 –1 0
1 2 3 4 5 t
Figure2.16
The functiong(t)is both piecewise continuous and periodic.
The square waveform is often used in electronic engineering, particularly in digital
electronic systems. One example is the clock signal that is generated to ensure that
all of the digital electronic circuits switch around the same time and so remain in
synchronisation.

68 Chapter 2 Engineering functions
EXERCISES2.3
1 Representthe followingintervals onthe real line:
(a) [1,3] (b) [2,4)
(c) (0,3.5) (d) [−2,0)
(e) (−1,1] (f) 2x<4
(g) 0<x<2 (h) −3x −1
(i) 0x<3
2 Describe the rule associated with the following
functions,sketch their graphs and state their domains
and ranges:
(a) f(x) =2x 2
(b)f(x)=x 2 −1
(c)g(t)=3t−4
(d)y(x) =x 3
0x
0t
(e)f(t)=0.5t+2 −2t10
(f)z(x)=3x−2
3 Iff(x)=5x+4,find
(a) f(3)
(b) f(−3)
(c) f(α)
(d) f(x+1)
(e) f(3α)
(f) f(x 2 )
4 Ifg(t) =5t 2 −4,find
(a) g(0)
(b) g(2)
(c) g(−3)
(d) g(x)
(e) g(2t −1)
3x8
5 Thereactance,X C ,offered by acapacitor is given by
X C = 1 ,where f isthe frequency ofthe applied
2πfC
alternating current, andC is the capacitance ofthe
capacitor. IfC = 10 −6 F,findX C when f = 50 Hz.
6 Classifythe functions in Question2asone-to-oneor
many-to-one.
7 Find the inverse ofthe followingfunctions:
(a) f(x)=x+4
(b) g(t) =3t +1
(c) y(x) =x 3
(d) h(t) = t −8
3
(e) f(t) = t −1
3
(f) h(x) =x 3 −1
(g) k(v)=7−v
(h) m(n) = 3 1 (1 −2n)
8 Givenf(t) =2t,g(t) =t −1andh(t) =t 2 write
expressions for
(a) f(g(t)) (b) f(h(t))
(c) g(h(t)) (d) g(f(t))
(e) h(g(t)) (f) h(f(t))
(g) f(f(t)) (h) g(g(t))
(i) h(h(t)) (j) f(g(h(t)))
(k) g(f(h(t))) (l) h(g(f(t)))
9 Givenf(t) =t 2 +1,g(t) =3t +2andh(t) = 1 t ,
write expressions for
(a) f(g(t)) (b) f(h(t))
(c) g(h(t)) (d) h(f(t))
(e) f(g(h(t)))
10 Given f(t) =2t,g(t) =2t +1,h(t) =1−3t,write
expressions forthe following:
(a) f −1 (t) (b)g −1 (t) (c)h −1 (t)
11 Givena(x) = 3x −2,b(x) = 2 x ,c(x)=1+1 x write
expressions for
(a)a −1 (x) (b)b −1 (x) (c)c −1 (x)
12 Givenf(t) =2t +3,g(t) =3tandh(t) = f(g(t))
write expressions for
(a) h(t)
(b) f −1 (t)
(c) g −1 (t)
(d) h −1 (t)
(e) g −1 (f −1 (t))
Whatdoyou noticeabout(d)and(e)?
13 Sketch the following functions:
{
t 0t3
(a) f(t) =
3 3<t4
{
2 −x 0x<1
(b) g(x) =
2 1x3
{
1 −t 0t1
(c) a(t) =
t−1 1<t2
⎧
⎪⎨
2 0 x1
(d) b(x) = 1 1<x2
⎪⎩
3−x 2<x3

2.3 Basic concepts of functions 69
14 Sketch
f(t)=
{
t 0t<2
5−2t 2t<3
Isthe function piecewise continuous orcontinuous?
State,ifthey exist, the position ofany discontinuities.
15 Thefunctionh(t)is defined by
{
2 −t 0t<2
h(t) =
2t−4 2t3
andh(t)has period 3.Sketchh(t)onthe interval
[0,6].
16 The functiong(t)is defined by
{
1 0 t1
g(t) =
2−t 1<t<2
andg(t)hasperiod 2.Sketchg(t)onthe interval
[−1,4].State any pointsofdiscontinuity.
Solutions
2 (a) Square the input and thenmultiply by2; domain
(−∞, ∞),range [0, ∞)
(b) Square the input, then subtract1; domain [0,∞),
range [−1,∞)
(c) Multiply input by3and subtract4; domain
[0, ∞),range [−4,∞)
(d) Cube the input;domain (−∞, ∞),range
(−∞,∞)
(e) Multiply input by0.5and then add2; domain
[−2,10], range [1,7]
(f) Multiply input by3and then subtract2; domain
[3,8],range [7,22]
3 (a) 19 (b) −11 (c) 5α +4
(d) 5x+9 (e) 15α+4 (f) 5x 2 +4
4 (a) −4 (b) 16 (c) 41
(d) 5x 2 −4 (e) 20t 2 −20t +1
5 3183ohms
8 (a) 2(t −1) (b) 2t 2 (c) t 2 −1
(d) 2t −1 (e) (t −1) 2 (f) 4t 2
(g) 4t (h) t −2 (i) t 4
(j) 2(t 2 −1) (k) 2t 2 −1 (l) (2t −1) 2
9 (a) 9t 2 1
+12t +5 (b)
t 2 +1
3
(c)
t +2 (d) 1
t 2 +1
9
(e)
t 2 + 12 +5
t
t
10 (a) (b) t −1
2 2
x +2 2
11 (a) (b)
3 x
12 (a) 6t +3 (b) t −3
2
(d) t −3 (e) t −3
6 6
13 SeeFigureS.1.
1 −t
(c)
3
1
(c)
x −1
t
(c)
3
6 (a) many-to-one (b) one-to-one
(c) one-to-one
(e) one-to-one
7 (a)f −1 (x)=x−4
(b) g −1 (t) = t −1
3
(c) y −1 (x) =x 1/3
(d) h −1 (t) =3t +8
(e) f −1 (t)=3t+1
(f) h −1 (x) = (x +1) 1/3
(g) k −1 (v) =7−v
(h) m −1 (n) = 1−3n
2
(d) one-to-one
(f) one-to-one
f
3
2
1
(a)
0
a
1
1
2 3 4 t
(c)
0 1 2 t
FigureS.1
(b)
(d)
g
2
1
0 1 2 3 x
b
2
1
0 1 2 3 x

70 Chapter 2 Engineering functions
14 Piecewise continuous; discontinuity att = 2. See
FigureS.2.
f
2
1
16 Discontinuitiesatt = 0,2.See FigureS.4.
g
1
–1 0 1 2 3 t
FigureS.2
15 See FigureS.3.
h
2
1
–1 0 1 2 3 4
FigureS.4
t
0 1 2 3 4 5 6
t
FigureS.3
2.4 REVIEWOFSOMECOMMONENGINEERINGFUNCTIONS
ANDTECHNIQUES
This section provides a catalogue of the more common engineering functions. The important
properties and definitions are included together with some techniques. It is intendedthatreaderswillrefertothissectionforrevisionpurposesandastheneedarises
throughout the rest of the book.
2.4.1 Polynomialfunctions
Apolynomial expressionhas the form
a n
x n +a n−1
x n−1 +a n−2
x n−2 +···+a 2
x 2 +a 1
x+a 0
wherenis a non-negative integer,a n
,a n−1
,...,a 1
,a 0
are constants andxis a variable.Apolynomial
function,P(x), has the form
P(x) =a n
x n +a n−1
x n−1 +a n−2
x n−2 + ··· +a 2
x 2 +a 1
x +a 0
(2.3)
Examples of polynomial functions include
P 1
(x)=3x 2 −x+2 (2.4)
P 2
(z) =7z 4 +z 2 −1 (2.5)
P 3
(t) =3t +9 (2.6)
P 4
(t) =6 (2.7)

2.4 Review of some common engineering functions and techniques 71
wherex,zandt are independent variables. It is common practice to contract the term
polynomial expression to polynomial. By convention, a polynomial is usually written
with the powers either increasing or decreasing. For example,
3x+9x 2 −x 3 +2
would be writtenas either
−x 3 +9x 2 +3x+2 or 2+3x+9x 2 −x 3
Thedegreeofapolynomialorpolynomialfunctionisthevalueofthehighestpower.
Equation(2.4)hasdegree2,Equation(2.5)hasdegree4,Equation(2.6)hasdegree1and
Equation(2.7)hasdegree0.Equation(2.3)hasdegreen.Polynomialswithlowdegrees
have special names (seeTable 2.1).
Table2.1
Polynomial Degree Name
ax 4 +bx 3 +cx 2 +dx +e 4 Quartic
ax 3 +bx 2 +cx +d 3 Cubic
ax 2 +bx +c 2 Quadratic
ax +b 1 Linear
a 0 Constant
Typical graphs of some polynomial functions are shown inFigure 2.17.
P(x)
Degree 2
Degree 3
x
Degree 0
Degree 1
Figure2.17
Some typical polynomials.
Engineeringapplication2.4
Ohm’slaw
RecallfromEngineeringapplication1.1thatthecurrentflowingthrougharesistoris
related tothe voltage applied across itby Ohm’s law. The equation is
V=IR
whereV = voltage across the resistor;
I = currentthrough the resistor;
R = resistance value of the resistor,which isaconstant
foragiven temperature.
➔

72 Chapter 2 Engineering functions
Note that the voltage is a linear polynomial function with I as the independent
variable.
This equation is only valid for a finite range of currents. If too much voltage is
appliedtotheresistor,thenthecurrentflowingthroughtheresistorbecomessufficient
forthe resistortooverheat and breakdown.
Engineeringapplication2.5
Anon-idealvoltagesource
An ideal voltage source has zero internal resistance and its output voltage,V, is independent
of the load applied to it; that is,V remains constant, independent of the
current it supplies. It is called an ideal voltage source because it is difficult to create
such a source in practice; it is in effect an abstraction that is useful when developing
engineering models of real electronic systems. A non-ideal voltage source has
aninternal resistance. Due tothis internal resistance, the output voltage from such a
source is reduced when current is drawn from the source. The voltage reduction increasesasmorecurrentisdrawn.Figure2.18showsanon-idealvoltagesource.Itis
modelledasanidealvoltagesourceinserieswithaninternalresistorwithresistance
R s
. The output voltage of the non-ideal voltage source is v o
while v R
is the voltage
drop across the internal resistor andiis the load current. Using Kirchhoff’s voltage
law,
V=v R
+v o
and hence by Ohm’s law,
V =iR s
+v o
v o
=V−iR s
Note thatV and R s
are constants and so the output voltage is a linear polynomial
function with independent variable i. The equation gives the output voltage across
theloadasafunctionofthecurrentthroughtheload.Theoutputcharacteristicforthe
non-idealvoltagesourceisobtainedbyvaryingtheloadresistorR L
andisplottedin
Figure2.19.Noticethattheoutputvoltageofthenon-idealvoltagesourcedecreases
as the load current increases and is equal in value to the ideal voltage source only
when there isno load current.
V
+
–
R s
v R
v o
i
R L
v o
V
gradient = –R s
i
non-ideal voltage source
Figure2.18
Anon-ideal voltage sourceconnected to aload
resistor,R L .
Figure2.19
Output characteristic ofanon-ideal
voltage source.

2.4 Review of some common engineering functions and techniques 73
Thisiswhyitiscalledanon-idealvoltagesource.Engineerswouldprefertohave
a source that maintained a constant voltage no matter how much current was drawn
but itisnotpossible tobuild such a source.
Engineeringapplication2.6
Windpowerturbines
Windturbinesareanimportantsourceofelectricalpower.Themostcommontype,
andtheoneswhichareusuallyfoundinoffshoreinstallations,resembleadesktopfan
and are called horizontal axis turbines. The wind driving a turbine blade consists
ofmanymoleculesofair,eachhavingatinyamountofmass.Thismasspassingthe
bladeareaeachsecondcarrieskineticenergy,whichisthesourceofthewindpower.
The wind power,P, can be calculated using the formula
P = 1 2 Mv2
whereM is the total mass of air per second passing the blade in kg s −1 and v is the
velocity of the air inms −1 .
The mass per second can be calculated by considering the area swept out by the
blade,A, the density ofthe air, ρ, and the velocity:
M=ρAv
This equation can besubstituted inthe power equation
P = 1 2 (ρAv)v2 = 1 2 ρAv3 (2.8)
Theavailablewindpowerthereforeincreaseswiththecubeofthevelocity.Notethat
the power isacubic polynomial function of the independent variable, v.
At 20 ◦ C the air density is approximately 1.204 kg m −3 . Consider the case of an
offshoreturbinethathasasweptareaof6362m 2 andaratedwindspeedof15ms −1 .
The maximum theoretical power atthe rated speed istherefore
P = 1 2 ρAv3 = 1 2 ×1.204 ×6362 ×153 = 12.93 MW
Theactualratedpowerofthedeviceisapproximately3MWbecauseotherphysical
processesandlosseshavetobeaccountedfor,yetEquation(2.8)remainsoneofthe
mostfundamental inthe studyof wind power.
Manyexcellentcomputersoftwarepackagesexistforplottinggraphsandthese,aswell
as graphics calculators, may be used to solve polynomial equations. The real roots of
theequationP(x) = 0aregivenbythevaluesoftheinterceptsofthefunctiony =P(x)
and thexaxis, because on thexaxisyiszero.
Figure 2.20 shows a graph of y = P(x). The graph intersects the x axis at x = x 1
,
x = x 2
and x = x 3
, and so the equation P(x) = 0 has real roots x 1
, x 2
and x 3
, that is
P(x 1
) =P(x 2
) =P(x 3
) = 0.

74 Chapter 2 Engineering functions
P(x)
x 1 x 2 x 3 x
Figure2.20
A polynomial function whichcuts
thexaxisat pointsx 1 ,x 2 andx 3 .
EXERCISES2.4.1
1 State the degree ofthe followingpolynomial (c) 3w −5w 2 +12w 4
expressions:
(d) 7x −x 2
(a) z 3 +2z 2 −8+13z
(e) 3(2t 2 −9t +1)
(b) t 2 −5t 5 +2−8t 3 (f) 2z(2z +1)(2z −1)
Solutions
1 (a) 3 (b) 5 (c) 4 (d) 2 (e) 2 (f) 3
TechnicalComputingExercises2.4.1
UseatechnicalcomputinglanguagesuchasMATLAB ® to
dothe following exercises.
1 (a) Ploty =x 3 andy = 4 −2x in the interval
[-3,3].Beaware that in MATLAB ® ,to carry out
operationsonindividual elements ofavector,
special notation is used.Forexample to multiply
each element in vectorabythe corresponding
element in vectorb(having the same dimension
asvectora)you would typea.*b.Other
functionssuch asraisingto apower alsorequire a
dotprefixifthey are to be carried outoneach
individual element, rather than the whole matrix.
Note thexcoordinate ofthe point ofintersection.
(b) Drawy =x 3 +2x −4. Note the coordinate of
the point where the curve cutsthexaxis.
Compare your answer with that from (a).Explain
your findings.
2 Plotthe followingfunctions:
(a) y=3x 3 −x 2 +2x+1
−2x2
(b) y=x 4 + x3
3 − 5x2 +x−1 −3x2
2
(c) y=x 5 −x 2 +2 −2x2
Henceestimatethe real roots of
0=3x 3 −x 2 +2x+1 −2x2
0=x 4 + x3
3 − 5x2
2 +x−1
−3x2
0=x 5 −x 2 +2 −2x2
3 Usetherootsfunctionin MATLAB ® orequivalent
to calculateamore accurate value forthe real roots
estimated in question 2 andconfirm your answers are
correct.
4 (a) Drawy = 2x 2 andy =x 3 +6 usingthe same
axes.Use yourgraphs to findapproximate
solutionstox 3 −2x 2 +6 = 0.
(b) Addthe liney = −3x +5 to yourgraph.State
approximatesolutionsto
(i) x 3 +3x+1=0
(ii)2x 2 +3x−5=0

2.4.2 Rationalfunctions
2.4 Review of some common engineering functions and techniques 75
Arational function,R(x), has the form
R(x) = P(x)
Q(x)
where P and Q are polynomial functions; P is the numerator and Q is the
denominator.
The functions
R 1
(x) = x +6
x 2 +1
R 2
(t) = t3 −1
2t+3
R 3
(z) = 2z2 +z−1
z 2 +3z−2
areallrational.Whensketchingthegraphofarationalfunction,y = f (x),itisusualto
drawupatableofxandyvalues.Indeedthishasbeencommonpracticewhensketching
anygraphalthoughtheuseofgraphicscalculatorsisnowreplacingthiscustom.Itisstill
usefultoanswer questions such as:
‘Howdoesthefunctionbehaveasxbecomeslargepositively?’
‘Howdoes thefunctionbehaveasxbecomeslarge negatively?’
‘Whatis thevalueof thefunctionwhenx = 0?’
‘At what valuesofxis thedenominator zero?’
Figure 2.21 shows a graph of the function y = 1+2x = 1 + 2. As x increases, the
x x
value ofyapproaches 2.We write thisas
y→2 as x→∞
and say ‘ytends to2asxtends toinfinity’.Also fromFigure 2.21, we see that
y→±∞ as x→0
Asx → ∞,thegraphgetsnearerandnearertothestraightliney = 2.Wesaythaty = 2
is an asymptote of the graph. Similarly,x = 0, that is theyaxis, is an asymptote since
the graph approaches the linex = 0 asx → 0.
If the graph of any function gets closer and closer to a straight line then that line is
calledanasymptote.Figure2.22illustratessomerationalfunctionswiththeirasymptotes
indicatedbydashedlines.InFigure2.22(a)theasymptotesarethehorizontalliney = 3
y
2
x
Figure2.21
Thefunction:y = 1+2x = 1 x x +2.

76 Chapter 2 Engineering functions
y
y
y
3
1
x
–2
x
–1
x
(a)
(b)
Figure2.22
Some examples offunctionswith their asymptotes:
(a)y= 3x+1
x
= 3 + 1 −1 +4x+2
;(b)y=x
x x +2 ;(c)y=x2 =x+3− 1
x +1 x +1 .
(c)
and the y axis, that is x = 0. In Figure 2.22(b) the asymptotes are the horizontal line
y = 1andtheverticallinex = −2;inFigure2.22(c)theyarey =x +3andthevertical
linex = −1.Theasymptotey =x+3,beingneitherhorizontalnorvertical,iscalledan
oblique asymptote. Oblique asymptotes occur only when the degree of the numerator
exceeds the degree of the denominator by one.
Weseethattheverticalasymptotesoccuratvaluesofxwhichmakethedenominator
zero. These values are particularly important to engineers and are known as the poles
of the function. The function shown in Figure 2.22(a) has a pole atx = 0; the function
showninFigure2.22(b)hasapoleatx = −2;andthefunctionshowninFigure2.22(c)
has a pole atx = −1.
Ifthegraphofafunctionapproachesastraightline,thelineisknownasanasymptote.
Asymptotesmaybehorizontal, vertical oroblique.
Values of the independent variable where the denominator is zero are called poles
ofthe function.
Example2.10 Sketch the rationalfunctiony =
x
x 2 +x−2 .
Solution For large values ofx, thex 2 term in the denominator has a much greater value than the
x inthe numerator. Hence,
y→0 as x→∞
y→0 as x→−∞
Therefore thexaxis, thatisy = 0,isanasymptote. Writingyas
x
y =
(x −1)(x +2)

2.4 Review of some common engineering functions and techniques 77
weseethefunctionhaspolesatx = 1andx = −2;thatis,thereareverticalasymptotes
atx = 1 andx = −2. Substitution into the function of a number of values ofxallows a
table tobe drawn up:
x −3 −2.5 −2.1 −1.9 −1.5 −1 0 0.5 0.9 1.1 1.5 2 3
y −0.75 −1.43 −6.77 6.55 1.20 0.50 0 −0.40 −3.10 3.55 0.86 0.50 0.30
The graph of the function can then be sketched as shown inFigure 2.23.
y
–2
1 x
Figure2.23
x
Thefunction:y =
x 2 +x−2 .
Engineeringapplication2.7
Equivalentresistance
RecallfromEngineeringapplication1.2thattheformulafortheequivalentresistance
oftwo resistors inparallel isgiven by:
1
R E
= 1 R 1
+ 1 R 2
Consider a circuit consisting of two resistors in parallel as shown in Figure 2.24.
Onehasaknown resistanceof1andtheother hasavariableresistance,R.The
equivalent resistance,R E
, satisfies
Hence,
1
= 1 R E
R + 1 1 = 1 +R
R
R E
=
R
1 +R
Thus the equivalent resistance is a rational function of R, with domain R 0.
The graph of this function is shown in Figure 2.25. When R = 0 we note
that R E
= 0, corresponding to a short circuit. As the value of R increases, that
isR → ∞,theequivalentresistanceR E
approaches1sothatR E
= 1isanasymptote.
➔

78 Chapter 2 Engineering functions
R E
1
R
1 V
R E
R
Figure2.24
Two resistorsin parallel.
Figure2.25
Theequivalent resistance,R E ,
increases asRincreases.
EXERCISES2.4.2
1 State the poles ofthe followingrational functions:
(a) y(x) = x +3
x −2
(c) y(t) = t2 +t+1
2t+3
3
(e) H(s) =
s 2 +6s+5
2s+7
(f) G(s) =
s 2 +3s−18
(g) x(t) = 9
t 3 −t
2t−6
(h) p(t) =
t 2 +10t+25
(b) y(x) = 2x+1
x +7
(d) X(s) = s +1
s 2 −4
2 Describethe horizontalasymptote ofeachofthe
following functions:
(a) y(x) =6+ 1 x
(c) y(r) =3− 2 5r
(e) r(v) = 2v−5
3v
(f) a(t) = t2 +2t+1
t 2
(g) m(s) = 10−2s−3s2
2s 2
(b) h(t) = 2 t −1
(d) v(t) = 6 +t
t
3 Describethe vertical asymptotes ofeachofthe
following functions:
(a) y(x) = 3x+1
x −2
(c) h(s) =
9
(s +2)(s −1)
(b) y(t) = 6t−3
4t+4
(d) G(t) = 1
t 2 −1
(f) y(x) =
(g) w(f)=
2x
x 2 −1
f +2
f 2 +f−6
(h) P(t) = t2 +t+1
t 2 +6t+9
(i) T(x) =
(j) Q(r) =
x3
2x−1
6 +r
r 2 −r−12
(e) H(s) =
s
s 2 −1
4 Describethe obliqueasymptote ofeachofthe
following functions:
(a) y(x)=x+3+ 1
x −1
(b) y(x)=2x−1+ 3
x +2
(c) y(x) = x 2 − 3 4 + 1
2x+7
(d) y(x)=3x−1+ 5
2x+2
(e) y(x)=2x−1+ x +2
x 2 −1
(f) y(x)=3x−2+ 4
2x−1
(g) y(x)=4−2x+ 3
2x+3

2.4 Review of some common engineering functions and techniques 79
5 Showthat
I(x) = 2x
3 − 7 9 + 23
9(3x +2)
can be expressed in the equivalent form
2x 2 −x+1
3x+2
Sketch the rational functionI(x)and state any
asymptotes.
6 Showthat
p(x) =2x + 1 2 + 9
2(2x +3)
can be writtenin the equivalent form
4x 2 +7x+6
2x+3
Sketch the rational function p(x)and stateany
asymptotes.
7 Show thatthe function
y(x)=x+ 7 −x
x 2 +3
can be expressed in the equivalent form
x 3 +2x+7
x 2 +3
Sketch the rational functiony(x) andstate any
asymptotes.
Solutions
1 (a) 2 (b) −7 (c) − 3 2
(d) −2,2 (e) −5, −1 (f) −6,3
(g) −1,0,1
(h) −5
2 (a)y=6 (b)h=−1
(c)y=3 (d) v=1
(e)r= 2 3
(f)a=1
4 (a) y=x+3 (b) y=2x−1
(c)y= x 2 − 3 4
(e) y=2x−1
(g) y=4−2x
5 x=− 2 2x
,I =
3 3 − 7 9
(d) y=3x−1
(f) y=3x−2
6 x=− 3 2 ,p=2x+1 2
(g)m=− 3 2
3 (a)x=2 (b)t=−1
(c) s=−2,1 (d) t =−1,1
(e) s=−1,1 (f) x=−1,1
(g) f =−3,2 (h) t =−3
(i) x=0.5 (j) r=−3,4
7 y=x
TechnicalComputingExercises2.4.2
1 Useatechnicalcomputing languageto plotthe
followingrational functions.Stateany asymptotes.
(2x +1)
(a) f(x) = −4x4
(x−3)
s
(b) g(s) = −3s3
(s+1)
z
(c) h(z) =
(z 2 −3z3
+1)
(d) y(x) = (x+1) −3x3
x
2x
(e) r(x) =
−3x3
(x −1)(x −2)
2 Plotthe functionsgiven in Question4in
Exercises2.4.2 for −10 x10.

80 Chapter 2 Engineering functions
2.4.3 Exponentialfunctions
Anexponentisanothernameforapowerorindex.Expressionsinvolvingexponentsare
called exponential expressions, for example 3 4 ,a b , andm n . In the exponential expressiona
x ,aiscalledthebase;xistheexponent.Exponentialexpressionscanbesimplified
and manipulated using the laws of indices. These laws aresummarized here.
a m a n =a m+n
a m
a n =am−n a 0 =1 a −m = 1
a m
(a m ) n =a mn
Example2.11 Simplify
a 3x a 2x
(a)
a 4x (b) a 2t (1 −a t ) +a 3t (c)
(a y ) 2
2a y
(d)
a −6z
a −2z
(e)
(2a 3r ) 2 a 2r
3a −5r
(f)
a x+y a y
a 2x
(g)
3a (x/y) a x
a y
a
Solution 3x a 2x
(a) = a5x
a 4x =ax
a4x (b) a 2t (1 −a t ) +a 3t =a 2t −a 3t +a 3t =a 2t
(c)
(d)
(e)
(f)
(g)
(a y ) 2
= a2y
2a y 2a = ay
y 2
a −6z
a −2z =a−6z−(−2z) =a −4z
(2a 3r ) 2 a 2r
= 4a6r a 2r
= 4a8r
3a −5r 3a −5r 3a
a x+y a y
=a x+2y−2x =a 2y−x
a 2x
3a (x/y) a x
= 3a (x/y)+x−y
a y
3
−5r
=
4a13r
Exponentialfunctions
Anexponential function, f (x), has the form
f(x)=a x
whereaisapositive constant called the base.
Some typical exponential functions are tabulated in Table 2.2 and are shown in Figure
2.26.Note from the graphsthatthese areone-to-onefunctions.
An exponential function is not a polynomial function. The powers of a polynomial
functionareconstants;thepowerofanexponentialfunction,thatistheexponent,isthe
variablex.

Table2.2
Values ofa x fora = 0.5, 2 and 3.
2.4 Review of some common engineering functions and techniques 81
x 0.5 x 2 x 3 x
(0.5) x
y
3 x
2 x
−3 8 0.125 0.037
−2 4 0.25 0.111
−1 2 0.5 0.333
0 1 1 1
1 0.5 2 3
2 0.25 4 9
3 0.125 8 27
Figure2.26
Some typical exponential functions.
x
Themostwidelyusedexponentialfunction,commonlycalledtheexponentialfunction,
is
f(x)=e x
where e is an irrational constant (e = 2.718281828...) commonly called the
exponential constant.
Most scientific calculators have values of e x available. The function is tabulated
in Table 2.3. The graph is shown in Figure 2.27. This particular exponential function
so dominates engineering applications that whenever an engineer refers to the exponential
function it almost invariably means this one. We will see later why it is so
important.
Asxincreasespositively,e x increasesveryrapidly;thatis,asx → ∞,e x → ∞.This
situationisknownasexponentialgrowth.Asxincreasesnegatively,e x approacheszero;
that is, asx → −∞, e x → 0. Thusy = 0 is an asymptote. Note that the exponential
function isnever negative.
Figure2.28showsagraphofe −x .Asxincreasespositively,e −x decreasestozero;that
is,asx → ∞,e −x → 0.Thisisknownasexponentialdecay.Thefunctionistabulated
inTable 2.4.
Table2.3
Thevalues ofthe exponential
function f (x) = e x for
variousvalues ofx.
x
e x
−3 0.050
−2 0.135
−1 0.368
0 1
1 2.718
2 7.389
3 20.086
e x 1
Figure2.27
Graph ofy = e x showing exponential growth.
x

82 Chapter 2 Engineering functions
Table2.4
The values ofthe exponential
function f (x) = e −x for
various values ofx.
e –x 1
x
e −x
−3 20.086
−2 7.389
−1 2.718
0 1
1 0.368
2 0.135
3 0.050
Figure2.28
Graph ofy = e −x showing exponential decay.
x
Engineeringapplication2.8
Dischargeofacapacitor
The capacitor is another of the three fundamental electronic components. It is a
device that is used to store electrical charge. It consists of two parallel conducting
platesseparatedbyaninsulatingmaterial,knownasadielectric.Abuildupofanet
positive charge on one plate and a net negative charge on the other plate creates an
electricfieldacrossthedielectric,allowingelectricalenergytobestoredthathasthe
potential to do useful work. The symbol for a capacitor is two parallel lines that are
perpendicular tothe conductors inthe circuit.
Consider thecircuitofFigure2.29.Beforetheswitchisclosed,thecapacitorhas
avoltageV acrossit.Supposetheswitchisclosedattimet = 0.Acurrentthenflows
in the circuit and the voltage, v, across the capacitor decays with time. The voltage
across the capacitor isgiven by
{
V t < 0
v =
Ve −t/(RC) t 0
ThequantityRC isknown asthetimeconstantofthecircuitand isusuallydenoted
byτ.So
{
V t < 0
v =
Ve −t/τ t 0
v
V
t larger
C
Figure2.29
Circuit to discharge acapacitor.
R
Figure2.30
Thecapacitor takes longerto discharge fora
largercircuit time constant, τ.
If τ is small, then the capacitor voltage decays more quickly than if τ is large.
This isillustratedinFigure 2.30.
t

2.4 Review of some common engineering functions and techniques 83
Engineeringapplication2.9
Thediodeequation
Adiodeisanelectronicdevicethatallowscurrenttoflowwitheaseinonedirection
andwithdifficultyintheother.Inplumbingtermsitistheequivalentofanon-return
valve.Itisconstructedbysandwichingtwodifferenttypesofsemiconductormaterial
together,knownasp-typeandn-typesemiconductors.Theresultantconstructionis
referred to as a p-n junction. These different types of semiconductors are created
by a procedure known as doping. The technical details are complex but it is one
of the most fundamental processes underpinning the electronics revolution that has
transformed society inrecent decades.
Asemiconductor diode can bemodelled by the equation
I=I s
(
e qV
nkT −1
)
whereV =applied voltage (V);
I =diode current (A);
I s
=reverse saturationcurrent (A);
k =1.38 ×10 −23 J K −1 ;
q =1.60 ×10 −19 C;
T =temperature (K);
n =idealityfactor.
This equation relates the current through the diode to the voltage across it. The idealityfactor,n,typicallyrangesbetween1and2dependingonhowthediodeismanufactured
and the type of semiconductor material used. We will consider the case
n = 1,whichcorrespondstothatofanidealdiode.Atroomtemperatureq/(kT ) ≈ 40
and sothe equation can be written as
I =I s
(e 40V −1)
Figure 2.31 shows a graph ofI againstV. Notice that for negative values ofV, the
equation may be approximated by
I≈−I s
since e 40V ≈ 0. The diode is said to be reverse saturated in this case. In reality,
I s
is usually quite small for a practical device, although its size has been exaggerated
in Figure 2.31. This model does not cater for the breakdown of the diode.
According to the model it would be possible to apply a very large reverse voltage
to a diode and yet only a small saturation current would flow. This illustrates an
importantpointthatnomathematicalmodelcoverseveryfacetofthephysicaldevice
orsystemitismodelling.Adifferentmodelwouldbeneededtodealwithbreakdown
characteristics.
➔

84 Chapter 2 Engineering functions
I
–I s V
Figure2.31
Atypical diode characteristic.
EXERCISES2.4.3
1 Simplify
(a)
(c)
(e)
e 2x
3e 3x
e x (e x +e 2x )
e 2x
(e 2t ) 3 (e 3t ) 4
e 10t
(b) e2t−3
e 2
(d) e−3 e −7
e 6 e −2
2 Consider theRC circuitofFigure2.29.Given an
initialcapacitor voltageof10Vplotthe variation in
capacitor voltage with time, usingthe same axes, for
the followingpairs ofcomponentvalues:
(a) R=1,C=1µF
(b) R=10,C=1µF
(c) R=3.3,C=1µF
(d) R=56,C=0.1µF
Calculate the time constant, τ,in eachcase.
3 Sketch the following functions,usingthe same axes:
y=e 2x y=e x/2 y=e −2x
for −3x3
4 Sketch agraph ofthe functiony = 1 −e −x for
x0.
Solutions
1 (a) e−x (b) e 2t−5 (c) 1 +e x 2 (a) 10 −6 (b) 10 −5 (c) 3.3 ×10 −6
3
(d) e −14 (e) e 8t (d) 5.6 ×10 −6
TechnicalComputingExercises2.4.3
1 Ploty = e kx fork = −3, −2, −1, 0, 1, 2, 3, for
−3x3.
2 Ploty =ke x fork = −3, −2, −1, 0, 1, 2, 3, for
−3x3.
3 Ploty=5−x 2 andy=e x for−3x3.Forwhich
valuesofxise x <5−x 2 ?
4 Ploty =x 4 andy = e x for −1 x9.Forwhich
valuesofxis(a)e x <x 4 ,(b)e x >x 4 ?

2.4.4 Logarithmfunctions
2.4 Review of some common engineering functions and techniques 85
Logarithms
The equation 16 = 2 4 may be expressed in an alternative form using logarithms. In
logarithmic form wewrite
log 2
16 = 4
and say ‘log to the base 2 of 16 equals 4’. Hence logarithms are nothing other than
powers. The logarithmic form isillustratedby more examples:
125 = 5 3 so log 5
125 = 3
64=8 2 solog 8
64=2
16=4 2 solog 4
16=2
1000 = 10 3 so log 10
1000 = 3
Ingeneral,
ifc =a b , thenb =log a
c
In practice, most logarithms use base 10 or base e. Logarithms using base e are called
naturallogarithms.Log 10
xandlog e
xareusuallyabbreviatedtologxandlnx,respectively.
Most scientific calculators have both logs to base 10 and logs to base e as preprogrammed
functions, usually denoted as log and ln, respectively. Some calculations
in communications engineering use base 2. Your calculator will probably not calculate
base 2 logarithms directly. We shall see how toovercome thisshortly.
Focusing on base 10 wesee that
ify=10 x then x=logy
Equivalently,
ifx=logy then y=10 x
Usingbase e we see that
ify=e x then x=lny
Equivalently,
ifx=lny then y=e x
Example2.12 Solve the equations
(a) 16 = 10 x (b) 30 = e x (c) logx = 1.5 (d) lnx = 0.75
Solution (a)
16 = 10 x
log16=x
x = 1.204
(c) logx = 1.5
x = 10 1.5
= 31.623
(b)
30 = e x
ln30=x
x = 3.401
(d) lnx = 0.75
x = e 0.75
= 2.117
Example2.13 Solve the equations
(a) 50 = 9(10 2x ) (b) 3e −(2x+1) = 10
(c) log(x 2 −1) =2 (d) 3ln(4x +7) =12

86 Chapter 2 Engineering functions
Solution (a) 50 = 9(10 2x ) (b) 3e −(2x+1) = 10
10 2x = 50 e −(2x+1) = 10
9
2x = log 50 3
−(2x +1) = ln 10 9
x = 1 2 log 50 3
9 = 0.372 2x=−ln 10 3 −1
(c) log(x 2 −1) = 2
x 2 −1=10 2 =100
x 2 = 101
x = ±10.050
2x = −2.204
x = −1.102
(d) 3ln(4x +7) = 12
4x+7=e 4
4x=e 4 −7
x = e4 −7
= 11.900
4
Logarithmicexpressionscanbemanipulatedusingthelawsoflogarithms.Theselaws
are identical for any base, but it is essential when applying the laws that bases are not
mixed.
log a
A +log a
B = log a
(AB)
( ) A
log a
A −log a
B = log a
B
nlog a
A = log a
(A n )
log a
a=1
Wesometimesneedtochangefromonebasetoanother.Thiscanbeachievedusingthe
following rule.
Inparticular,
Example2.14 Simplify
log a
X = log b X
log b
a
log 2
X = log 10 X
log 10
2 = log 10 X
0.3010
(a) logx +logx 3
(b) 3logx +logx 2
( 1
(c) 5lnx+ln
x)
(d) log(xy) +logx −2logy
( ) 4
(e) ln(2x 3 ) −ln + 1 x 2 3 ln27
Solution (a) Usingthe laws of logarithms we find
logx +logx 3 = logx 4

2.4 Review of some common engineering functions and techniques 87
(b) 3logx +logx 2 = logx 3 +logx 2 = logx
( ( ( 5
)
1 1 x
(c) 5lnx+ln =lnx
x)
5 5
+ln = ln =lnx
x)
4
x
(d) logxy+logx−2logy=logxy+logx−logy 2
( ) ( ) xyx x
2
=log = log
y 2 y
( ) 4
(e) ln(2x 3 ) −ln + ln27 ( ) 2x
3
=ln +ln27 1/3
x 2 3 4/x 2
( 2x 3 x 2 )
=ln +ln3
( x
5
=ln
2
4
)
+ln3=ln
( ) 3x
5
2
Example2.15 Findlog 2
14.
Solution Usingthe formulaforchangeofbasewehave
log 2
14 = log 10 14
log 10
2 = 1.146
0.301 = 3.807
Engineeringapplication2.10
Signalratiosanddecibels
Theratiobetweentwosignallevelsisoftenofinteresttoengineers.Forexample,the
outputandinputsignalsofanelectronicsystemcanbecomparedtoseeifthesystem
has increasedthe level ofasignal. Acommon case is anamplifier, where the output
signal is usually much larger than the input signal. This signal ratio, known as the
power gain, isoftenexpressed in decibels(dB)given by
( )
Po
power gain (dB) = 10log
P i
whereP o
isthepoweroftheoutputsignalandP i
isthepoweroftheinputsignal.The
term gain is used because ifP o
> P i
, then the logarithm function is positive, correspondingtoanincreaseinpower.IfP
o
<P i
thenthegainisnegative,corresponding
toadecrease inpower. Inthis situationthe signal issaidtobeattenuated.
The advantage of using decibels as a measure of gain is that if several electronic
systems are connected together then it is possible to obtain the overall system gain
in decibels by adding together the individual system gains. We will show this for
threesystems connected together, butthe development iseasilygeneralized tomore
systems. Let the power input to the first system be P i1
, and the power output from
➔

88 Chapter 2 Engineering functions
the third system be P o3
. Suppose the three are connected so that the power output
from system 1,P o1
, is used as input to system 2, that isP i2
=P o1
. The power output
from system 2,P o2
, is then used as input to system 3, that isP i3
= P o2
. We wish to
find the overall power gain, 10log(P o3
/P i1
). Now
P o3
= P o3P o2
P o1
P i1
P i3
P i2
P i1
becauseP i3
=P o2
andP i2
=P o1
. Therefore,
( ) ( )
Po3 Po3 P
10log = 10log o2
P o1
P i1
P i3
P i2
P i1
thatis
10log
(
Po3
P i1
)
( ) ( ) ( )
Po3 Po2 Po1
= 10log +10log +10log
P i3
P i2
P i1
usingthe laws of logarithms.
It follows that the overall power gain is equal to the sum of the individual power
gains. Often engineers are more interested in voltage gain rather than power gain.
The power of a signal is proportional to the square of its voltage. We definevoltage
gain (dB) by
( ) V
2
voltage gain (dB) = 10log
o
V 2
i
( )
Vo
= 20log
V i
Engineeringapplication2.11
TheuseofdBminradiofrequencyengineering
Inthepreviousengineeringapplicationitwasshownhowthedecibelcanbeusedto
express a ratio of the power of two signal levelsP o
andP i
. It is possible to specify
a fixed value for the input signalP i
. This is termed a reference level. When this is
done the decibel becomes an absolute quantity. The notation is normally changed
slightly to indicate the assumed reference level. For example, dBm is used as an
absolutemeasureofpowerinthefieldofradiofrequency(RF)engineering.Amobile
telephone handset, a microwave oven and a radar transmitter on an airfield are all
devices that might have their output power quoted in dBm. The definition of power
gain measured indBmisasfollows:
( )
Po
power gain (dBm) = 10log
10 −3
Herethe reference level chosen is1mWor 10 −3 W.
If a device is quoted as having an output power of 15 dBm we can convert this
into a power value inwatts as follows:
( )
Po
15 (dBm) = 10log
10 −3
Dividing both sides by 10
( )
Po
1.5 = log
10 −3

2.4 Review of some common engineering functions and techniques 89
And so
10 1.5 =
(
Po
10 −3 )
Therefore the actual power is
P o
= 10 1.5 ×10 −3 = 10 −1.5 = 0.0316 ∼ = 32 mW
ThisisthetypicalamountofRFpowerthatmightbetransmittedbyalaptopcomputer
with WiFicapability.
Engineeringapplication2.12
Attenuationinastep-indexopticalfibre
Fibre optical cables are used to guide high-bandwidth light signals generated by
lasers. One design is the step-index optical fibre. This consists of a glass core of
silicondioxidewithahighrefractiveindex,obtainedbydopingtheglasswiththeelementgermanium.Asurroundingsheathoflowerrefractiveindexglassensuresthat
nearly all of the light remains within the core. Transmission occurs by total internal
reflection atthe interfacebetween the two typesofglass.
The losses that do occur in the optical fibre can be described by the following
equation:
I(z) =I 0
e −αz
where the intensity of the light I(z) is a function of the distance along the fibre, z,
fromthelightsource.TheintensityofthelightsourceattheinsertionpointisI 0
and
α is anattenuation factor.
Note that α varies with the wavelength of the light and so the rate of attenuation
dependsonthecolourofthelight.Itispossibletoobtainplotsforthevariationofthe
attenuationfactorwithwavelengthfromthemanufacturers.Theattenuationfactoris
often expressed in units of dB km −1 . A typical value for a cable operating at a light
wavelength of1550 nmis0.3 dBkm −1 .
An alternative measure for gain/attenuation is that of the neper (Np). Like the
decibel this is also a dimensionless ratio that has a logarithmic form. Whereas the
decibel is defined in terms of base 10 logarithms, the neper is defined in terms of
natural logarithms, that is base e logarithms. The gain measured in Np is defined by
the following expression:
voltage gain (Np) = ln V o
V i
=lnV o
−lnV i
whereV o
is the signal value in volts after gain/attenuation andV i
is the reference
signal value involts.
It is possible to derive a conversion factor between Np and dB by considering
the case whenV o
is a factor of 10 greater thanV i
. Using the neper measurement this
corresponds to
voltage gain (Np) = ln 10V i
V i
=ln10
➔

90 Chapter 2 Engineering functions
Using the dB measurement for voltage gain (see Engineering application 2.10)
this corresponds to voltage gain (dB) =20log V o
V i
which equals 20log10. So,
ln10 (Np) = 20log10 (dB). Thus,
1Np= 20log10 dB ≈ 8.685 89 dB
ln 10
Using the value quoted earlier of an attenuation factor of 0.3 dB km −1 , this corresponds
to
0.3 ÷8.685 89 Np km −1 = 0.034539 Np km −1
It is more usual to quote values of Np m −1 and so this value becomes 3.4539 ×
10 −5 Np m −1 .
Engineeringapplication2.13
Referencelevels
We saw in Engineering application 2.11 that the suffix ‘m’ is used in dBm to indicatetheprovisionofaspecificreferencelevel.Alternativesuffixesareusedtodenote
other reference levels and quantities, which do not necessarily have to be related to
electricalpower.Forexample,whenmeasuringsoundpressure,P,inairtheconventionalreferencelevelforsoundpressureis20
µPar.m.s.Thisischosentocorrespond
totheapproximatethresholdofhumanhearingfora1kHzsinusoidalsignal.Theunit
forsoundpressureisthereforequotedwithreferencetoaninputpressureof20 µPa.
ThisiscommonlywrittenasdBre20 µPar.m.s.orusingtheshorthanddBSPL(dB,
soundpressure level). Inotherwords wehave
( )
P
soundpressure level (dBSPL) = 20log
20 ×10 −6
Asaconsequenceofthechoiceofthehumanhearingthresholdasthereferencelevel,
anegativevalueofdBSPLcorrespondstoasoundthatistooquiettobeheardbythe
averageperson;0dBSPLisasoundthatcanjustbeheardandanythingabovethisis
fullyaudible.Anofficemighthaveanambient(background)levelof30dBSPLand
a person talking to you at the next desk might produce 60 dB SPL, both quantities
being measured atyour hearing position.
Logarithmfunctions
Thelogarithm functionsaredefined by
f(x)=log a
x x>0
whereaisapositive constant called the base.
In particular the logarithm functions f (x) = logx and f (x) = lnx are shown in Figure2.32andsomevaluesaregiveninTable2.5.Thedomainofbothofthesefunctionsis
(0,∞)andtheirrangesare (−∞,∞).Weobservefromthegraphsthatthesefunctions

2.4 Review of some common engineering functions and techniques 91
Table2.5
Somevalues forlogarithm
functionslogx and lnx.
x logx lnx
y
ln x
log x
0.1 −1 −2.303
0.5 −0.301 −0.693
1 0 0
2 0.301 0.693
5 0.699 1.609
10 1 2.303
50 1.699 3.912
Figure2.32
Graphsoflnx and logx.
1 x
are one-to-one. It is important to stress that the logarithm functions, log a
x, are only
defined for positive values ofx. The following properties should be noted:
}
}
logx→∞ logx→−∞
asx→∞ asx→0
lnx→∞ lnx→−∞
log1=ln1=0
log10=1 lne=1
Connectionbetweenexponentialandlogarithmfunctions
Theexponentialfunction, f (x) =a x ,isaone-to-onefunctionandsoaninversefunction,
f −1 (x), exists.Recall
So
Now
f −1 (f(x))=x
f −1 (a x )=x
log a
(a x ) =xlog a
a
using laws of logarithms
=x since log a
a = 1
Hence the inverse of f (x) = a x is f −1 (x) = log a
x. By similar analysis the inverse of
f(x)=log a
xisf −1 (x) =a x .
The inverse of the exponential function, f (x) = a x , is the logarithm function, that
is f −1 (x) = log a
x.
Theinverseofthelogarithmfunction, f (x) = log a
x,istheexponentialfunction,
thatis f −1 (x) =a x .
In particular:
If f(x) =e x ,then f −1 (x) =lnx.
If f(x) = lnx,then f −1 (x) =e x .
If f(x) = 10 x ,then f −1 (x) =logx.
If f(x) = logx,then f −1 (x) =10 x .

92 Chapter 2 Engineering functions
x
1 1
2 64
3 729
4 4096
5 15625
6 46656
7 117649
8 262144
9 531441
10 1000000
y
log y
6
5
4
3
2
1
0
1
2 3 4 5 6 7 8 9 10 x
Figure2.33
Thefunctiony =x 6 plotted on alog--linear graph.
Useoflog–logandlog–linearscales
Suppose wewish toplot
y(x)=x 6
1x10
This may appear a straightforward exercise but consider the variation in the x and y
values. Asxvaries from 1 to 10, thenyvaries from 1 to 1000000, as tabulated above.
Severalofthesepointswouldnotbediscernibleonagraphandsoinformationwouldbe
lost.Thiscanbeovercomebyusingalogscalewhichaccommodatesthelargevariation
iny. Thus log yisplotted againstx, rather thanyagainstx. Note thatinthisexample
logy=logx 6 =6logx
so asxvaries from 1 to 10, logy varies from 0 to 6. A plot in which one scale is logarithmic
and the other is linear is known as a log--linear graph. Figure 2.33 shows logy
plottedagainstx.Ineffect,useofthelogscalehascompressedalargevariationintoone
which ismuch smaller and easier toobserve.
Example2.16 Considery = 7 x for −3 x 3.Plot a log--linear graph of this function.
Solution We have
and so
y=7 x
logy = log(7 x ) =xlog7 = 0.8451x
PuttingY = logy we haveY = 0.8451x which is the equation of a straight line passing
through the origin with gradient log7. Hence when logy is plotted againstxastraight
linegraphisproduced.ThisisshowninFigure2.34.Notethatbytakinglogs,therange
on the verticalaxis has been greatly reduced.
Aplotinwhichbothscalesarelogarithmicisknownasalog--logplot.Herelogyis
plotted against logx.

2.4 Review of some common engineering functions and techniques 93
Y
2.5
x y Y=logy
−3 0.003 −2.54
−2 0.020 −1.69
−1 0.143 −0.85
0 1 0
1 7 0.85
2 49 1.69
3 343 2.54
–3 –2 –1 1 2 3 x
–2.5
Figure2.34
Alog--linear plotofy = 7 x producesa
straightline graph.
Example2.17 Considery =x 7 for 1 x10. Plot a log--log graph of thisfunction.
Solution We have
and so
y=x 7
logy = log(x 7 ) = 7logx
We plot logy against logx in Figure 2.35 for a log--log plot. PuttingY = logy and
X = logx we haveY = 7X which isastraightlinethrough the originwith gradient 7.
x y X=logx Y=logy
1 1 0 0
2 128 0.301 2.107
3 2187 0.477 3.340
4 16384 0.602 4.214
5 78125 0.699 4.893
6 279936 0.778 5.447
7 823543 0.845 5.916
8 2097152 0.903 6.322
9 4782969 0.954 6.680
10 10000000 1 7
Y
7
1 X
Figure2.35
A log--logplotofy =x 7
produces astraightline graph.
Examples 2.16 and 2.17 illustrate the following general points:
Alog--linear plot ofy =a x produces a straightlinewith a gradient of loga.
Alog--log plot ofy =x n produces a straightlinewith a gradient ofn.

94 Chapter 2 Engineering functions
1
9
8
7
6
5
4
3
Second cycle
2
Logarithmic scale
1
9
8
7
6
5
4
3
First cycle
2
Linear
1
Figure2.36
Two-cycle log--lineargraph paper.
Useoflog–linearandlog–loggraphpaper
The requirement to take logarithms is a tedious process which can be avoided by using
specialgraphpaperscalledlog--lineargraphpaperandlog--loggraphpaper.Anexample
of log--linear graph paper isshown inFigure 2.36.
Note that on one axis the scale is uniform; this is the linear scale. On the other, the
scaleisnotuniformandismarkedincyclesfrom1to9.Thisisthelogarithmicscale.On
this scale values ofyare plotted directly, without first taking logarithms. On the graph
papershowninFigure2.36therearetwocyclesbutpapersarealsoavailablewiththree

2.4 Review of some common engineering functions and techniques 95
or more cycles. To decide which sort of graph paper is appropriate it is necessary to
examine the variation in size of the variable to be plotted measured in powers of 10. If,
for example,yvaries from 1 to 10, then paper with one cycle is appropriate. Ifyvaries
from 1 to 10 2 , two-cycle paper is necessary. If y varies from 10 −1 to 10 4 , then paper
with4 − (−1) = 5cycleswouldbeappropriate.Toseehowlog--linearpaperisusedin
practice, consider the following example.
Example2.18 During anexperiment the following pairs of data values wererecorded:
A B C D
x 0 1 5 12
y 4.00 5.20 14.85 93.19
It is believed thatyandxare related by the equationy = ab x . By plotting a log--linear
graph verify the relationship isof thisform and determineaandb.
Solution If the relationship isgiven byy =ab x , then taking logarithms yields
logy=loga+xlogb
So, plotting logy againstxshould produce a straight line graph with gradient logband
vertical intercept loga. The need to find logy is eliminated by plotting theyvalues directly
on a logarithmic scale. Examining the table of data we see that y varies from
approximately 10 0 to 10 2 so that two-cycle paper is appropriate. Values ofybetween 1
and 10 are plotted on the first cycle, and those between 10 and 100 are plotted on the
second.ThepointsareplottedinFigure2.37.Noteinparticularthatinthisexamplethe
‘1’atthestartofthesecondcyclerepresentsthevalue10,the‘2’representsthevalue20
and so on. From the graph, the straight line relationship between logy andxis evident.
It is therefore reasonable to assume that the relationship betweenyandxis of the form
y =ab x .
Tofindthegradientofthegraphwecanchooseanytwopointsontheline,forexample
Cand B. The gradient isthen
log14.85 −log5.20
5 −1
= log(14.85/5.20)
4
Recall thatlogbisthe gradient ofthe lineand so
= 0.1139
logb = 0.1139, thatisb= 10 0.1139 = 1.2999
The vertical intercept isloga. From the graph the vertical intercept islog4 so that
loga=log4,
thatisa=4
We conclude that the relationship betweenyandxisgiven byy = 4(1.3) x .

96 Chapter 2 Engineering functions
log y
D
(12, 93.19)
1
9
8
7
6
5
Second cycle
4
3
2
C (5, 14.85)
B (1, 5.20)
1
9
8
7
6
5
A (0, 4)
First cycle
4
3
2
0 1 2 3 4 5 6 7 8 9 10 11 12 x
Figure2.37
The log--lineargraph isastraightline.
1
Engineeringapplication2.14
Bodeplotofalinearcircuit
It is possible to find out a great deal about an electronic circuit by measuring how
it responds to a sinusoidal input signal. We will examine the sinusoidal functions in

2.4 Review of some common engineering functions and techniques 97
Section 3.4. The usual procedure is to apply a range of fixed-amplitude sinusoidal
signals with different frequencies in order to obtain information about the circuit or
system.InSection23.9wewillseethatifthecircuitislinearthen,afterithassettled
down, the output signal is also a sinusoidal signal of the same frequency but with a
different amplitude and phase (see Section 3.7 fordetails of these terms).
ABode plotconsists of two components:
(1) The ratio of the amplitudes of the output signal and the input signal is plotted
against frequency.
(2) Thephaseshiftbetweentheinputandoutputsignalsisplottedagainstfrequency.
A log scale is used for the frequency in order to compress its length; for example,
a typical frequency range is 0.1 to 10 6 Hz which corresponds to a range of −1 to 6
on a log scale. A log scale is also used for the ratio of the signal powers as this is
calculatedindecibels.Phaseshiftisplottedonalinearscale.Sothesignalamplitude
ratio versus frequency is a log--log graph and the phase shift versus frequency is a
linear-- log graph.
Anoperationalamplifierisanexampleofalinearcircuit.Inatechnicalcomputing
language such as MATLAB ® itisusually easytoproduce a Bode plot.
An example function which describes the behaviour ofsuch a device is:
A(f)= 100000
1 + j f 8
which gives the input-output voltage function. Do not worry about the meaning of
the ‘j’ in the equation for now. This will be covered in Chapter 9 when we discuss
complex numbers.
We could produce a Bode plot by typing:
f=1:1:10000;
semilogx(f, 20*log10(abs(100000./(1+j*f/8))));
The function semilogx plots a graph with a logarithmic scale on thex-axis. Chapter
21 examines Bode plotsinmore detail.
EXERCISES2.4.4
1 Evaluate
(a) log 2 8 (b) log 2 15
(c) log 16 50 (d) log 16 123
(e) log 8 23 (f) log 8 47
2 Simplify eachofthe following to asingle logterm:
(a) log7 +logx
(b) logx+logy+logz
(c) lny−ln3
(d) 2logy+logx
(e) ln(xy) +ln(y 2 )
(f) ln(x +y) −lny
(g) log(2x 2 ) +log(4x)
(h) 3logy+2logx
1
(i) 2 logx 4 −2logx
(j) 3lnx+2lny+4lnz
(k) logz−2logx+3logy
(l) logt 3 −log(2t) +2logt
3 Simplify each ofthe following to asingle log term:
(a) 3lnt−lnt
(b) 6logt 2 +4logt

98 Chapter 2 Engineering functions
(c) ln(3y 6 )−2ln3+lny
(d) ln(6x +4) −ln(3x +2)
(e) log(9x) ( 2
−log
2 3x)
4 Sketch graphs ofthe followingfunctions,usingthe
same axes:
y=ln(2x),
y=lnx 0<x10
Measurethe vertical distancebetweenthe graphs for
x = 1,x = 2 andx = 8. Canyou explain your
findings usingthe laws oflogarithms?
5 Solve the following equations:
(a) e x = 70 (b) e x = 1 3
(c) e −x = 1 (d) 3e x = 50
(e) e 3x = 50 (f) e 2x+3 = 300
(g) e −x+1 = 0.75 (h) 2ee 2x = 50
3
(i)
e x +1 = 0.6 (j) 3
= 0.6
ex+1 (k) (e x ) 3 = 200 (l) √ e 2x =2
(m) √ e 2x e x
+4=6 (n)
e x +2 = 0.7
(o) e 2x = 7e x (p) 2e −x = 9
(q) (e x +3) 2 =25 (r) (3e −x −6) 3 =8
(s) e 2x −3e x +2=0 (t) 2e 2x −7e x +3=0
(u) e x (5−e x )=6 (v) e x −7+ 12
e x = 0
6 Solve the following equations:
(a) 10 x = 30 (b) 10 x = 0.25
(c) 4(10 x ) = 20 (d) 10 2x = 90
(e) 10 3x−2 = 20 (f) 3(10 x+3 ) = 36
(g) 10 −3x = 0.02 (h) 7(10 −2x ) = 1.4
(i) 10 x−2 = 20 (j) 10 3x+1 = 75
4
(k)
10 x = 6 (l) (10−x ) 2 = 40
(m) √ 10 4x 10 −x
= 3 (n)
2 +10 −x = 1 2
(o) √ 10 2x +6 = 5 (p) 10 6x = 30(10 3x )
(q) (10 −x +2) 2 = 6 (r) 6(10 −3x ) = 10
(s) 10 2x −7(10 x ) +10 = 0
(t) 10 4x −8(10 2x ) +16 = 0
(u) 10 x −5 +6(10 −x ) = 0
(v) 4(10 2x ) −8(10 x ) +3 = 0
7 Solve
(a) logx = 1.6 (b) log2x = 1.6
(c) log(2 +x) = 1.6 (d) 2log(x 2 ) = 2.4
(e) log(2x −3) = 0.7
8 Solve
(a) lnx=2.4
(b) ln3x=4
(c) 2ln(2x
(
−1)
)
= 5 (d) ln(2x 2 ) = 4.5
x +1
(e) ln = 0.9
3
9 Solve
(a) e 3x = 21 (b) 10 −2x = 6.7
1
(c)
e −x +2 = 0.3 (d) 2e(x/2) −1 = 0
(e) 3(10 (−4x+6) ) = 17
(f) (e x−1 ) 3 +e 3x = 500
(g) √ 10 2x +100 = 3(10 x )
10 Calculate the voltagegain in decibelsofthe following
amplifiers:
(a) input signal = 0.1V, outputsignal =1V;
(b) input signal = 1 mV, outputsignal = 10V;
(c) input signal = 5 mV, outputsignal = 8 V;
(d) input signal = 60mV, outputsignal =2V.
11 An audio amplifierconsistsoftwo stages:a
preamplifierandamainamplifier. Given the
following data,calculatethe voltagegain in decibels
ofthe individual stagesandthe overall gain in
decibelsofthe audio amplifier:
preamplifier:
main amplifier:
input signal = 10mV,
outputsignal = 200 mV
input signal = 400 mV,
outputsignal = 3 V
12 ABluetooth radio system operating in Class3has a
maximumoutputpower of0dBm at2.45 GHz. Find
the maximumoutputpower in watts (W).
13 Amicrowave oven has an outputpower of1kW.
Expressthisfigurein dBm.
14 Express0dBSPLasasound pressure in
micropascals (µPa).
15 Acar audio speaker has an outputsound pressure
level of55dBSPLwhenmeasuredat adistanceof
1 m. Calculate the sound pressure atthatpoint in
pascal r.m.s.
16 An active sonarsystemfitted to aboatproducesa
sourcelevel of220 dBre 1 µPa at1m. Calculate the
sound pressure in kPa.

2.4 Review of some common engineering functions and techniques 99
17 Byusinglog--linear paper findthe relationship
betweenxandygiven the following table ofvalues:
x 1.5 1.7 3.2 3.9 4.3 4.9
y 8.5 9.7 27.6 44.8 59.1 89.6
18 Byusinglog--logpaper findthe relationshipbetween
x andygiven the following table ofvalues:
x 2.0 2.5 3.0 3.5 4.0 4.5
y 13.0 19.0 25.9 33.6 42.2 51.6
Solutions
1 (a) 3 (b) 3.9069
(c) 1.4110 (d) 1.7356
(e) 1.5079 (f) 1.8515
2 (a) log7x (b) logxyz
( )
y
(c) ln (d) logxy 2
3
( )
(e) ln(xy 3 x +y
) (f) ln
y
(g) log8x 3 (h) logx 2 y 3
(i) log1 = 0 (j) lnx 3 y 2 z 4
( ) ( )
y 3 z t 4 (k) log
x 2 (l) log
2
3 (a) 2lntorlnt 2 (b) 16logt
( )
y 7 (c) ln (d) ln2
3
( )
9x 3/2
(e) log
2
5 (a) 4.2485 (b) −1.0986
(c) 0 (d) 2.8134
(e) 1.3040 (f) 1.3519
(g) 1.2877 (h) 1.1094
(i) 1.3863 (j) 0.6094
(k) 1.7661 (l) 0.6931
(m) 1.7329 (n) 1.5404
(o) 1.9459 (p) −1.5041
(q) 0.6931 (r) −0.9808
(s) 0, 0.6931
(t) −0.6931,1.0986
(u) 0.6931,1.0986 (v) 1.0986,1.3863
6 (a) 1.4771 (b) −0.6021
(c) 0.6990 (d) 0.9771
(e) 1.1003 (f) −1.9208
(g) 0.5663 (h) 0.3495
(i) 3.3010 (j) 0.2917
(k) −0.1761 (l) −0.8010
(m) 0.2386 (n) −0.3010
(o) 0.6395 (p) 0.4924
(q) 0.3473 (r) −0.0739
(s) 0.3010,0.6990 (t) 0.3010
(u) 0.3010,0.4771 (v) −0.3010,0.1761
7 (a) 39.81 (b) 19.91 (c) 37.81
(d) ±3.98 (e) 4.01
8 (a) 11.02 (b) 18.20 (c) 6.59
(d) ±6.71 (e) 6.38
9 (a) 1.0148 (b) −0.4130 (c) −0.2877
(d) −1.3863 (e) 1.3117 (f) 2.0553
(g) 0.5485
10 (a) 20dB (b) 80 dB (c) 64.1 dB
(d) 30.46dB
11 Preamplifier gain = 26.02 dB, mainamplifiergain
= 17.50 dB,total gain = 43.52 dB
12 10 −3 W or1mW
13 60 dBm
14 20 µPa
15 0.011Paor11 mPa
16 10 5 or100 kPa
17 y=3(2 x )
18 y = 4x 1.7

100 Chapter 2 Engineering functions
TechnicalComputingExercises2.4.4
1 Useatechnical computing language to draw
y =log(kx)for0.5 x 50fork =1,2,3and4.
( 1
2 Drawy=lnxandy=ln for
x)
0.5 x20. What doyou observe? Can you explain
your observation usingthe laws oflogarithms?
3 Drawy=lnxandy=1− x 3 for
0.5 x4.Fromyourgraphs statean approximate
solution to
lnx=1− x 3
2.4.5 Thehyperbolicfunctions
The hyperbolic functions are y(x) = coshx, y(x) = sinhx, y(x) = tanhx, y(x) =
sechx, y(x) = cosechx and y(x) = cothx. Cosh is a contracted form of ‘hyperbolic
cosine’, sinh of‘hyperbolic sine’ and so on. We definecoshx and sinhx by
y(x) =coshx = ex +e −x
2
y(x) =sinhx = ex −e −x
2
Note:
cosh(−x) = e−x +e x
= coshx
2
sinh(−x) = e−x −e x
=−sinhx
2
so, for example, cosh1.7 = cosh(−1.7) and sinh(−1.7) = −sinh1.7. Clearly, hyperbolicfunctionsarenothingotherthancombinations
oftheexponential functionse x and
e −x .However,theseparticularcombinationsoccursofrequentlyinengineeringthatitis
worth introducing the coshx and sinhx functions. The remaining hyperbolic functions
are defined interms of coshx and sinhx.
y(x) =tanhx = sinhx
coshx = ex −e −x
e x +e −x
y(x) =sechx = 1
coshx = 2
e x +e −x
y(x) = cosechx = 1
sinhx = 2
e x −e −x
y(x) =cothx = coshx
sinhx = 1
tanhx = ex +e −x
e x −e −x
Values of the hyperbolic functions for various x values can be found from a scientific
calculator. UsuallyaHyp button followed by a Sin, Cos or Tan button isused.
Example2.19 Evaluate
(a) cosh3
(c) tanh1.6
(e) coth1
(b) sinh(−2)
(d) sech(−2.5)
(f) cosech(−1)

2.4 Review of some common engineering functions and techniques 101
Solution (a) cosh3 = 10.07
(b) sinh(−2) = −3.627
(c) tanh(1.6) = 0.9217
(d) sech(−2.5) = 1/cosh(−2.5) = 0.1631
(e) coth1 = 1/tanh1 = 1.313
(f) cosech(−1) = 1/sinh(−1) = −0.8509
Graphs of the functions sinhx, coshx and tanhx can be obtained using a graphics
calculator.
Hyperbolicidentities
Severalidentitiesinvolvinghyperbolicfunctionsexist.Theycanbeverifiedalgebraically
usingthe definitions given, and are listedfor reference.
cosh 2 x −sinh 2 x = 1
1 −tanh 2 x = sech 2 x
coth 2 x −1 = cosech 2 x
sinh(x ±y) = sinhxcoshy ±coshxsinhy
cosh(x ±y) = coshxcoshy ±sinhxsinhy
sinh2x = 2sinhxcoshx
cosh2x = cosh 2 x +sinh 2 x
cosh 2 x =
sinh 2 x =
cosh2x +1
2
cosh2x −1
2
Note also that
e x = coshx +sinhx
e −x = coshx −sinhx
Hence any combination of exponential terms may be expressed as a combination of
coshx and sinhx, and viceversa.
Example2.20 Express
(a) 3e x −2e −x intermsofcoshx and sinhx,
(b) 2sinhx +coshx intermsofe x and e −x .
Solution (a) 3e x −2e −x = 3(coshx +sinhx) −2(coshx −sinhx) = coshx +5sinhx.
(b) 2sinhx+coshx=e x −e −x + ex +e −x
= 3ex −e −x
.
2 2

102 Chapter 2 Engineering functions
Inversehyperbolicfunctions
The inverse hyperbolic functions use the familiar notation. Both y = sinhx and y =
tanhx are one-to-one functions and no domain restriction is needed for an inverse to be
defined.However,on (−∞,∞),y = coshxisamany-to-onefunction.Ifthedomainis
restricted to [0,∞) the resulting function is one-to-one and an inverse function can be
defined.
The inverse of the function sinhx is denoted by sinh −1 x. Here the −1 must not be
interpreted as a power but rather the notation we use for the inverse function. Similarly
the inverses of coshx and tanhx aredenoted by cosh −1 x and tanh −1 x respectively.
Values of sinh −1 x, cosh −1 x and tanh −1 x can be obtained using a scientific
calculator.
Example2.21 Evaluate
(a) cosh −1 (3.7) (b) sinh −1 (−2) (c) tanh −1 (0.5)
Solution Using a calculator we get
(a) 1.9827 (b) −1.4436 (c) 0.5493
Engineeringapplication2.15
Capacitancebetweentwoparallelwires
Althoughitmaynotseemobvious,asmallcapacitanceexistsbetweentwowiresthat
run close to each other and at certain frequencies this can be a significant factor for
someelectrical systems.
Themutualcapacitancepermetre,C,betweentwolongparallelwiresinaireach
havingaradiusrmetresandwiththewirecentresseparatedbyd metresiscalculated
using
πε
C = 0
cosh −1 (d/2r)
Thisexpressionincludesaninversehyperbolicfunction.Intheequationd>2r,otherwise
the wires would be overlapping. The constant ε 0
is a fundamental physical
constant called the permittivity of free space. It has an approximate value of
8.85 ×10 −12 Fm −1 .
Recall the general expression forthe hyperbolic function coshx:
coshx = ex +e −x
2
To derive the inverse of this hyperbolic function, we need to restrict the domain to
[0, ∞), that isx 0.We lety = coshx and then solve forxintermsofy:
y = ex +e −x
2
2y=e x +e −x
0=e x −2y+e −x

2.4 Review of some common engineering functions and techniques 103
Inorder tosolve this equation wefirst multiplyboth sides by e x , togive
0 =e x e x −e x 2y+e x e −x
Using the laws of indices,
0=e 2x −2e x y+1
We can now write this as a quadratic equation in e x and solve it using the standard
formula.This gives
(e x ) 2 −2y(e x )+1=0
e x = −b ± √ b 2 −4ac
2a
= −(−2y) ± √ (−2y) 2 −4(1)(1)
2(1)
= 2y ± √ 4y 2 −4
=y ± √ y
2
2 −1
Asx 0,the function e x 1 so wereject the negative rootand itfollows that
e x =y+ √ y 2 −1
from which we get
(
x=ln y + √ )
y 2 −1
Recall thaty =coshxand sox = cosh −1 y. Therefore
(
cosh −1 y = ln y + √ )
y 2 −1
Hencetheequationforthepairofwirescanalsobewrittenintermsoflogarithmsas
πε
C = [ 0
ln (d/2r) + √ ]
(d/2r) 2 −1
EXERCISES2.4.5
A word of warning about the inverse of coshx is needed. The calculator returns a
value of 1.9827 for cosh −1 (3.7). Note, however, that cosh(−1.9827) = 3.7. The value
−1.9827 is not returned by the calculator; only values on [0,∞) will be returned. This
isbecause the domainofy = coshx isrestrictedtoensureithas aninverse function.
1 Evaluate the following:
(a) sinh3 (b) cosh1.6
(c) tanh0.95 (d) sech1
(e) cosech2 (f) coth1.5
(g) cosh(−3) (h) cosech(−1.6)
(i) sinh(−2) (j) coth(−2.7)
(k) tanh(−1.4) (l) sech(−0.5)
2 Evaluate
(a) sinh −1 3
(b) cosh −1 2
(c) tanh −1 (−0.25)
3 Express
(a) 6e x +5e −x in terms ofsinhx andcoshx,
(b) 4e 2x −3e −2x in terms ofsinh2x andcosh2x,
(c) 2e −3x −5e 3x in terms ofsinh3x andcosh3x.
4 Express
(a) 4sinhx +3coshx in terms ofe x ande −x ,
(b) 3sinh2x −cosh2x in termsofe 2x ande −2x ,
(c) 3cosh3x −0.5sinh3x in terms ofe 3x ande −3x .
5 Expressae x +be −x ,whereaandbare constants,in
terms ofcoshx andsinhx.

104 Chapter 2 Engineering functions
6 Expressacoshx +bsinhx,whereaandbare
constants,in terms ofe x and e −x .
7 Show that the pointx = coshu,y = sinhulies onthe
curve
x 2 −y 2 =1
8 Provethehyperbolicidentitieslistedintheboxearlier
in thissection.
Solutions
1 (a) 10.0179 (b) 2.5775 (c) 0.7398
(d) 0.6481 (e) 0.2757 (f) 1.1048
(g) 10.0677 (h) −0.4210 (i) −3.6269
(j) −1.0091 (k) −0.8854 (l) 0.8868
2 (a) 1.8184 (b) 1.3170 (c) −0.2554
3 (a) 11coshx +sinhx (b) cosh2x +7sinh2x
(c) −3cosh3x −7sinh3x
4 (a) 3.5e x −0.5e −x
(b) e 2x −2e −2x
(c) 1.25e 3x +1.75e −3x
5 (a+b)coshx+(a−b)sinhx
6
a +b
2 ex + a −b
2 e−x
TechnicalComputingExercises2.4.5
1 Useatechnicalcomputing language to draw
(a)y = sinhx (b)y = coshx
(c)y=tanhxfor−5x5.
2 Draw graphs ofy = sinhx,y = coshx andy = ex
2 for
0 x5. What happens to the threegraphs asx
increases? Canyou explain thisalgebraically?
3 Draw
(a) y=sinh −1 x
(b) y=cosh −1 x
(c) y=tanh −1 x
−5x5
1x5
−1<x<1
2.4.6 Themodulusfunction
Themodulusofapositivenumberissimplythenumberitself.Themodulusofanegative
number is a positive number of the same magnitude. For example, the modulus of 3 is
3; the modulus of −3 is also 3. We enclose the number in vertical lines to show we are
findingits modulus,thus
|3|=3 |−3|=3
Mathematically wedefinethe modulusfunctionasfollows:
The modulus
{
functionisdefinedby
x x0
f(x)=|x|=
−x x<0
Figure2.38illustratesagraphof f (x) = |x|.Themodulusofaquantityisnevernegative.
Consider two points on thexaxis,aandb, asshown inFigure 2.39. Then
|a −b| = |b −a| = distancefromatob

2.4 Review of some common engineering functions and techniques 105
f (x)
a
0
b
x
|a – b|
Figure2.38
The function: f (x) = |x|.
Figure2.39
Distance fromatob=|a −b|.
Example2.22 Find the distance from
(a)x=2tox=9 (b)x=−2tox=9 (c)x=−2tox=−9
Solution (a) Distance = |2 −9| = | −7| = 7
(b) Distance=|−2−9|=|−11|=11
(c) Distance = | −2− (−9)| = |7| = 7
From the definition of the modulus function itfollows:
If |x| =a,then eitherx =aorx = −a.
If|x|<a,then−a<x<a.
If |x| >a, then eitherx >aorx < −a.
For example, if |x| = 4 then eitherx = 4 orx = −4. If |x| < 4 then −4 < x < 4; that
is,xlies between −4 and 4. If |x| > 4, then eitherx > 4 orx < −4; that is,xis either
greater than 4 orless than −4.
Example2.23 Describe the interval on thexaxis defined by
(a) |x| < 2
(b) |x| 3
(c) |x−1|<3
(d) |x+2|>1
Solution (a) |x| < 2 is the same statement as −2 < x < 2; that is,xis numerically less than 2.
Figure2.40illustratesthisregion.Notethattheregionisanopeninterval.Sincethe
pointsx = −2 andx = 2 arenot included, they areshown on the graph as ◦.
(b) If |x| 3theneitherx 3orx −3.ThisisshowninFigure2.41.Therequired
region of thexaxis has two distinct parts. Since the pointsx = 3 andx = −3 are
included inthe interval of interest,they are shown on the graph as •.
(c) |x −1| < 3 isequivalent to −3 <x −1 < 3,thatis −2 <x<4.
(d) |x+2|>1isequivalenttox+2>1orx+2<−1,thatisx>−1orx<−3.

106 Chapter 2 Engineering functions
–2
0
|x| < 2
Figure2.40
The quantity |x| < 2 is equivalent to −2 <x<2.
2
–3
0 3
|x| > 3 |x| > 3
Figure2.41
Thequantity |x| 3isequivalenttox 3orx −3.
Themodulusfunctioncan beusedtodescribe regionsinthex--yplane.
Example2.24 Sketch the region defined by
(a) |x| < 2and |y| < 1
(b) |x 2 +y 2 | 9
Solution (a) The region is a rectangle as shown in Figure 2.42. The boundary is not part of the
regionasstrictinequalitieswereusedtodefineit.Theregion |x| 2, |y| 1isthe
sameasthatinFigure 2.42 with the boundaryincluded.
(b) |x 2 +y 2 | 9isequivalentto −9 x 2 +y 2 9.Note,however,thatx 2 +y 2 isnever
negative and so the region isgiven by0 x 2 +y 2 9.
Let P(x,y) be a general point as shown in Figure 2.43. Then from Pythagoras’s
theorem,the distance fromPtothe originis √ x 2 +y 2 .So,
Then,
(distance fromorigin) 2 =x 2 +y 2 9
(distance fromorigin) 3
This describes a disc, centre the origin, of radius 3 (see Figure 2.44).
y
1
y
P (x, y)
y
3
–2
–1
2
x
O
r
x
–3
–3
3 x
Figure2.42
Theregion: |x| < 2 and |y| < 1.
Figure2.43
PointP(x,y) is ageneral point.
Figure2.44
Theregion: |x 2 +y 2 | 9.
Engineeringapplication2.16
Full-waverectifier
A rectifier is an electronic circuit that is used to convert an alternating voltage to
a direct voltage, that is, one that does not change in polarity with time. Usually
the alternating voltage has the shape of a sinusoidal function. We will examine the

2.4 Review of some common engineering functions and techniques 107
sinusoidalfunctionsinSection3.4.Afull-waverectifierisonethatmakesuseofthe
negativepartoftheinputsinusoidalsignalaswellasthepositivepart,byreversingits
polarity.Itcontrastswithahalf-waverectifierthatmerelydiscardsthenegativepart
of the input sinusoidal signal. This means that the half-wave circuit is not driven by
the alternating voltage supply at all during half of the cycle. In both types of circuit
a capacitor is usually connected across the output to store energy when the signal
approaches a peak. The capacitor then discharges into the circuit when the rectifier
output voltage falls. By this means, a relatively constant voltage is produced at the
power supply output.
Afullyrectifiedsinewave isthemodulusofthesinewave.ThecircuitforafullwaverectifierisshowninFigure2.45togetherwiththeinputandoutputwaveforms.
The input signal is v in
and the output signal is v o
. Ignoring the voltage drops across
the diodes gives
v o
= |v in
|
v in
v o
v o
t
v in
t
Figure2.45
A full-wave rectifier.
EXERCISES2.4.6
1 Sketchthe interval defined by
(a) |x|>4 (b) |y−1|3
(c) |t+6|3
(d) |t 2 −2|<7
(e) |t| < −1
Also, state the intervals withoutusingthe modulus
signs.
2 Sketchthe regionsdefined by
(a) |x| > 2, |y| < 3
(b) |x+2|<4,|y+1|<3
(c) |x 2 +y 2 −1|4
(d) |(x−1) 2 + (y+2) 2 | 1
3 Express the following intervalsusingmodulus
notation:
(a) −2x2
(c)1<y<3
(b) −4<t<4
(d) −6t0
Solutions
1 (a)x>4orx<−4 (b) −2y4
(c)t −3ort −9 (d) −3<t<3
(e) novalue oft satisfiesthis
3 (a) |x| 2 (b) |t| <4
(c) |y−2|<1
(d) |t+3|3

108 Chapter 2 Engineering functions
TechnicalComputingExercises2.4.6
1 Useatechnical computing language suchas
MATLAB ® to produce a plotofthe output of:
(a) afull wave rectifier circuit
(b) ahalfwave rectifier circuit
f (t)
2.4.7 Therampfunction
Theramp functionisdefined by
slope = c
f(t)=
{
ct t 0 cconstant
0 t<0
Figure2.46
Theramp function.
t
Its graph isshown inFigure 2.46.
Engineeringapplication2.17
Telescopedrivesignal
Inordertotrackthemotionofthestarslargetelescopesareusuallydrivenbyanelectricmotor.Thespeedofthismotoriscontrolledinorderthattheangularpositionof
thetelescopefollowsaspecifiedtrajectorywithtime.Thewholeassembly,including
telescope, gears, motor, controller and sensors, forms a position control system or
servo-system. This servo-system must be fed a signal corresponding to the desired
trajectoryofthesystem.OftenthistrajectoryisarampfunctionasillustratedinFigure2.47.Thedrivemotorisstartedattimet
=t 0
andthedesiredangularpositionof
the telescopeis θ.
u
Desired telescope
position
Drive motor started
t 0
t
Figure2.47
Tracking signal foratelescope.
2.4.8 Theunitstepfunction,u(t)
The unit stepfunctionisdefinedby
{ 1 t0
u(t) =
0 t<0

2.4 Review of some common engineering functions and techniques 109
u (t)
1
u (t – d)
1
Figure2.48
Theunit stepfunction.
t
Figure2.49
Graph ofu(t −d).
d
t
Its graph is shown in Figure 2.48. Note thatu(t) has a discontinuity att = 0. The point
(0, 1) is part of the function defined ont 0. This is depicted by •. The point (0, 0) is
notpartof the function defined ont < 0.We use ◦toillustratethis.
The position of the discontinuity may be shifted.
{ 1 td
u(t−d)=
0 t<d
The graph ofu(t −d) isshown inFigure 2.49.
Engineeringapplication2.18
RLCcircuit
Theinductorisanotherofthethreefundamentalelectroniccomponents.Itisadevice
that is used to store electrical energy in the form of a concentrated magnetic field. It
consistsofatightlywoundcoil,oftenaroundaferromagneticmaterial,whichhelps
to strengthen the magnetic field. It is usually depicted on a circuit diagram as a set
ofloops.Figure2.50showsaresistor,inductorandacapacitorconnectedinparallel.
This is usually known as an RLC circuit. The circuit also contains a switch, which
here is depicted as an arrow connecting two terminals with the switch in the open
position. When the switch is open the voltage v across terminals A and B is zero. If
theswitchisclosedatt = 0thevoltageacrossAandBisV S
,whereV S
isthesupply
voltage. Thus v can bemodelled by the function
{ 0 t<0
v(t) =
t0
V S
This can bewritten,usingthe unit stepfunction, as
v(t) =V S
u(t)
+
A
V S
–
L R C
B
Figure2.50
RLCcircuit.

110 Chapter 2 Engineering functions
Example2.25 Sketch the following functions:
(a) f =u(t−3)
(c) f =u(t−1)−u(t−3)
(e) f =e t u(t)
(b) f =u(t−1)
(d) f =u(t−3)−u(t−1)
Solution (a) See Figure 2.51. (b) See Figure 2.52.
(c) See Figure 2.53. (d) See Figure 2.54.
(e) See Figure 2.55.
f
1
f
1
Figure2.51
Thefunction: f =u(t −3).
3 t
Figure2.52
The function: f =u(t −1).
1
t
f
1
f
f
Figure2.53
Thefunction:
f =u(t−1)−u(t−3).
1
3
t
–1
1 3
Figure2.54
Thefunction:
f =u(t−3)−u(t−1).
t
1
Figure2.55
Thefunction: f = e t u(t).
t
EXERCISES2.4.8
1 Sketch the followingfunctions:
(a) f(t) =u(t −1)
(b) f(t) =u(t +1)
(c) f(t)=u(t−2)−u(t−6)
(d) f(t) =3u(t −1)
(e) f(t)=u(t+1)−u(t−1)
(f) f(t)=u(t−1)−u(t+1)
(g) f(t) =u(t +1)−2u(t −1)
(h) f(t) =2u(t +1)−u(t −1)
(i) f(t) =3u(t −1)−2u(t −2)
2 Aramp function, f (t),is defined by
{ 2t t0
f(t)=
0 t<0
Sketch the following for −1 t 3:
(a) f(t)
(b) u(t)f(t)
(c) u(t −1)f(t)
(d) u(t −1)f(t)−u(t −2)f(t)
Solutions
2 SeeFigureS.5.

2.4 Review of some common engineering functions and techniques 111
f
6
4
2
u(t – 1)f
6
4
2
u(t – 1)f – u(t –2)f
6
4
2
(a) (b)
–1
0
1
2 3 t
(c)
–1
0
1
2 3 t
(d)
–1
0
1
2 3 t
FigureS.5
2.4.9 Thedeltafunctionorunitimpulsefunction, δ(t)
Consider the rectangle function,R(t), shown in Figure 2.56. The base of the rectangle
ish, the height is 1/h and so the area is 1. Fort >h/2 andt < −h/2, the function is 0.
Ashdecreases, the base diminishes and the height increases; the area remains constant
at1.
Ashapproaches 0, the base becomes infinitesimally small and the height infinitely
large. The area remains at unity. The rectangle function is then called a delta function
or unit impulse function. Ithas a valueof0everywhere except atthe origin.
δ(t) = rectanglefunction ashapproaches0
We write thisconcisely as
δ(t) =R(t) as h →0
The position of the delta function may be changed from the origin tot =d. Consider a
rectangle function,R(t −d), shown in Figure 2.57.R(t −d) is obtained by translating
R(t)anamountd totheright.Again,lettinghapproach0producesadeltafunction,this
time centredont =d.
δ(t−d)=R(t−d) as h→0
We have seen that the delta function can be regarded as bounding an area 1 between
itselfand the horizontalaxis.More generallythe area bounded by the function
f(t) =kδ(t)
isk.Wesaythatkδ(t)representsanimpulseofstrengthkattheorigin,andkδ(t −d)is
animpulseofstrengthkatt =d.Itisoftenusefultodepictsuchanimpulsebyanarrow
where the height of the arrow gives the strength of the impulse. A series of impulses is
oftentermedan impulse train.
R (t)
1–
h
R (t – d)
1–
h
– h–
2
h–
2
t
d – h– d d + h–
2 2
t
Figure2.56
The rectangle function,R(t).
Figure2.57
Thedelayed rectangle function,R(t −d).

112 Chapter 2 Engineering functions
Example2.26 Atrainof impulses isgiven by
f(t) = δ(t)+3δ(t −1)+2δ(t −2)
Depict the traingraphically.
Solution Figure 2.58 shows the representation. In Section 22.8 we shall call such a function a
series of weighted impulseswhere the weights are1,3and 2,respectively.
f (t)
3
2
1
0 1 2 t
Figure2.58
A trainofimpulsesgiven by
f(t) = δ(t)+3δ(t −1)+2δ(t −2).
Engineeringapplication2.19
Impulseresponseofasystem
An impulse signal is sometimes used to test an electronic system. It can be thought
ofasgiving the systemavery harsh joltforavery shortperiodoftime.
It is not possible to produce an impulse function electronically as no practical
signalcanhaveaninfiniteheight.However,anapproximationtoanimpulsefunction
isoftenused,consistingofapulsewithalargevoltage,V,andshortduration,T.The
strength of such an impulse isVT. When this pulse signal is injected into a system
the output obtained isknown astheimpulse response ofthe system.
The approximation is valid provided the width of the pulse is an order of magnitudelessthanthefastesttimeconstantinthesystem.IfthevalueofT
requiredissmall
inordertosatisfythisconstraint,thenthevalueofV mayneedtobelargetoachieve
thecorrectimpulsestrength,VT.Oftenthiscanruleoutitsuseformanysystemsas
the value ofV would thenbelargeenoughtodistort the systemcharacteristics.
EXERCISES2.4.9
1 Sketch the impulsetrain given by
(a) f(t) = δ(t −1)+2δ(t −2)
(b) f(t) =3δ(t)+4δ(t −2)+δ(t −3)

Review exercises 2 113
REVIEWEXERCISES2
1 Statethe ruleand sketch the graphofeach ofthe
followingfunctions:
(a) f(x)=7x−2
(b) f(t)=t 2 −2 0t5
(c) g(x)=3e x +4 0x2
(d) y(t) = (e 2t −1)/2 t 0
(e) f(x)=x 3 +2x+5 −2x2
2 Statethe domainand range ofthe functions in
Question1.
3 Determine the inverse ofeach ofthe following
functions:
(a) y(x) =2x (b) f(t) =8t −3
(c) h(x) = 2x +1 (d)m(r)=1−3r
3
(e) H(s) = 3 s +2
(g) f(t) =e 2t
(i) g(v) = ln(v +1)
4 Iff(t)=e t find
(a) f(2t)
(b) f(x)
(c) f(λ) (d) f(t − λ)
(f) f(v)=lnv
(h) g(v) =lnv+1
(j) y(t) = 3e t−2
5 Ifg(t) =ln(t 2 +1)find(a)g(λ),(b)g(t − λ).
6 Sketchthe following functions:
{ 0 t<0
(a) f(t) =
0.5t t0
⎧
⎨4 t<0
(b) f(t) = t 0t<3
⎩
2t 3t4
(c) f(t)=|e t |
−3t 3
7 Thefunction f (x) isperiodicwith a periodof2, and
f(x)=|x|,−1x1.Sketchffor−3x3.
8 Givena(t) =3t,b(t) =t +3andc(t) =t 2 −3write
expressions
(a) b(c(t))
(c) a(b(t))
(e) a(b(c(t)))
(b) c(b(t))
(d) a(c(t))
(f) c(b(a(t)))
9 Sketchthe following functions,stating any
asymptotes:
(a) y(x) = 3 +x
x
(b) y(x) =
2x
x 2 −1
(c) y(x) =3−e −x (d) y(x) = ex +1
e x
10 Simplify each expression asfar aspossible.
(a) e 2x e 3x
(c) e 4 e 3 e
(e)
( )
e x 2
e −x
(b) e x e 2x e −3x
e x
(d)
e −x
( )
2
(f) ln3x +ln
x
(g) 3lnt+2lnt 2 (h) e lnx
(i) ln(e x ) (j) e lnx2
(k) e 0.5lnx2 (l) ln(e 2x )
(m) e 2lnx (n) ln(e 3 ) +ln(e 2x )
11 Solve the following:
(a) e 4x = 200 (b) e 3x−6 = 150
(c) 9e −x = 54 (d) e (x2) = 60
1
(e) = 0.1
6+e−x 12 Solve the following:
(a) 0.5lnt = 1.2
(b) ln(3t +2) = 1.4
(c) 3ln(t−1)=6
(d) log 10 (t 2 −1) = 1.5
(e) log 10 (lnt) = 0.5
(f) ln(log 10 t) = 0.5
13 Express eachofthe following in terms ofsinhx and
coshx:
(a) 7e x +3e −x (b) 6e x −5e −x
(c) 3ex −2e −x 1
(d)
2 e x +e −x
e x
(e)
1+e x
14 Express eachofthe following in terms ofe x ande −x :
(a) 2sinhx +5coshx
(b) tanhx +sechx
(c) 2coshx − 3 4 sinhx
1
(d)
sinhx −2coshx
(e) (sinhx) 2

114 Chapter 2 Engineering functions
15 Describe the interval onthexaxisdefined by
(a) |x| < 1.5 (b) |x| > 2
(c) |x+3|<7 (d) |2x|6
(e) |2x−1|5
16 Sketch
(a) f=|−2t| −3t3
(b) f=−|2t| −3t3
(c) 2δ(t)−δ(t −1)+3δ(t +1)
Solutions
1 (a) Multiply input by 7, then subtract2
(b) Square the input, thensubtract 2
(c) Calculate e x ,wherexisthe input, multiply by3
and thenadd 4
(d) Multiply by2, calculate exponential, subtract1
and thendivide by 2
(e) Cube input, addto twice the input and add 5
2 (a) Domain (−∞, ∞)range (−∞, ∞)
(b) Domain [0,5]range [−2,23]
(c) Domain [0,2]range [7,3e 2 +4]
(d) Domain [0,∞) range [0,∞)
(e) Domain [−2,2]range [−7,17]
3 (a) y −1 (x) = x 2
(b) f −1 (t) = t +3
8
(c) h −1 (x) = 3 2 (x−1)
(d) m −1 (r) = 1 −r
3
(e) H −1 (s) = 3
s −2
(f) f −1 (v) =e v
(g) f −1 (t) = 1 2 lnt
(h) g −1 (v) = e v−1
(i) g −1 (v) =e v (
−1
)
(j) y −1 t
(t) = ln +2
3
4 (a) e 2t (b) e x (c) e λ (d) e t−λ
5 (a) ln(λ 2 +1) (b) ln((t − λ) 2 +1)
8 (a) t 2
(b) (t+3) 2 −3ort 2 +6t+6
(c) 3(t +3)
(d) 3(t 2 −3)
(e) 3t 2
(f) (3t+3) 2 −3or9t 2 +18t+6
10 (a) e 5x (b) 1 (c) e 8
(d) e 2x (e) e 4x (f) ln6
(g) lnt 7 (h) x (i) x
(j) x 2 (k) x (l) 2x
(m) x 2 (n) 3 +2x
11 (a) 1.3246 (b) 3.6702
(c) −1.7918 (d) ±2.0234
(e) −1.3863
12 (a) 11.0232 (b) 0.6851
(c) 8.3891 (d) ±5.7116
(e) 23.6243 (f) 44.5370
13 (a) 10coshx +4sinhx
(b) coshx +11sinhx
coshx +5sinhx
(c)
2
1
(d)
2coshx
coshx +sinhx
(e)
1+coshx+sinhx
7
14 (a)
2 ex + 3 2 e−x
e x −e −x +2
(b)
e x +e −x
5e x +11e −x
(c)
8
−2
(d)
e x +3e −x
( )
e x −e −x 2
(e)
2
15 (a) −1.5 <x<1.5
(b) x>2andx<−2
(c) −10<x<4
(d) x3andx−3
(e) −2x3

3 Thetrigonometric
functions
Contents 3.1 Introduction 115
3.2 Degreesandradians 116
3.3 Thetrigonometricratios 116
3.4 Thesine,cosineandtangentfunctions 120
3.5 Thesincxfunction 123
3.6 Trigonometricidentities 125
3.7 Modellingwavesusingsint andcost 131
3.8 Trigonometricequations 144
Reviewexercises3 150
3.1 INTRODUCTION
ManycommonengineeringfunctionswerestudiedinSection2.4.However,thetrigonometricfunctionsaresoimportantthattheydeserveseparatetreatment.Oftenalternating
currents and voltages can be described using trigonometric functions, and they occur
frequently inthe solution of various kinds of equations.
Section3.3introducesthethreetrigonometricratios:sinθ,cosθ andtanθ.Theseare
definedasratiosofthesidesofaright-angledtriangle.Thedefinitionsarethenextended
so that angles of any magnitude may be considered. Section 3.5 introduces an important
function, sincx. This is really a combination of the familiar functionsxand sinx,
but because this combination occurs frequently in some engineering applications it deservesspecialmention.TrigonometricidentitiesareintroducedinSection3.6,themost
common ones being tabulated. These identities are useful in simplifying trigonometric
expressions. Graphs of the trigonometric functions are illustrated. The application
of these functions to the modelling of waveforms is an important section. The chapter
closes with the solution of trigonometric equations.

116 Chapter 3 The trigonometric functions
3.2 DEGREESANDRADIANS
Angles can be measured in units of either degrees or radians. The symbol for degree
is ◦ . Usually no symbol is used to denote radians and this is the convention adopted in
this book.
A complete revolution is defined as 360 ◦ or 2π radians. It is easy to use this fact to
convert between the two measures. We have
Note that
360 ◦ = 2π radians
1 ◦ = 2π
360 = π
180 radians
1 radian = 180
π
π
radians = 90◦
2
3π
2
π radians = 180 ◦
radians = 270◦
degrees ≈ 57.3◦
Yourcalculatorshouldbeabletoworkwithanglesmeasuredinbothradiansanddegrees.
Usually the Mode button allows you toselect the appropriate measure.
3.3 THETRIGONOMETRICRATIOS
Considertheangle θ intheright-angledtriangleABC,asshowninFigure3.1.Wedefine
the trigonometricratiossine, cosineandtangent asfollows:
sinθ=
cosθ=
tanθ=
sideopposite toangle
hypotenuse
sideadjacent toangle
hypotenuse
= BC
AC
= AB
AC
sideopposite toangle
sideadjacent toangle = BC
AB
C
a
A
u
B
Figure3.1
A right-angled triangle, ABC.

Note that
tanθ = BC
AB = BC
AC × AC
AB = sinθ
cos θ
3.3 The trigonometric ratios 117
Note that when θ reduces to0 ◦ the length of the sideBC reduces tozero and so
sin0 ◦ =0, tan0 ◦ =0
Also when θ reduces to0 ◦ the lengths of AB and ACbecome equal and so
cos0 ◦ =1
Similarlywhen θ approaches90 ◦ ,thelengthsofBCandACbecomeequalandthelength
of ABshrinks tozero.Hence
sin90 ◦ = 1, cos90 ◦ = 0
Note also that the length of AB shrinks to zero as θ approaches 90 ◦ , and so tanθ approaches
infinity. We write thisas
tanθ→∞
asθ→90 ◦
The trigonometric ratios of 30 ◦ , 45 ◦ and 60 ◦ occur frequently in calculations. They can
be calculated exactly by considering the right-angled triangles shown inFigure 3.2.
sin45 ◦ = 1 √
2
, cos45 ◦ = 1 √
2
, tan45 ◦ =1
sin30 ◦ = 1 2 , cos30◦ =
sin60 ◦ =
√
3
2 , tan30◦ = 1 √
3
√
3
2 , cos60◦ = 1 2 , tan60◦ = √ 3
Mostscientificcalculatorshavepre-programmedvaluesofsinθ,cosθ andtanθ.Anglescanbemeasuredindegreesorradians.Wewilluseradiansunlessstatedotherwise.
If welet ̸ ACB = α (see Figure 3.1),then
and
sinα= AB
AC =cosθ
cosα= BC
AC =sinθ
45°
2
1
45°
1
30°
2
3
60°
1
Figure3.2
Thetrigonometric ratiosfor30 ◦ ,
45 ◦ and60 ◦ canbe foundexactly
fromthese triangles.

118 Chapter 3 The trigonometric functions
But
α+θ= π 2
Hence,
( ) π
sinθ = cos
2 − θ
( ) π
cos θ = sin
2 − θ
Since θ is an angle in a right-angled triangle it cannot exceed π/2. In order to define
the sine, cosine and tangent ratios for angles larger than π/2 we introduce an extended
definition which isapplicable toangles of any size.
Consider an arm,OP, fixed atO, which can rotate (see Figure 3.3). The angle, θ, in
radians,betweenthearmandthepositivexaxisismeasuredanticlockwise.Thearmcan
beprojectedontoboththexandyaxes.TheseprojectionsareOAandOB,respectively.
Whetherthearmprojectsontothepositiveornegativexandyaxesdependsuponwhich
quadrant the armissituatedin. The length of the armOPisalways positive. Then,
sinθ= projectionofOPontoyaxis
OP
cosθ=
tanθ=
projection ofOPontoxaxis
OP
= OB
OP
= OA
OP
projection ofOP ontoyaxis
projection ofOPontoxaxis = OB
OA
In the first quadrant, that is 0 θ < π/2, both thexandyprojections are positive,
so sinθ, cos θ and tanθ are positive. In the second quadrant, that is π/2 θ < π, the
x projection,OA, is negative and theyprojection,OB, positive. Hence sinθ is positive,
and cosθ and tanθ are negative. Both thexandyprojections are negative for the third
quadrantandsosinθ andcosθ arenegativewhiletanθ ispositive.Finally,inthefourth
quadrant, thexprojection is positive and theyprojection is negative. Hence, sinθ and
tan θ are negative, and cosθ is positive (see Figure 3.4). The sign of the trigonometric
ratios can be summarized by Figure 3.5.
For angles greater than 2π, the armOP simply rotates more than one revolution before
coming to rest. Each complete revolution bringsOP back to its original position.
Second quadrant
P
y
B
u
First quadrant
A
O
x
Third quadrant
Fourth quadrant
Figure3.3
An arm,OP,fixed atO, which can rotate.

3.3 The trigonometric ratios 119
y
y
B
P
P
B
u
u
O A x
A
O
x
First quadrant
y
Second quadrant
y
A
O
u
x
O
u
A
x
P
B
B
P
Third quadrant
Fourth quadrant
Figure3.4
Evaluating the trigonometric ratios in each ofthe fourquadrants.
So,for example,
sin(8.76) = sin(8.76 −2π) = sin(2.477) = 0.617
cos(14.5) = cos(14.5 −4π) = cos(1.934) = −0.355
Negative angles are interpreted as a clockwise movement of the arm. Figure 3.6 illustrates
an angle of −2. Note that
sin(−2) = sin(2π −2) = sin(4.283) = −0.909
sinceananticlockwise movement ofOPof4.283radians would resultinthe armbeing
inthe same position as a clockwise movement of 2 radians.
y
sin u > 0
cos u < 0
tan u < 0
sin u > 0
cos u > 0
tan u > 0
y
sin u < 0
cos u < 0
tan u > 0
sin u < 0
cos u > 0
tan u < 0
x
O
u = –2
x
Figure3.5
Sign ofthe trigonometric ratios in
each ofthe four quadrants.
P
Figure3.6
Illustrationofthe angle θ = −2.

120 Chapter 3 The trigonometric functions
Thecosecant,secantandcotangentratiosaredefinedasthereciprocalsofthesine,
cosine and tangent ratios.
cosec θ = 1
sinθ
secθ = 1
cos θ
cotθ = 1
tanθ
Example3.1 An angle θ issuchthatsinθ > 0 and cosθ < 0.Inwhichquadrantdoes θ lie?
Solution FromFigure3.5weseethatsinθ > 0when θ isinthefirstandsecondquadrants.Also,
cosθ < 0when θ isinthesecondandthirdquadrants.Forbothsinθ > 0andcosθ < 0
thus requires θ to be in the second quadrant. Hence sin θ > 0 and cosθ < 0 when θ is
inthe secondquadrant.
EXERCISES3.3
1 Verify usingascientificcalculator that
(a)sin30 ◦ = sin390 ◦ = sin750 ◦
(b)cos100 ◦ = cos460 ◦ = cos820 ◦
(c)tan40 ◦ = tan220 ◦ = tan400 ◦
(d)sin70 ◦ = sin(−290 ◦ ) = sin(−650 ◦ )
(e)cos200 ◦ = cos(−160 ◦ ) = cos(−520 ◦ )
(f) tan150 ◦ = tan(−30 ◦ ) = tan(−210 ◦ )
2 Verify the following usingascientificcalculator. All
angles are in radians.
(a) sin0.7 = sin(0.7 +2π) = sin(0.7 +4π)
(b) cos1.4 = cos(1.4 +8π) = cos(1.4 −6π)
(c) tan1 = tan(1 + π) = tan(1 +2π) = tan(1 +3π)
(d) sin2.3 = sin(2.3 −2π) = sin(2.3 −4π)
(e) cos2 = cos(2 −2π) = cos(2 −4π)
(f) tan4 = tan(4 − π) = tan(4 −2π) = tan(4 −3π)
3 An angle θ is suchthat cos θ > 0 andtan θ < 0.In
whichquadrant does θ lie?
4 An angle α is suchthattan α > 0 andsin α < 0.In
whichquadrant does α lie?
Solutions
3 4thquadrant 4 3rd quadrant
3.4 THESINE,COSINEANDTANGENTFUNCTIONS
The sine, cosine and tangent functions follow directly from the trigonometric ratios.
These are defined to be f (x) = sinx, f (x) = cosx and f (x) = tanx. Graphs can be
constructedfromatableofvaluesfoundusingascientificcalculator.Theyareshownin
Figures3.7 to3.9. Notethatthese functions aremany-to-one.

3.4 The sine, cosine and tangent functions 121
f (x)
f (x)
1
1
p
– –2
p
–p
p
–
2
2p
3p
x
–p
p
2p
x
–1
–1
Figure3.7
Graphof f(x) = sinx.
Figure3.8
Graphof f(x) = cosx.
f (x)
0 p 2p x
Figure3.9
Graphof f(x) = tanx.
By shifting the cosine function to the right by an amount π/2 the sine function is
obtained. Similarly, shifting the sine function to the left by π/2 results in the cosine
function. This interchangeability between the sine and cosine functions is reflected in
their being commonly referred to as sinusoidal functions. Notice also from the graphs
two importantpropertiesofsinx and cosx:
sinx = −sin(−x)
cosx =cos(−x)
For example,
sin π (
3 = −sin − π )
3
and cos π (
3 = cos − π )
3
Note that f(x) = sinx is 0 whenx = 0,±π,±2π,±3π,..., ±nπ,.... Also f(x) =
cosxis0whenx=± π 2 ,±3π 2 ,±5π 2 ,..., ±(2n+1)π 2 ,....
3.4.1 Inversetrigonometricfunctions
Allsixtrigonometricfunctionshaveaninversebutwewillonlyexaminethoseofsinx,
cosxandtanx.Theinversefunctionsofsinx,cosxandtanxaredenotedsin −1 x,cos −1 x
and tan −1 x. This notation can and does cause confusion. The ‘−1’ in sin −1 x is sometimesmistakenlyinterpretedasapower.Wewrite
(sinx) −1 todenote1/sinx.Valuesof
sin −1 x, cos −1 x and tan −1 x can be found using a scientific calculator. Ify = sinx, then
x = sin −1 y, as in
sinx = 0.7654 x = sin −1 (0.7654) = 0.8717

122 Chapter 3 The trigonometric functions
sin x
1
–1 1 x
sin –1 x
– p –2
p –2 x
–1
Figure3.10
Thefunctionsinx isone-to-one [ ] ifthe
domainis restrictedto − π 2 , π .
2
–1
p –2
– p –2
Figure3.11
Theinverse sine
function, sin −1 x.
1
x
Figure3.12
A single input produces many
outputvalues.Thisisnotafunction.
Notethaty = sinxisamany-to-onefunction.Ifthedomainisrestrictedto[−π/2, π/2]
then the resulting function isone-to-one. This isshown inFigure 3.10.
Recall from Section 2.3 that a one-to-one function has a corresponding inverse. So
if the domain of y = sinx is restricted to [−π/2,π/2], then an inverse function exists.
A graph ofy = sin −1 x is shown in Figure 3.11. Without the domain restriction, a
one-to-many graph would result as shown in Figure 3.12. To denote the inverse sine
function clearly, we write
y=sin −1 x − π 2 y π 2
Example3.2 Useascientific calculatortoevaluate
(a) sin −1 (0.3169)
(b) sin −1 (−0.8061)
Solution (a) sin −1 (0.3169) = 0.3225
(b) sin −1 (−0.8061) = −0.9375
A word of warning about inverse trigonometric functions is needed. The calculator returns
a value of 0.3225 for sin −1 (0.3169). Note, however, that sin(0.3225 ± 2nπ) =
0.3169, n = 0,1,2,3,..., so there are an infinite number of values of x such that
sinx = 0.3169. Only one of these values is returned by the calculator. This is because
the domain ofy = sinx is restricted to ensure it has an inverse function. To ensure the
inverse functionsy = cos −1 x andy = tan −1 x can be obtained, restrictions are placed
on the domains of y = cosx and y = tanx. By convention, y = cosx has its domain
restrictedto[0,π] whereas withy = tanx the restrictionis (−π/2,π/2).

3.5 The sincxfunction 123
EXERCISES3.4
1 Evaluate the following:
(a) sin −1 (0.75) (b) cos −1 (0.625) (c) tan −1 3
(d) sin −1 (−0.9) (e) cos −1 (−0.75) (f) tan −1 (−3)
3 Sketchy = sinx andy = cosx forxin the interval
[−2π,2π].Markon the graphs the pointswhere
x=1,x=1.5,x=−3andx=2.3.
2 Showthat
sin −1 x +cos −1 x = π 2
Solutions
1 (a) 0.8481 (b) 0.8957 (c) 1.2490 (d) −1.1198 (e) 2.4189 (f) −1.2490
TechnicalComputingExercises3.4
1 Using atechnicalcomputing languagedrawy = sinx
andy = cosx for0 x4π onthe same axes. Use
your graphs to findapproximate solutionsto the
equationsinx = cosx.
2 Draw the graphs ofy = cos −1 x andy = tan −1 x.
3.5 THESINCxFUNCTION
Thesinefunctionisusedtodefineanotherimportantfunctionusedinengineering.The
cardinalsinefunction,sincx,occursfrequentlyinengineeringmathematicsinapplicationsrangingfromcommunications,powerelectronics,digitalsignalprocessing(d.s.p.)
and optical engineering. The standard sincxfunction without normalization is plotted
inFigure 3.13 and isdefined by
⎧
⎨ sinx
x≠0
sincx = x
⎩
1 x=0
It is necessary to separately define the value of the function at x = 0 because the
definitionofthefunction, sin x ,wouldcauseadivisionbyzeroatthispoint.Fortunately
x
it can be shown that asxapproaches 0, then sinx approaches 1, and so specifying the
x
valuesincx=1atx = 0isadequatetoresolvetheproblem.Notethatthesincfunction
without normalization is 0 whenx = ...,−3π, −2π, −π, π,2π, 3π,..., so that apart
fromx = 0 where sincx = 1 by definition, it has the same zero crossing points as that
ofsinx.
Some engineers use a different definition of the sincxfunction:
⎧
⎨ sinπx
x≠0
sincx = πx
⎩
1 x=0
This iscommonly termed thenormalized sinc function.

124 Chapter 3 The trigonometric functions
Sinc x
1
0.75
0.5
0.25
–4p –3p –2p –p p 2p 3p 4p
x
–0.25
Figure3.13
The standard sincfunction withoutnormalization.
Engineeringapplication3.1
Zerocrossingpointsofthenormalizedsinc(x)function
The sinc function often appears in communications engineering and signal processing.Ingeneral,asignalcanberegardedasbeingmadeofalargenumberofindividual
frequencycomponents.Inthecaseofasquarepulse,thegraphshowingtheamplitude
of these components against their frequencies takes the formof a sincfunction.
Find the zero crossing points of the normalized sincfunction,
⎧
⎨ sin(πx)
x≠0
sinc(x) = πx
⎩
1 x=0
Solution
It is evident from Figure 3.13 that the sinc function which has not been normalized
hasthesamezerocrossingpointsassinx,withtheexceptionofthecrossingatx = 0,
thatisatx=±π,±2π,±3π,...,±nπ,....
Thenormalizedsincfunctionis0whensin(πx)is0.Now,frompage121wesee
that
and so
sinx=0whenx=...−3π,−2π,−π,0,π,2π,3π,...
sinπx=0whenπx=...−3π,−2π,−π,0,π,2π,3π,...
and hence
sin(πx) =0whenx = −3,−2,−1,0,1,2,3...
Noting thatsinc(0) isdefined tohave a value of 1,then wesee that
sinc(x)=0whenx=...−3,−2,−1,1,2,3,...

3.6 Trigonometric identities 125
or
sinc(x)=0whenx=±nforn=1,2,3,...
The plot ofthe normalized sincfunction isshown inFigure 3.14.
Sinc x
1
0.75
0.5
0.25
–15 –10 –5 5 10
x
15
–0.25
Figure3.14
The sincfunctionwith normalization.
3.6 TRIGONOMETRICIDENTITIES
We have seen that tanθ = sinθ for all values of θ. We call this an identity. In an
cos θ
identity, the l.h.s. and the r.h.s. are always equal, unlike in an equation where the l.h.s.
and the r.h.s. are equal only for particular values of the variable concerned. Table 3.1
listssome common trigonometric identities.
Example3.3 Simplify
cotA
cosA
Solution We know that
tanA= sinA
cosA
and so
cotA= 1
tanA = cosA
sinA
Hence
cotA
cosA =cotA× 1
cosA
= cosA
sinA × 1
cosA
= 1
sinA
This may alsobewritten as cosecA.

126 Chapter 3 The trigonometric functions
Table3.1
Common trigonometricidentities.
tanA= sinA
cosA
sin(A ±B) =sinAcosB ±sinBcosA
cos(A ±B) =cosAcosB ∓sinAsinB
tan(A ±B) = tanA±tanB
1∓tanAtanB
2sinAcosB =sin(A +B) +sin(A −B)
2cosAcosB =cos(A +B) +cos(A −B)
2sinAsinB =cos(A −B) −cos(A +B)
sin 2 A+cos 2 A=1
1 +cot 2 A = cosec 2 A
tan 2 A+1=sec 2 A
cos2A=1−2sin 2 A=2cos 2 A−1=cos 2 A−sin 2 A
sin2A=2sinAcosA
sin 2 A = 1−cos2A
2
cos 2 A = 1+cos2A
2
Note: sin 2 Ais the notationused for (sinA) 2 .Similarly cos 2 A means (cosA) 2 .
Example3.4 Show that
tanA+cotA
maybewritten as
2
sin2A
Solution We have
and so
tanA= sinA
cosA ,
cotA=cosA sinA
tanA+cotA= sinA
cosA + cosA
sinA
= sin2 A +cos 2 A
sinAcosA
1
= usingthe identity sin 2 A +cos 2 A = 1
sinAcosA
2
=
2sinAcosA
= 2 using the identity sin2A = 2sinAcosA
sin2A

3.6 Trigonometric identities 127
Example3.5 Usethe( identities ) inTable 3.1( tosimplify ) the following expressions:
π 3π
(a) sin
2 + θ (b) cos
2 − θ
(c) tan(2π − θ) (d) sin(π − θ)
( π
)
Solution (a) The expression sin
2 + θ isof the form sin(A +B) and so we use the identity
sin(A +B) =sinAcosB +sinBcosA
PuttingA = π andB = θ weobtain
2
( π
)
sin
2 + θ = sin π 2 cos θ +sin θ cos π 2
We note thatsin π 2 = 1,cos π 2 =0andso
( π
)
sin
2 + θ =1cosθ +sinθ(0)
=cosθ
( ) 3π
(b) The expression cos
2 − θ has the form cos(A −B); hence we use the identity
cos(A −B) =cosAcosB +sinAsinB
PuttingA = 3π ,B = θ weobtain
2
( ) 3π
cos
2 − θ = cos 3π 2 cos θ +sin 3π 2 sinθ
Now cos 3π 2 = 0,sin 3π = −1 and so
2
( ) 3π
cos
2 − θ =0cosθ+(−1)sinθ
(c) We use the identity
= −sin θ
tan(A −B) = tanA−tanB
1+tanAtanB
SubstitutingA = 2π,B = θ we obtain
tan(2π − θ) =
tan2π −tan θ
1+tan2πtanθ
Since tan2π = 0 this simplifies to
tan(2π − θ) = −tanθ
1
(d) We use the identity
=−tanθ
sin(A −B) =sinAcosB −sinBcosA

128 Chapter 3 The trigonometric functions
SubstitutingA = π,B = θ, this thenbecomes
sin(π−θ)=sinπcosθ −sinθcosπ
Now sinπ = 0,cos π = −1 and so we obtain
sin(π−θ)=0cosθ −sinθ(−1)=sinθ
Example3.6 Simplify
(a) cos(π + θ)
(b) tan(π − θ)
(c) sin 3 B +sinBcos 2 B
(d) tanA(1 +cos2A)
Solution (a) Usingthe identityfor cos(A +B) withA = π,B = θ weobtain
cos(π+θ)=cosπ cosθ −sinπ sinθ
= (−1)cosθ − (0)sinθ
= −cos θ
(b) Usingthe identityfor tan(A −B) withA = π,B = θ weobtain
tan(π−θ)= tanπ−tanθ
1+tanπtanθ
= 0−tanθ
1+(0)tanθ
=−tanθ
(c) sin 3 B +sinBcos 2 B =sinB(sin 2 B +cos 2 B)
=sinB since sin 2 B +cos 2 B = 1
(d) Firstlywe notethat tanA = sinA . Also wehave from Table 3.1
cosA
cos 2 A = 1+cos2A
2
from which
Hence
1+cos2A=2cos 2 A
tanA(1 +cos2A) = sinA
cosA 2cos2 A
=2sinAcosA
= sin2A

3.6 Trigonometric identities 129
Example3.7 Show that
( ) ( ) A +B A −B
sinA+sinB=2sin cos
2 2
Solution Consider the identities
sin(C +D) = sinCcosD +sinDcosC
sin(C −D) = sinCcosD −sinDcosC
Byadding these identities weobtain
sin(C +D) +sin(C −D) =2sinCcosD
We now make the substitutionsC +D =A,C −D =Bfromwhich
C = A +B , D = A −B
2 2
Hence
( ) ( ) A +B A −B
sinA+sinB=2sin cos
2 2
TheresultofExample3.7isoneofmanysimilarresults.ThesearelistedinTable3.2.
Table3.2
Further trigonometricidentities
( A +B
sinA+sinB =2sin
2
( A −B
sinA−sinB =2sin
2
( A +B
cosA+cosB=2cos
2
( A +B
cosA−cosB=−2sin
2
)
cos
)
cos
)
cos
( ) A −B
2
( ) A +B
2
( ) A −B
2
( ) A −B
)
sin
2
Example3.8 Simplify
sin70 ◦ −sin30 ◦
cos50 ◦
Solution We note that the numerator, sin70 ◦ − sin30 ◦ , has the form sinA − sinB. Using the
identityfor sinA −sinBwithA = 70 ◦ andB = 30 ◦ wesee
( ) ( )
70
sin70 ◦ −sin30 ◦ ◦ −30 ◦ 70 ◦ +30 ◦
= 2sin cos
2 2
= 2sin20 ◦ cos50 ◦

130 Chapter 3 The trigonometric functions
Hence
sin70 ◦ −sin30 ◦
= 2sin20◦ cos50 ◦
cos50 ◦ cos50 ◦
= 2sin20 ◦
EXERCISES3.6
1 Usethe identities forsin(A ±B),cos(A ±B)and
tan(A ±B)to ( simplifythe ) following: ( )
(a) sin
θ − π 2
(b) cos
θ − π 2
(c) tan(θ + π) (d) sin(θ − π)
(e) cos(θ − π)
(g) sin(θ + π)
(
)
(i) sin 2θ + 3π 2
( )
π
(k) cos
2 + θ
(f) tan(θ −3π)
( )
(h) cos θ + 3π 2
( )
(j) cos
θ − 3π 2
2 Writedown the trigonometricidentityfortan(A
( )
+ θ).
BylettingA → π 2 showthat tan π
2 + θ can be
simplified to −cot θ.
3 (a) Bydividing the identity
sin 2 A +cos 2 A = 1 bycos 2 A showthat
tan 2 A +1 = sec 2 A.
(b) Bydividing the identity
sin 2 A +cos 2 A = 1 bysin 2 A showthat
1 +cot 2 A = cosec 2 A.
4 Simplify the followingexpressions:
(a) cosAtanA (b) sin θ cot θ
(c) tanB cosecB
( )
(d) cot2xsec2x
π
(e) tanθ tan
2 + θ (f) sin2t
cost
[Hint: seeQuestion 2.]
(g) sin 2 A +2cos 2 A
(h) 2cos 2 B −1
(i) (1 +cot 2 X)tan 2 X (j) (sin 2 A +cos 2 A) 2
(k) 1 2 sin2AtanA
(m)
sin2A
cos2A
(o) (tan 2 θ +1)cot 2 θ
5 Simplify
(a) sin110 ◦ −sin70 ◦
(b) cos20 ◦ −cos80 ◦
(c) sin40 ◦ +sin20 ◦
(d)
6 Show that
cos50 ◦ +cos40 ◦
√
2
sin60 ◦ +sin30 ◦
sin50 ◦ −sin40 ◦
isequivalent to
cos15 ◦
sin5 ◦
(l) (sec2 t −1)cos 2 t
(n)
sinA
sin2A
(p) cos2A +2sin 2 A
Solutions
1 (a) −cos θ (b) sin θ (c) tan θ
(d) −sin θ (e) −cos θ (f) tan θ
(g) −sin θ (h) sin θ (i) −cos2θ
(j) −sin θ (k) −sin θ
4 (a) sinA (b) cos θ (c) secB
(d) cosec 2x (e) −1 (f) 2sint
(g) 1 +cos 2 A (h) cos2B (i) sec 2 X
(j) 1 (k) sin 2 A (l) sin 2 t
(m) tan2A
(p) 1
(n)
1
2 secA (o) cosec 2 θ
5 (a) 0 (b) sin50 ◦ (c) cos10 ◦ (d) cos5 ◦

3.7 Modelling waves using sint and cost 131
3.7 MODELLINGWAVESUSINGSINt ANDCOSt
Examiningthegraphsofsinxandcosxrevealsthattheyhaveasimilarshapetowaves.
In fact, sine and cosine functions are often used to model waves and we will see in
Chapter 23 that almost any wave can be broken down into a combination of sine and
cosinefunctions.Themainwavesfoundinengineeringareonesthatvarywithtimeand
sot isoften the independent variable.
The amplitude of a wave is the maximum displacement of the wave from its mean
position.So,forexample,sint andcost haveanamplitudeof1,theamplitudeof2sint
is2,and the amplitude ofAsint isA(see Figure 3.15).
The amplitude ofAsint isA. The amplitude ofAcost isA.
AmoregeneralwaveisdefinedbyAcos ωt orAsin ωt.Thesymbol ω representsthe
angularfrequencyofthewave.Itismeasuredinradianspersecond.Forexample,sin3t
hasanangularfrequencyof3rads −1 .Ast increasesby1secondtheangle,3t,increases
by 3 radians. Note thatsint has anangular frequency of1rad s −1 .
The angular frequency ofy =Asin ωt andy =Acos ωt is ω radians per second.
The sine and cosine functions repeat themselves at regular intervals and so are periodic
functions. Looking at Figure 3.7 we see that one complete cycle of sint is completedevery
2πseconds.Thetimetaken tocomplete one fullcycle iscalled theperiod
andisdenotedbyT.Hencetheperiodofy = sint is2πseconds.Similarlytheperiodof
y = cost is2πseconds.Mathematicallythismeansthataddingorsubtractingmultiples
of 2πtot does notchange the sineor cosine of thatangle.
sint=sin(t±2nπ)
cost=cos(t±2nπ)
n=0,1,2,3,...
n=0,1,2,3,...
Inparticular we note that
sint = sin(t +2π)
cost = cos(t +2π)
We now considery =Asin ωt andy =Acos ωt. Whent = 0 seconds, ωt = 0 radians.
Whent = 2π ω seconds, ωt = ω2π = 2π radians. We can see that ast increases from 0
ω
f (t)
A
2
A sin t
2 sin t
–2
2p
t
–A
Figure3.15
The amplitude of f (t) =Asint isA.

132 Chapter 3 The trigonometric functions
to 2π seconds, the angle ωt increases from 0 to 2π radians. We know that as the angle
ω
ωt increases by 2π radians then Asin ωt completes a full cycle. Hence a full cycle is
completed in 2π ω seconds, thatisthe period ofy =Asin ωt is 2π ω seconds.
Ify =Asin ωt ory =Acos ωt, thenthe periodT is 2π ω .
Inparticular wenote thatthe period ofy =Asint andy =Acost is2π.
Closelyrelatedtotheperiodisthefrequencyofawave.Thefrequencyisthenumber
ofcyclescompletedin1second.Frequency ismeasuredinunitscalledhertz(Hz).One
hertz isone cycle per second. We have seen thaty =Asin ωt takes
2π
seconds tocomplete one cycle
ω
and soitwill take
1 second tocomplete ω 2π cycles
We use f as the symbol forfrequency and so
frequency, f = ω 2π
( ) 3
For example, sin3t has a frequency of Hz.
2π
Note that by rearrangement wemay write
ω=2πf
and sothe wavey =Asin ωt may alsobewritten asy =Asin2πft.
Fromthe definitions of period and frequency wecan see that
1
period =
frequency
that is
T = 1 f
We see that the period is the reciprocal of the frequency. Identical results apply for the
wavey =Acosωt.
A final generalization is to introduce a phase angle or phase, φ. This allows the
wave to be shifted along the time axis. It also means that either a sine function or a
cosine function can be usedtorepresent the same wave. So the general formsare
Acos(ωt + φ), Asin(ωt + φ)
Figure 3.16 depictsAsin(ωt + φ). Note from Figure 3.16 that the actual movement of
the wave along the time axis is φ/ω. Itis easytoshow thismathematically:
(
Asin(ωt +φ) =Asinω t + φ )
ω
The quantity φ iscalled thetimedisplacement.
ω

3.7 Modelling waves using sint and cost 133
f (t)
A
T = 1– 2p –v f =
f –
v
t
Figure3.16
The generalized waveAsin(ωt + φ).
The waves met in engineering are often termed signals or waveforms. There are
no rigid rules concerning the use of the words wave, signal and waveform, and often
engineers use them interchangeably. We will followthis convention.
Example3.9 Statethe amplitude, angular frequency and periodofeachofthe following waves:
(a) 2sin3t
1
(b)
(2t
2 cos + π )
6
Solution (a) Amplitude,A = 2,angular frequency, ω = 3,period,T = 2π ω = 2π 3 .
(b) Amplitude,A = 0.5, angular frequency, ω = 2,period,T = 2π ω = 2π 2 = π.
Example3.10 Statethe amplitude, period, phase angle and time displacement of
(a) 2sin(4t +1)
2cos(t −0.7)
(b)
( 3 ) 2t+1
(c) 4cos
3
( )
3 4t
(d)
4 sin 3
Solution (a) Amplitude = 2, period = 2π 4 = π , phase angle = 1 relative to 2sin4t, time
2
displacement = 0.25.
(b) Amplitude = 2 3 ,period = 2π,phaseangle = −0.7relativeto 2 3 cost,timedisplacement
= −0.7.
(c) Amplitude = 4, period = 2π
2/3 = 3π, phase angle = 1 ( ) 2t
3 relative to 4cos , time
3
displacement = 0.5.
(d) Amplitude = 3 2π
,period =
4 4/3 = 3π 2 ,phaseangle = 0relativeto 3 ( ) 4t
4 sin ,time
3
displacement = 0.

134 Chapter 3 The trigonometric functions
Engineeringapplication3.2
Alternatingcurrentwaveformsandtheelectricitysupply
Alternatingcurrentwaveformsareoftenfoundinengineering.Theelectricitysupply
to homes and businesses most often takes the form of an alternating current. This is
becauseitisfareasiertosupplyalternatingcurrentelectricitythandirectcurrentelectricitywhendistributingacrosslongdistances.Energylossesalongthesupplycables
can be reduced by transforming the electricity to high voltages prior to distribution
butelectricity transformersonly work with alternating currents.
Sineandcosinefunctionsareoftenusedtomodelalternatingcurrent(a.c.)waveforms.The
equations foran a.c. waveform are
I =I m
sin(ωt +φ) or I =I m
cos(ωt +φ)
whereI m
=maximumcurrent, ω =angularfrequencyand φ =phaseangle.Inpracticethefunctionscanbeshiftedalongthetimeaxisbygiving
φ anon-zerovalueand
so both the sine and the cosine function can be used to model any a.c. waveform;
which one isused isusually a matter of convenience.
The angular frequency, ω, can be written as ω = 2πf, where f is the frequency
of the waveform in Hertz (Hz). The frequency of the electricity supply in Europe
and across large parts of the world is 50 Hz, while in the Americas and in areas of
Asiaitis60Hz.Alternatingcurrentsuppliesarealsofoundonships,submarinesand
aircraftbut these often use 400 Hz as the operating frequency.
3.7.1 Combiningwaves
There are many situations in which engineers need to combine two or more waves
together to form a single wave. It is possible to make use of trigonometric identities
to calculate the resulting waveform when several waves are combined. Consider the
following example.
Engineeringapplication3.3
Combiningtwosinusoidalvoltagesignals
Two voltage signals, v 1
(t) and v 2
(t), have the following mathematical expressions:
v 1
(t) =3sint
v 2
(t) =2cost
(a) State the amplitude and angular frequency ofthe two signals.
(b) Obtain an expression forthe signal, v 3
(t), given by
v 3
(t) = v 1
(t)+2v 2
(t)
(c) Reduce the expression obtained in part (b) to a single sinusoid and hence state
its amplitude and phase.

3.7 Modelling waves using sint and cost 135
Solution
(a) v 1
(t) has an amplitude of 3 volts and an angular frequency ω = 1 rad s −1 . v 2
(t)
has an amplitude of 2 volts and an angular frequency ω = 1 rad s −1 . Note that
both of these signals have the same angular frequency.
(b) v 3
(t)=v 1
(t)+2v 2
(t)
=3sint+2(2cost)
=3sint+4cost
(c) We wish to write v 3
(t) in the formRsin(t + φ). The choice of sine is arbitrary.
We could have chosen cosine instead.Ris the amplitude of the single sinusoid
and φ isits phase angle.
Using the trigonometric identity sin(t + φ) = sintcos φ +sinφcost found
inTable 3.1 wecan write
Rsin(t +φ) =R(sintcosφ +sinφcost)
= (Rcosφ)sint +(Rsinφ)cost
Comparing thisexpression with thatfor v 3
(t) wenotethat, inorder tomake the
expressions identical,
Rcosφ =3 (3.1)
Rsinφ =4 (3.2)
Weneedtosolve(3.1)and(3.2)toobtainRand φ.Squaringeachequationgives
R 2 cos 2 φ=9
R 2 sin 2 φ = 16
Adding these equations together weobtain
R 2 cos 2 φ+R 2 sin 2 φ=9+16
R 2 (cos 2 φ +sin 2 φ) = 25
Usingthe identitycos 2 φ +sin 2 φ = 1 thissimplifies to
R 2 =25
R = 5
Next wedetermine φ. Dividing (3.2) by (3.1) wefind
Rsinφ
Rcosφ = 4 3
tanφ = 4 3
)
φ = tan −1 ( 4
3
From (3.1) and (3.2) we can see that both sinφ and cos φ are positive, and so
φ must lie in the first quadrant. Calculating tan −1( 4
3)
using a calculator gives
φ = 0.927 radians. So we can express v 3
(t) as
v 3
(t) =3sint +4cost =5sin(t +0.927)
Finally v 3
(t) has amplitude 5 volts and phase 0.927 radians.

136 Chapter 3 The trigonometric functions
Engineering application 3.3 illustrates an important property when combining together
two sinusoidal waves of the same angular frequency.
If two waves of equal angular frequency, ω, areadded the resultisawave of the
same angular frequency, ω.
Infactthisresultholdstruewhencombininganynumberofwavesofthesameangular
frequency.
Engineeringapplication3.4
Combiningtwosinusoidalcurrentsignals
Two current signals,i 1
(t) andi 2
(t), have the following mathematical expressions:
i 1
(t) =10sin4t
i 2
(t) =5cos4t
(a) State the amplitude and angular frequency ofthe two signals.
(b) Obtain an expression forthe signal,i 3
(t), given byi 3
(t) = 0.3i 1
(t) −0.4i 2
(t).
(c) Reduce the expression obtained in part (b) to a single sinusoid in the form
Rcos(4t + φ) and hence stateits amplitude and phase.
Solution
(a) i 1
(t)hasanamplitudeof10ampsandanangularfrequency ω = 4 rad s −1 .i 2
(t)
has an amplitude of 5 amps and an angular frequency ω = 4 rad s −1 . Note that
both signals have the same angular frequency.
(b) i 3
(t) =0.3i 1
(t) −0.4i 2
(t)
(c) Let
=0.3×10sin4t−0.4×5cos4t
=3sin4t−2cos4t
3sin4t −2cos4t =Rcos(4t +φ)
Then using the trigonometric identity given inTable 3.1
cos(A +B) =cosAcosB −sinAsinB
withA=4tandB=φwefind
3sin4t −2cos4t =Rcos(4t +φ)
Hence
=R(cos4tcosφ −sin4tsinφ)
= (Rcosφ)cos4t −(Rsinφ)sin4t
3=−Rsinφ (3.3)
−2=Rcosφ (3.4)
Bysquaring each equation and adding we obtain
9+4=R 2 (sin 2 φ+cos 2 φ)=R 2
so thatR = √ 13.

3.7 Modelling waves using sint and cost 137
From (3.3) and (3.4), both sinφ and cosφ are negative and so φ lies in the
third quadrant. Division of (3.3)by (3.4) gives
3
−2 = −Rsin φ
Rcosφ =−tanφ
tanφ = 1.5
Usingacalculatorandnotingthat φ liesinthethirdquadrantwefind φ = 4.124.
Finally
3sin4t−2cos4t = √ 13cos(4t +4.124)
Soi 3
(t) = √ 13cos(4t +4.124).Thereforei 3
(t)hasanamplitudeof √ 13amps
and a phase of4.124 radians.
Example3.11 Express0.5cos3t +sin3t asasingle cosinewave.
Solution Let
0.5cos3t +sin3t =Rcos(3t + φ)
Hence
=R(cos3tcosφ −sin3tsinφ)
= (Rcosφ)cos3t −(Rsinφ)sin3t
0.5=Rcosφ (3.5)
1=−Rsinφ (3.6)
Bysquaring and adding weobtain
1.25 =R 2
R = √ 1.25 = 1.1180
Division of (3.6)by (3.5) yields
2=−tanφ
(4 d.p.)
From (3.5), cosφ is positive; from (3.6), sin φ is negative; and so φ lies in the fourth
quadrant. Hence, using a calculator, φ = 5.1760. So
0.5cos3t +sin3t = 1.1180cos(3t +5.1760)
Example3.12 Ifacos ωt +bsinωt is expressed in the formRcos(ωt − θ) show thatR = √ a 2 +b 2
and tanθ = b a .
Solution Let
acosωt+bsinωt =Rcos(ωt−θ)
Then, using the trigonometric identity forcos(A −B), we can write
acosωt+bsinωt =Rcos(ωt−θ)
=R(cosωtcosθ +sinωtsinθ)
= (Rcosθ)cosωt+(Rsinθ)sinωt

138 Chapter 3 The trigonometric functions
Equating coefficients ofcosωt and then sinωt gives
a=Rcosθ (3.7)
b=Rsinθ (3.8)
Squaring these equations and adding gives
that is
a 2 +b 2 =R 2
R = √ a 2 +b 2
Division of (3.8)by (3.7) gives
b
a =tanθ
as required.
Note that this example demonstrates that adding two waves of angular frequency ω
forms another wave having the same angular frequency but with a modified amplitude
and phase.
3.7.2 Wavelength,wavenumberandhorizontalshift
The sine and cosine waves described earlier in this section hadt as their independent
variable because the waves commonly met in engineering vary with time. There are
occasionswheretheindependentvariableisdistance,xsay,andinthiscasesomeofthe
terminology changes.Consider the wave
y =Asin(kx+φ)
Asbefore,Aistheamplitudeofthewave.Thequantitykiscalledthewavenumber.It
playsthesameroleasdidtheangularfrequency, ω,whent wastheindependentvariable.
The length of one cycle of the wave, that is the wavelength, commonly denoted λ, is
related tokby the formula λ = 2π . The phase angle is φ and its introduction has the
k
effect of shiftingthe graph horizontally.
Example3.13 Figure 3.17 shows a graph ofy = sin2x.
(a) State the wave numberforthiswave.
(b) Findthe wavelength ofthe wave.
(c) State the phase angle.
Solution (a) Comparingy = sin2x withy = sinkx wesee thatthe wave number,k, is2.
(b) The wavelength, λ = 2π = π. Note by observing the graph that this result is
k
consistent inthat the distance required for one cycle of the wave is π units.
(c) Comparingy = sin2x with sin(kx + φ)we see thatthe phase angle, φ, is 0.

3.7 Modelling waves using sint and cost 139
y
1
–2p
–p
0
p
2p
x
–1
Figure3.17
A graphofthe wavey = sin2x.
(
Example3.14 Figure 3.18 shows a graph ofsin 2x + π )
.
3
(a) State the phase angle.
(b) BycomparingFigures3.17and3.18weseethattheintroductionofthephaseangle
has caused a horizontal shift of the graph (tothe left).Calculate thisshift.
y
1
–2p
–p 0
p 2p x
–1
p –6
Figure3.18
Agraph(
ofthe wave )
y=sin 2x + π .
3
(
Solution (a) By comparing sin 2x + π )
with sin(kx + φ) wesee thatthe phase angle is π 3
3 .
(
(b) By writingy = sin 2x + π ) (
as sin2 x + π )
we note that this isy = sin2x
3 6
shifted tothe left by a horizontal distance π 6 units.
The results of this example can be generalized. The wave y = Asin(kx + φ) can be
writteny = Asink(x + φ/k) so that a phase angle of φ introduces a horizontal shift of
length φ/k. (Compare thiswith the expression fortime displacement inSection3.7.)
Noting that λ = 2π k ,thenk=2π and we may write Asin(kx + φ) equivalently
( )
λ 2πx
as Asin
λ + φ . Again, using k = 2π λ , the horizontal shift, φ , may similarly be
k
written as
horizontal shift = φ k =
φ
2π/λ = φλ
2π

140 Chapter 3 The trigonometric functions
from which
phase angle = φ =
2π ×horizontal shift
λ
This result is important in the engineering application that follows because, more generally,
when any two waves arrive at a receiver it enables the difference in their phases,
φ, tobe calculated from knowledge of the horizontal shift between them.
Thepresenceof φ iny =Asin(kx + φ)causesahorizontal(left)shiftof φ k = φλ
2π .
Note that adding any multiple of 2π onto the phase angle φ will result in the same
graphbecauseoftheperiodicityofthesinefunction.Consequently,aphaseanglecould
(
be quoted as φ + 2nπ. For example, the wave sin 2x + π )
is the same wave as
3
sin
(2x + π )
3 +2π , sin
(2x + π )
3 +4π and so on. Normally, we would quote a value
of the phase thatwas lessthan 2πby subtracting multiples of 2πasnecessary.
Engineeringapplication3.5
Two-raypropagationmodel
Itisveryusefultobeabletomodelhowanelectromagneticwaveemittedbyatransmitter
propagates through space, in order to predict what signal is collected by the
receiver. This can be quite a complicated modelling exercise. One of the simplest
models is the two-ray propagation model. This model assumes that the signal received
consists of two main components. There is the signal that is sent direct from
the transmitter to the receiver and there is the signal that is received after being reflectedoff
the ground.
Figure 3.19 shows a transmitter with height above the groundh t
together with a
receiver with height above the groundh r
. The distance between the transmitter and
the receiver alongthe ground isd.
T
S
h t
Transmitter
O
Ground
d d
d r
d
A
R
Receiver
P
h r
Q
Figure3.19
Atransmitterandreceiver at
different heightsabove the
ground.
Note thatthe quantitiesh t
,h r
andd areall known.

3.7 Modelling waves using sint and cost 141
Waves can be considered to propagate between the transmitter and the receiver
in two ways. There is the direct route between transmitter and receiver. The direct
distance between transmitter and receiver is d d
. We obtain an expression for d d
in
terms of the known quantitiesh t
,h r
andd by considering the triangle RST. In this
triangle, RS =dand ST = TO − SO =h t
−h r
. Hence by Pythagoras’s theorem in
RST we have
and so
TR 2 = RS 2 +ST 2
d 2 d =d2 +(h t
−h r
) 2
d d
=
√
d 2 +(h t
−h r
) 2
Note thatd d
isexpressed intermsof the known quantitiesh t
,h r
andd.
ThereisalsoaroutewherebyawaveisreflectedoffthegroundatpointAbefore
arrivingatthereceiver.ThepointAonthegroundissuchthat ̸ TAOequals ̸ RAP.
The distance travelled in this case isd r
= TA +AR. We wish to find an expression
ford r
intermsoftheknownquantitiesh t
,h r
andd.Inordertosimplifythecalculation
of this distance we construct an isosceles triangle, TAQ, in which TA = QA and
̸ TAO = ̸ QAO. Note thatinthistriangle,TO = QO =h t
.
Then the distance travelled by thisreflected wave,d r
, is
distance travelled =d r
= TA +AR
=QA+AR
= QR
Consider now QSR. QR is the hypotenuse of this triangle. So by Pythagoras’s
theorem wehave
d 2 r
= QR 2 = SR 2 +SQ 2
WehaveSR=dandSQ=QO+SO=h t
+h r
.So
d 2 r
=d 2 +(h t
+h r
) 2
from which
√
d r
= d 2 +(h t
+h r
) 2
Now if the wavelength of the transmitted wave is λ then we can calculate the phase
differencebetweenthedirectwaveandthereflectedwave, φ,bynotingthedifference
in the distance travelled,d r
−d d
. Using the result for phase difference from Section
3.7.2 wehave
φ = 2π λ ×horizontal shift = 2π λ (d r −d d )
φ = 2π (√
)
d
λ
2 +(h t
+h r
) 2 −
√d 2 +(h t
−h r
) 2
⎛ √
φ = 2π ( )
√ ⎞
ht +h 2 ( )
⎝d 1 + r ht −h 2
−d 1 + r ⎠
λ d d
➔

142 Chapter 3 The trigonometric functions
Now the binomial expansion for √ (1 +x)is (see Section 6.4)
√
(1+x)=(1+x) 1/2 =1+ x 2 − x2
8 + x3
16 −···≈1+x 2 , if|x|<1
Usingthisexpansionintheexpressionfor φ andnotingthatthemoduliofboth(h t
+
h r
)/d and (h t
−h r
)/d arelessthan 1,we have
φ ≈ 2πd (1 + (h t +h r )2
−1− (h t −h )
r )2
λ 2d 2 2d 2
Expanding the bracketed termsgives
φ ≈ 2πd
(
h
2
t
+2h t
h r
+h 2 r −h2 t
+2h t
h r
−hr)
2
λ 2d 2
So
φ ≈ 4h t h r π
λd
This is a simplified approximation for the phase difference between the direct wave
andthereflectedwave.Notethatitdependsontheheightofthetransmitter,theheight
of the receiver and the distance between the transmitter and the receiver.
Thiscalculationisimportantbecauseundersomeconditionsthephasedifference
between the two paths means that the directed and reflected waves destructively
interfere.Inseverecasesthiscausesthesignaltodecreaseatthereceiverenoughso
that the communications link is lost. The effect is often termed multipath-induced
fading.
EXERCISES3.7
1 Statethe amplitude,angularfrequency, frequency,
phase angleandtime displacement ofthe following
waves:
(a) 3sin2t (b) 1 sin4t
2
(c) sin(t +1)
(d) 4cos3t (e) 2sin(t −3) (f) 5cos(0.4t)
(g) sin(100πt) (h) 6cos(5t +2) (i) 2 3 sin(0.5t)
(j) 4cos(πt −20)
2 Statethe period of
(a) 2sin7t (b) 7sin(2t +3)
(c) tan t 2
(e) cosec(2t −1)
(d) sec3t
( )
2t
(f) cot
3 +2
3 Avoltage sourceproduces atime-varying voltage,
v(t),given by
v(t) =15sin(20πt +4)
(a) Statethe amplitudeof v(t).
t 0
(b) Statethe angularfrequency of v(t).
(c) Statethe periodof v(t).
(d) Statethe phaseof v(t).
(e) Statethe time displacement of v(t).
(f) Statethe minimumvalue of v(t).
4 Asinusoidalfunctionhas an amplitude of 2 3 anda
periodof2. Stateapossibleform ofthe function.
5 Statethe phase angleandtime displacementof
(a) 2sin(t +3) relative to 2sint
(b) sin(2t −3) relative to sin2t
( )
t
(c) cos
2 +0.2 relative to cos t 2
(d) cos(2 −t)relative to cost
( )
3t+4
(e) sin relative to sin 3t
5 5
(f) sin(4 −3t) relative to sin3t

3.7 Modelling waves using sint and cost 143
(g) sin(2πt + π) relative to sin2πt
(h) 3cos(5πt −3) relative to 3cos5πt
( )
πt
(i) sin
3 +2 relative to sin πt
3
(j) cos(3π −t)relative to cost
6 Writeeachofthe followingin the form
Asin(3t+θ),θ 0:
(a) 2sin3t +3cos3t
(b) cos3t −2sin3t
(c) sin3t −4cos3t
(d) −cos3t −4sin3t
7 Writeeachofthe followingin the form
Acos(t−θ),θ 0:
(a) 2sint−3cost
(b) 9sint+6cost
(c) 4cost−sint
(d) 3sint
8 Writeeachofthe followingexpressions in the form
(i)Asin(ωt + θ),(ii)Asin(ωt − θ),
(iii)Acos(ωt + θ),(iv)Acos(ωt − θ)where θ 0:
(a) 5sint+4cost
(c) 4sin2t −6cos2t
(b) −2sin3t+2cos3t
(d) −sin5t −3cos5t
Solutions
1 (a) 3,2, 1 π ,0,0 (b) 1
2 ,4, 2 π ,0,0
(c) 1,1, 1 3
,1,1 (d) 4,3,
2π 2π ,0,0
(e) 2,1, 1
1
, −3, −3 (f) 5,0.4,
2π 5π ,0,0
(g) 1,100π,50,0,0 (h) 6,5, 5
2π ,2,0.4
(i)
2 (a)
(d)
2
3 ,0.5, 1
4π ,0,0
2π
7
2π
3
(b) π
(e) π
(j) 4,π,1 , −20, −20
2 π
(c) 2π
(f)
3π
2
3 (a) 15 (b) 20π (c) 0.1
1
(d) 4 (e) (f) −15
5π
2
4
3 sin(πt +k)or 2 cos(πt +k)
3
5 (a) 3, 3 (b) −3, − 3 2
4
(d) −2, −2 (e)
5 , 4 3
(g) π, 1 (h) −3, − 3
2 5π
(j) −3π,−3π
√
6 (a) 13sin(3t +0.9828)
(b) √ 5sin(3t +2.6779)
√
(c) 17sin(3t +4.9574)
(d) √ 17sin(3t +3.3866)
(c) 0.2, 0.4
(f) −0.858,−0.286
(i) 2, 6
π
√
7 (a) 13cos(t −2.5536)
√
(b) 117cos(t −0.9828)
√
(c) 17cos(t −6.0382)
( )
(d) 3cos t − π 2
8 (a) (i) √ 41sin(t +0.675)
(ii) √ 41sin(t −5.608)
(iii) √ 41cos(t +5.387)
(iv) √ 41cos(t −0.896)
( )
√
(b) (i) 8sin 3t + 3π 4
(ii) √ ( )
8sin 3t − 5π 4
( )
(iii) √ 8cos
(
3t + π 4
)
(iv) √ 8cos 3t − 7π 4
√
(c) (i) 52sin(2t +5.300)
(ii) √ 52sin(2t −0.983)
(iii) √ 52cos(2t +3.730)
(iv) √ 52cos(2t −2.554)
√
(d) (i) 10sin(5t +4.391)
(ii) √ 10sin(5t −1.893)
(iii) √ 10cos(5t +2.820)
(iv) √ 10cos(5t −3.463)

144 Chapter 3 The trigonometric functions
TechnicalComputingExercises3.7
1 Ploty=sin2tfor0t 2π.
2 Ploty=cos3tfor0t 3π.
( )
t
3 Ploty = sin for0t4π.
2
( )
2t
4 Ploty = cos for0t6π.
3
5 Ploty = sint +3cost for0 t 3π.Byreading
from yourgraph,statethe maximumvalue of
sint+3cost.
6 (a) Ploty=2sin3t−cos3tfor0t 2π.
Useyourgraph to findthe amplitudeof
2sin3t −cos3t.
(b) On the same axes ploty = sin3t.Estimate the
time displacement of2sin3t −cos3t.
3.8 TRIGONOMETRICEQUATIONS
Weexaminetrigonometricequationswhichcanbewritteninoneoftheformssinz =k,
cosz = k or tanz = k, wherezis the independent variable andkis a constant. These
equationsallhaveaninfinitenumberofsolutions.Thisisaconsequenceofthetrigonometric
functions being periodic. For example, sinz = 1 has solutions z = ..., −7π
2 ,
−3π
2 , π 2 , 5π 2 ,....Thesesolutionscouldbeexpressedasz = π 2 ±2nπ,n=0,1,2,....
Sometimesitisuseful,indeednecessary,tostateallthesolutions.Atothertimesweare
interestedonlyinsolutionsinsomespecifiedinterval,forexamplesolvingsinz = 1for
0 z2π. The following examples illustratethe method of solution.
Example3.15 Solve
(a) sint = 0.6105 for0 t 2π (b) sint = −0.6105 for0 t 2π.
Solution Figure 3.20 shows a graph ofy = sint, with horizontal lines drawn aty = 0.6105 and
y = −0.6105.
(a) From Figure 3.20 we see that there are two solutions in the interval 0 t 2π.
These aregiven bypoints Aand B. We have
sint = 0.6105
and so, usingascientific calculator, wehave
t = sin −1 (0.6105) = 0.6567
This is the solution at point A. From the symmetry of the graph, the second solution
is
t = π −0.6567 = 2.4849
This isthe solution atpointB. Therequired solutions aret = 0.6567,2.4849.
(b) Again from Figure 3.20 we see that the equation has two solutions in the interval
0 t 2π.ThesearegivenbythepointsCandD.Onesolutionliesintheinterval
π to 3π 2 ; the other solution liesinthe interval 3π 2
sint = −0.6105
to2π. We have

3.8 Trigonometric equations 145
sin t
1
0.6105
0 p –2 3p –2
Figure3.20
A B p C D 2p
t
–0.6105
–1
Aand Bare solution pointsfor
sint = 0.6105.Cand Dare
solution pointsforsint = −0.6105.
and so, using a calculator, wesee
t = sin −1 (−0.6105) = −0.6567
Although this value oft is a solution of sint = −0.6105 it is outside the range of
values of interest.Recall that
sint = sin(t +2π)
that is,adding 2πtoanangle does notchange the sineof the angle. Hence
t = −0.6567 +2π = 5.6265
is a required solution. This is the solution given by point D. From the symmetry of
Figure 3.20 the other solution is
t = π +0.6567 = 3.7983
This isthe solution atpoint C. The required solutions aret = 3.7983, 5.6265.
Example3.16 Solve
(a) cost = 0.3685 for0 t 2π
(b) cost = −0.3685 for0 t 2π
Solution Figure 3.21 shows a graph of y = cost between t = 0 and t = 2π together with
horizontallines aty = 0.3685andy = −0.3685.
(a) FromFigure3.21weseethatthereisasolutionofcost = 0.3685between0and π 2
and a solution between 3π 2
and 2π. These aregiven by points Aand D.Now
cost = 0.3685
and so, using a scientific calculator, wesee
t = cos −1 (0.3685) = 1.1934

146 Chapter 3 The trigonometric functions
cos t
1
0.3685
A
B
p
D
C 2p t
–0.3685
0 p –2 3p –2
Figure3.21
–1
A andDare solution pointsfor
cost = 0.3685.Band Care
solutionpointsforcost = −0.3685.
This is the solution between 0 and π , that is at point A. Using the symmetry of
2
Figure 3.21 the other solution atpoint Dis
t = 2π −1.1934 = 5.0898
The required solutions aret = 1.1934 and 5.0898.
(b) The graph in Figure 3.21 shows there are two solutions of cost = −0.3685. These
solutions areatpoints Band C. Given
cost = −0.3685
then usingascientific calculator we have
t = cos −1 (−0.3685) = 1.9482
This isthe solution given by pointB. By symmetrythe other solution atpointCis
t = 2π −1.9482 = 4.3350
The required solutions aret = 1.9482 and 4.3350.
Example3.17 Solve
(a) tant = 1.3100for0 t 2π
(b) tant = −1.3100 for0 t 2π
Solution Figure3.22showsagraphofy = tant fort = 0tot = 2πtogetherwithhorizontallines
y = 1.3100 andy = −1.3100.
(a) Thereisasolutionoftant = 1.3100between0and π andasolutionbetween πand
2
3π
. These aregiven by points Aand C.
2
tant = 1.3100
t = tan −1 (1.3100) = 0.9188

3.8 Trigonometric equations 147
tan t
2
1.3100
1
0
p
A –2 3p
B p C –2 D 2p t
–1
–1.3100
–2
Figure3.22
Aand Care solution pointsfortant = 1.3100. B andDare
solution pointsfortant = −1.3100.
This isthe solution between 0 and π given by point A.UsingFigure 3.22 wecan
2
see thatthe second solution isgiven by
t = π +0.9188 = 4.0604
This isgiven by point C.
(b) Figure 3.22 shows that there are two solutions of tant = −1.3100, one between π 2
and π,theotherbetween 3π 2 and2π.PointsBandDrepresentthesesolutions.Using
a scientific calculator we have
t = tan −1 (−1.3100) = −0.9188
This solution is outside the range of interest. Noting that the period of tant is π we
see that
t = −0.9188 + π = 2.2228
isasolution between π and π. This isgiven by pointB. The second solution is
2
t = −0.9188 +2π = 5.3644
This is the solution given by point D. The required solutions aret = 2.2228 and
5.3644.

148 Chapter 3 The trigonometric functions
Example3.18 Solve
(a) sin2t = 0.6105 for0t 2π
(b) cos(3t +2) = −0.3685for0 t 2π
Solution (a) Letz = 2t.Ast variesfrom0to2πthenzvariesfrom0to4π.Thustheproblemis
equivalent tosolving
sinz=0.6105
0z4π
From Example 3.15 the solutions between 0 and 2π are 0.6567 and 2.4849. Since
sinz has period 2π, then the solutions in the next cycle, that is between 2π and 4π,
arez = 0.6567 +2π = 6.9399 andz = 2.4849 +2π = 8.7681. Hence
z = 2t = 0.6567,2.4849,6.9399,8.7681
and so, tofour decimal places,
t = 0.3284,1.2425,3.4700,4.3840
(b) Letz = 3t +2. Ast varies from 0 to 2π thenzvaries from 2 to 6π +2. Hence the
problem isequivalent tosolving
cosz=−0.3685
2z6π+2
Solutions between 0 and 2π are given in Example 3.16 as z = 1.9482, 4.3350.
Notingthatcoszhasperiod2π,thensolutionsbetween2πand4πarez = 1.9482+
2π = 8.2314 andz = 4.3350 + 2π = 10.6182, solutions between 4π and 6π are
z = 1.9482+4π = 14.5146andz = 4.3350+4π = 16.9014andsolutionsbetween
6π and 8π arez = 1.9482 +6π = 20.7978 andz = 4.3350 +6π = 23.1846. The
solutions betweenz = 0 andz = 8πarethus
z = 1.9482,4.3350,8.2314,10.6182,14.5146,16.9014,20.7978,23.1846
Noting that 6π + 2 = 20.8496 we require values of z between 2 and 20.8496,
thatis
Finally
z = 4.3350,8.2314,10.6182,14.5146,16.9014,20.7978
t = z −2
3
= 0.7783,2.0771,2.8727,4.1715,4.9671,6.2659
Example3.19 Avoltage, v(t), isgiven by
v(t)=3sin(t+1)
t 0
Findthe first time thatthe voltage has a value of1.5 volts.
Solution We needtosolve 3sin(t +1) = 1.5,thatis
sin(t+1)=0.5
t0
Letz =t +1.Sincet 0 thenz 1.Theproblemisthus equivalent to
sinz=0.5
z1

Usingascientific calculator wehave
z = sin −1 (0.5) = 0.5236
3.8 Trigonometric equations 149
This solution is outside the range of interest. By reference to Figure 3.7 the next solution
is
z = π −0.5236 = 2.6180
This isthe first value ofzgreater than 1 such thatsinz = 0.5. Finally
t=z−1=1.6180
The voltage first has a value of 1.5 volts whent = 1.618 seconds.
EXERCISES3.8
1 Solvethe followingequations for0 t 2π:
(a) sint = 0.8426 (b) sint = 0.2146
(c) sint = 0.5681 (d) sint = −0.4316
(e) sint = −0.9042 (f) sint = −0.2491
2 Solvethe followingequations for0 t 2π:
(a) cost = 0.4243 (b) cost = 0.8040
(c) cost = 0.3500 (d) cost = −0.5618
(e) cost = −0.7423 (f) cost = −0.3658
3 Solvethe followingequations for0 t 2π:
(a) tant = 0.8493 (b) tant = 1.5326
(c) tant = 1.2500 (d) tant = −0.8437
(e) tant = −2.0612 (f) tant = −1.5731
4 Solvethe followingequations for0 t 2π:
(a)sin2t = 0.6347
(b)sin3t = −0.2516
( )
t
(c)sin = 0.4250
2
(d)sin(2t +1) = −0.6230
(e)sin(2t −3) = 0.1684
( )
t +2
(f) sin = −0.4681
3
5 Solve the following equations for0 t 2π:
(a) cos2t = 0.4234
( )
t
(b) cos = −0.5618
3
( )
2t
(c) cos = 0.6214
3
(d) cos(2t +0.5) = −0.8300
(e) cos(t −2) = 0.7431
(f) cos(πt −1) = −0.5325
6 Solve the following equations for0 t 2π:
(a) tan2t = 1.5234
( )
t
(b) tan = −0.8439
3
(c) tan(3t −2) = 1.0641
(d) tan(1.5t −1) = −1.7300
( )
2t+1
(e) tan = 1.0000
3
(f) tan(5t −6) = −1.2323
7 A time-varying voltage, v(t),has the form
v(t) =20sin(50πt +20)
t 0
Calculate the firsttime that the voltage has avalue of
(a)2volts (b)10volts (c)15volts
Solutions
1 (a) 1.0021,2.1395 (b) 0.2163,2.9253
(c) 0.6042,2.5374 (d) 3.5879,5.8369
(e) 4.2711,5.1537 (f) 3.3933,6.0314
2 (a) 1.1326,5.1506 (b) 0.6368,5.6464
(c) 1.2132,5.0700 (d) 2.1674,4.1158
(e) 2.4073,3.8759 (f) 1.9453,4.3379

150 Chapter 3 The trigonometric functions
3 (a) 0.7041, 3.8457 (b) 0.9927, 4.1343
(c) 0.8961, 4.0376 (d) 2.4408, 5.5824
(e) 2.0225, 5.1641 (f) 2.1370, 5.2786
4 (a) 0.3438,1.2270, 3.4854, 4.3686
(b) 1.1320,2.0096, 3.2264, 4.1040, 5.3208, 6.1984
(c) 0.8779,5.4053
(d) 1.4071,2.3053, 4.5487, 5.4469
(e) 1.5846,2.9862, 4.7262, 6.1278
(f) nosolutions
5 (a) 0.5668,2.5748, 3.7084, 5.7164
(b) nosolutions
(c) 1.3504
(d) 1.0250,1.6166, 4.1665, 4.7582
(e) 1.2669, 2.7331
(f) 0.9971, 1.6396, 2.9971, 3.6396, 4.9971, 5.6396
6 (a) 0.4950, 2.0658, 3.6366, 5.2073
(b) no solutions
(c) 0.9388, 1.9860, 3.0332, 4.0804, 5.1276, 6.1748
(d) 2.0633, 4.1577, 6.2521
(e) 0.6781, 5.3905
(f) 0.3939, 1.0222, 1.6505, 2.2788, 2.9071, 3.5355,
4.1638, 4.7921, 5.4204, 6.0487
7 (a) 1.2038 ×10 −2
(b) 9.3427 ×10 −3
(c) 7.2771 ×10 −3
TechnicalComputingExercises3.8
1 Ploty = sint for0 t 2π andy = 0.3500 using
the same axes. Useyour graphs to find approximate
solutionsto
sint =0.3500
0t 2π
2 Ploty = cost for0 t 2π andy = −0.5500using
the same axes. Useyour graphs to find approximate
solutionsto
cost+0.5500=0
0t2π
3 Ploty=sin(2t+1)andy=2sintfor0t 2π.
Useyourgraphs to stateapproximate solutionsto
sin(2t+1)=2sint
0t 2π
4 Ploty=2sin3tandy=3cos2tfor0t 2π.
Hence state approximate solutionsof
2sin3t =3cos2t
0t 2π
REVIEWEXERCISES3
1 Expressthe followingangles in radians:
(a) 45 ◦ (b) 72 ◦ (c) 100 ◦ (d) 300 ◦
(e) 440 ◦
2 Thefollowing angles are in radians. Expressthemin
degrees.
π 3π
(a) (b) 3π (c) (d)2 (e) 3.62
3 4
3 Statethe quadrant in whichthe angle α liesgiven
(a) sinα>0andtanα>0
(b) cosα>0andsinα<0
(c) tanα>0andcosα<0
(d) sinα<0andcosα<0
(e) tanα<0andcosα<0
4 Simplify the followingexpressions:
(a) sint cosect
sinx
(b)
tanx
cotA
(c)
cosA
secA
(d)
cosecA
(e) cotxtanx
5 Simplify the following expressions:
(a) cos 2 A +1+sin 2 A
2sinAcosA
(b)
cos 2 A −sin 2 A
√
(c) sec 2 x−1
(d) sintcost +
(e)
1
cosec 2 A −1
1
sect cosect

Review exercises 3 151
6 Simplify the following expressions:
(a) (sinx +cosx) 2 −1
(b) tanAsin2A +1+cos2A
sin4θ +sin2θ
(c)
cos2θ −cos4θ
(d) 4sinAcosAcos2A
sint
(e) ( )
t
cos
2
7 Statethe amplitude,angularfrequency, period,
frequency, phase andtime displacement ofthe
followingwaves:
(a) 2sin3t
(b) 4cos6t
(c) 0.7sin(2t +3)
(d) 0.1cos πt
(e) sin(50πt +20)
(f) 6cos(100πt −30)
( )
1
(g)
2 sin t
2
(h) 0.25cos(2πt +1)
8 Expressthe following in the form
Asin(ωt +φ),φ 0:
(a)6sin5t +5cos5t
(b)0.1sint −0.2cost
(c)7sin3t +6cos3t
( ) ( )
t t
(d)9cos −4sin
2 2
(e)3sin2t +15cos2t
9 Expressthe following in the form
Asin(ωt −φ),φ 0:
(a) 3sin4t +7cos4t
(b) 3cos2t −5sin2t
(c) 4sin6t −7cos6t
1
(d)
2 cost+2 3 sint
(e) 0.75sin(0.5t) −1.25cos(0.5t)
10 Expressthe following in the formAcos(ωt + φ),
φ0:
(a) 10sin3t +16cos3t
(b) −6sin2t −3cos2t
(c) sint−2cost
(d) 0.6cos4t +1.3sin4t
(e) cos7t −5sin7t
11 Express eachofthe following in the form
Acos(ωt −φ),φ 0:
(a) 2.3sin3t +6.4cos3t
(b) −sin2t −2cos2t
(c) 2cos9t +9sin9t
(d) 4sin4t
(
+5cos4t
) ( )
t t
(e) −6cos −2sin
2 2
12 Express eachofthe following in the form
Asin(ωt +φ),φ 0:
(a) sin(t +1) +cos(t +1)
(b) 2sin(2t +3) −3cos(2t +1)
(c) cos(3t −1) −2sin(3t +4)
(d) sin(t +1) +sin(t +3)
(e) cos(2t −1) +3cos(2t +3)
13 Reduceeachofthe followingexpressions to a single
wave andin each casestate the amplitudeandphase
angleofthe resultantwave:
(a) 2cosωt
(
+3sin
)
ωt
(b) cos
(c) 2sin
(d) 0.5sin
ωt + π 4
(
ωt + π 2
(
+sinωt
)
ωt − π 4
+4cos
)
(
+1.5sin
ωt + π 4
(
(e) 3sinωt +4sin(ωt +π)−2cos
)
ωt + π 4
(
)
ωt − π 2
14 Solve the following equations,stating allsolutions
between 0 and2π:
(a) sint = 0.5216
(b) sint = −0.3724
(c) cost = 0.9231
(d) cost = −1
(e) tant = 0.1437
(f) tant = −1
15 Solve the following equations,stating allsolutions
between 0 and2π:
(a) sin2t = 0.5421
(b) cos2t
(
=
)
−0.4687
t
(c) tan = −1.6235
2
(d) 2sin4t = 1.5
(e) 5cos2t =2
(f) 4tan2t =5
)

152 Chapter 3 The trigonometric functions
16 Avoltage, v(t),varieswith time,t,according to
v(t) = 240sin(100πt +30) t 0
Find the firsttime that v(t)has avalue of
(a) 240 (b) 0
(c) −240 (d) 100
17 Simplify asfar aspossible
(a) cos100 ◦ +cos80 ◦
(b) cos100 ◦ −cos80 ◦
sin50 ◦ +sin40 ◦
(c)
cos5 ◦
(d) sin80◦ −sin60 ◦
2sin10 ◦
Solutions
1 (a)
π
4
(b)
2π
5
(d) 5.2360 (e) 7.6794
(c) 1.7453
2 (a) 60 ◦ (b) 540 ◦ (c) 135 ◦
(d) 114.6 ◦ (e) 207.4 ◦
3 (a) 1st (b) 4th (c) 3rd
(d) 3rd (e) 2nd
4 (a) 1 (b) cosx (c) cosecA
(d) tanA (e) 1
5 (a) 2 (b) tan2A (c) tanx
(d) sin2t (e) tan 2 A
6 (a) sin2x (b) 2 (c) cot θ
( )
t
(d) sin4A (e) 2sin
2
7 (a) 2,3, 2π 3 , 3
2π ,0,0
(b) 4,6, π 3 , 3 π ,0,0
(c) 0.7,2, π, 1 π ,3, 3 2
(d) 0.1, π,2, 1 2 ,0,0
(e) 1,50π,0.04,25,20, 2
5π
(f) 6,100π,0.02,50, −30, − 3
10π
1
(g)
2 , 1 2 ,4π, 1
4π ,0,0
(h) 0.25,2π,1,1,1, 1
2π
√
8 (a) 61sin(5t +0.6947)
√
(b) 0.05sin(t +5.1760)
√
(c) 85sin(3t +0.7086)
(d)
(e)
9 (a)
( )
√ t 97sin
2 +1.9890
√
234sin(2t +1.3734)
√
58sin(4t −5.1173)
(b) √ 34sin(2t −3.6820)
√
(c) 65sin(6t −1.0517)
5
(d) sin(t −5.6397)
6
(e) 1.4577sin(0.5t −1.0304)
√
10 (a) 356cos(3t +5.7246)
(b) √ 45cos(2t +2.0344)
√
(c) 5cos(t +3.6052)
(d) √ 2.05cos(4t +5.1448)
√
(e) 26cos(7t +1.3734)
11 (a) 6.8007cos(3t −0.3450)
12 (a)
(b) √ 5cos(2t −3.6052)
√
(c) 85cos(9t −1.3521)
(d) √ 41cos(4t −0.6747)
( )
√ t
(e) 40cos
2 −3.4633
√
2sin(t +1.7854)
(b) 1.4451sin(2t +5.0987)
(c) 2.9725sin(3t +0.7628)
(d) 1.0806sin(t +2)
(e) 2.4654sin(2t +4.8828)
√ √
13 (a) 13sin(ωt +0.5880), amplitude = 13,phase
angle = 0.5880
(b) 0.765sin(ωt +1.1781),amplitude = 0.765,
phaseangle = 1.1781
(c) 5.596sin(ωt +2.10),amplitude = 5.596, phase
angle = 2.10

Review exercises 3 153
(d) 1.581sin(ωt +0.4636), amplitude = 1.581,
phaseangle = 0.4636
(e) −3sin ωt,amplitude = 3, phase angle = π
14 (a) 0.5487,2.5929 (b) 3.5232,5.9016
(c) 0.3947,5.8885
(e) 0.1427,3.2843
(d) 3.1416, i.e. π
3π
(f)
4 , 7π 4
15 (a) 0.2865,1.2843,3.4281,4.4259
(b) 1.0293,2.1123,4.1709,5.2539
(c) 4.2457
(d) 0.2120,0.5734,1.7828,2.1442,3.3536,3.7150,
4.9244,5.2858
(e) 0.5796,2.5620,3.7212,5.7035
(f) 0.4480,2.0188,3.5896,5.1604
16 (a) 9.5070 ×10 −3 (b) 4.5070 ×10 −3
(c) 1.9507 ×10 −2
17 (a) 0 (b) −2sin10 ◦
(c) √ 2
(d) cos70 ◦
(d) 5.8751 ×10 −3

4 Coordinatesystems
Contents 4.1 Introduction 154
4.2 Cartesiancoordinatesystem--twodimensions 154
4.3 Cartesiancoordinatesystem--threedimensions 157
4.4 Polarcoordinates 159
4.5 Somesimplepolarcurves 163
4.6 Cylindricalpolarcoordinates 166
4.7 Sphericalpolarcoordinates 170
Reviewexercises4 173
4.1 INTRODUCTION
The coordinates of a point describe its position. The most common coordinate system
is the x--y system: the first number, x, gives the distance along the x axis, the second
number, y, gives the distance along the y axis. However, this is not the only way to
describethepositionofapoint.Thischapteroutlinesseveralwaysinwhichtheposition
of a point can be described.
4.2 CARTESIANCOORDINATESYSTEM--TWODIMENSIONS
The Cartesian coordinate system is named after the French mathematician Descartes.
The system comprises two axes -- the x axis and the y axis -- which intersect at right
angles at the point O. The point O is called the origin. Figure 4.1 shows the Cartesian
coordinate system. By convention thexaxis is drawn horizontally. The positivexaxis
liestotherightoftheorigin,thenegativexaxisliestotheleftoftheorigin,thepositive
y axis liesabove the origin and the negativeyaxis lies below the origin.
Note that this coordinate system can be used only for locating points in a plane, that
isithas two dimensions.
Consider any point, P, in the plane. The horizontal distance of P from the y axis
is called the x coordinate. The vertical distance of P from the x axis is called the y
coordinate. Either coordinate can bepositive ornegative.

4.2 Cartesian coordinate system -- two dimensions 155
y
3
2
1
x coordinate
P
y coordinate
–4 –3 –2 –1 O 1 2 3 4
–1
x
–2
–3
Figure4.1
Cartesian coordinate system in
two dimensions.
When stating the coordinates of a point, by convention we always state thexcoordinate
first. Thus (3, 1) means that thexcoordinate is 3 and theycoordinate is 1. We
also write, for example, A(3, 1) to mean that the point whose coordinates are (3, 1) is
labelled A.InFigure 4.1 Phas coordinates (2, 3).
Example4.1 Plot the points whoseCartesiancoordinates are
(a) (4,2) (b) (−1,−3) (c) (−2,1)
Solution AsplottedinFigure 4.2
(a) Rhas coordinates (4,2).
(b) S has coordinates (−1,−3).
(c) Thas coordinates (−2,1).
Example4.2 Statethe coordinates ofthe points Aand B asshown inFigure 4.3.
Solution Ahas coordinates (−3,−1); Bhas coordinates (3,−2).
y
2
R
y
2
T
1
1
–4 –3 –2 –1 O 1 2 3 4
–1
–2
S
–3
x
–3 –2 –1 O 1 2 3
A
–1
–2
B
–3
x
Figure4.2
Figure4.3

156 Chapter 4 Coordinate systems
Engineeringapplication4.1
Electrodecoordinates
Often in science and engineering it is useful to place metallic conductors, known as
electrodes, inside a glass container from which air has been evacuated. For example,electricalvalves,
whicharealsoknown asvacuum tubes,areconstructed inthis
manner.Audioamplifiersthatmakeuseofvalvetechnologyhavemadeacomeback
inrecent years. Many musicians prefer the sound they generate.
Figure 4.4 shows two electrodes inside an evacuated glass envelope. State the
coordinates of the four points A,B, Cand D.
y
12
11
10
9
8
7
6
5
4
3
2
1
A
0 1 2 3 4 5 6 7 8 9 10 11 12
D
glass envelope
C
B
x
Figure4.4
Electrode coordinates.
Solution
A has coordinates (4,9).
B has coordinates (9,6).
C has coordinates (8,10).
D has coordinates (8,6).
Tosimulatetheelectricfieldbetweentheseelectrodes,andtoproduceaproperdesign
formanufacture,acomputermodelmaybeused.Suchamodelwouldinvolvetheuse
of mathematical techniques to solve the fields in the space between the electrodes.
However, the user would have to input the shape of the device. Coordinates would
be used to specify exactly where each part of the device is located. The coordinates
could easily be changed in the computer model to test the effects of changing the
sizeandpositionoftheelectrodes.Computermodelsspeedupthedesignprocessby
avoiding the need forearly-stage prototypes.

4.3 Cartesian coordinate system -- three dimensions 157
EXERCISES4.2
1 Plotthe following points:A(2,−2),B(−2,1),
C(−1,0),D(0, −2).
2 Statethe coordinatesofthe pointsU, V andWas
shown in Figure4.5.
3 A point Plies onthexaxis. State theycoordinate
ofP.
4 A point Qlies on theyaxis. Statethexcoordinate
ofQ.
y
y
2
2
1
V
B
1
C
–2 –1 O 1 2 3 4
–1
W
U –2
x
–2 –1 O
–1
–2
D
1
2
A
x
Figure4.5
FigureS.6
Solutions
1 FigureS.6 shows the pointsA, B,CandD.
2 U(−2, −2),V(4,1),W(3, −1)
3 0
4 0
4.3 CARTESIANCOORDINATESYSTEM--THREEDIMENSIONS
Manyengineeringproblemsrequiretheuseofthreedimensions.Figure4.6illustratesa
three-dimensional coordinate system. It comprises three axes,x,yandz. The axes are
all atright anglestoone another and intersectatthe origin, O.
The position of any point in three-dimensional space is given by specifying itsx,y
andzcoordinates.Byconventionthexcoordinateisstatedfirst,thentheycoordinateand
finally thezcoordinate. For example, P(2, 3, 4) has anxcoordinate of 2, aycoordinate
of 3 and azcoordinate of 4. It is illustrated in Figure 4.6. From the origin, P is located
by travelling 2 units in thexdirection, followed by 3 units in theydirection, followed
by 4 unitsinthezdirection.
Note that, aswith a two-dimensionalsystem,coordinates can benegative.
Example4.3 Plot the following points:A(1,−1,2), B(0,1,2), C(0,0,1).
Solution Figure 4.7 illustratesthe points A,Band C.
Wenowconsidertheequationofaplane.Anypointonthex--yplanehasazcoordinate
of 0. Hence the equation of thex--y plane isz = 0. Similarlyz = 1 represents a plane
parallel to the x--y plane but 1 unit above it. Point C in Figure 4.7 lies in the plane
z = 1.Allpoints inthis plane have azcoordinateof1.

158 Chapter 4 Coordinate systems
x
2
1
z
4
3
2
1
O
3
1
4
P (2, 3, 4)
2
2
3 y
Figure4.6
Cartesian coordinate systemin three
dimensions.
A
x
–1
1
z
2
1
C
1
B
2 y
Figure4.7
ThepointsA, Band Care
plotted in threedimensions.
AandBinFigure4.7lieintheplanez = 2.Allpointsinthisplanehaveazcoordinate
of 2.
EXERCISES4.3
1 Plotthe pointsA(2,0, −1),B(1, −1,1)and
C(−1,1,2).
2 Statethe equationofthe planepassingthrough
(4,7, −1),(3,0,−1) and(1,2, −1).
3 Statethe equationofthe planepassingthrough
(3,1,7),(−1,1,0)and(6,1, −3).
Solutions
1 FigureS.7 illustratesthe pointsA, B andC.
3 y=1
2 z=−1
B (1, –1, 1)
x
2
FigureS.7
–1
1
z
2
1
A (2, 0, –1)
–1
–1
1
C (–1, 1, 2)
2 y
y
O
u
r
x
P (x, y)
y
A
Figure4.8
Phas polar coordinatesrand θ.
x

4.4 Polar coordinates 159
4.4 POLARCOORDINATES
We have seen how the x and y coordinates of a point describe its location in the x--y
plane.Thereisanalternativewaytodescribethelocationofapoint.Figure4.8illustrates
a point Pinthex--yplane.
Phas Cartesian coordinates (x,y). Hence
OA=x,
AP=y
Consider the arm OP. The length of OP is the distance of P from the origin. We denote
thisbyr, that is
length of OP =r
Clearly,risnever negative, thatisr 0.
We note that the angle between the positivexaxis and OP is θ. The value of θ lies
between 0 and 2πradians or 0 ◦ to360 ◦ ifdegrees areused.
The values ofrand θ are known as the polar coordinates of P. Conventionally, the
valueofrisstatedfirst,thenthevalueof θ.Wecommonlywritethesepolarcoordinates
asr̸ θ.
Thevaluesofr and θ specifythepositionofapoint.Conventionally,positivevalues
of θ aremeasured anticlockwise from the positivexaxis.
Thepolarcoordinatesofapoint,P,arer̸ θ.ThevalueofristhedistanceofPfrom
the origin; the value of θ isthe angle between the positivexaxis and the armOP.
r0 0θ<2π(0 ◦ θ<360 ◦ )
Engineeringapplication4.2
Pickandplacerobot
Robotsarenowwidelyusedinfactoriesinordertoreducelabourcosts.Theyvaryin
complexity depending on their function. Almost all printed circuit boards found in
electronicdevicessuchascomputersandmobilesareassembledbyrobots.Figure4.9
shows a simple robot that can pick up a surface-mount electronic component in one
position and placeit inanotherposition.
It consists of a rotating arm the length of which can be extended and contracted.
On the end of the arm is a hand which can be closed to pick up a component and
opened toreleaseit.
r
zero datum
u
Figure4.9
A pickand place robot.
➔

160 Chapter 4 Coordinate systems
Itisnecessaryforacomputertocarryoutcalculationsinordertofindwhereacomponentislocatedandwheretoplaceacomponent.Decideuponasuitablecoordinate
systemtouse when carrying outthese calculations.
Solution
If we examine the geometry of the robot then we see that a polar coordinate system
wouldbethemostsuitable.Thecentreofthecoordinatesystemshouldbeontheaxis
ofrotation.Thelengthofthearmisthengivenbyr andtheorientationofthearmis
given by θ relative toan agreed zero datum mark.
Example4.4 Plot the points P, Qand R whose polar coordinates are
(a) 2,70 ◦
(b) 4,160 ◦
(c) 3,300 ◦
Solution Figure 4.10 shows the three points plotted.
y
y
P
Q
y
2
O
70°
x
4
O
160°
x
300°
O
x
3
(a) (b) (c)
R
Figure4.10
A point can be located bythe values ofits polar coordinates.
Consider the arm from the origin to the point. The value of r gives the length of this
arm. The value of θ gives the angle between the positivexaxis and the arm, measuring
anticlockwise from the positivexaxis.
Bystudying △OPA,shown inFigure 4.11, wecan see that
cosθ = x r
sinθ = y r
and sox =rcos θ (4.1)
and soy =rsin θ (4.2)
Hence if we know the values of r and θ, that is the polar coordinates of a point, we
can use Equations (4.1) and (4.2) to calculatexandy, the Cartesian coordinates of the
point.

4.4 Polar coordinates 161
y
O
u
r
x
Figure4.11
Thepolar coordinates arer, θ;the
Cartesian coordinates arex,y.
P
y
A
x
–3.4641
4
y
210°
–2
Figure4.12
The Cartesian coordinates canbe
calculated from the polar coordinates.
x
Example4.5 A point has polar coordinatesr = 4, θ = 210 ◦ . Calculate the Cartesian coordinates of
the point.Plot the point.
Solution TheCartesian coordinatesare given by
x =rcos θ = 4cos210 ◦ = −3.4641
y=rsinθ =4sin210 ◦ =−2
Figure 4.12 illustratesthe point.
We now look atthe problemofcalculatingthe polarcoordinates given the Cartesian
coordinates.Equations(4.1)and(4.2)canbearrangedsothatrand θ canbefoundfrom
the values ofxandy. Consider a typical pointPasshown inFigure 4.11.
The Cartesian coordinates are (x,y). Suppose that the values ofxandyare known.
The polar coordinates are r,θ; these values are unknown. By applying Pythagoras’s
theoremto △OPA wesee that
and so
r 2 =x 2 +y 2
r = √ x 2 +y 2
Note that since r is the distance from O to P it is always positive and so the positive
square rootistaken.
We now express θ in terms of the Cartesian coordinatesxandy. From Figure 4.11
wesee that
tanθ = y x
and hence
)
θ = tan −1 ( y
x
Insummary wehave
r = √ x 2 +y 2
θ = tan −1 ( y
x
)
( ) y
However, we need to exercise a little extra care before calculating tan −1 and
x
reading the resultfrom a calculator. As an illustrationnote that
tan40 ◦ = 0.8391 and tan220 ◦ = 0.8391

162 Chapter 4 Coordinate systems
and so tan −1 (0.8391) could be 40 ◦ or 220 ◦ . Similarly tan105 ◦ = −3.7321 and
tan285 ◦ = −3.7321 and so tan −1 (−3.7321)
( )
could be 105 ◦ or 285 ◦ . The value given
y
on your calculator when calculating tan −1 may not be the actual value of θ we
x
require. In order to clarify the situation it is always useful to sketch the Cartesian coordinates
and the angle θ before embarking on the calculation.
Example4.6 The Cartesian coordinates of P are (4, 7); those of Q are (−5,6). Calculate the polar
coordinates ofPand Q.
Solution Figure 4.13 illustratesthe situation forP.
Then
r = √ 4 2 +7 2 = √ 65 = 8.0623
Note from Figure 4.13 that P is in the first quadrant, that is θ lies between 0 ◦ and 90 ◦ .
Now
( ) ( ) y 7
tan −1 = tan −1
x 4
( ) 7
From a calculator, tan −1 = 60.26 ◦ . Since we know that θ lies between 0 ◦ and 90 ◦
4
then clearly 60.26 ◦ isthe required value.
The polar coordinates of Parer = 8.0623, θ = 60.26 ◦ .
Figure 4.14 illustratesthe situation forQ.
We have
r = √ (−5) 2 +6 2 = √ 61 = 7.8102
From Figure 4.14 wesee that θ lies between 90 ◦ and 180 ◦ . Now
( ) ( ) y 6
tan −1 = tan −1 = tan −1 (−1.2)
x −5
Acalculatorreturnsthevalueof −50.19 ◦ whichisclearlynottherequiredvalue.Recall
that tanθ isperiodic with period 180 ◦ . Hence the required angle is
180 ◦ + (−50.19 ◦ ) = 129.81 ◦
The polar coordinates of Qarer = 7.8102,θ = 129.81 ◦ .
y
P
y
r
7
Q 6
O
u
4
x
–5
O
u
x
Figure4.13
Pis in the firstquadrant.
Figure4.14
Qlies in the second quadrant.

4.5 Some simple polar curves 163
EXERCISES4.4
1 Given the polar coordinates, calculate the Cartesian
coordinatesofeach point.
(a) r=7,θ=36 ◦
(b) r = 10, θ = 101 ◦
(c) r = 15.7, θ = 3.7 radians
(d) r=1,θ= π 2 radians
2 Given the Cartesian coordinates, calculatethe polar
coordinatesofeachpoint.
(a) (7,11) (b) (−6,−12) (c) (0,15)
(d) (−4,6) (e) (4,0) (f) (−4,0)
Solutions
1 (a) 5.6631,4.1145 (b) −1.9081,9.8163
(c) 15, 90 ◦
(c) −13.3152, −8.3184 (d) 0, 1
(d) 7.2111,123.69 ◦
(e) 4,0 ◦
2 (a) r = 13.0384, θ = 57.53 ◦
(f) 4,180 ◦
4.5 SOMESIMPLEPOLARCURVES
UsingCartesiancoordinatestheequationy =mxdescribestheequationofalinepassing
through the origin. The equation of a line through the origin can also be stated using
polar coordinates. In addition, it is easy to state the equation of a circle using polar coordinates.
Equationofaline
Consider all points whose polar coordinates are of the formr̸ 45 ◦ . Note that the angle
θ is fixed at 45 ◦ but thatr, the distance from the origin, can vary. Asr increases, a line
at45 ◦ tothe positivexaxis istraced out.Figure 4.15 illustratesthis.
Thus, θ = 45 ◦ is the equation of a line starting at the origin, at 45 ◦ to the positive
x axis.
Ingeneral, θ = θ c
,where θ c
isafixedvalue,istheequationofalineinclinedat θ c
to
the positivexaxis, startingatthe origin.
Equationofacircle,centreontheorigin
Consider all points with polar coordinates 3̸ θ. Herer, the distance from the origin, is
fixed at 3 and θ can vary. As θ varies from 0 ◦ to 360 ◦ a circle, radius 3, centre on the
origin, issweptout.Figure 4.16 illustratesthis.
Ingeneralr =r c
wherer c
isafixedvalue,0 ◦ θ 360 ◦ describesacircleofradius
r c
, centred on the origin.
Example4.7 Draw the curve traced outbyr = 3,0 ◦ θ 180 ◦ .
Solution Hereris fixed at 3 and θ varies from 0 ◦ to 180 ◦ . As θ varies a semicircle is traced out.
Figure 4.17 illustratesthis.
AtP, θ =0 ◦ ,whileatQ, θ =180 ◦ .

164 Chapter 4 Coordinate systems
y
u = 45°
y
3
3
x
O
45°
Figure4.15
When θ isfixed andrvaries, a straight
line from the origin is traced out.
x
Figure4.16
Whenrisfixedand θ varies,acircleis
sweptout.
Q
y
3
O
Figure4.17
As θ varies from0 ◦ to 180 ◦ a
semicircle is traced out.
P
x
y
S
R
O
P
Q
Figure4.18
Surface forExample4.8.
x
Example4.8 Describethe surfacedefined by 1 r2,0 ◦ θ 90 ◦ .
Solution Herervariesfrom1to2and θ variesfrom0 ◦ to90 ◦ .Figure4.18illustratesthesurface
so formed.
AtP,r=1,θ=0 ◦ ;atQ,r=2,θ=0 ◦ ;atR,r=1,θ=90 ◦ ;atS,r=2,θ=90 ◦ .
We have seen some simple polar curves in Figures 4.16 and 4.17. In general a polar
curve isgiven by the equationr = f (θ),where the radiusrvaries with the angle θ.
Engineeringapplication4.3
Two-dimensionalantennaradiationpattern
Polarcurves areoftenusedtodepictradiation patternsfromantennas.Itisoftenthe
casethattheelectricfieldstrengthatafixeddistancefromanantennasuchasshown
inFigure 4.19 dependsupon the angle θ.
A typical expression for field strength at a particular angle θ could be
∣ cos( π cosθ)
2 ∣
∣.
sinθ

4.5 Some simple polar curves 165
u
Antenna
Figure4.19
Electric field strength atafixed distance from the
antenna depends upon the angle θ.
Byconsidering
cos(
r =
π cosθ)
2 ∣ sinθ ∣
we can use a polar curve to depict the field strength at any angle. Graph-plotting
softwareisavailableforproducingpolarplotssuchastheoneshowninFigure4.20,
which shows a typical angle, θ, and itsassociated radiusr.
The radius of the plot for a particular value of θ represents the transmitted field
strengthandsotheantennainFigure4.20hasnoradiationinthehorizontaldirection
and maximum radiation inthe vertical direction.
y
1
0.5
r
u
–0.5 0.5
x
–0.5
–1
Figure4.20
Thetransmitted field strength forthe antenna,
r,varies with the angularposition, θ.
EXERCISES4.5
1 Describethe curve defined byr = 2,0 ◦ θ 90 ◦ . 2 Describethe surfacedefined by0 r2,
30 ◦ θ 45 ◦ .

166 Chapter 4 Coordinate systems
Solutions
1 This isaquarter circle ofradius2asshown in
FigureS.8.
2 FigureS.9 illustratesthe surface. OPis setat30 ◦ to
thexaxis;OQis at45 ◦ to thexaxis. OP = OQ = 2.
y
y
Q
2
P
O
x
O
x
FigureS.8
FigureS.9
4.6 CYLINDRICALPOLARCOORDINATES
Consider the problem of studying the flow of water around a cylinder. A problem like
this would be studied by engineers when investigating the forces exerted by the sea on
the cylindrical supports of oil-rigs. It is often mathematically convenient to choose a
coordinate system that fits the shape of the object being described. It makes sense here
toselect a cylindricalcoordinate system.
Cylindrical polar coordinates comprise polar coordinates with the addition of a
vertical, or z, axis. Figure 4.21 illustrates a typical point, P, and its cylindrical polar
coordinates.
The point Q is in the x--y plane and lies directly below P. Q is the projection of P
onto thex--y plane.
Consider a point P in three-dimensional space, with Cartesian coordinates (x,y,z).
We can also describe the position of P using cylindrical polar coordinates. To do this,
the x and y coordinates are expressed as their equivalent polar coordinates, while the
z coordinate remains unaltered. Hence the cylindrical polar coordinates of a point have
theform (r,θ,z).
Recall thatris the length of the arm OQ (see Figure 4.21); that is, it is the distance
of a pointinthex--yplane from the origin, and sor 0.The angle θ is measured from
thepositivexaxistothearmOQandso θ hasvaluesbetween0 ◦ and360 ◦ or2πradians.
Finally, z is positive for points above the x--y plane and negative for points below the
x--yplane.Insummary
r0, 0θ<2π, −∞<z<∞
z
P
O
u
z
y
x
r
Q
Figure4.21
Thecylindrical polar coordinates ofPare (r,θ,z).

4.6 Cylindrical polar coordinates 167
We can relate the Cartesian coordinates, (x,y,z), to the cylindrical polar coordinates,
(r,θ,z). The following key point does this.
x=rcosθ r0
y=rsinθ 0θ <2π
z =z
Engineeringapplication4.4
Fluidflowalongapipe
Cylindrical polar coordinates provide a convenient framework for analysing liquid
flow down a pipe. The radial symmetry of a pipe makes it the natural choice. The
distance along the pipe is defined usingz. In order to utilize the radial symmetry of
thepipeitisnecessarytoalignthezaxiswiththecentreaxisofthepipe.Figure4.22
illustratesthearrangement.Distancefromthecentreofthepipeisdefinedbyr.The
angle θ isusedinconjunctionwithzandrtofixthepositionwithinthepipe.Atypical
problemthatmaybeanalysedisthevariationinfluidvelocitywithdistancefromthe
centreofthepipe.Forsmoothflow,liquidtendstotravelfasteratthecentreofapipe
than itdoes near the edge.
r
u
z
Figure4.22
Fluid flowalong apipe.
Pipes with metal walls are often used to guide electromagnetic waves, rather
than fluids, in high-powered microwave communications systems. They are termed
waveguides.Mathematicallyanalysingthewaveguide’spropagationmodesismade
much simplerby using cylindrical polar coordinates.
Example4.9 The Cartesian coordinates of P are (4,7,−6). State the cylindrical polar coordinates
ofP.
Solution We have
x=4, y=7, z=−6
Usingx = 4 andy = 7 the values ofrand θ are found to ber = 8.0623,θ = 60.26 ◦
(see Example 4.6). The z coordinate remains unchanged. Hence the cylindrical polar
coordinates are (8.0623,60.26 ◦ ,−6).

168 Chapter 4 Coordinate systems
Example4.10 Describe the figure defined by
(a) 1r2,θ=60 ◦ ,−1z1
(b)r=1,0 ◦ θ90 ◦ ,0z2
z
A
B
y
P
Q
x
P
C
Q
y
O
60°
x
D
Figure4.23
AtP,r=1;atQ,r=2.
Figure4.24
On the line AB,z = 1; onthe line CD,z = −1.
Solution (a) Consider therand θ coordinates first. Thercoordinate varies from 1 to 2 while θ
is fixed at 60 ◦ . This represents the line PQ as shown in Figure 4.23. At P the value
ofris 1; at Q the value ofris 2. The length of PQ is 1 and it is inclined at 60 ◦ to
thexaxis.
Now, we note thatzvaries from −1 to 1. We imagine the line PQ moving in the
z direction fromz = −1 toz = 1. This movement sweeps out a plane. Figure 4.24
illustratesthis.
(b) The r coordinate is fixed at r = 1. The θ coordinate varies from 0 ◦ to 90 ◦ . This
produces the quarter circle, AB, as shown in Figure 4.25. At A,r = 1, θ = 0 ◦ ; at
B,r=1,θ =90 ◦ .
z
D
y
C
B
B
A
y
O
A
x
x
Figure4.25
As θ varies from0 ◦ to 90 ◦ ,aquarter
circle is sweptout.
Figure4.26
Aszvaries from0to 2,the curveAB
sweepsoutthe curved surface.

4.6 Cylindrical polar coordinates 169
Examining the z coordinate, we see thatzvaries from 0 to 2. As z varies from
0 to 2, we imagine the curve AB sweeping out the curved surface as shown in Figure4.26.AtC,r
= 1, θ = 0 ◦ ,z = 2;atD,r = 1, θ = 90 ◦ ,z = 2.Thissurfaceis
partof a cylinder.
Iftherangeofvaluesofacoordinateisnotgivenitisunderstoodthatthatvariable
variesacrossallitspossiblevalues.Forexample,acurvemaybedescribedbyr = 1,
z = −2.Herethereisnomentionofthevaluesthat θ canhave.Itisassumedthat θ
can have its fullrange of values, thatis0 ◦ to360 ◦ .
Engineeringapplication4.5
Helicalantennas
The helix is a shape commonly found inengineering. For example, the springs used
inacar’ssuspensionoftenhave ahelicalshape.Helicalantennaswereinvented by
John Kraus in the 1940s and since then have been used extensively in a variety of
applications including space exploration, satellite communications and mobile telephony.
Developing a mathematical definition of a helix is essential to analysing its
electromagnetic properties.
Wecansetupacylindricalpolarcoordinatesystemwiththezaxisalignedwiththe
axis of the helix as shown in Figure 4.27. If we were to look at the helix along the
directionofthezaxis,allwewouldseewouldbeacircle.Wesaythattheprojectionof
the helix onto thex--y plane isacircle.
z
y
x
Figure4.27
Helix along thezaxis.
Suppose a particular helix can be defined parametrically by the Cartesian equations
x(t) =3cos2t, y(t) =3sin2t, z(t) =t
wheret is varied over a particular range in order to generate the finite length helix
required. By specifying a particular value oft these equations enable us to calculate
particular values ofx,yandzcorresponding toapoint on the helix.
➔

170 Chapter 4 Coordinate systems
We now develop the equation of the helix in cylindrical polar coordinates. By
comparing
x=3cos2t,y=3sin2t with x=rcosθ,y=rsinθ
(see Equations (4.1) and (4.2)), we haver = 3 and θ = 2t. Note thatr = 3 is the
equation of a circle, radius 3,centre the origin.
So, the projection of the helix onto thex--y plane is a circle of radius 3. Because
z(t) =t, the value ofzincreases as the parametert increases and the helix is traced
out.
Wecannowstateanalternativedefinitionofthehelixintermsofcylindricalpolar
coordinates:
r(t) =3, θ(t) =2t, z(t) =t
AsintheCartesiancase,specifyingavalueoftheparametert enablesustocalculate
particular values ofr, θ andzcorresponding to a point on the helix. This provides a
more elegant definition of the helix than that available using Cartesian coordinates.
Many problems require the use of these alternative coordinate systems in order to
simplifyanalysis.
EXERCISES4.6
1 Expressthe followingCartesian coordinates as
cylindrical polar coordinates.
(a) (−2,−1,4) (b) (0,3,−1) (c) (−4,5,0)
2 Expressthe followingcylindrical polar coordinates as
Cartesian coordinates.
(a) (3,70 ◦ ,7) (b) (1,200 ◦ ,6) (c) (5,180 ◦ ,0)
3 Describe the surface defined by
(a) z=0
(b) z=−1
(c) r=2,z=1
(d) θ =90 ◦ ,z=3
(e) r=2,0z4
Solutions
1 (a) ( √ 5,206.57 ◦ ,4)
(b) (3,90 ◦ ,−1)
(c) ( √ 41,128.66 ◦ ,0)
2 (a) (1.0261,2.8191,7)
(b) (−0.9397,−0.3420,6)
(c) (−5,0,0)
3 (a) thex--y plane
(b) aplaneparallelto thex--yplaneand1unit
below it
(c) acircle,radius2,parallelto thex--yplaneand
with centreat(0,0,1)
(d) aline 3 units above the positiveyaxisand
parallelto it
(e) thecurvedsurfaceofacylinder,radius2,height4
4.7 SPHERICALPOLARCOORDINATES
When problems involve spheres, for example modelling the flow of oil around a ball
bearing,itmaybeusefultousesphericalpolarcoordinates.Thepositionofapointis
given bythreecoordinates, (R,θ,φ). These areillustratedinFigure 4.28.

4.7 Spherical polar coordinates 171
z
z
P
f
R
O
x
u
y
x
y
Q
Figure4.28
Spherical polar coordinates are (R,θ, φ).
Consider a typical point, P. We look ateach of the three coordinates inturn.
The value ofRis the distance of the point from the origin; that is,Ris the length of
OP. Note thatR 0.
Let Q be the projection of P onto the x--y plane. Then θ is the angle between the
positive x axis and OQ. Thus, θ has the same definition as for polar and cylindrical
polar coordinates. Note that θ can have any value from0 ◦ to360 ◦ .
Consider the line OP. Then φ is the angle between the positive z axis and OP. The
angle φ can have values between 0 ◦ and 180 ◦ . When P is above thex--y plane, then φ
lies between 0 ◦ and 90 ◦ ; when P lies below the x--y plane, then φ is between 90 ◦ and
180 ◦ . When φ = 0 ◦ , then P is on the positivezaxis; when φ = 90 ◦ , P lies in thex--y
plane; when φ = 180 ◦ , Plieson the negativezaxis.
WecandetermineequationswhichrelatetheCartesiancoordinates, (x,y,z),andthe
spherical polar coordinates, (R,θ,φ).
Note that some books describe spherical polar coordinates as (R,φ,θ), that is the
definitions of θ and φ areinterchanged. Be aware of thiswhen reading other texts.
Consider △OPQ.Note that ̸ OQP isaright angle and so
OQ=OPsinφ=Rsinφ
OQ liesinthex--y plane and so
x=OQ cosθ =Rsinφcosθ
y=OQ sinθ =Rsinφsinθ
We also notethat
z=OPcosφ=Rcosφ
Insummarywe have
and
x=Rsinφcosθ
y=Rsinφsinθ
z=Rcosφ
R0, 0φ π(180 ◦ ) 0θ <2π(360 ◦ )

172 Chapter 4 Coordinate systems
Example4.11 Show that
R = √ x 2 +y 2 +z 2
Solution From Figure 4.28
x 2 +y 2 =OQ 2
From △OPQ
OP 2 = OQ 2 +PQ 2
ButOP=RandPQ=z,so
R 2 =x 2 +y 2 +z 2
and so
R = √ x 2 +y 2 +z 2
Example4.12 Describethe surfaceR = 4.
Solution We have R = 4 and θ and φ can vary across their full range of values. Such points
generate a sphereofradius 4,centredon the origin.
Engineeringapplication4.6
Three-dimensionalradiationpatternofahalf-wavedipole
Oneofthesimplesttypesofpracticalantennaisthehalf-wavedipole.Thisconsists
of two conductor elements stretched out along a straight line having a combined
lengthofapproximatelyhalfthewavelengthatthefrequencyofthea.c.signalthatis
tobetransmitted.Thesignalisappliedtotheantennaatthecentreofthearrangement
byafeedcable.Theelectricfieldstrengthproducedbytheantennaatafixeddistance
is usually expressed using a spherical coordinate system. The coordinates for the
antennaandtheoriginoftheradiationitselfareassumedtobelocatedattheantenna
feedpointandtheelectricfieldstrengthisrepresentedbytheradius,R.Plotsproduced
likethisareingeneraltermedradiationpatternsandareausefulwayofvisualizing
the amountofradiated field inagiven direction, (θ,φ), foraparticularantenna.
Thehalf-wave dipole patternisdescribedby the equation
( π
)
cos
R =K
2 cos φ sin φ
∣ ∣
whereRrepresentstheelectricfieldstrengthandK isaconstantforagivendistance
fromtheantennacentrepoint.Theequationforthissimpleantennadoesnotinvolve
θ,whichindicatesthatRdoesnotdependonit,hencethereisradialsymmetrytothe
pattern. This function isplotted inFigure 4.29.

Review exercises 4 173
z
z
Feeding lines
attached to
an a.c source
f
Antenna
elements
3D radiation
pattern
Figure4.29
The half-wave dipole antenna and itsradiation pattern in sphericalpolar coordinates.
EXERCISES4.7
1 Apoint has spherical polar coordinates (3,40 ◦ ,70 ◦ ).
Determine the Cartesian coordinates.
2 Apoint has Cartesian coordinates (1,2,3).Determine
the spherical polar coordinates.
3 Describe the surfaceR = 1,0 ◦ θ 360 ◦ ,
0 ◦ φ90 ◦ .
Solutions
1 (2.1595,1.8121, 1.0261)
3 A hemisphere ofradius1. Theflat sideis onthex--y
2 R = 3.7417, θ = 63.43 ◦ , φ = 36.70 ◦ plane.
TechnicalComputingExercises4.7
Useatechnical computing language such as
MATLAB ® to produce a plotsimilarto Engineering
application 4.3. You may findithelpful to firstgenerate
asetofnumbers0 t < 2π andthen use a builtin
functionsuch aspolarorpolarplotto generate a
graphofthe equation.
REVIEWEXERCISES4
1 Phas Cartesian coordinates (6, −3, −2).Calculate
the distance ofPfrom the origin.
2 Phas Cartesian coordinates (−4, −3).Calculate the
polar coordinates ofP.
3 Phas polar coordinates (5,240 ◦ ).Calculate the
Cartesiancoordinates ofP.
4 Describe the surface defined by
1r4,0 ◦ θ90 ◦ .
5 Calculate the cylindrical polar coordinates ofapoint
with Cartesian coordinates (−1,4,1).
6 Describe the surfacex = 0 in three dimensions.

174 Chapter 4 Coordinate systems
7 Apoint has Cartesian coordinates (−1,−1,2).
Calculate the sphericalpolar coordinatesofthe point.
8 Describe the surfaceR = 2,0 ◦ θ 180 ◦ ,
0 ◦ φ 180 ◦ .
9 Describe the three-dimensional surface defined
byx=y.
10 Thespheredefined byR = 2 intersects the plane
defined byz = 1. Describethe curve ofintersection.
Solutions
1 7
2 5, 216.87 ◦
3 −2.5, −4.3301
4 FigureS.10 illustratesthe surface.
y
5 (4.1231,104.04 ◦ ,1)
6 y--z plane
7 R = √ 6, θ = 225 ◦ , φ = 35.26 ◦
8 Ahemisphereofradius2.Theflat surfaceison the
x--z plane.
9 Thesurfaceis aplanegenerated bymoving the line
y =xup anddown thezaxis.
10 Acircle ofradius √ 3,centre (0,0,1),in the plane
z=1.
O 1 4
FigureS.10
x

5 Discretemathematics
Contents 5.1 Introduction 175
5.2 Settheory 175
5.3 Logic 183
5.4 Booleanalgebra 185
Reviewexercises5 197
5.1 INTRODUCTION
The term discrete is used to describe a growing number of modern branches of mathematics
involving topics such as set theory, logic, Boolean algebra, difference equations
and z transforms. These topics are particularly relevant to the needs of electrical and
electronic engineers. Set theory provides us with a language for precisely specifying a
great deal of mathematical work. In recent years this language has become particularly
important as more and more emphasis has been placed upon verification of software.
Boolean algebra finds its main use in the design of digital electronic circuits. Given
thataverylargeproportionofelectroniccircuitsaredigitalratherthananalogue,thisis
animportantareaofstudy.Digitalelectroniccircuitsconfinethemselvestotwoeffective
voltagelevelsratherthantherangeofvoltagelevelsusedbyanalogueelectroniccircuits.
Thesemakethemeasiertodesignandmanufactureastolerancesarenotsocritical.Digital
circuits are becoming more complex each year and one of the few ways of dealing
withthiscomplexityistousemathematics.Oneofthelikelytrendsforthefutureisthat
moreandmorecircuitdesignswillbeprovedtobecorrectusingmathematicsbeforethey
areimplemented.Differenceequationsandztransformsareofincreasingimportance
in fields such as digital control and digital signal processing. We shall deal with these
inChapter 22.
5.2 SETTHEORY
Asetisany collection of objects,things or states.

176 Chapter 5 Discrete mathematics
Theobjectsmaybenumbers,letters,daysoftheweek,or,infact,anythingunderdiscussion.Onewayofdescribingasetistolistthewholecollectionofmembersorelements
and enclose them inbraces { }.Consider the following examples.
A = {1,0}
B = {off, on}
C = {high, low}
D = {0,1,2,3,4,5,6,7,8,9}
the set ofbinary digits, one and zero
the set of possible statesof a two-statesystem
the set of effective voltage levels inadigital electronic
circuit
the set ofdigits used inthe decimal system
Notice that we usually use a capital letter to represent a set. To state that a particular
object belongs to a particular set we use the symbol ∈ which means ‘is a member of’.
So, forexample, wecan write
off∈B
3∈D
Likewise, /∈means ‘is notamember of’so that
low /∈B
5/∈A
are sensiblestatements.
Listingmembersofasetisfinewhentherearerelativelyfewbutisuselessifweare
dealingwithverylargesets.Clearly,wecouldnotpossiblywritedownallthemembers
of the set of whole numbers because there are an infinite number. To assist us special
symbols have been introduced tostand for some commonly used sets. These are
N the set of non-negative whole numbers, 0,1,2,3,...
N + the set of positive whole numbers, 1,2,3,...
Z the set of whole numbers, positive, negative and zero,
...−3,−2,−1,0,1,2,3...
R the set of all real numbers
R + the set of positive real numbers
R − the set of negative real numbers
Q the set of rational numbers
Note thatareal number isany number inthe interval (−∞,∞).
Another way of defining a set is to give a rule by which all members can be found.
Consider the following notation:
A={x:x∈Randx<2}
Thisreads‘Aisthesetofvalues ofxsuchthatxisamember ofthesetofrealnumbers
andxis less than 2’. ThusAcorresponds to the interval (−∞,2). Using this notation
other setscan be defined.
Note that
R + ={x:x∈Randx>0}
R − ={x:x∈Randx<0}

5.2 Set theory 177
Example5.1 Useset notation todescribe the intervals on thexaxis given by
(a) [0,2] (b) [0,2) (c) [−9,9]
Solution (a) {x :x∈Rand 0 x 2}
(b) {x:x∈Rand0 x < 2}
(c) {x:x∈Rand−9 x 9}
SometimesweshallbecontenttouseanEnglishdescriptionofasetofobjects,suchas
5.2.1 Equalsets
M is thesetof capacitors madebymachine M
N is thesetof capacitors madeby machineN
Q is thesetof faulty capacitors
Twosetsaresaidtobeequaliftheycontainexactlythesamemembers.Forexample,the
sets {9,5,2}and {5,9,2}areidentical.Theorderinwhichwewritedownthemembers
isimmaterial.The sets {2,2,5,9} and {2,5,9} are equal since repetition of elements is
ignored.
5.2.2 Venndiagrams
Venndiagramsprovideagraphicalwayofpicturingsetswhichoftenaidsunderstanding.
The sets are drawn as regions, usually circles, from which various properties can be
observed.
Example5.2 Suppose we are interested in discussing the set of positive whole numbers between 1
and10.LetA = {2,3,4,5}andB = {1,3,5,7,9}.TheVenndiagramrepresentingthese
sets is shown in Figure 5.1. The set containing all the numbers of interest is called the
universalset, E. Eisrepresentedbytherectangularregion.SetsAandBarerepresented
bytheinteriorsofthecirclesanditisevidentthat2,3,4and5aremembersofAwhile1,
3, 5, 7 and 9 are members ofB. It is also clear that 6 /∈A,6/∈B,8/∈A,8/∈ B. The
elements 3 and 5 are common toboth sets.
E
8
A
2
3
1
5
4 9
6
7
B
10
Figure5.1
Venn diagram forExample 5.2.
Theset containing all the members ofinterest iscalled the universal set E.
Itisusefultoaskwhethertwoormoresetshaveelementsincommon.Thisleadstothe
following definition.

178 Chapter 5 Discrete mathematics
5.2.3 Intersection
Given setsAandB, a new set which contains the elements common to bothAandBis
called theintersectionofAandB, written as
A∩B={x:x∈Aandx∈B}
In Example 5.2, we see thatA ∩B = {3,5}, that is 3 ∈A ∩B and 5 ∈A ∩B. If the set
A ∩BhasnoelementswesaythesetsAandBaredisjointandwriteA ∩B = ◦/,where
◦/denotes theempty set.
Aset with no elements iscalled an empty set and isdenoted by ◦/.
IfA∩B=◦/, thenAandBare disjoint sets.
5.2.4 Union
Given two setsAandB, the set which contains all the elements ofAand those ofBis
called theunionofAandB, written as
A∪B={x:x∈Aorx∈Borboth}
InExample5.2,A∪B = {1,2,3,4,5,7,9}.Wenotethatalthoughtheelements3and5
are common toboth setsthey are listedonly once.
5.2.5 Subsets
If all the members of a setAare also members of a setBwe sayAis a subset ofBand
writeA ⊂B. We have already metanumber ofsubsets.Convince yourself that
N⊂ZandZ⊂R
Example5.3 IfM representsthesetofallcapacitorsmanufacturedbymachineM,andM f
represents
the faulty capacitors made by machine M, then clearlyM f
⊂M.
5.2.6 Complement
If we are given a well-defined universal set E and a setAwithA ⊂ E, then the set of
membersof EthatarenotinAiscalledthecomplementofAandiswrittenasA.Clearly
A ∪A = E. There areno members inthe setA∩A, thatisA ∩A = ◦/.
Example5.4 A company has a number of machines which manufacture thyristors. We consider only
two machines, M and N. A small proportion made by each is faulty. Denoting the sets
offaultythyristorsmadebyMandNbyM f
andN f
,respectively,depictthissituationon
a Venn diagram.Describethe setsM f
∪N f
andM∪N.
Solution Let E be the universal set of all thyristors manufactured by the company. The Venn
diagram is shown in Figure 5.2. Note in particular that M ∩N = ◦/. There can be no
thyristorsintheintersectionsinceifathyristorismadebymachineMitcannotbemade

5.2 Set theory 179
E
M
N
M f
Nf
Figure5.2
Venn diagram forExample 5.4.
by machine N and vice versa. Thus M and N are disjoint sets. Also note M f
⊂ M,
N f
⊂N.ThesetM f
∪N f
isthesetoffaultythyristorsmanufacturedbyeithermachineM
or N.The setM∪N isthe set of thyristors made by machines other than M or N.
Wehaveseenhowtheoperations ∩, ∪canbeusedtodefinenewsets.Itisnotdifficult
to show that a number of laws hold, most of which are obvious from the inspection of
anappropriateVenn diagram.
5.2.7 Lawsofsetalgebra
For any setsA,B,C and a universal set E, we have the laws in Table 5.1. From these it
ispossible toprove the laws given inTable 5.2.
Table5.1
Thelaws ofset algebra.
}
A∪B=B∪A
Commutative laws
A∩B=B∩A
}
A∪(B∪C)=(A∪B)∪C
Associativelaws
A∩(B∩C)=(A∩B)∩C
}
A∩(B∪C)=(A∩B)∪(A∩C)
Distributivelaws
A∪(B∩C)=(A∪B)∩(A∪C)
}
A ∪◦/ =A
Identity laws
A∩E=A
⎫
A ∪A = E ⎪⎬
A ∩A = ◦/ Complement laws
A =A
⎪⎭
Table5.2
Laws derivable from Table 5.1.
}
A∪(A∩B)=A
Absorption laws
A∩(A∪B)=A
}
(A∩B)∪(A∩B)=A
Minimization laws
(A∪B)∩(A∪B)=A
}
A∪B=A ∩B
De Morgan’slaws
A∩B=A ∪B
5.2.8 Setsandfunctions
Ifwearegiventwosets,AandB,ausefulexerciseistoexaminerelationships,givenby
rules,betweentheelementsofAandtheelementsofB.Forexample,ifA = {0,1,4,9}

180 Chapter 5 Discrete mathematics
A
0
r : A — B
‘take plus or minus the square root of’
1
9
4
Figure5.3
Arelation between setsAandB.
2
0
1
3 –3
B
–1
–2
s : D — E
s : m — 3m + 1
D 1
0
1
2
3
4
5
4
7
10
13
16
19
22
Figure5.4
The relationsmaps elements ofDtoE.
E
andB = {−3,−2,−1,0,1,2,3}theneachelementofBisplusorminusthesquareroot
of some element ofA. We can depict thisas inFigure 5.3.
The rule, which, when given an element ofA, produces an element ofB, is called a
relation. If the ruleof the relation isgiven the symbolr wewrite
r:A→B
and say ‘the relation r maps elements of the set A to elements of the set B’. For the
example above, we can writer : 1 → ±1,r : 4 → ±2, and generallyr : x → ± √ x.
Thesetfromwhichwechooseourinputiscalledthedomain;thesettowhichwemap
is called the co-domain; the subset of the co-domain actually used is called the range.
As weshall see thisneed notbe the whole of the co-domain.
A relationrmaps elements of a setD, called the domain, to one or more elements
ofasetC, called the co-domain. We write
r:D→C
Example5.5 IfD = {0,1,2,3,4,5} andE = {1,4,7,10,13,16,19,22} and the relation with symbolsis
defined bys :D→E,s :m→3m +1, identify the domain and co-domain of
s. Draw a mapping diagramtoillustrate the relation.What isthe range ofs?
Solution Thedomainofsisthesetofvaluesfromwhichwechooseourinput,thatisD = {0,1,2,
3,4,5}.Theco-domainofsisthesettowhichwemap,thatisE = {1,4,7,10,13,16,
19,22}.Therules :m→3m+1enablesustodrawthemappingdiagram.Forexample,
s : 3 → 10 and so on. The diagram is shown in Figure 5.4. The range ofsis the subset
ofE actuallyused,inthiscase {1,4,7,10,13,16}. Wenotethatnotalltheelementsof
the co-domainareactuallyused.
The notation introduced is very similar to that for functions described in Section 2.3.
This is no accident. In fact, a function is a very special form of a relation. Let us recall
the definition ofafunction:
‘Afunctionis arulewhich whengivenaninputproducesasingleoutput’.
If we study the two relations r and s, we note that when relation r received input, it
couldproducetwooutputs.Onthemappingdiagramthisshowsupastwoarrowsleaving

5.2 Set theory 181
someelementsinA.Whenrelationsreceivedaninput,itproducedasingleoutput.This
shows up as a single arrow leaving each element in D. Hence the relation r is not a
function, whereas the relationsis. This leads to the following more rigorous definition
of a function.
A function f is a relation which maps each element of a set D, called the
domain, toasingleelement of a setC, called the co-domain. We write
f:D→C
Example5.6 IfM = {off, on},N = {0,1} and wedefine a relationr by
r:M→N
r:off→0
r:on→1
then the relationris a function since each element inM is mapped to a single element
inN.
Example5.7 IfP = {0,1} andQ = {high} and wedefine a relationr by
r:P→Q
r:1→high
thenrisnotafunction since each element inPisnot mapped toan element inQ.
All of the functions described in Chapter 2 have domains which are subsets of the real
numbers R.Theinputtoeachfunctionistheparticularvalueoftheindependentvariable
chosen from the domain and the output is the value of the dependent variable. When
dealing with continuous domains the graphs we have already considered replace the
mapping diagrams.
Example5.8 Find the domain,D, of the rational function f :D→Rgiven by
f:x→
3x
x −2
Solution Since no domain is given, we choose it to be the largest set possible. This is the set
of all real numbers except the value x = 2 at which point f is not defined. We have
D={x:x∈R,x≠2}.
EXERCISES5.2
1 Useset notationto describethe intervals onthexaxis
given by
(a) (−3,2) (b) [0,2] (c) [−2, −1)
(d) (3,6] (e) |x| < 1
2 Sketch the followingsets. [Hint:seeSection2.2 on
open andclosedintervals.]
(a) {x:x∈Rand2<x4}
(b) {x:x∈Rand−1x0}

182 Chapter 5 Discrete mathematics
(c) {x:x∈Rand0x<2}
(d) {x:x∈Rand1<x<3}
3 Using the definitions given in Section 5.2, state
whether each ofthe followingis trueorfalse:
(a) 7∈Z
(b) R − ∩Q=◦/
(c) 0.7 /∈Q (d) N + ∪Q=R +
(e) R − ∩N=◦/
(g) 5∈Q
(f) R∩Z=Z
(h) N⊂Q
4 IfA = {1,3,5,7,9,11} andB = {3,4,5,6}find
(a)A∩B
(b)A∪B
5 GivenA = {1,2,3,4,5,6},B = {2,4,6,8,10} and
C = {3,6,9} state the elements ofeach ofthe
following:
(a) A∩B
(c) A∩C
(e) A∩(B∪C)
(b)B∩C
(d)A∩B∩C
(f) B∪(A∩C)
6 Write outallthe members ofthe followingsets:
(a) A={x:x∈Nandx<10}
(b) B={x:x∈Rand0x10andxisdivisible
by3}
7 ThesetsA,BandC are given byA = {1,3,5,7,9},
B = {0,2,4,6} andC = {1,5,9} andthe universal
set,E={0,1,2,...,9}.
(a) Representthe sets onaVenn diagram.
(b) StateA ∪B.
(c) StateB ∩C.
(d) State E ∩C.
(e) StateA.
(f) StateB ∩C.
(g) StateB∪C.
8 UseVenn diagrams to illustrate the followingfor
general setsC andD:
(a)C∩D (b)C∪D (c)C∩D
(d) C∪D
(e)C∩D.
9 BydrawingVenn diagrams verifyDeMorgan’s laws
A∩B=A ∪ B and A∪B=A ∩B
10 ForsetsA = {0,1,2} andB = {3,4},draw amapping
diagramto illustrate the following relations.
Determine whichrelationsare functions.Forthose
thatare notfunctions,give reasonsforyourdecision.
(a) r:A→B,r:0→3,r:1→4,r:2→4
(b) s:A→B,s:0→3,s:0→4,s:1→3,
s:2→3
(c) t:A→B,t:0→3,t:1→4
11 IfA = {1,3,5,7} andB = {1,2,3,4},draw a
mapping diagramto illustrate the relationr :A→B,
whereristherelation‘isbiggerthan’.Israfunction?
Solutions
1 (a) {x:x∈Rand−3<x<2}
(b) {x:x∈Rand0x2}
(c) {x:x∈Rand−2x<−1}
(d) {x:x∈Rand3<x6}
(e) {x:x∈Rand−1<x<1}
2 SeeFigureS.11.
(a)
0 1 2 3 4 5 x (b) –1 0 1 x
(c)
0 1 2 3 x (d) 0 1 2 3 4 x
FigureS.11
3 (a)T (b)F (c)F (d)F
(e) T (f) T (g) T (h) T
4 (a) {3,5} (b) {1,3,4,5,6,7,9,11}
5 (a) {2,4,6} (b) {6} (c) {3,6} (d) {6}
(e) {2,3,4,6} (f) {2,3,4,6,8,10}
6 (a) {0,1,2,3,4,5,6,7,8,9}
(b) {0,3,6,9}
7 (a) SeeFigure S.12.
A
3
7
8
FigureS.12
1
5
9
B
C
0
2 4
6
(b) {0,1,2,3,4,5,6,7,9} (c) ◦/
(d) {1,5,9} (e) {0,2,4,6,8}
(f) {3,7,8} (g) {3,7,8}

5.3 Logic 183
C
D
C
D
(a)
(b)
C
D
C
D
(c)
(d)
(e)
C
D
FigureS.13
8 SeeFigure S.13.
10 (a)Function
(b)Not afunction since 0 is mapped to two elements
(c)Not afunction since 2 is notmapped to anything
11 r isnot afunction because 1 is notmapped to
anything.
5.3 LOGIC
In Section 5.4 we will examine Boolean algebra. This concerns itself with the manipulation
of logic statements and so is suitable for analysing digital logic circuits. In this
section we introduce the basic concepts of logic by means of logic gates as these form
the usual startingpoint for engineers studying this topic.
5.3.1 TheORgate
The OR gate is an electronic device which receives two inputs each in the form of a
binary digit, that is 0 or 1, and produces a binary digit as output, depending upon the
values of the two inputs.Itisrepresented by the symbolshown inFigure 5.5.
AandBarethetwoinputs,andF isthesingleoutput.Ashigh(1)orlow(0)voltages
are applied to A and B various possible outputs are achieved, these being defined by
means of a truth table as shown in Table 5.3. So, for example, if a low (0) voltage is
applied to A and a high (1) voltage is applied to B, the output is a high (1) voltage at
F. We note that a ‘1’ appears in the right-hand column of the truth table whenever A
orBtakes the value 1,hence the name OR gate. We use the symbol + torepresent OR.
BecauseitconnectsthevariablesAandB,ORisknownasalogicalconnective.Weshall
meet other logical connectives shortly. This connective is also known as adisjunction,
so thatA +Bissaidtobethe disjunction ofAandB.
Table5.3
The truth table for an OR gate
with inputsAandB.
A
B
Figure5.5
Symbol foran ORgate.
F = A + B
A B F=A+B
1 1 1
1 0 1
0 1 1
0 0 0

184 Chapter 5 Discrete mathematics
Table5.4
ThetruthtableforanANDgate
with inputsAandB.
A
B
Figure5.6
Symbol foran AND gate.
F = A . B
A B F=A·B
1 1 1
1 0 0
0 1 0
0 0 0
Table5.5
The truth table for
a NOTgate.
A
F = A
A
F=A
Figure5.7
Symbol foran inverter.
1 0
0 1
5.3.2 TheANDgate
It is possible to construct another electronic device called an AND gate which works
in a similar way except that the output only takes the value 1 when both inputs are 1.
ThesymbolforthisgateisshowninFigure5.6andthecompletetruthtableisshownin
Table 5.4. The logical connective AND is given the symbol · and is known as a conjunctionso
thatA·Bissaidtobethe conjunctionofAandB.
5.3.3 TheinverterorNOTgate
The inverter is a device with one input and one output and has the symbol shown in
Figure 5.7. It has a truth table defined by Table 5.5. If the input isA, then the output is
representedby the symbolA, known as the complement ofA.
5.3.4 TheNORgate
This gate is logically equivalent to a NOT gate in series with an OR gate as shown in
Figure 5.8. It is represented by the symbol shown in Figure 5.9 and has its truth table
defined inTable 5.6.
A
B
A + B
Figure5.8
A NOTgatein series with an OR gate.
A
B
Figure5.9
Symbol foraNOR gate.
F = A + B
F = A + B
Table5.6
ThetruthtableforaNORgate.
A B F =A +B
1 1 0
1 0 0
0 1 0
0 0 1

5.4 Boolean algebra 185
A
B
A . B
Figure5.10
ANOT gate in serieswith an AND gate.
A
B
Figure5.11
Symbol foraNAND gate.
F = A . B
F = A . B
Table5.7
Thetruthtable foraNAND gate.
A B F =A ·B
1 1 0
1 0 1
0 1 1
0 0 1
5.3.5 TheNANDgate
This gate is logically equivalent to a NOT gate in series with an AND gate as shown in
Figure5.10.ItisrepresentedbythesymbolshowninFigure5.11andhasthetruthtable
definedby Table 5.7.
Althoughwehaveonlyexaminedgateswithtwoinputsitispossibleforagatetohave
morethantwo.Forexample,theBooleanexpressionforafour-inputNANDgatewould
beF=A·B·C ·D while that of a four-input OR gate would beF = A +B+C +D,
whereA,B,C andDare the inputs, andF is the output. Logic gates form the building
blocks formore complicated digital electronic circuits.
5.4 BOOLEANALGEBRA
SupposeAandBarebinarydigits,thatis1or0.These,togetherwiththelogicalconnectives
+and · andalsothecomplementNOT,formwhatisknownasaBooleanalgebra.
The quantities A and B are known as Boolean variables. Expressions such as A +B,
A·BandAareknownasBooleanexpressions.MorecomplexBooleanexpressionscan
bebuiltupusingmoreBooleanvariablestogetherwithcombinationsof +, · andNOT;
forexample, wecan draw up a truth table forexpressions such as (A·B) + (C ·D).
Weshallrestrictourattentiontothelogicgatesdescribedinthelastsectionalthough
the techniques of Boolean algebra are more widely applicable. A Boolean variable can
only take the values 0 or 1. For our purposes a Boolean algebra is a set of Boolean
variables with the two operations · and +, together with the operation of taking the
complement, for which certainlaws hold.
5.4.1 LawsofBooleanalgebra
ForanyBooleanvariablesA,B,C,wehavethelawsinTable5.8.Fromtheseitispossible
toprovethelawsgiveninTable5.9.Youwillnoticethattheselawsareanalogoustothose
ofsetalgebraifweinterpret +as ∪, · as ∩,1astheuniversalset E,and0astheempty
set ◦/. In ordinary algebra, multiplication takes precedence over addition. In Boolean
algebra · takes precedence over +. So, for example, we can write the first absorption
lawwithout brackets, thatis
A+A·B=A
Similarly, the first minimization law becomes
A·B+A·B =A
We shall follow thisruleof precedence fromnow on.

186 Chapter 5 Discrete mathematics
Table5.8
Laws ofBoolean algebra.
}
A+B=B+A
Commutative laws
A·B=B·A
}
A+(B+C)=(A+B)+C
Associativelaws
A·(B·C)=(A·B)·C
}
A·(B+C)=(A·B)+(A·C)
Distributivelaws
A+(B·C)=(A+B)·(A+C)
}
A+0=A
Identity laws
A·1=A
⎫
A +A = 1⎪⎬
A ·A = 0 Complement laws
A =A
⎪⎭
Table5.9
Laws derived from the laws ofTable 5.8.
}
A+(A·B)=A
Absorption laws
A·(A+B)=A
}
(A·B)+(A·B)=A
Minimization laws
(A+B)·(A+B)=A
}
A+B=A ·B
DeMorgan’s laws
A·B=A +B
A+1=1
A·0=0
Example5.9 Findthe truth tableforthe BooleanexpressionA +B·C.
Solution WeconstructthetablebynotingthatA,BandC areBooleanvariables;thatis,theycan
takethevalues0or1.Thefirststageintheprocessistoformallpossiblecombinationsof
A,BandC,asshowninTable5.10.ThenwecompletethetablebyformingC,thenB·C
and finallyA +B·C, using the truth tables defined earlier. So, for example, whenever
C=1,C = 0. The complete process is shown in Table 5.11. Work through the table to
ensure you understand how itwas constructed.
Table5.10
The possible
combinations forthree
variables,A,BandC.
A B C
Table5.11
Thetruthtable forA +B·C.
A B C C B·C A+B·C
1 1 1
1 1 0
1 0 1
1 0 0
0 1 1
0 1 0
0 0 1
0 0 0
1 1 1 0 0 1
1 1 0 1 1 1
1 0 1 0 0 1
1 0 0 1 0 1
0 1 1 0 0 0
0 1 0 1 1 1
0 0 1 0 0 0
0 0 0 1 0 0

5.4 Boolean algebra 187
Table5.12
Truthtablefor (A +B)· (A +C).
A B C C A+B A+C (A+B)·(A+C)
1 1 1 0 1 1 1
1 1 0 1 1 1 1
1 0 1 0 1 1 1
1 0 0 1 1 1 1
0 1 1 0 1 0 0
0 1 0 1 1 1 1
0 0 1 0 0 0 0
0 0 0 1 0 1 0
5.4.2 Logicalequivalence
We know from the distributive laws ofBooleanalgebrathat
A+(B·C)=(A+B)·(A+C)
Let us construct the truth table for the r.h.s. of this expression (Table 5.12). If we now
observethefinalcolumnofTable5.12weseeitisthesameasthatofTable5.11.Wesay
thatA+(B·C)islogicallyequivalentto (A+B)· (A+C).Figures5.12and5.13show
thetwowaysinwhichtheselogicallyequivalentcircuitscouldbeconstructedusingOR
gates, AND gates and inverters. Clearly different electronic circuits can be constructed
toperform the same logical task. We shall shortlyexplore a wayof simplifyingcircuits
toreduce the number ofcomponents required.
A
B
C
C
B . C
A + (B . C)
Figure5.12
Circuit to implementA + (B ·C).
A
A + B
B
A
C
C
A + C
(A + B) . (A + C)
Figure5.13
Circuit to implement (A +B)· (A +C).
Engineeringapplication5.1
Booleanexpressionandtruthtableforanelectroniccircuit
Find the Boolean expression and truth table for the electronic circuit shown in Figure
5.14.
➔

188 Chapter 5 Discrete mathematics
Solution
BylabellingintermediatepointsinthecircuitweseethatX =A·B +C.Inorderto
obtain the truth table we form all possible combinations ofA,BandC, followed by
A·B,CandfinallyX =A·B+C.ThecompletecalculationisshowninTable5.13.
Table5.13
The truthtableforFigure5.14.
A B C A·B C X=A·B+C
A
B
C
Figure5.14
Circuit forEngineeringapplication 5.1.
X
1 1 1 1 0 1
1 1 0 1 1 1
1 0 1 0 0 0
1 0 0 0 1 1
0 1 1 0 0 0
0 1 0 0 1 1
0 0 1 0 0 0
0 0 0 0 1 1
Whatwewouldnowliketobeabletodoiscarryoutthereverseprocess:thatis,startwith
a truth table and find an appropriate Boolean expression so that the required electronic
device can beconstructed.
Example5.10 Given inputsA,BandC, find a Boolean expression forX as given by the truth table in
Table 5.14.
Solution To find an equivalent Boolean expression the procedure is as follows. Look down the
rows of the truth table and select those with an r.h.s. equal to 1. In this example, there
arefivesuchrows:1,2,4,6and8.Eachoftheserowsgivesrisetoatermintherequired
Booleanexpression.Eachtermisconstructedsothatithasavalue1fortheinputvalues
of that row. For example, for the input values of row 1, that is 1, 1, 1, we findA·B·C
has the value 1, whereas for the input values of row 2, that is 1, 1, 0, we findA·B·C
has the value 1. Carrying out this process for the other rows we find that the required
expression is
Table5.14
Thetruthtable fora
systemwithinputsA,B
andC andan outputX.
A B C X
1 1 1 1
1 1 0 1
1 0 1 0
1 0 0 1
0 1 1 0
0 1 0 1
0 0 1 0
0 0 0 1
X =A·B·C+A·B·C+A·B ·C +A·B·C +A ·B·C (5.1)
thatis,adisjunctionofterms,eachtermcorrespondingtooneoftheselectedrows.This
important expression is known as a disjunctive normal form (d.n.f.). We note that the
truthtableisthesameasthatofEngineeringapplication5.1whichhadBooleanexpression
(A·B)+C. The d.n.f. wehave justcalculated, whilecorrect,isnot the simplest.
Moregenerally,tofindtherequiredd.n.f.fromatruthtablewewritedownanexpression
ofthe form
( )+( )+···+( )
whereeachtermhasthevalue1fortheinputvaluesofthatrow.Wecouldnowconstruct
an electronic circuit corresponding to Equation (5.1) using a number of AND and OR
gatestogetherwithinverters,anditwoulddotherequiredjobinthesensethatthedesired
truth table would beachieved.

5.4 Boolean algebra 189
However, we know from Engineering application 5.1 that the much simpler expression
(A·B) +C has the same truth table and if a circuit were to be built corresponding
tothisitwouldrequirefewercomponents.Clearlywhatweneedisatechniqueforfinding
the simplest expression which does the desired job since the d.n.f. is not in general
the simplest. It is not obvious what is meant by ‘simplest expression’. In what follows
we shall be concerned with finding the simplest d.n.f. It is nevertheless possible that a
logically equivalent statement exists which would give a simpler circuit. Simplification
canbeachievedusingthelawsofBooleanalgebraasweshallseeinExample5.11and
Engineering application 5.3.
Engineeringapplication5.2
TheexclusiveORgate
WehavealreadylookedattheORgateinSection5.3.Thefullnameforthistypeof
ORgateistheinclusiveORgate.Itissocalledbecauseitgivesanoutputof1when
either or both inputs are 1. The exclusive OR gate only gives an output of 1 when
eitherbutnotbothinputsare1.ThetruthtableforthisgateisgiveninTable5.15and
its symbolisshown inFigure 5.15. Usingthe truthtable, the d.n.f. forthe gateis
F=A·B +A ·B
The exclusive OR often arises in the design of digital logic circuits. In fact, it is
possible tobuy integrated circuits thatcontain exclusive ORgates as basic units.
Table5.15
The truthtable foran
exclusive OR gate.
A B F
1 1 0
1 0 1
0 1 1
0 0 0
A
B
F = A . B + A . B
Figure5.15
Symbol foran exclusive OR gate.
Example5.11 UsethelawsofBooleanalgebragiveninTable5.8tosimplifythefollowingexpressions.
(a) A·B+A·B
(b) A+A ·B
(c) A+A ·B·C
(d) A·B·C+A·B·C+A·B·C+A·B ·C +A ·B·C
Solution (a) Using the distributive lawwe can write
A·B+A·B=A·(B+B)
Using the complement law,B +B = 1 and hence
A·B+A·B=A·1
=A using the identity law

190 Chapter 5 Discrete mathematics
HenceA·B+A·BsimplifiestoA.Notethatthisisthefirstminimizationlawgiven
inTable 5.9.
(b) A+A ·B=(A+A)·(A+B) by the distributive law
=1·(A+B)
by the complement law
=A +B using the identity law
(c) Note thatA +A·B·C can be written asA + (A·B)·C using the associative laws.
Then
A + (A·B)·C=(A+A·B)· (A +C) by the distributive law
=(A+B)· (A +C) usingpart(b)
=A +B ·C by the distributive law
(d) A·B·C+A·B·C+A·B·C+A·B ·C +A ·B·C
can be rearranged usingthe commutative lawtogive
A·B·C+A·B·C+A·B·C+A·B ·C +A ·B·C
This equals
A·B·(C+C)+A·B·(C+C) +A ·B·C by the distributive law
=A·B·1+A·B·1+A·B·C usingthe complement law
=A·B+A·B +A·B·C usingthe identitylaw
=A·(B+B) +A ·B·C usingthe distributive law
=A +A ·B·C usingthe complement and
identitylaws
Usingthe resultof part(c)this can be furthersimplified toA+B·C.
Engineeringapplication5.3
Designofabinaryfull-addercircuit
The binary adder circuit is a common type of digital logic circuit. For example, the
accumulator of a microprocessor is essentially a binary adder. The term full-adder
is used to describe a circuit which can add together two binary digits and also add
the carry-out digit from a previous stage. The outputs from the full-adder consist of
thesumvalueandthecarry-outvalue.Byconnectingtogetheraseriesoffull-adders
itispossibletoaddtogethertwobinarywords.(Abinarywordisagroupofbinary
digits, such as 0111 1010.) For example, adding together two 4-bit binary words
would requirefourfull-addercircuits.This isshown inFigure 5.16.
The inputs A0--A3 and B0--B3 hold the two binary words that are to be added.
TheoutputsS0--S3holdtheresultofcarryingouttheaddition.ThelinesC0--C3hold
the carry-out values from each of the stages. Sometimes there will be a carry-in to
stage 0 as a result of a previous calculation. Let us consider the design of stage 2 in
more detail. The design of the other stages will be identical. First of all a truth table
forthe circuitisderived. This isshown inTable 5.16.

5.4 Boolean algebra 191
S0
Stage
0
A0 B0
C0
S1
Stage
1
A1 B1
C1
S2
Stage
2
A2 B2
C2
S3
Stage
3
A3 B3
Figure5.16
Four full-addersconnected to allow two 4-bitbinarywordsto
be added.
C3
Table5.16
Truthtable forafull-adder.
C1 A2 B2 S2 C2
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1
1 1 0 0 1
1 1 1 1 1
Notice that there are three inputs to the circuit,C1, A2 and B2. There are also
twooutputsfromthecircuit,S2andC2.Writingexpressionsfortheoutputsind.n.f.
yields
S2 =C1 ·A2·B2+C1·A2·B2+C1·A2 ·B2+C1·A2·B2
C2 =C1·A2·B2+C1·A2·B2+C1·A2·B2+C1·A2·B2
Itisimportanttoreducetheseexpressionstoassimpleaformaspossibleinorderto
minimize the number of electronic gates needed to implement the expressions. So,
startingwithS2,
Let
S2 =C1 ·A2·B2+C1·A2·B2+C1·A2 ·B2+C1·A2·B2
=C1·(A2·B2+A2 ·B2) +C1·(A2·B2 +A2·B2)
by the distributive law (5.2)
X=A2·B2 +A2·B2 (5.3)
Notice this isanequation for an exclusive OR gate with inputsA2 andB2.
UsingEquation (5.3) wehave
X =A2·B2 +A2·B2
= (A2·B2)·(A2·B2) by DeMorgan’s law
= (A2 +B2)·(A2 +B2) by DeMorgan’s law
= (A2+B2)·(A2+B2) by the complement law
=A2·A2+A2 ·B2+B2·A2+B2·B2
by the distributive law
= 0 +A2 ·B2 +B2·A2 +0 by the complement law
=A2 ·B2 +B2·A2 by the identity law
=A2·B2+A2 ·B2
Itisnow possible towrite Equation (5.2) as
by the commutative law
S2=C1·X +C1·X
➔

192 Chapter 5 Discrete mathematics
This is the expression for an exclusive OR gate with inputsC1 andX. It is therefore
possible to obtainS2 with two exclusive OR gates which are usually available
asabasicbuildingblockonanintegratedcircuit.ThecircuitforS2isshowninFigure
5.17. Turning toC2, wehave
C2 =C1·A2·B2+C1·A2·B2+C1·A2·B2+C1·A2·B2
=A2·B2·(C1+C1)+C1·(A2·B2+A2·B2)
=A2·B2 +C1·X
by the complement law
by the distributive law
sinceC1 +C1 = 1, whereX is given by Equation (5.3). The output,X, has already
been generated to produceS2 but can be used again provided the exclusive OR gate
can stand feeding two inputs. Assuming this is so then the final circuit for the fulladder
isshown inFigure 5.18.
A2
B2
X
C1
S2
Figure5.17
Circuit to implementS2 =C1·X +C1·X,
whereX =A2 ·B2 +A2 ·B2.
A2
B2
C1
X
S2
A2 . B2
C1 . X
C2
Figure5.18
Circuit to implement the stage 2 full-adder.
Engineeringapplication5.4
Realizationoflogicgates
Logic gates are usually constructed from one or more transistors. The most commontechnologyusedisCMOS(complementarymetal-oxidesemiconductor)logic,
which was invented by Frank Wanlass and was perfected whilst he was working
at Fairchild Semiconductor in the 1960s. CMOS logic makes use of two different
typesoftransistorknownasPMOS(p-typemetal-oxidesemiconductor)andNMOS
(n-type metal-oxide semiconductor) transistors. Both types can be readily manufactured
in vast numbers on a single silicon wafer along with their associated wiring.
Thisprincipleistheconstructionalbasisofmodernmicroprocessorswhichmaycontainbillionsofindividualtransistors.Thetermvery-large-scaleintegration(VLSI)
was coined to refer to the practice of assembling such a large number of transistor
devices onasinglesilicon wafer.
Figure5.19showstheconstructionofasingleNANDgatemadefromitsindividual
transistors.

5.4 Boolean algebra 193
V DD
PMOS
Q1
D
D
PMOS
Q2
A
G
S
D
S
G
Y
G
S
D
NMOS
Q3
B
G
S
NMOS
Q4
0 V
Figure5.19
Internal construction ofasingle
CMOS NAND gate.
ThediagramshowstwoPMOStransistors,labelledQ1andQ2,connectedinparallel.
These are connected in series with two NMOS transistors, Q3 and Q4. There
arefoursourceterminalsalllabelledS,togetherwithfourdrainterminalslabelledD.
ThevoltageV DD
isthepositivevoltagesupplywhichisconnectedtothedrainonthe
fieldeffecttransistors.ThelabellingV DD
isaconventionoftenadoptedinthistypeof
circuit.Logiclevelsinacircuitlikethisarerepresentedbytakingalowvoltage,close
to 0V, to be a logic 0 and a high voltage, close toV DD
, to be a logic 1. The PMOS
transistors Q1 and Q2 each carry current between their source and drain terminals
only when a low voltage (logic 0) is connected at their gate terminal (labelled G).
The NMOS transistors Q3 and Q4 are a complementary type where current flows,
whichonlyoccurswhenahighvoltage,correspondingtologic1,ispresentedattheir
gates.ThuswhenbothAandBareatlogic0,Q1andQ2areswitchedonandQ3and
Q4 are switched off, hence the output isV DD
, which represents logic 1. This output
is still the same if either A or B, but not both, are at logic 1 because although Q3 or
Q4 will be turned on they are connected in series and individually have no effect. If
both A and B are at logic 1, then Q1 and Q2 are switched off, and Q3 and Q4 are
switched on, hence the output will be approximately 0V, which represents logic 0.
This behaviour isconsistentwith the truthtable given inTable 5.7.
All modern VLSI chips are designed using high-level design languages such as
VHDL(aspecializedcomputerlanguageforhardware)andthetransistordesignand
layoutisfullyautomated.Itisnowrarelynecessaryforthemicroprocessordesigner
toconsider individual transistors or even individual gates.
EXERCISES5.4
1 WriteBoolean expressionsforthe output,F,from the
electronic devices shown in Figure 5.20.
2 WriteBoolean expressionsforthe output from the
devices shown in Figure5.21.
3 Design electronic devices which produce the
following outputs:
(a) A +B (b) A·(B·C)
(d) A +B
(e)A·B+A ·B+B·C
(c) (C+D)·(A+B)

194 Chapter 5 Discrete mathematics
(a)
A
B
C
F
(b)
A
B
F
(c)
A
B
F
C
(d)
A
F
B
Figure5.20
(a)
A
F
B
(b)
A
B
F
C
Figure5.21
4 Draw upthe truthtablesforthe expressions given in
Question 3.
5 Usetruthtables to verifythat the following pairsof
expressions are logically equivalent:
(a) p+p·q·r+p·qandp+q
(b) (A +B)·(A+C)andA+B ·C
(c) p·q·r+p·q·r+p·q·randp·(r+q)
6 Simplify the followingBoolean expressions usingthe
laws ofBoolean algebra:
(a) A·A·A
(b) A·A·A
(c) A·A ·A
(d) (A+A)·(A+A)
(e) A +0
(f) (A+1)·(A+1)
(g) A +1
7 Simplify the following Boolean expressions usingthe
lawsofBoolean algebra:
(a) (A+A)·(B+B)
(b) A·(A+B+A·B)
(c) (A+A)·(A+C)

5.4 Boolean algebra 195
(d) A ·B·(A+B)
(e) A·(A +B)·B
(f) A·B·C+A·B·C
(g) A·B·C+A +B+C
8 Constructatruthtable showingA·BandA +B in
order to verifythe logicalequivalence expressed in
DeMorgan’s lawA·B=A +B.Carryout asimilar
exercise to verifyA+B=A·B.
9 LetB = 1 andthenB = 0 in the absorption laws,and
use the identitylaws to obtain (a)A +A =Aand
(b)A ·A =A.Verify your resultsusingtruthtables.
10 Derive Booleanexpressions andtruthtablesforthe
circuits shown in Figure5.22.
11 Simplify the following Booleanexpressions using
Booleanalgebra:
(a) A·B+A·B+B·C+A·B·C
(b) A·(C+A)+C·B+D+C+B·C+C·A
(c) A·B·C·D+A·B·C+A·B·C·D +
A ·B·C·D+A·B·C ·D
12 Thetruthvalues ofthe Booleanexpression,X,are
given in the followingtables. WriteX in disjunctive
A
B
C
D
E
A
B
(a)
X
normal form.Usethe laws ofBooleanalgebra to
simplifyyourexpressions.
(a)
(b)
A B X A B X
(c)
(d)
0 0 1
0 1 0
1 0 1
1 1 1
A B C X
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
A B C X
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 0
0 0 1
0 1 1
1 0 0
1 1 0
13 ExpressA·B +A·B usingonly conjunction (AND
gate) and negation (NOTgate).
C
D
X
E
F
Figure5.22
(b)
Solutions
1 (a) (A+B)·C (b) (A·B)·B
(c) (A·B)+B+C (d) A +B
2 (a) A ·B +A ·B which isthe same asA·B+A·B
(b) (A·B+A·B)·(B+C)+(A·B +A·B)·(B+C)
3 SeeFigureS.14.

196 Chapter 5 Discrete mathematics
A
B
A
B
C
(a)
C
D
B
A
A
B
(c)
A
B
C
(e)
(b)
FigureS.14
(d)
4 (a)
A B A +B
1 1 0
1 0 0
0 1 1
0 0 0
(d)
A B A +B
1 1 1
1 0 0
0 1 0
0 0 0
(b)
A B C A·(B·C)
1 1 1 0
1 1 0 1
1 0 1 1
1 0 0 1
0 1 1 1
0 1 0 1
0 0 1 1
0 0 0 1
(e)
A B C A·B+A ·B+B·C
1 1 1 1
1 1 0 1
1 0 1 0
1 0 0 0
0 1 1 0
0 1 0 1
0 0 1 1
0 0 0 1
(c)
A B C D (C+D)·(A+B)
1 1 1 1 1
1 1 1 0 1
1 1 0 1 1
1 1 0 0 0
1 0 1 1 1
1 0 1 0 1
1 0 0 1 1
1 0 0 0 0
0 1 1 1 0
0 1 1 0 0
0 1 0 1 0
0 1 0 0 0
0 0 1 1 1
0 0 1 0 1
0 0 0 1 1
0 0 0 0 0
6 (a)A (b)0 (c)0 (d)A
(e) A (f)1 (g)0
7 (a) A·B (b) A (c) A
(d) 0 (e) 0 (f) A·B
(g) A·B·C
10 (a) X =(A·B)·(C+D+E)
(b) X=A·B +A·B+C·D+E+F
11 (a) A+B·C
(b) A+B+C+D
(c) (A·B)+(A·B ·D) which can be further
simplifiedtoA · (B +D)

Review exercises 5 197
12 (a) A·B+A·B+A·B
(b) A·B +A·B
(c) A ·B·C +A ·B·C+A·B·C+A·B·C
(d) A·B·C+A·B·C+A·B·C
13 (A ·B)·(A·B)
REVIEWEXERCISES5
1 Classifythe followingastrue orfalse:
(a) R + ⊂ R
(c) 0.667∈Q
(e) −6∈Q
(g) N∩Q=N
(b) 0.667 ∈ R −
(d) N∪Z=R
(f)9/∈ R
(h) R − ∩Q=◦/
2 Useset notationto describethe followingintervals on
thexaxis.
(a) [−6,9] (b) (−1,1) (c) |x| < 1.7
(d) (0,2] (e) |x| > 1 (f) |x| 2
3 ThesetsA,BandC are given by
A = {1,2,6,7,10,11,12,13},B = {3,4,7,8,11},
C = {4,5,6,7,9,13} andthe universal set
E = {x :x∈N + ,1 x13}.Listthe elements of
the following sets:
(a) A∪B (b)B∪C (c)A∩B
(d) A∩(B∪C) (e)A∩B∩C
(f) A∪(B∩C) (g)C∪(B∩A)
4 Representthe setsA,BandC describedin Question3
onaVenn diagram.
5 List allthe elementsofthe followingsets:
(a) S={n:n∈Z,5n 2 50}
(b) S={m:m∈N,5m 2 50}
(c) S={m:m∈N,m 2 +2m−15=0}
6 IfA={n:n∈Z,−10n20},
B={m:m∈N,m>15}listthemembersofA∩B
andwrite down an expression forA ∪B.
7 WriteBooleanexpressions forthe output,F,from the
electronic devices shown in Figure5.23.
8 Simplify the followingBoolean expressions usingthe
laws ofBooleanalgebra:
(a) A·B·1
(c) 1+0
(e) (A+C)·(C+A)
(f) A·B+A +B (g) A·B
(h) A +B
(b) A+A·B+B
(d) D·(C+B)+C·D
(i) A·(B+C)·C
(j) (C+A+D)·(C·D+A)
9 Draw uptruthtables to verifythatA·B +B and
A ·B +A are logically equivalent.
A
B
F
(a)
C
A
B
F
(b)
(c)
A
B
C
D
Figure5.23
F

198 Chapter 5 Discrete mathematics
10 ExpressA +B+C +Dusingthe Boolean
connectives AND ( · )and negation.
11 Simplify the followingexpressions usingthe laws of
Boolean algebra:
(a) (A +A)·(B+B)
(b) A·B·A
(c) (A+B)·(A+B) (d) A ·A
(e) A·B·C·D·1·0
(g) B·C·1
(f) A·B·C·B
12 Reducethe following expressions usingBoolean
algebra:
(a) A+C+A·B·(B+C)
(b) (A+B)·B·C+(A+C)·(B+A·C)
(c) (A +B)·(A +C)
13 Thetruthtables ofthe Boolean expression,X,are
given in the followingtables. Writethe disjunctive
normal formofX in eachcase.
(a)
A B X
0 0 0
0 1 1
1 0 0
1 1 0
(b)
(c)
(d)
A B X
0 0 0
0 1 0
1 0 1
1 1 0
A B C X
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 1
A B C X
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 0
Solutions
1 (a)T (b)F (c)T (d)F
(e) T (f) F (g) T (h) F
2 (a) {x:x∈R,−6x9}
(b) {x:x∈R,−1<x<1}
(c) {x:x∈R,−1.7<x<1.7}
(d) {x:x∈R,0<x2}
(e) {x:x∈R,x>1orx<−1}
(f) {x:x∈R,x2orx−2}
3 (a) {1,2,3,4,6,7,8,10,11,12,13}
(b) {3,4,5,6,7,8,9,11,13}
(c) {7,11}
(d) {6,7,11,13}
(e) {7}
(f) {1,2,4,6,7,10,11,12,13}
(g) {4,5,6,7,9,11,13}
4 SeeFigureS.15.
A
1
2
11 3
10
12 6 7 13 4
FigureS.15
9
5
5 (a) S = {−7, −6, −5, −4, −3,3,4,5,6,7}
(b) S = {3,4,5,6,7}
(c) S = {3}
6 A ∩B = {16,17,18,19,20}
A∪B={m:m∈Z,m−10}
C
8
B

Review exercises 5 199
7 (a) (A·B)·C + (A·B)·C
(b) (A+B)·B
(c) (A·B)+(C+D)
8 (a) A·B (b) A+B (c) 1
(d) D·(C+B) (e) C (f) A ·B
(g) A+B (h) A·B (i) A ·C
(j) A
10 A·B·C ·D
11 (a) 1 (b) 0 (c) A+B (d) A
(e) 0 (f) 0 (g) B·C
12 (a) A +B ·C
(b) A·B+A·C+B·C
(c) A·(B+C)
13 (a) A ·B
(b) A·B
(c) A ·B·C +A·B·C+A·B·C+A·B·C
(d) A ·B·C+A·B·C+A·B ·C+A·B·C

6 Sequencesandseries
Contents 6.1 Introduction 200
6.2 Sequences 201
6.3 Series 209
6.4 Thebinomialtheorem 214
6.5 Powerseries 218
6.6 Sequencesarisingfromtheiterativesolutionofnon-linearequations 219
Reviewexercises6 222
6.1 INTRODUCTION
Muchofthematerialinthischapterisofafundamentalnatureandisapplicabletomany
differentareasofengineering.Forexample,ifcontinuoussignalsorwaveforms,suchas
those described in Chapter 2, are sampled at periodic intervals we obtain a sequence of
measuredvalues.Sequencesalsoarisewhenweattempttoobtainapproximatesolutions
of equations which model physical phenomena. Such approximations are necessary if
a solution is to be obtained using a digital computer. For many problems of practical
interest to engineers a computer solution is the only possibility. The z transform is an
exampleofaninfiniteserieswhichisparticularlyimportantinthefieldofdigitalsignal
processing. Signal processing is concerned with modifying signals in order to improve
them in some way. For example, the signals received from space satellites have to undergo
extensive processing in order to counteract the effects of noise, and to filter out
unwanted frequencies, before they can provide, say, acceptable visual images. Digital
signal processing is signal processing carried out using a computer. So, skill in manipulating
sequences and series is crucial. Later chapters will develop these concepts and
show examples of their use insolving real engineering problems.

6.2 Sequences 201
6.2 SEQUENCES
Asequenceisasetofnumbersorterms,notnecessarilydistinct,writtendownina
definite order.
For example,
1,3,5,7,9 and 1, 1 2 , 1 4 , 1 8 , 1
16
are both sequences. Sometimes we use the notation ‘...’ to indicate that the sequence
continues. For example, the sequence 1,2,3,...,20 is the sequence of integers from 1
to 20 inclusive. These sequences have a finite number of terms but we shall frequently
dealwithonesinvolvinganinfinitenumber ofterms.Toindicatethatasequencemight
go on for ever we can use the ... notation. Thus
and
2,4,6,8,...
1,−1,1,−1,...
can be assumed tocontinue indefinitely.
Ingeneral situations we shall writeasequence as
x[1],x[2],x[3],...
or more compactly,
x[k]
k=1,2,3,...
An alternative notation is
x 1
,x 2
,x 3
,...
Theformernotationisusuallyusedinsignalprocessingwherethetermsinthesequence
representthevalues ofthesignal.Thelatternotationarisesinthenumerical solutionof
equations. Hence both forms will be required. Often x[1] will be the first term of the
sequence although thisisnotalways the case.The sequence
...,x[−3],x[−2],x[−1],x[0],x[1],x[2],x[3],...
isusually writtenas
x[k]
k=...,−3,−2,−1,0,1,2,3,...
Acompletesequence,asopposedtoaspecifictermofasequence,isoftenwrittenusing
braces,for example
{x[k]} =x[1],x[2],...
although it is common to writex[k] for both the complete sequence and a general term
inthesequencewhenthereisnoconfusion,andthisistheconventionweshalladoptin
thisbook.
Asequencecanalsoberegardedasafunctionwhosedomainisasubsetofthesetof
integers.For example, the function defined by
x:N→R
x:k→ 3k
2

202 Chapter 6 Sequences and series
isthe sequence
x[0] = 0 x[1] = 3 2
x[2] = 3 x[3] = 9 2 ...
The values in the range of the function are the terms of the sequence. The independent
variableisk.FunctionsofthissortdifferfromthosedescribedinChapter2becausethe
independent variable is not selected from a continuous interval but rather isdiscrete. It
is,nevertheless,possibletorepresentx[k]graphicallyasillustratedinExamples6.1--6.3,
butinsteadofapiecewisecontinuouscurve,wenowhaveacollectionofisolatedpoints.
Example6.1 Graphthe sequences given by
{ 0 k <0
(a) x[k] = k=...,−3,−2,−1,0,1,2,...,thatisk∈Z
1 k0
{ 1 keven
(b) x[k] = k ∈ Z
−1 kodd
Solution (a) Fromthedefinitionofthissequence,thetermx[k]iszeroifk<0and1ifk 0.The
graph is obtained by plotting the terms of the sequence againstk(see Figure 6.1).
This sequenceisknown asthe unit stepsequence. We shall denotethisbyu[k].
(b) Thesequencex[k] isshown inFigure 6.2.
Example6.2 Graphthe sequencedefinedby
x[k] =
{ 1 k = 0
0 k≠0
k=...,−3,−2,−1,0,1,2,3,...
Solution From the definition, ifk = 0 thenx[k] = 1. Ifkis not equal to zero the corresponding
term in the sequence equals zero. Figure 6.3 shows the graph of this sequence which is
commonly called theKronecker delta sequence.
Example6.3 Thesequencex[k]isobtainedbymeasuringorsamplingthecontinuousfunction f (t) =
sint,t ∈ R, att = −2π, −3π/2, −π, −π/2, 0, π/2, π, 3π/2 and 2π. Write down the
termsofthissequenceand show themon a graph.
x[k]
1
x[k]
–4 –3 –2 –1 1
1
2 3 4 5
k
–4 –3 –2 –1 1
2 3 4 5
k
–1
Figure6.1
Theunitstepsequence.
Figure6.2 { 1 k even
Thesequencex[k] =
−1 k odd.

6.2 Sequences 203
sin t
1
x[k]
1
– 3p
–2p –2 –p – p –2 p
0 –2 p
3p –2
2p
t
–4 –3 –2 –1 1
Figure6.3
TheKronecker delta sequence.
2 3 4 5
k
–1
Figure6.4
Thefunction f (t) = sint with sampled points
shown.
x[k]
1
–4 –3 –2 –1 1 2 3 4 k Figure6.5
Sequence formed from sampling
–1
f(t) =sint.
Solution The function f (t) = sint, for −2π t 2π, is shown in Figure 6.4. We sample the
continuousfunctionattherequiredpoints.Thesamplevaluesareshownas •.Fromthe
graph we see that
x[k]=0,1,0,−1,0,1,0,−1,0
The graph ofx[k] isshown inFigure 6.5.
k=−4,−3,...,3,4
Sometimes it is possible to describe a sequence by a rule giving the kth term. For example,
the sequence for whichx[k] = 2 k ,k = 0,1,2,..., is given by 1,2,4,8.... On
occasions,arulegivesx[k]intermsofearliermembersofthesequence.Forexample,the
previoussequencecouldhavebeendefinedbyx[k] = 2x[k−1],x[0] = 1.Thesequence
isthensaidtobedefinedrecursivelyandthedefiningformulaiscalledarecurrencerelationordifferenceequation.Differenceequationsareparticularlyimportantindigital
signalprocessingand are dealtwith inChapter22.
Example6.4 Write down the termsx[k] fork = 0,...,7 ofthe sequencedefined recursively as
x[k] =x[k −2] +x[k −1]
wherex[0] = 1 andx[1] = 1.
Solution Thevalues ofx[0]andx[1]aregiven. Usingthe given recurrencerelationwefind
x[2] =x[0] +x[1] = 2
x[3] =x[1] +x[2] = 3
Continuing inthisfashion wefind the first eight termsofthe sequenceare
1,1,2,3,5,8,13,21
This sequenceis known asthe Fibonaccisequence.

204 Chapter 6 Sequences and series
6.2.1 Arithmeticprogressions
An arithmetic progression is a sequence where each term is found by adding a fixed
quantity, called thecommon difference, tothe previous term.
Example6.5 Writedownthefirstfivetermsofthearithmeticprogressionwherethefirsttermis1and
the common difference is3.
Solution The second term is found by adding the common difference, 3, to the first term, 1, and
so the second termis4.Continuing inthis way wecan construct the sequence
1,4,7,10,13,...
A more general arithmetic progression has first term a and common difference d,
that is
a,a+d,a+2d,a+3d,...
Itis easytosee thatthekth termis
a+(k−1)d
Allarithmeticprogressions can be writtenrecursively asx[k] =x[k −1] +d.
Arithmetic progression:a,a +d,a +2d,...
a = first term,d = common difference,kth term =a + (k −1)d
Example6.6 Findthe10thand20thtermsofthearithmeticprogressionwithafirstterm5andcommon
difference −4.
Solution Here a = 5 and d = −4. The kth term is 5 − 4(k − 1). Therefore the 10th term is
5 −4(9) = −31andthe20thtermis5−4(19) = −71.
6.2.2 Geometricprogressions
A geometric progression is a sequence where each term is found by multiplying the
previous termby a fixed quantity called thecommon ratio.
Example6.7 Write down the geometric progression whose first term is 1 and whose common ratio
is 1 2 .
Solution Thesecondtermisfoundbymultiplyingthefirstbythecommonratio, 1 2 ,thatis 1 1
2 ×1 =
. Continuing inthisway we obtain the sequence
2
1, 1 2 , 1 4 , 1 8 ,...

6.2 Sequences 205
Ageneralgeometricprogressionhasfirsttermaandcommonratiorandcantherefore
be writtenas
a,ar,ar 2 ,ar 3 ,...
anditiseasytoseethatthekthtermisar k−1 .Allgeometricprogressionscanbewritten
recursively asx[k] =rx[k −1].
Geometric progression:a,ar,ar 2 ,...
a = first term,r = common ratio,kth term =ar k−1
6.2.3 Moregeneralsequences
We have already metanumber of infinite sequences. For example,
(1) x[k]=2,4,6,8,...
(2) x[k] = 1, 1 2 , 1 4 ,...
In case (1) the terms of the sequence go on increasing without bound. We say the sequenceisunbounded.Ontheotherhand,incase(2)itisclearthatsuccessivetermsget
smallerandsmallerandask → ∞,x[k] → 0.Thenotionofgettingcloserandcloserto
afixedvalueisveryimportantinmathematicsandgivesrisetotheconceptofalimit.In
case(2)wesay‘thelimitofx[k]asktendstoinfinityis0’andwewritethisconciselyas
lim x[k] = 0
k→∞
We say that the sequence converges to 0, and because its terms do not increase without
bound wesay itisbounded.
More formally, we say that a sequencex[k] converges to a limitl if, by proceeding
far enough along the sequence, all subsequent terms can be made to lie as close tol as
wewish.Whenever a sequence isnotconvergent itissaidtobedivergent.
Itispossibletohavesequences whicharebounded butnevertheless donotconverge
toalimit. The sequence
x[k] = −1,1,−1,1,−1,1,...
clearly fails to have a limit ask → ∞ although it is bounded, that is its values all lie
within a given range. This particular sequence issaidtooscillate.
Itispossibletoevaluatethelimitofasequence,whensuchalimitexists,fromknowledgeofitsgeneralterm.Tobeabletodothiswecanmakeuseofcertainrules,theproofs
of which arebeyond the scope of thisbook, but which wenow state:
Ifx[k]andy[k]aretwosequencessuchthatlim k→∞
x[k] =l 1
,andlim k→∞
y[k] =l 2
,
wherel 1
andl 2
arefinite,then:
(1) Thesequencegiven byx[k] ±y[k] has limitl 1
±l 2
.
(2) Thesequencegiven bycx[k], wherecisaconstant, has limitcl 1
.
(3) Thesequencex[k]y[k] has limitl 1
l 2
.
(4) Thesequence x[k]
y[k] has limit l 1
providedl
l 2 ≠ 0.
2

206 Chapter 6 Sequences and series
Furthermore, wecan always assume that
lim
k→∞
1
= 0 forany constantm > 0
km Example6.8 Find,ifpossible, the limitofeachofthe following sequences,x[k].
(a) x[k] = 1 k
(b) x[k]=5
(c) x[k] = 3 + 1 k
(d) x[k] = 1
k +1
(e) x[k]=k 2
k=1,2,3,4,...
k=1,2,3,4,...
k=1,2,3,4,...
k=1,2,3,4,...
k=1,2,3,4,...
Solution (a) The sequencex[k] isgiven by
1, 1 2 , 1 3 , 1 4 ,...
Successivetermsgetsmallerandsmaller,andask → ∞,x[k] → 0.Byproceeding
far enough along thesequence wecan get asclose tothelimit0as wewish.Hence
1
lim x[k] = lim
k→∞ k→∞ k = 0
(b) The sequencex[k] isgiven by 5,5,5,5,.... This sequence has limit5.
(c) Thesequence3,3,3,3,...haslimit3.Thesequence1, 1 2 , 1 3 ,...haslimit0.Therefore,usingrule(1)
wehave
lim
k→∞ 3 + 1 k =3+0=3
The terms of the sequence x[k] = 3 + 1 k are given by 4,31 2 ,31 3 ,...,andby
proceeding far enough along we can make all subsequent terms lie as close to the
limit3as wewish.
(d) The sequencex[k] = 1
k +1 ,k=1,2,3,4,...,isgivenby
1
2 , 1 3 , 1 4 ,...
and has limit0.
(e) The sequence x[k] = k 2 ,k = 1,2,3,4,..., is given by 1,4,9,16,..., and
increases withoutbound. This sequence has no limit-- itisdivergent.
Example6.9 Given a sequencewith generaltermx[k] = k −1
k +1 , find lim k→∞ x[k].
Solution It is meaningless simply to writek = ∞ to obtain lim k→∞
x[k] = ∞ −1 , since such
∞ +1
a quantity is undefined. What we should do is try to rewritex[k] in a form in which we

6.2 Sequences 207
can sensibly letk → ∞. Dividing both numerator and denominator byk, we write
k −1 1− (1/k)
=
k +1 1+ (1/k)
Then, ask → ∞, 1/k → 0 so that
( ) 1− (1/k)
lim
k→∞
1+ (1/k)
= lim k→∞
(1 − (1/k))
lim k→∞
(1 + (1/k))
byrule (4)
= 1 1
= 1
Example6.10 Given a sequencewith generalterm
x[k] = 3k2 −5k+6
k 2 +2k+1
find lim k→∞
x[k].
Solution Dividing the numerator and denominator byk 2 introduces terms which tend to zero as
k → ∞, that is
3k 2 −5k+6
k 2 +2k+1 = 3 − (5/k) + (6/k2 )
1 + (2/k) + (1/k 2 )
Then ask → ∞, wefind
lim
k→∞ x[k] = 3 1 = 3
k
Example6.11 2
Examine the behaviour of
3k+1 ask→∞.
Solution k 2
3k+1 = k
3+ (1/k)
As k → ∞, 1/k → 0 so that the denominator approaches 3. On the other hand, as
k → ∞the numerator tends toinfinity so thatthissequence diverges toinfinity.
EXERCISES6.2
1 Graphthe sequences given by
(a)x[k]=k,k=0,1,2,3,...
{ 3 k = 2
(b)x[k] = k=0,1,2,3,...
0 otherwise
(c)x[k]=e −k ,k=0,1,2,3,...
2 Thesequencex[k] is obtainedbysampling
f (t) = cos(t +2),t ∈ R.Thesampling begins at
t = 0 andthereafter att = 1,2,3, . . ..Writedown
the firstsixterms ofthe sequence.
3 A sequence,x[k], is defined by
x[k] = k2
2 +k,k=0,1,2,3,...
State the firstfiveterms ofthe sequence.
4 Writedownthefirstfiveterms,andplotgraphs,ofthe
sequences given recursively by
(a) x[k] =
x[k −1]
, x[0]=1
2

208 Chapter 6 Sequences and series
(b) x[k] = 3x[k −1] −2x[k −2],
x[0] = 2, x[1] = 1
5 Arecurrence relation is defined by
x[n +1] =x[n] +10, x[0] = 1,
n=0,1,2,3,...
Findx[1],x[2],x[3]andx[4].
6 Asequence isdefined by meansofthe recurrence
relation
x[n +1] =x[n] +n 2 , x[0] = 1,
n=0,1,2,3,...
Write down the firstfive terms.
7 Considerthe difference equation
x[n +2] −x[n +1] = 3x[n],
n=0,1,2,3,...
Ifx[0] = 1 andx[1] = 2, findthe terms
x[2],x[3],...,x[6].
8 Write down the 10th and 19th termsofthe arithmetic
progressions
(a) 8,11,14, . . .
(b) 8,5,2,...
9 An arithmetic progressionis given by
b, 2b
3 , b 3 ,0,...
(a) State the sixth term.
(b) State thekth term.
(c) Ifthe 20th term has avalue of15, findb.
10 Writedown the 5thand10th terms ofthe geometric
progression 8,4,2, . . ..
11 Find the 10th and20th terms ofthe geometric
progression with firstterm 3 andcommon ratio 2.
12 Ageometric progression is given by
a,ar,ar 2 ,ar 3 ,...
If |(k +1)thterm | > |kthterm |and (k +1)thterm ×
kth term < 0, whichofthe following,ifany, mustbe
true?
(a)r>1 (b)a>1
(c) r < −1 (d) ais negative
(e) −1<r<1
13 Ageometricprogression has firstterma = 1. The
ninth termexceeds the fifthtermby 240. Find
possiblevalues forthe eighthterm.
14 Ifx[k] = 3k+2 find lim
k k→∞ x[k].
3k+2
15 Findlim k→∞
k 2 +7 .
16 Findthe limitsasktends to infinity,ifthey exist, of
the following sequences:
(a) x[k] =k 3
(b) x[k] = 2k+3
4k+2
(c) x[k] = k2 +k
k 2 +k+1
( ) 4
6k+7
17 Findlim k→∞ .
3k−2
18 Findlim k→∞ x[k], ifitexists, when
(a) x[k] = (−1) k
(b) x[k] = 2 − k
10
( ) k
1
(c) x[k] =
3
(d) x[k] = 3k3 −2k 2 +4
5k 3 +2k 2 +4
( ) 2k
1
(e) x[k] =
5
Solutions
2 cos2,cos3,cos4,cos5,cos6,cos7
3 0, 3 15
,4,
2 2 ,12
4 (a) 1, 1 2 , 1 4 , 1 8 , 1 16
5 11,21, 31,41
6 1,1,2,6,15
7 5, 11, 26,59, 137
(b)2,1, −1, −5, −13
8 (a) 10th term = 35, 19th term = 62
(b) 10th term = −19, 19th term = −46
9 (a)− 2b
3
10 1 2 , 1 64
11 1536,1572864
(b)
12 Only(c)must be true
b(4−k)
3
(c) − 45
16

6.3 Series 209
13 ±128
14 3
15 0
16 (a) Limitdoes not exist (b) 1 2
(c) 1
17 16
18 (a) Limit doesnot exist
(b) Limit doesnotexist
(c) 0 (d) 3 5
(e) 0
6.3 SERIES
Whenever the terms of a sequence are added together we obtain what is known as a
series.Forexample,ifweaddthetermsofthesequence1, 1 2 , 1 4 , 1 8 ,weobtaintheseries
S, where
S=1+ 1 2 + 1 4 + 1 8
This series ends after the fourth term and is said to be a finite series. Other series we
shall meet continue indefinitely and aresaidtobeinfinite series.
Given an arbitrarysequencex[k], we use thesigmanotation
n∑
S n
= x[k]
k=1
to mean the sum x[1] +x[2] + ··· +x[n], the first and last values of k being shown
below and above the Greek letter , which is pronounced ‘sigma’. If the first term of
the sequence isx[0] rather thanx[1] wewould write ∑ n
k=0 x[k].
6.3.1 Sumofafinitearithmeticseries
An arithmetic series isthe sum ofan arithmetic progression. Consider the sum
S=1+2+3+4+5
Clearly this sums to 15. When there are many more terms it is necessary to find a more
efficient way of adding them up. The equation forScan be writtenintwo ways:
and
S=1+2+3+4+5
S=5+4+3+2+1
If weadd these two equations together we get
2S=6+6+6+6+6
There arefive termsso that
thatis
2S=5×6=30
S=15
Now a general arithmeticseries withkterms can be writtenas
S k
=a+(a+d)+(a+2d)+···+(a+(k−1)d)
butrewritingthis back tofront, wehave
S k
=(a+(k−1)d)+(a+(k−2)d)+···+(a+d)+a

210 Chapter 6 Sequences and series
Addingtogetherthefirsttermineachseriesproduces2a+ (k−1)d.Addingthesecond
terms together produces 2a + (k − 1)d. Indeed adding together the ith terms yields
2a + (k −1)d.Hence,
2S k
=(2a+(k−1)d)+(2a+(k−1)d)+···+(2a+(k−1)d)
} {{ }
k times
that is
so that
2S k
=k(2a + (k −1)d)
S k
= k 2 (2a+(k−1)d)
This formula tells us the sum to k terms of the arithmetic series with first term a and
common differenced.
Sumofanarithmeticseries:S k
= k 2 (2a+(k−1)d)
Example6.12 Findthesumofthearithmeticseriescontaining30terms,withfirstterm1andcommon
difference 4.
Solution We wish tofindS k
:
S k
=1+5+9+···
} {{ }
30 terms
UsingS k
= k 2 (2a + (k −1)d)wefindS 30 = 30 (2 +29 ×4) =1770.
2
Example6.13 Find the sum of the arithmetic series with first term 1, common difference 3 and with
last term100.
Solution Wealreadyknowthatthekthtermofanarithmeticprogressionisgivenbya+(k−1)d.
Inthis casethe lasttermis100. We can use thisfacttofind the numberofterms.Thus,
that is
100=1+3(k−1)
3(k−1)=99
k−1=33
k=34
So thereare34 termsinthisseries.Thereforethe sum,S 34
, isgiven by
S 34
= 34 {2(1) + (33)(3)}
2
= 17(101)
= 1717

6.3.2 Sumofafinitegeometricseries
6.3 Series 211
A geometric series is the sum of the terms of a geometric progression. If we sum the
geometric progression 1, 1 2 , 1 4 , 1 8 , 1
16 wefind
S=1+ 1 2 + 1 4 + 1 8 + 1
16
(6.1)
Iftherehadbeenalargenumberoftermsitwouldhavebeenimpracticaltoaddthemall
directly. However, let us multiplyEquation (6.1) by the common ratio, 1 2 :
1
2 S = 1 2 + 1 4 + 1 8 + 1
16 + 1
32
so that, subtracting Equation (6.2)from Equation (6.1), we find
(6.2)
S − 1 2 S=1− 1
32
since mostterms cancel out. Therefore 1 2 S = 31
32 andsoS=31 16 = 115 16 .
We can apply this approach more generally: when we have a geometric progression
with first termaand common ratior, the sum toktermsis
S k
=a+ar+ar 2 +ar 3 +···+ar k−1
Multiplying byr gives
rS k
=ar+ar 2 +ar 3 +···+ar k−1 +ar k
Subtraction givesS k
−rS k
=a −ar k , so that
S k
= a(1−rk )
1 −r
providedr ≠ 1
This formula gives the sum to k terms of the geometric series with first term a and
common ratior.
Sumofageometricseries:S k
= a(1−rk )
1 −r
r≠1
6.3.3 Sumofaninfiniteseries
When dealing with infinite series the situation is more complicated. Nevertheless, it is
frequently the case that the answer to many problems can be expressed as an infinite
series.Incertaincircumstances,thesumofaseriestendstoafiniteanswerasmoreand
more terms are included and we say the series has converged. To illustrate this idea,
consider the graphical interpretation of the series 1 + 1 2 + 1 4 + 1 8 +···,asgivenin
Figure 6.6.
Start at0and move a length 1:total distance moved = 1
Move further, a length 1 2 : total distance moved = 11 2
Move further, a length 1 4 : total distance moved = 13 4

212 Chapter 6 Sequences and series
1
–
3
–
7
0 1 1 1 1–
2
2 4 8
Figure6.6
Graphical interpretation ofthe series
1 + 1 2 + 1 4 + 1 8 +···.
At each stage the extra distance moved is half the distance remaining up to the point
x = 2. It is obvious that the total distance we move cannot exceed 2 although we
can get as close to 2 as we like by adding on more and more terms. We say that the
series 1 + 1 2 + 1 + ··· converges to 2. The sequence of total distances moved, given
4
previously,
1,1 1 2 ,13 4 ,17 8 ,...
iscalled thesequence of partialsumsof the series.
For any given infinite series ∑ ∞
k=1x[k], wecan formthe sequence ofpartialsums,
1∑
S 1
= x[k] =x[1]
S 2
=
S 3
=
.
k=1
2∑
x[k] =x[1] +x[2]
k=1
3∑
x[k] =x[1] +x[2] +x[3]
k=1
If the sequence {S n
} converges to a limitS, we say that the infinite series has a sumS
or that it has converged toS. Clearly not all infinite series will converge. For example,
consider the series
1+2+3+4+5+···
The sequence of partial sums is 1,3,6,10,15,.... This sequence diverges to infinity
and sothe series 1 +2+3+4+5+··· is divergent.
It is possible to establish tests or convergence criteria to help us to decide whether
or not a given series converges or diverges, but for these you must refer to a more
advanced text.
6.3.4 Sumofaninfinitegeometricseries
In the case of an infinite geometric series, it is possible to derive a simple formula for
its sum when convergence takes place. We have already seen that the sum tok terms is
given by
S k
= a(1−rk )
1 −r
r≠1
What we must do is allowkto become large so that more and more terms are included
in the sum. Provided that −1 < r < 1, thenr k →0ask→∞.ThenS k
→
a
1 −r .
When thishappens we write
S ∞
=
a
1 −r

6.3 Series 213
whereS ∞
is known as the ‘sum to infinity’. Ifr > 1 orr < −1,r k fails to approach a
finitelimitask → ∞and the geometric series diverges.
Sum of aninfinite geometric series:S ∞
=
a
1 −r
−1<r<1
Example6.14 Findthe sum toktermsofthe following series and deducetheirsumstoinfinity:
(a) 1+ 1 3 + 1 9 + 1
27 +··· (b) 12+6+3+11 2 +···
Solution (a) This isageometric series with first term1and common ratio1/3.Therefore,
S k
= a(1−rk )
= 1(1 − (1/3)k )
= 3 ( ( 1 k )
1 −
1 −r 2/3 2 3)
Ask → ∞, (1/3) k → 0sothatS ∞
= 3/2.
(b) This isageometric series with first term12 and common ratio 1 2 . Therefore,
S k
= 24(1 − (1/2) k )
Ask → ∞, (1/2) k → 0sothatS ∞
= 24.Thiscould,ofcourse,havebeenobtained
directly from the formulafor the sum toinfinity.
EXERCISES6.3
1 An arithmeticseries has afirstterm of4andits 30th
term is1000.Find the sum to 30terms.
2 Find the sum to 20 termsofthe arithmeticserieswith
firstterma,andcommon differenced,given by
(a)a=4,d=3 (b)a=4,d=−3
3 Ifthe sumto 10terms ofan arithmeticseries is100
anditscommondifference,d,is −3,finditsfirstterm.
4 Thesum to 20terms ofan arithmeticseries is
identicalto the sum to 22 terms. Ifthe common
difference is −2, find the firstterm.
5 Find the sum to five terms ofthe geometricseries
with firstterm 1 andcommon ratio 1/3. Find the sum
to infinity.
6 Find the sum ofthe first20terms ofthe geometric
serieswith firstterm 3 andcommon ratio 1.5.
7 Find the sum ofthe arithmeticseries with firstterm 2,
common difference 2, andlast term50.
8 The sumto infinityofageometricseries isfour times
the firstterm.Find the common ratio.
9 The sumto infinityofageometricseries istwicethe
sum ofthe firsttwo terms. Findpossiblevalues ofthe
common ratio.
10 Express the alternatingharmonic series
1 − 1 2 + 1 3 − 1 + · · ·in sigma notation.
4
11 Writedown the firstsixterms ofthe series ∑ ∞
k=0 z −k .
12 Explain why ∑ ∞
k=1 x[k] is the same as ∑ ∞
n=1 x[n].
Further, explain why both canbe written as
∑ ∞k=0
x[k +1].
Solutions
1 15060
2 (a) 650 (b) −490
3 23.5
4 41

214 Chapter 6 Sequences and series
5 1.494,S ∞ = 3 2
6 19946
7 650
8
3
4
9 ± √ 1 2
∑
10 ∞ (−1) n+1
1 n
11 1+ 1 z + 1 z 2 + 1 z 3 + 1 z 4 + 1 z 5
TechnicalComputingExercises6.3
Most technicalcomputing languageshave built-in
functionsforgenerating geometricseries.MATLAB ®
has the functioncumprod whichcalculates the
cumulative productofthe numberspassed to its
input. Ittakes the first term andthensuccessively
multiplies the succeedingarguments in turnbythe
result.
Forexample, in MATLAB ® wecould type:
S=cumprod([1 0.5 0.5 0.5 0.5])
to produce the geometricseriesandstore itin a row
vectorS.
1.000000 0.500000 0.250000
0.125000 0.062500
1 Calculate the sum ofallofthe elementsin the finite
seriesgiven above.
2 Increase the number ofelementsin the series to 10
andnotethe difference in youranswer.
3 Compare the answers to the previous two exercises to
the exact equationgiven atthe endofSection6.3.2
witha = 1 andr = 0.5.Whatwould you expect to
happeniftherewere100 elementsin the series?
6.4 THEBINOMIALTHEOREM
It is straightforward to show that the expression (a + b) 2 can be written as a 2 +
2ab + b 2 . It is slightly more complicated to expand the expression (a + b) 3 to a 3 +
3a 2 b +3ab 2 +b 3 . However, it is often necessary to expand quantities such as (a +b) 6
or (a +b) 10 , say, and then the algebra becomes extremely lengthy. A simple technique
forexpandingexpressionsoftheform (a+b) n ,wherenisapositiveinteger,isgivenby
Pascal’s triangle.
Pascal’striangleisthetriangleofnumbersshowninFigure6.7,whereitisobserved
that every entry is obtained by adding the two entries on either side in the preceding
row, always starting and finishing a row with a ‘1’. You will note that the third row
down,1 2 1,givesthecoefficientsintheexpansionof (a +b) 2 = 1a 2 +2ab +1b 2 ,
whilethefourthrow,1 3 3 1,givesthecoefficientsintheexpansionof (a +b) 3 =
1a 3 +3a 2 b +3ab 2 +1b 3 .Furthermore,thetermsintheseexpansionsarecomposedof
decreasingpowersofaandincreasingpowersofb.Whenwecometoexpandthequantity
(a +b) 4 therowbeginning‘1 4’inthetrianglewillprovideuswiththenecessary
coefficientsintheexpansionandwemustsimplytakecaretoputinplacetheappropriate
powers ofaandb. Thus (a +b) 4 = 1a 4 +4a 3 b +6a 2 b 2 +4ab 3 +1b 4 .
1
1 1
1 2 1
1 3 3 1
1 4 + 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Figure6.7
Pascal’striangle.

6.4 The binomial theorem 215
Example6.15 UsePascal’s triangle toexpand (a +b) 6 .
Solution We look to the row commencing ‘1 6’, that is 1 6 15 20 15 6 1, because
a+bisraised tothe power 6.This row provides the necessary coefficients. Thus,
(a +b) 6 =a 6 +6a 5 b +15a 4 b 2 +20a 3 b 3 +15a 2 b 4 +6ab 5 +b 6
Example6.16 Expand (1 +x) 7 usingPascal’s triangle.
Solution Forming the rowcommencing ‘1 7’we select the coefficients
1 7 21 35 35 21 7 1
Inthisexample,a = 1 andb=xso that
(1 +x) 7 =1+7x +21x 2 +35x 3 +35x 4 +21x 5 +7x 6 +x 7
When it is necessary to expand the quantity (a + b) n for large n, it is clearly inappropriate
to use Pascal’s triangle since an extremely large triangle would have to be
constructed. However, it is frequently the case that in such situations only the first few
termsinthe expansion are required. This iswhere thebinomial theoremisuseful.
The binomial theorem states thatwhennisapositive integer
(a+b) n =a n +na n−1 b +
+···+b n
n(n −1)
a n−2 b 2 +
2!
n(n −1)(n −2)
a n−3 b 3
3!
Itisalsofrequently quoted for the case whena = 1 andb=x, so that
(1+x) n =1+nx+
n(n −1)
x 2 +
2!
n(n −1)(n −2)
x 3 +···+x n (6.3)
3!
Example6.17 Expand (1 +x) 10 uptothe terminx 3 .
Solution WecouldusePascal’striangletoanswerthisquestionandlooktotherowcommencing
‘1 10’buttofindthisrowconsiderablecalculationswouldbeneeded.Weshallusethe
binomialtheoreminthe formofEquation (6.3). Takingn = 10, wefind
(1+x) 10 =1+10x+ 10(9) x 2 + (10)(9)(8) x 3 +···
2! 3!
=1+10x+45x 2 +120x 3 +···
so that, up toand includingx 3 , the expansion is
1+10x+45x 2 +120x 3

216 Chapter 6 Sequences and series
Wehaveassumedintheforegoingdiscussionthatnisapositiveintegerinwhichcase
the expansion given by Equation (6.3) will eventually terminate. In Example 6.17 this
wouldoccurwhenwereachedtheterminx 10 .Itcanbeshown,however, thatwhennis
not a positive integer the function (1 +x) n and the expansion given by
(1+x) n =1+nx+
n(n −1)
x 2 +
2!
n(n −1)(n −2)
x 3 +··· (6.4)
3!
havethesamevalueprovided −1 <x<1.However,whennisnotapositiveintegerthe
seriesdoesnotterminateandwemustdealwithaninfiniteseries.Thisseriesconverges
when −1 < x < 1 and the expansion is then said to be valid. Whenxlies outside this
intervaltheinfiniteseriesdivergesandsobearsnorelationtothevalueof (1 +x) n .The
expansion isthen saidtobe invalid.
Thebinomialtheorem:
(1+x) n =1+nx+
n(n −1)
x 2 +
2!
n(n −1)(n −2)
x 3 +···
3!
−1<x<1
Example6.18 Usethebinomialtheoremtoexpand
Solution
the terminx 3 .
1
1 +x inascendingpowersofxuptoandincluding
1
1 +x canbewrittenas (1 +x)−1 .UsingthebinomialtheoremgivenbyEquation(6.4)
withn = −1, wefind
(1 +x) −1 = 1 + (−1)x + (−1)(−2) x 2 + (−1)(−2)(−3) x 3 +···
2! 3!
=1−x+x 2 −x 3 +···
provided −1 <x<1.Consequently,ifinfutureapplicationswecomeacrosstheseries
1 −x+x 2 −x 3 + ..., we shall be able to write it in the form (1 +x) −1 . This closed
formavoidstheuseofaninfiniteseriesandsoitiseasiertohandle.Weshallmakeuse
of this technique inChapter 22 when wemeet theztransform.
Example6.19 Obtain a quadraticapproximation to (1 −2x) 1/2 usingthe binomialtheorem.
Solution Using Equation (6.4)withxreplaced by −2x andn = 1 2 wehave
( 1
(1 −2x) 1/2 = 1 + (−2x) +
2)
(1/2)(−1/2) (−2x) 2 + ···
2!
=1−x− 1 2 x2 +···
provided that −1 < −2x < 1, that is − 1 2 < x < 1 . A quadratic approximation is
2
therefore 1 −x− 1 2 x2 .

6.4 The binomial theorem 217
EXERCISES6.4
1 Usethe binomial theorem to expand
(a) (1 +x) 3 (b) (1 +x) 4 (c)
(d)
(
1 − x 2) 5
(e)
(
2 + x 2) 5
(f)
2 UsePascal’striangleto expand (a +b) 8 .
(
(
1 + x 3
3 − x 4
3 UsePascal’striangleto expand (2x +3y) 4 .
4 Expand (a −2b) 5 .
5 Usethe binomial theoremto findthe expansion of
(3 −2x) 6 upto andincluding the term inx 3 .
6 Obtainthe firstfour terms in the expansion of
(
1 + 1 2 x ) 10
.
7 Obtainthe firstfive termsin the expansion of
(1 +2x) 1/2 .Statethe rangeofvalues ofxforwhich
the expansion is valid.Chooseavalue ofxwithin the
rangeofvalidity andcomputevalues ofyour
expansionforcomparison with the truefunction
values. (
8 Expand 1 + 1 ) −4
2 x in ascending powersofxupto
the term inx 4 ,stating the range ofvalues ofxfor
whichthe expansion is valid.
(
9 Expand 1 + 1 −1/2
in descending powers upto the
x)
fourthterm.
) 4
) 4
10 (a) Expand (1 +x 2 ) 4 .
(b) Expand (1 +1/x 2 ) 4 .
11 A function, f (x),isgiven by
( ) 1/2
f(x)=
1 + 1 x
(a) Obtain the firstfour termsin the expansion of
f (x)in descending powersofx.State the range
ofvalues ofxforwhich the expansion is valid.
(b) By writing f (x)in the form
f(x) =x −1/2 (1 +x) 1/2
obtain the firstfourterms in the expansion of
f (x)in ascending powersofx.State the range of
values ofxforwhich the expansion is valid.
12 The function,g(x),is defined by
g(x) =
1
(1+x 2 ) 4
(a) Obtain the firstfour termsin the expansion of
g(x) in ascending powersofx. Statethe range of
values ofxforwhich the expansion is valid.
(b) By rewritingg(x) in an appropriate form,obtain
the firstfour termsin the expansion ofg(x) in
descending powers ofx.State the range ofvalues
ofxforwhich the expansion isvalid.
Solutions
1 (a) 1+3x+3x 2 +x 3
3 16x 4 +96x 3 y +216x 2 y 2 +216xy 3 +81y 4
(b) 1+4x+6x 2 +4x 3 +x 4
4 a 5 −10a 4 b +40a 3 b 2 −80a 2 b 3 +80ab 4 −32b 5
(c) 1+ 4x
3 + 2x2
3 + 4x3
27 + x4
5 729 −2916x +4860x 2 −4320x 3
81
(d) 1− 5x
2 + 5x2
2 − 5x3
4 + 5x4
16 − x5 6 1+5x+ 45x2 +15x 3
32
4
(e) 32 +40x +20x 2 +5x 3 + 5x4
8 + x5 7 1+x− x2
32
2 + x3
2 − 5x4
8 valid for −1 2 <x< 1 2
(f) 81 −27x + 27x2 − 3x3
8 16 + x4 8 1−2x+ 5x2
256
2 − 5x3
2 + 35x4 valid for −2 <x<2
16
2 a 8 +8a 7 b +28a 6 b 2 +56a 5 b 3 +70a 4 b 4 +56a 3 b 5 + 9 1 − 1
28a 2 b 6 +8ab 7 +b 8 2x + 3
8x2 − 5
16x 3

218 Chapter 6 Sequences and series
(
)
10 (a) 1+4x 2 +6x 4 +4x 6 +x 8
(b) 1+ 4 x 2 + 6 x 4 + 4 x 6 + 1 (b) x −1/2 1 + x 2 − x2
8 + x3 valid for |x| < 1
16
x 8
12 (a) 1 −4x 2 +10x 4 −20x 6 validfor |x| < 1
11 (a) 1+ 1 2x − 1
8x 2 + 1
(
·)
16x 3 valid for |x| > 1 (b) x −8 1 − 4 x 2 + 10
x 4 − 20
x 6 · · valid for |x| > 1
6.5 POWERSERIES
Aparticularlyimportantclassofseriesareknownaspowerseriesandtheseareinfinite
series involving integer powers ofthe variablex. For example,
and
1+x+x 2 +x 3 +···
1+x+ x2
2 + x3
6 +···
arebothpowerseries.Notethatapowerseriescanberegardedasaninfinitepolynomial.
Many common functions can be expressed intermsof a power series,forexample
sinx=x− x3
3! + x5
− ··· x inradians
5!
which converges for any value ofx. For example,
sin(0.5) = 0.5 − (0.5)3
6
+ (0.5)5
120 −···
Taking justthe first threeterms,we find
sin(0.5) ≈ 0.5 −0.0208333 +0.0002604 = 0.4794271
as compared with the truevalue, sin0.5 = 0.4794255.
More generally, a power series is only meaningful if the series converges for the
particular value of x chosen. We define an important quantity known as the radius of
convergence,R, as the largest value for which anxchosen in the interval −R <x<R
causes the series toconverge.
Theopeninterval (−R,R)isknownastheintervalofconvergence.Testsforconvergenceofapowerseriesarethesubjectofmoreadvancedtexts.Furtherconsiderationwill
be given to power series in Chapter 18, but for future reference we give some common
expansions now:
sinx=x− x3
3! + x5
− ··· x inradians
5!
cosx=1− x2
2! + x4
− ··· x inradians
4!
e x =1+x+ x2
2! + x3
3! + x4
4! +···
all of which converge for any value ofx.

6.6 Sequences arising from the iterative solution of non-linear equations 219
Each of these series converges rapidly when x is small, and so can be used to obtain
usefulapproximations. Inparticular, wenote that
Ifxis smalland measured inradians
sinx≈x and cosx≈1− x2
2!
These formulae areknown as thesmall-angle approximations.
EXERCISES6.5
1 Thepower seriesexpansion ofe x is given by
e x =1+x+ x2
2! + x3
3! +···
andis valid foranyx.Take four terms ofthe series
whenx=0, 0.1,0.5and1, tocompare the sumtofour
termswith the value ofe x obtained fromyour
calculator. Comment upon the result.
2 Using the power series expansion forcosx:
(a) Writedown the power series expansion for
cos2x.
(b) Writedown the power series expansion for
cos(x/2).
Byconsidering the power series expansionfor
cos(−x) showthatcosx = cos(−x).
3 Byconsidering the power series expansionofe x find
∑ ∞k=0
1/k!.
4 Obtainacubicapproximation to e x sinx.
5 (a) Statethe power series expansion fore −x .
(b) Byusingyoursolution to (a)andthe expansion
fore x ,deduce the power seriesexpansions of
coshx andsinhx.
Solutions
1
x e x Sum to 4 terms
0 1 1
0.1 1.1052 1.1052
0.5 1.6487 1.6458
1 2.7183 2.6667
Values are in close agreementwhenxissmall.
2 (a) 1−2x 2 + 2x4
3
3 e
−··· (b) 1−x2
8 + x4
384 −···
4 x+x 2 + x3
3
5 (a) 1−x+ x2
2! − x3
3! +···
(b) coshx=1+ x2
2! + x4
4! +···,
sinhx=x+ x3
3! + x5
5! +···
6.6 SEQUENCESARISINGFROMTHEITERATIVESOLUTION
OFNON-LINEAREQUATIONS
Itisoften necessarytosolve equationsofthe form f (x) = 0.For example,
f(x)=x 3 −3x 2 +7=0, f(x)=lnx− 1 x = 0
To solve means to find values of x which satisfy the given equation. These values are
knownasroots.Forexample,therootsofx 2 −3x +2 = 0arex = 1andx = 2because

220 Chapter 6 Sequences and series
whenthesevaluesaresubstitutedintotheequationbothsidesareequal.Equationswhere
the unknown quantity, x, occurs only to the first power are called linear equations.
Otherwise an equation is non-linear. A simple way of finding the roots of an equation
f (x) = 0 istosketch a graph ofy = f (x)as shown inFigure 6.8.
The roots are those values of x where the graph cuts or touches the x axis.
Generally, thereisnoanalytical wayofsolvingtheequation f (x) = 0andsoitisoften
necessary to resort to approximate or numerical techniques of solution. An iterative
technique is one which produces a sequence of approximate solutions which may converge
to a root. Iterative techniques can fail in that the sequence produced can diverge.
Whetherornotthishappensdependsupontheequationtobesolvedandtheavailability
ofagoodestimateoftheroot.Suchanestimatecouldbeobtainedbysketchingagraph.
The technique we shall describe here is known as simple iteration. It requires that the
equation berewrittenintheformx =g(x). Anestimateoftherootismade,sayx 0
,and
this value is substituted into the r.h.s. ofx =g(x). This yields another estimate,x 1
. The
process isthen repeated. Formally weexpress this as
x n+1
=g(x n
)
This is a recurrence relation which produces a sequence of estimates x 0
,x 1
,x 2
,....
Under certain circumstances the sequence will converge to a root of the equation. It
is particularly simple to program this technique on a computer. A check would be built
into the program totestwhether ornot successive estimates areconverging.
Example6.20 Solve the equation f (x) = e −x −x = 0 by simple iteration.
Solution Theequationmustfirstbearrangedintotheformx =g(x),andsowewritee −x −x = 0
as
x=e −x
In this example we see that g(x) = e −x . The recurrence relation which will produce
estimates of the rootis
x n+1
= e −x n
Table6.1
Iterativesolutionof
e −x −x=0.
n
x n
f(x)
Figure6.8
A rootof f (x) = 0 occurswhere the
graph touches orcrossesthexaxis.
x
0 0
1 1
2 0.368
3 0.692
4 0.501
5 0.606
6 0.546
.
.
. 0.567

6.6 Sequences arising from the iterative solution of non-linear equations 221
Suppose weestimatex 0
= 0.Then
Then
x 1
=e −x 0 =e −0 =1
x 2
= e −x 1 = e −1 = 0.368
Theprocessiscontinued.ThecalculationisshowninTable6.1fromwhichweseethat
the sequence eventually converges to 0.567 (3 d.p.). We conclude thatx = 0.567 is a
rootof e −x −x = 0.
Notethatiftheequationtobesolvedinvolvestrigonometricfunctions,angleswillusuallybe
measured inradians and not degrees.
Itispossibletodeviseatesttocheckwhetheranygivenrearrangementwillconverge.
Fordetailsofthisyoushouldrefertoatextbookonnumericalanalysis.Thereareother
more sophisticated iterative methods for the solution of non-linear equations. One such
method, the Newton--Raphson method, isdiscussedinChapter 12.
EXERCISES6.6
1 Showthat the quadratic equationx 2 −5x −7 = 0 can
be written in the formx = √ 7+5x.Withx 0 =6
locate arootofthisequation.
2 Forthe quadraticequationofQuestion1showthat an
alternative rearrangement is
x = x2 −7
.Withx
5 0 = 0.6 findthe second
solution ofthisequation.
3 Forthe quadraticequationofQuestion1showthat
anotherrearrangement isx = 7 +5. Tryto
x
solvethe equation usingvarious initialestimates.
Investigatefurtheralternative arrangements ofthe
original equation.
4 Showthat onerecurrence relationforthe solution of
the equation
e x +10x−3=0
is
x n+1 = 3−ex n
10
Withx 0 = 0 locate aroot ofthe given equation.
5 (a) Show thatthe cubicequationx 3 +3x −5 = 0
can be writtenas
(i)x= 5−x3 ,
3
(ii) x = 5
x 2 +3 .
(b) By sketching agraph forvalues ofxbetween 0
and 3 obtain aroughestimate ofthe rootofthe
equation given in part (a).
(c) Determine which, ifeither, ofthe arrangements
in part (a)converges more rapidly to the root.
Solutions
1 6.14
2 −1.14
5 (b) 1.15
(c) Arrangement (ii)converges more rapidly
4 0.18

222 Chapter 6 Sequences and series
TechnicalComputingExercises6.6
Write acomputer program to implement the simple
iteration method. Bycomparing successive estimates
the program should check whether convergence is
taking place.
REVIEWEXERCISES6
1 Write down andgraph the firstfive terms ofthe
sequencesx[k] defined by
(a) x[k] = (−1) k , k =0,1,2,3,...
(b) x[k] = (−1)k
2k+1 , k=0,1,2,3,...
2 Find expressions forthekth termsofthe sequences
whose firstfiveterms are
(a) 1, 9, 17, 25,33
(b) −1, 1, −1, 1, −1
3 Forthe Fibonacci sequence
showthat
1,1,2,3,5,8,...
lim
k→∞
x[k +1]
x[k]
= 1 2
(
1 + √ )
5
[Hint:writex[k +1] =x[k] +x[k −1],form
x[k +1]/x[k] andtake limits.]
4 Usethe binomial theoremto expand (1 +x+x 2 ) 5 as
far asthe term inx 3 .
1
5 Usethe binomial theoremto expand
(3 +x) 3
in ascending powers ofxasfar asthe term inx 3 .
6 Thepower series expansion forln(1 +x) isgiven by
ln(1+x)=x− x2
2 + x3
3 − x4
4 +···
and isvalid for −1 <x 1. Take anumber ofvalues
ofxin thisinterval and obtain an approximate value
ofln(1 +x)by meansofthisseries.Compare your
answerswiththevaluesobtainedfromyourcalculator.
7 Bymultiplyingboth numerator anddenominatorof
√ √
k+1− k
by √ k+1+ √ k find
2
√ √
k+1− k
lim
k→∞ 2
8 Find lim
k→∞
(
) 3
3k−1
.
2k+7
2k 5 −3k 2
9 Find lim
k→∞ 7k 7 +2k .
10 Writedown the firsteight termsofthe series ∑ n
k=1 k.
Bynotingthat thisis an arithmeticseries showthat
n∑
k =
k=1
n(n +1)
2
11 Writedown the firstsixterms ofthe sequence defined
bythe recurrence relation
x[n +3] =x[n +2] −2x[n]
x[0] = 0 x[1] = 2 x[2] = 3
12 Findthe limit, ifitexists, ask → ∞ ofthe geometric
progression
when
a,ar,ar 2 ,...,ar k−1 ,...
(a) −1<r<1
(b) r>1
(c) r<−1
(d) r=1
(e) r=−1
13 An arithmeticseries has afirst termof4andthe 10th
termis 0.
(a) FindS 20 .
(b) IfS n = 0,findn.
14 Ageometricseries has
S 3 = 37
8
S 6 = 3367
512
Findthe firsttermand the common ratio.

Review exercises 6 223
15 (a) Write down the seriesgiven by ∑ 5
r=1 r 2 .
(b) The sumofthe squaresofthe firstnwhole
numbers can be foundfrom the formula
n∑
r 2 n(n +1)(2n +1)
=
6
r=1
Use thisformula to find
(i) ∑ 5
r=1 r 2 , (ii) ∑ 10
r=1 r 2 , (iii) ∑ 10
r=6 r 2 .
16 (a) Writedown the series given by ∑ 6
r=1 r 3 .
(b) The sumofthe cubesofthe firstnwhole
numbers canbe foundfrom the formula
(
n∑
r 3 =
n(n +1)
2
r=1
Use thisformula to find
) 2
(i) ∑ 6
r=1 r 3 , (ii) ∑ 12
r=1 r 3 , (iii) ∑ 12
r=7 r 3 .
17 Thethirdterm ofan arithmeticprogression is 18.The
fifthterm is28. Find the sum of20terms.
18 (a) Find an expression forthe general term in the
sequence
2,5,10,17,...
(b) Define the sequence in terms ofarecurrence
relation.
19 (a) Show thatthe equationx 3 +2x −14 = 0 canbe
rearrangedintothe formx = 3√ 14 −2x.With
x 0 = 2 use simple iteration to findarootofthe
equation.
(b) Rearrange the equation 0.8sinx −0.5x = 0 into
the formx =g(x).Withx 0 = 2 use simple
iteration to findarootofthe equation.
(c) Rearrange the equationx 3 = 2e −x intothe form
x =g(x).Withx 0 = 0usesimpleiterationtofind
a rootofthe equation.
20 Writeoutexplicitlythe firstfour terms ofthe series
∞∑
m=0
(−1) m x 2m
2 2m (m!) 2 .
21 Writeoutexplicitlythe firstfour terms ofthe series
∞∑
m=0
(−1) m x 2m+1
2 2m+1 m!(m +1)! .
Solutions
1 (a) 1,−1,1, −1,1 (b) 1, − 1 3 , 1 5 , −1 7 , 1 9
2 (a) 8k−7 k 1 (b) (−1) k k 1
4 1+5x+15x 2 +30x 3
(
+···)
1
5 1 −x+ 2x2
27 3 − 10x3
27
7 0
8
9 0
27
8
11 0, 2, 3, 3, −1, −7
12 (a) 0 (b) nolimit (c) nolimit
(d) a (e) nolimit
13 (a) − 40
9
(b) 19
14 2, 3 4
15 (a) 1+4+9+16+25
(b) (i) 55 (ii) 385 (iii) 330
16 (a) 1+8+27+64+125+216
(b) (i) 441 (ii) 6084 (iii) 5643
17 1110
18 (a) x[k]=k 2 +1,k=1,2,3,...
(b)x[k+1]=x[k]+2k+1
19 (a) 2.13
(b) x = 1.6sinx,1.6
(c)x= 3 √
2e −x ,0.93
20 1− 1 4 x2 + 1 64 x4 − 1
2304 x6 +···
21 1 2 x − 1 16 x3 + 1
384 x5 − 1
18432 x7 +···

7 Vectors
Contents 7.1 Introduction 224
7.2 Vectorsandscalars:basicconcepts 224
7.3 Cartesiancomponents 232
7.4 Scalarfieldsandvectorfields 240
7.5 Thescalarproduct 241
7.6 Thevectorproduct 246
7.7 Vectorsofndimensions 253
Reviewexercises7 255
7.1 INTRODUCTION
Certainphysicalquantitiesarefullydescribedbyasinglenumber:forexample,themass
of a stone, the speed of a car. Such quantities are called scalars. On the other hand,
some quantities are not fully described until a direction is specified in addition to the
number. For example, a velocity of 30 metres per second due east is different from a
velocityof30metrespersecondduenorth.Thesequantitiesarecalledvectorsanditis
importanttodistinguishthemfromscalars.There aremany engineering applications in
which vector and scalar quantities play important roles. For example, speed, potential,
work and energy are scalar quantities, while velocity, electric and magnetic forces, the
position of a robot and the state-space representation of a system can all be described
by vectors. A variety of mathematical techniques have been developed to enable useful
calculations tobe carriedout usingvectors and inthischapter these will be discussed.
7.2 VECTORSANDSCALARS:BASICCONCEPTS
Scalarsarethesimplestquantitieswithwhichtodeal;thespecificationofasinglenumberisallthatisrequired.Vectorsalsohaveadirectionanditisusefultoconsideragraphical
representation. Thus the line segment AB of length 4 in Figure 7.1 can represent

7.2 Vectors and scalars: basic concepts 225
A
a
B
A
C
AB
B
CD
D
Figure7.1
Avector, AB,oflength → 4.
Figure7.2
Two equal vectors.
a vector in the direction shown by the arrow on AB. This vector is denoted by AB.
→
→
Note that AB ≠ BA. → The vector AB → →
is directed from A to B, but BA is directed from
BtoA.
An alternative notation is frequently used: we denote AB → by a. This bold notation
is commonly used in textbooks but the notation a is preferable for handwritten work.
Wenowneedtorefertothediagramtoappreciate theintended direction.Thelengthof
the line segment represents the magnitude, or modulus, of the vector and we use the
notation | → AB|, |a| or simply a to denote this. Note that whereas a is a vector, |a| is a
scalar.
7.2.1 Negativevectors
The vector −a is a vector in the opposite direction to, but with the same magnitude as,
a. Geometrically itwill be → BA. Thus −aisthe same as → BA.
7.2.2 Equalvectors
Twovectorsareequaliftheyhavethesamemagnitudeanddirection.InFigure7.2vectors
→ CDand → ABareequaleventhoughtheirlocationsdiffer.Thisisausefulpropertyof
vectors:avectorcanbetranslated,maintainingitslengthanddirectionwithoutchanging
thevectoritself.Thereareexceptionstothisproperty.Forexample,weshallsoonmeet
position vectors which are used to locate specific fixed points in space. They clearly
cannot be moved around freely. Nevertheless most of the vectors we shall meet can be
translated freely, and assuch are often calledfree vectors.
7.2.3 Vectoraddition
Itisfrequentlyusefultoaddtwoormorevectorstogetherandtheadditionofvectorsis
definedbythetrianglelaw.ReferringtoFigure7.3,ifwewishtoaddABto → CD,
→ CDis
translateduntilCandBcoincide.Asmentionedearlier,thistranslationdoesnotchange
→
thevectorCD.Thenthesum,
→ AB+ CD,isdefinedbythevectorrepresentedbythethird
→
sideof the completed triangle, thatis → AD(Figure 7.4).Thus,
→
AB + → CD = → AD
→
Similarly,ifABisdenotedbya,
→ CDisdenotedbyb, ADisdenotedbyc,thenwehave
c=a+b
Wenotethattoaddaandbatriangleisformedusingaandbastwoofthesidesinsuch
a way that the head of one vector touches the tail of the other as shown in Figure 7.4.
The suma +bisthen represented by the third side.
→
→

226 Chapter 7 Vectors
A
Figure7.3
a
B
Two vectors, → ABand → CD.
C
b
Itispossible toprove the following rules:
D
A
Figure7.4
AB
a
B
c
AD
C
b
CD
Addition ofthe two vectors ofFigure 7.3using
the triangle law.
a +b = b +a vector addition iscommutative
a + (b +c) = (a +b) +c vector addition isassociative
D
ToseewhyitisappropriatetoaddvectorsusingthetrianglelawconsiderExamples7.1
and 7.2.
Engineeringapplication7.1
Routingofanautomatedvehicle
Automatedvehiclesareacommonfeatureofthemodernwarehouse.Sometimesthey
are guided by tracks set into the factory floor. Consider the case of an automated
vehiclecarryingelectricalcomponentsfromstoresatAtoworkersatCasillustrated
inFigure 7.5.
ThevehiclemayarriveatCeitherdirectly,orviapointB.ThemovementfromA
toBcanberepresentedbyavector → ABknownasadisplacementvector,whosemagnitude
is the distance between points A and B. Similarly, movement from B to C is
representedby → BC,andmovementdirectlyfromAtoCisrepresentedby → AC.Clearly,
sinceA,BandCarefixedpoints,thesedisplacementvectorsarefixedtoo.Sincethe
headofthevector → ABtouchesthetailofthevector → BCwearereadytousethetriangle
lawof vector addition tofind the combined effect of the two displacements:
→
AB + → BC = → AC
Thismeansthatthecombinedeffectofdisplacements → ABand → BCisthedisplacement
→
AC. We say that → AC istheresultant of → ABand → BC.
C
AC
BC
A
AB
B
Figure7.5
Routing ofan automated vehicle from storesat
Ato workersat C.

7.2 Vectors and scalars: basic concepts 227
In considering motion from point A to point B, the vector → AB is called a displacement
vector.
Engineeringapplication7.2
Resultantoftwoforcesactingonabody
Sometimesitisusefultobeabletocalculatetheresultantforcewhenseveraldifferent
forces act on a body. Consider the simplest case when there aretwo forces.
AforceF 1
of2Nactsverticallydownwards,andaforceF 2
of3Nactshorizontally
to the right, upon the body shown in Figure 7.6. Translating F 1
until its tail touches
the head of F 2
we can apply the triangle law to find the combined effect of the two
forces.This isasingle forceRknown as the resultant ofF 2
andF 1
. We write
R=F 2
+F 1
andsaythatRisthevectorsumofF 2
andF 1
.Theresultantforce,R,actsatanangle
of θ to the vertical, where tanθ = 3/2, and has a magnitude given by Pythagoras’s
theorem as √ 2 2 +3 2 = √ 13 N.
F 2
u
F 1
R
u
Figure7.6
Resultant force,R, producedbyavertical force,F 1 ,anda
horizontalforce,F 2 .
Engineeringapplication7.3
Resolvingaforceintotwoperpendiculardirections
Inthepreviousexamplewesawthattwoforcesactinguponabodycanbereplacedby
asingleforcewhichhasthesameeffect.Equivalentlywecanconsiderasingleforce
astwoforcesactingatrightanglestoeachother.ConsidertheforceFinFigure7.7.
It can be replaced by two forces, one of magnitude |F|cosθ and one of magnitude
|F|sin θ asshown.WesaythattheforceFhasbeenresolvedintotwoperpendicular
components.
For example, Figure 7.8 shows a force of 5 N acting at an angle of 30 ◦ to thex
axis. It can be resolved into two components. The first component is directed along
the x axis and has magnitude 5cos30 ◦ N. The second component is perpendicular
to the first and has magnitude 5sin30 ◦ N. These two components combine to have
the sameeffect asthe original5Nforce.
➔

228 Chapter 7 Vectors
F sin u
u
F
F cos u
Figure7.7
Aforce F can be resolved into two
perpendicular components.
5 sin 30°
30°
5
5 cos 30°
Figure7.8
The5Nforce can be resolved intotwo
perpendicular components.
x
Example7.1 Vectors p,q,r andsform the sidesofthe squareshown inFigure 7.9.Express
(a) pintermsof r
(b) sintermsof q
→
(c) diagonalBDintermsof qandr
Solution (a) The vector p has the same length as r but has the opposite direction. Therefore
p=−r.
(b) Vectorshasthesamelengthasqbuthastheoppositedirection.Therefores = −q.
(c) The head of BC → →
coincides with the tail of CD. Therefore, by the triangle law of
addition, the third sideoftriangle BCDrepresents the sum of → BCand → CD, thatis
→
BD = → BC + → CD
=q+r
7.2.4 Vectorsubtraction
Subtraction of one vector from another is performed by adding the corresponding negative
vector; thatis, ifweseeka −b, weforma + (−b).
Example7.2 ConsidertherectangleillustratedinFigure7.10withaandbasshown.Expressinterms
ofA,B, CorDthe vectors a −bandb −a.
B
p
q
C
r
a
B
C
A
s
Figure7.9
Thevectors p,q,r andsform the sides
ofasquare.
D
A
b
Figure7.10
Therectangle ABCD.
D

7.2 Vectors and scalars: basic concepts 229
B
B
AB
DB
BA
BD
A
DA
D
A
AD
D
Figure7.11
ThevectorsABand → DA.
→
Figure7.12
ThevectorsBAand → AD.
→
Solution We have
hence
b = → AD
−b = → DA
Then
a−b=a+(−b)
= → AB + → DA
= → DA + → AB by commutativity
→
DAand → ABareshown inFigure 7.11. Their sum isgiven by the triangle law, thatis
Similarly,
hence
a−b= → DA + → AB = → DB
a = → AB
−a = → BA
Then
b−a=b+(−a)
= → AD + → BA
= → BA + → AD
→
BAand → ADareshowninFigure7.12.Againtheirsumisgivenbythetrianglelaw,thatis
b−a= → BA + → AD = → BD
Example7.3 Referring to Figure 7.13, if r 1
and r 2
are as shown, find the vector b = → QP represented
by the third sideof the triangle OPQ.
Solution FromFigure 7.13 we note fromthe triangle law that
→
QP = → QO + → OP = → OP + → QO by commutativity.
→
ButQO = −r 2
andQP → = b, and so
b=r 1
−r 2

230 Chapter 7 Vectors
r 1
P
P
Q
P
PQ
Q
O
r 2
b
Q
p
O
q
r
R
O
p
q
Figure7.13
Thetwo vectors r 1 and r 2 .
Figure7.14
Thetetrahedron OPQR.
Figure7.15
Thetriangle OPQ.
Vectors do not necessarily lie in a two-dimensional plane. Three-dimensional vectors
are commonly used as isillustratedinthe following example.
Example7.4 OPQR is the tetrahedron shown in Figure 7.14. Let
→
ExpressPQ,
→ QRandRP → intermsofp, qandr.
Solution Consider the triangle OPQ shown in Figure 7.15. We note that
→
OP=p, → OQ=qand → OR=r.
→
OQ represents the third
sideofthetriangleformedwhenpand → PQareplacedheadtotail.Usingthetrianglelaw
we find
→
OP + → PQ = → OQ
Therefore,
→
PQ = → OQ − → OP
=q−p
→
Similarly, QR=r−qandRP=p−r.
→
7.2.5 Multiplicationofavectorbyascalar
Ifkis any positive scalar and a is a vector thenka is a vector in the same direction as a
butktimesaslong.Ifkisanynegative scalar,kaisavectorintheoppositedirectionto
a, andktimesas long.Bywayof example, studythe vectors inFigure 7.16. Clearly 2a
is twice as long as a but has the same direction. The vector 1 b is half as long as b but
2
has the same direction asb. Itispossible toprove the following rules.
For any scalarskandl, and any vectorsaand b:
k(a+b)=ka+kb
(k+l)a=ka+la
k(la) = (kl)a
Thevectorkaissaid tobeascalarmultiple ofa.

7.2 Vectors and scalars: basic concepts 231
A
a 1 –2 b
2a
Figure7.16
Scalar multiplication ofavector.
b
C
b
M
Figure7.17
Thetriangle ABC,with midpointM
ofside AB.
a
B
→
Example7.5 In a triangle ABC, M is the midpoint of AB. Let AB be denoted by a, and BC → by b.
ExpressAC,
→ →
CA andCMinterms →
ofaandb.
Solution The situationissketched inFigure 7.17. Usingthe triangle ruleforaddition wefind
→
AB + → BC = → AC
Therefore,
→
AC=a+b
Itfollows that → CA=− → AC = −(a +b).
Again by the triangle ruleapplied totriangle CMB we find
→
CM = CB → + BM
→
→
NowBM →
= 1 BA=− 1 aand so
2 2
(
→
CM=−b+ − 1 )
2 a
=−
(b+ 1 )
2 a
7.2.6 Unitvectors
Vectors which have length 1 are called unit vectors. If a has length 3, for example,
then a unit vector in the direction of a is clearly 1 a. More generally we denote the unit
3
vectorinthedirectionabyâ.Recallthatthelengthormodulusofais |a|andsowecan
write
â = a
|a|
Note that |a|and hence 1
|a| arescalars.

232 Chapter 7 Vectors
7.2.7 Orthogonalvectors
Iftheanglebetweentwovectorsaandbis90 ◦ ,thatisaandbareperpendicular,thena
and baresaidtobeorthogonal.
EXERCISES7.2
1 ForthearbitrarypointsA,B,C,DandE,findasingle
vector which isequivalent to
→
(a) DC + CB
→ →
(b) CE + DC
→
2 Figure7.18 shows acube. Letp = → AB,q= → AD and
r = → AE.Expressthe vectors representing → BD, → ACand
→
AGin terms of p,qand r.
F
G
3 Inatriangle ABC,Mis the midpoint ofBC,andNis
→
the midpoint ofAC.Show thatNM = 1 →
2AB.
4 Considerarectangle with verticesat E,F, GandH.
SupposeEF=pand → FG → = q.Expresseachofthe
vectorsEH,
→ → →
GH, FH andGE → in terms of p andq.
5 Ifaisan arbitraryvector, represent on adiagramthe
vectorsa, 1 2 a, −a ,3a, −3aand â.
4
B
p
E
r
A q
Figure7.18
C
D
H
6 Aparticleis positionedat the origin.Two forcesact
onthe particle.Thefirstforce has magnitude 7 N
andactsin the negativexdirection.The secondforce
has magnitude 12 Nandactsin theydirection.
Calculate the magnitude anddirectionofthe resultant
force.
7 Aforce of15Nacts atanangleof65 ◦ to thexaxis.
Resolvethisforce intotwo forces, onedirected along
thexaxisandthe other directed along the
yaxis.
Solutions
1 (a)
→
DB
(b)
→
DE
a
2 q−p,q+p,q+r+p
4 q,−p,q−p,−p−q
5 SeeFigureS.16.
6 13.9 Natan angleof59.7 ◦ to the negativexaxis
7 6.34 N, 13.59N
1– a
2
– –– a
4
FigureS.16
3a
–3a
7.3 CARTESIANCOMPONENTS
Consider thex--y plane in Figure 7.19. The general point P has coordinates (x,y). We
can join the origin to P by a vectorOP, → which is called the position vector of P, which
weoftendenotebyr.Themodulusofris |r| =r,andisthelengthofOP.Itispossible
→

y
j
O
i
r
Figure7.19
Thex--y plane with
point P.
P(x, y)
M
x
7.3 Cartesian components 233
to express r in terms of the numbers ofxandy. If we denote a unit vector along thex
axis by i, and a unit vector along theyaxis by j (we usually omit the ˆ here), then it is
clearfromthedefinitionofscalarmultiplicationthatOM =xi,andMP → =yj.Itfollows
fromthe triangle law of addition that
r = → OP = →
OM + → MP=xi+yj
Clearly the vectors i and j are orthogonal. The numbersxandyare the i and j componentsof
r. Furthermore, using Pythagoras’s theorem we can deduce that
r = √ x 2 +y 2
Alternative notations which aresometimes usefulare
(
r = OP → x
=
y)
→
and
r = → OP = (x,y)
( x
When written intheseforms iscalled a columnvector and (x,y) iscalled arow
y)
vector.Toavoidconfusionwiththecoordinates (x,y)weshallnotuserowvectorshere
but they will be needed in Chapter 26. We will also use the column vector notation for
moregeneralvectors, thus,
( a
ai+bj=
b)
Wesaidearlierthatavectorcanbetranslated,maintainingitslengthanddirectionwithoutchanging
the vector itself.While this is true generally, position vectors form animportant
exception. Position vectors are constrained to their specific position and must
always remaintiedtothe origin.
Example7.6 IfAisthepointwithcoordinates(5,4)andBisthepointwithcoordinates (−3,2)find
the position vectors ofAand B, and the vector → AB. Further, find | → AB|.
( 5
Solution The position vector of A is 5i +4j = , which we shall denote by a. The position
( ) 4)
−3
vector of B is −3i + 2j = , which we shall denote by b. Application of the
2
trianglelaw totriangleOAB (Figure 7.20)gives
thatis
→
OA + → AB = → OB
a + → AB=b

234 Chapter 7 Vectors
y
5
B(–3, 2)
4
3
2
b
1
–6 –5 –4 –3 –2 –1 0
a
1 2 3 4
A(5, 4)
5 6 x
Figure7.20
Points Aand Bin thex--yplane.
Therefore,
→
AB=b−a
= (−3i +2j) − (5i +4j)
=−8i−2j
Alternatively, intermsof column vectors
b−a=
=
( ) ( −3 5
−
2 4)
( ) −8
−2
Wenotethatsubtraction(andlikewiseaddition)ofcolumnvectorsiscarriedoutcomponentbycomponent.Tofind
| → AB|wemustobtainthelengthofthevector → AB.Referring
toFigure7.20,wenotethatthisquantityisthelengthofthehypotenuseofaright-angled
triangle with perpendicular sides 8 and 2.Thatis, | → AB| = √ 8 2 +2 2 = √ 68 = 8.25.
More generallywehave the following result:
Given vectors a = → OA=a 1
i+a 2
jandb= → OB =b 1
i +b 2
j (Figure7.21), then
and
→
AB=b−a=(b 1
−a 1
)i+(b 2
−a 2
)j
| AB|=|b−a|=|(b → 1
−a 1
)i+(b 2
−a 2
)j|
√
= (b 1
−a 1
) 2 +(b 2
−a 2
) 2
Example7.7 Ifa =
( ( ) 7 −2
andb= ,
3)
5
(a) find a +b, a −b, b +aandb −a, commentingupon the results
(b) find 2a −3b
(c) find |a −b|.

7.3 Cartesian components 235
Solution (a)
a 2
O
b
a
a 1
A
b – a
b 1
Figure7.21
Thequantity √
|b−a|= (b 1 −a 1 ) 2 + (b 2 −a 2 ) 2 .
a+b=
a−b=
b+a=
b−a=
( ( ) ( 7 −2 5
+ =
3)
5 8)
x
( ( 7 −2
− =
3)
5)
( ( −2 7
+ =
5)
3)
( ) 9
−2
( 5
8)
( ( ( −2 7 −9
− =
5)
3)
2)
A
x
z
O
{
P}
z
r
B
}
y
Figure7.22
Thequantity |r| = √ x 2 +y 2 +z 2 .
x
y
y
b 2
B
We note thataddition is commutative whereas subtraction isnot.
( ( ( ( ) ( )
7 −2 14 −6 20
(b) 2a−3b=2 −3 = − =
3)
5)
6)
15 −9
(c) |a−b|=|9i−2j|= √ 9 2 + (−2) 2 = √ 85
The previous development readily generalizes to the three-dimensional case. Taking
Cartesian axes x, y and z, any point in three-dimensional space can be represented by
givingx,yandzcoordinates(Figure7.22).Denotingunitvectorsalongtheseaxesbyi,
jandk, respectively, wecan write the vectorfrom OtoP(x,y,z) as
→
OP=r=xi+yj+zk=
⎛ ⎞
x
⎝y⎠
z
Thevectors i,jand kareorthogonal.
Example7.8 Ifr =xi +yj +zk show thatthe modulus ofrisr = √ x 2 +y 2 +z 2 .
Solution Recalling Figure 7.22 we first calculate the length of OB. Now OAB is a right-angled
triangle with perpendicular sides OA = x and AB = y. Therefore by Pythagoras’s

236 Chapter 7 Vectors
theoremOBhaslength √ x 2 +y 2 .Then,applyingPythagoras’stheoremtoright-angled
triangle OBP which has perpendicular sides OBand BP =z, we find
|r|=OP=
as required.
√
OB 2 +BP 2
= √ (x 2 +y 2 )+z 2
= √ x 2 +y 2 +z 2
Ifr=xi+yj+zk then |r|= √ x 2 +y 2 +z 2
Inthreedimensions we have the following general result:
Given vectors a = → OA=a 1
i+a 2
j+a 3
kandb= → OB =b 1
i+b 2
j+b 3
k,then
and
→
AB=b−a=(b 1
−a 1
)i+(b 2
−a 2
)j+(b 3
−a 3
)k
| AB|=|b−a|=|(b → 1
−a 1
)i+(b 2
−a 2
)j+(b 3
−a 3
)k|
√
= (b 1
−a 1
) 2 +(b 2
−a 2
) 2 +(b 3
−a 3
) 2
Example7.9 Ifa = 3i −2j +k andb = −2i +j−5k, find
(a) |a| (b) â (c) |b| (d) ˆb (e)b−a (f)|b−a|
Solution (a) |a| = √ 3 2 + (−2) 2 +1 2 = √ 14
(b) â = a
|a| = 1 √
14
(3i−2j+k)= 3 √
14
i − 2 √
14
j + 1 √
14
k
(c) |b| = √ (−2) 2 +1 2 + (−5) 2 = √ 30
(d) ˆb = b
|b| = √ 1 (−2i+j−5k)= √ −2 i + √ 1 j − √ 5 k
30 30 30 30
(e) b−a=−5i+3j−6k
(f) |b−a|= √ (−5) 2 +3 2 + (−6) 2 = √ 70
7.3.1 Thezerovector
A vector, all the components of which are zero, is called a zero vector and is denoted
by 0 to distinguish it from the scalar 0. Clearly the zero vector has a length of 0; it is
unusualinthatithas arbitrarydirection.

7.3 Cartesian components 237
Engineeringapplication7.4
Robotpositions
Positionvectorsprovideausefulmeansofdeterminingthepositionofarobot.There
aremanydifferenttypesofrobotbutacommontypeusesaseriesofrigidlinksconnected
together by flexible joints. Usually the mechanism is anchored at one point.
Atypical example isillustratedinFigure 7.23.
b
c
d
Y
a
p
X
Figure7.23
Atypical robot configuration with vectors representing
the robot links.
TheanchorpointisXandthetipoftherobotissituatedatpointY.Thefinallink
is sometimes called the hand of the robot. The hand often has rotating and gripping
facilitiesanditssizerelativetotherestoftherobotisusuallyquitesmall.Eachofthe
robotlinkscanberepresentedbyavector(seeFigure7.23).Thevectordcorresponds
tothehand.Acommonrequirementinroboticsistobeabletocalculatetheposition
of the tip of the hand to ensure it does not collide with other objects. This can be
achieved by defining a set of Cartesian coordinates with origin at the anchor point
of the robot, X. Each of the link vectors can then be represented in terms of these
coordinates. For example, inthe case of the robot inFigure 7.23:
a=a 1
i+a 2
j+a 3
k
b=b 1
i+b 2
j+b 3
k
c=c 1
i+c 2
j+c 3
k
d=d 1
i+d 2
j+d 3
k
Thepositionofthetipofthehandcanbecalculatedbyaddingtogetherthesevectors.
So,
p=a+b+c+d
=(a 1
+b 1
+c 1
+d 1
)i+(a 2
+b 2
+c 2
+d 2
)j+(a 3
+b 3
+c 3
+d 3
)k
7.3.2 Linearcombinations,dependenceandindependence
Supposewehavetwovectorsaandb.Ifweformarbitraryscalarmultiplesofthese,that
isk 1
a andk 2
b, and add these together, we obtain a new vector c where c = k 1
a+k 2
b.
The vector c is said to be a linear combination of a and b. Note that scalar multiplication
and addition of vectors are the only operations allowed when forming a linear
combination. Vector c is said to depend linearly on a and b. Of course we could also
write
a = 1 k 1
c − k 2
k 1
b providedk 1 ≠ 0

238 Chapter 7 Vectors
so thatadepends linearly oncandb. Providedk 2 ≠ 0,then
b = 1 k 2
c − k 1
k 2
a
so that b depends linearly on c and a. The set of vectors {a,b,c} is said to be linearly
dependent and any one of the set can be written as a linear combination of the other
two.Ingeneral, we have the following definition:
Asetofnvectors {a 1
,a 2
,...,a n
} islinearlydependentif the expression
k 1
a 1
+k 2
a 2
+···+k n
a n
=0
canbesatisfiedbyfindingscalarconstantsk 1
,k 2
,...,k n
,notallofwhicharezero.
If the only way we can make the combination zero is by choosing all thek i
s to be
zero,thenthe given setofvectors issaidtobe linearly independent.
Example7.10 Show thatthe vectors iandjarelinearly independent.
Solution Weformtheexpressionk 1
i +k 2
j = 0andtrytochoosek 1
andk 2
sothattheequationis
satisfied. Using columnvectors wehave
( ) ( ) ( 1 0 0
k 1
+k
0 2
=
1 0)
that is (k10 ) ( ) ( ) ( 0 k1 0
+ = =
k 2
k 2
0)
The only way we can satisfy the equation is by choosingk 1
= 0 andk 2
= 0 and hence
we conclude that the vectors i and j are linearly independent. Geometrically, we note
that sincethey are perpendicular, no scalar multiple ofican give j and viceversa.
Example7.11 Thevectors
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 5 13
⎝2⎠
⎝1⎠
⎝−1⎠
3 9 21
are linearlydependentbecause, forexample
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
5 1 13 0
3⎝1⎠ −2⎝2⎠ −1⎝−1⎠ = ⎝0⎠
9 3 21 0
EXERCISES7.3
1 PandQliein thex--y plane. Find → PQ,wherePis the
point with coordinates (5,1) andQisthe point with
coordinates (−1,4).Find | → PQ|.
2 AandBlie in thex--yplane. If Ais the point (3,4)
and Bis the point (1,−5) write down the vectors
→
OA, → OB and → AB.Find aunitvector in the direction
of → AB.

7.3 Cartesian components 239
3 Ifa=4i−j+3kandb=−2i+2j−k,findunit
vectors in the directions ofa,b andb −a.
4 Ifa = 5i −2j,b = 3i −7jandc = −3i +4j,express,
in termsof iand j,
a+b a+c c−b 3c−4b
Drawdiagrams to illustrate your results.Repeat the
calculations usingcolumn vectornotation.
5 Writedown aunit vectorwhich is parallel to the line
y=7x−3.
6 Find → PQwhere Pisthe point in three-dimensional
space with coordinates (4,1,3) andQisthe point
with coordinates (1,2,4).Find the distancebetween
PandQ. Further,findthe position vectorofthe point
dividing PQ in the ratio 1:3.
7 If P,QandRhave coordinates (3,2,1), (2,1,2) and
(1,3,3),respectively, use vectors to determine which
pair ofpointsare closestto each other.
8 Considerthe robot ofExample7.4. Thelinkvectors
have the following values:
a =12i+18j+k
b=6i−3j+8k
c =3i+2j−4k
d = 0.5i −0.2j +0.6k
Calculate the length ofeach ofthe links and the
position vector ofthe tip ofthe robot.
9 Show that the vectorsa = i +j andb = −i +jare
linearly independent.
10 Show that the vectorsi,j and kare linearly
independent.
Solutions
1
→
PQ=−6i+3j | → PQ| = 6.71
2 3i+4j,i−5j,−2i−9j
−1
unitvector: √ (2i +9j)
85
3
1
√ (4i −j +3k), 1 (−2i +2j −k),
26 3
1
√
61
(−6i +3j −4k)
4 8i −9j,2i +2j, −6i +11j,−21i +40j
5
6
1
√
50
(i +7j)
→
PQ = −3i +j+k,distancefrom Pto Q = 3.32,
1
(13i +5j +13k)
4
7 PandQ
8 21.66,10.44,5.39,0.81,21.5i +16.8j +5.6k
TechnicalComputingExercises7.3
To plotadisplacement vectorbetween points(1,2)and
(3,4)in MATLAB ® wecouldtype:
X=[1 3];
Y=[2 4];
plot(X,Y)
whichwould resultin aplotcontaining asingle line
between the two points.Notice the order in whichthe
coordinatesare passed to the plotfunction. Thefirst
entry in Xandthe firstentry in Yare forthe first
location andsoon. Ifwewantedto plotasecond
displacement vector from (3,4)to (5,6)wewould
type:
X=[1 3 5];
Y=[2 4 6];
plot(X,Y)
Now consider the problemofroutingan automated
vehicle between the following positionson afactory
floor:
Doorway: A =(0,0)
Resistorbin: B =(1,1)
Capacitorbin: C = (1,-0.5)
PCBrack: D =(2,0)
Assemblyline: E =(2.5, 0)
Forsimplicityyou mayassume that the robotwill
notcollidewith any objects ifittravels directly
betweenany ofthose two points.
1 Plotthe displacement vectors forthe robotifit
enters through the doorway andvisits the resistor
bin,the PCBrackandthenthe assemblyline. How
would thisactionbe writtenin vectornotation?

240 Chapter 7 Vectors
2 Calculate the distance travelled bythe robot.
3 Now consider the problem ofvisitingallofthe
positionsA, B,C,Dand Eonetime only, with the
robot travelling the least distance. The robot must
enter through the doorway and finishat the
assembly line. Bytrialand errorproduce a solution
to the problem, usingthe computer to plot each
trial. Expressyourfinal calculated solution in
vector notation.
7.4 SCALARFIELDSANDVECTORFIELDS
Imaginealargeroomfilledwithair.Atanypoint,P,wecanmeasurethetemperature, φ,
say. The temperature will depend upon whereabouts in the room we take the measurement.
Perhaps, close to a radiator the temperature will be higher than near to an open
window. Clearly the temperature φ isafunctionofthe position ofthe point.Ifwelabel
the point by its Cartesian coordinates (x,y,z), then φ will be a function of x,y and z,
that is
φ = φ(x,y,z)
Additionally, φ may be a function of time but for now we will leave this additional
complication aside. Since temperature is a scalar what we have done is define a scalar
ateach pointP(x,y,z) inaregion. This isanexample of ascalar field.
Alternatively,supposeweconsiderthemotionofalargebodyoffluid.Ateachpoint,
fluid will be moving with a certain speed in a certain direction; that is, each small fluid
elementhasaparticularvelocity,v,dependinguponwhereaboutsinthefluiditis.Since
velocity is a vector, what we have done is define a vector at each pointP(x,y,z). We
now have a vector function ofx,y andz, known as avector field. Let us write
v = (v x
,v y
,v z
)
so that v x
,v y
and v z
arethei,j and kcomponents respectively ofv, thatis
v=v x
i+v y
j+v z
k
We note that v x
, v y
and v z
will each bescalar functions ofx,y andz.
Engineeringapplication7.5
ElectricfieldstrengthEandelectricdisplacementD
Electrostaticsisthestudyoftheforceswhichstationarypositiveandnegativeelectric
charges exert upon one another. Consider Figure 7.24 which shows a single positive
charge placed at O. The presence of this charge gives rise to an electric force field
around it. Faraday introduced the idea of lines of force to help visualize the field.
At any point, P, there exists a vector which gives the direction and magnitude of
the electrostatic force at P. Because all the lines of force emerge radially from O,
the direction of the electrostatic force is radially outwards. It can be shown that the
magnitudeoftheforceisinverselyproportionaltothesquareofthedistancefromO.

7.5 The scalar product 241
O
E
Figure7.24
Acharge atOgives riseto an electric fieldE.
Ifasecondchargeisplacedinthisfielditexperiencesaforce.Animportantquantity
is the electric field strength, E. This is a vector field which describes the force
experienced by a unit charge.
A related quantity is the electric displacement, D, also called the electric flux
density, defined as D = εE, where ε is called the permittivity of the medium in
which the field islocated. Note thatDisascalar multiple ofE.
Engineeringapplication7.6
ElectrostaticpotentialV
An important electrostatic field is the electrostatic potentialV. This is an example
of a scalar field. The difference between the potential measured at any two points in
thefieldisequaltotheworkwhichneedstobedonetomoveaunitchargefromone
point to the other. Later, in Chapter 26, we will see that the scalar fieldV is closely
relatedtothe vectorfield E.
7.5 THESCALARPRODUCT
u
b
a
Figure7.25
Two vectorsaandb
separated by
angle θ.
Givenanytwovectorsaandb,therearetwowaysinwhichwecandefinetheirproduct.
Theseareknownasthescalarproductandthevectorproduct.Asthenamessuggest,the
resultoffindingascalarproductisascalarwhereastheresultoffindingavectorproduct
isavector. The scalar product of aandbiswritten as
a · b
Thisnotationgivesrisetothealternativenamedotproduct.Itisdefinedbytheformula
a·b=|a‖b|cosθ
where θ isthe angle between the two vectors asshown inFigure 7.25.

242 Chapter 7 Vectors
From the definition of the scalar product, it is possible to show that the following
rules hold:
a·b = b·a the scalar product iscommutative
k(a·b) = (ka·b) wherekisascalar
(a +b)·c = (a·c) + (b·c) thedistributiverule
Itisimportantatthisstagetorealizethatnotationisveryimportantinvectorwork.You
should not use a× to denote the scalar product because this is the symbol we shall use
for the vector product.
Example7.12 If a and b are parallel vectors, show that a·b = |a‖b|. If a and b are orthogonal show
that theirscalar product iszero.
Solution If a and b are parallel then the angle between them is zero. Therefore a·b =
|a‖b|cos0 ◦ = |a||b|. If a and b are orthogonal, then the angle between them is 90 ◦
and a·b = |a||b|cos90 ◦ = 0.
Similarly we can show that if a and b are two non-zero vectors for which a·b = 0,
thenaandbmustbeorthogonal.
If a andbareparallel vectors, a·b = |a‖b|.
If a andbareorthogonal vectors,a·b = 0.
Animmediateconsequenceofthepreviousresultisthefollowingusefulsetofformulae:
i·i=j·j=k·k=1
i·j=j·k=k·i=0
Example7.13 Ifa =a 1
i+a 2
j+a 3
kandb =b 1
i+b 2
j+b 3
kshowthata·b =a 1
b 1
+a 2
b 2
+a 3
b 3
.
Solution We have
a·b = (a 1
i+a 2
j+a 3
k)·(b 1
i+b 2
j+b 3
k)
=a 1
i·(b 1
i+b 2
j+b 3
k)+a 2
j·(b 1
i+b 2
j+b 3
k)
+a 3
k·(b 1
i+b 2
j+b 3
k)
=a 1
b 1
i·i+a 1
b 2
i·j+a 1
b 3
i·k+a 2
b 1
j·i+a 2
b 2
j·j+a 2
b 3
j·k
+a 3
b 1
k·i+a 3
b 2
k·j+a 3
b 3
k·k
=a 1
b 1
+a 2
b 2
+a 3
b 3
as required. Thus, given two vectors in component form their scalar product is the sum
of the products of corresponding components.
The resultdeveloped inExample 7.13 isimportant and should bememorized:
Ifa =a 1
i+a 2
j+a 3
kandb =b 1
i+b 2
j+b 3
k,
thena·b =a 1
b 1
+a 2
b 2
+a 3
b 3
.

7.5 The scalar product 243
Example7.14 Ifa = 5i −3j +2kand b = −2i +4j +kfind the scalar producta·b.
Solution Usingthe previous resultwe find
a·b=(5i−3j+2k)·(−2i+4j+k)
= (5)(−2) + (−3)(4) + (2)(1)
=−10−12+2
= −20
Example7.15 Ifa =a 1
i +a 2
j +a 3
k, find
(a) a·a (b) |a| 2
Solution (a) Using the previous resultwefind
a·a = (a 1
i+a 2
j+a 3
k)·(a 1
i+a 2
j+a 3
k)
=a 2 1 +a2 2 +a2 3
(b) From Example 7.8 we know that the modulus ofr =xi +yj +zk is √ x 2 +y 2 +z 2
and therefore
√
|a| = a 2 1 +a2 2 +a2 3
so that
|a| 2 =a 2 1 +a2 2 +a2 3
We notethe generalresultthat
a·a=|a| 2
Example7.16 Ifa = 3i +j−k andb = 2i +j+2k finda·band the angle betweenaandb.
Solution We have
a·b = (3)(2) + (1)(1) + (−1)(2)
=6+1−2
= 5
Furthermore, from the definition ofthe scalar producta·b = |a||b|cos θ. Now,
|a| = √ 9+1+1= √ 11 and |b| = √ 4+1+4=3
Therefore,
cosθ = a·b
|a‖b| = 5
3 √ 11
fromwhich wededuce that θ = 59.8 ◦ or 1.04 radians.

244 Chapter 7 Vectors
Engineeringapplication7.7
Workdonebyaforce
Ifaforceisappliedtoanobjectandtheobjectmoves,thenworkisdonebytheforce.
It is possible to use vectors to calculate the work done. Suppose a constant force F
is applied and as a consequence the object moves from A to B, represented by the
displacement vectors, asshown inFigure 7.26.
Wecanresolvetheforceintotwoperpendicularcomponents,oneparalleltosand
oneperpendiculartos.Theworkdonebyeachcomponentisequaltotheproductof
itsmagnitudeandthedistancemovedinitsdirection.Thecomponentperpendicular
to s will not do any work because there is no movement in this direction. For the
component alongs, that is |F|cos θ, we find
work done = |F|cos θ|s|
From the definition of the scalar product we see that the r.h.s. of this expression is
the scalar product ofFands.
F cos u
F
F sin u
u
A
AB = s
B
Figure7.26
ThecomponentofFinthedirectionofs
isFcosθ.
The work done by a constant force F which moves an object along the vector s is
equal tothe scalar productF·s.
Example7.17 A force F = 3i + 2j − k is applied to an object which moves through a displacement
s = 2i +2j +k. Find the work done by the force.
Solution The work done isequal to
F·s=(3i+2j−k)·(2i+2j+k)
=6+4−1
= 9 joules
Engineeringapplication7.8
Movementofachargedparticleinanelectricfield
Figure7.27showstwochargedplatessituatedinavacuum.PlateAhasanexcessof
positivecharge,whileplateBhasanexcessofnegativecharge.Suchanarrangement
gives risetoan electric field.An electricfield isan example of avector field.

–
–
–
–
–
–
–
–
–
–
7.5 The scalar product 245
A
+
+
+
+
+
+
+
+
+
+
X
E
ds
E
E
B
Figure7.27
Two charged platessituated in avacuum.
InFigure7.27theelectricfieldvectorEintheregionofspacebetweenthecharged
plateshasadirectionperpendiculartotheplatespointingfromAtoB.Themagnitude
of the electric field vector is constant in this region if end effects are ignored. If
a charged particle is required to move against an electric field, then work must be
donetoachievethis.Forexample,totransportapositivelychargedparticlefromthe
surfaceofplateBtothesurfaceofplateAwouldrequireworktobedone.Thiswould
lead toan increase inpotential of the charged particle.
If s represents the displacement andV the potential it is conventional to write δs
to represent a very small change in displacement, and δV to represent a very small
change inpotential.
Ifaunitpositivechargeismovedasmalldisplacement δsinanelectricfield(Figure
7.27) then the change inpotential δV isgiven by
δV =−E·δs (7.1)
This is an example of the use of a scalar product. Notice that although E and δs are
vector quantities the change inpotential, δV, isascalar.
ConsideragainthechargedplatesofFigure7.27.Ifaunitchargeismovedasmall
displacement along the plane X,then δs isperpendicular toE. So,
δV = −E·δs = −|E‖δs|cosθ
With θ = 90 ◦ ,wefind δV = 0.ThesurfaceXisknownasanequipotentialsurface
because movement of a charged particle along this surface does not give rise to a
change in its potential. Movement of a charge in a direction parallel to the electric
field gives risetothe maximum change inpotential, as forthis case θ = 0 ◦ .
EXERCISES7.5
1 Ifa=3i−7jandb=2i+4jfinda·b,b·a,a·a
andb·b.
2 Ifa=4i+2j−k,b=3i−3j+3kand
c=2i−j−k,find
(a) a·a (b) a·b
(c) a·c (d) b·c
3 Evaluate (−13i −5j) · (−3i +4j).
4 Find the angle between the vectorsp = 7i +3j +2k
andq=2i−j+k.
5 Find the angle between the vectors 7i +j and 4j −k.
6 Findtheanglebetweenthevectors4i −2jand3i −3j.
7 Ifa=7i+8jandb=5ifinda·ˆb.
⎛ ⎞ ⎛ ⎞
3 5
8 Ifr 1 = ⎝1⎠and r 2 = ⎝1⎠ findr 1 ·r 1 ,r 1 ·r 2
2 0
andr 2 ·r 2 .
9 Given thatp = 2qsimplify p ·q, (p +5q) ·q and
(q−p)·p.

246 Chapter 7 Vectors
10 Find the modulusofa = i −j−k,the modulusof
b = 2i +j+2kand the scalar producta ·b.Deduce
the angle betweenaand b.
11 Ifa=2i+2j−kandb=3i−6j+2k,find
|a|, |b|,a·b and the angle betweenaand b.
12 Useavector methodto showthat the diagonals ofthe
rhombusshown in Figure7.28 intersect at90 ◦ .
A
B
D
Figure7.28
TherhombusABCD.
C
13 Usethe scalar product to findatwo-dimensional
vector a =a 1 i +a 2 j perpendicular to the vector
4i −2j.
14 Ifa=3i−2j,b=7i+5jandc=9i−j,showthat
a·(b−c)=(a·b)−(a·c).
15 Findthe work done bythe forceF = 3i −j+k
in moving an object through adisplacement
s=3i+5j.
16 Aforce ofmagnitude 14Nactsin the direction
i +j+k upon an object. Itcauses the object to move
from point A(2,1,0)to point B(3,3,3).Find the
work done bythe force.
17 (a) Usethe scalar product to showthat the
component ofain the direction ofbisa · ˆb,
where ˆb isaunitvector in the directionofb.
(b) Findthe component of2i +3j in the directionof
i+5j.
Solutions
1 −22, −22,58,20
2 (a) 21 (b) 3 (c) 7 (d) 6
3 19
4 47.62 ◦
5 82.11 ◦
6 18.4 ◦
7 7
8 14,16, 26
9 2|q| 2 ,7|q| 2 , −2|q| 2
10 √ 3,3, −1,101.1 ◦
11 3,7, −8,112.4 ◦
13 c(i +2j),cconstant
15 4J
16 48.5 J
17 17/ √ 26
7.6 THEVECTORPRODUCT
The result of finding the vector product of a and b is a vector of length |a||b|sinθ,
where θ is the angle between a and b. The direction of this vector is such that it is
perpendicular to a and to b, and so it is perpendicular to the plane containing a and
b (Figure 7.29). There are, however, two possible directions for this vector, but it is
conventionaltochoosetheoneassociatedwiththeapplicationoftheright-handedscrew
rule. Imagine turning a right-handed screw in the sense from a towards b as shown. A
right-handedscrewisonewhich,whenturnedclockwise,entersthematerialintowhich
it is being screwed. The direction in which the screw advances is the direction of the
required vector product. The symbol we shall use to denote the vector product is ×.
Formally, wewrite
a×b=|a‖b|sinθê

7.6 The vector product 247
a 3 b
u
b
Plane containing
a and b
a
b
Figure7.29
a ×b is perpendicular to the plane containing
aand b.Theright-handed screw ruleallows
the direction ofa ×b to be found.
a
Figure7.30
Right-handedscrew ruleallowsthe
directionofb ×ato be found.
where ê is the unit vector required to define the appropriate direction, that is ê is a unit
vectorperpendicular toaandtobinasensedefined bytheright-handedscrewrule.To
evaluate b ×a we must imagine turning the screw from the direction of b towards that
of a. The screwwill advance as shown inFigure 7.30.
We notice immediately that a ×b ≠ b ×a since their directions are different. From
the definition of the vector product, itispossible toshow thatthe following rules hold:
a ×b = −(b ×a)
a × (b +c) = (a ×b) + (a ×c)
k(a ×b) = (ka) ×b =a× (kb)
the vector product is notcommutative
thedistributiverule
wherekisascalar
Example7.18 Ifaandbare parallel show thata ×b = 0.
Solution If a and b are parallel then the angle between a and b is zero. Therefore, a × b =
|a‖b|sin0ê= 0.Notethattheanswerisstillavector,andthatwedenotethezerovector
0i +0j +0k by0, todistinguish itfromthe scalar 0.Inparticular, wenote that
i×i=j×j=k×k=0
Ifaandbareparallel,thena ×b = 0.
Inparticular:
i×i=j×j=k×k=0
Example7.19 Show thati ×j = kand find expressions forj ×kandk ×i.
Solution We note that i and j are perpendicular so that |i ×j| = |i‖j|sin90 ◦ = 1. Furthermore,
theunitvectorperpendiculartoiandtojinthesensedefinedbytheright-handedscrew
rule is k. Therefore, i ×j = k as required. Similarly you should be able to show that
j×k=iandk×i=j.
i×j=k j×k=i k×i=j
j×i=−k k×j=−i i×k=−j

248 Chapter 7 Vectors
Example7.20 Simplify (a ×b) − (b ×a).
Solution Usethe resulta ×b = −(b ×a) toobtain
(a×b)−(b×a)=(a×b)−(−(a×b))
=(a×b)+(a×b)
=2(a×b)
Example7.21 (a) Ifa =a 1
i +a 2
j +a 3
kandb =b 1
i +b 2
j +b 3
k, show that
a×b = (a 2
b 3
−a 3
b 2
)i− (a 1
b 3
−a 3
b 1
)j+ (a 1
b 2
−a 2
b 1
)k
(b) Ifa=2i+j+3kandb=3i+2j+kfinda×b.
Solution (a) a×b=(a 1
i+a 2
j+a 3
k)× (b 1
i+b 2
j+b 3
k)
=a 1
i× (b 1
i+b 2
j+b 3
k)+a 2
j× (b 1
i+b 2
j+b 3
k)
+a 3
k× (b 1
i+b 2
j+b 3
k)
=a 1
b 1
(i×i)+a 1
b 2
(i×j)+a 1
b 3
(i×k)+a 2
b 1
(j×i)+a 2
b 2
(j×j)
+a 2
b 3
(j×k)+a 3
b 1
(k×i)+a 3
b 2
(k×j)+a 3
b 3
(k×k)
Using the results of Examples 7.18 and 7.19, we find that the expression for a ×b
simplifies to
a×b = (a 2
b 3
−a 3
b 2
)i− (a 1
b 3
−a 3
b 1
)j+ (a 1
b 2
−a 2
b 1
)k
(b) Usingthe resultof part(a) we have
a ×b = ((1)(1) − (3)(2))i − ((2)(1) − (3)(3))j + ((2)(2) − (1)(3))k
=−5i+7j+k
Verifyforyourself thatb ×a = 5i −7j −k.
7.6.1 Usingdeterminantstoevaluatevectorproducts
Evaluationofavectorproductusingthepreviousformulaisverycumbersome.Amore
convenient and easily remembered method is now described. The vectors a and b are
written inthe following pattern:
∣ i j k ∣∣∣∣∣
a 1
a 2
a 3
∣b 1
b 2
b 3
This quantity is called a determinant. A more thorough treatment of determinants is
given in Section 8.7. To find the i component of the vector product, imagine crossing
out the row and column containing i and performing the following calculation on what
isleft,thatis
a 2
b 3
−a 3
b 2
Theresultingnumberistheicomponentofthevectorproduct.Thejcomponentisfound
by crossing out the row and column containing j, performing a similar calculation, but
now changing the sign of the result.Thus thejcomponent equals
−(a 1
b 3
−a 3
b 1
)

7.6 The vector product 249
The k component is found by crossing out the row and column containing k and performingthe
calculation
We have
a 1
b 2
−a 2
b 1
Ifa =a 1
i+a 2
j+a 3
kandb =b 1
i+b 2
j+b 3
k,then
∣ i j k ∣∣∣∣∣
a×b=
a 1
a 2
a 3
∣b 1
b 2
b 3
= (a 2
b 3
−a 3
b 2
)i− (a 1
b 3
−a 3
b 1
)j+ (a 1
b 2
−a 2
b 1
)k
Example7.22 Find the vector product ofa = 2i +3j +7k andb = i +2j +k.
Solution The two given vectors arerepresented inthe following determinant:
i jk
237
∣121∣
Evaluating this determinant we obtain
a×b=(3−14)i−(2−7)j+(4−3)k=−11i+5j+k
Youwillfindthat,withpractice,thismethodofevaluatingavectorproductissimpleto
apply.
7.6.2 Applicationsofthevectorproduct
The following threeexamples illustrateapplications of the vector product.
Engineeringapplication7.9
MagneticfluxdensityBandmagneticfieldstrengthH
Itispossibletomodeltheeffectofmagnetismbymeansofavectorfield.Amagnetic
fieldwithmagneticfluxdensityBisavectorfieldwhichisdefinedinrelationtothe
forceitexertsonamovingchargedparticleplacedinthefield.ConsiderFigure7.31.
If a charge q moves with velocity v in a magnetic field B it experiences a force F
given by
F=qv×B
Notethatthisforceisdefinedusingavectorproduct.Theunitofmagneticfluxdensityistheweberpersquaremetre(Wbm
−2 ),ortesla(T).Thedirectionofthisforce
is at right angles to both v and B, in a sense defined by the right-handed screw rule.
Its magnitude, or modulus, is
F =qvBsinθ
➔

250 Chapter 7 Vectors
F
q
B
v
Figure7.31
Force,F,exerted on aparticle with charge,q,when moving with
velocity,v,in a magnetic field,B.
where θ is the angle between v and B, v is the modulus of v andBis the modulus
of B.
Note thatif θ = 90 ◦ , sin θ = 1,thenB = F qv .
Theseformulaeareusefulbecausetheycanbeusedtocalculatetheforcesexerted
on a conductor in an electric motor. They are also used to analyse electricity generators
in which the motion of a conductor in a magnetic field leads to movement of
charges within the conductor, thus generating electricity.
Arelatedquantityisthemagneticfieldstrength,orthemagneticfieldintensity,
H, defined from
B=µH
µ is called the permeability of a material and has units of webers per ampere per
metre (Wb A −1 m −1 ). The units of H are amperes per metre (A m −1 ). Confirm for
yourself thatthe units match on both sides of the equation.
Engineeringapplication7.10
Magneticfieldduetoamovingcharge
A chargeqmoving with velocity v gives rise to a magnetic field with magnetic flux
densityBinitsvicinity.Asaresultofthis,anothermovingchargeplacedinthisfield
will experienceamagnetic force.Themagnetic fluxdensityis given by
B = qµ 0
4πr 2 (v׈r)
where r is a vector from the charge to the point at which B is measured, and µ 0
is a
constant called the permeability of free space.
Thisequationcanbeusedtofindthemagneticfieldduetoacurrentinawire.Supposeasmallportionofwirehaslength
δsandcontainsacurrentI.Bywriting δsasa
vectoroflength δsinthedirectionofthewire,itcanbeshownthatthecorresponding
contribution tothe magnetic fluxdensityis given by
δB = µ 0 I (δs × ˆr)
4πr2 ThisistheBiot--Savartlaw.Techniquesofintegrationarerequiredinordertocompletethecalculation,
butusingthisitispossibletoshow,forexample,thatthemagnetic
flux density a distance r from a long straight wire has magnitude B = µ 0 I
2πr .

7.6 The vector product 251
Engineeringapplication7.11
TheHalleffectinasemiconductor
A frequent requirement in the semiconductor industry is to be able to measure the
density of holes in a p-type semiconductor and the density of electrons in an n-type
semiconductor. This can be achieved by using the Hall effect. We will consider the
case of a p-type semiconductor but the derivation for an n-type semiconductor is
similar.
area, A
z
y
x
B
L
I
hole
E H
+
V
–
B
p-type
semiconductor
+ –
Meter
V H
Figure7.32
Hall effect in ap-type semiconductor.
Consider the piece of semiconductor shown in Figure 7.32. A d.c. voltage,V, is
applied to the ends of the semiconductor. This gives rise to a flow of current composed
mainly of holes as they are the majority carriers for a p-type semiconductor.
This current can be represented by a vector pointing in thexdirection and denoted
by I. A magnetic field, B, is applied to the semiconductor in the y direction. The
moving holes experience a force, F B
, per unit volume, caused by the magnetic field
given by
F B
= 1 A I ×B
where A is the cross-sectional area of the semiconductor. This causes the holes to
drift in thezdirection and so causes an excess of positive charge to appear on one
side of the semiconductor. This gives rise to a voltage known as the Hall voltage,
V H
. As this excess charge builds up it creates an electric field, E H
, in the negativez
direction,whichinturnexertsanopposingforceontheholes.Thisforceisgivenby
F E
= qp 0
E H
whereq = elementary charge = 1.60 ×10 −19 C, and p 0
= density of
holes (holes per cubic metre). Equilibrium is reached when the two forces are equal
inmagnitude, that is |F B
| = |F E
|. Now,
|F E
| =qp 0
|E H
| |F B
| = |I×B|
A
= IB A
➔

252 Chapter 7 Vectors
In equilibrium the magnitude of the electric field, |E H
|, is constant and so we can
write |E H
| = V H
, whereListhe widthof the semiconductor. Hence,
L
IB
A =qp H
0V
L
so that
p 0
= BIL
V H
qA
So, by measuring the value of the Hall voltage, it is possible to calculate the density
of the holes, p 0
, inthe semiconductor.
EXERCISES7.6
1 Evaluate
i jk
(a)
312
∣ 214 ∣
i jk
(c)
010
∣ 104 ∣
(b)
∣
(d)
∣
∣
jk
∣
ij k
−1 2 −3
−4 0 1
i
3 52
−3 −1 4
2 Ifa=i−2j+3kandb=2i−j−k,find
(a) a×b
(b) b×a
3 Ifa=i−2jandb=5i+5kfinda×b.
4 Ifa=i+j−k,b=i−jandc=2i+kfind
(a) (a×b)×c
(b) a×(b×c)
5 Ifp=6i+7j−2kandq=3i−j+4kfind|p|,|q|
and |p ×q|.Deduce the sineofthe anglebetween p
and q.
6 Forarbitraryvectors p andqsimplify
(a) (p+q)×p
(b) (p+q)×(p−q)
7 Ifc=i+jandd=2i+k,findaunitvector
perpendicular to bothcandd.Further, findthe sineof
the anglebetweencandd.
8 A, B,Care pointswith coordinates (1,2,3), (3,2,1)
and (−1,1,0),respectively. Find aunitvector
perpendicular to the planecontaining A, B andC.
9 Ifa=7i−2j−5kandb=5i+j+3k,findavector
perpendicular toaand b.
10 Ifa=7i−j+k,b=3i−j−2kand
c = 9i +j−3k, showthat
a×(b+c)=(a×b)+(a×c)
11 (a) Thearea,A,ofaparallelogramwith baseb
andperpendicular heighthisgiven byA =bh.
Showthat ifthe two non-parallelsidesofthe
parallelogramare representedby the vectorsa
andb,thenthe areais alsogiven by
A=|a×b|.
(b) Findthe area ofthe parallelogramwith sides
representedby2i +3j +k and3i +j−k.
12 Thevolume,V,ofaparallelepipedwithsidesa,band
cisgivenbyV = |a · (b ×c)|.Findthevolumeofthe
parallelepiped with sides 3i +2j +k,2i +j+k and
i+2j+4k.
13 Suppose a forceFactsthrough the pointPwith
position vector r.Themoment aboutthe origin, M,
ofthe force isameasureofthe turningeffect ofthe
force andis given by M = r ×F.Aforce of4Nacts
in the directioni +j+k,andthrough the point with
coordinates (7,1,3).Findthe momentofthe force
about the origin.
14 Inthe theory ofelectromagnetic waves an important
quantityassociatedwith the flow ofelectromagnetic
energy isthe Poyntingvector S.This isdefined as
S = E ×Hwhere Eis the electric field strength and
Hthe magnetic field strength. Suppose thatin a plane
electromagnetic wave
( ) 2πz
E=E 0 cos
λ − ωt j

7.7 Vectors ofndimensions 253
and
H = − 2πE ( )
0 2πz
cos
λωµ 0 λ − ωt i
where λ, ω,µ 0 andE 0 are constants. Findthe
Poyntingvector andconfirm thatthe directionof
energy flow isthezdirection.
Solutions
1 (a) 2i−8j+k (b) 2i+13j+8k
(c) 4i −k (d) 22i −18j +12k
8
1
√
27
(i−5j+k)
2 (a) 5i+7j+3k (b) −5i−7j−3k
3 −10i −5j +10k
4 (a) −i−3j+2k (b) i−j
5 √ 89, √ 26,48.01,0.9980
6 (a) q×p (b) 2q×p
7
1
√
6
(i −j−2k),0.775
9 −i−46j+17k
11 (b) √ 90 = 9.49
12 5
13
4
√
3
(−2i −4j +6k)
( )
14 2πE2 0
cos 2 2πz
λωµ 0 λ − ωt k,
which is avector in thezdirection.
TechnicalComputingExercises7.6
Vectorandscalarproductsarereadilycalculatedinmost
technicalcomputing languages.Forexample in
MATLAB ® wecouldcalculatethe scalar product of
a=i−2j+3kandb=2i−j−kbytyping:
a=[1 -2 3];
b=[2 -1 -1];
dot(a,b)
producingthe answer:
ans= 1
andthe vector product bythentyping:
cross(a,b)
producingthe answer:
ans =
5 7 3
1 ConfirmusingMATLAB ® ,orasimilarlanguage,that
i×j=k.
2 (a) Selecttwo different vectorsaand b and
calculate |a ×b| 2 + (a.b) 2 .
(b) Calculate |a| 2 |b| 2
(c) Whatare yourobservations about the results
from (a)and(b)? Verifyyourconclusions by
changing the vectorsaandbandrepeating
the calculations.
7.7 VECTORSOFnDIMENSIONS
The examples we have discussed have all concerned two- and three-dimensional vectors.
Our understanding has been helped by the fact that two-dimensional vectors can
be drawn in the plane of the paper and three-dimensional vectors can be visualized in
thethree-dimensionalspaceinwhichwelive.However,therearesomesituationswhen

254 Chapter 7 Vectors
the generalization to higher dimensions is appropriate, but no convenient geometrical
interpretation is available. Nevertheless, many of the concepts we have discussed are
still applicable. For example, wecan introduce the four-dimensional vectors
⎛ ⎞
⎛ ⎞
3 1
a = ⎜1
⎟
⎝2
⎠ and b= ⎜0
⎟
⎝3
⎠
4 1
It is natural to define the magnitude, or norm, of a as √ 3 2 +1 2 +2 2 +4 2 = √ 30 and
the scalar product of a and b as a·b = (3)(1) + (1)(0) + (2)(3) + (4)(1) = 13. An
n-dimensional vector will havencomponents. Operations such as addition, subtraction
and scalar multiplication aredefined inanobvious way.
It is also possible to define a set of variables as a vector. This turns out to be a useful
way of modelling a physical system. The system is described by means of a vector
which consists of an ordered set of variables sufficient to describe the state of the system.
Such a vector is called a state vector. This concept is explored in more detail in
Chapter 20.
Engineeringapplication7.12
Meshcurrentvector
When analysing a complex circuit it can be convenient to assign a current to each
smallindependentloopwithinthecircuit.Eachofthesecurrentsisknownasamesh
current. It is possible to collect these individual currents together to form a vector
quantity. Consider the following example.
A circuit as shown in Figure 7.33 has a set of mesh currents {I 1
,I 2
,I 3
,I 4
} from
whichwecan formacurrentvector
⎛ ⎞
I 1
I = ⎜I 2 ⎟
⎝I 3
⎠
I 4
Nogeometricalinterpretationispossiblebutneverthelessthisvectorprovidesauseful
mathematical way of handling the mesh currents. We shall see how vectors such
asthesecan bemanipulatedinSection8.12.
I 1 I 2 I 3 I 4
Figure7.33
Acircuitwith meshcurrents
shown.

Review exercises 7 255
EXERCISES7.7
1 If
⎛ ⎞
1
1
a =
⎜0
⎟
⎝1
⎠
1
⎛ ⎞
3
2
and b=
⎜1
⎟
⎝0
⎠
1
findthe norm ofa, the norm ofband a·b.Further,
findthe norm ofa −b.
2 Two non-zero vectors are mutually orthogonal iftheir
scalar product iszero. Determine which ofthe
following are mutually orthogonal.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 2 3
a = ⎜ 2
⎟
⎝ 4 ⎠ b = ⎜1
⎟
⎝0
⎠ c = ⎜0
⎟
⎝1
⎠
−1 0 0
⎛ ⎞ ⎛ ⎞
0 3
d = ⎜0
⎟
⎝7
⎠ e = ⎜ 0
⎟
⎝ −2 ⎠
2 −5
Solutions
1 2, √ 15,6, √ 7 2 aande,bandd
REVIEWEXERCISES7
1 Finda·banda×bwhen
(a) a=7i−j+k,b=3i+2j+5k
(b) a=6i−6j−6k,b=i−j−k.
2 Foratriangle ABC, express assimplyaspossiblethe
vector → AB + → BC + → CA.
3 Ifa=7i−j+2kandb=8i+j+k,find|a|,|b|and
a ·b. Deducethe cosineofthe anglebetweenaand b.
4 Ifa=6i−j+2kandb=3i−j+3k,find
|a|,|b|, |a ×b|.Deducethe sineofthe anglebetween
aandb.
5 Ifa=7i+9j−3kandb=2i−4j,findâ, ˆb, ̂ a×b.
6 Bycombining the scalar andvector productsother
types ofproductscan be defined. The triplescalar
productforthreevectors isdefined as (a ×b)·c
whichisascalar.Ifa=3i−j+2k,b=2i−2j−k,
c=3i+j,finda×band(a×b)·c.Showthat
(a×b)·c=a·(b×c).
7 Thetriplevectorproduct isdefined by (a ×b) ×c.
Find the triple vectorproduct ofthe vectors given in
Question6. Alsofind a ·c,b·c andverifythat
(a·c)b−(b·c)a=(a×b)×c
Further,finda × (b ×c) andconfirm that
a×(b×c)≠(a×b)×c.
8 Show thatthe vectorsp = 3i −2j +k,
q=2i+j−4kandr=i−3j+5kformthethree
sides ofaright-angledtriangle.
9 Find aunitvector parallelto the liney = 7x −3.Find
a unitvector paralleltoy = 2x +7.Use the scalar
product to findthe anglebetweenthese two lines.
10 An electric chargeqwhichmoves with avelocity v
produces amagnetic fieldBgiven by
B = µq v× ˆr
4π |r| 2 where µ = constant
FindBifr=3i+j−2kandv=i−2j+3k.
11 InatriangleABC, denote → ABbyc, → AC byband → CB
by a.Usethe scalar product to provethe cosine rule:
a 2 =b 2 +c 2 −2bccosA.
12 Evaluate
ij k
(a)
−4 0 −3
∣ 71 4 ∣
(b)
i jk
825
∣100∣
13 Find the area ofthe parallelogram with sides
representedby3i +5j −k and i +3j −k.
14 Find the anglebetween the vectors 7i +2j and i −3j.
15 Find aunitvector in the directionofthe linejoining
the points (2,3,4) and (7,17,1).

256 Chapter 7 Vectors
16 Show that the vectorsi −j and −3i −3j are
perpendicular.
17 Find the norm ofeach ofthe vectors
⎛ ⎞ ⎛ ⎞
7 2
⎜ 2
⎟
⎝ −1⎠ and ⎜ 1
⎟
⎝ 0 ⎠
2 −4
18 (a)Usethe scalar product to findthe value ofthe
scalar µ sothat i +j+µk isperpendicular to the
vector i +j+k.
(b)Usethevectorproductandtheresultsfrompart(a)
to findamutually perpendicular setofunitvectors
ˆv 1 , ˆv 2 and ˆv 3 , where ˆv 1 is inclined equally to the
vectorsi,jand k.
19 ThepointsA, Band C have coordinates (2,−1, −2),
(4,−1,−3)and (1,3,−1).
(a) Writedown the vectors → ABand → AC.
(b) Using the vectorproduct findaunitvector
which isperpendicular to the plane containing
A, B and C.
(c) IfDisthe point with coordinates (3,0,1),use
the scalar product to find the perpendicular
distance fromDto the plane ABC.
20 Thecondition forvectorsa,band cto be coplanar
(i.e.they liein the same plane) isa · (b ×c) = 0.
(a) Show that the vectorsa = 4i +5j +6k,
b=6i−3j−3kandc=−i+2j+2kare
notcoplanar.
(b) Givend = −i +2j + λk,findthe value of λ so
thata, b anddare coplanar.
21 Points Aand Bhave position vectorsaand b
respectively. Show that the position vector ofan
arbitrarypoint on the line AB isgiven by
r = a + λ (b −a) forsome scalar λ.Thisis the
vector equationofthe line.
22 Usevector methods to show that the threemedians of
any triangle intersect atacommon point (called the
centroid).
23 Usethe vectorproduct to findthe area ofatriangle
with vertices atthe pointswith coordinates (1,2,3),
(4,−3,2),and (8,1,1).
Solutions
1 (a) 24, −7i −32j +17k (b) 18, 0
2 0
3 √ 54, √ 66,57,0.9548
4 √ 41, √ 19, √ 154,0.4446
5
1
√
139
a,
1
√
20
b,
6 5i+7j−4k,22
1
√
2296
(−12i −6j −46k)
7 (a×b)×c=4i−12j−16k
a·c=8,b·c=4
a×(b×c)=−2i−22j−8k
1 1
9 √ (i +7j), √ (i +2j),18.4 ◦
50 5
10
µq
56 √ (i +11j +7k)
14π
12 (a) 3i−5j−4k
(b) 5j −2k
13 √ 24 = 4.90
14 87.5 ◦
15
1
√
230
(5i +14j −3k)
17 √ 58, √ 21
18 (a) µ=−2
(b) ˆv 1 = 1 √
3
(1,1,1), ˆv 2 = 1 √
6
(1,1,−2),
ˆv 3 = 1 √
2
(−1,1,0)
19 (a) (2,0,−1), (−1,4,1) (b)
1
(4,−1,8) (c) 3
9
20 (a) a · (b ×c) = 9 ≠ 0 andhence the vectors are not
coplanar
(b) λ = 31/14
23 1 2√
1106

8 Matrixalgebra
Contents 8.1 Introduction 257
8.2 Basicdefinitions 258
8.3 Addition,subtractionandmultiplication 259
8.4 Usingmatricesinthetranslationandrotationofvectors 267
8.5 Somespecialmatrices 271
8.6 Theinverseofa2×2matrix 274
8.7 Determinants 278
8.8 Theinverseofa3×3matrix 281
8.9 Applicationtothesolutionofsimultaneousequations 283
8.10 Gaussianelimination 286
8.11 Eigenvaluesandeigenvectors 294
8.12 Analysisofelectricalnetworks 307
8.13 Iterativetechniquesforthesolutionofsimultaneousequations 312
8.14 Computersolutionsofmatrixproblems 319
Reviewexercises8 321
8.1 INTRODUCTION
Matrices provide a means of storing large quantities of information in such a way that
each piece can be easily identified and manipulated. They permit the solution of large
systemsoflinearequationstobecarriedoutinalogicalandformalwaysothatcomputer
implementation follows naturally. Applications of matrices extend over many areas of
engineering including electrical network analysis and robotics.

258 Chapter 8 Matrix algebra
An example of an extremely large electrical network is the national grid in Britain.
Theequationsgoverningthisnetworkareexpressedinmatrixformforanalysisbycomputerbecausesolutionsarerequiredatregularintervalsthroughoutthedayandnightin
order to make decisions such as whether or not a power station should be connected to,
or removed from, the grid.
Toobtainthetrajectoryofarobotitisnecessarytoperformmatrixcalculationstofind
thespeedatwhichvariousmotorswithintherobotshouldoperate.Thisisacomplicated
problemasitisnecessarytoensurethatarobotreachesitsrequireddestinationanddoes
not collide with another object during its movement.
8.2 BASICDEFINITIONS
Amatrix isarectangular pattern or arrayof numbers.
For example,
⎛
A = ⎝
⎞
123
456⎠ B =
−124
( ) 1 2 −2
340.5
C = ( 1 −1 1 )
areallmatrices.Notethatweusuallyuseacapitallettertodenoteamatrix,andenclose
the array of numbers in brackets. To describe the size of a matrix we quote its number
of rows and columns in that order so, for example, an r ×s matrix has r rows and s
columns. We say the matrix hasorderr×s.
Anr×smatrix hasrrows andscolumns.
Example8.1 DescribethesizesofthematricesA,BandCatthestartofthissection,andgiveexamples
of matrices of order 3 ×1,3 ×2 and 4 ×2.
Solution Ahas order 3 ×3,Bhas order 2 ×3 andC has order 1 ×3.
⎛ ⎞
⎛ ⎞
−1
1 2
⎝−2⎠ isa3 ×1 matrix ⎝3 4⎠ isa3 ×2 matrix
3
5 6
and ⎛ ⎞
−1 −1
⎜ ⎜⎝ −1 2
⎟
2 −0.5⎠ 1 0
isa4 ×2 matrix
More generally, ifthe matrixAhasmrows andncolumns wecan write
⎛
⎞
a 11
a 12
... a 1n
a 21
a 22
... a 2n
A = ⎜
⎝
.
.
. .. .
⎟
⎠
a m1
a m2
... a mn

8.3 Addition, subtraction and multiplication 259
wherea ij
representsthenumberorelementintheithrowand jthcolumn.Amatrixwith
a singlecolumn can alsobe regarded asacolumn vector.
Theoperationsofaddition,subtractionandmultiplicationaredefineduponmatrices
and these areexplained inSection 8.3.
8.3 ADDITION,SUBTRACTIONANDMULTIPLICATION
8.3.1 Matrixadditionandsubtraction
Twomatricescanbeadded(orsubtracted)iftheyhavethesameshapeandsize,thatisthe
sameorder.Theirsum(ordifference)isfoundbyadding(orsubtracting)corresponding
elements as the following example shows.
Example8.2 If
A =
( ) 1 5 −2
31 1
and B=
( ) 120
−1 1 4
findA+BandA−B.
Solution
( ) ( ) ( )
1 5 −2 120 2 7 −2
A +B= + =
31 1 −1 1 4 22 5
( ) ( ) ( )
1 5 −2 120 0 3 −2
A−B= − =
31 1 −1 1 4 4 0 −3
Ontheotherhand,thematrices
they have different orders.
( ) ( 1 2 3
and cannotbeaddedorsubtractedbecause
0 1 1)
Example8.3 IfC =
( ) a b
andD=
c d
( ) e f
showthatC+D=D+C.
g h
Solution C+D =
D+C =
( ) ( ) ( )
a b e f a +e b+f
+ =
c d g h c+g d+h
( ) ( ) ( )
e f a b e +a f+b
+ =
g h c d g+c h+d
Nowa +e is exactly the same ase +a because addition of numbers is commutative.
The same observation can be made ofb + f,c +gandd+h. HenceC +D =D +C.
The addition of these matrices is therefore commutative. This may seem an obvious
statementbut weshall shortlymeet matrix multiplication which isnotcommutative, so
ingeneral commutativity should not be simplyassumed.

260 Chapter 8 Matrix algebra
The resultobtained inExample 8.3 istruemore generally:
Matrix addition iscommutative, thatis
A+B=B+A
Itis alsoeasytoshow that
Matrix addition isassociative, that is
A+(B+C)=(A+B)+C
8.3.2 Scalarmultiplication
GivenanymatrixA,wecanmultiplyitbyanumber,thatisascalar,toformanewmatrix
of the same order asA. This multiplication is performed by multiplying every element
ofAby the number.
Example8.4 If
⎛ ⎞
1 3
A = ⎝−2 1⎠
0 1
find 2A, −3Aand 1 2 A.
⎛ ⎞ ⎛ ⎞
1 3 2 6
Solution 2A=2⎝−2 1⎠ = ⎝−4 2⎠
0 1 0 2
⎛ ⎞
1 3
⎛ ⎞
−3 −9
−3A = −3⎝−2 1⎠ = ⎝ 6 −3⎠
0 1 0 −3
and
⎛
1
⎛ ⎞
1
2 A = 1 1 3
2
⎝−2 1⎠ =
−1
2 ⎜
0 1 ⎝
0
3
⎞
2
1
2⎟
⎠
1
2

8.3 Addition, subtraction and multiplication 261
Ingeneral wehave
⎛
⎞ ⎛
⎞
a 11
a 12
... a 1n
ka 11
ka 12
... ka 1n
a 21
a 22
... a 2n
IfA= ⎜
⎝
.
.
. ..
. ..
⎟
⎠ thenkA = ka 21
ka 22
... ka 2n
⎜
⎝
.
.
. ..
. ..
⎟
⎠
a m1
a m2
... a mn
ka m1
ka m2
... ka mn
8.3.3 Matrixmultiplication
Matrix multiplication is defined in a special way which at first seems strange but is in
factveryuseful.IfAisa p ×qmatrixandBisanr×smatrixwecanformtheproduct
AB only ifq =r; that is, only if the number of columns inAis the same as the number
of rows inB. The product isthen a p ×smatrixC, thatis
C=AB where Aisp×q
Bisq×s
Cisp×s
Example8.5 GivenA = ( 4 2 ) andB=
Solution Ahas size1 ×2
Bhassize2×3
( ) 3 7 6
can the productAB beformed?
52 −1
Because the number of columns inAis the same as the number of rows in B, we can
formtheproductAB.Theresultingmatrixwillhavesize1 ×3becausethereisonerow
inAand three columns inB.
SupposewewishtofindABwhenA = ( 4 2 ) andB=
( 3
7)
.Ahassize1 ×2andBhas
size2 ×1andsowecanformtheproductAB. Theresultwillbea1 ×1matrix,thatis
a singlenumber. We perform the calculation as follows:
AB = ( 4 2 )( )
3
=4×3+2×7=12+14=26
7
Note that we have multiplied elements in the row ofAwith corresponding elements in
the column ofB, and added the results together.
⎛ ⎞
Example8.6 FindCD whenC = ( 192 ) 2
andD= ⎝6⎠.
8
⎛ ⎞
Solution CD = ( 192 ) 2
⎝6⎠ =1×2+9×6+2×8=2+54+16=72
8
Let us now extend this idea to general matrices A and B. Suppose we wish to findC
whereC =AB.Theelementc 11
isfoundbypairingeachelementinrow1ofAwiththe

262 Chapter 8 Matrix algebra
Example8.7 IfA =
correspondingelementincolumn1ofB.Thepairsaremultipliedtogetherandthenthe
resultsareaddedtogivec 11
.Similarly,tofindtheelementc 12
,eachelementinrow1of
Aispairedwiththecorrespondingelementincolumn2ofB.Again,thepairedelements
are multiplied together and the results are added to form c 12
. Other elements ofC are
foundinasimilarway.Ingeneraltheelementc ij
isfoundbypairingelementsintheith
rowofAwiththoseinthe jthcolumnofB.Thesearemultipliedtogetherandtheresults
are added togivec ij
. Consider the following example.
( ) 1 2
andB=
4 3
Solution We can form the product
( ) ( ) 1 2 5
C=AB=
4 3 −3
( ) 5
find, ifpossible, the matrixC whereC =AB.
−3
↑ ↑
2×2 2×1
because the number of columns inA, that is 2, is the same as the number of rows inB.
The size of the product is found by inspecting the number of rows in the first matrix,
which is 2, and the number of columns in the second, which is 1. These numbers give
the number of rows and columns respectively inC. ThereforeC will be a 2 ×1 matrix.
To find the elementc 11
we pair the elements in the first row ofAwith those in the
first column ofB, multiplyand then add these together. Thus
c 11
=1×5+2×−3=5−6=−1
Similarly,tofindtheelementc 21
wepairtheelementsinthesecondrowofAwiththose
inthe first column ofB, multiply and then add these together. Thus
c 21
=4×5+3×−3=20−9=11
The complete calculation iswritten as
( )( ) ( )
1 2 5 1 ×5+2×−3
AB = =
4 3 −3 4×5+3×−3
( ) 5 −6
=
20−9
( ) −1
=
11
IfAis a p ×q matrix andBis aq ×s matrix, then the productC = AB will be a
p ×s matrix. To findc ij
we take theith row ofAand pair its elements with the jth
column ofB. The paired elements aremultiplied together and added toformc ij
.

Example8.8 IfB =
( ) 123
andC=
456
⎛ ⎞
1
⎝2⎠ findBC.
4
8.3 Addition, subtraction and multiplication 263
Solution B has order 2 ×3 andC has order 3 ×1 so clearly the productBC exists and will have
order 2 ×1.BC isformed as follows:
( ) ⎛ ⎞
( ) ( )
123
BC = ⎝1
2⎠ 1 ×1+2×2+3×4 17
=
=
456 4×1+5×2+6×4 38
4
Note that the order of the product, 2 ×1, can be determined at the start by considering
the orders ofBandC.
Example8.9 FindABwhere
⎛ ⎞
1 2
⎛ ⎞
−1
A = ⎝ 3 4⎠ and B= ⎝ 2⎠
−1 0
−1
Solution A andBhave orders 3 × 2 and 3 × 1, respectively, and consequently the product,AB,
cannot beformed.
Example8.10 Given
⎛ ⎞
1 11
A = ⎝2 10⎠ and
⎛ ⎞
0 31
B= ⎝4−10⎠
3−12 2 21
find, ifpossible,ABandBA, and comment upon the result.
Solution A andBboth have order 3 × 3 and the productsAB andBA can both be formed. Both
will have order 3 ×3.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 11 0 31 6 42
AB = ⎝2 10⎠
⎝4−10⎠ = ⎝4 52⎠
3−12 2 21 0145
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
0 31 1 11 922
BA = ⎝4−10⎠
⎝2 10⎠ = ⎝234⎠
2 21 3−12 934
ClearlyAB andBA are not the same. Matrix multiplication is not usually commutative
and wemustpay particular attention tothisdetailwhen weareworking withmatrices.
IngeneralAB ≠BA and so matrix multiplication isnot commutative.
In the product AB we say that B has been premultiplied by A, or alternatively A has
beenpostmultiplied byB.

264 Chapter 8 Matrix algebra
Example8.11 Given
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1−12 −1 29 415
A = ⎝ 3 02⎠ B = ⎝ 1 00⎠ C = ⎝231⎠
−1 35 3−21 015
findBC,A(BC),AB and (AB)C, commenting upon the result.
⎛
−1
⎞ ⎛ ⎞
2 9 415
⎛ ⎞
0 14 42
Solution BC = ⎝ 1 00⎠
⎝231⎠ = ⎝4 1 5⎠
3 −2 1 015 8 −2 18
⎛ ⎞ ⎛ ⎞
1 −1 2 0 14 42
⎛
12 9
⎞
73
A(BC) = ⎝ 3 02⎠
⎝4 1 5⎠ = ⎝16 38 162⎠
−1 3 5 8 −218 52 −21 63
⎛ ⎞ ⎛
1 −1 2 −1
⎞
29
⎛
4 −2
⎞
11
AB = ⎝ 3 02⎠
⎝ 1 00⎠ = ⎝ 3 2 29⎠
−1 3 5 3 −21 19 −12 −4
⎛
4 −2
⎞ ⎛ ⎞
11 415
⎛
12 9
⎞
73
(AB)C = ⎝ 3 2 29⎠
⎝231⎠ = ⎝16 38 162⎠
19 −12 −4 015 52 −21 63
We note thatA(BC) = (AB)C so thatinthiscase matrix multiplication isassociative.
The resultobtained inExample 8.11 isalsotrueingeneral:
Matrix multiplication isassociative:
(AB)C =A(BC)
EXERCISES8.3
1 Evaluate
( ) ( )
−1
(a) 1 4
4
(b) ( 3 7 ) ( )
3
−4
( ( ) 5 2 9
(c)
1 3)
1
( ) ( ) 1 4 5 2
(d)
−1 3 −3 4
( ) ( ) 5 1 6 −1
(e)
29 6 −29 5
⎛ ⎞
( )
1
(f) 124 ⎝3⎠
2
⎛ ⎞
(g) ( 5 −1 3 ) 3 3
⎝2 6⎠
4 1
(h) ( 1 9 ) ( )
111
222
( )( ) 1 3 2 1
(i)
−1 0 5 0
( )( ) 3 0 0 3
(j)
0 1 9 0
( ) ( ( )
1 1 2 −7 1
2 IfA= ,B = ,C = ,
3 4 1)
0 4
D = ( 321 ) ( ) 23 4
andE= find,ifpossible,
12 −1
(a)A+D,C−AandD−E
(b) AB,BA,CA,AC,DA,DB,BD,EB,BE andAE
(c) 7C, −3D andkE,wherekisascalar.

8.3 Addition, subtraction and multiplication 265
3 Plotthe pointsA,B,C with position vectors given by
( ( ( 1 2 2
v 1 = v
0)
2 = v
0)
3 =
3)
respectively. Treating these vectors asmatrices of
order 2 ×1 findthe productsMv 1 ,Mv 2 ,Mv 3 when
( ) 1 0
(a) M=
0 −1
( ) 0 1
(b)M=
1 0
( ) 0 −1
(c) M=
1 0
Ineach casedraw adiagram to illustrate the effect
uponthe vectors ofmultiplication bythe matrix.
4 FindABandBA where
⎛ ⎞
13 2
A = ⎝−10 4⎠
51 −1
⎛ ⎞
521
B = ⎝034⎠
135
5 Given thatA 2 meansthe product ofamatrixAwith
( )
itself,findA 2 4 2
whenA = .FindA
1 3
3 .
( ) ( ) 1 3 2 1
6 IfA= ,B = findAB,BA,A +B
−2 4 −4 5
and (A +B) 2 .Show that
(A+B) 2 =A 2 +AB+BA+B 2
Whyis (A +B) 2 notequaltoA 2 +2AB +B 2 ?
7 Find, ifpossible,
⎛ ⎞⎛
⎞
10 00 1
(a) ⎜00 −10
⎟⎜1
⎟
⎝01 00⎠⎝2⎠
00 01 1
⎛ ⎞⎛
⎞
0010 2
(b) ⎜ 0100
⎟⎜5
⎟
⎝−1000⎠⎝5⎠
0001 1
( ) ⎛ 2 1 36
8 Find ⎝1
3 −5⎠.
2 −57
6 7
⎛ ⎞
1
9 Given the vectorv = ⎝2⎠calculate the vectors
3
obtained whenvispremultiplied by the following
matrices:
⎛ ⎞ ⎛ ⎞
62 9
(a) ⎝ 13 2⎠
−12 −3
⎛ ⎞
131
(c) ⎝926⎠
280
⎛ ⎞
683
(e) ⎜964
⎟
⎝539⎠
252
⎞
(b)
(d)
−103
⎝ 719⎠
134
( ) 312
654
Solutions
1 (a) 15 (b) −19
( ) ( )
47
−7 18
(c) (d)
12
−14 10
( ) 1 0
(e) (f) 15
0 1
(g) (25 12) (h) (19 19 19)
(i)
( ) 17 1
−2 −1
(j)
( ) 0 9
9 0
( ) −8 0
2 (a) A +Ddoes notexist, ,
−3 0
D −E does not exist
( ) 3
(b) ,BAdoes notexist,
10
(c)
3 (a)
(c)
( ) −4 −3
,
12 16
( ) −7 5
,
−21 19
DAdoes notexist,DBdoesnot exist,
( ) 642
,
321
EBdoes notexist,BE doesnot exist,
( ) 3 53
10178
( ) ( )
−49 7 2k 3k 4k
, (−9−6−3),
0 28 k2k −k
( ( ( ) ( ( ( 1 2 2 0 0 3
, , (b) , ,
0)
0)
−3 1)
2)
2)
( ( ( 0 0 −3
, ,
1)
2)
2)

266 Chapter 8 Matrix algebra
4
5
6
⎛ ⎞ ⎛ ⎞
71723 81617
⎝−1 10 19⎠,
⎝17 4 8⎠
2410 4 23 8 9
( ) ( ) 18 14 86 78
,
7 11 39 47
( ) ( )
−10 16 0 10
, ,
−20 18 −14 8
( ) ( ) 3 4 −15 48
,
−6 9 −72 57
⎛ ⎞
1
7 (a) ⎜−2
⎟
⎝ 1⎠
1
( ) 46 29
8
29 78
⎛ ⎞
37
9 (a) ⎝ 13⎠
−6
(d)
( ) 11
28
(b)
⎛
⎜
⎝
⎞
5
5
⎟
⎠
−2
1
⎛ ⎞
8
(b) ⎝36⎠
19
⎛ ⎞
31
(e) ⎜33
⎟
⎝38⎠
18
(c)
⎛ ⎞
10
⎝31⎠
18
TechnicalComputingExercises8.3
Technical computing languages such asMATLAB ® are
usuallydesigned to perform operationson scalars,
vectorandmatrices.However,ashasbeendemonstrated
in thischapter, different mathematical rules apply when
workingwith vectors and matrices.Consider the
followingquantities:
a = −2 (a scalar quantity)
b = 7 (a scalar quantity)
( ) 1 3
c = , (a 2 by2matrix)
2 4
( ) 5 7
d = , (a 2 by 2 matrix)
6 8
⎛ ⎞
1 4
e = ⎝2 5⎠, (a 3 by2matrix)
3 6
( ) 135
f = , (a 2 by3matrix)
246
Load these asvariables in atechnical computing
language and thenattempt to perform the following
calculations andanswer the questions.
1. Find the product ofthe scalarsaandb.Isthisa
valid calculation?
2. Find the product ofthe scalar aand the matrixc.Is
thiscalculation valid? Similarlyfindthe product of
the matrixcandscalara.Do these calculations
produce the same result?
3. Find the productsofthe matricescandd,that iscd
anddc. Are these calculations valid and dothey
produce the same result?
4. Find the productsofthe matriceseand f,that isef
and fe.Are these calculations valid and dothey
produce the same result?
5. Find the productsofthe matricesd ande,that isde
anded. Are these calculations valid and dothey
produce the same result?
6. Findc 2 .Isthiscalculation valid?
7. Finde 2 .Isthiscalculation valid?
Solutions
1 Yes, the calculation is valid. Itis ascalar multiplied
byascalar.InMATLAB ® :
a*b
ans = -14
2 Yes, the calculations are valid and produce the same
result.
a*c
ans =
-2 -6
-4 -8

8.4 Using matrices in the translation and rotation of vectors 267
c*a
ans =
-2 -6
-4 -8
Multiplication ofascalar byamatrixis
commutative.Itdoes notmatter in which order the
calculation is performed. Each element in the matrix
ismultiplied bythe scalar value.
3 Thecalculations are valid but they do notproduce the
same result becausematrixmultiplication isnot
commutative.
c*d
ans =
23 31
34 46
>> d*c
ans =
19 43
22 50
Thisresult emphasizes the need forcare when
manipulating matrixequations.
4 Bothcalculations are valid but the resultsare not the
same.
e*f
ans =
f*e
ans =
9 19 29
12 26 40
15 33 51
22 49
28 64
5 de isnot valid asthere are 2 columns ind and 3 rows
ine. Itshould generate an error in atechnical
computing language. Itisworth observing the error
message andunderstanding why thishappensfor
future reference.
ed isvalid,
e*d
ans =
29 39
40 54
51 69
6 Itis valid. We takethisto mean the matrix
multiplication ofcbyitself:
c*c
ans =
7 15
10 22
Notice that thisisdifferent to taking the squareof
each element withinc. Ifwe wanted to dothat it
would be necessaryto use the ‘dot’ notation:
c.^2
ans =
1 9
4 16
Thecommandc.ˆ2means‘takeeach elementofcand
raiseitto the power of2.There is asetofoperators
within MATLAB ® which allow multiplication of
matrices onan element byelement basis.Care has to
be taken as(forexample)c ∗dandc. ∗dare both
valid operationswhich generate a3by 3 result but the
elements ofthe matrixare different. Inother words,it
could give an incorrect result but MATLAB ® would
not giveany error messages.
7 Itis notvalid. Itis only possibleto multiply amatrix
by itselfifthere are the same number ofcolumns as
rows.
8.4 USINGMATRICESINTHETRANSLATIONANDROTATION
OFVECTORS
There are many applications in engineering that require vectors to be manipulated into
newconfigurations.Onemechanismforachievingthisisbytheuseofmatrices.Wewill
introduce this topic through the example of robotics. However, it is important to note
thatthe ideas aremore generally applicable inengineering.

268 Chapter 8 Matrix algebra
Engineeringapplication8.1
Robotcoordinateframes
InChapter7wesawthatvectorsprovideausefultoolfortheanalysisoftheposition
ofrobots.Byassigningavectortoeachofthelinksthepositionvectorcorresponding
tothetipoftherobot canbecalculated (Figure8.1).Inpracticetheinverse problem
ismorelikely:calculatethelinkvectorstoachieveaparticularpositionvector.Usually
a desired position for the tip of the robot is known and link positions to achieve
this are required. The problem is made more complicated because the position of a
link depends upon the movements of all the joints between it and the anchor point.
Thesolutionofthisproblemcanbequitecomplicated,especiallywhentherobothas
several links. One way forward is to define the position of a link by its own local
set of coordinates. This is usually termed a coordinate frame because it provides a
frameofreferenceforthelink.Matrixoperationscanthenbeusedtorelatethecoordinateframes,thusallowingalinkpositiontobedefinedwithrespecttoaconvenient
coordinateframe.Acommonrequirementistobeabletorelatethelinkpositionsto
a world coordinate frame. If a robot is being used in conjunction with other machinesthentheworldcoordinateframemayhaveanoriginsomedistanceawayfrom
the robot. The advantage of defining link coordinate frames is that the position of a
linkiseasilydefinedwithinitsowncoordinateframeandthemovementofcoordinate
frames relative toeach other can be expressed by means ofmatrix equations.
b
c
d
a
p
p = a + b + c + d
Figure8.1
Arobotwithlinksrepresentedbyvectors
a, b, cand d.
8.4.1 Translationandrotationofvectors
An introduction to the mathematics involved in analysing the movement of robots can
be obtained by examining the way in which vectors can be translated and rotated using
matrix operations.
Consider the pointPwith position vector given by
⎛ ⎞
x
r = ⎝y⎠
z
In order to translate and rotate this vector it is useful to introduce an augmented vector
Vgiven by
⎛ ⎞
x
V = ⎜y
⎟
⎝z⎠ (8.1)
1

8.4 Using matrices in the translation and rotation of vectors 269
Itisthen possible todefine several matrices:
⎛
⎞
1 0 0 0
Rot(x,θ) = ⎜0 cosθ −sinθ 0
⎟
⎝0 sinθ cosθ 0⎠ (8.2)
0 0 0 1
⎛
⎞
cosθ 0 sinθ 0
Rot(y,θ) = ⎜ 0 1 0 0
⎟
⎝−sinθ 0 cosθ 0⎠ (8.3)
0 0 0 1
⎛
⎞
cosθ −sinθ 0 0
Rot(z,θ) = ⎜sinθ cosθ 0 0
⎟
⎝ 0 0 1 0⎠ (8.4)
0 0 0 1
⎛ ⎞
100a
Trans(a,b,c) = ⎜010b
⎟
⎝001c⎠ (8.5)
0001
Matrices (8.2) to (8.4) allow vectors to be rotated by an angle θ around axesx,yandz,
respectively. For example, the product Rot(x,θ)V has the effect of rotating r through
anangle θ about thexaxis. Matrix (8.5)allows a vector tobetranslatedaunits inthex
direction,bunits intheydirection andcunits inthezdirection.
It is possible to combine these matrices to calculate the effect of several operations
on a vector. In doing so, it is important to maintain the correct order of operations as
matrix multiplication isnon-commutative.
For example, the position of a vector that has first been translated and then rotated
about thexaxis can be defined by
V new
= Rot(x,θ)Trans(a,b,c)V old
Afewexamples will help toclarifythese ideas.
Example8.12 Rotate the vector
⎛ ⎞
1
r = ⎝1⎠
2
through 90 ◦ about thexaxis.
⎛ ⎞
⎛ ⎞ 1
1
Solution r old
= ⎝1⎠ V old
= ⎜1
⎟
⎝2⎠
2
1
Rot(x,90 ◦ ) =
⎛
⎞ ⎛ ⎞
1 0 0 0 10 00
⎜0 cos90 ◦ −sin90 ◦ 0
⎟
⎝0 sin90 ◦ cos90 ◦ 0⎠ = ⎜00−10
⎟
⎝01 00⎠
0 0 0 1 00 01

270 Chapter 8 Matrix algebra
V new
=
So,
⎛ ⎞
1
r new
= ⎝−2⎠
1
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
10 00 1 1
⎜00−10
⎟ ⎜1
⎟
⎝01 00⎠
⎝2⎠ = ⎜−2
⎟
⎝ 1⎠
00 01 1 1
Example8.13 Translate the vector
⎛ ⎞
1
r = ⎝3⎠
2
by
⎛ ⎞
1
⎝2⎠
3
and then rotate by 90 ◦ about theyaxis.
⎛ ⎞
⎛ ⎞ 1
1
Solution r old
= ⎝3⎠ V old
= ⎜3
⎟
⎝2⎠
2
1
To translate r old
, weform
⎛ ⎞
1001
Trans(1,2,3) = ⎜0102
⎟
⎝0013⎠
0001
Then
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1001 1 2
Trans(1,2,3)V old
= ⎜0102
⎟ ⎜3
⎟
⎝0013⎠
⎝2⎠ = ⎜5
⎟
⎝5⎠
0001 1 1
To rotate by 90 ◦ about theyaxis we require
⎛
⎞ ⎛ ⎞
cos90 ◦ 0 sin90 ◦ 0 0010
Rot(y,90 ◦ ) = ⎜ 0 1 0 0
⎟
⎝−sin90 ◦ 0 cos90 ◦ 0⎠ = ⎜ 0100
⎟
⎝−1000⎠
0 0 0 1 0001
⎛ ⎞
2
The vector ⎜5
⎟
⎝5⎠ ispremultiplied by thismatrix togive
1

8.5 Some special matrices 271
V new
= Rot(y,90 ◦ )Trans(1,2,3)V old
⎛ ⎞⎛
⎞ ⎛ ⎞
0010 2 5
= ⎜ 0100
⎟⎜5
⎟
⎝−1000⎠⎝5⎠ ⎜ 5
⎟
⎝−2⎠
0001 1 1
Hence
⎛ ⎞
5
r new
= ⎝ 5⎠
−2
8.5 SOMESPECIALMATRICES
8.5.1 Squarematrices
Amatrixwhichhasthesamenumberofrowsascolumnsiscalledasquarematrix.Thus
⎛ ⎞
123
( )
⎝−101⎠ −130
isasquare matrix, while isnot
241
321
8.5.2 Diagonalmatrices
Some square matrices have elements which are zero everywhere except on the leading
diagonal (top-lefttobottom-right). Such matrices aresaidtobediagonal. Thus
⎛ ⎞
⎛ ⎞
10 0 ( )
3000
⎝02 0⎠
1 0
⎜0200
⎟
0 b ⎝0010⎠
00−1
0000
areall diagonal matrices, whereas
⎛ ⎞
124
⎝010⎠
301
isnot.
8.5.3 Identitymatrices
Diagonal matrices which have only ones on their leading diagonals, for example
⎛ ⎞
( ) 100
1 0
and ⎝010⎠
0 1
001
arecalledidentity matrices and aredenoted by the letterI.
( ) ( )
1 0 2 44
Example8.14 FindIAwhereI = andA= and comment upon the result.
0 1 3−10
( )( ) ( )
1 0 2 44 2 44
Solution IA =
=
0 1 3−10 3−10

272 Chapter 8 Matrix algebra
TheeffectofpremultiplyingAbyIhasbeentoleaveAunaltered.Theproductisidentical
tothe originalmatrixA, and this iswhyI iscalled an identitymatrix.
In general, if A is an arbitrary matrix and I is an identity matrix of the appropriate
size,then
IA=A
IfAisasquare matrix thenIA =AI =A.
8.5.4 Thetransposeofamatrix
IfAisanarbitrarym×nmatrix,arelatedmatrixisthetransposeofA,writtenA T ,found
by interchanging the rows and columns ofA. Thus the first row ofAbecomes the first
column ofA T and so on.A T isann ×mmatrix.
( ) 1 −1
Example8.15 IfA = findA
2 4
T .
( )
Solution 1 2
A T =
−1 4
( ) 426
Example8.16 IfA = findA
187
T and evaluateAA T .
⎛ ⎞
4 1
( ) ⎛ ⎞
1
Solution A T = ⎝2 8⎠ 426
AA T = ⎝4
2 8⎠ =
187
6 7
6 7
( ) 56 62
62 114
8.5.5 Symmetricmatrices
IfasquarematrixAanditstransposeA T areidentical,thenAissaidtobeasymmetric
matrix.
⎛ ⎞
5−4 2
Example8.17 IfA = ⎝−4 6 9⎠findA T .
2 913
⎛ ⎞
5−4 2
Solution A T = ⎝−4 6 9⎠
2 913
whichisclearlyequaltoA.HenceAisasymmetricmatrix.Notethatasymmetricmatrix
issymmetrical about its leading diagonal.
8.5.6 Skewsymmetricmatrices
If a square matrixAissuchthatA T = −A thenAissaidtobe skewsymmetric.

Example8.18 IfA =
Solution We haveA T =
( ) 0 5
, findA
−5 0
T and deduce thatAisskew symmetric.
8.5 Some special matrices 273
( ) 0 −5
which isclearly equal to −A. HenceAisskew symmetric.
5 0
EXERCISES8.5
1 IfA=
(a)findA T ,
(b)findB T ,
(c)findAB,
( ) 3 1
andB=
2 6
(d)find (AB) T ,
( ) −1 4
3 8
(e)deduce that (AB) T =B T A T .
⎛ ⎞
x
2 Treating the columnvectorx = ⎝y⎠asa3×1
z
matrix, findIxwhereI is the 3 ×3 identitymatrix.
( ) a b
3 IfA= showthatAA
c d
T is asymmetric
matrix.
⎛ ⎞ ⎛ ⎞
213 1 −70
4 IfA= ⎝ 421⎠andB=
⎝0 25⎠
−132 3 45
findA T ,B T ,ABand (AB) T .
Deduce that (AB) T =B T A T .
5 Determine the type ofmatrixobtained when two
diagonal matrices are multiplied together.
Solutions
( ) 3 2
1 (a)
1 6
( ) 0 20
(c)
16 56
⎛ ⎞
x
2 ⎝y⎠
z
4
( ) −1 3
(b)
4 8
( ) 0 16
(d)
20 56
⎛ ⎞ ⎛ ⎞
24 −1 103
⎝12 3⎠,
⎝−724⎠,
31 2 055
6 IfA=
( ) a b
is skew symmetric, showthat
c d
a =d = 0,that is the diagonal elementsare zero.
( ) 1 13
7 IfA=
15 7
(a) findA T ,
(b) find (A T ) T ,
(c) deduce that (A T ) T is equal toA.
( ) 9 4
8 IfA=
3 2
(a) findA +A T andshowthat thisisasymmetric
matrix,
(b) findA −A T andshowthat thisisaskew
symmetric matrix.
9 The sumofthe elements onthe leading diagonal
ofasquarematrixis known asitstrace. Find the
trace of
( ) ( )
7 2 0 9
(a) (b)
−1 5 −1 0
⎛ ⎞ ⎛ ⎞
(c)
100
⎝010⎠
001
(d)
7 2 1
⎝8 2 3⎠
9 −1 −4
⎛ ⎞ ⎛ ⎞
11 020 11 7 5
⎝ 7 −20 15⎠,
⎝ 0 −20 21⎠
5 2125 20 15 25
5 Diagonal matrix
( ) 1 15
7 (a)
13 7
( ) 18 7
8 (a)
7 4
( ) 1 13
(b)
15 7
( ) 0 1
(b)
−1 0
9 (a)12 (b)0 (c)3 (d)5

274 Chapter 8 Matrix algebra
8.6 THEINVERSEOFA2×2MATRIX
Whenwearedealingwithordinarynumbersitisoftennecessarytocarryouttheoperationofdivision.Thus,forexample,ifweknowthat3x
= 4,thenclearlyx = 4/3.Ifwe
are given matricesAandC and know that
AB=C
how do we findB? Itmight be tempting towrite
B = C A
Unfortunately, this would be entirely wrong since division of matrices is not defined.
However, given expressions like AB = C it is often necessary to be able to find the
appropriateexpressionforB.Thisiswhereweneedtointroducetheconceptofaninverse
matrix.
IfAisasquare matrix and we can find another matrixBwith the property that
AB=BA=I
thenBissaid tobe theinverseofAand iswrittenA −1 , thatis
AA −1 =A −1 A =I
IfBis the inverse ofA, thenAis also the inverse ofB. Note thatA −1 does not mean a
reciprocal; there is no such thing as matrix division.A −1 is the notation we use for the
inverse ofA.
Multiplyingamatrix byits inverse yieldsthe identitymatrixI, thatis
AA −1 =A −1 A =I
Example8.19 IfA =
SinceAisasquarematrix,A −1 isalsosquareandofthesameorder,sothattheproducts
AA −1 andA −1 A can be formed. The term ‘inverse’ cannot be applied to a matrix which
isnotsquare.
( ) 2 1
show thatthe matrix
3 2
( ) 2 −1
isthe inverse ofA.
−3 2
Solution Forming the products
( )( ) ( )
2 1 2 −1 1 0
=
3 2 −3 2 0 1
( )( ) ( )
2 −1 2 1 1 0
=
−3 2 3 2 0 1
wesee that
( ) 2 −1
is the inverse ofA.
−3 2

8.6.1 Findingtheinverseofamatrix
For 2 ×2 matrices a simple formula exists tofind the inverse of
( ) a b
A =
c d
This formulastates
IfA=
( ) a b
thenA
c d
−1 =
1
ad−bc
8.6 The inverse of a 2 × 2 matrix 275
( ) d −b
.
−c a
Example8.20 IfA =
( ) 3 5
findA
1 2
−1 .
Solution Clearlyad −bc = 6 −5 = 1,so that
A −1 = 1 ( ) ( ) 2 −5 2 −5
=
1 −1 3 −1 3
The solution should always be checked by formingAA −1 .
Example8.21 IfA =
( ) 1 5
findA
2 4
−1 .
Solution Herewehavead −bc = 4 −10 = −6.Therefore
⎛
A −1 = 1 ( ) − 2 5
⎞
4 −5 ⎜
=
3 6⎟
−6 −2 1
⎝
1
⎠
3 −1 6
Example8.22 IfA =
( ) 2 4
findA
1 2
−1 .
1
Solution Thistime,ad −bc = 4 −4 = 0,sowhenwecometoform
ad−bc wefind1/0which
isnotdefined. We cannot form the inverse ofAinthiscase; itdoes notexist.
Clearly not all square matrices have inverses. The quantity ad − bc is obviously the
important determining factor since only if ad −bc ≠ 0 can we find A −1 . This quantity
is therefore given a special name: the determinant ofA, denoted by |A|, or detA.
Given any 2 × 2 matrix A, its determinant, |A|, is the scalar ad − bc. This is easily
remembered as
[product of ց diagonal] −[product of ւ diagonal]

276 Chapter 8 Matrix algebra
( )
a b
If A is the matrix , we write its determinant as
c d
∣ a b
c d∣ . Note that the straight
lines || indicate that we are discussing the determinant, which is a scalar, rather than
the matrix itself. If the matrixAis such that |A| = 0, then it has no inverse and is said
tobe singular. If |A| ≠ 0 thenA −1 exists andAissaidtobe non-singular.
Asingular matrixAhas |A| = 0.
Anon-singular matrixAhas |A| ≠ 0.
Example8.23 IfA =
( ) 1 2
andB=
5 0
( ) −1 2
find |A|, |B|and |AB|.
−3 1
Solution |A| =
∣ 1 2
5 0∣ = (1)(0) − (2)(5) = −10
|B| =
∣ −1 2
−3 1∣ = (−1)(1) − (2)(−3) =5
( )( ) ( )
1 2 −1 2 −7 4
AB =
=
5 0 −3 1 −5 10
|AB| = (−7)(10) − (4)(−5) = −50
We note that |A||B| = |AB|.
The resultobtained inExample 8.23 istruemore generally:
IfAandBaresquare matrices of the same order, |A||B| = |AB|.
8.6.2 Orthogonalmatrices
A non-singular square matrixAsuch thatA T = A −1 is said to be orthogonal. Consequently,
ifAisorthogonalAA T =A T A =I.
Example8.24 Find the inverse ofA =
( ) 0 −1
. Deduce thatAisan orthogonal matrix.
1 0
Solution From the formula forthe inverse of a 2 ×2 matrix wefind
A −1 = 1 ( ) ( ) 0 1 0 1
=
1 −1 0 −1 0
This isclearly equal tothe transpose ofA. HenceAisan orthogonal matrix.
To findthe inversesoflargermatricesweshallneedtostudydeterminantsfurther.This
isdone inSection8.7.

8.6 The inverse of a 2 × 2 matrix 277
EXERCISES8.6
1 IfA=
( ) 5 6
findA
−4 8
−1 .
2 Find the inverse, ifitexists,ofeach ofthe following
matrices:
( ) ( ) ( )
1 0 −1 0 2 3
(a) (b) (c)
0 1 0 −1 4 1
( ) ( ) ( )
−1 0 6 2 −6 2
(d) (e) (f)
−1 7 9 3 9 3
⎛ ⎞
1 1
(g) ⎜2
2
⎝
0 1 ⎟
⎠
2
( ) ( ) 3 0 7 8
3 IfA= andB=
−1 4 4 3
find |AB|, |BA|.
( ) a b
4 IfA= ,B =
c d
findAB, |A|, |B|, |AB|.
( ) e f
g h
Verifythat |AB| = |A||B|.
( ) 1 2
5 IfA= findA
3 4
−1 .
Find values ofthe constantsaandb
such thatA +aA −1 =bI.
( ) ( )
1 1 2 1
6 IfA= andB=
0 3 −1 3
findAB, (AB) −1 ,B −1 ,A −1 andB −1 A −1 .
Deduce that (AB) −1 =B −1 A −1 .
7 Given that the matrix
⎛
⎞
cos ωt −sin ωt 0
M = ⎝sin ωt cos ωt 0⎠
0 0 1
is orthogonal, findM −1 .
( ) a b
8 (a) IfA= andkis ascalar constant,
c d
showthat the inverse ofthe matrixkA
is 1 k A−1 .
( ) 1 1
(b) Findthe inverse of andhence write
1 0
⎛ ⎞
1 1
down the inverse of ⎜3
3
⎟
⎝1
3 0 ⎠ .
Solutions
1
64
( ) 8 −6
4 5
1
⎛
( ) ( ) 1 0 −1 0
− 1 ⎞
3
2 (a) (b) (c) ⎜ 10 10
0 1 0 −1 ⎝ 2
− 1 ⎟
⎠
5 5
⎛
( ) −1 0
(d)
− 1 − 1 ⎞
1
1 (e) No inverse (f) ⎜ 12 18
⎟
⎝ 1 1 ⎠
7 7
( )
4 6
2 −2
(g)
0 2
3 −132, −132
4
( ) ae +bg af+bh
,
ce+dg cf+dh
ad−bc,eh−fg,
(ad −bc)(eh − fg)
( ) −2 1
5 3
2 −1 a=−2,b=5
2
( ) 1 4
6 AB= , (AB)
−3 9
−1 = 1 ( ) 9 −4
,
21 3 1
B −1 = 1 ( ) 3 −1
,A
7 1 2
−1 = 1 ( ) 3 −1
3 0 1
⎛
⎞
cos ωt sinωt 0
7 ⎝−sin ωt cos ωt 0⎠
0 0 1
( ) ( ) 0 1 0 3
8 (b) ,
1 −1 3 −3

278 Chapter 8 Matrix algebra
8.7 DETERMINANTS
⎛ ⎞
a 11
a 12
a 13
IfA= ⎝a 21
a 22
a 23
⎠, the value of itsdeterminant, |A|, isgiven by
a 31
a 32
a 33
∣ ∣ ∣ ∣ ∣ ∣ ∣∣∣ a
|A| =a 22
a ∣∣∣ ∣∣∣ 23
a
11
−a 21
a ∣∣∣ ∣∣∣ 23
a
a 32
a 12
+a 21
a ∣∣∣ 22
33
a 31
a 13
33
a 31
a 32
If we choose an element ofA, a ij
say, and cross out its row and column and form the
determinant of the four remaining elements, this determinant is known as theminor of
the elementa ij
.
Amoment’s study will therefore reveal thatthe determinant ofAisgiven by
|A| = (a 11
×its minor) − (a 12
×its minor) + (a 13
×its minor)
This method of evaluating a determinant isknown asexpansion along the first row.
Example8.25 Find the determinant of the matrix
⎛ ⎞
121
A = ⎝−134⎠
512
Solution The determinant ofA, written as
121
−134
∣ 512∣
isfound by expanding along itsfirst row:
|A|=1
∣ 3 4
∣ ∣ ∣∣∣ 1 2∣ −2 −1 4
∣∣∣ 5 2∣ +1 −1 3
5 1∣
= 1(2) −2(−22) +1(−16)
= 2+44−16
= 30
Example8.26 Find the minors ofthe elements 1 and 4 inthe matrix
⎛ ⎞
723
B = ⎝103⎠
042
Solution To find the minor of 1 delete its row and column to form the determinant
∣ 2 3
4 2∣ . The
required minor istherefore 4 −12 = −8.
Similarly, the minor of 4 is
∣ 7 3
1 3∣ =21−3=18.

8.7 Determinants 279
In addition to finding the minor of each element in a matrix, it is often useful to find a
related quantity -- the cofactor of each element. The cofactor is found by imposing on
the minor a positive or negative sign depending upon its position, that is a place sign,
according tothe following rule:
+ − +
− + −
+ − +
Example8.27 If
⎛ ⎞
3 27
A = ⎝9 10⎠
3−12
find the cofactors of 9 and 7.
Solution The minor of 9 is
2 7
∣−1 2∣ = 4 − (−7) = 11, but since its place sign is negative, the
required cofactor is −11.
The minor of 7 is
∣ 9 1
3 −1∣ = −9 − 3 = −12. Its place sign is positive, so that the
required cofactor issimply −12.
8.7.1 Usingdeterminantstofindvectorproducts
Determinants can also be used to evaluate the vector product of two vectors. If a =
a 1
i +a 2
j +a 3
k and b = b 1
i +b 2
j +b 3
k, we showed in Section 7.6 that a × b is the
vector defined by
a×b = (a 2
b 3
−a 3
b 2
)i+ (a 3
b 1
−a 1
b 3
)j+ (a 1
b 2
−a 2
b 1
)k
If weconsider the expansion of the determinant given by
∣ i jk ∣∣∣∣∣
a 1
a 2
a 3
∣b 1
b 2
b 3
we find the same result. This definition is therefore a convenient mechanism for evaluating
a vector product.
Ifa =a 1
i+a 2
j+a 3
kandb =b 1
i+b 2
j+b 3
k,then
∣ i jk ∣∣∣∣∣
a×b=
a 1
a 2
a 3
∣b 1
b 2
b 3

280 Chapter 8 Matrix algebra
Example8.28 Ifa=3i+j−2kandb=4i+5kfinda×b.
Solution We have
ij k
a×b =
31−2
∣40 5∣
=5i−23j−4k
8.7.2 Cramer’srule
A useful application of determinants is to the solution of simultaneous equations. Consider
the case of three simultaneous equations inthreeunknowns:
a 11
x+a 12
y+a 13
z = b 1
a 21
x+a 22
y+a 23
z = b 2
a 31
x+a 32
y+a 33
z = b 3
Cramer’s rulestates thatx,yandzaregiven by the following ratios of determinants.
Cramer’srule:
∣ ∣ ∣ b 1
a 12
a ∣∣∣∣∣
13
a 11
b 1
a ∣∣∣∣∣
13
a 11
a 12
b ∣∣∣∣∣ 1
b 2
a 22
a 23
a 21
b 2
a 23
a 21
a 22
b 2
∣b 3
a 32
a ∣ 33
a
x =
∣ 31
b 3
a ∣ 33
a
y =
∣ 31
a 32
b
z =
3
∣ a 11
a 12
a ∣∣∣∣∣ 13
a 11
a 12
a ∣∣∣∣∣ 13
a 11
a 12
a ∣∣∣∣∣ 13
a 21
a 22
a 23
a 21
a 22
a 23
a 21
a 22
a 23
∣a 31
a 32
a ∣ 33
a 31
a 32
a ∣ 33
a 31
a 32
a 33
Note that in all cases the determinant in the denominator is identical and its elements
arethecoefficientsonthel.h.s.ofthesimultaneousequations.Whenthisdeterminantis
zero, Cramer’s method will clearly fail.
Example8.29 Solve
3x+2y−z =4
2x−y+2z =10
x−3y−4z =5
Solution We find
4 2−1
10−1 2
∣ 5−3−4∣
x =
3 2 −1
2−1 2
∣1−3−4∣
= 165
55 = 3
Verifyfor yourself thaty = −2 andz = 1.

8.8 The inverse of a 3 × 3 matrix 281
EXERCISES8.7
∣ ∣ 1 Find
∣ 4 6
∣∣∣∣∣ 13 4
∣∣∣∣∣ 2 8∣ , 672
21 0
35 −1∣ and 143
−114∣ .
2 Find
cos ωt sin ωt
∣−sin ωt cos ωt∣ .
∣ 500
∣∣∣∣∣ 900
3 Evaluate
632
∣457∣ and 070
008∣ .
⎛ ⎞
2 −17
4 IfA= ⎝0 84⎠,find |A|and |A T |.
3 64
Comment uponyour result.
5 UseCramer’s ruleto solve
(a) 2x−3y+z=0
5x+4y+z=10
2x−2y−z=−1
(b) 3x+y=−1
2x−y+z=−1
5x+5y−7z=−16
(c) 4x+y+z=13
2x−y=4
x+y−z=−3
(d) 3x+2y=1
x+y−z=1
2x+3z=−1
6 Given
⎛ ⎞
37 6
A = ⎝−21 0⎠
42 −5
(a) find |A|
(b) findthe cofactors ofthe elements ofrow 2,that is
−2,1, 0
(c) calculate
−2 × (cofactorof −2)
+1 × (cofactorof1)
+0 × (cofactorof0).
What doyou deduce?
7 Ifa=7i+11j−2kandb=6i−3j+kfinda×b.
8 Find a ×bwhen
(a) a=3i−j,b=i+j+k
(b) a=2i+j+k,b=7k
(c) a=−7j−k,b=−3i+j
Solutions
1 20,33,39
2 1
3 55,504
4 −164, −164 Note |A| = |A T |
5 (a) x=y=z=1
(b) x=−1,y=2,z=3
(c) x=2,y=0,z=5
(d) x=1,y=−1,z=−1
6 (a) −133 (b) 47, −39,22 (c) −133
7 5i−19j−87k
8 (a) −i−3j+4k
(b) 7i −14j
(c) i+3j−21k
8.8 THEINVERSEOFA3×3MATRIX
Given a 3×3 matrix,A, its inverse isfound as follows:
(1) Find the transpose ofA, by interchanging the rows and columns ofA.
(2) ReplaceeachelementofA T byitscofactor;byitsminortogetherwithitsassociated
place sign. The resulting matrix isknown as theadjoint ofA, denoted adj(A).

282 Chapter 8 Matrix algebra
(3) Finally, the inverse ofAisgiven by
A −1 = adj(A)
|A|
Example8.30 Findthe inverse of
⎛ ⎞
1−20
A = ⎝ 3 15⎠
−1 23
⎛ ⎞
13−1
Solution A T = ⎝−21 2⎠
05 3
ReplacingeachelementofA T byits cofactor, wefind
⎛ ⎞
−7 6 −10
adj(A) = ⎝−14 3 −5⎠
70 7
ThedeterminantofAisgiven by
|A|=1
∣ 1 5
∣ ∣ ∣∣∣ 2 3∣ − (−2) 3 5
∣∣∣ −1 3∣ +0 3 1
−1 2∣
Therefore,
= (1)(−7) + (2)(14)
= 21
A −1 = adj(A)
|A|
⎛ ⎞
= 1 −7 6 −10
⎝−14 3 −5⎠
21
70 7
Note thatthis solution should be checked by formingAA −1 togiveI.
It is clear that should |A| = 0 then no inverse will exist since then the quantity 1/|A| is
undefined.Recallthatsuchamatrix issaidtobesingular.
Forany squarematrixA, the following statements are equivalent:
|A|=0
Aissingular
Ahas no inverse

8.9 Application to the solution of simultaneous equations 283
EXERCISES8.8
1 Find adj(A), |A| and, ifitexists,A −1 ,if
⎛ ⎞
2 −3 1
(a) A= ⎝5 4 1⎠
2 −2 −1
⎛ ⎞
3 1 0
(b) A= ⎝2 −1 1⎠
5 5 −7
⎛ ⎞
2 −1 4
(c) A= ⎝5 −2 9⎠
3 2 −1
⎛ ⎞
10 −5 −4
2 IfP= ⎝−5 10 −3⎠,find adj(P)and |P|.
−4 −3 8
DeduceP −1 .
Solutions
1 (a) |A| = −43
⎛ ⎞
−2 −5 −7
adj(A) = ⎝ 7 −4 3 ⎠ ,
−18 −2 23
⎛ ⎞
A −1 = 1 25 7
⎝ −74 −3 ⎠
43
18 2 −23
(b) |A| = 25
⎛ ⎞
2 7 1
adj(A) = ⎝19 −21 −3⎠,
15 −10 −5
⎛ ⎞
A −1 = 1 2 7 1
⎝19 −21 −3⎠
25
15 −10 −5
(c) |A| = 0
⎛
⎞
−16 7 −1
adj(A) = ⎝ 32 −14 2 ⎠
16 −7 1
A −1 does notexist.
2 |P| = 230
⎛ ⎞
71 52 55
adj(P) = ⎝52 64 50⎠
55 50 75
⎛ ⎞
P −1 = 1 71 52 55
⎝52 64 50⎠
230
55 50 75
8.9 APPLICATIONTOTHESOLUTIONOFSIMULTANEOUS
EQUATIONS
Thematrixtechniqueswehavedevelopedallowthesolutionofsimultaneousequations
tobefound inasystematic way.
Example8.31 Useamatrix methodtosolve the simultaneousequations
2x+4y=14
x−3y=−8
(8.6)
Solution We first notethatthe systemofequationscan bewritteninmatrix form asfollows:
( ( ) 2 4 x
=
1 −3)
y
( ) 14
−8
(8.7)

284 Chapter 8 Matrix algebra
To understand this expression it is necessary that matrix multiplication has been fully
mastered, for, by multiplying outthe l.h.s.,wefind
( ( ) ( )
2 4 x 2x +4y
=
1 −3)
y 1x−3y
and the form (8.7)follows immediately.
We can writeEquation (8.7) as
AX=B (8.8)
( )
( ( )
2 4 x 14
whereAisthe matrix ,X isthe matrix andBis the matrix .
1 −3 y)
−8
( x
In order to findX = it is now necessary to makeX the subject of the equation
y)
AX =B. We can premultiply Equation (8.8) byA −1 , the inverse ofA, provided such an
inverse exists,togive
A −1 AX =A −1 B
Then, noting thatA −1 A =I, we find
that is
IX =A −1 B
X =A −1 B
usingthepropertiesoftheidentitymatrix.WehavenowmadeX thesubjectoftheequationasrequiredandweseethattofindX
wemustpremultiplyther.h.s.ofEquation(8.8)
by the inverse ofA.
Inthiscase
A −1 = 1 ( ) −3 −4
−10 −1 2
( ) 3/10 2/5
=
1/10 −1/5
and
A −1 B =
=
( ) ( ) 3/10 2/5 14
1/10 −1/5 −8
( 1
3)
( x
that is,X = =
y)
( 1
3)
, so thatx = 1 andy = 3 isthe required solution.
IfAX =BthenX =A −1 BprovidedA −1 exists.
Thistechniquecanbeappliedtothreeequationsinthreeunknownsinananalogousway.

8.9 Application to the solution of simultaneous equations 285
Example8.32 Express the following equations inthe formAX =Band hence solve them:
3x+2y−z =4
2x−y+2z =10
x−3y−4z =5
Solution Usingthe rules of matrix multiplication, wefind
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
3 2−1 x 4
⎝2−1 2⎠
⎝y⎠ = ⎝10⎠
1−3−4 z 5
whichisintheformAX =B.ThematrixAiscalledthecoefficientmatrixandissimply
the coefficients ofx,yandzinthe equations. As before,
AX=B
A −1 AX = A −1 B
IX =X=A −1 B
We musttherefore find the inverse ofAinorder tosolve the equations.
To invertAweuse the adjoint. If
⎛ ⎞
3 2−1
A = ⎝2−1 2⎠
1−3−4
then
⎛ ⎞
3 2 1
A T = ⎝ 2−1−3⎠
−1 2−4
and you should verify thatadj(A) isgiven by
⎛ ⎞
10 11 3
adj(A) = ⎝ 10 −11 −8⎠
−5 11 −7
The determinant ofAis found by expanding along the first row:
|A|=3
∣ −1 2
∣ ∣ ∣∣∣ −3 −4∣ −2 2 2
∣∣∣ 1 −4∣ −1 2 −1
1 −3∣
Therefore,
= (3)(10) − (2)(−10) − (1)(−5)
=30+20+5
= 55
A −1 = adj(A)
|A|
⎛ ⎞
= 1 10 11 3
⎝ 10 −11 −8⎠
55
−5 11 −7

286 Chapter 8 Matrix algebra
Finally, the solutionX isgiven by
⎛ ⎞ ⎛ ⎞⎛
⎞
x
X = ⎝y⎠ =A −1 B = 1 10 11 3 4
⎝10 −11 −8⎠⎝10⎠
z
55
−5 11 −7 5
⎛ ⎞
3
= ⎝−2⎠
1
that is,the solution isx = 3,y = −2 andz = 1.
EXERCISES8.9
1 Byexpressing the following equations in matrixform
andfinding an inverse matrix, solve
(a) 4x−2y=14
2x+ y=5
(b) 2x−2y=0
x+3y=−8
(c) 8x+3y=59
−2x+ y=−13
2 Solve the following equationsAX =Bbyfinding
A −1 ,ifitexists.
( ( ) ( ) 6 3 x 12
(a) =
5 2)
y 9
(b)
(c)
(d)
(e)
(f)
( ) ( ) 4 4 x 20
=
1 3)(
y 11
( ) ( 2 −1 x −4
=
3 2)(
y 1)
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
4 13 x 20
⎝2 −14⎠
⎝y⎠ = ⎝20⎠
0 15 z 20
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
4 13 x 15
⎝2 −14⎠
⎝y⎠ = ⎝12⎠
0 15 z 17
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
4 13 x 0
⎝2 −14⎠
⎝y⎠ = ⎝0⎠
0 15 z 0
Solutions
1 (a)x=3,y=−1
(b) x=−2,y=−2
(c)x=7,y=1
2 (a) x=1,y=2 (b) x=2,y=3
(c) x=−1,y=2 (d) x=2,y=0,z=4
(e) x=1,y=2,z=3 (f) x=y=z=0
8.10 GAUSSIANELIMINATION
An alternative technique for the solution of simultaneous equations is that of Gaussian
eliminationwhich weintroduce bymeansofthe following trivialexample.
Example8.33 UseGaussian eliminationtosolve
2x+3y=1
x+y=3

Solution Firstconsider the equations with a steppattern imposed as follows:
2x+3y=1 (1)
x +y =3 (2)
8.10 Gaussian elimination 287
Ouraimwillbetoperformvariousoperationsontheseequationstoremoveoreliminate
all the values underneath the step. You will probably remember from your early work
on simultaneous equations that in order to eliminate a variable from an equation, that
equationcanbemultipliedbyanysuitablenumberandthenaddedtoorsubtractedfrom
another equation. In this example we can eliminate thexterm from below the step by
multiplying the second equation by 2 and subtracting the first equation. Since the first
equationisentirelyabovethestepweshallleaveitasitstands.Thiswholeprocesswill
be writtenas follows:
R 1
R 2
→2R 2
−R 1
2x+3y=1
0x −y =5
(8.9)
where the symbolR 1
means that Equation (1) is unaltered, andR 2
→ 2R 2
−R 1
means
that Equation (2) has been replaced by 2 × Equation (2) -- Equation (1). All this may
seem to be overcomplicating a simple problem but a moment’s study of Equation (8.9)
will reveal why this ‘stepped’ form is useful. Because the value under the step is zero
wecan read offyfrom the lastline, thatis −y = 5,so that
y=−5
Knowingywe can then move up tothe first equation and substituteforytofindx.
2x +3(−5) =1
x = 8
This laststage isknown asback substitution.
Beforeweconsideranother example,let usnotesome important points:
(1) It is necessary to write down the operations used as indicated previously. This aids
checkingand provides a record ofthe working used.
(2) Theoperationsallowedtoeliminate unwanted variables are:
(a) any equation can bemultipliedby any non-zeroconstant;
(b) any equation can beadded toorsubtractedfromany other equation;
(c) equationscan beinterchanged.
It is often convenient to use matrices to carry out this method, in which case the operationsallowedarereferredtoasrowoperations.Theadvantageofusingmatricesisthat
it is unnecessary to write downx,y(and laterz) each time. To do this, wefirst form the
augmentedmatrix:
( ) 231
113
( )
( 2 3
1
so called because the coefficient matrix is augmented by the r.h.s. matrix .
1 1
3)
Itistobeunderstoodthatthisnotationmeans2x+3y = 1,andsoon,sothatwenolonger
write down x and y. Each row of the augmented matrix corresponds to one equation.

288 Chapter 8 Matrix algebra
The aim, as before, istocarryout rowoperations on the stepped form
( )
231
1 1 3
inordertoobtainvaluesofzerounderthestep.Clearlytoachievetherequiredform,the
row operations we performed earlier arerequired, thatis
( )
R 1 2 31
R 2
→2R 2
−R 1 0 −1 5
The last line means 0x −1y = 5, that isy = −5, and finally back substitution yieldsx,
as before.
This technique has other advantages in that it allows us to observe other forms of
behaviour.Weshallseethatsomeequationshaveauniquesolution,somehavenosolutions,whileothers
have an infinite number.
Example8.34 UseGaussian eliminationtosolve
2x+3y=4
4x+6y=7
Solution Inaugmentedmatrix form wehave
( )
234
4 6 7
We proceed toeliminate entries under the step:
( )
R 1 23 4
R 2
→R 2
−2R 1 0 0 −1
Study of the last line seems to imply that 0x + 0y = −1, which is clearly nonsense.
When this happens the equations have no solutions and we say that the simultaneous
equations areinconsistent.
Example8.35 UseGaussian eliminationtosolve
x+y=0
2x+2y=0
Solution Inaugmentedmatrix form wehave
( )
110
2 2 0
Eliminating entries under the stepwefind
( )
R 1 110
R 2
→R 2
−2R 1 0 0 0
This last line implies that 0x + 0y = 0. This is not an inconsistency, but we are now
observing a third type of behaviour. Whenever this happens we need to introduce what
are called free variables. The first row starts off with a non-zero x. There is now no
row which starts off with a non-zero y. We therefore say y is free and choose it to be

8.10 Gaussian elimination 289
anything we please, thatis
y = λ
λisour free choice
Then back substitution occurs as before:
x+λ=0
x=−λ
The solution is therefore x = −λ, y = λ, where λ is any number. There are thus an
infinite number of solutions, forexample
or
x=−1 y=1
x = 1 2
y = − 1 2
and so on.
Observation of the coefficient matrices in the last two examples shows that they have a
determinantofzero.Wheneverthishappensweshallfindtheequationseitherareinconsistentorhaveaninfinitenumberofsolutions.Weshallnextconsiderthegeneralization
ofthismethodtothreeequationsinthreeunknowns.
Example8.36 Solve byGaussian elimination
x−4y−2z =21
2x+y+2z =3
3x+2y−z = −2
Solution We first formthe augmentedmatrix and add the steppedpatternasindicated:
⎛
⎞
1−4−2 21
⎝ 2 1 2 3 ⎠
3 2 −1 −2
The aim is to eliminate all numbers underneath the steps by carrying out appropriate
rowoperations.Thisshouldbecarriedoutbyeliminatingunwantednumbersinthefirst
column first. We find
R 1
R 2
→R 2
−2R 1
R 3
→R 3
−3R 1
⎛
⎝
1−4−2 21
0 9 6−39
0 14 5 −65
We have combined the elimination of unwanted numbers in the first column into one
stage.We now remove unwanted numbers inthe second column:
⎛
⎞
R 1 1−4 −2 21
R 2
⎜ 0 9 6 −39 ⎟
R 3
→R 3
− 14 ⎝
9 R 2 0 0 − 13 ⎠
3 −13 3
and the elimination is complete. AlthoughR 3
→R 3
+ 14 4 R 1
would eliminate the 14, it
would reintroduce a non-zero term into the first column. It is therefore essential to use
the second row and not the first to eliminate this element. We can now read offzsince
⎞
⎠

290 Chapter 8 Matrix algebra
thelastequationstates0x +0y − 13 3 z = −13 ,thatisz = 1.Backsubstitutionofz = 1
3
in the second equation gives y = −5 and, finally, substitution of z and y into the first
equation givesx = 3.
Example8.37 Solve the following equationsby Gaussian elimination:
x−y+z =3
x+5y−5z =2
2x+y−z =1
Solution Forming the augmented matrix,wefind
⎛ ⎞
1−1 13
⎝ 1 5−52⎠
2 1 −1 1
Then, as before, we aim to eliminate all non-zero entries under the step. Starting with
those inthe first column, wefind
⎛
R 1
R 2
→R 2
−R 1
⎝ 1 −1 1 3
⎞
0 6−6−1⎠
R 3
→R 3
−2R 1 0 3 −3 −5
Then,
R 1
R 2
R 3
→2R 3
−R 2
⎛
⎝ 1 −1 1 3
0 6−6−1
0 0 0 −9
Thislastlineimpliesthat0x+0y+0z = −9,whichisclearlyinconsistent.Weconclude
that thereareno solutions.
⎞
⎠
You will see from Examples 8.36 and 8.37 that not only have all entries under the step
beenreducedtozero,butalsoeachsuccessiverowcontainsmoreleadingzerosthanthe
previousone.Wesaythesystemhasbeenreducedtoechelonform.Moregenerallythe
systemhasbeenreducedtoechelonformiffori < jthenumberofleadingzerosinrow
j islarger thanthe number inrowi. Consider Example 8.38.
Example8.38 Solve the following equationsby Gaussian elimination:
2x−y+z =2
−2x+y+z =4
6x−3y−2z = −9
Solution Forming the augmented matrix,wehave
⎛
⎝ 2 −1 1 2
⎞
−2 1 1 4 ⎠
6 −3 −2 −9

Eliminating the unwanted values inthe first column, wefind
⎛
R 1
R 2
→R 2
+R 1
⎝ 2 −1 1 2
⎞
0 0 2 6 ⎠
R 3
→R 3
−3R 1 0 0 −5 −15
8.10 Gaussian elimination 291
Entriesunderthesteparenowzero.Toreducethematrixtoechelonformwemustensure
each successive row has more leading zeros than the row before.We continue:
⎛
R 1
R 2
⎝ 2−112
⎞
0 026⎠
R 3
→ 2R 3
+5R 2 0 0 0 0
which isnow inechelon form.
In this form there is a row which starts off with a non-zeroxvalue, that is the first
row, there is a row which starts off with a non-zerozvalue, but no row which starts off
withanon-zeroyvalue.Therefore,wechooseytobethefreevariable,y = λsay.From
thesecondrowwehavez = 3andfromthefirst2x −y+z = 2,sothat2x = λ−1,that
isx = (λ −1)/2.
Engineeringapplication8.2
TheVandermondeMatrix
Many data storage devices, for example the Blu-ray disc , and data transmission
standards, for example WiMAX ® , have built-in techniques to reduce the effects of
errors which occur during normal use. These errors originate from noise and interference
and they can result in the loss of data. One such technique is the use of an
error-correctingcodetoprovideameasureofprotectionagainstdataerrors.Errorcorrecting
codes are used to encode the data when it is stored or transmitted. The
data is then decoded when it is read or received. In the case of a transmission line,
theerror-correctingcodemakesitpossibletodetecterrorsinthereceivedsignaland
tomake corrections,so thatthe errors arenotsubsequently retransmitted.
Oneimportantclassoferror-correctingcodesarebasedonReed-Solomoncodes,
details ofwhicharebeyond the scopeofthistext.However animportantmathematical
concept usedinReed-Solomoncodes is thatofaVandermondematrix.
The Vandermonde matrix can be illustrated by considering the problem of representing
a signal by a polynomial. For example, to approximate a signal f (t) by a
second-degree polynomialwewrite
f(t)≈a 0
+a 1
t+a 2
t 2
wherea 0
,a 1
anda 2
are coefficients of the polynomial which must be found. These
are found by forcing the original signal f (t) and its polynomial approximation to
agree at three different values oft, sayt 0
,t 1
andt 2
. This gives rise to the following
systemofequations:
⎛
1t 0
t 2 ⎞ ⎛ ⎞ ⎛ ⎞
0 a
⎜
⎝1t 1
t1
2 ⎟ 0
f(t 0
)
⎠ ⎝a 1
⎠ = ⎝f(t 1
) ⎠
1t 2
t2
2 a 2
f(t 2
)
➔

292 Chapter 8 Matrix algebra
ThecoefficientmatrixisknownasaVandermondematrix.Supposewewishtofind
asecond-degreepolynomialapproximationtothesignal f (t) = cos πt
2 forvaluesof
t between −1 and 1. We can do this by making the approximating polynomial and
the original signal equal at three points, sayt = −1,t = 0 andt = 1. The equations
tobe solved arethen
⎛ ⎞⎛
⎞ ⎛ ⎞ ⎛ ⎞
1−11 a 0
f(−1) 0
⎝1 00⎠⎝a 1
⎠ = ⎝ f(0) ⎠ = ⎝1⎠
1 11 a 2
f(1) 0
It is straightforward to solve this system of equations by Gaussian elimination and
obtaina 0
=1,a 1
=0 anda 2
= −1. Therefore the second-degree polynomial which
approximates f (t) = cos πt
2 is1−t2 .
8.10.1 Findingtheinversematrixusingrowoperations
AsimilartechniquecanbeusedtofindtheinverseofasquarematrixAwherethisexists.
Suppose wearegiven the matrixAand wish tofind its inverseB. Thenweknow
AB=I
that is,
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
a 11
a 12
a 13
b 11
b 12
b 13
100
⎝a 21
a 22
a 23
⎠ ⎝b 21
b 22
b 23
⎠ = ⎝010⎠
a 31
a 32
a 33
b 31
b 32
b 33
001
We form the augmentedmatrix
⎛
⎞
a 11
a 12
a 13
1 0 0
⎝a 21
a 22
a 23
0 1 0⎠
a 31
a 32
a 33
0 0 1
Nowcarryoutrowoperationsonthismatrixinsuchawaythatthel.h.s.isreducedtoa
3 ×3identitymatrix.Thematrixwhichthenremainsonther.h.s.istherequiredinverse.
Example8.39 Findthe inverse of
⎛ ⎞
−18 −2
A = ⎝−6 49 −10⎠
−4 34 −5
by rowreduction tothe identity.
Solution We form the augmentedmatrix
⎛
⎞
−1 8 −2 100
⎝−649−10 010⎠
−434 −5 001
We now carry out row operations on the whole matrix to reduce the l.h.s. to an identity
matrix. This means we must eliminate all the elements off the diagonal. Work through
the following calculationyourself:

⎛
R 1
R 2
→R 2
−6R 1
⎝
R 3
→R 3
−4R 1
⎞
−18−2 100
01 2−610⎠
02 3−401
8.10 Gaussian elimination 293
Thishasremovedalltheoff-diagonalentriesincolumn1.Toremovethoseincolumn2:
⎛
⎞
R 1
→R 1
−8R 2
−10−18 49−80
R 2
⎝ 01 2−6 10⎠
R 3
→R 3
−2R 2
00 −1 8−21
To remove those incolumn 3:
⎛
⎞
R 1
→R 1
−18R 3
−1 0 0 −95 28 −18
R 2
→R 2
+2R 3
⎝ 01 0 10−3 2⎠
R 3
00−1 8−2 1
We mustnow adjust the ‘−1’ entries toobtain the identitymatrix:
⎛
⎞
R 1
→ −R 1
100 95−28 18
R 2
⎝ 010 10 −3 2⎠
R 3
→ −R 3
001−8 2−1
Finally, the required inverse isthe matrix remaining on the r.h.s.:
⎛ ⎞
95 −28 18
⎝ 10 −3 2⎠
−8 2 −1
You should check thisresultby evaluatingAA −1 .
EXERCISES8.10
1 Solvethe followingequations byGaussian
elimination:
(a) 2x−3y=32
3x+7y=−21
(b) 2x+y−3z=−5
x−y+2z=12
7x−2y+3z=37
(c) x+y−z=1
3x−y+5z=3
7x+2y+3z=7
(d) 2x+y−z=−9
3x−2y+4z=5
−2x−y+7z=33
(e) 4x+7y+8z=2
5x+8y+13z=0
3x+5y+7z=1
2 Use Gaussian elimination to solve
x+y+z=7
x−y+2z=9
2x+y−z=1
3 Find the inversesofthe following matrices usingthe
technique ofExample 8.39:
( ) 4 1
(a)
3 2
⎛ ⎞
421
(b) ⎝ 034⎠
−113
⎛ ⎞
103
(c) ⎝ 215⎠
−721

294 Chapter 8 Matrix algebra
Solutions
1 (a) x=7,y=−6
(b) x=3,y=−5,z=2
(c) x=1−µ,y=2µ,z=µ
(d) x=−3,y=1,z=4
(e) Inconsistent
2 x=2,y=1,z=4
3 (a)
(b)
(c)
1
5
( ) 2 −1
−3 4
⎛ ⎞
5 −5 5
⎝−4 13 −16⎠
3 −6 12
⎛ ⎞
−9 6 −3
⎝−37 22 1⎠
11 −2 1
1
15
1
24
8.11 EIGENVALUESANDEIGENVECTORS
We are now in a position to examine the meaning and calculation of eigenvalues and
theircorrespondingeigenvectors.
8.11.1 Solutiontosystemsoflinearhomogeneousequations
Recall that an equation is linear when the variables occur only to the first power. For
example,
2x+3y=1 (1)
isalinear equation but
2x 2 +3y=1 (2)
isanon-linearequation due tothe term2x 2 .
Equation (1) is called inhomogeneous. When the r.h.s. of a linear equation is 0 then
the equation ishomogeneous.Forexample,
2x+3y=0 and 7x−3y=0
are both homogeneous. This section looks at the solution of systems of linear homogeneous
equations.
Consider the simultaneouslinear homogeneousequations
ax+by =0
cx+dy =0
where a, b, c and d are constants. Clearly x = 0, y = 0 is a solution. It is called the
trivial solution. Non-trivial solutions are solutions other thanx = 0, y = 0. We now
study the system to find conditions on a, b, c and d under which non-trivial solutions
exist.
Fordefinitenessweconsidertwo caseswith values ofa,b,candd given.
Case1
3x−5y =0
6x−7y =0

8.11 Eigenvalues and eigenvectors 295
Solving Case 1,forexample by Gaussian elimination, leads tox = 0,y = 0 as the only
possible solution.Thus, the only solution isthe trivialsolution.
Case2
x+y=0
2x+2y =0
This systemwas solved inExample 8.35 usingGaussian elimination toyield
x=−λ,
y=λ
where λ is any number and y is a free variable. Thus there are an infinite number of
solutions.Notethatinthissystemthesecondequation,2x+2y = 0,isamultipleofthe
first equation,x+y = 0.The second equation istwice the first equation.
We now return tothe system
ax+by =0
cx+dy =0
Asseen,dependinguponthevaluesofa,b,candd thesystemhaseitheronlythetrivial
solution or an infinite number of non-trivial solutions. For there to be non-trivial solutionsthesecondequationmustbeamultipleofthefirst.Whenthisisthecase,thencis
a multiple ofaandd is the same multiple ofb, thatis
c = αa, d = αb forsomevalueof α
Inthiscase, consider the quantityad −bc:
ad −bc = a(αb) −b(αa)
= αab− αab
= 0
Hence the condition for non-trivial solutions to exist is thatad −bc = 0. Writing the
systeminmatrix formgives
( ) ( a b x 0
=
c d)(
y 0)
or
where
AX=0
A =
( ) ( ( a b x 0
, X = , 0 =
c d y)
0)
We note that ad −bc is the determinant of A, so non-trivial solutions exist when the
determinant ofAis zero;thatis, whenAisasingular matrix.
Insummary:

296 Chapter 8 Matrix algebra
Consider the system
AX=0
If |A| = 0,the systemhas non-trivial solutions.
If |A| ≠ 0,the systemhas only the trivialsolution.
Example8.40 Decide which of the following systems of equations has non-trivial solutions:
(a) 3x+7y=0
2x−y=0
(b) 2x+y=0
6x+3y=0
Solution (a) We writethe systemas
( ) ( 3 7 x 0
=
2 −1)(
y 0)
Let
Then
A =
( ) 3 7
2 −1
|A| = 3(−1) −2(7) = −17
Since the determinant ofAisnon-zero, the systemhas only the trivialsolution.
(b) We writethe systemas
( ( ) ( 2 1 x 0
=
6 3)
y 0)
Let
A =
( ) 2 1
6 3
fromwhich |A| = 2(3) −1(6) = 0.SincethedeterminantofAis0,thesystemhas
non-trivial solutions.
Example8.41 Determine which of the following systems ofequations has non-trivial solutions:
(a) 2x+y−3z=0
x−3y+2z=0
5x−8y+3z=0
(b) 2x+y−3z=0
x−3y+2z=0
5x−7y+3z=0

8.11 Eigenvalues and eigenvectors 297
Solution (a) We have
AX=0
where
⎛ ⎞ ⎛ ⎞
2 1−3 x
A = ⎝1−3 2⎠ X = ⎝y⎠
5−8 3 z
Evaluation of |A|shows that |A| = 0 and so the systemhas non-trivial solutions.
(b) Here
⎛ ⎞
2 1−3
A = ⎝1−3 2⎠
5−7 3
from which |A| = −7.Since |A| ≠ 0 the systemhas only the trivialsolution.
EXERCISES8.11.1
1 Explain what ismeant by the trivial solution ofa
systemoflinear equations and what is meantby a
non-trivial solution.
2 Determine which ofthe following systemshas
non-trivial solutions:
(a) x−2y=0
3x−6y=0
(b) 3x+y=0
9x+2y=0
(c) 4x−3y=0
−4x+3y=0
(d) 6x−2y=0
2x − 2 3 y = 0
(e) y=2x
x=3y
3 Determine whichofthe following systemshave
non-trivial solutions:
(a) x+2y− z=0
3x+ y+2z=0
x+y =0
(b) 2x−3y−2z=0
3x+ y−3z=0
x−7y− z=0
(c) x+2y+3z=0
4x−3y− z=0
6x+ y+3z=0
(d) x+ 3z=0
x−y =0
y+2z=0
Solutions
2 (a),(c)and(d)have non-trivialsolutions. 3 (a)and(b)have non-trivialsolutions.
8.11.2 Eigenvalues
We will explain the meaning of the term eigenvalue by means of an example. Consider
the system
2x+y =λx
3x+4y=λy

298 Chapter 8 Matrix algebra
where λ is some unknown constant. Clearly these equations have the trivial solution
x = 0,y = 0.The equations may be written inmatrix formas
( ) ( 2 1 x x
= λ
3 4)(
y y)
or, usingthe usual notation,
AX=λX
We now seek values of λ so that the system has non-trivial solutions. Although it is
tempting to write (A − λ)X = 0 this would be incorrect sinceA−λ is not defined:A
is a matrix and λ is constant. Hence to progress we need to write the r.h.s. ( in a slightly
x
differentway.Tohelpusdothisweusethe2 ×2identitymatrix,I.Now λ maybe
y)
expressed as
( ) 1 0 x
λ
0 1)(
y
( x
since multiplying by the identity matrix leaves it unaltered. So λX may be written
y)
as λIX. Hence wehave
AX = λIX
which can bewritten as
AX−λIX =0
(A−λI)X =0
Notethattheexpression (A − λI)isdefinedsincebothAand λI aresquarematrices
of the same size.
WehaveseeninSection8.11.1thatforAX = 0tohavenon-trivialsolutionsrequires
|A| = 0.Hence for
(A−λI)X=0
to have non-trivial solutions requires
|A−λI|=0
Now
( ) ( ) 2 1 1 0
A−λI = − λ
3 4 0 1
( ) ( ) 2 1 λ 0
= −
3 4 0 λ
( ) 2 − λ 1
=
3 4−λ
So the condition |A − λI| = 0 gives
∣ 2 − λ 1
3 4−λ∣ = 0

Itfollows that
so that
(2−λ)(4−λ)−3 =0
λ 2 −6λ+5 =0
(λ−1)(λ−5) =0
λ=1 or 5
8.11 Eigenvalues and eigenvectors 299
Thesearethevaluesof λwhichcausethesystemAX = λX tohavenon-trivialsolutions.
They arecalledeigenvalues.
The equation
|A−λI|=0
which when written out explicitly is the quadratic equation in λ, is called the characteristicequation.
Example8.42 Find values of λfor which
x+4y = λx
2x+3y = λy
has non-trivial solutions.
Solution We write the systemas
where
AX=λX
A =
( ) ( 1 4 x
, X =
2 3 y)
To have non-trivialsolutions we require
|A−λI|=0
Now
( ) ( ) 1 4 1 0
A−λI = − λ
2 3 0 1
( ) ( ) 1 4 λ 0
= −
2 3 0 λ
( ) 1 − λ 4
=
2 3−λ
Hence
|A−λI| = (1−λ)(3−λ)−8
= λ 2 −4λ−5
To have non-trivialsolutions we require
λ 2 −4λ−5 =0
(λ+1)(λ−5) =0

300 Chapter 8 Matrix algebra
which yields
λ=−1 or 5
The given system has non-trivial solutions when λ = −1 and λ = 5. These are the
eigenvalues.
If A is a 2 × 2 matrix, the characteristic equation will be a polynomial of degree 2,
that is a quadratic equation in λ, leading to two eigenvalues. IfAis a 3 ×3 matrix, the
characteristicequationwillbeapolynomialofdegree3,thatisacubic,leadingtothree
eigenvalues. In general ann×n matrix gives rise to a characteristic equation of degree
n and hence toneigenvalues.
The characteristic equation of a square matrixAisgiven by
|A−λI|=0
Solutions of this equation are the eigenvalues of A. These are the values of λ for
whichAX = λX has non-trivial solutions.
Example8.43 Determine the characteristic equation and eigenvalues, λ, inthe system
( ( ) ( 3 1 x x
= λ
−1 5)
y y)
Solution InthisexampletheequationshavebeenwritteninmatrixformwithA =
characteristic equation is given by
|A−λI| =0
( ) ( )∣ 3 1 1 0 ∣∣∣
∣ − λ = 0
−1 5 0 1
∣ 3 − λ 1
−1 5−λ∣ = 0
(3−λ)(5−λ)+1 =0
λ 2 −8λ+16 =0
( ) 3 1
.The
−1 5
The characteristic equation is λ 2 − 8λ + 16 = 0. Solving the characteristic equation
gives
λ 2 −8λ+16 =0
(λ−4)(λ−4) =0
λ = 4 (twice)
There isone repeated eigenvalue, λ = 4.

8.11 Eigenvalues and eigenvectors 301
Example8.44 Find the eigenvalues λinthe system
( ( ) ( 4 1 x x
= λ
3 2)
y y)
Solution We form the characteristic equation, |A − λI| = 0.Now
( ) 4 − λ 1
A−λI=
3 2−λ
Then
|A−λI|=(4−λ)(2−λ)−3=λ 2 −6λ+5
Solving the characteristic equation, λ 2 −6λ +5 = 0,gives
λ=1 or 5
There aretwo eigenvalues, λ = 1, λ = 5.
The process of finding the characteristic equation and eigenvalues of a matrix has
been illustrated using 2 × 2 matrices. This same process can be applied to a square
matrix of any size.
Example8.45 Find (a)the characteristic equation (b)the eigenvalues ofAwhere
⎛ ⎞
1 2 0
A = ⎝−1−1 1⎠
3 2−2
Solution (a) We need tocalculate |A − λI|. Now
⎛
⎞
1−λ 2 0
A−λI= ⎝ −1 −1−λ 1 ⎠
3 2 −2−λ
and ∣∣∣∣∣∣ 1 −λ 2 0
∣ ∣ ∣∣∣
−1 −1−λ 1
3 2 −2−λ∣ = (1−λ) −1 − λ 1
∣∣∣ 2 −2−λ∣ −2 −1 1
3 −2−λ∣
= (1 − λ)[(−1 − λ)(−2 − λ) −2]
−2[−1(−2 − λ) −3]
Upon simplification thisreduces to −λ 3 −2λ 2 + λ +2.Hence
yields
|A−λI|=0
−λ 3 −2λ 2 +λ+2=0
which maybe written as
λ 3 +2λ 2 −λ−2=0
The characteristic equation is λ 3 +2λ 2 − λ −2 = 0.

302 Chapter 8 Matrix algebra
(b) The characteristic equation issolved toyield the eigenvalues:
λ 3 +2λ 2 −λ−2=0
Factorizing yields
(λ+2)(λ+1)(λ−1)=0
from which λ = −2,−1,1.
The eigenvalues are λ = −2,−1,1.
EXERCISES8.11.2
1 Calculate (i) the characteristic equation (ii) the
eigenvalues ofthe systemAX = λX,
whereAisgiven by
( ) ( )
5 6 −3 4
(a) (b)
(c)
2 1
( 7 −2
1 4
)
(d)
−4 5
( ) 1 3
4 −1
2 Calculate (i) the characteristic equation (ii) the
eigenvalues ofthe following 3 ×3 matrices:
⎛ ⎞
1 −12
(a) ⎝−3 −23⎠
2 −11
(b)
(c)
(d)
(e)
⎛ ⎞
10 −1
⎝31 4⎠
02 2
⎛ ⎞
21 2
⎝−11 −1⎠
83 0
⎛ ⎞
−2 62
⎝ 0 34⎠
3 −35
⎛ ⎞
3 −21
⎝ 2 −43⎠
16 −41
Solutions
1 (a) (i) λ 2 −6λ−7=0
(ii) λ = −1,7
(b) (i) λ 2 −2λ+1=0
(ii) λ = 1 (twice)
(c) (i) λ 2 −11λ+30=0
(ii) λ=5,6
(d) (i) λ 2 −13=0
(ii) λ = − √ 13, √ 13
2 (a) (i) −λ 3 +7λ+6=0
(ii) λ = −2, −1,3
(b) (i) λ 3 −4λ 2 −3λ+12=0
(ii) λ = − √ 3, √ 3,4
(c) (i) λ 3 −3λ 2 −10λ+24=0
(ii) λ = −3,2,4
(d) (i) λ 3 −6λ 2 +5λ=0
(ii) λ=0,1,5
(e) (i) λ 3 −13λ+12=0
(ii) λ = −4,1,3
TechnicalComputingExercises8.11.2
Thecalculationofeigenvalues andeigenvectors is
usuallyperformedby abuilt-infunction. Forexample,
in MATLAB ® the functioneigcan used to calculate
the eigenvalues.
To produce asolution to question 1.(a)in the previous
exercise wewould type:
A=[56;21];
eig(A)

8.11 Eigenvalues and eigenvectors 303
which produces:
ans =
7
-1
which isavectorcontaining the eigenvalues.
UseMATLAB ® orasimilar language to confirm the
restofthe eigenvalues given in Exercises8.11.2.
8.11.3 Eigenvectors
We have studied the system
AX=λX
anddeterminedthevaluesof λforwhichnon-trivialsolutionsexist.Thesevaluesof λare
calledeigenvaluesofthesystem,or,moresimply,eigenvaluesofA.Foreacheigenvalue
thereisanon-trivialsolution of the system.This solution iscalled aneigenvector.
Example8.46 Find the eigenvectors of
where
AX=λX
A =
( ) ( 4 1 x
andX=
3 2 y)
Solution We seek solutions ofAX = λX which may be written as
(A−λI)X=0
The eigenvalues werefound inExample 8.44 tobe λ = 1,5.
Firstlywe consider λ = 1.The systemequation becomes
(A−λI)X =0
(A−I)X =0
[( ) ( ( ) ( 4 1 1 0 x 0
− =
3 2 0 1)]
y 0)
( ) ( 3 1 x 0
=
3 1)(
y 0)
Writtenas individual equations wehave
3x+y =0
3x+y =0
Clearlythereisonlyoneequationwhichisrepeated.Aslongasy = −3xtheequationis
satisfied.Thusthereareaninfinitenumberofsolutionssuchasx = 1,y = −3;x = −5,
y = 15;and so on. Generally we write
x=t,y=−3t
for any numbert. Thus the eigenvector corresponding to λ = 1 is
( ( ) ( ) x t 1
X = = =t
y)
−3t −3

304 Chapter 8 Matrix algebra
Notethattheeigenvectorhasbeendeterminedtowithinanarbitraryscalar,t.Thusthere
isan infinity of solutions corresponding to λ = 1.
We now consider λ = 5 and seek solutions of the systemequation:
(A−λI)X =0
[( ) ( ( ) ( 4 1 1 0 x 0
−5 =
3 2 0 1)]
y 0)
( ( ) ( −1 1 x 0
=
3 −3)
y 0)
Written as individual equations we have
−x+y =0
3x−3y =0
Wenotethatthesecondequationissimplyamultipleofthefirstsothatinessencethere
isonlyoneequation.Solving −x+y = 0givesy =xforanyx.Sowewritex ( =t,y =t.
1
Hencetheeigenvectorcorrespondingto λ = 5isX =t .Againtheeigenvectorhas
1)
been determined towithin an arbitraryscaling constant.
Sometimesthearbitraryscalingconstantsarenotwrittendown;itisunderstoodthat
they arethere. Insuch a case wesay the eigenvectors of the systemare
( )
( 1 1
X = and X=
−3 1)
Example8.47 Determine the eigenvectors of
( ( ) ( 3 1 x x
= λ
−1 5)
y y)
Solution InExample8.43wefoundthatthereisonlyoneeigenvalue, λ = 4.Weseekthesolution
of (A−λI)X =0.Withλ =4wehave
( ) ( ) ( )
3 1 1 0 −1 1
A−λI= −4 =
−1 5 0 1 −1 1
Hence (A − λI)X = 0is the same as
( ( ) ( −1 1 x 0
=
−1 1)
y 0)
Thus there isonly one equation, namely
−x+y=0
which has an infinity of solutions:x =t,y =t. Hence thereisone eigenvector:
( 1
X =t
1)

8.11 Eigenvalues and eigenvectors 305
The concept ofeigenvectors iseasilyextended tomatrices ofhigher order.
Example8.48 Determine the eigenvectors of
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 2 0 x x
⎝−1−1 1⎠
⎝y⎠ = λ ⎝y⎠
3 2−2 z z
The eigenvalues werefound inExample 8.45.
Solution FromExample8.45theeigenvaluesare λ = −2,−1,1.Weconsidereacheigenvaluein
turn.
λ=−2
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 2 0 x x
⎝−1−1 1⎠
⎝y⎠ = −2⎝y⎠
3 2−2 z z
⎡⎛
⎞ ⎛ ⎞⎤⎛
⎞ ⎛ ⎞
1 2 0 100 x 0
⎣⎝−1−1 1⎠ +2⎝010⎠⎦
⎝y⎠ = ⎝0⎠
3 2−2 001 z 0
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
320 x 0
⎝−111⎠
⎝y⎠ = ⎝0⎠
320 z 0
We note that the first and lastrows areidentical. So wehave
3x +2y =0
−x+y +z=0
Solving these equations gives
x=t, y=− 3 2 t, z=5 2 t
Hence the corresponding eigenvector is
⎛ ⎞
1
− 3 X =t ⎜ 2⎟
⎝ ⎠
5
2
λ=−1
We have
⎡⎛
⎞
1 2 0
⎛ ⎞⎤⎛
⎞
100 x
⎣⎝−1−1 1⎠ + ⎝010⎠⎦
⎝y⎠ =
3 2−2 001 z
⎛ ⎞ ⎛ ⎞
22 0 x
⎝−10 1⎠
⎝y⎠ =
32−1 z
⎛ ⎞
0
⎝0⎠
0
⎛ ⎞
0
⎝0⎠
0

306 Chapter 8 Matrix algebra
Thus we have
2x+2y =0
−x +z=0
3x+2y −z=0
Wenotethatthethirdequationcanbederivedfromthefirsttwoequations,bysubtracting
thesecondequationfromthefirst.Ifyoucannotspotthistheequationsshouldbesolved
by Gaussian elimination. Ineffect we have only two equations:
2x+2y =0
−x +z =0
Solving these givesx =t,y = −t,z =t. The eigenvector is
⎛ ⎞
1
X =t ⎝−1⎠
1
λ = 1
We have
⎡⎛
⎞
1 2 0
⎛ ⎞⎤
⎛ ⎞
100 x
⎛ ⎞
0
⎣⎝−1−1 1⎠ − ⎝010⎠⎦
⎝y⎠ = ⎝0⎠
3 2−2 001 z 0
Thus we have
2y =0
−x−2y+ z=0
3x +2y−3z=0
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
0 2 0 x 0
⎝−1−2 1⎠
⎝y⎠ = ⎝0⎠
3 2−3 z 0
From the first equation,y = 0;puttingy = 0 into the other equations yields
−x+z =0
3x−3z =0
Here the second equation can be derived from the first by multiplying the first by −3.
Solving, wehavex =t,z =t. So the eigenvector is
⎛ ⎞
1
X =t ⎝0⎠
1
EXERCISES8.11.3
1 Calculate the eigenvectors ofthe matrices given in
Question 1 ofExercises8.11.2.
2 Calculate the eigenvectors ofthe matrices given in
Question2ofExercises8.11.2.

8.12 Analysis of electrical networks 307
Solutions
( ) )
1
1
1 (a)t ,t(
1
−1
3
( )
1
1
(c) t ,t(
1
1)
2
( ) (
1
(d) t
(
− 1 + √ )
,t
13 /3
( 1
(b) t
1)
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 1 1
2 (a)t⎝5⎠,t
⎝12⎠,t
⎝0⎠
1 5 1
⎛ ⎞ ⎛ ⎞
1 1
(b)t ⎝−5.0981⎠,t
⎝ 0.0981⎠,t
2.7321 −0.7321
)
(
1 √13−1 )
/3
⎛ ⎞
1
⎝−3⎠
−3
⎛ ⎞
1
− 1 ⎛ ⎞ ⎛ ⎞
1 1
(c)t ⎜ 3⎟
⎝
− 7 ⎠ ,t ⎝−2⎠,t
⎝−0.8⎠
1 1.4
3
⎛ ⎞
1 ⎛ ⎞ ⎛ ⎞
4
1 1
(d)t ⎜ 9⎟
⎝
− 1 ⎠ ,t ⎝ 0.6⎠,t
⎝ 1⎠
−0.3 0.5
3
⎛ ⎞
1 ⎛ ⎞ ⎛ ⎞
19
1 1
(e)t ⎜ 6 ⎟
⎝
− 2 ⎠ ,t ⎝4⎠,t
⎝2⎠
6 4
3
8.12 ANALYSISOFELECTRICALNETWORKS
Matrix algebra is very useful for the analysis of certain types of electrical network. For
suchnetworksitispossibletoproduceamathematicalmodelconsistingofsimultaneous
equations which can be solved using Gaussian elimination. We will consider the case
whenthenetworkconsistsofresistorsandvoltagesources.Thetechniqueissimilarfor
othertypesofnetwork.
In order to develop this approach, it is necessary to develop a systematic method
for writing the circuit equations. The method adopted depends on what the unknown
variables are. A common problem is that the voltage sources and the resistor values are
knownanditisdesiredtoknowthecurrentvaluesineachpartofthenetwork.Thiscan
beformulatedasamatrix equation. Given
where
V=RI ′
V = voltage vectorforthe network
I ′ = currentvectorforthe network
R = matrix ofresistor values
theproblemistocalculateI ′ whenV andRareknown.I ′ isusedtoavoidconfusionwith
the identitymatrix.
Anysizeofelectricalnetworkcanbeanalysedusingthisapproach.Wewilllimitthe
discussion to the case whereI ′ has three components, for simplicity. The extension to
larger networks is straightforward. Consider the electrical network of Figure 8.2. Mesh
currentshavebeendrawnforeachoftheloopsinthecircuit.Ameshisdefinedasaloop
that cannot contain a smaller closed current path. For convenience, each mesh current
is drawn in a clockwise direction even though it may turn out to be in the opposite
directionwhenthecalculationshavebeenperformed.Thenetcurrentineachbranchof
thecircuitcanbeobtainedbycombiningthemeshcurrents.Thesearetermedthebranch

308 Chapter 8 Matrix algebra
R 1 R 2
+
E 1
I 1
–
R 4
R 5
R 3
R 6 R 7
I 2 I 3
–
E 2 E 3
+ R 8
+
–
Figure8.2
An electrical network with meshcurrents
shown.
currents.Theconceptofameshcurrentmayappearslightlyabstractbutitdoesprovide
aconvenientmechanismforanalysingelectricalnetworks.Wewillexamineanapproach
that avoids the use of mesh currents later inthis section.
ThenextstageistomakeuseofKirchhoff’svoltagelawforeachofthemeshesinthe
network.Thisstatesthatthealgebraicsumofthevoltagesaroundanyclosedloopinan
electrical network iszero. Therefore the sum of the voltage risesmustequal the sum of
voltage drops. When applying Kirchhoff’s voltage law it is important to use the correct
sign for a voltage source depending on whether or notitis‘aiding’ a mesh current.
For mesh 1
E 1
= I 1
R 1
+I 1
R 2
+ (I 1
−I 3
)R 4
+ (I 1
−I 2
)R 3
E 1
= I 1
(R 1
+R 2
+R 4
+R 3
)+I 2
(−R 3
)+I 3
(−R 4
)
For mesh 2
−E 2
−E 3
= I 2
R 5
+ (I 2
−I 1
)R 3
+ (I 2
−I 3
)R 6
+I 2
R 8
−E 2
−E 3
= I 1
(−R 3
)+I 2
(R 5
+R 3
+R 6
+R 8
)+I 3
(−R 6
)
For mesh 3
E 3
= (I 3
−I 2
)R 6
+ (I 3
−I 1
)R 4
+I 3
R 7
E 3
= I 1
(−R 4
)+I 2
(−R 6
)+I 3
(R 6
+R 4
+R 7
)
These equations can be written inmatrix form as
⎛ ⎞ ⎛
⎞ ⎛ ⎞
E 1 R 1 +R 2 +R 4 +R 3 −R 3 −R 4 I 1
⎝−E 2 −E 3
⎠ = ⎝ −R 3 R 5 +R 3 +R 6 +R 8 −R 6
⎠ ⎝I 2
⎠
E 3
−R 4 −R 6 R 6 +R 4 +R 7 I 3
Engineeringapplication8.3
Calculationofmeshandbranchcurrentsinanetwork
Consider the electrical network of Figure 8.3. It has the same structure as that of
Figure8.2butwithactualvaluesforthevoltagesourcesandresistors.Branchcurrents
aswellasmeshcurrentshavebeenshown.Calculatethemeshcurrentsandhencethe
branch currents forthe network.

8.12 Analysis of electrical networks 309
2 V I 4 V
a
3 V
+
– I b
3 V
I 1
1 V
I c
5 V I 2 2 V I 3
3 V
I d – I e +
2 V 4 V
+ 4 V –
I f
Figure8.3
Theelectrical network ofFigure 8.2
with values forthe source voltagesand
resistorsadded.
Solution
We have already obtained the equations for this network. Substituting actual values
for the resistorsand voltage sources gives
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
3 10 −3 −1 I 1
⎝−2−4⎠ = ⎝−3 14 −2⎠
⎝I 2
⎠
4 −1−2 6 I 3
This is now in the formV =RI ′ . We shall solve these equations by Gaussian elimination.
Forming the augmented matrix, we have
⎛
⎞
10−3−1 3
⎝−3 14 −2 −6⎠
−1−2 6 4
Then
and
Hence,
R 1
R 2
→ 10R 2
+3R 1
R 3
→ 10R 3
+R 1
R 1
R 2
R 3
→ 131R 3
+23R 2
Similarly,
I 3
= 4460
7200 = 0.619 A
I 2
=
Finally,
I 1
=
−51 +23(0.619)
131
3 +0.619 +3(−0.281)
10
⎛
⎞
10 −3 −1 3
⎝0 131 −23 −51⎠
0 −23 59 43
⎛
⎞
10 −3 −1 3
⎝0 131 −23 −51⎠
0 0 7200 4460
= −0.281 A
= 0.278 A
➔

310 Chapter 8 Matrix algebra
The branch currents arethen
I a
=I 1
=278mA
I b
= I 2
−I 1
= −281 −278 = −559 mA
I c
=I 3
−I 1
=619−278=341mA
I d
= −I 2
= 281 mA
I e
= I 2
−I 3
= −281 −619 = −900 mA
I f
=I 3
=619mA
An alternative approach to analysing an electrical network is to use the node voltage
method which is often simply called nodal analysis. This technique is fundamental to
manycomputerprogramswhichareusedtosimulateelectricalcircuits,suchasSPICE.
We introduce thistechnique by means of an example.
Engineeringapplication8.4
Analysinganelectricalnetworkusingthenodevoltagemethod
The node voltage method utilises the notion that ‘islands’ of equal potential lie between
electrical components and sources.The procedure isas follows:
(1) Pick a reference node. In order to simplify the equations this is usually chosen
tobethenodewhichiscommontothelargestnumberofvoltagesourcesand/or
the largestnumber of branches.
(2) Assignanodevoltagevariabletoalloftheothernodes.Iftwonodesareseparated
solelybyavoltagesourcethenonlyoneofthenodesneedbeassignedavoltage
variable. The node voltages areall measured with respect tothe reference node.
(3) At each node, write Kirchhoff’s current law in terms of the node voltages. Note
that once the node voltages have been calculated it is easy to obtain the branch
currents.
We will again examine the network of Figure 8.2, but this time use the node voltage
method.ThenetworkisshowninFigure8.4withnodevoltagesassignedandbranch
currentslabelled.Thereferencenodeisindicatedbyusingtheearthsymbol.Writing
Kirchhoff’s current lawfor each node, weobtain:
node a
node b
I a
=I a
V b
+E 1
−V a
R 1
= V a −V d
R 2
V b
R 2
+E 1
R 2
−V a
R 2
=V a
R 1
−V d
R 1
V a
(R 1
+R 2
)−V b
R 2
−V d
R 1
=E 1
R 2
I a
+I b
+I d
=0
V b
+E 1
−V a
+ V b −V c
R 1
R 3
+ V b +E 2 −V e
R 5
= 0

8.12 Analysis of electrical networks 311
R 1 V R a 2
a I
+
a
E 1
–
I
R 3 b
V R c 4 I
V c
b
b c d
R 5 R 6 R 7
V d
I d –
I e +
I f
E 2 E 3
+ R 8 –
e
V e
Figure8.4
The network ofFigure8.2with node
voltages labelled.
Rearrangement yields
that is,
V b
R 3
R 5
+ E 1
R 3
R 5
−V a
R 3
R 5
+V b
R 1
R 5
−V c
R 1
R 5
+V b
R 1
R 3
+ E 2
R 1
R 3
−V e
R 1
R 3
=0
V a
R 3
R 5
− V b
(R 1
R 3
+R 1
R 5
+R 3
R 5
)+V c
R 1
R 5
+V e
R 1
R 3
node c
= E 1
R 3
R 5
+E 2
R 1
R 3
I b
=I c
+I e
so that
V b
−V c
R 3
= V c −V d
R 4
+ V c −E 3
R 6
that is,
node d
node e
V b
R 4
R 6
−V c
R 4
R 6
=V c
R 3
R 6
−V d
R 3
R 6
+V c
R 3
R 4
−E 3
R 3
R 4
V b
R 4
R 6
−V c
(R 4
R 6
+R 3
R 6
+R 3
R 4
)+V d
R 3
R 6
= −E 3
R 3
R 4
I a
+I c
=I f
V a
−V d
+ V c −V d
R 2
R 4
= V d
R 7
V a
R 4
R 7
−V d
R 4
R 7
+V c
R 2
R 7
−V d
R 2
R 7
=V d
R 2
R 4
V a
R 4
R 7
+V c
R 2
R 7
−V d
(R 4
R 7
+R 2
R 7
+R 2
R 4
) =0
I d
=I d
V b
+E 2
−V e
R 5
= V e
R 8
V b
R 8
−V e
(R 5
+R 8
) = −E 2
R 8
➔

312 Chapter 8 Matrix algebra
These equations can be writteninthe matrix formAV =B, whereAisthe matrix
⎛
⎞
R 1 +R 2 −R 2 0 −R 1 0
R 3 R 5 −R 1 R 3 −R 1 R 5 −R 3 R 5 R 1 R 5 0 R 1 R 3
⎜ 0 R
⎝ 4 R 6 −R 4 R 6 −R 3 R 6 −R 3 R 4 R 3 R 6 0 ⎟
R 4 R 7 0 R 2 R 7 −R 4 R 7 −R 2 R 7 −R 2 R 4 0 ⎠
0 R 8 0 0 −R 5 −R 8
and
⎛ ⎞
⎛
⎞
V a
E 1
R 2
V b
E 1
R 3
R 5
+E 2
R 1
R 3
V =
⎜V c ⎟ and B=
⎜ −E 3
R 3
R 4 ⎟
⎝V d
⎠
⎝ 0 ⎠
V e
−E 2
R 8
TheequationswouldgenerallybesolvedbyGaussianeliminationtoobtainthenode
voltages and hence the branch currents.
Using the component values from Engineering application 8.3, these equations
become
⎛
⎞ ⎛ ⎞ ⎛ ⎞
6 −4 0 −2 0 V a
12
15−31 10 0 6
V b
⎜ 0 2 −11 6 0
⎟ ⎜V c ⎟
⎝ 3 0 12 −19 0⎠
⎝V d
⎠ = 57
⎜−12
⎟
⎝ 0⎠
0 4 0 0 −9 V e
−8
Use of a computer package avoids the tedious arithmetic associated with Gaussian
elimination and yields
V a
= 2.969 V b
= 0.5250 V c
= 2.200 V d
= 1.858 V e
= 1.122
Itisthen straightforward tocalculate the branch currents:
I a
= V a −V d
=278mA I
R b
= V b −V c
= −558 mA
2
R 3
I c
= V c −V d
R 4
=342mA I d
= V e
R 8
= 281 mA
I e
= V c −E 3
R 6
= −900 mA I f
= V d
R 7
= 619 mA
Compare these answers withthose of Engineering application 8.3.
It is possible to analyse electrical networks containing more complex elements
suchascapacitors,inductors,activedevices,etc.,usingthesameapproach.Theequationsaremorecomplicatedbutthetechniqueisthesame.Oftenitisnecessarytouse
iterative techniques in view of the size and complexity of the problem. These are
examined inthe following section.
8.13 ITERATIVETECHNIQUESFORTHESOLUTION
OFSIMULTANEOUSEQUATIONS
The techniques met so far for the solution of simultaneous equations are known as direct
methods, which generally lead to the solution after a finite number of stages in
the calculation process have been carried out. An alternative collection of techniques is

8.13 Iterative techniques for the solution of simultaneous equations 313
available and these are known as iterative methods. They generate a sequence of approximate
solutions which may converge to the required solution, and are particularly
advantageous when large systems of equations are to be solved by computer. We shall
studytwo such techniques here: Jacobi’s method and Gauss--Seidel iteration.
Example8.49 Solve the equations
2x+y =4
x−3y = −5
usingJacobi’s iterative method.
Solution We first rewrite the equations as
2x =−y+4
−3y =−x−5
and then as
x = − 1 2 y +2
(8.10)
y = 1 3 x + 5 3
Jacobi’smethodinvolves‘guessing’asolutionandsubstitutingtheguessinther.h.s.of
the equations in(8.10).Suppose we guessx = 0,y = 0.Substitution then gives
x = 2
y = 5 3
We now use these values as estimates of the solution and resubstitute into the r.h.s. of
Equation (8.10).This time wefind
x = − 1 2
( 5
3)
+2 = 1.1667 (tofour decimal places)
y = 1 3 (2) + 5 3 = 2.3333
(tofour decimal places)
The whole process is repeated in the hope that each successive application oriteration
will give an answer close to the required solution, that is successive iterates will converge.
In order to keep track of the calculations, we label the initial guessx (0) ,y (0) , the
resultof the first iterationx (1) ,y (1) and so on. Generally, we find
x (n+1) = − 1 2 y(n) +2
y (n+1) = 1 3 x(n) + 5 3

314 Chapter 8 Matrix algebra
Table8.1
Iterates produced byJacobi’smethod.
Iterationno. (n) x (n) y (n)
0 0 0
1 2.0000 1.6667
2 1.1667 2.3333
3 0.8333 2.0556
4 0.9722 1.9444
5 1.0278 1.9907
6 1.0047 2.0093
7 0.9954 2.0016
8 0.9992 1.9985
9 1.0008 1.9997
10 1.0002 2.0003
TheresultsofsuccessivelyapplyingtheseformulaeareshowninTable8.1.Thesequence
of values ofx (n) seems toconverge to1while thatofy (n) seems toconverge to2.
Clearly this sort of approach is simple to program and iterative techniques such as
Jacobi’s method are best implemented on a computer. When writing the program a test
should be incorporated so that after each iteration a check for convergence is made by
comparing successive iterates. In many cases, even when convergence does occur, it is
slow and so other techniques are used which converge more rapidly. The Gauss--Seidel
method is attractive for this reason. It uses the most recent approximation to x when
calculatingyleadingtoimprovedratesofconvergenceasthefollowingexampleshows.
Example8.50 Usethe Gauss--Seidel method tosolve the equations ofExample 8.49.
Solution As before wewrite the equations inthe form
x = − 1 2 y +2
y = 1 3 x + 5 3
Withx (0) = 0,y (0) = 0 as our initial guess,wefind
x (1) = − 1 2 (0)+2=2
To findy (1) weuse the mostrecent approximation toxavailable, thatisx (1) :
y (1) = 1 3 (2) + 5 3 = 2.3333
Generally, wefind
x (n+1) = − 1 2 y(n) +2
y (n+1) = 1 3 x(n+1) + 5 3

8.13 Iterative techniques for the solution of simultaneous equations 315
Table8.2
Iterates produced bythe Gauss--Seidel
method.
Iterationno. (n) x (n) y (n)
0 0 0
1 2.0000 2.3333
2 0.8334 1.9445
3 1.0278 2.0093
4 0.9954 1.9985
5 1.0008 2.0003
6 0.9999 2.0000
andtheresultsofsuccessivelyapplyingtheseformulaeareshowninTable8.2.Asbefore,
we see that the sequence x (n) seems to converge to 1 and y (n) seems to converge to 2,
although more rapidly than before.
Both ofthese techniques generalize tolarger systems of equations.
Example8.51 Perform three iterations of Jacobi’s method and three iterations of the Gauss--Seidel
method tofind anapproximate solution of
−8x+y+z=1
x−5y+ z =16
x+ y−4z =7
with aninitial guess ofx =y=z=0.
Solution Werewritethesystemtomakex,yandzthesubjectofthefirst,secondandthirdequation,
respectively:
x = 1 8 y + 1 8 z − 1 8
y = 1 5 x + 1 5 z − 16
5
(8.11)
z = 1 4 x + 1 4 y − 7 4
To apply Jacobi’s method we substitute the initial guessx (0) = y (0) = z (0) = 0 into
the r.h.s. of Equation (8.11) to obtainx (1) ,y (1) andz (1) , and then repeat the process. In
general,
x (n+1) = 1 8 y(n) + 1 8 z(n) − 1 8
y (n+1) = 1 5 x(n) + 1 5 z(n) − 16 5
z (n+1) = 1 4 x(n) + 1 4 y(n) − 7 4

316 Chapter 8 Matrix algebra
We find
x (1) = − 1 8 = −0.1250
y (1) = − 16 5 = −3.2000
z (1) = − 7 4 = −1.7500
Then,
x (2) = 1 8 (−3.2000) + 1 8 (−1.7500) − 1 8 = −0.7438
y (2) = 1 5 (−0.1250) + 1 5 (−1.7500) − 16 5 = −3.5750
z (2) = 1 4 (−0.1250) + 1 4 (−3.2000) − 7 4 = −2.5813
Finally,
x (3) = 1 8 (−3.5750) + 1 8 (−2.5813) − 1 8 = −0.8945
y (3) = 1 5 (−0.7438) + 1 5 (−2.5813) − 16 5 = −3.8650
z (3) = 1 4 (−0.7438) + 1 4 (−3.5750) − 7 4 = −2.8297
To apply the Gauss--Seidel iteration to Equation (8.11), the most recent approximation
isusedateach stage leading to
x (n+1) = 1 8 y(n) + 1 8 z(n) − 1 8
y (n+1) = 1 5 x(n+1) + 1 5 z(n) − 16 5
z (n+1) = 1 4 x(n+1) + 1 4 y(n+1) − 7 4
Startingfromx (0) =y (0) =z (0) = 0,wefind
x (1) = − 1 8 = −0.1250
y (1) = 1 5 (−0.1250) + 1 5 (0) − 16 5 = −3.2250
z (1) = 1 4 (−0.1250) + 1 4 (−3.2250) − 7 4 = −2.5875

8.13 Iterative techniques for the solution of simultaneous equations 317
Table8.3
Comparisonofthe Jacobi andGauss--Seidel methods.
Iteration
no. (n)
Jacobi’smethod Gauss--Seidel
x (n) y (n) z (n) x (n) y (n) z (n)
0 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
1 −0.1250 −3.2000 −1.7500 −0.1250 −3.2250 −2.5875
2 −0.7438 −3.5750 −2.5813 −0.8516 −3.8878 −2.9348
3 −0.8945 −3.8650 −2.8297 −0.9778 −3.9825 −2.9901
4 −0.9618 −3.9448 −2.9399 −0.9966 −3.9973 −2.9985
5 −0.9856 −3.9803 −2.9767 −0.9995 −3.9996 −2.9998
6 −0.9946 −3.9924 −2.9915 −0.9999 −3.9999 −3.0000
7 −0.9980 −3.9972 −2.9968
8 −0.9992 −3.9990 −2.9988
9 −0.9997 −3.9996 −2.9996
Then,
x (2) = 1 8 (−3.2250) + 1 8 (−2.5875) − 1 8 = −0.8516
y (2) = 1 5 (−0.8516) + 1 5 (−2.5875) − 16 5 = −3.8878
Finally,
z (2) = 1 4 (−0.8516) + 1 4 (−3.8878) − 7 4 = −2.9348
x (3) = 1 8 (−3.8878) + 1 8 (−2.9348) − 1 8 = −0.9778
y (3) = 1 5 (−0.9778) + 1 5 (−2.9348) − 16 5 = −3.9825
z (3) = 1 4 (−0.9778) + 1 4 (−3.9825) − 7 4 = −2.9901
For completeness, further iterations areshown inTable 8.3.
AsexpectedtheGauss--SeidelmethodconvergesmorerapidlythanJacobi’s.Thisis
generally the case because ituses the mostrecently calculated values ateach stage.
Unfortunately, as with all iterative methods, convergence is not guaranteed. However,
it can be shown that if the matrix of coefficients is diagonally dominant, that is each
diagonal element is larger in modulus than the sum of the moduli of the other elements
inits row, thenthe Gauss--Seidel methodwill converge.
Engineeringapplication8.5
Findingapproximatesolutionsforthenodevoltagesofan
electricalnetwork
When trying to analyse complicated electrical networks it is frequently necessary
to resort to computer-based methods in order to find the voltages and currents. For ➔

318 Chapter 8 Matrix algebra
example, large networks for the distribution of electricity from power stations to
points of consumption can be very complicated to analyse due to the large number
of loops involved. This example illustrates the method by making use of the circuit
given in Engineering application 8.3. However, it should be borne in mind that this
approach isof mostvalue when analysing much larger networks.
Use Jacobi’s method and the Gauss--Seidel method to obtain approximate solutions
for the node voltages of the electrical network examined in Engineering
application 8.3.
Solution
The node voltage equations are
6V a
−4V b
−2V d
= 12 3V a
+12V c
−19V d
= 0
15V a
−31V b
+10V c
+6V e
= 57
2V b
−11V c
+6V d
= −12
These can be rearranged togive
4V b
−9V e
= −8
V a
= 2V b +V d +6
3
V b
= 15V a +10V c +6V e −57
31
V c
= 2V b +6V d +12
11
V d
= 3V a +12V c
19
V e
= 4V b +8
9
The results of applying Jacobi’s method with aninitial guess of
V (0)
a
=V (0)
b
=V (0)
c
=V (0)
d
=V (0)
e
= 0
are shown in Table 8.4. Convergence was achieved to within 0.001 after 44 iterations.TheresultsofapplyingtheGauss--SeidelmethodareshowninTable8.5.Convergence
was achieved to within 0.001 after 21 iterations. Clearly the Gauss--Seidel
method converges more rapidly than Jacobi’s method.
Table8.4
Node voltages derived from Jacobi’smethod.
Iterationno.(n)
V (n)
a
V (n)
b
V (n)
c
V (n)
d
V (n)
e
0 0.0000 0.0000 0.0000 0.0000 0.0000
1 2.0000 −1.8387 1.0909 0.0000 0.8889
.
20 2.8710 0.4802 2.1379 1.8265 1.0738
.
44 2.9679 0.5243 2.1990 1.8578 1.1215

8.14 Computer solutions of matrix problems 319
Table8.5
Node voltages derived from the Gauss--Seidel method.
Iterationno.(n)
V (n)
a
V (n)
b
V (n)
c
V (n)
d
V (n)
e
0 0.0000 0.0000 0.0000 0.0000 0.0000
1 2.0000 −0.8710 0.9326 0.9048 0.5018
.
20 2.9665 0.5226 2.1985 1.8569 1.1212
21 2.9674 0.5233 2.1989 1.8573 1.1215
EXERCISES8.13
1 Performthree iterationsofthe methods ofJacobiand
Gauss--Seidel to obtain approximate solutionsofthe
following.Ineach case, use an initialguess of
x (0) =y (0) =z (0) =0
(a) 4x+y+z=−1
x+6y+2z=0
x+2y+4z=1
(b) 5x+y−z=4
x−4y+z=−4
2x+2y−4z=−6
(c) 4x+y+z=17
x+3y−z=9
2x−y+5z=1
Solutions
1 (a) Jacobi
x 1 = −0.2500,y 1 = 0,z 1 = 0.2500
x 2 = −0.3125,y 2 = −0.0417,z 2 = 0.3125
x 3 = −0.3177,y 3 = −0.0521,z 3 = 0.3490
Gauss--Seidel
x 1 = −0.2500,y 1 = 0.0417,z 1 = 0.2917
x 2 = −0.3333,y 2 = −0.0417,z 2 = 0.3542
x 3 = −0.3281,y 3 = −0.0634,z 3 = 0.3637
(b) Jacobi
x 1 = 0.8000,y 1 = 1,z 1 = 1.5000
x 2 = 0.9000,y 2 = 1.5750,z 2 = 2.4000
x 3 = 0.9650,y 3 = 1.8250,z 3 = 2.7375
Gauss--Seidel
x 1 = 0.8000,y 1 = 1.2000,z 1 = 2.5000
x 2 = 1.0600,y 2 = 1.8900,z 2 = 2.9750
x 3 = 1.0170,y 3 = 1.9980,z 3 = 3.0075
(c) Jacobi
x 1 = 4.2500,y 1 = 3,z 1 = 0.2000
x 2 = 3.4500,y 2 = 1.6500,z 2 = −0.9000
x 3 = 4.0625,y 3 = 1.5500,z 3 = −0.8500
Gauss--Seidel
x 1 = 4.2500,y 1 = 1.5833,z 1 = −1.1833
x 2 = 4.1500,y 2 = 1.2222,z 2 = −1.2156
x 3 = 4.2484,y 3 = 1.1787,z 3 = −1.2636
8.14 COMPUTERSOLUTIONSOFMATRIXPROBLEMS
Theuseoftechnicalcomputinglanguageshasvastlyimprovedtheabilityofengineersto
calculate the solutions to complex engineering problems. As has been demonstrated by
theexamplesgiveninpreviouschapters,highlevelmathematicalfunctionsareprovided
such as the ability to calculate the roots of an equation and the eigenvalues of a matrix.
In electrical engineering problems it is frequently necessary to solve matrix equations.
Consider again Example 8.32. The equations areof the form
AX=B

320 Chapter 8 Matrix algebra
where
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
3 2−1 4 x
A = ⎝2−1 2⎠ B = ⎝10⎠ X = ⎝y⎠
1−3−4 5 z
Inordertosolvetheseequationsinatechnicalcomputinglanguagethematricesneedto
be loaded into memory. InMATLAB ® wewould type
A = [3,2,−1;2,−1,2;1,−3,−4]
B = [4;10;5]
We note that commas are used to separate entries on the same row and semicolons are
used toseparate different rows.
ThevalueofX canbeobtainedbythemethodofmatrixinversion.Thecodeforthisis
X =inv(A) ∗B
Alternatively,X can be obtained by Gaussian elimination. The code for thisis
X=A\B
Gaussian elimination is the usual method of solving systems of linear equations in
MATLAB ® because it is computationally efficient. If we run this three-line program
using MATLAB ® then weobtain
A = [3,2,−1;2,−1,2;1,−3,−4]
B = [4;10;5]
X=A\B
X = 3.0000
−2.0000
1.0000
and sox = 3.0000, y = −2.0000, z = 1.0000.
We see that MATLAB ® is a powerful tool for carrying out matrix calculations. Part
ofitspowerderivesfromtheextremelyhigh-levelnatureofitscommands.Acommand
suchasinv(A)wouldrequiretypically50linesifwritteninanormalhigh-levellanguage
such as Cor FORTRAN.
Engineeringapplication8.6
SolutionofanelectricalnetworkusingMATLAB®
Consider again the electrical network of Engineering application 8.3. Write the
MATLAB ® code tosolve thisnetwork.
Solution
TheequationsareoftheformV =RI ′ whereV isthevoltagevectorofthenetwork,
RisthematrixofresistorvaluesandI ′ isthecurrentvectorofthenetwork.Weknow
V andRand we wish to obtainI ′ . Using the values of Engineering application 8.3
wecan write
➔

Review exercises 8 321
V = [3;−6;4]
R = [10,−3,−1;−3,14,−2;−1,−2,6]
IPRIME = inv(R) ∗V
Running thisprogram gives
IPRIME = 0.2778
−0.2806
0.6194
REVIEWEXERCISES8
1 Evaluate the following products:
(a) ( −4 1 )( )
2
(b) ( −4 1 )( )
3
7
6
(c) ( −4 1 )( )
( )
2 3 2 −3 −1
(d)
7 6 4 5)(
1
⎛ ⎞
(e) ( 2 −1 0 ) 131 ( )
⎝240⎠
2 1
(f)5
3 6
107
( )
1 732
(g)
2 101
2 Simplify
∣ cosh θ sinh θ
sinh θ cosh θ∣ .
⎛ ⎞
0 23
3 Given thatA = ⎝2 00⎠findA −1 andA 2 .Show
1 −10
thatA 2 +6A −1 −7I = 0, whereI denotes the 3 ×3
identitymatrix.
⎛ ⎞
2 −14
4 IfA= ⎝1 00⎠find
1 −20
(a) |A|
(b) adj(A)
(c) A −1
5 Find the inverse ofthe matrix
⎛ ⎞
1 −20
⎝ 3 15⎠
−1 23
Hencesolvethe equations
x−2y=3
3x+y+5z=12
−x+2y+3z=3
6 Use Gaussian elimination to solve
x+2y−3z+2w=2
2x+5y−8z+6w=5
3x+4y−5z+2w=4
7 Use Jacobi’smethodto obtain a solution ofAX =B
to three decimalplaces where
⎛ ⎞ ⎛ ⎞
10 1 0 1
A = ⎝ 1 10 1 ⎠ and B= ⎝2⎠
0 1 10 1
8 Use amatrixmethodto solve
2x+y−z=3
x−y+2z=1
3x+4y+3z=2
9 Consider the Vandermonde matrix
⎛
1aa 2 ⎞
V = ⎝1bb 2 ⎠
1cc 2
(a) Find detV andshowthat itcanbe written as
(a −c)(a −b)(c −b).
(b) Show thatifa,bandcare all different, thenthe
Vandermonde matrixis non-singular.
10 (a) Thesignal f (t) = sin πt isto be approximated
2
byathird-degree polynomial forvalues oft
between −1 and 2.Byforcingthe original signal
and itsapproximating polynomial to agree at
t=−1,t=0,t=1andt=2,findthis
approximation. [Hint: seeEngineering
application 8.2.]
(b) Useagraphics calculator orgraph plotting
package to compare the graphs of f (t)andits
approximating polynomial.

322 Chapter 8 Matrix algebra
( ) 1 −2
11 SupposeA = .
1 4
( −2
(a) FindAv when v = .
2)
( −2
(b) Find 3vwhen v = .Deduce thatAv = 3v.
2)
( −µ
(c) FindAv when v = forany constant µ.
µ)
Deduce thatAv = 3v.
12 Usethe Gauss--Seidel method to find an approximate
solution of
(a) 5x +3y = −34
2x−7y=93
(b) 3x+y+z=6
2x+5y−z=5
x−3y+8z=14
13 Determine which ofthe followingsystemshave
non-trivial solutions.
(a) 2x−y=0
3x−1.5y=0
(b) 6x+5y=0
5x+6y=0
(c) −x−4y=0
2x+8y=0
(d) 7x−3y=0
1.4x −0.6y = 0
(e) −4x+5y=0
3x−4y=0
14 Determine which ofthe followingsystemshave
non-trivial solutions.
(a) 3x−2y+2z=0
x−y+z=0
2x+2y−z=0
(b) x+3y−z=0
4x−y+2z=0
6x+5y=0
(c) x+2y−z=0
x−3z=0
5x+6y−9z=0
15 ThematrixAis defined by
( ) 3 2
A =
−3 −4
(a) Determine the characteristic equation ofA.
(b) Determine the eigenvalues ofA.
(c) Determine the eigenvectors ofA.
(d) Formanew matrixM whose columns are the two
eigenvectors ofA.M iscalled amodal matrix.
(e) Show thatM −1 AM is adiagonal matrix,D,with
the eigenvalues ofAonits leading diagonal.Dis
called thespectral matrixcorresponding to the
modal matrixM.
16 (a) Show that the matrix
( ) 5 2
A =
−2 1
has only one eigenvalue and determine it.
(b) Calculate the eigenvector ofA.
17 ThematrixH isgiven by
⎛ ⎞
4 −11
H = ⎝−2 40⎠
−4 31
(a) Findthe eigenvalues ofH.
(b) Determine the eigenvectors ofH.
(c) Formanew matrixM whose columns are the
threeeigenvectors ofH.M iscalled a modal
matrix.
(d) ShowthatM −1 HM isadiagonal matrix,D, with
the eigenvalues ofH onits leading diagonal.Dis
called thespectral matrixcorresponding to the
modal matrixM.
Solutions
1 (a) −1 (b) −6
( −5
(c) (−1 −6) (d)
1)
(e) (0 2 2)
⎛ ⎞
7 3
(g) ⎜2
2 1
⎝1
2 0 1 ⎟
⎠
2
(f)
( ) 10 5
15 30
2 1
⎛ ⎞
3 A −1 = 1 0 3 0
⎝0 3 −6⎠
6
2 −2 4
⎛ ⎞
7 −30
A 2 = ⎝ 0 46⎠
−2 23

Review exercises 8 323
4 (a) −8
⎛
(b) ⎝
⎞
0 −80
0 −44⎠
−2 31
⎛ ⎞
0 8 0
1
(c) ⎝0 4 −4⎠
8
2 −3 −1
5
⎛
1
⎝
21
−7 6 −10
−14 3 −5
70 7
x=1,y=−1,z=2
⎞
⎠
6 x=2µ−λ,y=1+2λ−2µ,z=λ,w=µ
7 x = 0.082,y = 0.184,z = 0.082
8 x = 1.462,y = −0.308,z = −0.385
12 (a) x=1,y=−13
(b)x=1,y=1,z=2
13 (a),(c)and(d)have non-trivialsolutions.
14 (b)and(c)have non-trivialsolutions.
15 (a) λ 2 +λ−6=0
(b) ( λ=−3,2 ) ( ) 1 1
(c) ,
−3 −0.5
16 (a) λ=3
( ) 1
(b)
−1
17 (a) λ=2,3,4
⎛
(b) ⎝
⎞ ⎛ ⎞ ⎛ ⎞
1 1 0
1⎠,
⎝2⎠,
⎝1⎠
−1 1 1
10 (a) f(t) = 4 3 t − 1 3 t3
( ( −6 −6
11 (a) (b)
6)
6)
(c)
( ) −3µ
3µ

9 Complexnumbers
Contents 9.1 Introduction 324
9.2 Complexnumbers 325
9.3 Operationswithcomplexnumbers 328
9.4 Graphicalrepresentationofcomplexnumbers 332
9.5 Polarformofacomplexnumber 333
9.6 Vectorsandcomplexnumbers 336
9.7 Theexponentialformofacomplexnumber 337
9.8 Phasors 340
9.9 DeMoivre’stheorem 344
9.10 Lociandregionsofthecomplexplane 351
Reviewexercises9 354
9.1 INTRODUCTION
Complex numbers often seem strange when first encountered but it is worth persevering
with them because they provide a powerful mathematical tool for solving several
engineeringproblems.Oneofthemainapplicationsistotheanalysisofalternatingcurrent
(a.c.) circuits. Engineers are very interested in these because the mains supply is
itself a.c., and electricity generation and transportation are dominated by a.c. voltages
and currents.
A great deal of signal analysis and processing uses mathematical models based on
complex numbers because they allow the manipulation of sinusoidal quantities to be
undertakenmoreeasily.Furthermore,thedesignoffilterstobeusedincommunications
equipment relies heavily on their use.
One area of particular relevance is control engineering -- so much so that control
engineers often prefer to think of a control system in terms of a ‘complex plane’ representation
rather than a ‘time domain’ representation. We will develop these concepts in
this and subsequent chapters.

9.2 Complex numbers 325
9.2 COMPLEXNUMBERS
We have already examined quadratic equations such as
x 2 −x−6=0 (9.1)
andhavemettechniquesforfindingtherootsofsuchequations.Theformulaforobtainingthe
roots of a quadratic equationax 2 +bx +c = 0 is
x = −b ± √ b 2 −4ac
2a
Applying thisformulatoEquation (9.1), wefind
(9.2)
x = +1 ± √ (−1) 2 −4(1)(−6)
2
= 1 ± √ 25
2
= 1 ±5
2
so thatx = 3 andx = −2 are the two roots. However, if we try to apply the formula to
the equation
we find
2x 2 +2x+5=0
x = −2 ± √ −36
4
A problem now arises in that we need to find the square root of a negative number. We
knowfromexperiencethatsquaringbothpositiveandnegativenumbersyieldsapositive
result; thus,
6 2 = 36 and (−6) 2 = 36
so that there is no real number whose square is −36. In the general case, if
ax 2 +bx +c = 0, we see by examining the square root in Equation (9.2) that this problemwillalwaysarisewheneverb
2 −4ac < 0.Nevertheless,itturnsouttobeveryuseful
to invent a technique for dealing with such situations, leading to the theory of complex
numbers.
To make progress weintroduce a number, denoted j,with the property that
j 2 =−1
We have already seen that using the real number system we cannot obtain a negative
numberbysquaringareal numbersothenumberjisnotreal-- wesay itisimaginary.
Thisimaginarynumberhasaveryusefulroletoplayinengineeringmathematics.Using
it we can now formally write down an expression for the square root of any negative
number. Thus,
√
−36 =
√
36 × (−1)
= √ 36×j 2
= 6j

326 Chapter 9 Complex numbers
Returning tothe solution of the quadratic equation 2x 2 +2x +5 = 0,wefind
x = −2 ± √ −36
4
= −2±6j
4
= −1±3j
2
We have found two roots, namelyx = − 1 2 + 3 2 jandx=−1 2 − 3 j.These numbers are
2
calledcomplexnumbersandweseethattheyaremadeupoftwoparts--arealpartand
animaginarypart.Forthefirstcomplexnumbertherealpartis − 1 2 andtheimaginary
part is 3 2 . For the second complex number the real part is −1 and the imaginary part is
2
− 3 2 .Inamoregeneralcaseweusuallyusetheletterztodenoteacomplexnumberwith
realpartaandimaginarypartb,soz =a +bj.Wewritea = Re(z)andb= Im(z),and
denote the set of all complex numbers by C. Note thata,b∈ Rwhereasz ∈ C.
z =a +bj,zisamemberofthe setofcomplex numbers, that isz ∈ C
a =Re(z) b =Im(z)
Complex numbers which have a zero imaginary part are purely real and hence all real
numbers arealsocomplex numbers, thatis R ⊂ C.
Example9.1 Solve the quadraticequation 2s 2 −3s +7 = 0.
Solution Using the formulaforsolvingaquadraticequation wefind
s = −(−3) ± √ (−3) 2 −4(2)(7)
2(2)
= 3 ± √ −47
4
= 3 ± √ 47j
4
= 0.75 ±1.71j
Usingthe factthat j 2 = −1 wecan develop other quantities.
Example9.2 Simplify the expression j 3 .
Solution We have
j 3 =j 2 ×j
=(−1)×j
= −j

9.2 Complex numbers 327
9.2.1 Thecomplexconjugate
Ifz =a +bj,wedefineitscomplexconjugatetobethenumberz =a −bj;thatis,we
change the sign of the imaginary part.
Example9.3 Write down the complex conjugates of
(a) −7+j (b) 6−5j (c) 6 (d) j
Solution Tofindthecomplexconjugatesofthegivennumberswechangethesignoftheimaginary
parts. Apurelyreal number has animaginarypart0.We find
(a) −7 −j (b) 6 +5j (c) 6,thereisno imaginarypart toalter (d) −j
We recall that the solution of the quadratic equation 2x 2 +2x +5 = 0 yielded the two
complexnumbers − 1 2 + 3 2 jand−1 2 − 3 2 j,andnotethattheseformacomplexconjugate
pair. This illustratesamore general result:
When the polynomial equationP(x) = 0 has real coefficients, any complex roots
will always occur incomplex conjugatepairs.
Consider the following example.
Example9.4 Show that the equationx 3 −7x 2 +19x −13 = 0 has a root atx = 1 and find the other
roots.
Solution IfweletP(x) =x 3 −7x 2 +19x −13,thenP(1) =1−7 +19 −13 =0sothatx =1
is a root. This means thatx −1 must be a factor ofP(x) and so we can expressP(x) in
the form
P(x)=x 3 −7x 2 +19x−13=(x−1)(αx 2 +βx+γ)
=αx 3 +(β−α)x 2 +(γ−β)x−γ
where α, β and γ arecoefficientstobedetermined.Comparingthecoefficientsofx 3 we
find α = 1. Comparing the constant coefficients we find γ = 13. Finally, comparing
coefficients ofxwefind β = −6,and hence
P(x) =x 3 −7x 2 +19x −13 = (x −1)(x 2 −6x +13)
Theothertwo rootsofP(x) = 0 arefoundby solvingthe quadraticequation
x 2 −6x+13=0,thatis
x = 6 ± √ 36−52
2
= 6 ± √ −16
2
=3±2j
and again we note that the complex roots occur as a complex conjugate pair. This illustratesthegeneralresultgiveninSection1.4thatannth-degreepolynomialhasnroots.

328 Chapter 9 Complex numbers
EXERCISES9.2
1 Solve the following equations:
(e) cos ωt +jsin ωt (f) cos ωt −jsin ωt
complex numbers.
1
(c)z=0 (d)z= 1 +j
(a) −11 −8j (b) 5 +3j (c)
2 j (d) −17 2
(f)z=j (e)z=j 2
(a)x 2 +1=0 (b)x 2 +4=0
(g) −0.333j +1
(c)3x 2 +7=0 (d)x 2 +x+1=0 4 Recallfrom Chapter2thatthe polesofarational
(e) x2
2 −x+2=0
functionR(x) =P(x)/Q(x)are those values ofxfor
whichQ(x) = 0.Findany poles of
(f)−x 2 −3x−4=0
x 3x 3
(g)2x 2 (a) (b)
+3x+3=0
x −3 x 2 (c)
+1 x 2 +x+1
(h)x 2 +3x+4=0
2 Solve the cubicequation
5 Solvethe equations 2 +2s +5 = 0.
6 Expressasacomplex number
3x 3 −11x 2 +16x−12=0
(a) j 4 (b) j 5 (c) j 6
given thatoneofthe rootsisx = 2. 7 StateRe(z)andIm(z)where
3 Writedown the complex conjugates ofthe following (a)z=7+11j (b)z=−6+j
Solutions
1 (a) ±j (b) ±2j
(c) ± √ 7/3j (d) −1/2 ± ( √ 3/2)j
(e) 1± √ 3j (f) −3/2 ± ( √ 7/2)j
(g) −3/4 ± ( √ 15/4)j (h) −3/2 ± ( √ 7/2)j
2 2,5/6±( √ 47/6)j
3 (a) −11 +8j (b) 5 −3j (c) − 1 2 j
(d) −17 (e) cos ωt −jsin ωt
(f) cos ωt +jsin ωt (g) 0.333j +1
4 (a)x=3 (b)x=±j
(c)x = −1/2 ± ( √ 3/2)j
5 s=−1±2j
6 (a) 1 (b) j (c) −1
7 (a) 7,11 (b) −6,1 (c) 0, 0
(d)
1
2 , 1 2
(e) 0, 1 (f) −1,0
9.3 OPERATIONSWITHCOMPLEXNUMBERS
Twocomplexnumbersareequalifandonlyiftheirrealpartsareequalandtheirimaginaryparts
areequal.
Example9.5 Findxandyso thatx+6jand 3 −yjrepresentthe samecomplex number.
Solution If both quantitiesrepresentthe same complex number wehave
x+6j=3−yj
Sincethe real parts mustbeequal wecan equatethem,thatis
x = 3

Similarly, wefind, by equating imaginary parts
6=−y
so thaty = −6.
9.3 Operations with complex numbers 329
Theoperationsofaddition,subtraction,multiplicationanddivisioncanallbeperformed
on complex numbers.
9.3.1 Additionandsubtraction
Toaddtwocomplexnumberswesimplyaddtherealpartsandaddtheimaginaryparts;
tosubtractacomplexnumberfromanotherwesubtractthecorrespondingrealpartsand
subtractthe corresponding imaginary parts as shown inExample 9.6.
Example9.6 Ifz 1
= 3 −4jandz 2
= 4 +2jfindz 1
+z 2
andz 1
−z 2
.
Solution
z 1
+z 2
=(3−4j)+(4+2j)
=(3+4)+(−4+2)j
=7−2j
z 1
−z 2
=(3−4j)−(4+2j)
=(3−4)+(−4−2)j
=−1−6j
9.3.2 Multiplication
Wecanmultiplyacomplexnumberbyarealnumber.Boththerealandimaginaryparts
of the complex number aremultiplied by the real number. Thus 3(4 −6j) = 12 −18j.
To multiply two complex numbers we use the factthat j 2 = −1.
Example9.7 Ifz 1
= 2 −2jandz 2
= 3 +4j, findz 1
z 2
.
Solution
z 1
z 2
= (2 −2j)(3 +4j)
Removing brackets wefind
z 1
z 2
=6−6j+8j−8j 2
=6−6j+8j+8 usingj 2 =−1
=14+2j
Example9.8 Ifz = 3 −2jfindzz.
Solution Ifz = 3 −2jthen itsconjugate isz = 3 +2j. Therefore,
zz = (3 −2j)(3 +2j)
=9−6j+6j−4j 2
=9−4j 2
= 13
We see thatthe answer isareal number.

330 Chapter 9 Complex numbers
Whenever we multiplyacomplex number by its conjugate the answer is a real number.
Thusifz=a+bj
zz = (a +bj)(a −bj)
=a 2 +baj−abj−b 2 j 2
=a 2 +b 2
Ifz=a+bjthenzz=a 2 +b 2
9.3.3 Division
To divide two complex numbers it is necessary to make use of the complex conjugate.
We multiply both the numerator and denominator by the conjugate of the denominator
and then simplifythe result.
Example9.9 Ifz 1
= 2 +9jandz 2
= 5 −2jfind z 1
z 2
.
Solution We seek 2+9j . The complex conjugate of the denominator is 5 + 2j, so we multiply
5−2j
bothnumeratoranddenominatorbythisquantity.Theeffectofthisistoleavethevalue
of z 1
unaltered since wehave only multiplied by 1.Therefore,
z 2
z 1
= 2+9j (2 +9j) (5 +2j)
=
z 2
5−2j (5 −2j) (5 +2j)
=
10 +45j +4j +18j2
25+4
=
−8 +49j
29
= − 8
29 + 49
29 j
The multiplication of two conjugates in the denominator allows a usefulsimplification.
Weseethattheeffectofmultiplyingbytheconjugateofthedenominatoristomakethe
denominator of the solution purely real.
Ifz 1
=x 1
+jy 1
andz 2
=x 2
+jy 2
thenthequotient z 1
z 2
isfoundbymultiplyingboth
numerator and denominator by the conjugate of the denominator, thatis
z 1
z 2
= x 1 +jy 1
x 2
+jy 2
= x 1 +jy 1
x 2
+jy 2
× x 2 −jy 2
x 2
−jy 2
= x 1 x 2 +y 1 y 2 +j(x 2 y 1 −x 1 y 2 )
x 2 2 +y2 2

9.3 Operations with complex numbers 331
EXERCISES9.3
1 Ifz 1 =3+2jandz 2 =4−8jfind
(a) z 1 +z 2 (b) z 1 −z 2 (c) z 2 −z 1
2 Expressthe following in the forma +bj:
1
(a) (b) −2
1 +j j
1
(c)
j + 1 j
(d)
2 −j 1 +j
3
(e)
3+2j + 1
5 −j
3 Expressthe following in the forma +bj:
2
(a) (b) −2+3j
1 −j j
(c)3j(4 −2j)
(e) 5+3j
2+2j
(b) (7 −2j)(5 +6j)
4 Find aquadraticequationwhose rootsare 1 −3j and
1+3j.
5 If (x +jy) 2 =3+4j,findxandy,wherex,y ∈ R.
6 Find the real andimaginaryparts of
(a)
2
4 +j − 3
2 −j
(b) j 4 −j 5
(c) 1 j +j (d) 1
j 3 −3j
Solutions
1 (a)7−6j (b) −1+10j (c) 1−10j 4 x 2 −2x+10=0
1
2 (a)
2 − 1 2 j (b) 2j (c) 2 5 − 4 5 j 5 x=2,y=1;x=−2,y=−1
1
(d)
2 + 1 23 −11j
j (e) 6 (a) − 62
2 26
85 , −61 85
(b) 1, −1
3 (a) 1+j (b) 3+2j (c) 6+12j
(d) 47 +32j (e) 2 − 1 (c) 0, 0
2 j (d) 0, 1 4
TechnicalComputingExercises9.3
1 Many technicalcomputing languagesallow the user
toinputandmanipulatecomplex numbers.Investigate
howthe complex numbera +bjisinput to the
software to whichyou have access.
2 Usetechnicalcomputing software to simplifythe
followingcomplex numbers:
(a) (1 +j) 5 3j
(b)
(1 −2j) 7
4
(c)
(3 +j) 3 − 7
(2 −3j) 2
3 Solvethe polynomial equations
(a) x 3 +7x 2 +9x+63=0
(b) x 4 +15x 2 −16=0
YoumayrecallfromChapter1thatin MATLAB ® the
rootsfunctioncan be used to solvepolynomial
equationswith real roots.This functioncan also
calculatecomplex roots.
4 Use apackageto findthe real andimaginaryparts of
15+j
7−5j .
5 Acommonrequirementincontroltheoryistofindthe
poles ofarationalfunction. Useapackageto findthe
poles ofthe function
0.5
G(s) =
s 3 +2s 2 +0.2s +0.0556
To solvethisproblem automatically mayrequire the
use ofspecialist toolboxesin MATLAB ® orGNU
Octave which are not partofthe core program.For
example, the Signal Processingtoolboxin
MATLAB ® has the functiontf2zp which convertsa
transfer function expressed asaratio ofpolynomials
to pole/zero form.

332 Chapter 9 Complex numbers
7
6
z 2 = –3+5j
Imaginary axis
5
4
3
2
z 1 = 7+2j
1
Imaginary axis
y
b
Figure9.1
Argand diagram.
a + bj
(a, b)
a x
Real axis
–4 –3 –2 –1 1 2 3 4 5 6 7
–1
Real axis
z 3 = –1–2j
–2
–3
–4
–5
z 4 = –4j
Figure9.2
Argand diagram forExample9.10.
9.4 GRAPHICALREPRESENTATIONOFCOMPLEXNUMBERS
Given a complex numberz = a +bj we can obtain a useful graphical interpretation of
it by plotting the real part on the horizontal axis and the imaginary part on the vertical
axis and obtain a unique point in thex--y plane (Figure 9.1). We call thexaxis the real
axis and theyaxis the imaginary axis, and the whole picture an Argand diagram. In
this context, thex--y plane isoften referredtoasthe complex plane.
Example9.10 Plotthecomplexnumbersz 1
= 7 +2j,z 2
= −3 +5j,z 3
= −1 −2jandz 4
= −4jonan
Argand diagram.
Solution TheArgand diagramisshown inFigure 9.2.
EXERCISES9.4
1 Plotthe followingcomplex numbersonan Argand
diagram:
(a) z 1 =−3−3j
(b) z 2 =7+2j
(c) z 3 =3
(d) z 4 = −0.5j
(e) z 5 = −2
2 (a) Plotthe complex numberz = 1 +jon anArgand
diagram.
(b) Simplify the complex number j(1 +j)andplot
the result onyour Arganddiagram.Observe that
theeffectofmultiplyingthecomplexnumberbyj
isto rotatethe complex numberthrough an angle
of π/2 radiansanticlockwise aboutthe origin.

9.5 Polar form of a complex number 333
Solutions
1 SeeFigure S.17.
2 (a) SeeFigure S.18. (b) j(1 +j) = −1 +j
Imaginary
Imaginary
–2
7+2j
–1+j 1
1+j
–0.5j
3
Real
–3–3j
FigureS.17
–1
FigureS.18
1
Real
9.5 POLARFORMOFACOMPLEXNUMBER
ItisoftenusefultoexchangeCartesiancoordinates (a,b)forpolarcoordinatesrand
θ asdepictedinFigure 9.3.
FromFigure 9.3 wenotethat
cosθ = a r
sinθ = b r
and so,
a=rcosθ
b=rsinθ
Furthermore,
tanθ = b a
UsingPythagoras’stheoremweobtainr = √ a 2 +b 2 .Byfindingrand θ wecanexpress
the complex numberz =a +bj in polar formas
z=rcosθ+jrsinθ =r(cosθ+jsinθ)
whichweoftenabbreviatetoz =r̸ θ.Clearly,risthe‘distance’ofthepoint (a,b)from
the origin and is called the modulus of the complex numberz. The modulus is always
a non-negative number and is denoted |z|. The angle is conventionally measured from
the positivexaxis. Angles measured in an anticlockwise sense are regarded as positive
whilethosemeasuredinaclockwisesenseareregardedasnegative.Theangle θ iscalled
the argument ofz, denoted arg(z). Since adding or subtracting multiples of 2π from θ
will result in the ‘arm’ in Figure 9.3 being in the same position, the argument can have
many values. Usuallywe shall choose θ tosatisfy −π < θ π.
Cartesian form:z =a +bj
Polarform:z =r(cosθ +jsinθ) =r̸
|z|=r= √ a 2 +b 2
θ
a=rcosθ b=rsinθ tanθ = b a

334 Chapter 9 Complex numbers
y
y
b
r
u
a
(a, b)
x
7– 4
p
p –4 –
1
2
–1 (1, –1)
x
Figure9.3
Polar andCartesian formsofacomplex
number.
Figure9.4
Argand diagram depictingz = 1 −jin
Example 9.11.
Note that
r̸
(−θ) =r(cos(−θ) +jsin(−θ))
=r(cosθ −jsinθ)
=z
Ifz=a+bjthenz=a−bjandz =r̸
(−θ).
Example9.11 Depict the complex numberz = 1 − j on an Argand diagram and convert it into polar
form.
Solution Therealpartofzis1andtheimaginarypartis −1.Wethereforeplotapointinthex--y
plane withx = 1 andy = −1 asshown inFigure 9.4.
FromFigure9.4weseethatr = √ 1 2 + (−1) 2 = √ 2and θ = −45 ◦ or −π/4radians.
Thereforez = 1 −j = √ 2̸ (−π/4).
To express a complex number in polar form it is essential to draw an Argand diagram
and notsimplyquoteformulae, asthe following example will show.
Example9.12 Expressz = −1 −j inpolar form.
Solution If we use the formula |z| =r= √ a 2 +b 2 ,wefindthatr = √ 2. Using tan θ =b/a, we
find that tanθ = −1/ −1 = 1 so that you may be tempted to take θ = π/4. Figure 9.5
√shows the Argand diagram and it is clear that θ = −3π/4. Therefore,z = −1 − j =
2̸ −3π/4,and we see the importance ofdrawing an Argand diagram.
y
–1
2
5p —4
3p —4 –
–1
x
Figure9.5
Argand diagramdepictingz = −1 −jin Example 9.12.

9.5.1 Multiplicationanddivisioninpolarform
9.5 Polar form of a complex number 335
ThepolarformmayseemmorecomplicatedthantheCartesianformbutitisoftenmore
useful. For example, suppose wewant tomultiplythe complex numbers
We find
z 1
=r 1
(cosθ 1
+jsinθ 1
) and z 2
=r 2
(cosθ 2
+jsinθ 2
)
z 1
z 2
=r 1
(cosθ 1
+jsinθ 1
)r 2
(cosθ 2
+jsinθ 2
)
=r 1
r 2
{(cosθ 1
cosθ 2
−sinθ 1
sinθ 2
) +j(sinθ 1
cosθ 2
+sinθ 2
cosθ 1
)}
which can be written as
r 1
r 2
{cos(θ 1
+ θ 2
) +jsin(θ 1
+ θ 2
)}
using the trigonometric identities of Section 3.6. This is a new complex number which,
ifwecomparewiththegeneralformr(cosθ +jsin θ),weseehasamodulusofr 1
r 2
and
an argument of θ 1
+ θ 2
. To summarize: to multiply two complex numbers we multiply
their moduliand add their arguments, thatis
z 1
z 2
=r 1
r 2̸ (θ 1
+θ 2
)
Example9.13 Ifz 1
= 3̸ π/3 andz 2
= 4̸ π/6 findz 1
z 2
.
Solution Multiplyingthemoduliwefindr 1
r 2
= 12,andaddingtheargumentswefind θ 1
+ θ 2
=
π/2.Thereforez 1
z 2
= 12̸ π/2.
Asimilardevelopmentshowsthattodividetwocomplexnumberswedividetheirmoduli
and subtracttheirarguments,that is
z 1
z 2
= r 1
r 2
̸ (θ 1
−θ 2
)
Example9.14 Ifz 1
= 3̸ π/3 andz 2
= 4̸ π/6 findz 1
/z 2
.
Solution Dividing the respective moduli, we find r 1
/r 2
= 3/4 and subtracting the arguments,
π/3 − π/6 = π/6.Hencez 1
/z 2
= 0.75̸ π/6.
EXERCISES9.5
1 Markonan Arganddiagrampointsrepresenting
z 1 =3−2j,z 2 =−j,z 3 =j 2 ,z 4 =−2−4jand
z 5 = 3. Find the modulus andargumentofeach
complex number.
2 Expressthe following complex numbersin polar
form:
(a) 3 −j (b) 2 (c) −j (d) −5 +12j
3 Find the modulusandargumentof(a)z 1 = − √ 3 +j
and (b)z 2 = 4 +4j. Hence expressz 1 z 2 andz 1 /z 2 in
polar form.
4 Express √ 2̸ π/4,2̸ π/6and2̸ −π/6 in Cartesian
forma +bj.
5 Prove the result z 1
z 2
= r 1
r 2
̸ θ 1 −θ 2 .

(
336 Chapter 9 Complex numbers
6 Expressz = 1 ,where ω andC are real constants,
jωC
in the forma +bj.Plotzonan Argand diagram.
7 Ifz 1 = 4(cos40 ◦ +jsin40 ◦ )andz 2 = 3(cos70 ◦ +
jsin70 ◦ ),expressz 1 z 2 andz 1 /z 2 in polar form.
8 Simplify
( √ 2̸ (5π/4)) 2 (2̸ (−π/3)) 2
2̸ (−π/6)
Solutions
1 |z 1 | = √ 13,arg(z 1 ) = −0.5880
|z 2 | = 1,arg(z 2 ) = −π/2
|z 3 | = 1,arg(z 3 ) = π
|z 4 | = √ 20,arg(z 4 ) = −2.0344
|z 5 | = 3,arg(z 5 ) = 0
2 (a) √ 10̸ −0.3218 (b) 2̸ 0
(c)1̸ −π/2 (d) 13̸ 1.9656
3 (a) 2,5π/6
(b) 4 √ 2,π/4
z 1 z 2 =8 √ 2̸ 13π/12
√
2
z 1 /z 2 =
4 ̸ 7π/12
4 1+j, √ 3+j, √ 3 −j
6 − j
ωC
7 z 1 z 2 = 12(cos110 ◦ +jsin110 ◦ )
8 4
z 1 /z 2 = 4 3 (cos30◦ −jsin30 ◦ )
9.6 VECTORSANDCOMPLEXNUMBERS
It is often convenient to represent complex numbers by vectors in the x--y plane. Figure
9.6(a) shows the complex numberz = a + jb. Figure 9.6(b) shows the equivalent
vector. Figure 9.7 shows the complex numbersz 1
= 2 +j andz 2
= 1 +3j.
If we now evaluate z 3
= z 1
+z 2
we find z 3
= 3 + 4j which is also shown. If we
form a parallelogram, two sides of which are the representations ofz 1
andz 2
, we find
thatz 3
isthediagonaloftheparallelogram.Ifweregardz 1
andz 2
asvectorsintheplane
we see that there is a direct analogy between the triangle law of vector addition (see
Section7.2.3)and the addition ofcomplex numbers.
y
4
y
b
z = a + jb
y
b
(
a
b
3
2
z 2
z 3
a x a x
1
z 1
(a)
Figure9.6
Thecomplex numberz =a +jbandits equivalent vector.
(b)
0 1 2 3 4
Figure9.7
Vector addition in the complex plane.
x

9.7 The exponential form of a complex number 337
y
B
z 2
OC
z 1
BA
A
C
y
5
4
3
2
1
B
z 2
z 1 + z 2
z 1 – z 2
z 1
A
C
O
x
O 1
2 3 4 5 6 7 8 9 10
x
Figure9.8
Vector addition and subtraction.
Figure9.9
Diagram forExample9.15.
More generally, if z 1
and z 2
are any complex numbers represented on an Argand
diagram by the vectors OA ⃗ and OB ⃗ (Figure 9.8) then upon completing the parallelogram
OBCA, the sum z 1
+ z 2
is represented by the vector OC. ⃗ We can also obtain
a representation of the difference of two complex numbers in the following way. If
wewrite
z 3
=z 1
−z 2
=z 1
+(−z 2
)
andnotethatifz 2
isrepresentedbythevectorOB,then ⃗ −z 2
isrepresentedbythevector
−OB ⃗ = BO ⃗ = CA. ⃗ The complex number z 1
is represented by OA ⃗ = BC, ⃗ so that
z 1
+ (−z 2
) = BC ⃗ + CA ⃗ = BA. ⃗ Thus the difference z 1
− z 2
is represented by the
diagonal BA. To summarize, the sum and difference of z 1
and z 2
are represented by
the two diagonals of the parallelogram OBCA.
Example9.15 Represent z 1
= 6 + j and z 2
= 3 + 4j, and their sum and difference, on an Argand
diagram.
Solution WedrawvectorsOAand ⃗ OBrepresentingz ⃗ 1
andz 2
,respectively (Figure9.9).Thenwe
completetheparallelogramOACBasshown.Thesumz 1
+z 2
isthenrepresentedby ( ) OC ⃗
andthedifferencez 1
−z 2
byBA.Itiseasytoseethatthevector ⃗ BA ⃗ 3
(
= ,thatisit
−3
representsthecomplexnumber3 −3j,whileOC ⃗ 9
= ,thatisitrepresents9 +5j,so
5)
thatz 1
+z 2
=9+5jandz 1
−z 2
=3−3j.
9.7 THEEXPONENTIALFORMOFACOMPLEXNUMBER
You will recall from Chapter 6 that many functions possess a power series expansion,
thatisthefunctioncanbeexpressedasthesumofasequenceoftermsinvolvinginteger
powers ofx. For example,
e x =1+x+ x2
2! + x3
3! +···

338 Chapter 9 Complex numbers
andthisrepresentationisvalidforanyrealvalueofx.Theexpressiononther.h.s.is,of
course, an infinite sum but its terms get smaller and smaller, and as more are included,
the sum weobtain approaches e x . Other examples ofpower series include
and
sinx=x− x3
3! + x5
−··· (9.3)
5!
cosx=1− x2
2! + x4
−··· (9.4)
4!
whicharealsovalidforanyrealvalueofx.Itisusefultoextendtherangeofapplicability
of these power series by allowing x to be a complex number. That is, we define the
function e z tobe
e z =1+z+ z2
2! + z3
3! +···
andtheorybeyondthescopeofthisbookcanbeusedtoshowthatthisrepresentationis
valid for all complex numbersz.
We have already seen thatwecan express a complex number inpolar form:
z=r(cosθ+jsinθ)
Using Equations (9.3) and (9.4) wecan write
z =r
{(1 −···)
− θ2
2! + θ4
+j
(θ −···)}
− θ3
4! 3! + θ5
5!
=r
(1 ···)
+jθ− θ2
2! −jθ3 3! + θ4
4! +jθ5 5!
Furthermore, wenote thate jθ can be written as
so that
e jθ =1+jθ+ j2 θ 2
2!
+ j3 θ 3
3!
+···
=1+jθ− θ2
2! −jθ3 3! +···
z=r(cosθ+jsinθ)=re jθ
This is yet another form of the same complex number which we call the exponential
form. We see that
e jθ =cosθ+jsinθ (9.5)
Itis straightforward toshowthat
e −jθ =cosθ−jsinθ

9.7 The exponential form of a complex number 339
Therefore,ifz =r(cos θ −jsin θ)wecanequivalentlywritez =re −jθ .Thetwoexpressions
for e jθ and e −jθ are known as Euler’s relations. From these it is easy to obtain the
following usefulresults:
cosθ = ejθ +e −jθ
2
sinθ = ejθ −e −jθ
2j
Example9.16 WesawinSection3.7thatawaveformcanbewrittenintheform f (t) =Acos(ωt +φ).
Consider the complex numbere j(ωt+φ) . We can use Euler’s relations towrite
e j(ωt+φ) = cos(ωt + φ) +jsin(ωt + φ)
and hence,
f(t) =ARe(e j(ωt+φ) )
EXERCISES9.7
1 Find the modulusandargumentof
(a) 3e jπ/4 (b) 2e −jπ/6 (c) 7e jπ/3
2 Find the real andimaginaryparts of
(a) 5e jπ/3
(c) 11e jπ
(b) e j2π/3
(d) 2e −jπ
3 Expressz = 6(cos30 ◦ +jsin30 ◦ ) in exponential
form.Plotzonan Arganddiagramandfind itsreal
andimaginaryparts.
4 If σ, ω,T ∈ R,findthe real andimaginaryparts of
e (σ+jω)T .
5 Expressz = e 1+jπ/2 in the forma +bj.
6 Express −1 −jin the formre jθ .
7 Express
(a) 7 +5j and
1
(b)
2 − 1 jin exponential form.
3
8 Expressz 1 = 1 −jand
z 2 = 1 +j √
3 −j
in the formre jθ .
Solutions
1 (a) 3, π/4 (b) 2,−π/6 (c) 7,π/3
2 (a) 2.5, 4.3301 (b) −0.5, 0.8660 (c) −11,0
(d) −2, 0
3 6e (π/6)j ,Re(z) = 5.1962,Im(z) = 3
4 Real part: e σT cos ωT, imaginarypart: e σT sin ωT
5 ej
6 √ 2e (−3π/4)j
7 (a) √ 74e 0.62j (b) 0.60e −0.59j
8 √ 2e (−π/4)j ,
1
√
2
e (5π/12)j

340 Chapter 9 Complex numbers
9.8 PHASORS
Electrical engineers are often interested in analysing circuits in which there is an a.c.
power supply. Almost invariably the supply waveform is sinusoidal and the resulting
currentsandvoltageswithinthecircuitarealsosinusoidal.Forexample,atypicalvoltage
isof the form
v(t) =Vcos(ωt +φ) =Vcos(2πft +φ) (9.6)
whereV is the maximum or peak value, ω is the angular frequency, f is the frequency,
and φ is the phase relative to some reference waveform. This is known as the time
domain representation. Each of the voltages and currents in the circuit has the same
frequency as the supply but differs inmagnitude and phase.
In order to analyse such circuits it is necessary to add, subtract, multiply and divide
these waveforms. If the time domain representation is used then the mathematics
becomes extremely tedious. An alternative approach is to introduce a waveform representation
known as a phasor. A phasor is an entity consisting of two distinct parts: a
magnitude and an angle. It is possible to represent a phasor by a complex number in
polarform.Thefixedmagnitudeofthiscomplex number corresponds tothemagnitude
of the phasor and hence the amplitude of the waveform. The argument of this complex
number, φ, corresponds to the angle of the phasor and hence the phase angle of the
waveform. Figure 9.10 shows a phasor for the sinusoidal waveform of Equation (9.6).
The time dependency of the waveform is catered for by rotating the phasor anticlockwise
at an angular frequency, ω. The projection of the phasor onto the real axis
gives the instantaneous value of the waveform. However, the main interest of an engineer
is in the phase relationships between the various sinusoids. Therefore the phasors
are‘frozen’atacertainpointintime.Thismaybechosensothatt = 0oritmaybechosensothataconvenientphasor,knownasthereferencephasor,alignswiththepositive
realaxis.Thisapproachisvalidbecausethephaseandmagnituderelationshipsbetween
the various phasors remain the same at all points in time once a circuit has recovered
from any initial transients caused by switching.
Sometextbooksrefertophasorsasvectors.Thiscanleadtoconfusionasitispossible
todividephasorswhereasdivisionofvectorsbyothervectorsisnotdefined.Inpractice
this is not a problem as phasors, although thought of as vectors, are manipulated as
complex numbers, whichcan bedivided. We willavoidthese conceptual difficulties by
introducing a different notation. We will denote a phasor byṼ, which corresponds to
V ̸ φ incomplexnumbernotation(seeFigure9.10).Thus,forexample,acurrenti(t) =
Icos(ωt+φ)wouldbewrittenĨinphasornotationandI̸ φincomplexnumbernotation.
Many engineers use the root mean square (r.m.s.) value of a sinusoid as the magnitude
of a phasor. The justification for this is that it represents the value of a d.c. signal
that would dissipate the same amount of power in a resistor as the sinusoid. For
y
v
V
~
V
f
x
Figure9.10
Illustration ofthe phasorṼ =V̸ φ where ω =angular
frequency with which the phasor rotates.

9.8 Phasors 341
example, in the case of a current signal, Irms 2 R is the average power dissipated by the
sinusoidIcos(ωt +φ)inaresistorR.Forthecaseofasinusoidalsignalther.m.s.value
ofthesignalis1/ √ 2timesthepeakvalueofthesignal(seeSection15.3,Example15.4,
foraproof of this).We will not adopt thisapproach butitisacommon one.
We start by examining the phasor representation of individual circuit elements. In
order todo thiswe need a phasor form of Ohm’s law. This is
Ṽ
=ĨZ (9.7)
where Ṽ is the voltage phasor, Ĩ is the current phasor and Z is the impedance of an
element or group of elements and may be a complex quantity. Note that phasors and
complex numbers are mixed together in the same equation. This is a common practice
because phasors areusually manipulated as complex numbers.
9.8.1 Resistor
Experimentally it can be shown that if an a.c. voltage is applied to a resistor then the
currentisinphasewiththevoltage.Theratioofthemagnitudeofthetwowaveformsis
equal to the resistance,R. So, givenĨ = I̸ 0,Z=R̸ 0, using Equation (9.7) we have
Ṽ=IR̸ 0.This isillustratedinFigure 9.11.
9.8.2 Inductor
Foraninductorweknowfromexperimentthatthevoltageleadsthecurrentbyaphaseof
π/2, and so the phase angle of the impedance is π/2. We also know that the magnitude
oftheimpedanceisgivenby ωL.So,givenĨ =I̸ 0,Z=ωL̸ π/2,usingEquation(9.7)
we haveṼ =IωL̸ π/2. An alternative way of representingZ for an inductor is to use
the Cartesian form, that is
(
Z=ωLe jπ/2 = ωL cos π )
2 +jsinπ 2
= jωL
Thisisusefulwhenphasorsneedtobeaddedandsubtracted.Thephasordiagramforan
inductor isillustratedinFigure 9.12.
9.8.3 Capacitor
For a capacitor it is known that the voltage lags the current by a phase of π/2 and the
magnitude of the impedance is given by 1
ωC . So givenĨ = I̸ 0,Z= 1
ωC ̸ −π/2, we
y
y
p –2
~
I
V ~
x
~
I x
V ~
Figure9.11
Phasordiagram foraresistor.
Figure9.12
Phasordiagram foran inductor.

342 Chapter 9 Complex numbers
have, using Equation (9.7),Ṽ = I
ωC ̸ −π/2. Alternatively,
Z = e−jπ/2
ωC = 1 (
cos π )
ωC 2 −jsinπ 2
= − j
ωC
Engineers often prefer torewrite this lastexpression as
1
jωC
The phasor diagram forthe capacitor isillustratedinFigure 9.13.
Wehaveshownhowphasorscanbemultipliedbyacomplexnumber;divisionisvery
similar.Addition ofphasors will nowbeillustrated;subtraction issimilar.Consider the
circuit shown in Figure 9.14 in which a resistor, capacitor and inductor are connected
in series and fed by an a.c. source. As this is a series circuit, the current through each
element,Ĩ,isthesamebyKirchhoff’scurrentlaw.ByKirchhoff’svoltagelawthevoltage
rise produced by the supply,Ṽ S
, must equal the sum of the voltage drops across the
elements. Therefore,
Ṽ S
=ṼR +ṼC +ṼL
Note that this is an addition of phasors so that the voltage drops across the elements
do not necessarily have the same phase. The phasor diagram for the circuit is shown in
Figure 9.15, inthiscase with |ṼL | > |ṼC |;Ĩ isthe reference phasor.
Notethatthephasoradditionoftheelementvoltagesgivestheoverallsupplyvoltage
for a particular supply current. If the magnitude of these element voltage phasors is
known then it is possible to calculate the supply voltage graphically. In practice it is
easier to convert the polar form of the phasors into Cartesian form and use algebra to
analyse the circuit.
Now,
Ṽ R
=ĨR̸
Ṽ L
=ĨωL̸
Ṽ C
=
0 =ĨR
π/2 =ĨjωL
Ĩ Ĩ
ωC ̸ −π/2 =
jωC
y
~ I x
–
p –2
V ~
v
~
I
R
~
V R
~
~
~ V S
C V C
L
~
V L
y
~
V L
~
V C
~
V S
~ ~
I V x R
Figure9.13
Phasordiagram foracapacitor.
Figure9.14
RLC circuit.
Figure9.15
Phasordiagram forthe circuit in
Figure 9.14.

9.8 Phasors 343
Therefore,
Ṽ S
=ṼR +ṼC +ṼL
Ĩ
=ĨR +ĨjωL +
jωC
(
=Ĩ R +jωL+ 1 )
jωC
Therefore the impedance of the circuit is Z = R + jωL + 1 . We can calculate the
jωC
frequency forwhich the impedance of the circuithas minimum magnitude:
Now
Z=R+jωL+ 1
jωC
=R+jωL− j
(
=R+j
|Z| =
√
R 2 +
ωC
ωL − 1
ωC
)
(
ωL − 1 ) 2
ωC
and so, as ω varies |Z| isaminimum when
ωL − 1
ωC = 0
ω 2 = 1
√
LC
1
ω =
LC
This minimum value is |Z| = R. Examining Figure 9.15 it is clear that the minimum
impedance occurs whenṼL andṼC have the same magnitude, in which caseṼS has no
imaginary component. The frequency at which this occurs is known as the resonant
frequencyof the circuit.
Engineeringapplication9.1
ThePoyntingvector
An electromagnetic wave freely travelling in space has electric and magnetic field
componentswhichoscillateatrightanglestoeachotherandtothedirectionofpropagation.
This type of wave is known as a transverse wave. Figure 9.16 illustrates
anelectromagneticwavetravellinginfreespace.ThePoyntingvectorisusedtodescribetheenergyfluxassociatedwithanelectromagneticwave.IthasunitsofWm
−2
andisapowerperunitarea,thatisapowerdensity.Ifthetotalpowerassociatedwith
anelectromagneticwavefrontisrequiredthenthePoyntingvectorcanbeintegrated
over anareaofinterest.
ThePoynting vector, S, forelectricfield,E, and magnetic field, H,is definedas
S=E×H
➔

344 Chapter 9 Complex numbers
Electric field
Magnetic field
x
y
z
Figure9.16
An electromagnetic wave travelling in freespace.
where E and H are vector quantities. Note that S is also a vector quantity as this
expression isavector product.
It is possible to specify magnitude and phase of the field quantities using complex
numbers; that is, in terms of phasors. Consider the case when E and H are the
following:
E=(2+j0.5)i+0j+0k
H = 0i + (3.5 +j0.25)j +0k
HereCartesiancoordinateshavebeenusedtodefinethetwofieldquantitiesandiisa
unitvectorinthexdirection,jisaunitvectorintheydirectionandkisaunitvector
inthezdirection.Examiningthesetwotermswenotethattheelectricfieldisaligned
inthexdirection and the magnetic field isaligned intheydirection.
Calculating the Poynting vector we obtain
i j k
S=E×H=
2+j0.5 0 0
∣0 3.5 +j0.25 0 ∣
= 0i −0j + (2 +j0.5)(3.5 +j0.25)k = (6.875 +j2.25)k
Weseethattheonlynon-zerocomponentisalignedinthezdirection.Thisisconsistentwithourunderstandingofatransversewaveinthatthedirectionofpropagation
is at right angles to the two field components. This implies that the energy flow of
the electromagnetic wave isinthezdirection.
9.9 DEMOIVRE’STHEOREM
AveryimportantresultincomplexnumbertheoryisDeMoivre’stheoremwhichstates
that ifn ∈ N,
(cosθ +jsinθ) n = cosnθ +jsinnθ (9.8)
Example9.17 VerifyDe Moivre’s theorem whenn = 1 andn = 2.
Solution Whenn = 1,the theorem states:
(cosθ +jsinθ) 1 =cos1θ +jsin1θ

which isclearly true,and the theorem holds. Whenn = 2,wefind
(cosθ +jsinθ) 2 = (cosθ +jsinθ)(cosθ +jsinθ)
=cos 2 θ+jsinθcosθ+jcosθsinθ+j 2 sin 2 θ
=cos 2 θ −sin 2 θ +j(2sinθcosθ)
9.9 De Moivre’s theorem 345
Recallingthetrigonometricidentitiescos2θ = cos 2 θ − sin 2 θ andsin2θ = 2sin θ cos θ,
wecan write the previous expression as
cos2θ +jsin2θ
Therefore,
(cosθ +jsinθ) 2 =cos2θ +jsin2θ
and DeMoivre’s theorem has been verified whenn = 2.
The theorem also holds whennis a rational number, that isn = p/q where pandqare
integers.Thus wehave
(cosθ +jsinθ) p/q =cos p q θ+jsinp q θ
Inthisform itcan be used toobtain roots of complex numbers. For example,
√
3
cosθ +jsinθ =(cosθ +jsinθ) 1/3 =cos 1 3 θ+jsin1 3 θ
Insuchacasetheexpressionobtainedisonlyoneofthepossibleroots.Additionalroots
can be found as illustratedinExample 9.18.
De Moivre’s theorem is particularly important for the solution of certain types of
equation.
Example9.18 Findallcomplex numberszwhich satisfy
z 3 =1 (9.9)
Solution The solution of this equation is equivalent to finding the solutions ofz = 1 1/3 ; that is,
findingthecuberootsof1.Sinceweareallowingztobecomplex,thatisz ∈ C,wecan
write
z=r(cosθ+jsinθ)
Then,usingDeMoivre’s theorem,
z 3 =r 3 (cosθ +jsinθ) 3
=r 3 (cos3θ +jsin3θ)
Wenextconverttheexpressiononther.h.s.ofEquation(9.9)intopolarform.Figure9.17
shows the number1 = 1 +0jon anArgand diagram.
From the Argand diagram we see that its modulus is 1 and its argument is 0, or possibly
±2π, ±4π,..., thatis2nπ wheren ∈ Z.Consequently, wecan write
1 =1(cos2nπ +jsin2nπ) n ∈ Z

346 Chapter 9 Complex numbers
z 2
1
y
y
1 x
2p —3
4p —3
1 z 1
1
z 3
x
Figure9.17
The complex numberz = 1 +0j.
Figure9.18
Solutionsofz 3 = 1.
Using the polar form, Equation (9.9) becomes
r 3 (cos3θ +jsin3θ) = 1(cos2nπ +jsin2nπ)
Comparing both sides of thisequation we see that
and
r 3 =1 thatis r=1sincer∈R
3θ = 2nπ thatis θ = 2nπ/3 n ∈ Z
Apparently θ can take infinitely many values, but, as we shall see, the corresponding
complex numbers aresimplyrepetitions. Whenn = 0,we find θ = 0,so that
z=z 1
=1(cos0+jsin0)=1
isthe first solution.Whenn = 1 wefind θ = 2π/3,so that
(
z=z 2
=1 cos 2π )
3 +jsin2π = − 1 √
3
3 2 +j 2
isthe second solution.Whenn = 2 we find θ = 4π/3,so that
(
z=z 3
=1 cos 4π )
3 +jsin4π = − 1 √
3
3 2 −j 2
is the third solution. If we continue searching for solutions using larger values ofnwe
find that we only repeat solutions already obtained. It is often useful to plot solutions
on an Argand diagram and this is easily done directly from the polar form, as shown in
Figure 9.18. We note thatthe solutions are equally spaced atangles of 2π/3.
Example9.19 Findthe complex numberszwhichsatisfyz 2 = 4j.
Solution Sincez ∈ Cwewritez =r(cos θ +jsin θ). Therefore,
z 2 =r 2 (cosθ +jsinθ) 2
=r 2 (cos2θ +jsin2θ)

9.9 De Moivre’s theorem 347
y
2
5p —4
p –4
x
2
Figure9.19
Solution ofz 2 = 4j.
by De Moivre’s theorem. Furthermore, 4j has modulus 4 and argument π/2 + 2nπ,
n ∈ Z, thatis
4j = 4{cos(π/2 +2nπ) +jsin(π/2 +2nπ)}
n ∈ Z
Therefore,
r 2 (cos2θ +jsin2θ) = 4{cos(π/2 +2nπ) +jsin(π/2 +2nπ)}
Comparing both sides of this equation, we see that
and
r 2 =4 andso r=2
2θ=π/2+2nπ andso θ=π/4+nπ
Whenn = 0wefind θ = π/4,andwhenn = 1wefind θ = 5π/4.Usinglargervaluesof
n simply repeats solutions already obtained. These solutions are shown in Figure 9.19.
We note that in this example the solutions are equally spaced at intervals of 2π/2 = π.
InCartesian form,
z 1
= 2 √
2
+j 2 √
2
= √ 2(1+j) and z 2
=− 2 √
2
−j 2 √
2
= − √ 2(1 +j)
Ingeneral, thenrootsofz n =a +jbare equallyspacedatangles2π/n.
Once the technique for solving equations like those in Examples 9.18 and 9.19 has
beenmastered,engineersfinditsimplertoworkwiththeabbreviatedformr̸ θ.Example
9.19 reworked inthisfashion becomes
Letz=r̸ θ, then z 2 =r 2̸ 2θ
Furthermore, 4j = 4̸ π/2 +2nπ, and hence ifz 2 = 4j, we have
( ) π
2 +2nπ
r 2̸
fromwhich
2θ=4̸
r 2 =4 and 2θ= π 2 +2nπ
as before.Rework Example 9.18 foryourself usingthisapproach.

348 Chapter 9 Complex numbers
Engineeringapplication9.2
Characteristicimpedance
In a later application we shall explain the concept of the characteristic impedance
ofanelectricaltransmissionlineinmoredetail.Fornow,wecanassumethatitisan
important electricalparameter represented by the equation
√
R+jωL
Z 0
=
G+jωC
whereR,L,GandCaretransmissionlineparametersand ωistheangularfrequency,
all of which are real numbers. If R and G are zero, as they would be for an ideal
transmissionline, thenZ 0
iswholly real because:
√ √
jωL L
Z 0
=
jωC = C
Alternatively, ifGorRis included then the equation involves taking the square root
of a complex number.
Given values ofR,G,L,C and ω we can write
R+jωL
G+jωC
in Cartesian form (see Example 9.9) and hence in polar form (see Section 9.5).
Application of De Moivre’s Theorem will then allow us to calculate the required
square roots tofindZ 0
.
Another application of De Moivre’s theorem is the derivation of trigonometric
identities.
Example9.20 UseDeMoivre’s theoremtoshow that
cos3θ =4cos 3 θ−3cosθ
and
sin3θ =3sinθ−4sin 3 θ
Solution We know that
(cosθ +jsinθ) 3 =cos3θ +jsin3θ
Expanding the l.h.s. wefind
cos 3 θ+3jcos 2 θsinθ−3cosθsin 2 θ−jsin 3 θ =cos3θ+jsin3θ
Equating the real parts gives
cos 3 θ −3cos θ sin 2 θ = cos3θ (9.10)
and equating the imaginarypartsgives
3cos 2 θ sinθ −sin 3 θ = sin3θ (9.11)

9.9 De Moivre’s theorem 349
Now, writing sin 2 θ = 1 −cos 2 θ in Equation (9.10) and cos 2 θ = 1 −sin 2 θ in Equation
(9.11),wefind
cos3θ =cos 3 θ −3cosθ(1−cos 2 θ)
=cos 3 θ+3cos 3 θ−3cosθ
=4cos 3 θ−3cosθ
sin3θ = 3(1 −sin 2 θ)sinθ −sin 3 θ
as required.
=3sinθ−4sin 3 θ
This technique allows trigonometric functions of multiples of angles to be expressed
in terms of powers. Sometimes we want to carry out the reverse process and express a
power interms ofmultiple angles. Consider Example 9.21.
Example9.21 Ifz = cos θ +jsin θ show that
z + 1 z =2cosθ
z−1 z =2jsinθ
and find similarexpressions forz n + 1 z n andzn − 1 z n.
Solution Consider the complex number
z=cosθ+jsinθ
UsingDe Moivre’s theorem,
1
z =z−1 = (cosθ +jsinθ) −1 = cos(−θ) +jsin(−θ)
Butcos(−θ) = cosθ andsin(−θ) = −sinθ,sothatifz = cosθ +jsinθ
1
z =cosθ−jsinθ
Consequently,
z + 1 z =2cosθ and z−1 z =2jsinθ
Moreover,
so that
z n = cosnθ +jsinnθ and z −n = cosnθ −jsinnθ
z n + 1 z n =2cosnθ and zn − 1 z n = 2jsinnθ
z n + 1 z n =2cosnθ and zn − 1 z n = 2jsinnθ

350 Chapter 9 Complex numbers
Example9.22 Show thatcos 2 θ = 1 (cos2θ +1).
2
Solution The formulae obtained in Example 9.21 allow us to obtain expressions for powers of
cosθ andsinθ.Since2cosθ =z+ 1 , squaring both sides wehave
z
(
2 2 cos 2 θ = z + 1 ) 2
=z 2 +2+ 1 z z 2
=
(z 2 + 1 )
+2
z 2
Butz 2 + 1 z =2cos2θ,so 2
2 2 cos 2 θ =2cos2θ+2
and therefore,
cos 2 θ = 1 2 cos2θ + 1 2
as required.
= 1 (cos2θ +1)
2
EXERCISES9.9
1 Express (cos θ +jsinθ) 9 and (cos θ +jsinθ) 1/2 in
the formcosnθ +jsinnθ.
2 UseDeMoivre’stheoremto simplify
(a) (cos3θ +jsin3θ)(cos4θ +jsin4θ)
cos8θ +jsin8θ
(b)
cos2θ −jsin2θ
3 Solve the equations
(a) z 3 +1=0
(b) z 4 =1+j
4 Find 3 √ 2 +2j and displayyoursolutionsonan
Argand diagram.
5 Prove the following trigonometricidentities:
(a) cos4θ =8cos 4 θ−8cos 2 θ+1
(b) 32sin 6 θ = 10 −15cos2θ +6cos4θ −cos6θ
6 Solvethe equationz 4 +25 = 0.
7 Findthe fifthrootsofjanddepictyoursolutionson
an Arganddiagram.
8 Show thatcos 3 θ = 1 (cos3θ +3cosθ).
4
9 Show thatsin 4 θ = 1 (cos4θ −4cos2θ +3).
8
10 Expresscos5θ in termsofpowers ofcos θ.
11 Expresssin5θ in terms ofpowersofsin θ.
12 Given e jθ = cos θ +jsin θ,proveDeMoivre’s
theoremin the form
(e jθ ) n =cosnθ+jsinnθ
Solutions
1 cos9θ +jsin 9θ,cos(θ/2) +jsin(θ/2)
2 (a) cos 7θ +jsin 7θ
(b) cos 10θ +j sin 10θ
3 (a)1̸ π/3 +2nπ/3 n = 0,1,2
(b) 2 1/8̸ π/16 +nπ/2 n = 0,1,2,3
4 8 1/6̸ π/12 +2nπ/3 n = 0,1,2

9.10 Loci and regions of the complex plane 351
6 √ 5̸ π/4+nπ/2 n=0,1,2,3
7 1̸ π/10 +2nπ/5 n = 0,1,2,3,4
10 16cos 5 θ−20cos 3 θ+5cosθ
11 16sin 5 θ−20sin 3 θ+5sinθ
9.10 LOCIANDREGIONSOFTHECOMPLEXPLANE
Regionsofthecomplexplanecanoftenbeconvenientlydescribedbymeansofcomplex
numbers. For example, the points that lie on a circle of radius 2 centred at the origin
(Figure 9.20) represent complex numbers all of which have a modulus of 2. The argumentsareanyvalueof
θ, −π < θ π.Wecandescribeallthepointsonthiscircleby
the simple expression
|z|=2
thatis,allcomplexnumberswithmodulus2.Wesaythatthelocus(orpath)ofthepoint
z is a circle, radius 2, centred at the origin. The interior of the circle is described by
|z| < 2 whileitsexterior isdescribed by |z| > 2.
Similarlyallpointslyinginthefirstquadrant(shadedinFigure9.21)havearguments
between 0 and π/2.This quadrantistherefore describedby the expression:
0 < arg(z) < π/2
y
y
2
z
y
2
x
x
p –4
arg z = p – 4
x
Figure9.20
A circle, radius2, centred atthe
origin.
Figure9.21
First quadrant ofthex--y plane.
Figure9.22
Locusofpointssatisfying
arg(z) = π/4.
Example9.23 Sketch the locus ofthe pointsatisfyingarg(z) = π/4.
Solution The set of points with arg(z) = π/4 comprises complex numbers whose argument is
π/4.Allthesecomplex numbers lie onthe lineshown inFigure 9.22.
Example9.24 Sketch the locus ofthe pointsatisfying |z −2| = 3.
Solution First mark the fixed point 2 on the Argand diagram labelling it ‘A’ (Figure 9.23). Consider
the complex numberzrepresented by the point P. From the vector triangle law of
addition
⃗OA + ⃗ AP = ⃗ OP
⃗AP = ⃗ OP − ⃗ OA

352 Chapter 9 Complex numbers
y
y
z
P
z such that |z – 2| = 3
1 2 3 4 5
x
A
0 1 2
x
Figure9.23
Pointszand2 +0j.
Figure9.24
Locusofpointssatisfying |z −2| = 3.
Recall from Section 9.6 that the graphical representation of the sum and difference of
vectors in the plane, and the sum and difference of complex numbers, are equivalent.
Sincevector ⃗ OPrepresentsthecomplexnumberz,andvector ⃗ OArepresentsthecomplex
number2, ⃗ AP = ⃗ OP− ⃗ OAwillrepresentz−2.Therefore |z−2|representsthedistance
betweenAandP.Wearegiventhat |z−2| = 3,whichthereforemeansthatPcanbeany
point such that its distance from A is 3. This means that P can be any point on a circle
ofradius3centredatA(2,0).ThelocusisshowninFigure9.24. |z −2| < 3represents
the interior of the circle while |z −2| > 3 represents the exterior. Alternatively we can
obtain the same result algebraically: given |z −2| = 3 and also thatz =x +jy, we can
write
|z−2|=|(x−2)+jy|=3
that is
√
(x−2)2 +y 2 =3
or
(x−2) 2 +y 2 =9
Generally,theequation (x−a) 2 +(y−b) 2 =r 2 representsacircleofradiusrcentredat
(a,b), soweseethat (x −2) 2 +y 2 = 9representsacircleofradius3centredat (2,0),
as before.
Example9.25 Usethe algebraicapproachtofind the locusofthe pointzwhichsatisfies
|z−1|= 1 2 |z−j|
Solution Ifz =x +jy, then we have
|(x−1)+jy|= 1 |x +j(y −1)|
2
Therefore,
(x−1) 2 +y 2 = 1 4 {x2 + (y−1) 2 }

9.10 Loci and regions of the complex plane 353
and so
thatis
4(x−1) 2 +4y 2 =x 2 +(y−1) 2
3x 2 −8x+3y 2 +2y+3=0
Bycompleting the square thismay be writteninthe form
(
3 x −
3) 4 2 (
+3 y + 1 ) 2
= 8 3 3
thatis
(
x − 4 2 (
+ y +
3) 1 ) 2
= 8 3 9
which represents a circle of radius √ ( 4
8/3 centred at
3 3)
,−1 .
EXERCISES9.10
1 Sketchthe locidefined by
(a) arg(z) = 0
(b) arg(z) = π/2
(c) arg(z −4) = π/4
(d) |2z| = |z −1|
2 Sketchthe regionsdefined by
(a) Re(z) 0
(b) Im(z) < 3
(c) |z| > 3
(d) 0 arg(z) π/2
(e) |z+2|3
(f) |z+j|>3
(g) |z−1|<|z−2|
3 Ifs = σ +jω sketchthe regionsdefined by
(a) σ0
(b)σ0
(c) −2ω2
Solutions
1 SeeFigure S.19.
arg(z) = 0
arg(z) = p/2
arg(z – 4) = p/4
2_
3
– 1_ 1_ 3 3
2|z| = |z – 1|
4
(a)
(b)
(c)
(d)
FigureS.19

354 Chapter 9 Complex numbers
2 See FigureS.20.
y
y
y
3
Im(z) , 3
3
|z| . 3
y
0 arg(z) p/2
Re(z) 0
x
x
x
x
(a)
(b)
(c)
(d)
3
y
y
2
|z+j| . 3
y
|z –1| , |z –2|
–2
x
|z+2| 3
–1
3
x
1 2
x
(e)
(f)
(g)
FigureS.20
3 See FigureS.21.
v v v
2
s > 0
–2 < v < 2
s < 0
s s s
–2
(a) (b) (c)
FigureS.21
REVIEWEXERCISES9
1 Show that
1
cosθ−jsinθ =cosθ+jsinθ.
2 Expressin Cartesian form
5+4j
(a)
5−4j
1
(c)
2+3j + 1
2−3j
1
(b)
2+3j
1
(d)
x−jy
3 Find the modulusandargumentof −j, −3,1 +j,
cosθ+jsinθ.
4 Markonan Arganddiagramvectors corresponding to
the followingcomplex numbers: −3 +2j, −3 −2j,
cosπ+jsinπ.
5 Expressin the forma +bj:
(a) (cosθ +jsinθ) 6
(c)
cosθ+jsinθ
cosφ+jsinφ
6 Ifz ∈ C,showthat
(a) z+z =2Re(z)
(c) zz = |z| 2
(b)
1
(cosθ +jsinθ) 3
(b) z −z =2jIm(z)
7 Show thate jωt +e −jωt = 2cosωt andfindan
expression fore jωt −e −jωt .
8 Express1 +e 2jωt in the forma +bj.

Review exercises 9 355
9 Sketch the region in the complex plane described by
|z+2j|<1.
10 Expresse (1/2−6j) in the forma +bj.
11 Solvethe equationz 4 +1 = j √ 3.
12 Expresss 2 +6s +13intheform (s −a)(s −b)
wherea,b∈ C.
13 Express2s 2 +8s +11intheform2(s −a)(s −b)
wherea,b∈ C.
Solutions
2 (a)
(d)
9
41 + 40
41 j (b) 2
13 − 3 13 j (c) 4
13
y
x 2 +y 2j
x
x 2 +y 2 +
3 | −j| = 1,arg(−j) = −π/2; |−3| = 3,arg(−3) = π;
|1+j|= √ 2,arg(1+j) = π/4,|cosθ +jsinθ| =1,
arg(cosθ +jsinθ)=θ
5 (a) cos 6θ +jsin6θ (b) cos 3θ −jsin 3θ
(c) cos(θ −φ)+jsin(θ −φ)
7 e jωt −e −jωt = 2jsin ωt
8 1+cos2ωt+jsin2ωt
10 1.5831 +0.4607j
11 2 1/4̸ π/6+nπ/2 n=0,1,2,3
12 [s − (−3 +2j)][s − (−3 −2j)]
[ ( √ )][ (
6
13 2 s − −2 +
2 j s− −2 −
√
6
2 j )]

10 Differentiation
Contents 10.1 Introduction 356
10.2 Graphicalapproachtodifferentiation 357
10.3 Limitsandcontinuity 358
10.4 Rateofchangeataspecificpoint 362
10.5 Rateofchangeatageneralpoint 364
10.6 Existenceofderivatives 370
10.7 Commonderivatives 372
10.8 Differentiationasalinearoperator 375
Reviewexercises10 385
10.1 INTRODUCTION
Differentiation is a mathematical technique for analysing the way in which functions
change. In particular, it determines how rapidly a function is changing at any specific
point.Asthefunctioninquestionmayrepresentthemagneticfieldofamotor,thevoltage
acrossacapacitor,thetemperatureofachemicalmix,etc.,itisoftenimportanttoknow
howquicklythesequantitieschange.Forexample,ifthevoltageonanelectricalsupply
network is falling rapidly because of a short circuit, then it is necessary to detect this
in order to switch out that part of the network where the fault has occurred. However,
the system should not take action for normal voltage fluctuations and so it is important
to distinguish different types and rates of change. Another example would be detecting
a sudden rise in the pressure of a fermentation vessel and taking appropriate action to
stabilize the pressure.
Differentiationwillbeintroducedinthischapter.Weshallderiveaformulawhichcan
be used to find the rate of change of a function. To avoid always having to resort to the
formulaengineersoftenuseatableofderivatives;suchatableisgiveninSection10.7.
The chapter closes with a discussion of an important property of differentiation -- that
of linearity.

10.2 Graphical approach to differentiation 357
y (t)
1 2 3 4 5 6 7 8 9 10 11 12 t
Figure10.1
Thefunctiony(t)hasdifferentratesof
change over different regionsoft.
10.2 GRAPHICALAPPROACHTODIFFERENTIATION
Differentiation is concerned with the rate at which a function is changing, rather than
the actual change itself. We can explore the rate of change of a function by examining
Figure 10.1. There are several regions to this curve corresponding to different intervals
oft. In the interval [0, 5] the function does not change at all. The rate of change of y
is zero. Fromt = 5 tot = 7 the function increases slightly. Thus the rate of change of
y ast increases is small. Sinceyis increasing, the rate of change ofyis positive. From
t = 7 tot = 8 there is a rapid rise in the value of the function. The rate of change ofy
islargeandpositive.Fromt = 8tot = 9thevalueofydecreasesveryrapidly.Therate
ofchangeofyislargeandnegative.Finallyfromt = 9tot = 12thefunctiondecreases
slightly. Thus the rate ofchangeofyissmalland negative.
The aim of differential calculus is to specify the rate of change of a function precisely.Itisnotsufficienttosay‘therateofchangeofafunctionislarge’.Werequirean
exactvalueorexpressionfortherateofchange.Beforebeingabletodothisweneedto
introduce two conceptsconcerningthe rate ofchangeofafunction.
10.2.1 Averagerateofchangeofafunctionacrossaninterval
ConsiderFigure10.2.Whent =t 1
,thefunctionhasavaluey(t 1
).ThisisdenotedbyA
on the curve. Whent =t 2
, the function has a value ofy(t 2
). This point is denoted by B
on the curve. The function changes by an amounty(t 2
) −y(t 1
) over the interval [t 1
,t 2
].
Thusthe average rate ofchangeofthe function over the intervalis
changeiny
change int = y(t 2 ) −y(t 1 )
t 2
−t 1
The straight line joining A and B is known as a chord. Graphically,y(t 2
) −y(t 1
) is the
verticaldistanceandt 2
−t 1
thehorizontaldistancebetweenAandB,sothatthegradient
of the chord AB isgiven by
BC
AC = y(t 2 ) −y(t 1 )
t 2
−t 1
Thegradientorslopeofalineisameasureofitssteepnessandlinesmayhavepositive,
negative orzero gradients as shown inFigure 10.3.
Thus the gradient of the chord AB corresponds to the average rate of change of the
function between Aand B. To summarize:
TheaveragerateofchangeofafunctionbetweentwopointsAandBisthegradient
ofthe chord AB.

358 Chapter 10 Differentiation
y (t)
y (t 2 )
B
y
Positive
gradient
y (t 1 )
A
t 1
(t 2 – t 1 )
Figure10.2
Average rate ofchange across an interval.
u
t 2
C
y (t 2 ) – y (t 1 )
t
Zero
gradient
Negative
gradient
Figure10.3
Linescan have different gradients.
t
y (t)
B 1
Chord AB 1
Chord AB 2
A
B 2
Tangent at A
10.2.2 Rateofchangeofafunctionatapoint
t
Figure10.4
Point B ismoved nearer to Ato improve
accuracy.
Consider again Figure 10.2. Suppose we require the rate of change of the function at
pointA.WecanusethegradientofthechordABasanapproximationtothisvalue.IfB
isclosetoAthentheapproximationisbetterthanifBisnotsoclosetoA.Thereforeby
moving B nearer to A it is possible to improve the accuracy of this approximation (see
Figure 10.4).
Suppose the chord AB is extended as a straight line on both sides of AB, and B is
moved closer and closer to A until both points eventually coincide. The straight line
becomesatangenttothecurveatA.Thisisthestraightlinethatjusttouchesthecurve
atA.However,therateofchangeofthistangent,thatisitsgradient,stillcorrespondsto
the rate of change of the function, but now it is the rate of change of the function at the
point A.To summarize:
The rate of change of a function at a point A on the curve is the gradient of the
tangenttothe curve atpointA.
We have still to address the question of how the gradients of chords and tangents are
found.This requires a knowledge oflimits whichisthe topic ofthe next section.
10.3 LIMITSANDCONTINUITY
The concept of a limit is crucial to the development of differentiation. We write
t →c to denote thatt approaches, or tends to, the value ofc. Note carefully that this is
distinct from statingt = c. Ast tends tocwe consider the value to which the function
approachesand call thisvalue thelimit ofthe functionast →c.

10.3 Limits and continuity 359
f (t)
5
y = 1 – x
y
4
3
y = x +1
2
–3 2 t
Figure10.5
Thecurvef(t) =t 2 +2t −3.
–1
Figure10.6
Asx → 0,y → 1,even though
y(0) =3.
1
0
3 x
Example10.1 Ift → 2,whatvalue does
approach?
f(t)=t 2 +2t−3
Solution Figure 10.5 shows a graph of f (t). Clearly, whethert = 2 is approached from the l.h.s.
orther.h.s.thefunctiontendsto5.Thatis,ift → 2,then f (t) → 5.Wenotethatthisis
thevalueof f (2).Informallywearesayingthatast getsnearerandnearertothevalue2,
so f (t)gets nearerand nearer to5.This isusuallywritten as
lim(t 2 +2t−3)=5
t→2
where ‘lim’ is an abbreviation of limit. In this example, the limit of f (t) ast → 2 is
simply f (2), butthisisnottrueforall functions.
Example10.2 Figure 10.6 illustratesy(x) defined by
⎧
⎨1−x x<0
y(x) = 3 x=0
⎩
x+1 x>0
Evaluate:
(a) lim x→3
y (b) lim x→−1
y (c) lim x→0
y
Solution We notethat thisfunctionispiecewise continuous.Ithas a discontinuityatx = 0.
(a) Weseekthelimitofyasxapproaches3.Asxapproaches3,wewillbeonthatpart
of the function defined byx > 0, that isy(x) =x +1. Asx → 3, theny → 3 +1,
that isy → 4.So
lim y = 4
x→3
(b) Whenxapproaches −1, we will be on that part of the function defined byx < 0,
thatisy(x) = 1 −x.Soasx → −1,theny → 1 − (−1),thatisy → 2.Hence
lim y = 2
x→−1
(c) Asxapproaches0whatvaluedoesyapproach?Notethatwearenotevaluatingy(0)
which actually has a value of 3. We simply ask the question ‘What value isynear

360 Chapter 10 Differentiation
whenxisnear,butdistinctfrom,0?’FromFigure10.6weseeyisnearto1,thatis
lim y = 1
x→0
Example10.3 The functiony(x) isdefined by
⎧
⎨0 x0
y(x) = x 0<x2
⎩
x−2 x>2
(a) Sketch the function.
(b) State the limitofyasxapproaches (i)3,(ii)2,(iii)0.
Solution (a) ThefunctionisshowninFigure10.7.Notethatthefigurehasthreeparts;eachpart
corresponds toapartinthe algebraic definition.
(b) (i) Asx → 3,the relevant partof the function isy(x) =x −2.Hence
lim y = 1
x→3
(ii) Supposex < 2 and gradually increases, approaching the value 2. Then, from
the graph, we see that y approaches 2. Now, suppose x > 2 and gradually
decreases,tendingto2.Inthiscaseyapproaches0.Hence,wecannotfindthe
limitofyasxtends to2.The lim x→2
ydoes notexist.
(iii) Asxtends to 0,ytends to 0. This is true whetherxapproaches 0 from below,
thatisfrom the left,orfrom above, thatisfrom the right.So,
lim y = 0
x→0
y
y = x
y = x – 2
y = 0
2
x
Figure10.7
Thefunctionyhas different limitsasx → 2
from the left and the right.
Itisappropriateatthisstagetointroducetheconceptofleft-handandright-handlimits.
ReferringtoExample10.3,weseethatasxapproaches2fromtheleft,thatisfrombelow,
then y approaches 2. We say that the left-hand limit of y as x tends to 2 is 2. This is
written as
lim
x→2 −y = 2
Similarly, the right-hand limit ofyis obtained by lettingxtend to 2 from above. In this
case,yapproaches0.This iswritten as
lim
x→2 +y = 0
Consider a point at which the left-hand and right-hand limits are equal. At such a point
wesay ‘thelimitexists atthatpoint’.

10.3 Limits and continuity 361
Thelimitofafunction, atapointx =a, existsonly iftheleft-handand right-hand
limits areequal there.
10.3.1 Continuousanddiscontinuousfunctions
Afunction f iscontinuousatthe point wherex =a, if
lim f =f(a)
x→a
that is, the limit value matches the function value at a point of continuity. A function
which is not continuous is discontinuous. In Example 10.3, the function is continuous
atx = 0 because
lim y=0=f(0)
x→0
butdiscontinuousatx = 2becauselim x→2
ydoesnotexist.InExample10.2,thefunction
isdiscontinuousatx = 0becauselim x→0
y = 1buty(0) = 3.Theconceptofcontinuity
corresponds to our natural understanding of a break in the graph of the function, as
discussed inChapter 2.
Afunction f iscontinuous atapointx =aifand only if
lim x→a
f = f(a)
thatis, the limitof f exists atx =aand isequal to f (a).
EXERCISES10.3
1 Thefunction, f (t),is defined by
⎧
⎨1 0t2
f(t)= 2 2<t3
⎩
3 t>3
Sketchagraphof f (t)andstatethefollowinglimitsif
they exist:
(a) lim t→1.5 f
(b) lim t→2 + f
(c) lim t→3 f
(d) lim t→0 + f
(e) lim t→3 − f
2 The functiong(t)is defined by
⎧
0 t<0
⎪⎨
t
g(t) =
2 0t3
2t+3 3<t 4
⎪⎩
12 t>4
(a) Sketchg.
(b) State any pointsofdiscontinuity.
(c) Find, ifthey exist,
(i) lim t→3 g
(ii) lim t→4 g
(iii) lim t→4 −g
Solutions
1 (a) 1 (b) 2 (c) notdefined
(d) 1 (e) 2
2 (b)t=4
(c) (i) 9 (ii) not defined (iii) 11

362 Chapter 10 Differentiation
10.4 RATEOFCHANGEATASPECIFICPOINT
WesawinSection10.2thattherateofchangeofafunctionatapointisthegradientof
thetangenttothecurveatthatpoint.Also,wecanthinkofatangentatAasthelimitof
an extended chord AB as B → A. We now put these two ideas together to find the rate
of change of a function atapoint.
Example10.4 Given y = f (x) = 3x 2 + 2, obtain estimates of the rate of change of y at x = 3 by
considering the intervals
(a) [3,4] (b) [3,3.1] (c) [3,3.01]
Solution (a) Consider Figure 10.8.
y(3) = 3(3) 2 +2 =29
y(4) = 3(4) 2 +2 =50
Let Abe the point (3, 29) on the curve. Let Bbe the point (4, 50).Then
average rate of change over the interval [3,4] =
=
change iny
change inx
y(4) −y(3)
4 −3
= 50−29
4 −3 = 21
ThisisthegradientofthechordABandisanestimateofthegradientofthetangent
atA.Thatis, the rate of change atAisapproximately 21.
(b) y(3.1) = 30.83 and so,
average rate of change over the interval [3,3.1] =
This isamore accurate estimateof the rate of change atA.
(c) y(3.01) = 29.1803 and so,
30.83 −29
3.1−3
29.1803 −29
average rate of change over the interval [3,3.01] =
3.01 −3
= 18.03
= 18.3
y
50
B
29
A
3 4
x
Figure10.8
Thefunction:y = 3x 2 +2.

10.4 Rate of change at a specific point 363
ThisisanevenbetterestimateoftherateofchangeatA.HenceatA,ifxincreases
by 1 unit then y increases by approximately 18 units. This corresponds to a steep
upward slope atA.
Example10.4illustratestheapproachofestimatingtherateofchangeatapointbyusing
the‘shrinkinginterval’method.Bytakingsmallerandsmallerintervals,betterandbetter
estimates of the rate of change of the function atx = 3 can be obtained. However, we
eventuallywanttheintervalto‘shrink’tothepointx = 3.Weintroduceasmallchange
orincrementofxdenotedby δxandconsidertheinterval[3,3+δx].Byletting δxtend
tozero,the interval [3,3 + δx] effectively shrinks tothe pointx = 3.
Example10.5 Find the rate of change ofy = 3x 2 +2 atx = 3 by considering the interval [3,3 + δx]
and letting δx tend to0.
Solution Whenx = 3,y(3) = 29. Whenx = 3 + δx then
So,
y(3 + δx) = 3(3 + δx) 2 +2
=3(9+6δx+(δx) 2 )+2
= 3(δx) 2 +18δx +29
average rate of change ofyacross [3,3 + δx] =
change iny
change inx
= (3(δx)2 +18δx +29) −29
δx
= 3(δx)2 +18δx
δx
=
δx(3δx +18)
δx
=3δx+18
We now let δx tend to0,so thatthe intervalshrinks toapoint:
rate of change ofywhenxis3 = lim
δx→0
(3δx +18) =18
We have found the rate of change ofyat a particular value ofx, rather than across an
interval.We usually write
( ) y(3 + δx) −y(3)
rate of change ofywhenxis3 = lim
δx→0 δx
( 3(δx) 2 )
+18δx
= lim
= lim (3δx +18)
δx→0 δx
δx→0
= 18

364 Chapter 10 Differentiation
EXERCISES10.4
1 Find the rate ofchange ofy = 3x 2 +2 at
(a) x = 4 by considering the interval[4,4 + δx]
(b) x = −2byconsideringtheinterval[−2,−2 + δx]
(c) x = 1 byconsidering the interval[1 − δx,1 + δx]
2 Find the rate ofchange ofy = 1/x atx = 2.
3 To determine the rate ofchange ofy =x 2 −x
atx = 1 the interval [1,1 + δx] could be used.
Equally the intervals [1 − δx,1]or
[1 − δx,1 + δx] could be used.Show that the
same answer resultsregardless ofwhich interval is
used.
4 Findtherateofchangeofy(x) = 2 −x 2 at
(a) x = 3,byconsidering the interval[3,3 + δx]
(b) x = −5,byconsidering the interval
[−5,−5 + δx]
(c) x = 1,byconsidering the interval
[1 − δx,1 + δx].
5 Findthe rate ofchange ofy(x) = x
x +3 atx=3by
considering the interval [3,3 + δx].
Solutions
1 (a) 24 (b) −12 (c) 6
4 (a) −6 (b) 10 (c) −2
2 −0.25
3 1
5
1
12
10.5 RATEOFCHANGEATAGENERALPOINT
Example 10.5 shows that the rate of change of a function at a particular point can be
found. We will now develop a general terminology for the method. Suppose we have a
function ofx,y(x). We wish to find the rate of change ofyat a general value ofx. We
beginbyfindingtheaveragerateofchangeofy(x)acrossanintervalandthenallowthe
intervaltoshrinktoasinglepoint.Considertheinterval[x,x + δx].Atthebeginningof
theintervalyhasavalueofy(x).Attheendoftheintervalyhasavalueofy(x + δx)so
that the changeinyisy(x + δx) −y(x), whichwedenoteby δy(see Figure 10.9).So,
average rate ofchangeofy = changeiny
change inx
=
y(x + δx) −y(x)
δx
= δy
δx
Now let δx tend to0,so thatthe interval shrinks toapoint.Then
( ) ( )
y(x + δx) −y(x) δy
rate of change ofy = lim
= lim
δx→0 δx
δx→0 δx
To see how weproceed toevaluate this limitconsider Example 10.6.

10.5 Rate of change at a general point 365
y
dy = y(x + dx) – y(x)
y(x + dx)
y(x)
x
x + dx
dx
x
Figure10.9
Therate ofchange ofyat apoint is foundby letting δx → 0.
Example10.6 Findtherateofchangeofy(x) = 2x 2 +3x.Calculatetherateofchangeofywhenx = 2
and whenx = −3.
Solution Giveny(x) = 2x 2 +3x
then
Hence
So,
y(x + δx) = 2(x + δx) 2 +3(x + δx)
= 2x 2 +4xδx +2(δx) 2 +3x +3δx
y(x + δx) −y(x) =2(δx) 2 +4xδx +3δx
( ) y(x + δx) −y(x)
rate ofchangeofy = lim
δx→0 δx
= lim
δx→0
( 2(δx) 2 +4xδx +3δx
δx
)
= lim
δx→0
(2δx+4x+3)=4x+3
Whenx = 2,therateofchangeofyis4(2) +3 = 11.Whenx = −3,therateofchange
ofyis4(−3) +3 = −9.Apositiverateofchangeshowsthatthefunctionisincreasing
at that particular point. A negative rate of change shows that the function is decreasing
atthatparticular point.
( ) δy
Therateofchangeofyiscalledthederivativeofy.Wedenote lim by dy
δx→0 δx dx .This
ispronounced ‘deeyby deex’.
rate ofchangeofy(x) = dy ( ) y(x + δx) −y(x)
dx = lim
δx→0 δx
Note thatthe notation dy δy
means lim
dx δx→0 δx .

366 Chapter 10 Differentiation
then
Usingthe previous example, if
y(x) =2x 2 +3x
dy
dx =4x+3
dy
isoften abbreviated toy′
dx
y ′ is pronounced ‘y dash’ or ‘y prime’. To stress thatyis the dependent variable andx
the independent variable we often talk of ‘the rate of change ofywith respect tox’, or,
more compactly, ‘the rate of change of y w.r.t. x’. The process of finding y ′ from y is
calleddifferentiation.Thisshrinkingintervalmethodoffindingthederivativeiscalled
differentiation from first principles. We know that the derivative dy is the gradient
dx
of the tangent to the function at a point. It is also the rate of change of the function. In
manyexamples,theindependentvariableist andweneedtofindtherateofchangeofy
withrespecttot;thatis,find dy
dt .Thisisalsooftenwrittenasy′ althoughẏ,pronounced
‘ydot’,isalsocommon.Thereadershouldbeawareofbothnotations.Finally,y ′ isused
to denote the derivative ofywhatever the independent variable may be. So dy
dz , dy
dr and
dy
dw could all be represented byy′ .
Example10.7 Findthe gradientofthe tangent toy =x 2 atA(1,1),B(−1,1) andC(2,4).
Solution We havey =x 2 and so
y(x + δx) = (x + δx) 2
and hence
Then
=x 2 +2xδx + (δx) 2
y(x + δx) −y(x) =x 2 +2xδx + (δx) 2 −x 2
dy
dx
= 2xδx + (δx) 2
= gradient of a tangent tocurve
( ) y(x + δx) −y(x)
= lim
δx→0 δx
)
= lim
δx→0
( (x + δx) 2 −x 2
δx
= lim
δx→0
(2x + δx) =2x
( ) 2xδx + (δx)
2
= lim
δx→0 δx

10.5 Rate of change at a general point 367
At (1, 1), dy
dy
= 2 = gradient of tangent at A. At (−1,1),
dx dx
tangent atB. At (2, 4), dy = 4 = gradient of tangent atC.
dx
= −2 = gradient of
Suppose we wish to evaluate the derivative, dy
dx , at a specific value ofx, sayx 0
. This is
denoted by
dy
dx (x =x 0 )
dy
or more compactly by
dx (x 0 ) ory′ (x 0
)
An alternative notation is
dy
∣
dx
∣
x=x0
or
dy
∣
dx
∣
x0
So,for Example 10.7 wecould have written
dy
dx (1)=2
dy
dx (x=−1)=−2
dy
dx∣ =4 y ′ (2)=4
x=2
Example10.8 Refer to Figure 10.10. By considering the gradient of the tangent at the points A, B, C,
Dand Estatewhether dy ispositive, negative orzero atthese points.
dx
Solution AtAandCthetangentisparalleltothexaxisandso dy
dx iszero.AtBandEthetangent
has a positive gradient and so dy is positive. At D the tangent has a negative gradient
dx
and thus dy
dx isnegative.
y
E
B
A
D
C
x
Figure10.10
Graph forExample 10.8.
AswesawinChapter3,functionsareusedtorepresentphysicallyimportantquantities
suchasvoltageandcurrent.Whenthecurrentthroughcertaindeviceschanges,thiscan
give rise to voltages, the magnitudes of which are proportional to the rate of change of
the current. Consequently differentiation is needed to model these effects as illustrated
inEngineeringapplications10.1 and 10.2

368 Chapter 10 Differentiation
Engineeringapplication10.1
Voltageacrossaninductor
The voltage, v, across an inductor with inductance, L, is related to the current, i,
through the inductor by
v =L di
dt
Figure10.11showstherelationshipbetweenmagneticfluxlinespassingthroughthe
coils of aninductor and the current flowing through the inductor.
i
Figure10.11
Schematic diagram ofan
inductorshowing the
relationshipbetween magnetic
fluxlinesand current.
This relationship is a quantification of Faraday’s law which states that the voltage
induced in a coil is proportional to the rate of change of magnetic flux through it. If
the current in a coil is changing then this corresponds to a change in the magnetic
flux through the coil. Note that if di is large then v is large. This is why care has
dt
to be taken when abruptly switching off the current to an inductor because it causes
high voltages tobe generated.
Engineeringapplication10.2
Currentthroughacapacitor
The current,i, through a capacitor with capacitance,C, is related to the voltage, v,
acrossthe capacitorby
i =C dv
dt
Displacement current i D
Conduction current i
i
Figure10.12
Schematic diagram ofacapacitor
showingthe conduction current and
the displacement current.
It may appear confusing to talk of a current flow through a capacitor as no actual
charge flows through the capacitor apart from that caused by any leakage current.
Insteadthereisabuild-upofchargeontheplatesofthecapacitor.Thisinturngives
risetoavoltageacrossthecapacitor.Ifthecurrentflowislargethentherateofchange
ofthisvoltagewillbelarge.Thecurrentflowthroughthecapacitorwiresistermeda

10.5 Rate of change at a general point 369
conductioncurrentwhilethatbetweenthecapacitorplatesiscalledadisplacement
current (see Figure 10.12).
Thedisplacementcurrentcanbethoughtofasavirtualcurrentthatflowsbetween
the plates due to the build-up of positive charge on one plate and negative charge
on the other plate. It doesn’t correspond to a real current flowing between the plates
as they are separated by an insulating material.
We can relate the smallchanges δx and δy tothe derivative dy
dx .
If δx isvery smallyet still finitewe can statethat
dy
dx ≈ δy
(10.1)
δx
Thisresultallowsanimportantapproximationtobemade.FromEquation(10.1)wesee
thatifasmallchange, δx,ismadetotheindependentvariable,thecorrespondingchange
inthe dependent variable isgiven by the following formula:
δy ≈ dy
dx δx
Example10.9 Ify =x 2 estimate the changeinycausedbychangingxfrom3to3.1.
Solution Ify =x 2 then dy = 2x. The approximate change inthe dependent variable isgiven by
dx
δy ≈ dy δx =2xδx
dx
Takingx = 3 and δx = 0.1 we have
δy ≈ (2)(3)(0.1) = 0.6
We conclude that at the point wherex = 3 a change inxto 3.1 causes an approximate
change of 0.6 inthe value ofy.
EXERCISES10.5
1 Calculatethe gradientofthe functionsat the specified
points.
(a) y = 2x 2 at(1,2)
(b) y=2x−x 2 at(0,0)
(c) y=1+x+x 2 at(2,7)
(d) y=2x 2 +1at(2,9)
2 Afunction,y,is such that dy is constant.
dx
What can you say about the function,y?
3 Forwhichgraphs in Figure10.13isthe derivative
always (a) positive or(b)negative?
4 Find the derivative ofy(x)whereyis
(a) x 2 (b) −x 2 +2x
5 Differentiatey = 2x 2 +9; thatis,find dy
dx .
What is the rate ofchange ofywhenx = 3, −2, 1,0?
6 Find the rate ofchange ofy = 4t −t 2 .Whatis the
value of dy whent = 2?
dt
7 Ify=x 3 −3x 2 +xthen
dy
dx =3x2 −6x+1
Estimate the change inyasxchanges from
(a) 2 to 2.05
(b) 0 to 0.025
(c) −1 to −1.05

370 Chapter 10 Differentiation
y
y
y
y
x
x
x
(i) (ii) (iii) (iv)
Figure10.13
x
Solutions
1 (a)4 (b)2 (c)5 (d)8
2 yislinearinx,thatisy =ax +b
3 (i) always negative (ii) always positive
(iii) always positive (iv) always negative
4 (a) 2x (b) −2x +2
5 4x,12,−8,4,0
6 4−2t,0
7 (a) 0.05 (b) 0.025 (c) −0.5
10.6 EXISTENCEOFDERIVATIVES
So far we have seen that the derivative, dy , of a function,y(x), may be viewed either
dx
algebraically orgeometrically.
( )
dy y(x + δx) −y(x)
dx = lim
δx→0 δx
dy
= rate of change ofy
dx
= gradient of the graph ofy
We now discuss briefly the existence of dy . For some functions the derivative does not
dx
exist at certain points and we need to be able to recognize such points. Consider the
graphs shown in Figure 10.14. Figure 10.14(a) shows a function with a discontinuity at
y
y
(a)
a
x
Figure10.14
(a) The graphhas a discontinuity atx =a. (b)The graphhas a cuspat
x=a.
(b)
a
x

10.6 Existence of derivatives 371
x = a. The function shown in Figure 10.14(b) is continuous but has a cusp or corner
atx = a. In both cases it is impossible to draw a tangent atx = a, and so dy does not
dx
existatx =a.Itisimpossibletodrawatangent toacurve atapointwhere thecurve is
not smooth. Note from Figure 10.14(b) that continuity is not sufficient to guarantee the
existence of a derivative.
Example10.10 Sketch the following functions. State the values oft for which the derivative does not
exist.
(a) y=|t| (b) y=tant (c) y=1/t
Solution (a) Thegraphofy = |t|isshowninFigure10.15(a).Acornerexistsatt = 0andsothe
derivative doesnotexist here.
(b) A graph ofy = tant is shown in Figure 10.15(b). There is a discontinuity in tant
when t = ... −3π/2,−π/2,π/2,3π/2,.... No derivative exists at these
points.
(c) Figure 10.15(c) shows a graph of y = 1/t. The function has one discontinuity at
t = 0,and so the derivative doesnotexist here.
y
y = tan t
y
y
y = |t|
–
p
–
p
–
2
2
t
y = 1_ t
t
(a)
t
(b)
Figure10.15
(a) There is acorner att = 0; (b)tant has discontinuities; (c)y = 1/t has a discontinuity att = 0.
(c)
EXERCISES10.6
1 Sketchthe functionsanddetermineany pointswhere
aderivative does notexist.
(a) y= 1
t −1
(b) y=|sint|
(c) y=e t
(d) y = |1/t|
{ 1 t0
(e) The unitstepfunctionu(t) =
0 t<0
{ ct t0
(f) The ramp function f (t) =
0 t<0

372 Chapter 10 Differentiation
Solutions
1 See FigureS.22.
(a) No derivative exists for t = 1
y
(b) No derivative exists for t = nπ
|sin t|
(c) Derivative exists for all values of t
e t
1
t
–2π–π
0 π 2π3π
t
0
t
(d) No derivative exists for t = 0
1
–
t
FigureS.22
0 t
(e) No derivative exists for t = 0
u(t)
1
t
(f) No derivative exists for t = 0, although
the function is continuous here
f(t)
t
10.7 COMMONDERIVATIVES
Itistimeconsumingtofindthederivative ofy(x) usingthe‘shrinkinginterval’method
(often referred to as differentiation from first principles). Consequently the derivatives
of commonly used functions are listed for reference in Table 10.1. It will be helpful
to memorize the most common derivatives. Note that a, b and n are constants. In
the trigonometric functions, the quantity ax +b, being an angle, must be measured in
radians.
A shorter table of the more common derivatives is given on the inside back cover of
this bookforeasyreference.
Example10.11 UseTable 10.1 tofindy ′ when
(a) y = e −7x
(b)y=x 5
(c) y = tan(3x −2)
(d) y = sin(ωx + φ)
(e)y= √ 1 x
(f)y= 1 x 5
(g) y = cosh −1 5x

10.7 Common derivatives 373
Table10.1
Derivatives ofcommonly usedfunctions.
Function,y(x) Derivative,y ′
constant 0
x n
e x
e −x
e ax
lnx
sinx
cosx
sin(ax +b)
cos(ax +b)
tan(ax +b)
cosec(ax +b)
sec(ax +b)
cot(ax +b)
sin −1 (ax +b)
nx n−1
e x
−e −x
ae ax
1
x
cosx
−sinx
acos(ax +b)
−asin(ax +b)
asec 2 (ax +b)
−acosec(ax +b)cot(ax +b)
asec(ax +b)tan(ax +b)
−acosec 2 (ax +b)
a
√
1−(ax+b) 2
Function,y(x) Derivative,y ′
cos −1 (ax +b)
tan −1 (ax +b)
sinh(ax +b)
cosh(ax +b)
tanh(ax +b)
cosech(ax +b)
sech(ax +b)
coth(ax +b)
sinh −1 (ax +b)
cosh −1 (ax +b)
tanh −1 (ax +b)
−a
√
1−(ax+b) 2
a
1+(ax+b) 2
acosh(ax +b)
asinh(ax +b)
asech 2 (ax +b)
−acosech(ax +b)×
coth(ax +b)
−asech(ax +b)×
tanh(ax +b)
−acosech 2 (ax +b)
a
√
(ax+b) 2 +1
a
√
(ax+b) 2 −1
a
1−(ax+b) 2
Solution (a) FromTable 10.1,wefind thatif
y = e ax then y ′ =ae ax
Inthis case,a = −7 and so if
y = e −7x then y ′ = −7e −7x
(b) FromTable 10.1,wefind thatif
y =x n then y ′ =nx n−1
Inthis case,n = 5 and so if
y =x 5 then y ′ = 5x 4
(c) Ify = tan(ax+b)theny ′ =asec 2 (ax+b).Inthiscase,a = 3andb = −2.Henceif
y =tan(3x −2) then y ′ =3sec 2 (3x −2)
(d) Ify = sin(ax +b)theny ′ =acos(ax +b).Herea = ωandb = φ,andsoif
(e) Note that
y = sin(ωx + φ) then y ′ = ωcos(ωx + φ)
1
√ x
= x −1/2 . From Table 10.1 we find that ify = x n theny ′ = nx n−1 . In
this case,n = −1/2 and so if
y = 1 √ x
then y ′ = − 1 2 x−3/2

374 Chapter 10 Differentiation
(f) Note that 1 x 5 =x−5 . UsingTable 10.1, we find thatify =x −5 theny ′ = −5x −6 .
(g) FromTable 10.1, ify = cosh −1 (ax +b) then
y ′ =
a
√
(ax+b)2 −1
Inthiscase,a = 5 andb= 0.Hence, if
y = cosh −1 5x then y ′ =
5
√
25x2 −1
Example10.12 Differentiatey(t) = e t .
Solution We note that the independent variable ist. However, Table 10.1 can still be used. From
Table 10.1, wefind
dy
dt =et =y
We note that the derivative of e t is again e t . This is the only function which reproduces
itself upon differentiation.
EXERCISES10.7
1 UseTable 10.1 to findy ′ given:
(a) y=t 2 (b) y=t 9
(c) y=t −3 (d) y=t
(e) y= 1 (f) y= 1
t t 2
(g) y = e 3t (h) y = e −3t
(i) y= 1
e 5t (j) y =t 1/2
(k) y =sin(2t +3) (l) y =cos(4 −t)
( )
t
(m) y = tan
2 +1 (n) y = cosec(3t +7)
(o) y =cot(1−t)
(q) y =sin −1 (t +π)
(p) y =sec(2t −π)
(r) y = π
(s) y = tan −1 (−2t −1) (t) y = cos −1 (4t −3)
(u) y = tanh(6t)
( )
(v) y = cosh(2t +5)
t +3
(w) y = sinh
2
(x) y = sech(−t)
( )
2t
(y) y = coth
3 − 1 2
(z) y = cosh −1 (t +3)
2 Find dy
dx when
(a) y= 1 √ x
(b) y = e 2x/3
(c) y = e −x/2 ( )
2x−1
(e) y = cosec
3
(f) y = tan −1 (πx +3)
(g) y = tanh(2x +1)
(h) y = sinh −1 (−3x)
(i) y = cot(ωx + π)
1
(j) y=
sin(5x +3)
(l) y= 1
cos3x(
)
x −1
(n) y = cosech
2
(o) y = tanh −1 (
2x+3
7
)
(d) y = lnx
ω constant
(k) y = cos3x
(m) y = tan(2x + π)

10.8 Differentiation as a linear operator 375
Solutions
1 (a) 2t (b) 9t 8
(c) −3t −4 (d) 1
(e) −t −2 (f) −2t −3
(g) 3e 3t (h) −3e −3t
(i) −5e −5t (j) 0.5t −1/2
(k) 2cos(2t +3) (l) sin(4 −t)
(m) 0.5sec 2 (t/2 +1)
(n) −3cosec(3t +7)cot(3t +7)
(o) cosec 2 (1 −t)
(p) 2sec(2t − π)tan(2t − π)
1
(q) √ (r) 0
1−(t+π) 2
2
4
(s) −
1 + (−2t −1) 2 (t) −√ 1−(4t−3) 2
(u) 6sech 2 6t
( )
(v) 2sinh(2t +5)
t +3
(w) 0.5cosh
2
(x) sech(−t)tanh(−t)
(y) − 2 3 cosech2 (
2t
3 − 1 2
)
(z)
1
√
(t+3) 2 −1
2 (a) −0.5x −3/2 2
(b)
3 e2x/3
(c) −0.5e −x/2 (d) 1/x
(e) − 2 3 cosec (
2x−1
3
)
cot
(
2x−1
3
π
(f)
1+(πx+3) 2 (g) 2sech 2 (2x +1)
3
(h) −√ (i) −ωcosec 2 (ωx + π)
9x 2 +1
(j) −5cosec(5x +3)cot(5x +3)
(k) −3sin3x
(l) 3sec3xtan3x (m) 2sec 2 (2x + π)
( ) ( )
x −1 x −1
(n) −0.5cosech coth
2 2
(o)
2
[ ( ) 2 ]
2x+3
7 1 −
7
)
10.8 DIFFERENTIATIONASALINEAROPERATOR
Inmathematicallanguagedifferentiationisalinearoperator.Thismeansthatifwewish
to differentiate the sum of two functions we can differentiate each function separately
and thensimplyadd the two derivatives, thatis
derivative of (f +g) = derivative of f +derivative ofg
This isexpressed mathematically as
d
dx (f+g)=df dx + dg
dx
We can regard d as the operation of differentiation being applied to the expression
dx
whichfollows it.Thepropertiesofalinear operator alsomake thehandling ofconstant
factorseasy.Todifferentiatekf,wherekisaconstant,wetakektimesthederivativeof
f,thatis
derivative of (kf) =k ×derivative of f
Mathematically, wewould state:
d
dx (kf)=kdf dx

376 Chapter 10 Differentiation
Table 10.1 together with these two linearity properties allow us to differentiate some
quite complicated functions.
Example10.13 Differentiate
(a) 3x 2 (b) 9x (c) 7 (d) 3x 2 +9x +7
Solution (a) Lety = 3x 2 , then
dy
dx = d dx (3x2 )
= 3 d dx (x2 ) using linearity
= 3(2x) fromthe table
= 6x
(b) Lety = 9x, then
dy
dx = d dx (9x)
= 9 d (x) using linearity
dx
= 9
(c) Lety = 7,theny ′ = 0.
(d) Lety=3x 2 +9x+7
dy
dx = d dx (3x2 +9x +7)
dy
dx = 3 d dx (x2 )+9 d dx (x) + d (7) usinglinearity
dx
=6x+9
Engineeringapplication10.3
Fluidflowintoatank
If fluid is being poured into a tank at a rate of q m 3 s −1 , then this will result in an
increase in volume,V, of fluid in the tank. The arrangement is illustrated in Figure
10.16. Therate ofincrease involume, dV m 3 s −1 , isgiven by
dt
dV
=q
dt
This relationship follows from the principle of conservation of mass. If q is large,
then dV is large, which corresponds to the fluid volume in the tank increasing at a
dt

10.8 Differentiation as a linear operator 377
fast rate. Consequently, the height of the fluid,h, also increases at a fast rate. If the
cross-sectional area ofthe tank,A, isconstant, thenV =Ah. Therefore,
dV
= d (Ah) =Adh
dt dt dt
because differentiation isalinear operator. So,
A dh
dt
=q
q
h
V
Figure10.16
A closedtankcontaining avolume offluidV.
Example10.14 UseTable 10.1 and the linearityproperties ofdifferentiation tofindy ′ where
(a) y=3e 2x
(b) y = 1/x
(c) y=3sin4x
(d) y = sin2x −cos5x
(e) y=3lnx
(f) y=ln2x
(g) y=3x 2 +7x−5
Solution (a) Ify = 3e 2x , then
dy
dx = d dx (3e2x ) = 3 d dx (e2x ) usinglinearity
= 3(2e 2x ) usingTable 10.1
=6e 2x
(b) Ify =x −1 ,then
y ′ = −1x −2 from Table 10.1
= − 1 x 2
(c) Ify = 3sin4x, then
dy
dx = d dx (3sin4x) =3 d dx (sin4x) usinglinearity
= 3(4cos4x) usingTable 10.1
= 12cos4x

378 Chapter 10 Differentiation
(d) The linearityproperties allowus todifferentiate each termindividually. If
y = sin2x −cos5x
then
dy
dx = d dx (sin2x) − d dx (cos5x)
=2cos2x+5sin5x
(e) Ify = 3lnx,then
dy
dx = d dx (3lnx)=3d dx (lnx)
= 3 x
using linearity
using Table 10.1
(f) Ify = ln2x, then
y = ln2 +lnx
usinglaws of logarithms
and so
dy
dx = 0 + 1 x = 1 x
(g) Each termisdifferentiated:
y ′ =6x+7−0=6x+7
since ln2 isconstant
Engineeringapplication10.4
Dynamicresistanceofasemiconductordiode
RecallfromEngineeringapplication2.9thatasemiconductordiodecanbemodelled
by the equation
I =I S
(e qV
nkT −1)
where V is the applied voltage, I is the diode current, I S
is the reverse saturation
current, k is the Boltzmann constant, q is the charge on the electron, and T is the
temperature of the diode junction in Kelvin. The parameter n is the ideality factor
whichisusuallyconsideredtobeaconstantforagiventypeofsemiconductorjunction,
dependent primarily on the design and manufacture ofthe diode.
Examiningthediodeelectricalcharacteristics,showninFigure2.31,wenotethat
there are two main electrical regimes. The first occurs when the applied voltage,V,
isnegativeandthecurrent,I,islimitedtoanegativesaturationcurrent,I s
,apartfrom
a small region around the origin. The second occurs when the applied voltage,V, is
positive and the current,I, takes on the shape of an exponential curve, apart from a
smallregion around the origin.
Wewillconcentrateonthesecondcaseandfurthermorewewillassumethate qV
nkT
is very much greater than 1. This is written concisely as e qV
nkT >> 1. This is a good
approximationapartfromthesmallregionclosetotheorigin.Theoriginalexpression
forIcan then be writtenas
I≈I S
e qV
nkT
where the symbol ≈means ‘is approximately equal to’.

10.8 Differentiation as a linear operator 379
For a fixed temperature the only variables in this equation are I andV. We can
differentiate thisequation togive
dI
dV ≈I q
S
nkT e qV
nkT
Back substituting the approximate expressionI ≈I S
e qV
nkT into this equation yields
dI
dV ≈I q
S
nkT e qV q
nkT ≈
nkT I
Inverting this equation gives
dV
dI ≈ nkT
qI
Itisconstructiveatthisstagetocomparethisresultwiththestaticequivalent,V =IR,
from which
V
I =R
The quantity dV
dI can be thought of as a dynamic value of the diode resistance,r d ,
which varies depending on the value of the current in the diode. It is valid for a
positive voltageV that is sufficiently large to ensure the diode is in the exponential
portion of the electrical characteristic.
Engineeringapplication10.5
Operatingpointforanidealsemiconductordiode
In Engineering application 2.18 we examined the mathematical model of a semiconductordiodeandthenconsideredthespecialisedcaseofanidealdiode.Atroom
temperaturetheequationdescribingthebehaviourofitselectricalcharacteristicswas
I =I s
(e 40V −1)
where I is the diode current,V is the applied voltage and I s
is the constant reverse
saturation current. Sometimes when a diode is used in a circuit it may be biased
to operate in a certain region of its I--V characteristic. This means that its use is
restricted to a certain voltage range. The point around which it operates is known as
its operating point. This isillustratedinFigure 10.17.
Deviationsfromthisoperatingpointmaybesmallincertaincases.Ifso,theyare
known assmall-signalvariations andarecausedbysmalla.c.voltagesbeingsuperimposedonthemaind.c.biasvoltage.Incalculatinghowthediodewillreacttosuch
small-signalvoltages,theslopeofthediodecharacteristicaroundtheoperatingpoint
is more relevant than the overall ratio of current to voltage. Provided the deviations
from the operating point are not large, the tangent to theI--V curve at the operating
point provides an adequate model for how the diode will behave. The slope of the
curve can beobtained by differentiating the diodeequation.If
then
I =I s
(e 40V −1) =I s
e 40V −I s
dI
dV = 40I s e40V
➔

380 Chapter 10 Differentiation
I
Bias current
Operating
point
dV
dI
Bias
voltage
Tangent
approximation
V
Figure10.17
Diode characteristic showing
operating point and tangent
approximation.
sinceI s
isconstant.
It is usual to write small changes in current and voltage as δI and δV. Therefore,
since
δI
δV ≈ dI
dV
δI ≈ 40I s
e 40V δV
Thisexpressionallowsthechangeindiodecurrent, δI,tobeestimatedgivenachange
indiodevoltage, δV,providedtheoperatingpointisknownandthechangesaresmall.
Example10.15 Findthe derivative ofy = e −t +t 2 , when
(a)t=1
(b)t=0
Solution
y=e −t +t 2
y ′ = −e −t +2t
(a) Whent = 1,y ′ = −e −1 +2 = 1.632, thatisy ′ (1) = 1.632.
(b) Whent = 0,y ′ = −1,thatisy ′ (0) = −1.
Engineeringapplication10.6
Obtainingalinearmodelforasimplefluidsystem
Consider the fluid system illustrated in Figure 10.18. The pump is driven by a d.c.
motor. The pump/motor can be modelled by a linear relationship in which the fluid
flowrate,q i
, isproportionaltothe control voltage, v in
, thatis
q i
=k p
v in

10.8 Differentiation as a linear operator 381
v in
Tank Valve
Pump/motor
q i
h
p
q o
Figure10.18
Afluidsystem comprising pump, tank
andvalve.
where k p
= pump/motor constant, v in
= control voltage (V), q i
= flow rate into
the tank (m 3 s −1 ). The valve has a non-linear characteristic given by the quadratic
polynomial
p = 20000q 2 o
where p = pressure at the base of the tank (Nm −2 ), q o
= flow out of the tank
(m 3 s −1 ).Thefluidbeingusediswaterwhichhasadensity ρ = 998kg m −3 .Assume
g = 9.81 ms −2 and thatk p
= 0.03 m 3 s −1 V −1 . Carry outthe following:
(a) Calculate theflowrateoutofthetank,q o
,andthecontrol voltage, v in
,when the
systemisinequilibrium and the height ofthe water inthe tank,h, is0.25 m.
(b) Obtainalinearmodelforthesystem,validforsmallchangesaboutawaterheight
of0.25m.Usethismodel tocalculatethenewwaterheightandflowrateoutof
the tank when the control voltage isincreased by 0.4 V.
Solution
(a) The pressure atthe bottomof the tank isgiven by
p = ρgh = 998 ×9.81 ×0.25 = 2448
The flowrate through the valve isgiven by
q 2 o = p
20000
√ √
p 2448
q o
=
20000 = 20000 = 0.350 m3 s −1
Nowifthesystemisinequilibrium,thentheheightofthewaterinthetankmust
have stabilized toaconstant value. Therefore,
and so
q i
=q o
= 0.350
v in
= q i
k p
= 0.350
0.03 = 11.7 V
(b) Before answering this part it is worth examining what is meant by a linear
model. Figure 10.19 shows the valve characteristic together with a linear approximationaroundanoutputflowrateofq
o
= 0.350m 3 s −1 .Thiscorresponds
toawater height of 0.25 m.
➔

382 Chapter 10 Differentiation
p (N m –2 )
2448
0.35
q o (m 3 s –1 )
Figure10.19
Relationshipbetween pressureacross
valve (p)and flowthrough valve (q o ).
Alinearmodelforthevalveisoneinwhichtherelationshipbetweenpand
q o
isapproximated by the straightlinewhich forms a tangent tothe curve atthe
operatingpoint.Theoperatingpointisthepointaroundwhichthemodelisvalid.
Itisclearthatifthestraightlineapproximationisusedforpointsthatarealarge
distance from the operating point, then the linear model will not be very accurate.
However, for small changes around the operating point the approximation
isreasonablyaccurate.Clearlyadifferentoperatingpointwillrequireadifferent
linearapproximation.Inordertoobtainthegradientofthislineitisnecessaryto
differentiate the function relating valve pressure tovalve flow. So,
p = 20000q 2 o
dp
dq o
= 40000q 0
At the operating pointq o
= 0.350. Therefore,
∣
dp ∣∣∣qo
= 40000 ×0.350 = 14000
dq o =0.350
This value is the gradient of the tangent to the curve at the operating point.
Smallchangesaroundanoperatingpointareusuallyindicatedbythenotation δ.
Therefore,
δp
≈ dp ∣ ∣∣∣qo
= 14000
δq o
dq o =0.350
δp = 14000δq o
(10.2)
Note that equality has been assumed for the purposes of the linear model. It is
easy to relate a change in pump flow to a change in control voltage because the
relationship islinear and so a linear approximation isnotrequired. So,
q i
=k p
v in
Differentiating thisequation w.r.t. v in
yields
dq i
dv in
=k p
Inthis case
δq i
δv in
= dq i
dv in
=k p

10.8 Differentiation as a linear operator 383
and has a constant value independent of the operating point:
δq i
=k p
δv in
(10.3)
Therelationshipbetweenpressureatthebottomofthetankandthewaterheight
isalsolinear.
Since p = ρgh we have dp dp
= ρg. Because is a constant wecan write
dh dh
δp
δh = dp
dh = ρg
δp= ρgδh (10.4)
Thechangeincontrolvoltage, δv in
,isgivenas0.4.Wealsoknowthatk p
= 0.03.
Therefore, using Equation (10.3),we have
δq i
= 0.03 ×0.4 = 0.012
Now if time is allowed for the system to reach equilibrium with this increased
inputflow,then δq o
= δq i
.Inotherwords,theoutputflowincreasesbythesame
amount as the input flow and the water height once again stabilizes to a fixed
value. Therefore,
δq o
= δq i
=0.012
UsingEquation (10.2),wefind
δp =14000 δq o
=14000 ×0.012 =168
UsingEquation (10.4),weget
δh = δp
ρg = 168
998 ×9.81 = 0.0172
Therefore the new water height is 0.25 +0.0172 = 0.267 m to three significant
figures. The new water flowrate is0.35 +0.012 = 0.362 m 3 s −1 .
Torecap,alltheelementsofthefluidsystemwerelinearapartfromthevalve.
Byobtainingalinearmodelforthevalve,validforvaluesclosetotheoperating
point, it was possible to calculate the effect of changing the control voltage to
themotor.Itisimportanttostressthatthelinearmodelforthevalveisonlygood
forsmallchangesaroundtheoperatingpoint.Inthiscasetheincreaseincontrol
voltage was approximately 3%. The model would not have been very good for
predictingtheeffectofa50%increaseincontrolvoltage.Linearmodelsofnonlinear
systems are particularly useful when several components are non-linear,
astheyaremucheasiertoanalyse.Weexaminetheseconceptsinmoredetailin
Chapter 18.
EXERCISES10.8
1 Differentiate the following functions:
(a) y = 4x 3 −5x 2
(b) y = 3sin(5t) +2e 4t
(c) y = sin(4t) +3cos(2t) −t
(d) y = tan(3z)
(e) y =2e 3t +17−4sin(2t)
(f) y= 1
t 3 + cos5t
2

384 Chapter 10 Differentiation
(g)y= 2w3
3 + e4w
2
(h)y= √ ( √x )
x+ln .[Hint: √ x =x 1/2 ,and use the
laws oflogarithms.]
2 Evaluate the derivatives ofthe functionsat the given
value:
(a) y=2t+9+e t/2 t=1
(b) y= t2 −4t+6
t = 2
3
(c) y=sint+cost t=1
( t
(d) y=3e 2t −2sin t = 0
2)
(e) y = 5tan(2x) + 1
e 2x x=0.5
(f) y =3lnt +sin(3t) t =0.25
3 Find dx
dt , if
(a) x = e ωt (b) x = e −ωt
where ω isaconstant.
4 Find the derivative of
(a) y = 3sin −1 (2t) −5cos −1 (3t)
(b) y= 1 2 tan−1 (t +2) +4cos −1 (2t −1)
( )
t −3
(c) y = 2sinh(3t −1) −4cosh
2
cosech(4t) +3sech(6t)
(d) y=
2
( ) ( )
(e) y = 2sinh −1 t +1
−3cosh −1 1 −t
2 2
(f) y = 3tanh −1 (2t +3) −2tanh −1 (3t +2)
5 Afunction,y(t),is given by
y(t) = t3 3 − 5t2
2 +4t+1
(a) Find dy
dt .
(b) Forwhich values oft isthe derivative zero?
6 Findthe equationofthe tangent to the curve
y(x)=x 3 +7x 2 −9
atthe point (2,27).
7 Findvalues oft in the interval [0,π] forwhichthe
tangent tox(t) = sin2t has zerogradient.
8 Findthe rate ofchangeof
when
(a) t=0
(b) t=3
z(t) =2e t/2 −t 2
Solutions
1 (a) 12x 2 −10x (b) 15cos5t +8e 4t
(c) 4cos4t −6sin2t −1
(d) 3sec 2 3z
(e) 6e 3t −8cos2t (f) − 3
t 4 −2.5sin5t
(g) 2w 2 +2e 4w (h) 0.5x −1/2 + 1 2x
2 (a) 2.8244 (b) 0 (c) −0.3012
(d) 5 (e) 33.5194 (f) 14.195
3 (a) ωe ωt (b) −ωe −ωt
4 (a)
(b)
6
√ + 15
√
1−4t 2 1−9t 2
1
2[1+ (t +2) 2 ] − 4
√ t(1−t)
(c) 6cosh(3t −1) −2sinh t −3
2
(d) −2cosech4tcoth4t −9sech6ttanh6t
(e)
(f)
1
√ (1
2 (t+1) ) 2
+1
+
6
1−(2t+3) 2 − 6
1−(3t+2) 2
5 (a) t 2 −5t+4 (b) 1,4
6 y=40x−53
7 π/4,3π/4
8 (a) 1 (b) −1.5183
3
2√ (1
2 (1−t) ) 2
−1

Review exercises 10 385
REVIEWEXERCISES10
1 Findtherateofchangeof f(x) = 5 +3x 2 atx = 2by
considering each ofthe intervals [2 − δx,2],
[2,2 + δx],[2 − δx,2 + δx].Showthatthesame
resultis obtained in each case.
2 Useatable ofderivatives and the linearity rules to
differentiate the following:
(a) y=4x 2 +6x−11
(b)y=−x 2 +2x−10
(c) y =x 1/3 −x 1/4
(d) y=5cos4x−3cos2x
(e) y = sin −1 (4x +3)
(f) y= √ 2x 4 − 5
3x 2
(g) y =t 3/2 +cost
(h)y=t 2 −14t+8
(i) y=5lnt+sin4t
(j) y= 1 2 x − 1 3 x2
(k)y= 2t2
3 +e2t
3 Find the equationofthe tangent toy =x 2 +7x −4 at
the point on the graph wherex = 2.
4 Find the rate ofchange of f (t) = 2cost +3sint at
t=1.
5 At any timet,the voltage, v, acrossan inductorof
inductanceLis relatedto the current,i,through the
inductorby v =L di
dt .
(a) Findan expression forthe voltage when
i = 5cosωt where ω isthe constantangular
frequency.
(b) Findan expression forthe voltage when the
current takesthe formofasinewave with
amplitude 10 andperiod 0.01 seconds.
6 Use the shrinking interval methodto findthe rate of
change of f (t) = sint att = 0 by consideringthe
interval [0, δt]. [Hint:use the trigonometricidentities
in Section3.6 andthe small-angle approximation in
Section6.5.]Usetheshrinkingintervalmethodtofind
the rate ofchange of f (t) = sint at ageneral point.
7 Giveny(t) = 3 +sin2t,findthe averagerate of
change ofyast variesfrom 0 to 2.
8 Explain the essentialdifference between δy dy
and
δx dx .
9 Findy ′ forthe followingfunctions:
(a) y =2e −t +6cos(t/2)
(b) y = (−t +2) 2
10 Using derivatives, estimatethe changeinyasx
changes from 1.5 to 1.55 wherey = 2e 2x +x 3 .
Solutions
1 12
2 (a) 8x+6
(b) −2x +2
1
(c) 3 x−2/3 − 1 4 x−3/4
(d) −20sin4x +6sin2x
4
(e) √
1−(4x+3) 2
(f) 2 √ 2x + 10
3x 3
3
(g) 2 t1/2 −sint
(h) 2t −14
(i)
5
t +4cos4t
(j)
(k)
1
2 − 2 3 x
4
3 t+2e2t
3 y=11x−8
4 −0.062
5 (a) −5ωLsin ωt (b) 2000πLcos200πt
7 −0.3784
( )
9 (a) −2e −t t
−3sin
2
10 4.355
(b) 2t−4

11 Techniquesof
differentiation
Contents 11.1 Introduction 386
11.2 Rulesofdifferentiation 386
11.3 Parametric,implicitandlogarithmicdifferentiation 393
11.4 Higherderivatives 400
Reviewexercises11 404
11.1 INTRODUCTION
InthischapterwedevelopthetechniquesofdifferentiationintroducedinChapter10so
thatratesofchangeofmorecomplicatedfunctionscanbefound.Weintroducerulesfor
differentiatingproductsandquotientsoffunctions.Thechainruleisusedfordifferentiating
functions of functions. We explain what is meant by defining functions implicitly
andparametricallyandshowhowthesecanbedifferentiated.Thetechniqueoflogarithmic
differentiation allows complicated products of functions to be simplified and then
differentiated.Finallyderivativesoffunctionscanthemselvesbedifferentiated.Thisinvolvestheuseofhigherderivativeswhichareexplainedtowardstheendofthechapter.
11.2 RULESOFDIFFERENTIATION
Therearethreeruleswhichenableustodifferentiatemorecomplicatedfunctions.They
are (a) the product rule, (b) the quotient rule, (c) the chain rule. Traditionally they are
written withxas the independent variable but apply in an analogous way for other independent
variables.
11.2.1 Theproductrule
Asthenamesuggests,thisruleallowsustodifferentiateaproductoffunctions,suchas
xsinx,t 2 cos2t and e z lnz.

11.2 Rules of differentiation 387
The product rulestates: if
then
y(x) =u(x)v(x)
dy
dx = du
dx v +udv dx =u′ v+uv ′
To apply this rule one of the functions in the product must be chosen to beu, and the
other, v. Beforewe can apply the ruleweneed tocalculateu ′ and v ′ .
Example11.1 Findy ′ given
(a) y=xsinx
(b)y=t 2 e t
Solution (a) y = xsinx = uv. Chooseu = x and v = sinx. Thenu ′ = 1, v ′ = cosx. Applying
the productruletoyyields
y ′ =sinx+xcosx
(b)y=t 2 e t =uv.Chooseu =t 2 andv =e t .Thenu ′ =2tandv ′ = e t . Applying the
product ruletoyyields
y ′ =2te t +t 2 e t
Engineeringapplication11.1
Dampedsinusoidalsignal
Acommonfunctionfoundinengineeringisthedampedsinusoidalsignal.Thisconsists
of a negative exponential function multiplied by a sinusoid. A typical example
is
f (t) = e −0.1t cost
The graph of this function is shown in Figure 11.1. This function approximates the
wayacarbodyreactswhenthecardrivesoveralargebumpintheroad.Fortunately,
thecarshockabsorbersensurethattheoscillationsreduceinamplitudequitequickly.
When sketching such a function it can be useful to think of the exponential term,
and its mirror image around the time axis, providing an envelope that contains the
signal. When values of the sinusoid are 1 then the signal touches the positive part
of the envelope and when values of the sinusoid are −1 then the signal touches the
negative partofthe envelope.
Therateofchangeofthissignalwithrespecttotimecanbefoundbydifferentiating
f (t)usingtheproductrule.Todosowenotethat f (t)isaproductofu(t) = e −0.1t
and v(t) = cost.Recastingtheformulafordifferentiatingaproductintermsoftwe
have
f(t) =u(t)v(t)
➔

388 Chapter 11 Techniques of differentiation
df
= du
dt dt v +udv dt
Now wehave
du
= −0.1e −0.1t
dt
dv
=−sint
dt
So
df
= du
dt dt v +udv dt
df
= −0.1e −0.1t cost +e −0.1t (−sint)
dt
Rearranging gives
df
dt
= −e −0.1t (0.1cost +sint)
1
f(t)
0.8
0.6
0.4
0.2
–0.2
–0.4
–0.6
–0.8
–1
p 2p 3p 4p 5p 6p 7p 8p 9p 10p 11p 12p 13p 14p 15p
t
Figure11.1
Adamped sinusoidalsignal.
11.2.2 Thequotientrule
This rule allows us to differentiate a quotient of functions, such as
and e−3z
z 2 −1 .
x
sinx , t2 −t+3
t +2

11.2 Rules of differentiation 389
The quotient rulestates: when
y(x) = u(x)
v(x)
then
( ) ( ) du dv
v −u
y ′ dx dx
=
= vu′ −uv ′
v 2 v 2
Example11.2 Findy ′ given
(a)y= sinx
x
t2
(b)y=
2t+1
e2t
(c)y=
t 2 +1
Solution (a)y= sinx = u x v .Sou=sinx,v=xandu′ =cosx,v ′ = 1. Using the quotient rule
the derivative ofyisfound:
(b)y=
y ′ = xcosx−sinx
x 2
t2
2t+1 = u v .Sou=t2 ,v=2t+1andu ′ =2t,v ′ = 2.Hence,
y ′ = (2t +1)2t − (t2 )(2) 2t(t +1)
=
(2t +1) 2 (2t +1) 2
(c)y=
e2t
t 2 +1 .Sou=e2t ,v=t 2 +1andu ′ = 2e 2t , v ′ = 2t. Application of the
quotient ruleyields
y ′ = (t2 +1)2e 2t −e 2t 2t
= 2e2t (t 2 −t+1)
(t 2 +1) 2 (t 2 +1) 2
11.2.3 Thechainrule
Thisrulehelpsustodifferentiatecomplicatedfunctions,whereasubstitutioncanbeused
to simplify the function. Supposey = y(z) andz = z(x). Thenymay be considered as
afunctionofx.Forexample,ify =z 3 −zandz = sin3x,theny = (sin3x) 3 −sin(3x).
Suppose weseekthe derivative, dy . Note that the derivative w.r.t.xissought.
dx
Thechain rule states:
dy
dx = dy
dz × dz
dx

390 Chapter 11 Techniques of differentiation
Example11.3 Giveny =z 6 wherez =x 2 +1 find dy
dx .
Solution If y = z 6 and z = x 2 + 1, then y = (x 2 + 1) 6 . We recognize this as the composition
y(z(x)) (see Section 2.3.6). Now y = z 6 and so dy
dz = 6z5 . Alsoz = x 2 +1 and so
dz
= 2x. Usingthe chain rule,
dx
dy
dx = dy
dz × dz
dx =6z5 2x =12x(x 2 +1) 5
Example11.4 Ify = ln(3x 2 +5x +7), find dy
dx .
Solution Weuseasubstitutiontosimplifythegivenfunction:letz = 3x 2 +5x+7sothaty = lnz.
Since
dy
y=lnz then
dz = 1 z
Also
z=3x 2 +5x+7
Using the chain rulewefind
dy
dx = dy
dz × dz
dx
= 1 z ×(6x+5)
=
6x+5
3x 2 +5x+7
andso
dz
dx =6x+5
Note that in the previous answer the numerator is the derivative of the denominator.
This result is true more generally and can be applied when differentiating the natural
logarithm ofany function:
Wheny = ln f(x)then dy
dx = f ′ (x)
f(x)
Example11.5 Findy ′ when
(a) y =ln(x 5 +8) (b) y =ln(1−t) (c) y =8ln(2−3t)
(d) y =ln(1 +x)
(e) y =ln(1 +cosx)
Solution Ineachcase weapplythe previousrule.
(a) Ify = ln(x 5 +8),theny ′ =
5x4
x 5 +8 .

(b) Ify = ln(1 −t),theny ′ = −1
1 −t = − 1
1 −t = 1
t −1 .
(c) Ify =8ln(2−3t),theny ′ =8 −3
2−3t = − 24
2−3t = 24
3t−2 .
(d) Ify =ln(1 +x)theny ′ = 1
1 +x .
(e) Ify = ln(1 +cosx)theny ′ = −sinx
1+cosx = − sinx
1+cosx .
11.2 Rules of differentiation 391
Example11.6 Differentiate
(a) y = 3e sinx
(b) y = (3t 2 +2t −9) 10
(c)y= √ 1 +t 2
Solution Inthese examples we mustformulate the functionzourselves.
(a) Letz(x) = sinx. Theny(z) = 3e z so dy
dz = 3ez ;z(x) = sinx and so dz
dx = cosx.
The chain ruleisused tofind dy
dx .
dy
dx = dy
dz × dz
dx = 3ez cosx = 3e sinx cosx
(b) Letz(t) = 3t 2 +2t −9.Theny(z) =z 10 , dy
dz = 10z9 , dz = 6t +2.Usingthechain
dt
rule dy
dt isfound:
dy
dt = dy
dz × dz
dt = 10z9 (6t +2) = 20(3t +1)(3t 2 +2t−9) 9
(c) Letz(t) =1+t 2 .Theny= √ z =z 1/2 , dy
dz = 1 2 z−1/2 and dz
dt = 2t.Usingthechain
rule,we obtain
dy
dt = dy
dz × dz
dt = 1 2 z−1/2 2t = √ t t
= √ z 1 +t
2
EXERCISES11.2
1 Usethe product ruleto differentiate the following
functions:
(a) y=sinxcosx
(b) y=lnttant
(c) y = (t 3 +1)e 2t
(d) y= √ xe x
(e) y=e t sintcost
(f) y = 3sinh2tcosh3t
(g) y=(1+sint)tant
(h) y = 4sinh(t +1)cosh(1 −t)

392 Chapter 11 Techniques of differentiation
2 Usethe quotient ruleto findthe derivatives ofthe
following:
cosx tant
(a) (b)
sinx lnt
(c)
(e)
(g)
e 2t
t 3 +1
(d)
x 2 +x+1
1+e x (f)
1+e t
1+e 2t
3x 2 +2x−9
x 3 +1
sinh2t
cosh3t
3 Usethe chainruleto differentiate the following:
(a) (t 3 +1) 100 (b) sin 3 (3t +2)
(c) ln(x 2 +1) (d) (2t +1) 1/2
(e) 3 √ cos(2x −1)
(g) (at +b) n ,aandbconstants
(f)
1
t +1
4 Differentiate eachofthe followingfunctions:
(a) y=5sinx
(c) y = 5e sinx
(e) y = (t 3 +4t) 15
(g)y=
sinx
4cosx+1
(b) y=5e x sinx
(d)y= 5sinx
e −x
(f) y = 7e −3t2
5 Forwhichvalues oft is the derivative ofy(t) = e −t t 2
zero?
6 Findtherateofchangeofyatthespecifiedvaluesoft.
(a) y = sin 1 t = 1
t
(b) y=(t 2 −1) 17 t=1
(c) y =sinh(t 2 ) t =2
(d) y= 1+t+t2
1 −t
(e) y=
et
tsint
t = 1
t = 2
7 Findthe equationofthe tangent to
8 Differentiate
y(x) = e 3x (1 −x) atthepoint (0,1)
(a) y=lnx
(b) y=ln2x
(c) y = lnkx,k constant
(d) y =ln(1+t)
(e) y =ln(3+4t)
(f) y =ln(5 +7sinx)
Solutions
1 (a) cos 2 x −sin 2 1
x (b)
t tant+lntsec2 t
e x (−x 2 +x)+2x+1
(e)
( )
(e x +1) 2
√x
(c) e 2t (2t 3 +3t 2 +2) (d) e x 1 +
2 √ 2cosh2tcosh3t −3sinh2tsinh3t
x
(f)
(cosh3t) 2
(e) e t (2cos 2 t +sintcost −1)
−e 3t −2e 2t +e t
(f) 3[2cosh2tcosh3t +3sinh2tsinh3t]
(g)
(e 2t +1) 2
(g) (1+sint)sec 2 t +sint
3 (a) 300t 2 (t 3 +1) 99
(h) 4[cosh(t +1)cosh(1 −t)
−sinh(t +1)sinh(1 −t)]
(b) 9sin 2 (3t +2)cos(3t +2)
2 (a) −cosec 2 x
2x
(c)
x 2 +1
lntsec 2 t − (tant)/t
(b)
(lnt) 2
(d) (2t +1) −1/2
e 2t (2t 3 −3t 2 −3sin(2x −1)
+2)
(e) √
(c)
cos(2x −1)
(t 3 +1) 2
−3x 4 −4x 3 +27x 2 (f) −(t +1) −2
+6x +2
(d)
(x 3 +1) 2 (g) an(at +b) n−1

11.3 Parametric, implicit and logarithmic differentiation 393
4 (a) 5cosx
5 0,2
(e) 15(t 3 +4t) 14 (3t 2 +4)
1 1 1
8 (a) (b) (c)
(f) −42te −3t2
x x x
4+cosx
1 4 7cosx
(g)
(d) (e) (f)
(4cosx +1) 2 1 +t 3+4t 5+7sinx
(b) 5e x (cosx +sinx)
6 (a) −0.5403 (b) 0 (c) 109.2
(c) 5e sinx cosx
(d) 2 (e) −2.0742
(d) 5e x (cosx +sinx)
7 y=2x+1
11.3 PARAMETRIC,IMPLICITANDLOGARITHMIC
DIFFERENTIATION
11.3.1 Parametricdifferentiation
In some circumstances bothyandxdepend upon a third variable,t. This third variable
is often called a parameter. By eliminatingt, y can be found as a function of x. For
example, ify = (1 +t) 2 andx = 2t then, eliminatingt, we can writey = (1 +x/2) 2 .
Hence,ymay be considered as a function ofx, and so the derivative dy can be found.
dx
However, sometimes the elimination oft is difficult or even impossible. Consider the
exampley=sint+t,x=t 2 +e t .Inthiscase,itisimpossibletoobtainyintermsofx.
The derivative dy can stillbe found usingthe chain rule.
dx
dy
dx = dy
dt × dt
dx = dy / dx
dt dt
Finding dy by thismethod isknown asparametric differentiation.
dx
Example11.7 Giveny = (1 +t) 2 ,x = 2t find dy
dx .
Solution Byeliminatingt, we see
(
y = 1 + x ) 2
=1+x+ x2
2 4
and so
dy
dx = 1 + x 2
Parametricdifferentiationisanalternativemethodoffinding dy
dx whichdoesnotrequire
the elimination oft.
dy
dt =2(1+t)
dx
dt = 2

394 Chapter 11 Techniques of differentiation
Using the chain rule,we obtain
dy
dx = dy / dx
dt dt = 2(1+t) =1+t=1+ x 2 2
Example11.8 Giveny = e t +t,x =t 2 +1,find dy using parametric differentiation.
dx
Solution
dy
dt =et +1
dx
dt = 2t
Hence,
dy
dx = dy / dx
dt dt = et +1
2t
In this example, the derivative is expressed in terms oft. This will always be the case
whent has not been eliminated betweenxandy.
Example11.9 Ifx = sint +cost andy =t 2 −t+1find dy (t =0).
dx
Solution
Hence,
dy
dt =2t−1
dy
dx = 2t−1
cost−sint
Whent = 0, dy
dx = −1
1 = −1.
dx
dt =cost−sint
11.3.2 Implicitdifferentiation
Suppose wearetold that
y 3 +x 3 =5sinx+10cosy
Althoughydependsuponx,itisimpossibletowritetheequationintheformy = f (x).
We say y is expressed implicitly in terms of x. The form y = f (x) is an explicit
expression for y in terms of x. However, given an implicit expression for y it is still
possible to find dy dy
. Usually will be expressed in terms of bothxandy. Essentially,
dx dx
the chain ruleisused when differentiating implicit expressions.
Whencalculating dy
dx weneedtodifferentiateafunctionofy,asopposedtoafunction
ofx. For example, we may need tofind d dx (y4 ).
Example11.10 Find
d
(a)
dx (y4 )
(b)
d
dx (y−3 )

11.3 Parametric, implicit and logarithmic differentiation 395
Solution (a) We make a substitution and letz = y 4 so that the problem becomes that of finding
dz
. Now, usingthe chain rule,
dx
dz
dx = dz
dy × dy
dx
Ifz =y 4 then dz
dy = 4y3 and so
dz
dx = 4y3dy dx
We conclude that
d
dx (y4 ) = 4y 3dy
dx
(b) Wemakeasubstitutionandletz =y −3 sothattheproblembecomesthatoffinding
dz
dxİfz =y−3 then dz
dy = −3y−4 and so
d
dx (y−3 ) = dz
dx = dz
dy × dy
dx = −3y−4dy dx
Example11.11 Find d dt (lny).
Solution Weletz = lnysothattheproblembecomesthatoffinding dz
dt .Ifz=lnythendz dy = 1 y
and so usingthe chain rule
dz
dt = dz
dy × dy
dt = 1 dy
y dt
weconclude that
d
dt (lny) = 1 dy
y dt
Examples11.10and 11.11illustrate the generalformula:
d
dx (f(y))=df dy × dy
dx .
This issimplythe chain ruleexpressed inaslightly different form.

396 Chapter 11 Techniques of differentiation
Example11.12 Find d dx (y).
Solution We have
d dy
(y) =
dx dy × dy
dx = dy
dx
Example11.13 Given
find dy
dx .
x 3 +y=1+y 3
Solution Consider differentiation of the l.h.s.w.r.t.x.
d
dx (x3 +y)= d dx (x3 )+ dy
dx =3x2 + dy
dx
Now consider differentiation of the r.h.s.w.r.t.x.
d
dx (1+y3 )= d dx (1) + d dx (y3 )= d dx (y3 )
d
We note from the formulafollowing Example 11.11 that
dx (y3 ) = 3y 2dy . So finally,
dx
3x 2 + dy
dx = 3y2dy dx
from which
dy
dx = 3x2
3y 2 −1
Note that dy isexpressed intermsofxandy.
dx
Example11.14 Find dy
dx given
(a) lny=y−x 2
(b) x 2 y 3 −e y = e 2x
Solution (a) Differentiating the given equation w.r.t.xyields
1dy
y dx = dy
dx −2x
from which
dy
dx = 2xy
y −1

11.3 Parametric, implicit and logarithmic differentiation 397
(b) Consider d dx (x2 y 3 ). Using the product rule wefind
d
dx (x2 y 3 ) = d dx (x2 )y 3 +x 2 d dx (y3 ) = 2xy 3 +x 2 3y 2dy
dx
Consider d dx (ey ).Letz = e y so dz
dy = ey . Hence,
d
dx (ey )= dz
dx = dz
dy
dy
dx = eydy dx
So, upon differentiating, the equation becomes
So,
2xy 3 +x 2 3y 2dy
dx −eydy dx = 2e2x
dy
dx (3x2 y 2 −e y ) = 2e 2x −2xy 3
from which
dy
dx = 2e2x −2xy 3
3x 2 y 2 −e y
11.3.3 Logarithmicdifferentiation
Thetechniqueof logarithmicdifferentiationisusefulwhenweneedtodifferentiatea
cumbersomeproduct.The methodinvolves takingthe naturallogarithm ofthe function
tobedifferentiated.This isillustratedinthe following examples.
Example11.15 Given thaty =t 2 (1−t) 8 find dy
dt .
Solution Theproductrulecouldbeusedbutwewilldemonstrateanalternativetechnique.Taking
the natural logarithm of both sides ofthe given equation yields
lny = ln(t 2 (1−t) 8 )
Usingthe laws of logarithms we can write thisas
lny=lnt 2 +ln(1−t) 8
=2lnt+8ln(1−t)
Both sides of thisequation are now differentiated w.r.t.t togive
d
dt (lny) = d dt (2lnt)+ d (8ln(1 −t))
dt
The evaluation of d (lny) has already been found inExample 11.11, and so
dt
1dy
y dt = 2 t − 8
1 −t

398 Chapter 11 Techniques of differentiation
Hence
(
dy 2
dt =y t − 8 )
1 −t
Finally, replacingybyt 2 (1 −t) 8 wehave
(
dy 2
dt =t2 (1−t) 8 t − 8 )
1 −t
Example11.16 Given
find dy
dx .
y =x 3 (1+x) 9 e 6x
Solution Theproductrulecouldbeused.However,wewilluselogarithmicdifferentiation.Taking
the natural logarithm of the equation and applying the laws of logarithms produces
lny = ln(x 3 (1 +x) 9 e 6x ) =lnx 3 +ln(1 +x) 9 +lne 6x
lny=3lnx+9ln(1+x)+6x
This equation isnow differentiated:
and so
1dy
y dx = 3 x + 9
1 +x +6
(
dy 3
dx =y x + 9 )
1 +x +6
= 3x 2 (1 +x) 9 e 6x +9x 3 (1 +x) 8 e 6x +6x 3 (1 +x) 9 e 6x
Example11.17 Ify = √ 1 +t 2 sin 2 t findy ′ .
Solution Taking logarithms we find
lny=ln( √ 1 +t 2 sin 2 t)
= ln √ 1 +t 2 +ln(sin 2 t)
= 1 2 ln(1 +t2 )+2ln(sint)
Differentiation yields
1
y y′ = 1 2t
21 +t +2cost 2 sint
( ) t
y ′ =y
1 +t +2cott 2
= √ 1 +t 2 sin 2 t
( t
1 +t 2 +2cott )

11.3 Parametric, implicit and logarithmic differentiation 399
EXERCISES11.3
1 Find each ofthe following:
(d) x=cos2t y=3t
d d d
(a)
dx (x5 ) (b)
dx (y5 ) (c)
dt (y5 )
(e) x= 3 y=e 2t
t
(f) x=e
d d
(d)
dx (y2 ) (e)
dx (5y) (f) d
t −e −t y=e t +e −t
dx (y) 5 Use logarithmicdifferentiationto find the derivatives
d
ofthe followingfunctions:
(g)
dx (y2 +y 3 )
(a) y =x 4 e x (b) y = 1 x e−x
2 Find eachofthe following:
(c) z =t 3 (1+t) 9 (d) y =e x sinx
d d
(a)
dx (ey ) (b)
dx (siny) (e) y =x 7 sin 4 x
d
(c)
dx (cos2y) (d) d
6 Use logarithmicdifferentiationto find the derivatives
dx (e−3y )
ofthe followingfunctions:
d
(e)
dx (√ d
y) (f)
dx (x+y) (a) z =t 4 (1−t) 6 (2+t) 4
d
(g) (cos(x +y))
dx (h) d
dx (lny)
(b)y= (1 +x2 ) 3 e 7x
(2 +x) 6
(c) x = (1+t) 3 (2+t) 4 (3+t) 5
d
(i)
dx (ln7y)
(d)y= (sin4 t)(2 −t 2 ) 4
3 Find dy
dx given
(1+e t ) 6
(e) y =x 3 e x sinx
(a) 2y 2 −3x 3 =x+y
(b) √ y + √ x=x 2 +y 3
7 Ifx=t+t 2 +t 3 andy=sin2t,find dy whent = 1.
dx
√
(c) 2x+3y=1+e x
8 Givenx=1+t 6 andy=1−t 2 ,find:
(d) y= ex√ 1 +x
(a) the rate ofchange ofxw.r.t.t whent = 2
x 2
(b) the rate ofchange ofyw.r.t.t whent = 1
(e) 2xy 4 =x 3 +3xy 2
(c) the rate ofchange ofyw.r.t.xwhent = 1
(f) sin(x+y)=1+y
(d) the rate ofchange ofxw.r.t.ywhent = 2
(g) ln(x 2 +y 2 ) = 2x −3y
(h) ye 2y =x 2 e x/2
9 Find the equations ofthe tangentsto
4 Find dy
dx ,given
y 2 =x 2 +6y
whenx = 4.
(a) x=t 2 y=1+t 3
(b) x=sint y=e t
10 Given the implicitfunction3x 2 +y 3 =yfindan
(c) x=(1+t) 3 y=1+t 3 expression for dy
dx .
Solutions
1 (a) 5x 4 (b) 5y 4dy
dx
(c) 5y 4dy
dt
(d) 2y dy
dx
(e) 5 dy
dx
(g) (2y +3y 2 ) dy
dx
(f)
dy
dx

400 Chapter 11 Techniques of differentiation
2 (a) e ydy
dx
(c) −2sin2y dy
dx
1
(e)
2 y−1/2dy dx
(
(g) −sin(x +y)
(h)
3 (a)
(b)
1 dy
y dx
1+9x 2
4y−1
(4x √ x−1) √ y
(1−6y 2√ y) √ x
(b) cosy dy
dx
(d) −3e −3ydy
dx
(f) 1+ dy
dx
)
1 + dy
dx
1 dy
(i)
y dx
2
(c)
3 (ex√ 2x+3y−1)
(
1
(d) y
2(1 +x) +1− 2 )
x
(e)
(f)
(g)
3x 2 +3y 2 −2y 4
2xy(4y 2 −3)
cos(x +y)
1 −cos(x +y)
2(x 2 +y 2 −x)
3x 2 +3y 2 +2y
5 (a) (4x 3 +x 4 )e x
( )
(b) −e −x 1
x 2 + 1 x
( )
(c) t 3 (1+t) 9 3
t + 9
1 +t
(d) e x sinx(1 +cotx)
( )
(e) x 7 sin 4 7
x
x +4cotx
(
)
4
6 (a)z
t − 6
1 −t + 4
2 +t
(
)
6x
(b) y
1+x 2 +7− 6
2 +x
(
)
3
(c) x
1 +t + 4
2 +t + 5
3 +t
(
)
(d) y 4cott −
8t
2−t 2 − 6et
1+e t
( )
(e) x 3 e x 3
sinx
x +1+cotx
7 −0.1387
8 (a) 192 (b) −2 (c) − 1 3
(h)
e (x/2−2y) x(x +4)
2(1 +2y)
4 (a) 3t/2 (b)
(d)
−3
2sin2t
e t
cost
(c)
(e) −2e 2t t 2 /3 (f)
( ) 2
t
1 +t
e t −e −t
e t +e −t
(d)
dx
dy = −3t4 ;whent =2, dx
dy = −48
9 y = −4x+6 ,y = 4x+24
5 5
10 y ′ = 6x
1−3y 2
11.4 HIGHERDERIVATIVES
The derivative, y ′ , of a function y(x) is more correctly called the first derivative of y
w.r.t.x.Sincey ′ itselfisafunctionofx,thenitisoftenpossibletodifferentiatethistoo.
Thederivative ofy ′ iscalled the secondderivative ofy:
secondderivative ofy = d dx
( ) dy
dx
which iswrittenas d2 y
dx 2 or more compactly asy′′ .

11.4 Higher derivatives 401
Example11.18 Ify(x) = 3x 2 +8x +9,findy ′ andy ′′ .
Solution y ′ =6x+8 y ′′ = d dx (6x+8)=6
Example11.19 Ify(t) = 2sin3t, findy ′ andy ′′ .
Solution y ′ = 6cos3t y ′′ = −18sin3t
Thefirst and secondderivatives w.r.t.time,t, arealsodenotedbyẏandÿ.
Example11.20 Findy ′′ given
1+xy=x 2 +y 2
Solution Theequation isdifferentiated implicitly toobtain dy
dx :
0 +y+xy ′ =2x+2yy ′
y ′ = 2x−y
x−2y
Thequotientruleisnowusedwithu = 2x −yand v =x −2y.Thederivativesofuand
v are
Then,
du
dx =u′ =2−y ′
dv
dx = v′ =1−2y ′
y ′′ = (x −2y)(2 −y′ ) − (2x −y)(1 −2y ′ )
(x −2y) 2
This issimplified to
y ′′ = 3xy′ −3y
(x −2y) 2
Replacingy ′ by 2x−y and simplifying yields
x−2y
y ′′ = 6(x2 −xy+y 2 )
(x −2y) 3
Note that it is possible to simplify this further by observing thatx 2 +y 2 = 1 +xy as
given. Therefore,
y ′′ =
6(1 +xy −xy)
(x −2y) 3 =
6
(x −2y) 3
Justasthefirstderivativemaybedifferentiatedtoobtainthesecondderivative,sothe
second derivative may be differentiated to find the third derivative and so on. A similar

402 Chapter 11 Techniques of differentiation
notation is used. The third derivative is written d3 y
dx 3 ory′′′ ory (3) . The fourth derivative
iswritten d4 y
dx 4 oryiv ory (4) . The fifth derivative iswritten d5 y
dx 5 oryv ory (5) .
Example11.21 Findthe first five derivatives ofz(t) = 2t 3 +sint.
Solution z ′ =6t 2 +cost z iv =sint
z ′′ =12t−sint z v =cost
z ′′′ =12 −cost
Example11.22 Calculate the values ofxforwhichy ′′ = 0,giveny =x 4 −x 3 .
Solution y=x 4 −x 3 y ′ =4x 3 −3x 2 y ′′ =12x 2 −6x
Puttingy ′′ = 0 gives
Hence
12x 2 −6x =0 andso 6x(2x −1) =0
x=0, 1 2
The first and second derivatives can be used to describe the nature of increasing and
decreasing functions. In Figure 11.2(a, b) the tangents to the curves have positive gradients,
that isy ′ > 0. As can be seen, asxincreases the value of the function increases.
Conversely, in Figure 11.2(c, d) the tangents have negative gradients (y ′ < 0) and as
x increases the value of the function decreases. The sign of the first derivative tells us
whetheryis increasing or decreasing. However, the curves in (a) and (b) both showy
increasingbut, clearly, thereisadifference inthe wayychanges.
ConsideragainFigure11.2(a).ThetangentsatA,BandCareshown.Asxincreases
thegradientofthetangentincreases,thatisy ′ increasesasxincreases.Sincey ′ increases
as x increases then the derivative of y ′ is positive, that is y ′′ > 0. (Compare with: y
increaseswhenitsderivativeispositive.)SoforthecurveshowninFigure11.2(a),y ′ > 0
andy ′′ > 0.
For that shown in Figure 11.2(b) the situation is different. The value ofy ′ decreases
asxincreases, as can be seen by considering the gradients of the tangents at A, B and
C, thatisthe derivative ofy ′ mustbenegative. For thiscurvey ′ > 0 andy ′′ < 0.
A function is concave down wheny ′ decreases and concave up wheny ′ increases.
HenceFigure11.2(a)illustratesaconcaveupfunction;Figure11.2(b)illustratesaconcavedownfunction.Thesignofthesecondderivativecanbeusedtodistinguishbetween
concave upand concave down functions.
Consider now the functions shown in Figure 11.2(c) and Figure 11.2(d). In both (c)
and(d),yisdecreasingandsoy ′ < 0.In(c)thegradientofthetangentbecomesincreasinglynegative;thatis,itisdecreasing.Hence,forthefunctionin(c)y
′′ < 0.Conversely,
for the function in (d) the gradient of the tangent is increasing asxincreases, although
it is always negative, that isy ′′ > 0. So for the function in (c)y ′ < 0 andy ′′ < 0; that

11.4 Higher derivatives 403
y
C
B
y
C
A
y
B
A
y
(a)
A
B
x
A
(b)
x
(c)
C
x
(d)
B
C
x
Figure11.2
(a)yis concave up (y ′ > 0,y ′′ > 0);(b)yisconcave down (y ′ > 0,y ′′ < 0);(c)yis concave down (y ′ < 0,y ′′ < 0);(d)y
is concave up (y ′ < 0,y ′′ > 0).
is, the function is concave down. For the function in (d)y ′ < 0 andy ′′ > 0; that is, the
function isconcave up. Insummary, we can state:
Wheny ′ > 0,yisincreasing.Wheny ′ < 0,yisdecreasing.
Wheny ′ isincreasingthe functionisconcave up. Inthiscasey ′′ > 0.
Wheny ′ isdecreasing the functionisconcave down. Inthiscasey ′′ < 0.
EXERCISES11.4
An easy way of determining the concavity of a curve is to note that as the curve is
traced from left to right, an anticlockwise motion reveals that the curve is concave up.
Aclockwisemotionmeansthatthe curve is concave down.
Aswillbeseeninthenextchapter,higherderivativesareusedtodeterminethelocationandnatureofimportantpointscalledmaximumpoints,minimumpointsandpoints
ofinflexion.
1 Calculate dy
dt and d2 y
dt 2 given
(a) y=t 2 +t
2 If
(b) y=2t 3 −t 2 +1
(c) y = sin2t
(d) y = sinkt
(e) y=2e 3t −t 2 +1
(f) y=
t
t +1
(g) y=4cos t 2
(h) y =e t t
(i) y = sinh4t
(j) y = sin 2 t
k constant
y=2x 3 +3x 2 −12x+1
findvalues ofxforwhichy ′′ = 0.
3 If dy
dt =3t2 +t,find
(a)
d 2 y
dt 2
(b)
d 3 y
dt 3
4 Find values oft atwhichy ′′ = 0,where
y = t3 3 − 7t2
2 +12t−1
5 Determine whether the following functionsare
concave up orconcave down.
(a) y=e t (b) y=t 2 (c) y=1+t−t 2
6 Determine the interval onwhichy =t 3 is
(a) concave up,
(b) concave down.
7 Evaluatey ′′ atthe specifiedvalue oft.
(a) y=2cost−t 2 t=1
(b) y= sint+cost
2
t=π/2
(c) y=(1+t)e t t=0
8 Find d2 y
dx 2 givenxy +x2 =y 2 .
9 Find dx
dt whenx3 + x t =t2 +x 2 t.

404 Chapter 11 Techniques of differentiation
Solutions
1 (a) 2t+1,2
(b) 6t 2 −2t,12t −2
(c) 2cos2t, −4sin2t
(d) kcoskt,−k 2 sinkt
(e) 6e 3t −2t,18e 3t −2
1
(f)
2
(t +1) 2,− (t +1) 3
(g) −2sin(t/2),−cos(t/2)
(h) e t (t +1),e t (t +2)
(i) 4cosh4t,16sinh4t
(j) 2sintcost,2cos2t
2 − 1 2
3 (a) 6t+1 (b) 6
4
7
2
5 (a) concave up
(b) concave up
(c) concave down
6 (a) concave upon (0,∞)
(b) concave down on (−∞,0)
7 (a) −3.08 (b) − 1 2
8
9
10y 2 −10x 2 −10yx
(2y −x) 3
x+2t 3 +x 2 t 2
3x 2 t 2 +t −2xt 3
(c) 3
REVIEWEXERCISES11
1 Differentiate eachofthe followingfunctions:
(a) y = sin(5 +x) 2
(b) y = e 2sinx
(c) y = (4x +7) 5
(d) y =x 4 sin3x
(e) y=
e4x
x 3 +11
(f) y=x 2 tanx
(g) y= cos3x
x 2
(h) y = e −x cos5x
(i) y = lncos4x
(j) y = sin2tcos2t
(k) y= 1
x 2 +1
2 Find dy in each ofthe following cases:
dx
(a) y= x3 sin2x
cosx
(b) y =x 3 e −x tanx
(c) y= xe5x
sinx
(d) x 2 +3xy+y 2 =5
(e) 9=3x 3 +2xy 2 −y
3 Ifx= 5+3t
1 −t
andy= 2 −t dy
find
1 −t dx and d2 y
dx 2.
4 Ifx=4(1+cosθ)andy=3(θ−sinθ)find dy
dx .
5 Ifx=3cos 2 θandy=2sin 2 θfind dy
dx and d2 y
dx 2.
6 Show thaty = e −4x sin8x satisfiesthe equation
y ′′ +8y ′ +80y = 0.
7 Differentiatey =x x .
8 Giveny =x 3 e 2x find
dy d 2 y
(a) (b)
dx dx 2
(c)
d 3 y
dx 3
9 Uselogarithmicdifferentiationto find dx
dt given
(a) x =te t sint
(b) x =t 2 e −t cos3t
(c) x =t 2 e 3t sin4tcos3t
10 Show thatify(t) =Asin ωt +Bcos ωt,where ω isa
constant, then
y ′′ +ω 2 y=0

Review exercises 11 405
Solutions
1 (a) 2(5 +x)cos(5 +x) 2
(b) 2cosxe 2sinx
(c) 20(4x +7) 4
(d) 3x 4 cos3x +4x 3 sin3x
(e)
2 (a)
e 4x (4x 3 −3x 2 +44)
(x 3 +11) 2
(f) x 2 sec 2 x +2xtanx
−3xsin3x −2cos3x
(g)
x 3
(h) −e −x (5sin5x +cos5x)
(i) −4tan4x
(j) 2cos4t
−2x
(k)
(x 2 +1) 2
cosx(2x 3 cos2x +3x 2 sin2x) +x 3 sin2xsinx
cos 2 x
whichsimplifiesto 2x 3 cosx +6x 2 sinx
(b) e −x (x 3 sec 2 x −x 3 tanx +3x 2 tanx)
(c)
e 5x (5xsinx +sinx −xcosx)
sin 2 x
3
(d) − 2x+3y
3x+2y
9x 2 +2y 2
(e)
1−4xy
1
8 ,0
4 − 3(1−cosθ)
4sinθ
5 − 2 3 ,0
7 x x (lnx+1)
8 (a) e 2x (2x 3 +3x 2 )
(b) 2xe 2x (2x 2 +6x +3)
(c) 2e 2x (4x 3 +18x 2 +18x +3)
9 (a) e t (tcost+(t+1)sint)
(b) −e −t ((t 2 −2t)cos3t +3t 2 sin3t)
(
)
(c) t 2 e 3t 2 4cos4t
sin4tcos3t +3+ − 3sin3t
t sin4t cos3t

12 Applicationsof
differentiation
Contents 12.1 Introduction 406
12.2 Maximumpointsandminimumpoints 406
12.3 Pointsofinflexion 415
12.4 TheNewton--Raphsonmethodforsolvingequations 418
12.5 Differentiationofvectors 423
Reviewexercises12 427
12.1 INTRODUCTION
In this chapter the techniques of differentiation are used to solve a variety of problems.
It is possible to use differentiation to find the maximum or minimum values of a function.
For example, it is possible to find the maximum power transferred from a voltage
source to a load resistor, as we shall show later in the chapter. Differentiation is also
usedintheNewton--Raphsonmethodofsolvingnon-linearequations.Suchanequation
needs to be solved to calculate the steady-state values of current and voltage in a series
diode--resistorcircuit.
Finallyweshowhowvectorscanbedifferentiated.Thisformsanintroductiontothe
important topic of vector calculus which is discussedinChapter 26.
12.2 MAXIMUMPOINTSANDMINIMUMPOINTS
Consider Figure 12.1. A and B are important points on the curve. At A the function
stops increasing and starts to decrease. At B it stops decreasing and starts to increase.
A is a local maximum, B is a local minimum. Note that A is not the highest point on
the curve, nor B the lowest point. However, for that part of the curve near to A, A is
the highest point. The word ‘local’ is used to stress that A is maximum in its locality.
Similarly, Bisthe lowestpoint initslocality.

12.2 Maximum points and minimum points 407
y
A
y
A
(a)
B
Figure12.1
Thefunctionyhas a local maximum at Aand alocal minimumat B.
x
(b)
B
x
In Figure 12.1(a) tangents drawn at A and B would be parallel to thexaxis and so
atthesepoints dy
dx iszero.However,inFigure12.1(b)therearecornersatAandB.Itis
impossibletodraw tangents atthese points and so dy does notexistatthese points.
dx
Hence, when searching for maximum and minimum points we need only examine
thosepoints atwhich dy dy
is zero,or does not exist.
dx dx
Points at which dy is zero are known as turning points or stationary values of the
dx
function.
At maximumand minimumpoints either:
dy
(i) does notexist, or
dx
dy
(ii)
dx = 0
To distinguish between maximum and minimum points we can study the sign of dy
dx on
eithersideofthepoint.AtmaximumpointssuchasA,yisincreasingimmediatelytothe
leftofthepoint,anddecreasingimmediatelytotheright.Thatis, dy
dx ispositiveimmediatelytotheleft,and
dy
dx isnegativeimmediatelytotheright.Atminimumpointssuchas
B,yisdecreasingimmediatelytotheleftofthepoint,andincreasingimmediatelytothe
right.Thatis, dy
dy
isnegativeimmediatelytotheleft,and
dx dx ispositiveimmediatelytothe
right.Thisso-calledfirst-derivativetestenablesustodistinguishmaximafromminima.
This testcan be used even when the derivative does notexistatthe pointinquestion.
Thefirst-derivative testtodistinguish maximafromminima:
To the left ofamaximumpoint, dy dy
ispositive; tothe right,
dx dx isnegative.
To the left of a minimum point, dy
dy
isnegative; tothe right,
dx dx ispositive.

408 Chapter 12 Applications of differentiation
Example12.1 Determinethepositionandnatureofallmaximumandminimumpointsofthefollowing
functions:
(a)y=x 2
(b)y=−t 2 +t+1
(c)y= x3
3 + x2
2 −2x+1
(d) y = |t|
Solution (a) Ify =x 2 , then by differentiation dy
dx = 2x.
Recall thatatmaximum and minimum points either
(i)
(ii)
dy
does notexist, or
dx
dy
= 0.We mustcheck both of these conditions.
dx
The function 2x exists for all values of x, and so we move to examine any points
where dy = 0.So, wehave
dx
dy
dx =2x=0
Theequation2x = 0hasonesolution,x = 0.Weconcludethataturningpointexists
atx = 0. Furthermore, from the given functiony =x 2 , we see that whenx = 0 the
valueofyisalso0,soaturningpointexistsatthepointwithcoordinates (0,0).To
determine whether this point is a maximum or minimum we use the first-derivative
test and examine the sign of dy on either side ofx = 0. To the left ofx = 0,xis
dx
clearly negative and so 2x is also negative. To the right ofx = 0,xis positive and
so2xisalsopositive.Henceyhasaminimumatx = 0.Agraphofy =x 2 showing
thisminimum isgiven inFigure 12.2.
(b) Ify=−t 2 +t + 1, theny ′ = −2t + 1 and this function exists for all values oft.
Solvingy ′ = 0 we have
−2t+1=0 andsot= 1 2
Weconcludethatthereisaturningpointatt = 1 2 .Theycoordinatehereis − ( 1
2) 2
+
1
2 +1=11 4 . We now inspect the sign ofy′ to the left and to the right oft = 1 2 . A
little to the left, say att = 0, we see thaty ′ = −2(0) +1 = 1 which is positive. A
littletotheright,sayatt = 1,weseethaty ( ′ = −2(1) +1 = −1whichisnegative.
1
Hence thereisamaximum atthe point
2 4)
,11 .
Agraph ofthe function isshown inFigure 12.3.

12.2 Maximum points and minimum points 409
y
y = x 2
y
y = –t 2 + t + 1
1–
2
t
x
Figure12.2
Thefunctionyhas a minimumatx = 0.
Figure12.3
Thefunctionyhas amaximum att = 1 2 .
(c) Ify= x3
3 + x2
2 −2x +1,theny′ =x 2 +x−2andthisfunctionexistsforallvalues
ofx. Solvingy ′ = 0 wefind
x 2 +x−2=0
(x−1)(x+2)=0
x=1,−2
Therearethereforetwoturningpoints,oneatx = 1andoneatx = −2.Weconsider
each inturn.
Atx = 1,weexaminethesignofy ′ totheleftandtotherightofx = 1.Alittleway
to the left, say atx = 0, we see thaty ′ = −2 which is negative. A little to the right,
sayatx = 2,weseethaty ′ = 2 2 +2−2 = 4whichispositive.Sothepointwhere
x = 1 isaminimum.
Atx = −2, we examine the sign ofy ′ to the left and to the right ofx = −2. A little
way to the left, say atx = −3, we see thaty ′ = (−3) 2 + (−3) − 2 = 4 which is
positive.Alittletotheright,sayatx = −1,weseethaty ′ = (−1) 2 +(−1)−2 = −2
which isnegative. So the point wherex = −2 isamaximum.
Agraph of the function is shown inFigure 12.4.
(d) Recall that the modulus functiony = |t| isdefined as follows:
{ −t t0
y=|t|=
t t>0
A graph of this function was given in Figure 10.13(a) and this should be looked at
before continuing. Note that dy = −1 fort negative, and dy = +1 fort positive.
dt
dt
Thederivativeisnotdefinedatt = 0becauseofthecornerthere.Therearenopoints
when dy = 0. Because the derivative is not defined att = 0 this point requires
dt
further scrutiny. To the left oft = 0, dy dy
< 0; to the right,
dt dt >0andsot=0isa
minimum point.
y
y = —
x 3
+ —
x 2
– 2x + 1
3 2
–2
1
x
Figure12.4
Thefunctionyhas a maximum atx = −2
and aminimum atx = 1.

410 Chapter 12 Applications of differentiation
y
dy
—
dx > 0
dy
—
dx = 0
Figure12.5
The derivative dy decreases on
dx
passingthrough amaximum point.
dy
—
dx < 0
x
y
dy
—
dx < 0
dy
—
dx = 0
dy
—
dx > 0
Figure12.6
Thederivative dy increases onpassing
dx
through aminimum point.
x
Rather than examine the sign of y ′ on both sides of the point, a second-derivative
test may be used. On passing through a maximum pointy ′ changes from positive to 0
to negative, as shown in Figure 12.5. Hence,y ′ is decreasing. Ify ′′ is negative then this
indicates y ′ is decreasing and the point is therefore a maximum point. Conversely, on
passingthroughaminimumpoint,y ′ increases,goingfromnegativeto0topositive(see
Figure 12.6). Ify ′′ ispositive theny ′ isincreasing and this indicates a minimum point.
So,havinglocatedthepointswherey ′ = 0,welookatthesecondderivative,y ′′ .Thus
y ′′ > 0 implies a minimum point;y ′′ < 0 implies a maximum point. Ify ′′ = 0, then we
must return to the earlier, more basic test of examiningy ′ on both sides of the point. In
summary:
Thesecond-derivative test todistinguish maximafrom minima:
Ify ′ = 0 andy ′′ < 0 atapoint,thenthisindicates thatthe pointisamaximum
turningpoint.
Ify ′ = 0 andy ′′ > 0 atapoint,thenthisindicates thatthe pointisaminimum
turningpoint.
Ify ′ = 0 andy ′′ = 0 atapoint,the second-derivative testfailsand you must
use the first-derivative test.
Example12.2 Usethesecond-derivativetesttofindallmaximumandminimumpointsofthefunctions
inExample 12.1.
Solution (a) Given y = x 2 then y ′ = 2x and y ′′ = 2. We locate the position of maximum and
minimum points by solvingy ′ = 0 and so such a point exists atx = 0. Evaluating
y ′′ atthispointweseethaty ′′ (0) = 2whichispositive.Usingthesecond-derivative
testweconcludethatthe pointisaminimum.
(b) Giveny = −t 2 +t + 1 theny ′ = −2t + 1 andy ′′ = −2. Solvingy ′ = 0 we find
t = 1 ( ) 1
2 . Evaluatingy′′ at this point we findy ′′ = −2 which is negative. Using
2
the second-derivative testwe conclude thatt = 1 isamaximum point.
2
(c) Given y = x3
3 + x2
2 −2x+1,theny′ =x 2 +x−2andy ′′ =2x+1.y ′ =0
at x = 1 and x = −2. At x = 1, y ′′ = 3 which is positive and so the point

12.2 Maximum points and minimum points 411
is a minimum. At x = −2, y ′′ = −3 which is negative and so the point is a
maximum. ⎧
⎪⎨ −1 t<0
(d) y ′ = 1 t>0
⎪⎩
undefined att = 0
Since y ′ (0) is undefined, we use the first-derivative test. This was employed in
Example 12.1.
Engineeringapplication12.1
Risetimeforasecond-orderelectricalsystem
Consider the electrical system illustrated in Figure 12.7. The input voltage, v i
, is
appliedtoterminalsa--b.Theoutputfromthesystemisavoltage, v o
,measuredacross
the terminals c--d. The easiest way to determine the time response of this system to
a particular input istouse the technique of Laplace transforms (seeChapter 21).
When a step input is applied to the system, the general form of the response dependsonwhetheraquantitycalledthedampingratio,
ζ,issuchthat ζ > 1,ζ = 1or
ζ < 1.Thequantity ζ itselfdependsuponthevaluesofL,CandR.Thisisillustrated
inFigure12.8.Ifthedampingratio, ζ < 1,then v o
overshootsitsfinalvalueandthe
systemissaidtobe underdamped. For thiscase itcan beshown that
(
v o
=U −Ue −αt cos(βt)+ αsin(βt) )
for t > 0 (12.1)
β
whereU isthe height ofastepinputapplied att = 0,and
α = R 2L
(12.2)
ω r
= 1 √
LC
resonant frequency (12.3)
β =
√
ωr 2 − α2 naturalfrequency (12.4)
Engineers are often interested in knowing how quickly a system will respond to a
particular input. For many systems this is an important design criterion. One way of
characterizingthespeedofresponseofthesystemisthetimetakenfortheoutputto
reach a certain level in response to a step input. This is known as the rise time and
is often defined as the time taken for the output to rise from 10% to 90% of its final
value.However,bylookingattheunderdampedresponseillustratedinFigure12.8it
is clear that the time,t m
, required for the output to reach its maximum value would
also provide an indicator of system response time. As the derivative of a function is
zero atamaximum point itispossible tocalculate thistime.
Differentiating Equation (12.1) and usingthe product rule,
dv o
dt
= d dt
( (U −Ue −αt cos(βt)+ αsin(βt)
β
))
( ) e −αt αsin(βt)
t > 0
=0−U d dt (e−αt cos(βt)) −U d dt
β
➔

412 Chapter 12 Applications of differentiation
= −U(−αe −αt cos(βt) −e −αt βsin(βt))
( −αe −αt )
αsin(βt)
−U
+ e−αt αβcos(βt)
β
β
(
)
= −Ue −αt −αcos(βt) − βsin(βt) − α2 sin(βt)
+ αcos(βt)
β
( )
=Ue −αt β + α2
sin(βt)
β
At a turning point dv o
= 0.Hence
dt
( β
Ue −αt 2 + α 2 )
sin(βt) =0
β
y o (t)
z < 1
a
R
L
c
U
z = 1
y i
b
Figure12.7
Asecond-order electrical system.
C
d
y o
t m
z > 1
Figure12.8
Responseofasecond-order system to a step
input.
This occurs when sin(βt) = 0, which corresponds tot = kπ/β, k = 0,1,2....
It is now straightforward to calculate t m
, once β has been calculated, using Equations
(12.2), (12.3) and (12.4) for particular values ofR,L andC. You may like to
show that the turning point corresponding to k = 1 is a maximum by calculating
d 2 v o
dt 2 and carrying outthe second-derivative test.
Itispossibletocheckwhetherornotasystemisunderdampedusingthefollowing
formulae:
ζ = R R c
damping ratio (12.5)
t
√
L
R c
=2
C
criticalresistance (12.6)
LetuslookataspecificcasewithtypicalvaluesL = 40mH,C = 1 µF,R = 200.
UsingEquations (12.5) and (12.6),wefind
√ √
L
R c
=2
C = 2 4 ×10 −2
= 400
1 ×10−6 ζ = R R c
= 200
400 = 0.5

12.2 Maximum points and minimum points 413
and therefore the systemisunderdamped because ζ < 1.
Also,
ω r
= 1 √
LC
= 5000
α = R 2L = 200
= 2500
2×4×10−2 √
β = ωr 2 −α2 = √ 5000 2 −2500 2 = 4330
Finally,
t m
=
π
4330 = 7.26 ×10−4 = 726 µs
We conclude forthis case that the risetimeis726 µs.
Engineeringapplication12.2
Maximumpowertransfer
ConsiderthecircuitofFigure12.9inwhichanon-idealvoltagesourceisconnected
toavariableloadresistorwithresistanceR L
.ThesourcevoltageisV anditsinternal
resistanceisR S
.CalculatethevalueofR L
whichresultsinthemaximumpowerbeing
transferred from the voltage source to the load resistor. This is an essential piece of
information for engineers involved in the design of power systems. Often an importantdesignconsiderationistotransferthemaximumamountfromthepowersource
tothe pointwhere the power isbeingconsumed.
i
R S
Non-ideal
voltage
source
+
V
–
R L
Figure12.9
Maximum power transferoccurs whenR L =R S .
Solution
Letibethecurrentflowinginthecircuit.UsingKirchhoff’svoltagelawandOhm’s
law gives
V =i(R S
+R L
)
LetPbethe power developed inthe loadresistor.Then,
P=i 2 R L
=
V2 R L
(R S
+R L
) 2 ➔

414 Chapter 12 Applications of differentiation
ClearlyPdependsonthevalueofR L
.Differentiatingw.r.t.R L
andusingthequotient
rule,weobtain
dP
=V 21(R S +R L )2 −R L
2(R S
+R L
)
dR L
(R S
+R L
) 4
=V 2 (R S +R L )−2R L
(R S
+R L
) 3
=V 2 R S
−R L
(R S
+R L
) 3
Equating dP
dR L
tozero toobtain the turning pointgives
thatis,
V 2 R S
−R L
(R S
+R L
) 3 = 0
R L
=R S
So a turning point occurs when the load resistance equals the source resistance. We
need tocheck ifthisisamaximum turning point,so
dP
dR L
=V 2 R S
−R L
(R S
+R L
) 3
d 2 P
dR 2 L
=V 2 −1(R S +R L )3 − (R S
−R L
)3(R S
+R L
) 2
(R S
+R L
) 6
=V 2 −(R S +R L )−3(R S −R L )
(R S
+R L
) 4
=V 2 2R L −4R S
(R S
+R L
) 4
= 2V 2 (R L −2R S )
(R S
+R L
) 4
WhenR L
= R S
, this expression is negative and so the turning point is a maximum.
Therefore, maximum power transfer occurs when the load resistance equals the
source resistance.
EXERCISES12.2
1 Locatethe position ofany turningpointsofthe
following functionsanddeterminewhether they are
maxima orminima.
(a) y=x 2 −x+6 (b) y=2x 2 +3x+1
(c) y=x−1 (d) y=1+x−2x 2
(e) y=x 3 −12x (f) y=7+3x
2 Locateandidentify allturningpointsof
(a) y= x3
3 −3x2 +8x+1
(b) y =te −t
(c) y=x 4 −2x 2

12.3 Points of inflexion 415
Solutions
1 (a)
( 1
2 ,53 4)
,minimum
(b) (−0.75,−0.125),minimum
(c) none
(d) (0.25,1.125),maximum
(e) (2,−16) minimum, (−2,16) maximum
(f) none
2 (a)
(
2, 23 ) (
maximum, 4, 19 )
minimum
3 3
(b) (1,0.368),maximum
(c) (0,0) maximum, (1,−1) minimum, (−1,−1)
minimum
TechnicalComputingExercises12.2
Computerlanguagessuch asMATLAB ® are matrix
orientatedanddonotalways providethe ability to
differentiate functions.Others such asWolfram
Mathematica andMaplesoftMaple have thiscapability by
default. Ifyou are attempting the followingexercisesin
MATLAB ® you mayrequire the Symbolic Math Toolbox
whichis an add-on forthe main program.
(a) Useatechnicalcomputinglanguagetofindy ′ and
y ′′ wheny = e −0.2t cost.
(b) Solvey ′ = 0 andhence locate any turningpoints
in the interval [0,6]anddeterminetheirtype.
(c) Plotagraphofyandcheck the position of
the turningpointswith the resultsobtainedin
part (b).
12.3 POINTSOFINFLEXION
Recall from Section 11.4 that when the gradient of a curve, that is y ′ , is increasing,
the second derivative y ′′ is positive and the curve is said to be concave up. When the
gradient is decreasing the second derivative y ′′ is negative and the curve is said to be
concave down. A point at which the concavity of a curve changes from concave up to
concave down orviceversa iscalled a point ofinflexion.
Apointofinflexionisapointonacurvewheretheconcavitychangesfromconcave
up to concave down or vice versa. It follows that y ′′ = 0 at such a point or, in
exceptionalcases,y ′′ doesnotexist.
Figure 12.10(a) shows a graph for which a point of inflexion occurs at the point
markedA.Notethatatthispointthegradientofthegraphiszero.Figure12.10(b)shows
a graph with points of inflexion occurring at A and B. Note that at these points the
gradientofthe graph isnotzero.
Tolocateapointofinflexionwemustlookforapointwherey ′′ = 0ordoesnotexist.
We mustthenexamine the concavity ofthe curve oneither sideofsuchapoint.

416 Chapter 12 Applications of differentiation
y
y
A
A
B
(a)
x
Figure12.10
(a) There isapoint ofinflexion atA; (b)there are pointsofinflexion at A
and B.
(b)
y
x
y = x 4 x
y
y = x 3
y'' > 0
y'' = 0
x
y' < 0, y'' > 0 y' > 0, y'' > 0
y'' < 0
Figure12.11
The second derivative,y ′′ ,changes
sign atx = 0.
Figure12.12
Thederivative,y ′ ,changessignatx = 0,but
y ′′ remainspositive.
Example12.3 Locateany points ofinflexion ofthe curvey =x 3 .
Solution Given y = x 3 , then y ′ = 3x 2 and y ′′ = 6x. Points of inflexion can only occur where
y ′′ = 0 or does not exist. Clearlyy ′′ exists for allxand is zero whenx = 0. It is possible
that a point of inflexion occurs whenx = 0 but we must examine the concavity of
the curve on either side. To the left ofx = 0,xis negative and soy ′′ is negative. Hence
to the left, the curve is concave down. To the right of x = 0, x is positive and so y ′′
is positive. Hence to the right, the curve is concave up. Thus the concavity changes at
x = 0.Weconcludethatx = 0isapointofinflexion.AgraphisshowninFigure12.11.
Note thatatthis pointofinflexiony ′ = 0 too.
Acommonerroristostatethatify ′ =y ′′ = 0thenthereisapointofinflexion.This
isnotalways true;considerthe next example.
Example12.4 Locateall maximumpoints,minimum points and points ofinflexion ofy =x 4 .
Solution y ′ = 4x 3 y ′′ = 12x 2

12.3 Points of inflexion 417
y ′ = 0 atx = 0. Alsoy ′′ = 0 atx = 0 and so the second-derivative test is of no help
in determining the position of maximum and minimum points. We return to examiney ′
on both sides ofx = 0. To the left ofx = 0,y ′ < 0; to the righty ′ > 0 and sox = 0 is
a minimum point. Figure 12.12 illustrates this. Note that at the pointx = 0, the second
derivativey ′′ is zero. However,y ′′ is positive both to the left and to the right ofx = 0;
thusx = 0 isnot a point of inflexion.
Example12.5 Find any maximum points,minimum points and points of inflexion ofy =x 3 +2x 2 .
Solution Giveny =x 3 +2x 2 theny ′ = 3x 2 +4xandy ′′ = 6x +4.Letusfirstfindanymaximum
andminimumpoints.Thefirstderivativey ′ iszerowhen3x 2 +4x =x(3x+4) = 0,that
iswhenx=0orx=− 4 3 . Using the second-derivative test we findy′′ (0) = 4 which
(
corresponds to a minimum point. Similarly,y ′′ − 4 )
= −4 which corresponds to a
3
maximum point.
We seek points of inflexion by looking for points wherey ′′ = 0 and then examining
the concavity on either side.y ′′ = 0 whenx = − 2 3 .
Sincey ′′ is negative whenx < − 2 3 , theny′ is decreasing there, that is the function is
concavedown.Also,y ′′ ispositivewhenx > − 2 3 andsoy′ isthenincreasing,thatisthe
functionisconcave up. Hence thereisapointofinflexion whenx = − 2 .Thegraph of
3
y =x 3 +2x 2 isshown inFigure 12.13.
y
Figure12.13
– 4–
3
– 2–
3
x
There isamaximum atx = − 4 ,aminimum at
3
x = 0 and apoint ofinflexion atx = − 2 3 .
FromExamples12.4 and 12.5 wenotethat:
(1) The conditiony ′′ = 0 is not sufficient to ensure a point is a point of inflexion.
The concavity of the function on either side of the point wherey ′′ = 0 must be
considered.
(2) At a pointofinflexion itis notnecessarytohavey ′ = 0.
(3) At a pointofinflexiony ′′ = 0 ory ′′ doesnotexist.

418 Chapter 12 Applications of differentiation
EXERCISES12.3
1 Locate the maximumpoints,minimum pointsand (e) y=t 5
pointsofinflexion of
(f) y=t 6
(a) y=3t 2 +6t−1
(g) y=x 4 −2x 2
(b) y=4−t−t 2
(h) z=t+ 1
(c) y= x3
3 − x2
t
2 +10
(i) y=x 5 − 5x3
(d) y= x3
3 + x2
3
2 −20x+7 (j) y =t 1/3
Solutions
1 (a) (−1,−4) minimum
(b) (−0.5,4.25) maximum
(c) (0,10) maximum,
( 1
2 , 119
12
(
(d) 4, − 131
3
(
1, 59
6
)
point ofinflexion
)
minimum,
)
minimum, (−5,77.83) maximum,
(− 1 2 ,17.08 )
point ofinflexion
(e) (0,0) point ofinflexion
(f) (0,0) minimum
(g) (0,0) maximum, (1,−1) minimum,
( ) ( )
1
(−1,−1) minimum, √ , − 5 , −√ 1 , − 5
3 9 3 9
pointsofinflexion
(h) (1,2) minimum, (−1, −2)maximum
(
(i) 1, − 2 ) (
minimum, −1, 2 )
maximum,
3 3
( ) ( )
1
(0,0), √ , − 7
2 12 √ , −√ 1 7
,
2 2 12 √ are
2
alsopointsofinflexion
(j) (0,0) point ofinflexion
TechnicalComputingExercises12.3
(a) Useatechnicalcomputing language such as
MATLAB ® to produce agraph ofy = 3t 1/5 .
(b) Fromyour graphfind the position ofany maxima,
minimaorpointsofinflexion.
12.4 THENEWTON--RAPHSONMETHODFORSOLVING
EQUATIONS
We often needtosolve equationssuchas
f(x)=2x 4 −x 3 +x 2 −10=0
f(t) =2e −3t −t 2 = 0
f(t)=t−sint=0
The Newton--Raphson technique is a method of obtaining an approximate solution, or
root, ofsuchequations.Itinvolves the use ofdifferentiation.

12.4 The Newton--Raphson method for solving equations 419
f (x)
y = f (x)
B
^
x
C A
x 2 x 1 x
Figure12.14
Thetangent at Bintersects thexaxisat C.
Suppose we wish to find a root of f (x) = 0. Figure 12.14 illustrates the curvey =
f (x). Roots of the equation f (x) = 0 correspond to where the curve cuts the x axis.
One such root is illustrated and is labelledx = ˆx. Suppose we know thatx = x 1
is an
approximate solution. Let A be the point on thexaxis wherex = x 1
and let B be the
pointonthecurvewherex =x 1
.ThetangentatBisdrawnandcutsthexaxisatCwhere
x =x 2
.Clearlyx =x 2
isabetterapproximationto ˆxthanx 1
.Wenowfindx 2
intermsof
the known value,x 1
.
Hence,
AB = distance of Babove thexaxis = f (x 1
)
CA=x 1
−x 2
gradient of lineCB = AB
CA = f(x 1 )
x 1
−x 2
ButCB isatangent tothe curve atx =x 1
and so has gradient f ′ (x 1
). Hence,
f ′ (x 1
) = f(x 1 )
x 1
−x 2
x 1
−x 2
= f(x 1 )
f ′ (x 1
)
and therefore,
x 2
=x 1
− f(x 1 )
f ′ (x 1
)
(12.7)
Equation (12.7) is known as the Newton--Raphson formula. Knowing an approximate
root of f (x) = 0, that is x 1
, the Newton--Raphson formula enables us to calculate an
improved approximate root,x 2
.
Example12.6 Given thatx 1
= 7.5 is an approximate root of e x −6x 3 = 0, use the Newton--Raphson
techniquetofind animproved value.
Solution x 1
=7.5
f(x)=e x −6x 3 f(x 1
)=−723
f ′ (x)=e x −18x 2 f ′ (x 1
) =796

420 Chapter 12 Applications of differentiation
Using the Newton--Raphson technique the value ofx 2
isfound:
x 2
=x 1
− f(x 1 )
f ′ (x 1
) =7.5−(−723) = 8.41
796
An improved estimate of the root of e x − 6x 3 = 0 isx = 8.41. To two decimal places
the trueanswer isx = 8.05.
The Newton--Raphson technique can be used repeatedly as illustrated in
Example12.7.Thisgeneratesasequenceofapproximatesolutionswhichmayconverge
tothe required root. Eachapplication ofthe methodisknown asan iteration.
Example12.7 A root of 3sinx = x is near to x = 2.5. Use two iterations of the Newton--Raphson
technique tofind a more accurateapproximation.
Solution Theequation mustfirst bewritten inthe form f (x) = 0,thatis
Then
Then
f(x)=3sinx−x=0
x 1
= 2.5
f(x)=3sinx−x f(x 1
)=−0.705
f ′ (x)=3cosx−1 f ′ (x 1
) = −3.403
x 2
=2.5− (−0.705)
(−3.403) = 2.293
The process isrepeated withx 1
= 2.293 as the initial approximation:
Then
x 1
= 2.293 f (x 1
) = −0.042 f ′ (x 1
) = −2.983
x 2
= 2.293 − (−0.042)
(−2.983) = 2.279
Using two iterations of the Newton--Raphson technique, we obtainx = 2.28 as an improved
estimateof the root.
Example12.8 An approximaterootof
Solution We have
x 3 −2x 2 −5=0
isx = 3. By using the Newton--Raphson technique repeatedly, determine the value of
the rootcorrecttotwo decimalplaces.
Hence
x 1
=3
f(x)=x 3 −2x 2 −5 f(x 1
)=4
f ′ (x)=3x 2 −4x f ′ (x 1
)=15
x 2
=3− 4
15 = 2.733

12.4 The Newton--Raphson method for solving equations 421
Animprovedestimateofthevalueoftherootis2.73(2d.p.).Themethodisusedagain,
takingx 1
= 2.73 asthe initial approximation:
x 1
= 2.73 f(x 1
) = 0.441 f ′ (x 1
) = 11.439
x 2
= 2.73 − 0.441
11.439 = 2.691
An improved estimateisx = 2.69 (2d.p.). The method isused again:
So
x 1
= 2.69 f (x 1
) = −0.007 f ′ (x 1
) = 10.948
x 2
= 2.69 − (−0.007)
10.948 = 2.691
Thereisnochangeinthevalueoftheapproximaterootandsototwodecimalplacesthe
rootof f(x) =0isx =2.69.
The calculation can be performed inMATLAB ® usingtherootsfunction:
roots([1 -2 0 -5])
which will produce the real root of 2.69, agreeing with our numerical calculation using
the Newton–Raphson method. The two complex roots of the equation will also be returned.Technicalcomputinglanguagesmakeuseofavarietyofnumericalmethodsand
to some extent the user has to take it on trust that they are correctly implemented and
testedso thatthey always produce the correctresult.
Theprevious examples illustrate the generalNewton--Raphson formula.
Ifx = x n
is an approximate root of f (x) = 0, then an improved estimate,x n+1
, is
given by
x n+1
=x n
− f(x n )
f ′ (x n
)
TheNewton--Raphsonformulaiseasytoprograminaloopstructure.Exitfromtheloop
is usually conditional upon |x n+1
−x n
| being smaller than some prescribed very small
value. This condition shows that successive approximate roots are very close to each
other.
Engineeringapplication12.3
Seriesdiode--resistorcircuit
Consider the circuit of Figure 12.15(a). A diode is in series with a resistor with resistanceR.ThevoltageacrossthediodeisdenotedbyV
andthecurrentthroughthe
diode is denoted byI. TheI--V relationship for the diode is non-linear and is given
by
I =I s
(e 40V −1)
where I s
is the reverse saturation current of the diode. Given that the supply voltage,V
s
,is2V,I s
is10 −14 AandRis22k,calculatethesteady-statevaluesofIandV.
➔

422 Chapter 12 Applications of differentiation
I
I
+
–
R
V
V s —R
V s
V s V
Point at which both
the diode and the
resistor equations
are satisfied
(a)
(b)
Figure12.15
Asimple non-linear circuit: (a)seriesdiode--resistorcircuit;(b)resistorload line
superimposed on diode characteristic.
Solution
There are several ways to solve this problem. A difficulty exists because the diode
I--V relationship is non-linear. One possibility is to draw a load line for the resistor
superimposedonthediodeI--V characteristic,asshowninFigure12.15(b).Theload
lineisanequationfortheresistorcharacteristicwrittenintermsofthevoltageacross
the diode,V, and the current through the diode,I. Itisgiven by
V s
−V=IR
I = − 1 R V +V s
R
This is a straight line with slope − 1 R and vertical intercept V s
.WhenV =0,I=
V
R
s
. This corresponds to all of the supply voltage being dropped across the resistor.
R
WhenV = V s
,I = 0. This corresponds to all of the supply voltage being dropped
across the diode. Therefore, these two limits correspond to the points within which
the circuit must operate. The solution to the circuit can be obtained by determining
the intercept of the diode characteristic and the load line. This is possible because
both the resistor characteristic and diode characteristic are formulated in terms ofV
andI,andsoanysolutionmusthavethesamevaluesofI andV forbothcomponents.
If an accurate graph is used, it is possible to obtain a reasonably good solution. An
alternative approach is to use the Newton--Raphson technique. Combining the two
component equations gives
−V +V s
=RI s
(e 40V −1)
NowR =2.2×10 4 ,I s
=10 −14 ,V s
=2andso
−V +2 = 2.2 ×10 4 ×10 −14 (e 40V −1)
Now, define f (V) by
f(V)=2.2×10 −10 (e 40V −1)+V −2
We wishtosolve f (V) = 0.We have
f ′ (V ) = 2.2 ×10 −10 ×40e 40V +1 = 8.8 ×10 −9 e 40V +1
Choose aninitial guess ofV 1
= 0.5:
V 2
=V 1
− f(V 1 )
f ′ (V 1
) =0.5−2.2 ×10−10 (e 20 −1) +0.5 −2
= 0.7644
8.8 ×10 −9 e 20 +1

12.5 Differentiation of vectors 423
Withanequationofthiscomplexity,itisbettertouseacomputeroraprogrammable
calculator. Doing so gives
V 5
= 0.6895,...,V 10
= 0.5770,...,V 14
= 0.5650
which is accurate tofourdecimal places.
Itisusefultocheckthesolutionbyindependentlycalculatingthecurrentthrough
the diode using the two different expressions. So,
I = 10 −14 (e 40×0.5650 −1) = 6.53 ×10 −5 A
I = − 0.5650
2.2 ×10 + 2
4 2.2 ×10 = 6.53 4 ×10−5 A
and therefore the solution iscorrect.
EXERCISES12.4
1 Usethe Newton--Raphson technique to findthe value
ofarootofthe following equations correctto two
decimalplaces. An approximateroot,x 1 ,isgiven in
eachcase.
(a)2cosx=x 2 x 1 =0.8
(b)3x 3 −4x 2 +2x−9=0 x 1 =2
(c)e x/2 −5x=0 x 1 =6
(d)lnx= 1 x
x 1 = 1.6
(e)sinx + 2x
π =1 x 1 =0.6
2 Explain circumstances in whichthe Newton--Raphson
technique mayfail to converge to arootof f (x) = 0.
Solutions
1 (a) 1.02 (b) 1.85 (c) 7.15 (d) 1.76 (e) 0.64
12.5 DIFFERENTIATIONOFVECTORS
Consider Figure 12.16. If r represents the position vector of an object and that object
moves along a curve C, then the position vector will be dependent upon the time,t. We
writer = r(t)toshowthedependenceupontime.Supposethattheobjectisatthepoint
P with position vector r at timet and at the point Q with position vector r(t + δt) at
the later timet + δt as shown in Figure 12.17. Then PQ ⃗ represents the displacement
vectoroftheobjectduringtheintervaloftime δt.Thelengthofthedisplacementvector
represents the distance travelled while its direction gives the direction of motion. The
average velocity during the time fromt tot + δt is the displacement vector divided by
the time interval δt, thatis
PQ
average velocity = ⃗ r(t +δt)−r(t)
=
δt δt
The instantaneous velocity,v, isgiven by
r(t +δt)−r(t)
v=lim
= dr
δt→0 δt dt

424 Chapter 12 Applications of differentiation
y
r
P
C
r(t)
P
PQ
Q
Figure12.16
Position vector ofapoint Pona
curve C.
x
r(t + dt)
Figure12.17
VectorPQrepresents ⃗ the displacement of
the object during the time interval δt.
Now,sincethexandycoordinatesoftheobjectdependuponthetime,wecanwritethe
position vectorras
r(t) =x(t)i +y(t)j
Therefore,
so that
r(t +δt) =x(t +δt)i+y(t +δt)j
x(t +δt)i+y(t +δt)j−x(t)i−y(t)j
v(t) = lim
δt→0 δt
{ }
x(t +δt)−x(t) y(t +δt)−y(t)
= lim i + j
δt→0 δt
δt
= dx
dt i + dy
dt j
often abbreviated to v =ṙ =ẋi+ẏj. Recall the dot notation for derivatives w.r.t.
time which is commonly used when differentiating vectors. So the velocity vector is
the derivative of the position vector with respect to time. This result generalizes in an
obvious way tothree dimensions. If
then
r(t) =x(t)i +y(t)j +z(t)k
ṙ(t) =ẋ(t)i+ẏ(t)j+ż(t)k
The magnitude of the velocity vector gives the speed of the object. We can define the
acceleration inasimilarway:
a = dv
dt = d2 r
= ¨r =ẍi+ÿj+¨zk
dt2 Inmoregeneralsituations,wewillnotbedealingwithpositionvectorsbutotherphysical
quantities such as time-dependent electricor magnetic fields.
Example12.9 Ifa = 3t 2 i +cos2tj, find
(a) da
(b)
da
dt ∣dt
∣
(c) d2 a
dt 2

Solution (a) Ifa = 3t 2 i +cos2tj, then differentiation with respect tot yields
(b)
(c)
da
dt =6ti−2sin2tj
da
∣dt
∣ = √ (6t) 2 + (−2sin2t) 2 =
d 2 a
dt 2 =6i−4cos2tj
√
36t 2 +4sin 2 2t
12.5 Differentiation of vectors 425
It is possible to differentiate more complicated expressions involving vectors provided
certain rules are adhered to. If a and b are vectors andcis a scalar, all functions
oftimet, then
d
(ca) =cda
dt dt + dc
dt a
d
dt (a+b)=da dt + db
dt
d
(a·b)=a·db
dt dt + da
dt ·b
d
dt (a×b)=a×db dt + da
dt ×b
Example12.10 Ifa = 3ti −t 2 jandb=2t 2 i +3j,verify
d
(a) (a·b)=a·db
dt dt + da
dt ·b (b) d
dt (a×b)=a×db dt + da
dt ×b
Solution (a) a·b = (3ti −t 2 j)· (2t 2 i +3j)=6t 3 −3t 2
Also
So,
d
dt (a·b)=18t2 −6t
da
dt =3i−2tj
a· db
dt
+b· da
dt
db
dt
=4ti
=(3ti−t 2 j)·(4ti) + (2t 2 i+3j)·(3i−2tj)
= 12t 2 +6t 2 −6t=18t 2 −6t
We have verified d (a·b)=a·db
dt dt + da
dt ·b.
i j k
(b) a×b=
3t −t 2 0
∣2t 2 3 0∣
=(9t+2t 4 )k
d
dt (a×b)=(9+8t3 )k

426 Chapter 12 Applications of differentiation
Also,
a × db
dt
i j k
=
3t −t 2 0
∣4t 0 0∣
= 4t 3 k
∣ ∣∣∣∣∣ i j k
da
dt ×b = 3 −2t 0
2t 2 3 0∣
=(9+4t 3 )k
and so
a × db
dt + da
dt ×b=4t3 k +(9+4t 3 )k=(9+8t 3 )k = d dt (a×b)
as required.
EXERCISES12.5
1 Ifr =3ti+2t 2 j+t 3 k,find
dr d 2 r
(a) (b)
dt dt 2
2 GivenB =te −t i+costjfind
dB d 2 B
(a) (b)
dt dt 2
3 Ifr =4t 2 i+2tj−7kevaluaterand dr
dt
4 Ifa=t 3 i−7tk,andb=(2+t)i+t 2 j−2k,
(a) finda ·b
(b) find da
dt
whent = 1.
(c) find db
dt
(d) showthat d dt (a·b)=a·db
dt + da
dt ·b.
5 Givenr =sinti+costj,find
(a)r (b)r (c) |r|
Show thatthe position vectorandvelocityvector are
perpendicular.
6 Show r = 3e −t i + (2 +t)jsatisfies
¨r +ṙ =j
7 Givena=t 2 i−(4−t)j,b=i+tjshow
( ) ( )
d
(a)
dt (a×b)= a × db da
+
dt dt ×b
(b)
d
(a·b)=a·db da
+b ·
dt dt dt
Solutions
1 (a) 3i+4tj+3t 2 k
(b) 4j+6tk
2 (a) (−te −t +e −t )i −sintj
(b) e −t (t −2)i −costj
3 4i+2j−7k,8i+2j
4 (a) t(t 3 +2t 2 +14) (b) 3t 2 i−7k
(c) i+2tj
5 (a) costi−sintj (b) −sinti−costj (c) 1

Review exercises 12 427
REVIEWEXERCISES12
1 Determine the position ofall maximum points,
minimumpointsand pointsofinflexion of
(a)y=2t 3 −21t 2 +60t+9
(b)y =t(t 2 −1)
2 InSection 9.8 weshowed that the impedance ofan
LCRcircuitcan be written as
( )
Z=R+j ωL − 1
ωC
(a) Find |Z|.
(b) Foragiven circuit,R,L andC are constants,and
ω can be varied. Find d|Z|
dω .
(c) Forwhat value of ω will |Z| have amaximum or
minimum value? Does thisvalue give a
maximum orminimum value of |Z|?
3 Usetwo iterationsofthe Newton--Raphson technique
to findan improved estimateofthe rootof
t 3 =e t
givent = 1.8 is an approximate root.
4 Given
a=(t 2 +1)i−j+tk
b=2tj−k
find
da
(a)
dt
(c)
d
dt (a·b)
(b)
db
dt
(d) d
dt (a×b)
5 Use two iterationsofthe Newton--Raphson method
to find animproved estimateofthe rootof
sint=1− t ,0 t π,givent=0.7isan
2
approximate root.
6 Determine the position ofallmaximumpoints,
minimum pointsandpointsofinflexion of
(a) y = e −x2
(b) y =t 3 e −t
(c) y=x 3 −3x 2 +3x−1
(d) y=e x +e −x
(e) y=|t|−t 2
Solutions
( 7
1 (a) (2,61) maximum, (5,34) minimum,
2 , 95 )
2
point ofinflexion
( ) ( )
1
(b) √ ,− 2
3 3 √ minimum, −√ 1 2
,
3 3 3 √ 3
maximum, (0,0) point ofinflexion
√
2 (a) R 2 +ω 2 L 2 − 2L
C + 1
ω 2 C 2
(b)
ωL 2 −1/ω 3 C 2
√
R 2 + ω 2 L 2 −2L/C +1/(ω 2 C 2 )
(c) ω = 1 √
LC
producesaminimumvalue ofZ
3 1.859,1.857
4 (a) 2ti+k
(b) 2j
(c) −3
(d) −4ti+2tj+ (6t 2 +2)k
5 0.705, 0.705
6 (a) (0,1)is amaximum
√
2
Points ofinflexion whenx = ±
2
(b) (0,0)point ofinflexion, (3,1.34) maximum.
Further pointsofinflexion whent = 4.73,1.27
(c) (1,0)point ofinflexion
(d) (0,2)minimum
(e) (0.5,0.25) maximum, (−0.5,0.25) maximum,
minimum at (0,0)(y ′ doesnot existhere)

13 Integration
Contents 13.1 Introduction 428
13.2 Elementaryintegration 429
13.3 Definiteandindefiniteintegrals 442
Reviewexercises13 453
13.1 INTRODUCTION
When a function, f (x), is known we can differentiate it to obtain the derivative, df
dx .
Thereverseprocessistoobtainthefunction f (x)fromknowledgeofitsderivative.This
process iscalledintegration. Thus,integration isthe reverse of differentiation.
A problem related to integration is to find the area between a curve and thexaxis.
At first sight it may not be clear that the calculation of area is connected to integration.
This chapter aims to explain the connection. An area can have various interpretations.
For example, the area under a graph of power used by a motor plotted against time
representsthetotalenergyusedbythemotorinaparticulartimeperiod.Theareaunder
a graph of current flow into a capacitor against time represents the total charge stored
by the capacitor.
Circuits to carry out integration are used extensively in electronics. For example, a
circuittodisplaythetotaldistancetravelledbyacarmayhaveaspeedsignalasinputand
may integrate this signal to give the distance travelled as output. Integrator circuits are
widely used in analogue computers. These computers can be used to model a physical
system and observe its response to a range of inputs. The advantage of this approach is
thatthesystemparameterscanbevariedinordertoseewhateffecttheyhaveonsystem
performance.Thisavoidstheneedtobuildtheactualsystemandallowsdesignideasto
be explored relatively quickly and cheaply by an engineer.

13.2 Elementary integration 429
13.2 ELEMENTARYINTEGRATION
Consider the following problem: given dy = 2x, findy(x). Differentiation of the functiony(x)
= x 2 +c, wherecis a constant, yields dy = 2x for anyc. Thereforey(x) =
dx
dx
x 2 +cisasolutiontotheproblem.Asccanbeanyconstant,thereareaninfinitenumber
of different solutions. The constantcis known as aconstant of integration. Inthis example,thefunctionyhasbeenfoundfromaknowledgeofitsderivative.
Wesay2xhas
beenintegrated,yieldingx 2 +c.Toindicatetheprocessofintegrationthesymbols ∫ and
dx are used. The ∫ sign denotes that integration is to be performed and the dx indicates
thatxisthe independent variable. Returning tothe previous problem, we write
dy
dx = 2x
∫
y = 2xdx=x 2 +c
↑ ↑ ↑
symbols for
integration
constant of integration
Ingeneral, if
dy
dx =f(x)
then
∫
y = f(x)dx
Consider a simple example.
Example13.1 Given dy
dx =cosx−x,findy.
Solution We need to find a function which, when differentiated, yields cosx −x. Differentiating
sinx yields cosx, while differentiating −x 2 /2 yields −x. Hence,
∫
y =
(cosx−x)dx=sinx− x2
2 +c
wherecis the constant of integration. Usually brackets are not used and the integral is
written simplyas ∫ cosx−xdx.
Thefunctiontobeintegratedisknownastheintegrand.InExample13.1theintegrand
iscosx−x.

430 Chapter 13 Integration
Example13.2 Find d dx
( x
n+1
n +1 +c )
and hence deduce that ∫ x n dx= xn+1
n +1 +c.
Solution From Table 10.1 wefind
( )
d x
n+1
dx n +1 +c = d ( ) x
n+1
+ d dx n +1 dx (c) usingthe linearity
of differentiation
= 1 d
n +1dx (xn+1 ) + d dx (c)
= 1
n +1 {(n +1)xn } +0
=x n
again usingthe
linearityof differentiation
usingTable10.1
Consequently, reversing the process wefind
∫
x n dx= xn+1
n +1 +c
as required. Note that this result is invalid if n = −1 and so this result could not be
applied tothe integral ∫ (1/x)dx.
Table 13.1 lists several common functions and their integrals. Although the variable
x is used throughout Table 13.1, we can use this table to integrate functions of other
variables, forexamplet andz.
Example13.3 UseTable 13.1 tointegratethe following functions:
(a) x 4
(b) coskx, wherekisaconstant
(c) sin(3x +2)
(d) 5.9
(e) tan(6t −4)
(f) e −3z
1
(g)
x 2
(h) cos100nπt, wherenisaconstant
Solution (a) From Table 13.1, we find ∫ x n dx= xn+1
n +1 +c,n ≠ −1. To find ∫ x 4 dxletn=4;
weobtain
∫
x 4 dx= x5
5 +c
(b) FromTable 13.1, wefind ∫ cos(ax)dx = sin(ax) +c. Inthiscasea =kand so
a
∫
coskxdx = sinkx +c
k

13.2 Elementary integration 431
Table13.1
Theintegrals ofsome common functions.
f(x)
∫
f(x)dx
f(x)
∫
f(x)dx
k, constant kx +c
x n x n+1
+c n≠−1
n +1
x −1 = 1 x
ln|x|+c
e x e x +c
e −x −e −x +c
e ax
sinx
sinax
sin(ax +b)
cosx
cosax
e ax
a +c
−cosx+c
−cosax
+c
a
−cos(ax +b)
+c
a
sinx+c
sinax
a
+c
cos(ax +b)
tanx
tanax
tan(ax +b)
cosec(ax +b)
sec(ax +b)
sin(ax +b)
+c
a
ln|secx|+c
ln|secax|
+c
a
ln|sec(ax +b)|
+c
a
1
{ln|cosec(ax +b)
a
−cot(ax +b)|} +c
1
{ln|sec(ax +b)
a
+tan(ax +b)|} +c
cot(ax +b)
1
{ln|sin(ax +b)|} +c
a
1
√
a2 −x 2 sin −1 x a +c
1
a 2 +x 2 1
a tan−1 x a +c
Notethata,b,nandcareconstants.Whenintegratingtrigonometricfunctions,
anglesmustbeinradians.
(c) FromTable13.1,wefind ∫ −cos(ax +b)
sin(ax+b)dx =
a
andb= 2,and so
∫
−cos(3x +2)
sin(3x +2)dx = +c
3
+c.Inthiscasea = 3
(d) From Table 13.1, we find thatifkisaconstant then ∫ k dx =kx +c.Hence,
∫
5.9 dx = 5.9x +c
(e) In this example, the independent variable ist but nevertheless from Table 13.1 we
can deduce
∫
ln|sec(at +b)|
tan(at +b)dt = +c
a
Hence witha = 6 andb=−4,we obtain
∫
ln|sec(6t −4)|
tan(6t −4)dt = +c
6

432 Chapter 13 Integration
(f) TheindependentvariableiszbutfromTable13.1wecandeduce ∫ e az dz = eaz
a +c.
Hence, takinga = −3 we obtain
∫
e −3z dz = e−3z
−3 +c=−e−3z 3 +c
(g) Since 1 x 2 =x−2 ,wefind
∫ 1
x 2 dx = ∫
x −2 dx = x−1
−1 +c=−1 x +c
(h) Whenintegratingcos100nπt withrespecttot,notethat100nπisaconstant.Hence,
usingpart (b)wefind
∫
cos100nπt dt = sin100nπt
100nπ +c
13.2.1 Integrationasalinearoperator
Integration, like differentiation, is a linear operator. If f and g are two functions of x,
then
∫ ∫ ∫
f+gdx= fdx+ gdx
This states that the integral of a sum of functions is the sum of the integrals of the individual
functions.IfAisaconstant and f a functionofx, then
∫
∫
Afdx=A
f dx
Thus,constantfactors can betakenthroughthe integral sign.
IfAandBareconstants,and f andgarefunctions ofx, then
∫
∫
Af+Bgdx=A
∫
fdx+B
gdx
These three properties are all consequences of the fact that integration is a linear operator.
Note that the first two are special cases of the third. The properties are used in
Example 13.4.
Example13.4 Use Table 13.1 and the properties of a linear operator to integrate the following
expressions:
(a) x 2 +9 (d) (t +2) 2 (g) 3sin4t
(b) 3t 4 − √ t
(c)
1
x
(e)
(f) 4e 2z
1
+z (h) 4cos(9x +2)
z
(i) 3e 2z

Solution (a)
sinx+cosx
(j)
2
(k) 2t −e t
∫
∫
x 2 +9dx=
∫
x 2 dx+
( ) z −1
(l) tan
2
(m) e t +e −t
9 dx usinglinearity
13.2 Elementary integration 433
(n) 3sec(4x −1)
(o) 2cot9x
(p) 7cosec(π/3)
(b)
+9x +c usingTable 13.1
3
Note thatonly a single constant of integration isrequired.
∫
3t 4 − √ ∫ ∫
tdt=3 t 4 dt − t 1/2 dt using linearity
= x3
( ) t
5
=3 − t3/2
+c using Table 13.1
5 3/2
= 3t5
5 − 2t3/2
+c
3
(c)
∫ 1
dx = ln |x| +c using Table 13.1.
x
Sometimesitisconvenienttousethelawsoflogarithmstorewriteanswersinvolving
logarithms. For example, we can write ln|x| +c as ln|x| +ln|A| wherec = ln |A|.
This enables us towritethe integralas
∫ 1
dx = ln|Ax|
x
(d)
∫
∫
(t+2) 2 dt=
t 2 +4t+4dt= t3 3 +2t2 +4t+c
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
∫ 1
z +zdz=ln|z|+z2 2 +c
∫
4e 2z dz = 4e2z
2 +c=2e2z +c
∫
3sin(4t)dt = − 3cos4t +c
4
∫
4sin(9x +2)
4cos(9x +2)dx = +c
9
∫
3e 2z dz = 3e2z
2 +c
∫ sinx+cosx
dx = −cosx+sinx +c
2 2
∫
2t−e t dt=t 2 −e t +c
∫ ( )
( )∣ z −1
tan dz=2ln
z −1 ∣∣∣
2 ∣ sec +c
2

434 Chapter 13 Integration
∫
(m)
∫
(n)
∫
(o)
∫
(p)
e t +e −t dt=e t −e −t +c
3sec(4x −1)dx = 3 ln|sec(4x −1) +tan(4x −1)| +c
4
2cot9xdx= 2 9 ln|sin9x|+c
7cosec(π/3) dx = {7cosec(π/3)}x +c
as cosec(π/3) isaconstant
Engineeringapplication13.1
Distancetravelledbyaparticle
The speed, v, of a particle is the rate of change of distance,s, with respect to timet,
thatis v = ds .Thespeedattimet isgivenby3+2t.ThisisillustratedinFigure13.1.
dt
Note that the speed of the particle is increasing linearly with time. Find the distance
intermsoft.
Solution
We aregiven that
v = ds
dt =3+2t
and arerequired tofinds. Therefore,
∫
s = 3+2tdt=3t+t 2 +c
Notethatthespeedoftheparticleismodelledbyalinearexpressionint.Thismeans
thatthespeedincreasesbythesameamountineachsubsequentsecond.Ontheother
hand,thedistance,s,ismodelledbyaquadraticint.Therefore,thedistancetravelled
ineachsubsequentsecondincreases.Figure13.1illustratesthespeed-timegraphand
Figure 13.2 illustratesthe distance-time graph.
y
s
3
c
O
Figure13.1
Graphofspeed ofthe particle, v,against
time,t.
t
O
Figure13.2
Graphofdistancetravelled by the
particle,s,against time,t.
t

13.2 Elementary integration 435
Engineeringapplication13.2
Voltageacrossacapacitor
Recall from Engineering application 10.2 that the current, i, through a capacitor
depends upon time,t, and isgiven by
i =C dv
dt
where v is the voltage across the capacitor andC is the capacitance of the capacitor.
Derive an expression for v.
Solution
If
i =C dv then
dt
Therefore,
∫ i
v =
C dt = 1 ∫
C
i dt
dv
dt
= i C
usinglinearity
Notethatwhereasthecapacitance,C,isconstant,thecurrent,i,isnotandsoitcannot
betakenthroughtheintegralsign.Inordertoperformtheintegrationweneedtoknow
i as a function oft.
13.2.2 Electronicintegrators
Often there is a requirement in engineering to integrate electronic signals. Various circuits
are available to carry out this task. One of the simplest circuits is shown in
Figure 13.3. The input voltage is v i
, the output voltage is v o
, the voltage drop across
theresistoris v R
andthecurrentflowinginthecircuitisi.ApplyingKirchhoff’svoltage
lawyields
v i
=v R
+v o
(13.1)
For the resistor with resistance,R,
v R
=iR (13.2)
For the capacitorwith capacitance,C,
i =C dv o
dt
Combining Equations (13.1) to(13.3) yields
(13.3)
v i
=RC dv o
+ v
dt o
(13.4)
In general, v i
will be a time-varying signal consisting of a range of frequencies. For
the case where v i
is sinusoidal we can specify a property of the capacitor known as the
capacitive reactance,X c
, given by
X c
= 1
2πfC

436 Chapter 13 Integration
C
R
i
R 1 i 1 i
y f
1
R 2 i y2
2
–
R
i A
y 3 i
X 3 3
+
y i
y R
C
y o
y o
Figure13.3
Simple integrator.
Figure13.4
Summingintegrator usingan operational
amplifier(op amp).
where f =frequencyofthesignal(Hz).ItcanbeseenthatX c
decreaseswithincreasing
frequency, f. For frequencies whereX c
is small compared withR, most of the voltage
droptakesplaceacrosstheresistor.Inotherwords, v o
issmallcomparedwith v R
.ExaminingEquation(13.1)forthecasewhenX
c
isverymuchlessthanR(writtenasX c
≪R),
and v o
≪ v R
, itcan be seen thatEquation (13.4) simplifies to
v i
=RC dv o
dt
(13.5)
Thisequationisonlyvalidfortherangeoffrequencies forwhichX c
≪R.Rearranging
Equation (13.5) yields
dv o
dt
= v i
RC
v o
= 1
RC
∫
v i
dt
The output voltage from the circuit is an integrated version of the input voltage with a
1
scaling factor
RC .
Anelectronicintegratorwhichperformsbettercanbemadefromanoperationalamplifier.
The circuit for an operational amplifier integrator is given in Figure 13.4. The
main advantage of this circuit is the low output impedance and high input impedance,
makingitusefulforelectroniccontrolapplicationsandanaloguesignalprocessing.The
function of the operational amplifier is to amplify the potential difference between the
invertingandnon-invertinginputs.Thesearelabelled −and +respectivelyinthecircuit
diagram.Usuallythegainoftheamplifierisextremelyhigh,soevenasmallvoltagedifferencebetweenthetwoinputswillgiveaverylargeoutputvoltage,whichislimitedby
the voltage supply attached to the device. Notice in this circuit that there is a capacitor
connectedfromtheoutputbacktotheinput.Thiscapacitorprovidesnegativefeedback.
This means that a proportion of the output voltage is fed back to the input and this in
turnservestoreducetheoutput.Asaconsequenceofthistheoverallgainofthecircuit
is limited and the amplifier reaches an equilibrium state where the voltage at point X is
the same as at the non-inverting input, which is connected to earth. For this reason the
point Xinthe circuitis sometimes referredtoas avirtual earth.
Assuming point Xisatzero volts, and using Ohm’s law, gives
i 1
= v 1
R 1
i 2
= v 2
R 2
i 3
= v 3
R 3
(13.6)

13.2 Elementary integration 437
4.00
3.00
2.00
1.00
y 0
0
–1.00
–2.00
–3.00
–4.00
290 291 292 293 294 295 296 297
t (ms)
y 1
298 299 300
Figure13.5
Output froman operational
amplifier integrator circuit with
v 1 being asquare wave input
andv 2 =v 3 =0.
Assumingi A
isnegligible, then
i f
=i 1
+i 2
+i 3
(13.7)
For the capacitor,
i f
=−C dv o
dt
(13.8)
Thenegativesignisaresultofthedirectionchosenfori f
.CombiningEquations(13.6)--
(13.8) yields
v 1
+ v 2
+ v 3
= −C dv o
R 1
R 2
R 3
dt
∫
v
v o
=− 1
R 1
C + v 2
R 2
C + v 3
R 3
C dt
The circuit therefore acts as an integrator. The minus sign in the integration is a result
of the circuit design which isknown as aninverting circuit. Atypical output from this
circuitisgiven inFigure 13.5.
13.2.3 Integrationoftrigonometricfunctions
The trigonometric identities given in Table 3.1 together with Table 13.1 allow us to
integrate a number of trigonometric functions.
Example13.5 Evaluate
(a) ∫ cos 2 t dt
(b) ∫ sin 2 t dt

438 Chapter 13 Integration
Solution Powersoftrigonometricfunctions,forexamplesin 2 t,donotappearinthetableofstandardintegrals.Whatwemustattempttodoisrewritetheintegrand
toobtainastandard
form.
(a) FromTable 3.1
cos 2 t = 1+cos2t
2
(b)
and so
∫ ∫ 1+cos2t
cos 2 t dt = dt
2
∫ ∫ 1 cos2t
=
2 dt + dt
2
= t 2 + sin2t +c
4
∫ ∫
sin 2 t dt = 1 −cos 2 t dt using the trigonometric identities
∫
=
∫
1dt−
cos 2 t dt
( t
= t −
2 + sin2t )
+c
4
= t 2 − sin2t
4
+k
using linearity
using part(a)
Example13.6 Find
(a) ∫ sin2tcost dt
(b) ∫ sinmtsinnt dt, wheremandnareconstants withm ≠n
Solution (a) Usingthe identitiesinTable3.1 wefind
2sinAcosB =sin(A +B) +sin(A −B)
hence sin2tcost can bewritten 1 (sin3t +sint). Therefore,
2
∫
∫ 1
sin2tcost dt = (sin3t +sint)dt
2
= 1 ( )
−cos3t
−cost +c
2 3
= − 1 6 cos3t − 1 2 cost+c
(b) Usingthe identity2sinAsinB = cos(A −B) −cos(A +B), we find
sinmtsinnt = 1 {cos(m −n)t −cos(m +n)t}
2

Therefore,
∫
∫ 1
sinmtsinnt dt = {cos(m −n)t −cos(m +n)t}dt
2
= 1 { }
sin(m −n)t sin(m +n)t
− +c
2 m −n m +n
13.2 Elementary integration 439
EXERCISES13.2
1 Integratethe following expressions usingTable 13.1:
1 1
(e) √ (f)
(a) x 10 (b) 9 (c) x 1.5
0.25 −x 2 0.01 + v 2
(d) √ 1 1
1
(g)
t (e) (f) −3.2
10 6 +r 2 (h)
10+t 2
z
1 1
1 1
(g)
x 3 (h) √ (i) (x 2 ) 3
(i) √ (j)
t 2−x 2 1
(j) x −2 (k) t 1/3
9 +x2
5 Integratethe followingexpressions:
2 Integratethe following expressions usingTable 13.1:
(a) e 5x (b) e 6x (c) e −3t (a)3+x+ 1 x
1
(d)
e x (e) e 0.5t (f) e z/3 (b)e 2x −e −2x
1
(g)
e 4t (h) e −2.5x
(c)2sin3x +cos3x
( )
t
(d)sec(2t + π) +cot
3 Integratethe following expressions usingTable 13.1:
2 − π
( )
(a) sin4x (b) sin9t
( )
( )
t
(e)tan +cosec(3t − π)
x 2t
2
(c) sin (d) sin
2 5
(e) cos7x (f) cos(−3x)
(f) sinx + x 3 + 1 e
( )
x
5t
1
(g) cos (h) tan9x
(g)
3
cos(3x)
( ) 2
(i) cosec2x (j) sec5t
(h) t + 1 (k) cot8y (l) cos(5t +1)
t
(m) tan(3x +4) (n) sin(3t − π)
1
( ) (i)
x
3e 2x
(o) cosec(5z +2) (p) sec
2 +1 (j) tan(4t −3) +2sin(−t −1)
( )
2t
(k)1+2cot3x
(q) sin
3 −1 (r) cot(5 −x)
( ) ( )
t t
(l) sin −3cos
(s) cosec(π −2x)
2 2
4 UseTable 13.1 to integrate the following expressions: (m)(t −2) 2
1 1
(n)3e −t −e −t/2
(a)
1+x 2 (b) √
1−x 2
(o)7 −7x 6 +e −x
1 1
(c) √ (d)
(p) (k +t) 2 k constant
4−x 2 9+z 2 (q)ksint −coskt k constant

440 Chapter 13 Integration
1
(r)
25+t 2
1
(s) √
25−t 2
(t)
6
1+x 2 + 1+x2
6
6 Theacceleration,a, ofaparticleis the rate ofchange
ofspeed, v,withrespecttotimet,thatisa = dv
dt .The
speed ofthe particle is the rate ofchange of
distance,s,that is v = ds .Ifthe acceleration isgiven
dt
by1+ t ,find expressionsforspeed anddistance.
2
7 Thespeed, v,ofaparticlevaries with time according
to
v(t) =2−e −t
(a) Obtainan expression forthe distancetravelled by
the particle.
(b) Calculate the distancetravelled bythe particle
betweent = 0 andt = 3.
8 Bywritingsinhax andcoshax in terms ofthe
exponential functionfind
∫
(a) sinhax dx
∫
(b) coshax dx
(c) Useyour results from (a)and(b)to find
∫
3sinh2x +cosh4x dx.
9 Acapacitor ofcapacitance10 −2 Fhas acurrenti(t)
through itwhere
i(t) =10−e −t
Find an expression forthe voltageacross the
capacitor.
10 Integratethe following:
1 1
(a)
x 2 (b)
+4 2x 2 +4
3
(c)
2x 2 +1
2
(e) √
4−x 2
1
(g) √
1 − (x 2 /2)
(d)
(f)
1
√
9−x 2
−7
√
2−3x 2
11 Bywriting 3 +x in the form 3 x x +1,
∫ 3 +x
find dx.
x
12 (a) Express
x 2 +2x+1
x(x 2 +1)
asitspartial fractions.
∫ x 2 +2x+1
(b) Hence find
x(x 2 dx.
+1)
13 Thevelocity, v,ofaparticleisgiven by
v=2+e −t/2
(a) Given distance,s,and v are relatedby ds
dt = v
findan expression fordistance.
(b) Acceleration isthe rate ofchange ofvelocity
with respect tot.Determine the acceleration.
14 (a) Usethe product ruleofdifferentiationto verify
d
dx (xe2x ) = e 2x +2xe 2x
(b) Hence show
∫
xe 2x dx = xe2x
2 − e2x
4 +c
15 Integrate
2−t 2
(a)
t 3
(c)
(e)
cos4x
sin4x
4+e 2t
(b)
e 3t
1
(d)
2sin3x
2+x 2
1+x 2 (f) sin 2 t +cos 2 t
Solutions
x 11
1 (a) +c (b) 9x+c
11
2 2
(c)
5 x2.5 +c (d)
3 t1.5 +c
(e) ln |z| +c (f) −3.2x +c
(g) − 1
2x 2 +c (h) 2√ t +c
x 7
(i)
7 +c (j) −1 x +c
3
(k)
4 t4/3 +c

13.2 Elementary integration 441
2 (a)
e 5x
5 +c (b) e 6x
6 +c
(c) − e−3t
3 +c (d) −e−x +c
(e) 2e 0.5t +c (f) 3e z/3 +c
(g) − e−4t
4
3 (a) − cos4x +c
4
(b) − cos9t +c
9 ( )
x
(c) −2cos +c
2
(d) − 5 2 cos (
(e)
sin7x
7
+c (h) −2e−2.5x
5
+c
2t
5
)
+c
(f) − 1 sin(−3x) +c
3 ( )
3
(g)
5 sin 5t
+c
3
ln|sec9x|
(h) +c
9
1
(i) (ln|cosec2x −cot2x|) +c
2
1
(j) (ln|sec5t +tan5t|)+c
5
1
(k)
8 ln|sin8y|+c
1
(l) sin(5t +1) +c
5
(m) 1 ln|sec(3x +4)| +c
3
(n) − 1 cos(3t −π)+c
3
1
5
+c
(o) ln|cosec(5z +2) −cot(5z +2)| +c
( ) ( )∣ (p) 2ln
∣ sec x
2 +1 x ∣∣∣∣
+tan
2 +1 +c
( )
(q) − 3 2 cos 2t
3 −1 +c
(r) −ln|sin(5 −x)| +c
(s) − 1 ln|cosec(π −2x) −cot(π −2x)| +c
2
4 (a) tan −1 x +c
(b) sin −1 x +c
( )
(c) sin −1 x
+c
2
(d)
( )
1 z
3 tan−1 +c
3
(e) sin −1 (2x) +c
(f) 10tan −1 (10v) +c
( )
(g) 10 −3 tan −1 x
10 3 +c
(h)
( )
1
√ tan −1 t
√ +c
10 10
( )
(i) sin −1 x
√ +c
2
(j) 3tan −1 (3x) +c
5 (a) 3x+ x2
2 +ln|x|+c
(b) e2x
2 + e−2x +c
2
(c) − 2 sin3x
cos(3x) + +c
3 3
(d) 0.5ln|sec(2t + π) +tan(2t + π)|
( )∣ +2ln
∣ sin t ∣∣∣∣
2 − π +c
( )∣ (e) 2ln
∣ sec t ∣∣∣∣
+ 1 ln|cosec(3t − π)
2 3
−cot(3t −π)|+c
(f) −cosx+ x2
6 −e−x +c
(g) 1 3 ln|sec3x+tan3x|+c
(h) t3 3 +2t−1 t +c
(i) − e−2x +c
6
1
(j) ln|sec(4t −3)| +2cos(−t −1) +c
4
(k)x+ 2 3 ln|sin3x|+c
( ) ( )
t t
(l) −2cos −6sin +c
2 2
(m) t3 3 −2t2 +4t+c
(n) −3e −t +2e −t/2 +c
(o)7x−x 7 −e −x +c
(p) k 2 t +kt 2 + t3 3 +c
(q) −kcost − sinkt
k
+c

442 Chapter 13 Integration
( )
1 t
(r)
5 tan−1 +c
5
( )
(s) sin −1 t
+c
5
(t) 6tan −1 x + x 6 + x3
18 +c
6 Speed:t + t2 4 +c,distance: t2 2 + t3
12 +ct+d
7 (a) 2t +e −t +c (b) 5.0498
8 (a) coshax +c
a
(b) sinhax +c
a
(c) 3 2 cosh2x + 1 sinh4x +c
4
9 100(10t +e −t ) +c
( )
1 x
10 (a)
2 tan−1 +c
2
( )
1
(b)
2 √ x
2 tan−1 √ +c
2
(c)
3
√
2
tan −1 ( √ 2x)+c
( )
(d) sin −1 x
+c
3
( )
(e) 2sin −1 x
+c
2
(√ )
−7 3
(f) √ sin −1 √ x +c
3 2
(g) √ ( )
2sin −1 x
√ +c
2
11 3ln|x|+x+c
12 (a)
1
x + 2
x 2 +1
(b) ln|x| +2tan −1 x +c
13 (a) 2t −2e −t/2 +c (b) − 1 2 e−t/2
15 (a) −t −2 −ln|t|+c
(b) − 4 3 e−3t −e −t +c
1
(c)
4 ln|sin4x|+c
1
(d) ln|cosec3x −cot3x| +c
6
(e) x+tan −1 x+c
(f) t+c
13.3 DEFINITEANDINDEFINITEINTEGRALS
Alltheintegrationsolutionssofarencounteredhavecontainedaconstantofintegration.
Such integrals are known as indefinite integrals. Integration can be used to determine
the area under curves and this gives rise todefinite integrals.
To estimate the area under y(x), it is divided into thin rectangles. The sum of the
rectangularareasisanapproximationtotheareaunderthecurve.Severalthinrectangles
will give a better approximationthanafewwide ones.
ConsiderFigure13.6wheretheareaisapproximatedbyalargenumberofrectangles.
Suppose each rectangle has width δx. The area of rectangle 1 is y(x 2
)δx, the area of
rectangle2isy(x 3
)δxandsoon.LetA(x n
)denotethetotalareaunderthecurvefromx 1
tox n
. Then,
A(x n
) ≈ sum ofthe rectangularareas =
n∑
y(x i
)δx
i=2
Lettheareabeincreasedbyextendingthebasefromx n
tox.ThenA(x)isthetotalarea
under the curve fromx 1
tox(see Figure 13.7). Then,
increaseinarea = δA =A(x) −A(x n
) ≈y(x)δx

13.3 Definite and indefinite integrals 443
y
y
n
1 2 3 4 n–1
dx
x 1 x 2
dx
x 3 x 4 x 5
x n
x
x 1
x n
x
x
Figure13.6
The area isapproximated by (n −1) rectangles.
Figure13.7
Thearea isextended byadding an extrarectangle.
So,
δA
δx ≈y(x)
Inthe limitas δx → 0,we get
( ) δA
lim = dA
δx→0 δx dx =y(x)
Since differentiation isthe reverse of integration, we can write
∫
A = y(x)dx
To denote the limits of the area being considered we place values on the integral sign.
Thearea under the curve,y(x), betweenx =aandx =bisdenotedas
∫ x=b
x=a
ydx
ormorecompactlyby
∫ b
a
ydx
Theconstantsaandbareknownasthelimitsoftheintegral:lowerandupper,respectively.Sinceanareahasaspecificvalue,suchanintegraliscalledadefiniteintegral.The
area under the curve up to the vertical linex = b isA(b) (see Figure 13.8). Similarly
A(a) is the area up to the vertical linex = a. So the area betweenx = a andx = b is
A(b) −A(a), asshown inFigure 13.9.
Thearea betweenx =aandx =bis given by
Area =
∫ b
a
ydx =A(b) −A(a)
The integral is evaluated at the upper limit, b, and at the lower limit, a, and the
difference between thesegives the required area.

444 Chapter 13 Integration
y
y
y
A(b)
a
A(a)
Figure13.8
Thearea depends onthe limitsaandb.
x
b
x
a
Figure13.9
Thearea betweenx =aandx =bis
A(b) −A(a).
b
x
The expressionA(b) −A(a) isoften writtenas [A(x)] b a . Similarly[x2 +1]
2 3 isthe value
ofx 2 +1atx=3lessthevalueofx 2 +1atx=2.Thus
[x 2 +1] 3 2 =(32 +1)−(2 2 +1)=5
Ingeneral
[f(x)] b a =f(b)−f(a)
Note that since
∫ b
a
ydx =A(b) −A(a)
then, interchanging upper and lower limits,
∫ a
that is,
b
ydx =A(a) −A(b) = −{A(b) −A(a)}
∫ b
ydx=−
∫ a
a b
ydx
Interchanging the limitschanges the sign of the integral.
The evaluation of definite integrals isillustratedinthe following examples.
Example13.7 Evaluate
(a)
∫ 2
1
x 2 +1dx
(b)
∫ 1
∫ 2
Solution (a) LetI stand for x 2 +1dx.
I =
∫ 2
1
1
2
x 2 +1dx
[ ] x
x 2 3 2
+1dx=
3 +x 1
(c)
∫ π
0
sinx dx
Theintegralisnowevaluatedattheupperandlowerlimits.Thedifferencegivesthe
value required.
( ) ( ) 2
3 1
3
I =
3 +2 −
3 +1 = 8 3 +2− 4 3 = 10 3

13.3 Definite and indefinite integrals 445
(b) Because interchanging thelimitsofintegration changes thesignoftheintegral,we
(c)
find
∫ 1
x 2 +1dx=−
∫ π
0
∫ 2
2
1
x 2 +1dx=− 10 3
sinx dx = [−cosx] π 0 =(−cosπ)−(−cos0)=1−(−1)=2
Figure 13.10 illustratesthisarea.
sin x
0 p
x
Figure13.10
The area isgiven byadefinite integral.
Note that:
(1) Theintegratedfunctionisevaluatedattheupperandlowerlimits,andthedifference
found.
(2) No constantofintegration is needed.
(3) Any anglesaremeasured inradians.
Example13.8 Findthe area underz(t) = e 2t fromt = 1 tot = 3.
∫ 3 ∫ 3
[ e
Solution Area = zdt= e 2t 2t
dt =
1 1 2
[ ] [ ] e
6 e
2
= − = 198
2 2
] 3
1
If the evaluation of an area by integration yields a negative quantity this means that
some or all of the corresponding area is below the horizontal axis. This is illustrated in
Example 13.9.
Example13.9 Findthe area bounded byy =x 3 and thexaxisfromx = −3tox = −2.
Solution Figure 13.11illustratesthe required area.
∫ −2
[ ] x
x 3 4 −2
dx=
−3 4
−3
{ } { }
(−2)
4 (−3)
4
= − = − 65 4 4 4
The area is65/4 square units; the negative sign indicates thatitisbelow thexaxis.

446 Chapter 13 Integration
y
y = x 3
y
y = sin x
–3
–2
0 x
–p
p
x
Figure13.11
Areas below thexaxisare classed asnegative.
Figure13.12
Thepositive and negative areas cancel
each other out.
Example13.10 (a) Sketchy = sinx forx = −πtox = π.
(b) Calculate
∫ π
−π
sinxdx and comment onyourfindings.
(c) Calculatetheareaenclosedbyy = sinxandthexaxisbetweenx = −πandx = π.
Solution (a) Agraph ofy = sinx betweenx = −πandx = π isshown inFigure 13.12.
(b)
∫ π
−π
sinx dx = [−cosx] π −π = −cos(π) +cos(−π) = 0
Examining Figure 13.12 we see that the positive and negative contributions have
cancelledeachotherout;thatis,theareaabovethexaxisisequalinsizetothearea
below thexaxis.
(c) From(b)theareaabovethexaxisisequalinsizetotheareabelowthexaxis.From
Example 13.7(c) the area above the x axis is 2. Hence the total area enclosed by
y=sinxandthexaxisfromx=−πtox=πis4.
Ifanareacontainspartsbothaboveandbelowthehorizontalaxisthencalculatingan
integral will give the net area. If the total area is required, then the relevant limits must
first befound.Asketchofthe function oftenclarifiesthe situation.
Example13.11 Findthe area containedbyy = sinx fromx = 0 tox = 3π/2.
Solution Figure13.13illustratestherequiredarea.Fromthisweseethattherearepartsbothabove
and below thexaxisand the crossover pointoccurswhenx = π.
∫ π
0
∫ 3π/2
π
sinx dx = [−cosx] π 0
=−cosπ+cos0=2
sinx dx = [−cosx] 3π/2
π
( ) 3π
=−cos +cosπ=−1
2

13.3 Definite and indefinite integrals 447
y
0
p
3p ––2 y = sin x
x
Figure13.13
Thepositive andnegative areas are calculated
separately.
The total area is 3 square units. Note, however, that the single integral over 0 to 3π/2
evaluates to1;thatis, itgives the net value of 2 and −1.
∫ 3π/2
0
( ) 3π
sinx dx = [−cosx] 3π/2
0
=−cos +cos0=1
2
Theneedtoevaluatetheareaunderacurveisacommonrequirementinengineering.
Oftentherateofchangeofanengineeringvariablewithtimeisknownanditisrequired
to calculate the value of the engineering variable. This corresponds to calculating the
area under a curve.
Engineeringapplication13.3
Energyusedbyanelectricmotor
Consider a small d.c. electric motor being used to drive an electric screwdriver. The
amount of power that is supplied to the motor by the battery depends on the load on
the screwdriver. Therefore the power supplied to the screwdriver is a function that
varies with time.Figure 13.14shows a typicalcurve ofpower versus time.Now,
P = dE
dt
whereP = power (W),E = energy (J). Therefore, to calculate the energy used by
the motor between timest 1
andt 2
, wecan write
E =
∫ t2
t 1
Pdt
This is equivalent to evaluating the area under the curve, P(t), between t 1
and t 2
,
which is shown as the shaded region inFigure 13.14.
Power (watts)
t 1
t 2
Time (s)
Figure13.14
Shaded area representsthe energy usedto
drivethe motor duringthe time interval
t 1 tt 2 .

448 Chapter 13 Integration
Engineeringapplication13.4
Capacitanceofacoaxialcable
A coaxial cable has an inner conductor with a diameter of 1.02 mm and an outer
conductorwithaninternaldiameterof3mm,asshowninFigure13.15.Theinsulator
separatingthetwoconductorshasarelativepermittivityof1.55.Letuscalculatethe
capacitance ofthe cable per metre length.
Outer conductor
Cable sheath
r
a
b
1.02 mm 3 mm
Insulator
Inner conductor
Figure13.15
Cross-sectionofacoaxial
cable.
Before solving this problem it is instructive to derive the formula for the capacitance
of a coaxial cable. Imagine that the inner conductor has a charge of +Q per
metrelength and thatthe outer conductor has a charge of −Qper metrelength. Furtherassumethecableislongandacentralsectionisbeinganalysedinorderthatend
effects can be ignored.
Consideranimaginarycylindricalsurface,radiusrandlengthl,withintheinsulator(ordielectric).Gauss’stheoremstatesthattheelectricfluxoutofanyclosedsurfaceisequaltothechargeenclosedbythesurface.Inthiscase,becauseofsymmetry,
theelectricfluxpointsradiallyoutwardsandsonofluxisdirectedthroughtheends
of the imaginary cylinder; that is, end effects can be neglected. The curved surface
area of the cylinder is2πrl. Therefore, using Gauss’stheorem
D×2πrl=Ql
whereD =electricflux density.
WhenaninsulatorordielectricispresentthenD = ε r
ε 0
E,whereE istheelectric
field strength, ε r
is the relative permittivity, ε 0
is the permittivity of free space and
has a value of 8.85 ×10 −12 Fm −1 . Therefore,
thatis,
ε r
ε 0
E2πrl =Ql
E =
Q
2πε r
ε 0
r
(13.9)
This equation gives a value for the electric field within the dielectric. In order to
calculatethecapacitanceofthecableitisnecessarytocalculatethevoltagedifference
between thetwoconductors. LetV a
representthevoltageoftheinnerconductor and
V b
the voltage of the outer conductor.

13.3 Definite and indefinite integrals 449
The electric field is a measure of the rate of change of the voltage with position.
Inotherwords,ifthevoltageischangingrapidlywithpositionthenthiscorresponds
to a large magnitude of the electric field. This is illustrated in Figure 13.16. The
magnitude of the electric field at point A is larger than at point B. As a positive
electricfield,E,correspondstoadecreaseinvoltage,V,withpositiontherelationship
betweenE andV, ingeneral, is
E = − dV
(13.10)
dr
This expression can be used to calculate the voltage difference arising as a result of
an electric field. In practice, this is a simplified equation and is only valid provided
r is in the same direction as the electric field. If this is not the case, then a modified
vector formof Equation (13.10) isneeded.
V
A
B
r
Figure13.16
Thegradient ofthe curve, dV , isproportional to the
dr
magnitude ofthe electric field.
In the case of the coaxial cable, E is in the same direction as r and so
Equation (13.10) can be used to calculate the voltage difference between the two
conductors. From Equation (13.10)
dV
dr = −E
Therefore the voltage atanarbitrarypoint,r, isgiven by
∫
V = − E dr
Consequently, the voltage difference between pointsr =bandr =aisgiven by
V b
−V a
=−
∫ b
a
E dr
= − Q ∫ b
1
dr using Equation (13.9)
2πε r
ε 0
r
= − Q
2πε r
ε 0
[lnr] b a
= − Q
2πε r
ε 0
ln( b
a
a
)
Thisgivesthevoltageoftheouterconductorrelativetotheinnerone.Thusthevoltage
of the inner conductor relative tothe outer one is
V a
−V b
=
Q ) b
ln(
2πε r
ε 0
a
➔

450 Chapter 13 Integration
More generally, capacitance is defined asC = Q/V, whereV is the voltage difference.
Therefore
C =
Q = 2πε r ε 0
(
V a
−V b b
ln
a)
Note that this is the capacitance per unit length of cable. Using ε r
= 1.55, a =
5.1 ×10 −4 m,b= 1.5 ×10 −3 m, weget
2π ×1.55 ×8.85 ×10−12
C = ( ) = 7.99 ×10 −11 ≈ 80 pF m −1
1.5 ×10
−3
ln
5.1 ×10 −4
Engineeringapplication13.5
Characteristicimpedanceofacoaxialcable
Acommonlyquotedparameterofacoaxialcableisitscharacteristicimpedance,Z 0
.
Thecharacteristicimpedanceistheratioofthevoltagetothecurrentforapropagating
wave travelling on an electrical transmission line in the absence of reflections.
ThevalueofZ 0
iseasytoselectatthedesignstagebycarefullychoosingthedimensionsaandbtogether
with the type of insulating material within the cable. The two
mostcommoncharacteristicimpedancesforflexiblecablesareapproximately50
and 75 . The main reason for selecting 50 is that it represents a good compromise
between the ability to handle high power and the minimization of losses that
occur in thermoplastic dielectrics. The value of 75 is mainly considered optimal
for situations of low power transmission and where losses are the most important
consideration. Often these 75 cables are of the air-dielectric type where the inner
conductor is supported by a spacer rather than a solid plastic dielectric. An example
of an application for a 75 cable is the connection from a rooftop TV antenna to a
TV set.
Itcanbeshownfromfundamentaltransmissionlinetheorythatthecharacteristic
impedanceofaloss-freecableis
√
L
Z 0
=
C
Itcan be shown thatthe expression for the inductance of a coaxial cable isgiven by
L = µ ( )
0 b
2π ln a
As shown inEngineering application 13.4, the expression for the capacitance is
C = 2πε r ε 0
( b
ln
a)

13.3 Definite and indefinite integrals 451
Substituting forLandC inthe equation forZ 0
( )
µ 0 b Z 0
=
2π ln a
/ ( ) = 1 √ ( )
µ0 b
ln
√2πε r
ε 0 b 2π ε r
ε 0
a
ln
a
For the cable defined in Engineering application 13.4, and substituting for the permeability
of free space, µ 0
= 4π ×10 −7 ,
Z 0
= 1 √
( )
µ 0 1.5 ×10
−3
2π 1.55 ×8.85 ×10 ln =52
−12 0.51 ×10 −3
13.3.1 Useofadummyvariable
Consider the following integrals,I 1
andI 2
:
I 1
=
∫ 1
Then,
[ ] t
3 1
I 1
= =
3
0
0
[ x
3
I 2
=
3
t 2 dt I 2
=
] 1
0
=
∫ 1
0
x 2 dx
( 1
3)
−(0)= 1 3
( 1
3)
−(0)= 1 3
So clearlyI 1
= I 2
. The value ofI 1
does not depend upont, and the value ofI 2
does not
depend uponx. Ingeneral,
I =
∫ b
a
f(t)dt =
∫ b
a
f(x)dx
Because the value ofI is the same, regardless of what the integrating variable may be,
wesayxandt are dummy variables. Indeed we could write
I =
∫ b
a
f(z)dz=
∫ b
a
f(r)dr=
Thenz,randyaredummy variables.
∫ b
a
f(y)dy
EXERCISES13.3
1 Evaluate the following integrals:
∫ 3 ∫ 4
(a) x 3 1
dx (b)
1 1 x dx
∫ 1 ∫ 1
(c) 2 dx (d) e x dx
0
−1
∫ π/3
(e) sint dt
0
∫ π/2
(g) cos3t dt
0
∫ 1.2
(i) tanx dx
1
(f)
(h)
∫ π
sin(t +3)dt
0
∫ 2
cos πt dt
1

452 Chapter 13 Integration
2 Evaluate the followingintegrals:
∫ 1
(a) t 2 +1dt
0
∫ 1 1
(b)
0 1+t 2 dt
∫ 2
(c) 3e 2x −2e 3x dx
1
∫ 2
(d) (x +1)(x +2)dx
1
∫ 2
(e) 2sin4t dt
0
∫ ( ) π t
(f) 4cos dt
0 2
3 Evaluate the followingintegrals:
∫ 1
(a) t 2 +0.5t−6dt
0
∫ 3 3
(b)
2 2x + 2x
3 dx
∫ 2
(c) e −2x −3e −x dx
1
∫ 2
(d) 3sin(4t − π) +5cos(3t + π/2)dt
0
∫ 1
(e) 2tantdt
0
∫ 1.5 2
(f)
0 1+x 2 − 1
√ dx
4−x 2
4 Calculate the areabetween f (t) = cost andthet axis
ast variesfrom
(a) 0 to π (b) 0 to π 4 2
3π
(c) toπ (d) 0toπ
4
5 Calculate the total area between f (t) = cos2t andthe
t axisast variesfrom
(a) 0 to π 4
(b)
π
4 to π 2
(c) 0 to π 2
6 Calculate the area enclosed by the curvesy =x 2 and
y=x.
7 Calculate the area enclosed by the graphs ofthe
functionsy =t 2 +5 andy = 6.
8 Evaluate the following definiteintegrals:
∫ 1.5 1
(a)
1 t + 1 e t + 1
sint dt
∫ 4 5
(b)
1 8+3x 2 dx
∫ 1
(c) sinhx dx
−1
∫ 1
(d) coshx dx
−1
9 Calculate the area betweeny = 2tant andthet axis
for −1 t 1.4.
10 Findthe areabetweeny = sint,y = cost andthey
axis, fort 0.
11 Thevelocity, v,ofaparticleisgiven by
v=(1+t) 2
Findthe distancetravelled bythe particlefromt = 1
tot = 4; thatis,evaluate ∫ 4
1 vdt.
12 Evaluate the area under the functionx = 1/t for
1t10.
13 Evaluate
∫ 2
(a) x 1.4 dx
3
∫ π
(c) sinxcosx dx
0
∫ 1
(b) (e t ) 2 dt
0
∫ 2
(d) sinh 2 x dx
1
Solutions
1 (a) 20 (b) 1.3863 (c) 2
(d) 2.3504 (e) 0.5 (f) −1.9800
(g) − 1 3
4
2 (a)
3
53
(d)
6
(h) 0 (i) 0.3995
(b)
π
4
(e) 0.5728 (f) 8
(c) −184.75
3 (a)− 65
12
(b) 2.275 (c) −0.639
(d) −0.9255 (e) 1.2313 (f) 1.1175
4 (a) 0.7071 (b) 1
5 (a)
(c) 0.7071 (d) 2
1
2
(b)
1
2
(c) 1

Review exercises 13 453
6
7
1
6
4
3
8 (a) 1.0839 (b) 0.6468
(c) 0 (d) 2.3504
9 4.7756
10 0.4142
11 39
12 2.3026
13 (a) −3.6202 (b) 3.1945
(c) 0 (d) 5.4158
REVIEWEXERCISES13
1 Find the following integrals:
∫
∫ 2
(a) 3x 2 +xdx (b)
t +2t+2dt
∫
∫ √t 1
(c) (1 +z)(1 −z)dz (d) − √t dt
(e)
∫
2 Given
x 2/3 +4x 3 dx
dy
dx =x2 +sinx+cos2x+1
findan expression fory(x).
3 Find the following integrals:
∫
(a) 3e x + 3 e x dx
∫
(b) (1+e x )(1−e −x )dx
∫
(c) 2e 4t +1 dt
∫
(d) e x (1+e x )dx
∫
(e) 4e −t −e −2t dt
4 Find the integrals
∫
(a) sin2x +cos2x dx
∫
(b) 2sint−costdt
∫ ( )
t
(c) 4tan dt
2
∫
(d) sin(π −z) +cos(π −2z)dz
∫
(e) tan(t + π)dt
∫ ( )
t
(f) 2sin3t+2sin dt
3
5 Find the following integrals:
∫
(a) cosec(3t + π)dt
∫ ( )
x
(b) sec
2 +1 dx
∫ ( )
π +t
(c) cot dt
2
∫ ( )
y
(d) 3cosec
3 −2 dy
∫ 1
(e) cot(π −2z)dz
2
∫ 2
(f) sec(2t − π)dt
3
6 Find the following integrals:
∫
∫
4
(a) √ dv (b) 1−v 2
(c)
(e)
(g)
∫
∫
∫
∫
1
49+t 2 dt (d)
2
√
1−4t 2 dt
3
x 2 +3 dx
(f) ∫
1
2 √ 1−v dv 2
1
50+2t 2 dt
1
√
36 −9x 2 dx
7 The speed, v(t),ofaparticleis given by
v(t) =t +e −t
(a) Findthe distancetravelled by the particle.
(b) Calculate the distancetravelled betweent = 1
andt=3.
8 The capacitanceofacapacitor is 0.1F.The current,
i(t),through the capacitor is given by
i(t) = 50sinπt
Derive an expression forthe voltageacross the
capacitor.

454 Chapter 13 Integration
9 Byusingsuitable trigonometric identities find
∫
(a) sin 2 2t +cos 2 2t dt
∫
(b) sin2tcos2t dt
∫
1
(c)
sin2t dt
∫ cos2t
(d)
sin2t dt
∫ sin2t
(e)
cos2t dt
10 Byexpressing
2x 2 +x+2
x 3 +x
asits partialfractions, find
∫ 2x 2 +x+2
x 3 dx
+x
11 Evaluate the followingdefiniteintegrals:
∫ 1
(a) 2t 2 +t 3 dt
0
∫ 3 2
(b)
1 x − x 2 dx
∫ 1
(c) 7−t+7t 2 dt
0
∫ 2
(d) (z +1)(z +2)dz
0
∫ 4 √
(e) x dx
1
∫ −1 x −2
(f) dx
−2 x
12 Evaluate the followingdefiniteintegrals:
∫ 1
∫ 1
(a) e 3x+1 dx (b) e 2t −e t +1dt
0
−1
∫ 2
∫ 1
(c) (e z −1) 2 dz (d) e −2x +e 2x dx
1
0
∫ 0
(e) x+e x dx
−1
13 Evaluate the followingintegrals:
∫ π/2
(a) 2sin3t dt
0
∫ π
(b) sin2t −cos2t dt
π/2
∫ π/4
(c) tant+tdt
0
∫ ( ) ( )
π/4 t t
(d) sin +cos dt
0 2 2
∫ 0.1
(e) tan3t dt
0
14 Evaluate the following definiteintegrals:
∫ 2π/k
(a) sinkt dt
0
∫ 2π/k
(b) coskt dt
0
wherekis aconstant.
15 Evaluate the following definiteintegrals:
∫ 0.5
(a) cosec(2x +1)dx
0
∫ 0.1
(b) sec3t dt
0
∫ π/4
(c) tan(x + π)dx
−π/4
16 Evaluate the following definiteintegrals:
∫ 2 3
(a) √ dx
0 9−x 2
∫ 1 2
(b)
−1 9+x 2 dx
∫ 1 1
(c) √ dx
0 8−2x 2
∫ 3 1
(d)
1 10+4t 2 dt
17 (a) Calculate the area enclosed bythe curvey =x 3 ,
thexaxisandx = 2.
(b) Calculate the area enclosed bythe curvey =x 3 ,
theyaxisandy = 8.
18 Findthe total area between f (t) =t 2 −4 andthet
axisonthe followingintervals:
(a) [−4, −3] (b) [−3, −1]
(c) [0,3]
(d) [−3,3]
19 Calculate the area enclosed by the curve
y=x 2 −x−6andthexaxis.
20 Calculate the total area betweeny =x 2 −3x −4 and
thexaxisonthe followingintervals:
(a) [−2, −1] (b) [−2,1]
(c) [2,4] (d) [2,5]
21 Calculate the area enclosed by the curvey =x 3 and
theliney =x.

Review exercises 13 455
22 Calculate the area enclosed byy =x 2 +4 and
y=12−x 2 .
23 Calculate the area enclosed byy = sinx andy = 2x
π .
Solutions
1 (a) x 3 + x2
2 +c
2
(b) 2ln|t|+t 2 +2t+c
(c) z− z3 3 +c
2
(d)
3 t3/2 −2t 1/2 +c
3
(e)
5 x5/3 +x 4 +c
x 3
3 −cosx+1 2 sin2x+x+c
3 (a) 3e x −3e −x +c
(b) e x +e −x +c
(c)
e 4t
2 +t+c
(d) e x + e2x
2 +c
(e) −4e −t + e−2t
2 +c
4 (a) − 1 2 cos2x + 1 2 sin2x+c
(b) −2cost−sint+c
( )∣ (c) 8ln
∣ sec t ∣∣∣∣
+c
2
(d) cos(π −z) − 1 sin(π −2z) +c
2
(e) ln|sec(t +π)|+c
( )
(f) − 2 3 cos3t −6cos t
+c
3
1
5 (a) ln|cosec(3t + π) −cot(3t + π)| +c
3 ( ) ( )∣ (b) 2ln
∣ sec x
2 +1 x ∣∣∣∣
+tan
2 +1 +c
( )∣ (c) 2ln
∣ sin π +t ∣∣∣∣
+c
2
( ) ( )∣ (d) 9ln
∣ cosec y
3 −2 y ∣∣∣∣
−cot
3 −2 +c
(f)
(e) − 1 ln|sin(π −2z)| +c
4
1
ln|sec(2t −π)+tan(2t −π)|+c
3
6 (a) 4sin −1 v+c
(b) 1 2 sin−1 v +c
( )
1 t
(c)
7 tan−1 +c
7
( )
1 t
(d)
10 tan−1 +c
5
(e) sin −1 (2t) +c
( )
1 x
(f)
3 sin−1 +c
2
(g) √ ( )
3tan −1 x
√ +c
3
7 (a) t2 2 −e−t +c (b) 4.3181
8 − 500
π cosπt+c
9 (a) t+c
(b) − cos4t +c
8
1
(c) ln|cosec2t −cot2t|+c
2
1
(d)
2 ln|sin2t|+c
1
(e)
2 ln|sec2t|+c
10 2ln|x|+tan −1 x+c
11
11 (a)
12
38
(d)
3
(b) 0.1972
(e)
14
3
(c)
53
6
(f) 2.3863
12 (a) 17.2933 (b) 3.2765 (c) 15.2630
(d) 3.6269 (e) 0.1321
2
13 (a)
3
(b) −1 (c) 0.6550
(d) 0.9176 (e) 0.0152
14 (a) 0 (b) 0

456 Chapter 13 Integration
15 (a) 0.5238 (b) 0.1015 (c) 0
16 (a) 2.1892 (b) 0.4290
(c) 0.3702 (d) 0.0825
17 (a) 4 (b) 12
18 (a)
25
3
(b) 4
(c)
23
3
(d)
46
3
19 125
6
20 (a)
21 0.5
22 64
3
17
6
23 0.4292
(b)
61
6
(c)
22
3
(d)
61
6

14 Techniques
ofintegration
Contents 14.1 Introduction 457
14.2 Integrationbyparts 457
14.3 Integrationbysubstitution 463
14.4 Integrationusingpartialfractions 466
Reviewexercises14 468
14.1 INTRODUCTION
The previous chapter showed us how to integrate functions which matched the list of
standard integrals given in Table 13.1. Clearly, it is impossible to list all possible functions
in the table and so some general techniques are required. Integration techniques
maybeclassifiedasanalytical,thatisexact,ornumerical,thatisapproximate.Wewill
now study threeanalytical techniques:
(1) integration by parts;
(2) integration by substitution;
(3) integration usingpartialfractions.
14.2 INTEGRATIONBYPARTS
This technique is used to integrate a product, and is derived from the product rule for
differentiation. Let u and v be functions of x. Then the product rule of differentiation
states:
d du
(uv) =
dx dx v +udv dx
Rearranging we have
u dv
dx = d dx (uv)−vdu dx

458 Chapter 14 Techniques of integration
Integrating thisequation yields
∫
u dv ∫ ∫ d
dx dx = dx (uv)dx−
v du
dx dx
Recognizing thatintegration and differentiation areinverse processes allows
∫ d(uv)
dx
dx
tobe simplified touv. Hence,
∫
( ) ∫ dv
u dx=uv−
dx
( du
v
dx
)
dx
This isthe formulafor integration by parts.
∫
Example14.1 Find
xsinxdx.
Solution We recognize the integrand as a product of the functions x and sinx. Let u = x,
dv du
=sinx. Then =1, v = −cosx. Using the integration by parts formula we get
dx dx
∫
∫
xsinxdx=x(−cosx)− (−cosx)1dx
=−xcosx+sinx+c
Whendealingwithdefiniteintegralsthecorrespondingformulaforintegrationbyparts
is
∫ b
a
( ∫ dv b
( ) du
u
)dx =[uv] b a
dx
− v dx
dx
a
Example14.2 Evaluate
Solution We let
Then
∫ 2
0
u=x
xe x dx
and
dv
dx = ex
du
=1 and v=ex
dx

14.2 Integration by parts 459
Usingthe integration by parts formula fordefinite integrals we have
∫ 2
∫ 2
xe x dx = [xe x ] 2 0 − e x·1dx
0
0
= 2e 2 −[e x ] 2 0
=2e 2 −[e 2 −1]
=e 2 +1
Sometimes integration by parts needs tobe used twice,as the next example illustrates.
Example14.3 Evaluate
∫ 2
0
x 2 e x dx
Solution We let
Then
u=x 2
and
dv
dx = ex
du
=2x and v=ex
dx
Usingthe integration by parts formula wehave
Now
∫ 2
0
∫ 2
0
∫ 2
0
∫ 2
x 2 e x dx = [x 2 e x ] 2 0 − 2xe x dx
=4e 2 −2
0
∫ 2
0
xe x dx
xe x dx has been evaluated usingintegration by parts inExample 14.2. So
x 2 e x dx = 4e 2 −2[e 2 +1] = 2e 2 −2 = 12.78
The next example illustrates a case in which the integral to be found reappears after
repeated application ofintegration by parts.
Example14.4 Find
∫
e t sintdt

460 Chapter 14 Techniques of integration
Solution We let
and so
u=e t
and
dv
dt
=sint
du
dt =et and v=−cost
Applying integration by parts yields
∫
∫
e t sintdt =−e t cost− (−cost)e t dt +c
∫
=−e t cost+
e t costdt +c (14.1)
We now apply integration by parts to ∫ e t costdt. We let
Then
u=e t
and
dv
dt
=cost
du
dt =et and v =sint
So
∫ ∫
e t costdt =e t sint−
e t sintdt (14.2)
Substituting Equation (14.2) into Equation (14.1) yields
∫
∫
e t sintdt =−e t cost+e t sint− e t sintdt+c
Rearranging the equation gives
∫
2 e t sintdt =−e t cost+e t sint+c
from which we see that
∫
e t sintdt = −et cost+e t sint+c
2
Example14.5 Evaluate
∫ 2
0
x n e x dx
forn=3,4,5.
Solution The integral may be evaluated by using integration by parts repeatedly. However, this
is slow and cumbersome. Instead it is useful to develop a reduction formula as is now
illustrated.
Letu =x n and dv
dx = ex . Then du
dx =nxn−1 and v = e x .

Usingintegration by parts we have
∫ 2
0
Writing
I n
=
wesee that
Hence
I n−1
=
∫ 2
x n e x dx = [x n e x ] 2 0 − nx n−1 e x dx
∫ 2
0
∫ 2
=2 n e 2 −n
x n e x dx
0
x n−1 e x dx
0
∫ 2
0
x n−1 e x dx
14.2 Integration by parts 461
I n
= 2 n e 2 −nI n−1
(14.3)
Equation (14.3) iscalled areduction formula.
We have already evaluatedI 1
, thatis
I 1
=e 2 +1
∫ 2
Usingthe reduction formula withn = 2 gives
∫ 2
0
x 2 e x dx =I 2
=2 2 e 2 −2I 1
= 4e 2 −2(e 2 +1)
=2e 2 −2
Note that thisisinagreement with Example 14.3.
Withn = 3 the reduction formulayields
∫ 2
0
x 3 e x dx =I 3
=2 3 e 2 −3I 2
Withn = 4 we have
∫ 2
0
= 8e 2 −3(2e 2 −2)
=2e 2 +6
x 4 e x dx =I 4
=2 4 e 2 −4I 3
Withn = 5 we have
∫ 2
= 16e 2 −4(2e 2 +6)
=8e 2 −24
0
xe x dx inExample 14.2, and found
0
x 5 e x dx =I 5
=2 5 e 2 −5I 4
= 32e 2 −5(8e 2 −24)
= 120 −8e 2

462 Chapter 14 Techniques of integration
EXERCISES14.2
1 Useintegration by partsto findthe following:
∫
∫
(a) xsin(2x)dx (b) te 3t dt
(c)
(e)
∫
xcosxdx
∫ x
e x dx
(d)
∫
( )
v
2vsin dv
2
2 Useintegration by partsto find
∫
(a) tlntdt
∫
(b) lntdt
∫
(c) t n lntdt (n ≠ −1)
∫
(d) tsin(at +b)dt a,bconstants
∫
(e) te at+b dt a,bconstants
3 Evaluate the followingdefiniteintegrals:
∫ 1
∫ π/2
(a) xcos2xdx (b) xsin2xdx
0
0
∫ 1
∫ 3
(c) te 2t dt (d) t 2 lntdt
−1
1
∫ 2 2x
(e) dx
0 e2x 4 Find
∫
(a) t 2 e 2t dt
(b)
(c)
∫
t 2 cos3tdt
∫ ( )
t 2 t
sin dt
2
5 Evaluate the following definiteintegrals:
∫ 2
(a) t 2 e t dt
0
∫ 1
(b) t 2 sintdt
−1
∫ 1
(c) t 2 cos3tdt
0
6 Obtainareduction formulafor
∫
I n = t n e kt dt n,k constants
∫
Hencefind
∫
t 2 e 3t dt,
∫
t 3 e 3t dt and
t 4 e 3t dt.
7 Useintegration byparts twiceto obtain areduction
formulafor
∫ π/2
I n = t n sintdt
0
∫ π/2 ∫ π/2
Hencefind t 3 sintdt, t 5 sintdt
0 0
∫ π/2
and t 7 sintdt.
0
8 Useintegration byparts to find
∫ π/2
e 2x cosxdx
0
Solutions
sin2x
1 (a) − xcos2x
4 2
( )
(b) e 3t t
3 − 1 +c
9
+c
(c) cosx+xsinx+c
( ) ( )
v v
(d) 8sin −4vcos +c
2 2
(e) −e −x (x +1) +c
2 (a)
t 2 lnt
2
− t2 4 +c
(b) tlnt−t+c
(c)
(d)
(lnt)t n+1
−
tn+1
n +1 (n +1) 2 +c
sin(at +b)
a 2
−
(e) e at+b (
t
a − 1 a 2 )
+c
tcos(at +b)
a
+c

14.3 Integration by substitution 463
3 (a) 0.1006 (b) 0.7854 (c) 1.9488
4 (a)
(d) 6.9986 (e) 0.4542
(b)
e 2t (2t 2 −2t +1)
4
+c
)
2
(t 2
9 tcos3t+ 3 − 2 sin3t+c
27
( ) ( )
t
(c) 8tsin −2(t 2 t
−8)cos +c
2 2
5 (a) 12.7781 (b) 0 (c) −0.1834
6 I n = tn e kt
− n k k I n−1 , e3t (9t 2 −6t +2)
,
27
e 3t (9t 3 −9t 2 +6t −2)
,
27
e 3t (27t 4 −36t 3 +36t 2 −24t +8)
81
7 I n =n
8 4.2281
{(
π
2
) n−1
− (n −1)I n−2
}
1.4022,2.3963,4.5084
14.3 INTEGRATIONBYSUBSTITUTION
∫
Example14.6 Find
This technique is the integral equivalent of the chain rule. It is best illustrated by
examples.
(3x +1) 2.7 dx.
Solution Letz = 3x + 1, so that dz
dx
becomes
∫
z 2.71 3 dz = 1 ∫
z 2.7 dz = 1 ( z
3.7
3 3 3.7
dz
= 3, that is dx = . Writing the integral in terms ofz, it
3
)
+c = 1 (3x +1) 3.7
+c
3 3.7
Example14.7 Evaluate
∫ 3
Solution Let v =t 2 so dv
dt
2
dt = 1 2t dv
tsin(t 2 )dt.
= 2t, thatis
When changing the integral from one in terms oft to one in terms of v, the limits must
alsobe changed. Whent = 2, v = 4;whent = 3, v = 9.Hence, the integral becomes
∫ 9
4
sinv
2 dv = 1 2 [−cos v]9 4 = 1 [−cos9 +cos4] = 0.129
2
Sometimesthe substitution can involve a trigonometric function.

464 Chapter 14 Techniques of integration
Example14.8 Evaluate
∫ 2
1
sint cos 2 tdt.
Solution Putz = cost so that dz = −sint, that is sintdt = −dz. Whent = 1,z = cos1; when
dt
t = 2,z = cos2.Hence
∫ 2
∫ cos2
[ ] z
sint cos 2 tdt =− z 2 3 cos2
dz=−
3
∫
Example14.9 Find
1
e
tanx
cos 2 x dx.
cos1
cos1
= cos3 1 −cos 3 2
3
Solution Putz = tanx. Then dz
dx = sec2 x,dz =
dx
cos 2 x . Hence,
∫ e
tanx
∫
cos 2 x dx = e z dz=e z +c=e tanx +c
∫
Example14.10 Find
Integration by substitution allows functions ofthe form df/dx
f
3x 2 +1
x 3 +x+2 dx.
= 0.0766
tobe integrated.
Solution Putz =x 3 +x+2,then dz
dx = 3x2 +1,thatisdz = (3x 2 +1)dx. Hence,
∫
3x 2 ∫
+1 dz
x 3 +x+2 dx = z =ln|z|+c=ln|x3 +x+2|+c
∫ df/dx
Example14.11 Find dx.
f
Solution Putz = f.Then dz
dx = df
dx , thatis
dz = df
dx dx.
Hence,
∫ df/dx
f
dx =
∫ dz
z =ln|z|+c=ln|f|+c
and so
∫
df/dx
f
dx=ln|f|+c

14.3 Integration by substitution 465
The resultofExample 14.11 isparticularly important.
∫ df/dx
dx=ln|f|+c
f
∫ 4
3t
Example14.12 2 +2t
Evaluate
2 t 3 +t 2 +1 dt.
Solution The numerator isthe derivative ofthe denominator and so
∫ 4
2
3t 2 +2t
t 3 +t 2 +1 dt = [ln|t3 +t 2 +1|] 4 2 =ln81−ln13=1.83
Example14.13 Find
∫
(a)
∫
(b)
∫
(c)
4
5x−7 dx
t
t 2 +1 dt
e t/2
e t/2 +1 dt
Solution The integrands arerewritten so thatthe numerator isthe derivative of the denominator.
∫
4
(a)
5x−7 dx = 4 ∫
5
5 5x−7 dx = 4 5 ln|5x−7|+c
∫
t
(b)
t 2 +1 dt = 1 ∫
2t
2 t 2 +1 dt
(c)
∫
= 1 2 ln|t2 +1|+c
e t/2 ∫ 1
e t/2 +1 dt=2 2 et/2
e t/2 +1 dt =2ln|et/2 +1|+c
EXERCISES14.3
1 Usethe given substitutionsto findthe following
integrals:
∫
(a) (4x+1) 7 dx, z=4x+1
∫
(b) t 2 sin(t 3 +1)dt, z =t 3 +1
∫
(c) 4te −t2 dt, z=t 2
∫
(d) (1−z) 1/3 dz, t=1−z
∫
(e) cost(sin 5 t)dt, z =sint
2 Evaluate the following definiteintegrals:
∫ 2
∫ π/2
(a) (2t +3) 7 dt (b) sin2tcos 4 2tdt
1
0
∫ 1
∫ 2
(c) 3t 2 √
e t3 dt (d) 4+3xdx
0
0
(
∫ √x )
2 sin
(e) √ dx
1 x
3 Find the area betweeny =x(3x 2 +2) 4 andthexaxis
fromx=0tox=1.

466 Chapter 14 Techniques of integration
4 Find
∫
(a)
∫
(c)
∫
(e)
4x+1
2x 2 +x+3 dx
3
9−2x dx
1
tlnt dt
(b) ∫
(d) ∫
2sin2x
cos2x+7 dx
t
t 2 +1 dt
5 Evaluate the following:
∫ 1
∫
2
π/2
(a)
0 (1 +3x) 2 dx (b) sint √ costdt
0
∫ 2
∫
3 +x 2
(c)
1 x 2 +6x+1 dx (d) e √ t
√ dt
1 t
∫ 2
(e) xsin(π −x 2 )dx
0
Solutions
1 (a)
(4x +1) 8
32
+c (b) − cos(t3 +1)
3
+c
(c) −2e −t2 +c (d) − 3 4 (1 −z)4/3 +c
(e)
1
6 sin6 t+c
2 (a) 3.3588 ×10 5 (b) 0.2 (c) 1.7183
(d) 5.2495 (e) 0.7687
3 103.098
4 (a) ln(2x 2 +x+3)+c
(b) −ln(cos2x +7) +c
(c) − 3 ln(9 −2x) +c
2
1
(d)
2 ln(t2 +1)+c
(e) ln(lnt) +c
5 (a) 0.5 (b) 0.6667 (c) 0.3769
(d) 2.7899 (e) 0.8268
14.4 INTEGRATIONUSINGPARTIALFRACTIONS
The technique of expressing a rational function as the sum of its partial fractions has
been covered in Section 1.7. Some expressions which at first sight look impossible to
integrate mayinfactbeintegrated when expressed astheirpartialfractions.
Example14.14 Find
∫
(a)
∫
(b)
1
x 3 +x dx
13x−4
6x 2 −x−2 dx
Solution (a) Firstexpress the integrand inpartialfractions:
Then,
1
x 3 +x = 1
x(x 2 +1) = A x + Bx+C
x 2 +1
1=A(x 2 +1)+x(Bx+C)
Equating the constant terms: 1 =Aso thatA = 1.
Equating the coefficients ofx: 0 =C so thatC = 0.
Equating the coefficients ofx 2 : 0 =A +B and henceB = −1.

14.4 Integration using partial fractions 467
(b)
Then, ∫ ∫
1 1
x 3 +x dx = x − x
x 2 +1 dx
=ln|x|− 1 2 ln|x2 +1|+c=ln
x
∣√ x2 +1
∣ +c
∫
∫
13x−4
6x 2 −x−2 dx = 13x−4
(2x +1)(3x −2) dx
∫
=
= 3 2
3
2x+1 + 2
3x−2 dx
∫
2
2x+1 dx + 2 ∫
3
using partialfractions
3
3x−2 dx
= 3 2 ln|2x+1|+2 3 ln|3x−2|+c
∫ 1
4t
Example14.15 3 −2t 2 +3t−1
Evaluate dt.
0 2t 2 +1
Solution Usingpartialfractions we maywrite
4t 3 −2t 2 +3t−1 t
=2t−1+
2t 2 +1 2t 2 +1
Hence,
∫ 1
0
4t 3 −2t 2 +3t−1
2t 2 +1
dt =
=
∫ 1
0
2t−1+
[
t
2t 2 +1 dt = t 2 −t + 1 ] 1
4 ln |2t2 +1|
0
[1 −1+ 1 ]
4 ln3 −
[0 −0+ 1 ]
4 ln1 = 0.275
EXERCISES14.4
1 Bywriting the integrand asits partialfractions find
∫ x +3
(a)
x 2 +x dx
∫ t −3
(b)
t 2 −1 dt
∫
8x+10
(c)
4x 2 +8x+3 dx
∫ 2t 2 +3t+3
(d) dt
2(t +1)
∫ 2x 2 +x+1
(e)
x 3 +x 2 dx
2 Evaluate the following integrals:
∫ 3 5x+6
(a)
1 2x 2 +4x dx
∫ 1 3x+5
(b)
0 (x +1)(x +2) dx
∫ 2 3−3x
(c)
1 2x 2 +6x dx
∫ 0 4x+1
(d)
−1 2x 2 +x−6 dx
∫ 3 x 2 +2x−1
(e)
2 (x 2 +1)(x −1) dx

468 Chapter 14 Techniques of integration
3 Find the area betweeny =
axisfromx=0tox=1.
4 Usepartialfractions to find
∫ 3 3x+2
(a)
2 x 2 −1 dx
4x+7
4x 2 and thex
+8x+3
(b)
(c)
(d)
∫
t +3
t 2 +2t+1 dt
∫ 2t 2 +3t+1
t 3 dt
+t
∫
6t+3
2t 2 −5t+2 dt
Solutions
1 (a) 3lnx−2ln(x+1)+c
(b) 2ln(t+1)−ln(t−1)+c
1
(c)
2 ln(2x +3) + 3 ln(2x +1) +c
2
t(t +1)
(d) ln(t +1) + +c
2
(e) 2ln(x +1) − 1 x +c
2 (a) 2.1587
(b) 1.7918
(c) −0.09971
(d) 0.1823
(e) 0.9769
3 1.2456
4 (a) 1.8767
(b) ln(t +1) − 2
t +1 +c
(c) 3tan −1 t + 1 2 ln(t2 +1)+lnt+c
(d) 5ln(t−2)−2ln(2t−1)+c
REVIEWEXERCISES14
1 Usethe given substitutionto findthe following
integrals:
∫ 1
(a) (9t+2) 10 dt z=9t+2
0
∫ 5
(b) (−t+1) 6 dt z=−t+1
3
∫ 3
(c) (4x−1) 27 dx z=4x−1
∫6
√3t+1dt
(d) z=3t+1
∫
(e) (9y−2) 17 dy z=9y−2
∫ 2 3
(f) dz
0 (2z +5) 6
∫
y=2z+5
(g) t 2 sin(t 3 )dt z =t 3
∫
(h) x 2 e x3 +1 dx z =x 3 +1
∫ 0.5
(i) sin(2t)e cos(2t) dt z = cos(2t)
∫0
π
(j) sintcos 2 tdt z=cost
∫0
(k) cost √ sintdt z=sint
2 Useintegration byparts to find
∫ π/2
∫ π/2
(a) e 2x cosxdx (b) e 2x sinxdx
0
0
3 Find
∫ lnt
dt
t
using
(a)integrationby parts
(b)the substitutionz = lnt.
4 TheintegralI n isgiven by
∫ π/2
I n = sin n θ dθ
0
(a) StateI n−2 .
(b) Show
I n = n −1
n I n−2
(c) EvaluateI 0 ,I 1 ,I 2 andI 3 .
5 Evaluate
∫
t 2
(a)
t 3 +1 dt

Review exercises 14 469
∫ π/3
(b) sintcostdt
0
∫ 3 4
(c) dt
1 e2t ∫
3x−7
(d)
(x −2)(x −3)(x −4) dx
∫
e x
(e)
e x +1 dx
6 Evaluate
∫ 1 3
(a)
0 (e t ) 2 +sintcos2 tdt
∫ 1
(b) 4t 2 e t3 +t(1+t 2 ) 12 dt
0
∫ π
(c) sin 2 ωt +cos 2 ωt + ωdt
0
ω constant
∫ 2 1+t+t 2
(d)
1 t(1+t 2 ) dt
∫ 1
(e) (t+e t )sintdt
0
∫ 3 1+4x
(f)
1 2x +4x 2 dx
7 Calculatethe area undery(x) = 1 +2e2x
x+e 2x fromx=1
tox=3.
8 Evaluate the following integrals:
(a)
∫
(−2t +0.1) 4 dt
∫
(b) (1+x)sinxdx
∫ 2
(c) xsin(1 +x)dx
1
∫ 6 t
(d) √
3 t 2 +1 dt
∫
1
(e)
t 3 +2t 2 +t dt
∫ 5 1
(f)
1 1+e tdt
9 Find
∫
(a)
cost
10+sint dt
∫
1
(c)
t(1+lnt) dt
∫ 1+lnx
(e)
xlnx dx
∫
10 Find
x 3 e x2 dx.
∫ 2sintcost
(b)
1+sin 2 dt
t
∫
(d) 1
e t (1+e −t ) dt
Solutions
1 (a) 2.8819 ×10 9
(b) 2322
(c) −1.2 ×10 36
(d) 2 9 (3t +1)3/2 +c
(9y −2) 18
(e) +c
162
(f) 9.092 ×10 −5
(g) − 1 3 cos(t3 )+c
+1
(h) ex3 +c
3
(i) 0.5009
2
(j)
3
(k) 2 3 (sint)3/2 +c
2 (a) 4.2281 (b) 9.4563
3
(lnt) 2
+c
2
∫ π/2
4 (a) I n−2 = sin n−2 θ dθ
0
π
(c)
2 ,1,0.7854,0.6667
1
5 (a)
3 ln|t3 +1|+c
3
(b)
8
(c) 0.2657
(d) − 1 2 ln|x−2|−2ln|x−3|+5 2 ln|x−4|+c
(e) ln|e x +1|+c
6 (a) 1.5778 (b) 317.3 (c) π(1 + ω)
(d) 1.0149 (e) 1.2105 (f) 0.9730
7 3.8805
(−2t +0.1)5
8 (a) − +c
10
(b) −(1+x)cosx+sinx+c
(c) 0.7957

470 Chapter 14 Techniques of integration
(d) 2.9205
(e) ln|t|−ln|t+1|+ 1
t +1 +c
(f) 0.3066
9 (a) ln(sint +10) +c
(b) ln(sin 2 t +1) +c
(c) ln(1+lnt)+c
(d) −ln(1+e −t )+c
(e) ln(xlnx) +c
10 ex2 (x 2 −1)
2
+c

15 Applicationsof
integration
Contents 15.1 Introduction 471
15.2 Averagevalueofafunction 471
15.3 Rootmeansquarevalueofafunction 475
Reviewexercises15 479
15.1 INTRODUCTION
Currents and voltages often vary with time. Engineers may wish to know the average
valueofsuchacurrentorvoltageoversomeparticulartimeinterval.Theaveragevalue
of a time-varying function is defined in terms of an integral. An associated quantity is
therootmeansquare(r.m.s.)valueofafunction.Ther.m.s.valueofacurrentisusedin
the calculation of the power dissipated by a resistor.
15.2 AVERAGEVALUEOFAFUNCTION
Suppose f (t)isafunction defined ona t b. The area,A, under f isgiven by
A =
∫ b
a
f dt
A rectangle with base spanning the interval [a,b] and heighthhas an area ofh(b −a).
Supposetheheight,h,ischosensothattheareaunder f andtheareaoftherectangleare
equal. This means
h(b−a)=
h =
∫ b
a
f dt
∫ b
a f dt
b −a
Thenhiscalledtheaverage valueofthefunctionacrosstheinterval[a,b] andisillustrated
inFigure 15.1:
average value =
∫ b
a f dt
b −a

472 Chapter 15 Applications of integration
f (t)
h
a b t
Figure15.1
Thearea underthe curve fromt =atot =bandthe
area ofthe rectangle are equal.
Example15.1 Findthe average value of f (t) =t 2 across
(a) [1,3]
(b) [2,5]
Solution (a) average value =
(b) average value =
∫ 3
1 t2 dt
3 −1 = 1 [ ] t
3 3
2 3
1
∫ 5
2 t2 dt
5 −2 = 1 [ ] t
3 5
3 3
2
= 13
3
= 13
Example15.1showsthatiftheintervalofintegrationchangesthentheaveragevalueof
a functioncan change.
Engineeringapplication15.1
Saw-toothwaveform
Recall from Engineering application 2.2 that engineers frequently make use of the
saw-toothwaveform. Consider the saw-toothwaveform shown inFigure 15.2.
y
5
–2 0 2
4 6 8 t
Figure15.2
Asaw-tooth waveform.
Calculate the average value of thiswaveform over a complete period.

15.2 Average value of a function 473
Solution
We first need to obtain an equation for the waveform. We choose the interval
0t<2.
The general equation for a straightlineis
v=mt+c
Whent=0thenv=0.So
0=0+c
c = 0
Whent=2thenv=5.So
Hence
5 =m(2)
m=2.5
v = 2.5t
The average value isgiven by
∫ 2
[ 2.5t
2
] 2
0
v av
= 1 2.5tdt = 1 2 0 2 2
v av
= 1 ( ) 2.5×4
−0 = 10 2 2 4 =2.5V
Engineeringapplication15.2
Athyristorfiringcircuit
Figure 15.3 shows a simple circuit to control the voltage across a load resistor,R L
.
This circuit has many uses, one of which is to adjust the level of lighting in a room.
Thecircuithasana.c.powersupplywithpeakvoltage,V S
.Themaincontrolelement
is the thyristor. This device is similar in many ways to a diode. It has a very high
resistance when it is reverse biased and a low resistance when it is forward biased.
However,unlikeadiode,thislowresistancedependsonthethyristorbeing‘switched
on’ by the application of a gate current. The point at which the thyristor is switched
on can be varied by varying the resistor,R G
. Figure 15.4 shows a typical waveform
ofthe voltage, v L
, across the loadresistor.
The point at which the thyristor is turned on in each cycle is characterized by
the quantity αT, where 0 α 0.25 and T is the period of the waveform. This
restrictionon αreflectsthefactthatifthethyristorhasnotturnedonwhenthesupply
voltage has peaked inthe forward direction thenitwill never turnon.
Calculate the average value of the waveform over a period and comment on the
result.
➔

474 Chapter 15 Applications of integration
y L
~
V S
y L R L
Thyristor Gate R G
V S
aT aT aT
t
Figure15.3
A thyristorfiringcircuit.
Figure15.4
Load voltage waveform.
Solution
The average value of load voltage is
∫
1 T
v
T L
dt= 1 ∫ T/2
( ) 2πt
V
T S
sin dt
T
0
= V S
T
αT
T
2π
[
−cos
( 2πt
T
)] T/2
αT
= V S
(1 +cos2πα)
2π
If α = 0, then the average value is V S
/π, the maximum value for this circuit. If
α = 0.25,thentheaveragevalueisV S
/2πwhichillustratesthatdelayingtheturning
on ofthe thyristor reduces the average value of the load voltage.
EXERCISES15.2
1 Calculate the averagevalue ofthe given functions
across the specified interval:
(a) f (t) = 1 +t across [0,2]
(b) f (x) = 2x −1 across [−1,1]
(c) f (t) =t 2 across [0,1]
(d) f (t) =t 2 across [0,2]
(e) f(z) =z 2 +zacross[1,3]
2 Calculate the averagevalue ofthe given functions
over the specified interval:
(a) f (x) =x 3 across[1,3]
(b) f(x) = 1 x across[1,2]
(c) f(t) = √ t across [0,2]
3
(d) f (z) =z 3 −1 across [−1,1]
(e) f(t) = 1 across[−3, −2]
t2 Calculate the average value ofthe following:
[ ]
(a) f (t) = sint across
0, π 2
(b) f (t) = sint across[0, π]
(c) f (t) = sin ωt across [0,π]
[ ]
(d) f (t) = cost across
0, π 2
(e) f (t) = cost across [0,π]
(f) f (t) = cos ωt across [0, π]
(g) f (t) = sin ωt +cos ωt across [0,1]
4 Calculate the average value ofthe following
functions:
(a) f(t) = √ t +1 across[0,3]
(b) f (t) = e t across[−1,1]
(c) f (t) = 1 +e t across[−1,1]

15.3 Root mean square value of a function 475
Solutions
1 (a) 2 (b) −1 (c)
1
3
(d)
2 (a) 10 (b) 0.6931 (c) 0.9428
1
(d) −1 (e)
6
2 2
3 (a) (b)
π π
4
3
(e)
19
3
1
(c)
πω [1 −cos(πω)] (d) 2
π
sin(πω)
(e) 0 (f)
πω
1+sinω−cosω
(g)
ω
14
4 (a) (b) 1.1752 (c) 2.1752
9
15.3 ROOTMEANSQUAREVALUEOFAFUNCTION
If f (t)isdefinedon [a,b], theroot mean square (r.m.s.)value is
r.m.s. =
√ ∫b
a (f(t))2 dt
b −a
Example15.2 Findthe r.m.s. valueof f (t) =t 2 across[1, 3].
Solution r.m.s. =
√ ∫ 3
1 (t2 ) 2 dt
3 −1
=
√ ∫ 3
1 t4 dt
2
=
√ √
[t 5 /5]
1
3 242
=
2 10 = 4.92
Example15.3 Calculate the r.m.s. valueof f (t) =Asint across [0,2π].
√ ∫ 2π
A
0 2 sin 2 tdt
Solution r.m.s. =
2π
√
A ∫ 2 2π
(1 −cos2t)/2dt
0
=
2π
√
[
A
=
2
t − sin2t ] 2π
4π 2
0
√
A2 2π
=
4π = √ A = 0.707A
2
Thus the r.m.s.value is0.707 ×the amplitude.

476 Chapter 15 Applications of integration
Example15.4 Calculate the r.m.s.value of f (t) =Asin(ωt + φ) across [0,2π/ω].
√ ∫ 2π/ω
A
0
Solution 2 sin 2 (ωt + φ)dt
r.m.s. =
2π/ω
√
∫
A
=
2 ω 2π/ω
1−cos2(ωt +φ)dt
4π 0
√
[
]
A
=
2 ω sin2(ωt + φ) 2π/ω
t −
4π 2ω
0
=
√
(
A 2 ω 2π
4π ω
−
sin2(2π + φ)
2ω
+ sin2φ )
2ω
Now sin2(2π + φ) = sin(4π + 2φ) and since sin(t + φ) has period 2π we see that
sin(4π +2φ) = sin2φ. Hence,
√ √
A2 ω 2π A
r.m.s. =
4π ω = 2
2 = √ A = 0.707A
2
Notethatsin(ωt + φ)hasperiod2π/ω.TheresultofExample15.4illustratesageneral
result:
The r.m.s. value of any sinusoidal waveform taken across an interval of length one
periodis
0.707 ×amplitudeofthe waveform
Root mean square value is an effective measure of the energy transfer capability of a
time-varyingelectricalcurrent.Toseewhythisisso,considerthefollowingEngineering
application.
Engineeringapplication15.3
Average power developed across a resistor by a time-varying
current
Consideracurrenti(t)whichdevelopsapowerp(t)inaloadresistorR.Thiscurrent
flowsfromtimet =t 1
totimet =t 2
.LetP av
betheaveragepowerdissipatedbythe
resistor during the time interval [ ]
t 1
,t 2 . We require that total energy transfer,E,be
the same inboth cases. So wehave
E =P av
(t 2
−t 1
)=
∫ t2
t 1
p(t)dt

15.3 Root mean square value of a function 477
Now
and so
p(t) = (i(t)) 2 R
P av
(t 2
−t 1
)=
∫ t2
t 1
i 2 Rdt
Ifwenowconsidertheaveragepowerdissipatedbytheresistortobetheresultofan
effective currentI eff
then wehave
I 2 eff R(t 2 −t 1 )= ∫ t2
I 2 eff (t 2 −t 1 )= ∫ t2
t 1
i 2 Rdt
t 1
i 2 dt
∫ t2
i 2 dt
I 2 eff = t 1
t 2
−t 1
∫ t2
√
i
t 2 dt
I eff
=
1
t 2
−t 1
We see that the equivalent direct current is the r.m.s. value of the time-varying
current.
Engineeringapplication15.4
Averagevalueandr.m.s.valueofaperiodicwaveform
Consider the periodicwaveform shown inFigure 15.5.
Thecurrenti(t) is
i(t) =20e −t 0 t <10, period T =10
(a) Calculate the average valueofthe currentover a complete period.
(b) Calculate the r.m.s. valueofthe currentover a complete period.
i(t)
20
0 10 20
30 t
Figure15.5
Waveform forEngineering application
15.4.
➔

478 Chapter 15 Applications of integration
Solution
(a) As the waveform isperiodic, we need only consider the interval 0 t < 10.
I av
= 1 T
∫ T
∫ t2
√
i
t 2 dt
(b) I eff
=
1
t 2
−t 1
Here
0
i(t)dt = 1
10
∫ 10
0
20e −t dt
t 1
=0 t 2
=10 i(t)=20e −t
Firstwe evaluate
So
∫ t2
t 1
i 2 dt=
∫ 10
0
[ e
−2t
= 400
−2
√
200.00
I eff
=
10−0
= 4.4721 A
= 1
10 [−20e−t ] 10
0
= 1
10 (−20e−10 +20e 0 )
= 20
10 (1 −e−10 ) = 2 ×0.99995 = 2.000A
400e −2t dt
] 10
= 400
−2 (e−20 −e 0 )
= 200(1 −e −20 ) = 200.00
0
EXERCISES15.3
1 Calculate the r.m.s.values ofthe functions in
Question1in Exercises15.2.
2 Calculate the r.m.s.values ofthe functions in
Question2in Exercises15.2.
3 Calculate the r.m.s.values ofthe functionsin
Question3in Exercises15.2.
4 Calculate the r.m.s.values ofthe functionsin
Question4in Exercises15.2.
Solutions
1 (a) 2.0817 (b) 1.5275 (c) 0.4472
(d) 1.7889 (e) 6.9666
2 (a) 12.4957 (b) 0.7071 (c) 1
(d) 1.0690 (e) 0.1712
3 (a) 0.7071
(b) 0.7071
√
1
(c)
2 − sinπωcosπω
2πω

Review exercises 15 479
(d) 0.7071
(e) 0.7071
√
1
(f)
2 + sinπωcosπω
2πω
(g)
√
1 + sin2 ω
ω
4 (a) 1.5811 (b) 1.3466 (c) 2.2724
REVIEWEXERCISES15
1 Find the average value ofthe following functions
acrossthe specified interval:
(a) f (t) = 3 −t across[0,4]
(b) f(t) =t 2 −2across[1,3]
(c) f(t)=t+ 1 across [1,4]
t
(d) f(t) = √ t +1 across [0,4]
(e) f (t) =t 2/3 across [0,1]
2 Calculatethe average value ofthe following:
[ ]
(a) f (t) = 2sin2t across 0, π 2
[ ]
(b) f (t) =Asin4t across
0, π 2
(c) f (t) = sint +cost across [0,π]
( ) [ ]
t
(d) f(t) =cos across 0, π 2 2
(e) f (t) = sintcost across [0, π]
3 Calculate the averagevalue ofthe following
functions:
(a) f (t) =Ae kt across [0,1]
(b) f(t) = 1
e 3t across [0,2]
(c) f (t) = 3 −e −t across [1,3]
(d) f (t) = e t +e −t across[0,2]
(e) f (t) =t +e t across[0,2]
4 Find the average andr.m.s.values of
Acost +Bsint across
(a) [0,2π]
(b) [0, π]
5 Find the r.m.s.values ofthe functions
in Question1.
6 Find the r.m.s.values ofthe functionsin
Question2.
7 Find the r.m.s.values ofthe functions
in Question3.
Solutions
1 (a) 1
7
(b)
3
(c) 2.9621
7 3
(d) (e)
3 5
4
2
2 (a) (b) 0 (c)
π
π
(d) 0.9003
(
(e) 0
)
e k −1
3 (a)A
k
(b) 0.1663
(c) 2.8410 (d) 3.6269
(e) 4.1945
4 (a) average = 0, r.m.s. =
√
A 2 +B 2
2
(b) average = 2B π ,r.m.s. = √
A 2 +B 2
5 (a) 1.5275 (b) 3.2965 (c) 3.0414
(d) 2.3805 (e) 0.6547
6 (a) 1.4142 (b)
2
A
√
2
(c) 1
(d) 0.9046 (e) 0.3536
√
e 2k −1
7 (a)A (b) 0.2887
2k
(c) 2.8423 (d) 3.9554
(e) 4.8085

16 Furthertopicsin
integration
Contents 16.1 Introduction 480
16.2 Orthogonalfunctions 480
16.3 Improperintegrals 483
16.4 Integralpropertiesofthedeltafunction 489
16.5 Integrationofpiecewisecontinuousfunctions 491
16.6 Integrationofvectors 493
Reviewexercises16 494
16.1 INTRODUCTION
Thischapterexaminessomefurthertopicsinintegration.Orthogonalfunctionsareintroduced
in Section 16.2. These functions are used extensively in Fourier analysis
(see Chapter 23). Some integrals have one or two infinite limits of integration, or have
an integrand which becomes infinite at particular points in the interval of integration.
Such integrals are termed ‘improper’ and require special treatment. They are used extensivelyinthetheoryofLaplaceandFouriertransforms.TheDiracdeltafunction,
δ(t),
hasbeenintroducedinChapter2.TheintegralpropertiesareexaminedinSection16.4.
The chapter concludes with the integration of piecewise continuous functions and the
integration of vectors.
16.2 ORTHOGONALFUNCTIONS
Two functions f (x)andg(x) aresaidtobe orthogonal over the interval [a,b] if
∫ b
a
f(x)g(x)dx =0
Toshowthattwofunctionsareorthogonalwemustdemonstratethattheintegraloftheir
product over the interval of interest iszero.

[
Example16.1 Show that f (x) =xandg(x) =x −1 areorthogonal on 0, 3 ]
.
2
16.2 Orthogonal functions 481
Solution
] 3/2
∫ 3/2 ∫ 3/2
[ x
x(x−1)dx= x 2 3
−xdx=
0
0 3 − x2
2
0
[
Hence f andgare orthogonal over the interval 0, 3 ]
.
2
= 9 8 − 9 8 = 0
Clearlyfunctionsmaybeorthogonaloveroneintervalbutnotorthogonaloverothers.
For example,
∫ 1
0
x(x−1)dx≠0
and soxandx −1 arenotorthogonal over [0,1].
Example16.2 Show f (t) = 1,g(t) = sint andh(t) = cost aremutually orthogonal over [−π,π].
Solution We are required toshow thatany pairoffunctions isorthogonal over [−π,π].
∫ π
−π
1sintdt =[−cost] π −π = −cosπ +cos(−π)
∫ π
−π
= −(−1) + (−1) =0
1costdt =[sint] π −π = sinπ −sin(−π) =0
Usingthe trigonometricidentitysin2A = 2sinAcosA, wecan write
∫ π
−π
sintcostdt =
∫ π
−π
[
1 cos(2t)
2 sin(2t)dt = − 4
cos(2π) −cos(−2π)
= − = 0
4
] π
−π
Hence the functions 1,sint, cost form an orthogonal set over [−π,π].
ThesetofExample 16.2 maybeextendedto
{1,sint,cost,sin(2t),cos(2t),sin(3t),cos(3t),...,sin(nt),cos(nt)}
n ∈ N
Example16.3 Verifythat {1,sint,cost,sin(2t),cos(2t),...} formsanorthogonal setover [−π,π].
Solution Supposen,m ∈ N. We must show that all combinations of 1,sinnt, sinmt, cosnt and
cosmt are orthogonal.
∫ π
[ −cos(nt) −cos(nπ) +cos(−nπ)
1sin(nt)dt =
= = 0
n n
−π
] π
−π

482 Chapter 16 Further topics in integration
Inasimilarmanner, itiseasytoshow
∫ π
1cos(nt)dt = 0
−π
Also,usingthe trigonometric identities inSection 3.6
∫ π
=
−πcos(nt)sin(mt)dt 1 ∫ π
sin(n +m)t −sin(n −m)tdt
2 −π
We have seen that ∫ π
−πsinnt dt = 0 for any n ∈ N. Noting that (n +m) ∈ N and
(n −m) ∈ N,weseethat
∫ π
−π
sin(n +m)t −sin(n −m)tdt =0
Itis left as anexercise for the reader toshow that
∫ π
−π
∫ π
−π
sinntsinmtdt =0 n≠m
cosntcosmtdt =0
n≠m
The functions thus form an orthogonal set across [−π,π].
TheresultofExample 16.3 can beextended:
{1,sint,cost,sin2t,cos2t,...}isanorthogonalsetoveranyintervaloflength2π.
More generally:
{
1,sin
( 2πt
T
)
,cos
( 2πt
T
)
,sin
( 4πt
T
)
,cos
( 4πt
T
) }
,...
isanorthogonalsetoveranyintervaloflengthT.Inparticular,thesetisorthogonal
[
over [0,T] and − T ]
2 ,T .
2
These results areused extensively inFourier analysis.
Example16.4 Find
(a)
(b)
∫ π
−π
∫ π
−π
sin 2 (nt)dt n∈ Z n ≠0
cos 2 (nt)dt n∈ Z n ≠0

Solution (a) We use the trigonometric identity
(b)
sin 2 nt = 1 −cos2nt
2
toget
∫ π
sin 2 (nt)dt =
−π
∫ π
−π
1 −cos2nt
2
dt =
using the orthogonal properties of cos(nt).
∫ π
−π
cos 2 (nt)dt =
∫ π
−π
1 −sin 2 (nt)dt
=2π−π=π
∫ π
−π
16.3 Improper integrals 483
1
2 dt=π
It is a simple extension to show that integrating sin 2 (nt) or cos 2 (nt) over any interval
of length 2π yields the same result, namely π. It is also possible to extend the result of
Example 16.4 toshow
∫ T/2
−T/2
sin 2 ( 2nπt
T
)
dt =
Finally, integrating sin 2 ( 2nπt
T
the same result,thatis T 2 .
∫ T/2
−T/2
)
( ) 2nπt
cos 2 dt = T T 2
)
and cos 2 ( 2nπt
T
n∈Z
n≠0
over any interval of lengthT gives
EXERCISES16.2
1 Show f (x) =x 2 andg(x) = 1 −xareorthogonal
[ ]
across
0, 4 3
.
2 Show f(x) = 1 x andg(x) =x2 are orthogonal over
[−k,k].
3 (a)Showf(t)=1−tandg(t)=1+tare
orthogonalover [0, √ 3].
(b)Findanotherinterval over which f (t)andg(t)are
orthogonal.
4 Show f (t) = e t andg(t) = 1 −e −2t are orthogonal
across [−1,1].
5 Show f(x) = √ xandg(x)=1− √ x are orthogonal
[ ]
on 0, 16 .
9
Solutions
3 (b) [− √ 3,0]
16.3 IMPROPERINTEGRALS
There aretwo caseswhen evaluation ofanintegral needs specialcare:
(1) one,orboth,ofthe limits ofanintegral areinfinite;
(2) the integrand becomesinfiniteatone,ormore,points ofthe interval ofintegration.

484 Chapter 16 Further topics in integration
If either (1) or (2) is true the integral is called an improper integral. Evaluation of
improper integrals involves the use of limits.
∫ ∞
1
Example16.5 Evaluate
2 t dt. 2
Solution
∫ ∞
2
1
[−
t dt = 1 ] ∞ 2 t
2
To evaluate − 1 atthe upper limitwe consider lim
t −1 . Clearly the limitis0.Hence,
t→∞ t
∫ ∞
1
(−
t dt=0− 1 )
= 1 2 2 2
2
∫ 1
Example16.6 Evaluate e 2x dx.
−∞
Solution
∫ 1
−∞
[ e
e 2x 2x
dx =
2
] 1
−∞
We need toevaluate lim
x→−∞
∫ 1
−∞
e 2x
. This limitis 0.So,
2
[ ] e
e 2x 2x 1
dx = = e2
2
−∞
2 −0=3.69
Engineeringapplication16.1
Capacitorsinseries
Engineers are often called upon to simplify an electronic circuit in order to make
it easier to analyse. One of the arrangements frequently met is that of two or more
capacitors connected together in series. It is useful to be able to replace this configuration
by a single capacitor with a capacitance value equivalent to that of the
original capacitors in series. Derive an expression for the equivalent capacitance of
two capacitors connected togetherinseries (see Figure 16.1).
Solution
In Engineering application 13.2 we obtained an expression for the voltage across a
capacitor. Thiswas
v = 1 ∫
idt
C

16.3 Improper integrals 485
i
C 1 C 2
y 1
y
y 2
Figure16.1
Two capacitors
connected in series.
This can be written as a definite integral to give the voltage expression across the
capacitor atageneral pointintime,t. The expression is
v = 1 C
∫ t
−∞
idt
Now consider the situation depicted in Figure 16.1. Writing an equation for each of
the capacitors gives
v 1
= 1 C 1
∫ t
−∞
idt
v 2
= 1 C 2
∫ t
By Kirchhoff’s voltage law, v = v 1
+ v 2
and so
v = 1 ∫ t
idt+ 1 ∫ t
( 1
idt = + 1 ) ∫ t
idt
C 1 −∞ C 2 −∞ C 1
C 2 −∞
−∞
idt
Therefore,thetwocapacitorscanbereplacedbyanequivalentcapacitance,C,given
by
so that
1
C = 1 C 1
+ 1 C 2
= C 1 +C 2
C 1
C 2
C = C 1 C 2
C 1
+C 2
The result is easily generalised for more than two capacitors. For example, for three
capacitors wehave
1
C = 1 + 1 + 1 C 1
C 2
C 3
so that
C
C = 1
C 2
C 3
C 2
C 3
+C 1
C 3
+C 1
C 2
You may wish toprove this result.
∫ ∞
2
Example16.7 Evaluate
3 2t+1 − 1 t dt.
Solution
∫ ∞
3
2
2t+1 − 1 t dt =[ln|2t +1| −ln |t|]∞ 3
[
] = ln
2t+1
∞ [
∣ t ∣ = ln
∣ 2 + 1 3
t ∣
[ (
= lim ln 2 + 1 )]
−ln 7
t→∞ t 3
] ∞
3
=ln2−ln 7 3 = ln 6 7 = −0.1542

486 Chapter 16 Further topics in integration
Example16.8 Evaluate
∫ ∞
1
sintdt.
Solution
∫ ∞
1
sintdt =[−cost] ∞ 1
Nowlim t→∞
(−cost)doesnotexist,thatisthefunctioncost doesnotapproachalimit
ast → ∞, and so the integral cannot be evaluated. We say the integraldiverges.
∫ 1
1
Example16.9 Evaluate √ dx.
0 x
1
Solution Theintegrand, √ ,becomesinfinitewhenx = 0,whichisintheintervalofintegration.
x
The point x = 0 is ‘removed’ from the interval. We consider
slightly greater than 0,and then letb→0 + . Now,
∫ 1
b
Then,
∫ 1
0
1
√ x
dx = [2 √ x] 1 b =2−2√ b
∫
1 1
√ dx = lim x b→0 + b
1
√ x
dx = lim
b→0 +(2−2√ b)=2
The improper integral exists and has value 2.
∫ 1
b
1
√ x
dx where b is
∫ 2
1
Example16.10 Determinewhether the integral dx exists ornot.
0 x
Solution As in Example 16.9 the integrand is not defined atx = 0, so we consider
b >0andthenletb →0 + .
So,
∫ 2
b
1
x dx = [ln|x|]2 b =ln2−lnb
(∫ 2
)
1
lim
b→0 b x dx = lim(ln2−lnb)
b→0
Since lim b→0
lnbdoes notexistthe integral diverges.
∫ 2
b
1
dx for
x
∫ 2
1
Example16.11 Evaluate dx ifpossible.
−1 x
Solution We ‘remove’
∫
the point x = 0 where the integrand becomes infinite and consider two
b
∫
1
2
integrals:
−1 x dx wherebis slightly smaller than 0, and 1
dx wherecis slightly
c x ∫ 2
largerthan0.Iftheseintegralsexistasb→0 − andc → 0 + 1
then
x dxconverges.If
−1

∫ 2
either of the integrals failstoconverge then
∫ b
−1
1
dx = ln |b| −ln |−1| = ln|b|
x
∫ b
1
lim dx = lim
b→0 − −1 x b→0 −(ln|b|)
This limitfailstoexist and so
∫ 2
−1
−1
1
dx diverges.
x
16.3 Improper integrals 487
1
dx diverges. Now,
x
Engineeringapplication16.2
Energystoredinacapacitor
Acapacitorprovidesausefulmeansofstoringenergy.Thisenergycanbedischarged
by connecting the capacitor in series with a load resistor and closing a switch. The
stored electrical energy is converted into heat energy as a result of electrical current
flowing through the resistor. Consider the circuit shown in Figure 16.2 which
consists of a capacitor,C F, connected in series with a resistor with valueRand
isolatedbymeansofaswitch,S.Wewishtocalculatetheamountofenergystoredin
the capacitor. The switch is closed att = 0 and a current,i, flows in the circuit. We
have already seen in Chapter 2 that for such a case the time-varying voltage across
the capacitordecaysexponentially and isgiven by
v =Ve −t/RC
So,usingOhm’s law
i = v R = Ve−t/RC
R
S
i
C
R
Figure16.2
The capacitor is dischargedbyclosingthe switch.
Now the effect of closing the switch is to allow the energy stored in the capacitor to
be dissipated in the resistor. Therefore, if the total energy dissipated in the resistor
is calculated then this will allow the energy stored in the capacitor to be obtained.
However, the energy dissipation rate, that is power dissipated, is not a constant for
theresistorbutdependsonthecurrentflowingthroughit.Thetotalenergydissipated,
E, isgiven by
E =
∫ ∞
0
P(t)dt
whereP(t) is the power dissipated in the resistor at timet. This equation has been
discussedinEngineeringapplication 13.3.Now,
P=i 2 R= RV2 e −2t/RC
R 2
= V2 e −2t/RC
R
➔

488 Chapter 16 Further topics in integration
Now
E =
∫ ∞
0
= V2 RC
−2R
V 2 e −2t/RC
R
lim
t→∞ e−2t/RC = 0
[
e
−2t/RC ] ∞
0
dt = V2
R
∫ ∞
0
e −2t/RC dt
and so the energy storedinthe capacitor isgiven by
E = CV2
2
Example16.12 Find
∫ ∞
0
e −st sintdt s>0
Solution Using integration by parts, withu = e −st and dv = sint, wehave
dt
∫ ∞
0
e −st sintdt = [ −e −st cost ] ∫ ∞
∞
−s e −st costdt
0
Considerthefirsttermonther.h.s.Weneedtoevaluate [ −e −st cost ] ast → ∞andwhen
t = 0.Notethat −e −st cost → 0ast → ∞becausewearegiventhatsispositive.When
t = 0, −e −st cost evaluates to −1, and so
∫ ∞
0
e −st sintdt =1−s
∫ ∞
0
0
e −st costdt
Integrating by parts for a second time yields
∫ ∞
{ [e
e −st sintdt =1−s
−st sint ] ∫ ∞
∞
+s 0
0
∫ ∞
=1−s 2 e −st sintdt
0
0
}
e −st sintdt
because [ e −st sint ] ∞
0 evaluatestozeroatbothlimits.Atthisstagethereadermightsuspectthatwehavegonearoundinacircleandstillneedtoevaluatetheoriginalintegral.
However, somealgebraic manipulation yields the required result.We have
∫ ∞
0
∫ ∞
e −st sintdt +s 2 e −st sintdt = 1
(1+s 2 )
0
∫ ∞
0
∫ ∞
0
e −st sintdt = 1
e −st sintdt = 1
1+s 2

16.4 Integral properties of the delta function 489
EXERCISES16.3
1 Evaluate, ifpossible,
∫ ∞
(a) e −t dt
0
∫ ∞
(b) e −kt dt k is aconstant,k > 0
0
∫ ∞ 1
(c)
1 x dx
∫ ∞ 1
(d)
1 x 2 dx
∫ 3 1
(e)
1 x −2 dx
2 Evaluate the following integrals wherepossible:
∫ 4
∫
3 3
(a)
0 x −2 dx (b) 1
0 x −1 + 1
x −2 dx
∫ 2
∫
1
∞
(c)
0 x 2 −1 dx (d) sin3tdt
0
∫ 3
(e) xe x dx
−∞
3 Find
∫ ∞
e −st costdt s>0
0
Solutions
1
1 (a) 1 (b) (c) does not exist
k
(d) 1 (e) does not exist
2 (a) does notexist (b) does notexist
3
(c) does notexist
(e) 2e 3
s
s 2 +1
(d) doesnot exist
16.4 INTEGRALPROPERTIESOFTHEDELTAFUNCTION
Thedeltafunction, δ(t −d),wasintroducedinChapter2.Thefunctionisdefinedtobea
rectanglewhoseareais1inthelimitasthebaselengthtendsto0andastheheighttends
toinfinity.Sometimesweneedtointegratethedeltafunction.Inparticular,weconsider
the improper integral
∫ ∞
−∞
δ(t −d)dt
Theintegral gives the area under the functionand this isdefinedtobe1.Hence,
∫ ∞
−∞
δ(t−d)dt =1
InChapter21 weneedtoconsiderthe improper integral
∫ ∞
−∞
f(t)δ(t −d)dt
where f (t) is some known function of time. The delta function δ(t −d) is zero everywhere
except att =d. Whent =d, then f (t)has a value f (d).Hence,
∫ ∞
−∞
f(t)δ(t −d)dt =
∫ ∞
−∞
f(d)δ(t −d)dt = f(d)
∫ ∞
−∞
δ(t −d)dt

490 Chapter 16 Further topics in integration
∫ ∞
since f (d)is a constant. But
∫ ∞
−∞
∫ ∞
−∞
The result
∫ ∞
−∞
−∞
f(t)δ(t −d)dt = f(d)
∫ ∞
−∞
δ(t−d)dt=1
f(t)δ(t −d)dt=f(d)
f(t)δ(t −d)dt = f(d)
δ(t −d)dt =1andhence
is known as the sifting property of the delta function. By multiplying a function, f (t),
by δ(t −d) and integrating from −∞ to ∞ we siftfrom the function the value f (d).
Example16.13 Evaluate the following integrals:
(a)
∫ ∞
−∞
Solution (a) We use
∫ ∞
t 2 δ(t −2)dt
−∞
(b)
∫ ∞
f(t)δ(t −d)dt = f(d)
withf(t) =t 2 andd = 2.Hence
∫ ∞
−∞
0
e t δ(t −1)dt
t 2 δ(t −2)dt=f(2)=2 2 =4
(b) We note thatthe expression e t δ(t −1) is0everywhere except att = 1.Hence
∫ ∞
0
e t δ(t−1)dt =
∫ ∞
−∞
e t δ(t −1)dt
EXERCISES16.4
Using ∫ ∞
f(t)δ(t −d)dt = f(d)
−∞
with f (t) = e t andd = 1 gives
∫ ∞
0
e t δ(t−1)dt =f(1)=e 1 =e
1 Evaluate
∫ ∞
(a) e t δ(t)dt
−∞
∫ ∞
(b) e t δ(t −4)dt
−∞
∫ ∞
(c) e t δ(t +3)dt
−∞
∫ ∞
(d)4 t 2 δ(t −3)dt
−∞

16.5 Integration of piecewise continuous functions 491
∫ ∞ (1+t)δ(t −1)
(e) dt
−∞ 2
∫ ∞
(f) e −kt δ(t)dt
−∞
∫ ∞
(g) e −kt δ(t −a)dt
−∞
∫ ∞
(h) e −k(t−a) δ(t −a)dt
−∞
2 Evaluate the following integrals:
∫ ∞
(a) (sint)δ(t −2)dt
−∞
∫ ∞
(b) e −t δ(t +1)dt
−∞
∫ 0
(c) e −t δ(t +3)dt
−∞
∫ 10
(d) x 3 δ(x−2)dx
0
∫ 1
(e) x 2 δ(x+2)dx
−1
3 Evaluate the following:
∫ ∞
(a) t(sin2t)δ(t −3)dt
−∞
∫ ∞
(b) δ(t+1)−δ(t−1)dt
0
∫ ∞
(c) δ(t −d)dt
0 ∫ a
(d) δ(t −d)dt
−a
Solutions
1 (a) 1 (b) 54.5982
(c) 4.9787 ×10 −2 (d) 36
(e) 1 (f) 1
(g) e −ak (h) 1
2 (a) 0.9093
(b) 2.7183
(c) 20.0855
(d) 8
(e) 0
3 (a) −0.8382
(b) −1
(c) 1 ifd 0,0otherwise
(d) 1 if −a d a,0otherwise
16.5 INTEGRATIONOFPIECEWISECONTINUOUSFUNCTIONS
Integration of piecewise continuous functions is illustrated in Example 16.14. If a discontinuity
occurs within the limits of integration then the interval is divided into subintervals
so thatthe integrand iscontinuouson eachsub-interval.
Example16.14 Given
f(t)=
{
2 0 t<1
t 2
evaluate ∫ 3
0 f(t)dt.
1<t3
Solution The function, f (t), is piecewise continuous with a discontinuity att = 1. The function
is illustrated in Figure 16.3. The discontinuity occurs within the limits of integration.
We split the interval ofintegration atthe discontinuitythus:
∫ 3
0
f(t)dt =
∫ 1
0
f(t)dt+
∫ 3
1
f(t)dt

492 Chapter 16 Further topics in integration
f (t)
2
1 3 t
Figure16.3
Thefunction f (t) =
{ 2 0t<1
t 2 1<t 3.
On the intervals (0, 1) and (1, 3), f (t)iscontinuous, so
∫ 3
0
fdt=
=
∫ 1
0
[
2t
2dt+
] 1
0
∫ 3
1
t 2 dt
[ ] t
3 3
+ = 2 +
3
1
[
9 − 1 ]
= 32 3 3
EXERCISES16.5
1 Given
⎧
⎪⎨
3 −1t<1
f(t)= 2t 1t2
⎪⎩
t 2 2<t3
evaluate
∫ 1 ∫ 1.5
(a) f dt (b) f dt
−1
−0.5
∫ 2.5 ∫ 3
(c) f dt (d) f dt
0
−1
2 Given
⎧
⎨3t
0t<3
g(t) = 15−2t 3t<4
⎩
6 4t6
evaluate
∫ 2 ∫ 4
(a) g(t)dt (b) g(t)dt
0 2
∫ 5
∫ 6
(c) g(t)dt (d) g(t)dt
3
0
∫ 4.5
(e) g(t)dt
3.5
3 Givenu(t)isthe unitstepfunction, evaluate
∫ 4
(a) u(t)dt
0
∫ 2
(b) u(t)dt
−3
∫ 4
(c) 2u(t +1)dt
−2
∫ 2
(d) tu(t)dt
−1
∫ 4
(e) e kt u(t −3)dt k constant
0
Solutions
1 (a) 6 (b) 5.75
(c) 8.5417 (d) 15.3333
2 (a) 6 (b) 15.5 (c) 14
(d) 33.5 (e) 6.75
3 (a)4 (b)2
(c) 10 (d) 2
e 4k −e 3k
(e)
k

16.6 Integration of vectors 493
16.6 INTEGRATIONOFVECTORS
Ifavectordependsupontimet,itisoftennecessarytointegrateitwithrespecttotime.
Recall that i, j and k are constant vectors and are treated thus in any integration. Hence
the integral
∫
I = (f(t)i+g(t)j+h(t)k)dt
issimplyevaluated as three scalar integrals, and so
( ∫ ) ( ∫ ) ( ∫ )
I = f(t)dt i + g(t)dt j + h(t)dt k
Example16.15 Ifr = 3ti +t 2 j + (1 +2t)k, evaluate ∫ 1
0 rdt.
Solution
∫ 1
0
( ∫ ) (
1 ∫ ) (
1 ∫ )
1
rdt= 3tdt i + t 2 dt j + 1+2tdt k
0
[ 3t
2
=
2
0
] 1 [ ] t
3 1
i + j +
0
3
0
0
[
t +t 2 ] 1
0k = 3 2 i + 1 3 j+2k
EXERCISES16.6
1 Givenr = 3sinti−costj+(2−t)k,evaluate ∫ π
0 rdt.
2 Given v = i −3j +k,evaluate
(a)
∫ 1
vdt
0
(b)
∫ 2
vdt
0
3 Thevector, a,is defined by
a=t 2 i+e −t j+tk
Evaluate
∫ 1
(a) adt
0
(b)
∫ 3
adt
2
(c)
∫ 4
adt
1
4 Letaandbbe two three-dimensional vectors. Isthe
following true?
∫ t2
∫ t2
∫ t2
adt× bdt = a×bdt
t 1 t 1 t 1
Recallthat ×denotes the vectorproduct.
Solutions
1 6i +1.348k
2 (a) i−3j+k (b) 2i−6j+2k
3 (a)0.333i +0.632j +0.5k
4 no
(b)6.333i +0.0855j +2.5k
(c)21i +0.3496j +7.5k

494 Chapter 16 Further topics in integration
REVIEWEXERCISES16
1 Show f (x) =x n andg(x) =x m areorthogonalon
[−k,k] ifn +mis an odd number.
2 Show f (t) = sint andg(t) = cost are orthogonal on
[a,a +nπ].
3 Show f (t) = sinht andg(t) = cosht are orthogonal
over [−k,k].
4 Determine the values ofkforwhich ∫ ∞
0 t k dt exists.
5 Evaluate ifpossible
∫ ∞
(a) u(t)dt
∫0
∞
(b) e −1000t dt
−∞
∫ 0 1
(c)
−2 x +1 dx
∫ ∞
(d) u(t)e −t dt
−∞
∫ 2 1
(e)
−2 x 2 −1 dx
(Note thatu(t)is the unitstepfunction.)
6 Find the values ofkforwhich ∫ ∞
0 e kt dt exists.
7 Evaluate
∫ ∞
(a) (t 2 +1)δ(t −1)dt
−∞
∫ ∞
(b) te t δ(t −2)dt
−∞
∫ ∞
(c) (t 2 +t +2)δ(t −1)dt
−∞
∫ ∞ δ(t +2)
(d)
−∞ t 2 +1 dt
∫ ∞
(e) δ(t +1)δ(t +2)dt
−∞
8 (a) Evaluate
∫ ∞
(t 2 +1)δ(2t)dt
−∞
[Hint: substitutez = 2t.]
(b) Show
∫ ∞
f(t)δ(nt)dt = 1 f(0), n>0
−∞ n
9 Evaluate
∫ ∞
(a) (1+t)δ(t −2)dt
0
∫ ∞
(b) tδ(1+t)dt
−∞
∫ ∞
(c) (1+t)δ(−t)dt
−∞
∫ 0
(d) δ(t−6)+δ(t+6)dt
−∞
∫ 4k
(e) δ(t +k)+δ(t +3k)+δ(t +5k)dt
−2k
k > 0
10 Thefunctiong(t)ispiecewise continuous and
defined by
{ 2t 0t1
g(t) =
3 1<t2
Evaluate
∫ 1
(a) g(t)dt
0
∫ 1.5
(b) g(t)dt
0
∫ 1.7
(c) g(t)dt
0.5
∫ 2
(d) g(t)dt
1
∫ 1.5
(e) g(t)dt
1.3
11 Given
⎧
⎨1+t −1t<3
f(t)= t−1 3t4
⎩
0 otherwise
evaluate
∫ 2
(a) f(t)dt
−1
∫ 4
(b) f(t)dt
−1
∫ 5
(c) f(t)dt
0
∫ ∞
(d) f(t)δ(t −2)dt
−∞
∫ ∞
(e) f(t)u(t)dt
−∞

Review exercises 16 495
12 Given
14 Given
⎧
⎨t 2 v(t) =2ti+(3−t 2 )j+t 3 k
−2t2
h(t) = t
⎩
3 find
2<t3
∫ 1
4 3<t5
(a) vdt
0
evaluate
∫
∫ 2
2
(b) vdt
(a) h(t)dt
1
−1
∫
∫ 2
2.5
(c) vdt
(b) h(t)dt
0
0
∫ 4
15 Evaluate the following integrals:
(c) h(t)+1dt
∫
2
3
∫ 4
(a) |t|dt
−1
(d) h(t +1)dt
∫
0
2
∫ 2
(b) |t+2|dt
−3
(e) h(2t)dt
∫
−1
2
13 Ifa =ti−2tj+3tk,find ∫ 1
0 adt. (c) |3t −1|dt
0
Solutions
4 k<−1
5 (a) does notexist
(b) does notexist
(c) does notexist
(d) 1
(e) does notexist
6 k<0
7 (a) 2 (b) 14.7781 (c) 4
(d) 0.2 (e) 0
8 (a) 1 2
9 (a) 3 (b) −1 (c) 1
(d) 1 (e) 1
5
10 (a) 1 (b)
2
(d) 3 (e) 0.6
(c) 2.85
11 (a) 4.5 (b) 10.5 (c) 10
(d) 3 (e) 10
12 (a) 3
(b) 8.4323
(c) 22.25
(d) 26.5833
(e) 12.7917
13 0.5i −j+1.5k
14 (a) i+ 8 3 j + 1 4 k
(b) 3i + 2 3 j + 15
4 k
(c) 4i + 10
3 j+4k
15 (a) 5 (b) 8.5 (c)
13
3

17 Numericalintegration
Contents 17.1 Introduction 496
17.2 Trapeziumrule 496
17.3 Simpson’srule 500
Reviewexercises17 505
17.1 INTRODUCTION
Many functions, for example sinx 2 and ex
, cannot be integrated analytically. Integrationofsuchfunctionsmustbeperformednumerically.Thissectionoutlinestwosimple
x
numericaltechniques--thetrapeziumruleandSimpson’srule.Moresophisticatedones
exist and there are many excellent software packages available which implement these
methods.
17.2 TRAPEZIUMRULE
We wish to find the area undery(x), fromx = a tox = b, that is we wish to evaluate
∫ b
ydx. The required area is divided intonstrips, each of widthh. Note that the width,
a
h, of each strip is given byh = b −a . Each strip is then approximated by a trapezium.
n
AtypicaltrapeziumisshowninFigure17.1.Theareaofthetrapeziumis 1 2 h[y i +y i+1 ].
Summing the areas of all the trapezia will yield an approximation tothe total area:
area oftrapezia = h 2 (y 0 +y 1 )+h 2 (y 1 +y 2 )+···+h 2 (y n−1 +y n )
= h 2 (y 0 +2y 1 +2y 2 +2y 3 +···+2y n−1 +y n )

17.2 Trapezium rule 497
y
{ {
y 0 y n
y i y i +1
x i
a
x i +1 b
h
{
{
x
Figure17.1
Each stripis approximated by atrapezium.
If the number of strips is increased, that ishis decreased, then the accuracy of the approximation
is increased.
Table17.1
Example17.1 Usethe trapezium ruletoestimate
(a) a strip widthof0.2
∫ 1.3
0.5
e (x2) dx using
(b) a strip widthof0.1.
Solution (a) Table 17.1 lists values ofxand correspondingvalues ofe (x2) .
We notethath = 0.2.Usingthe trapeziumrulewefind
x
0.5
y=e (x2 )
1.2840 =y 0
sum ofareasoftrapezia = 0.2 {1.2840 +2(1.6323) +2(2.2479)
2
+2(3.3535) +5.4195}
0.7 1.6323 =y 1
= 2.117
0.9 2.2479 =y 2
Table17.2
1.1 3.3535 =y 3 Hence
1.3 5.4195 =y 4 ∫ 1.3
x y=e (x2 )
0.5
e (x2) dx ≈ 2.117
x 0 = 0.5 1.2840 =y 0
x 1 = 0.6 1.4333 =y 1
x 2 = 0.7 1.6323 =y 2
x 3 = 0.8 1.8965 =y 3
x 4 = 0.9 2.2479 =y 4
x 5 = 1.0 2.7183 =y 5
x 6 = 1.1 3.3535 =y 6
x 7 = 1.2 4.2207 =y 7
x 8 = 1.3 5.4195 =y 8
(b) Table 17.2 listsxvalues and correspondingvalues ofe (x2) .
Using Table17.2, wefind
Hence
sum ofareasoftrapezia = 0.1 {1.2840 +2(1.4333) +2(1.6323) + ···
2
+2(4.2207) +5.4195}
∫ 1.3
0.5
e (x2) dx ≈ 2.085.
= 2.085
Dividingtheinterval[0.5,1.3]intostripsofwidth0.1resultsinamoreaccurateestimate
thanusingstripsofwidth0.2.

498 Chapter 17 Numerical integration
Engineeringapplication17.1
Distancetravelledbyarocket
A rocket is released and travels at a variable speed v. A motion sensor on the rocket
measures this speed and the value is sampled by an onboard computer at 1 second
intervals. The computer is required to calculate the distance travelled by the rocket
andrelaythevaluetoagroundstationatregularintervals.Table17.3recordsvalues
of the measurements taken by the computer during the first 10 seconds of flight.
Assumingthatthecomputerusesthetrapeziumruletoestimatethedistancetravelled
by the rocket, calculate the value that the computer will relay to the ground station
after 10 seconds.
Table17.3
Time(s) Speed(ms −1 )
0 10.1
1 17.2
2 24.4
3 29.2
4 34.6
5 41.2
6 50.9
7 57.8
8 60.3
9 61.2
10 62.1
Solution
We know that v = ds where v =speed ands = distance travelled intimet. So
dt
s =
∫ t
0
We estimate the value of this integral using the trapezium rule. We choose a strip of
widthh = 1 as this is the time interval at which data is collected by the computer.
Therefore
s =
∫ 10
0
vdt
vdt≈ 1 {10.1 +2(17.2 +24.4 +29.2 +34.6
2
+41.2 +50.9 +57.8 +60.3 +61.2) +62.1}
= 1 2 (825.8)
= 412.9
The distance travelled after 10 seconds isapproximately 412.9 m.
17.2.1 Errorduetothetrapeziumrule
Suppose we wish to estimate ∫ b
f dx using the trapezium rule. The interval [a,b] is
a
dividedintonequal strips, eachofwidthh = b −a
n .

17.2 Trapezium rule 499
The difference between the estimated value of the integral and the true value of the
integral istheerror. So
error = estimated value −truevalue
Weareabletofindthemaximumvalueof |error|.Firstlywecalculatethesecondderivative
of f, that is f ′′ . Suppose that |f ′′ | is never greater than some value,M, throughout
the interval [a,b], thatis
|f ′′ | M forallxvalues on [a,b]
WesaythatM isanupperboundfor |f ′′ |on[a,b].Thentheerrorduetothetrapezium
ruleissuch that
|error | (b−a) h 2 M
12
The expression (b−a) h 2 M is an upper bound for the error. Note that the error
12
depends upon h 2 : if the strip width, h, is halved the error reduces by a factor of 4; if
hisdivided by 10 the erroris divided by 100.
Example17.2 Find an upper bound for the error in the estimates calculated in Example 17.1. Hence
find upper and lower boundsforthe truevalue of ∫ 1.3
0.5 e(x2) dx.
Solution We have f = e (x2) . Then
f ′ = 2xe (x2 )
and f ′′ = 2e (x2) (1+2x 2 )
We note that f ′′ is increasing on [0.5,1.3] and so its maximum value is obtained at
x = 1.3. Thus the maximum value of f ′′ on [0.5,1.3] is 2e (1.3)2 [1 + 2(1.3) 2 ]. Noting
that2e (1.3)2 [1 +2(1.3) 2 ] = 47.47weseethat48isanupperboundfor f ′′ on[0.5,1.3],
thatisM = 48.
We notethata = 0.5,b= 1.3.Thus
|error | (0.8)h2 (48)
= 3.2h 2
12
(a) InExample 17.1(a),h = 0.2 and so
| error | 3.2(0.2) 2 = 0.128
Thus an upper bound for the error is0.128. We have
−0.128 error 0.128
The estimated value of the integral is2.117 and so
that is
2.117 −0.128 truevalue of integral 2.117 +0.128
1.989
∫ 1.3
0.5
e (x2) dx 2.245
An upper bound for ∫ 1.3
0.5 e(x2) dx is2.245 and a lower bound is 1.989.

500 Chapter 17 Numerical integration
(b) InExample 17.1(b),h = 0.1 and so
|error | 3.2(0.1) 2 = 0.032
Hence an upper bound forthe erroris 0.032. Now
−0.032 error 0.032
The estimated value of the integral is2.085 and so
2.085 −0.032
thatis
2.053
∫ 1.3
0.5
∫ 1.3
0.5
e (x2) dx 2.117
e (x2) dx 2.085 +0.032
EXERCISES17.2
1 Estimate the followingdefinite integrals usingthe
trapezium rule:
∫ 1
(a) sin(t 2 )dt useh = 0.2
0
Solutions
∫ 1.2 e x
(b)
1 x dx
use five strips
1 (a) 0.3139 (b) 0.5467
TechnicalComputingExercises17.2
1 (a) Plotthe second derivative of f (t) = sin(t 2 ) for
0 t 1. Useyour graph to findan upperbound
for f ′′ (t)for0 t 1. Hence findan upper
bound forthe error in Question1(a) in
Exercises17.2.Stateupperandlower boundsfor
the integral given in the question.
(b) Repeat(a)with f (x) = ex
x for1x1.2.
Hence findan upper bound forthe errorin
Question1(b) in Exercises17.2. Stateupper and
lower bounds forthe integral given in the
question.
17.3 SIMPSON’SRULE
Inthetrapeziumrulethecurvey(x)isapproximatedbyaseriesofstraightlinesegments.
InSimpson’srulethecurveisapproximatedbyaseriesofquadraticcurvesasshownin
Figure 17.2. The area is divided into an even number of strips of equal width. Consider
thefirstpairofstrips.AquadraticcurveisfittedthroughthepointsA,BandC.Another

y
A
B
C
D
E
17.3 Simpson’s rule 501
quadraticcurveisfittedthroughthepointsC,DandE.Aftersomeanalysisanexpression
forapproximating the area isfound.
Simpson’s rulestates:
Figure17.2
InSimpson’s rulean
even number ofstrips
isused. Thecurve is
approximated by
quadratic curves.
x
area ≈ h 3 (y 0 +4y 1 +2y 2 +4y 3 +2y 4 +···+2y n−2 +4y n−1 +y n )
= h 3 {y 0 +4(y 1 +y 3 +···)+2(y 2 +y 4 +···)+y n }
wherenisan even number.
Example17.3 Estimate ∫ 1.3
0.5 e(x2) dx usingSimpson’srulewith (a)fourstrips, (b)eight strips.
Solution (a) We use Table 17.1 and notethath = 0.2.UsingSimpson’s rulewehave
estimatedvalue = 0.2
3 {1.2840+4(1.6323+3.3535)+2(2.2479)+5.4195}
= 2.0762
Using Simpson’s rulewe have found ∫ 1.3
0.5 e(x2) dx ≈ 2.0762.
(b) We use Table 17.2 and note thath = 0.1:
estimated value = 0.1 {1.2840 +4(1.4333 +1.8965 +2.7183 +4.2207)
3
+2(1.6323 +2.2479 +3.3535) +5.4195}
= 2.0749
Using Simpson’s rulewe have found ∫ 1.3
0.5 e(x2) dx ≈ 2.0749.
Example17.4 Estimate ∫ 2
1
√
1 +x3 dx using Simpson’s rulewith 10 strips.
Solution With10 stripsh = 0.1. UsingTable 17.4, wefind
estimatedvalue ≈ 0.1
3 {1.4142+4(1.5268+1.7880+2.0917+2.4317+2.8034)
+2(1.6517 +1.9349 +2.2574 +2.6138) +3.000}
= 2.130
In some cases the numerical values are not derived from a function but from actual
measurements.Numericalmethods can still beappliedinanidentical manner.

502 Chapter 17 Numerical integration
Table17.5
3.0000 =y 10
Table17.4
x 10 = 2.0
Using Simpson’s rule to estimate
∫ 2
√ Measurementsusedto
1 1+x 3 dx.
estimatethe integral.
x y= √ 1+x 3 t Measurement (f )
x 0 = 1.0 1.4142 =y 0
0 4
x 1 = 1.1 1.5268 =y 1
1 4.7
x 2 = 1.2 1.6517 =y 2
2 4.9
x 3 = 1.3 1.7880 =y 3
3 5.3
x 4 = 1.4 1.9349 =y 4
4 6.0
x 5 = 1.5 2.0917 =y 5
5 5.3
x 6 = 1.6 2.2574 =y 6
6 5.9
x 7 = 1.7 2.4317 =y 7
x 8 = 1.8 2.6138 =y 8
x 9 = 1.9 2.8034 =y 9
Example17.5 Measurements of a variable, f, were made at 1 second intervals and are given in
Table 17.5. Estimate ∫ 6
f dt using
0
(a) the trapezium rule
(b) Simpson’srule.
Solution (a) Thesum ofthe areasofthe trapezia is
1
[4 +2(4.7 +4.9 +5.3 +6.0 +5.3) +5.9] = 31.2
2
(b) The area has been divided into sixstripsand so Simpson’s rule can beapplied:
approximate value of integral = 1 [4 +4(4.7 +5.3 +5.3)
3
+2(4.9 +6.0) +5.9] = 31.0
Engineeringapplication17.2
Energydissipationinaresistor
Aresistorisbeingusedtodissipateenergyfromavariabled.c.supply.Acalculation
is needed of how much energy has been dissipated over a period of time. Table 17.6
contains values of current,I, through the resistor, and voltage,V, across the resistor
forthefirst100secondssinceelectricalpowerwasfirstapplied.Calculatetheenergy
dissipation during this time period using Simpson’s rule with a step interval of 10
seconds.
Solution
Theenergy dissipated,E, isgiven by
E =
∫ t
0
Pdt

17.3 Simpson’s rule 503
Table17.6
Time(s) Voltage(V) Current(A)
0 50.5 10.1
10 101.0 20.2
20 67.5 13.5
30 80.5 16.1
40 92.0 18.4
50 96.0 19.2
60 78.5 15.7
70 82.0 16.4
80 90.5 18.1
90 107.0 21.4
100 86.0 17.2
Table17.7
Time(s) Voltage(V) Current(A) Power(W)
0 50.5 10.1 510.05
10 101.0 20.2 2040.20
20 67.5 13.5 911.25
30 80.5 16.1 1296.05
40 92.0 18.4 1692.80
50 96.0 19.2 1843.20
60 78.5 15.7 1232.45
70 82.0 16.4 1344.80
80 90.5 18.1 1638.05
90 107.0 21.4 2289.80
100 86.0 17.2 1479.20
wherePis the power. AlsoP =IV, and so
E =
∫ t
0
IV dt
We first need toevaluatePas shown inTable 17.7.
Then using Simpson’s rulewithh = 10 wehave
E =
∫ 100
0
IVdt≈ 10 {510.05 +4(2040.20 +1296.05 +1843.20 +1344.80
3
+2289.80) +2(911.25 +1692.80 +1232.45 +1638.05)
+1479.20}
= 160648.5 J
= 160.649 kJ
The energy dissipated istherefore approximately 160.6 kJ.
17.3.1 ErrorduetoSimpson’srule
Simpson’sruleprovidesanestimatedvalueofadefiniteintegral.Thedifferencebetween
theestimatedvalueandthetrue(exact)valueistheerror.Justaswiththetrapeziumrule,
wecan calculateanupper bound forthiserror.
We need to calculate the fourth derivative of f, that is f (4) . Suppose |f (4) | is never
greaterthansome value,M, throughoutthe interval [a,b], thatis
|f (4) | M forallxvalues on [a,b]
ClearlyM isanupperboundfor |f (4) |on[a,b].TheerrorduetoSimpson’sruleissuch
that
|error | (b−a)h4 M
180
Note that the error isproportional toh 4 .

504 Chapter 17 Numerical integration
Example17.6 Find an upper bound for the error in the estimates calculated in Example 17.3. Hence
find upper and lower bounds for ∫ 1.3
0.5 e(x2) dx.
Solution Here wehave f = e (x2) . Calculating the fourthderivative gives
f (4) = 4e (x2) (4x 4 +12x 2 +3)
Weseekanupperboundfor |f (4) |on[0.5,1.3].Wenotethat f (4) increasesasxincreases
and soits maximum value inthe interval occurs whenx = 1.3:
f (4) (1.3) = 752.319 < 753
Hence 753 is anupper bound for |f (4) |.
Inthis examplea = 0.5 andb=1.3 and so
| error | (0.8)h4 (753)
180
(a) Hereh = 0.2 and so
= 3.347h 4
|error | 3.347(0.2) 4 = 0.0054
An upper bound forthe erroris0.0054. Now
−0.0054 error 0.0054
The estimated value of the integral is2.0762 and so
2.0762 −0.0054
∫ 1.3
0.5
e (x2) dx 2.0762 +0.0054
thatis
2.0708
∫ 1.3
0.5
e (x2) dx 2.0816
(b) Hereh = 0.1 and so
|error | 3.347(0.1) 4 = 3.347 ×10 −4
An upper bound forthe erroris3.347 ×10 −4 . Now
−3.347 ×10 −4 error 3.347 ×10 −4
Noting thatthe estimated value of the integral is2.0749 we have
2.0749 −3.347 ×10 −4
∫ 1.3
0.5
e (x2) dx 2.0749 +3.347 ×10 −4
thatis
2.0746
∫ 1.3
0.5
e (x2) dx 2.0752

Review Exercises 17 505
EXERCISES17.3
1 Estimate the values ofthe following integrals using
Simpson’s rule:
∫ 3
(a) ln(x 3 +1)dx use10strips
2
∫ 2.6 √
(b) te t dt use eightstrips
1
2 Evaluate,usingthe trapeziumruleandSimpson’s
rule,
∫ 1
(a) (x 2 +1) 3/2 dx use four strips
0
∫ 1.6 sin2t
(b) dt use sixstrips
1 t
Solutions
1 (a) 2.7955 (b) 15.1164
2 (a) trapeziumrule: 1.5900,Simpson’s rule: 1.5681
(b) trapeziumrule:0.2464,Simpson’s rule:
0.2460
TechnicalComputingExercises17.3
1 (a) Plotthe fourthderivative of f (x) =
ln(x 3 +1)for2 x3.Useyourgraph
to findan upper bound for f (4) (x)for2x3.
Hence findan upperbound forthe error in
Question1(a) in Exercises17.3. Stateupper and
lower boundsforthe integral given in the
question.
(b) Repeat (a)with f (t) = √ te t for1 t 2.6.
Hence findan upper bound forthe error in Question1(b)
in Exercises17.3. Statelower and upper
boundsforthe integralgiven in the question.
2 Use MATLAB ® orasimilar technicalcomputing
language to findupper andlower boundsforthe
integrals in Question2in Exercises17.3.
REVIEWEXERCISES17
1 Iff(t)= √ t 2 +1find ∫ 2
1 f (t)dt using
(a) the trapezium rulewithh = 0.25
(b) Simpson’s ruleusingeightstrips.
2 Estimate the followingdefiniteintegrals usingthe
trapeziumrulewith sixstrips:
∫ 4 √
∫ 0.6
(a) x 3 +1dx (b) sin(t 2 )dt
1
0
∫ 0.8
∫ 0.3
(c) e (t2) 3
dt (d)
0.2
0 x 3 +2 dx
∫ 2.5 e t
(e) dt
1 t
3 Estimate the definiteintegrals in Question2using
Simpson’s rulewith sixstrips.
4 Estimate the followingdefiniteintegrals usingthe
trapeziumrule with eightstrips:
∫ 4
(a) cos( √ t)dt
0
∫ 6
(b) (t 2 +1) 3/2 dt
−2
∫ 3
(c) e √x dx
1
∫ 0.8
(d) tan(t 2 )dt
0
∫ 5
(e) ln(x 2 +1)dx
3

506 Chapter 17 Numerical integration
Solutions
1 (a) 1.8111
(b) 1.8101
2 (a) 12.9113
(b) 0.07227
(c) 0.80860
(d) 0.44845
(e) 5.19384
3 (a) 12.87181
(b) 0.071334
(c) 0.80643
(d) 0.44849
(e) 5.17879
4 (a) 0.81058
(b) 370.25
(c) 8.27686
(d) 0.18390
(e) 5.63032
TechnicalComputingExercises17
1 UseMATLAB ® orasimilartechnical computing
language to findupper boundsforthe errorsin
Questions1to 4 in Review exercises17. Hence find
upperand lower boundsforthe integrals given in
these questions.

18 Taylorpolynomials,
Taylorseriesand
Maclaurinseries
Contents 18.1 Introduction 507
18.2 Linearizationusingfirst-orderTaylorpolynomials 508
18.3 Second-orderTaylorpolynomials 513
18.4 Taylorpolynomialsofthenthorder 517
18.5 Taylor’sformulaandtheremainderterm 521
18.6 TaylorandMaclaurinseries 524
Reviewexercises18 532
18.1 INTRODUCTION
Often the value of a function and the values of its derivatives are known at a particular
pointandfromthisinformationitisdesiredtoobtainvaluesofthefunctionaroundthat
point.TheTaylorpolynomialsandTaylorseriesallowengineerstomakesuchestimates.
Oneapplicationofthisisinobtaininglinearizedmodelsofnon-linearsystems.Thegreat
advantageofalinearmodelisthatitismucheasiertoanalysethananon-linearone.Itis
possibletomakeuseoftheprincipleofsuperposition:thisallowstheeffectsofmultiple
inputstoasystemtobeconsideredseparately,andtheresultantoutputtobeobtainedby
summing the individual outputs.
Oftenasystemmaycontainonlyafewcomponentsthatarenon-linear.Bylinearizing
theseitisthenpossibletoproducealinearmodelforthesystem.Wesawanexampleof
thiswhenweanalysedafluidsysteminEngineeringapplication10.6.Althoughelectricalsystemsareoftenlinear,mechanical,thermalandfluidsystems,orsystemscontainingamixtureofthese,arelikelytocontainsomenon-linearcomponents.Unfortunately
itmaynotbepossibletoobtainasufficientlyaccuratelinearmodelforeverynon-linear
systemas weshall see inthischapter.
Taylorpolynomialsofhigherdegreecanbefoundwhichapproximatetoagivenfunction.ThisisdealtwithinSections18.3and18.4.Thedifferencebetweenagivenfunction
andthecorrespondingTaylorpolynomialiscoveredinSection18.5.Thechaptercloses
with a treatment of Taylor and Maclaurin series.

508 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
18.2 LINEARIZATIONUSINGFIRST-ORDERTAYLOR
POLYNOMIALS
Suppose we know that y is a function of x and we know the values of y and y ′ when
x =a,thatisy(a)andy ′ (a)areknown.Wecanusey(a)andy ′ (a)todeterminealinear
polynomial which approximates toy(x). Let this polynomial be
p 1
(x) =c 0
+c 1
x
We choose the constantsc 0
andc 1
so that
p 1
(a) =y(a)
p ′ 1 (a) =y′ (a)
that is, the values of p 1
and its first derivative evaluated atx = a match the values ofy
and itsfirst derivative evaluated atx =a. Then,
p 1
(a) =y(a) =c 0
+c 1
a
p ′ 1 (a) =y′ (a) =c 1
Solving forc 0
andc 1
yields
Thus,
c 0
=y(a) −ay ′ (a)
c 1
=y ′ (a)
p 1
(x) =y(a) −ay ′ (a) +y ′ (a)x
p 1
(x) =y(a) +y ′ (a)(x −a)
p 1
(x) isthefirst-order Taylor polynomialgenerated byyatx =a.
y
O
Q
Figure18.1
Graphical representation
ofafirst-orderTaylor
polynomial.
a
x
The function,y(x), isoften referred toas thegenerating function. Note that p 1
(x) and
its first derivative evaluated atx =a agree withy(x) and its first derivative evaluated at
x=a.
First-order Taylor polynomials can also be viewed from a graphical perspective.
Figure 18.1 shows the function, y(x), and a tangent at Q where x = a. Let the equation
ofthe tangent atx =abe
p(x)=mx+c
The gradient of the tangent is, by definition, the derivative ofyatx = a, that isy ′ (a).
So,
p(x) =y ′ (a)x +c
The tangent passes through the point (a,y(a)), and so
y(a) =y ′ (a)a +c
that isc =y(a) −y ′ (a)a. The equation of the tangent isthus
p(x) =y ′ (a)x +y(a) −y ′ (a)a
p(x) =y(a) +y ′ (a)(x −a)

18.2 Linearization using first-order Taylor polynomials 509
This is the first-order Taylor polynomial. We see that the first-order Taylor polynomial
issimplythe equation of the tangent toy(x) wherex =a.
Clearly,forvaluesofxneartox =athevalueofp 1
(x)willbeneartoy(x);p 1
(x)isa
linearapproximationtoy(x).Intheneighbourhoodofx =a,p 1
(x)closelyapproximates
y(x), but being linear isamuch easier function todeal with.
Example18.1 Afunction,y, and its first derivative areevaluated atx = 2.
y(2)=1 y ′ (2)=3
(a) State the first-order Taylor polynomial generated byyatx = 2.
(b) Estimatey(2.5).
Solution (a) p 1
(x) =y(2) +y ′ (2)(x −2) =1+3(x −2) = −5 +3x
(b) We use the first-order Taylor polynomial toestimatey(2.5):
p 1
(2.5) = −5 +3(2.5) = 2.5
Hence,y(2.5) ≈ 2.5.
Example18.2 Find a linear approximation toy(t) =t 2 neart = 3.
Solution We require the equation of the tangent toy = t 2 att = 3, that is the first-order Taylor
polynomial aboutt = 3.Note thaty(3) = 9 andy ′ (3) = 6.
p 1
(t) =y(a) +y ′ (a)(t −a) =y(3) +y ′ (3)(t −3)
=9+6(t−3)
=6t−9
Att = 3, p 1
(t) and y(t) have an identical value. Near tot = 3, p 1
(t) and y(t) have
similarvalues, forexample p 1
(2.8) = 7.8,y(2.8) = 7.84.
18.2.1 Linearization
It is a frequent requirement in engineering to obtain a linear mathematical model of a
systemwhichisbasicallynon-linear.Mathematicallyandcomputationallylinearmodels
arefareasiertodealwiththannon-linearmodels.Themainreasonforthisisthatlinear
modelsobeytheprincipleofsuperposition.Itfollowsthatiftheapplication,separately,
of inputs u 1
(t) and u 2
(t) to the system produces outputs y 1
(t) and y 2
(t), respectively,
thentheapplicationofaninputu 1
(t) +u 2
(t)willproduceanoutputy 1
(t) +y 2
(t).This
isonly trueforlinear systems.
The value of this principle is that the effect of several inputs to a system can be calculated
merely by adding together the effects of the individual inputs. This allows the
effect of simple individual inputs to the system to be analysed and then combined to
evaluate the effect of more complicated combinations of inputs. A few examples will
help clarifythese points.

510 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Engineeringapplication18.1
Ad.c.electricalnetwork
Consider the d.c. network of Figure 18.2. This network is a linear system. This is
because the voltage/current characteristic of a resistor is linear provided a certain
voltage isnot exceeded. Recall Ohm’s lawwhich isgiven by
V=IR
whereV is the voltage across the resistor,I is the current through the resistor andR
istheresistance.Thismakestheanalysisofthenetworkrelativelyeasy.Thevoltage
sources,V 1
,V 2
,V 3
,V 4
, can be thought of as the inputs to the system. It is possible to
analyse the effect of each of these sources separately, for example the voltage drop
across the resistor,R 5
, resulting from the voltage sourceV 1
, and then combine these
effects to obtain the total effect on the system. The voltage drop acrossR 5
when all
sources are considered would be the sum of each of the voltage drops due to the
individual sourcesV 1
,V 2
,V 3
andV 4
.
R V 1 2 V 3
+ – + –
–
R 2 R 3 V 4 R 4 R 5
+
+ –
V R 6 R 1 7
Figure18.2
Ad.c. electrical network.
Engineeringapplication18.2
Agravityfeedwatersupply
Consider the water supply network of Figure 18.3. The network consists of three
sourcereservoirsandaseriesofconnectingpipes.Wateristakenfromthenetworkat
twopoints,S 1
andS 2
.Inapracticalnetwork,reservoirsareusuallyseveralkilometres
away from the points at which water is taken from the network and so the effect of
pressure drops along the pipes is significant. For this reason many networks require
pumps toboostthe pressure.
The main problem with analysing this network is that it is non-linear. This is because
the relationship between pressure drop along a pipe and water flow through a
pipeisnotlinear:adoublingofpressuredoesnotleadtoadoublingofflow.Forthis
R 1
P 1 P 2
R 2 R 3
Figure18.3
P 3 P 4 P 5 P 6
S 1 S 2
Awater supply network.

18.2 Linearization using first-order Taylor polynomials 511
reason, it is not possible to use the principle of superposition when analysing the
network. For example, if the effect of the pressure at S 1
for a given flow rate was
calculatedseparatelyforeachoftheinputstothesystem--thereservoirsR 1
,R 2
,R 3
--
then the effect of all the reservoirs could not be obtained by adding the individual
effects. It is not possible to obtain a linear model for this system except under very
restrictive conditions and so the analysis of water networks isvery complicated.
Having demonstrated the valueoflinear models itisworthanalysing howand when
a non-linear system can be linearized. The first thing to note is that many systems may
containamixtureoflinearandnon-linearcomponentsandsoitisonlynecessarytolinearizecertainpartsofthesystem.AsystemofthistypehasbeenstudiedinEngineering
application 10.6. Thereforelinearization involves deciding which components ofasystemarenon-linear,decidingwhetheritwouldbevalidtolinearizethecomponentsand,
ifso, then obtaining linearized models of the components.
Consider again Figure 18.1. Imagine it illustrates a component characteristic. The
actual component characteristic is unimportant for the purposes of this discussion. For
instance, it could be the pressure/flow relationship of a valve or the voltage/current
relationship of an electronic device. The main factor in deciding whether a valid linear
model can be obtained is the range of values over which the component is required to
operate. If an operating point Q were chosen and deviations from this operating point
were small then it is clear from Figure 18.1 that a linear model -- corresponding to the
tangent tothe curve atpointQ-- would be an appropriate model.
Obtaining a linear model is relatively straightforward. It consists of calculating the
first-order Taylor polynomial centred around the operating point Q.This isgiven by
p 1
(x) =y(a) +y ′ (a)(x −a)
Then p 1
(x)isusedasthelinearizedmodelofthecomponentwithcharacteristicy(x).It
isvalidprovidedthatitisonlyusedforvaluesofxsuchthat |x−a|issufficientlysmall.
Asstated, p 1
(x)isalsotheequationofthetangenttothecurveatpointQ.Therangeof
values for which the model is valid depends on the curvature of the characteristic and
the accuracy required.
Engineeringapplication18.3
Powerdissipationinaresistor
The power dissipated in a resistor varies with the current. Derive a linear model for
thispowervariationvalidforanoperatingpointof0.5A.Theresistorhasaresistance
of 10 .
Solution
For a resistor
where
P=I 2 R
P = power dissipated (W)
I = current (A)
R = resistance ().
➔

512 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
The first-order polynomial, valid around an operating pointI = 0.5,is
p 1
(I) =P(0.5) +P ′ (0.5)(I −0.5)
NowP(0.5) = (0.5) 2 10 = 2.5,P ′ (0.5) = 2(0.5)(10) = 10,andso
p 1
(I) =2.5 +10(I −0.5) =10I −2.5
It is interesting to compare this linear approximation with the true curve for values
ofI around the operating point. Table 18.1 shows some typical values. Figure 18.4
shows a graph of the power dissipated in the resistor,P, against the current flowing
intheresistor,I.Noticethatthelinearapproximationisquitegoodwhenclosetothe
operating pointbut deteriorates further away.
Table18.1
A comparison oflinear approximations with truevalues.
I (A) TruevalueofP (W) ApproximatevalueofP (W)
0.5 2.5 2.5
0.499 2.49001 2.49
0.501 2.51001 2.51
0.49 2.401 2.4
0.51 2.601 2.6
0.4 1.6 1.5
0.6 3.6 3.5
1.0 10 7.5
P
10
8
6
4
Operating point
P = 10 I 2
p 1 (I) = 10 I – 2.5
2
0
0.2 0.4 0.6 0.8 1
I
Figure18.4
We see that the tangential approximation is
good when close to the operating point.
EXERCISES18.2
1 Calculate the first-orderTaylor polynomial generated
byy(x) = e x about
(a) x=0 (b) x=2 (c) x=−3
2 Calculate the first-orderTaylor polynomial generated
byy(x) = sinxabout
(a) x=0 (b) x=1 (c) x=−0.5
3 Calculate the first-orderTaylor polynomial generated
byy(x) = cosxabout
(a)x=0
(b)x=1
(c)x = −0.5

18.3 Second-order Taylor polynomials 513
4 (a) Find alinear approximation, p 1 (t),toh(t) =t 3
aboutt = 2.
(b) Evaluateh(2.3)and p 1 (2.3).
5 (a) Findalinearapproximation, p 1 (t),toR(t) = 1 t
aboutt = 0.5.
(b) EvaluateR(0.7)and p 1 (0.7).
Solutions
1 (a) p 1 (x)=x+1
(b) p 1 (x) =e 2 (x −1)
(c) p 1 (x) =e −3 (x +4)
2 (a) p 1 (x) =x
(b) p 1 (x) = 0.5403x +0.3012
(c) p 1 (x) = 0.8776x −0.0406
4 (a) p 1 (t) =12t −16
(b) h(2.3) = 12.167,p 1 (2.3) = 11.6
5 (a) p 1 (t) = −4t +4
(b) R(0.7) = 1.4286,p 1 (0.7) = 1.2
3 (a) p 1 (x) =1
(b) p 1 (x) = −0.8415x +1.3818
(c) p 1 (x) = 0.4794x +1.1173
18.3 SECOND-ORDERTAYLORPOLYNOMIALS
Suppose that in addition to y(a) and y ′ (a), we also have a value of y ′′ (a). With this
informationasecond-orderTaylorpolynomialcanbefound,whichprovidesaquadratic
approximationtoy(x). Let
p 2
(x) =c 0
+c 1
x+c 2
x 2
We require
p 2
(a) =y(a)
p ′ 2 (a) =y′ (a)
p ′′ 2 (a) =y′′ (a)
thatis,thepolynomialanditsfirsttwoderivativesevaluatedatx =amatchthefunction
and its first two derivatives evaluated atx =a. Hence
p 2
(a) =c 0
+c 1
a +c 2
a 2 =y(a) (18.1)
p ′ 2 (a) =c 1 +2c 2 a =y′ (a) (18.2)
p ′′ 2 (a) = 2c 2 =y′′ (a) (18.3)
Solvingforc 0
,c 1
andc 2
yields
c 2
= y′′ (a)
from Equation (18.3)
2
c 1
=y ′ (a) −ay ′′ (a) fromEquation (18.2)
c 0
=y(a) −ay ′ (a) + a2
2 y′′ (a) from Equation (18.1)

514 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Hence,
Finally,
p 2
(x) =y(a) −ay ′ (a) + a2
2 y′′ (a)
+{y ′ (a) −ay ′′ (a)}x + y′′ (a)
2 x2
p 2
(x) =y(a) +y ′ (a)(x −a) +y ′′ (x −a)2
(a)
2
p 2
(x) isthesecond-order Taylor polynomialgenerated byyaboutx =a.
Example18.3 Giveny(1) = 0,y ′ (1) = 1,y ′′ (1) = −2,estimate
(a) y(1.5)
(b) y(2)
(c) y(0.5)
usingthe second-order Taylor polynomial.
Solution Thesecond-order Taylor polynomialis p 2
(x):
p 2
(x) =y(1) +y ′ (1)(x −1) +y ′′ (x −1)2
(1)
2
(x −1)2
=x−1−2 =x−1−(x−1) 2 =−x 2 +3x−2
2
We use p 2
(x) as an approximation toy(x).
(a) The value ofy(1.5) isapproximated by p 2
(1.5):
y(1.5) ≈ p 2
(1.5) = 0.25
(b) The value ofy(2) isapproximated by p 2
(2):
y(2) ≈ p 2
(2) =0
(c) The value ofy(0.5) isapproximated by p 2
(0.5):
y(0.5) ≈ p 2
(0.5) = −0.75
Example18.4 (a) Calculatethe second-order Taylor polynomial, p 2
(x), generatedby
y(x)=x 3 +x 2 −6aboutx=2.
(b) Verifythaty(2) = p 2
(2),y ′ (2) = p ′ 2 (2)andy′′ (2) = p ′′ 2 (2).
(c) Comparey(2.1) and p 2
(2.1).
Solution (a) We needtocalculatey(2),y ′ (2) andy ′′ (2). Now
and so
y(x)=x 3 +x 2 −6,y ′ (x)=3x 2 +2x,y ′′ (x)=6x+2
y(2) =6,y ′ (2) =16,y ′′ (2) =14

18.3 Second-order Taylor polynomials 515
The required second-order Taylor polynomial, p 2
(x), isthus given by
p 2
(x) =y(2) +y ′ (2)(x −2) +y ′′ (x −2)2
(2)
2
(x −2)2
=6+16(x−2)+14
2
=6+16x−32+7(x 2 −4x+4)
=7x 2 −12x+2
(b) Using (a)we can see that
and so
Hence
(c) We have
p 2
(x) =7x 2 −12x +2,p ′ 2 (x) = 14x −12,p′′ 2
(x) =14
p 2
(2) =6,p ′ 2 (2) = 16,p′′ 2
(2) =14
y(2) = p 2
(2),y ′ (2) = p ′ 2 (2),y′′ (2) = p ′′ 2 (2)
y(2.1) = (2.1) 3 + (2.1) 2 −6 = 7.671
p 2
(2.1) = 7(2.1) 2 −12(2.1) +2 = 7.67
Clearly there is a very close agreement between values of y(x) and p 2
(x) near to
x=2.
Engineeringapplication18.4
Quadraticapproximationtoadiodecharacteristic
In Engineering application 10.5 we derived a linear approximation to a diode characteristic
suitable for small signal variations around an operating point. Sometimes
it is not possible to use a linear approximation because the variations are too large
to maintain sufficient accuracy. Even so, an approximate model may be desirable.
In general, a higher order Taylor polynomial will give a more accurate model than
that of a lower order polynomial. We will consider a quadratic model for a diode
characteristic. TheV −I characteristic of typical diode at room temperature can be
modelledby the equation
I =I(V) =I s
(e 40V −1)
Given anoperating point,V a
, the second-order Taylor polynomialis
p 2
(V)=I(V a
)+I ′ (V a
)(V −V a
)+I ′′ (V a
) (V −V a )2
2
Now
I ′ (V) = 40I s
e 40V I ′′ (V) = 1600I s
e 40V ➔

516 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
so
p 2
(V) =I s
(e 40V a −1) +40Is e 40V a (V −Va ) +1600I s
e 40V (V −V
a a
) 2
2
Thecoefficientsneedtobecalculatedonlyonce.Afterthatthecalculationofacurrent
value only involves evaluating a quadratic.
EXERCISES18.3
1 (a) Obtainthe second-order Taylor polynomial,
p 2 (x),generatedbyy(x) = 3x 4 +1aboutx = 2.
(b) Verifythaty(2) = p 2 (2),y ′ (2) = p ′ 2 (2)and
y ′′ (2) = p ′′ 2 (2).
(c) Evaluate p 2 (1.8)andy(1.8).
2 (a) Calculate the second-order Taylor polynomial,
p 2 (x),generated byy(x) = sinx aboutx = 0.
(b) Calculate the second-order Taylor polynomial,
p 2 (x),generated byy(x) = cosx aboutx = 0.
(c) Compare your resultsfrom (a) and(b)with the
small-angle approximations given in Section6.5.
3 Afunction,y(x),issuchthaty(−1) = 3,y ′ (−1) = 2
andy ′′ (−1) = −2.
(a) Statethe second-order Taylor polynomial
generated byyaboutx = −1.
(b) Estimatey(−0.9).
4 Afunction,y(x),satisfiesthe equation
y ′ =y 2 +x
y(1)=2
(a) Estimatey(1.3)usingafirst-orderTaylor
polynomial.
(b) Bydifferentiatingthe equationwith respect tox,
obtain an expression fory ′′ .Hence evaluate
y ′′ (1).
(c) Estimatey(1.3)usingasecond-order Taylor
polynomial.
5 Afunction,x(t),satisfiesthe equation
ẋ=x+ √ t+1
x(0)=2
(a) Estimatex(0.2)usingafirst-orderTaylor
polynomial.
(b) Differentiate the equation w.r.t.t and hence
obtain an expression forẍ.
(c) Estimatex(0.2)usingasecond-order Taylor
polynomial.
6 Afunction,h(t),isdefined by
h(t) = sin2t +cos3t
Obtainthe second-order Taylor polynomial generated
byh(t)aboutt = 0.
7 Thefunctionsy 1 (x),y 2 (x)andy 3 (x)are defined by
y 1 (x) =Ae x ,y 2 (x) =Bx 3 +Cx,
y 3 (x) =y 1 (x) +y 2 (x)
(a) Obtainasecond-order Taylor polynomial for
y 1 (x)aboutx = 0.
(b) Obtainasecond-order Taylor polynomial for
y 2 (x)aboutx = 0.
(c) Obtainasecond-order Taylor polynomial for
y 3 (x)aboutx = 0.
(d) Canyou draw any conclusionsfrom your
answers to (a),(b)and(c)?
Solutions
1 (a) p 2 (x) = 72x 2 −192x +145
(c) p 2 (1.8) = 32.68,y(1.8) = 32.4928
2 (a) p 2 (x) =x (b) p 2 (x) =1− x2
2
3 (a) p 2 (x) = −x 2 +4
(b) p 2 (−0.9) = 3.19.This isan approximation to
y(−0.9).
4 (a) 3.5
(b) y ′′ = 2yy ′ +1,y ′′ (1) = 21
(c) 4.445
5 (a) 2.6
1
(b) ẍ=ẋ +
2 √ t +1
(c) 2.67

18.4 Taylor polynomials of the nth order 517
6 p 2 (t) =1+2t −4.5t 2
( )
7 (a)A1 +x+ x2
2
(b) Cx
(c) A+(A+C)x+ Ax2
2
TechnicalComputingExercises18.3
Many technicalcomputing languageshave the
capability ofproducingaTaylor polynomial ofa
function.Insome languagesthisis available asa
defaultwhereasin others, such asMATLAB ® ,thisis
offered via a toolboxfunctionwhichmayneedto be
loaded orpurchased separately. Ifthe software you
are usingdoes not have the capability to produce the
Taylor seriesautomaticallyyou maychoose to dothis
partmanuallyandthenplot the resultsusingthe
technicalcomputing language.
1 (a) Calculate the second-order Taylor polynomial,
p 2 (x),generatedbyy(x) =x 3 aboutx = 0.
(b) Drawy(x)andp 2 (x)for −2 x2.
2 (a) Calculate the second-order Taylor polynomial,
p 2 (x),generated byy(x) = sinx aboutx = 0.
(b) Drawy(x)andp 2 (x)for −2 x 2.
3 (a) Calculate the second-order Taylor polynomial,
( )
1
p 2 (x),generated byy(x) = sin aboutx = 3.
x
(b) Drawy(x)andp 2 (x)for1 x 5.
4 (a) Calculate the second-order Taylor polynomial,
p 2 (x),generated byy(x) = e cosx aboutx = 0.
(b) Drawy(x)andp 2 (x)for −2 x 2.
18.4 TAYLORPOLYNOMIALSOFTHEnTHORDER
If we know y and its first n derivatives evaluated at x = a, that is y(a), y ′ (a),
y ′′ (a), ...,y (n) (a), thenthenth-orderTaylorpolynomial, p n
(x), maybewrittenas
p n
(x)=y(a) +y ′ (a)(x −a) +y ′′ (x −a)2
(a) +y (3) (x −a)3
(a)
2! 3!
+···+y (n) (x −a)n
(a)
n!
This provides an approximation toy(x). The polynomial and its firstnderivatives evaluated
atx = a match the values ofy(x) and its firstnderivatives evaluated atx = a,
thatis
p n
(a) =y(a)
p ′ n (a) =y′ (a)
p ′′ n (a) =y′′ (a)
.
p (n)
n (a) =y(n) (a)

518 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Example18.5 Giveny(0) = 1,y ′ (0) = 1,y ′′ (0) = 1,y (3) (0) = −1,y (4) (0) = 2,obtainafourth-order
Taylor polynomial generated byyaboutx = 0.Estimatey(0.2).
Solution Inthis examplea = 0 and hence
p 4
(x) =y(0) +y ′ (0)x +y ′′ (0) x2
2! +y(3) (0) x3
3! +y(4) (0) x4
4!
=1+x+ x2
2 − x3
6 + x4
12
The Taylor polynomial can be used toestimatey(0.2):
p 4
(0.2) = 1 +0.2 + (0.2)2
2
y(0.2) ≈ 1.2188
− (0.2)3
6
+ (0.2)4
12
= 1.2188
Example18.6 (a) Calculatethe first-,second-,third-and fourth-orderTaylor polynomialsgenerated
byy(x) = e x aboutx = 0.
(b) Ploty = e x and the Taylor polynomials for −2 x2.
Solution (a) We have
and
y(x) =y ′ (x) =y ′′ (x) =y (3) (x) =y (4) (x) =e x
y(0) =y ′ (0) =y ′′ (0) =y (3) (0) =y (4) (0) =1
Thus the Taylor polynomials, p 1
(x),p 2
(x),p 3
(x) and p 4
(x), are given by
p 1
(x) =y(0)+y ′ (0)x =1+x
p 2
(x) =y(0) +y ′ (0)x +y ′′ (0) x2
2 =1+x+x2 2
p 3
(x) =y(0) +y ′ (0)x +y ′′ (0) x2
2 +y(3) (0) x3
3! =1+x+x2 2 + x3
6
p 4
(x) =y(0) +y ′ (0)x +y ′′ (0) x2
2 +y(3) (0) x3
3! +y(4) (0) x4
4!
=1+x+ x2
2 + x3
6 + x4
24
(b) The graphs ofy = e x and the Taylor polynomials are shown in Figure 18.5. Note
that the Taylor polynomials become better and better approximations to e x as the
order of the polynomial increases.
Example18.7 Given thatysatisfiesthe equation
y ′′ −(y ′ ) 2 +2y=x 2 (18.4)
and alsothe conditions
y(0)=1 y ′ (0)=2

18.4. Taylor polynomials of the nth order 519
y(x), p i (x)
8
7
6
5
4
3
2
p
1 2 (x)
0 y(x)
p 1 (x)
–1
–2.0 –1.5 –1.0 –0.5 0
x
(a)
y(x)
p 2 (x)
p 1 (x)
0.5 1.0 1.5 2.0
y(x), p i (x)
Figure18.5
Agraph ofy = e x andfour Taylor polynomials.
8
7
6
5
4
3
2
1
p 4 (x)
y(x)
0 p 3 (x)
–1
–2.0 –1.5 –1.0 –0.5 0
(b)
x
p 4 (x)
y(x)
p 3 (x)
0.5 1.0 1.5 2.0
use a third-order Taylor polynomial toestimatey(0.5).
Solution To write down the third-order Taylor polynomial about x = 0 we require y(0), y ′ (0),
y ′′ (0) andy (3) (0). FromEquation (18.4),
So,
y ′′ (x) =x 2 + {y ′ (x)} 2 −2y(x) (18.5)
y ′′ (0) =0+ {y ′ (0)} 2 −2y(0) =2
To findy (3) (0), Equation (18.5) is differentiated w.r.t.x:
y (3) (x) = 2x +2y ′ (x)y ′′ (x) −2y ′ (x)
y (3) (0) = 2y ′ (0)y ′′ (0) −2y ′ (0) =4
The Taylor polynomial may now be written as
p 3
(x) =y(0) +y ′ (0)x +y ′′ (0) x2
2 +y(3) (0) x3
6
=1+2x+x 2 + 2x3
3
We use p 3
(x) as anapproximation toy(x):
thatis,
p 3
(0.5) = 1 +1+0.25 +0.0833 = 2.333
y(0.5) ≈ 2.333
EXERCISES18.4
1 Afunction,y(x),hasy(0) = 3,y ′ (0) = 1,
y ′′ (0) = −1andy (3) (0) =2.
(a) Obtainathird-orderTaylor polynomial, p 3 (x),
generated byy(x) aboutx = 0.
(b) Estimatey(0.2).
2 Afunction,h(t),hash(2) = 1,h ′ (2) = 4,
h ′′ (2) = −2,h (3) (2) =1andh (4) (2) =3.
(a) Obtainafourth-orderTaylor polynomial, p 4 (t),
generated byh(t)aboutt = 2.
(b) Estimateh(1.8).

520 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
3 Giveny(x) = sinx,obtain the third-, fourth-and
fifth-orderTaylor polynomials generated byy(x)
aboutx = 0.
4 Giveny(x) = cosx, obtain the third-, fourth- and
fifth-orderTaylor polynomials generated byy(x)
aboutx = 0.
5 (a) Giveny(x) = sin(kx),k a constant, obtain the
third-, fourth-and fifth-orderTaylor polynomials
generated byy(x)aboutx = 0.
(b) Writedown the third-, fourth-and fifth-order
Taylor polynomials generated byy = sin(−x)
aboutx = 0.
6 (a) Giveny(x) = cos(kx),k aconstant, obtain the
third-, fourth-and fifth-orderTaylor polynomials
generated byy(x)aboutx = 0.
(b) Writedown the third-, fourth-and fifth-order
Taylor polynomials generated byy = cos(2x)
aboutx = 0.
7 If p n (x) isthenth-order Taylor polynomial generated
byy(x) aboutx = 0, showthat p n (kx) is thenth-order
Taylor polynomial generated byy(kx) aboutx = 0.
8 Thefunction,y(x),satisfiesthe equation
y ′′ =y+3x 2 y(1) =1,y ′ (1) =2
(a) Obtain athird-orderTaylor polynomial generated
byyaboutx = 1.
(b) Estimatey(1.3)using(a).
(c) Obtain afourth-orderTaylor polynomial
generated byyaboutx = 1.
(d) Estimatey(1.3)using(c).
9 Afunction,y(x),satisfiesthe equation
y ′′ +y 2 =x 3
y(0) =1,y ′ (0) = −1
(a) Estimatey(0.25)usingathird-orderTaylor
polynomial.
(b) Estimatey(0.25)usingafourth-orderTaylor
polynomial.
10 Afunction,y(x),hasy(1) = 3,y ′ (1) = 6,y ′′ (1) = 1
andy (3) (1) = −1.
(a) Estimatey(1.2)usingathird-orderTaylor
polynomial.
(b) Estimatey ′ (1.2)usingan appropriate
second-order Taylor polynomial.
[Hint: define anew variable,z,given byz =y ′ .]
Solutions
1 (a) 3+x− x2
2 + x3
3
2 (a)
t 4 8 − 5t3
6 +t2 +6t− 31
3
(b) 0.1589
3 p 3 (x)=x− x3
3! ,
p 4 (x) =x− x3
3! ,
p 5 (x) =x− x3
3! + x5
5!
4 p 3 (x)=1− x2
2! ,
p 4 (x) =1− x2
2! + x4
4! ,
p 5 (x) =1− x2
2! + x4
4!
(b) 3.1827
5 (a) p 3 (x) =kx− k3 x 3
3! ,
p 4 (x) =kx− k3 x 3
3! ,
p 5 (x) =kx− k3 x 3
3!
(b) p 3 (x) = −x + x3
3! ,
p 4 (x) = −x+ x3
3! ,
+ k5 x 5
5!
p 5 (x) = −x+ x3
3! − x5
5!
6 (a) p 3 (x) =1− k2 x 2
2! ,
p 4 (x) =1− k2 x 2
2!
p 5 (x) =1− k2 x 2
2!
+ k4 x 4
4! ,
+ k4 x 4
4!

18.5 Taylor’s formula and the remainder term 521
(b) p 3 (x) =1−2x 2 ,
p 4 (x) =1−2x 2 + 2x4
3 ,
p 5 (x) =1−2x 2 + 2x4
3
4x 3
8 (a)
3 −2x2 +2x− 1 3
(b) 1.816
5x 4
(c)
12 − x3
3 + x2
2 + x 3 + 1 12
(d) 1.8194
9 (a) 0.7240 (b) 0.7240
10 (a) 4.2187 (b) 6.18
TechnicalComputingExercises18.4
1 Drawy(x) = 1 for1 x8. On the same axesdraw
x
the fourth-orderTaylor polynomial generated byy(x)
aboutx = 3.
3 Drawy(x) = lnx for0.5 x10.On the same axes,
draw the third-, fourth- andfifth-orderTaylor
polynomials generatedbyy(x)aboutx = 1.
2 Drawy = tanx for −1 x1. Using the same axes,
drawthe fifth-orderTaylor polynomial generatedby
y(x)aboutx = 0.
18.5 TAYLOR’SFORMULAANDTHEREMAINDERTERM
So far we have found Taylor polynomials of orders 1, 2, 3, 4 and so on. Example 18.6
suggeststhatthegeneratingfunction,y(x),andtheTaylorpolynomialsareincloseagreementforvalues
ofxneartothe pointwherex =a. Itisreasonable toask:
‘HowaccuratelydoTaylorpolynomialsgeneratedbyy(x)atx =aapproximateto
y atvaluesofxotherthana?’
‘IfmoreandmoretermsareusedintheTaylorpolynomialwillthisproduceabetter
and better approximation toy?’
To answerthesequestionsweintroduceTaylor’s formulaand the remainder term.
Suppose p n
(x) is an nth-order Taylor polynomial generated by y(x) about x = a.
ThenTaylor’s formulastates:
y(x) = p n
(x) +R n
(x)
whereR n
(x) iscalled theremainderofordernand isgiven by
remainder ofordern =R n
(x) = y(n+1) (c)(x −a) n+1
for somenumbercbetweenaandx.
(n +1)!
Theremainderofordernisalsocalledtheerrorterm.Theerrortermineffectgives
thedifferencebetweenthefunction,y(x),andtheTaylorpolynomialgeneratedbyy(x).
ForaTaylorpolynomialtobeacloseapproximationtothegeneratingfunctionrequires
the error termtobesmall.

522 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Example18.8 Calculate the error term of order 5 due to y(x) = e x generating a Taylor polynomial
aboutx = 0.
Solution Heren = 5 and son +1 = 6.Inthisexamplea = 0.We see thaty = e x and so
y ′ (x) =y ′′ (x) = ··· =y (5) (x) =y (6) (x) =e x
and soy (6) (c) = e c . The remainder termoforder 5,R 5
(x), isgiven by
R 5
(x) = ec x 6
6!
forsome numbercbetween 0 andx
Example18.9 (a) Calculatethe fourth-orderTaylor polynomial, p 4
(x), generatedbyy(x) = sin2x
aboutx = 0.
(b) State the fourth-order errorterm,R 4
(x).
(c) Calculate anupper bound forthis errortermgiven |x| < 1.
(d) Comparey(0.5) and p 4
(0.5).
Solution (a) y(x) =sin2x, y(0) = 0
y ′ (x)=2cos2x, y ′ (0) =2
y ′′ (x)=−4sin2x, y ′′ (0) =0
y (3) (x)=−8cos2x,
y (3) (0) = −8
y (4) (x)=16sin2x, y (4) (0) =0
Thefourth-orderTaylor polynomial, p 4
(x), is
p 4
(x) =y(0) +y ′ (0)x +y ′′ (0) x2
2! +y(3) (0) x3
3! +y(4) (0) x4
4!
=0+2x+0− 8x3
6 +0
=2x− 4x3
3
(b) Wenotethaty (5) (x) = 32cos2xandsoy (5) (c) = 32cos(2c).Theerrorterm,R 4
(x),
isgiven by
R 4
(x) =y (5) (c) x5
5!
= 32cos(2c)x5
120
= 4
15 cos(2c)x5 wherecisanumber between 0 andx
(c) In order to calculate an upper bound for this error term we note that |cos(2c)| 1
forany value ofc. Hence an upper bound forR 4
(x) isgiven by
|R 4
(x)|
∣
4x 5
15
∣

We know |x| < 1, and so 4x5
15
18.5 Taylor’s formula and the remainder term 523
4
is never greater than . Hence an upper bound for
15
the error term is 4 15 . If we use p 4
(x) to approximatey = sin2x the error will be no
greater than 4 provided |x| < 1.
15
(d) We letx = 0.5.
y(0.5) = sin1 = 0.8415
p 4
(0.5) = 2(0.5) − 4 3 (0.5)3 = 0.8333
Thedifferencebetweeny(0.5)andp 4
(0.5)canneverbegreaterthananupperbound
of the errortermevaluated atx = 0.5. This isverified numerically.
and
y(0.5) −p 4
(0.5) = 0.8415 −0.8333 = 0.0082
|R 4
(0.5)| 4
15 (0.5)5 = 0.0083
EXERCISES18.5
1 Thefunction,y(x),is given byy(x) = sinx.
(a) Calculate the fifth-orderTaylor polynomial
generated byyaboutx = 0.
(b) Find an expression forthe remainder term of
order 5.
(c) Statean upper bound foryour expression in (b).
2 RepeatQuestion1withy(x) = cosx.
3 Thefunctiony(x) = e x maybe approximated by the
quadraticexpression1 +x+ x2
2 .Findanupperbound
forthe error term given |x| < 0.5.
4 (a) Find the third-orderTaylor polynomial generated
byh(t) = 1 aboutt = 2.
t
(b) State the error term.
(c) Find an upper bound forthe errorterm given
1t4.
5 The functiony(x) =x 5 +x 6 isapproximated bya
third-orderTaylor polynomial aboutx = 1.
(a) Find an expression forthe third-ordererror term.
(b) Find an upper boundforthe errorterm given
0x2.
Solutions
1 (a) x− x3
3! + x5
5!
(b) R 5 =− (sinc)x6
6!
whereclies between 0 andx
x 6
(c)
∣6!
∣
2 (a) 1− x2
2! + x4
4!
(b) R 5 =− (cosc)x6
6!
whereclies between 0 andx
x 6
(c)
∣6!
∣

524 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
3 3.435 ×10 −2
4 (a) − t3
16 + t2 2 − 3t
2 +2
(t −2) 4
(b)
c 5 whereclies between 2 andt
(c) 0.5
5 (a) 5c(1 +3c)(x −1) 4 forcbetween1andx
(b) 70
18.6 TAYLORANDMACLAURINSERIES
We have seen that y and its first n derivatives evaluated at x = a match p n
(x) and its
firstnderivatives evaluated atx = a. We have studied the difference betweeny(x) and
p n
(x), thatisthe errorterm,R n
(x).
AsmoreandmoretermsareincludedintheTaylorpolynomial,weobtainaninfinite
series,called aTaylorseries. We denotethisinfiniteseries by p(x).
Taylor series:
p(x) =y(a) +y ′ (a)(x −a) +y ′′ (x −a)2
(a)
2!
+···+y (n) (x −a)n
(a) +···
n!
+y (3) (x −a)3
(a)
3!
ForsomeTaylorseries,thevalueoftheseriesequalsthevalueofthegeneratingfunction
for every value ofx. For example, the Taylor series for e x , sinx and cosx equal the
valuesofe x ,sinxandcosxforeveryvalueofx.However,somefunctionshaveaTaylor
series which equals the function only for a limited range of x values. Example 18.14
gives a case of a function which equals its Taylor series only when −1 <x<1.
To determine whether a Taylor series, p(x), is equal to its generating function,y(x),
we need to examine the error term of ordern, that isR n
(x). We examine this error term
as more and more terms are included in the Taylor polynomial, that is as n tends to
infinity. If this error term approaches 0 asnincreases, then the Taylor series equates to
thegeneratingfunction,y(x).Sometimestheerrortermapproaches 0asnincreasesfor
all values ofx, sometimes it approaches 0 only whenxlies in some specified interval,
say, for example, (−1,1). Hence wehave:
IfR n
(x) → 0 asn → ∞ for all values ofx, then the Taylor series, p(x), and the
generatingfunction,y(x), are equal forallvalues ofx.
IfR n
(x) → 0asn → ∞forvaluesofxintheinterval (α,β),thentheTaylor
series,p(x),andthegeneratingfunction,y(x),areequalforxvaluesontheinterval
(α,β). For values ofxoutside the interval (α,β), the values of p(x) andy(x) are
different.
By examining the error terms associated withy = e x ,y = sinx andy = cosx it is
possible to show that these errors all approach 0 asn → ∞ for all values ofx. Hence
the functionsy = e x ,y = sinx andy = cosx are allequal totheircorrespondingTaylor
series forall values ofx.

18.6 Taylor and Maclaurin series 525
Aspecial,commonlyused,caseofaTaylorseriesoccurswhena = 0.Thisisknown
as theMaclaurin series.
Maclaurin series:
p(x) =y(0) +y ′ (0)x +y ′′ (0) x2
2! +y(3) (0) x3
3! +···+y(n) (0) xn
n! +···
Example18.10 Determinethe Maclaurinseries fory = e x .
Solution Inthisexampley(x) = e x and clearlyy ′ (x) = e x too.Similarly,
y ′′ (x) =y (3) (x) = ··· =y (n) (x) =e x
forall values ofn. Evaluating atx = 0 yields
and so
y(0)=y ′ (0)=y ′′ (0)=···=y (n) (0)=1
p(x)=1+x+ x2
2! + x3
3! +···+xn n! +···
Asmentionedearlier,theseriesandthegeneratingfunctionareequalforallvaluesofx.
Hence,
e x =1+x+ x2
2! + x3
3! +···
forall values ofx, that is
∞∑
e x x n
=
n!
n=0
Example18.11 Obtain the Maclaurinseries fory(x) = sinx.
Solution y(x)=sinx, y(0)=0
y ′ (x)=cosx, y ′ (0)=1
y ′′ (x)=−sinx, y ′′ (0)=0
y (3) (x)=−cosx,
y (3) (0)=−1
y (4) (x)=sinx, y (4) (0)=0
p(x) =y(0) +y ′ (0)x + y′′ (0)x 2
2!
=x − x3
3! + x5
5! − x7
7! +···
∞∑ (−1) i x 2i+1
=
(2i +1)!
i=0
+ y(3) (0)x 3
3!
+···

526 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Since the generating function and series are equal forall values ofxwehave
∞∑ (−1) i x 2i+1
sinx= =x − x3
(2i +1)! 3! + x5
5! − x7
7! +···
i=0
Example18.12 Obtain the Maclaurinseries fory(x) = cosx.
Solution y(x)=cosx, y(0)=1
y ′ (x)=−sinx, y ′ (0)=0
y ′′ (x)=−cosx,
y ′′ (0)=−1
y (3) (x)=sinx, y (3) (0)=0
y (4) (x)=cosx, y (4) (0)=1
and soon. Therefore
p(x) =y(0) +y ′ (0)x +y ′′ (0) x2
2! +y(3) (0) x3
3! +y(4) (0) x4
4! +···
= 1 − x2
2! + x4
4! −···
∞∑ (−1) i x 2i
=
(2i)!
i=0
Since the series and the generating function areequal for all values ofxthen
∞∑ (−1) i x 2i
cosx= = 1 − x2
(2i)! 2! + x4
4! − x6
6! +···
i=0
FromExamples18.10, 18.11and 18.12wenotethreeimportantMaclaurinseries:
e x =1+x+ x2
2! + x3
3! + x4
+ ··· forall values ofx
4!
sinx=x− x3
3! + x5
5! − x7
+ ··· forall values ofx
7!
cosx=1− x2
2! + x4
4! − x6
+ ··· forall values ofx
6!
Example18.13 Findthe Maclaurinseries forthe following functions:
( x
(a)y = e 2x (b)y = sin3x (c)y = cos
2)
Solution We use the previously stated series.
(a) We note that
e z =1+z+ z2
2! + z3
3! + z4
4! +···

18.6 Taylor and Maclaurin series 527
Substitutingz = 2x we obtain
(b) We note that
e 2x =1+2x+ (2x)2
2!
+ (2x)3
3!
+ (2x)4
4!
=1+2x+2x 2 + 4x3
3 + 2x4
3 +···
sinz=z− z3
3! + z5
5! − z7
7! +···
By putting z = 3x we obtain
(c) We note that
sin3x=3x− (3x)3
3!
+ (3x)5
5!
− (3x)7
7!
+···
+···
=3x− 9 2 x3 + 81
40 x5 − 243
560 x7 +···
cosz=1− z2
2! + z4
4! − z6
6! +···
Puttingz = x we obtain
2
( x
cos = 1 −
2) (x/2)2 + (x/2)4 − (x/2)6 +···
2! 4! 6!
= 1 − x2
8 + x4
384 − x6
46080 +···
Example18.14 Determinethe Maclaurinseries fory(x) = 1
1 +x .
Solution The value ofyand its derivatives atx = 0 arefound.
y(x) = 1
1 +x , y(0)=1
y ′ (x) =
−1
(1 +x) 2, y′ (0)=−1
y ′′ 2!
(x) =
(1 +x) 3, y′′ (0) =2!
y ′′′ (x) = −3!
(1 +x) 4, y′′′ (0) = −3!
.
.
y (n) (x) = (−1)n n!
(1 +x) n+1, y(n) (0) = (−1) n n!
Hence usingthe formulafor the Maclaurin series wefind
p(x) = 1 −1(x) +2! x2
2! −3!x3 3! +···+ (−1)n n! xn
n! +···
=1−x+x 2 −x 3 +x 4 −···+(−1) n x n +···

528 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
1
Itcan beshown thatthisseries converges to for |x| < 1.Hence,
1 +x
1
∞∑
1 +x =1−x+x2 −x 3 +···= (−1) n x n for |x| < 1
0
1
Forvaluesofxoutside (−1,1)thevaluesof
1 +x and ∑ (−1) n x n aresimplynotequal;
try evaluating the l.h.s. and r.h.s.with,say,x = −2.
Example18.15 Findthe Taylor series fory(x) = e −x aboutx = 1.
Solution y=e −x , y(1)=e −1
y ′ = −e −x ,
y ′′ =e −x ,
and soon. Hence,
y ′ (1) = −e −1
y ′′ (1) =e −1
e −x = e −1 − (e −1 −1
(x −1)2
)(x −1) +e
2!
{
= e −1 (x −1)2
1 −(x−1)+ −
2!
∑ ∞
e −x = e −1 n
(x −1)n
(−1)
n!
0
−1
(x −1)3
−e
3!
}
(x −1)3
3!
+···
+···
Example18.16 Findthe Maclaurinseries fory(x) =xcosx.
Solution TheMaclaurinseries, p(x), forcosx is
cosx=p(x)=1− x2
2! + x4
4! − x6
6! +···
So the Maclaurin series forxcosx isxp(x), that is
xcosx=xp(x)=x− x3
2! + x5
4! − x7
6! +···
An alternative form of Taylor series is often used in numerical analysis. We know
that the Taylor series fory(x) generatedaboutx =aisgiven by
y(x) =y(a) +y ′ (a)(x −a) +y ′′ (x −a)2
(a) +y (3) (x −a)3
(a) +···
2! 3!
Replacingabyx 0
we obtain
y(x) =y(x 0
) +y ′ (x 0
)(x −x 0
) +y ′′ (x 0
) (x−x 0 )2
2!
If we now letx−x 0
=h, wesee that
+y (3) (x 0
) (x−x 0 )3
3!
y(x 0
+h) =y(x 0
) +y ′ (x 0
)h +y ′′ (x 0
) h2
2! +y(3) (x 0
) h3
3! +···
To interpret thisformof Taylor series werefer toFigure 18.6.
+···

18.6 Taylor and Maclaurin series 529
y
y(x)
x 0 x 0 +h
x
Figure18.6
Thevaluey(x 0 +h) can be estimated usingvalues ofy
and itsderivatives atx =x 0 .
Ifyand its derivatives are known whenx =x 0
, then the Taylor series can be used to
findyatanearbypoint,wherex =x 0
+h.ThisformofTaylorseriesisusedinnumerical
methods of solving differential equations.
Engineeringapplication18.5
Electricfieldofanelectrostaticdipole
Engineersoftenneedtocalculatetheelectricalfieldduetoseveralstationaryelectric
charges.Theseareknownaselectrostaticproblems.Oneofthesimplestelectrostatic
configurations is that of two charges of opposite polarity separated by a distance,d.
Thisarrangementisknownasanelectrostaticdipole.ItisillustratedinFigure18.7
forcharges +Q and −Q.
The origin of the x axis is located midway between the two charges so that the
charge −Q has coordinate −d/2 and the charge +Qhas coordinated/2.
d
–Q 0 +Q
x
Figure18.7
Electrostatic dipole with two point charges of −Q
and +Qcoulombs respectively.
We wish to calculate the combined electric field of the two charges as a function
of the distance along thexaxis.
The electricfield,E, ofasinglecharge isgiven by
E =
q
4πε 0
r 2
where q is the charge in coulombs, ε 0
is the permittivity of free space (a constant
of approximately 8.85 ×10 −12 F m −1 ), andris the distance from the charge to the
measuring point.
The electricfield atpointxdue tothe left-hand charge is
E LEFT
=
−Q
(
4πε 0
x + d ) 2
2
and forthe right-hand charge the electric field atpointxis
E RIGHT
=
+Q
(
4πε 0
x − d ) 2
2
➔

530 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Thetotalelectricfield,E T
,canbeobtainedbyaddingthetwocontributionstogether:
E T
=E LEFT
+E RIGHT
=
⎡
= Q ⎣
4πε 0
1
(
x − d 2
−Q Q
(
4πε 0
x + d ) 2
+ (
4πε
2 0
x − d ) 2
2
) 2
−
1
⎤
(
x + d ) 2 ⎦
2
It is possible to gain further insight into the properties of the electrostatic dipole by
carrying out a power series expansion. To prepare the equation we take a factor of
1/x 2 outside of the brackets
E T
=
⎡
Q ⎣
4πε 0
x 2
1
(
1 − d 2x
) 2
−
1
(
1 + d 2x
) 2
⎤
⎦
1
Now consider the Maclaurin series of
(1−α) 2.Following
the process explained in
the previous examples, the first five termsare calculated:
1
y(α) =
(1−α) 2, y(0)=1
y ′ 2!
(α) =
(1−α) 3, y′ (0)=2!
y ′′ 3!
(α) =
(1−α) 4, y′′ (0) =3!
y (3) 4!
(α) =
(1−α) 5, y(3) (0) =4!
y (4) 5!
(α) =
(1−α) 6, y(4) (0) =5!
1
So, = 1+2!α+3! α2
(1−α) 2 2! +4!α3 3! +5!α4 4! +··· = 1+2α+3α2 +4α 3 +
5α 4 +···.
Following a similarprocess for the Maclaurin series of 1/(1 + α) 2 ,
1
y(α) =
(1+α) 2, y(0)=1
y ′ (α) = −2!
(1+α) 3, y′ (0) = −2!
y ′′ 3!
(α) =
(1+α) 4, y′′ (0) =3!
y (3) (α) = −4!
(1+α) 5, y(3) (0) = −4!
y (4) 5!
(α) =
(1+α) 6, y(4) (0) =5!
1
weobtain =1+ (−2!)α +3!α2
(1+α)
2
2! + (−4!)α3 3! +5!α4 4! +···
=1−2α+3α 2 −4α 3 +5α 4 +···.
These two expansions may be used to approximate the total electric field by substituting
α =d/(2x).

18.6 Taylor and Maclaurin series 531
The total electricfield isthus
E T
=
Q
+4α 3 +5α 4 +···)
4πε 0
x 2[(1+2α+3α2
−(1−2α+3α 2 −4α 3 +5α 4 +···)]
E T
=
Q +···]
4πε 0
x 2[4α+8α3
Providing x is much larger than d, then d/(2x), that is α is small, and the higher
order terms in the series become increasingly small. Thus we can approximate the
field using only the first term,
E T
∼ =
Q
4πε 0
x 2[4α] =
Q
4πε 0
x 24 d 2x =
Qd
2πε 0
x 3
It can now be seen that the total electric field of the dipole decays asx −3 rather than
x −2 , as would be the case for a singlepoint charge.
EXERCISES18.6
1 Usethe Maclaurin series forsinx to write down the
Maclaurinseries forsin5x.
2 Usethe Maclaurin series forcosx to write down the
Maclaurinseries forcos3x.
3 Usethe Maclaurin series fore x to write down the
Maclaurinseries for 1 e x.
4 Find the Taylor seriesfory(x) = √ x aboutx = 1.
5 (a) Findthe Maclaurin seriesfor
y(x) =x 2 +sinx.
(b) Deduce the Maclaurin series for
y(x) =x n +sinx forany positiveintegern.
6 (a) Obtainthe Maclaurinseries fory(x) =xe x .
(b) Statethe range ofvalues ofxforwhichy(x)
equals itsMaclaurin series.
7 Find the Taylor series fory(x) =x +e x aboutx = 1.
8 Find the Maclaurin seriesfory(x) = ln(1 +x).
Solutions
1 5x− (5x)3
3!
2 1 − (3x)2
2!
+ (5x)5
5!
+ (3x)4
4!
3 1−x+ x2
2! − x3
3! + x4
4! −···
(x −1)2
8
4 1 + x −1 −
2
5(x −1)4
− +···
128
5 (a) x+x 2 − x3
3! + x5
5! − x7
−···
(b) x n +x− x3
3! + x5
5! − x7
7! +···
− (3x)6 +··· 6 (a) x+x 2 + x3
6!
2! + x4
3! + x5
4! +···
(b) Forallvalues ofx
7 p(x) = (1 +e)x
(x −1)3
{
+ (x −1)
16
2 (x −1)3
+e + +
2! 3!
7! +··· 8 x − x2
2 + x3
3 − x4
4 + x5
5 −···
(x −1)4
4!
+···
}

532 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
REVIEWEXERCISES18
1 Afunction, f (t),and itsfirstand second derivatives (a) Calculatey ′ (1),y ′′ (1)andy (3) (1).
dy
dx + y2
2 =xy y(1)=2 sin(kx)
(c) lim 1 −cos(kx)
are evaluated att = 3. Find the second-order Taylor
polynomial generated by f aboutt = 3.
(b) Statethe third-orderTaylor polynomial generated
byy(x)aboutx = 1.
f(3)=2 f ′ (3)=−1 f ′′ (3)=1 (c) Estimatey(1.25).
2 Afunction,s(t),andits firstderivative are evaluated
att =3;s(3) =4,s ′ (3) = −1.
(a) Find the first-orderTaylor polynomial generated
bysatt=3.
(b) Estimates(2.9)ands(3.2).
(c) Ifadditionally weknows ′′ (3) = 1, use a
second-order Taylor polynomial to estimate
s(2.9)ands(3.2).
3 Obtain alinear approximation to
y(t) =at 4 +bt 3 +ct 2 +dt +earoundt =1.
4 Find a quadratic approximation toy =x 3 nearx = 2.
5 Find linear approximations to
8 (a) Findthe third-orderTaylor polynomial generated
byy(x) = e −x aboutx = 1.
(b) Statethe third-ordererror term.
(c) Findan upperbound forthe error termgiven
|x| < 1.
9 (a) Findaquadraticapproximation toy(x) = sin 2 x
aboutx = 0.
(b) Statethe remainder termoforder2.
(c) Statean upper boundforthe remainderterm
given |x| < 0.5.
10 (a) Findacubicapproximation toy(x) =xcosx
aboutx = 0.
(b) Statethe error termoforder3.
(a) z(t) =e t neart =1
(c) Statean upper boundforthe errorterm given
(b) w(t) =sin3t neart =1
|x| < 0.25.
(c) v(t) =e t +sin3tneart =1
6 Thefunction,y(x),satisfiesthe equation
y ′′ +xy ′ −3y=x 2 +1
11 Giventhaty(x) =x 2 ,
(a) Calculate the Taylor seriesofy(x) aboutx =a.
(b) Calculate the Maclaurin series ofy(x).
y(0) =1 y ′ (0) =2
12 Findthe Maclaurinexpansion ofy(x) = ln(2 +x) up
(a) Evaluatey ′′ (0).
to andincluding the terminx 4 .
(b) Differentiate the equation.
(c) Evaluatey (3) (0).
13 Byconsideringthe Maclaurinexpansions ofsin(kx)
andcos(kx),kconstant,evaluate ifpossible
(d) Write down acubic approximation fory(x).
sin(kx)
(a) lim
(e) Estimatey(0.5).
x→0 x
7 Given thatysatisfiesthe equation
cos(kx) −1
(b) lim x→0 x
Solutions
1
t 2 2 −4t+19 2
2 (a) −t +7 (b) 4.1, 3.8 (c) 4.105, 3.82
3 (4a+3b+2c+d)t−3a−2b−c+e
4 6x 2 −12x+8
5 (a) et (b) −2.97t +3.11
(c) −0.25t +3.11
6 (a) 4 (b) y ′′′ +xy ′′ −2y ′ = 2x
(c) 4 (d) 1+2x+2x 2 + 2 3 x3
(e) 2.5833

Review exercises 18 533
7 (a) 0, 2, −2 (b) − x3
3 +2x2 −3x+ 10
3
(c) 2.0573
{
}
8 (a)e −1 − x3
6 +x2 − 5x
2 + 8 3
(b) e−c (x −1) 4
4!
(c) 2e
3
9 (a) x 2
whereclies between 1 andx
(b) − 2sin(2c)x3 forcbetween 0 andx
3
(c) 0.07
10 (a) x− x3
2!
(b) (4sinc +ccosc) x4 forcbetween 0 andx
4!
(c) 2 ×10 −4
11 (a) x 2 (b) x 2
12 ln2+ x 2 − x2
8 + x3
24 − x4
64 +···
13 (a) k (b) 0 (c) notdefined

19 Ordinarydifferential
equationsI
Contents 19.1 Introduction 534
19.2 Basicdefinitions 535
19.3 First-orderequations:simpleequationsandseparationofvariables 540
19.4 First-orderlinearequations:useofanintegratingfactor 547
19.5 Second-orderlinearequations 558
19.6 Constantcoefficientequations 560
19.7 Seriessolutionofdifferentialequations 584
19.8 Bessel’sequationandBesselfunctions 587
Reviewexercises19 601
19.1 INTRODUCTION
Thesolutionofproblemsconcerningthemotionofobjects,theflowofchargedparticles,
heat transport, etc.,often involves discussion of relations of the form
d 2 x
dt 2 +6dx dt +2x=3t or dq
dt +8q=sint
In the first equation,xmight represent distance. For this case dx is the rate of change
dt
of distance with respect to timet, that is speed, and d2 x
represents acceleration. In the
dt2 second equation,qmight be charge and dq the rate of flow of charge, that is current.
dt
These are examples of differential equations, so called because they are equations involving
the derivatives of various quantities. Such equations arise out of situations in
whichchangeisoccurring.Tosolvesuchadifferentialequationmeanstofindthefunctionx(t)orq(t)whenwearegiventhedifferentialequation.Solutionstotheseequations
maybeanalyticalinthatwecanwritedownananswerintermsofcommonelementary
functions such as e t , sint and so on. Alternatively, the problem may be so difficult that
only numerical methodsare available, which produce approximate solutions.

19.2 Basic definitions 535
In engineering, differential equations are most commonly used to model dynamic
systems. These are systems which change with time. Examples include an electronic
circuit with time-dependent currents and voltages, a chemical production line in which
pressures, tank levels, flow rates, etc., vary with time, and a semiconductor device in
which hole and electron densities change with time.
19.2 BASICDEFINITIONS
Inordertosolveadifferentialequationitisimportanttoidentifycertainfeatures.Recall
fromChapter2thatinafunctionsuchasy =x 2 +3xwesayxistheindependentvariable
andyis the dependent variable since the value ofydepends upon the choice we have
made forx.
Inadifferential equation suchas
dy
dx −2y=3x2
x isthe independent variable, andyisthe dependent variable.
Similarly, forthe differential equation
dx
dt +7x=et
t isthe independent variable andxisthe dependent variable.
We see thatthe variable being differentiated isthe dependent variable.
Before classifying differential equations, wewill derive one.
Engineeringapplication19.1
AnRC chargingcircuit
Consider the RC circuit of Figure 19.1. Suppose we wish to derive a differential
equation which models the circuit so that we can determine the voltage across the
capacitor at any time,t. Clearly there are two different cases corresponding to the
switchbeingopenandtheswitchbeingclosed.Wewillconcentrateonthelatterand
for convenience assume that the switch is closed att = 0. From Kirchhoff’s voltage
law wehave
v S
=v R
+v C
thatis v R
=v S
−v C
where v S
isthevoltageofthesupply, v C
isthevoltageacrossthecapacitor,and v R
is
the voltageacrossthe resistor.Using Ohm’slaw forthe resistor thengives
i = v S −v C
R
whereiisthecurrentflowinginthecircuitaftertheswitchisclosed.Forthecapacitor
i =C dv C
dt
Combining these equations gives
C dv C
dt
= v S −v C
R
➔

536 Chapter 19 Ordinary differential equations I
y R
y S
+
–
i
R
C
y C
Figure19.1
AnRC charging circuit.
and hence
RC dv C
+v
dt C
=v S
This isthedifferential equation which models the variation involtageacross thecapacitorwithtime.Here
v C
isthedependentvariableandt istheindependentvariable,
and when we are required to solve this differential equation we must attempt to find
v C
as a function oft.
19.2.1 Order
Differential equations which have features in common are often grouped together and
givencertainclassificationsanditisusuallythecasethatappropriatemethodsofsolution
depend upon the classifications. Someimportantterminology isnow given.
Theorder ofadifferential equation isthe orderofits highestderivative.
Example19.1 State the orderof
(a) d2 y
dx 2 + dy
dx =x
dx
(b)
dt = (xt)5
Solution (a) The highest derivative is d2 y
dx2, a second derivative. The order istherefore two.
(b) The only derivative appearing is dx , a first derivative. The order istherefore one.
dt
19.2.2 Linearity
Recall that in a differential equation such as dy
dx +3y =x2 , the independent variable is
x and the dependent variable isy.
Adifferential equation issaidtobelinearif:
(1) the dependent variable and itsderivatives occur tothe first power only,
(2) thereare no products involving the dependent variable with its derivatives, and
(3) there are no non-linear functions of the dependent variable such as sine, exponential,etc.
Ifanequationisnotlinear,thenitissaidtobenon-linear.Notefrom(2)thataproduct
of terms involving the dependent variable such asy dy is non-linear. Note from (3) that
dx
the existence of termssuch asy 2 , siny and e y causes anequation tobe non-linear.

19.2 Basic definitions 537
Note also that the conditions for linearity are conditions on the dependent variable.
The linearity of a differential equation is not determined or affected by the presence of
non-linear terms involving the independent variable.
The distinction between a linear and a non-linear differential equation is important
because the methods of solution depend upon whether an equation is linear or nonlinear.
Furthermore, it is usually the case that a linear differential equation is easier to
solve.
Example19.2 Decide whether or notthe following equations are linear:
(a) sinx dy
dx +y=x
dx
(b)
dt +x=t3
d 2 y
(c)
dx 2 +y2 =0
dy
(d)
dx +siny=0
Solution In (a), (c) and (d) the dependent variable isy, and the independent variable isx. In (b)
the dependent variable isxand the independent variable ist.
(a) This equation islinear.
(b) Thisequationislinear.Itdoesnotmatterthatthetermint,theindependentvariable,
israised tothe power 3.
(c) This equation isnon-linear, the non-linearity arising through the termy 2 .
(d) This equation isnon-linear, the non-linearity arising through the termsiny.
19.2.3 Thesolutionofadifferentialequation
The solution of a differential equation is a relationship between the dependent and independent
variables such that the differential equation is satisfied for all values of the
independentvariable over a specified domain.
Example19.3 Verifythaty = e x isasolution ofthe differential equation
dy
dx =y
Solution If y = e x then dy
dx = ex . For all values of x, we see that dy
dx =yandsoy=ex isa
solution.Note alsothatthisequation isfirst order and linear.
Therearefrequentlymanydifferentfunctionswhichsatisfyadifferentialequation;that
is,therearemanysolutions.Thegeneralsolutionembracesalloftheseandallpossible
solutionscan beobtained from it.

538 Chapter 19 Ordinary differential equations I
Example19.4 Verifythaty =Ce x is a solution of dy =y, whereC is any constant.
dx
Solution If y = Ce x , then dy
dx = Cex . Therefore, for all values of x, dy = y and the equation
dx
is satisfied for any constantC;C is called an arbitrary constant and by varying it, all
possible solutions can be obtained. For example, by choosingC to be 1, we obtain the
solution of the previous example. Infact,y =Ce x isthe general solution of dy
dx =y.
More generally, to determine C we require more information given in the form of a
condition.Forexample,ifwearetoldthat,atx = 0,y = 4thenfromy =Ce x wehave
4=Ce 0 =C
so thatC = 4. Therefore y = 4e x is the solution of the differential equation which
additionally satisfies the condition y(0) = 4. This is called a particular solution. In
general, application of conditions to the general solution yields the particular solution.
Toobtainaparticularsolution,thenumberofgivenindependentconditionsmustbethe
same asthe number ofconstants.
Consider the following example.
Example19.5 Consider the second-order differential equation
d 2 y
dx 2 +y=0
The general solution of this equation can be shown tobe
y=Acosx+Bsinx
whereAandBarearbitraryconstants.
Find the particular solution which satisfies the conditions
(a) whenx = 0,theny = 0,and
(b) whenx = 3π 2 ,theny=1.
Solution Wenotethatbecausethegeneralsolutionhastwoarbitraryconstants,AandB,thentwo
conditions are necessary to obtain a particular solution. Applying the first condition to
the general solution gives
0=Acos0+Bsin0
=A
ThereforeA = 0,andthesolutionreducestoy =Bsinx.Applyingthesecondcondition
we find
1=Bsin 3π 2

19.2 Basic definitions 539
fromwhich
1 =B(−1)
B=−1
The particular solution becomesy = −sinx.
Sometimes the conditions involve derivatives.
Example19.6 ForthedifferentialequationofExample19.5findtheparticularsolutionwhichsatisfies
the conditions
(a) whenx = 0,theny = 0,and
(b) whenx = 0,then dy
dx = 5.
Solution Application ofthe first condition tothe general solutiony =Acosx +Bsinx gives
0=Acos0+Bsin0
=A
ThereforeA = 0 and the solution becomesy = Bsinx. To apply the second condition
wemustdifferentiatey:
dy
dx =Bcosx
Then applying the second condition weget
5=Bcos0
=B
so thatB = 5.Finallythe required particular solution isy = 5sinx.
In this example both conditions have been specified atx = 0, and are often referred
toas initialconditions.
EXERCISES19.2
1 Verifythaty = 3sin2xisasolutionof d2 y
dx 2 +4y=0.
2 Verifythat 3e x ,Axe x ,Axe x +Be x ,whereA,Bare
constants,allsatisfythe differentialequation
d 2 y
dx 2 −2dy dx +y=0
3 Verifythatx =t 2 +Alnt +Bis asolution of
t d2 x
dt 2 + dx
dt = 4t
4 Verify thaty =Acosx +Bsinx satisfiesthe
differentialequation
d 2 y
dx 2 +y=0
Verify alsothaty =Acosx andy =Bsinx each
individually satisfythe equation.
5 Ify =Ae 2x is the general solution of dy = 2y, find
dx
the particular solution satisfyingy(0) = 3.What is
the particular solution satisfying dy
dx =2whenx=0?

540 Chapter 19 Ordinary differential equations I
6 Identifythe dependent andindependent variables of
the followingdifferential equations. Give the order of
the equations and state which are linear.
(a)
(b)
(c)
dy
dx +9y=0
( )( )
dy d 2 y
dx dx 2 +3 dy
dx = 0
d 3 x
dt 3 +5dx dt =sinx
7 Show thatx(t) = 7cos3t −2sin2t isasolution of
d 2 x
dt 2 +2x=−49cos3t+4sin2t
8 Thegeneral solution ofthe equation
d 2 x
dt 2 −3dx +2x = 0 is given by
dt
x=Ae t +Be 2t
Findthe particularsolution whichsatisfiesx = 3 and
dx
dt =5whent=0.
9 Thegeneral solution of d2 y
dx 2 −2dy dx +y=0is
y =Axe x +Be x .Findthe particular solution
satisfyingy(0) = 0, dy (0) =1.
dx
10 Thegeneral solution of d2 x
dt 2 = −ω2 xis
x =Ae jωt +Be −jωt ,where j 2 = −1.Verify that this
isindeed asolution. What is the particular solution
satisfyingx(0) = 0, dx (0) = 1? Expressthe general
dt
solution and the particular solution in terms of
trigonometric functions.
Solutions
5 y(x) =3e 2x ,y(x) =e 2x
6 (a) y isthe dependentvariable;xisthe independent
variable;firstorder, linear
(b) y isthe dependentvariable;xisthe independent
variable;second order,non-linear
(c) x is the dependent variable;t is the independent
variable;thirdorder,non-linear.
8 x=e t +2e 2t
9 y=xe x
10 particularsolution: sin ωt/ω;general solution:
(A +B)cosωt + (A −B)jsinωt
19.3 FIRST-ORDEREQUATIONS:SIMPLEEQUATIONS
ANDSEPARATIONOFVARIABLES
19.3.1 Simpleequations
Thesimplest first-orderequationstodealwith arethose ofthe form
dy
dx =f(x)
where the r.h.s. is a function of the independent variable only. No special treatment is
necessary and direct integration yieldsyas a function ofx, thatis
∫
y = f(x)dx

19.3 First-order equations: simple equations and separation of variables 541
Example19.7 Find the general solution of dy
dx = 3cos2x.
Solution Given that dy
dx = 3cos2x, theny = ∫ 3cos2xdx = 3 sin2x +C. This is the required
2
general solution.
If dy
dx =f(x)theny= ∫
f(x)dx.
19.3.2 Separationofvariables
When the function f on the r.h.s. of the equation depends upon both independent and
dependentvariablestheapproachofSection19.3.1isnotpossible.However,first-order
equations which can be writteninthe form
dy
= f(x)g(y) (19.1)
dx
forman important class known asseparable equations. For example,
dy
dx = 3x2 e −4y
isaseparable equation forwhich
f(x) = 3x 2 and g(y) = e −4y
To obtain a solution wefirst divide both sides of Equation (19.1) byg(y) togive
1 dy
g(y) dx =f(x)
Integrating both sides with respecttoxyields
∫ ∫ ∫
1 dy
g(y) dx dx = 1
g(y) dy = f(x)dx
Theequationisthensaidtobeseparated.Ifthelasttwointegralscanbefound,weobtain
arelationshipbetweenyandx,althoughitisnotalwayspossibletowriteyexplicitlyin
termsofxasthe following examples will show.
Separationofvariables:
Thesolution of dy = f (x)g(y) isfound from
dx
∫ ∫
1
g(y) dy = f(x)dx

542 Chapter 19 Ordinary differential equations I
Example19.8 Solve dy
dx = e−x
y .
Solution Here f (x) = e −x andg(y) = 1 . Multiplication through byyyields
y
y dy
dx = e−x
Integration of both sides with respect toxgives
∫ ∫
ydy= e −x dx
so that
y 2
2 =−e−x +C
Note that the constants arising from the two integrals have been combined to give a
single constant,C. Finally we can rearrange this expression togiveyinterms ofx:
that is,
y 2 =−2e −x +2C
y=± √ D−2e −x
whereD = 2C.
It is important to stress that the constant of integration must be inserted at the stage at
which the integration is actually carried out, and not simply added to the answer at the
end.
Example19.9 Solve dy
dx = 3x2 e −y subjecttoy(0) = 1.
Solution Hereg(y) = e −y and f (x) = 3x 2 . Separating the variables and integrating wefind
∫ ∫
e y dy= 3x 2 dx
so thate y =x 3 +C. Imposing the initial conditiony(0) = 1 wefind
e 1 =(0) 3 +C
so thatC = e.Therefore,
e y =x 3 +e
Note that since the exponential function is always positive, the solution will be valid
only forx 3 +e > 0.Taking natural logarithms gives the particular solution explicitly:
y =ln(x 3 +e)

19.3 First-order equations: simple equations and separation of variables 543
Example19.10 Solve
dx
dt = t2 +1
x 2 +1
Solution Separating the variables and integrating we find
∫ ∫
x 2 +1dx= t 2 +1dt
Therefore,
x 3
3 +x = t3 3 +t+C
which is the general solution. Here we note that x has not been obtained explicitly in
terms oft, although we have found a relationship between x andt which satisfies the
differential equation. To obtain the value of x at any givent it would be necessary to
solve the cubic equation.
Sometimes,equationswhicharenotimmediatelyseparablecanbereducedtoseparable
formby anappropriatesubstitution asthe following example shows.
Example19.11 Bymeansofthe substitutionz = y , solve the equation
x
dy
dx = y2
x + y +1 (19.2)
2 x
Solution Ifz = y theny = zx. Because the solution,y, is a function ofxthe variablezdepends
x
uponxalso. The product rulegives dy
dx =z+xdz , so thatEquation (19.2) becomes
dx
z +x dz
dx =z2 +z+1
thatis,
x dz
dx =z2 +1
Thisnewequationhasindependentvariablexanddependentvariablez,andisseparable.
We find
∫
∫
dz dx
z 2 +1 = x
so thattan −1 z = ln|x| +C. WritingC = ln |D|wehave
tan −1 z = ln|x| +ln|D| = ln|Dx|
so that z = tan(ln|Dx|). Returning to the original variables we see that the general
solution is
y =zx =xtan(ln|Dx|)

544 Chapter 19 Ordinary differential equations I
Engineeringapplication19.2
RLcircuitwithstepinput
Writedownthedifferentialequationgoverningthecurrent,i,flowingintheRLcircuit
shown in Figure 19.2 when a stepvoltage of magnitudeE isapplied tothe circuitat
t = 0.Solvethisdifferentialequationtoobtaini(t).Assumethatwhent = 0,i = 0.
Solution
Applying Kirchhoff’s voltage law and Ohm’s law tothe circuitwefind
iR+L di =E
dt
for t0
thatis,
L di
dt =E−iR
so that
∫ ∫
L
E−iR di = dt
R
L
i
E/R
i
Closed
at t = 0
+ –
E
Figure19.2
Astepvoltage isapplied to the circuitat
t=0.
0.63E/R
t
Figure19.3
Responseofthe circuitofFigure 19.2 to
astepinput.
NoteinparticularthatinthisequationL,E andRareconstantsandsothevariablesi
andt have been separated. If the applied voltage,E, varied with time this would not
havebeenthecasesincethel.h.s.wouldcontaintermsdependentupont.Integrating,
wefind ∫
∫
L −L
di =
E−iR R
−R
E−iR di=−L R ln(E−iR)=t+C
Tofindtheconstantofintegration,C,aconditionisrequired.Thephysicalcondition
i = 0 att = 0 provides this. Applyingi = 0 whent = 0 we find
C = − L R lnE
Substituting thisvalue gives
− L R ln(E−iR)=t−L R lnE
Thus,
L
R [lnE−ln(E−iR)]=t
t

19.3 First-order equations: simple equations and separation of variables 545
so that
L
R ln E
E−iR =t
Then
E
ln
E−iR = Rt
L
hence
E
E−iR = eRt/L
Rearranging toobtainE gives
hence
and so
so
E =(E−iR)e Rt/L
E =Ee Rt/L −iRe Rt/L
iRe Rt/L =E ( e Rt/L −1 )
iR=E ( 1 −e −Rt/L)
Finally we have
i = E ( )
1 −e
−Rt/L
R
ThegraphofthiscurrentagainsttimeisshowninFigure19.2.Wenotethatast → ∞,
i → E R .Therateatwhichthecurrentincreasestowardsitsfinalvaluedependsupon
thevaluesofthecomponentsRandL.Itiscommontodefineatimeconstant, τ,for
the circuit. Inthis case τ = L and the equation forthe current can be written as
R
i = E R
(
1 −e
−t/τ )
The smaller the value of τ, the more rapidly the current reaches its final value. It is
possibletoestimateavalueof τ fromalaboratorytestcurvebynotingthatafterone
time constant, thatist = τ,ihas reached 1 −e −1 ≈ 0.63 of its final value.
EXERCISES19.3
1 Find the general solution ofthe followingequations:
(a)
(c)
dy
dx =3
dy
dx = 2x
(b) dx
dt = 5
(d) dy
dt = 6t
(e)
(g)
(i)
dy
dx = 8x2
dy
dx = x2
y
dx
dt = et
x
(f)
(h)
(j)
dx
dt =3t3
dx
dt = t3
x 2
dy
dx = e−2x
y 2

546 Chapter 19 Ordinary differential equations I
dy
(k)
dx = 6sinx dx
(l)
y dt = 9cos4t
dy
x 2
(b)
dx = −ky, k constant
dx
(m)
dt = 3cos2t+8sin4t
dy
(c)
x 2 dx =y2
+x
(d) y dy
2 Find the particular solution ofthe following
dx =sinx
equations:
(e) y dy
dx
dx =x+2
(a) =3t, x(0) =1
dt (f) x 2dy
dy
(b)
dx = 6x2
dx =2y2 +yx
y , y(0)=1
dx
(g)
dy
(c)
dt = 3sint
dt = t4
x 5
, y(0)=2
y
5 Findthe general solutionsofthe followingequations:
dy
(d)
dx = e−x
y , y(0)=3
dx
(a)
dt =xt (b) dy
dx = x y
dx
(e)
dt = 4sint+6cos2t , x(0)=2 (c) t dx
x
dt =tanx
3 Find the general solution of dx
dx
= lnt.Find the (d)
dt dt = x2 −1
t
particular solution satisfyingx(1) = 1.
6 Findthe general solution ofthe equation
4 Find the general solutionsofthe following equations: dx
=t(x −2).Findthe particular solution which
dy
(a)
dx =kx, k constant dt
satisfiesx(0) = 5.
Solutions
1 (a) y=3x+c (b) x=5t+c 3 tln|t|−t+c,tln|t|−t+2
(c) y=x 2 +c (d) y=3t 2 +c
(e) y= 8 3 x3 +c (f) x= 3 4 (a) kx2
2 +c
4 t4 +c
(b)Ae −kx
y 2
(g)
2 = x3
3 +c (h) x 3
3 = t4 4 +c (c) − 1
x +c
x 2 y 3 (i)
2 =et e−2x
+c (j) =c − (d)y 2 =2(c −cosx)
3 2
y 2
(k)
2 =c−6cosx (l) x 3
3 = 9 (e)y 2 =x 2 +4x+c
4 sin4t+c x
(f)
x 3
(m)
3 + x2
2 = 3 A−2ln|x|
2 sin2t−2cos4t+c
2 (a) x= 3 (g) x6
2 t2 +1 (b) y 2 =4x 3 6 = t5 5 +c
+1
5 (a) x=Ae t2 /2 (b) y 2 =x 2 +c
y 2
(c)
2 =5−3cost (d) y2 =11−2e −x
(c) x = sin −1 (kt) (d) x = 1+At2
x 2
1−At 2
(e)
2 =3sin2t−4cost+6 6 2+Ae t2 /2 ,2+3e t2 /2

19.4 First-order linear equations: use of an integrating factor 547
19.4 FIRST-ORDERLINEAREQUATIONS:USEOFAN
INTEGRATINGFACTOR
Inthissection wedevelop a method for solvingfirst-order linear differential equations.
19.4.1 Exactequations
Consider the differential equation
dy
dx = 3x2
This can be solved very easily by simplyintegrating both sidestogive
∫
y = 3x 2 dx
=x 3 +c
wherecis the constant of integration. An equation which can be solved by integrating
both sides is said to be an exact equation. A more complicated example which, nevertheless,can
besolved inthe same way is
d
(xy) = 3x2
dx
Integrating both sides wefind
∫
xy = 3x 2 dx
=x 3 +c
and dividing through byxgives the general solution
y=x 2 + c x
The differential equation we have justsolved isan exact equation.
Example19.12 Solve the equation
d (
x 2 y ) =cosx
dx
Solution Integrating both sides wefind
∫
x 2 y = cosxdx
so that
=sinx+c
y = sinx
x 2 + c x 2

548 Chapter 19 Ordinary differential equations I
Consider again the differential equation of the previous example:
d
dx (x2 y) =cosx
Using the product rulefordifferentiation we can expand the l.h.s. as follows:
d
dx (x2 y) =x 2dy
dx +2xy
Doing this, the differential equation can bewritten inthe equivalent form
x 2dy
dx +2xy=cosx
Suppose we had posed the question in this form. A method of solving this equation
would be to recognize that the equation is exact and that the l.h.s. could be written as
d
dx (x2 y).
Itiseasy torecognize an exact equation because itwill always take the form
µ dy
dx +µ′ y=f(x)
where µ is some function ofx. That is, the coefficient ofyis the derivative of the coefficient
of dy
dx . When thisisthe case the l.h.s. can be written d dx (µy).
Example19.13 The following equations are exact. Note in each case that the coefficient of y is the
derivative ofthe coefficient of dy . Solve them.
dx
(a) x 3dy
dx +3x2 y =e 2x
(b) cosx dy − (sinx)y =1
dx
Solution (a) The equation can be written
d
dx (x3 y) =e 2x
and so, upon integrating,
∫
x 3 y = e 2x dx
so that
= e2x
2 +c
y = e2x
2x 3 + c x 3
(b) The equation can be written
d
((cosx)y) =1
dx

and so, upon integrating,
so that
(cosx)y=x+c
y =
19.4 First-order linear equations: use of an integrating factor 549
x
cosx + c
cosx
EXERCISES19.4.1
1 Eachofthe followingequations is exact.Solvethem.
(a) x 2dy
dx +2xy=x3
(b)
1 dy
x 2 dx − 2 x 3y=5x3
(c) e x (
y + dy
dx
)
=cosx
Solutions
1 (a) y= x2
4 + C x 2 (b) y= 5x6
4 +Cx2 (c) y=e −x sinx+Ce −x
19.4.2 Apreliminaryresultinvolvingseparationofvariables
Consider the following differential equation forthe dependentvariable µ:
dµ
dx
= µP (19.3)
wherePissome function ofxonly. Usingseparation ofvariables wehave
1 dµ
µ dx =P
and integrating both sides
∫ 1
µ dµ = ∫
Pdx
∫
lnµ= Pdx
µ = e ∫ Pdx
Inthisdevelopmenttheconstantofintegrationhasbeenomitted.Thereasonforthiswill
be apparent in what follows. So the solution of Equation (19.3), for any functionP(x),
isµ=e ∫ Pdx .
For example, if P(x) = 1 dµ
, then Equation (19.3) is
x dx = µ and its solution is
x
µ=e ∫ (1/x)dx = e lnx =x.

550 Chapter 19 Ordinary differential equations I
19.4.3 First-orderlinearequations
First-order linear differential equations can always bewritten inthe ‘standard’ form
dy
+P(x)y =Q(x) (19.4)
dx
wherePandQare both functions of the independent variable,x, only. In some cases,
either of these may be simplyconstants.
An example of suchan equation is
dy
+3xy = 7x2
dx
Comparing this with the standard form in Equation (19.4), note that P(x) = 3x and
Q(x) = 7x 2 . As a second example consider
dy
dx − 2y
x =4e−x
HereP(x) = − 2 x (note inparticular the minus sign) andQ(x) = 4e−x .
Finally, inthe equation
dy
dx −5y=sinx
note thatP(x) issimplythe constant −5.
Variables other thanyandxmay be used. So, forexample,
dx
dt +8tx=3t2 −5t
isafirst-orderlinearequationintheformofEquation(19.4)butwithindependentvariablet
and dependent variablex. HereP(t) = 8t andQ(t) = 3t 2 −5t.
In what follows it will be important that you can distinguish between the dependent
and independent variables, and alsothatyou areable toidentifythe functionsPandQ.
Equations such as these arise naturally when modelling many engineering applications.Forexample,theequationwhichdeterminesthecurrentflowinaseriesRLcircuit
when the applied voltage takes the form of a ramp isgiven by
di
dt + R L i = t L
This is a first-order linear equation in whichP(t) is the constant R L andQ(t) = t L . You
will learnhow tosolve such equations inthe following section.
19.4.4 Theintegratingfactormethod
All first-order linear equations, even when they are not exact, can be made exact by
multiplying them through by a function known as an integrating factor. As we have
seen, the solution then follows by performing an integration. For example, the linear
equation
dy
dx + 3 x y = e2x
x 3
is not exact, but multiplying it through by x 3 produces the differential equation in
Example 19.13(a) which isexact.

19.4 First-order linear equations: use of an integrating factor 551
Consider again a first-order linear equation instandard form:
dy
+Py=Q (19.5)
dx
where P and Q are functions of x only. We are aiming to solve this and express the
dependentvariableyintermsoftheindependentvariablex.Theaimistomultiply(19.5)
through by a function µ to make the equation exact. That is, so that the l.h.s. can be
written inthe form
d
dx (µy)
At thisstage the function µisnot known. Multiplying (19.5) through by µyields
µ dy +µPy=µQ (19.6)
dx
If the l.h.s. istoequal
d
(µy) = µdy
dx dx + µPy
d
(µy) then wemusthave
dx
Expanding the l.h.s. using the product rulegives
µ dy
dx +ydµ dx = µdy dx + µPy
which simplifies to
y dµ
dx = µPy
and consequently
dµ
dx = µP
Usingthe resultinSection 19.4.2 wesee that thisequation has solution
µ = e ∫ Pdx
The function µ is called an integrating factor and is a function of x only. With this
choice of µ, the l.h.s. of(19.6) isthe same as d (µy) and hence (19.6) can bewritten
dx
d
(µy) = µQ
dx
This exact equation can be solved by integration togive
∫
µy = µQdx
and consequently
y = 1 ∫
µQdx
µ

552 Chapter 19 Ordinary differential equations I
Insummary:
Given any first-order linear equation instandard form
dy
dx +Py=Q
wherePandQarefunctions ofx, the integrating factor µ isgiven by
µ = e ∫ Pdx
and the solution of the equation isthen obtained from
∫
µy = µQdx
It is important when working on a particular differential equation to rewrite the standard
formulae in the correct form before use. Essentially this means using the correct
dependentandindependentvariablesintheequations.Forexample,ifxisthedependent
variable andt isthe independent variable then the equations areas follows:
Theintegrating factorfor
dx
dt +Px=Q
wherePandQarefunctions oft, isgiven by
µ = e ∫ Pdt
and the solution of the equation isobtained from
∫
µx = µQdt
These results areillustratedinthe examples which follow.
Example19.14 Solve the differential equation dy
dx + y = 1 using the integrating factor method.
x
Solution Referring tothe standard first-order linear equation
dy
+P(x)y =Q(x)
dx
weseethatP(x) = 1 andQ(x) = 1.Usingthe previous formula for µ(x), we find
x
µ(x) = e ∫ (1/x)dx
= e lnx
=x

19.4 First-order linear equations: use of an integrating factor 553
Then fromthe first key pointon page 552 with µ =xandQ = 1 wehave
∫
xy = xdx= x2
2 +C
and finallyy = x 2 +C x
isthe required general solution.
Engineeringapplication19.3
RLcircuitwithrampinput
Recall from Section 2.4.7 that a standard ramp function has value 0 fort < 0 and a
valuect fort 0. This is shown in Figure 2.46. A voltage ramp signal with a value
c = 1 is applied to an RL circuit. The arrangement is shown in Figure 19.4. The
differential equation governing the currentflow,i(t), inthiscircuitisgiven by
iR+L di =t for t0 i(0)=0
dt
ShowthatthisequationcanbewrittenintheformofEquation(19.4).Henceusethe
integrating factor method tofindi(t).
i
y (t)
R
0
t
L
Figure19.4
A ramp input signal applied to aseriesRL
circuit.
Solution
This equation can bewritten as
di
dt + R L i = t for t0
L
which is a first-order linear equation.
Noteinthiscasethattheindependentvariableist andthedependentvariableisi.
Instandard form, wehave
di
+P(t)i =Q(t)
dt
whereP(t) = R L andQ(t) = t . The integrating factor, µ, isgiven by
L
µ = e ∫ P(t)dt
= e ∫ (R/L)dt
= e Rt/L ➔

554 Chapter 19 Ordinary differential equations I
Then
ie Rt/L = 1 L
= 1 L
∫
te Rt/L dt
( te
Rt/L
R/L − ∫ e
Rt/L
R/L dt )
= teRt/L
R − L R 2 eRt/L +K
whereK isthe constant of integration. The general solution is
i = t R − L R 2 +Ke−Rt/L
When t = 0, i = 0. This gives the initial condition required to find K. Applying
i(0) = 0 givesK =L/R 2 so thatthe particular solution is
i = t R + L R 2 (
e −Rt/L −1 )
In many engineering applications the terms transient response, steady-state response,
zero-input response, and zero-state response arise through the solution of differential
equations.These termsare explained inthe following example.
Engineeringapplication19.4
RC circuit:zero-inputresponseandzero-stateresponse
The differential equation which is used to model the variation in voltage across a
capacitor, v C
(t), inaseriesRC circuitwasderived inEngineeringapplication 19.1:
RC dv C
+v
dt C
=v S
(t)
where v S
(t) is the applied voltage. Suppose that this applied voltage takes the form
v S
(t) = V cos ωt. Suppose also that when the switch is closed att = 0, the initial
voltage across the capacitor isV 0
, thatis v C
(0) =V 0
.
(a) Show that this equation can be written in the form of Equation (19.4); that is, it
is a first-order linear equation.
(b) Usetheintegratingfactormethodtoobtaintheparticularsolutionofthisequation
which satisfies the given initial condition.
(c) Obtain the particular solution of the equation subject instead to the initial condition
v C
(0) = 0. It corresponds to the response of the system when there is no
initialenergyinthecircuit.Thissolutionisreferredtoasthezero-stateresponse.
(d) Obtainthesolutionoftheequationinthecasewhenthesupplyvoltageisidenticallyzero,
v S
(t) = 0,subjecttothegivencondition v C
(0) =V 0
.Thissolutionis
often referred to as the zero-input response, and represents the response of the
systemwhen the input iszero.

19.4 First-order linear equations: use of an integrating factor 555
(e) Showthatthesolutionin(b)canbewrittenasthesumofthezero-stateresponse
and the zero-input response.
(f) Identifythe transient and steady-state termsinthe solution topart(e).
Solution
(a) The given differential equation can berewritten as
dv C
+ 1
dt RC v C = 1
RC v S (t)
Comparing the form of this equation with (19.4) we see that it is a first-order
linearequation,withindependentvariablet,anddependentvariable v C
,inwhich
P(t) = 1
RC andQ(t) = 1
RC v S (t).
(b) The integrating factor isgiven by
µ = e ∫ (1/RC)dt = e t/(RC)
Itfollows from the second key point on page 552 that
e t/(RC) v C
= V ∫
e t/(RC) cosωtdt
RC
Thisintegralcanbeevaluatedusingintegrationbypartstwicefollowingthetechnique
inExample 14.4. You should verify that
∫
e t/(RC) cosωtdt = R2 C 2 e t/(RC) [
cos ωt
]
ωsinωt +
R 2 C 2 ω 2 +1 RC
Then
e t/(RC) v C
= VRCet/(RC)
R 2 C 2 ω 2 +1
+constant of integration
[
ωsinωt +
cos ωt
]
+K
RC
from which
VRC
[
v C
= ωsinωt + cosωt ]
+Ke −t/(RC)
R 2 C 2 ω 2 +1 RC
Thisisthegeneralsolutionofthedifferentialequation.Applyingtheinitialcondition
gives us a value forK. Whent = 0, v C
=V 0
, so
[ ]
VRC 1
V 0
=
+K
R 2 C 2 ω 2 +1 RC
from which
Finally,
K=V 0
−
v C
=
V
R 2 C 2 ω 2 +1
VRC
[
ωsinωt+
R 2 C 2 ω 2 +1
cos ωt
] (
+ V
RC 0
−
)
V
e −t/(RC)
R 2 C 2 ω 2 +1
This is the particular solution which satisfies the given initial condition. It tells
us the voltage across the capacitor as a function of timet.
➔

556 Chapter 19 Ordinary differential equations I
(c) To find the particular solution subject to the condition v C
(0) = 0 we need only
replaceV 0
by0intheparticularsolutionobtainedinpart(b).Hencethezero-state
response is
v C
(t) =
VRC
[
ωsin ωt +
R 2 C 2 ω 2 +1
cos ωt
]
−
RC
V
R 2 C 2 ω 2 +1 e−t/(RC)
(d) In the case of zero input we have v S
(t) = 0. In part (b) we solved the equation
with v S
(t) =V cos ωt.Hence,puttingV = 0inthesolutiontopart(b)willyield
the solution when v S
(t) = 0.So,
v C
(t) =V 0
e −t/(RC)
Alternatively we can note thatwhen v S
(t) = 0 the originalequation becomes
dv C
+ 1
dt RC v C =0
subject to v C
(0) =V 0
. This can be solved usingseparation of variables. So
1 dv C
= − 1
v C
dt RC
Integration yields
∫ 1
v C
dv C
=−∫
1
RC dt
lnv C
=− t
RC +k
v C
= e −t/(RC)+k
= e k e −t/(RC)
=Ae −t/(RC)
(k a constant)
where A is the constant e k . Applying the initial condition v C
(0) = V 0
we find
V 0
=Aand so finally
as before.
v C
(t) =V 0
e −t/(RC)
(e) Inspectionofpart(b)showsthatitisthesumofthezero-inputresponseandthe
zero-state response:
zero-input zero-state response
response
{ }} { { }} {
v C
(t)=V 0
e −t/(RC) +
VRC [
cos ωt
]
V
ωsinωt+ −
R 2 C 2 ω 2 +1 RC R 2 C 2 ω 2 +1 e−t/(RC)
(f) In this example, the terms involving e −t/(RC) tend to zero ast increases, and are
termed transients. Once the system has settled down their contribution will not
be important. The remaining terms represent the longer term behaviour of the
systemor the so-calledsteady state.
Hence the transient termsare
V 0
e −t/(RC) V
−
R 2 C 2 ω 2 +1 e−t/(RC)

19.4 First-order linear equations: use of an integrating factor 557
and the steady-state terms are
[
ωsinωt +
VRC
R 2 C 2 ω 2 +1
cos ωt
]
RC
EXERCISES19.4.4
1 Find the general solution ofthe followingequations:
dy
(a)
dx +y=1 (b) dy
dx +2y=6
(c)
(e)
dx
dt +6x=4
dy
dx =6y+9
(d) dy
dx −3y=2
(f) dx
dt =3x−8
2 Find the particular solution ofthe following
equations:
(a) dy
dx +4y=7, y(0)=1
(b) dx
dt −x=4,
(c) dy
dt =3y+2,
(d) dy
dx =4y−8,
x(0)=2
y(0)=2
y(1)=2
3 Find the general solution of dx
dt =2x+4t.Whatis
the particular solution which satisfiesx(1) = 2?
4 Find the general solution of dy
dx +y=2x+5.
5 Solve dx
dt =t−tx,x(0)=0.
6 Useanintegratingfactortoobtainthegeneralsolution
ofiR+L di = sin ωt,whereR,Land ω areconstants.
dt
7 Solvex dy
dx +y=x4 .
8 Use an integrating factorto findthe general solution
oft dx
dt +x=3t.
9 Find the general solution of dx +2xt =t.Findthe
dt
particular solution satisfyingthe condition
x(0) = −1.
10 Find the general solution of
tẋ+3x= et
t 2
Solutions
1 (a) y=1+ce −x
(b) y=3+ce −2x
(c) x= 2 3 +ce−6t
(d) y=ce 3x − 2 3
(e) y=ce 6x − 3 2
(f) x= 8 3 +ce3t
2 (a)y= 7 4 − 3 4 e−4x
(b)x=6e t −4
(c)y= 8 3 e3t − 2 3
(d)y=2
3 −2t−1+ce 2t ,−2t−1+5e 2(t−1)
4 2x+3+ce −x
5 1−e −t2 /2
6
7
8
9
L((R/L)sinωt − ωcosωt)
R 2 +L 2 ω 2
x 4
5 + c x
3t
2 + c t
1
2 − 3 2 e−t2
10 et +c
t 3
+ce −Rt/L

558 Chapter 19 Ordinary differential equations I
19.5 SECOND-ORDERLINEAREQUATIONS
The general form ofasecond-order linear ordinary differential equation is
p(x) d2 y
dx +q(x)dy +r(x)y = f(x) (19.7)
2 dx
where p(x),q(x),r(x) and f (x)arefunctions ofxonly.
An important relative of thisequation is
p(x) d2 y
dx +q(x)dy +r(x)y = 0 (19.8)
2 dx
whichisobtainedfromEquation(19.7)byignoringthetermwhichisindependentofy.
Equation (19.8) is said to be a homogeneous equation -- all its terms contain y or its
derivatives. Equation (19.7) issaidtobeinhomogeneous.
For example,
x 2d2 y
dx 2 +xdy dx + (x2 −1)y =e −x
isaninhomogeneoussecond-orderlinearequationinwhichp(x) =x 2 ,q(x) =x,r(x) =
x 2 −1 and f (x) = e −x . The associated homogeneous equation is
x d2 y
dx +xdy 2 dx +(x2 −1)y=0
The following properties of linear equations are necessary for finding solutions of
second-order linear equations.
19.5.1 Property1
Ify 1
(x) andy 2
(x) are any two linearly independent solutions of a second-order homogeneous
equation then the general solution,y H
(x), is
y H
(x) =Ay 1
(x) +By 2
(x)
whereA,Bare constants.
We see that the second-order linear ordinary differential equation has two arbitrary
constantsinitsgeneralsolution.Thefunctionsy 1
(x)andy 2
(x)arelinearlyindependent
ifone isnotsimplyamultiple ofthe other.
19.5.2 Property2
Lety P
(x) be any solution of an inhomogeneous equation. Lety H
(x) be the general solution
of the associated homogeneous equation. The general solution of the inhomogeneous
equation isthengiven by
y(x) =y H
(x) +y P
(x)
Inotherwords,tofindthegeneralsolutionofaninhomogeneousequationwemustfirst
findthegeneralsolutionofthecorrespondinghomogeneousproblem,andthenaddtoit
any solutionofthe inhomogeneousequation.
Thefunctiony H
(x)isknownasthecomplementaryfunctionandy P
(x)iscalledthe
particularintegral.Clearlythecomplementaryfunctionofahomogeneousproblemis

19.5 Second-order linear equations 559
thesameasitsgeneralsolution;weshalloftenwritey(x)forboth.Ifconditionsaregiven
theyareappliedtothegeneralsolutionoftheinhomogeneousequationtodetermineany
unknown constants. This yields the particular solution satisfyingthe given conditions.
Example19.15 Verify that y 1
(x) = x and y 2
(x) = 1 both satisfy d2 y
= 0. Write down the general
dx2 solution of thisequation and verifythatthis indeed satisfies the equation.
Solution Ify 1
(x) = x then dy 1
y 1
dx =1andd2 = 0, so thaty
dx 2 1
satisfies d2 y
dx = 0.Ify 2 2
(x) = 1,
then dy 2
y 2
dx =0andd2 = 0, so thaty
dx 2 2
satisfies d2 y
= 0. From Property 1, the general
dx2 solution of d2 y
dx =0is 2
y H
(x) =Ax +B(1)
=Ax+B
To verifythat thissatisfies the equation proceed as follows:
dy H
dx =A
d 2 y H
dx 2 = 0
and soy H
(x) satisfies d2 y
dx 2 = 0.
Example19.16 Given
d 2 y
+y=x (19.9)
dx2 (a) Show thaty H
= Acosx +Bsinx is a solution of the corresponding homogeneous
equation.
(b) Verifythaty P
=xisaparticular integral.
(c) Verifythaty H
+y P
does indeed satisfythe inhomogeneous equation.
Solution (a) Ify H
=Acosx +Bsinx, then
y ′ H =−Asinx+Bcosx
y′′ H =−Acosx−Bsinx
We see immediately thaty H
+y ′′ H = 0 so thaty H
is a solution of the homogeneous
equation.
(b) Ify P
= x theny ′ P = 1 andy′′ P
= 0. Substitution into the inhomogeneous equation
shows thaty P
satisfies Equation (19.9),thatisy P
=xisaparticular integral.
(c) Writing
we have
y=Acosx+Bsinx+x
y ′ =−Asinx+Bcosx+1
y ′′ =−Acosx−Bsinx

560 Chapter 19 Ordinary differential equations I
Substitution into the l.h.s. of Equation (19.9) gives
(−Acosx−Bsinx)+(Acosx+Bsinx+x)
which equals x, and so the complementary function plus the particular integral is
indeed a solution ofthe inhomogeneous equation, as required by Property 2.
19.6 CONSTANTCOEFFICIENTEQUATIONS
Wenowproceedtostudyindetailthosesecond-orderlinearequationswhichhaveconstant
coefficients. The general formof such an equation is
a d2 y
dx +bdy +cy = f(x) (19.10)
2 dx
wherea,b,careconstants. The homogeneous formof Equation (19.10) is
a d2 y
dx +bdy +cy=0 (19.11)
2 dx
Equationsofthisformariseintheanalysisofcircuits.Considerthefollowingexample.
Engineeringapplication19.5
TheLCRcircuit
WritedownthedifferentialequationgoverningthecurrentflowingintheseriesLCR
circuitshown inFigure 19.5.
Solution
UsingKirchhoff’s voltage lawand the individual laws foreachcomponent wefind
L di ∫
dt +iR+1 idt = v(t)
C
If thisequation isnow differentiated w.r.t.t wefind
L d2 i
dt +Rdi 2 dt + 1 C i = dv(t)
dt
Thisisaninhomogeneoussecond-orderdifferentialequation,withtheinhomogeneity
arising from the term dv . When the circuit componentsL,RandC are constants
dt
wehavewhatistermedalineartime-invariantsystem,andthedifferentialequation
then has constant coefficients.
L C R
i
y (t)
Figure19.5
AnLCRcircuit.

19.6 Constant coefficient equations 561
A linear time-invariant system has components whose properties do not vary with
time and as such can be modelled by a linear constant coefficient differential
equation.
19.6.1 Findingthecomplementaryfunction
AsstatedinProperty2(Section19.5.2),findingthegeneralsolutionofay ′′ +by ′ +cy = f
is a two-stage process. The first task is to determine the complementary function. This
is the general solution of the corresponding homogeneous equation, that isay ′′ +by ′ +
cy = 0.We now focus attention on the solution of such equations.
Example19.17 Verify that y 1
= e 4x and y 2
= e 2x both satisfy the constant coefficient homogeneous
equation
d 2 y
dx −6dy +8y=0 (19.12)
2 dx
Writedown the general solution of thisequation.
Solution Ify 1
= e 4x , differentiation yields
dy 1
dx =4e4x
and similarly,
d 2 y 1
dx 2 = 16e4x
Substitution into Equation (19.12) gives
16e 4x −6(4e 4x )+8e 4x =0
so that y 1
= e 4x is indeed a solution. Similarly if y 2
= e 2x , then dy 2
dx = 2e2x and
d 2 y 2
dx 2 = 4e2x . Substitution into Equation (19.12) gives
4e 2x −6(2e 2x )+8e 2x =0
so that y 2
= e 2x is also a solution of Equation (19.12). Now e 2x and e 4x are linearly
independent functions. So,from Property 1 we have
y H
(x)=Ae 4x +Be 2x
as the general solution of Equation (19.12).
Example19.18 Findvalues ofkso thaty = e kx isasolution of
d 2 y
dx − dy
2 dx −6y=0
Hence statethe general solution.

562 Chapter 19 Ordinary differential equations I
Solution As suggested we try a solution of the formy = e kx . Differentiating we find dy
dx =kekx
and d2 y
dx 2 =k2 e kx . Substitution into the given equation yields
that is,
k 2 e kx −ke kx −6e kx =0
(k 2 −k−6)e kx =0
The only way this equation can be satisfied forall values ofxisif
that is,
k 2 −k−6=0 (19.13)
(k−3)(k+2)=0
so thatk = 3 ork = −2. That is to say, ify = e kx is to be a solution of the differential
equationkmustbe either 3 or −2. We therefore have found two solutions
y 1
(x) = e 3x and y 2
(x) = e −2x
These two functions are linearly independent and we can therefore apply Property 1 to
give the general solution:
y H
(x) =Ae 3x +Be −2x
Equation (19.13) fordeterminingkiscalled theauxiliary equation.
Example19.19 Findthe auxiliary equation ofthe differential equation
a d2 y
dx 2 +bdy dx +cy=0
Solution We try a solution of the formy = e kx so that dy
dx =kekx and d2 y
dx = 2 k2 e kx . Substitution
into the given differential equation yields
that is,
ak 2 e kx +bke kx +ce kx =0
(ak 2 +bk+c)e kx =0
Since thisequation istobe satisfied for all values ofx, then
ak 2 +bk+c=0
isthe required auxiliary equation.

19.6 Constant coefficient equations 563
The auxiliary equation ofa d2 y
dx 2 +bdy dx +cy=0is
ak 2 +bk+c=0
Solution of this quadratic equation gives the values of k which we seek. Clearly the
nature of the roots will depend upon the values ofa,bandc. Ifb 2 > 4ac the roots will
be real and distinct. The two values ofkthus obtained,k 1
andk 2
, will allow us to write
down two independent solutions:
y 1
(x) =e k 1 x
y 2
(x) =e k 2 x
and so the general solution of the differential equation will be
y(x) =Ae k 1 x +Be k 2 x
Iftheauxiliaryequationhasreal,distinctrootsk 1
andk 2
,thecomplementaryfunction
will be
y(x) =Ae k 1 x +Be k 2 x
On the other hand, if b 2 = 4ac the two roots of the auxiliary equation will be equal
andthismethodwillthereforeonlyyieldoneindependentsolution.Inthiscase,special
treatmentisrequired.Ifb 2 <4acthetworootsoftheauxiliaryequationwillbecomplex,
that isk 1
andk 2
will be complex numbers. The procedure for dealing with such cases
will become apparentinthe following examples.
Example19.20 Findthe generalsolutionof
d 2 y
dx 2 +3dy dx −10y=0
Find the particular solution which satisfies the conditionsy(0) = 1 andy ′ (0) = 1.
Solution By lettingy = e kx , so that dy
dx =kekx and d2 y
dx = 2 k2 e kx , the auxiliary equation is found
tobe
k 2 +3k−10=0
Therefore,
(k−2)(k+5)=0
sothatk = 2andk = −5.Thusthereexisttwosolutions,y 1
= e 2x andy 2
= e −5x .From
Property 1 we can writethe general solution as
y=Ae 2x +Be −5x

564 Chapter 19 Ordinary differential equations I
To find the particular solution we mustnow imposethe given conditions:
y(0)=1 gives 1=A+B
y ′ (0)=1 gives 1=2A−5B
from which A = 6 7 andB = 1 . Finally, the required particular solution is
7
y = 6 7 e2x + 1 7 e−5x .
Example19.21 Findthe generalsolution of
d 2 y
dx 2 +4y=0
Solution As before, lety = e kx so that dy
dx = kekx and d2 y
dx = 2 k2 e kx . The auxiliary equation is
easily found tobe
that is
so that
k 2 +4=0
k 2 =−4
k=±2j
that is,we have complex roots.The two independent solutions ofthe equation arethus
y 1
(x) = e 2jx and y 2
(x) = e −2jx
so thatthe general solution can be written inthe form
y(x) =Ae 2jx +Be −2jx
However, in cases such as this, it is usual to rewrite the solution in the following way.
Recall from Chapter 9 thatEuler’s relations give
and
so that
e 2jx = cos2x +jsin2x
e −2jx = cos2x −jsin2x
y(x) =A(cos2x +jsin2x) +B(cos2x −jsin2x)
If we now relabel the constants such that
A+B=C and Aj−Bj=D
we can write the general solution inthe form
y(x) =Ccos2x +Dsin2x

19.6 Constant coefficient equations 565
Engineeringapplication19.6
Oscillatingmass--springsystem
Consider the simple mechanical problem of a mass resting on a smooth frictionless
table. A spring is attached to the mass and an adjacent anchor point as shown in
Figure 19.6. The spring is capable of being compressed as well as stretched. When
the spring isneither compressed nor stretched the mass islocated atx = 0.
m
0
x
Figure19.6
Mass--spring system.
If the mass is pulled in thexdirection and let go, it will oscillate about thex = 0
position. We wish to find the position of the mass,x, as a function of time,t. Differential
equations are needed todescribe thisproblem fully.
Newton’s second law states that if a forceF is applied to a body of massmthen
themotionofthebodyisgovernedbyF =ma,whereaistheacceleration.Applying
Newton’s second law, and noting thata = d2 x
dt2,we obtain
F =m d2 x
dt 2
The force, F, is provided by the spring. The force exerted by a spring is given by
Hooke’slaw,whichstatesthattheforceisproportionaltotheextensionorcompression
of the spring,
F=−kx
wherekisthespringconstantforthespringinuse.Theminussignisrequiredsothat
when the spring is stretched (x > 0) the force is in the negativexdirection. When
the spring iscompressed (x < 0) the force isinthe positivexdirection.
If the table is sufficiently smooth and if there are no other external forces acting,
thenm d2 x
dt 2 = −kx.
Therefore thedifferentialequationthatgoverns motioninthe system is
m d2 x
dt 2 +kx=0
We write thisas
d 2 x
dt 2 + k m x = 0
Let k m = ω2 ,giving
d 2 x
dt 2 +ω2 x=0
➔

566 Chapter 19 Ordinary differential equations I
Usingthe technique illustratedinExample 19.21 we obtain
x(t) =Acosωt +Bsinωt
whereA,Bareconstants. Note this solution may alsobe expressed as a singlewave
x =Ccos(ωt +φ)
whereC and φ are constants, using the technique described in Section 3.7.1, Combining
waves.
We note that this solution is sinusoidal and oscillates with time. It gives the position
of the mass at a given point in time. The constantsC and φ depend on the
position and velocity of the mass when it is released. These are known as theinitial
conditionsofthedifferentialequation(seeExample19.6onpage539).Itisintuitive
thatanequationthatdescribesthepositionofthemassatagiventimemusttakethese
into account.
Mathematically describing or modelling systems like this one using differential
equationsisanextremelyimportantdiscipline.Mathematicalmodelsofmechanical,
electricalandothersubsystemscanbelinkedtogetherandasaresultwholesystems
can be accurately characterized.
Example19.22 Givenay ′′ +by ′ +cy = 0,writedowntheauxiliaryequation.Iftherootsoftheauxiliary
equation arecomplex and aredenoted by
k 1
=α+βj
k 2
=α−βj
show thatthe general solution is
y(x) =e αx (Acosβx+Bsinβx)
Solution Substitution ofy = e kx into the differential equation yields
(ak 2 +bk+c)e kx =0
and so
ak 2 +bk+c=0
This isthe auxiliary equation. Ifk 1
= α + βj,k 2
= α − βj thenthe general solution is
y =Ce (α+βj)x +De (α−βj)x
whereC andDarearbitraryconstants. Usingthe laws ofindices thisisrewritten as
y=Ce αx e βjx +De αx e −βjx =e αx (Ce βjx +De −βjx )
Then, usingEuler’s relations, weobtain
y =e αx (Ccosβx+Cjsinβx+Dcosβx−Djsinβx)
=e αx {(C +D)cosβx+ (Cj−Dj)sinβx}

19.6 Constant coefficient equations 567
WritingA =C +DandB =Cj −Dj,wefind
y =e αx (Acosβx+Bsinβx)
This isthe required solution.
Iftheauxiliaryequationhascomplexroots α +βjand α −βj,thenthecomplementaryfunction
is
y =e αx (Acosβx+Bsinβx)
Note that Example 19.21 isaspecial case of Example 19.22 with α = 0 and β = 2.
Example19.23 Find the general solution ofy ′′ +2y ′ +4y = 0.
Solution The auxiliary equation isk 2 +2k +4 = 0.This equation has complex roots given by
k = −2 ± √ 4−16
2
= −2 ± √ 12j
2
=−1± √ 3j
Using the result of Example 19.22 with α = −1 and β = √ 3 we find the general
solution is
y = e −x (Acos √ 3x+Bsin √ 3x)
Example19.24 Theauxiliaryequationofay ′′ +by ′ +cy = 0isak 2 +bk +c = 0.Supposethisequation
has equal rootsk =k 1
. Verifythaty =xe k 1 x isasolution ofthe differential equation.
Solution We have
y =xe k 1 x y ′ =e k 1 x (1+k 1
x) y ′′ =e k 1 x (k 2 1 x +2k 1 )
Substitutioninto the l.h.s. ofthe differential equation yields
e k x{ 1 a ( k 2 1 x +2k )
1 +b(1+k1 x)+cx } =e k x{( 1 ak 2 1 +bk 1 +c) x +2ak 1
+b }
Butak1 2 +bk 1 +c = 0 sincek 1
satisfies the auxiliary equation. Also,
k 1
= −b ± √ b 2 −4ac
2a
butsincetherootsareequal,thenb 2 −4ac = 0andhencek 1
= − b
2a .So2ak 1 +b=0.
We conclude thaty = xe k 1 x is a solution ofay ′′ +by ′ +cy = 0 when the roots of the
auxiliary equation areequal.

568 Chapter 19 Ordinary differential equations I
If the auxiliary equation has two equal roots,k 1
, the complementary function is
y =Ae k 1 x +Bxe k 1 x
Example19.25 Obtain the general solution of the equation
d 2 y
dx 2 +8dy dx +16y=0
Solution As before, a trialsolution of the formy = e kx yields an auxiliary equation:
k 2 +8k+16=0
This equation factorizes so that
(k+4)(k+4)=0
and we obtain equal roots, that is k = −4 (twice). If we proceed as before, writing
y 1
(x) = e −4x , y 2
(x) = e −4x , it is clear that the two solutions are not independent. To
apply Property 1 we need to find a second independent solution. Using the result of
Example 19.24 we conclude that, because the roots of the auxiliary equation are equal,
the second independent solution isy 2
=xe −4x . The general solution isthen
y(x) =Ae −4x +Bxe −4x
EXERCISES19.6.1
1 Obtainthe general solutions,that isthe
complementary functions,ofthe following
homogeneous equations:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
d 2 y
dx 2 −3dy dx +2y=0
d 2 y
dx 2 +7dy dx +6y=0
d 2 x
dt 2 +5dx dt +6x=0
d 2 y
dt 2 +2dy dt +y=0
d 2 y
dx 2 −4dy dx +4y=0
d 2 y
dt 2 + dy
dt +8y=0
d 2 y
dx 2 −2dy dx +y=0
(h)
(i)
(j)
(k)
(l)
d 2 y
dt 2 + dy
dt +5y=0
d 2 y
dx 2 + dy
dx −2y=0
d 2 y
dx 2 +9y=0
d 2 y
dx 2 −2dy dx = 0
d 2 x
dt 2 −16x=0
2 Findthe auxiliaryequation forthe differential
equation
L d2 i
dt 2 +Rdi dt + 1 C i = 0
Hence write down the complementary function.
3 Findthe complementaryfunctionofthe equation
d 2 y
dx 2 + dy
dx +y=0

19.6 Constant coefficient equations 569
Solutions
1 (a) y=Ae x +Be 2x
(b)y=Ae −x +Be −6x
(c) x =Ae −2t +Be −3t
(d)y=Ae −t +Bte −t
(e) y=Ae 2x +Bxe 2x
(f) y = e −0.5t (Acos2.78t +Bsin2.78t)
(g)y=Ae x +Bxe x
(h) y = e −0.5t (Acos2.18t +Bsin2.18t)
(i) y=Ae −2x +Be x
(j) y=Acos3x+Bsin3x
(k)y=A+Be 2x
(l) x=Ae 4t +Be −4t
2 Lk 2 +Rk+ 1 C =0 i(t) =Aek 1 t +Be k 2 t
where √
R 2 C −4L
−R ±
C
k 1 ,k 2 =
2L
√ √ )
3x 3x
3 e
(Acos −x/2 +Bsin
2 2
19.6.2 Findingaparticularintegral
We stated in Property 2 (Section 19.5.2) that the general solution of an inhomogeneous
equation is the sum of the complementary function and a particular integral. We have
seenhowtofindthecomplementaryfunctioninthecaseofaconstantcoefficientequation.
We shall now deal with the problem of finding a particular integral. Recall that
theparticularintegralisanysolutionoftheinhomogeneous equation.Thereareanumber
of advanced techniques available for finding such solutions but these are beyond
the scope of this book. We shall adopt a simpler strategy. Since any solution will do
we shall try to find such a solution by a combination of educated guesswork and trial
and error.
Example19.26 Findthe generalsolutionofthe equation
d 2 y
dx − dy
2 dx −6y=e2x (19.14)
Solution ThecomplementaryfunctionforthisequationhasalreadybeenshowninExample19.18
tobe
y H
=Ae 3x +Be −2x
We shall attempt to find a solution of the inhomogeneous problem by trying a function
of the same form as that on the r.h.s. In particular, let us tryy P
(x) = αe 2x , where α is a
constant that we shall now determine. Ify P
(x) = αe 2x then dy P
dx = 2αe2x and d2 y P
=
dx 2
4αe 2x . Substitution inEquation (19.14) gives
thatis,
4αe 2x −2αe 2x −6αe 2x = e 2x
−4αe 2x = e 2x

570 Chapter 19 Ordinary differential equations I
sothaty P
willbeasolutionif αischosensothat −4α = 1,thatis α = − 1 4 .Thereforethe
particularintegralisy P
(x) = − e2x
4 .FromProperty2thegeneralsolutionoftheinhomogeneous
equation is found by summing this particular integral and the complementary
function
y(x) =Ae 3x +Be −2x − 1 4 e2x
Example19.27 Obtain a particularintegral ofthe equation
d 2 y
dx 2 −6dy dx +8y=x
Solution In the last example, we found that a fruitful approach was to assume a solution in the
same form as that on the r.h.s. Suppose we assume a solutiony P
(x) = αx and proceed
to determine α. This approach will actually fail, but let us see why. Ify P
(x) = αx then
dy P
y P
dx =αandd2 = 0.Substitution into the differential equation yields
dx2 0−6α+8αx=x
and α ought now to be chosen so that this expression is true for allx. If we equate the
coefficients ofxwe find 8α = 1 so that α = 1 , but with this value of α the constant
8
terms are inconsistent. Clearly a particular integral of the form αx is not possible. The
problemarisesbecausedifferentiationoftheterm αxproducesconstanttermswhichare
unbalanced on the r.h.s.So, wetry a solution of the form
y P
(x)=αx+β
with α, β constants. Proceeding as before, dy P
dx = α, d2 y P
= 0. Substitution in the
dx
differential equation now gives
2
0−6α+8(αx+β)=x
Equating coefficients ofxwefind
8α=1 (19.15)
and equating constant termswe find
−6α+8β=0 (19.16)
From Equation (19.15), α = 1 and then from Equation (19.16)
8
( 1
−6 +8β=0
8)
so that
8β = 3 4

thatis,
β = 3
32
The required particular integral isy P
(x) = x 8 + 3
32 .
19.6 Constant coefficient equations 571
Experienceleadstothe trialsolutionssuggestedinTable 19.1.
Table19.1
Trial solutionsto find the particular
integral.
f (x)
constant
polynomial inx
ofdegreer
coskx
sinkx
ae kx
Trialsolution
constant
polynomial inx
ofdegreer
acoskx +bsinkx
acoskx +bsinkx
αe kx
Example19.28 Findaparticular integralforthe equation
d 2 y
dx 2 −6dy dx +8y=3cosx
Solution We shall try a solution ofthe form
y P
(x)=αcosx+βsinx
Differentiating,we find
dy P
dx =−αsinx+βcosx
d 2 y P
dx 2 =−αcosx−βsinx
Substitution into the differential equation gives
(−αcosx−βsinx)−6(−αsinx+βcosx)+8(αcosx+βsinx)
=3cosx
Equating coefficients of cosxwefind
−α−6β+8α=3 (19.17)
whilethose of sinx give
−β+6α+8β=0 (19.18)

572 Chapter 19 Ordinary differential equations I
Solving Equations (19.17) and (19.18) simultaneously we find α = 21
85 andβ=−18 85 ,
so thatthe particular integral is
y P
(x) = 21
85 cosx−18 85 sinx
Engineeringapplication19.7
AnLC circuitwithsinusoidalinput
The differential equation governing the flow of current in a seriesLC circuit when
subject toanapplied voltage v(t) =V 0
sinωt is
L d2 i
dt + 1 2 C i = ωV 0cos ωt
Derive this equation and then obtain its general solution.
Solution
Kirchhoff’s voltage law and the component laws give
L di
dt + 1 ∫
idt =V
C 0
sinωt
To avoid processes of differentiation and integration in the same equation let us differentiate
thisequation w.r.t.t. This yields
L d2 i
dt + 1 2 C i = ωV 0cos ωt
as required.
The homogeneous equation isL d2 i
dt + i 2 C = 0. Lettingi = ekt we find the auxiliary
equation isLk 2 + 1 C =0sothatk=± √ j . Therefore, using the result of
LC
Example 19.22, with α = 0 and β = 1 √
LC
, the complementary function is
t t
i=Acos√ +Bsin√ LC LC
Tofindaparticularintegral,tryi =Ecos ωt+F sin ωt,whereE andF areconstants.
We find
di
= −ωEsinωt + ωFcosωt
dt
d 2 i
dt 2 = −ω2 Ecosωt −ω 2 Fsinωt
Substitution into the inhomogeneous equation yields
L(−ω 2 Ecosωt − ω 2 Fsinωt) + 1 C (Ecosωt +Fsinωt) =V 0 ωcosωt
Equating coefficients of sinωt gives
−ω 2 LF + F C = 0

19.6 Constant coefficient equations 573
Equating coefficients of cosωt gives
−ω 2 LE + E C =V 0 ω
CV
sothatF = 0andE = 0
ω
. It follows that the particular integral is
1−ω 2 LC
i =
CV 0 ω cos ωt. Finally, the general solution is
1−ω 2 LC
t t
i=Acos√ +Bsin√ + CV 0ωcos ωt
LC LC 1−ω 2 LC
The terms zero-input response and zero-state response were introduced in Section 19.4
inconnectionwithfirst-orderequations.Thisterminologyisidenticalwhenwedealwith
second-order equationsasthe following example illustrates.
Engineeringapplication19.8
ParallelRLC circuit
Figure19.7showsaparallelRLCcircuitwhichhasacurrentsourcei S
(t).Theinductor
current,i, can befound bysolvingthe second-order differential equation
LC d2 i
dt + L di
2 Rdt +i=i S (t) fort0
It would be a useful exercise for you to derive this equation. Note that this equation
issecond order, and inhomogeneous due tothe source termi S
(t).
i S (t)
R L C
i
Figure19.7
AparallelRLC circuit.
SupposeL = 10 H,R = 10 ,C = 0.1 F andi S
(t) = e −2t and that the initial
conditions arei = 1 and di
dt =2whent=0.
(a) Obtain the solution of this equation subject to the given initial conditions and
hence statethe inductor currenti.
(b) Obtain the zero-input response. This isthe solution wheni S
(t) = 0.
(c) Obtain the zero-state response. This is the solution of the inhomogeneous equation
subject to the conditionsi = 0 and di = 0 att = 0. It corresponds to there
dt
being no initial energy inthe circuit.
(d) Show that the solution in (a) is the sum of the zero-input response and the zerostateresponse.
➔

574 Chapter 19 Ordinary differential equations I
Solution
(a) With the given parameter values the differential equation becomes
d 2 i
dt + di
2 dt +i=e−2t
Itisfirstnecessarytofind thecomplementary function. Lettingi = e kt , theauxiliaryequation
isk 2 +k+1 = 0 which has complex solutions
k = − 1 √
3
2 ± 2 j
The complementary function istherefore
√ √ ) 3 3
i =e
(Acos
−t/2 2 t+Bsin 2 t
For a particular integral we try a solution of the form i = αe −2t . Substitution
into the differential equation gives
so that
4αe −2t −2αe −2t + αe −2t = e −2t
3α=1, thatis α= 1 3
Hence a particular integral isi = 1 3 e−2t . The general solution is the sum of the
complementary function and the particular integral:
√ √ ) 3 3
i =e
(Acos
−t/2 2 t+Bsin 2 t + 1 3 e−2t
We now apply the initial conditions to find the constantsAandB. Giveni = 1
whent = 0 means
1=A+ 1 so that A = 2 3 3
To apply the second condition weneed tofind di
dt :
( √ √ √ √ )
di 3 3 3 3
dt = e−t/2 −
2 Asin 2 t + 2 Bcos 2 t
− 1 2 e−t/2 (Acos
Given di = 2 whent = 0 means
dt
√
3
2 =
2 B − 1 2 (A) − 2 3
√
3
2 =
2 B − 1 ( ) 2
− 2 2 3 3
√
3
2 t+Bsin √
3
2 t )
− 2 3 e−2t
2 =
3 =
√
3
2 B −1
√
3
2 B

19.6 Constant coefficient equations 575
B = 6 √
3
= 2 √ 3
Finally, the required particular solution which gives the current through the
inductor atany time is
( √ √ )
2 3 3
i =e −t/2 3 cos 2 t+2√ 3sin
2 t + 1 3 e−2t
(b) The zero-input response is obtained by ignoring the source term i S
(t). From
(a)we see thatthis isjustthe complementary function:
√ √ ) 3 3
i =e
(Acos
−t/2 2 t+Bsin 2 t
Applying the given initial conditions to this solution givesA = 1 andB = √ 5 3
and so the zero-input response is
√
3
i =e
(cos
−t/2 2 t + √ 5 √ ) 3
sin
3 2 t
(c) Thezero-stateresponseoftheinhomogeneousequationisfoundbyapplyingthe
conditionsi = 0 and di = 0 att = 0 tothe general solution already obtained in
dt
(a). It is straightforward to show thatA = − 1 3 andB= √ 1 . So the zero-state
3
response is
(
i =e −t/2 − 1 √
3
3 cos 2 t + √ 1
√ ) 3
sin
3 2 t + 1 3 e−2t
(d) Inspectionofthepreviousworkingshowsthattheparticularsolutionobtainedin
(a)isthe sum of the zero-state response and zero-input response.
19.6.3 Inhomogeneoustermappearsinthecomplementaryfunction
In some examples, terms which form part of the complementary function also appear
in the inhomogeneous term. This gives rise to an additional complication. Consider
Example 19.29.
Example19.29 Considertheequationy ′′ −y ′ −6y = e 3x .Itisstraightforwardtoshowthatthecomplementaryfunctionis
y=Ae 3x +Be −2x
Findaparticular integraland deducethe generalsolution.
Solution Suppose we try to find a particular integral by using a trial solution of the form
y P
= αe 3x . Substitutioninto the l.h.s. ofthe inhomogeneous equation yields
9αe 3x −3αe 3x −6αe 3x whichsimplifies to0
so that αe 3x is clearlynotasolution of the inhomogeneousequation. Thereason is that
e 3x is part of the complementary function and so causes the l.h.s. to vanish. To obtain a

576 Chapter 19 Ordinary differential equations I
particularintegralinsuchacase,wecarryouttheprocedurerequiredwhentheauxiliary
equation has equal roots.Thatis, we tryy P
= αxe 3x . We find
y ′ =αe 3x (3x+1)
y ′′ =αe 3x (9x+6)
Substitution into the inhomogeneous equation yields
αe 3x (9x +6) − αe 3x (3x +1) −6αxe 3x =e 3x
Mosttermscancel, leaving
5αe 3x = e 3x
sothat α = 1 5 .Finally,therequiredparticularintegralisy P =xe3x 5 .Thegeneralsolution
istheny=Ae 3x +Be −2x + xe3x
5 .
Engineeringapplication19.9
Transmissionlines
A transmission line is an arrangement of electrical conductors for transporting
electromagnetic waves. Although this definition could be applied to most electrical
cables,itisusuallyrestrictedtocablesusedtotransporthigh-frequencyelectromagneticwaves.Thereareseveraldifferenttypesoftransmissionlines.Themostfamiliar
one is the coaxial cable which is used to carry the signal from a television aerial to
atelevisionset(seeFigure19.8).Whenahigh-frequencywaveisbeingcarriedbya
cableseveraleffectsbecomeimportantwhichcanusuallybeneglectedwhendealing
with a low-frequency wave. These are:
(1) the capacitance,C, between the two conductors,
(2) the series inductance,L, ofthe two conductors,
(3) theleakagecurrentthroughtheinsulationlayerthatseparatesthetwoconductors.
The electrical parameters of a coaxial cable are evenly distributed along its length.
Thisistruefortransmissionlinesingeneralandsoitisusualtospecifyperunitlength
values for the parameters. When constructing a mathematical model of a transmission
line it is easier to think in terms of lumped components spanning a distance δz
and then allow δz to tend to zero (see Figure 19.9). The leakage between the two
conductors is conventionally modelled by a conductance, G, as this simplifies the
mathematics and avoids confusion with the line resistance,R. Note thatC,G,Land
Rareper unit length values forthe transmissionline.
×
Outer conductor
Inner conductor
Protective coating
Insulator
Figure19.8
Acoaxial cable.

19.6 Constant coefficient equations 577
For many transmission lines of interest the signal that is being carried varies sinusoidally
with time. Therefore the voltage and current depend on both position along
theline,z,andtime,t.However,itiscommontoseparatethetimedependencefrom
the voltage and current expressions in order to simplify the analysis. It must be remembered
that any voltage and current variation with position has superimposed
uponitasinusoidalvariationwithtime.Therefore,ignoringthetime-dependentelement
we write the voltage as v and the current asi, knowing that they are functions
ofz.
y
i
i + di
di
Cdz
y L
Ldz
Gdz
y R
Rdz
y + dy
i C
iG
z
dz
Figure19.9
A section ofatransmissionline.
ConsiderthecircuitofFigure19.9whichrepresentsasectionofthetransmission
line oflength δz. Applying Kirchhoff’s voltage law tothe circuityields
v+δv−v+v L
+v R
=0
δv=−v L
−v R
where v L
is the voltage across the inductor and v R
is the voltage across the resistor.
Using the individual component laws for the inductor and resistorgives
δv = −ijωLδz −iRδz
= −i(R +jωL)δz
Notethat δihasbeenignoredbecauseitissmallcomparedtoi.Nowconsidertheparallelcombinationofthecapacitorandresistor(withunitsofconductance).Applying
Kirchhoff’s current law tothiscombination yields
δi=i C
+i G
wherei C
isthecurrentthroughthecapacitorandi G
isthecurrentthroughtheresistor.
Using the individual component laws for the capacitor and resistor gives
δi = −vjωCδz − vGδz
= −v(G +jωC)δz
Dividing these two circuitequations by δzyields
δv
δz
= −i(R +jωL)
δi
δz = −v(G+jωC)
➔

578 Chapter 19 Ordinary differential equations I
Inordertomodelacontinuoustransmissionlinewithevenlydistributedparameters,
δz isallowed totend tozero. Inthe limitthe two circuitequations become
dv
= −i(R +jωL) (19.19)
dz
di
= −v(G+jωC) (19.20)
dz
Differentiating Equation (19.19) yields
d 2 v
= −(R +jωL)di
dz2 dz
Substituting for di fromEquation (19.20) yields
dz
d 2 v
= (R +jωL)(G +jωC)v
dz2 This isusually writtenas
d 2 v
dz 2 = γ2 v where γ 2 = (R +jωL)(G +jωC) (19.21)
This is the differential equation that describes the variation of the voltage, v, with
position,z,alongthetransmissionline.Thegeneralsolutionofthisequationiseasily
shown tobe
v=v 1
e −γz +v 2
e γz (19.22)
where v 1
and v 2
areconstantsthatdependontheinitialconditionsforthetransmission
line. It is useful to write γ = α +jβ thus separating the real and imaginary parts of
γ. Equation (19.22) can then be writtenas
v = v 1
e −αz e −jβz + v 2
e αz e jβz (19.23)
The quantity v 1
e −αz e −jβz represents the forward wave on the transmission line. It
consists of a decaying exponential multiplied by a sinusoidal term. The decaying
exponentialrepresentsagradualattenuationofthewavecausedbylossesasittravels
along the transmission line. The quantity v 2
e αz e jβz represents the backward wave
produced by reflection. Reflection occurs if the transmission line is not matched
withitsload.Asthewaveistravellingintheoppositedirectiontotheforwardwave,
e αz stillrepresents an attenuation butinthiscase an attenuation aszdecreases.
Alosslesslineisoneinwhichtheattenuationisnegligible.Thiscasecorresponds
toα=0,andsoγ =jβ.Ifγ =jβthenγ 2 = −β 2 so that, from Equation (19.21),
(R + jωL)(G + jωC) must be real and negative. We see that this is the case when
R = 0 andG = 0. This agrees with what would be expected in practice as it is the
resistive and conductive termsthatlead toenergy dissipation.
Engineeringapplication19.10
Voltagereflectioncoefficient
Consider a lossless transmission line of the type already described in Engineering
application19.9.Weknowingeneralthattheforwardwaveatpositionzisgivenby
v 1
e −αz e −jβz . For the lossless line this simplifies to v 1
e −jβz . The reverse term, by the
samereasoning,is v 2
e jβz .

19.6 Constant coefficient equations 579
A definition used regularly when analysing transmission lines is the voltage
reflection coefficient. Usually this is denoted by ρ and can be defined either at a
specific position on a lineor as a function ofdistance
ρ(z) = v 2 ejβz
v 1
e −jβz = v 2
v 1
e 2jβz
Itisadimensionlessquantityandistheratioofthereversetoforwardwavecomponents.
Inmanysystemsencounteredinradio-frequency(RF)engineeringitisdesirable
to minimize the relative amplitude of the backward wave component and hence the
reflection coefficient. An example of this can be found in the transmit circuit of a
mobilehandsetwhereatransmissionlinecarryingthesignalisattachedtoanantenna.
Duringtransmission,forwardwavespropagatetowardstheantennaterminals.Atthis
point they are either radiated from or dissipated within the antenna, or they reflect
back.Ifreflectedbacktheymayreturntotheamplifiercircuitwhichgeneratedthem,
and are wasted in the form of heat energy. As a consequence, battery life can be
reduced due to wasted power. Hence the minimization of the reflection coefficient,
usually by carefully designing the antenna, at the working frequency of the handset
isan important activity.
Fortheantennaitisdesirablefor ρ(z)tobeassmallaspossibleandfor v 1
≫ v 2
.
Note that in a system such as this |ρ(z)| < 1 and that ρ(z) is in general a complex
number.
We now consider a transmission line of length l with characteristic impedance
Z 0
terminated at z = l with a load having impedance Z L
. The setup is shown in
Figure 19.10.
z = 0
y 2 e jbz
Z 0
y 1 e –jbz
z = l
Z L
z
Figure19.10
Lossless transmissionline with
termination.
The total voltage at the load end of the line, v L
, is the sum of the forward and
backward wave componentsatz = l, thatis
v L
= v 1
e −jβl +v 2
e jβl
Using the definition ofthe reflection coefficient atz = l,
ρ(l) = v 2
v 1
e 2jβl
we can rewrite the voltage atthe load as
v L
= v 1
e −jβl e jβl
+ρ(l)v 1
e = v j2βl 1 e−jβl + ρ(l)v 1
e −jβl = v 1
e −jβl [1 + ρ(l)]
Thecurrentattheload,i L
,canbefoundfromfirstprinciplesbyasimilaranalysisto
the voltage, butwill be statedhere for simplicity
i L
= 1 Z 0
(v 1
e −jβl − v 2
e jβl )
➔

580 Chapter 19 Ordinary differential equations I
This equation can alsobe written intermsof the reflection coefficient
i L
= v 1 e−jβl
[1 − ρ(l)]
Z 0
The voltage and currentatthe load arerelated by the simple equation
Z L
= v L
i L
and hence
Z L
= v 1 e−jβl [1 + ρ(l)]
v 1
e −jβl
[1 − ρ(l)]
Z 0
=Z 0
[1 + ρ(l)]
[1 − ρ(l)]
Itcan be shown by further manipulation that
ρ(l) = Z L −Z 0
Z L
+Z 0
Thereflectioncoefficientattheloadisthereforedependentonlyonthecharacteristic
impedance of the line, which is usually known, and the load impedance. Some RF
measuringinstrumentssuchasnetworkanalyserscanmeasuretheamountofforward
and backward waves. From thisthey areable todetermine the load impedance.
Engineeringapplication19.11
Standingwavesontransmissionlines
Thedifferentialequationsthatmodelvoltageandcurrentwavesontransmissionlines
alsodescribethepresenceof standingwaves.Astandingwave issocalledbecause
it appears to remain stationary in space. Standing waves are an effect caused by the
interference pattern created when two waves propagate in opposite directions in the
sametransmissionmedium.
Thepreviousexamplesuggestedthatminimizingbackwardwavesisoftendesired.
In this case the standing wave component on the line is also minimized. In order to
studythisinmoredetailwewishtoplotthevoltagestandingwavepatternforagiven
loadimpedance,Z L
.
Recall from Equation (19.23) (with α = 0 for a lossless line) that the voltage at
any pointzon the lineisgiven by
v(z) = v 1
e −jβz +v 2
e jβz
This voltage is the sum of the forward and reflected waves. The modulus of this
function can be plotted withzas the independent variable. Once the amplitude v 1
is
known, then the amplitude v 2
can be determined using the reflection coefficient at
the load
ρ(l) = v 2
v 1
e j2βl
from which v 2
= v 1
ρ(l)e −j2βl so that
v(z) = v 1
e −jβz + v 1
ρ(l)e −j2βl e jβz

19.6 Constant coefficient equations 581
which simplifies to
v(z) = v 1
e −jβz[ 1 + ρ(l)e j2β(z−l)]
The modulus of v(z) is now plotted against z for several different values of the
reflection coefficient at the load ρ(l). Consider a transmission line of total length
l = 10 m. Another quantity weneed toknow inorder toplot the voltage on the line
is β, which is the phase constant. For a transmission line of given construction and
ataparticular frequency, β isconstantand represents thephase change per metreof
transmissionline. Herewe take β = π/2 rad m −1 .
Case1: ρ(l)= −1
Consider the case when the transmission line is terminated in a short circuit. The
reflection coefficient atthe load is
ρ(l) = Z L −Z 0
Z L
+Z 0
= 0−Z 0
0+Z 0
= −1
Taking the amplitude v 1
= 1, which represents a 1 V peak sine wave, and plotting
the modulus of v(z), gives the resultshown inFigure 19.11.
y(z)
2
1
0
2 4 6 8 10 z
Figure19.11
Voltage standing wave pattern forashort-circuitloadwith a 1 Vinput wave.
Noticethatthevoltagepeakonthelineis2Vwhereasthevoltageinputwasonly
1V.Thisisduetotheforward-going1Vinputwavereachingtheendofthelineand
reflectingbackuponitself.Atsomevaluesofzitconstructivelyinterfereswithitself
giving double the input; atothers itdestructively interferesgiving zero volts.
Case2: ρ(l)=1
Asimilareffectisseenforanopen-circuitload.HerebyconsideringZ L
→ ∞,ρ(l)
can be shown to equal +1, giving rise to the standing wave pattern shown in
Figure 19.12. Note here that the voltage maximum is at the load, unlike the case
ofthe shortcircuitwhere the minimum wasatthe load.
y(z)
2
1
0
2 4 6 8 10 z
Figure19.12
Voltage standing wave pattern foran open-circuit loadwith a1Vinput wave.
➔

582 Chapter 19 Ordinary differential equations I
Case3: ρ(l)=0
Consider now the case in which the load impedance is equal to the characteristic
impedance, thatis
Z L
=Z 0
This time, ρ(l) = 0 and hence v(z) = v 1
e −jβz . The magnitude of this complex
exponential is shown plotted in Figure 19.13 and is a horizontal straight line with
|v(z)| = 1 for all values ofz.
y(z)
2
1
0
2 4 6 8 10 z
Figure19.13
Voltage onatransmissionline with a reflection coefficient of ρ(l) = 0.
No standing wave component is present and the line is said to be matched. This
is the ideal case for power transfer because it represents the case in which all of the
input wave isdelivered tothe load and nothing isreflected back.
Case4:ageneralcase
Oftenitisdifficulttoarrangeforaperfectlymatchedtransmissionline.MoregenerallythepeakvoltageonthelineappearsliketheoneshowninFigure19.14.Herean
assumed reflection coefficient of ρ(l) = 0.187 −j 0.015 isused.
y(z)
2
1
0
2 4 6 8 10 z
Figure19.14
Voltage onatransmissionline with a reflection coefficient of ρ(l) = 0.187 − j0.015.
The standing wave component is still present although it is of smaller magnitude
than seen for the open- or short-circuit loads. Most of the input wave is transferred
tothe load but a proportion isreflected back and interferes.
The standing wave pattern of a transmission line can be measured by a special
piece of apparatus known as a slotted line. The device consists of a length of transmissionlinewithaslotintowhichasmallprobeisinserted.Theprobeisusedtofind
thevoltageasafunctionofdistanceasitismovedalongthelinetowardsanunknown
load. The equations presented above can be used with the results of the slotted line
experiment todetermine the reflection coefficient ofthe load.

19.6 Constant coefficient equations 583
EXERCISES19.6.3
1 Find the general solution ofthe followingequations:
(a) d2 x
dt 2 −2dx dt −3x=6
(b) d2 y
dx 2 +5dy dx +4y=8
(c) d2 y
dt 2 +5dy dt +6y=2t
(d) d2 x
dt 2 +11dx dt +30x=8t
(e) d2 y
dx 2 +2dy dx +3y=2sin2x
d 2 y
(f)
dt 2 + dy
dt +y=4cos3t
(g) d2 y
dx 2 +9y=4e8x
(h) d2 x
dt 2 −16x=9e6t
2 Find aparticular integral forthe equation
d 2 x
dt 2 −3dx dt +2x=5e3t
3 Find aparticular integral forthe equation
d 2 x
dt 2 −x=4e−2t
4 Obtainthe general solution of
y ′′ −y ′ −2y=6.
5 Obtainthe general solution ofthe equation
d 2 y
dx 2 +3dy dx +2y=10cos2x
Find the particularsolution satisfying
y(0) =1, dy (0) =0.
dx
6 Find aparticularintegralforthe equation
d 2 y
dx 2 + dy
dx +y=1+x
7 Find the general solution of
d 2 x
(a)
dt 2 −6dx dt +5x=3
d 2 x
(b)
dt 2 −2dx dt +x=et
8 Forthe circuitshown in Figure19.15showthat
RCL d2 i 2
dt 2 +Ldi 2
+Ri
dt 2 =E(t)
IfL=1mH,R=10,C=1Fand
E(t) = 2sin100πt,findthe complementary function.
9 Find the general solution of
i
d 2 i
dt 2 +8di +25i = 48cos3t −16sin3t
dt
R
Figure19.15
i 1
i 2
E(t)
C
L
Solutions
1 (a) x=Ae −t +Be 3t −2
(f) y = e −0.5t (Acos0.866t +Bsin0.866t) −
0.438cos3t +0.164sin3t
(b) y=Ae −x +Be −4x +2
(c) y=Ae −2t +Be −3t + t 3 − 5 (g) y =Acos3x +Bsin3x +0.0548e 8x
18
(d) x =Ae −6t +Be −5t +0.267t −0.0978 (h) x=Ae 4t +Be −4t + 9
20 e6t
(e) y = e −x [Asin √ 2x+Bcos √ 2x] − 8 17 cos2x − 2 x=2.5e 3t
2
17 sin2x 3 x = 4 3 e−2t

584 Chapter 19 Ordinary differential equations I
4 Ae 2x +Be −x −3
5 Ae −2x +Be −x + 3 2 sin2x − 1 2 cos2x,
3
2 e−2x + 3 2 sin2x − 1 2 cos2x
6 x
7 (a) Ae t +Be 5t + 3 5
(b) Ae t +Bte t + 1 2 t2 e t
8 Ae −11270t +Be −88730t
9 e −4t (Asin3t +Bcos3t)+
14sin3t +18cos3t
13
19.7 SERIESSOLUTIONOFDIFFERENTIALEQUATIONS
Wenowintroduceatechniqueforfindingsolutionsofdifferentialequationsthatcanbe
∞∑
expressed in the form of a power seriesy = a m
x m =a 0
+a 1
x +a 2
x 2 + ....Here,
a m
,m = 0,1,2,3... areconstants whosevalues needtobefound.
With the given power series form of y, we can differentiate term by term to obtain
power series for dy
dx and d2 y
dx2. By substituting these series into certain classes of differentialequationwecandeterminetheconstantsa
m
,form = 0,1,2,3...andthusobtain
the power series solution fory. This technique of representing a solution in the form of
a power series paves the way for the introduction of an important family of differential
equations,knownasBessel’sequations,whichcanbesolvedbythismethod.Thepower
series solutions so formed are known as Bessel functions which will be introduced in
Section 19.8.
Consider the following example:
Example19.30 The general solution of the differential equation d2 y
+y = 0 was shown in Example
dx2 19.16 tobe
Solution We let
y=Acosx+Bsinx
whereAandBarearbitraryconstants.Obtainthisresultbylookingforasolutionofthe
∞∑
equation inthe form ofapower seriesy = a m
x m .
y =
m=0
m=0
∞∑
a m
x m =a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 +...
m=0
from which, by differentiating with respecttox,
and
∞
dy
dx = ∑
ma m
x m−1 =a 1
+2a 2
x+3a 3
x 2 +4a 4
x 3 +5a 5
x 4 +...
m=0
d 2 y
dx = ∑ ∞
(m −1)ma 2 m
x m−2 = 2a 2
+ (2)(3)a 3
x + (3)(4)a 4
x 2 + (4)(5)a 5
x 3 + ...
m=0

19.7 Series solution of differential equations 585
These expressions for y and d2 y
are substituted into the given differential equation
dx2 d 2 y
dx +y=0.Thus 2
(2a 2
+ (2)(3)a 3
x + (3)(4)a 4
x 2 + (4)(5)a 5
x 3 + ...) +
(a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 +...) =0.
We now use the technique of equating coefficients. This was first introduced in
Section 1.7 Partial fractions. Equating the constant terms:
2a 2
+a 0
=0 sothat a 2
=− 1 2 a 0 .
Equating coefficients ofx:
(2)(3)a 3
+a 1
=0 sothat a 3
= − 1
(2)(3) a 1 .
Equating coefficients ofx 2 :
(3)(4)a 4
+a 2
=0 sothat a 4
= − 1
(3)(4) a 2 = 1
(2)(3)(4) a 0 .
Equating coefficients ofx 3 :
(4)(5)a 5
+a 3
=0 sothat a 5
= − 1
(4)(5) a 3 = 1
(2)(3)(4)(5) a 1 .
Note that by using factorial notation we can express these coefficients concisely as:
a 2
=− 1 2! a 0 , a 3 =−1 3! a 1 , a 4 = 1 4! a 0 , a 5 = 1 5! a 1 .
Byequatinghigherordertermsinthesamewayfurthercoefficientsa m
canbedetermined
∞∑
and the power seriesy = a m
x m can then be written out explicitly. Observe that all
m=0
coefficientsa m
are multiples ofa 0
whenmis even, and multiples ofa 1
whenmis odd.
We can therefore collect terms together as follows:
∞∑
y = a m
x m
m=0
=a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 +...
=a 0
+a 1
x− 1 2! a 0 x2 − 1 3! a 1 x3 + 1 4! a 0 x4 + 1 5! a 1 x5 +...
=a 0
(1− 1 2! x2 + 1 4! x4 −...)+a 1
(x− 1 3! x3 + 1 5! x5 − ...) (19.24)
ObservethattheseriesinthebracketsinEquation19.24arethepowerseriesexpansions
of cosx and sinx as discussed inSection 18.6. Hence wecan writethe solution as
y=a 0
cosx+a 1
sinx.
We see that this is simply the general solution of the differential equation as given
in the question but where the coefficients a 0
and a 1
are the arbitrary constants. Had
we not recognised the series expansions for the sine and cosine functions we would
still have obtained the general solution as expressed in the form of the power series in
Equation 19.24.

586 Chapter 19 Ordinary differential equations I
This method of solution iswidely applicable and involves the following steps:
• given a differential equation, we seek a solution in the form of the power series
∞∑
y = a m
x m ,
m=0
• thispower series issubstitutedinto the differential equation
• the coefficientsa m
are then found by the technique of equating coefficients.
Once the values of a m
are found, the solution follows in the form of a power series.
Some differential equations and their solutions have such importance in engineering
and scientific applications that the series obtained are used to introduce new functions
in much the same way as the sine and cosine functions appeared in Equation 19.24.
We shall see in the following section how an important class of functions known as
Besselfunctionsarisesinthisway.Thereisofcoursethequestionofwhetheranypower
series expansion converges but such discussion is beyond the scope of the introductory
treatment given here.
EXERCISES19.7
1 Thegeneral solution ofthe equation d2 y
dx 2 −y=0is
y =Ae x +Be −x whereAandBarearbitraryconstants.
Obtain thisresultby lookingforasolution ofthe
∞∑
equation in the form ofapower seriesy = a m x m .
2 Thegeneral solution ofthe equation
m=0
d 2 y
dx 2 −3dy dx +2y = 0 isy =Aex +Be 2x whereAand
Barearbitraryconstants.Obtainthisresultbylooking
forasolution ofthe equation in the formofapower
∞∑
seriesy = a m x m .
m=0
3 Verifyusingthemethodofseparationofvariablesthat
the generalsolutionof dy
dx = 2xyisy =Kex2 whereK
isan arbitraryconstant. Obtainthe same resultby
seeking apower seriessolution ofthe differential
∞∑
equationin the formy = a m x m .
m=0
4 Obtainthe first fournon-zeroterms in the power
seriessolution ofthe initialvalue problem
dy
dx +xy=0,y(0)=1.
5 Obtainthe power series solution ofthe equation
d 2 y
dx 2 +xy = 0,upto termsinvolvingx7 . This
differentialequation isknown asanAiryequation.
Solutions
4 y=1− x2
2 + x4
8 − x6
48 +.... 5
y=a 0
(
1 − x3
6 + x6
180 −... )
(
)
+a 1 x − x4
12 + x7
504 −... .

19.8 Bessel’s equation and Bessel functions 587
19.8 BESSEL’SEQUATIONANDBESSELFUNCTIONS
Alineardifferentialequationthathasimportantapplicationsinthetheoryofwavepropagation,
particularly when using cylindrical or spherical polar coordinates, is known as
Bessel’sequation.InfactthetermBessel’sequationrepresentsawholefamilyofequations
having the form
x 2d2 y
dx +xdy 2 dx +(x2 −v 2 )y=0.
Herexandyaretheindependentvariableanddependentvariablerespectively.Thethird
quantity v,termedaparameter,isknownastheorderoftheequation. 1 Aswevarythe
parameter v we obtain the family of differential equations. Note that v can be any real
number. In Sections 19.8.1 and 19.8.2 we will solve Bessel’s equation when v = 0 and
v = 1.Intheengineeringapplicationswhichfollowlaterinthechapteryouwillseethat
v may be other integers including those that are negative. Observe that the equation is
linearandhomogeneous,butitdoesnothaveconstantcoefficients:thecoefficientof d2 y
dx 2
isx 2 , a function of the independent variable. In what follows we show that solutions of
Bessel’sequationcanbefoundintheformofinfinitepowerserieswhichwecallBessel
functions.
Bessel functions have many applications in physics and engineering. In electrical
engineeringtheyarecommonlyusedinthefieldofcommunications.Theyareimportant
in finding the bandwidth and frequency content of some radio signals. They are also
essentialintheanalysisofpropagatingmodesincylindricalwaveguidesandcylindrical
cavityresonators,bothofwhichareimportantdevicesusedinhighfrequencymicrowave
engineering(seeEngineeringapplication19.12).Besselfunctionsarealsoutilisedinthe
design of filters.
19.8.1 Bessel’sequationoforderzero
To introduce the solution of Bessel’s equation and thereby introduce a Bessel function
the following example deals with the specialcase when v = 0.
Example19.31 By seeking a power series solution in the form
∞∑
a m
x m obtain the first four non-zero
termsinapower series solution ofBessel’s equation inthe specialcase when v = 0.
Solution When v = 0 Bessel’s equation is:
x 2d2 y
dx +xdy 2 dx +x2 y=0.
which fornon-zeroxbecomes
x d2 y
dx + dy +xy=0. (19.25)
2 dx
m=0
1 Note that in this context the word ‘order’ means something different from the order of a differential
equation, which wasdefined in 19.2.1.

588 Chapter 19 Ordinary differential equations I
As was done inSection 19.7 we seekapower series solution inthe form
y =
∞∑
a m
x m =a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 +...
m=0
The strategy is to substitutey, and its first and second derivatives into Equation 19.25
and choose valuesa m
so that the equation is satisfied.yand its first two derivatives can
be written:
y=a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 +...
dy
dx =a 1 +2a 2 x+3a 3 x2 +4a 4
x 3 +5a 5
x 4 +...
d 2 y
dx 2 =2a 2 +6a 3 x+12a 4 x2 +20a 5
x 3 +...
Substitution intox d2 y
dx + dy +xy = 0gives
2 dx
(2a 2
x +6a 3
x 2 +12a 4
x 3 +20a 5
x 4 ...) + ( a 1
+2a 2
x+3a 3
x 2 +4a 4
x 3 ... )
+ (a 0
x+a 1
x 2 +a 2
x 3 +a 3
x 4 +a 4
x 5 ...) =0
We now equate coefficients on both sides of this equation starting with the constant
term:
a 1
=0.
Equating coefficients ofx:
4a 2
+a 0
=0 sothat a 2
=− 1 4 a 0 .
Equating coefficients ofx 2 :
9a 3
+a 1
=0 sothat a 3
=− 1 9 a 1 = 0 sincea 1 = 0.
Equating coefficients ofx 3 :
16a 4
+a 2
=0 sothat a 4
=− 1 (
16 a 2 = − 1 ) (
− 1 )
a
16 4 0
= 1
64 a 0 .
Itiseasytoverifythatallsubsequenttermsa m
,whenmisodd,arezero.Youshouldverify,inasimilarway,thata
6
= − 1
2304 a 0 .Whenmiseven,alla m aremultiplesofa 0 .We
can now write down a power series that satisfies Bessel’s differential
Equation 19.25:
(
y=a 0
1 − 1 4 x2 + 1
64 x4 − 1 )
2304 x6 +...
(19.26)
We have succeeded in obtaining a power series solution of Bessel’s Equation 19.25
wherea 0
isan arbitraryconstant.
As we have already noted, a 0
is an arbitrary constant. However, historically its value
is chosen to be 1 in Equation 19.26. The resulting power series is used to define a

19.8 Bessel’s equation and Bessel functions 589
J o (x)
1
0
–0.4
10 20 30
x
Figure19.16
TheBesselfunction ofthe firstkind oforder
zero,J 0 (x).
function called a Bessel function of the first kind of order zero that is conventionallydenotedJ
0
(x). Itispossible toshow thatitcan beexpressed concisely usingsigma
notation as
J 0
(x) =
∞∑
m=0
(−1) m x 2m
2 2m (m!) 2 . (19.27)
You shouldverifythisbywritingoutthefirstfewtermsofthisinfiniteseries.Note that
this method of solving the differential equation produces a power series solution and is
usedtointroduceanewfunction,J 0
(x).ItisnotpracticaltotrytodrawagraphofJ 0
(x)
by hand but this can be obtained readily using a technical computing language such as
MATLAB ® .Figure19.16showsagraphobtainedinthisway.ObservethatJ 0
hassome
similar characteristics to a cosine wave: its value whenx = 0,J 0
(0), equals 1, and the
functionoscillates,albeitwithdecreasingamplitude.Thevaluesofxwherethefunction
crosses the horizonal axis are known as zeros of the Bessel function and these values
arise in important applications as we shall see in Engineering application 19.12: The
modes of a cylindrical microwave cavity.
WhenaspecificvalueofaBesselfunctionisrequireditisusualpracticetolookthis
up rather than work directly with the power series definition. Both printed tables and
computer packages are available forthispurpose.
Recall from Section 19.5 that the general solution of a second order linear equation
requires two independent solutions. The function J 0
(x) is one such solution. Another
independent solution is called a Bessel function of the second kind of order
zero and is denoted byY 0
(x). Derivation of this function is beyond the scope of this
book but, like J 0
(x), values ofY 0
(x) can be obtained from tables and by using computer
software. The general solution of Bessel’s equation of order zero can then be
written
y =AJ 0
(x) +BY 0
(x)
whereAandBare arbitraryconstants.
19.8.2 Bessel’sequationoforderone
Here we adopt the previous approach to obtain a solution of Bessel’s equation of order
one, and thereby introduce the Bessel function of the first kind of order one.

590 Chapter 19 Ordinary differential equations I
Example19.32 By seeking a power series solution in the form
∞∑
a m
x m obtain the first three non-zero
m=0
terms inthe solution of Bessel’s equation inthe case when v = 1:
x 2d2 y
dx 2 +xdy dx + (x2 −1)y = 0. (19.28)
Solution As before weseek a power series solution inthe form
∞∑
y = a m
x m =a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 ...
m=0
The strategy is to substitutey, and its first and second derivatives into Equation 19.28
and choose valuesa m
so that the equation is satisfied.yand its first two derivatives can
be written:
y=a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 ...
dy
dx =a 1 +2a 2 x+3a 3 x2 +4a 4
x 3 +5a 5
x 4 ...
d 2 y
dx 2 =2a 2 +6a 3 x+12a 4 x2 +20a 5
x 3 ...
Substitution intox 2d2 y
dx 2 +xdy dx + (x2 −1)y =0gives
2a 2
x 2 +6a 3
x 3 +12a 4
x 4 +20a 5
x 5 ... + ( a 1
x +2a 2
x 2 +3a 3
x 3 +4a 4
x 4 +5a 5
x 5 ... )
Removing the brackets
+ (x 2 −1)(a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 ...) =0
2a 2
x 2 +6a 3
x 3 +12a 4
x 4 +20a 5
x 5 ...+ ( a 1
x +2a 2
x 2 +3a 3
x 3 +4a 4
x 4 +5a 5
x 5 ... )
+a 0
x 2 +a 1
x 3 +a 2
x 4 +a 3
x 5 +a 4
x 6 +a 5
x 7 ...
−(a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 ...) =0
We now equate coefficients on both sides of this equation starting with the constant
term:
−a 0
= 0 so that a 0
= 0.
Equating coefficients ofx:
a 1
−a 1
= 0
sothata 1
isarbitrary.
Equating coefficients ofx 2 :
2a 2
+2a 2
+a 0
−a 2
= 0 sothat 3a 2
= −a 0
, from which a 2
= 0.
Equating coefficients ofx 3 :
6a 3
+3a 3
+a 1
−a 3
= 0 sothat 8a 3
= −a 1
, from which a 3
= − 1 8 a 1 .
It is easy to verify that all subsequent terms a m
when m is even, are zero. Equating
coefficients ofx 5 :
20a 5
+5a 5
+a 3
−a 5
= 0 so that 24a 5
= −a 3

fromwhich
a 5
=− 1
24 a 3 = 1
192 a 1
using a 3
= − 1 8 a 1 .
19.8 Bessel’s equation and Bessel functions 591
Note that all non-zero coefficients are multiples ofa 1
. We can now write down the first
threenon-zero terms inthe power series solution:
(
y=a 1
x − 1 8 x3 + 1 )
192 x5 −...
(19.29)
Herea 1
isanarbitraryconstant.Wehavesucceededinobtainingapowerseriessolution
of Bessel’s Equation 19.28.
Although a 1
is an arbitrary constant, historically its value is chosen to equal 1 2 in
Equation19.29andtheresultingpowerseriesisusedtodefineafunctioncalledaBessel
functionofthefirstkindoforderonethatisconventionallydenotedJ 1
(x).Itispossible
toshow thatitcan beexpressed concisely using sigmanotation as
J 1
(x) =
∞∑
m=0
(−1) m x 2m+1
2 2m+1 m!(m +1)! . (19.30)
You should verify this by writing out the first few terms of this infinite series. Figure
19.17 shows a graph of J 1
(x) obtained using computer software. Observe that the
graph is similar to that of the sine function but has decreasing amplitude. Here,
J 1
(0) =0.
J 1 (x)
0.5
0
10 20 30
x
–0.3
Figure19.17
The Besselfunction ofthe firstkind of
order 1,J 1 (x).
A second independent solution is called a Bessel function of the second kind of
orderoneandisdenotedbyY 1
(x).Derivationofthisfunctionisbeyondthescopeofthis
bookbut,likeJ 1
(x),valuesofY 1
(x)canbeobtainedfromtablesandbyusingcomputer
software. The general solution ofBessel’s equation of order one can thenbe written
y =AJ 1
(x) +BY 1
(x)
whereAandBare arbitraryconstants.

592 Chapter 19 Ordinary differential equations I
EXERCISES19.8.2
1 Write outexplicitly the firstfiveterms in the series
expansion ofJ 0 (x) andthe firstfour terms in the
expansion ofJ 1 (x).BydifferentiatingJ 0 (x) verify
that d dx (J 0 (x)) = −J 1 (x).
2 (Method ofFrobenius) Substituteapower series in
∞∑
the more general formy = a m x m+r ,whereris to
m=0
be determined, intoBessel’sequationoforder 1:
x 2d2 y
dx 2 +xdy dx +(x2 −1)y=0.
(i) By equatingcoefficients ofx r show thatr
satisfies theindicialequationr 2 −1 = 0 and
hence deducer = ±1.
(ii) Continuethe solution forthe caser = 1 to
obtain a seriessolution ofthe differential
equation. Findthe first 3 non-zeroterms.
(iii) Confirm that the solution is equivalent to that
obtained previously in Equation19.29.
19.8.3 Bessel’sequationandBesselfunctionsofhigherorder
Tocopewithmoregeneralcases,includingthoseofanon-integerorder,theusualpractice
is to seek solutions in the form a m
∞∑
x m+r where r is an unknown real number
m=0
that is chosen during the course of the calculation in order to generate a solution (see
Exercise 19.8.2, question 2). For details on how to proceed in such cases you should
consultanadvanced texton differential equations.
When the order, v, is a positive integer, it is conventional to label it asninstead. A
solution of Bessel’s equation of positive integer ordernis a Bessel function of the first
kind ofordern, denotedJ n
(x), and can beshown tobe
J n
(x) =
∞∑
m=0
(−1) m x 2m+n
2 2m+n m!(m +n)! , forn=0,1,2,...
Graphs of these functions forn = 0,1,2,3,4 are shown inFigure 19.18.
Whilst to this point we have only discussed cases where the order is non-negative,
in fact Bessel functions are defined for any order v, −∞ < v < ∞. Later, in Engineering
application 19.13 we shall make use of Bessel functions with negative integer
order. In such cases, careful inspection of the above definition shows that it will fail.
When n is negative, then m + n will be negative for some values of m and in these
case (m +n)! is not defined. For a complete treatment, and one which is beyond the
scope of this book, it is necessary to define a new function - the gamma function Ŵ(x)
- which can be thought of as a generalisation of factorials to deal with negative arguments.
Nevertheless, this is possible and it can be shown that Bessel functions with
negativeintegerorderarerelatedtotheircounterpartswithpositiveintegerorderbythe
result
J −n
(x) = (−1) n J n
(x).
(See Exercise 19.8.3)

n 5 2
n 5 3n
5 4
19.8 Bessel’s equation and Bessel functions 593
J n (x)
1
n 5 0
n 5 1
0
2 4 6 8 10 12 x
–0.4
Figure19.18
BesselfunctionsJ n (x),forn = 0 . . .4.
Engineeringapplication19.12
Modesofacylindricalmicrowavecavity
Circularcross-sectionwaveguidesandcylindricalmicrowavecavitiesarebothcomponents
that are found in high power microwave systems such as radar transmitters.
Bessel functions are used in the mathematical description of electromagnetic wave
behaviour in such devices. By way of example, consider the microwave cavity depictedinFigure
19.19.
r
l
Figure19.19
Acylindrical microwave cavity.
➔

594 Chapter 19 Ordinary differential equations I
Thewallsofmicrowavecavitiesareusuallymadeofmetalswithgoodconductivity
such as copper, steel or aluminium but in some applications the internal surfaces
areplatedwithpreciousmetalssuchasgoldtopreventoxidationwhichwouldreduce
the surface conductivity. When used at high power levels the conductivity is important.Lowconductivity
canleadtolossofpower andheatingofthecavityduetothe
currents flowing on the internal walls. The space inside the cavity can be empty or
it can contain a dielectric insulating material such as PTFE. The analysis here will
consider an empty cavity.
The purpose of the device is to act as a narrow band filter which selects a specific
frequency. In fact, there is a set of frequencies at which the microwave cavity
will work, known as the resonant frequencies. The electromagnetic waves inside
the cavity are governed by the propagation modes within the cylinder. These modes
are described mathematically by Bessel functions. It can be shown that the resonant
frequencies, f mnp
, associated with the transverse magnetic modes of the cavity are
given by
f mnp
= c
2π
√ (Xmn
r
) 2 ( pπ
+
l
) 2
wherem,nand paremodeindexnumbers,0,1,2,...andX mn
denotesthen-thzero
oftheBesselfunctionofthefirstkindoforderm,J m
(x),wherecisthespeedofwave
propagation inthe medium.
The zeros can be observed in Figure 19.18 as the points where the graph crosses
thexaxis.UsuallysoftwaresuchasMATLAB ® hasthecapabilitytocalculatenumericallythezerosofafunction.IfweobservethatthefirstzeroofJ
0
(x)isapproximately
x = 2.5 (see Figure 19.18) we can type the following at the MATLAB ® command
prompt:
X = fzero(@(x)besselj(0,x),2.5)
which gives the resultX = 2.4048.
The second zero ofJ 0
(x) isclose to5.5,thus wetype,
X = fzero(@(x)besselj(0,x),5.5)
which gives the resultX = 5.5201.
The first zero ofJ 1
(x) isapproximately 4,so,
X = fzero(@(x)besselj(1,x),4)
which gives the resultX = 3.8317.
Table19.2showsasummaryoftheresultsofapplying thismethodform = 0,1,
and 2.
Consideramicrowavecavityresonatormadefromanaluminiumfizzydrinkscan
with dimensionsr = 33 mm and l = 120 mm. The transverse magnetic modes are
labelledTM nmp
andthefirstcommonlyusedmodeisTM 011
.Theresonantfrequency
would be

19.8 Bessel’s equation and Bessel functions 595
Table19.2
Besselfunction zeros foramicrowave
cavity.
m X m1 X m2 X m3
0 2.4048 5.5201 8.6537
1 3.8317 7.0156 10.173
2 5.1356 8.4172 11.620
f 011
= 3×108
2π
√ (2.4048 ) 2
+
0.033
( ) 1 × π 2
= 3.70GHz
0.120
EXERCISES19.8.3
1 WehaveseenthatforpositiveintegerordertheBessel
functionsofthe firstkind are defined by
∞∑ (−1) m x 2m+n
J n (x) =
2 2m+n m!(m +n)! ,
m=0
for n=0,1,2,...
(a)Now suppose that weare interestedin Bessel
functionswithnegativeintegerorder.Replacenby
−nin thisdefinition to showthat
∞∑ (−1) m x 2m−n
J −n (x) =
2 2m−n m!(m −n)! ,
m=0
withn=1,2,...
(b)Now make the assumption that (m −n)!,which
appears in the denominator ofeach term,becomes
infinite whenm = 0,1,2, . . . ,n −1.Itis possible
to prove thisusingproperties ofthe Gamma
functionbutisbeyondthescopeofthisbook.With
thisassumption showthat
∞∑ (−1) m x 2m−n
J −n (x) =
2
m=n
2m−n m!(m −n)! ,
withn=1,2,...
(c)Byintroducing anew dummy variable,s,where
m =s +n, showthat
∞
J −n (x) = (−1) n ∑ (−1) s x 2s+n
2 2s+n m!s!
s=0
for n=0,1,2,...
and hence deduceJ −n (x) = (−1) n J n (x).
19.8.4 Besselfunctionsandthegeneratingfunction
A different set of modes known asTE modes may also be calculated using Bessel
functions. Rather than using zeros of the Bessel functions directly it is necessary to
usethederivativesoftheBesselfunctions,atopicbeyondthescopeofthisintroductorytreatment.
SeveralimportantpropertiesofBesselfunctionsusedinphysicsandengineeringapplications
can be deduced from knowledge of the so-called generating function. In this
sectionweintroducethisgeneratingfunctionanduseittoderivetheJacobi-Angeridentitieswhich
wethen employ inEngineering application 19.13.
Firstly, recall the power series expansion of e x introduced inSection 6.5.
∞
e x =1+x+ x2
2! + x3
3! + x4
4! +...= ∑ 1
n! xn . (19.31)
n=0

596 Chapter 19 Ordinary differential equations I
Recallalsothatthereisnothingspecialaboutlabellingthedummyvariablen.Anyother
letter could have been used and the statement of the expansion would be equivalent.
Now replacexfirstly by xt
2 and then by − x in Equation 19.31 to produce the following
expansions where we have usedr andsasthe dummy
2t
variables:
and
e xt
2 =
e − x 2t =
∞∑
r=0
∞∑
s=0
1
r!
( xt
) r
2
1
(
− x ) s
s! 2t
Multiplying these two expansions together wefind
e xt
2 e
− x 2t = e x 2 (t−1 t ) =
∞∑
r=0
1
r!
( xt
) r ∑ ∞
1
(
− x ) s
2 s! 2t
s=0
or, more explicitly,
[
]
e x 2 (t−1 t ) = 1 + xt
1!2 + x2 t 2
2!2 2 + x3 t 3
3!2 3 + x4 t 4
4!2 4 +... ×
[
]
1 − x
1!2t + x2
2!2 2 t 2 − x3
3!2 3 t 3 + x4
4!2 4 t 4 −...
(19.32)
Now imagine multiplying out the brackets on the right and observe that the resulting
expression can be considered as the sum or difference of terms involving powers oft,
that is, t, t 2 ,...t n ... and t −1 ,t −2 ,...t −n ,.... In fact we can express Equation 19.32
concisely as
e x 2 (t−1 t ) =
∞∑
n=−∞
y n
(x)t n (19.33)
where eachy n
(x) is a function ofxthat we can regard as the coefficient of the termt n
in the expansion. For example, let us focus first on the those terms in Equation 19.32
whichwhenmultipliedoutresultinmultiplesoft 0 ,aswillhappenbyformingproducts
such as
( )( )
1 1 ,
( xt
) (
− x )
,
1!2 1!2t
( )( )
x 2 t 2 x 2
2!2 2 2!2 2 t 2 ,
(
x 3 t 3
3!2 3 )(
− x3
3!2 3 t 3 )
Then together, these termswill formy 0
(x)t 0 inEquation 19.33, that is
y 0
(x)=1−
or more concisely,
y 0
(x) =
∞∑
n=0
x2
(1!) 2 2 + x 4
2 (2!) 2 (2 2 ) − x 6
2 (3!) 2 (2 3 ) +... 2
and soon.
(−1) n x 2n
(n!) 2 2 2n (19.34)
Comparing Equation 19.34 with Equation 19.27 we see thaty 0
(x) is simply the Bessel
function oforder zero,J 0
(x).

19.8 Bessel’s equation and Bessel functions 597
In a similar manner, suppose we focus attention on the terms in Equation 19.32 that
when multiplied outresultint 1 as will happen when weform products such as
( xt
)
(1) ,
1!2
( x 2 t 2 ) (
− x )
,
2!2 2 1!2t
( x 3 t 3
3!2 3 )( x
2
2!2 2 t 2 )
and so on.
Then, together these will formy 1
(x)t 1 inEquation 19.33, so that
y 1
(x) = x
1!2 − x3
2!2 3 + x5
3!2!2 5 ...
or more concisely,
y 1
(x) =
∞∑
m=0
(−1) m x 2m+1
(m +1)!m!2 2m+1 (19.35)
Comparing Equation 19.35 with 19.30 we see thaty 1
(x) is simply the Bessel function
of order one,J 1
(x).
Inthesameway,itispossibletoshowthatalltheothercoefficientsy n
(x)inEquation
19.33 are the Bessel functions J n
(x). The function e x 2 (t−1 t ) in Equation 19.33 is called
a generating function for the Bessel functions because the coefficient oft n in the expansion
isthe Bessel functionJ n
(x). Thus the expansion isused togenerate the Bessel
functions.We have the following important result:
Generatingfunctionforthe Bessel functions:
∞∑
e x 2 (t−1 t ) = J n
(x)t n (19.36)
n=−∞
Ifthefunctionontheleftisexpandedinpowersoft thenthecoefficientoft n isthe
Bessel function of ordern.
EXERCISES19.8.4
1 Usethe generating functionto show thatifnis an
even integer, thenthe BesselfunctionJ n (x)is an even
function, thatisJ n (−x) =J n (x).Ifnisan odd
integer, show thatJ n (x) isan odd function, thatis
J n (−x) = −J n (x).
19.8.5 TheJacobi-Angeridentities
Two results which relate to Bessel functions and which are important in communicationsengineering
are known as the Jacobi-Anger identities. We first statethem before
deriving them from the generatingfunction.

598 Chapter 19 Ordinary differential equations I
Jacobi-Anger identities
∞∑
e jxcos θ = J n
(x)j n e jnθ
n=−∞
∞∑
e jxsinθ = J n
(x)e jnθ
n=−∞
To obtain the first identity we let t = je jθ , so that t n = j n e jnθ , in Equation 19.36.
Noting that
t − 1 t = jejθ − 1
je jθ = jejθ +je −jθ = 2jcos θ
(where we have used the Euler relation cosθ = ejθ +e −jθ
, Section 9.7, p 339) then the
2
left hand sideof Equation 19.36 is
e x 2 (t−1 t ) = e x 2 (2jcosθ) = e jxcosθ
and sofrom Equation 19.36 we have
∞∑
e jxcosθ = J n
(x)j n e jnθ .
n=−∞
Inasimilarfashion, welett = e jθ , sothatt n = e jnθ , inEquation 19.36. Noting that
t − 1 t =ejθ − 1
e jθ = ejθ −e −jθ = 2jsin θ
(where we have used the Euler relation sinθ = ejθ −e −jθ
, Section 9.7 p 339) then the
2j
left hand sideof Equation 19.36 is
e x 2 (t−1 t ) = e x 2 (2jsinθ) = e jxsin θ
and sofrom Equation 19.36 we have
∞∑
e jxsinθ = J n
(x)e jnθ .
n=−∞
Engineeringapplication19.13
UseofBesselfunctionsinfrequencymodulation
A common type of modulation scheme used in electrical engineering is frequency
modulation(FM).Thepurposeofmodulationistoprepareasignalfortransmission,
such as when a signal is to be sent from a transmitter to a remote receiver during a
radiobroadcast.Atthereceiverthesignalisrecoveredfromitsmodulatedformbya
device called a demodulator.
Even though the process of electronically producing an FM signal is fairly
straightforward, the mathematics underpinning frequency modulation can become
complicatedandrequirestheuseofBesselfunctionsandaJacobi-Angeridentity.In

19.8 Bessel’s equation and Bessel functions 599
order to explain how FM works it is helpful to start with the electronic process by
which the FM signal is generated and to then derive the properties of the modulated
signal mathematically.
The FM signal is usually produced by a device called a voltage controlled oscillator
(VCO). The VCO generates a sinusoidal signal on its output, the frequency
of which is controlled by a voltage on its signal input. If we were to connect a d.c.
voltage to the input then we would observe a simple sine wave on the output. The
frequencyofthesinewavecouldbechangedbychangingthed.c.voltage.Inananalogueradiobroadcastsystemhoweverwewishtotransmitatime-varyingsignaland
this, rather thanad.c. voltage, isused as inputtothe VCO.
Consider the case of a sinusoidal modulating signal, sin(2πf m
t), that we wish
totransmitsuchasthatshowninFigure19.20.Here f m
isthefrequencyofthismodulating
signal. Another sinusoidal signal, termed the carrier, is used to carry the
modulatedsignal.Thefrequencyofthecarrierisdenoted f c
andisthefrequencyyou
wouldtunetoonaradiodialtoreceivetheradiostation.Letussuppose,forexample,
that the carrier isAcos(2πf c
t) as illustratedinFigure 19.21.
VCO input voltage
1
–1
Figure19.20
A sinusoidalmodulating signal: the signal we wishto transmit.
Time, t
Carrier signal
A
–A
Time, t
Figure19.21
A sinusoidalcarrier signal.
Itis usefultodefine a term, β, which isknown as themodulation index:
β = f
f m
where f isthemaximumchangeinthefrequencyoftheoutputfromtheVCOwhen
the modulating signal is applied. In other words the largest instantaneous frequency
of the carrier is f c
+ f and the smallest is f c
− f. It may seem intuitive that the
frequencieswithintheoutputsignalwouldbebetween f c
+f and f c
−f,butthe
situation is actually more complex. The changing signal produces a different set of
output frequencies. Consider the following analysis.
We can describe this modulation process mathematically as
v(t) =Acos [ 2πf c
t + βsin(2πf m
t) ]
➔

600 Chapter 19 Ordinary differential equations I
and thisoutput waveform isshown inFigure 19.22.
VCO output y(t)
A
–A
Time, t
Figure19.22
Theoutput signal.
We may rewrite thisequation using phasor notation as:
v(t) =Re
[Ae j[2πf c t+βsin(2πf t)]] m
= Re [ Ae j2πf c t e jβsin(2πf m t)]
where Re means the real part of what follows. Notice that the second exponential
∞∑
term is in the same form as the Jacobi-Anger identity, e jxsinθ = J n
(x)e jnθ , and
n=−∞
can be expanded outinterms ofBessel functions of the first kind.Thus,
[ ∞
]
∑
v(t) =Re Ae j2πf c t J n
(β)e jn2πf m t
n=−∞
This can be rearranged by combining the exponentials:
[
]
∞∑
v(t) =Re A J n
(β)e jn2πf m t e j2πf c t
v(t) =Re
[
A
n=−∞
∞∑
n=−∞
J n
(β)e j2π(f c +nf m )t ]
Convertingbackfromthephasornotation,andforsimplicitytakingA = 1,weobtain:
v(t) =J 0
(β)cos(2πf c
t)
+J −1
(β)cos[2π(f c
− f m
)t] +J 1
(β)cos[2π(f c
+ f m
)t]
+J −2
(β)cos[2π(f c
−2f m
)t] +J 2
(β)cos[2π(f c
+2f m
)t]
+J −3
(β)cos[2π(f c
−3f m
)t] +J 3
(β)cos[2π(f c
+3f m
)t]
+ ...
We sawinSection 19.8.3 thatforinteger values ofn,J −n
(x) = (−1) n J n
(x), hence,
v(t) =J 0
(β)cos(2πf c
t)
−J 1
(β)cos[2π(f c
− f m
)t] +J 1
(β)cos[2π(f c
+ f m
)t]
+J 2
(β)cos[2π(f c
−2f m
)t] +J 2
(β)cos[2π(f c
+2f m
)t]
−J 3
(β)cos[2π(f c
−3f m
)t] +J 3
(β)cos[2π(f c
+3f m
)t]
+ ...
WecanseethattheoutputfromtheVCOisactuallymadeupfromasetofsinusoidal
waveformswithfrequenciesf c
−nf m
,...,f c
−3f m
,f c
−2f m
,f c
−f m
,f c
,f c
+f m
,f c
+
2f m
, f c
+ 3f m
,..., f c
+nf m
. This is known as the frequency content of the signal
andwillbediscussedinmoredetailinChapters23&24.Theequationsuggeststhat
a broadcast station with a carrier at f c
would have an infinite number of other fre-

Review exercises 19 601
quencies, known as side bands associated with it. The amplitudes of the side bands
are given by the values of the Bessel functions. When these are evaluated, the values
ofJ n
(β) generally diminish with increasingnhence only the sidebands near to
the carrier frequency have significant amplitude. Therefore frequencies that are a
long way away from f c
can be safely ignored when designing the communications
system.
TabulatedvaluesofBesselfunctionsareavailablesimilartothoseshowninTable
19.3. It can be seen how the signal amplitudes generally decrease with increasing
order of the Bessel function. A larger value of β also implies a larger number of
side bands are significant and need to be considered. The use of Bessel functions
whendesigningradiocommunicationssystemsisveryimportantbecauseeachradio
channel has an allocated bandwidth outside of which it should not stray. To do so
would create the possibility of different radio channels interfering with each other.
Thismathematicalmethodallowstheamplitudesofthesidebandstobecalculatedto
ensuretheyaresmallenoughsoastonotinterferesignificantlywithotherchannels.
Table19.3
Table ofBesselfunctionsJ n (β)ofinteger order,n,fordifferent values of
modulation index, β.
β n=0 n=1 n=2 n=3 n=4 n=5 n=6 n=7
0.00 1.00 – – – – – – –
0.25 0.98 0.12 – – – – – –
0.50 0.94 0.24 0.03 – – – – –
0.75 0.86 0.35 0.07 – – – – –
1.00 0.77 0.44 0.11 0.02 – – – –
1.25 0.65 0.51 0.17 0.04 – – – –
1.50 0.51 0.56 0.23 0.06 0.01 – – –
1.75 0.37 0.58 0.29 0.09 0.02 – – –
2.00 0.22 0.58 0.35 0.13 0.03 – – –
2.25 0.08 0.55 0.40 0.17 0.05 0.01 – –
2.50 -0.05 0.50 0.45 0.22 0.07 0.02 – –
2.75 -0.16 0.43 0.47 0.26 0.10 0.03 – –
3.00 -0.26 0.34 0.49 0.31 0.13 0.04 0.01 –
3.25 -0.33 0.24 0.48 0.35 0.17 0.06 0.02 –
3.50 -0.38 0.14 0.46 0.39 0.20 0.08 0.03 –
3.75 -0.40 0.03 0.42 0.41 0.24 0.10 0.04 0.01
4.00 -0.40 -0.07 0.36 0.43 0.28 0.13 0.05 0.02
REVIEWEXERCISES19
1 Find the general solution ofthe followingequations:
(a)
(c)
dx
dt =2x (b) (1+t)dx
dt = 3
dy
dx =y2 cosx
3
2 Solve dy = 2,subjecttoy(0) = 3.
dx
Find the general solution oftẋ+x=2t.
4 Solve dx
dt +2x=e2t cost
(a)byusingan integrating factor
(b)byfinding itscomplementary function anda
particular integral.

602 Chapter 19 Ordinary differential equations I
5 Find the general solution ofy ′′ +16y =x 2 .
6 Find the particular solution of
y ′′ +3y ′ −4y =e x ,y(0) =2,y ′ (0) =0.
7 Aparticle moves in astraightline such that its
displacement from the origin Oisx,wherexsatisfies
the differentialequation
d 2 x
dt 2 +16x=0
(a) Findthe general solution ofthisequation.
( ) ( )
π
π
(b) Ifx = −12, andẋ = 20,findthe
4 4
displacement ofthe particlewhent = π 2 .
8 Usean integrating factorto solvethe differential
equation
dx
+xcott =cos3t
dt
Solutions
1 (a) x=Ae 2t
(b) x=3ln|1+t|+c
1
(c) y=
A−sinx
2 y=2x+3
3 x=t+ c t
4 x = e2t (4cost +sint)
17
+ce −2t
5 y=Acos4x+Bsin4x+ x2
16 − 1
128
6 y = 39ex +11e −4x
25
+ xex
5
7 (a) Asin4t +Bcos4t (b) 12
cos2t − 1
8 sint;x(t) =
2 cos4t+c
4sint

20 Ordinarydifferential
equationsII
Contents 20.1 Introduction 603
20.2 Analoguesimulation 603
20.3 Higherorderequations 606
20.4 State-spacemodels 609
20.5 Numericalmethods 615
20.6 Euler’smethod 616
20.7 ImprovedEulermethod 620
20.8 Runge--Kuttamethodoforder4 623
Reviewexercises20 626
20.1 INTRODUCTION
Engineerssometimesfinditconvenienttorepresentadifferentialequationusinganelectronic
circuit. This is known as analogue simulation. Analogue controllers are also frequentlyusedincontrolsystemsandthesearebasedonsimilarcircuits.Complicatedengineering
systems need to be modelled using a set of differential equations. This leads
tothetopicofstate-spacemodelling.Suchmodelsfindwidespreaduseintheaerospace
industries as well as in other areas that have complex systems. Numerical techniques
are very important as they allow differential equations to be solved using computers.
Often this is the only way to obtain a solution of a differential equation, owing to its
complexity. State-space models arealmost always solved using computers.
20.2 ANALOGUESIMULATION
Itispossibletosolvedifferentialequationsusingelectroniccircuitsbasedonoperational
amplifiers. The advantage of this approach is the ease with which the coefficients of
the differential equation can be adjusted and the effect on the solution observed. The

604 Chapter 20 Ordinary differential equations II
y ic
y 1
10
y 1
1
y 2
1
y o
y 2
1
y o
y 3
1
y 3
10
Figure20.1
An integrator.
Figure20.2
A summer.
technique is known as analogue simulation. Prior to the mass availability of powerful
digital computers this was a common approach in engineering design because it providedtheengineerwithanautomatedmethodtosolvearangeofmathematicalmodels.
Special-purposecomputers,knownasanaloguecomputers,werehistoricallyusedwhich
had the electronic circuits already incorporated, thus making iteasier tosimulate a particular
differential equation.
Three basic types of circuit are required to enable ordinary differential equations
with constant coefficients to be simulated. The first type is an integrator which has already
been discussed in Section 13.2. The usual symbol for such a circuit is shown in
Figure 20.1.
If an analogue computer is used it is common for there to be a gain of only 1 or 10
ontheinputvoltage.Againabove10isbetterachievedbylinkingtogethertwocircuits
in series. In addition there is usually a facility to allow initial conditions to be set. The
equation forthe circuitof Figure 20.1 is
v o
=−
∫ t
0
(10v 1
+ v 2
+ v 3
)dt − v ic
(20.1)
where v ic
indicatestheinitialoutputvoltage.Notethatthegainisusuallywrittenonthe
input line.
The second type of circuit is the summer. It has the symbol shown in Figure 20.2.
The equation forthe circuitof Figure 20.2 is
v o
= −(v 1
+ v 2
+10v 3
) (20.2)
Finally, a circuit is required to allow gains to be varied. This is simply a potentiometer
andisillustratedtogetherwithitssymbolinFigure20.3.Theequationforthecircuitof
Figure 20.3 issimply
v o
=kv i
0k1 (20.3)
The potentiometer only allows the gain to be varied between 0 and 1. If a variable gain
greater than 1 is required, the potentiometer must be placed in series with a circuit that
increases gain, for example a one-input summer withagain of 10.
Thesecircuitscanbecombinedtoobtainthesolutionsofdifferentialequations.This
isbestillustratedbymeansofanexample.Considerthegeneralformofasecond-order
differential equation with constant coefficients:
a d2 y
dt +bdy +cy=f(t) (20.4)
2 dt
The general approach to obtaining the circuit for a particular differential equation is to
assume a certain point in the circuit corresponds to a particular term and then arrange

20.2 Analogue simulation 605
y i
y i y o
k
y o
d
a —
2 y
dt 2
1—
a
d 2 y
—
dt 2
dy
– —
1 dt 1
y
Figure20.3
Apotentiometer.
Figure20.4
Firststage in synthesizing Equation(20.4).
toconnectthatpointtothecorrectlysynthesizedvalue.Itismorestraightforwardifthe
assumed point corresponds to the highest derivative. Rearranging Equation (20.4) we
find
a d2 y
dt =f(t)−bdy −cy (20.5)
2 dt
Assuminga d2 y
dt isalreadyavailable,apotentiometercanbeusedtoobtain d2 y
2 dt 2.Integratorscanthenbeusedtoobtainthevariables
dy
dt andy.ThisisshowninFigure20.4.The
nextstageistoobtaintheexpressioncorrespondingtother.h.s.ofEquation(20.5).This
requiresasummertoaddtogethertheindividualtermsandpotentiometerstoallowindividualcoefficientstobeobtained.Aninvertorisalsorequiredtoobtainthecorrectsign
forthevariable dy
dt .Thissimplyconsistsofasummerwithoneinputandagainof1.The
final circuit is shown in Figure 20.5. The three potentiometers allow different values of
a,bandctobeobtained.Ifgainsgreaterthan1wererequiredthenextrasummerswould
be needed, or alternatively the input gains of the existing summers could be adjusted.
The input to the circuit, f (t), can be simulated by applying a signal at the appropriate
point in the circuit. The output of the circuit, y(t), corresponds to the solution of the
differential equation. It may be necessary to include initial conditions corresponding to
y(0)and dy
dt (0).
Although analogue computers remain an interesting historical development in the
simulation of control systems it is now much more common to use digital computers.
Thesystemtobesimulatedcanberepresentedeitherdirectlyintheformofmathematical
equations or by block diagrams similar to that shown in Figure 20.5. The response to
various inputs can be readily obtained.
f(t)
1
–f(t)
1
1
1
a
b dy — dt
d 2 y
—
dt 2
1
1—
a
–b dy — dt
d 2 y
—
dt 2
– dy —
1 dt 1 y
b
cy
c
Figure20.5
Thecomplete circuit to synthesize Equation(20.4).

606 Chapter 20 Ordinary differential equations II
20.3 HIGHERORDEREQUATIONS
Inthissectionweshallconsidersecond-andhigherorderequationsandshowhowthey
can be represented as a set of simultaneous first-order equations. The main reason for
doingthisisthatwhenacomputersolutionisrequireditisusefultoexpressanequation
in this form. Details of the analytical solution of such systems are not considered here
although one technique isdiscussed inSection 21.10.
It is possible to express a second-order differential equation as two first-order equations.Thus
ifwehave
d 2 ( )
y dy
dx = f 2 dx ,y,x (20.6)
we can introduce the new dependent variablesy 1
andy 2
such thaty 1
= y andy 2
= dy
dx .
Equation (20.6) thenbecomes
dy 1
dx =y 2
dy 2
dx = f(y 2 ,y 1
,x) (20.7)
These first-order simultaneous differential equations are often referred to as coupled
equations.
Example20.1 Expressthe equation
d 2 y
dx 2 −7dy dx +3y=0
as a set of first-order equations.
Solution Lettingy 1
= y, andy 2
= dy
dx , we find dy 2
dx = d2 y
dx2. Therefore the differential equation
becomes
dy 1
dx =y 2
dy 2
dx −7y 2 +3y 1 =0
We note thatthese equations can alsobewritten as
dy 1
dx =y 2
dy 2
dx = −3y 1 +7y 2
or, inmatrix form,
( ) ( ) ( ) y
′
1
0 1 y1
=
−3 7 y 2
y ′ 2
where ′ denotes d dx .

20.3 Higher order equations 607
Higher order differential equations can be reduced to a set of first-order equations in a
similarway.
Example20.2 Express the equation
d 3 x
dt 3 −7d2 x
dt 2 +3dx dt +2x=0
as a set of first-order equations.
Solution Lettingx 1
=x,x 2
= dx
dt andx 3 = d2 x
dt 2 wefind
dx 1
dt
dx 2
dt
=x 2
=x 3
dx 3
dt
−7x 3
+3x 2
+2x 1
=0
isthe set of first-order equations representing the given differential equation.
Example20.3 (a) Expressthe coupledfirst-order equations
dy 1
dx =y 1 +y 2
dy 2
dx = 4y 1 −2y 2
as a second-order ordinary differential equation, and obtain its general solution.
(b) Express the given equations inthe form
( ) ) y
′
1
y1
=A(
y 2
y ′ 2
whereAisa2 ×2 matrix.
Solution (a) Differentiating the second of the given equations we have
d 2 y 2
dx = 4dy 1
2 dx −2dy 2
dx
and then, using the first, we find
d 2 y 2
dx =4(y 2 1 +y 2 )−2dy 2
dx
But fromthe second given equation 4y 1
= dy 2
dx +2y 2
, and therefore
d 2 y 2
dx = dy 2
2 dx +2y 2 +4y 2 −2dy 2
dx

608 Chapter 20 Ordinary differential equations II
thatis,
d 2 y 2
dx 2 + dy 2
dx −6y 2 =0
Writingyfory 2
wefind
d 2 y
dx 2 + dy
dx −6y=0
To solve thiswe lety = e kx toobtain
k 2 +k−6=0
(k−2)(k+3)=0
Therefore,
k=2,−3
The general solution is theny = y 2
= Ae 2x +Be −3x . It is straightforward to show
thaty 1
satisfiesthesamesecond-orderdifferentialequationandhencehasthesame
general solution, butwith different arbitraryconstants.
(b) The first-order equations can be written as
( ) ( ) ( ) y
′
1
1 1 y1
y ′ =
2
4 −2 y 2
( ) 1 1
ThereforeA = .
4 −2
EXERCISES20.3
1 Expressthe followingequations asasetoffirst-order
equations:
(a) d2 y
dx 2 +2dy dx +3y=0
(b) d2 y
dx 2 +8dy dx +9y=0
(c) d2 x
dt 2 +4dx dt +6x=0
(d) d2 y
dt 2 +6dy dt +7y=0
(e) d3 y
dx 3 +6d2 y
dx 2 +2dy dx +y=0
(f) d3 x
dt 3 +2d2 x
dt 2 +4dx dt +2x=0
2 Expressthe followingcoupled first-orderequations as
a single second-order differentialequation:
(a)
dy 1
dx =y 1 +y 2 ,dy 2
dx = 2y 1 −2y 2
(b) dy 1
dx =2y 1 −y 2 ,dy 2
dx =4y 1 +y 2
(c)
dx 1
dt
(d) dy 1
dt
3 Express
=3x 1 +x 2 , dx 2
dt
=2y 1 +4y 2 , dy 2
dt
dy 1
dt
= 2x 1 −3x 2
= 2y 1 +6y 2 and
= 6y 1 −7y 2
dy 2
= −2y
dt 1 −5y 2
asasingle second-order equation. Solve thisequation
and hence findy 1 andy 2 .Expressthe equations in the
formy ′ =Ay.

20.4 State-space models 609
Solutions
1 (a)
(b)
(c)
(d)
(e)
(f)
dy 1
dx =y dy 2
2
dx = −3y 1 −2y d 2 y
2 2 (a)
dx 2 + dy
dx −4y=0
dy 1
dx =y dy 2
2
dx = −9y 1 −8y 2
d 2 y
(b)
dx 2 −3dy dx +6y=0
dx 1 dx
=x 2
dt 2 = −6x
dt 1 −4x 2
d 2 x
(c)
dy 1 dy dt 2 −11x=0
=y 2
dt 2 = −7y
dt 1 −6y 2
d 2 y
(d)
dy 1
dx =y dy 2
2
dx =y dt 2 +5dy dt −38y=0
3
3 y ′′ +3y ′ +2y=0
dy 3
dx = −y 1 −2y 2 −6y 3
y 1 (t) =Ae −2t +Be −t
dx 1 dx
=x 2
y
dt 2 =x 2 (t) = − 2
dt 3
3 Ae−2t − 1 2 Be−t
( ) ( )
y ′ (y1 ) 2 6
dx 1
3
= −2x
dt 1 −4x 2 −2x 3 y 2
′ =
−2 −5 y 2
20.4 STATE-SPACEMODELS
Thereareseveralwaystomodellineartime-invariantsystemsmathematically.Oneway,
which we have already examined, is to use linear differential equations with constant
coefficients. A second method is to use transfer functions which will be discussed in
Chapter 21. A third type of model is the state-space model. The state-space technique
is particularly useful for modelling complex engineering systems in which there are
several inputs and outputs. It also has a convenient form for solution by means of a
digital computer.
Thebasisofthestate-spacetechniqueistherepresentationofasystembymeansofa
setoffirst-ordercoupleddifferentialequations,knownasstateequations.Thenumber
of first-order differential equations required to model a system defines the order of the
system. For example, if three differential equations are required then the system is a
third-ordersystem.Associatedwiththefirst-orderdifferentialequationsareasetofstate
variables, the samenumber astherearedifferential equations.
The concept of a state variable lies at the heart of the state-space technique. A system
is defined by means of its state variables. Provided the initial values of these state
variables are known, it is possible to predict the behaviour of the system with time by
means of the first-order differential equations. One complication is that the choice of
statevariablestocharacterizeasystemisnotunique.Manydifferentchoicesofasetof
state variables for a particular system are often possible. However, a system of ordern
only requires n state variables to specify it. Introducing more state variables than this
only introducesredundancy.
Thechoiceofstatevariablesforasystemis,tosomeextent,dependentonexperience
buttherearecertainguidelinesthatcanbefollowedtoobtainavalidchoiceofvariables.
One thing that is particularly important is that the state variables are independent of

610 Chapter 20 Ordinary differential equations II
y C
C
R
i
~
L
Figure20.6
Thecircuitcould be modelled using v C andiasthe state
variables.
each other. For example, for the electrical system of Figure 20.6, the choice of v C
and
3v C
wouldleadtothetwostatevariablesbeingdependentoneachother.Avalidchoice
wouldbe v C
andiwhicharenotdirectlydependentoneachother.Thisisasecond-order
systemand so only requires two statevariables tomodel itsbehaviour.
Many engineering systems may have a high order and so require several differential
equations to model their behaviour. For this reason, a standard way of laying out these
equations has evolved to reduce the chance of making errors. There is also an added
advantage in that the standard layout makes it easier to present the equations to a digital
computer for solution. Before introducing the standard form, an example will be
presented toillustratethe statevariable method.
Engineeringapplication20.1
State-spacemodelforaspring-mass-dampersystem
ConsiderthemechanicalsystemillustratedinFigure20.7.Amassrestsonafrictionlesssurfaceandisconnectedtoafixedwallbymeansofanidealspringandanideal
damper. Aforce, f (t), isapplied tothe mass and the position ofthe mass is w.
K
B
M
w
f(t)
Figure20.7
Asecond-order mechanicalsystem.
By considering the forces acting on the mass,M, it is possible to devise a differential
equation that models the behaviour of the system. The force produced by the
spring isKw (K is the spring stiffness) and opposes the forward motion. The force
produced by the damper isB dw , where B is the damping coefficient, and this also
dt
opposestheforwardmotion.Sincethemassisconstant,Newton’ssecondlawofmotion
states that the net force on the mass is equal to the product of the mass and its
acceleration. Thus wefind
f(t)−B dw −Kw=M d2 w
(20.8)
dt dt 2
Note that this is a second-order differential equation and so the system is a secondorder
system.
Inordertoobtainastate-spacemodelthestatevariableshavetobechosen.There
areseveralpossiblechoices.Anobviousoneistheposition, w,ofthemass.Asecond

20.4 State-space models 611
statevariableisrequiredasthesystemissecondorder.Inthiscasethevelocityofthe
masswillbechosen.Thevelocityofthemassisnotdirectlydependentonitsposition
and so the two variables are independent. Another possible choice would have been
dw
−w.Thismayseemaclumsychoicebutforcertainproblemssuchchoicesmay
dt
lead tosimplifications inthe statevariable equations.
Itiscustomarytousethesymbolsx 1
,x 2
,x 3
,...torepresentvariablesforreasons
that will become clear shortly. So,
x 1
=w
(20.9a)
x 2
= dw
(20.9b)
dt
Because of the particular choice of state variables, it is easy to obtain the first of
thefirst-orderdifferentialequations--thusillustratingtheneedforexperiencewhen
choosing statevariables.
Differentiating Equation (20.9a) gives
dx 1
= dw =x
dt dt 2
This isthe first stateequation although itisusually written as
ẋ 1
=x 2
where ẋ 1
denotes dx 1
. The second of the first-order equations is obtained by rearranging
Equation
dt
(20.8):
d 2 w
= − K dt 2 M w − B dw
+ 1 f (t) (20.10)
M dt M
However, differentiating Equation (20.9b) we get
d 2 w
dt 2 =ẋ 2
Then, usingEquation (20.10) weobtain
ẋ 2
=− K M x 1 − B M x 2 + 1 M f(t)
Finally, itisusual toarrange the stateequations inaparticular way:
ẋ 1
= + x 2
ẋ 2
=− K M x 1 − B M x 2 + 1 M f(t)
w=x 1
Note that it is conventional to relate the output of the system to the state variables.
Assume that for this system the required output variable is the position of the mass,
that is w.
Itisstraightforward torewrite these equations inmatrix form:
) ( )( ) ( )
(ẋ1 0 1 x1 0
=
+ f(t)
ẋ2 −K/M −B/M x 2
1/M
w = ( 1 0 ) ( )
x 1
x 2

612 Chapter 20 Ordinary differential equations II
More generally, the standard form ofthe stateequations for a linear system isgiven by
ẋ(t) =Ax(t) +Bu(t)
y(t) =Cx(t) +Du(t)
For a systemwithnstatevariables,rinputs and poutputs:
x(t) is ann-component column vector representing the states of thenth-order system.
It is usuallycalled thestatevector.
u(t) is an r-component column vector composed of the input functions to the
system. It is usuallycalled theinputvector.
y(t) is a p-component column vector composed of the defined outputs of the
system. It is referred toas theoutputvector.
A = (n ×n)matrix, knownas the statematrix.
B = (n ×r)matrix, knownas the inputmatrix.
C = (p ×n)matrix, knownas the outputmatrix.
D = (p ×r)matrix, knownas the directtransmission matrix.
A,B,C andDhave constant elements if the system is time invariant. When presented
in this form the equations appear to be extremely complicated. In fact, the problem is
only one of notation. The general nature of the notation allows any linear system to be
specified but in many cases the matrices are zero or have simple coefficient values. For
example,thematrixDisoftenzeroforasystemasitisunusualtohave directcoupling
betweentheinputandtheoutputofasystem.Inthisformatitwouldbestraightforward
topresent the equations toadigital computer for solution.
Engineeringapplication20.2
Anarmature-controlledd.c.motor
Deriveastate-spacemodelforanarmature-controlledd.c.motorconnectedtoamechanicalloadwithcombinedmomentofinertiaJ,andviscousfrictioncoefficientB.
The arrangement isshown inFigure 20.8.
v a
= applied armature voltage
i a
= armature current
R a
= armature resistance
L a
= armature inductance
e b
= back e.m.f.of the motor
ω m
= angular speed of the motor
T = torque generated by the motor.
Solution
Letusassumethatthesysteminputisthearmaturevoltage, v a
,andthesystemoutput
istheangularspeedofthemotor, ω m
.Thisisasecond-ordersystemandsotwostate
variablesarerequired.Wewillchoosethearmaturecurrentandtheangularspeedof
the motor. So,
x 1
=i a
x 2
=ω m

20.4 State-space models 613
R a L a
T
i a
Fixed field
y a
e b
v m
J
Bv m
Figure20.8
An armaturecontrolled
d.c. motor.
The next stage is to obtain a mathematical model for the system. Using Kirchhoff’s
voltage law and the component laws for the resistor and inductor we obtain, for the
armature circuit,
v a
=i a
R a
+L a
di a
dt
+e b
Now for a d.c. motor the back e.m.f. is proportional to the speed of the motor and is
given by e b
=K e
ω m
,K e
constant. So,
di
v a
=i a
R a
+L a
a
+K
dt e
ω m
di a
= − R a
i
dt L a
− K e
ω
a
L m
+ 1 v
a
L a
(20.11)
a
Let us now turn to the mechanical part of the system. If G is the net torque about
theaxisofrotationthentherotationalformofNewton’ssecondlawofmotionstates
G = J dω
dω
, whereJ is the moment of inertia, and is the angular acceleration. In
dt dt
this example, the torques are that generated by the motor,T, and a frictional torque
Bω m
which opposes the motion, so that
T−Bω m
=J dω m
dt
For a d.c. motor, the torque developed by the motor is proportional to the armature
current and isgiven byT =K T
i a
, whereK T
isaconstant. So,
K T
i a
−Bω m
=J dω m
dt
dω m
dt
= K T
J i a − B J ω m
(20.12)
Equations (20.11) and (20.12) are the state equations for the system. They can be
arranged inmatrix form togive
( ) ( )( ) ( )
˙i a −Ra /L
= a
−K e
/L a ia 1/La
+ v
˙ω m
K T
/J −B/J ω m
0 a
( ) ( )
x1 ia
Alternatively the notation x = = andu = v
x 2
ω a
can be used. However,
m
whenthereisnoconfusionitisbettertoretaintheoriginalsymbolsbecauseitmakes
iteasier tosee ataglance what the statevariables are.
➔

614 Chapter 20 Ordinary differential equations II
Finally,anoutputequationisneeded.Inthiscaseitistrivialastheoutputvariable
isthe same as one of the statevariables. So,
ω m
= ( 0 1 )( )
i a
ω m
Engineeringapplication20.3
State-spacemodelforacoupled-tankssystem
Derive a state-space model for the coupled tank system shown in Figure 20.9. The
tankshavecross-sectionalareasA 1
andA 2
,valveresistancesR 1
andR 2
,fluidheights
h 1
and h 2
, input flows q 1
and q 2
. Additionally, there is a flow, q o
, out of tank 2.
Assumethatthevalvescanbemodelledaslinearelementsandletthedensityofthe
fluid inthe tanks be ρ. Letq i
be the intermediate flowbetween the two tanks.
q 1
q 2
h 1
q i
qo
h 2
Tank 1, R 1 Tank 2, R 2 Figure20.9
area A 1 area A 2 Coupled tanks.
Solution
A convenient choice of state variables is the height of the fluid in each of the tanks,
although a perfectly acceptable choice would be the volume of fluid in each of the
tanks.For tank 1,conservation of mass gives
q 1
−q i
=A 1
dh 1
dt
Fortheresistanceelement,R 1
,thepressuredifferenceacrossthevalveisequaltothe
productoftheflowthroughthevalveandthevalveresistance.Thiscanbethoughtof
asafluidequivalentofOhm’slaw.Notethatatmosphericpressurehasbeenignored
as itisthe same on both sides of the valve. So,
ρgh 1
−ρgh 2
=q i
R 1
q i
= ρg (h
R 1
−h 2
) (20.13)
1
Combining these two equations gives
dh 1
= − ρg h
dt R 1
A 1
+ ρg h
1
R 1
A 2
+ 1 q
1
A 1
(20.14)
1
For tank 2,
q i
+q 2
−q o
=A 2
dh 2
dt
ρgh 2
=R 2
q o
(20.15)

20.5 Numerical methods 615
Combining these equations and usingEquation (20.13) toeliminateq i
andq o
gives
dh 2
= ρg ( ρg
h
dt R 1
A 1
− + ρg )
h
2
R 1
A 2
R 2
A 2
+ 1 q
2
A 2
(20.16)
2
Equations (20.14) and (20.16) are the state-space equations for the system and can
be written inmatrix formas
⎛
−ρg ρg
⎞ ⎛ ⎞
1
)
(ḣ1
R 1
A 1
R 1
A 1
( )
0
h1
A ( )
1 = ⎜
ḣ 2 ⎝ ρg
− ρg − ρg
⎟ +
q1
⎠ h ⎜ ⎟
2 ⎝ 1 ⎠
q 2
0
R 1
A 2
R 1
A 2
R 2
A 2 A 2
Note that in this case the input vector is two-dimensional as there are two inputs to
the system.The output equation isgiven by
q o
= ( ) ( )
h
0 ρg/R 1
2
h 2
and isobtained directly from Equation (20.15).
20.5 NUMERICALMETHODS
All the techniques we have so far met for solving differential equations are known as
analytical methods, and these methods give rise to a solution in terms of elementary
functionssuchassinx,e x ,x 3 ,etc.Inpractice,mostengineeringproblemsinvolvingdifferential
equations are too complicated to be solved easily using analytical techniques.
It is therefore frequently necessary to make use of computers. One such approach is to
use analogue simulation, which we discussed in Section 20.2. Another, more common,
approachistouse digital computers.
There are a variety of computer software packages available for solving differential
equations,includingMATLAB ® whichhasalreadybeenintroduced.Oftenthedetailof
howthecomputersolvestheequationishiddenfromtheuseranditusesoneofarange
of numericalmethods.Wewillexaminesomeofthemethodsthatmightbeappliedby
the computer inthissection.
Whenusingnumericalmethodsitisimportanttonotethattheyresultinapproximate
solutions to differential equations and solutions are only calculated at discrete intervals
oftheindependentvariable,typicallyxort.Weshallbeginbyexaminingthefirst-order
differential equation
dy
dx =f(x,y)
subject to the initial condition y(x 0
) = y 0
. Usually the solution is obtained at equally
spaced values of x, and we call this spacing the step size, denoted by h. By choosing
a suitable value ofhwe can control the accuracy of the approximate solution obtained.
Weshallwritey n
forthisapproximatesolutionatx =x n
,whereaswewritey(x n
)forthe
true solution atx = x n
. Generally these values will not be the same although we try to
ensure the difference issmallfor obvious reasons.
The simplest numerical method forthe solution of the differential equation
dy
dx =f(x,y)

616 Chapter 20 Ordinary differential equations II
with initial condition
y(x 0
) =y 0
isEuler’s method which weshall study inthe next section.
20.6 EULER’SMETHOD
You will recall, from Chapter 12, that given a functiony(x), the quantity dy
dx represents
thegradientofthatfunction.Soif dy
dx = f (x,y)andweseeky(x),weseethatthedifferential
equation tells us the gradient of the required function. Given the initial condition
y =y 0
whenx =x 0
wecanpicturethissinglepointasshowninFigure20.10.Moreover,
we know the gradient of the solution here. Because dy
dx = f(x,y)weseethat
dy
dx∣ =f(x 0
,y 0
)
x=x0
which equals the gradient of the solution at x = x 0
. Thus the exact solution passes
through (x 0
,y 0
)andhasgradient f (x 0
,y 0
)there.Wecandrawastraightlinethroughthis
point with the required gradient to approximate the solution as shown in Figure 20.10.
Thisstraightlineapproximatesthetruesolution,butonlynear (x 0
,y 0
)because,ingeneral,
the gradient is not constant but changes. So, in practice we only extend it a short
distance,h,alongthexaxistowherex =x 1
.Theycoordinateatthispointisthentaken
asy 1
.Wenowdevelopanexpressionfory 1
.Thestraightlinehasgradient f (x 0
,y 0
)and
passes through (x 0
,y 0
). Itcan beshown thatits equation istherefore
y=y 0
+(x−x 0
)f(x 0
,y 0
)
Whenx =x 1
theycoordinate isthen given by
y 1
=y 0
+(x 1
−x 0
)f(x 0
,y 0
)
and sincex 1
−x 0
=hwe find
y 1
=y 0
+hf(x 0
,y 0
)
Thisequationcanbeusedtofindy 1
.Wethenregard (x 1
,y 1
)asknown.Fromthisknown
point the whole process isthen repeated using the formula
y i+1
=y i
+hf(x i
,y i
)
y
y(x 1 )
y 1
True solution
Tangent line
approximation
y 0
x 0 x 1 x
0
Figure20.10
Approximation usedin Euler’smethod.

20.6 Euler’s method 617
and we can therefore generate a whole sequence of approximate values ofy. Naturally,
theaccuracyofthesolutionwilldependuponthestepsize,h.Infact,forEuler’smethod,
theerrorincurredisroughlyproportionaltoh,sothatbyhalvingthestepsizeweroughly
halve the error.
An alternative way of deriving Euler’s method is to use a Taylor series expansion.
Recall from Chapter 18 that
y(x 0
+h) =y(x 0
) +hy ′ (x 0
) + h2
2! y′′ (x 0
)+···
If wetruncate afterthe second termwe find
thatis,
y(x 0
+h) ≈y(x 0
) +hy ′ (x 0
)
y 1
=y 0
+hy ′ (x 0
)=y 0
+hf(x 0
,y 0
)
sothatEuler’smethodisequivalenttotheTaylorseriestruncatedafterthesecondterm.
Example20.4 UseEuler’s methodwithh = 0.25 toobtain a numericalsolution of
dy
dx = −xy2
subject toy(0) = 2, giving approximate values ofyfor 0 x 1. Work throughout
tothreedecimal places and determine the exact solution for comparison.
Solution We need to calculatey 1
,y 2
,y 3
andy 4
. The correspondingxvalues arex 1
= 0.25,x 2
=
0.5,x 3
= 0.75 andx 4
= 1.0.Euler’s method becomes
We find
y i+1
=y i
+0.25(−x i
y 2 i ) with x 0 =0 y 0 =2
y 1
= 2 −0.25(0)(2 2 ) = 2.000
y 2
= 2 −0.25(0.25)(2 2 ) = 1.750
y 3
= 1.750 −0.25(0.5)(1.750 2 ) = 1.367
y 4
= 1.367 −0.25(0.75)(1.367 2 ) = 1.017
The exact solution can be found by separating the variables:
∫ ∫ dy
y = − xdx
2
so that
− 1 y = −x2 2 +C
Imposingy(0) = 2 givesC = − 1 so that
2
Finally,
− 1 y = −x2 2 − 1 2
y = 2
x 2 +1

618 Chapter 20 Ordinary differential equations II
Table20.1
ComparisonofnumericalsolutionbyEuler’s
method with exact solution.
i x i y i y(x i )
numerical exact
0 0.000 2.000 2.000
1 0.250 2.000 1.882
2 0.500 1.750 1.600
3 0.750 1.367 1.280
4 1.000 1.017 1.000
Table20.2
The solution to Example 20.5
byEuler’smethod.
i x i y i
0 1.00 1.00
1 1.20 1.20
2 1.40 1.40
3 1.60 1.60
4 1.80 1.80
5 2.00 2.00
Table 20.1 summarizes the numerical and exact solutions. In this example only one
correct significant figure is obtained. In practice, for most equations a very small step
size is necessary which means that computation is extremely time consuming. An improvement
to Euler’s method, which usually yields more accurate solutions, is given in
Section 20.7.
Example20.5 Obtain a solution forvalues ofxbetween 1 and 2 of
dy
dx = y x
subject toy = 1whenx = 1usingEuler’smethod.Useastepsizeofh = 0.2,working
throughouttotwodecimalplacesofaccuracy.Compareyouranswerwiththeanalytical
solution. Comment upon the approximate and exact solutions.
Solution Here we have f (x,y) = y x , y 0 = 1, x 0
= 1 and h = 0.20. With a step size of 0.2,
x 1
= 1.2,x 2
= 1.4,...,x 5
= 2.0. We need to calculatey 1
,y 2
,...,y 5
. Euler’s method,
y i+1
=y i
+hf (x i
,y i
), reduces to
( )
yi
y i+1
=y i
+0.2
x i
Therefore,
Similarly,
y 1
=y 0
+0.2
y 2
=y 1
+0.2
(
y0
(
y1
x 0
)
( 1
=1+0.2 = 1.20
1)
) ( ) 1.20
= 1.20 +0.2 = 1.40
x 1
1.20
ContinuinginasimilarfashionweobtaintheresultsshowninTable20.2.Toobtainthe
analytical solution we separate the variables togive
∫ ∫ dy dx
y = x
so that
lny=lnx+lnD=lnDx
Therefore,
y=Dx

20.6 Euler’s method 619
Whenx = 1,y = 1 so thatD = 1, and the analytical solution is thereforey = x. We
see that in this example the numerical solution by Euler’s method produces the exact
solution. This will always be the case when the exact solution is a linear function and
exact arithmeticisemployed.
EXERCISES20.6
1 Using Euler’s method estimatey(3) given
y ′ = x +y
x
y(2) =1
Useh = 0.5 andh = 0.25. Solvethisequation
analytically and compare your numerical solutions
with the true solution.
2 Findy(0.5)ify ′ =x +y,y(0) = 0.Useh = 0.25and
h = 0.1.Find the truesolution forcomparison.
3 Use Euler’s methodto find v(0.01)given
10 −2dv +v =sin100πt v(0) =0
dt
Takeh = 0.005 andh = 0.002. Find the analytical
solution forcomparison.
Solutions
1 Exact:y =xln |x| −0.1931x
x i y i y i y
(h = 0.5) (h = 0.25) (exact)
2.00 1.0000 1.0000 1.0000
2.25 -- 1.3750 1.3901
2.50 1.7500 1.7778 1.8080
2.75 -- 2.2056 2.2509
3.00 2.6000 2.6561 2.7165
3 Exact:
sin100πt − πcos100πt + πe−100t
v =
π 2 +1
t i v i v
(h = 0.005) (exact)
0 0.0000 0.0000
0.005 0.0000 0.2673
0.010 0.5000 0.3954
2 Exact:y=−x−1+e x
x i y i y
(h = 0.25) (exact)
0 0.0000 0.0000
0.25 0.0000 0.0340
0.50 0.0625 0.1487
x i y i y
(h = 0.1) (exact)
t i v i v
(h = 0.002) (exact)
0 0.0000 0.0000
0.002 0.0000 0.0569
0.004 0.1176 0.1919
0.006 0.2843 0.3354
0.008 0.4176 0.4178
0.010 0.4517 0.3954
0 0.0000 0.0000
0.1 0.0000 0.0052
0.2 0.0100 0.0214
0.3 0.0310 0.0499
0.4 0.0641 0.0918
0.5 0.1105 0.1487

620 Chapter 20 Ordinary differential equations II
20.7 IMPROVEDEULERMETHOD
YouwillrecallthatEuler’smethodisobtainedbyfirstfindingtheslopeofthesolutionat
(x 0
,y 0
) and imposing a straight line approximation (often called a tangent line approximation)
there as indicated in Figure 20.10. If we knew the gradient of the solution at
x =x 1
in addition to the gradient atx =x 0
, a better approximation to the gradient over
the whole interval might be the mean of the two. Unfortunately, the gradient atx = x 1
which is
dy
dx∣ =f(x 1
,y 1
)
x=x1
cannot be obtained until we knowy 1
. What we can do, however, is use the value ofy 1
obtained by Euler’s method in estimating the gradient atx = x 1
. This gives rise to the
improved Euler method:
y 1
=y 0
+h×(average of gradients atx 0
andx 1
)
{ } f(x0 ,y
=y 0
+h× 0
)+f(x 1
,y 1
)
2
=y 0
+ h 2 {f(x 0 ,y 0 )+f(x 1 ,y 0 +hf(x 0 ,y 0 ))}
Then, knowingy 1
the whole process isstartedagain tofindy 2
, etc. Generally,
y i+1
=y i
+ h 2 {f(x i ,y i )+f(x i+1 ,y i +hf(x i ,y i ))}
It can be shown that, like Euler’s method, the improved Euler method is equivalent to
truncating the Taylor series expansion, inthis case afterthe third term.
Example20.6 Use the improved Euler method to solve the differential equationy ′ = −xy 2 ,y(0) = 2,
inExample20.4.Asbeforetakeh = 0.25,butworkthroughouttofourdecimalplaces.
Solution Here f(x,y) = −xy 2 ,y 0
= 2,x 0
= 0.Wefind f(x 0
,y 0
) = f(0,2) = −0(2 2 ) = 0.The
improved Euler method
yields
y 1
=y 0
+ h 2 {f(x 0 ,y 0 )+f(x 1 ,y 0 +hf(x 0 ,y 0 ))}
y 1
=2+ 0.25 {0 + f(0.25,2)}
2
Now f (0.25,2) = −0.25(2 2 ) = −1,so that
y 1
= 2 +0.125(−1) = 1.875
WeshallsetoutthecalculationsrequiredinTable20.3.Youshouldworkthroughthenext
fewstagesyourself.Comparingthevaluesobtainedinthiswaywiththeexactsolution,
we see that, over the interval of interest, our solution is usually correct to two decimal
places (Table 20.4).

20.7 Improved Euler method 621
Table20.3
Applying the improved Eulermethod to Example 20.6.
i x i y i f(x i ,y i ) y i +hf(x i ,y i ) f(x i+1 ,y i y i+1
+hf(x i ,y i ))
0 0 2 0 2 −1 1.8750
1 0.25 1.8750 −0.8789 1.6553 −1.3700 1.5939
2 0.5 1.5939 −1.2703 1.2763 −1.2217 1.2824
3 0.75 1.2824 −1.2334 0.9741 −0.9489 1.0096
Table20.4
Comparison of the improved Euler
method with the exact solution.
i x i y i y(x i )
0 0.000 2.0000 2.0000
1 0.2500 1.8750 1.8824
2 0.5000 1.5939 1.6000
3 0.7500 1.2824 1.2800
4 1.0000 1.0096 1.0000
Example20.7 Apply both Euler’s method and the improved Euler method tothe solution of
dy
dx =2x y=1whenx=0
for 0 x 0.5 using h = 0.1. Compare your answers with the analytical solution.
Work throughout tothreedecimal places.
Solution We have dy
dx = 2x,x 0 = 0,y 0
= 1.Therefore, usingEuler’s method we find
y i+1
=y i
+hf(x i
,y i
)
Therefore,
=y i
+0.1(2x i
)
=y i
+0.2x i
y 1
=y 0
+0.2x 0
=1+ (0.2)(0) =1
y 2
=y 1
+0.2x 1
= 1 + (0.2)(0.1) = 1.02
The complete solution appears in Table 20.5. Check the values given in the table for
yourself.Usingthe improved Euler method we have
y i+1
=y i
+ h 2 {f(x i ,y i )+f(x i+1 ,y i +hf(x i ,y i ))}
=y i
+0.05(2x i
+2x i+1
)
=y i
+0.1(x i
+x i+1
)
Table20.5
ThesolutionofExample20.7usingEuler’smethod,
the improved Eulermethod and the exact solution.
x i y i (Euler) y i (improved y(x i )
Euler) (exact)
0 1.000 1.000 1.000
0.1 1.000 1.010 1.010
0.2 1.020 1.040 1.040
0.3 1.060 1.090 1.090
0.4 1.120 1.160 1.160
0.5 1.200 1.250 1.250

622 Chapter 20 Ordinary differential equations II
Therefore,
y 1
=y 0
+0.1(x 0
+x 1
) = 1 +0.1(0 +0.1) = 1.01
y 2
=y 1
+0.1(x 1
+x 2
) = 1.01 +0.1(0.1 +0.2) = 1.04
and so on. The complete solution appears in Table 20.5. Check this for yourself. The
analytical solution is
∫
y = 2xdx=x 2 +c
Applyingtheconditiony(0) = 1givesc = 1,andsoy =x 2 +1.Valuesofthisfunction
arealsoshowninthetableandwenotethemarkedimprovementgivenbytheimproved
Euler method.
EXERCISES20.7
1 Apply the improved Eulermethod to Question 1 in
Exercises20.6.
2 Apply the improved Eulermethod to Question 2 in
Exercises20.6.
3 Apply the improved Eulermethod to Question 3 in
Exercises20.6.
Solutions
1
x i
y i (h = 0.5) y i (h = 0.25) y(exact)
3
t i v i (h = 0.005) v(exact)
2
2.00 1.0000 1.0000 1.0000
2.25 -- 1.3889 1.3901
2.50 1.8000 1.8057 1.8080
2.75 -- 2.2476 2.2509
3.00 2.7017 2.7123 2.7165
x i y i (h = 0.25) y(exact)
0 0.0000 0.0000
0.25 0.0313 0.0340
0.50 0.1416 0.1487
0 0.0000 0.0000
0.005 0.2500 0.2673
0.010 0.2813 0.3954
t i v i (h = 0.002) v(exact)
0 0.0000 0.0000
0.002 0.0588 0.0569
0.004 0.1903 0.1919
0.006 0.3273 0.3354
0.008 0.4032 0.4178
0.010 0.3777 0.3954
x i y i (h = 0.1) y(exact)
0 0.0000 0.0000
0.1 0.0050 0.0052
0.2 0.0210 0.0214
0.3 0.0492 0.0499
0.4 0.0909 0.0918
0.5 0.1474 0.1487

20.8 Runge--Kutta method of order 4 623
20.8 RUNGE--KUTTAMETHODOFORDER4
The‘Runge--Kutta’methods arealargefamilyofmethods usedforsolvingdifferential
equations. The Euler and improved Euler methods are special cases of this family. The
orderofthemethodreferstothehighestpowerofhincludedintheTaylorseriesexpansion.Weshallnowpresentthefourth-orderRunge--Kuttamethod.Thederivationofthis
isbeyond the scope ofthis book.
To solve the equation dy
dx = f (x,y) subject toy = y 0 whenx = x 0
we generate the
sequence ofvalues,y i
, from the formula
where
y i+1
=y i
+ h 6 (k 1 +2k 2 +2k 3 +k 4 ) (20.17)
k 1
= f(x i
,y i
)
(
k 2
=f x i
+ h )
2 ,y i +h 2 k 1
(
k 3
=f x i
+ h )
2 ,y i +h 2 k 2
k 4
=f(x i
+h,y i
+hk 3
)
Example20.8 UsetheRunge--Kuttamethodtosolve dy
dx = −xy2 ,for0 x 1,subjecttoy(0) = 2.
Useh = 0.25 and work tofour decimal places.
Solution We shall show the calculations required to findy 1
. You should follow this through and
then verifythe results inTable 20.6. The exact solution is shown forcomparison.
f(x,y)=−xy 2 h=0.25 x 0
=0 y 0
=2
Takingi = 0 inEquation (20.17), wehave
k 1
= f(x 0
,y 0
) = −0(2) 2 =0
k 2
= f (0.125,2) = −0.125(2) 2 = −0.5
(
k 3
=f 0.125,2 + 0.25 )
2 (−0.5) = f (0.125,1.9375)
= −0.125(1.9375) 2 = −0.4692
k 4
= f (0.25,2 +0.25(−0.4692)) = f (0.25,1.8827)
Therefore,
= −0.25(1.8827) 2 = −0.8861
y 1
=2+ 0.25 (0 +2(−0.5) +2(−0.4692) + (−0.8861)) = 1.8823
6

624 Chapter 20 Ordinary differential equations II
Table20.6
Comparison of the Runge--Kutta method
with exact solutions.
i x i y i y(x i )
0 0.0 2.0 2.0
1 0.25 1.8823 1.8824
2 0.5 1.5999 1.6000
3 0.75 1.2799 1.2800
4 1.0 1.0000 1.0000
Table20.7
The solution to
Example 20.9.
x i
y i
0.0 0.0
0.2 0.0213
0.4 0.0897
0.6 0.2112
Example20.9 Usethe fourth-order Runge--Kutta method toobtain a solution of
Solution We have
dy
dx =x2 +x−y
subject toy = 0 whenx = 0, for 0 x 0.6 withh = 0.2. Work throughout to four
decimal places.
where
y i+1
=y i
+ h 6 (k 1 +2k 2 +2k 3 +k 4 )
k 1
= f(x i
,y i
)
(
k 2
=f x i
+ h )
2 ,y i +h 2 k 1
(
k 3
=f x i
+ h )
2 ,y i +h 2 k 2
k 4
=f(x i
+h,y i
+hk 3
)
Inthisexample, f (x,y) =x 2 +x−yandx 0
= 0,y 0
= 0.Thefirststageinthesolution
isgiven by
k 1
=f(x 0
,y 0
)=0
k 2
= f(0.1,0) = 0.11
k 3
= f(0.1,0 + (0.1)(0.11)) = f(0.1,0.011) = 0.099
k 4
= f (0.2,0 + (0.2)(0.099)) = f (0.2,0.0198) = 0.2202
Therefore,
y 1
=0+ 0.2 (0 +2(0.11) +2(0.099) +0.2202) = 0.0213
6
which, in fact, is correct to four decimal places. Check the next stage for yourself. The
complete solution isshown inTable 20.7.

20.8.1 Higherorderequations
20.8 Runge--Kutta method of order 4 625
Thetechniquesdiscussedforthesolutionofsinglefirst-orderequationsgeneralizereadilytohigherorderequationssuchasthosedescribedinSection20.3andSection20.4.It
isobviousthatacomputersolutionisessentialandthereareawidevarietyofcomputer
packages available tosolve such equations.
EXERCISES20.8
1 Findy(0.4)ify ′ = (x +y) 2 andy(0) = 1usingthe
Runge--Kutta method oforder 4. Take (a) h = 0.2
and(b) h = 0.1.
2 Repeat Question 1 in Exercises20.6 usingthe
fourth-orderRunge--Kutta method.
3 Repeat Question 2 in Exercises20.6 usingthe
fourth-orderRunge--Kutta method.
4 Repeat Question 3 in Exercises20.6 usingthe
fourth-orderRunge--Kutta method.
Solutions
1 (a)
2
x i
y i
0 1.0000
0.2 1.3085
0.4 2.0640
(b)
x i
y i
0 1.0000
0.1 1.1230
0.2 1.3085
0.3 1.5958
0.4 2.0649
x i y i y i y
(h = 0.5) (h = 0.25) (exact)
2.00 1.0000 1.0000 1.0000
2.25 -- 1.3900 1.3901
2.50 1.8078 1.8079 1.8080
2.75 -- 2.2507 2.2509
3.00 2.7163 2.7164 2.7165
4
x i y i y
(h = 0.1) (exact)
0.0 0.0000 0.0000
0.1 0.0052 0.0052
0.2 0.0214 0.0214
0.3 0.0499 0.0499
0.4 0.0918 0.0918
0.5 0.1487 0.1487
t i v i v
(h = 0.005) (exact)
0 0.0000 0.0000
0.005 0.2675 0.2673
0.010 0.3959 0.3954
3
x i y i y
(h = 0.25) (exact)
0.00 0.0000 0.0000
0.25 0.0340 0.0340
0.50 0.1487 0.1487
t i v i v
(h = 0.002) (exact)
0 0.0000 0.0000
0.002 0.0569 0.0569
0.004 0.1919 0.1919
0.006 0.3352 0.3354
0.008 0.4178 0.4178
0.010 0.3954 0.3954

626 Chapter 20 Ordinary differential equations II
REVIEWEXERCISES20
1 UseEuler’smethodwithh = 0.1 to estimatex(0.4)
given dx
dt =x2 −2xt,x(0) =2.
2 Thegeneralization ofEuler’smethodto the two
coupledequations
dy 1
dx = f(x,y 1 ,y 2 )
dy 2
dx =g(x,y 1 ,y 2 )
is given by
y 1(i+1)
= y 1(i)
+hf(x i ,y 1(i)
,y 2(i)
)
y 2(i+1)
= y 2(i)
+hg(x i ,y 1(i)
,y 2(i)
)
Given the coupled equations
dy 1
dx =xy 1 +y dy 2
2
dx =xy 2 +y 1
estimatey 1 (0.3)andy 2 (0.3),ify 1 (0) = 1 and
y 2 (0) = −1.Takeh = 0.1.
3 Expressthe followingequations asasetoffirst-order
equations:
(a)
d 2 y
dt 2 +6dy dt +8y=0
(b) 3 d2 x
dt 2 +5dx dt +4x=0
(c) 4 d3 x
dt 3 +8d2 x
dt 2 +6dx dt +5x=0
4 Expressthe followingcoupledfirst-order equationsas
asingle second-order differentialequation:
(a)
(b)
(c)
dy 1
dx =3y 1 +4y 2 ,2dy 2
dx = 6y 1 +8y 2
dx 1
dt
dy 1
dt
=6x 1 −5x 2 ,2 dx 2
dt
=8y 1 +4y 2 ,2 dy 2
dt
= 4x 1 −3x 2
= 4y 1 −6y 2
Solutions
1 4.7937
4 (a)
2 0.7535, −0.7535
dy
3 (a) 1 dy (b)
=y 2
dt 2 = −8y
dt 1 −6y 2
dx
(b) 1
=x
dt 2 3 dx 2
(c)
= −4x
dt 1 −5x 2
dx
(c) 1 dx
=x 2
dt 2 =x
dt 3
4 dx 3
= −5x
dt 1 −6x 2 −8x 3
d 2 y
dx 2 −7dy dx = 0
d 2 x
dt 2 − 9 dx
2 dt +x=0
d 2 y
dt 2 −5dy dt −32y=0

21 TheLaplacetransform
Contents 21.1 Introduction 627
21.2 DefinitionoftheLaplacetransform 628
21.3 Laplacetransformsofsomecommonfunctions 629
21.4 PropertiesoftheLaplacetransform 631
21.5 Laplacetransformofderivativesandintegrals 635
21.6 InverseLaplacetransforms 638
21.7 UsingpartialfractionstofindtheinverseLaplacetransform 641
21.8 FindingtheinverseLaplacetransformusingcomplexnumbers 643
21.9 Theconvolutiontheorem 647
21.10 Solvinglinearconstantcoefficientdifferentialequationsusing
theLaplacetransform 649
21.11 Transferfunctions 659
21.12 Poles,zerosandthesplane 668
21.13 Laplacetransformsofsomespecialfunctions 675
Reviewexercises21 678
21.1 INTRODUCTION
TheLaplacetransformisusedtosolvelinearconstantcoefficientdifferentialequations.
This is achieved by transforming them to algebraic equations. The algebraic equations
are solved, then the inverse Laplace transform is used to obtain a solution in terms of
the original variables. This technique can be applied to both single and simultaneous
differentialequationsandsoisextremelyusefulgiventhatdifferentialequationmodels
arecommon as we sawinChapters 19 and 20.
TheLaplacetransformisalsousedtoproducetransferfunctionsfortheelementsofan
engineeringsystem.Thesearerepresentedindiagrammaticformasblocks.Thevarious

628 Chapter 21 The Laplace transform
blocksofthesystem,correspondingtothesystemelements,areconnectedtogetherand
the result is a block diagram for the whole system. By breaking down a system in this
way it is much easier to visualize how the various parts of the system interact and so
a transfer function model is complementary to a time domain model and is a valuable
way of viewing an engineering system. Transfer functions are useful in many areas of
engineering, butareparticularly important inthe design of control systems.
21.2 DEFINITIONOFTHELAPLACETRANSFORM
Let f (t) be a function of time t. In many real problems only values of t 0 are of
interest.Hence f (t)is given fort 0,and forallt < 0, f (t) istaken tobe 0.
The Laplace transform of f (t)isF(s), defined by
F(s) =
∫ ∞
0
e −st f (t)dt
Note that to find the Laplace transform of a function f (t), we multiply it by e −st and
integrate between the limits 0 and ∞.
TheLaplacetransformchanges,ortransforms,thefunction f (t)intoadifferentfunctionF(s).Notealsothatwhereas
f (t)isafunctionoft,F(s)isafunctionofs.Todenote
the Laplace transform of f (t) we write L{f (t)}. We use a lower case letter to represent
the time domain function and an upper case letter to represent thesdomain function.
Thevariablesmayberealorcomplex.Astheintegralisimproperrestrictionsmayneed
tobe placed onstoensure thatthe integral does not diverge.
Example21.1 Find the Laplace transformsof
(a) 1
(b) e −at
∫ ∞
[ ] e
Solution (a) L{1} = e −st −st ∞
1dt = = 1
0 −s
0
s =F(s)
This transformexists provided the real partofs,Re(s), ispositive.
(b) L{e −at } =
=
∫ ∞
0
∫ ∞
0
e −st e −at dt
e −(s+a)t dt
[ e
−(s+a)t ∞
=
−(s +a)]
0
( )
1
= 0 −
−(s +a)
= 1
s +a =F(s)
This transformexists provideds +a > 0.
Throughoutthe chapteritisassumed thatshas a valuesuchthatall integrals exist.

21.3 Laplace transforms of some common functions 629
21.3 LAPLACETRANSFORMSOFSOMECOMMON
FUNCTIONS
Determining the Laplace transform of a given function, f (t), is essentially an exercise
inintegration.Inordertosaveeffortalook-uptableisoftenused.Table21.1listssome
common functions and their corresponding Laplace transforms.
Table21.1
TheLaplace transformsofsome common functions.
Function, f (t) Laplacetransform,F(s) Function, f (t) Laplacetransform,F(s)
1
t
1
s
1
s 2
t 2 2
s 3
t n n!
s n+1
e at 1
s −a
e −at 1
s +a
e −at cosbt
sinhbt
coshbt
e −at sinhbt
e −at coshbt
tsinbt
t n e −at n!
(s +a) n+1 tcosbt
sinbt
cosbt
e −at sinbt
b
s 2 +b 2
s
s 2 +b 2
u(t)unitstep
u(t −d)
b
(s+a) 2 +b 2 δ(t) 1
δ(t −d)
s +a
(s+a) 2 +b 2
b
s 2 −b 2
s
s 2 −b 2
b
(s+a) 2 −b 2
s +a
(s+a) 2 −b 2
2bs
(s 2 +b 2 ) 2
s 2 −b 2
(s 2 +b 2 ) 2
1
s
e −sd
s
e −sd
Example21.2 UseTable 21.1 todetermine the Laplace transformofeachofthe following functions:
(a) t 3 (b) t 7
(c) sin4t (d) e
( −2t
t
(e) cos (f) sinh3t
2)
(g) cosh5t
(i) e −t sin2t
(h) tsin4t
(j) e 3t cost

630 Chapter 21 The Laplace transform
Solution From Table 21.1 wefind the results listedinTable 21.2.
Table21.2
f(t) F(s) f(t) F(s) f(t) F(s)
(a) t 3 6
s 4
(b) t 7 7!
s 8
(c) sin4t
4
s 2 +16
(d) e −2t 1
s +2
(e) cos(t/2)
(f) sinh3t
(g) cosh5t
s
s 2 +0.25
3
s 2 −9
s
s 2 −25
(h) tsin4t
(i) e −t sin2t
(j) e 3t cost
8s
(s 2 +16) 2
2
(s+1) 2 +4
s −3
(s−3) 2 +1
EXERCISES21.3
1 Determine the Laplacetransformsofthe following
functions:
(a) sin6t
( )
2t
(c) sin
3
(b) cos4t
( )
4t
(d) cos
3
(e) t 4 (f) t 2 t 3
(g) e −3t
(i)
1
e 4t
(h) e 3t
(j) tcos3t
(k) tsint
cos7t
(m)
e 5t
(o) cosh5t
(q) e −2t cosh7t
(s) e 7t cosh9t
(l) e −t sin3t
(n) sinh6t
(p) e −3t sinh4t
(r) e 4t sinh3t
2 ShowfromthedefinitionoftheLaplacetransformthat
L{u(t −d)} = e−sd ,d>0
s
Solutions
6
1 (a)
s 2 +36
6
(c)
9s 2 +4
24
(e)
s 5
1
(g)
s +3
(i)
1
s +4
(b)
(d)
(f)
(h)
(j)
s
s 2 (k)
+16
9s
9s 2 (m)
+16
120
s 6
(o)
1
s −3
(q)
s 2 −9
(s 2 +9) 2 (s)
2s
(s 2 +1) 2 (l)
s +5
(s+5) 2 +49
s
s 2 −25
s +2
(s+2) 2 −49
s −7
(s−7) 2 −81
(n)
(p)
(r)
3
(s+1) 2 +9
6
s 2 −36
4
(s+3) 2 −16
3
(s−4) 2 −9

21.4 Properties of the Laplace transform 631
21.4 PROPERTIESOFTHELAPLACETRANSFORM
There are some useful properties of the Laplace transform that can be exploited. They
allow us to find the Laplace transforms of more difficult functions. The properties we
shall examine are:
(1) linearity;
(2) shift theorems;
(3) final value theorem.
21.4.1 Linearity
Let f andgbe two functions oft and letkbe a constant which may benegative. Then
L{f +g}= L{f}+L{g}
L{kf}=kL{f}
The first property states that to find the Laplace transform of a sum of functions, we
simply sum the Laplace transforms of the individual functions. The second property
says that if we multiply a function by a constant k, then the corresponding transform
is also multiplied by k. Both of these properties follow directly from the definition of
the Laplace transform and linearity properties of integrals, and mean that the Laplace
transformisalinearoperator.UsingthelinearitypropertiesandTable21.1,wecanfind
the Laplace transformsof more complicated functions.
Example21.3 Find the Laplace transforms ofthe following functions:
(a) 3 +2t (b) 5t 2 −2e t
Solution (a) L{3 +2t} = L{3} + L{2t}
= 3L{1} +2L{t}
= 3 s + 2 s 2
(b) L{5t 2 −2e t }= L{5t 2 } + L{−2e t }
= 5L{t 2 } −2L{e t }
= 10
s − 2
3 s −1
With a littlepractice, someofthe intermediatesteps maybeexcluded.
Example21.4 Findthe Laplace transforms ofthe following:
(a) 5cos3t+2sin5t−6t 3
Solution (a) L{5cos3t +2sin5t −6t 3 } =
(b) −e −t + 1 (sint +cost)
2
5s
s 2 +9 + 10
s 2 +25 − 36
s 4

632 Chapter 21 The Laplace transform
{
(b) L −e −t + sint+cost }
= −1
2 s +1 + 1
2(s 2 +1) + s
2(s 2 +1)
= −1
s +1 + 1 +s
2(s 2 +1)
21.4.2 Firstshifttheorem
If L{f(t)} =F(s)then
L{e −at f (t)} =F(s +a)
aconstant
We obtainF(s +a) by replacing every ‘s’ inF(s) by ‘s +a’. The variableshas been
shifted by anamounta.
Example21.5 (a) UseTable 21.1 tofind the Laplace transform of
f(t)=tsin5t
(b) Usethe first shift theoremtowritedown
L{e −3t tsin5t}
Solution (a) FromTable 21.1 wehave
L{f(t)} = L{tsin5t} =
10s
(s 2 +25) 2 =F(s)
(b) Fromthe first shift theorem witha = 3 we have
L{e −3t f(t)} =F(s +3)
10(s +3)
=
((s +3) 2 +25) 2
=
10(s +3)
(s 2 +6s +34) 2
Example21.6 TheLaplace transformofafunction, f (t), isgiven by
F(s) = 2s+1
s(s +1)
State the Laplace transformof
(a) e −2t f (t)
(b) e 3t f (t)
Solution (a) Usethe first shift theorem witha = 2.
L{e −2t f(t)} =F(s +2)
=
=
2(s+2)+1
(s+2)(s+2+1)
2s+5
(s +2)(s +3)

(b) Usethe first shift theorem witha = −3.
L{e 3t f(t)} =F(s−3)
=
=
2(s−3)+1
(s−3)(s−3+1)
2s−5
(s −3)(s −2)
21.4 Properties of the Laplace transform 633
21.4.3 Secondshifttheorem
If L{f(t)} =F(s)then
L{u(t −d)f(t −d)} =e −sd F(s) d >0
The function, u(t − d)f (t − d), is obtained by moving u(t)f (t) to the right by an
amount d. This is illustrated in Figure 21.1. Note that because f (t) is defined to be
0fort < 0,then f (t −d) = 0fort <d.Itmayappearthatu(t −d)isredundant.However,
it is necessary for inversion of the Laplace transform, which will be covered in
Section21.6.
u(t) f(t)
u(t – d) f(t – d)
(a)
t
Figure21.1
Shifting the functionu(t)f (t)to the rightbyan amountd yields the function
u(t−d)f(t−d).
(b)
d
t
Example21.7 Given L{f (t)} =
2s ,find L{u(t −2)f(t −2)}.
s +9
Solution Usethe second shift theorem withd = 2.
L{u(t −2)f(t −2)} = 2se−2s
s +9
Example21.8 TheLaplace transform ofafunctionis e−3s
. Find the function.
s2 Solution Theexponentialterminthetransformsuggeststhatthesecondshifttheoremisused.Let
e −sd F(s) = e−3s
s 2

634 Chapter 21 The Laplace transform
sothatd =3andF(s) = 1 s2. If we let
L{f(t)} =F(s) = 1 s 2
then f (t) =t and so f (t −3) =t −3.Now, usingthe second shift theorem
L{u(t −3)f(t −3)} = L{u(t −3)(t −3)} = e−3s
s 2
Hence the required function isu(t −3)(t −3) as shown inFigure 21.2.
f(t)
3 t
Figure21.2
Thefunction: f(t) =u(t −3)(t −3).
21.4.4 Finalvaluetheorem
This theorem applies only to real values of s and for functions, f (t), which possess a
limitast → ∞.
Thefinal valuetheorem states:
limsF(s) = lim f(t)
s→0 t→∞
Some care is needed when applying the theorem. The Laplace transform of some functions
exists only for Re(s) > 0 and for these functions taking the limit ass → 0 is not
sensible.
Example21.9 Verifythe final value theoremfor f (t) = e −2t .
Solution ∫ ∞
0
e −st e −2t dt exists provided Re(s) > −2 and so
F(s) = 1
s +2
Re(s) > −2
Since weonly require Re(s) > −2, itispermissibletolets → 0.
s
limsF(s) =lim
s→0 s→0 s +2 = 0
Furthermore
lim
t→∞ e−2t = 0
So lim s→0
sF(s) = lim t→∞
f (t)and the theorem isverified.

21.5 Laplace transform of derivatives and integrals 635
EXERCISES21.4
1 Find the Laplace transformsofthe following
functions:
(a)3t 2 −4
(c)2−t 2 +2t 4
( )
(e) 1 3 sin3t −4cos t
2
(f) 3t 4 e 5t +t
(g)sinh2t +3cosh2t
(h)e −t sin3t +4e −t cos3t
(b) 2sin4t+11−t
(d) 3e 2t +4sint
2 TheLaplacetransform of f (t)is given as
F(s) = 3s2 −1
s 2 +s+1
Find the Laplace transformof
(a)e −t f (t) (b) e 3t f (t) (c) e −t/2 f (t)
3 Given
find
L{f(t)} =
4s
s 2 +1
(a) L{u(t −1)f(t −1)}
(b) L{3u(t −2)f(t −2)}
{
}
(c) L u(t−4) f(t−4)
2
4 Find the final value ofthe followingfunctionsusing
the final value theorem:
(a) f(t) =e −t sint
(b) f(t)=e −t +1
(c) f(t) =e −3t cost +5
Solutions
1 (a)
(c)
(e)
(g)
6
s 3 − 4 s
2
s − 2 s 3 + 48
s 5
1
s 2 +9 − 16s
4s 2 +1
3s+2
s 2 −4
(b)
(d)
(f)
(h)
8
s 2 +16 + 11
s − 1 s 2
3
s −2 + 4
s 2 +1
72
(s −5) 5 + 1 s 2
4s+7
(s+1) 2 +9
2 (a)
(c)
3 (a)
3s 2 +6s+2
s 2 +3s+3
12s 2 +12s −1
4s 2 +8s+7
4se −s
s 2 (b)
+1
(b)
12se −2s
s 2 +1
4 (a)0 (b)1 (c)5
3s 2 −18s +26
s 2 −5s+7
(c)
2se −4s
s 2 +1
21.5 LAPLACETRANSFORMOFDERIVATIVES
ANDINTEGRALS
In later sections we shall use the Laplace transform to solve differential equations. In
order to do this we need to be able to find the Laplace transform of derivatives of functions.Let
f (t)beafunctionoft,and f ′ and f ′′ thefirstandsecondderivativesof f.The
Laplace transform of f (t)isF(s). Then
L{f ′ } =sF(s) − f(0)
L{f ′′ } =s 2 F(s)−sf(0)−f ′ (0)
where f(0) and f ′ (0) are the initial values of f and f ′ . The general case for the
Laplace transform ofannthderivative is
L{f (n) } =s n F(s) −s n−1 f(0) −s n−2 f ′ (0)−···−f (n−1) (0)

636 Chapter 21 The Laplace transform
Another usefulresultis
{ ∫ }
t
L f(t)dt = 1 s F(s)
0
Example21.10 The Laplace transform ofx(t) isX(s). Givenx(0) = 2 andx ′ (0) = −1, write expressionsforthe
Laplace transformsof
(a) 2x ′′ −3x ′ +x (b) −x ′′ +2x ′ +x
Solution L{x ′ } =sX(s) −x(0) =sX(s) −2
L{x ′′ } =s 2 X(s)−sx(0)−x ′ (0) =s 2 X(s)−2s+1
(a) L{2x ′′ −3x ′ +x} = 2(s 2 X(s) −2s +1) −3(sX(s) −2) +X(s)
= (2s 2 −3s+1)X(s)−4s+8
(b) L{−x ′′ +2x ′ +x} = −(s 2 X(s) −2s +1) +2(sX(s) −2) +X(s)
= (−s 2 +2s+1)X(s)+2s−5
Engineeringapplication21.1
Voltageacrossacapacitor
Thevoltage, v(t), acrossacapacitorofcapacitanceC isgiven by
v(t) = 1 C
∫ t
0
i(t)dt
Taking Laplace transforms yields
V(s)= 1 Cs I(s)
whereV(s) = L{v(t)}andI(s) = L{i(t)}.
Engineeringapplication21.2
Frequencyresponseofasystem
The Laplace variablesis sometimes referred to as the generalized or complex frequency
variable. It consists of a real and an imaginary part, wheres = σ + jω. If
onlythesinusoidalsteady-stateresponsetoasinewaveinputforasystemisrequired
then we can obtain this by putting σ = 0 into the expression for s and sos = jω.
Thusitispossibletomakethissubstitutioninanytransferfunctiongivenintermsof
the Laplace variablesto obtain the sinusoidal steady-state frequency response. We
willnotprovethisassertionhereforreasonsofbrevity.Insteadwewilldemonstrate
its usefulness.

21.5 Laplace transform of derivatives and integrals 637
Consider the Laplace transformof theequation forthe voltage across acapacitor
of capacitanceC. Recall fromEngineering application 21.1 that
V(s)= 1 Cs I(s)
Rearrangingthisexpressiongivesaformulafortheimpedanceofthecapacitor,Z(s),
inthesdomain
Z(s) = V(s)
I(s) = 1 Cs
Thisformoftheequationcanbeusedtodeterminethebehaviourofthecapacitorin
a variety of situations where the voltages and currents can take a number of forms,
for example step inputs, sine waves, triangular waves, etc. However, if we are only
interested in the sinusoidal steady-state response of the system when all transients
have decayed, then itispossible tosubstitutes = jω into the expression forZ(s)
Z(jω) = V(jω)
I(jω) = 1
jωC
This form of the impedance is the one regularly used in a.c. circuit theory. It gives
theimpedanceofthecapacitorintermsofthecapacitanceandtheangularfrequency
ω,where ω = 2πf.(Thisresulthasalreadybeendiscussedinthecontextofphasors
on page 342).
Thesamesubstitutioncan bemade toobtain thefrequency responseofasystem,
givenitstransferfunction.Thesystemfrequencyresponseissomethingthatcanusuallybeobtainedeasilybyexperiment.Allthatisrequiredisasignalsourceofaknown
amplitude and a means of measuring the output amplitude and phase relative to the
original signal. Measurements of this sort can reveal some of the properties of the
systeminquestionreflectingthecloserelationshipbetweenthetransferfunctionand
the frequency response.
EXERCISES21.5
1 TheLaplacetransform ofy(t)isY(s),y(0) = 3,
y ′ (0) = 1. Find the Laplacetransformsofthe
followingexpressions:
(a) y ′ (b) y ′′
(c) 3y ′′ −y ′
(d) y ′′ +2y ′ +3y
(e) 3y ′′ −y ′ +2y
(f) −4y ′′ +5y ′ −3y
(g) 3 d2 y
dt 2 +6dy dt +8y
(h) 4 d2 y
dt 2 −8dy dt +6y
2 Given the Laplacetransform of f (t)isF(s),
f(0)=2,f ′ (0)=3andf ′′ (0) = −1,findthe
Laplace transformsof
(a) 3f ′ −2f
(c) f ′′′
(b) 3f ′′ − f ′ + f
(d) 2f ′′′ − f ′′ +4f ′ −2f
3 (a) IfF(s) = L{f(t)} = ∫ ∞
0 e −st f (t)dt,showusing
integration byparts that
(i) L{f ′ (t)} =sF(s)−f(0)
(ii) L{f ′′ (t)} =s 2 F(s)−sf(0)−f ′ (0)
(b) IfF(s) = L{f (t)}prove that
L{e −at f(t)} =F(s +a)
Deduce L{te −t },given L{t} = 1 s 2.

638 Chapter 21 The Laplace transform
Solutions
1 (a) sY−3
(b) s 2 Y−3s−1
(c) 3s 2 Y −sY −9s
(d) (s 2 +2s+3)Y−3s−7
(e) (3s 2 −s +2)Y −9s
(f) (−4s 2 +5s −3)Y +12s −11
(g) (3s 2 +6s +8)Y −9s −21
(h) (4s 2 −8s +6)Y −12s +20
2 (a) (3s−2)F−6
(b) (3s 2 −s+1)F−6s−7
(c) s 3 F−2s 2 −3s+1
(d) (2s 3 −s 2 +4s−2)F−4s 2 −4s−3
1
3 (b)
(s +1) 2
21.6 INVERSELAPLACETRANSFORMS
As mentioned in Section 21.5, the Laplace transform can be used to solve differential
equations. However, before such an application can be put into practice, we must study
theinverseLaplacetransform.Sofarinthischapterwehavebeengivenfunctionsoftand
foundtheirLaplacetransforms.Wenowconsidertheproblemoffindingafunction f (t),
havingbeengiventheLaplacetransform,F(s).ClearlyTable21.1andthepropertiesof
Laplace transformswill helpus todothis. If L{f (t)} =F(s)wewrite
f(t) = L −1 {F(s)}
and call L −1 the inverse Laplace transform. Like L, L −1 can be shown to be a linear
operator.
Example21.11 Findthe inverse Laplace transformsofthe following:
(a)
2
s 3
(b)
16
s 3
(c)
s
s 2 +1
(d)
1
s 2 +1
(e)
s +1
s 2 +1
Solution (a) We need to find a function of t which has a Laplace transform of
Table 21.1 wesee L −1 { 2
s 3 }
=t 2 .
{ } { } 16 2
(b) L −1 = 8L −1 = 8t 2
s 3 s 3
{ } s
(c) L −1 =cost
s 2 +1
{ } 1
(d) L −1 =sint
s 2 +1
{ } s +1
(e) L −1 s 2 +1
{ } { } s 1
= L −1 + L −1 =cost+sint
s 2 +1 s 2 +1
2
s3. Using
In parts (a), (c) and (d) we obtained the inverse Laplace transform by referring directly
to the table. In (b) and (e) we used the linearity properties of the inverse transform, and
then referredtoTable 21.1.

21.6 Inverse Laplace transforms 639
Example21.12 Find the inverse Laplace transformsof the following functions:
10 (s+1)
(a) (b) (c)
(s +2) 4 (s+1) 2 +4
{ }
Solution (a) L −1 10
= 10 { }
6
(s +2) 4 6 L−1 (s +2) 4
15
(s−1) 2 −9
= 5t3 e −2t
3
{ } { }
(b) L −1 s +1
= L −1 s +1
= e −t cos2t
(s+1) 2 +4 (s+1) 2 +2 2
{ } { }
(c) L −1 15
= 5L −1 3
= 5e t sinh3t
(s−1) 2 −9 (s−1) 2 −3 2
The function is written to match exactly the standard forms given in Table 21.1, with
possibly a constant factor being present. Often the denominator needs to be written in
standard form asillustratedinthe next example.
Example21.13 Findthe inverse Laplace transformsofthe following functions:
(a)
s +3
s 2 +6s+13
(b)
2s+3
s 2 +6s+13
Solution (a) By completing the square we can write
(c)
s 2 +6s+13=(s+3) 2 +4=(s+3) 2 +2 2
s −1
2s 2 +8s+11
(b)
Hence we may write
{ } { }
L −1 s +3
= L −1 s +3
= e −3t cos2t
s 2 +6s+13 (s+3) 2 +2 2
2s+3
s 2 +6s+13 = 2s+3
(s+3) 2 +2 = 2s+6
2 (s+3) 2 +2 − 3
2 (s+3) 2 +2 2
(
(
= 2
s +3
(s+3) 2 +2 2 )
− 3 2
)
2
(s+3) 2 +2 2
The expressions in brackets are standard forms so their inverse Laplace transforms
can befound fromTable 21.1.
{ }
L −1 2s+3
= 2e −3t cos2t − 3e−3t sin2t
s 2 +6s+13
2
(c) We write the expression usingstandard forms:
s −1
2s 2 +8s+11 = 1 s −1
2s 2 +4s+5.5 = 1 s −1
2 (s +2) 2 +1.5
= 1 (
)
s +2
2 (s +2) 2 +1.5 − 3
(s +2) 2 +1.5
= 1 (
s +2
2 (s +2) 2 +1.5 − √ 3
√ ) 1.5
1.5 (s +2) 2 +1.5

640 Chapter 21 The Laplace transform
Havingwrittentheexpressionintermsofstandardforms,theinverseLaplacetransform
can now be found:
{ }
L −1 s −1
= 1 (
e −2t cos √ 1.5t − √ 3 e −2t sin √ )
1.5t
2s 2 +8s+11 2 1.5
In every case the given function ofsis written as a linear combination of standard
formscontained inTable 21.1.
EXERCISES21.6
1 Byusingthe standard formsin Table 21.1 findthe
inverseLaplacetransformsofthefollowingfunctions:
1 2
(a) (b)
s s 3
8 1
(c) (d)
s s +2
6 5
(e) (f)
s +4 s −3
7 2
(g) (h)
s +8 s 2 +4
s 6s
(i)
s 2 (j)
+9 s 2 +4
20 2
(k)
s 2 (l)
+16 (s +1) 2
8 9
(m)
(s +2) 2 (n)
(s +3) 3
e −3s 4e −6s
(o) (p)
s s
(q) e −9s (r) 4e −8s
4 s
(s)
s 2 (t)
−16 s 2 −9
12 6s
(u)
s 2 (v)
−9 s 2 −8
2
(w)
(s+1) 2 −4
2
(y)
(s+3) 2 −4
s +3
(x)
(s+3) 2 −4
6s
(z)
s 2 −5
2 Findthe inverse Laplacetransformsofthe following
functions:
3 4
(a) (b)
2s s − 1 s 3
30
1
(c)
s 2 (d)
3(s +2)
3s−7 s −6
(e)
s 2 (f)
+9 s −4
s +4 5
(g)
(s+4) 2 (h)
+1 (s+4) 2 +1
(i)
(k)
(m)
6s+17
(s+4) 2 +1
(j)
0.5
(s +0.5) 2 (l)
6s+9
s 2 +2s+10
(n)
s
s 2 +2s+7
s +5
s 2 +8s+20
7s+3
s 2 +4s+8
Solutions
1 (a) 1 (b) t 2
(o) u(t −3) (p) 4u(t −6)
(c) 8 (d) e −2t
(q) δ(t −9) (r) 4δ(t −8)
(e) 6e −4t (f) 5e 3t
(s) sinh4t (t) cosh3t
(g) 7e −8t (h) sin2t
(u) 4sinh3t (v) 6cosh √ 8t
(i) cos3t (j) 6cos2t
(w) e −t sinh2t (x) e −3t cosh2t
(k) 5sin4t (l) 2te −t
(y) e −3t sinh2t (z) 6cosh √ 5t
(m) 8te −2t 9
(n)
2 t2 e −3t

21.7 Using partial fractions to find the inverse Laplace transform 641
2 (a)
3
2
(c) 30t
(b) 4− t2 2
(d)
(e) 3cos3t − 7 3 sin3t
(f) δ(t)−2e 4t
(g) e −4t cost
(h) 5e −4t sint
e −2t
3
(i) 6e −4t cost −7e −4t sint
(j) e −t cos √ 6t − 1 √
6
e −t sin √ 6t
(k)
te −t/2
2
(l) e −4t cos2t + 1 2 e−4t sin2t
(m) 6e −t cos3t +e −t sin3t
(n) 7e −2t cos2t − 11
2 e−2t sin2t
21.7 USINGPARTIALFRACTIONSTOFINDTHEINVERSE
LAPLACETRANSFORM
TheinverseLaplacetransformofafractionisoftenbestfoundbyexpressingitasitspartialfractions,andfindingtheinversetransformofthese.(SeeSection1.7foratreatment
ofpartialfractions.)
Example21.14 Findthe inverse Laplace transform of
4s−1 6s+8
(a) (b)
s 2 −s s 2 +3s+2
Solution (a) We express 4s−1
s 2 −s
as its partialfractions:
4s−1
s 2 −s = 4s−1
s(s −1) = A s + B
s −1
Using the technique of Section 1.6 wefind thatA = 1,B = 3 and hence
4s−1
s 2 −s = 1 s + 3
s −1
The inverse Laplace transform of each partialfraction isfound:
( ) ( )
1 3
L −1 = 1, L −1 =3e t
s s −1
Hence
L −1 ( 4s−1
s 2 −s
)
=1+3e t
(b) The expression isexpressed as its partialfractions. This was done inSection 1.7.
6s+8
s 2 +3s+2 = 2
s +1 + 4
s +2
The inverse Laplace transform of each partialfraction isnoted:
( )
( )
2
4
L −1 = 2e −t L −1 = 4e −2t
s +1 s +2

642 Chapter 21 The Laplace transform
and so
( ) 6s+8
L −1 s 2 +3s+2
=2e −t +4e −2t
Example21.15 Findthe inverse Laplace transformof 3s2 +6s+2
s 3 +3s 2 +2s .
Solution Following fromthe work inSection 1.7.1,we have
3s 2 +6s+2
s 3 +3s 2 +2s = 1 s + 1
s +1 + 1
s +2
The inverse Laplace transform of the partialfractions iseasily found:
{ } 3s
L −1 2 +6s+2
=1+e −t +e −2t
s 3 +3s 2 +2s
Example21.16 Findthe inverse Laplace transformof
Solution From Example 1.33 wehave
3s 2 +11s +14
s 3 +2s 2 −11s−52 .
3s 2 +11s +14
s 3 +2s 2 −11s−52 = 2
s −4 + s +3
s 2 +6s+13
We find the inverse Laplace transformsof the partialfractions:
{ } 2
L −1 =2e 4t
s −4
{ }
L −1 s +3
= e −3t cos2t
s 2 +6s+13
So
{ } 3s
L −1 2 +11s +14
= 2e 4t +e −3t cos2t
s 3 +2s 2 −11s−52
EXERCISES21.7
1 Expressthe followingaspartialfractions andhence
findtheir inverse Laplacetransforms:
(a)
(b)
(c)
5s+2
(s +1)(s +2)
3s+4
(s +2)(s +3)
4s+1
(s +3)(s +4)
(d)
(e)
(f)
(g)
6s−5
(s +5)(s +3)
4s+1
s(s +2)(s +3)
7s+3
s(s +3)(s +4)
6s+7
s(s +2)(s +4)

21.8 Finding the inverse Laplace transform using complex numbers 643
(h)
(i)
(j)
8s−5
(s +1)(s +2)(s +3)
3s+5
(s +1)(s 2 +3s +2)
2s−8
(s +2)(s 2 +7s +6)
2 Expressthe following fractions aspartialfractions
andhence findtheir inverse Laplacetransforms:
(a)
(b)
3s+3
(s −1)(s +2)
5s
(s +1)(2s −3)
(c) 2s+5
s +2
s 2 +4s+4
(d)
s 3 +2s 2 +5s
1 −s
(e)
(s +1)(s 2 +2s +2)
s +4
(f)
s 2 +4s+4
(g)
(h)
2(s 3 −3s 2 +s−1)
(s 2 +4s +1)(s 2 +1)
3s 2 −s+8
(s 2 −2s +3)(s +2)
Solutions
1 (a) 8e −2t −3e −t
(b) 5e −3t −2e −2t
(c) 15e −4t −11e −3t
35
(d)
2 e−5t − 23
2 e−3t
1
(e)
6 + 7 2 e−2t − 11
3 e−3t
1
(f)
4 +6e−3t − 25
4 e−4t
7
(g)
8 + 5 4 e−2t − 17
8 e−4t
(h) 21e −2t − 29
2 e−3t − 13
2 e−t
(i) e −t +2te −t −e −2t
(j) 3e −2t −e −6t −2e −t
2 (a) 2e t +e −2t
(b) e −t + 3e3t/2
2
(c) 2δ(t) +e −2t
(d) 1 5 (4 +e−t cos2t + 11 2 e−t sin2t)
(e) e −t (2−2cost −sint)
(f) e −2t (1 +2t)
(
(g) e −2t 3cosh √ 3t − √ 8 sinh √ )
3t −cost
3
(h) e t (cos √ 2t + √ 2sin √ 2t) +2e −2t
21.8 FINDINGTHEINVERSELAPLACETRANSFORMUSING
COMPLEXNUMBERS
InSections21.6and21.7wefoundtheinverseLaplacetransformusingstandardforms
and partial fractions. We now look at a method of finding inverse Laplace transforms
usingcomplexnumbers.Essentiallythemethodisoneusingpartialfractions,butwhere
all the factors in the denominator are linear -- that is, there are no quadratic factors. We
illustrate the methodusingExample21.13.
Example21.17 Findthe inverse Laplace transformsofthe following functions:
(a)
s +3
s 2 +6s+13
(b)
2s+3
s 2 +6s+13
(c)
s −1
2s 2 +8s+11

644 Chapter 21 The Laplace transform
Solution (a) We first factorize the denominator. To do this we solves 2 +6s +13 = 0 using the
formula
s = −6 ± √ 36 −4(13)
2
= −6 ± √ −16
2
= −6±4j
2
=−3±2j
It then follows that the denominator can be factorized as (s − a)(s − b) where
a = −3 +2jandb=−3 −2j. Then, usingthe partialfractions method
s +3
s 2 +6s+13 = s +3
(s −a)(s −b) = A
s −a + B
s −b
The unknown constantsAandBcan now be found:
s+3=A(s−b)+B(s−a)
Puts=a=−3+2j
2j =A(−3 +2j −b) =A(4j)
A = 1 2
Equate the coefficients ofs
1=A+B
B = 1 2
So,
s +3
s 2 +6s+13 = 1 ( 1
2 s −a + 1 )
s −b
The inverse Laplace transform can now be found:
{ }
L −1 s +3
= 1 { 1
s 2 +6s+13 2 L−1 s −a + 1 }
s −b
= 1 2 (eat +e bt ) = 1 2 (e(−3+2j)t +e (−3−2j)t )
= 1 2 e−3t (e 2jt +e −2jt )
= 1 2 e−3t (cos2t +jsin2t +cos2t −jsin2t)
= e −3t cos2t

21.8 Finding the inverse Laplace transform using complex numbers 645
(b)
2s+3
s 2 +6s+13 = 2s+3
(s −a)(s −b) = A
s −a + B
s −b
wherea = −3 +2jandb= −3 −2j. Hence,
2s+3=A(s−b)+B(s−a)
Puts=a=−3+2j
−3 +4j =A(4j)
A=1+0.75j
Equate the coefficients ofs
Hence,
2=A+B
B=1−0.75j
2s+3
s 2 +6s+13 = 1 +0.75j + 1 −0.75j
s −a s −b
(c)
Taking the inverse Laplace transform yields
{ }
L −1 2s+3
= (1 +0.75j)e at + (1 −0.75j)e bt
s 2 +6s+13
= (1 +0.75j)e (−3+2j)t + (1 −0.75j)e (−3−2j)t
s −1
2s 2 +8s+11 = 1 (
2
= (1 +0.75j)e −3t (cos2t +jsin2t)
+ (1 −0.75j)e −3t (cos2t −jsin2t)
= e −3t (2cos2t −1.5sin2t)
)
= 1 (
2
s −1
s 2 +4s+5.5
s −1
(s −a)(s −b)
)
wherea = −2 + √ 1.5j,b = −2 − √ 1.5j.Applyingthemethodofpartialfractions
produces
Hence,
s −1
(s −a)(s −b) = A
s −a + B
s −b
s−1=A(s−b)+B(s−a)
By lettings =a, thens =binturn, gives
A=0.5+ √ 1.5j B = 0.5 − √ 1.5j
Hence we may write
s −1
(s −a)(s −b) = 0.5 + √ 1.5j
+ 0.5 − √ 1.5j
s −a s −b

646 Chapter 21 The Laplace transform
Taking the inverse Laplace transformyields
Hence
{ }
L −1 s −1
= (0.5 + √ 1.5j)e at + (0.5 − √ 1.5j)e bt
(s −a)(s −b)
= (0.5 + √ 1.5j)e (−2+√ 1.5j)t
+ (0.5 − √ 1.5j)e (−2−√ 1.5j)t
= e −2t {(0.5 + √ 1.5j)e √ 1.5jt
+ (0.5 − √ 1.5j)e −√ 1.5jt }
= e −2t {(0.5 + √ 1.5j)(cos √ 1.5t +jsin √ 1.5t)
+ (0.5 − √ 1.5j)(cos √ 1.5t −jsin √ 1.5t)}
= e −2t (cos √ 1.5t −2 √ 1.5sin √ 1.5t)
{ }
L −1 s −1
= e−2t
2s 2 +8s+11 2 (cos √ 1.5t −2 √ 1.5sin √ 1.5t)
EXERCISES21.8
AsseenfromExample21.17,whencomplexnumbersareallowed,allthefactorsinthe
denominatorarelinear.Theunknown constantsareevaluatedusingparticularvaluesof
s orequating coefficients.
1 Expressthe followingexpressions aspartialfractions,
usingcomplex numbers ifnecessary. Hence findtheir
inverse Laplacetransforms.
(a)
3s−2
s 2 +6s+13
(b)
2s+1
s 2 −2s+2
(c)
(e)
s 2
(s 2 /2)−s+5
2s+3
−s 2 +2s−5
(d)
s 2 +s+1
s 2 −2s+3
Solutions
3
1 (a) 2 +11j/4
+
s −a
3
2 −11j/4
s −b
wherea=−3+2j,b=−3−2j,
e −3t[ 3cos2t − 11
2 sin2t ]
(b) 1 −3j/2 + 1 +3j/2
s −a s −b
wherea=1+j,b=1−j,
e t (2cost+3sint)
(c) 2 +8j/3 + 2 −8j/3 +2
s −a s −b
wherea=1+3j,b=1−3j,
(
e t 4cos3t − 16 )
3 sin3t +2δ(t)
3
(d)1+ 2 −j/(2 √ 2)
+
s −a
(e)
3
2 +j/(2√ 2)
s −b
wherea=1+ √ 2j,b=1− √ 2j,
)
δ(t)+e t (3cos √ 2t + 1 √
2
sin √ 2t
−1 +5j/4
s −a
+
−1 −5j/4
s −b
wherea=1+2j,b=1−2j,
e t (−2cos2t − 5 2 sin2t)

21.9 The convolution theorem 647
21.9 THECONVOLUTIONTHEOREM
Let f (t) andg(t) be two piecewise continuous functions. The convolution of f (t)
andg(t), denoted (f ∗g)(t), isdefined by
(f ∗g)(t)=
∫ t
0
f(t −v)g(v)dv
Example21.18 Find the convolution of2t andt 3 .
Solution f (t) = 2t,g(t) =t 3 , f(t −v) =2(t −v),g(v) = v 3 .Then
2t∗t 3 =
∫ t
0
2(t −v)v 3 dv =2
∫ t
0
tv 3 −v 4 dv
[ tv
4
= 2
4 − v5
5
= t5
10
] t
0
[ ] t
5
= 2
4 − t5 5
Itcan beshown that
f∗g=g∗f
butthe proof isomitted. Instead, thispropertyisillustratedby anexample.
Example21.19 Show that f ∗g=g ∗ f where f (t) = 2t andg(t) =t 3 .
Solution f ∗g= 2t ∗t 3 =t 5 /10 by Example 21.18. Fromthe definitionofconvolution
(g∗f)(t)=
∫ t
0
g(t −v)f(v)dv
We haveg(t) =t 3 ,sog(t −v) = (t −v) 3 ,and f(v) =2v.Therefore
g∗f=t 3 ∗2t=
=
= 2
∫ t
0
∫ t
0
∫ t
(t −v) 3 2vdv
(t 3 −3t 2 v +3tv 2 −v 3 )2vdv
0
t 3 v −3t 2 v 2 +3tv 3 −v 4 dv
[ t 3 v 2
] t
= 2
2 −t2 v 3 + 3tv4
4 − v5
= t5
5
0
10

648 Chapter 21 The Laplace transform
For any functions f (t)andg(t)
f∗g=g∗f
21.9.1 Theconvolutiontheorem
Let f (t) and g(t) be piecewise continuous functions, with L{f (t)} = F(s) and
L{g(t)} = G(s). The convolution theorem allows us to find the inverse Laplace transform
of a product oftransforms,F(s)G(s):
L −1 {F(s)G(s)} = (f ∗g)(t)
Example21.20 Use the convolution theorem to find the inverse Laplace transforms of the following
functions:
1 3
(a) (b)
(s +2)(s +3) s(s 2 +4)
Solution (a) LetF(s) = 1
s +2 ,G(s) = 1
s +3 .
Then f(t) = L −1 {F(s)} = e −2t ,g(t) = L −1 {G(s)} = e −3t .
{ }
L −1 1
= L −1 {F(s)G(s)} = (f ∗g)(t)
(s +2)(s +3)
=
=
∫ t
0
∫ t
0
e −2(t−v) e −3v dv =
e −2t e −v dv
∫ t
0
e −2t e 2v e −3v dv
= e −2t [−e −v ] t 0 = e−2t (−e −t +1) = e −2t −e −3t
(b) LetF(s) = 3 s ,G(s) = 1
s 2 +4 .Thenf(t) =3,g(t) = 1 sin2t. So,
2
{ }
L −1 3
= L −1 {F(s)G(s)}
s(s 2 +4)
=(f ∗g)(t)
∫ t
∫ t
= 3 sin2v dv = 3 sin2vdv
0 2 2 0
= 3 [ −cos2v
= 3 2 2 4 (1−cos2t)
] t
0

21.9 Solving linear constant coefficient differential equations 649
EXERCISES21.9
1 Find
(a) e −2t ∗e −t
(b) t 2 ∗e −3t
2 Find f ∗gwhen
(a) f =1,g=t
(b) f =t 2 ,g=t
(c) f =e t ,g=t
(d) f =sint,g=t
Ineach caseverify that
L{f}×L{g}=L{f ∗g}.
3 IfF(s) = 1
s −1 ,G(s) = 1 s andH(s) = 1
2s+3
use the convolution theoremto find the inverse
Laplace transformsof
(a) F(s)G(s)
(b) F(s)H(s)
(c) G(s)H(s)
4 Use the convolution theoremto determine the inverse
Laplace transformsof
(a)
1
s 2 (s+1)
(b)
1
(s +3)(s −2)
(c)
1
(s 2 +1) 2
Solutions
1 (a) e −t −e −2t
t 2 (b)
3 − 2t
9 + 2 27 − 2e−3t
27
t 2 t 4
2 (a) (b)
2 12
(c) −t−1+e t (d)t−sint
3 (a)e t −1 (b) et −e −3t/2
4 (a) t−1+e −t (b) e2t −e −3t
(c) sint−tcost
2
5
5
(c) 1 −e−3t/2
3
21.10 SOLVINGLINEARCONSTANTCOEFFICIENT
DIFFERENTIALEQUATIONSUSINGTHELAPLACE
TRANSFORM
SofarwehaveseenhowtofindtheLaplacetransformofafunctionoftimeandhowto
find the inverse Laplace transform. We now apply this to finding the particular solution
ofdifferentialequations.Theinitialconditionsareautomaticallysatisfiedwhensolving
an equation using the Laplace transform. They are contained in the transform of the
derivative terms.
TheLaplacetransformoftheequationisfound.Thistransformsthedifferentialequationintoanalgebraicequation.Thetransformofthedependentvariableisfoundandthen
the inverse transform iscalculated toyield the required particular solution.
Example21.21 Solve the differential equation
dx
dt +x=0
x(0)=3
usingLaplace transforms.

650 Chapter 21 The Laplace transform
Solution The Laplace transformof each termisfound:
L(x) =X(s),
( ) dx
L =sX(s)−x(0)=sX(s)−3
dt
Note that we write X(s) to draw attention to the fact that X is a function of s; X(s)
does not mean X multiplied by s. Taking the Laplace transform of the equation
yields
sX(s)−3+X(s)=0
The equation isrearranged forX(s):
sX(s)+X(s) =3
(s+1)X(s) =3
X(s) = 3
s +1
Taking the inverse Laplace transform of both sides of the equation gives
x(t) =3e −t
Example21.22 Solve
dx
dt +x=9e2t
x(0)=3
using the Laplace transform.
Solution The Laplace transformof both sides of the equation isfound:
sX(s) −x(0) +X(s) = 9
s −2
sX(s)−3+X(s) = 9
s −2
(s +1)X(s) = 9
s −2 +3
3(s +1)
(s +1)X(s) =
s −2
X(s) = 3
s −2
Taking the inverse Laplace transform yields
x(t) =3e 2t

21.10 Solving linear constant coefficient differential equations 651
Engineeringapplication21.3
RLcircuitwithrampinput
Recall from Engineering application 19.3 the differential equation for an RL circuit
with a unit ramp input isgiven by
iR+L di
dt
=t t0 i(0)=0
Usethe Laplace transform tosolve thisequation.
Solution
Let
L(i(t)) =I(s)
Then
( ) di
L =sI(s)−i(0) =sI(s)−0 =sI(s)
dt
From Table 21.1 wesee that
L(t) = 1 s 2
We note thatRandLare constants. We can now take the Laplace transform of the
given equation. This gives
I(s)R +LsI(s) = 1 s 2
This equation issolved forI(s):
I(s)(R +Ls) = 1 s 2
1
I(s) =
s 2 (R +Ls)
InordertotaketheinverseLaplacetransformandhencefindi(t)weexpressI(s)as
the sum of itspartial fractions. The expression
1
s 2 (R +Ls)
has a repeated linear factor,s 2 inthe denominator, giving risetopartialfractions
A
s + B s 2
The linear factor,R+Ls, gives risetoapartialfraction of the form
Hence
C
R+Ls
I(s) =
1
s 2 (R +Ls) = A s + B s + C
2 R+Ls
➔

652 Chapter 21 The Laplace transform
The unknown constants, A, B andC, need to be expressed in terms of the known
constantsRandL. Multiplying the above equation bys 2 (R +Ls) yields
1 =As(R +Ls) +B(R +Ls) +Cs 2 (1)
Lettings = 0 inEquation (1)gives
1=BR
andsoB= 1 R .
Wenotethatwhens = − R L thenR+Lsis0.Hencelettings = −R L inEquation(1)
gives
(
1 =C − R ) 2
fromwhichC = L2
L R 2
Comparing the coefficient ofson both sides of Equation (1) gives
0=AR+BL
A = − BL
R = − L R 2
Substituting the expressions forA,BandC into the expression forI(s)gives
I(s) = − L
R 2 s + 1
Rs 2 + L 2
R 2 (R +Ls)
InreadinessfortakinginverseLaplacetransformswewritethefinaltermasfollows:
L 2
R 2 (R +Ls) = L 2
R 2 L (R/L +s) = L
R 2 (R/L +s)
Hence
I(s) = − L
R 2 s + 1
Rs + L
2 R 2 (R/L +s)
Taking the inverse Laplace transformyields
i(t) = − L R 2 + t R + L R 2 e−(R/L)t t 0
This may berearranged as
i(t) = t R + L R 2 (e−(R/L)t −1)
t 0
Example21.23 Solve
y ′′ −y=−t 2 y(0) =2, y ′ (0) =0
using the Laplace transform.
Solution Let L(y) =Y(s). Thenusingthe resultstated inSection21.5 wehave
L(y ′′ ) =s 2 Y(s) −sy(0) −y ′ (0) =s 2 Y(s) −2s

21.10 Solving linear constant coefficient differential equations 653
We also notethat
L(−t 2 ) = − 2 s 3
The Laplace transform of the differential equation is taken. This yields
s 2 Y(s)−2s−Y(s)=− 2 s 3
The equation isrearranged forY(s):
(s 2 −1)Y(s) = − 2 s 3 +2s
= 2(s4 −1)
s 3
= 2(s2 −1)(s 2 +1)
s 3
Bydividing the equation by (s 2 −1) weobtain
Y(s)= 2(s2 −1)(s 2 +1)
s 3 (s 2 −1)
= 2(s2 +1)
s 3
= 2s2
s 3 + 2 s 3 = 2 s + 2 s 3
Taking the inverse Laplace transform gives
y(t)=2+t 2
Example21.24 Solve
y ′′ +y ′ −2y=−2 y(0)=2,y ′ (0)=1
Solution Let L(y) =Y(s). Then
L(y ′ ) =sY(s) −y(0) =sY(s) −2
L(y ′′ ) =s 2 Y(s) −sy(0) −y ′ (0) =s 2 Y(s) −2s −1
We also notethat
L(−2) = − 2 s
We can now take the Laplace transform of the differential equation toget
and so
s 2 Y(s)−2s−1+sY(s)−2−2Y(s)=− 2 s
(s 2 +s−2)Y(s)=− 2 s +2s+3
(s −1)(s +2)Y(s) = 2s2 +3s−2 (2s −1)(s +2)
=
s s
Dividing the equation by (s −1)(s +2) gives
Y(s)=
(2s −1)(s +2)
s(s −1)(s +2) = 2s−1
s(s −1)

654 Chapter 21 The Laplace transform
The expression,
Y(s)= 1 s + 1
s −1
2s−1
, iswritten as itspartialfractions:
s(s −1)
Taking the inverse Laplace transform yields
y(t)=1+e t
Example21.25 Solve
x ′′ +2x ′ +2x =e −t x(0) =x ′ (0) =0
usingLaplace transforms.
Solution Taking the Laplace transform ofboth sides:
s 2 X(s) −sx(0) −x ′ (0) +2(sX(s) −x(0)) +2X(s) = 1
s +1
Therefore,
(s 2 +2s +2)X(s) = 1 sincex(0) =x ′ (0) =0
s +1
1
X(s)=
(s +1)(s 2 +2s +2) = 1
(s +1)(s −a)(s −b)
wherea = −1 +j,b = −1 −j.Usingpartialfractions gives
X(s)= 1
s +1 − 1 ( 1
2 s −a + 1 )
s −b
Taking the inverse Laplace transform
x(t) =e −t − 1 2 (eat +e bt )
=e −t − 1 2 (e(−1+j)t +e (−1−j)t )
=e −t − e−t
2 (ejt +e −jt )
=e −t − e−t
2 (cost+jsint+cost−jsint)
= e −t −e −t cost
Example21.26 Solve
x ′′ −5x ′ +6x=6t−4 x(0)=1,x ′ (0)=2
Solution TheLaplace transformofboth sides ofthe equation isfound.Let L{x} =X(s).
s 2 X(s) −sx(0) −x ′ (0) −5(sX(s) −x(0)) +6X(s) = 6 s 2 − 4 s
(s 2 −5s +6)X(s) = 6 s 2 − 4 s +s−3=s3 −3s 2 −4s+6
s 2

X(s)= s3 −3s 2 −4s+6
s 2 (s 2 −5s+6)
21.10 Solving linear constant coefficient differential equations 655
= A s + B s + C
2 s −2 + D
s −3
= s3 −3s 2 −4s+6
s 2 (s −2)(s −3)
The constantsA,B,C andDare evaluated inthe usual way:
X(s)= 1 6s + 1 s 2 + 3
2(s −2) − 2
3(s −3)
Taking the inverse Laplace transform yields
x(t) = 1 6 +t + 3 2 e2t − 2 3 e3t
Engineeringapplication21.4
Dischargeofacapacitorthroughaloadresistor
InEngineeringapplication2.8,weexaminedthevariationinvoltageacrossacapacitor,C,
when it was switched in series with a resistor,R, at timet = 0. We stated a
relationshipforthetime-varyingvoltage, v,acrossthecapacitor.Provethisrelationship.
Refer tothe example fordetails ofthe circuit.
Solution
Firstwemustderiveadifferentialequationforthecircuit.UsingKirchhoff’svoltage
law and denoting the voltageacrossthe resistorby v R
weobtain
v+v R
=0
Using Ohm’slaw and denoting the currentinthe circuitbyiweobtain
v+iR=0
For the capacitor,
i =C dv
dt
Combining these equations gives
v+RC dv = 0
dt
We now take the Laplace transformof this equation. Using L{v} =V(s)weobtain
V(s) +RC(sV(s) − v(0)) =0
V(s)(1 +RCs) =RCv(0)
V(s)= RCv(0)
1+RCs = v(0)
1
RC +s
Taking the inverse Laplace transform ofthe equation yields
v = v(0)e −t/(RC)
t 0
This isequivalent tothe relationship statedinEngineering application 2.8.

656 Chapter 21 The Laplace transform
Engineeringapplication21.5
Electronicthermometermeasuringoventemperature
Many engineering systems can be modelled by a first-order differential equation.
Thetimeconstantisameasureoftherapiditywithwhichthesesystemsrespondto
a change in input. Suppose an electronic thermometer is used to measure the temperature
of an oven. The sensing element does not respond instantly to changes in
the oven temperature because it takes time for the element to heat up or cool down.
Provided the electronic circuitry does not introduce further time delays then the differential
equation thatmodels the thermometer isgiven by
τ dv m
+v
dt m
=v o
where v m
=measuredtemperature, v o
=oventemperature, τ =timeconstantofthe
sensor. For convenience the temperature is measured relative to the ambient room
temperature, which forms a ‘baseline’ for temperature measurement.
Suppose the sensing element of an electronic thermometer has a time constant
of 2 seconds. If the temperature of the oven increases linearly at the rate of 3 ◦ Cs −1
startingfromanambientroomtemperatureof20 ◦ Catt = 0,calculatetheresponseof
the thermometer to the changing oven temperature. State the maximum temperature
error.
Solution
Taking Laplace transforms ofthe equation gives
τ(sV m
(s) − v m
(0)) +V m
(s) =V o
(s)
v m
(0) = 0 as the oven temperature and sensor temperature are identical att = 0.
Therefore,
τsV m
(s) +V m
(s) =V o
(s)
V m
(s) = V o (s)
1+τs
(21.1)
Forthisexample,theinputtothethermometerisatemperaturerampwithaslopeof
3 ◦ Cs −1 . Therefore, v o
= 3t fort 0:
V o
(s) = L{v o
(t)} = 3 s 2 (21.2)
Combining Equations (21.1) and (21.2) yields
V m
(s) =
3
s 2 (1+τs) = 3
s 2 (1 +2s)
Then using partialfractions, wehave
V m
(s) = 3 s 2 − 6 s + 12
1+2s
Taking the inverse Laplace transformyields
since τ = 2
v m
=3t−6+6e −0.5t
t0

21.10 Solving linear constant coefficient differential equations 657
This response consists of threeparts:
(1) a decaying transient, 6e −0.5t , which disappears with time;
(2) a ramp,3t, with the same slope as the oven temperature;
(3) a fixed negative temperature error, −6.
Therefore,afterthetransienthasdecayedthemeasuredtemperaturefollowstheoven
temperaturewithafixednegativeerror.Itisinstructivetoobtainthetemperatureerror
by an alternative method. Given thatthe temperature error is v e
, then
and
v e
=v m
−v o
V e
(s) =V m
(s) −V o
(s) (21.3)
Combining Equations (21.1) and (21.3) yields
V e
(s) = V o (s)
1+τs −V o (s)= −τsV o (s)
1+τs
Combining Equations (21.2) and (21.4) yields
V e
(s) = −τs 3
1+τss = −3τ
2 s(1+τs)
The final value theorem can beused tofind the steady-state error:
[ ] −3τs
lim v
t→∞
e
(t) =lim sV e
(s) = lim =−3τ=−6
s→0 s→0 s(1+τs)
(21.4)
that is, the steady-state temperature error is −6 ◦ C. It is important to note that the
final value theorem can only be used if it is known that the time function tends to a
limitast → ∞. Inmany cases engineers know thisisthe case fromexperience.
The Laplace transform technique can also be used to solve simultaneous differential
equations.
Example21.27 Solve the simultaneousdifferential equations
x ′ +x+ y′
=1 x(0)=y(0)=0
2
x ′
2 +y′ +y=0
Solution Take the Laplace transforms of both equations:
sY(s) −y(0)
sX(s) −x(0) +X(s) + = 1 2 s
sX(s) −x(0)
+sY(s)−y(0)+Y(s)=0
2
These arerearranged togive
(s +1)X(s) + sY(s) = 1 2 s
sX(s)
+(s+1)Y(s)=0
2

658 Chapter 21 The Laplace transform
These simultaneous algebraic equations need to be solved for X(s) and Y(s). By
Cramer’s rule(see Section 8.7.2)
∣ s +11/s
s/2 0 ∣ −1/2
Y(s)=
∣ s +1 s/2
=
(s+1) 2 −s 2 /4
s/2 s+1∣
= − 1 { }
1
2 3s 2 /4+2s+1
= − 1 { }
4
2 3s 2 +8s+4
= − 1 { }
4
2 (3s +2)(s +2)
Using partialfractions wefind
Y(s)=− 1 { 3
2 3s+2 − 1 }
s +2
= − 1 { 1
− 1 }
2 s + 2 s +2
3
and hence
Similarly,
and so
y(t) = 1 2 (e−2t −e −2t/3 )
X(s)=
4(s +1)
s(3s +2)(s +2) = 1 s − 1
2 ( 1
) −
s + 2 2(s +2)
3
x(t)=1− 1 2 (e−2t/3 +e −2t )
EXERCISES21.10
1 UseLaplace transformsto solve
(a) x ′ +x=3,
x(0) =1
(b) 3 dx
dt +4x=2,
x(0) =2
(c) 2 dy
dt +4y=1,
y(0) =4
(d) 4 dy
dt +8y=7,
y(0) =6
(e) y ′′ +y ′ +y=1,
y(0) =1,y ′ (0) =3
2 UseLaplacetransformsto solve
(a) x ′′ +x =2t,
x(0) =0,x ′ (0) =5
(b) 2x ′′ +x ′ −x =27cos2t +6sin2t,
x(0) = −1,x ′ (0) = −2

21.11 Transfer functions 659
(c) x ′′ +x ′ −2x=1−2t,
x(0) =6,x ′ (0) = −11
(d) x ′′ −4x = 4(cos2t −1),
x(0) =1,x ′ (0) =0
(e) x ′ −2x−y ′ +2y=−2t 2 +7,
x ′
2 +x+3y′ +y=t 2 +6
x(0) =3,y(0) =6
(f) x ′ +x+y ′ +y=6e t ,
x ′ +2x−y ′ −y=2e −t
x(0) =2,y(0) =1
3 Using Laplace transformsfindthe particular
solution of
d 2 y
dt 2 −5dy dt −6y=14e−t
satisfyingy = 3 and dy
dt =8whent=0.
Solutions
1 (a) x(t)=3−2e −t
(b) 3 2 e−4t/3 + 1 2
1
(c)
4 + 15 4 e−2t
(d) y(t) = 7 8 + 41
8 e−2t
(e) y(t) = 1 +3.464e −0.5t sin0.866t
2 (a) 3sint+2t
(b) −3cos2t +2e −t
(c) t +6e −2t
(d)
e −2t
4 + e2t
4 − 1 2 cos2t+1
(e) x=t 2 +3,y=6−t
(f) x(t) = 6et
5 +2e−t − 6e−3t/2 ,
5
y(t) = 6e−3t/2
5
3 −2te −t + 8e−t
7
−2e −t + 9et
5
+ 13e6t
7
21.11 TRANSFERFUNCTIONS
It is possible to obtain a mathematical model of an engineering system that consists of
one or more differential equations. This approach was introduced in Section 20.4. We
have already seen that the solution of differential equations can be found by using the
Laplace transform. This leads naturally to the concept of a transfer function which will
bedeveloped inthissection.Consider the differential equation
dx(t)
+x(t) = f(t) x(0) =x
dt
0
(21.5)
andassumethatitmodelsasimpleengineeringsystem.Then f (t)representstheinputto
thesystemandx(t)representstheoutput,orresponseofthesystemtotheinput f (t).For
reasonsthatwillbeexplainedbelowitisnecessarytoassumethattheinitialconditions
associatedwiththedifferentialequationarezero.InEquation(21.5)thismeanswetake
x 0
tobe zero.Taking the Laplace transform of Equation (21.5) yields
sX(s)−x 0
+X(s) =F(s)
(1 +s)X(s) =F(s) assumingx 0
=0
so that
X(s)
F(s) = 1
1 +s
The function, X(s)/F(s), is called a transfer function. It is the ratio of the Laplace
transform of the output to the Laplace transform of the input. It is often denoted by

660 Chapter 21 The Laplace transform
G(s). Therefore, forEquation (21.5),
G(s) = 1
1 +s
The concept of a transferfunction isvery usefulinengineering. Itprovides a simple
algebraic relationship between the input and the output. In other words, it allows the
analysisofdynamicsystemsbasedonthedifferentialequationtoproceedinarelatively
straightforward manner. Earlier we noted that it was necessary to assume zero initial
conditions in order to form the transfer function. Without such an assumption, the relationship
between the input and the output would have been more complicated. What
is more, the relationship would vary depending on how much energy is stored in the
system at t = 0. By assuming zero initial conditions, the transfer function depends
purely on the system characteristics. Such an approach, whilst very convenient, does
have its limitations. The transfer function approach is useful for analysing the effect of
aninputtothesystem.However,ifonerequirestheeffectofthecombinationofasystem
inputand initial conditions,thenitisnecessarytocarryoutafullsolution ofthedifferential
equation as we did in Section 21.10. In practice there are many cases where the
simplified treatment provided by the transfer function is perfectly acceptable. An alternativeapproachisprovidedbystate-spacemodels,whichweexaminedinSection20.4.
Theseprovideanaturalwayofhandlinginitialconditions.Inordertosolveastate-space
model wherethereareinitialconditions,allthatisnecessaryistouseanon-zeroinitial
valueofthestatevector.RecallthatinSection20.4weonlysetupthestate-spacemodels
and did not solve them. However, the method for the standard solution of a state-space
model can befound inmany advanced textbooks on control and systems engineering.
Example21.28 Findthetransferfunctionsofthefollowingequationsassumingthat f (t)representsthe
inputandx(t) represents the output:
(a)
(b)
dx(t)
dt
d 2 x(t)
dt 2
−4x(t) =3f(t), x(0) =0
+3 dx(t)
dt
−x(t) = f(t),
dx(0)
dt
=0, x(0)=0
Solution (a) Taking Laplace transforms ofthe differential equation gives
sX(s) −x(0) −4X(s) =3F(s)
(s −4)X(s) =3F(s) asx(0) =0
X(s)
F(s) =G(s) = 3
s −4
(b) Taking Laplace transforms ofthe differential equation gives
s 2 X(s) −sx(0) − dx(0) +3(sX(s) −x(0)) −X(s) =F(s)
dt
(s 2 +3s −1)X(s) =F(s) as dx(0) =0 and x(0)=0
X(s)
F(s) =G(s) = 1
s 2 +3s−1
dt
Whencreatingamathematicalmodelofanengineeringsystemitisoftenconvenientto
thinkofthevariableswithinthesystemassignalsandelementsofthesystemasmeans

21.11 Transfer functions 661
R(s)
G(s)
Y(s)
Figure21.3
Therelationship
Y(s) =G(s)R(s)holds
forasingle block.
by which these signals are modified. The word signal is used in a very general sense
and is not restricted to, say, voltage. On this basis each of the elements of the system
can be modelled by a transfer function. A transfer function defines the relationship between
an input signal and an output signal. The relationship is defined in terms of the
Laplace transforms of the signals. The advantage of this is that the rules governing the
manipulation of transfer functions are then of a purely algebraic nature. Consider
Figure 21.3. If
then
R(s) = L{r(t)} = Laplace transform of the input signal
Y(s) = L{y(t)} = Laplace transformof the output signal
G(s) = transferfunction
Y(s) =G(s)R(s)
Transfer functions are represented schematically by rectangular blocks, while signals
are represented as arrows. Engineers often speak of the time domain and the s domain
in order to distinguish between the two mathematical representations of an engineering
system. However, it is important to emphasize the equivalence between the
two domains.
Oftenwhenconstructingamathematicalmodelofasystemusingtransferfunctions,
it is convenient first to obtain transfer functions of the elements of the system and then
combinethem.Beforetheoveralltransferfunctioniscalculatedablockdiagramisdrawn
which shows the relationship between the various transfer functions. Block diagrams
consistof threebasic components. These are shown inFigure 21.4.
R(s)
G(s)
Y(s)
R(s)
+
–
R(s) – X(s)
Y(s)
Y(s)
X(s)
(a) (b) (c)
Figure21.4
Thethree components ofblock diagrams. (a)Abasic block; the block contains a
transferfunction which relatesthe input and outputsignals. (b)Asummingpoint.
(c) Atake-off point.
We have already examined the basic block which is governed by the relationship
Y(s) =G(s)R(s).Asummingpointaddstogethertheincomingsignalstothesumming
point and produces an outgoing signal. The polarity of the incoming signals is denoted
bymeansofapositiveornegative sign.Therecanbeseveral incomingsignalsbutonly
one outgoing signal. A take-off point is a point where a signal is tapped. This process
of tapping the signal has no effect on the signal value; that is, the tap does not load the
original signal. There are several rules governing the manipulation of block diagrams.
Only two will beconsidered here.
21.11.1 Rule1.Combiningtwotransferfunctionsinseries
Consider Figure 21.5. The following relationships hold:
X(s) =G 1
(s)R(s)
Y(s) =G 2
(s)X(s)
Y(s)

662 Chapter 21 The Laplace transform
R(s)
G 1 (s)
X(s)
G 2 (s)
Y(s)
R(s)
G 1 (s) G 2 (s)
Y(s)
Figure21.5
Two blocksin series.
Figure21.6
Figure21.5issimplifiedtoasingleblock.
EliminatingX(s) from these equations yields
Y(s) =G 1
(s)G 2
(s)R(s)
Y(s)
R(s) =G 1 (s)G 2 (s)
Finally the overall transfer function, G(s), is given by G 1
(s)G 2
(s) as shown in
Figure 21.6.
Fortwo transferfunctions inseries the overall transferfunction isgiven by
G(s) =G 1
(s)G 2
(s)
21.11.2 Rule2.Eliminatinganegativefeedbackloop
Consider Figure 21.7 which shows a negative feedback loop. It is so called because the
outputsignalis‘fedback’andsubtractedfromtheinputsignal.Suchloopsarecommon
inavarietyofengineeringsystems.ThequantitiesX 1
(s)andX 2
(s)representintermediate
signals in the system. We wish to obtain an overall transfer function for this system
relatingY(s)andR(s). For the two transferfunctions the following hold:
Y(s) =G(s)X 2
(s) (21.6)
X 1
(s) =H(s)Y(s) (21.7)
For the summingpoint
X 2
(s) =R(s) −X 1
(s) (21.8)
Combining Equations(21.7)and (21.8) gives
X 2
(s) =R(s) −H(s)Y(s) (21.9)
Combining Equations(21.6)and (21.9) gives
Y(s) =G(s)(R(s) −H(s)Y(s)) =G(s)R(s) −G(s)H(s)Y(s)
Y(s)(1 +G(s)H(s)) =G(s)R(s)
Y(s)=
G(s)R(s)
1 +G(s)H(s)
Theoverall transferfunctionforanegative feedback loop isgiven by
Y(s)
R(s) = G(s)
1 +G(s)H(s)

21.11 Transfer functions 663
R(s) +
–
X 2 (s)
G(s)
Y(s)
X 1 (s)
H(s)
R(s)
G(s)
1 + G(s)H(s)
Y(s)
Figure21.7
Block diagram foranegative feedback loop.
Figure21.8
Simplifiedblock diagram foranegative
feedback loop.
The simplified block diagram for a negative feedback loop isshown inFigure 21.8.
A complicated engineering system may be represented by many differential equations.
The output from one part of the system may form the input to another part. Consider
the following example.
Example21.29 Asystemismodelled bythe differential equations
x ′ +2x=f(t) (21.10)
2y ′ −y =x(t) (21.11)
In Equation (21.10) the input is f (t) and the output is x(t). In Equation (21.11), x(t)
is the input andy(t) is the final output of the system. Find the overall system transfer
functionassumingzero initial conditions.
Solution The output from Equation (21.10) isx(t); this forms the input to Equation (21.11). The
blockdiagramsforEquations(21.10)and(21.11)arecombinedintoasingleblockdiagramasshown
inFigure 21.9.
UsingRule 1,the overall systemtransferfunction can thenbefound:
G(s) = Y(s)
F(s) = 1
(s +2)(2s −1)
This transferfunction relatesY(s)andF(s)(see Figure 21.10).
F(s)
1
X(s)
1
Y(s)
s + 2
2s – 1
F(s)
1
(s + 2)(2s – 1)
Y(s)
Figure21.9
Combinedblock diagram for
Equations(21.10)and (21.11).
Figure21.10
Theoverall systemtransfer function.
Example21.30 Asystemisrepresentedby the differential equations
2x ′ −x=f(t)
y ′ +3y =x(t)
z ′ +z=y(t)
The initial input is f (t) and the final output is z(t). Find the overall system transfer
function, assumingzeroinitial conditions.

664 Chapter 21 The Laplace transform
Solution The transfer function for each equation is found and combined into one block diagram
(see Figure 21.11). The three blocks are simplified to a single block as shown in Figure
21.12. The overall systemtransfer function is
G(s) = Z(s)
F(s) = 1
(2s −1)(s +3)(s +1) .
F(s)
1
2s – 1
X(s)
1
s + 3
Y(s)
1
s + 1
Z(s)
F(s)
1
(2s – 1)(s + 3)(s + 1)
Z(s)
Figure21.11
Block diagram forthe system given in
Example 21.30.
Figure21.12
Simplified block diagram for
Example 21.30.
Engineeringapplication21.6
Transportlaginasystemtosupplyfueltoafurnace
Transportlagisatermusedtodescribethetimedelaythatmaybepresentincertain
engineering systems. A typical example would be a conveyor belt feeding a furnace
with coal supplied by a hopper (see Figure 21.13). The amount of fuel supplied to
the furnace can be varied by varying the opening at the base of the hopper but there
is a time delay before this changed quantity of fuel reaches the furnace. The time
delaydependsonthespeedandlengthoftheconveyor.Mathematically,thefunction
describing the variation in the quantity of fuel entering the furnace is a time-shifted
version of the function describing the variation in the quantity of fuel placed on the
conveyor (see Figure 21.14).
Let
u(t)q(t) = quantity offuelplacedon the conveyor,
whereu(t) isthe unit stepfunction,
Coal
Conveyor velocity, y
l
Conveyor
Furnace
Figure21.13
Coalisfed intothe furnacevia the conveyor
belt.

21.11 Transfer functions 665
Fuel variation
(mass of fuel
per unit length
of conveyor,
kgm –1 )
d
q(t) q(t – d)
Q(s)
e –sd
Q d (s)
Time (s)
Figure21.14
Thetimedelay,d,isintroducedbytheconveyorbelt.
Figure21.15
Block diagram forthe
conveyor belt.
u(t −d)q(t −d) = quantity of fuelentering the furnace,
d = time delay introduced by the conveyor,
l = length of conveyor (m),
v = speed of conveyor (ms −1 ).
The input to the conveyor is u(t)q(t). The output from the conveyor is
u(t −d)q(t −d). If the conveyor is moving at a constant speed then v = l d and
sod= l . The transfer function that models the conveyor belt can be obtained by
v
using the second shift theorem:
Q d
(s) = L{u(t −d)q(t −d)} = e −sd L{q(t)} = e −sd Q(s)
The transfer function forthe conveyor isshown inFigure 21.15.
Transport lags can cause difficulty when trying to control a system because of
the delay between taking a control action and its effect being felt. In this example,
increasingthequantityoffuelontheconveyordoesnotleadtoanimmediateincrease
in fuel entering the furnace. The difficulty of controlling the furnace temperature
increases as the transport lag introduced by the conveyor increases.
Engineeringapplication21.7
Positioncontrolsystem
There are many examples of position control systems in engineering, for example
controlofthepositionofaplotterpen,andcontrolofthepositionofaradiotelescope.
Thecommon termforthesesystems is servo-systems.
Consider the block diagram of Figure 21.16 which represents a simple servosystem.
The actual position of the motor is denoted by a
(s) in the s domain and
θ a
(t)inthetimedomain.Thedesiredpositionisdenotedby d
(s)and θ d
(t),respectively.
The system is a closed loop with negative feedback. The difference between
the desired and the actual position generates an error signal which is fed to a controllerwithgainK.Theoutputsignalfromthecontrollerisfedtoaservo-motorand
its associated drive circuitry. The aim of the control system is to maintain the actual
position of the motor at a value corresponding to the desired position. In practice, if ➔

666 Chapter 21 The Laplace transform
Q d (s)
+
–
K
0.5
s(s + 1)
Controller Servo-motor +
drive amplifier
Q a (s)
Figure21.16
Position control system.
anewdesiredpositionisrequestedthenthesystemwilltakesometimetoattainthis
new position. The engineer can choose a value of the controller gain to obtain the
besttypeofresponsefromthecontrolsystem.Wewillexaminetheeffectofvarying
K on the response of the servo-system.
We can use Rules 1 and 2 to obtain an overall transfer function for the system.
The forward transfer function is
G(s) = 0.5K by Rule 1
s(s +1)
The overall transferfunction a (s) isobtained by Rule 2 withH(s) = 1.So,
d
(s)
0.5K
a
(s)
d
(s) = s(s +1) 0.5K
1 + 0.5K(1) =
s(s +1) +0.5K = 0.5K
s 2 +s+0.5K
s(s +1)
Let us now examine the effect of varying K. We will consider three values,
K = 0.375,K = 0.5,K = 5, and examine the response of the system to a unit step
input ineach case.
ForK = 0.375
a
(s)
d
(s) = 0.1875
s 2 +s+0.1875
With d
(s) = 1 s , then
a
(s) =
usingpartialfractions. So,
0.1875
(s 2 +s+0.1875)s = 1 s + 0.5
s +0.75 − 1.5
s +0.25
θ a
(t) = 1 +0.5e −0.75t −1.5e −0.25t t 0
This is shown in Figure 21.17. Engineers usually refer to this as an overdamped
response. The response does notovershoot the final value.
ForK = 0.5
a
(s)
d
(s) = 0.25
s 2 +s+0.25
a
(s) =
0.25
s(s 2 +s +0.25) = 1 s − 1
s+0.5 − 0.5
(s +0.5) 2
θ a
(t) = 1 −e −0.5t −0.5te −0.5t t 0

21.11 Transfer functions 667
u a (t)
K = 5
K = 0.5
K = 0.375
t
Figure21.17
Time response forvariousvalues ofK.
This is shown in Figure 21.17 and is termed acritically damped response. It corresponds
tothe fastestrisetime of the systemwithoutovershooting.
ForK=5
a
(s)
d
(s) = 2.5
s 2 +s+2.5
2.5
a
(s) =
s(s 2 +s +2.5)
Rearranging toenable standard formstobe inverted gives
a
(s) = 1 s −
s+0.5
(s +0.5) 2 +1.5 − 0.5
2 (s +0.5 2 ) +1.5 2
θ a
(t) = 1 −e −0.5t cos1.5t − 1 3 e−0.5t sin1.5t
The trigonometric terms can be expressed as a single sinusoid using the techniques
given inSection 3.7.Thus,
θ a
(t) = 1 −1.054e −0.5t sin(1.5t +1.249) for t 0
ThisisshowninFigure21.17andistermedanunderdampedresponse.Thesystem
overshoots its final value.
Inapracticalsystemitiscommontodesignforsomeovershoot,provideditisnot
excessive,asthisenablesthedesiredvaluetobereachedmorequickly.Itisinteresting
tocomparethesystemresponseforthethreecaseswiththenatureoftheirrespective
transfer function poles. For the overdamped case the poles are real and unequal, for
the critically damped case the poles are real and equal, and for the underdamped
casethepolesarecomplex.Engineersrelyheavilyonpolepositionswhendesigning
a system to have a particular response. By varying the value of K it is possible to
obtain a range of system responses and corresponding pole positions.
EXERCISES21.11
1 Find the transfer functionforeachofthe following
equationsassuming zeroinitialconditions:
(a) x ′′ +x=f(t)
(b) 2 d2 x
dt 2 + dx
dt −x=f(t)
(d)
(c) 2 d2 y
dt 2 +3dy +6y =p(t)
dt
3 d3 y y
dt 3 +6d2 dt 2 +8dy +4y =g(t)
dt
(e) 6 d2 y
dt 2 +8dy dt +4y=3df dt +4f,
f isthe systeminput
(f) 6x ′′′ +2x ′′ −x ′ +4x = 4f ′′ +2f ′ +6f,fisthe
systeminput

668 Chapter 21 The Laplace transform
Solutions
1 (a)
(c)
1
s 2 +1
1
2s 2 +3s+6
1
(b)
2s 2 +s−1
1
(d)
3s 3 +6s 2 +8s+4
(e)
(f)
3s+4
6s 2 +8s+4
4s 2 +2s+6
6s 3 +2s 2 −s+4
21.12 POLES,ZEROSANDTHEsPLANE
Mosttransferfunctionsforengineeringsystemscanbewrittenasrationalfunctions:that
is, asratiosoftwo polynomials ins, with a constantfactor,K:
G(s) =K P(s)
Q(s)
P(s) is of order m, and Q(s) is of order n; for a physically realizable system m < n.
HenceG(s) may bewritten as
G(s) = K(s−z 1 )(s−z 2 )...(s−z m )
(s−p 1
)(s−p 2
)...(s−p n
)
The values ofsthat makeG(s) zero are known as the system zeros and correspond to
the roots ofP(s) = 0, that iss =z 1
,z 2
,...,z m
. The values ofsthat makeG(s) infinite
are known as the system poles and correspond to the roots of Q(s) = 0, that is s =
p 1
,p 2
,...,p n
. As we have seen, poles may be real or complex. Complex poles always
occur incomplex conjugate pairs whenever the polynomialQ(s)has real coefficients.
Engineersfinditusefultoplotthesepolesandzerosonansplanediagram.Acomplex
planeplotisusedwith,conventionally,arealaxislabelof σ andanimaginaryaxislabel
of jω. Poles are marked as crosses and zeros are marked as small circles. Figure 21.18
shows ansplane plot for the transfer function
G(s) =
3(s −3)(s +2)
(s+1)(s+1+3j)(s+1−3j)
Thebenefitofthisapproachisthatitallowsthecharacterofalinearsystemtobedetermined
by examining thesplane plot. In particular, the transient response of the system
can easilybe visualized by the number and positions of the systempoles and zeros.
jv
3 3
s plane
3
–2–1 3 s
3–3
Figure21.18
Polesand zerosplotted forthe transferfunction:
G(s) =
3(s −3)(s +2)
(s+1)(s+1+3j)(s+1−3j) .

21.12 Poles, zeros and thesplane 669
Example21.31 Find the poles of
s −2
(s 2 +2s +5)(s +1) .
Solution The denominator isfactorized into linear factors:
(s 2 +2s+5)(s+1)= (s−p 1
)(s−p 2
)(s−p 3
)
where p 1
= −1 +2j, p 2
= −1 −2j, p 3
= −1.The poles are −1 +2j,−1 −2j,−1.
IfX(s)= 1
s−p 1
where the pole p 1
isgiven bya +bj,then
x(t) =e p 1 t =e at e btj = e at (cosbt +jsinbt)
Hence the real part of the pole,a, gives rise to an exponential term and the imaginary
part,b,givesrisetoanoscillatoryterm.Ifa < 0theresponse,x(t),willdecreasetozero
ast→∞.
ConsidertheLaplacetransforminExample21.25.Therearethreepoles: −1, −1+j
and −1 − j. The real pole is negative, and the real parts of the complex poles are also
negative.Thisensurestheresponse,x(t),decreaseswithtime.Theimaginarypartofthe
complex poles gives rise to the oscillatory term, cost. The characteristics of poles and
the corresponding responses arenow discussed.
GivenasystemwithtransferfunctionG(s),inputsignalR(s)andoutputsignalC(s),
then
C(s)
R(s) =G(s)
thatis,
and so
C(s) =G(s)R(s)
C(s) = K(s−z 1 )(s−z 2 )...(s−z m )R(s)
(s−p 1
)(s−p 2
)...(s−p n
)
The poles and zeros of the system are independent of the input that is applied. All that
R(s)contributes tothe expression forC(s)isextra poles and zeros.
Consider the case whereR(s) = 1 , corresponding toaunit stepinput:
s
C(s) = K(s−z 1 )(s−z 2 )...(s−z m )
(s−p 1
)(s−p 2
)...(s−p n
)s
= A 1
+ A 2
+···+
A n
+ B 1
s−p 1
s−p 2
s−p n
s
whereA 1
,A 2
,...,A n
andB 1
areconstants. Taking inverse Laplace transformsyields
c(t) =A 1
e p 1 t +A 2
e p 2 t +···+A n
e p n t +B 1
If the system is stable then p 1
,p 2
,...,p n
will have negative real parts and their contribution
toc(t) will vanish ast → ∞.
Theresponsecausedbythesystempolesisoftencalledatransientresponsebecause
itdecreaseswithtimeforastablesystem.Thecomponent ofthetransientresponsedue
to a particular pole is often termed its transient. Notice that the form of the transient
responseisindependentofthesysteminputandisdeterminedbythenatureofthesystem
poles. It is now possible to derive a series of rules relating the transient response of the
systemtothe positions of the systempoles inthesplane.

670 Chapter 21 The Laplace transform
jv
s plane
Transient
response
3
3
s
t
Figure21.19
A polewith anegative real part leads to a decaying transientwhile a
pole with a zeroreal partleads to atransientthat doesnotdecay with
time.
jv
s plane
Transient
response
3
3
s
t
21.12.1 Rule1
Figure21.20
The furtherapole isfrom the imaginary axis, the quicker the decay of
its transient.
Thepolesmaybeeitherrealorcomplexbutforaparticularpoleitisnecessaryforthereal
parttobenegativeifthetransientcausedbythatpoleistodecaywithtime.Otherwisethe
transient response will increase with time and the system will be unstable, a condition
engineers usually design to avoid. In simple terms this means that the poles of a linear
system must all lie in the left half of the s plane for stability. Poles on the imaginary
axisleadtomarginal stabilityasthetransientsintroduced by suchpoles donotgrowor
decay. This isillustratedinFigure 21.19.
21.12.2 Rule2
21.12.3 Rule3
The further a pole is to the left of the imaginary axis the faster its transient decays (see
Figure 21.20). This is because its transient contains a larger negative exponential term.
Forexample,e −5t decaysfasterthane −2t .Thepolesneartotheimaginaryaxisaretermed
the dominant poles as their transients take the longest to decay. It is quite common for
engineers to ignore the effect of poles that are more than five or six times further away
from the imaginary axis than the dominant poles.
For a real system, poles with imaginary components occur as complex conjugate pairs.
The transient resulting from this pair of poles has the form of a sinusoidal term

21.12 Poles, zeros and thesplane 671
3
jv
s plane
Transient
response
3
s
t
Figure21.21
Apairofstable complex poles gives riseto a decaying sinusoid
transient.
multiplying an exponential term. For a pair of stable poles, that is negative real part,
the transient can be sketched by drawing a sinusoid confined within a ‘decaying exponentialenvelope’(seeFigure21.21).Thereasonforthisisthatwhenthesinusoidalterm
has a value of 1 the transient touches the decaying exponential. When the sinusoidal
term has a value of −1 the transient touches the reflection in thet axis of the decaying
exponential. The larger the imaginary component of the pair of poles, the higher the
frequency ofthe sinusoidal term.
It should now be clear how useful a concept thesplane plot is when analysing the
response of a linear system. A complex system may have many poles and zeros but by
plotting them on the s plane the engineer begins to get a feel for the character of the
system. The form of the transients relating to particular poles or pairs of poles can be
obtainedusingtheaboverules.Themagnitudeofthetransients,thatisthevaluesofthe
coefficientsA 1
,A 2
,...,A n
, depends on the systemzeros.
Itcanbeshownthathavingazeroneartoapolereducesthemagnitudeofthetransient
relatingtothatpole.Engineersoftendeliberatelyintroducezerosintoasystemtoreduce
the effect of unwanted poles. If a zero coincides with the pole, it cancels it and the
transient corresponding tothatpole iseliminated.
Engineeringapplication21.8
Asymptotic Bode plot of a transfer function with real poles
andzeros
Recall that the Bode magnitude plot of a simple linear circuit was examined in
Engineeringapplication2.14.Itispossibletoconstructaplotofamorecomplicated
transferfunctionby following somesimple steps.
Hereweconsideratransferfunction with only real polesand zeros
( ) ( )···( )
s −z1 s −z2 s −zm
G(s) =K
s ( ) ( )···( )
s −p 1 s −p2 s −pn
wherez 1
,z 2
... are the systemzeros and p 1
,p 2
... are the system poles.
Thefirststageistoobtainanexpressionforthefrequencyresponseofthesystem
by substitutings = jω
( )( ) )
jω −z1 jω −z2 ···(
jω −zm
G (jω) =K
jω ( )( ) )
jω −p 1 jω −p2 ···(
jω −pn
➔

672 Chapter 21 The Laplace transform
We note that ( ( )
) jω
jω −z 1 can be written as −z1 +1 . Applying this process
−z 1
toeach bracket inthe numerator and denominator yields
( )( )···( )
−z1 −z2 −zm
G (jω) =K ( )( ) )
−p1 −p2 ···(
−pn
( ) ( ) ( )
jω jω jω
+1 +1 ... +1
−z
× 1
−z 2
−z
( ) ( ) ( m
)
jω jω jω
jω +1 +1 ... +1
−p 1
−p 2
−p n
The Bode plot isalogarithmic plot of20 log 10
|G(jω)|.So we wishtoplot
( ) ( ) )
−z1 −z2 ···(
−zm
∣ ∣∣∣∣
20log 10
|G (jω)| = 20log 10
K ( )( ) )
−p1 −p2 ···(
−pn
( ) ( ) ( )
jω jω jω
+1 +1 ... +1
−z
· 1
−z 2
−z
( m ) ( ) ( )
jω jω jω
jω +1 +1 ... +1
∣
−p 1
−p 2
−p n
Using the laws of logarithms (page 86) this can be rewritten as
20log 10
|G (jω)|
∣ ∣∣∣∣
( ) ( ) )
−z1 −z2 ···(
−zm = 20log 10
K ( )( ) )
−p1 −p2 ···(
−pn ∣ +20log jω
10 ∣ +1
−z ∣
1
∣ ∣∣∣ jω
+20log 10
+1
−z ∣ +···+20log jω
10∣
+1
2
−z ∣
m
∣ ∣∣∣ jω
−20log 10
|jω| −20log 10
+1
−p ∣
1
∣ ∣∣∣ jω
−20log 10
+1
−p ∣ −···−20log jω
10∣
+1
2
−p ∣
n
Inthisform the contribution ∣ of individual poles and zeros can bestudied.
∣∣∣∣
( ) ( ) )
−z1 −z2 ···(
−zm Theterm20log 10
K ( )( ) )
−p1 −p2 ···(
−pn ∣ isconstant.Ithasnodependenceon
frequency, ω.
Theterm20log 10
|jω|hasavalueof0at ω = 1.At ω = 10ithasavalueof20,at
ω = 100 it has a value of 40, and so on. At ω = 0.1 it has a value of −20, at 0.01, a
value of −40, and so on. If we plot this term by itself using a logarithmic frequency
scaleitwouldthereforebeastraightlinepassingthroughthepointwhere ω = 1with
a gradient of 20 dBfor ∣ each decade, or factor of 10, increase infrequency.
∣∣∣ jω
Theterm20log 10
+1
−z ∣ canbeexaminedbyconsideringtheinfluenceatdif-
1
∣ ∣ ferent frequencies. If
ω ∣∣∣ ∣∣∣ jω
∣ ≪ 1, then 20log
z 10
+1
1
−z ∣ ≈ 20log 10
|1| = 0. So the
1
termhas very littleinfluence.

21.12 Poles, zeros and thesplane 673
∣ ∣ Now consider
ω ∣∣∣ ∣∣∣ jω
∣ = 10. Then 20log
z 10
+1
1
−z ∣ ≈ 20log 10
|j10| = 20. It can
∣ 1 be shown by successively considering
ω ∣∣∣
∣ = 100, 1000, 10000,... that for every
z 1
decademultiplicationinfrequency(forfixedz 1
)thetermincreasesbyanother20dB.
Graphically,itapproximatestoastraightlinethatincreasesat20dBforeverydecade
(×10)changeinfrequency.Thepointatwhichthisstraightlineintersectsthe ω axis
iscalled thebreakpoint. ∣ ∣∣∣ jω
The term −20log 10
+1
−p ∣ and others like it have similar properties to those
1
discussed for the term involving zeros. The only difference is the straight line falls
by 20 dBfor every decade multiplication infrequency away from the breakpoint.
By considering the addition of the terms discussed above we can produce an approximation
to the Bode plot. These approximate lines are asymptotes to which the
actual plots tend as one moves away from the breakpoints. Hence the term asymptoticBode
diagramisused.
Consider a practical example of finding the poles and zeros, and sketching the
transfer function
G(s) =
5000 (s +3)
s (s +100)(s +500)
Firstlywenote the position of the poles and zeros by observing that
G(s) =
5000 (s − (−3))
s (s − (−100)) (s − (−500))
There is a single zero atz 1
= −3. There are three poles at p 1
= 0,p 2
= −100,
p 3
= −500. The pole at0isusually termed thepole at the origin.
The substitutions = jω is made togive
G(jω) =
5000 (jω +3)
jω (jω +100)(jω +500)
The equation isrearranged as
G(jω) =
5000 ×3
100 ×500 ·
( jω
3 +1 )
( ) ( ) jω jω
jω
100 +1 500 +1
The constant termcontributes a gain of
∣ ∣∣∣∣
( )( ) )
−z1 −z2 ···(
−zm 20log 10
K ( )( )···( )
−p1 −p2 −pn ∣ = 20log 5000 ×3
10 ∣100 ×500∣ = 20log 10 |0.3|
The zero contributes a gain of
∣ ∣ ∣∣∣ jω
20log 10
+1
−z ∣ = 20log jω ∣∣∣
10 ∣
1
3 +1
This is approximated by a straight line on the Bode diagram starting at ω = 3 and
increasing at20dB per decade.
➔

674 Chapter 21 The Laplace transform
The pole atthe origincontributes a gain of
−20log 10
|jω|
This is plotted as a straight line with no start or end points, with a slope of −20dB
per decade passing through ω = 1.
The two other poles contribute a gain of
∣ ∣ ∣∣∣ jω
−20log 10
+1
−p ∣ = −20log jω ∣∣∣
10 ∣
1
100 +1
and
∣ ∣ ∣∣∣ jω
−20log 10
+1
−p ∣ = −20log jω ∣∣∣
10 ∣
2
500 +1
respectively.Thefirstoftheseisapproximatedbyalinewithslope−20dBperdecade
startingat ω = 100, the second has the same gradient and startsat ω = 500.
TheindividualcontributionstotheoverallplotareshowninFigure21.22.Notice
thestartingpointsoftheasymptotes--thebreakpoints--andtheircorrespondenceto
the positions of the poles and zeros.
60
40
20 log 10 jv
3
dB
20
0
–20
–40
–20 log 10 |jv|
20 log 10 |0.3|
+ 1 –20 log 10 jv
500 + 1
–20 log 10 jv
100 + 1
–60
–80
0.1 1 10 100 1000 10 4
Figure21.22
Contributionsofthe individual terms to the Bode plot.Dottedlinesare the asymptotic
approximations.
v
The asymptotes can now be added together to form a combined graph which is
shown in Figure 21.23. Also plotted on the same graph is an exact Bode plot of the
transferfunction obtained usingacomputer.
Notice that the largest error in the approximation occurs at the breakpoints,
whereas the two graphs agree well at other frequencies. The reason for the error
is most apparent when we consider the case where ω = −z 1
, for example. The gain
atthisbreak frequency according tothe lineplottedshould be
20log 10
∣ ∣∣∣ −jz 1
−z 1
∣ ∣∣∣
= 20log 10
|j| = 0

21.13 Laplace transforms of some special functions 675
0
20
dB
40
60
80
0.1 1 10 100 1000 10 4
v
Figure21.23
Comparisonbetween the asymptotic Bodeplotand the exact frequency response. Thedotted
line isthe result ofaddingthe asymptotic approximations together at each frequency point.
whereas the exact value is
∣ ∣∣∣ −jz
20log 1
10
+1
−z ∣ = 20log 10 |j +1| = 20log 10 ∣ √ 2∣ ∼ = −3.01
1
Fortunately this is the worst-case error for a single pole or zero and if a more
accuratesketchisrequireditispossibletousethisfacttoimproveontheasymptotic
approximationofthetransferfunction.EngineersfinditusefultosketchaBodeplot
inordertogeta‘feel’forthefrequencyresponseofasystem.Ifamoreaccurateplot
isneeded then itispossible touse a computer package toobtain it.
21.13 LAPLACETRANSFORMSOFSOMESPECIALFUNCTIONS
In this section we apply the Laplace transform to the delta function and periodic functions.These
functions wereintroducedinChapter2.
21.13.1 Thedeltafunction, δ(t−d)
Recallthe integral propertyofthe deltafunction(see Section16.4),
∫ ∞
−∞
f(t)δ(t −d)dt = f(d)
The Laplace transform follows from this integral property. Let f (t) = e −st so that
f(d) = e −sd .Then
thatis,
∫ ∞
−∞
e −st δ(t −d)dt =
L{δ(t −d)} = e −sd
∫ ∞
0
e −st δ(t −d)dt = f(d) =e −sd

676 Chapter 21 The Laplace transform
The Laplace transformof δ(t) follows by settingd = 0:
L{δ(t)} =1
Example21.32 For a particular circuititcan be shown that the transferfunction,G(s), isgiven by
G(s) = V o (s)
V i
(s) = 1
s +2
where V i
(s) and V o
(s) are the Laplace transforms of the input and output voltages
respectively.
(a) Find v o
(t) when v i
(t) = δ(t).
(b) Usethe convolution theorem tofind v o
(t) when
{
e
−t
t0
v i
(t) =
0 t<0
Solution (a) We aregiven the transfer function
G(s) = V o (s)
V i
(s) = 1
s +2
When v i
(t) = δ(t),V i
(s) = 1andhence
and
V o
(s) = 1
s +2
{ } 1
v o
(t) = L −1 = e −2t
s +2
Thisisknownastheimpulseresponseofthesystem,g(t).Iftheimpulseresponse,
g(t),isknownthentheresponse, v o
(t),toanyotherinput, v i
(t),canbeobtainedby
convolution, thatis v o
(t) =g(t) ∗ v i
(t).
(b) The response toan input, v i
(t) =u(t)e −t , isgiven by
v o
(t) =g(t)∗v i
(t) =
=
∫ t
0
∫ t
0
e −2λ e −(t−λ) dλ
∫ t
= e −t e −λ dλ
0
[ ] e
= e −t −λ t
−1
0
=e −t (1−e −t )
= e −t −e −2t
g(λ)v i
(t − λ)dλ

21.13.2 Periodicfunctions
21.13 Laplace transforms of some special functions 677
Recallthedefinitionofaperiodicfunction, f (t).GivenT > 0, f (t)isperiodicif f (t) =
f (t +T ) for all t in the domain. If f (t) is periodic and we know the values of f (t)
over a period, then we know the values of f (t) over its entire domain. Hence, it seems
reasonable that the Laplace transform of f (t) can be found by studying an appropriate
integral over an interval whose length is just one period. This is indeed the case and
formsthe basis of the following development.
Let f (t)be periodic, with periodT. The Laplace transformof f (t)is
L{f(t)} =F(s) =
=
∫ T
0
∫ ∞
0
e −st f (t)dt
∫ 2T ∫ 3T
e −st f(t)dt + e −st f(t)dt + e −st f(t)dt+···
T
2T
Lett =xinthefirstintegral,t =x +T inthesecond,t =x +2T inthethirdandsoon.
F(s) =
∫ T
0
∫ T
+
0
∫ T
e −sx f(x)dx +
0
e −s(x+T ) f(x+T)dx
e −s(x+2T ) f(x+2T)dx+···
Since f isperiodic with periodT then
So,
f(x)=f(x+T)=f(x+2T)=···
F(s) =
∫ T
0
∫ T
∫ T
e −sx f(x)dx + e −sT e −sx f(x)dx + e −2sT e −sx f(x)dx + ···
0
0
=(1+e −sT +e −2sT +···)
∫ T
0
e −sx f(x)dx
We recognize the terms in brackets as a geometric series whose sum to infinity
1
is . Hence,
1 −e−sT F(s) =
∫ T
0 e−st f (t)dt
1 −e −sT
Example21.33 Awaveform, f (t), isdefinedasfollows:
f(t)=
{ 2 0<t1.25
0 1.25<t1.5
and f (t)isperiodicwith periodof1.5.Findthe Laplace transformofthe waveform.

678 Chapter 21 The Laplace transform
Solution L{f (t)} =
=
∫ 1.5
e −st f (t)dt
0
1 −e −1.5s
∫ 1.25
2e −st dt + ∫ 1.5
0 1.25 0e−st dt
1 −e −1.5s
= 2(1 −e−1.25s )
s(1 −e −1.5s )
EXERCISES21.13
1 Find the Laplacetransformsof
(a) 3u(t) + δ(t) (b) −6u(t) +4δ(t)
(c) 3u(t −2)+δ(t −2)
(d) u(t −3)−δ(t −4)
1
(e) 2 u(t −4)+3δ(t −4)
whereu(t)is the unitstepfunction.
2 Find the inverse Laplace transformsof
(a) 2 s −1 (b) 2
3s + 1 2
(d) 4s−3
s
3 Aperiodicwaveform is defined by
{
t 0t1
f(t)=
2−t 1<t2
(c) 3−2s
s
andhas aperiod of2.
(a) Sketch two cycles of f (t).
(b) Find L{f (t)}.
4 Awaveform, f (t),is defined by
{
2 0t1.5
f(t)=
−2t+5 1.5<t<2.5
andhas aperiod of2.5.
(a) Sketch f (t)on [0,5].
(b) Find L{f (t)}.
5 Ifthe impulseresponse ofanetwork isg(t) = 10e −4t
findthe outputwhenthe input is f (t) = e −t cos2t,
t0.
Solutions
3
1 (a)
s +1 (b) −6 s +4
3e −2s
(c) +e −2s (d) e−3s −e −4s
s s
e −4s
(e) +3e −4s
2s
2 (a)2u(t)−δ(t) (b) 2u(t) + δ(t)
3 2
3 (b)
4 (b)
5
(c)3u(t) −2δ(t)
1−e −s
s 2 (1+e −s )
2(s +e −2.5s −e −1.5s )
s 2 (1 −e −2.5s )
(30cos2t +20sin2t)e −t
13
(d) 4δ(t) −3u(t)
− 30e−4t
13
REVIEWEXERCISES21
1 GivenF(s) = s +1
s 2 isthe Laplace transformof
+2
f (t),findthe Laplace transformsofthe following:
(a) f(t)
e 2t
(b) 3e 2t f(t)
(c)2e −t (f(t)+1) (d) u(t−1)f(t−1)
(e)4u(t −3)f(t −3)
(f) e −2t u(t −2)f(t −2)

Review exercises 21 679
2 Find the Laplace transformsofthe following:
(a) IfV o (s) = L{v o (t)}andV i (s) = L{v i (t)},show
seriesRC network are relatedbythe differential (d) Findi(t)if
equation
{
2e
CR dv −2t t 0
o
v(t) =
+v
dt o =v i 0 otherwise
(a) 2tsin3t (b) 1 2 (−3tcos2t)
that the transfer function, V o (s) , isgiven by
V i (s)
1
(c) e −t u(t) (d) e −t u(t −1)
sCR+1 .
(e) 3t 2 u(t −1) (f) e −t δ(t −2)
(b) IfC = 0.1 µF,R = 100 k find,and sketch a
3 Find the inverse Laplacetransformsof
graph ofthe response, v o (t),when
2s+3
{
(a)
5volts t0
(s +1)(s +2)
v i (t) =
0 t<0
4s
(b)
s 2 −9
(c) Using the component values given in (b)findthe
(c) 2(s3 +4s 2 +4s +64)
response when the input is aunitimpulse, δ(t).
(s 2 +4)(s 2 +16)
(d) Using the component values given in (b)use the
(d)e −2s 6 s 4 (e) e−ss +1
convolution theoremto determine the response
s 2
when v i (t) = 5e −100t .
4 Solvethe followingdifferentialequations usingthe 7 Express the squarewave
Laplacetransformmethod:
{
1 0<t<a
(a) x ′′ +2x ′ −3x =2t, x(0) =1 x ′ (0) =2 f(t)=
−1 a <t < 2a period 2a
(b) x ′′ −2x ′ +5x =cost, x(0) =0 x ′ (0) =1 in terms ofunitstepfunctions. Hencededuce its
(c) ẋ+ẏ+x+y= 3 2 (1+t)
Laplace transform.
ẋ−2ẏ+x+2y=2t
8 Consider the circuitshown in Figure 21.24.
x(0) =0 y(0) =0
(d) ẋ−ẏ+x=−1,x(0)=0 y(0)=2
1–
3 F 1 H 4 V
2ẋ−ẏ + y 2 −x=1
5 Thecurrent,i(t),in a seriesLC circuitis governed by
i(t)
L di
dt + 1 ∫ t
idt = v(t)
y(t)
C 0
Figure21.24
where v(t)is the applied voltage.
(a) Assumingzero initialconditions showthat
(a) Show that
LsI + 1 Cs I =V(s)
∫
di t
dt +4i+3 idt = v(t)
0
(b) If v(t) = δ(t)show that
(b) Assumingzero initialconditions, showthat
i(t) = 1 L cos t
√ s
LC
I(s) =
(s+1)(s+3) V(s)
6 Theinput andoutputvoltages, v i (t)and v o (t),ofa (c) Findi(t)if v(t) = δ(t).

680 Chapter 21 The Laplace transform
Solutions
s +3
1 (a)
s 2 +4s+6
3(s −1)
(b)
s 2 −4s+6
(c)
(d)
(e)
2(2s 2 +5s +5)
(s +1)(s 2 +2s +3)
e −s (s +1)
s 2 +2
4e −3s (s +1)
s 2 +2
e −2(s+2) (s +3)
(f)
s 2 +4s+6
12s
2 (a)
(s 2 +9) 2
(b) − 3(s2 −4)
2(s 2 +4) 2
(c)
1
s +1
(d) e−(s+1)
s +1
( )
(e) 3e −s 2
s 3 + 2 s 2 + 1 s
(f) e −2(s+1)
3 (a) e −t +e −2t
(b) 2e −3t +2e 3t
(c) 2(2sin2t +cos4t)
(d) u(t −2)(t −2) 3
(e) u(t −1)t
4 (a) − 11e−3t
36
(b) e t [ − 1 5
+ 7et
4 − 2t
3 − 4 9
cos2t +
13
20 sin2t ]
+ 1 5 cost− 1 10 sint
(c) x=t,y=t/2
(d) x=e t −1,y=2e t
6 (b) v o (t) =5−5e −100t
7
8 (c)
(c) 100e −100t
(d) 500te −100t
1 −e −as
s(1 +e −as )
3e −3t
2
− e−t
2
(d) −e −t +4e −2t −3e −3t

22 Differenceequations
andtheztransform
Contents 22.1 Introduction 681
22.2 Basicdefinitions 682
22.3 Rewritingdifferenceequations 686
22.4 Blockdiagramrepresentationofdifferenceequations 688
22.5 Designofadiscrete-timecontroller 693
22.6 Numericalsolutionofdifferenceequations 695
22.7 Definitionoftheztransform 698
22.8 Samplingacontinuoussignal 702
22.9 TherelationshipbetweentheztransformandtheLaplacetransform 704
22.10 Propertiesoftheztransform 709
22.11 Inversionofztransforms 715
22.12 Theztransformanddifferenceequations 718
Reviewexercises22 720
22.1 INTRODUCTION
Difference equations are the discrete equivalent of differential equations. The terminology
is similar and the methods of solution have much in common with each other.
Differenceequationsarisewheneveranindependentvariablecanhaveonlydiscretevalues.
They are of growing importance in engineering in view of their association with
discrete-timesystems based on the microprocessor.
InChapter21theLaplacetransformwasshowntobeausefultoolforthesolutionof
ordinary differential equations, and for the construction of transfer functions in circuit
analysis,controltheory,etc.Generally,Laplacetransformmethodsapplywhenthevariables
being measured are continuous. Theztransform plays a similar role for discrete

682 Chapter 22 Difference equations and the z transform
systemstothatplayedbytheLaplacetransformforcontinuousones.Inthischapteryou
will be introduced to theztransform and one of its applications -- the solution of linear
constant coefficient difference equations. The z transform is of increasing importance
as more and more engineering systems now contain a microprocessor or computer and
sohaveoneormorediscrete-timecomponents.Forexample,mostindustrialcontrollers
nowhaveanembeddedmicroprocessor,andoverallcontrolofafactoryisoftenbymeans
ofasupervisoryprocesscontrolcomputer.Manyfactoriesalsohaveaproductioncontrol
computer toschedule production.
22.2 BASICDEFINITIONS
Before classifying difference equations wewill first derive an example of one.
Suppose a microprocessor system is being used to capture and analyse images. The
number of instruction cycles,i, of the microprocessor needed to process an image depends
on the number of pixels, n, that the image is broken down into. Clearly n is a
non-negative integer, that is n ∈ N. Since i depends upon n we write i = i[n]. The
squarebracketsnotationreflectsthefactthatnisadiscretevariable(seeSection6.2).If
there arenitemsof data, the number of instruction cycles used isi[n].
Suppose that if there aren + 1 items of data, the number of instruction cycles used
increases by 10n +1.Then,
i[n+1]=i[n]+10n+1
Thisisanexampleofadifferenceequation.Thedependentvariableisi;theindependent
variable isn.
Puttingn = 0 inthe difference equation gives
i[1] =i[0] +10(0) +1 =1
Similarly,i[2] = 12,i[3] = 33,i[4] = 64andsoon.Weseethatthedifferenceequation
gives risetoasequence of values.
There are strong similarities between difference and differential equations. The important
point to note is that with difference equations, the independent variable is discrete,
not continuous. In the above example,nis the number of pixels; it can have only
integer values. This discrete property of the independent variable is an essential and
distinguishing feature of difference equations. Much of the terminology of differential
equations isapplied, withidentical meaning, todifference equations.
22.2.1 Dependentandindependentvariables
Consider a simple difference equation:
x[n +1] −x[n] = 10
The dependent variable isx; the independent variable isn. Inthe difference equation
y[k+1]−y[k]=3k+5
the dependent variable isyand the independent variable isk.

22.2 Basic definitions 683
22.2.2 Thesolutionofadifferenceequation
A solution is obtained when the dependent variable is known for each value of interest
of the independent variable. Thus the solution takes the form of a sequence. There are
frequently many different sequences which satisfy a difference equation; that is, there
aremanysolutions.Thegeneralsolutionembracesalloftheseandallpossiblesolutions
can be obtained from it.
Example22.1 Showx[n] =A2 n , whereAisaconstant, isasolution of
x[n +1] −2x[n] = 0
Solution x[n] =A2 n x[n +1] =A2 n+1 = 2A2 n
Hence,
x[n +1] −2x[n] = 2A2 n −2A2 n = 0
Hencex[n] =A2 n isasolutionofthegivendifferenceequation.Infactx[n]isthegeneral
solution.
If additionally we are given a condition, sayx[0] = 3, the constantAcan be found. If
x[n] =A2 n , thenx[0] =A2 0 =Aand hence
A = 3
The solution is thus x[n] = 3(2 n ). This is the specific solution and satisfies both the
difference equation and the given condition.
22.2.3 Linearandnon-linearequations
Anequationislinearifthedependentvariableoccursonlytothefirstpower.Ifanequation
isnot linear itisnon-linear. For example,
3x[n +1] −x[n] = 10
y[n+1]−2y[n−1]=n 2
kz[k +2] +z[k] =z[k −1]
arealllinearequations.Notethatthepresenceofthetermn 2 doesnotmaketheequation
non-linear, sincenisthe independent variable. However,
(x[n +1]) 2 −x[n] = 10
y[k+1]= √ y[k] +1
areboth non-linear. Also,
z[n +1]z[n] =n 2 +100
sinx[n] =x[n −1]
arenon-linear.Theproducttermz[n +1]z[n]andthetermsinx[n]arethecausesofthe
non-linearity.

684 Chapter 22 Difference equations and the z transform
22.2.4 Order
Theorderisthedifferencebetweenthehighestandlowestargumentsofthedependent
variable. The equation
3x[n +2] −x[n +1] −7x[n] =n
issecond order because the difference betweenn+2 andnis2.
x[n +1]x[n −1] = 7x[n −2]
isthirdorderbecausethedifferencebetweenn +1andn −2is3.Ingeneral,thehigher
the order of an equation, the more difficultitistosolve.
Insomedifferenceequationsthedependentvariableoccursonlyonce.Theseareclassified
as zero order. Engineers refer to them as non-recursive difference equations because
calculation of the value of the dependent variable does not require knowledge of
thepreviousvalues.Incontrast,differenceequationsoforder1orgreaterarereferredto
asrecursivedifferenceequationsbecausetheirsolutionrequiresknowledgeofprevious
values of the dependent variable. The difference equation
x[n]=n 2 +n+1
has zero order and so isanon-recursive difference equation.
22.2.5 Homogeneousandinhomogeneousequations
Themeaningsofhomogeneousandinhomogeneousasappliedtolineardifferenceequationsareanalogoustothosemeaningswhenappliedtodifferentialequations.Todecide
whether a linear equation is homogeneous or inhomogeneous it is written in standard
form, with all the dependent variable terms on the l.h.s. Any remaining independent
variable termsare writtenon the r.h.s.For example,
3nx[n +1] −2n 3 =x[n −1]
iswritten as
3nx[n +1] −x[n −1] = 2n 3 (22.1)
If the r.h.s. is 0, the equation is homogeneous; otherwise it is inhomogeneous. Equation
(22.1) isinhomogeneous but
3nx[n+1]−x[n−1]=0
ishomogeneous.
Engineeringapplication22.1
Signalprocessingusingamicroprocessor
In engineering, an increasing number of products contain a microprocessor or computer
which is solving one or more difference equations. The input to the microprocessorisasequenceofsignalvalues,inmanycasesformedasaresultofsamplinga
continuous inputsignal.Theoutput fromthemicroprocessorisasequence ofsignal
values whichmaybesubsequently convertedintoacontinuous signal.Forexample,

22.2 Basic definitions 685
an inhomogeneous difference equation could be of the form
y[n] −2y[n −1] = 0.1s[n] +0.2s[n −1] −0.5s[n −2]
where y[n] is the output sequence or dependent variable and s[n] is the input sequence.Notethattheinputsequencecanstillbethoughtofastheindependentvariable
but instead of being expressed analytically in terms ofnit arises as a result of
sampling. The corresponding homogeneous equation is
y[n] −2y[n −1] = 0
22.2.6 Coefficient
Thetermcoefficientreferstothecoefficientofthedependentvariable.InEquation(22.1)
the coefficients are 3nand −1.
Example22.2 (a) State the order ofeach of the following equations (i)--(vii).
(b) State whether each equation islinear or non-linear.
(c) For each linear equation, statewhether itishomogeneous or inhomogeneous.
(i) 2x[n] −3nx[n −1] +x[n −2] +n 2 = 0
1
(ii) (x[n +1] −x[n −1]) =x[n]
3
(iii) z[n +2](2n −z[n −1]) =n +1
7x[n −1]
(iv)
x[n −2] = n +1
n −1
(v) w[n +3]w[n +1] =n 3 −1
(vi) y[n +2] +2y[n +1] = 6s[n +2] −2s[n +1] +s[n]whereyisthedependent
variable
(vii) x[k+3]−2x[k+2]+x[k] =e[k+2]−e[k]wherexisthedependentvariable.
Solution (a) (i) Second order
(ii) Second order
(iii) Third order
(iv) Firstorder
(v) Second order
(vi) Firstorder
(vii) Third order
(b) (i) Linear
(ii) Linear
(iii) Non-linear
(iv) Linear
(v) Non-linear
(vi) Linear
(vii) Linear
(c) Equations (iii) and (v) are non-linear. The linear equations are written in standard
form:
(i) 2x[n] −3nx[n −1] +x[n −2] = −n 2
(ii) x[n +1] −3x[n] −x[n −1] = 0

686 Chapter 22 Difference equations and the z transform
(iv) 7(n −1)x[n −1] − (n +1)x[n −2] =0
(vi) and (vii)are already instandard form.
Hence wefind the following:
(i) Inhomogeneous
(ii) Homogeneous
(iv) Homogeneous
(vi) Inhomogeneous
(vii) Inhomogeneous
EXERCISES22.2
1 Foreach ofthe followingequations (a)--(e): (c)y[n −2] +y[n −1] +y[n] =n 2
(i) Statethe order ofthe equation. (ii)State whether
each equation islinear ornon-linear. (iii)Foreach
linear equation, state whether itis homogeneous or
inhomogeneous.
(a)n(3n +x[n]) =x[n −1]
2z[k −4]
(b)
z[k −3] =z[k−2]
(d) √ n +x[n] =x[n −2] +e[n], wherexis the
dependent variable.
(e) (2w[n−1]+1) 2 = w[n−2]+s[n−1]−s[n−2],
where w isthe dependent variable.
Solutions
1 (a) First order; linear;inhomogeneous
(b) Second order; non-linear
(c) Second order; linear;inhomogeneous
(d) Second order; non-linear
(e) Firstorder; non-linear
22.3 REWRITINGDIFFERENCEEQUATIONS
Sometimes an equation or expression can be written in different ways. At first sight, it
may appear there are two independent equations when in fact there is only one. Thus
weneedtobeabletorewriteequationssothatcomparisonscanbemade.Whengeneral
solutions of equations are to be found, usually the equation is first written in a standard
form. Soonceagain thereisaneedtorewrite equations.
Example22.3 Rewrite the equation sothat the highestargumentofthe dependentvariable isn +1.
x[n+3]−x[n+2]=2n
x[2]=7
Solution Thehighestargumentinthegivenequationisn +3;thismustbereducedby2ton +1.
To do thisnisreplaced byn −2.Theequation becomes
x[n +1] −x[n] =2(n −2) x[2] =7
Note,however,thattheinitialcondition,x[2] = 7,isnotchanged.Thisissimplystating
thatxhas a valueof7when the independentvariablehas a value of2.

22.3 Rewriting difference equations 687
Example22.4 Write the following equations so that the highest argument of the dependent variable is
n+2.
(a) 3x[n +4] −2nx[n +2] = (n −1) 2 x[3] = 6 x[4] = −7
(b) z[n−2]+z[n−1]+z[n]=1+n z[0]=1 z[1]=0
Solution (a) Thehighestargument,n+4,mustbereplacedbyn+2,thatisnisreplacedbyn−2
throughout the equation.
3x[n +2] −2(n −2)x[n] = (n −3) 2 x[3] = 6 x[4] = −7
(b) The highest argument,n, isincreased ton +2,thatisnisreplaced byn+2.
z[n]+z[n+1]+z[n+2]=n+3 z[0]=1 z[1]=0
Example22.5 Write the following equations so that the highest argument of the dependent variable
isk:
s[k +2] −2s[k +1] +s[k]
(a) a[k +2] =
15
whereaisthe dependent variable.
(b) a[k +3] = l[k+3]+l[k+2]+l[k+1]+l[k]+l[k−1]
5
whereaisthe dependent variable.
Solution (a) The highest argument,k+2, must be replaced byk, that iskis replaced byk−2
throughout the equation:
a[k] =
s[k] −2s[k −1] +s[k −2]
15
(b) The highest argument,k+3, must be replaced byk, that iskis replaced byk−3
throughout the equation:
a[k] = l[k]+l[k−1]+l[k−2]+l[k−3]+l[k−4]
5
EXERCISES22.3
1 Writeeachequation sothat the highestargument of
the dependent variableisasspecified:
(a) p[k] −3p[k +1] = p[k −2],highestargumentof
the dependent variableisto bek+2.
(b) R[n−1]−R[n−2]−R[n−3]=n,R[0]=1,
R[1] = −2, highestargumentofthe dependent
variableisto ben.
(c) q[t] +tq[t −1] = 3q[t +1],q[1] = 0,
q[2] = −2,highestargumentofthe dependent
variableis to bet −1.
(d) T[m]+(m−1)T[m−2]=m 2 ,
T[0] =T[1] = 1,highestargumentofthe
dependent variableis to bem +2.
(e) y[k +1] −y[k +3] = (s[k +2] −s[k +4])/2,
whereyisthe dependentvariable andthe highest
argumentofthe dependent variableisto bek.

688 Chapter 22 Difference equations and the z transform
Solutions
1 (a) p[k+1]−3p[k+2]=p[k−1]
(b) R[n]−R[n−1]−R[n−2]=n+1
R[0] = 1,R[1] = −2
(c) q[t −2] + (t −2)q[t −3] =3q[t −1]
q[1] = 0,q[2] = −2
(d) T[m+2]+(m+1)T[m]= (m+2) 2
T[0]=T[1]=1
(e) y[k −2] −y[k] = s[k−1]−s[k+1]
2
22.4 BLOCKDIAGRAMREPRESENTATIONOFDIFFERENCE
EQUATIONS
Manyengineeringsystemscanbemodelledbymeansofdifferenceequations.Itispossibletorepresentadifferenceequationpictoriallybymeansofablockdiagram.Theuse
ofablockdiagramrepresentationhelpsanengineertovisualizeasystemandmayoften
behelpfulinsuggestingtherequiredhardwareorsoftwaretoimplementaparticulardifferenceequation.Thisisparticularlyimportantinareassuchasdigitalsignalprocessing
and digital controlengineering.
Before discussing block diagrams it is necessary to review the topic of sampling.
Difference equations operate on discrete-time data and therefore a continuous signal
needstobesampledbeforeuse.Inthemostcommonformofsampling,asampleistaken
at regular intervals,T. A continuous signal and the sequence produced by sampling it
areshowninFigure22.1.Someauthorswritethesequenceasx[nT]toindicatethatthe
sequencehasbeenobtainedbysamplingacontinuouswaveformatintervalsT.Wewill
notuse thisconvention butsimplyrefertothe sampledsequenceasx[n].
Several components are used in a block diagram. The delay block is shown in
Figure22.2.Theeffectofthiselementistodelaythesequencebyonesamplinginterval,
T. For example,if
x[n]=6,4,3,−2,0,2,5,0,...
n=0,1,2,...
wecan write thisasx[0] = 6,x[1] = 4,x[2] = 3,x[3] = −2,..., and then
x[n−1]=0,6,4,3,−2,0,2,5,0,...
n=0,1,2,...
x(t)
x[n]
(a)
t
Figure22.1
(a) Continuoussignal; (b)sequence produced asaresult ofsampling.
(b)
T
n

22.4 Block diagram representation of difference equations 689
x[n] x[n – 1]
T
Figure22.2
Adelay block delaysasequence
byatime interval,T.
x[n] x[n – 2]
T
T
Figure22.3
Two delay blocks in series.
x[n]
x[n]
p[n]
x[n] + a
1
kx[n]
3
q[n]
p[n] + q[n] + r[n]
S
a
k
r[n]
Figure22.4
Adding aconstant
to a sequence.
Figure22.5
Scalingasequence
byaconstant.
Figure22.6
Adding sequences together
usingasummer.
Notethatx[n −1]isundefinedwhenn = 0andsothisisassignedavalueof0.Adelay
of two sampling intervals results inthe sequencex[n −2]as shown inFigure 22.3.
Another block diagram element represents the addition of a constant to a sequence
and is shown in Figure 22.4. A sequence can be scaled by a constant. This is shown in
Figure 22.5. Finally, sequences can be added together using a summer. This is shown
inFigure 22.6.
Engineeringapplication22.2
Discrete-timefilter
The term filter is used by engineers to describe an electronic device that selectively
processessignalsofdifferentfrequencies.So,forexample,alowpassfilterisafilter
that allows low frequency signals to pass with little attenuation of their amplitude.
However, high frequency signals tend to be severely attenuated. In contrast, a high
passfilterattenuateslowfrequencysignalswhileallowinghighfrequencysignalsto
pass with relative ease. A discrete-time filter or, as more commonly termed, a digital
filter is an electronic device that uses digital methods to selectively filter digital
signals.
A simple example of a discrete-time filter is one described by the difference
equation
y[n] −ay[n −1] =x[n]
wherex[n]istheinputsequence,y[n]istheoutputsequenceandaisaconstant.Ifa
is positive then the filter behaves as a low-pass filter which rejects high frequencies
but allows low frequencies to pass. If a is negative then the filter behaves as a
high-pass filter. A block diagram for the filter is shown in Figure 22.7. Note that
this is a recursive filter because calculation ofy[n] requires knowledge of previous
values of the output sequence. Note also that the block diagram contains a feedback
path. This isafeatureofrecursive difference equations.
➔

690 Chapter 22 Difference equations and the z transform
x[n]
S
y[n]
ay[n – 1]
3 T
a
y[n – 1]
Figure22.7
Discrete-time filter.
Engineeringapplication22.3
Developmentofadifferenceequationtocalculatethe
accelerationofanobject
Engineersoftenneedtocalculatetheaccelerationofanobjectwhenallthatisknown
isthepositionofthatobject.Thiscanbeachievedbydevelopingadifferenceequation
that can be solved at regular time intervals by a computer. It is first necessary to
use a device that converts the position signal into a form suitable for analysis by a
computer. Usually the position signal will be continuous and therefore in analogue
form.Itisthereforenecessarytouseananalogue-to-digitalconvertertotransform
the analogue signal into a digital signal. Consider the following problem. Develop
a difference equation to convert a digital position signal into a digital acceleration
signal given that this position signal varies with time. Derive an associated block
diagramforthisdifference equation.
Lets =position, v =speed anda =acceleration:
v = ds a = dv = d2 s
dt dt dt 2
Therefore, in order to obtain an acceleration signal the position signal must be differentiatedtwice.ForasmalltimeintervalT,thederivative,y(t),ofasignalx(t)can
be approximated by
y(t) ≈ x(t)−x(t−T)
T
This follows directly from the definition of differentiation. If the signalx(t) is sampled
to givex[n] then the process of differentiation is represented by the difference
equation
x[n] −x[n −1]
y[n] =
T
Figure 22.8 shows a block diagram for the differentiator. It is important to note that
this difference equation is only an approximation to the process of differentiation.
This could be implemented using special-purpose hardware or by software on a microprocessor.
Itfollows thatthe speed of the object isgiven by
s[n] −s[n −1]
v[n] =
T
The problem of finding the acceleration, a[n], can now be solved by coupling two
differentiators together as shown in Figure 22.9. An alternative approach to this

22.4 Block diagram representation of difference equations 691
x[n]
x[n]
S
3
y[n]
T
3
–x[n – 1]
1
— T
–1
Figure22.8
Block diagram ofadifferentiator.
s[n]
y[n]
s[n]
3
y[n]
3
a[n]
T 3
—T
1
3
—T
1
–s[n – 1] –y[n – 1]
Figure22.9
Two differentiators in series.
–1 –1
problemistoobtainadifferenceequationfortheprocessoffindingasecondderivative.
Given that
s[n] −s[n −1]
v[n] = (22.2)
T
and
v[n] − v[n −1]
a[n] = (22.3)
T
then substituting Equation (22.2) into Equation (22.3) gives
a[n] =
(s[n] −s[n −1]) − (s[n −1] −s[n −2])
T 2
s[n] −2s[n −1] +s[n −2]
a[n] =
T 2
TheblockdiagramforthisdifferenceequationisshowninFigure22.10.Notethatthe
difference equation is non-recursive and so there are no feedback paths in the block
diagram.Theoutputsequenceisa[n]andtheinputsequenceiss[n].Theindependent
variable isn.
–2
T
3
–2s[n – 1]
s[n]
T
T
s[n]
S
s[n – 2]
3
1—
T 2
a[n]
Figure22.10
a[n] is obtained by
calculating the second
derivative ofs[n].

692 Chapter 22 Difference equations and the z transform
Engineeringapplication22.4
Development of a moving average filter to calculate the height
ofacidinachemicaltank
A common problem in chemical engineering is that the height of liquid in a tank is
difficulttomeasureaccuratelybecauseoftheliquidswillingaround.Onesolutionto
thisproblemistouseamovingaveragefiltertosmoothoutanyvariationsinheight
obtainedfromthetankpositionsensor.Akeydesigndecisionishowmanypastdata
points to use when constructing the moving average filter. In practice this depends,
amongst other things, on how much the liquid moves around. It is important not to
usetoomanypointsorsignificantvariationsinheightmaynotbepickedupintime.
However,iftoofewpointsareusedthentheoutputfromthefiltermaybetoospikyto
beofuse.Considerthecaseofamovingaveragefilterthatmakesuseofthefivemost
recentlymeasuredvaluesoftheleveloftheliquidinthetank.Formulateadifference
equation tocarryout thisaveraging and draw an equivalent block diagram.
Leta[n] represent the average level of the acid in the tank and letl[n] represent
the sampled values of the level measurements received from the transducer. Then,
a[n] = l[n]+l[n−1]+l[n−2]+l[n−3]+l[n−4]
5
Theblockdiagramforthisdifference equationisshowninFigure22.11. Theaction
of taking a moving average of sampled values is equivalent to passing the sampled
valuesthroughalow-passfilterbecauseitfiltersouthigh-frequencyvariationsinthe
sampledvalues.Thisprocessistermeddigitalfilteringordigitalsignalprocessing.
T
l[n – 1]
l[n]
l[n]
T
T
l[n – 2]
S
3
a[n]
T
T
T
T
T
T
l[n – 3]
T
l[n – 4]
1—
5
Figure22.11
Block diagram ofamoving averager.
EXERCISES22.4
1 Design adigital filterbasedontaking a moving
average ofthe last threevalues ofasampledsignal.
2 Acomputerisfedasignalrepresentingthevelocityof
an objectasafunctionoftime.Prior to enteringthe
computer the signal is sampled usingan
analogue-to-digitalconverter. Derive adifference
equationandassociatedblock diagramto obtain the
accelerationofthe objectasafunctionoftime.

22.5 Design of a discrete-time controller 693
Solutions
1 SeeFigure S.23.
2 a[n] =
v[n] − v[n −1]
.SeeFigure S.24.
T
x[n]
T
S 3 a[n]
y[n]
S 3
a[n]
FigureS.23
T
T
1–
3
FigureS.24
T
3
–1
1 –
T
22.5 DESIGNOFADISCRETE-TIMECONTROLLER
Figure22.12showsablockdiagramofasingleloopindustrialcontrolsystem.Thecontrol
system employs feedback to compare the desired value of a process variable with
the actual value. Any difference between the two generates an error signal e(t). This
signalisprocessedbythecontrollertoproduceacontrollersignalm(t).Anamplifieris
oftenpresenttomagnifythissignaltomakeitsuitablefordrivingtheplantthatisbeing
controlled.
Thecontrollercanbeimplementedbymeansofananalogueelectroniccircuit.However,
digital computers are being used increasingly as controllers. The signalse(t) and
m(t) are both continuous in time and so it is necessary to samplee(t) before it can be
used by the computer and to reconstruct the signal generated by the computer to producethecontrolleroutput,m(t).Thearrangementforimplementingadigitalcontroller
isshown inFigure 22.13.
r(t) +
e(t)
Controller
m(t)
Amplifier
Plant
c(t)
–
Sensor
Figure22.12
Asingle loop industrial controlsystem.
e(t)
Analogue-todigital
converter
Digital
computer
Digital-toanalogue
converter
m(t)
Figure22.13
Block diagramofadigital controller.

694 Chapter 22 Difference equations and the z transform
The most common type of controller used in industry is the proportional/integral/
derivative (p.i.d.)controller. Itcanbeshown thattheanalogue formofthiscontrolleris
modelled by the equation
m(t) =K p
e(t)+K i
∫ t
0
e(t)dt +K d
de(t)
dt
(22.4)
whereK p
,K i
andK d
are constants. In order to implement a discrete-time (digital) controlleritisnecessarytoconvertthisequationintoanequivalentdifferenceequation.The
approximation for the process of differentiation has already been examined in
Engineering application 22.3, and isgiven by
de(t)
dt
≈
e[n] −e[n −1]
T
Thereareseveralpossiblewaysofapproximatingtheprocessofintegration.Onemethod
is illustrated in Figure 22.14. Here the area under the curve is approximated by a series
of rectangles, each of widthT. If the approximate area under the curve fromt = 0 to
t =nT isdenoted byx[n], then
x[n] =x[n −1] +Te[n] (22.5)
The discrete form ofEquation (22.4) can now be formulated. Itisgiven by
m[n] =K p
e[n] +K i
x[n] +K d
e[n] −e[n −1]
T
(22.6)
Equations (22.5) and (22.6) form a set of equations toimplement a discrete form of the
p.i.d.controlleronadigitalcomputerormicroprocessor.Thesetwoequationsaretermed
coupled difference equations because both are needed to calculate m[n]. In addition,
Equation(22.5)isrecursive.Aflowchartforimplementingtheseequationsisshownin
Figure 22.15.
Initialization
Time to
sample?
YES
NO
Get e[n]
e(t)
Calculate
m[n]
Output m[n]
t
Figure22.14
Approximating the areaunder the curve by a
seriesofrectangles.
Figure22.15
Flow chartforap.i.d. controller.

22.6 Numerical solution of difference equations 695
EXERCISES22.5
1 Atransduceris used to measure the speed ofamotor
car.Design adigital filterto calculate the distance
travelled bythe car.
2 Draw a block diagram forEquations(22.5)
and (22.6).
Solutions
1 s[n] =s[n −1] +Tv[n]. SeeFigure S.25.
y[n]
3 S
s[n]
T
T
FigureS.25
22.6 NUMERICALSOLUTIONOFDIFFERENCEEQUATIONS
Having seen how difference equations are formulated we now proceed to methods of
solution. The numerical method illustrated may be applied to all classes of difference
equation.
Example22.6 Given
x[n+1]−x[n]=n
x[0]=1
determinex[1],x[2] andx[3].
Solution Thetermsinthe equation areevaluated forvarious values ofn:
n = 0
x[1] −x[0] = 0
x[1] = 1
n = 1
x[2] −x[1] = 1
x[2] = 2
n = 2
x[3] −x[2] = 2
x[3] = 4
Example22.7 Determinex[4]given
2x[k +2] −x[k +1] +x[k] = −k 2 x[0] = 1 x[1] = 3

696 Chapter 22 Difference equations and the z transform
Solution k =0 k =1
2x[2] −x[1] +x[0] = 0 2x[3] −x[2] +x[1] = −1
x[2] = 1 x[3] = − 3 2
k =2
2x[4] −x[3] +x[2] = −4
x[4] = − 13 4
Asthepreviousexamplesillustrate,todetermineauniquesolutiontoafirst-orderequation
requires one initial condition; for a second-order equation, two initial values are
required.
Engineeringapplication22.5
Numericalsolutionoftheoutputfromadigitallow-passfilter
Recall from Engineering application 22.2 that a low-pass filter is a filter that is designedtoattenuatehighfrequencysignals.Asimpledifferenceequationforadigital
filter isgiven by:
y[n] −ay[n −1] =x[n]
Ifaispositive thenthe filter isalow-passfilter. We will choosea = 0.5 and so
y[n] = 0.5y[n −1] +x[n]
Letusexaminetheresponseofthisfiltertoaunitstepinputappliedatn = 0,thatis
{ 0 n<0
x[n] = n ∈ Z
1 otherwise
Assume the output of the filter is zero prior to the application of the step input, that
isy[n] = 0 forn −1.Fromthe difference equation,wefind
Similarly,
y[0] = 0.5y[−1] +x[0]
= 0.5(0) +1
= 1
y[1] = 0.5y[0] +x[1]
= 0.5(1) +1
= 1.5
andsoon.Whennumericallysolvingadifferenceequation,itisoftenusefultoform
a table with intermediateresults. Table 22.1 shows suchatable.
Figure 22.16 shows the input and output sequences superimposed on the same
graph. The sequence values have been joined to illustrate their trends. Note that the
input signal reaches its final value immediately whereas the output signal takes several
sample intervals to reach its final value. Engineers often refer to this process as
‘smoothing’theinputsignal.WewillseeinChapter23thatrapidlychangingsignals

22.6 Numerical solution of difference equations 697
Table22.1
Numerical solution of a difference
equation.
n x[n] y[n −1] y[n]
−1 0 0 0
0 1 0 1
1 1 1 1.5
2 1 1.5 1.75
3 1 1.75 1.88
4 1 1.88 1.94
5 1 1.94 1.97
6 1 1.97 1.98
7 1 1.98 1.99
8 1 1.99 2.00
9 1 2.00 2.00
10 1 2.00 2.00
2
1
0 12345678910
Figure22.16
Input and output sequences forthe
low-pass filter.
y[n]
x[n]
n
tend to be richer in high frequencies than those that change more slowly. The effect
of the low-pass filter is to filter out these high frequencies and so the output from
the filter changes more slowly than the input. In Chapter 23 we will also examine a
continuous low-pass filter which does the same jobforcontinuous signals.
Numerical solution of difference equations is often the only feasible method of obtainingasolutionformanypracticalengineeringsystems.Thisisbecausetheinputterms
are usually obtained as a result of sampling an input signal and so cannot be expressed
analytically. However, in some cases it is possible to express the input analytically and
then an analytical solution to the difference equation may be feasible. While analytical
methods analogous to those applied to differential equations are available,ztransform
techniquesaremorepopularwithengineersandsotheseareintroducedinthefollowing
sections.
EXERCISES22.6
1 Given
x[n+2]+x[n+1]−x[n]=2
x[0] = 3 x[1] = 5
findx[2],x[3],x[4]andx[5].
2 If
z[n]z[n −1] =n 2 z[1] = 7
findz[2],z[3]andz[4].
3 Determinex[2]andx[3]given
(a) 2x[n +2] −5x[n +1] = 4n,x[1] = 2
(b) 6x[n] −x[n −1] +2x[n −2]
=n 2 −n,x[0]=1,x[1]=2
(c) 3x[n −1] +x[n −2] −9x[n −3]
= (n−1) 2 ,x[0] =3,x[1] =2
4 Calculate the firstfive termsofthe following
difference equations with the given initialconditions:
(a) x[n +1] −x[n] = 2,x[0] = 3
(b) x[n +2] +x[n +1] −x[n] = 4,
x[0] = 5,x[1] = 7
(c) x[p+2]−x[p+1]+2x[p] =2,
x[0] = 1,x[1] = 1

698 Chapter 22 Difference equations and the z transform
Solutions
1 0, 7, −5, 14
4
2
7 , 63
4 , 64
63
3 (a) 5, 29
2
(b)
1
3 , 7 18
(c)
29
3 , 52
9
4 (a) 3,5,7, 9,11
(b) 5,7,2, 9, −3
(c) 1,1,1, 1,1
TechnicalComputingExercises22.6
1 Writeacomputer program to implementthe
difference equations given in Engineeringapplication
22.5. Verifythat your resultsagreepreciselywith
those given in Table 22.1.
2 Modifyyour program sothat the input sequence is
x[n] = [0 1 0 0 0 0 0 0 0 0 0 0].Theoutput
is known astheimpulseresponse ofthe system.
Observe the shape ofthe plot andcompareitto the
output sequence in the Example.
3 Now change your program sothatitwill generate a
total of100 samples ofoutputandchange the input
sequence to:
x[n] = sin(2πn/100)
You should observe asinewave on the outputy(n).
Makeanoteofits amplitude.
4 Investigatewhat happensto the amplitudeofthe
outputwhenyou increasethe frequency ofthe input
sinewave. The frequency can be increased by
reducing the value ofthe denominatorin the sine
function, forexample,
x[n] = sin(2πn/10)
5 Does the result agreewith the description ofthe
operation ofthe discrete-time filterdescribedin the
example?
22.7 DEFINITIONOFTHEzTRANSFORM
Supposewehaveasequence f[k],k ∈ N.Suchasequencemayhavearisenbysampling
a continuoussignal. We define itsztransformtobe
∞∑
F(z) = Z{f[k]} = f[k]z −k (22.7)
k=0
Weseefromthedefinitionthattheztransformisaninfiniteseriesformedfromtheterms
ofthe sequence.Explicitly, wehave
Z{f[k]}=f[0]+ f[1]
z
+ f[2]
z 2 + f[3]
z 3 +···
Inmostengineeringapplicationswedonotactuallyneedtoworkwiththeinfiniteseries
since it is often possible to express this in a closed form. The closed form is generally
valid for values ofzwithin a region known as the radius of absolute convergence as
will become apparent from the following examples.

22.7 Definition of the z transform 699
Example22.8 Find theztransformof the sequence defined by
{
1 k=0
f[k]=
0 k≠0
Solution Z{f[k]} =
This sequence is sometimes called the Kronecker delta sequence, often denoted by
δ[k].
∞∑
f[k]z −k
k=0
=f[0]+ f[1]
z
=1 + 0 z + 0 z 2 + 0 z 3 +···
=1
HenceF(z) = 1.
+ f[2]
z 2 + f[3]
z 3 +···
Example22.9 Findtheztransformofthe sequencedefined by
f[k]=1
k∈N
Solution Z{f[k]} =
This istheunit stepsequence, oftendenotedbyu[k].
∞∑
f[k]z −k
k=0
=1 + 1 z + 1 z 2 + 1 z 3 +···
This is a geometric progression with first term 1 and common ratio 1 . The progression
z
converges if |z| > 1 inwhich case the sum toinfinity is
1
1−1/z = z
z −1
We see thatF(z)has the convenient closed form
F(z) =
for |z| > 1.
z
z −1
Note that the process of taking theztransform converts the sequence f[k] into the continuousfunctionF(z).
Example22.10 Findtheztransformofthesequencedefinedby f[k] =k,k ∈ N.Thissequenceiscalled
theunit ramp sequence.

700 Chapter 22 Difference equations and the z transform
Solution F(z) = Z{f[k]} =
∞∑
k=0
kz −k
= 1 z + 2 z 2 + 3 z 3 +···
= 1 z
{1 + 2 z + 3 z 2 +···}
Ifweusethebinomialtheorem(Section6.4)toexpress
(
1− 1 z
) −2
asaninfiniteseries,
we find that
(
1 − 1 ) −2
= 1 + 2 z z + 3 ∣ ∣∣∣
z + ··· provided 1
2 z ∣ < 1,that is |z| > 1
Using thisresultwe see that Z{f[k]}can be written as
F(z) = 1 (
1 − 1 ) −2
z z
and so
F(z) = 1 z
1
(1 −1/z) = z
for |z| > 1
2 (z −1)
2
Example22.11 Findtheztransform ofthe sequencedefinedby
f[k] =Ak
A constant
Solution We find
F(z) = Z{f[k]} =
=A
=
∞∑
Akz −k
k=0
∞∑
k=0
kz −k
Az
(z −1) 2 usingExample 22.10
In the same way as has been done for Laplace transforms, we can build up a library of
sequences and theirztransforms. Some common examples appear in Table 22.2. Note
that inTable22.2aandbareconstants.
Example22.12 UseTable 22.2 tofind theztransformsof
(a) sin 1 2 k
(b) e3k cos2k
Solution Directly fromTable 22.2 wefind
(a) Z{sin 1 2 k} = zsin 1 2
z 2 −2zcos 1 2 +1
(b) Z{e 3k cos2k} =
z 2 −ze 3 cos2
z 2 −2ze 3 cos2 +e 6

22.7 Definition of the z transform 701
Table22.2
Theztransformsofsome common functions.
f[k] F(z) f[k] F(z)
{ 1 k=0
δ[k] =
0 k≠0
{ 1 k0
u[k] =
0 k<0
k
e −ak
a k
ka k
k 2 a k
1 k 2 z(z +1)
(z −1) 3
z
z −1
z
(z −1) 2 sinak
z
z−e −a
z
z −a
k 3 z(z 2 +4z +1)
(z −1) 4
cosak
e −ak sinbk
az
(z −a) 2 e −ak cosbk
az(z +a)
(z −a) 3
zsina
z 2 −2zcosa+1
z(z −cosa)
z 2 −2zcosa+1
ze −a sinb
z 2 −2ze −a cosb +e −2a
z 2 −ze −a cosb
z 2 −2ze −a cosb +e −2a
EXERCISES22.7
1 Using the definition oftheztransform, find
closed-form expressions fortheztransformsofthe
followingsequences f[k] where
(a) f[0]=0,f[1]=0,f[k]=1fork 2
{ 0 k=0,1,...,5
(b) f[k] =
4 k>5
(c) f[k]=3k,k 0
(d) f[k]=e −k ,k=0,1,2,...
(e) f[0]=1,f[1]=2,f[2]=3,f[k]=0,k 3
(f) f[0]=3,f[k]=0,k ≠0
{ 2 k0
(g) f[k] =
0 k<0
2 UseTable 22.2 to find theztransformsof
(a) cos3k (b) e k (c) e −2k cosk
(d) e 4k sin2k (e) 4 k (f) (−3)
( ) ( )
k
kπ kπ
(g) sin (h) cos
2 2
3 Find, fromTable 22.2, the sequences whichhave the
followingztransforms:
z 2z 3z
(a) (b) (c)
z +4 2z−1 3z+1
z z
(d)
z−e 3 (e)
z 2 +1
4 Byconsidering the Taylor seriesexpansion ofe 1/z ,
find theztransformofthe sequence
f[k]= 1 k!
k 0
Solutions
1 (a)
(c)
1 4
(b)
z(z −1) z 5 (z−1)
3z ez
(z −1) 2 (d)
ez−1
(e)
2 (a)
z 2 +2z+3
z 2 (f) 3 (g)
z(z −cos3)
z 2 −2zcos3+1
(b)
z
z −e
2z
z −1

702 Chapter 22 Difference equations and the z transform
(c)
(d)
(e)
z 2 −ze −2 cos1
z 2 −2ze −2 cos1 +e −4
ze 4 sin2
z 2 −2ze 4 cos2 +e 8
z z
(f)
z −4 z +3
(g)
z
z 2 +1
(h)
z 2
z 2 +1
3 (a) (−4) k (b) (1/2) k (c) (−1/3) k
(d) e 3k
4 e 1/z
(e) sin(πk/2)
22.8 SAMPLINGACONTINUOUSSIGNAL
WehavealreadyintroducedsamplinginSection22.4.Wenowreturntothetopic.Most
of the signals that are encountered in the physical world are continuous in time. This
meansthattheyhaveasignallevelforeveryvalueoftimeoveraparticulartimeinterval
of interest. An example is the measured value of the temperature of an oven obtained
usinganelectronicthermometer.Thistypeofsignalcanbemodelledusingacontinuous
mathematical function in which for each value oft there is a continuous signal level,
f (t). Several engineering systems contain signals whose values are important only at
particularpointsintime.Thesepointsareusuallyequallyspacedandseparatedbyatime
interval, T. Such signals are referred to as discrete time, or more compactly, discrete
signals. They are modelled by a mathematical function that is only defined at certain
points in time. An example of a discrete system is a digital computer. It carries out
calculationsatfixed intervals governed by anelectronic clock.
Supposewehaveacontinuoussignal f (t),definedfort 0,whichwesample,that
is measure, at intervals of time, T. We obtain a sequence of sampled values of f (t),
thatis f[0], f[1], f[2],..., f[k],....Returningtotheexampleoftheoventemperature
signal, a discrete signal with a time interval of 5 seconds can be obtained by noting
the value of the electronic thermometer display every 5 seconds. Some textbooks use
the notation f[kT] as a reminder that the sequence has been obtained by sampling at
an interval T. We will not use this notation as it can become clumsy. However, it is
important to note that changing the value ofT changes theztransform as we shall see
in Example 22.13. It can be shown that sampling a continuous signal does not lose the
essence of the signal provided the sampling rate is sufficiently high, and it is in fact
possible to recreate the original continuous signal from the discrete signal, if required.
Itisoftenconvenienttorepresentadiscretesignalasaseriesofweightedimpulses.The
strengthofeachimpulseisthelevelofthesignalatthecorrespondingpointintime.We
write
f ∗ (t) =
∞∑
f[k]δ(t −kT) (22.8)
k=0
the * indicating that f (t) has been sampled. This representation is discussed in
Appendix I. This is a useful mathematical way of representing a discrete signal as the
propertiesoftheimpulsefunctionlendthemselvestoavaluethatonlyexistsforashort
interval of time. In practice, no sampling method has zero sampling time but provided
the sampling time is much smaller than the sampling interval, then this is a valid mathematical
model ofadiscrete signal.

22.8 Sampling a continuous signal 703
Wecanapplytheztransformdirectlytoacontinuousfunction, f (t),ifweregardthe
function as having been sampled at discrete intervals of time. Consider the following
example.
Example22.13 (a) Find theztransformof f (t) = e −t sampled att = 0,0.1,0.2,....
(b) Find theztransform of f (t) = e −t sampled att = 0,0.01,0.02,....
(c) Express the sequences obtained in(a)and (b) as series of weighted impulses.
Solution (a) The sequence of sampled values is
that is,
1,e −0.1 ,e −0.2 ,...
e −0.1k k ∈ NandT = 0.1
From Table 22.2 wefind theztransform ofthis sequence is
(b) The sequence of sampled values is
that is,
1,e −0.01 ,e −0.02 ,...
z
z −e −0.1.
e −0.01k k ∈ N and T = 0.01
z
From Table 22.2 we find the z transform of this sequence is
z −e−0.01. We note
that modifying the sampling interval,T, alters theztransform even though we are
dealing with the same function f (t).
(c) WhenT = 0.1 we have, from Equation (22.8),
∞∑
f ∗ (t) = e −0.1k δ(t −0.1k)
k=0
=1δ(t)+e −0.1 δ(t −0.1)+e −0.2 δ(t −0.2)+···
Notethatanadvantageofexpressingthesequenceasaseriesofweightedimpulsesis
thatinformationconcerningthetimeofoccurrenceofaparticularvalueiscontained
inthe corresponding δ term.
WhenT = 0.01, we have
∞∑
f ∗ (t) = e −0.01k δ(t −0.01k)
k=0
= 1δ(t) +e −0.01 δ(t −0.01) +e −0.02 δ(t −0.02) + ···
Example22.14 (a) Findtheztransform ofthe continuousfunction f (t) = cos3t sampledatt =kT,
k∈N.
(b) WritedownthefirstfourtermsofthesampledsequencewhenT = 0.2,andexpress
the sequenceasaseries ofweighted impulses.

704 Chapter 22 Difference equations and the z transform
Solution (a) The sampled sequence is
f[k] = cos3kT = cos((3T)k)
The z transform of this sequence can be obtained directly from Table 22.2 from
which we have
Z{cosak} =
Writinga = 3T wefind
Z{cos3kT} =
z(z −cosa)
z 2 −2zcosa+1
z(z −cos3T)
z 2 −2zcos3T +1
(b) WhenT = 0.2,the first four termsare
1 cos0.6 cos1.2 cos1.8
From Equation (22.8), the sequence of sampled values can be represented as the
following series of weighted impulses:
f ∗ (t) =
=
∞∑
cos3kTδ(t −kT)
k=0
∞∑
cos0.6kδ(t −0.2k)
k=0
= δ(t) +cos0.6δ(t −0.2) +cos1.2δ(t −0.4) +cos1.8δ(t −0.6) + ···
22.9 THERELATIONSHIPBETWEENTHEzTRANSFORM
ANDTHELAPLACETRANSFORM
We have defined theztransform quite independently of any other transform. However,
there is a close relationship between the z transform and the Laplace transform, the z
being regarded as the discrete equivalent of the Laplace. This can be seen from the following
argument.
Ifthecontinuoussignal f (t)issampledatintervalsoftime,T,weobtainasequence
of sampled values f[k], k ∈ N. From Section 22.8 we note that this sequence can be
regarded asatrainofimpulses.
f ∗ (t) =
∞∑
f[k]δ(t −kT)
k=0
Taking the Laplace transform,wehave
L{f ∗ (t)} =
=
∫ ∞
0
∑ ∞
e −st f[k]δ(t −kT)dt
∞∑
f[k]
k=0
k=0
∫ ∞
0
e −st δ(t −kT)dt

22.9 The relationship between the z transform and the Laplace transform 705
assuming that it is permissible to interchange the order of summation and integration.
NotingfromTable21.1thattheLaplacetransformofthefunction δ(t −kT )ise −skT ,we
can write
L{f ∗ (t)} =
∞∑
f[k]e −skT (22.9)
k=0
Now, making the change of variablez = e sT , we have
L{f ∗ (t)} =
∞∑
f[k]z −k
k=0
whichisthedefinitionoftheztransform.Theexpression L{f ∗ (t)}iscommonlywritten
asF ∗ (s).
In Appendix I it is shown that the continuous function f (t) can be approximated by
multiplying the function f ∗ (t) by the sampling intervalT. Correspondingly,TF ∗ (s)is
an approximation to the Laplace transform of f (t). This is illustrated in the following
example.
Example22.15 Considerthefunction f (t) =u(t)e −t whichhasLaplacetransformF(s) = 1
s +1 .Suppose
f (t) is sampled at intervals T to give the sequence f[k] = e −kT , for
k=0,1,2....
(a) UseTable 22.2 tofind theztransform of f[k].
(b) Make the change of variablez = e sT toobtainF ∗ (s).
(c) ShowthatprovidedthesampleintervalT issufficientlysmall,TF ∗ (s)approximates
the Laplace transformF(s).
Solution (a) From Table 22.2 wefind Z{f[k]} =F(z) =
(b) Lettingz = e sT givesF ∗ (s) =
e sT givesF ∗ (s) =
1
1−e −T(1+s).
z
z−e −T.
e sT
. Dividing numerator and denominator by
e sT −e−T (c) Usingthepowerseriesexpansionfore x wecanwritee x = 1 +x+ x2
2! +···.Sowe
can approximate e −T(1+s) for sufficiently smallT as1 −T(1 +s). Hence
F ∗ (s) ≈
=
1
1−(1−T(1+s))
1
T(1+s)
andsoTF ∗ (s) ≈ 1 , thatisthe Laplace transform of f (t).
1 +s
We have illustrated the connection between the two transforms and shown how the z
transformcan beregarded as the discrete equivalent of the Laplace transform.

706 Chapter 22 Difference equations and the z transform
22.9.1 Mappingthesplanetothezplane
When designing an engineering system it is often useful to consider ansplane representationofthesystem.Thecharacteristicsofasystemcanbequicklyidentifiedbythe
positionsofthepolesandzerosaswesawinChapter21.Engineerswilloftenmodifythe
system characteristics by introducing new poles and zeros or by changing the positions
ofexistingones.Unfortunately,itisnotconvenienttousethesplanetoanalysediscrete
systems. For a sampled signalEquation (22.9) yields
∞∑
F ∗ (s)=L{f ∗ (t)} = f[k]e −skT
k=0
ThecontinuoussignalsandsystemsthatwereanalysedinChapter21hadLaplacetransformsthatweresimpleratiosofpolynomials
ins.Thiswasoneofthemainreasonsfor
usingLaplacetransformstosolvedifferentialequations;theproblemwasreducedtoone
ofreasonablystraightforwardalgebraicmanipulation.HerewehaveaLaplacetransform
thatisverycomplicated.Infactitcanhaveaninfinitenumberofpolesandzeros.Tosee
this consider the following example.
( ) πt
Example22.16 The continuous signal f (t) = cos is sampled at 1 second intervals starting from
2
t=0.
(a) Find the Laplace transformof the sampled signal f ∗ (t).
(b) Show thatF ∗ (s)has an infinity of poles.
(c) Find theztransformof the sampled signaland show thatthis has justtwo poles.
( ) πt
Solution (a) The continuous signal f (t) = cos sampled at 1 second intervals gives rise to
2
the sequence 1,0, −1,0,1,0, −1,.... Consequently, fromEquation (22.9)
∞∑
∞∑
F ∗ (s)=L{f ∗ (t)} = f[k]e −skT = f[k]e −sk sinceT = 1
thatis,
k=0
k=0
F ∗ (s)=1+0−e −2s +0+e −4s +0−e −6s +···
This is a geometric progression with common ratio −e −2s and hence its sum to
infinity is
thatis,
1
1 − (−e −2s )
F ∗ 1
(s) =
1 +e −2s
(b) PolesofF ∗ (s)willoccurwhen1 +e −2s = 0.Writings = σ +jω weseethatpoles
will occur when e −2(σ+jω) = −1. Since −1 can be written as e j(2n−1)π ,n ∈ Z (see
Chapter 9), wesee thatpoles will occur when
e −2σ−2jω = e j(2n−1)π

22.9 The relationship between the z transform and the Laplace transform 707
jv
5p —2
3
s plane
3p —2
p —2
3
3
– p —2
–3p —2
3
3
s
jy
1 3
z plane
–5p —2
3
Figure22.17
Thesampled signal has an infinite
number ofpoles.
–1 3
Figure22.18
There are two polesatz = ±j.
x
thatis,when σ = 0and ω = −(2n −1)π/2.Thusthereexistaninfinitenumberof
poles occurring when
s=−(2n−1) π 2 j
n∈Z
Some of these areillustratedinFigure 22.17.
(c) Theztransformof the sampled signalis
Z{f ∗ (t)}=1+ 0 z − 1 z 2 + 0 z 3 + 1 z 4 +···
=
1
1 − (−1/z 2 )
= z2
z 2 +1
which has justtwo poles atz = ±j asshown inFigure 22.18.
It is possible to show in general that the result of sampling a continuous signal is
to convert each simple pole of the Laplace transform into an infinite set of poles. Suppose
that the Laplace transform of the signal f (t) can be broken down by using partial
fractions into a series of n + 1 terms with simple poles a 0
,a 1
,...,a n
. For simplicity,
repeatingpoleswill notbeconsideredbut the proof forsuchacase issimilar. So,
F(s) = A 0
s−a 0
+ A 1
s−a 1
+ A 2
s−a 2
+···+
Inthe time domain thiscorresponds to
f(t) =A 0
e a 0 t +A 1
e a 1 t +···+A n
e a n t
A n
s−a n

708 Chapter 22 Difference equations and the z transform
Now we consider the Laplace transformof the sampled signal f ∗ (t):
F ∗ (s)=L{f ∗ (t)} =
=
∞∑
f[k]e −skT
k=0
∞∑
(A 0
e a 0 kT +A 1
e a 1 kT + ··· +A n
e a n kT )e −skT
k=0
∑ ∞
∑ ∞
=A 0
e −kT(s−a 0 ) +A 1
e −kT(s−a 1 ) +···
k=0
+A n
∞
∑
k=0
e −kT(s−a n )
Now each of the summations can be converted into a closed form. For example,
∞∑
e −kT(s−a 0 ) =1+e −T(s−a 0 ) +e −2T(s−a 0 ) +e −3T(s−a 0 ) +···
k=0
Therefore,
F ∗ (s) =
1
=
1−e −T(s−a 0 )
k=0
A 0
1−e + A 1
−T(s−a 0 ) 1−e +···+ A n
−T(s−a 1 ) 1−e −T(s−a n )
It is possible to show that for each simple pole in F(s) there is now an infinite set of
poles. Consider the pole ats =a 0
. This contributes the term
A 0
1−e −T(s−a 0 )
toF ∗ (s). This term has poles whenever 1 − e −T(s−a 0 ) = 0, that is e −T(s−a 0 ) = 1. This
corresponds toT(s −a 0
) = 2πmj,m ∈ Z. Therefore,
T(s−a 0
)=2πmj
s−a 0
= 2πm
T j
s=a 0
+ 2πm
T j
m∈Z
The effect of sampling is to introduce an infinite set of poles. Each one is equal to the
pole of the original continuous signal but displaced by an imaginary component. This
is illustrated in Figure 22.19 for a real polea 0
. However, the proof is equally valid for
a complex conjugate pair of poles but the diagram is more cluttered and has, therefore,
not been shown.
Clearly discrete systems are not amenable tosplane design techniques. Fortunately,
thezplane can be used for analysing discrete systems in the same way that thesplane
can be used when analysing continuous systems. It is possible to map points from thes
plane to thezplane using the relationz = e sT which gives rise to the definition of thez
transform as described previously.

22.10 Properties of the z transform 709
jv
s plane
8p
3 a 0 + —– j T
6p
3 a 0 + —– j T
4p
3 a 0 + —– j T
2p
3 a 0 + —– j T
3 a 0
s
jv
s plane jy z plane
2p
3 a 0 – —– j T
4p
3 a 0 – —– j T
6p
3 a 0 – —– j T
8p
3 a 0 – —– j T
Figure22.19
The effect ofsampling isto introduce an
infinite set ofpoles.
s
Figure22.20
Theleft half ofthesplane maps to the inside ofthe unit
circle ofthezplane.
1
1
x
The great advantage of thezplane is that it eliminates the problem of infinitely repeatingpolesandzeroswhenanalysingdiscretesystems.Thiscanbeillustratedbyconsideringhow
the imaginary axis of thesplane maps tothezplane.
Referring to Figure 22.20, we see that s = σ + jω. On the imaginary axis σ = 0
andthereforez = e sT = e jωT .As ω variesbetween − π T and π T ,thelocusofzisacircle
of radius 1, centred at the origin (see Section 9.10). As ω is increased from 0 to π T the
upper half of the unit circle is traced out, while as ω is decreased from 0 to − π T , the
lower half of the unit circle is traced out. Increasing ω above π T
or decreasing it below
− π T leadstoaretracingoftheunitcircle.Inotherwords,therepeatedsplanepointsare
superimposedontopofeachother.Thisisthereasonwhythezplaneapproachismuch
simplerthan thesplane approach when analysing discrete systems.
Theztransforms of discrete signals and systems are, in many cases, simple ratios of
polynomials. We shall see shortly that this means the process of analysing difference
equations which model these signals and systems is reduced to relatively simple algebraicmanipulations.
22.10 PROPERTIESOFTHEzTRANSFORM
Because of the relationship between the two transforms we would expect that many of
the properties of the Laplace transform would be mirrored by properties of theztransform.
This isindeedthe caseand someoftheseproperties aregiven now. These are:
(1) linearity;
(2) shift theorems;
(3) the complex translation theorem.

710 Chapter 22 Difference equations and the z transform
22.10.1 Linearity
If f[k] andg[k] are two sequences then
Z{f[k] +g[k]} = Z{f[k]} + Z{g[k]}
This statement simply says that to find theztransform of the sum of two sequences we
canaddtheztransformsofthetwosequences.Ifcisaconstant,whichmaybenegative,
and f[k]isasequence, then
Z{cf[k]} =cZ{f[k]}
Together these two properties mean thattheztransform isalinear operator.
Example22.17 Find theztransform of e −k +k.
Solution From Table 22.2 wehave
Z{e −k z
} =
z−e −1
and
z
Z{k} =
(z −1) 2
Therefore,
Z{e −k +k} =
z
z−e −1 +
z
(z −1) 2
Example22.18 Findtheztransform of3k.
Solution FromTable 22.2 wehave
z
Z{k} =
(z −1) 2
Therefore,
Z{3k} =3Z{k} =3×
z
(z −1) = 3z
2 (z −1) 2
Example22.19 Findtheztransform ofthe function f (t) = 2t 2 sampledatt =kT,k ∈ N.
Solution Thesequenceofsampledvalues is
f[k] = 2(kT) 2 = 2T 2 k 2
TheztransformofthissequencecanbereaddirectlyfromTable22.2usingthelinearity
properties. We have
Z{2T 2 k 2 } =2T 2 Z{k 2 } = 2T2 z(z +1)
(z −1) 3

22.10 Properties of the z transform 711
22.10.2 Firstshifttheorem
If f[k]isasequence andF(z)isitsztransform,then
Z{f[k+i]} =z i F(z)−(z i f[0]+z i−1 f[1]+···+zf[i−1]),i ∈ N + (22.10)
Inparticular, ifi = 1 we have
Z{f[k +1]} =zF(z) −zf[0] (22.11)
Ifi=2wehave
Z{f[k+2]}=z 2 F(z)−z 2 f[0]−zf[1]
Example22.20 The sequence f[k]isdefined by
{ 0 k=0,1,2,3
f[k]=
1 k=4,5,6,...
Writedown the sequence f[k +1] and verify that
Z{f[k+1]}=zF(z)−zf[0]
whereF(z)istheztransform of f[k].
Solution The graph of f[k] is illustrated in Figure 22.21. The sequence f[k + 1] is defined as
follows:
Whenk = 0, f[k +1] = f[1]which is0.
Whenk = 1, f[k +1] = f[2]which is0.
Whenk = 2, f[k +1] = f[3]which is0.
Whenk = 3, f[k +1] = f[4]which is1,and soon.
Consequently,
{ 0 k=0,1,2
f[k+1]=
1 k=3,4,5,...
as illustrated in Figure 22.22. We see that the graph of f[k + 1] is simply that of f[k]
shiftedoneplacetotheleft.Moregenerally, f[k+i]isthesequence f[k]shiftediplaces
tothe left.Now
f [k]
1
f [k + 1]
1
0 1 2 3 4 5 6 7 8 9 k
Figure22.21
The sequence ofExample 22.25.
0 1 2 3 4 5 6 7 8 9 k
Figure22.22
AshiftedversionofthesequenceofFigure22.21.

712 Chapter 22 Difference equations and the z transform
F(z) = Z{f[k]} =
=
∞∑
f[k]z −k
k=0
∞∑
k=4
z −k
= 1 z 4 + 1 z 5 + 1 z 6 +···
= 1 z 4 (
1 + 1 z + 1 z 2 +···)
= 1 z 4 1
1−1/z
= 1 z 4 z
z −1
= 1 z 3 1
z −1
The same argument shows that
Z{f[k+1]}= 1 z 2 1
z −1
Itthen follows that
Z{f[k+1]}= 1 z 2 1
z −1 =zF(z)−zf[0]
since f[0] = 0.This illustratesthe first shift theorem.
22.10.3 Secondshifttheorem
Thefunction f (t)u(t)isdefined by
{ f(t) t0
f(t)u(t) =
0 t<0
whereu(t)istheunitstepfunction.Thefunction f (t−iT)u(t−iT ),whereiisapositive
integer,representsashifttotherightofisampleintervals.Supposethisshiftedfunction
issampled; thenweobtain
f[k−i]u[k−i]
k∈N
Thesecondshift theoremstates:
Z{f[k−i]u[k−i]} =z −i F(z) i ∈ N +
whereF(z)istheztransform of f[k].

22.10 Properties of the z transform 713
Example22.21 The functiont u(t) is sampled at intervals T = 1 to give k u[k]. This sample is then
shifted tothe right by one sampling interval togive
(k −1)u[k −1]
Find itsztransform.
Solution Figure 22.23(a) shows t u(t) and Figure 22.23(b) shows the sampled function. Figures
22.23(c) and (d) show (t − 1)u(t − 1) and (k − 1)u[k − 1], respectively. From
Table 22.2, we have
z
Z{k} =
(z −1) 2
and so, fromthe second shift theorem withi = 1,wehave
Z{(k −1)u[k −1]} =z −1 z
(z −1) = 1
2 (z −1) 2
2
2
1
1
1
1
1
t
1 2 3 k
1 2 3 t
(a) (b) (c) (d)
1 2 3 k
Figure22.23
GraphsforExample 22.21:(a)t u(t);(b)ku[k];(c) (t −1)u(t −1);(d) (k −1)u[k −1].
Example22.22 Find theztransform of the unit step functionu(t) and the shifted unit stepu(t − 2T),
sampledatintervals ofT seconds.
Solution If the function u(t) is sampled at intervals T then we are concerned with finding the
z transform of the sequence u[k]. This has been derived earlier: Z{u[k]} =
z
z −1 . If
u(t −2T) issampled, we have
{
1 k=2,3,4,...
u[k−2]=
0 otherwise
Therefore, by the second shift theorem,
Z{u[k −2]} =z −2 Z{u[k]}
=z −2 z
z −1
=
1
z(z −1)

714 Chapter 22 Difference equations and the z transform
Example22.23 Find the sequence whoseztransformis
Solution
1
z −1 = 1 z
z z
z −1 =z−1
z −1
From Table 22.2 wehave
Z{u[k]} =
z
z −1
So from the second shift property, wehave
Z{u[k −1]} =z −1 z
z −1
1
z −1 .
The required sequence isthereforeu[k −1].
1
Example22.24 Findthe sequencewhoseztransformis
z 2 (z −1) 2.
1
Solution The expression does not appear in the table of transforms, but we observe
z 2 (z −1)
2
that
1
z 2 (z −1) = 1 z
2 z 3 (z −1) 2
z
and does appear. Itfollows from Table 22.2 that
(z −1)
2
z
Z{k} =
(z −1) 2
From the second shift property,z −3 z
istheztransform of (k −3)u[k −3].
(z −1)
2
22.10.4 Thecomplextranslationtheorem
Z{e −bk f[k]} =F(e b z) whereF(z)istheztransform of f[k].
Example22.25 Given thattheztransform ofcos(ak) is
z(z −cosa)
z 2 −2zcosa+1
find theztransform of e −2k cos(ak).
Solution Since b = 2, the complex translation theorem states that we replace z by e 2 z in the z
transformF(z).
F(e 2 z) =
e2 z(e 2 z −cosa)
e 4 z 2 −2e 2 zcosa+1
istherefore the required transform.

22.11 Inversion of z transforms 715
EXERCISES22.10
1 UseTable 22.2 to find theztransformsof
(a)3(4) k +7k 2 ,k 0
(b)3e −k sin4k−k,k 0
2 Find theztransformsofthe following continuous
functionssampledatt =kT,k ∈ N:
(a) t 2 (b) 4t (c) sin2t
(d) u(t −4T )
(e) e 3t
3 Find theztransformof (k −3)u[k −3]bydirectuse
ofthe definition oftheztransform.Hence verifythe
resultofExample 22.24.
4 Provethat theztransformofe −at f (t)isF(e aT z).
5 Prove the firstandsecond shifttheorems.
6 Use the complex translation theoremto findthez
transformsof
(a) ke −bk
(b) e −k sink
7 If f[k] = 4(3) k find Z{f[k]}.Usethefirstshift
theorem to deduce Z{f[k +1]}.Show that
Z{f[k+1]}−3Z{f[k]}=0.
8 Write down the firstfive terms ofthe sequence
definedby f[k] = 4(2) k−1 u[k −1],k 0.Finditsz
transformdirectly, andalso byusingthe second shift
theorem.
Solutions
1 (a)
(b)
2 (a)
(c)
(e)
3z 7z(z +1)
1
+
z −4 (z −1) 3
3
z 2 (z −1) 2
3ze −1 sin4
z 2 −2ze −1 cos4 +e −2 − z
ze b
(z −1) 2 6 (a)
(ze b −1) 2
T 2 z(z +1) 4Tz
ezsin1
(z −1) 3 (b)
(z −1) 2 (b)
e 2 z 2 −2ezcos1+1
zsin2T 1
z 2 (d)
4z
−2zcos2T +1 z 3 (z−1) 7
z −3 , 12z
z −3
z
z−e 3T 8 0, 4,8,16,32.
4
z −2
22.11 INVERSIONOFzTRANSFORMS
Just as it is necessary to invert Laplace transforms we need to be able to invertztransforms
and as before we can make use of tables of transforms, partial fractions and the
shift theorems.Invery complicatedcasesmore advanced techniquesarerequired.
Example22.26 IfF(z) = z +3 ,find f[k].
z −2
Solution Note that wecan writeF(z)as
z +3
z −2 = z
z −2 + 3
z −2
Thereasonforthischoiceisthatquantitieslikethefirstonther.h.s.appearinTable22.2.
Fromthis table we find
Z{2 k } = z
z −2

716 Chapter 22 Difference equations and the z transform
and write
{ } z
Z −1 z −2
= 2 k
where Z −1 denotes the inverseztransform.Also,
3
z −2 = 3 z
zz −2
= 3z −1 z
z −2
Using the second shift property wesee that
Z{3(2) k−1 u[k −1]} = 3 z
so that
{ } z +3
Z −1 z −2
z
z −2
{ } { z
= Z −1 + Z −1
z −2
3z −1
}
z
z −2
= 2 k +3(2) k−1 u[k −1]
{ 1 k=0
=
2 k +3(2) k−1 k=1,2,...
{ 1 k=0
=
2(2) k−1 +3(2) k−1 k = 1,2,...
{ 1 k=0
=
5(2) k−1 k =1,2,...
Often it is necessary to split a complicated expression into several simpler ones, using
partialfractions,beforeinversioncanbecarriedout.Also,ifweexaminetheztransform
table, Table 22.2, wenoticethat nearlyall ofthe entrieshave azterminthe numerator.
For this reason it is convenient to divide a complicated expression byzbefore splitting
it into partialfractions. We illustrate this techniqueby meansofanexample.
Example22.27 Findthe sequencewhoseztransformis
F(z) =
2z 2 −z
(z −5)(z +4)
Solution The first stage istodivide the expression forF(z)byz. We have
F(z) =
F(z) =
F(z)
z
=
2z 2 −z
(z −5)(z +4)
z(2z −1)
(z −5)(z +4)
2z−1
(z −5)(z +4)

22.11 Inversion of z transforms 717
We now split the r.h.s. expression into partial fractions using the standard techniques
discussed inSection 1.7.This gives
F(z)
z
= 1
z −5 + 1
z +4
Multiplying byzgives
F(z) =
z
z −5 + z
z +4
It is now possible to invert this expression for F(z) using the z transform table,
Table 22.2. This gives
f[k] =5 k + (−4) k
22.11.1 Directinversion
Sometimes it is possible to invert a transform F(z) directly by reading off the coefficients.
Example22.28 Find f[k]if
F(z)=1+z −1 +z −2 +z −3 +···
Solution Usingthe definitionoftheztransform wesee that
f[k]=1,1,1,...
Occasionallyitispossible torewriteF(z)toobtain the required form.
Example22.29 Use the binomial theorem to expand
sequence withztransformF(z) =
Solution Usingthe binomial theorem, wehave
(
1 − 1 ) −3 (
=1+(−3) − 1 z z
(
1 − 1 ) −3
up to the term 1 z z4. Hence find the
z 3
(z −1) 3.
)
+ (−3)(−4)
2!
+ (−3)(−4)(−5) (
− 1 3! z
= 1 + 3 z + 6 z + 10
2 z + 15
3 z +··· 4
provided |z| > 1.Since
z 3 ( ) z −1 −3 (
F(z) =
(z −1) = = 1 − 1 ) −3
3 z z
(
− 1 ) 2
z
) 3
+ (−3)(−4)(−5)(−6)
4!
(
− 1 z
) 4
+···

718 Chapter 22 Difference equations and the z transform
we have
F(z)=1+ 3 z + 6 z 2 + 10
z 3 + 15
z 4 +···
ThusF(z) can beinverted directly togive
f[k] = 1,3,6,10,15,... thatis,f[k] =
(k +2)(k +1)
2
k 0
EXERCISES22.11
1 Find the inverseztransformsofthe following:
4z z 2 +2z
(a) (b)
z −4 3z 2 −4z−7
z +1 2z 3 +z
(c)
(z −3)z 2 (d)
(z−3) 2 (z−1)
2 Find the inverseztransformsof
(a)
2z
(z −2)(z −3)
(c) 1− 2 z +
z
(z −3)(z −4)
(b)
(d)
3 Express
(e)
ez
(ez −1) 2 4 If
z 2
(z 2 − 1 9 )
2z 2
(z −1)(z −0.905)
F(z) =
(z +1)(2z −3)(z −2)
z 3
in partial fractions andhence obtain its inversez
transform.
F(z) =
find f[k].
10z
(z −1)(z −2)
Solutions
1 (a) 4(4 k )
(b)
13(7/3) k
30
− (−1)k
10
(c) u[k −2] 4(3)k−2 − δ[k −2]
3
May be written as:
(d)
3 k−2 u[k −2] +3 k−3 u[k −3]
19k(3 k )
6
+ 3u[k]
4
2 (a) −2(2 k ) +2(3 k )
+ 5(3k )
4
(b) e −k k
(c) δ[k] −2δ[k −1] +4 k −3 k
(1/3) k + (−1/3) k
(d)
2
(e) 21.05u[k] −19.05(0.905) k
3 2 − 5 z − 1 z 2 + 6 z 3
f[0]=2,f[1]=−5,f[2]=−1,
f[3]=6,f[k]=0 k4
4 10(2 k −u[k])
22.12 THEzTRANSFORMANDDIFFERENCEEQUATIONS
InChapter21wesawhowusefultheLaplacetransformcanbeinthesolutionoflinear,
constantcoefficient,ordinarydifferentialequations.Similarlytheztransformhasarole
toplayinthe solutionofdifference equations.

22.12 The z transform and difference equations 719
Example22.30 Solve the difference equationy[k +1] −3y[k] = 0,y[0] = 4.
Solution Taking theztransformof both sides of the equation we have
Z{y[k +1] −3y[k]} = Z{0} = 0
since Z{0} = 0.Usingthe properties of linearitywefind
Z{y[k +1]} −3Z{y[k]} = 0
Usingthe first shift theorem on the first of the terms on the l.h.s. weobtain
zZ{y[k]} −4z −3Z{y[k]} = 0
Writing Z{y[k]} =Y(z), thisbecomes
(z −3)Y(z) =4z
so that
Y(z)=
4z
z −3
Thefunctiononther.h.s.istheztransformoftherequiredsolution.Invertingthis,from
Table 22.2 we findy[k] = 4(3) k .
Higherorderequationsaretreatedinthe same way.
Example22.31 Solve the second-order difference equation
y[k +2] −5y[k +1] +6y[k] = 0 y[0] = 0 y[1] = 2
Solution Takingtheztransformofbothsidesoftheequationandusingthepropertiesoflinearity
wehave
Z{y[k +2]} −5Z{y[k +1]} +6Z{y[k]} = 0
Fromthe first shift theoremwehave
where
z 2 Y(z) −z 2 y[0] −zy[1] −5(zY(z) −zy[0]) +6Y(z) = 0
Y(z) = Z{y[k]}
Substitutingvalues forthe conditionsgives
Hence
so that
z 2 Y(z) −0z 2 −2z −5(zY(z) −0z) +6Y(z) = 0
(z 2 −5z +6)Y(z) =2z
Y(z) = Z{y[k]} =
z 2 Y(z) −5zY(z) +6Y(z) = 2z
2z
(z −2)(z −3)

720 Chapter 22 Difference equations and the z transform
Dividing both sides byzgives
Y(z)
z
=
2
(z −2)(z −3)
Expressing the r.h.s.inpartialfractions yields
Y(z)
z
Y(z)=
= 2
z −3 − 2
z −2
2z
z −3 − 2z
z −2
Inverting gives the solution tothe difference equation:
y[k] = 2(3 k ) −2(2 k )
EXERCISES22.12
1 Useztransformsto solvethe following difference
equations:
(a) x[k +1] −3x[k] = −6,x[0] = 1
(b) 2x[k +1] −x[k] = 2 k ,x[0] = 2
(c) x[k +1] +x[k] = 2k +1,x[0] = 0
(d) x[k +2] −8x[k +1] +16x[k] = 0,
x[0] = 10,x[1] = 20
(e) x[k +2] −x[k] = 0,x[0] = 0,x[1] = 1
2 Solvethe difference equation
x[k +2] −3x[k +1] +2x[k] = δ[k]
subjectto the conditionsx[0] =x[1] = 0.
3 Solvethe difference equation
y[k +2] +3y[k +1] +2y[k] = 0
subjectto the conditionsy[0] = 0,y[1] = 1.
4 Solvethe difference equation
x[k +2] −7x[k +1] +12x[k] =k
subjectto the conditionsx[0] = 1,x[1] = 1.
Solutions
1 (a) x[k] = 3 −2(3 k )
2 k
(b)
3 + 5(1/2)k
3
(c) k (d) 10(4 k ) −5(k4 k )
(e)
u[k] − (−1) k
2
May be expressed asx[k] =
{ 0 k even
1 kodd
2 (2 k−1 −1)u[k −1]
3 (−1) k − (−2) k
4
k
6 + 5 11
u[k] +
36 4 (3)k − 17
9 (4)k
REVIEWEXERCISES22
1 State
(i) the order
(ii) the independentvariable
(iii) the dependentvariable
(iv) whether linearornon-linear
foreachofthe following equations:
(a) x[n] +x[n −2] = 6
(b) y[k+1]+ky[k−1]−k=0
(c) (y[z] +1)y[z +1] =z 2
(d) z[n] −z[n −1] =n 2 z[n −2]

Review exercises 22 721
(e) q[k +3] + √ q[k+2]=q[k]−1
Foreach linearequation state whether itis
homogeneous orinhomogeneous.
2 Given
3(x[n +1]) 2 −2x[n] =n 2 x[0] = 2
findx[1],x[2]andx[3].
3 Rewrite eachequationsothatthe highestargumentof
the dependent variableisasspecified.
(a) 3ny[n +1] −y[n −1] =n 2 ,highest argumentof
the dependent variableisto ben.
(b) z[k +2] + (3 +k/2)z[k] = √ kz[k −1],highest
argumentofthedependentvariableistobek+1.
(c) x[3]x[n] −x[2]x[n −1] = (n +1) 2 ,highest
argumentofthedependentvariableistoben +1.
4 Find f[k]if
F(z) =
z(1 −a)
(z −1)(z −a)
5 Find the inverseztransform of
(a)
(b)
3z(z +2)
(z −2)(z −3) 2
z 2 +3z
3z 2 +2z−5
6 Thesequence δ[k −i]istheKroneckerdeltasequence
shiftediunits to the right. Finditsztransform.
7 Show thatsinak canbe writtenas
Given that
show that
e akj −e −akj
2j
Z{e −ak } =
Z{sinak} =
z
z−e −a
zsina
z 2 −2zcosa+1
Solutions
1 (a) Second order,n,x,linear, inhomogeneous (b) z[k +1] +
(3 + 1 )
2 (k−1) z[k −1]
(b) Second order,k,y,linear, inhomogeneous
= √ k−1z[k−2]
(c) First order,z,y, non-linear
(c) x[3]x[n +1] −x[2]x[n] = (n +2) 2
(d) Second order,n,z,linear, homogeneous
4 u[k] −a k
(e) Third order,k,q,non-linear
5 (a) 12(2 k ) −12(3 k ) +5k3 k
2 1.1547,1.0503,1.4260
u[k] − (−5/3) k /3
(b)
3 (a) 3(n −1)y[n] −y[n −2] = (n −1) 2 2
1
6
z i

23 Fourierseries
Contents 23.1 Introduction 722
23.2 Periodicwaveforms 723
23.3 Oddandevenfunctions 726
23.4 Orthogonalityrelationsandotherusefulidentities 732
23.5 Fourierseries 733
23.6 Half-rangeseries 745
23.7 Parseval’stheorem 748
23.8 Complexnotation 749
23.9 Frequencyresponseofalinearsystem 751
Reviewexercises23 755
23.1 INTRODUCTION
The ability to analyse waveforms of various types is an important engineering skill.
Fourieranalysisprovidesasetofmathematicaltoolswhichenabletheengineertobreak
down a wave into its various frequency components. It is then possible to predict the
effect a particular waveform may have from knowledge of the effects of its individual
frequency components. Often an engineer finds it useful to think of a signal in terms of
its frequency components rather than in terms of its time domain representation. This
alternative view is called a frequency domain representation. It is particularly useful
when trying to understand the effect of a filter on a signal. Filters are used extensively
inmanyareasofengineering.Inparticular,communicationengineersusetheminsignal
receptionequipmentforfilteringoutunwantedfrequenciesinthereceivedsignal;thatis,
removing the transmission signal to leave the audio signal. We shall begin this chapter
by reviewing the essential properties of waves before describing how breaking down
into frequency components isachieved.

23.2 Periodic waveforms 723
23.2 PERIODICWAVEFORMS
Inthischapterweshallbeconcernedwithperiodicfunctions,especiallysineandcosine
functions. Let us recall some important definitions and properties ( already discussed in
Section 3.7. The function f (t) = Asin(ωt + φ) = Asin ω t + φ )
is a sine wave of
ω
amplitudeA, angular frequency ω, frequency ω 2π , periodT = 2π ω
and phase angle φ.
The time displacement isdefined tobe φ . These quantities areshown inFigure 23.1.
ω
SimilarremarkscanbemadeaboutthefunctionAcos(ωt + φ)andtogetherthesine
andcosinefunctionsformaclassoffunctionsknownassinusoidsorharmonics.Itwill
beparticularlyimportantforwhatfollowsthatyouhavemasteredtheskillsofintegrating
these functions. The following results can be found inTable 13.1 (see page 431):
∫
sinnωtdt = − cosnωt
nω +c ∫
cosnωtdt = sinnωt
nω +c
forn=±1,±2,...
Sometimesafunctionoccursasthesumofanumberofdifferentsineorcosinecomponents
such as
f (t) = 2sin ω 1
t +0.8sin2ω 1
t +0.7sin4ω 1
t (23.1)
The r.h.s.ofEquation (23.1) isalinear combinationof sinusoids.
Note in particular that the angular frequencies of all components in Equation (23.1)
are integer multiples of the angular frequency ω 1
. Functions like these can easily be
plottedusingagraphicscalculatororcomputergraph-plottingpackage.Thecomponent
with the lowest frequency, or largest period, is 2sinω 1
t. The quantity ω 1
is called the
fundamentalangularfrequencyandthiscomponentiscalledthefundamentalorfirst
harmonic. The component with angular frequency 2ω 1
is called the second harmonic
andsoon.Inwhatfollowsallangularfrequenciesareintegermultiplesofthefundamental
angular frequency as in Equation (23.1). A consequence of this is that the resulting
function, f (t), is periodic and has the same frequency as the fundamental. Some harmonicsmaybemissing.Forexample,inEquation(23.1)thethirdharmonicismissing.
Insomecases the first harmonic may be missing.For example, if
f(t) =cos2ω 1
t +0.5cos3ω 1
t +0.4cos4ω 1
t + ···
f(t)
A
2p
T = —
v , period
Amplitude
– f v
t
–A
Figure23.1
The function: f (t) =Asin(ωt + φ).

724 Chapter 23 Fourier series
all angular frequencies are integer multiples of the fundamental angular frequency ω 1
,
which is missing. Nevertheless, f (t) has the same angular frequency as the fundamental.
A common value for ω 1
is 100π as this corresponds to a frequency of 50 Hz, the
frequency of the UK mains supply.
Example23.1 Describe the frequency and amplitude characteristics of the different harmonic components
of the function
f (t) = cos20πt +0.6cos60πt −0.2sin140πt
Solution Thefundamentalangularfrequencyis20πarisingthroughthetermcos20πt.Thiscorresponds
to a frequency of 10 Hz. This term has amplitude 1. The second, fourth, fifth
andsixthharmonicsaremissing,whilethethirdandseventhhaveamplitudes0.6and0.2,
respectively.
Example23.2 If f (t) = 2sint + 3cost, express f (t) as a single sinusoid and hence determine its
amplitude and phase.
Solution Both terms have angular frequency ω = 1. Recalling the trigonometric identity (Section
3.7)
Rcos(ωt−θ)=acosωt+bsinωt
whereR = √ a 2 +b 2 ,tanθ = b a , we see that in this caseR = √ 3 2 +2 2 = √ 13 and
tanθ = 2 , that is θ = 0.59 radians. Therefore we can express f (t)inthe form
3
f(t)= √ 13cos(t −0.59)
We see immediately that this is a sinusoid of amplitude √ 13 and phase angle −0.59
radians.
Example23.3 Findthe amplitudeand phase ofthe fundamentalcomponent ofthe function
f(t) =0.5sinω 1
t +1.5cosω 1
t +3.5sin2ω 1
t −3cos3ω 1
t
Solution Contributions to the fundamentalcomponent -- that is, that with the lowest frequency --
come from the terms 0.5sin ω 1
t and 1.5cos ω 1
t only. To find the amplitude and phase
wemustexpress theseasasinglecomponent.Usingthe trigonometricidentity
Rcos(ωt−θ)=acosωt+bsinωt
whereR = √ a 2 +b 2 ,tanθ = b a , wefind
R = √ 1.5 2 +0.5 2 = 1.58
tanθ = 0.5
1.5 = 1 thatis, θ = 0.32 radians
3
Therefore the fundamental can bewritten1.58cos(ω 1
t −0.32) and has amplitude1.58
and phase angle −0.32 radians.

23.2 Periodic waveforms 725
f(t)
1
f(t)
1
–1 0 1 2 3 t
Figure23.2
Graph forExample 23.4.
–1 0 1
Figure23.3
Graph forExample 23.5.
2 3 4 5 6 t
Many other periodic functions arise in engineering applications as well as the more familiar
harmonic waves. Remember, to be periodic the function values must repeat at
regularintervalsknownastheperiod,T.Theangularfrequency ω isgivenby ω = 2π
T .
To describe a periodic function mathematically it is sufficient to give its equation over
one full period and state that period. From this information the complete graph can be
drawn as Examples 23.4 and 23.5 show.
Example23.4 Sketch the graph ofthe periodicfunctiondefined by
f(t)=t 0t<1 period1
Solution To proceed we first sketch the graph on the given interval 0 t < 1 (Figure 23.2), and
thenusethefactthatthefunctionrepeatsregularlywithperiod1tocompletethepicture.
Example23.5 Write down a mathematical expression for the function whose graph is shown in
Figure 23.3.
Solution We first note that the interval over which the function repeats itself is 2; that is, period
= 2. It is then sufficient to describe the function over any interval of length 2. The
simplest interval to take is 0 t < 2. We note in this example that a single formula is
insufficienttodescribethefunctionfor0 t < 2sincedifferentbehaviourisexhibited
inthetwointervals0 t < 1and1 t < 2.For0 t < 1thefunctionisarampwith
slope1andpassesthroughtheorigin,thatisithasequation f (t) =t.For1 t < 2the
function value remains constant at 1. Therefore this periodic function can be described
by the expression
{
t 0t<1 period2
f(t)=
1 1t<2
EXERCISES23.2
1 Describethe frequency andamplitudecharacteristics
ofthedifferentharmoniccomponentsofthefollowing
waveforms:
(a)f(t) =3sin100πt −4sin200πt
+0.7sin300πt
(b) f (t) = sin40t −0.5cos120t
+0.3cos240t
Use agraph-plottingcomputer packageorgraphics
calculator to graphthese waveforms.

726 Chapter 23 Fourier series
2 Expresseach ofthe following functionsasasingle
sinusoidandhence findtheir amplitudes and phases.
(a) f(t)=2cost−3sint
(b) f(t) =0.5cost +3.2sint
(c) f(t) =3cos3t
(d) f(t)=2cos2t+3sin2t
3 Sketch the graphs ofthe following functions:
(a) f(t) =t 2 ,−1t 1,period2
{
0 0 t < π/2 period π
(b) f(t) =
sint π/2tπ
{
−t −2t<0 period3
(c) f(t) =
t 0t<1
f(t)
f(t)
1
1
–2 –1 1 2 t –2 –1 0 1 2 t
(a)
(b)
f(t)
1
–3 –2 1 3 4 t
(c)
Figure23.4
4 Write down mathematical expressions to describe the
functionswhose graphs are shown in Figure23.4.
Solutions
1 (a) Fundamental frequency is50 Hz, amplitude3.
Second harmonic has frequency of100 Hz,
amplitude4. Thirdharmonic has frequency of
150 Hz,amplitude0.7.
(b) Fundamental frequency is 20 Hz, amplitude 1.
π
Second harmonic is missing.Third harmonic has
frequency of 60 Hz, amplitude 0.5. Fourth and
π
fifth harmonics are missing.Sixth harmonic has
frequency of 120 Hz, amplitude 0.3.
π
√ √
2 (a) 13cos(t +0.983);amplitude = 13,
phase = 0.983
(b) 3.24cos(t +4.867);amplitude = 3.24,
phase = 4.867
(c) 3cos3t;amplitude = 3,phase = 0
√ √
(d) 13cos(2t −0.983);amplitude = 13,
phase = −0.983
{
1 0 t1
4 (a) f(t)=
0 1<t<2 period =2
{
2t 0t1/2
(b) f(t) =
2−2t 1/2<t<1
period =1
{
0 0 t1
(c) f(t) =
t/2−1/2 1<t<3
period =3
23.3 ODDANDEVENFUNCTIONS
Thefunctionssint andcost eachpossesscertainpropertieswhichcanbegeneralizedto
otherfunctions.Figure23.5showsthegraphof f (t) = cost.Itisobviousfromthegraph
that the function value at a negativet value, say − π , will be the same as the function
4
value at the corresponding positivet value, in this case + π . This is true because the
4
graph is symmetrical about the vertical axis. We can therefore state that for any value
oft,cos(−t) =cost.

23.3 Odd and even functions 727
f(t)
1
–2p
–p
– p –4 p –4
p
2p
t
Figure23.5
Thefunction: f (t) = cost.
f(t)
f(t)
3
(a)
Figure23.6
Examples ofeven functions:(a) f (t) =t 2 ;(b) f (t) = 3.
t
(b)
t
More generally, any function with the property that f (t) = f (−t) for any value of
its argument,t, issaidtobe aneven function.
If f (t) = f (−t) then f isaneven function.
In particular, the set of functions cosnωt, for any integern, is even. The graphs of all
even functions are symmetrical about the vertical axis -- or equivalently, the graph on
the left of the origin can be obtained by reflecting in the vertical axis that on the right.
Someother examples ofeven functions areshown inFigure 23.6.
Sketchingagraphshowsuptherequiredsymmetryimmediately.However,evenfunctions
can be identified by an algebraic approach as shown in Example 23.6. Given any
function f (t), weexamine f (−t)tosee if f (t) = f (−t).
Example23.6 Show that f (t) =t 2 is even.
Solution We can argue asfollows. If
f(t)=t 2
then
f(−t) = (−t) 2
=t 2
=f(t)
so that f (t)iseven, by definition.

728 Chapter 23 Fourier series
Example23.7 Test whether or notthe function f (t) = 4t 3 iseven.
Solution If
f(t)=4t 3
then
f(−t) = 4(−t) 3
= −4t 3
=−f(t)
so that f (−t) isnotequal to f (t)and therefore the given function isnot even.
Let us turn now to the graph of f (t) = sint in Figure 23.7. It is obvious from the
graph that the function value at a negativet value, say − π , will not be the same as the
4
function value at the corresponding positivet value, in this case + π . This graph is not
4
symmetricalabouttheverticalaxis.However,wecanstatesomethingelse.Thefunction
value at a negativet value ( is minus the function value at the corresponding positivet
value.Forexample,sin − π ) ( ) π
=−sin .Infact,forallvaluesoft wecanstatethat
4 4
sin(−t) = −sint. More generally, any function with the property that f (−t) = −f (t)
for all values of its argument,t, issaidtobe anodd function.
If f (−t) = −f (t) then f isanoddfunction.
In particular, the set of functions sinnωt, for any integer n, is odd. In Example 23.7,
f(t)=4t 3 , we found that f (−t) = −f (t) so this function is odd. The graph of an odd
functioncanbeobtainedbyreflectionfirstinthehorizontalaxisandtheninthevertical
axis. Somemoreexamples ofoddfunctions areshown inFigure 23.8.
Therearesomefunctionsthatareneitheroddnoreven--forexample,theexponential
function (see Chapter2).
Example23.8 Showthatanyfunction, f (t),canbeexpressedasthesumofanoddcomponentandan
even component.
Solution We can write
f(t)=f(t)+ f(−t)
2
Rearranging gives
− f(−t)
2
= f(t)
2 + f(t)
2 + f(−t) − f(−t)
2 2
f(t)= f(t)+f(−t)
2
+ f(t)−f(−t)
2

23.3 Odd and even functions 729
f(t)
t
f(t)
1
(a)
f(t)
1
– p –4
p –4
t
0
–1
t
–1
(b)
Figure23.7
The function: f (t) = sint.
Figure23.8
Examples ofodd functions:
(a)f(t) =t; { 1 t>0
(b)f(t) =
−1 t<0.
Now it is easy to check that the first term on the r.h.s. is even and the second term is
odd, so that we have expressed f (t) as the sum of an even and an odd component as
required.
Example23.9 Showthattheproductoftwoevenfunctionsisitselfanevenfunction.Determinewhether
the product of two odd functions is even or odd. Is the product of an even function and
anoddfunctioneven orodd?
Solution If f (t) andg(t) are even then f (−t) = f (t) andg(−t) = g(t). LetP(t) = f (t)g(t) be
the product of f andg. Then
P(−t) = f (−t)g(−t)
by definition
= f (t)g(t) since f andgare even
=P(t)
Therefore,P(−t) = P(t) and so the product f (t)g(t) is itself an even function. On the
otherhand, if f (t)andg(t) areboth oddwefind
P(−t) = f(−t)g(−t)
= (−f(t))(−g(t))
= f(t)g(t)
=P(t)
so thatthe product f (t)g(t)iseven.

730 Chapter 23 Fourier series
f(t)
f(t)
–a 0
a
t
Figure23.9
A typicalodd function, f (t).
–a 0
Figure23.10
Atypical even function, f (t).
a
t
If f (t)iseven andg(t) isodd, we find
P(−t) = f(−t)g(−t)
= f(t)(−g(t))
= −f(t)g(t)
= −P(t)
so the product is an odd function. These rules are obviously analogous to the rules for
multiplying positive and negative numbers.
Theresults ofExample 23.9 are summarizedthus:
(even) × (even) = even
(odd) ×(odd) =even
(even) × (odd) =odd
23.3.1 Integralpropertiesofevenandoddfunctions
Consider a typical odd function, f (t), such as that shown in Figure 23.9. Suppose we
wish to evaluate ∫ a
f (t)dt where the interval of integration [−a,a] is symmetrical
−a
abouttheverticalaxis.RecallfromChapter13thatadefiniteintegralcanberegardedas
theareaboundedbythegraphoftheintegrandandthehorizontalaxis.Areasabovethe
horizontalaxisarepositivewhilethosebelowarenegative.Weseethatbecausepositive
andnegativecontributionscancel,theintegralofanoddfunctionoveranintervalwhich
issymmetricalaboutthe vertical axiswill bezero.
Example23.10 Evaluate ∫ π
−π tcosnωtdt.
Solution The functiont is odd. The function cosnωt is even and hence the producttcosnωt is
odd. The interval [−π, π] is symmetrical about the vertical axis and hence the required
integral iszero.
Considernowatypicalevenfunction, f (t),suchasthatshowninFigure23.10.Suppose
we wish to evaluate ∫ a
f (t)dt. Clearly the area bounded by the graph and thet axis in
−a

23.3 Odd and even functions 731
the interval [−a,0] is the same as the corresponding area in the interval [0,a]. Hence
wecan write
∫ a
−a
f(t)dt =2
∫ a
0
f(t)dt
Example23.11 Evaluate ∫ π
−π tsintdt.
Solution Thefunctionst andsint arebothodd,andhencetheirproductiseven.Therefore,using
the factthatthe integrand iseven we can write
∫ π
−π
tsintdt=2
So,integrating by parts,
2
∫ π
0
∫ π
0
tsintdt
∫ π
)
tsintdt=2
([−tcost] π 0 + costdt
0
= 2((−πcosπ) − (0) +[sint] π 0 )
= 2π
EXERCISES23.3
1 Determine byinspection whether each ofthe (c)f(t) =t 3 ;
(b)f(t) = t 2 +1;
functionsin Figure23.11 isodd, even orneither. (d) f(t) =cost +0.1t 2 .
f(t)
f(t)
2 Byusingthe properties ofodd andeven functions
developed in Example23.9 statewhether the
following are odd, even orneither:
(a) t 3 sin ωt (b) tcos2t
t
t (c) sint sin4t (d) cos ωt sin2ωt
(e) e t sint
(a)
(b)
f(t)
f(t) 3 Evaluate the following integrals usingthe integral
propertiesofodd andeven functionswhere
appropriate:
t
t (a) ∫ 5
−5 t3 dt (b) ∫ 5
−5 t3 cos3tdt
(c)
(d)
(c) ∫ π
−π t2 sintdt (d) ∫ 2
−2 tcosh3tdt
Figure23.11
(a)The function: f (t) = −t;
(e) ∫ 1
−1 |t|dt (f) ∫ 1
−1 t|t|dt

732 Chapter 23 Fourier series
Solutions
1 (a) odd (b) neither
(c) odd (d) even
2 (a) even (b) odd (c) even
(d) odd (e) neither
3 (a)0 (b)0
(c) 0 (d) 0
(e) 1 (f) 0
23.4 ORTHOGONALITYRELATIONSANDOTHER
USEFULIDENTITIES
RecallfromChapter16thattwofunctions f (t)andg(t)aresaidtobeorthogonalonthe
intervala t bif
∫ b
a
f(t)g(t)dt =0
Example23.12 Show that the functions cosmωt and cosnωt withm,npositive integers andm ≠n are
orthogonal on the interval − π ω t π ω .
Solution We mustevaluate
∫ π/ω
−π/ω
cosmωt cosnωtdt
Usingthetrigonometricidentity2cosAcosB = cos(A +B) +cos(A −B),wefindthe
integral becomes
∫
1 π/ω
cos(m +n)ωt +cos(m −n)ωtdt
2 −π/ω
= 1 2
[ sin(m +n)ωt
(m +n)ω
]
sin(m −n)ωt π/ω
+
(m −n)ω
−π/ω
= 0
since sin(m ±n)π = 0 for all integersm,n. It was necessary to requirem ≠ n since
otherwise the second quantity inbrackets becomes undefined.
Hence cosmωt and cosnωt are orthogonal on the given interval.
AnumberofotherfunctionsregularlyappearinginworkconnectedwithFourieranalysis
are orthogonal. The main results together with some other useful integral identities are
given inTable 23.1.

23.5 Fourier series 733
Table23.1
Someusefulintegral identities.
∫ T
sin 2nπt dt = 0 forallintegersn
0 T
∫ T
0
∫ T
0
∫ T
0
∫ T
0
∫ T
0
cos 2nπt
T
cos 2nπt
T
cos 2mπt
T
sin 2mπt
T
sin 2mπt
T
dt =0
dt=T n=0
cos 2nπt
T
sin 2nπt
T
cos 2nπt
T
n=1,2,3,...
{ 0 m≠n
dt =
T/2 m=n≠0
{ 0 m≠n
dt =
T/2 m=n≠0
dt = 0
forall integersmandn
23.5 FOURIERSERIES
We have seen that the functions sinωt, sin2ωt, sin3ωt,...,cos ωt, cos2ωt,... are
periodic.Furthermore,linearcombinationsofthemarealsoperiodic.Theyarealsoconvenient
functions to deal with because they can be easily differentiated, integrated, etc.
Theyalsopossessanotherveryusefulproperty--thatof completeness.Thismeansthat
almost any periodic function can be expressed as a linear combination of them and no
additional functions are required to do this. In other words, they can be used as buildingblockstoconstructperiodicfunctionssimplybyaddingparticularmultiplesofthem
together.
We shall see, for example, that the sawtooth waveform with period 2π, shown in
Figure 23.12, isgiven by the particular combination
f(t)=2
(sint− 1 2 sin2t + 1 3 sin3t − 1 4 sin4t + 1 sin5t−···)
5
Thisisaninfiniteserieswhichcanbeshowntoconvergeforalmostallvaluesoft tothe
function f.Thismeansthatifanyvalueoft issubstitutedintotheinfiniteseriesandthe
series is summed, the result will be the same as the value of the sawtooth function at
thatvalueoft.Thereisanexception:ift isoneofthepointsofdiscontinuitytheinfinite
series will converge tothe mean of the values toitsleft and right,that is0.
Toobtainafeelforwhatishappening considerFigure23.13. Graphs(a),(b)and(c)
showtheeffectofincludingmoreandmoretermsintheseries.Asmoretermsaretaken
f(t)
p
–p
p
3p
t
Figure23.12
Sawtooth waveform.

734 Chapter 23 Fourier series
f(t)
f(t)
f(t)
(a)
t
(b)
Figure23.13
Fourier synthesisofasawtooth waveform: (a) f (t) = 2sint;
(b)f(t) =2(sint − 1 2 sin2t + 1 3 sin3t);
(c)f(t) =2(sint − 1 2 sin2t + 1 3 sin3t − 1 4 sin4t + 1 5 sin5t).
t
(c)
t
weseethattheseriesapproachesthedesiredsawtoothwaveform.Theprocessofadding
together sinusoids to form a new periodic function is called Fourier synthesis. We see
thatthesawtoothwaveformhasbeenexpressedasaninfiniteseriesofharmonicwaves,
sint beingthefundamentalorfirstharmonic,andtherestbeingwaveswithfrequencies
thatareintegermultiplesofthefundamentalfrequency.Thisinfiniteseriesiscalledthe
Fourier series representation of f (t) and what we have succeeded in doing is to break
down f (t) into its component harmonic waveforms. In this example, only sine waves
were required to construct the function. More generally we shall need both sine and
cosine waves.
Suppose the function f (t) is defined in the interval 0 < t < T and is periodic with
periodT. Then, under certainconditions, its Fourier series isgiven by
f(t)= a 0
2 + ∞
∑
orequivalently
n=1
(
a n
cos 2nπt
T
+b n
sin 2nπt )
T
f(t)= a ∞
0
2 + ∑
(a n
cosnωt +b n
sinnωt)
n=1
(23.2)
where a n
and b n
are constants called the Fourier coefficients. These are given by the
formulae
a 0
= 2 T
a n
= 2 T
b n
= 2 T
∫ T
0
∫ T
0
∫ T
0
f(t)dt (23.3)
f(t)cos 2nπt
T
f(t)sin 2nπt
T
dt fornapositive integer (23.4)
dt fornapositive integer (23.5)
The term a 0
represents the mean value or d.c. component of the waveform (see Section15.2).ThederivationoftheseformulaeappearsinExample23.16.Itisimportantto
2

23.5 Fourier series 735
point out that the integrals in Equations (23.3), (23.4) and (23.5) can be evaluated over
anycompleteperiod,forexamplefromt = − T 2 tot =T 2 .Prudentchoiceoftheinterval
ofintegrationcanoftensaveeffort.Theexpressionappearinginther.h.s.oftheFourier
representation, Equation (23.2), is an infinite series. We list conditions, often called the
Dirichlet conditions, sufficient for the series to converge to the value of the function
f (t). The integral ∫ |f (t)|dt over a complete period must be finite, and f (t) may have
no more than a finite number of discontinuities in any finite interval. Fortunately, most
signals of interest to engineers satisfy these conditions. At a point of discontinuity the
Fourier series converges to the average of the two function values at either side of the
discontinuity.
Example23.13 Findthe Fourier series representation ofthe function with periodT = 1 given by
50
{
1 0 t<0.01
f(t)=
0 0.01 t < 0.02
Solution Thefunction f (t)isshowninFigure23.14.Notethat f (t)isdefinedtobezerobetween
t = 0.01 and t = 0.02. This means we need only consider 0 t < 0.01. Using
Equations (23.3)--(23.5) wefind
∫ 0.02
a 0
= 100 f(t)dt =100
0
a n
= 100
b n
= 100
∫ 0.01
0
∫ 0.01
0
Noting thatcosnπ = (−1) n wefind
b n
= 1
nπ (1 − (−1)n )
∫ 0.01
0
= 100[t] 0.01
0
= 1
1dt+100
∫ 0.02
[ sin100nπt
cos100nπtdt = 100
100nπ
0.01
] 0.01
= 0
[ −cos100nπt
sin100nπtdt = 100
100nπ
0
0dt
] 0.01
= − 1 (cosnπ −cos0)
nπ
0
f(t)
1
1 –––
100
T =
1 ––
50
t
Figure23.14
Graph forExample 23.13.

736 Chapter 23 Fourier series
Ifnisevenb n
=0.Ifnisoddb n
= 2 . Therefore the Fourier series representation of
nπ
f(t)is
f(t)= 1 2 + 2 (
sin100πt + sin300πt + sin500πt )
+···
π 3 5
The average value of the waveform is 1 . This is the zero frequency component or d.c.
2
value. We notethat inthisexample only odd harmonics arepresent.
Example23.14 Findthe Fourier series representation of f (t) = 1 +t, −π <t π, period2π.
Solution As usual we sketch f (t) first as this often provides insight into what follows (see Figure
23.15). HereT = 2π, ω = 1, and for convenience we shall consider the period of
integration tobe[−π,π]. UsingEquation (23.3)wefind
a 0
= 1 ∫ π
π −π1+tdt= 1 [ ] π
t + t2
π 2 −π
= 1 ) ))
((π + π2
−
(−π + π2
π 2 2
= 1 π (2π)
= 2
Similarly, using Equation (23.4) wefind
a n
= 1 π
∫ π
−π
(1+t)cosntdt
Integrating by parts gives
a n
= 1 ([
(1+t) sinnt
π n
= 1 π
] π
−π
( [ ] cosnt π )
0 +
n 2 −π
= 1 (cosnπ −cos(−nπ))
πn2 ∫ π
)
sinnt
− dt
−π n
since sin±nπ = 0
f(t)
1
–p
p
t
Figure23.15
Graph forExample23.14.

23.5 Fourier series 737
but cos(−nπ) = cosnπ and hence a n
= 0, for all positive integers n. Using Equation
(23.5) we find
∫ π
b n
= 1 (1+t)sinntdt
π −π
= 1 ([
−(1+t) cosnt
π n
= 1 π
] π
−π
∫ π
)
cosnt
+ dt
−π n
(
−(1+π) cosnπ +(1−π) cos(−nπ) [ ] sinnt π )
+
n n n 2 −π
= 1 πn (−2πcosnπ)
since sin ±nπ = 0.Hence,
b n
=− 2 n cosnπ=−2 n (−1)n
Wefindb 1
=2,b 2
=−1,b 3
= 2 ,.... Thus the Fourier series representation is given
3
fromEquation (23.2) as
f(t)=1+2sint−sin2t+ 2 3 sin3t+···
which wecan writeconcisely as
f(t)=1−
∞∑
n=1
2
n (−1)n sinnt
Example23.15 Find the Fourier series representation of the function with period 2π defined by
f(t)=t 2 ,0<t2π.
Solution Asusualwesketch f (t),asshowninFigure23.16.HereT = 2πandweshallintegrate,
forconvenience, over the interval [0,2π]. UsingEquation (23.3)wefind
a 0
= 1 π
∫ 2π
0
t 2 dt = 1 π
[ t
3
3
] 2π
0
= 8π2
3
f(t)
4p 2
2p
t
Figure23.16
Graph forExample 23.15.

738 Chapter 23 Fourier series
Using Equation (23.4) wehave
a n
= 1 π
∫ 2π
0
t 2 cosntdt
Integrating by parts, wefind
a n
= 1 π
= − 2
nπ
([
t 2sinnt
n
∫ 2π
0
] 2π
0
−
tsinntdt
∫ 2π
= − 2 ([
−t cosnt ] 2π
nπ n
0
= − 2 ( −2πcos2nπ
+
nπ n
0
2t sinnt
n
)
dt
∫ 2π
cosnt
+
0 n
)
[ sinnt
n 2 ] 2π
0
)
dt
= 4 n 2
Hencea 1
= 4,a 2
= 1,a 3
= 4 ,.... Similarly,
9
b n
= 1 π
∫ 2π
0
t 2 sinntdt
= 1 ([
−t 2cosnt
] 2π
+
π n
0 0
= 1 (− 4π2
π n cos2nπ + 2 n
= 1 π
= 1 π
∫ 2π
(− 4π2
n + 2 ([ tsinnt
n n
(− 4π2
n − 2 [
− cosnt
n 2 n
∫ 2π
0
] 2π
0
2t cosnt
n
] 2π
0
)
dt
)
tcosntdt
∫ 2π
))
sinnt
− dt
0 n
)
= − 4π n
Thus b 1
= −4π, b 2
= −2π,.... Finally, the required Fourier series representation is
given by
f(t)= 4π2
3 + (
4cost+cos2t+ 4 9 cos3t+···)
(
−π 4sint+2sin2t+ 4sin3t
3
)
+···

23.5 Fourier series 739
Example23.16 Obtain the expressions for the Fourier coefficients a 0
, a n
and b n
in Equations (23.3),
(23.4) and (23.5).
Solution Assumethat f (t)can be expressed inthe form
f(t)= a ∞
0
2 + ∑
(
a n
cos 2nπt +b
T n
sin 2nπt )
T
n=1
(23.6)
Multiplying Equation (23.6) through by cos 2mπt and integrating from 0 toT wefind
T
∫ T
0
f(t)cos 2mπt
T
dt =
∫ T
a 0
2
0
∫ T
+
0
2mπt
cos dt
T
∞∑
(
a n
cos 2nπt
T
n=1
+b n
sin 2nπt )
cos 2mπt dt
T T
If we now assume that it is legitimate to interchange the order of integration and
summation weobtain
∫ T
0
f(t)cos 2mπt
T
dt =
∫ T
a 0
0
+
2
∞∑
n=1
2mπt
cos dt
T
∫ T
0
(
a n
cos 2nπt +b
T n
sin 2nπt
T
)
cos 2mπt dt
T
The first integral on the r.h.s.iseasily shown tobe zero unlessm = 0. Furthermore, we
can use the previously found orthogonality properties (Table 23.1) to show that the rest
of the integrals on the r.h.s. vanish except for the case whenn = m, in which case the
r.h.s.reduces to a m T
2 . Consequently,
a m
= 2 T
∫ T
0
f(t)cos 2mπt
T
dt
forall positive integersm
as required. Whenm = 0 all terms on the r.h.s.except the first vanish and weobtain
so that
∫ T
0
a 0
= 2 T
f(t)dt =
∫ T
0
∫ T
0
= a 0 T
2
f(t)dt
a 0
2 dt
To obtain the formula for the b n
multiply Equation (23.6) through by sin 2mπt and
T
integrate from 0 toT.
∫ T
0
f(t)sin 2mπt
T
dt =
∫ T
a 0
2
0
∫ T
+
0
2mπt
sin dt
T
∞∑
(
a n
cos 2nπt
T
n=1
+b n
sin 2nπt )
sin 2mπt dt
T T

740 Chapter 23 Fourier series
Again assuming that it is legitimate to interchange the order of integration and summation,
weobtain
∫ T
0
f(t)sin 2mπt
T
dt =
∫ T
a 0
0
+
2
∞∑
n=1
2mπt
sin dt
T
∫ T
0
(
a n
cos 2nπt +b
T n
sin 2nπt
T
)
sin 2mπt dt
T
The first integral on the r.h.s. is easily shown to be zero. Furthermore, we can use the
properties given in Table 23.1 to show that the rest of the integrals on the r.h.s. vanish
exceptforthecasewhenn =m,inwhichcasether.h.s.reducesto b m T
2 .Hencewefind
b m
= 2 T
as required.
∫ T
0
f(t)sin 2mπt
T
dt
23.5.1 Fourierseriesofoddandevenfunctions
LetusnowconsiderwhathappenswhenwedetermineFourierseriesoffunctionswhich
are either oddoreven.
Example23.17 Findthe Fourier series forthe functionwith period2π definedby
⎧
⎪⎨ 0 −π<t<−π/2
f(t)= 4 −π/2tπ/2
⎪⎩
0 π/2<t<π
Solution As usual we sketch the function first (Figure 23.17). Inspection of Figure 23.17 shows
that the Dirichlet conditions (page 735) are satisfied. We note from the graph that the
function is symmetrical about the vertical axis; that is, it is an even function. We shall
see shortly that this fact has important implications for the Fourier series representation.
For convenience we consider the period of integration to be
Equation (23.4) becomes
[
− T 2 ,T 2
]
. Hence
a n
= 2 T
∫ T/2
−T/2
f(t)cos 2nπt
T
dt
for all positive integersn
Inthis example the periodT equals 2π. The formulafora n
then simplifies to
a n
= 1 π
∫ π
−π
f(t)cosntdt
The interval of integration is fromt = −π tot = π. However, a glance at the graph
shows that the function is zero outside the interval − π 2 t π , and takes the value 4
2

23.5 Fourier series 741
y
4
– 3π –– –π π –2
π –2
2
–
π
3π ––2
t
Figure23.17
Graph forExample 23.17.
inside. The integral thus reduces to
a n
= 1 π
∫ π/2
−π/2
4cosntdt
= 4 π
∫ π/2
−π/2
cosntdt
[ sinnt
] π/2
= 4 π n
−π/2
= 4 [sin nπ (
nπ 2 −sin − nπ 2
= 8
nπ sinnπ 2
)]
We obtaina 1
= 8 π ,a 2 =0,a 3 =−8 3π , etc. We finda 0
using Equation (23.3), again
[
integrating over − T ]
2 ,T :
2
a 0
= 2 T
= 1 π
= 4
∫ T/2
−T/2
∫ π/2
−π/2
f(t)dt
4dt = 4 π [t]π/2 −π/2
Similarly, tofind the Fourier coefficients,b n
, we use Equation (23.5):
b n
= 2 T
∫ T/2
−T/2
which reduces to
∫ π/2
f(t)sin 2nπt
T
dt
b n
= 1 4sinntdt
π −π/2
= 4 [ ] −cosnt π/2
π n
−π/2
= 4 [
−cos nπ ]
nπ 2 +cosnπ 2
= 0

742 Chapter 23 Fourier series
that is, all the Fourier coefficients,b n
, are zero. Finally we can gather together all our
results and writedown the Fourier series representation of f (t):
f(t)=2+ 8 π cost− 8
3π cos3t + 8
5π cos5t−···
In this example we see that there are no sine terms at all. In fact, whenever a function
is even its Fourier series will possess no sine terms. To see this we note thatb n
can be
found from
b n
= 2 T
∫ T/2
−T/2
f(t)sin 2nπt
T
dt
Since f (t) is even and sin 2nπt is odd, the product f (t)sin 2nπt is odd also. Now the
T
T
integral of an odd function on an interval which is symmetrical about the vertical axis
wasshowninSection23.3tobezero.Hencewhenever f (t)isevenwecanimmediately
assumeb n
= 0 for alln.
Correspondingly, when a function is odd its Fourier series will contain no cosine or
constant terms.This isbecause the product
f(t)cos 2nπt
T
isodd alsoand so the integral
a n
= 2 T
∫ T/2
−T/2
f(t)cos 2nπt
T
dt
will equal zero. We conclude that when f (t) is odd,a n
= 0 for alln. These facts can
oftenbeusedtosavetimeandeffort.KnowingthefunctioninExample23.17waseven
before we started the Fourier analysis, we could have assumed that theb n
would all be
zero.
Example23.18 FindtheFourierseriesrepresentationofthesawtoothwaveformdescribedatthebeginning
ofthissection (see Figure 23.12).
Solution This function is defined by f (t) =t, −π <t < π, and has periodT = 2π. It is an odd
function and hencea n
= 0 foralln. To find theb n
wemustevaluate
b n
= 1 π
= 1 π
∫ π
−π
tsinntdt
{[ −tcosnt
n
] π
−π
= 1 {−πcosnπ − πcosnπ}
nπ
∫ π
}
cosnt
+ dt
−π n
since the lastintegral vanishes. Therefore,
b n
=− 2 n cosnπ=−2 n (−1)n

23.5 Fourier series 743
We conclude thatb 1
= 2,b 2
= −1,b 3
= 2 ,.... Therefore f (t)has Fourier series
3
f(t)=2
{sint − 1 2 sin2t + 1 3 sin3t − 1 4 sin4t + 1 sin5t−···}
5
Engineeringapplication23.1
Spectrumofapulsewidthmodulationcontrolledsolarcharger
Pulse width modulated (PWM) signals are found in a wide variety of electronics
applications. APWMsignalissimplyarectangular wave with a fixed timeperiod,
T,wherethe‘on’timeofthesignal,knownasthemarktime,m,canbevaried.The
ratio of the ‘on’ time,m, of the signal to the period,T, is known as the duty cycle,
denotedbyd:
duty cycle (d) = ‘on’time
period = m T
The frequency of the PWM signalis f = 1 T .
Solarchargecontrollers,whichoftenusepulsewidthmodulation,aredevicesthat
protect the batteries in the system from becoming overcharged or from becoming
fully discharged, both of which can cause permanent damage. Pulse width modulation
is used to vary the charge rate to the batteries. Charge controllers and other
devicesthatemploypulsewidthmodulationgeneratehigh-frequencyharmonicsdue
totheir switching.
Consider the PWM signal shown in Figure 23.18. It has an amplitude ofAvolts.
Ithas periodT and mark timem, so thatthe duty cycle isd = m T .
Note that the function is even because of the symmetry about the vertical axis.
This means that the Fourier series will contain only cosine terms, that is all theb n
values arezero. Thea n
values can beshown (see Exercises 23.5, Question 8) tobe
a n
= 2A
nπ sin(nπd) T t
m
A
m
T
Figure23.18
A PWM signal ofperiodT andduty cyclem/T.
➔

744 Chapter 23 Fourier series
Notethatthevalueofa n
istheamplitudeofthenthharmonic.Further,thefrequency
of the nth harmonic is nf, that is integer multiples of the frequency of the PWM
signal.Agraphof |a n
|against fnisoftenreferredtoasthefrequencyspectrum.This
concept isdiscussed further inChapter 24.
Consider the specific case of a 12 V signal at a frequency of f = 300 Hz.
Figures 23.19(a)--(d) show the frequency spectra when d = 0.01,0.15,0.30 and
0.50 respectively.
It can be seen that the frequency spectrum becomes narrower for higher values
of the duty cycle. The sinc function (see Section 3.5) forms the envelope for the
spectrum, which is most clearly evident for d = 0.15. For low values of the duty
cycletheharmonicsmayextenduptoradiofrequenciesandhencethereisapotential
forinterference tooccur.
Note thatwhend = 0.5
a n
= 2A ( nπ
)
nπ sin 2
( nπ
)
When n is even the value of sin is zero and so all the even harmonics are
2
absent. In other words, when d = 0.5 the PWM signal possesses only odd
harmonics.
a n (V)
—
—
0.24
0.22
0.20
0.18
0.16
0.14
a n (V) 3.5
—
—
3.0
2.5
2.0
1.5
1.0
0.5
a n (V)
—
—
6
(a)
2 4 6 8
f (kHz)
a n (V)
—
—
(b)
2 4 6 8
f (kHz)
5
6
4
3
4
2
1
2
(c)
2 4 6 8
f (kHz)
Figure23.19
Spectrum for(a)d = 0.01, (b)d = 0.15,(c)d = 0.30,(d)d = 0.50.
(d)
2 4 6 8
f (kHz)

23.6 Half-range series 745
EXERCISES23.5
1 Find the Fourier seriesrepresentation ofthe function
{
0 −5 <t < 0 period10
f(t)=
1 0<t<5
2 Find the Fourier seriesrepresentation ofthe function
{
−t −π <t < 0 period2π
f(t)=
0 0<t<π
3 Find the Fourier seriesrepresentation ofthe function
f(t)=t 2 +πt −π<t <π period2π
4 Find the Fourier seriesrepresentation ofthe function
{
−4 −π <t 0 period 2π
f(t)=
4 0<t<π
5 Find the Fourier seriesrepresentation ofthe function
{
2(1+t) −1<t0 period2
f(t)=
0 0<t<1
6 Find the Fourier seriesrepresentation ofthe function
with period2π given by
{
t 2 0t<π
f(t)=
0 πt<2π
7 Find the Fourier seriesrepresentation ofthe function
f(t)=2sint 0<t <2π
period2π
8 Forthe signal in Engineering application 23.1, show
that
a n = 2A
nπ sin(nπd)
Solutions
1
2
3
1
2 + 2 πt
sin
π 5 + 2 3πt
sin
3π 5 + 2 5πt
sin
5π 5
+ 2
4
7πt
sin
7π 5 · · ·
π
4 − 2 π cost−sint+1 2 sin2t− 2
5
9π cos3t−1 3 sin3t...
π 2
3 +2πsint−4cost−πsin2t+cos2t+ 6
2π
3 sin3t − 4 9 cos3t···
8
(2sint sin5t+···)
+ 3 2 sin3t + 5 2
π
1
{
2 +2 (2/π)cosπt−sinπt− 1 2 sin2πt
+(2/9π)cos3πt − 1 3 sin3πt+···}/π
π 2
6 + π2 −4
sint−2cost− π π 2 sin2t + 1 2 cos2t
+ 9π2 −4
27π
sin3t − 2 9 cos3t+···
7 2sint
23.6 HALF-RANGESERIES
Sometimes an engineering function is not periodic but is only defined over a finite
interval, 0 < t < T say, as shown in Figure 23.20. In cases like this Fourier analysis
can still be useful. Because the region of interest is only that between t = 0
2
and
t = T we may choose to define the function arbitrarily outside the interval. In particular,
we can make our choice so that the resulting function is periodic, with periodT.
2
Thereismorethanonewaytoproceed.Forexample,wecanreflecttheabovefunction
in the vertical axis and then repeat it periodically so that the result is the periodic even
function shown in Figure 23.21. We have performed what is called a periodic extension
of the given function. Note that within the interval of interest nothing has altered

746 Chapter 23 Fourier series
y
y
0 T –2
t
0 T –2
t
Figure23.20
Function defined over interval0 <t < T 2 .
Figure23.21
An even periodic extension.
y
y
0 T –2
t
0 T –2
t
Figure23.22
An odd periodicextension.
Figure23.23
Aperiodic extension that is neither even nor odd.
but we have now achieved our objective of finding a periodic function. We can find
the Fourier series of this periodic function and within the interval of interest this will
converge to the required function. What happens outside this interval is not important.
Moreover, since the periodic function is even the Fourier series will contain no sine
terms.
An alternative periodic extension is that shown in Figure 23.22, which has been obtainedbyreflectinginboththeverticalandt
axesbeforerepeatingitperiodicallytogive
a periodic odd function. Its Fourier series will contain no cosine terms and within the
interval of interest will converge tothe function required.
AthirdalternativeperiodicextensionisshowninFigure23.23.However,thisextensionisneitheroddnorevenandsoithasnoneofthedesirablepropertiesoftheothertwo.
Whicheverextensionwechoose,theresultingFourierseriesonlygivesarepresentation
of the original function in the interval 0 < t < T and as such is termed a half-range
2
Fourier series. Similarly we have the terminology half-range sine series for a series
containingonlysinetermsandhalf-rangecosineseriesforaseriescontainingonlycosine
terms. The Fourier series formulae then simplify to give the following half-range
formulae:
Half-range sineseries:
a 0
= 0,a n
= 0
b n
= 4 T
∫ T/2
0
fornapositive integer
f(t)sin 2nπt
T
and f (t)isgiven by
∞∑
f(t)= b n
sin 2nπt
T
n=1
dt fornapositive integer (23.7)

23.6 Half-range series 747
Half-range cosine series:
a 0
= 4 T
a n
= 4 T
b n
= 0
∫ T/2
0
∫ T/2
and then f (t)isgiven by
0
f(t)dt (23.8)
f(t)cos 2nπt
T
fornapositive integer
f(t)= a ∞
0
2 + ∑
a n
cos 2nπt
T
n=1
dt fornapositive integer (23.9)
Example23.19 BydefininganappropriateperiodicextensionofthefunctionillustratedinFigure23.24,
find the half-range cosineseries representation.
Solution The function illustrated in Figure 23.24 is given by the formula f (t) =t for 0 <t < π
and is undefined outside this interval. Since the cosine series is required an even periodic
extension must be formed. This is illustrated in Figure 23.25. Taking T = 2π in
Equations(23.8) and (23.9),wefinda 0
anda n
.
∫ π
a 0
= 2 tdt
π 0
= 2 [ ] t
2 π
π 2
0
= π
∫ π
a n
= 2 tcosntdt
π 0
= 2 {[ ] tsinnt π ∫ π
}
sinnt
− dt
π n
0 0 n
[ cosnt
= 2 π
n 2 ] π
0
y
π
y
π
π
t
–2π
–π 0
π
2π
t
Figure23.24
Graph forExample 23.19.
Figure23.25
GraphforExample23.19.

748 Chapter 23 Fourier series
Now cosnπ = (−1) n , so that
a n
= 2 ( (−1)
n
− 1 )
π n 2 n 2
n=1,2,...
Of course, all theb n
are zero.Therefore the half-range cosine series is
f(t)= π 2 − 4 π cost− 4
9π cos3t...
and thisseries converges tothe given function within the interval 0 <t < π.
EXERCISES23.6
1 Graph anappropriate periodicextension of
f(t)=3t
0<t<π
andhence findits half-range cosineseries
representation.
2 Find the half-range sineseriesrepresentation ofthe
functiongiven in Example 23.19.
3 Find the half-range cosineseries representingthe
function
f(t)=sint
0<t<π
4 Graph anappropriateperiodicextension of
f(t)=e t
0<t<1
andfinditshalf-range cosine series.
5 Findthe half-rangesineseries representation of
f(t)=2−t,0t2.
Solutions
( )
3π
1
2 + 6 ∞∑ cosnπ−1
π n 2 cosnt
1
( )
∞∑ cosnπ
2 −2 sinnt
n
1
( )
2
3
π − 2 ∞∑ cosnπ+1
π n 2 cosnt
−1
2
( )
∞∑ ecosnπ−1
4 e−1+2
n 2 π 2 cosnπt
+1
1
5
4
π
∞∑
1
sin(nπt/2)
n
23.7 PARSEVAL’STHEOREM
Ifthefunction f (t)isperiodicwithperiodT andhasFouriercoefficientsa n
andb n
,then
Parseval’s theoremstates:
2
T
∫ T
0
(f(t)) 2 dt = 1 2 a2 0 + ∞
∑
n=1
( )
a
2
n +b2 n
Itis frequently usefulinpower calculations as the following example shows.

23.8 Complex notation 749
Engineeringapplication23.2
Averagepowerofasignal
Find the average power developed across a 1 resistor by a voltage signal with
period 2πgiven by
v(t) =cost − 1 3 sin2t + 1 2 cos3t
Solution
We note that v(t) is periodic with period T = 2π; v(t) is already expressed as a
Fourierserieswitha 1
= 1,a 3
= 1 2 andb 2 =−1 3 .AllotherFouriercoefficientsare0.
The instantaneous power is (v(t)) 2 and hence the average power over one period is
given by
∫ 2π
P av
= 1 (v(t)) 2 dt
2π 0
Therefore, using Parseval’s theorem wefind
P av
= 1 ( (
1 2 + − 1 2 ( 1 2 )
+ = 0.68 W
2 3)
2)
23.8 COMPLEXNOTATION
AnalternativenotationforFourierseriesinvolvingcomplexnumbersisavailablewhich
leadsnaturallyintothemoregeneraltopicofFouriertransforms.RecallfromChapter9
the Euler relations
e ±jθ =cosθ±jsinθ
fromwhich wecan obtain expressions forcosθ and sinθ:
cosθ = ejθ +e −jθ
2
sinθ = ejθ −e −jθ
2j
which enable us torewrite the Fourier representation
f(t)= a ∞
0
2 + ∑
(
a n
cos 2nπt +b
T n
sin 2nπt )
T
as
n=1
f(t)= a ∞
0
2 + ∑ e
(a j2nπt/T +e −j2nπt/T
n
2
n=1
= a ∞
0
2 + ∑
(
an −jb n
2
n=1
e j2nπt/T + a n +jb n
2
+b n
e j2nπt/T −e −j2nπt/T
2j
e −j2nπt/T )
)

750 Chapter 23 Fourier series
which wecan write equivalently as
∞∑
f(t)= c n
e j2nπt/T
n=−∞
where
c n
= a n −jb n
c
2 −n
= a n +jb n
n=1,2,...
2
andc 0
=a 0
/2. Itcan be shown thatthe Fourier coefficients,c n
, arethen given by
c n
= 1 T
∫ T/2
−T/2
f (t)e −j2nπt/T dt
The integral can also be evaluated over any complete period as convenient. Further, if
we writeT = 2π
ω 1
then thiscomplex formcan be expressed as
where
f(t)=
c n
= ω 1
2π
∞∑
n=−∞
∫ π/ω1
c n
e jnω 1 t
−π/ω 1
f (t)e −jnω 1 t dt
Example23.20 FindthecomplexFourierseriesrepresentationofthefunctionwithperiodT definedby
{
1 −T/4 <t <T/4
f(t)=
0 otherwise
Solution We find
c n
= 1 T
∫ T/4
−T/4
1e −j2nπt/T dt
= 1 T
[ e
−j2nπt/T
−j2nπ/T
] T/4
−T/4
Therefore,
= −1
2nπ j (e−jnπ/2 −e jnπ/2 )
= 1 ( ) e jnπ/2 −e −jnπ/2
nπ 2j
= 1
nπ sinnπ 2
f(t)=
∞∑
n=−∞
1
nπ sinnπ 2 ej2nπt/T

23.9 Frequency response of a linear system 751
The observant reader will note that the expressions forc n
appear invalid whenn = 0,
since the denominator is then zero. We can computec 0
in either of two ways. Using an
integral expression we see that
c 0
= 1 T
∫ T/4
−T/41dt = 1 2
Also,using a Taylor series expansion itispossible toshow
[ ]
1 e
−j2nπt/T T/4
lim
= 1
n→0T
−j2nπ/T
−T/4
2
giving a consistentresult.
EXERCISES23.8
1 Find the complex Fourier seriesrepresentation of
(a) f(t) =
{
1 0 <t<2
0 2<t<4
period 4
(b) f(t)=e t −1<t<1 period2
{
Asinωt 0<t <π/ω
(c) f(t) =
0 π/ω<t<2π/ω
period 2π/ω
2 If
∞∑
f(t)= c n e j2nπt/T
−∞
show thatthe coefficients,c n ,are given by
c n = 1 ∫ T
f (t)e −j2nπt/T dt
T 0
[Hint: multiply both sidesbye −j2mπt/T and integrate
over [0,T].]
Solutions
1 (a)
(b)
∞∑ j
(cosnπ −1)ejnπt/2
2nπ
−∞
(
)
1
∞∑ e −jnπ+1 −e −1+jnπ
e jnπt
2 1−jnπ
−∞
(c)
∞∑ A(1 +e −jnπ )e jnωt
2π(1 −n
−∞
2 )
23.9 FREQUENCYRESPONSEOFALINEARSYSTEM
Linear systems have the property that the response to several inputs being applied to
thesystemcanbeobtainedbyaddingtheeffectsoftheindividualinputs.Anotheruseful
property of linear systems is that if a sinusoidal input is applied to the system then the
output will also be a sinusoid of the same frequency but with modified amplitude and
phase.This isillustratedinFigure 23.26.
InSection9.7wesawthatsinusoidalsignalscanberepresentedbycomplexnumbers
and that an a.c. electrical circuit can be analysed using complex numbers. This is true

752 Chapter 23 Fourier series
A o
f
v
A i
t
Linear
system
t
Figure23.26
The response ofalinear systemto a sinusoidalinput is alsosinusoidal.
for linear systems in general. It is possible to define a complex frequency function,
G(jω), where ω is the frequency of the input; G relates the output and the input of a
linear system.
If a sinewave ofamplitudeA i
isapplied tothe systemthen the amplitude,A o
, of the
output isgiven by
A o
= |G(jω)|A i
The phase shift, φ, isgiven by
φ = ̸
G(jω)
Note that A o
and φ depend upon ω. It is important to note that G(jω) is a frequencydependent
function. Although the notation for G(jω) may seem slightly odd it arises
becauseonemethodofobtainingthefrequencyfunctionforalinearsystemistosubstitutes
= jω inthe Laplace transform transferfunction,G(s), of the system.
Itisnowpossibletoanalysetheeffectofapplyingageneralizedperiodicwaveformto
alinearsystem.ThefirststageistocalculatetheFouriercomponentsoftheinputwaveform.Theamplitudeandphaseshiftofeachoftheoutputcomponentsisthencalculated
usingG(jω). Finally, the output components are added to obtain the output waveform.
This is only possible because of the additive nature of linear systems. An example will
help toclarifythese points.
Engineeringapplication23.3
Analoguelow-passfilter
Recall from Engineering application 22.2 that a low-pass filter is a filter that has a
tendencytoattenuatehighfrequencysignals.Ifthefilterconsistsofelectroniccomponentsandoperatesonanaloguesignalsdirectlythenitisknownasananaloguefilter.Asimplecircuitthatactsasananaloguelow-passfilterisshowninFigure23.27.
Derive the frequency response for this circuit and draw graphs of its amplitude and
phase characteristics.
Consider the circuit of Figure 23.27. Using Kirchhoff’s voltage law and Ohm’s
lawweobtain
v i
=iR+v o
For the capacitor,
v o
=
i
jωC

23.9 Frequency response of a linear system 753
Eliminatingiyields
v i
= v o
jωCR+v o
= v o
(1+jωRC)
v o
v i
=
1
1 +jωRC
(23.10)
Equation (23.10) relates the output of the system to the input of the system.
Therefore,
G(jω) =
1
1 +jωRC
Itis convenient toconvertG(jω) into polar form:
1̸ 0
G(jω) = √
1 + (ωRC)
2̸
=
1
√ ̸
1 + (ωRC)
2
tan −1 ωRC
−tan −1 ωRC
Therefore,
|G(jω)| =
1
√
1+ (ωRC)
2
(23.11)
̸ G(jω) = −tan −1 ωRC (23.12)
The amplitude and phase characteristics for the circuit of Figure 23.27 are shown in
Figure 23.28. These show the variation of |G(jω)| and ̸ G(jω) with angular
frequency ω.
|G(jv)|
1
0.707
0
1/RC
v
R
G(jv)
1
v
y i
Figure23.27
Circuit forEngineering
application 23.3.
i C y o
– p –4
– p –2
Figure23.28
Amplitude andphase characteristics forthe circuit
ofFigure23.27.
Note that the circuit is a low-pass filter; it allows low frequencies to pass easily
and rejects high frequencies. The cut-off point of the filter, that is the point at
which significant frequency attenuation begins to occur, can be varied by changing ➔

754 Chapter 23 Fourier series
thevaluesofRandC.ThequantityRC isusuallyknownasthetimeconstantforthe
system.Consider the case whenRC = 0.3. Equations (23.11) and (23.12) reduce to
|G(jω)| =
1
√
1 +0.09ω
2
(23.13)
̸ G(jω) = −tan −1 0.3ω (23.14)
Letusexaminetheresponseofthissystemtoasquarewaveinputwithfundamental
angular frequency 1 and amplitude 1. This waveform is shown in Figure 23.29(a).
We note that T = 2π. The waveform function is odd and so will not contain any
cosine Fourier components. It has an average value of 0 and so will not have a zero
frequencycomponent;thatis,therewillbenod.c.component.Thereforecalculating
the Fourier components reduces toevaluating
f(t)=
∞∑
n=1
b n
sin 2πnt
T
b n
= 2 T
∫ T/2
−T/2
f(t)sin 2πnt
T
dt
forpositive integersn
SinceT = 2π, wefind
b n
= 1 (∫ 0 ∫ π
)
−1sinntdt + sinntdt
π −π 0
= 1 ([ [ ] cosnt −cosnt π )
+
π n n
0
] 0
−π
= 1 (cos0 −cosπn −cosπn +cos0)
nπ
= 1 (2 −2cosπn)
nπ
= 2 (1 −cosπn)
nπ
y i
y o
t
t
(a)
Figure23.29
(a) Input to low-pass filter;(b)outputfrom low-pass filter.
(b)

̸
̸
̸
Review exercises 23 755
The values of the first few coefficients are
b 1
= 2 π (1−cosπ)= 4 π
b 2
= 2
2π (1−cos2π)=0
b 3
= 2 4
(1 −cos3π) =
3π 3π
The next stage is toevaluate the gain and phase changes of the Fourier components.
Using Equations (23.13) and (23.14):
n = 1
ω 1
=1
|G(jω 1
)| =
1
√
1+0.09×1
= 0.96
n = 3
n = 5
G(jω 1
) = −tan −1 0.3 = −16.7 ◦
ω 3
=3
|G(jω 3
)| =
1
√
1+0.09×9
= 0.74
G(jω 3
) = −tan −1 0.9 = −42.0 ◦
ω 5
=5
|G(jω 5
)| =
1
√
1+0.09×25
= 0.55
G(jω 5
) = −tan −1 1.5 = −56.3 ◦
It is clear that high-frequency Fourier components are attenuated and phase shifted
more than low-frequency Fourier components. The effect is to produce a rounding
of the rising and falling edges of the square wave input signal. This is illustrated
in Figure 23.29(b). The output signal has been obtained by adding together the
attenuated and phase-shifted output Fourier components. This is possible because
the systemislinear.
REVIEWEXERCISES23
1 Find the half-range Fourier sineseries representation
off(t)=tsint,0tπ.
2 Find the half-range sineseries representationof
f(t)=cos2t,0tπ.
3 Find (a) the half-range sineseries,and(b)the
half-rangecosine seriesrepresentation ofthe function
defined in the interval [0, τ]by
⎧
4t
⎪⎨ τ
f(t)= ( )
4
⎪⎩ 1 − t 3 τ
0t τ 4
τ
4 tτ
4 Find the Fourier series representation ofthe function
with periodT defined by
{
V(constant) |t| <T/6
f(t)=
0 T/6|t|T/2

756 Chapter 23 Fourier series
5 Theoutput from ahalf-wave rectifier isgiven by
{
Isinωt 0<t <T/2
i(t) =
0 T/2<t<T
and isperiodic with periodT = 2π ω .
Find its Fourier seriesrepresentation.
6 Find the complex Fourier series representation ofthe
functionwith periodT = 0.02defined by
{
V(constant) 0 t < 0.01
v(t) =
0 0.01 t < 0.02
7 Find the Fourier seriesrepresentation ofthe function
with period8given by
{
2 −t 0<t<4
f(t)=
t−6 4<t<8
8 Ther.m.s.voltage, v r.m.s. ,ofaperiodicwaveform,
v(t),with periodT, isgiven by
√
∫
1 T
v r.m.s. = (v(t))
T
2 dt
0
If v(t)has Fourier coefficientsa n andb n show, using
Parseval’stheorem,that
√
1
v r.m.s. =
4 a2 0 + 1 ∞∑
(an 2 2
+b2 n )
n=1
9 If f (t)has Fourier series
f(t)= a ∞ (
0
2 + ∑
a n cos 2nπt
T
n=1
)
+b n sin 2nπt
T
prove Parseval’stheorem.
[Hint: multiply both sidesby f (t)to obtain
(f(t)) 2 = a 0 f(t)
(
∞∑
+ a
2 n f(t)cos 2nπt
T
n=1
)
+b n f(t)sin 2nπt
T
and integrate both sidesover the interval[0,T]using
Equations(23.3)--(23.5).]
10 Findthe half-rangecosine seriesandthe half-range
sineseriesforthe function
f(t)=sinhπt 0<t <1
Solutions
1
2
π
∞ 2 − ∑ 4n(1 +cosnπ)
π(n +1) 2 (n −1) 2 sin(nt)
2
2
π
3 (a)
∞∑
1
n(1 −cosnπ)
(n 2 −4)
32 ∑ ∞
3π 2
1
sin(nt)
sin(nπ/4)sin(nπt/τ)
n 2
5
6
7
I
π + I 2 sin ωt − I π
v
2π
∞∑
−∞
8 ∑ ∞
π 2
1
j cosnπ−1
n
1−cosnπ
n 2
∞∑
2
cosnπ+1
n 2 −1
e 100nπjt
cos
(
)
nπt
4
cosnωt
4
(b)
1
2 + 8 π 2
×
v
3 + 2v π
∞∑
1
∞∑
1
4cos(nπ/4) −cosnπ −3
3n 2
sin(nπ/3)cos(2nπt/T)
n
cos
(
)
nπt
τ
10 cosineseries: 1 (coshπ −1)
π
+
∞∑
n=1
2
π(1+n 2 ) ((−1)n coshπ −1)cosnπt
sineseries:
∞∑ 2n
−
π(1+n 2 ) (−1)n sinh πsinnπt
n=1

24 TheFouriertransform
Contents 24.1 Introduction 757
24.2 TheFouriertransform--definitions 758
24.3 SomepropertiesoftheFouriertransform 761
24.4 Spectra 766
24.5 Thet−ω dualityprinciple 768
24.6 Fouriertransformsofsomespecialfunctions 770
24.7 TherelationshipbetweentheFouriertransformandthe
Laplacetransform 772
24.8 Convolutionandcorrelation 774
24.9 ThediscreteFouriertransform 783
24.10 Derivationofthed.f.t. 787
24.11 Usingthed.f.t.toestimateaFouriertransform 790
24.12 Matrixrepresentationofthed.f.t. 792
24.13 Somepropertiesofthed.f.t. 793
24.14 Thediscretecosinetransform 795
24.15 Discreteconvolutionandcorrelation 801
Reviewexercises24 821
24.1 INTRODUCTION
Wehaveseenthatalmostanyperiodicsignalcanberepresentedasalinearcombination
ofsineandcosinewavesofvariousfrequenciesandamplitudes.Allfrequenciesareintegermultiplesofthefundamental.However,manypracticalwaveformsarenotperiodic.
Examples are pulse signals and noise signals. The function shown in Figure 24.1 is an
example of a non-periodic signal.

758 Chapter 24 The Fourier transform
WeshallnowseehowFouriertechniquescanstillbeusefulbyintroducingtheFourier
transformwhichisusedextensivelyincommunicationsengineeringandsignalprocessing.Forexample,itcanbeusedtoanalysetheprocessesof
modulation,whichinvolves
superimposinganaudiosignalontoacarriersignal,anddemodulation,whichinvolves
removing the carrier signaltoleave the audio signal.
24.2 THEFOURIERTRANSFORM--DEFINITIONS
f(t)
1
0
a
Figure24.1
Anon-periodic
function.
b
t
Under certain conditions it can be shown that a non-periodic function, f (t), can be
expressed notasthe sum ofsineand cosinewaves but asanintegral. Inparticular,
where
f(t)=
Provided
∫ ∞
0
A(ω) = 1 π
A(ω)cosωt +B(ω)sinωtdω (24.1)
∫ ∞
−∞
f(t)cosωtdt and B(ω) = 1 π
∫ ∞
(1) f(t)andf ′ (t) are piecewise continuous inevery finiteinterval, and
(2) ∫ ∞
|f (t)|dt exists
−∞
−∞
f (t)sinωtdt (24.2)
then the above Fourier integral representation of f (t) holds. At a point of discontinuity
of f (t) the integral representation converges to the average value of the right- and lefthand
limits. As with Fourier series, an equivalent complex representation exists which
is, infact, more commonly used:
Fourier integral representationof f (t):
where
f(t)= 1
2π
F(ω) =
∫ ∞
∫ ∞
−∞
−∞
F(ω)e jωt dω (24.3)
f (t)e −jωt dt (24.4)
Thereisnouniversalconventionconcerningthedefinitionoftheseintegralsandanumber
of variants are still correct. For instance, some authors write the factor in the
1
2π
1
second integral rather than the first, while others place a factor √ in both, giving
2π
some symmetry to the equations. There is also variation in the location of the factors
e −jωt ande jωt .Weshallusedefinitions(24.3)and(24.4)throughoutbutitisimportantto
be aware of possible differences when consulting other texts.
Equations (24.3) and (24.4) form what is called a Fourier transform pair. The
Fourier transform of f (t) is F(ω) which is sometimes written F{f (t)}. Similarly
f (t) in Equation (24.3) is the inverse Fourier transform of F(ω), usually denoted
F −1 {F(ω)}.

24.2 The Fourier transform -- definitions 759
The Fourier transform of f (t)is defined tobe
F{f(t)}=F(ω) =
∫ ∞
−∞
f (t)e −jωt dt
YouwillalsonotethesimilaritybetweenEquation(24.4)andthedefinitionoftheLaplace
transformof f (t):
L{f(t)} =
∫ ∞
0
f (t)e −st dt (24.5)
We see that, apart from the limits of integration, the substitution jω = s in Equation(24.4)resultsintheLaplacetransforminEquation(24.5).ThereisindeedanimportantrelationshipbetweenthetwotransformswhichweshalldiscussinSection24.7.We
notethatEquation(24.3)providesaformulafortheinverseFouriertransformofF(ω),
although the integral isfrequently difficulttoevaluate.
Example24.1 Find the Fourier transform of the function f (t) = u(t)e −t , whereu(t) is the unit step
function.
f(t)
1
Figure24.2
Graph ofu(t)e −t .
Solution The functionu(t)e −t is shown in Figure 24.2. Using Equation (24.4), its Fourier transformis
given by
t
thatis,
F(ω) =
=
=
∫ ∞
−∞
∫ ∞
0
∫ ∞
0
f (t)e −jωt dt
e −t e −jωt dt since f (t) = 0 fort < 0
e −(1+jω)t dt
[ e
−(1+jω)t ∞
=
−(1+jω)]
0
= 1
1+jω
F(ω) = 1
1+jω
sincee −(1+jω)t → 0ast → ∞
Example24.2 Use Equation (24.3) to find the Fourier integral representation of the function defined
by
{ 1 −1t1
f(t)=
0 |t|>1
Solution UsingEquation (24.3)wefind
f(t)= 1
2π
∫ ∞
−∞
F(ω)e jωt dω

760 Chapter 24 The Fourier transform
where
F(ω) =
=
∫ ∞
−∞
∫ 1
−1
f (t)e −jωt dt
1e −jωt dt
[ ] e
−jωt 1
=
−jω
−1
= e−jω −e jω
−jω
= ejω −e −jω
jω
Using Euler’s relation (Section 9.6)
sinθ = ejθ −e −jθ
2j
we find
F(ω) = 2sinω
ω
so that
f(t)= 1 ∫ ∞
2sinω
2π ω
−∞
since f (t)iszero outside [−1,1]
ejωt dω
istherequiredintegralrepresentation.NotethatF(ω) = 2sinω istheFouriertransform
ω
of f (t).Thefunction sin ω
ω
occursfrequentlyandisoftenreferredtoasthesincfunction
(see Section 3.5).
As with Laplace transforms, tables have been compiled for reference. Such a table of
common transforms appearsinTable 24.1.
EXERCISES24.2
1 Find the Fourier transformsof
{ 1/4 |t| 3
(a)f(t) =
0 |t|>3
⎧
1 − t 0t2
⎪⎨ 2
(b)f(t) =
1 + t −2t0
⎪⎩ 2
0 otherwise
{ e −αt t0 α>0
(c)f(t) =
e αt t<0
{ e
(d)f(t) =
−t cost t 0
0 t<0
(e) f (t) =u(t)e −t/τ
where τ isaconstant
2 Find
(a) the Fourier transform, and
(b) the Laplace transformof
f(t) =u(t)e −αt α >0
Show thatmakingthe substitutions = jω in the
Laplacetransform of f resultsin the Fourier
transform.
{ 1 |t|2
3 Iff(t)=
0 otherwise
andg(t) = e jt ,find F{f (t)g(t)}.

24.3 Some properties of the Fourier transform 761
Table24.1
CommonFourier transforms.
f(t)
f(t) =Au(t)e −αt ,α >0
{ 1 −αtα
f(t)=
0 otherwise
F(ω)
A
α+jω
2sinωα
ω
f(t) =A constant 2πAδ(ω)
( )
f(t) =u(t)A A πδ(ω)− j ω
f(t) = δ(t) 1
f(t) = δ(t −a)
f(t) =cosat
f(t) =sinat
f(t) =e −α|t| ,α>0
{ 1 t>0
f(t) =sgn(t) =
−1 t<0
f(t)= 1 t
e −jωa
π(δ(ω +a)+δ(ω −a))
π
(δ(ω −a)−δ(ω +a))
j
2α
α 2 +ω 2
2
jω
−jπsgn(ω)
f(t) =e −at2 √ π
a e−ω2 /4a
Solutions
1 (a)
(d)
sin3ω
(b)
2ω
1+jω
(1+jω) 2 +1
1−cos2ω
ω 2
τ
(e)
1+jωτ
(c)
2α
α 2 +ω 2
2 (a)
3
1
α+jω
2sin2(1−ω)
1 − ω
(b)
1
s + α
24.3 SOMEPROPERTIESOFTHEFOURIERTRANSFORM
A number of the properties of Laplace transforms that we have already discussed hold
forFourier transforms. We considerlinearityand two shifttheorems.
24.3.1 Linearity
If f andgarefunctions oft andkisaconstant, then
F{f +g}=F{f}+F{g}
F{kf}=kF{f}

762 Chapter 24 The Fourier transform
Both of these properties follow directly from the definition and linearity properties of
integrals, and mean that F isalinear operator.
Example24.3 Find F{u(t)e −t +u(t)e −2t }.
Solution We sawinExample 24.1 that
F{u(t)e −t } = 1
1+jω
Furthermore,
F{u(t)e −2t } =
Therefore,
=
∫ ∞
−∞
∫ ∞
0
u(t)e −2t e −jωt dt
e −(2+jω)t dt
[ e
−(2+jω)t ∞
=
−(2+jω)]
0
= 1
2+jω
F{u(t)e −t +u(t)e −2t } = 1
1+jω + 1
2+jω
by linearity
= 2+jω+1+jω
(1+jω)(2+jω)
=
3+2jω
2−ω 2 +3jω
24.3.2 Firstshifttheorem
IfF(ω)isthe Fourier transform of f (t),then
F{e jat f (t)} =F(ω −a)
whereaisaconstant
Example24.4 (a) Show thatthe Fourier transform of
{ 3 −2t2
f(t)=
0 otherwise
isgivenbyF(ω) = 6sin2ω .
ω
(b) Usethe first shift theorem tofind the Fourier transform of e −jt f (t).
(c) VerifythefirstshifttheorembyobtainingtheFouriertransformofe −jt f (t)directly.

∫ 2
[ ] e
Solution (a) F(ω)=3 e −jωt −jωt 2
dt = 3
−2 −jω
−2
( e −2jω −e 2jω )
=3
−jω
( e 2jω −e −2jω )
=6
2jω
= 6 ω sin2ω
24.3 Some properties of the Fourier transform 763
(b) Wehave F{f(t)} =F(ω) = 6sin2ω . Using the first shift theorem witha = −1
ω
we have
(c) e −jt f (t) =
F{e −jt f(t)} =F(ω+1) = 6 sin2(ω +1)
ω +1
{
3e
−jt
−2t2
0 otherwise
So toevaluate its Fourier transformdirectly wemustfind
F{e −jt f(t)} =3
as required.
∫ 2
−2
e −jt e −jωt dt
∫ 2
= 3 e −(1+ω)jt dt
−2
[ e
−(1+ω)jt 2
= 3
−j(1+ω)]
−2
= 6
1 + ω
( ) e 2(1+ω)j −e −2(1+ω)j
2j
= 6
1 + ω sin2(1+ω)
Example24.5 UsethefirstshifttheoremtofindthefunctionwhoseFouriertransformis
given that F{u(t)e −mt } =
Solution Fromthe given resultwe have
Now
1
m+jω ,m>0.
F{u(t)e −3t } = 1
3+jω =F(ω)
1
3+j(ω−2) =F(ω−2)
1
3+j(ω−2) ,

764 Chapter 24 The Fourier transform
Therefore, fromthe first shift theorem witha = 2 wehave
F{e 2jt u(t)e −3t } =
1
3+j(ω−2)
Consequently the function whose Fourier transform is
1
3+j(ω−2) isu(t)e−(3−2j)t .
Example24.6 Findthe Fourier transformof
{ e
−3t
t0
f(t)=
e 3t t<0
Deducethe functionwhoseFourier transformisG(ω) =
∫ ∞
Solution F(ω) = f (t)e −jωt dt
−∞
Now
=
=
∫ 0
−∞
∫ 0
−∞
e 3t e −jωt dt +
e (3−jω)t dt +
∫ ∞
0
∫ ∞
0
e −3t e −jωt dt
e −(3+jω)t dt
[ e
(3−jω)t 0 [ e
−(3+jω)t ∞
= +
3−jω]
−∞
−(3+jω)]
0
= 1
3−jω + 1
3+jω
= 6
9+ω 2
G(ω) =
6
10+2ω+ω = 6
2 (ω+1) 2 +9 =F(ω+1)
6
10+2ω+ω 2.
Then, using the first shift theorem F(ω + 1) will be F{e −jt f (t)}; that is, the required
function is
{ e
(−3−j)t
t 0
g(t) =
e (3−j)t t < 0
24.3.3 Secondshifttheorem
IfF(ω)isthe Fourier transform of f (t)then
F{f(t −α)} =e −jαω F(ω)

24.3 Some properties of the Fourier transform 765
Example24.7 Giventhatwhen f (t) =
{ 1 |t|1
0 |t|>1
tofind the Fourier transform of
g(t) =
{ 1 1t3
0 otherwise
Verifyyour resultdirectly.
2sinω
,F(ω) =
ω
,applythesecondshifttheorem
g(t)
1
1 2 3
t
Figure24.3
The function
{ 1 1t3
g(t) =
0 otherwise.
Solution The functiong(t) is depicted in Figure 24.3. Clearlyg(t) is the function f (t)translated
2unitstotheright,thatisg(t) = f (t−2).NowF(ω) = 2sinω istheFouriertransform
ω
of f (t).Therefore, by the second shift theorem
F{g(t)} = F{f(t −2)} =e −2jω F(ω) = 2e−2jω sinω
ω
To verifythis resultdirectly wemustevaluate
∫ ∞ ∫ 3
[ ] e
F{g(t)} = g(t)e −jωt dt = e −jωt −jωt 3
dt =
−∞ 1 −jω
1
as required.
= e−3jω −e −jω
−jω
= 2e−2jω sinω
ω
= e −2jω ( e −jω −e jω
−jω
)
EXERCISES24.3
1 Provethe first shift theorem.
2 Find the Fourier transform of
{
1−t 2 |t|1
f(t)=
0 |t|>1
Usethe first shift theoremto deduce the Fourier
transformsof
{
e
(a)g(t) =
3jt (1−t 2 ) |t|1
0 |t|>1
{
e
(b)h(t) =
−t (1−t 2 ) |t|1
0 |t|>1
3 Find the inverse Fourier transformsof
(a)
(b)
1
(ω+7)j+1
2
1+2(ω−1)j
4 Prove the second shift theorem.

766 Chapter 24 The Fourier transform
5 Given F{u(t)e −t } = 1 ,use the second shift
1+jω
theorem to find
F{u(t +4)e −(t+4) }
Verify your result bydirect integration.
6 Find,usingthe secondshift theorem,
{ }
F −1 6e −4jωsin2ω
ω
Solutions
2
4cosω
−ω 2 + 4sinω
ω 3
(a)
(b)
−4cos(ω −3) 4sin(ω −3)
(ω −3) 2 +
(ω −3) 3
−4cos(ω −j) 4sin(ω −j)
(ω −j) 2 +
3 (a) u(t)e −t e −7jt (b) e jt u(t)e −t/2
e 4jω
5
1+jω
6 3 for2 t 6,0otherwise
(ω −j) 3
24.4 SPECTRA
IntheFourieranalysisofperiodicwaveformswestatedthatalthoughawaveformphysicallyexistsinthetime(orspatial)domainitcanberegardedascomprisingcomponents
with a variety of temporal (or spatial) frequencies. The amplitude and phase of these
components are obtained from the Fourier coefficients a n
and b n
. This is known as a
frequency domain description. Plots of amplitude against frequency and phase against
frequency are together known as the spectrum of a waveform. Periodic functions have
discrete or line spectra; that is, the spectra assume non-zero values only at certain frequencies.
Only a discrete set of frequencies is required to synthesize a periodic waveform.Ontheotherhandwhenanalysingnon-periodicphenomenaviaFouriertransform
techniqueswefindthat,ingeneral,acontinuousrangeoffrequenciesisrequired.Instead
ofdiscretespectrawehavecontinuousspectra.ThemodulusoftheFouriertransform,
|F(ω)|,givesthespectrumamplitudewhileitsargumentarg(F(ω))describesthespectrum
phase.
Example24.8 InExample 24.2 the Fourier transformof
f(t)=
{ 1 |t|1
0 |t|>1
wasfoundtobeF(ω) = 2sinω . Sketch the spectrum of f (t).
ω
Solution F(ω) is purely real. The spectrum of f (t) is depicted by plotting |F(ω)| against ω as
sinω
illustrated in Figure 24.4. Note that lim = 1. (See Review exercises in
ω→0 ω
Chapter 18.)

24.4 Spectra 767
uF(v)u
2
–4p –3p –2p –p p 2p 3p 4p
v
Figure24.4 { 1 |t|1
Spectrum of f (t) =
0 |t|>1.
Engineeringapplication24.1
Amplitudemodulation
Amplitudemodulationisatechniquethatallowsaudiosignalstobetransmittedas
electromagnetic radio waves. The maximum frequency of audio signals is typically
10 kHz. If these signals were to be transmitted directly then it would be necessary
to use a very large antenna. This can be seen by calculating the wavelength of an
electromagnetic wave offrequency 10 kHzusingthe formula
c=fλ
Herecisthevelocityofanelectromagneticwaveinavacuum (3 ×10 8 ms −1 ), f is
the frequency of the wave and λ is its wavelength, and hence λ = c = 30000 m.
f
It can be shown that an antenna must have dimensions of at least one-quarter of the
wavelengthofthesignalbeingtransmittedifitistobereasonablyefficient.Clearlya
verylargeantennawouldbeneededtotransmita10kHzsignaldirectly.Thesolution
istohaveacarriersignalofamuchhigherfrequencythantheaudiosignalwhichis
usuallytermedthemodulationsignal.Thisallowstheantennatobeareasonablesize
as a higher frequency signal has a smaller wavelength. The arrangement for mixing
the two signals isshown inFigure 24.5.
x(t)
3
x(t)cosv c t
Antenna
cosv c t
Figure24.5
Amplitudemodulation.
Let us now derive an expression for the frequency spectrum of an amplitudemodulated
signalgiven by
φ(t) =x(t)cosω c
t
where ω c
is the angular frequency of the carrier signal and x(t) is the modulation
signal. Now φ(t) =x(t)cos ω c
t can bewritten as
φ(t) =x(t) ejω c t +e −jω c t
2
➔

768 Chapter 24 The Fourier transform
uX(v)u
uF(v)u
(a)
v
(b)
–v c
Figure24.6
Amplitude modulation. (a)Spectrum ofthe modulation signal; (b)spectrum ofthe
amplitude-modulated signal.
Taking the Fourier transformand using the first shifttheorem yields
{ x(t)(e
jω t c +e −jω c t }
)
F{φ(t)} = (ω) = F
2
{ e
jω t } {
c x(t) e
−jω t }
c x(t)
= F + F
2 2
v c
v
= 1 2 (X(ω−ω c )+X(ω+ω c ))
whereX(ω) = F{x(t)}, the frequency spectrum of the modulation signal.
Let us consider the case where the frequency spectrum, |X(ω)|, has the profile
shown in Figure 24.6(a). The frequency spectrum of the amplitude-modulated signal,
|(ω)|, is shown in Figure 24.6(b). All of the frequencies of the amplitudemodulatedsignalaremuchhigherthanthefrequenciesofthemodulationsignalthus
allowing a much smaller antenna to be used to transmit the signal. This method of
amplitude modulation is known as suppressed carrier amplitude modulation because
the carrier signal is modulated to its full depth and so the spectrum of the
amplitude-modulated signalhas no identifiable carrier component.
EXERCISES24.4
1 Show thatthe Fourier transformofthe pulse
{ t+1 −1<t<0
f(t)=
1−t 0<t<1
can be written as
Plotagraph ofthe spectrum of f (t)for
−2π ω 2π.Writedown an integral expression
forthe pulse whichwould result ifthe signal f (t)
were passed through a filterwhicheliminatesall
angularfrequenciesgreaterthan2π.
Solutions
F(ω) = 2 ω 2 (1−cosω)
1
∫
1 2π
2(1−cosω) ejωt
2π −2π ω 2 dω
24.5 THEt−ω DUALITYPRINCIPLE
We have, fromthe definition ofthe Fourier integral,
f(t)= 1 ∫ ∞
F(ω)e jωt dω (24.6)
2π
−∞

The t−ω duality principle 769
f(t)
1
F(v)
2 sint
——— t
2pf(v)
2p
–1 1
t
v
t
–1 1
v
Figure24.7
Illustrating thet--ω duality principle.
Figure24.8
Illustrating thet--ω duality principle.
where
F(ω) =
∫ ∞
−∞
f (t)e −jωt dt (24.7)
is the Fourier transform of f (t). In Equation (24.6), ω is a dummy variable so, for example,
Equation (24.6) could be equivalently written as
f(t)= 1
2π
∫ ∞
−∞
F(z)e jzt dz (24.8)
Then, from Equation (24.8),replacingt by −ω wefind
f(−ω) = 1
2π
∫ ∞
−∞
F(z)e −jωz dz = 1
2π
∫ ∞
−∞
F(t)e −jωt dt
which werecognize as 1 timesthe Fourier transformofF(t).
2π
We have the following result:
IfF(ω)isthe Fourier transformof f (t)then
f(−ω)is 1 ×(theFourier transform ofF(t))
2π
which isknown as thet--ω duality principle.
We have seen inExample 24.2 that if
{ 1 |t|1
f(t)=
0 |t|>1
thenF(ω) = 2sinω
ω
.ThisisdepictedinFigure24.7.Fromthedualityprinciplewecan
immediately deduce that
{ } 2sint
F =2πf(−ω)=2πf(ω)
t
since f is an even function (Figure 24.8). Unfortunately it is very difficult to verify this
result in most cases because while one of the integrals is relatively straightforward to
evaluate, the other is usually very difficult. However, we can use the result to derive a
number of new Fourier transforms.
Example24.9 GiventhattheFouriertransformofu(t)e −t is
the transform of
1
1+jt .
1
1+jω usethedualityprincipletodeduce

770 Chapter 24 The Fourier transform
1
Solution We know F(ω) =
1+jω is the Fourier transform of f (t) = u(t)e−t . Therefore
2π(u(−ω)e ω 1
)is the Fourier transform of
1+jt .
24.6 FOURIERTRANSFORMSOFSOMESPECIALFUNCTIONS
WesawinSection24.4thattheFouriertransformtellsusthefrequencycontentofasignal.IfweweretofindtheFouriertransformofasignalcomposedofonlyonefrequency
component, for example f (t) = sint, we would hope that the exercise of finding the
Fourier transform would resultinaspectrum containing thatsinglefrequency.
Unfortunately if we try to find the Fourier transform of, say, f (t) = sint problems
arisesince the integral
∫ ∞
−∞
sinte −jωt dt
cannot be evaluated in the usual sense because sint oscillates indefinitely as |t| → ∞.
In particular, Condition (2) of Section 24.2 fails since ∫ ∞
|sint|dt diverges. There are
−∞
many other functions which give rise to similar difficulties, for instance the unit step
function, polynomials and so on. All these functions fail to have a Fourier transform
in its usual sense. However, by making use of the delta function it is possible to make
progress even with functions like these.
24.6.1 TheFouriertransformof δ(t −a)
Example24.10 Usethe propertiesofthe deltafunction todeduceits Fourier transform.
Solution Bydefinition
F{δ(t −a)} =
∫ ∞
−∞
δ(t −a)e −jωt dt
Next, recall the following propertyofthe delta function
∫ ∞
−∞
f(t)δ(t −a)dt = f(a) (24.9)
foranyreasonablywell-behavedfunction f (t).UsingEquation(24.9)with f (t) = e −jωt
wehave
F{δ(t −a)} =
∫ ∞
−∞
e −jωt δ(t −a)dt = e −jωa
In particular, ifa = 0 wehave F{δ(t)} = 1.This resultisdepictedinFigure 24.9.

24.6 Fourier transforms of some special functions 771
f(t)
1
F(v)
1
F(t)
1
2pd(v)
2p
t
v
t
v
Figure24.9
F{δ(t)} =1.
Figure24.10
F{1} = 2πδ(ω).
Example24.11 Apply thet--ω dualityprinciple tothe previous result.Interpretthe resultphysically.
Solution We have f (t) = δ(t) andF(ω) = 1.Thedualityprinciple tells us that
f (−ω) = δ(−ω) whichequals 1
2π F{1}
thatis,
F{1} =2πδ(ω)
(since δ(−ω) = δ(ω)). This is illustrated in Figure 24.10. PhysicallyF(t) = 1 can be
regardedasad.c.waveform.Thisresultconfirmsthatad.c.signalhasonlyonefrequency
component, namely zero.
Example24.12 Given that F{δ(t −a)} = e −jωa find F{e −jta }.
Solution We have f (t) = δ(t −a),F(ω) = e −jωa . Applyingthet--ω duality principle wefind
f(−ω) = δ(−ω −a) = 1
2π F{e−jta }
Therefore
F{e −jta } = 2πδ(−ω −a)
= 2πδ(−(ω +a))
=2πδ(ω +a)
since δ(ω)isan even function.
24.6.2 Fouriertransformsofsomeperiodicfunctions
From Example 24.12 we have F{e −jta } = 2πδ(ω +a) and also, replacing a by −a,
F{e jta } = 2πδ(ω −a). Addingthesetwo expressions wefind
F{e −jta }+ F{e jta } =2π(δ(ω +a)+δ(ω −a))
Recallingthe linearityproperties of F wecan write
F{e −jta +e jta } =2π(δ(ω +a)+δ(ω −a))
and usingEuler’srelations wefind
F{cosat} = π(δ(ω +a)+δ(ω −a))

772 Chapter 24 The Fourier transform
f(t) = cos at
F(v)
p
t
–a a
v
Figure24.11
The spectrum ofcosat.
We see that the spectrum of cosat consists of single lines at ω = ±a corresponding to
a singlefrequency component (Figure 24.11).
Example24.13 Find F{sinat}.
Solution Subtractingthepreviousexpressionsfor F{e jta }and F{e −jta }andusingEuler’srelations
wefind
F{e jta }− F{e −jta } =2π(δ(ω −a)−δ(ω +a))
that is,
{ e jta −e −jta }
F = π (δ(ω −a)−δ(ω +a))
2j j
so that
F{sinat} = π (δ(ω −a)−δ(ω +a))
j
24.7 THERELATIONSHIPBETWEENTHEFOURIER
TRANSFORMANDTHELAPLACETRANSFORM
Wehavealreadynoted(Section24.2)thesimilaritybetweentheLaplacetransformand
the Fourier transform.Let usnow look atthis a littlemoreclosely. We have
F{f(t)} =
∫ ∞
−∞
f (t)e −jωt dt and L{f (t)} =
∫ ∞
0
f (t)e −st dt
InthedefinitionoftheLaplacetransform,theparametersiscomplexandwemaywrite
s=σ+jω,sothat
L{f(t)} =
∫ ∞
0
f(t)e −σt e −jωt dt
Thus an additional factor, e −σt , appears in the integrand of the Laplace transform. For
σ > 0thisrepresentsanexponentiallydecayingfactor,thepresenceofwhichmeansthat
theintegralexistsforawidervarietyoffunctionsthanthecorrespondingFourierintegral.
Example24.14 Find,ifpossible,
(a) the Laplace transform
(b) the Fourier transform
of f (t) =u(t)e 3t . Comment uponthe result.

24.7 The relationship between the Fourier transform and the Laplace transform 773
Solution (a) Either by integration, orfrom Table 21.1, we find
L{u(t)e 3t } = 1
s −3
provideds > 3
(b) F{u(t)e 3t } =
∫ ∞
0
e 3t e −jωt dt =
∫ ∞
0
[ e
(3−jω)t ∞
e (3−jω)t dt = .
3−jω]
0
Now, as t → ∞, e 3t → ∞, so that the integral fails to exist. Clearly, u(t)e 3t has a
Laplace transform butno Fourier transform.
Suppose f (t)isdefined tobe0fort < 0.Thenits Fourier transform becomes
F{f(t)} =
∫ ∞
0
f (t)e −jωt dt
and its Laplace transform is
L{f(t)} =
∫ ∞
0
f (t)e −st dt
ByreplacingsbyjωintheLaplacetransformweobtaintheFouriertransformof f (t)if
itexists.CaremustbetakenheresincewehaveseenthattheFouriertransformmaynot
exist forafunctionthatnevertheless has a Laplace transform.
Example24.15 Findthe Laplace transforms of
(a) u(t)e −2t
(b) u(t)e 2t
Lets = jω and comment uponthe result.
Solution (a) L{u(t)e−2t } = 1
s +2 .
(b) L{u(t)e 2t } = 1
s −2 .
Replacingsby jω in (a) gives
Now
F{u(t)e −2t } = 1
jω+2
1
jω+2 . Similarly, replacingsby jω in (b) gives 1
jω−2 .
sothatreplacingsbyjωintheLaplacetransformresultsintheFouriertransform.However,
F{u(t)e 2t } does not exist and even though we can lets = jω in the Laplace transformand
obtain , wecannot interpretthis asaFourier transform.
1
jω−2
TheFouriertransformdoespossesscertainadvantagesovertheLaplacetransform.While
the Laplace transform can only be applied to functions which are zero fort < 0, the

774 Chapter 24 The Fourier transform
Fourier transform is applicable to functions with domain −∞ < t < ∞. In some applications
where, for example, t represents not time but a spatial variable, it is often
necessary towork with negative values.
The inverse Fourier transformisgiven by
F −1 {F(ω)} = 1 ∫ ∞
F(ω)e jωt dω (24.10)
2π −∞
ThecorrespondinginverseLaplacetransformrequiresadvancedtechniquesinthetheory
of complex variables which are beyond the scope of this book. The existence of Equation
(24.10) is not quite as advantageous as it may seem because it is often difficult to
perform the required integration analytically.
24.8 CONVOLUTIONANDCORRELATION
Convolution is an important technique in signal and image processing. It provides a
means of calculating the response or output of a system to an arbitrary input signal if
the impulse response is known. The impulse response is the response of the system to
animpulsefunction.Convolvingtheinputsignalandtheimpulseresponseresultsinthe
response to the arbitrary input. Correlation is a second important technique. It can be
used to determine the time delay between a transmitted signal and a received signal as
might occur inradarorsonar detection equipment.
24.8.1 Convolutionandtheconvolutiontheorem
If f (t) andg(t) are two real piecewise continuous functions, their convolution, which
wedenoteby f ∗g, isdefinedasfollows:
Theconvolution of f (t)andg(t):
f∗g=
∫ ∞
−∞
f(λ)g(t − λ)dλ
f ∗gisitselfafunctionoft, andtoshowthisexplicitly wesometimes write (f ∗g)(t).
Note that convolution is an integral with respect to the dummy variable λ. In general,
as can be seen from the limits of integration, λ varies from −∞ to ∞, but in particular
caseswewillseethatthisintervalofintegrationcanbereduced.Intheexampleswhich
follow the precisemeaning ofthe terms f (λ)andg(t − λ) will become apparent.
Because convolution is commutative, that is f ∗g = g ∗ f, the convolution can be
defined equivalently as
∫ ∞
−∞
f(t − λ)g(λ)dλ
Incaseswhen f (t)andg(t)arezerofort < 0thisexpressionreducestothatdefinedfor
the Laplace transform inSection21.9.

24.8 Convolution and correlation 775
Theconvolution theoremstatesthatconvolutioninthetimedomaincorresponds to
multiplication inthe frequency domain:
The convolution theorem:
If F{f(t)} =F(ω)and F{g(t)} =G(ω)then
F{f ∗g} =F(ω)G(ω)
Thistheoremgivesusatechniqueforcalculatingtheconvolutionoftwofunctionsusing
the Fourier transform,since
f∗g=F −1 {F(ω)G(ω)}
So,itis possible tofind the convolution of f (t)andg(t), by
(1) finding the corresponding Fourier transforms,F(ω)andG(ω),
(2) multiplying these together toformF(ω)G(ω),
(3) finding the inverse Fourier transform which then yields f ∗g.
This is a process often used for finding convolutions using a computer as will be explained
inSection 24.15.
Agraphicalrepresentationofconvolutionisusefulasitcanthrowlightontheunderlyingprocessandhelpustodetermineappropriatelimitsofintegration.Wewillillustrate
thisinthe following example.
Example24.16 (a) Usingthe definition ofconvolution, calculate the convolution f ∗gwhen
f (t) =u(t)e −t andg(t) =u(t)e −2t , whereu(t) isthe unit stepfunction.
(b) Verifythe convolution theorem for these functions.
Solution (a) The convolution of f andgisgiven by
1
f(t)=u(t)e –t
g(t) = u(t)e –2t t
Figure24.12
Graphsof
f(t) =u(t)e −t and
g(t) =u(t)e −2t .
f∗g=
∫ ∞
−∞
f(λ)g(t − λ)dλ
Graphs of f (t)andg(t) areshown inFigure 24.12.
Evaluatingaconvolutionintegralcanbedifficult,sowewilldevelopthesolutioninstages.Firstofallitisnecessarytobeclearaboutthemeaningofthedifferent
termsintheintegrand.Notethatif f (t) =u(t)e −t then f (λ) =u(λ)e −λ .Similarly
g(λ) =u(λ)e −2λ .Thefunctiong(−λ)isfoundbyreflectingg(λ)intheverticalaxis
asshowninFigure24.13(a).Insignalprocessingterminologythisisalsoknownas
folding.Thefoldedgraphcanbetranslatedadistancet totheleftortotherightby
changingtheargumentofgtog(t −λ).Ift isnegativethegraphmovestotheleftas
shown in Figure 24.13(b) whereas ift is positive it moves to the right as shown in
Figure 24.13(c). In Figure 24.13 we have superimposed the graphs of f (λ), shown
dashed, andg(t − λ), fort being negative, zero and positive. Where the graphs do
notoverlap,theproduct f (λ)g(t −λ),andhencetheconvolution,mustbezero.We
examineseparatelythedomainst < 0andt 0correspondingtowherethegraphs
do notoverlap and where they do overlap respectively.

776 Chapter 24 The Fourier transform
g(–l)
f( l)= u( l)e – l
(a)
l
g(t – l) when t < 0
(b)
t
l
g(t – l) when t > 0
(c)
t
l
Figure24.13
Thefunctiong(t − λ)forvariousvalues
oft.
Whent<0
Whent < 0there isclearlyno overlap and itfollows that f ∗g= 0.
Whent0
Whent 0thereisoverlapforvaluesof λbetween0andt,thatiswhen0 λ t,
and hence
f∗g=
=
∫ t
0
∫ t
0
e −λ e −2(t−λ) dλ
e −λ e −2t e 2λ dλ
∫ t
= e −2t e λ dλ
0
= e −2t [e λ ] t 0
= e −2t (e t −1)
= e −t −e −2t
Finally the complete expression for the convolution is
(f ∗g)(t)=
This may alsobe writtenas
{ e −t −e −2t whent 0
0 whent<0
(f ∗g)(t) =u(t)(e −t −e −2t )
(b) UsingTable 24.1 the Fourier transforms of f andgare
F(ω) = 1
1+jω
G(ω) = 1
2+jω

and theirproduct is
F(ω)G(ω) =
1
(1+jω)(2+jω)
24.8 Convolution and correlation 777
The Fourier transform of the convolution is, usinglinearityand Table 24.1,
F{u(t)(e −t −e −2t )} = 1
1+jω − 1
2+jω
= (2+jω)−(1+jω)
(1+jω)(2+jω)
1
=
(1+jω)(2+jω)
(24.11)
whichisthesameasEquation(24.11).Wehaveshownthat F{f ∗g} =F(ω)G(ω)
and sothe convolution theorem has been verified.
Example24.17 (a) Usingthe definition ofconvolution find the convolution, f ∗g, ofthe ‘top-hat’
function
{ 1 −1t1
f(t)=
0 otherwise
and the functiong(t) =u(t)e −t , whereu(t) isthe unit stepfunction.
(b) Verifythe convolution theoremforthesefunctions.
Solution (a) Thefunctions f (t)andg(t) are shown inFigure 24.14.
Theconvolution of f (t)andg(t) isgiven by
f∗g=
∫ ∞
−∞
f(λ)g(t − λ)dλ
Note that since g(t) = u(t)e −t , it follows that g(λ) = u(λ)e −λ as shown in Figure
24.15(a). The function g(−λ) is found by reflecting, or folding, g(λ) in the
vertical axis. This folding is shown in Figure 24.15(b). The folded graph can be
translated a distancet to the left or to the right by changing the argument of g to
g(t − λ).Ift isnegativethegraphinFigure24.15(b)movestotheleft,whereasift
ispositive itmoves tothe right.Study Figures24.15(c--g)toobserve this.
Convolution is the integral of the product of f (λ) andg(t − λ). We have superimposed
f (λ) on the graphs in Figure 24.15. For values of λ where the graphs do
notoverlap, thisproduct mustbezero.
Inspectionofthegraphsshowsthatwhent islessthan −1(Figure24.15(c))there
isno overlap and hence f (λ)g(t − λ) = 0.So:
ift<−1
f∗g=0
f(t)
1
g(t) =
u(t) e – t
–1 1
t
t
Figure24.14
The‘top-hat’ function f (t),
andg(t) =u(t)e −t .

778 Chapter 24 The Fourier transform
1
g( l)
(a)
l
g(– l)
1
(b)
l
g(t– l ) when t< –1
1
(c)
t
–1
1
l
g(t– l ) when t = –1
(d)
–1
1
l
g(t– l ) when – 1ø t ø 1
(e)
–1
t
1
l
g(t– l )whent= 1
(f)
–1
1
l
g(t– l ) when t . 1
(g)
–1
1
t
l
Figure24.15
Thefunctiong(t − λ) forvarious
values oft.

24.8 Convolution and correlation 779
Whent isgreaterthan −1butlessthan1,asinFigure24.15(e),thereisanoverlap,
andhenceanon-zeroproduct.Thisoccursforvaluesof λbetween −1andt,thatis
in the interval −1 λ t. Within this interval f (λ) = 1 andg(t − λ) = e −(t−λ) .
So:
if −1t<1
f∗g=
∫ t
−1
∫ t
e −(t−λ) dλ = e −t e λ dλ = e −t [e λ ] t −1 = e−t [e t −e −1 ] = 1 −e −1−t
−1
Whent is greater than 1 the graphs overlap, but only for values of λ between −1
and 1,thatisfor −1 λ < 1.So:
ift>1
f∗g=
∫ 1
−1
e −(t−λ) dλ = e −t [e λ ] 1 −1 = e−t (e 1 −e −1 )
Putting all these results together
⎧
⎨0 t<−1
(f ∗g)(t)= 1−e −1−t −1t<1
⎩
e −t (e 1 −e −1 ) t 1
(b) Using Table 24.1 the Fourier transformsof f andgare
F(ω) = 2sinω
ω
and theirproduct is
F(ω)G(ω) =
2sinω
ω(1+jω)
G(ω) = 1
1+jω
(24.12)
Since the convolution is defined differently on each part of the domain, then the
Fouriertransformoftheconvolutionmustbefoundbyintegratingovereachpartof
the domain separately. You will alsoneed torecall that
So,
sinω= ejω −e −jω
2j
F{(f ∗g)(t)} =
=
=
=
∫ ∞
−∞
∫ −1
−∞
∫ ∞
+
1
∫ 1
−1
∫ 1
−1
(f ∗g)(t)e −jωt dt
0 ·e −jωt dt +
∫ 1
−1
e −t (e 1 −e −1 )e −jωt dt
(1 −e −1−t )e −jωt dt
e −jωt −e −1−t−jωt dt + (e 1 −e −1 )
∫ ∞
e −jωt −e −1 e −t(1+jω) dt+(e 1 −e −1 )
1
e −t e −jωt dt
∫ ∞
1
e −t(1+jω) dt
[ ] e
−jωt 1 [ e
= −e −1 −t(1+jω) 1 [ e
+(e
−jω
−1
−(1+jω)] 1 −e −1 −t(1+jω) ∞
)
−1
−(1+jω)]
1

780 Chapter 24 The Fourier transform
( )
= e−jω
−jω + ejω e
−(1+jω)
jω −e−1 −(1+jω)
( ) ( )
e
+e −1 (1+jω) e
+(e 1 −e −1 −(1+jω)
)
−(1+jω) 1+jω
Thefirsttwotermssimplifyto 2sinω
ω
.Bysimplifyingandrearrangingtheremainder
becomes − ejω −e −jω
. Putting all thistogether we find
1+jω
F{(f ∗g)(t)} = 2sinω
ω
= 2sinω
ω
= 2sinω
ω(1+jω)
− ejω −e −jω
1+jω
− 2jsin ω
1+jω
whichisthesameasEquation(24.12).Wehaveshownthat F{f ∗g} =F(ω)G(ω)
and so the convolution theorem has been verified.
24.8.2 Correlationandthecorrelationtheorem
If f (t) and g(t) are two piecewise continuous functions their correlation (also called
their cross-correlation), whichwedenoteby f ⋆g, isdefined asfollows:
Thecorrelationof f (t)andg(t):
f⋆g=
∫ ∞
−∞
f(λ)g(λ −t)dλ
Thisformulaisthesameasthatforconvolutionexceptfortheimportantdifferencethat
the functiongisnotfolded; thatis, wehaveg(λ −t) hererather thang(t − λ).
It is possible to show that the correlation of f (t) andg(t) can be written in the alternative
form
f⋆g=
∫ ∞
−∞
f(t + λ)g(λ)dλ
Sometextsusethisform,butnotethatitistheargumentof f whichist + λ(seeQuestion
8 inExercises 24.8).
We can now statethecorrelation theorem.
Thecorrelationtheorem:
If F{f(t)} =F(ω)and F{g(t)} =G(ω)then
F{f ⋆g} =F(ω)G(−ω)

24.8 Convolution and correlation 781
Example24.18 (a) Usingthe definition ofcorrelation, calculate the correlation of f (t) =u(t)e −t and
g(t) =u(t)e −2t , whereu(t) isthe unit stepfunction.
(b) Verifythe correlation theorem for these functions.
Solution (a) The correlation of f andgisgiven by
f⋆g=
∫ ∞
−∞
f(λ)g(λ −t)dλ
1
f(t)= u(t) e –t
g(t) = u(t) e – 2t
Figure24.16
Graphsof f (t) =u(t)e −t
andg(t) =u(t)e −2t .
t
Graphs of f (t)andg(t) areshown inFigure 24.16.
InFigure24.17wehavesuperimposedthegraphsof f (λ),showndashed,and
g(λ −t), fort being negative, and then positive. Where the graphs do not overlap,
the product f (λ)g(λ −t), and hence the correlation, iszero.
Whent<0
Whent < 0 the graphs overlap for 0 λ < ∞. Hence
f⋆g=
=
∫ ∞
0
∫ ∞
0
f(λ)g(λ −t)dλ
e −λ e −2(λ−t) dλ
∫ ∞
= e 2t e −3λ dλ
0
= e 2t [ e
−3λ
−3
] ∞
0
= 1 3 e2t
f( l)= u( l)e –
l
(a)
l
g( l – t) when t < 0
(b)
t
l
(c)
g( l – t )when t > 0
t
l
Figure24.17
Graphsof f(λ),andg(λ −t)for(b)t < 0,
(c)t>0.

782 Chapter 24 The Fourier transform
Whent0
Whent 0 the graphs overlap fort λ < ∞. Hence
f⋆g=
=
∫ ∞
t
∫ ∞
t
f(λ)g(λ −t)dλ
e −λ e −2(λ−t) dλ
∫ ∞
= e 2t e −3λ dλ
= e 2t [ e
−3λ
−3
t
= e 2te−3t
3
= 1 3 e−t
] ∞
t
Finally the complete expression forthe correlation is
⎧
1
⎪⎨
(f ⋆g)(t)=
3 e2t whent < 0
⎪⎩ 1
3 e−t whent 0
(b) UsingTable 24.1 the Fourier transforms of f andgare
F(ω) = 1 G(ω) = 1
1+jω 2+jω
and hence
F(ω)G(−ω) =
1
(1+jω)(2−jω)
Furthermore, taking the Fourier transformof f ⋆g, obtained inpart(a), we have
∫ 0
∫ ∞
1
F{f⋆g}=
−∞ 3 e2t e −jωt 1
dt +
0 3 e−t e −jωt dt
[ e
t(2−jω) 0 [ e
t(−1−jω) ∞
=
+
3(2−jω)]
−∞
3(−1 −jω)]
0
=
1
3(2−jω) + 1
3(1+jω)
= 3+3jω+6−3jω
9(2−jω)(1+jω)
=
1
(2−jω)(1+jω)
We have shown that F{f ⋆g} = F(ω)G(−ω) and so the correlation theorem has
been verified.

24.9 The discrete Fourier transform 783
EXERCISES24.8
1 Find the convolution of
{ 2
f(t)= 3 t 0t3
0 otherwise
and
{ 4 −1t3
g(t) =
0 otherwise
2 Theconvolution ofafunctionwith itselfis known as
autoconvolution.Find the autoconvolution f ∗ f
when
{ 1 −1t1
f(t)=
0 otherwise
3 Find the correlationof f (t) = 1 for −1 t 1 and
zerootherwise, andg(t) =u(t)e −t .Verify the
correlationtheoremforthese functions.
4 Prove thatconvolution iscommutative, thatis
f ∗g=g ∗ f.Note thatcorrelationis not.
5 Show thatif f (t)andg(t) are both zero,fort < 0,
then (f ∗g)(t) = ∫ t
0 f(λ)g(t − λ)dλ.
6 Prove the convolution theorem.
7 Show that f (t) ⋆g(t) = f (t) ∗g(−t).Deduce thata
correlationcanbeexpressedintermsofaconvolution.
Showalsothat f(t) ∗g(t) = f(t) ⋆g(−t).
8 Prove thatthe correlationintegral
f⋆g= ∫ ∞
−∞ f (λ)g(λ −t)dλcanalsobe writtenin
the form ∫ ∞
−∞g(λ)f(t + λ)dλ.
9 Prove the correlationtheorem.
Solutions
⎧
0 t−1
4
⎪⎨ 3 (t+1)2 −1<t2
1 f∗g= 12 2<t3
12 − 4 ⎪⎩ 3 (t−3)2 3<t6
0 t>6
⎧
⎪⎨
t+2 −2t0
2 f∗f= 2−t 0<t2
⎪⎩
0 otherwise
⎧
⎪⎨
e t (e 1 −e −1 ) t <−1
3 f⋆g= 1−e
⎪⎩
t−1 −1t1
0 t>1
24.9 THEDISCRETEFOURIERTRANSFORM
FormostpracticalengineeringproblemsrequiringtheevaluationofaFouriertransform
it is necessary to use a computer and so some form of approximation is needed. In this
section we show how a function f (t), with Fourier transformF(ω), can be sampled at
intervalsofT togiveasequenceofvalues f (0), f (T), f (2T)....ThediscreteFourier
transform (d.f.t.) takes such a sequence and processes it to produce a new sequence
whichcanbethoughtofasasampledversionofF(ω).Thed.f.t.isimportantasitisthe
basis of most signal and image processing methods. The use of a computer is essential
because ofthe enormousnumberofcalculationsrequired.
We start by stating the transform and provide a simple example to show how it is
calculated. The interested reader should refer to Section 24.10 for a derivation which
shows the relationshipbetween the Fourier transform and the d.f.t.

784 Chapter 24 The Fourier transform
24.9.1 Definitionofthed.f.t.
We consider a sequence ofN terms, f[0], f[1], f[2],..., f[N −1].
Thed.f.t.ofasequence f[n],n = 0,1,2,...,N −1,isanothersequenceF[k],also
havingN terms,defined by
F[k] =
∑N−1
n=0
f[n]e −2jnkπ/N
fork=0,1,2,...,N−1
We writeF[k] = D{f[n]}todenote the d.f.t. of the sequence f[n].
There are a number of variations of this definition and you need to be aware that
different authors may use slightly different formulae. This can cause confusion when
thed.f.t.isfirstmet.Inparticular,someauthorsincludeafactor 1 N inthedefinition,and
others use a positive exponential term instead of the negative one given above. What is
crucial isthatyou know which formula isbeing used and you apply itconsistently.
We now give an example toshow the application of the formula.
Example24.19 Findthe d.f.t. ofthe sequence f[n] = 1,2,−5,3.
Solution We use the formula
D{f[n]} =F[k] =
∑N−1
n=0
f[n]e −2jnkπ/N
fork=0,1,2,...,N−1
Herethe number ofterms,N, isfour:
3∑
F[k] =
n=0
f[n]e −2jnkπ/4
fork=0,1,2,3
So,whenk = 0,
3∑
F[0] = f[n]e 0
n=0
= 1
Whenk = 1,
3∑
F[1] =
=1+2+(−5)+3
n=0
f[n]e −2jnπ/4
= 1 +2e −2jπ/4 + (−5)e −2j2π/4 +3e −2j3π/4
=1+2e −πj/2 −5e −πj +3e −3jπ/2
= 1 +2(−j) −5(−1) +3j
=6+j

24.9 The discrete Fourier transform 785
Whenk = 2,
3∑
F[2] = f[n]e −2jn2π/4
n=0
3∑
= f[n]e −πjn
n=0
= 1 +2e −jπ + (−5)e −2jπ +3e −3jπ
= 1 +2(−1) −5(1) +3(−1)
= −9
Finally, whenk = 3,
3∑
F[3] = f[n]e −2jn3π/4
n=0
3∑
=
n=0
f[n]e −3jnπ/2
= 1 +2e −3jπ/2 + (−5)e −3jπ +3e −9jπ/2
= 1 +2(j) −5(−1) +3(−j)
=6−j
So, the d.f.t. of the sequence 1,2,−5,3 isthe sequence 1,6 +j,−9,6 −j.
You will note from this simple example that a significant amount of calculation is necessary
even when there are just four points in the sampled sequence. This is why it is
necessary to use a computer program to find the transform.Even then, the computation
can be very time consuming. For this reason, much effort has gone into developing fast
algorithms known as fast Fourier transforms (f.f.t.s). For details of these algorithms
youshouldrefertooneofthemanyspecialisttextsonthesubject.Youshouldalsoinvestigatewhichpackagesareavailabletoenableyoutoperformthissortofcalculation.For
example, the computer package MATLAB ® has built-in commands for finding a d.f.t.
using an f.f.t. For example, the MATLAB ® command fft(f) calculates the d.f.t. of a
sequencefas shown below:
f=[1 2 -5 3];
y=fft(f)
y =
1.0000 6.0000+ 1.0000i -9.0000 6.0000- 1.0000i
InMATLAB ® either j or i can beused ininput torepresent √ −1,for example
1+5j
or
1+5i

786 Chapter 24 The Fourier transform
EXERCISES24.9.1
and they are both equivalent. However when variables are printed, i is always used.
Hence both of these inputs iftyped atthe command line would give
ans= 1+5i
this should beborne inmindwhen viewing the results of the f.f.t.
1 Usethe definition to findthe d.f.t. ofthe sequences
f[n] = 1,2,0, −1 andg[n] = 3,1, −1,1.
2 Calculate the d.f.t. ofthe sequence f[n] = 5,−1,2.
3 Useatechnical computing language suchas
MATLAB ® to verifyyour answers to Questions1
and 2.
4 Fromthe definition ofthe d.f.t. showthat if f[n] is a
sequence ofreal numberswith d.f.t.F[k], then
∑ N−1
n=0 f[n]e2jnkπ/N =F[k] where the overline
denotes the complex conjugate ofF[k].
5 Forasequence ofcomplex numbersF[k], letF[k]
represent the sequence obtained by taking the
complex conjugate ofeach termin the sequence.
Show that if f[n]is asequence ofreal numbers,then
F[N−k]=F[k]fork=0,1,2,...,N/2ifNis
even,andfork =0,1,2,...,(N −1)/2ifN isodd.
Thiscan be seen in Example 24.20 whereN = 4 and
F[3] =F[1].
Solutions
1 F[k]=2,1−3j,0,1+3j.G[k]=4,4,0,4 2 6,4.5 +2.5981j,4.5 −2.5981j
24.9.2 Theinversed.f.t.
Just as there is an inverse Fourier transform which transformsF(ω) back to f (t), there
isaninverse d.f.t. whichconvertsF[k] backto f[n].
Theinverse d.f.t. ofthe sequenceF[k],
k = 0,1,2,...,N −1,isthe sequence f[n], alsohavingN terms,given by
D −1 {F[k]} = f[n] = 1 N
∑
N−1
F[k]e 2jknπ/N
k=0
n=0,1,2,...,N−1
Example24.20 Using the definition, find the inverse d.f.t. of the sequence F[k], for k = 0,1,2,3,
given by
F[k] = −4,1,0,1
Solution Inthis sequencetherearefourtermsand soN = 4.Fromthe definition
D −1 {F[k]} = f[n] = 1 3∑
F[k]e 2jknπ/4 n=0,1,2,3
4
= 1 4
k=0
3∑
k=0
F[k]e jknπ/2
n=0,1,2,3
= 1 4 (−4 +1ejnπ/2 +0e jnπ +1e 3jnπ/2 )

24.10 Derivation of the d.f.t. 787
Then, takingn = 0,1,2,3 gives the sequence:
f[n]=− 1 2 ,−1,−3 2 ,−1
EXERCISES24.9.2
1 Usethe definition to findthe inverse d.f.t. ofthe
sequenceF[k] = 6, −2,2, −2.
2 Investigatewhether you have access to acomputer
packagewhichwill calculatean inverse d.f.t. Use the
packageto verifyyour answer to Question1.
3 Prove that f[n] = 1 ∑ N−1
N k=0 F[k]e2jknπ/N is indeed
the inverse ofF[k] = ∑ N−1
m=0 f[m]e−2jkmπ/N by
substituting the expression forF[k] fromthe second
formula intothe first,interchanging the orderof
summation andsimplifyingthe result.
Solutions
1 f[n]=1,1,3,1
24.10 DERIVATIONOFTHED.F.T.
24.10.1 Somepreliminaryresults
A number of results discussed earlier in this book will be required in the development
whichfollows. To assistinthisdevelopment weremind youofthesenow.
Euler’s relations have beendiscussedinSection9.7.Recallthat
e −jθ =cosθ−jsinθ
and inparticular that
e −2nπj = cos2nπ −jsin2nπ
Furthermore, whennisaninteger cos2nπ = 1 and sin2nπ = 0 and so e −2nπj = 1.
Thefollowing integral propertyofthe delta, orimpulse, functionwill beneeded:
∫ ∞
−∞
f(t)δ(t −a)dt = f(a)
So,multiplyinganyfunction, f (t),bythedeltafunction δ(t−a),representinganimpulse
occurring att = a, and integrating, results in f (a). This sifting property of the delta
function, so called because it sifts out the value of f (t) at the location of the impulse,
has beendiscussedinSection16.4. Inparticular, when f (t) = e −jωt wenote
∫ ∞
−∞
e −jωt δ(t −a)dt = e −jωa
When a continuous function f (t), which is defined for t 0, is evaluated at times
t = 0,T,2T,...,nT,... we obtain the sequence of values f(0),f(T),f(2T),...,
f (nT),..., which we denote more concisely by the sequence f[0], f[1], f[2],...,
f[n],…. This sequence can be expressed in the form of a continuous function, ˜f(t),

788 Chapter 24 The Fourier transform
inthe following way:
∞∑
˜f(t)=T f[n]δ(t −nT)
n=0
Notethatwhereas f (t)istheoriginalcontinuousfunctionoft, ˜f (t)isanapproximation
obtainedusingonlythesampledvalues.Thisrepresentationisdiscussedingreaterdetail
inAppendix I.
24.10.2 Derivation
Thisderivationisbaseduponthefactthatafunction f (t),definedfort 0,andhaving
been sampled atintervals ofT, can be represented by the function ˜f (t) where
∞∑
˜f(t)=T f[n]δ(t −nT)
n=0
In any real problem it is only possible to sample a signal over a finite time interval.
Suppose we obtainN samples of the signal at timest = 0,T,2T,..., (N −1)T. Then
thesampledsignalcanberepresentedbyamendingthelimitsofsummationandwriting
˜f(t)=T
∑N−1
n=0
f[n]δ(t −nT)
Taking the Fourier transform of both sides gives
F{ ˜f(t)}=T
∫ ∞
−∞
∑N−1
e −jωt
n=0
f[n]δ(t −nT)dt
If we make the assumption that it is permissible to interchange the order of integration
and summation wefind
F{ ˜f(t)}=T
=T
∑N−1
f[n]
∫ ∞
n=0
−∞
∑N−1
f[n]e −jωnT
n=0
e −jωt δ(t −nT )dt
where we have used the sifting property of the delta function. The quantity on the r.h.s.
isacontinuousfunctionof ωderivedusingthevaluesinthesequence f[n].Writethisas
F(ω)andnotethat ˜ F(ω)isanapproximationtotheFouriertransformF(ω) ˜
= F{f (t)}.
It can also be thought of as a Fourier transform for sequences. An important point to
note about this function of ω is that it is periodic with period 2π . This is proved in the
T
following example.
Example24.21 Show thatthe function
∑N−1
F(ω) ˜ =T
n=0
f[n]e −jωnT
isperiodic with period 2π
T .

(
Solution ConsiderF
˜ ω + 2π )
for any ω.
T
(
F˜ ω + 2π )
∑N−1
−j(ω+2π/T )nT
=T f[n]e
T
n=0
=T
=T
∑N−1
n=0
∑N−1
n=0
Now wesaw inSection 24.10.1 that
f[n]e −jωnT e −j2π T nT
f[n]e −jωnT e −2nπj
24.10 Derivation of the d.f.t. 789
usingthe laws of indices
e −2nπj =cos2nπ−jsin2nπ=1
Finally wehave
(
F˜ ω + 2π )
=T
T
∑N−1
n=0
f[n]e −jωnT
which equalsF(ω).
˜ (
We have shown that F ˜ ω + 2π )
= F(ω) ˜ for any ω. HenceF(ω) ˜ is periodic with
T
period 2π
T .
For computational purposes it is necessary to sampleF(ω) ˜ at discrete values. Suppose
we choose to sample at N points over the interval 0 ω < 2π . By choosing this
T
intervalwearesamplingoveracompleteperiodinthefrequencydomain.Sowechoose
the sample points given by ω =k 2π
(
fork = 0,1,2,...,(N −1).WritingF˜
k 2π )
NT NT
asF(ω ˜
k
)wehave
∑N−1
F(ω ˜
k
)=T
n=0
N−1
2π
−jk
f[n]e NT nT
∑
=T f[n]e −2jnkπ/N (24.13)
n=0
Thislastsum,withoutthefactorT,isusuallytakenasthedefinitionofthed.f.t.of f[n],
writtenF[k]. To denote the d.f.t. of the sequence f[n] wewill write D{f[n]}.
Thed.f.t. ofthe sequence f[n],n = 0,1,2,...,N −1,isthe sequencegiven by
D{f[n]} =F[k] =
∑N−1
n=0
f[n]e −2jnkπ/N
fork=0,1,2,...,N−1

790 Chapter 24 The Fourier transform
When we use this transform to approximate the Fourier transform of the function f (t)
sampled at intervalsT, it follows from Equation (24.13) that we must multiply by the
factor T to obtain our approximation. An example of this is given in the next
section.
24.11 USINGTHED.F.T.TOESTIMATEAFOURIERTRANSFORM
One application of the d.f.t. is to estimate the continuous Fourier transform of a signal
f (t). The following example shows how this can bedone.
Example24.22 The signal f (t) = u(t)e −2t , where u(t) is the unit step function, is shown in
Figure 24.18(a). Its Fourier transform, which can be found from Table 24.1, is
F(ω) = 1 and a graph of |F(ω)|isshown inFigure 24.18(b).
2+jω
Suppose we obtain 16 sample values of f (t) at intervals ofT = 0.1 fromt = 0 to
t = 1.5.
(a) Obtain the d.f.t. of the sampled sequence.
(b) Usethe d.f.t. toestimatethe trueFourier transform values.
(c) Plot a graph tocompare values of |F[k]| and |F(ω)|.
Solution (a) If f (t) =u(t)e −2t is sampled att =nT, that ist = 0,0.1,0.2,...,1.5, we obtain
the sequence f[n] = e −2nT = e −0.2n forn = 0,1,2,...,15. This is the sequence
given inthe second column of Table 24.2.
(b) Thecalculationofthed.f.t.isfartoolaborioustobedonebyhand.Insteadwehave
used the MATLAB ® command fft() to do the calculation and the results, F[k],
have been placed inthe third column.
(c) From Section 24.10 we know that the d.f.t. samples at intervals of 2π in the frequencydomain.SoforcomparisonthefourthcolumnshowsthetruevaluesofF(ω)
NT
1 f(t) = u(t)e – 2 t t
4 + 2 30
u F( v) u =
1
v
0.5
(a)
1
(b)
Figure24.18
The signal f (t) =u(t)e −2t and itsFourier transform.
–30
v

24.11 Using the d.f.t.to estimate a Fourier transform 791
Table24.2
n,k f[n] = e −0.2n F[k] F(ω k ) 0.1F[k]
0 1.0000 5.2918 0.5000 0.5292
1 0.8187 1.4835 −1.9082j 0.1030 −0.2022j 0.1484 −0.1908j
2 0.6703 0.7882 −1.0837j 0.0304 −0.1196j 0.0788 −0.1084j
3 0.5488 0.6311 −0.6952j 0.0140 −0.0825j 0.0631 −0.0695j
4 0.4493 0.5743 −0.4702j 0.0080 −0.0626j 0.0574 −0.0470j
5 0.3679 0.5485 −0.3159j 0.0051 −0.0504j 0.0548 −0.0316j
6 0.3012 0.5355 −0.1964j 0.0036 −0.0421j 0.0536 −0.0196j
7 0.2466 0.5293 −0.0944j 0.0026 −0.0362j 0.0529 −0.0094j
8 0.2019 0.5274 0.0020 −0.0317j 0.0527
9 0.1653 0.5293 +0.0944j 0.0016 −0.0282j 0.0529 +0.0094j
10 0.1353 0.5355 +0.1964j 0.0013 −0.0254j 0.0536 +0.0196j
11 0.1108 0.5485 +0.3159j 0.0011 −0.0231j 0.0548 +0.0316j
12 0.0907 0.5743 +0.4702j 0.0009 −0.0212j 0.0574 +0.0470j
13 0.0743 0.6311 +0.6952j 0.0008 −0.0196j 0.0631 +0.0695j
14 0.0608 0.7882 +1.0837j 0.0007 −0.0182j 0.0788 +0.1084j
15 0.0498 1.4835 +1.9082j 0.0006 −0.0170j 0.1484 +0.1908j
also sampled at intervals of 2π . Recall that to obtain an estimate of the Fourier
NT
transform of f (t) we must multiply the d.f.t. values byT = 0.1. This is shown in
the fifth column of the table.
Figure24.19showsgraphsof |0.1F[k]|and |F(ω)|.Valuesof |F[k]|obtainedbeyond
k = 8aremirrorimagesoftheearliervalues.ThisisbecauseF[N−k] =F[k]asproved
inQuestion5inExercises24.9.1.Thisisaphenomenon ofthed.f.t.andarisesbecause
ofitsperiodicityandsymmetryproperties,furtherdetailsofwhicharebeyondthescope
of this book. The interested reader is referred to a text on signal processing for further
details.
0.6
0.5
0.4
F ( k) = F
0.3
(
2 pk
u v u
N T )
0.1uF[k]u
0.2
0.1
0 2 4 6 8 10 12 14
k
Figure24.19
Comparisonoftruevalues ofaFourier
transformand approximate values obtained
usingad.f.t.

792 Chapter 24 The Fourier transform
24.12 MATRIXREPRESENTATIONOFTHED.F.T.
Whenitisnecessarytodevelopcomputercodeforperformingad.f.t.,anunderstanding
of the following matrix representation isuseful.
We have seen thatthe d.f.t. ofasequence f[n] isgiven by
F[k] =
∑N−1
n=0
f[n]e −2jnkπ/N
Considertheterme −2jnkπ/N whichcanbewrittenusingthelawsofindicesas (e −2jπ/N ) nk .
DefineW = e −2jπ/N so that
F[k] =
∑N−1
n=0
f[n]W nk
Note thatW does not depend uponnor k. For a fixed value of N we can calculateW
which isthen a constant. Writingout thissum explicitly we find
F[k] = f[0]W 0 + f[1]W k + f[2]W 2k + f[3]W 3k +···
+f[N−1]W (N−1)k
Writingthisout foreachkwefind
fork=0,1,2,...,N−1
F[0] = f[0]W 0 + f[1]W 0 + f[2]W 0 + f[3]W 0 +···+f[N−1]W 0
F[1] = f[0]W 0 + f[1]W 1 + f[2]W 2 + f[3]W 3 +···+f[N−1]W N−1
F[2] = f[0]W 0 + f[1]W 2 + f[2]W 4 + f[3]W 6 +···+f[N−1]W 2(N−1)
. = . .
F[N −1] = f[0]W 0 + f[1]W N−1 + f[2]W 2(N−1)
+ f[3]W 3(N−1) +···+f[N−1]W (N−1)(N−1)
These equations can be written inmatrix form asfollows:
⎛ ⎞ ⎛
⎞ ⎛
F[0] W 0 W 0 W 0 ... W 0
F[1]
W 0 W 1 W 2 ... W N−1
F[2]
=
W 0 W 2 W 4 ... W 2(N−1)
⎜
⎝
⎟ ⎜
. ⎠ ⎝
.
.
.
.
⎟ ⎜
. ⎠ ⎝
F[N −1] W 0 W N−1 W 2(N−1) ...W (N−1)(N−1)
whereW = e −2jπ/N .
f[0]
f[1]
f[2]
.
f[N−1]
⎞
⎟
⎠
Example24.23 (a) Find the matrix representing a three-point d.f.t.
(b) Usethe matrix tofind the d.f.t. of the sequence f[n] = 4,−7,11.
Solution (a) HereN = 3 and soW = e −2πj/3 . The required matrix is
⎛
⎞
1 1 1
⎝1 e −2πj/3 e −4πj/3 ⎠
1 e −4πj/3 e −8πj/3

24.13 Some properties of the d.f.t. 793
(b)
which usingthe Cartesian form can be written
⎛
⎞
1 1 1
⎜
1 − 1 √
3
⎝ 2 −j − 1 √
3
2 2 +j 2
1 − 1 √
3
2 +j − 1 √ ⎟
⎠ 3
2 2 −j 2
⎛
⎞
⎛ ⎞ 1 1 1
F[0]
⎝F[1]
⎠ = ⎜
1 − 1 √
3
F[2]
⎝ 2 −j − 1 √ ⎛ ⎞ ⎛ ⎞
3
2 2 +j 4 8
2
1 − 1 √
3
2 +j − 1 √ ⎟ ⎝−7⎠ = ⎝2 +15.5885j⎠
⎠ 3
2 2 −j 11 2 −15.5885j
2
So, the required d.f.t. isthe sequence
8,2 +15.5885j,2 −15.5885j
EXERCISES24.12
1 Using the definition ofthe d.f.t. show thatthe matrix
whichcan be used to implement the d.f.t. ofa
four-point signal is
⎛
⎞
1 1 1 1
W = ⎜1 −j −1 j
⎟
⎝1 −1 1 −1⎠
1 j −1 −j
Use thismatrix to findthe d.f.t. ofthe sequences
f[n] = {1,2,0, −1} andg[n] = {3,1, −1,1}.
Solutions
1 F[k]=2,1−3j,0,1+3j.G[k]=4,4,0,4
24.13 SOMEPROPERTIESOFTHED.F.T.
24.13.1 Periodicity
Thed.f.t.F[k] isperiodic with periodN, thatis
F[k+N] =F[k]
This is why we take onlyN terms in the frequency domain. Calculating further values
will only reproducethe earlier ones.Theinverse d.f.t. isalsoperiodicwith periodN.
24.13.2 Linearity
The d.f.t. is a linear transform. This means that the d.f.t. of the sequenceaf[n] +bg[n]
whereaandbareconstants isaF[k] +bG[k].

794 Chapter 24 The Fourier transform
24.13.3 Parseval’stheorem
If D{f[n]} =F[k] and D{g[n]} =G[k] then
∑N−1
n=0
f[n]g[n] = 1 ∑N−1
F[k]G[k]
N
k=0
where the overline indicates the complex conjugate.
24.13.4 Rayleigh’stheorem
This theorem isobtained fromParseval’s theorem by lettingg[n] = f[n].
∑N−1
|f[n]| 2 = 1 ∑N−1
|F[k]| 2
N
n=0 k=0
Example24.24 VerifyRayleigh’s theoremforthe sequence f[n] = 5,4.
Solution HereN = 2.We first calculateF[k].
F[k] =
1∑
n=0
f[n]e −jnkπ
= f[0] + f[1]e −jkπ
= 5 +4e −jkπ
Whenk =0,wehaveF[0] =5+4e 0 =9.
Whenk=1,wehaveF[1]=5+4e −jπ =5−4=1.
For brevity weoften writeF[k] = 9,1.
Then ∑ 1
n=0 |f[n]|2 =5 2 +4 2 =41.
Also, ∑ 1
k=0 |F[k]|2 = 9 2 +1 2 = 82.
We note that41 = 1 ×82 and this verifies Rayleigh’s theorem.
2
EXERCISES24.13
1 (a)Obtainthe d.f.t.sof f[n] = 1,2,2,1 and
g[n] = 2,1,1,2.
(b)Verify Rayleigh’stheoremforeachofthese
sequences.
2 Suppose f[n] = 3,1,5,4 andg[n] = 2, −1,9,5.
(a) Show that ∑ 3
n=0 f[n]g[n] = 70.
(b) Obtainthe d.f.t.sF[k] andG[k].
(c) CalculateF[k]G[k].
(d) HenceverifyParseval’stheoremforthese
sequences.
3 Show thatifthe d.f.t. of f[n] isF[k] thenthe d.f.t. of
f[n −i] is e −2πjki/N F[k]. This isknown asthe shift
theorem.
4 ProveParseval’stheorem.
5 ProveRayleigh’stheorem.

24.14 The discrete cosine transform 795
Solutions
1 (a) F[k]=6,−1−j,0,−1+j.
G[k]=6,1+j,0,1−j
(b) ∑ |f[n]| 2 = 10, ∑ |F[k]| 2 = 40.
∑ |g[n]| 2 = 10, ∑ |G[k]| 2 = 40
2 (b) F[k] = 13,−2 +3j,3,−2 −3j.
G[k] = 15,−7 +6j,7,−7 −6j
(c) 195,32 −9j,21,32 +9j
24.14 THEDISCRETECOSINETRANSFORM
There isanalternative method tothatofthe d.f.t.fortransformingaset oftimedomain
samples of the signal f (t) into the frequency domain,F[ω]. Itis known as the discrete
cosinetransformord.c.t.WehavealreadyseenthatthediscreteFouriertransformtakes
N samples, f[0], f[1], f[2], ..., f[N − 1] and producesN output samples in the frequencydomainF[0],F[1],F[2],
...,F[N −1].Thed.c.t.iscalculatedinasimilarway.
However, thed.c.t.ofasetofsampleshas specialpropertieswhichmakeitparticularly
suitable for image and audio processing. These will become apparent in the following
examples.
A detailed derivation will not be presented here for the d.c.t., although much of the
mathematics of the d.c.t. and the d.f.t. is very similar. It should also be mentioned that
very efficient methods for computing the d.c.t. are possible; this is beyond the scope of
thisbook which will focus on the importantproperties of the d.c.t. as a transformation.
24.14.1 Definitionofthed.c.t.anditsinverse
We consider a sequence ofN terms, f[0], f[1], f[2], ..., f[N −1].
Thed.c.t.ofasequence f[n],n = 0,1,2,...,N −1,isanothersequenceF[k],also
havingN terms,defined by
F[k] = √ 1 ∑N−1
[ ( π
f[n]cos n + 1 ) ]
k fork=0,1,2, ... ,N−1
N N 2
n=0
Thereareseveralvariantsofthedefinitionofthed.c.t.Theonegivenaboveisprobably
themostcommonlyusedandisusuallytermedd.c.t.-II.Themainmathematicaldifference
between this and the d.f.t. is that in the d.c.t. only real numbers are produced in
F[k], assuming real-valued samples in f[n]. The same cannot be said for the d.f.t. becauseevenwithreal-valuedsamplesin
f[n]theoutputF[k]ingeneralcontainssamples
which have both a real and imaginary part.
The d.c.t. and its variants are popular choices for image and audio compression.
Compression is the process of reducing the volume of data associated with a signal by
removing redundant, partially redundant or repeated data. We will see in the following
examples thatthe d.c.t. isvery usefulforcompressing certaintypes of signal.
Inordertomakeuseofthed.c.t.intheseapplicationsitisnecessarytousetheinverse
operationtorecovertheoriginalsamples.Theinverseofd.c.t.-IIisanotherd.c.t.termed
d.c.t.-III.

796 Chapter 24 The Fourier transform
The inverse d.c.t. of a sequenceF[k],k = 0,1,2,...,N −1, is another sequence
f[n],alsohavingN terms,defined by
f[n] = √ 1 ∑N−1
[ ( π
{F[0] +2 F[k]cos
N N k n + 1 )] }
2
k=1
forn=0,1,2, ... ,N−1
Again,aderivationofthisinversewillnotbepresentedhere.However,wewilldemonstrate
the inverse property for a specific example.
Example24.25 (a) Findthe d.c.t.,F[k], ofthe sequence f[n] = 2,4,6.
(b) Apply the inverse d.c.t. to F[k] and show that the original sequence, f[n], is
obtained.
Solution (a) We use the formula
F[k] = √ 1 ∑N−1
N
n=0
[ π
f[n] cos
N
Herethe number of terms,N, isthree.
Whenk=0
F[0] = 1 √
3
2∑
Whenk=1
n=0
(
n + 1 ) ]
k
2
[ ( π
f[n]cos n + 1 ) ]
×0
3 2
= 1 √
3
[2cos0 +4cos0+6cos0]= 12 √
3
F[1] = 1 √
3
2∑
= √ 1 [
3
Whenk=2
n=0
[ ( π
f[n]cos n + 1 ) ]
×1
3 2
]
fork=0,1,2, ... ,N−1
2cos π 6 +4cosπ 2 +6cos5π 6
√ )]
3
= 1 [√ √ ]
√ 3−3 3 = −2
2 3
= √ 1
√ (
3
[2 ×
3 2 +0+6× −
F[2] = 1 √
3
2∑
= √ 1 [
3
n=0
[ ( π
f[n]cos n + 1 ) ]
×2
3 2
]
2cos π 3 +4cosπ+6cos5π 3
= √ 1 [
2 × 1
3 2 +4× (−1) +6× 1 ]
=1−4+3=0
2
So the d.c.t. of the sequence 2,4,6is 12 √
3
, −2, 0.

24.14 The discrete cosine transform 797
(b) We now apply the inverse d.c.t. toF[k].
The inverse d.c.t. is
f[n] = √ 1 ∑N−1
[ ( π
{F[0] +2 F[k]cos
N N k n + 1 )] }
2
k=1
forn=0,1,2, ... ,N−1
Whenn=0
{
f[0] = √ 1 12
2∑
√ +2
3 3
k=1
= √ 1 [ 12
√ +2× (−2) ×cos
3
[ 3
= √ 1 12
√ −4×
3 3
[ ( π
F[k]cos
3 k 0 + 1 )] }
2
( π
) ]
+2×0
6
√
3
2 +0 ]
=4−2=2
Whenn=1
{
f[1] = √ 1 12
2∑
√ +2
3 3
= 1 √
3
[ 12
k=1
√ +2× (−2) ×cos
3
]
= 4
= 1 √
3
[ 12
√
3
−4×0+0
[ ( π
F[k]cos
3 k 1 + 1 )] }
2
( π
) ]
+2×0
2
Whenn=2
{
f[2] = √ 1 12
2∑
√ +2
3 3
k=1
[ ( π
F[k]cos
3 k 2 + 1 )] }
2
( ) ] 5π
+2×0
6
= √ 1 [ 12
√ +2× (−2) ×cos
3
[ 3
(
= √ 1
√ ) ]
12 3
√ −4× − +0 =4+2=6
3 3 2
So f[n] = 2,4,6, which was the originalsequence as expected.
If a d.f.t. had been performed on the sequence f[n], the output,F[k], would have been
the sequence of complex terms 12, −3 + j1.7321,−3 − j1.7321. Compare this with
the d.c.t. which isF[k] = 12 √
3
,−2, 0. Note that this latter sequence contains only realvalued
terms.

798 Chapter 24 The Fourier transform
Engineeringapplication24.2
Effectoftruncatingthed.c.t.andd.f.t.ofasetofsamples
Now we turn our attention to a longer sequence and consider the effect of deleting
samples inthe d.c.t.
In this example, we produce both a d.c.t. and a d.f.t. of a given set of samples
andthensetsomeofthesamplesinthelatterpartofthesequencetozero.Following
this we carry out an inverse d.c.t. to return an approximation to the original sample
set.
We consider the sequence of samples
f[n]=0,1,2,3,4,5,6,7,8,9
Usingthe methods described inthe previous example wecan show that the d.c.t. is
F DCT
[k] = 14.2302,−6.3815,0.0000,−0.6835,0.0000,
−0.2236,0.0000,−0.0904,0.0000,−0.0254
ThenotationF DCT
hasbeenusedtodistinguishthissequencefromthed.f.t.sequence
thatitwill be compared to. The d.f.t. is
F DFT
[k] = 45.0000,−5.0000 +j15.3884,−5.0000 +j6.8819,−5.0000
+j3.6327,−5.0000 +j1.6246,−5.0000,−5.0000 −j1.6246,
−5.0000 −j3.6327,−5.0000 −j6.8819,−5.0000 −j15.3884
Bydefinition,applyingtheappropriateinverseoperationstoeithersetofdatawould
recover the originalinput samples.
If instead we delete samples from the end of each sequence by setting the values
to zero and then perform the inversion then the results are different. In this example
weset the lastfive samples tobezero. So weperform the inversion on
F DCT
[k] = 14.2302,−6.3815,0.0000,−0.6835,0.0000,0,0,0,0,0
F DFT
[k] = 45.0000,−5.0000 +j15.3884,−5.0000 +j6.8819,−5.0000
+j3.63271,−5.0000 +j1.6246,0,0,0,0,0
The results of the inversion are shown plotted in Figure 24.20. The values obtained
from the inverse d.c.t. are all real values. The values plotted on the graph for the
inversed.f.t.arethemagnitudesofthecomplexvaluesthatarereturned.Thisisnecessary
because now samples have been removed from the d.f.t., it can no longer be
guaranteed thatthe solution consists of wholly real numbers.
Notice that the samples in both cases are not exactly the same as the ones we
startedwith.Thisisexpectedbecauseinsettingsomeofthesamplevaluestozeroin
thed.c.t.wehavedestroyedsomedata.Infactwehavedestroyedhalfofthesamples
in this example. However, notice also that the d.c.t. appears to perform much better
than the d.f.t. in recovering the input data. The reason the d.c.t. looks qualitatively
better than the d.f.t. is that the information content or energy is concentrated in the
lowest order samples. Hence when we set the later samples to zero they affect the
d.c.t. far lessthan the d.f.t.

24.14 The discrete cosine transform 799
f(n)
8
6
4
2
2 4 6 8
Figure24.20
Lossycompression usingthe d.c.t. andthe d.f.t. Markers:circle = original data; square =
d.c.t. derived values; diamond =d.f.t. derived values.
n
Thisisanexampleof lossycompression,whichisusedextensivelyinaudio,image
and video compression. Because some of the samples are known to be zero they do
not need to be stored and hence the file or data storage requirement is much lower. An
approximation of the original data can be recovered even though not all of the original
d.c.t. samples are provided. It can be shown that the more samples we retain, the better
thequalityofthereproduction.Itiswellknownthatthed.c.t.performswellonstraight
line data and is used extensively in compression standards for images such as JPEG. In
image data which is viewed qualitatively a certain amount of data loss is tolerable, and
thiswill be illustratedinthe following example.
Engineeringapplication24.3
Two-dimensionald.c.t.andimagecompression
Sofarwehaveonlyseenthed.c.t.appliedtoaone-dimensional dataset.Thiscould
be used to handle the compression of audio data, for example. When applying the
d.c.t. to a two-dimensional problem such as an image a slightly different form is
required:
F[k,l] = √ 1 ∑N−1
NM
n=0
{ M−1 ∑
[ ( π
f[n,m]cos m + 1 )
l] } [ ( π
cos n + 1 ) ]
k
M 2 N 2
m=0
forn=0,1,2, ... ,N−1
m=0,1,2, ... ,M−1
This formula can be applied to a 2D data set. We take the greyscale image given
in Figure 24.21 as the source data. Each pixel in the image has assigned to it a
numerical brightness value and itistothese values that weapply the d.c.t.
➔

800 Chapter 24 The Fourier transform
f(0, 0)
f(124, 124)
Figure24.21
Source imageford.c.t.
application.
We use the transform given above to produce an N ×M matrix containing the
real d.c.t. values,F[k,l]. Figure 24.22 is a visualization ofF[k,l]. Dark pixels represent
large values, white pixels represent zero or close to zero elements. Notice the
concentration of largevalues close toF[0,0], which isinthe top-left corner.
F(0, 0)
Figure24.22
Thed.c.t. matrix.
F(124, 124)
Inthenextstepwecompresstheimagebysettingthearrayelementstozeroatthe
right-handsideandbottom.Sincethecoefficientsaresmallinthisregiontheeffects
on the image after inversion are minor, unless a significant number of values on the
right-hand sideand bottom are set tozero.
The results of applying the inverse transform are shown in Figure 24.23. We can
seethattheimageontherightwhichcontainsthelowestnumberofd.c.t.coefficients
hasmoredistortionduetothecompression.Suchdistortionsareoftentermedvisual
artefacts. The advantage of having a highly compressed file is that it requires less
space for data storage.
In practice, the image is often broken into smaller blocks with the d.c.t. being
carried out on these blocks. If a highly compressed image is viewed on a computer
atalowresolutionitisoftenpossibletoseetheseblocks,whichtypicallymeasure8
by 8 pixels.

24.15 Discrete convolution and correlation 801
Figure24.23
Effect ofsetting elements to zero in the d.c.t.matrix. Moving from left to right agreater
number ofelements in the matrix are setto zero, resulting in more imageartefacts.
24.15 DISCRETECONVOLUTIONANDCORRELATION
As stated at the beginning of Section 24.8, convolution and correlation are important
techniquesinsignalandimageprocessing.Inthissectionwewilldescribetheirdiscrete
representations. Both ofthese can be implemented efficiently usingthe d.f.t.
24.15.1 Linearconvolution
The linear convolution of two real sequences f[n] andg[n] is another sequence,h[n]
say, which we denote by f ∗g, which isdefined as follows:
Linear convolution of f[n] andg[n]:
∞∑
h[n]=f∗g= f[m]g[n−m] forn=...−3,−2,−1,0,1,2,3,...
m=−∞
Notice the similaritybetween this definition and thatof convolution defined inthe context
of the continuous Fourier transform in Section 24.8. Notice also that in this formulathesequencegisfoldedand
shifted.Thiswillbeillustratedintheexample which
follows.
Frequently the sequences being considered will be finite.
Assume now that f[n] is a finite sequence ofN 1
terms, so that all terms other than
f[0], f[1],..., f[N 1
−1] arezero.
Suppose also that g[n] is a finite sequence of N 2
terms, with all terms other than
g[0],g[1],...,g[N 2
−1] being zero.
Itcanbeshownthatthesequenceh[n]willhavelengthN 1
+N 2
−1,andtheconvolution
sum simplifies tothe following:
The linear convolution of the two finite sequences f[n] andg[n] isdefined as
n∑
h[n]=f∗g= f[m]g[n−m] forn=0,1,2,3,...,(N 1
+N 2
−2)
m=0

802 Chapter 24 The Fourier transform
The terms of this sequence can be calculated by brute force as the following example
will show, although in practice a convolution can be calculated much more efficiently
using a d.f.t.
Example24.26 Suppose f[n]isthe sequence 3,9,2,−1, andg[n] isthe sequence −4,8,5.
(a) Find the linear convolutionh[n] = f ∗gusing the previous formula.
(b) Develop a graphical interpretation of thisprocess.
Solution (a) Notethatboth f andgarefinitesequences oflength4and3respectively. Theconvolution
f ∗gwill be a sequence of length 4 +3−1 = 6.By definition,
h[n] =
n∑
f[m]g[n−m]
m=0
forn=0,1,2,...,5
The first terminh[n] isobtained by lettingn = 0:
h[0] =
0∑
f[m]g[0 −m]
m=0
= f[0]g[0]
= (3)(−4)
= −12
The second termisobtained by lettingn = 1:
h[1] =
1∑
f[m]g[1 −m]
m=0
= f[0]g[1] + f[1]g[0]
= (3)(8) + (9)(−4)
=24−36
= −12
Subsequent terms are calculated in a similar manner. You should obtain these for
yourself toensure thatyou understand the process.The complete sequence is
h[n] = f ∗g= −12,−12,79,65,2,−5
(b) The graphical interpretation is developed along the same lines as was done for the
continuous convolution inSection 24.8.
The sequences f[m]andg[m] areshown inFigure 24.24.
The sequence g[−m] is found by reflection in the vertical axis, that is folding as
shown in Figure 24.25(a). Then the folded graph can be translatednunits to the left or
therightbychangingtheargumentofgfromg[m]tog[n −m].Ifnispositivethegraph
moves to the right. Study Figures 24.25(b--g) to observe this. Convolution is the sum
of products of f[m] with g[n −m]. The graph of f[m] is superimposed. We are only
interested in values ofnfor which the graphs overlap -- otherwise each product is zero.
For each value ofnthe superimposed graphs make it easy to see which values must be

24.15 Discrete convolution and correlation 803
f [m]
g[m]
10
8
6
4
2
–2
–4
1 2
m
10
8
6
4
2
–2
–4
Figure24.24
Thesequences f[m] = 3,9,2, −1 andg[m] = −4,8,5.
1 2
m
multipliedtogetherandadded.Ifrequired,atablecanbeconstructedwhichsummarizes
all the necessary information as shown below.
m −2 −1 0 1 2 3 4 5
f[m] -- -- 3 9 2 −1 -- --
n=0 g[−m] 5 8 −4 -- -- -- -- --
n=1 g[1−m] -- 5 8 −4 -- -- -- --
n=2 g[2−m] -- -- 5 8 −4 -- -- --
n=3 g[3−m] -- -- -- 5 8 −4 -- --
n=4 g[4−m] -- -- -- -- 5 8 −4 --
n=5 g[5−m] -- -- -- -- -- 5 8 −4
EXERCISES24.15.1
1 Given f[n] = 1,2,2,1 andg[n] = 2,1,1,2, findthe
linearconvolution f ∗g.
2 (a) Thelinearconvolution ofthe sequences
3,9,2, −1 and −4,8,5 was obtainedin
Example24.27. Show thatthisconvolution is
equivalent to multiplying the two polynomials
3+9x+2x 2 −x 3 and−4+8x+5x 2 .
(b) Byusingpolynomial multiplicationfind the
linearconvolution ofthe sequences f[n] = 9, −8
andg[n] = 1,2, −4.
(c) Use the polynomial methodto findthe linear
convolution of f[n] = 9, −1,3 and
g[n] = 7,2, −4.
3 Find the linearconvolution of f[n] = 1, −1,1,3 and
g[n] = 7,2,0,1.
4 Prove from the definition thatlinearconvolution is
commutative, that is f ∗g=g ∗ f.
Solutions
1 2,5,7,8,7,5,2
3 7, −5,5,24,5,1,3
2 (b) 9,10, −52,32 (c) 63,11, −17,10, −12
24.15.2 Circularconvolution
InthissectionweconsiderperiodicsequenceshavingperiodN.Wecanselectoneperiod
forexaminationbylookingattheterms f[0], f[1], f[2],..., f[N−1],say.Forexample,
the sequence
f[n]=...−7,11,2,−7,11,2,−7,11,2...

804 Chapter 24 The Fourier transform
(a)
(b)
(c)
(d)
(e)
g[ – m]
10
6 8
4
2
– 4
g[ – m]
10
6 8
4
2
– 4
g[ 1– m]
10
6 8
4
2
– 4
g[ 2 – m]
10
6 8
4
2
– 4
g[ 3 – m]
10
6 8
4
2
– 4
1 2
1 2
1 2
1 2
1 2
m
m
m
m
m
sum of products:
(3)( – 4) = –12
sum of products:
(3)(8) + (9)( – 4) = –12
sum of products:
(3)(5) + (9)(8) + (2)( – 4) = 79
sum of products:
(9)(5) + (2)(8) + ( – 1)( – 4) = 65
(f)
(g)
g[ 4 – m]
10
6 8
4
2
– 4
g[ 5 – m]
10
6 8
4
2
– 4
1 2
1 2
m
m
sum of products:
(2)(5) + (– 1)(8) = 2
sum of products:
(– 1)(5) = –5
Figure24.25
The effect oftranslatingg[−m] bynunits,forn = 0,1, . . . ,5.

24.15 Discrete convolution and correlation 805
f [0]= f [3]
–7
f [0]
–7
11 2
f [1]= f [4] f [2]= f [5]
(a)
11 2
f [–2] f [–1]
(b)
Figure24.26
Aperiodic sequence can be visualized
bylisting its termsaround acircle.
isaperiodicsequencewithperiodN = 3.Wecanselecttheterms f[0], f[1], f[2],thatis
terms −7,11,2, and use these tostudythe entire sequence.
Suchasequencecanberepresentedgraphicallybylistingitstermsaroundacircleas
shown in Figure 24.26. Doing this allows us to calculate further terms in the sequence
as we require them. Rotation anticlockwise represents increasingnas in (a). Rotation
clockwise represents decreasingnas in(b).
Suppose f[n] andg[n] are two periodic sequences having periodN. Their circular
convolution,orperiodicconvolution,isthesequenceh[n],whichwedenoteby f ○∗ g
and which isdefined as follows:
The circular convolution of two periodic sequences each of periodN isdefined as
h[n]=f○∗g=
∑N−1
m=0
f[m]g[n−m] forn=0,1,2,...,N−1
The sequence is periodic with period N and so we can state it for n = 0,1,2,...,
N−1.
Example24.27 (a) Calculate the circular convolution,h[n] = f ○∗ g, of the two periodic sequences
f[n] = 9,−1,3 andg[n] = 7,2,−4.
(b) Develop a graphical representation of thisprocess.
Solution (a) The sequenceg[n] isdepicted inFigure 24.27.
We use the formulagiven above. InthisExample,N = 3.First letn = 0.
g[0]
2∑
h[0] = f[m]g[0 −m]
7
2 – 4
g[1] g[2]
g[–2]
g[–1]
Figure24.27
Theperiodic sequence
g[n] = 7,2, −4.
m=0
= f[0]g[0] + f[1]g[−1] + f[2]g[−2]
= (9)(7) + (−1)(−4) + (3)(2)
= 73
Next letn = 1.
2∑
h[1] = f[m]g[1 −m]
m=0
= f[0]g[1] + f[1]g[0] + f[2]g[−1]
= (9)(2) + (−1)(7) + (3)(−4)
= −1

806 Chapter 24 The Fourier transform
n=0 n=1 n=2
7
2
–4
9
9
9
–4
– 1 3
2
7
– 1 3
–4
2
– 1 3
7
Sums of products:
(9)(7) + ( –1)(– 4) + (3)(2) = 73
(9)(2) + ( –1)(7) + (3)( – 4) = –1
(9)( – 4) + ( –1)(2) + (3)(7) = –17
Figure24.28
The sequences f[m]drawn around the inner circle,andg[n −m] drawn around the outer circle,
forn=0,1,2.
Finally, letn = 2.
h[2] =
2∑
f[m]g[2 −m]
m=0
= f[0]g[2] + f[1]g[1] + f[2]g[0]
= (9)(−4) + (−1)(2) + (3)(7)
= −17
So the circular convolution f ○∗ g = 73,−1,−17.
If required, a table can be constructed which summarizes all the necessary informationaswasshowninExample24.26.Thesequencesmustbeextendedtoshow
their periodicity, and this time we are only interested in generating the convolution
sequence over one period, namely thatsection of the table form = 0,1,2.
m −2 −1 0 1 2 3 4 5
f[m] −1 3 9 −1 3 9 −1 3
n=0 g[−m] −4 2 7 −4 2 7 −4 2
n=1 g[1−m] 7 −4 2 7 −4 2 7 −4
n=2 g[2−m] 2 7 −4 2 7 −4 2 7
(b) A graphical representation can be developed by listing the fixed sequence f[m],
for m = 0,1,2, anticlockwise around an inner circle as shown in Figure 24.28.
We list g[−m] around an outer circle but do so clockwise to take account of the
folding.Byrotatingtheoutercircleanticlockwiseweobtaing[1−m]andg[2−m].
Bymultiplyingneighbouringtermsandaddingweobtaintherequiredconvolution.
The resultis73,−1,−17 as obtained inpart(a).
In the following section you will see how circular convolution can be performed using
the d.f.t.

24.15 Discrete convolution and correlation 807
EXERCISES24.15.2
1 Calculate the circular convolution of f[n] = 1, −1,1,3 andg[n] = 7,2,0,1.
Solutions
1 12, −4,8,24
24.15.3 The(circular)convolutiontheorem
Suppose f[n]andg[n]areperiodicsequencesofperiodN.Supposefurtherthatthed.f.t.s
of f[n] andg[n] forn = 0,1,2,...,N −1 are calculated and are denoted byF[k] and
G[k]. The convolution theorem states that the d.f.t. of the circular convolution of f[n]
andg[n] isequal tothe product of the d.f.t.sof f[n]andg[n].
The convolution theorem:
D{f ○∗ g} =F[k]G[k]
This is important because it provides a technique for finding a circular convolution. It
follows fromthe theorem that
f ○∗ g = D −1 {F[k]G[k]}
So,tofind the circular convolution of f[n] andg[n] weproceed as follows:
(1) Find the corresponding d.f.t.s,F[k] andG[k].
(2) Multiply these together toobtainF[k]G[k].
(3) Find the inverse d.f.t. togive f ○∗ g.
Whilst this procedure may seem complicated, it is nevertheless an efficient way of calculating
a convolution.
Example24.28 Usethe convolution theorem tofind f ○∗ gwhen f[n] = 5,4 andg[n] = −1,3.
Solution Firstwefind the corresponding d.f.t.s,F[k] andG[k]. Using
F[k] =
∑N−1
n=0
withN = 2 gives
F[0] =
f[n]e −2jnkπ/N
1∑
f[n] =9, F[1] =
n=0
1∑
f[n]e −jnπ = 5 +4e −jπ = 1
n=0

808 Chapter 24 The Fourier transform
and similarly,
and so
G[0] =
1∑
g[n] = 2, G[1] =
n=0
F[k] =9,1
G[k] =2,−4
1∑
g[n]e −jnπ = −1 +3e −jπ = −4
These transformsaremultiplied together, termby term, togive
n=0
H[k] =F[k]G[k] = (9)(2), (1)(−4) = 18,−4
Finally the inverse d.f.t. of the sequence 18,−4, isfound using
togive
and so
h[n] = 1 N
∑
N−1
H[k]e 2jnkπ/N
k=0
(f ○∗ g)[0] = 1 2 (18−4)=7
(f ○∗ g)[1] = 1 2 (18) + 1 2 (−4ejπ ) = 11
f○∗g=7,11
TheconvolutioncouldalsobeevaluateddirectlyusingthetechniqueofSection24.15.2.
You should trythistoconfirm the resultobtained usingthe theorem.
Example24.29 Usetheconvolutiontheoremandacomputerpackagewhichcalculatesd.f.t.stofindthe
circularconvolution ofthe sequences f[n] = 1,2,−1,7 andg[n] = −1,3,2,−5.
Solution You will need access to a computer package such as MATLAB ® to work through this
example.Thed.f.t.of f[n]canbecalculatedeitherdirectly,whichislaborious,orusing
the MATLAB ® commandfft().
F=fft([1 2 -1 7])
ans =
9.0000 2.0000+ 5.0000i -9.0000 2.0000- 5.0000i
HenceF[k] = 9,2 +5j,−9,2 −5j.
Similarly,
G=fft([-1 3 2 -5])
ans =
-1.0000 -3.0000- 8.0000i 3.0000 -3.0000+ 8.0000i
HenceG[k] = −1,−3 −8j,3,−3 +8j. Then, the product of these d.f.t.s is calculated
by multiplyingcorrespondingtermstogether:
F[k]G[k] = −9,34 −31j,−27,34 +31j

24.15 Discrete convolution and correlation 809
InMATLAB ® ,
product = F.*G
product =
-9.0000 34.0000-31.0000i -27.0000 34.0000+31.0000i
Finally, taking the inverse d.f.t.,using the MATLAB ® commandifft(), gives
ifft([-9 34-31i -27 34+31i])
ans =
8 20 -26 -11
This isthe circular convolution of f[n]andg[n], thatis
f[n] ○∗ g[n] = 8,20,−26,−11
This example has illustrated how circular convolution can be achieved through the use
of the d.f.t.
Theconvolutiontheoremappliestocircularconvolutionbutnotlinearconvolution.However,bymodifyingtheprocedureslightly,thelinearconvolutionoftwofinitesequences
can alsobe found.
If f[n] is a finite sequence of lengthN 1
andg[n] is a finite sequence of lengthN 2
we
know from Section 24.15.1 that their linear convolution is a sequence of length N 1
+
N 2
−1.
Firstweextendboththesequences f[n]andg[n]tomakeeachhavelengthN 1
+N 2
−1.
This extension is done by adding zeros. This process is known as ‘padding’ with zeros.
Then the d.f.t.s of the padded sequences are calculated to give F[k] and G[k]. It can
be shown that the linear convolution of the original sequences is equal to the circular
convolution of the padded sequences. Hence the linear convolution f ∗gis then found
from
f ∗g = D −1 {F[k]G[k]}
Consider the following example.
Example24.30 If f[n] is the finite sequence 7,−1 andg[n] is the finite sequence 4,2,−7 use circular
convolution with padded zeros toobtain the linear convolution f ∗g.
Solution Their linear convolution isasequence of length 2 +3−1 = 4.
We pad f andgtogive sequences of length 4.
f[n] = 7,−1,0,0 andg[n] = 4,2,−7,0
Then, either by direct calculation or by using a computer package you can verify that
F[k]=6,7+j,8,7−j and G[k]=−1,11−2j,−5,11+2j

810 Chapter 24 The Fourier transform
Then
F[k]G[k] = −6,79 −3j,−40,79 +3j
Then taking the inverse d.f.t. gives
D −1 {F[k]G[k]} = 28,10,−51,7
whichistherequiredlinearconvolution.Youshouldverifythisbycalculatingthelinear
convolution directly.
Engineeringapplication24.4
Convolutionreverb
Convolutioncanbeusedonanaudiosignaltosimulatetheechoingorreverberation
of a real room or space. An application of such a method can be found in recording
studioswhereitmaybedesirabletosimulatetheacousticsofalargeroomsuchasa
concert hall on a recording.
The first stage involves capturing the impulse response of the room to be simulated.Animpulsivesoundsourcesuchasapistolorasmallexplosioncanbeusedto
exciteabroadrangeoffrequencies.Itcanbeshownthatashort-durationimpulsehas
a very broad spectrum. A popular alternative is to use a sinusoidal signal source of
time-varyingfrequency.Inthelattercasetheoutputoftheconvolutionprocess,h[n],
is known because the spectrum can be measured directly. The second stage consists
ofapplyingtheinverseprocess,termeddeconvolution,tothisdatainordertoobtain
the impulse response of the room.
When the impulse response has been found, whichever method has been used, it
becomes a case of convolving the impulse response of the room or space with the
signal. It is necessary to carry out a linear convolution process to obtain the desired
effect.Wedonotwishtousecircularconvolutionbecausethesignalisnotperiodic.
However, we may wishtomake use of the circular convolution theorem
f ○∗ g =D −1 {D{f[n]} · D{g[n]}}
Recall, however, that this theorem relates only to circular convolution, not to linear
convolution.Wemaystillmakeuseoftheconvolutiontheoremifwepadthesignals
withzerovaluesinordertopreventunwantedoverlap.Thisisbestillustratedbyuse
of anexample.
Theproblemofcalculatingthecircularconvolutionh[n] = f ○∗ gforthesignals
f[n] = 9, −1,3andg[n] = 7,2, −4hasalreadybeenexplained(seeExample24.27).
Theresultofh[n] = 73,−1,−17couldhavebeenobtainedeithergraphicallyorusingtheconvolution
theorem.Thelinearconvolution ish[n] = 63,11, −17,10, −12
and can be found using the direct method, and this result is presented here for reference
(see Exercises 24.15.1, Question 2(c)).
To use the convolution theorem it is first necessary to take the original signals
and pad them with zero values so that the overall length of each is equal to the sum

24.15 Discrete convolution and correlation 811
of the two − 1. So in the simple example, both signals start with length 3, and after
padding they musthave a length of 3 +3−1 = 5.The padded signals are
f ′ [n] =9,−1,3,0,0
g ′ [n] = 7, 2, −4, 0, 0
The calculation of the first term in the output signal is shown graphically in Figure
24.29.
7
n = 0
0
–1
9
0
2
0
3 0
–4
Figure24.29
The padded signals f ′ [m] (inner circle)andg ′ [m] (outer
circle).
The first element ofh[n] is (9)(7) = 63. Notice how the padding is sufficient to
ensurethatallofthevaluesinthelinearconvolutioncanbeproducedwithoutoverlap.
In this position all of the other terms are multiplied by 0 and hence can be removed
from the calculation. By rotating the outer circle anticlockwise four more times the
other values can be calculated. It can be seen that this method is exactly equivalent
tocarrying out the linear convolution.
Hence wecan usethe convolution theoremon thepadded signals f ′ [n],g ′ [n]and
f ○∗ g= D −1 {D{f ′ [n]} · D{g ′ [n]}}
Thismethodissometimespreferredbecausethed.f.t.valuesofthesignalsarereadily
and quickly calculated on a computer.
The signals may be padded with additional zeros if desired with no effect on the
result. In order to optimize the computation speed for the d.f.t. it is often desirable
to use the f.f.t. algorithm mentioned earlier. One restriction of the f.f.t. is that the
numberofsamplesintheinputsignalmustbeapowerof2,soithastocontain2,4,
8,16,32,... samples.Sincepaddingispossiblewithoutaffectingtheresultthenthe
f.f.t.can be used.
As an example we consider adding reverb to a signalg[n], which is a 30 second
duration piece of music played on an electric guitar. The signalg[n] is very ‘clean’,
obtained bypluggingtheguitardirectlyintotherecording equipment. Asaresult,it
contains no reverberant room effects that would have been present if a microphone
andamplifierhadbeenusedinstead.Animpulseresponse f[n]isobtainedbybursting
a balloon in an environment to be simulated. The two signals prior to introducing
padding areshown inFigure 24.30.
The two padded Fourier transformed signals are multiplied together and the inversed.f.t.takentoproducethesignalinFigure24.31.Thesignalh[n]hasaduration
of 32 s due to the effects of linear convolution. When the signal is played through
speakersithasadistinctiveechoorreverbeffectthatisnotpresenting[n].Thereare
visible differences inthe signalifitis examined carefully.

812 Chapter 24 The Fourier transform
f [n]
Scale:
1s
g[n]
Figure24.30
Two audio signals:a2simpulseresponse ofan environment to be simulated, f[n], anda30s
music trackto be processed,g[n]. Bothsignals are sampled atarate of44100 samples per
second.
Figure24.31
The output signalh[n] = f ○∗ gincluding reverb.
EXERCISES24.15.3
1 Thecircular convolution of f[n] = 1, −1,1,3 and
g[n] = 7,2,0,1 wascalculated in Question 1 in
Exercises24.15.2. Verify the convolution theorem for
these sequences.
2 Usecircular convolution and padding with zerosto
obtain the linear convolution of f[n] = 9,0,1 and
g[n] = 5,4,5,2,1.Further, verifythe convolution
theorem forthese sequences.
3 Provethe circular convolution theorem.
Solutions
1 F[k] =4,4j,0,−4j.G[k] =10,7−j,4,7+j 2 45,36,50,22,14,2,1
24.15.4 Linearcross-correlation
The linear cross-correlation of two real sequences f[n] andg[n] is another sequence,
c[n] say, which wedenote by f ⋆g, which isdefined asfollows:
Linear cross-correlation of f[n]andg[n]:
∞∑
c[n]=f⋆g= f[m]g[m−n] forn=...−3,−2,−1,0,1,2,3...
m=−∞

24.15 Discrete convolution and correlation 813
Note the similarity between this definition and that of linear convolution defined in
Section 24.15.1. In the formula for cross-correlation the sequence g is not folded. If
the two sequences are finite and of length N, that is f[n] and g[n] are non-zero only
when 0 nN−1, there are 2N −1 terms in the cross-correlation sequence and the
formulacan be written as follows:
Linear cross-correlation of two finite sequences f[n] andg[n]:
and
c[n]=f⋆g=
c[n]=f⋆g=
∑N−1
m=n
N+n−1
∑
m=0
f[m]g[m−n]
f[m]g[m−n]
forn=0,1,2,...,N−1
forn=0,−1,−2,...,−(N−1)
Example24.31 Suppose f[n] = 7,2,−3 andg[n] = 1,9,−1. Assume both sequences f andgstart at
n=0.
(a) Find the linear cross-correlationc[n] = f ⋆gusing the formulae above.
(b) Develop a graphical interpretation of thisprocess.
Solution (a) Both f and g are finite sequences of length N = 3. Their cross-correlation is a
sequencec[n], forn = −2,−1,0,1,2, of length 5.
Usingthe formulae above withn = −2 gives
c[−2] =
0∑
f[m]g[m +2]
m=0
= f[0]g[2]
= (7)(−1)
= −7
Whenn = −1 we have
c[−1] =
1∑
f[m]g[m +1]
m=0
= f[0]g[1] + f[1]g[2]
= (7)(9) + (2)(−1)
= 61
The remaining terms are calculated in a similar fashion. You should calculate one
or two terms yourself toverifythat the fullsequence is
c[n] = −7,61,28,−25,−3
n = −2,−1,0,1,2
(b) The graphical interpretation is developed along the same lines as for linear convolution
in Example 24.26. Figure 24.32 shows the sequence f[m], form = 0,1,2,
denoted by the symbols ◦. Also shown isthe sequenceg[m] denoted by •.

814 Chapter 24 The Fourier transform
(a)
(b)
(c)
(d)
(e)
g[m ]
g[m – 1]
g[m – 2]
g[m + 1]
g[m + 2]
10
6 8
4
2
– 4
10
6 8
4
2
– 4
10
6 8
4
2
– 4
10
6 8
4
2
– 4
10
6 8
4
2
– 4
m
m
m
m
m
Figure24.32
The effect oftranslatingg[m].
n = 0
sum of products:
(7)(1) + (2)(9) + (–3)(–1) = 7 + 18 + 3 = 28
n = 1
sum of products:
(2)(1) + (–3)(9) = 2 – 27 = –25
n = 2
sum of products:
(–3)(1) = –3
n=–1
sum of products:
(7)(9) + (2)( –1) = 63 – 2 = 61
n= –2
sum of products:
(7)( –1) = –7
The graph ofg[m] can be translatednunits to the left or right by changing the
argument ofgfromg[m] tog[m −n]. Ifnis positive the graph moves to the right.
Study the figure to observe this. Correlation is the sum of products of f[m] and
g[m − n]. For each value of n, the graph of f[m] is superimposed. We are only
interested in values ofnfor which the graphs overlap -- otherwise each product is
zero.Foreachvalueofnthesuperimposedgraphsmakeiteasytoseewhichvalues
must be multiplied together and added. Placing these results in order confirms the
resultobtained inpart(a), thatisthe correlation is −7,61,28,−25,−3.
Whenasequenceiscross-correlatedwithitselftheprocessisknownasautocorrelation.
Autocorrelationisusedtosearchforpossibleperiodicitiesinsignals,becauseifasignal
isperiodicwithperiodN itsautocorrelationsequencewillshowpeaksatintervalsofN.
EXERCISES24.15.4
1 Find the linearcross-correlationofthe sequences
f[n] = 4,5,9 andg[n] = 3,1,1.
2 Show thatthe linearcross-correlationof f[n] andg[n]
can be written in the alternative form
f⋆g= ∑ ∞
m=−∞ f[m +n]g[m].

24.15 Discrete convolution and correlation 815
3 There are variants ofthe definition ofcorrelation.
Showthat if f ⋆gisredefined to be
∑ ∞m=−∞
f[m −n]g[m] then the corresponding
correlation ofthe sequences in Question1is
27,24,26,9,4.
4 Find the linear autocorrelation ofthe sequence
f[n]=3,2,1.
Solutions
1 4,9,26,24,27 4 3,8,14,8,3
24.15.5 Circularcross-correlation
The circular cross-correlation of two periodic sequences of period N is defined in a
similarmanner totheir circular convolution. Itisasequencec[n] oflengthN.
The circular cross-correlation of two periodic sequences, f[n] and g[n], each of
periodN, isdefined as
c[n]=f○⋆ g =
∑N−1
m=0
f[m]g[m−n]
forn=0,1,2,...,N−1
Whenasequenceiscross-correlatedwithitselftheprocessisknownasautocorrelation.
Example24.32 (a) Find the circular autocorrelation of the sequence f[n] = 3,2,1 using the formula.
(b) Develop a graphical method forperforming thiscalculation.
Solution (a) HereN = 3.From the definition
2∑
c[n]=f○⋆ f = f[m]f[m−n]
m=0
forn=0,1,2
2∑
c[0] = f[m]f[m]
m=0
= (3)(3) + (2)(2) + (1)(1)
= 14
2∑
c[1] = f[m]f[m−1]
m=0
= f[0]f[−1]+f[1]f[0]+f[2]f[1]
= (3)(1) + (2)(3) + (1)(2)
= 11

816 Chapter 24 The Fourier transform
n=0 n=1 n=2
3
1
2
3
3
3
2
2 1
1
3
2 1
2
1
2 1
3
Sums of products:
(3)(3) + (2)(2) + (1)(1) = 14 (3)(1) + (2)(3) + (1)(2) = 11 (3)(2) + (2)(1) + (1)(3) = 11
Figure24.33
Graphical calculation ofthe autocorrelation of f[n] = 3,2,1.
c[2] =
2∑
f[m]f[m−2]
m=0
= (3)(2) + (2)(1) + (1)(3)
= 11
Hencec[n] = 14,11,11.
(b) The graphical method involves listing the sequence f[m], m = 0,1,2, around an
inner circle. Around an outer circle we listitagain. This method is identical to that
used in Example 24.27 for circular convolution, but because now there is no folding,
the sequence on the outer circle is not reversed. The calculation can be seen in
Figure 24.33.
24.15.6 (Circular)correlationtheorem
For real sequences f[n]andg[n] the correlation theorem states:
D{f ○⋆ g} =F[k]G[k]
whereG[k] denotes the complex conjugate ofG[k].
This provides a technique forcalculating a correlation usingthe d.f.t.
Example24.33 Findthe circularcorrelation f ○⋆ gwhen f[n] = 8,−9,3,2 andg[n] = 11,4,−1,−5.
Solution Either directly from the definition of the d.f.t., or by using a computer package, we can
show that
F[k] = 4,5 +11j,18,5 −11j G[k] = 9,12 −9j,11,12 +9j
Theconjugate ofG[k] isG[k] = 9,12 +9j,11,12 −9j. Then
F[k]G[k] = 36,−39 +177j,198,−39 −177j

24.15 Discrete convolution and correlation 817
Eitherdirectly,orusingacomputerpackage,takingtheinversed.f.t.yieldsthesequence
39,−129,78,48
which is the required circular correlation. You may like to verify this by directly calculating
f ○⋆ g.
Itfollowsfromthetheoremthatthecircularautocorrelationsequencehasthefollowing
property:
D{f ○⋆ f}=F[k]F[k]
Example24.34 Usethe d.f.t. tofind the circularautocorrelationofthe sequence f[n] = 8,3,−1,2.
Solution Itisstraightforward buttedioustoshow thatthe d.f.t.of f[n] is
F[k]=12,9−j,2,9+j
Then,the conjugate ofF[k] isF[k]=12,9+j,2,9−j,and
F[k]F[k] = 144,82,4,82
Finally,takingtheinversed.f.t.givesthesequence78,35,−4,35whichyoucanverify
isthe circular autocorrelation of f[n] = 8,3,−1,2.
Engineeringapplication24.5
Useofcorrelationinradar
Both sonar and radar operate by transmitting a signal which bounces off a distant
target and returns to a receiver. The main difference is that sonar uses sound waves
whereasradarusesradiowaves.Thetimetakenforthesignaltoreturntothereceiver
having reached the target can be used to deduce,d, the distance between the transmitter
and the target. In both radar and sonar, the received signal is typically very
much smaller than the transmitted signal and can contain a lot of unwanted noise.
High-gain electronic amplifiers may be required to make the signal large enough to
be processed and these introduce even more noise. In addition the receiver experiences
interference from other sources which has to be separated from the returning
signal. Correlation techniques are very useful for retrieving a signal from the noise
and the interference.
We will consider a radar system. The signals transmitted are usually modulated
usingahigh-frequencycarriersignal(seeEngineeringapplication24.1).Forourpurposesthe
presence ofacarrier signalisnotimportant.
However, the high-frequency component of the signal can be removed
electronically prior to the signals being sampled by a process called
demodulation.
➔

818 Chapter 24 The Fourier transform
The received and demodulated signal is represented by f[n] and the transmitted
signal prior to modulation is represented byg[n]. Consider the case whereg[n] is a
square pulse comprising 10 samples. Let the time between each sample beT. Both
signalsareshownplottedinFigure24.34.Notethatthisisanidealizedcaseinwhich
thereis no noise on the received signal.
1.0
0.8
0.6
g[n]
0.4
f [n]
0.2
20 40 60 80 100 120 140
Figure24.34
Plotoftransmitted and received digital signals foraradar system.
n
The returned signal is delayed due to the time taken for it to travel to a distant
target and back. Examining Figure 24.34 we note that the returned signal is delayed
by 65 samples when compared with the transmitted signal. This equates to a time
delayof65T.Weknowthespeedthatthetransmittedpulsetravelsisc,thespeedof
light.Using
distance =time × velocity
gives total distance travelled = 65T ×c. But the total distance travelled is twice the
distance between the transmitter and the target,thatis2d. Henced = 65Tc
2 .
We now confirm the number of samples delay by calculating the circular crosscorrelation
of f[n] andg[n]. We wish to calculatec[n] = f ○∗ gand plot this on a
graphforanalysis.Thiscanbedoneeitherdirectlybycalculatingthecircularcrosscorrelation
or by using the circular correlation theorem. Owing to the large number
ofcalculationsinvolvedtheyarenotpresentedhere.Normallysuchaprocesswould
be carriedout usingacomputer.
TheresultisasequenceofsampleswhichareplottedinFigure24.35.Thelargest
value of c[n] occurs at 65, which corresponds to the delay in samples between the
two signals. It agrees with our initial observation where the difference was quite
straightforward tosee by inspecting the two signals.

24.15 Discrete convolution and correlation 819
c[n]
2.0
1.5
1.0
0.5
20 40 60 80 100 120 140
Figure24.35
Cross-correlationfunctionc[n] forthe signals f[n] andg[n].
n
We now consider a more realistic case in which there is random noise on the returnedsignalandcarryoutthesamecalculationthatwedidpreviously.Figure24.36
shows the two signals. Although the return signal is still present it is no longer possible
to accurately determine where it actually starts and finishes by just looking at
the graph.
1.0
0.8
0.6
g[n]
0.4
f [n]
0.2
20 40 60 80 100 120 140
Figure24.36
Plotoftransmittedandreceiveddigitalsignalsforaradarsysteminwhichnoiseispresenton
the returnedsignal.
n
➔

820 Chapter 24 The Fourier transform
Carryingoutthecircularcross-correlationonthesesignalsgivestheresultshown
in Figure 24.37. The highest peak still occurs at 65 time intervals, allowing the distance
of the remoteobject tobe accurately calculated.
c[n]
2.0
1.5
1.0
0.5
20 40 60 80 100 120 140 n
Figure24.37
Cross-correlationfunctionc[n] forthe signals f[n]andg[n] forthe case where f[n]contains
additive noise.
Thecross-correlationfunctionallowsthedistanceoftheremoteobjecttobecalculatedevenwhenthereisalargequantityofnoiseonthereturnedsignal.Itistherefore
a very useful signal processing operation for radar systems. Similar considerations
apply when working with sonar signals.
EXERCISES24.15.6
1 Find the circular cross-correlationof f[n] = 2,3, −1
andg[n] = 8,7,1. Verify the correlation theorem.
2 Fromthe definition,prove the correlationtheorem.
3 There are variants ofthe definition ofcorrelation.
∑
Show that if f ○⋆ gis redefined to be
N−1
m=0 f[m −n]g[m] then the corresponding
correlation theorem states D{f ○⋆g}=F[k]G[k].
Solutions
1 36,19, 9

Review exercises 24 821
REVIEWEXERCISES24
1 Find the Fourier transformsof
{
1−t 2 |t|<1
(a)f(t) =
0 otherwise
{ sint |t|<π
(b)f(t) =
0 otherwise
{ 1 0<t<τ
(c)f(t) =
0 otherwise
{ e −αt t>0
(d)f(t) =
−e αt α >0
t<0
2 Find the Fourier integral representations of
{ 3t |t|<2
(a) f(t) =
0 otherwise
⎧
⎨0 t<0
(b) f(t) = 6 0<t<2
⎩
0 t>2
3 (a) IfF(ω) = F{f(t)},showthat
F{f ′ (t)} = jωF(ω).
(b) IfF(ω) = F{f(t)},showthat
F{f (n) (t)} = (jω) n F(ω).Theseresultsenableus
to calculate the Fourier transformsofderivatives
offunctions.
4 IfF(ω) = F{f(t)},showthat
( )
F{f(at)} = 1 a F ω
.
a
5 IfF(ω)is the Fourier transform of f (t)show that
(a) F(0) = ∫ ∞
−∞ f(t)dt.
(b) f(0) = 1 ∫ ∞
F(ω)dω.
2π −∞
(c) Show that if f (t)isan even function,
F(ω) =2 ∫ ∞
0 f(t)cosωtdt.
6 Theconvolution theoremgiven in Section24.8.1
representsconvolution in the time domain.
Convolution can alsobe performedin the frequency
domain,in whichcasethe equivalent convolution
theoremis
F{f(t)g(t)} = 1
2π [F(ω)∗G(ω)]
Provethe convolution theorem in thisform.
7 (a) Given that the Fourier transform of f (t) = e −α|t|
2α
is
α 2 use thet--ω duality principleto find
+ω2 the Fourier transformof
1
α 2 +t 2.
(b) Fromatable oftransformswrite down the
Fourier transformofcosbt.
(c) Use the convolution theoremobtained in
Question 6 to find the Fourier transformof
cosbt
α 2 +t 2,forb>0.
(d) Using the result in part(c)evaluate the integral
∫ ∞ cosbt
−∞ α 2 +t 2dt
8 Find the d.f.t. ofthe sequence f[n] = {3,3,0,3}.
Verify Rayleigh’stheoremforthissequence.
9 Findthelinearconvolutionofthetwofinitesequences
f[n] = 3, −1, −7 andg[n] = 4,0, 1 2 .
10 The signum functionis defined to be
⎧
⎨ 1 t>0
sgn(t) = −1 t<0
⎩
0 t=0
This functioncan be representedasthe exponential
functione −ǫt ,ift > 0, andas −e ǫt ,ift < 0,in the
limitas ǫ → 0.
(a) Show that
∫ ∞
sgn(t)e −jωt dt = 2
−∞ jω
(b) Usethet--ω duality principle to showthat
∫ ∞
−∞
1
πt e−jωt dt = −jsgn(ω)
(c) Usethe second resultin part (b), the convolution
theorem and the integral propertiesofthe delta
function to showthat
1
∗cos(αt) = sin(αt)
πt
11 Show that f[n] ○⋆ g[n] = f[n] ○∗ g[−n] andhence
deduce thatacorrelationcanbe expressed in termsof
a convolution.

822 Chapter 24 The Fourier transform
Solutions
1 (a)
4(sinω −ωcosω)
ω 3
(b) − 2jsinωπ
1−ω 2
sinωτ j(cosωτ −1)
(c) +
ω ω
(d) −
2jω
α 2 +ω 2
∫
1 ∞ 6
2 (a)
2π −∞ ω 2 (2jωcos2ω −jsin2ω)ejωt dω
(b)
7 (a)
(c)
(d)
∫
1 ∞ 6
2π −∞ ω [sin2ω + (cos2ω −1)j]ejωt dω
π
α e−α|ω|
π
2α [e−α|ω+b| +e −α|ω−b| ]
πe −αb
α
8 F[k] =9,3,−3,3
9 12, −4, −26.5, −0.5, −3.5

25 Functionsofseveral
variables
Contents 25.1 Introduction 823
25.2 Functionsofmorethanonevariable 823
25.3 Partialderivatives 825
25.4 Higherorderderivatives 829
25.5 Partialdifferentialequations 832
25.6 TaylorpolynomialsandTaylorseriesintwovariables 835
25.7 Maximumandminimumpointsofafunctionoftwovariables 841
Reviewexercises25 846
25.1 INTRODUCTION
Inengineeringtherearemanyfunctionswhichdependuponmorethanonevariable.For
example,thevoltageonatransmissionlinedependsuponpositionalongthetransmission
lineaswellastime.Theheight ofliquidinatankdepends upon theflowratesintoand
outof the tank. We shall examine some more examples inthis chapter.
Wehavealreadydiscusseddifferentiationoffunctionsofonevariable.However,we
alsoneedtobeabletodifferentiatefunctionsoftwoormorevariables.Thisisachieved
by allowing one variable to change at a time, holding the others fixed. Differentiation
under these conditions iscalledpartial differentiation.
25.2 FUNCTIONSOFMORETHANONEVARIABLE
InChapter10wesawhowtodifferentiateafunctiony(x)withrespecttox.Manystandard
derivatives were listed and some techniques explained. Sinceyis a function ofx
wecallythedependentvariableandxtheindependentvariable.Thefunctionydepends
upon the one variablex. Consider the following example.
The area,A, ofacircle depends only upon the radius,r, and isgiven by
A(r) = πr 2

824 Chapter 25 Functions of several variables
The rate of change of area w.r.t. the radius is dA = 2πr. In practice functions often
dr
dependuponmorethanonevariable.Forexample,thevolume,V,ofacylinderdepends
upon the radius,r, and the height,h, and isgiven by
V = πr 2 h
whereV isthedependentvariable;r andhareindependentvariables.V isafunctionof
the two independent variables,r andh. We writeV =V(r,h).
Engineeringapplication25.1
Electricalpotentialinsideacathoderaytube
In recent years the cathode ray tube has declined in popularity as a screen display
device. It is now rarely used in consumer electronics as flat screen televisions have
becomemorepopular.However,itisstillsometimesusedinthespecialistelectronics
sector and instruments such as cathode ray oscilloscopes are still commonly in use
inmany laboratories.
Withinacathoderaytubetheelectricalpotential,V,willvarywithspatialposition
and time.Given Cartesiancoordinatesx,yandz, wecan write
V =V(x,y,z,t)
toshow thisdependence. Note thatV isafunctionoffourindependentvariables.
Engineeringapplication25.2
Powerdissipatedinavariableresistor
Thepower,P,dissipatedinavariableresistordependsupontheinstantaneousvoltage
acrossthe resistor, v, and the resistance,r. Itisgiven by
P = v2
r
Hence we may writeP =P(v,r) to show this dependence. The power is a function
of two independent variables.
Asanotherexampleofafunctionofmorethanonevariableconsiderathree-dimensional
surface as shown in Figure 25.1. The height,z, of the surface above thex--y plane depends
upon thexandycoordinates, that isz = z(x,y). If we are given values ofxand
y, then z(x,y) can be evaluated. This value of z is the height of the surface above the
point (x,y). We write, for example,z(3,−1) for the value ofzevaluated whenx = 3
and y = −1. The dependent variable, z, is a function of the independent variables
x andy.
Some important features are shown in Figure 25.1. The value of z at a maximum
point is greater than the values of z at nearby points. Point A is such a point. As you

25.3 Partial derivatives 825
D
C
A
z = z(x,y)
x
z
y
E
B
Figure25.1
Theheight ofthe surface
above thex--yplane isz.
move away from A the value of z decreases. A minimum point is similarly defined.
At a minimum point, the value ofzis smaller than thezvalue at nearby points. This is
illustratedbypointB.PointCillustratesasaddlepoint.Atasaddlepoint,zincreasesin
one direction, axis D on the figure, but decreases in the direction of axis E. These axes
are at right angles to each other. The term ‘saddle’ is descriptive as a horse saddle has
a similar shape. Maximum points, minimum points and saddle points are considered in
greater depth inSection 25.7.
25.3 PARTIALDERIVATIVES
Consider
z =z(x,y)
thatis,zisafunctionoftheindependent variablesxandy.Wecandifferentiatezeither
w.r.t.x,orw.r.t.y.Weneedsymbolstodistinguishbetweenthesetwocases.Whenfinding
the derivative w.r.t. x, the other independent variable, y, is held constant and only
x changes. Similarly when differentiating w.r.t. y, the variable x is held constant. We
write ∂z
∂x todenotedifferentiationofzw.r.t.xforaconstanty.Itiscalledthefirstpartial
derivative of z w.r.t. x. Similarly, the first partial derivative of z w.r.t. y is denoted ∂z
∂y .
Referring to the surfacez(x,y), ∂z gives the rate of change ofzmoving only in thex
∂x
direction, and henceyisheld fixed.
Ifz =z(x,y), thenthe firstpartialderivatives ofzare
∂z
∂x
and
∂z
∂y
If we wish to evaluate a partial derivative, say ∂z
∂x , at a particular point (x 0 ,y 0
), we
indicate thisby
∂z
∂x (x 0 ,y 0 ) or ∂z
∣
∂x
∣
(x0 ,y 0
)
justas wedid forfunctions of one variable.

826 Chapter 25 Functions of several variables
Example25.1 Givenz(x,y) =x 2 y +sinx +xcosyfind ∂z ∂z
and
∂x ∂y .
Solution To find ∂z we differentiate z w.r.t. x, treating y as a constant. Note that since y is a
∂x
constant then soiscosy.
∂z
∂x =2xy+cosx+cosy
Infinding ∂z
∂y ,xand hencex2 and sinx areheld fixed, thus
∂z
∂y =x2 −xsiny
Example25.2 Givenz(x,y) = 3e x −2e y +x 2 y 3
(a) findz(1,1)
(b) find ∂z ∂z
and
∂x ∂y whenx=y=1.
Solution (a) z(1,1) = 3e 1 −2e 1 +1 = 3.718
∂z
(b)
∂x =3ex +2xy 3
∂z
∂y =−2ey +3x 2 y 2
Whenx =y=1,then
∂z
∂x =3e+2=10.155
∂z
∂y =−2e+3=−2.437
Atthepoint(1,1,3.718)onthesurfacedefinedbyz(x,y),theheightofthesurface
abovethex--yplaneisincreasinginthexdirection,anddecreasingintheydirection.
Note thatwecould alsowrite
∂z
∂z
∂x∣ = 10.155 and
(1,1)
∂y∣ = −2.437
(1,1)
or
∂z
∂x (1,1) = 10.155 and ∂z
(1,1) = −2.437
∂y
Both notations areincommon use.

25.3 Partial derivatives 827
Engineeringapplication25.3
Eddycurrentlosses
Eddycurrentsarecirculatingcurrentsthatariseinironcoresofelectricalequipment
as a resultof ana.c. magnetic field. They lead toenergy lossesgiven by
where
P e
=k e
f 2 B 2 max
P e
= eddy currentlosses (Wper unit mass),
B max
= maximum value of the magnetic field wave (T),
f = frequency of the magnetic field wave (Hz),
k e
= a constant thatdepends upon factors such as the lamination thickness of
the ironcore.
Calculate ∂P e
∂f , and ∂P e
∂B max
.
Solution
∂P e
∂f
=2k e
fB 2 max
and
∂P e
∂B max
=2k e
f 2 B max
Example25.3 IfV(x,y) = sin(xy),find ∂V
∂x
and
∂V
∂y .
Solution To find ∂V
∂x we treat y as a constant. Recalling that d
(sinkx) = kcoskx we find
dx
∂V
∂V
∂V
=ycos(xy).Tofind wetreatxas a constant. Thus
∂x ∂y ∂y =xcos(xy).
Example25.4 Findthe first partialderivatives ofzwhere
(a) z(x,y) =yxe x
(b) z(x,y) =x 2 sin(xy)
Solution (a) To find ∂z we must treatyas a constant. However, the differentiation of the factor
∂x
xe x will require use of the product rule.We find
∂z
∂x =y ∂ ∂x (xex )
=y((1)e x +xe x )
=ye x (x+1)

828 Chapter 25 Functions of several variables
To find ∂z the variablexisheld constant and so
∂y
∂z
∂y =xex
(b) Observe the product term in the variable x which means we shall need to use the
product rule.We find
∂z
∂x = 2xsin(xy) +x2 (ycos(xy)) = 2xsin(xy) +x 2 ycos(xy)
To find ∂z wetreatxas a constant and find
∂y
∂z
∂y =x2 (xcos(xy)) =x 3 cos(xy)
EXERCISES25.3
(c) y=te t +x (d) y=3x 2 e 2t +t 3 e −x
1 Find the firstpartialderivatives of
(a) y=te x (b) y=x 2 e −t h=1.
(a) z=2x+3y (b) z=x−y+x 2 (e) y = e xt (f) y = e 2x+3t
(c) z=x 2 +y 2 +3 (d) z=xy−1 5 Findthe firstpartialderivatives ofzwhere
(e)z =x 2 y+xy 2
(a)z= √ x 2 +y 2 x
(b) z=
(f)z=x 3 +2x 2 y−4xy 2 +2y 3
2x+3y
( )
x
2 Find the firstpartialderivatives of
(c) z = sin (d) z = e 3xy
(a)z= x (b)z= √ y
xy
y
(e) z = ln(2x −3y) (f) z = ln(xy)
( )
(c)z= y2
x 3 +2 (d)z= 1 y
(g) z =xln(xy) (h) z =xln
xy
x
√
(e)z= 3x2 y
y − (f)z= √ xy−3(x+y) 6 Evaluate the first partialderivatives of f atx = 1,
x
y=2.
3 Find the firstpartialderivatives of
(a) f =3x 2 y−2xy (b) f = x +y
(a) z=2sinx+3cosy (b) z=xsiny
y
(c) z=xtany−y 2 (c) f = 2sin(3x +2y) (d) f = 2e xy
sinx (d) z=xysiny
(e) f =ylnx+ln(xy) (f) f = √ 3x+y
(e) z =sin(x +y) (f) z =4cos(4x −6y)
(g)z= siny
7 Given
x
f(r,h) =2r 2 h− √ rh
4 Find the firstpartialderivatives ofy.Note thatyisa
functionofxandt.
evaluate the firstpartial derivatives of f whenr = 2,
Solutions
∂z ∂z
1 (a) = 2,
∂x ∂y = 3
(b) 1 +2x,−1
(c) 2x,2y
(d) y,x
(e) 2xy +y 2 ,x 2 +2xy
(f) 3x 2 +4xy −4y 2 ,2x 2 −8xy +6y 2

25.4 Higher order derivatives 829
∂z
2 (a)
∂x = 1 y , ∂z
∂y = − x y 2
y
(b)
2 √ x , √ x
(c) − 3y2
x 4 , 2y
x 3
(d) − 1
x 2 y ,− 1
xy 2
(e)
(f)
√
6x y
y + x 2 , −3x2 y 2 − 1
2x √ y
1
2√ y
x −3, 1 2
√ x
y −3
∂z ∂z
3 (a) =2cosx,
∂x ∂y =−3siny
(b) siny,xcosy
(c) tany −y 2 cosx,xsec 2 y −2ysinx
(d) ysiny,xsiny+xycosy
(e) cos(x +y),cos(x +y)
(f) −16sin(4x −6y),24sin(4x −6y)
(g) − siny
x 2 , cosy
x
∂y
4 (a)
∂x =tex , ∂y
∂t = ex
(b) 2xe −t ,−x 2 e −t
(c) 1,e t (1+t)
(d) 6xe 2t −t 3 e −x ,6x 2 e 2t +3t 2 e −x
(e) te xt ,xe xt
(f) 2e 2x+3t ,3e 2x+3t
5 (a)
(b)
(c)
∂z
∂x =
x ∂z
√
x 2 +y2, ∂y =
3y
(2x +3y) 2 , − 3x
(2x +3y) 2
( ( )
1
y cos x
), − x y y 2 cos x
y
(d) 3ye 3xy ,3xe 3xy
(e)
(f)
2
2x−3y ,− 3
2x−3y
1
x , 1 y
(g) 1 +ln(xy), x y
( )
y
(h) ln −1, x x y
y
√
x 2 +y 2
∂f ∂f
6 (a) (1,2) =8,
∂x ∂y (1,2)=1
(b) 0.5, −0.25
7
(c) 4.5234, 3.0156
(d) 29.5562, 14.7781
(e) 3,0.5
(f) 0.6708, 0.2236
∂f
∂f
(2,1) = 7.6464, (2,1) = 6.5858
∂r ∂h
25.4 HIGHERORDERDERIVATIVES
Just as functions of one variable have second and higher derivatives, so do functions of
several variables. Consider
z =z(x,y)
Thefirstpartialderivativesofzare ∂z ∂z
and
∂x ∂y .Thesecondpartialderivativesarefound
bydifferentiatingthefirstderivatives.Wecandifferentiatefirstpartialderivativeseither
w.r.t.xor w.r.t.ytoobtain various second partialderivatives:
differentiating ∂z
∂x w.r.t.xproduces ∂ ( ) ∂z
∂x ∂x
differentiating ∂z
∂x w.r.t.yproduces ∂ ( ) ∂z
∂y ∂x
= ∂2 z
∂x 2
= ∂2 z
∂y∂x

830 Chapter 25 Functions of several variables
differentiating ∂z
∂y w.r.t.xproduces ∂ ( ) ∂z
∂x ∂y
differentiating ∂z
∂y w.r.t.yproduces ∂ ( ) ∂z
∂y ∂y
= ∂2 z
∂x∂y
= ∂2 z
∂y 2
For mostcommon functions, the mixed derivatives ∂2 z
∂y∂x and ∂2 z
∂x∂y areequal.
Example25.5 Given
z(x,y) = 3xy 3 −2xy +sinx
find allsecondpartialderivatives ofz.
Solution
∂z
∂x =3y3 −2y+cosx
∂ 2 z
∂x = ∂ ( ) ∂z
=−sinx
2 ∂x ∂x
∂ 2 z
∂x∂y = ∂ ( ) ∂z
=9y 2 −2
∂x ∂y
Note that
∂ 2 z
∂x∂y = ∂2 z
∂y∂x .
∂z
∂y = 9xy2 −2x
∂ 2 z
∂y = ∂ ( ) ∂z
= 18xy
2 ∂y ∂y
∂ 2 z
∂y∂x = ∂ ( ) ∂z
=9y 2 −2
∂y ∂x
Example25.6 Given
H(x,t) =3x 2 +t 2 +e xt
verify that ∂2 H
∂x∂t = ∂2 H
∂t∂x .
Solution
∂H
∂x =6x+text
∂ 2 H
∂t∂x = ∂ ∂t
∂ 2 H
∂x∂t = ∂ ∂x
∂H
∂t
=2t+xe xt
( ) ∂H
= ∂ ∂x ∂t (6x+text )= e xt +txe xt
( ) ∂H
= ∂ ∂t ∂x (2t+xext )=e xt +xte xt
Third,fourthandhigherderivativesarefoundinasimilarwaytofindingsecondderivatives.
The third derivatives are found by differentiating the second derivatives and
so on.

25.4 Higher order derivatives 831
Example25.7 Find all third derivatives of f (r,s) = sin(2r) −3r 4 s 2 .
Solution
∂f
∂r = 2cos(2r) −12r3 s 2
∂ 2 f
∂r 2 = −4sin(2r) −36r2 s 2
∂ 2 f
∂r∂s = −24r3 s
∂f
∂s = −6r4 s
∂ 2 f
∂s 2 = −6r4
The third derivatives are ∂3 f
∂r , ∂ 3 f
3 ∂r 2 ∂s , ∂ 3 f
∂r∂s and ∂3 f
, and these are found by differentiating
the second derivatives.
2 ∂s3 ∂ 3 f
∂r = ∂ ( ) ∂ 2 f
= −8cos(2r) −72rs 2
3 ∂r ∂r 2
∂ 3 f
∂r 2 ∂s = ∂ ( ∂ 2 )
f
= −72r 2 s
∂r ∂r∂s
∂ 3 f
∂r∂s = ∂ ( ∂ 2 )
f
= −24r 3
2 ∂r ∂s 2
∂ 3 f
∂s = ∂ ( ∂ 2 )
f
= 0
3 ∂s ∂s 2
Note that the mixed derivatives can be calculated inavariety ofways.
∂ 3 f
∂r 2 ∂s = ∂ ( ∂ 2 )
f
= ∂ ( ∂ 2 )
f
∂r ∂r∂s ∂s ∂r 2
∂ 3 f
∂r∂s = ∂ ( ) ∂ 2 f
= ∂ ( ) ∂ 2 f
2 ∂r ∂s 2 ∂s ∂r∂s
EXERCISES25.4
1 Calculateall second derivatives of v where
v(h,r) =r 2√ h
2 Find the second partialderivatives of f given
(a) f =x 2 y+y 3 (b) f =2x 4 y 3 −3x 3 y 5
(c) f=4 √ xy 2 (d) f = x2 +1
y
(e) f= 3x3 √ y
(f) f=4 √ xy
3 Find allsecond partialderivatives of
(a) z =xe 2y
(c) z =xcos(2x +3y)
(e) z = e x siny
(g) z = e xy
(b) z = 2sin(xy)
(d) z =ysin(4xy)
(f) z = e 3x−y
4 Find allsecond partialderivatives of
(a) z = (3x −2y) 20
(c) z = sin(x 2 +y 2 )
1
(e) z=
3x−2y
(b) z = √ 2x+5y
(d) z = ln(2x +5y)
5 Find allthirdpartialderivatives ofzwhere
z(x,y) =
x2
y +1
6 Evaluate all secondpartialderivatives of f atx = 2,
y=1.
(a) f =ye x (b) f =sin(2x −y)
( )
y
(c) f=ln
x

832 Chapter 25 Functions of several variables
Solutions
1
2 (a)
∂ 2 v
∂h 2 = − r2
4h 3/2 , ∂2 v
∂h∂r = r √
h
, ∂2 v
∂r 2 = 2√ h
∂ 2 f
∂x 2 = 2y, ∂2 f
∂x∂y = 2x, ∂2 f
∂y 2 = 6y
(b) 24x 2 y 3 −18xy 5 ,
24x 3 y 2 −45x 2 y 4 ,
12x 4 y −60x 3 y 3
(c) −x −3/2 y 2 ,4x −1/2 y,8 √ x
(d) 2 y , −2x y 2 , 2(x2 +1)
y 3
(e) 18xy −1/2 , − 9 2 x2 y −3/2 , 9 4 x3 y −5/2
(f) −x −3/2 y 1/2 ,x −1/2 y −1/2 , −x 1/2 y −3/2
∂ 2 z
3 (a)
∂x 2 = 0, ∂2 z
∂x∂y =2e2y , ∂2 z
∂y 2 = 4xe2y
(b) −2y 2 sin(xy),
4 (a)
2cos(xy) −2xysin(xy),
−2x 2 sin(xy)
(c) −4sin(2x +3y) −4xcos(2x +3y),
−3sin(2x +3y) −6xcos(2x +3y),
−9xcos(2x +3y)
(d) −16y 3 sin(4xy),
8ycos(4xy) −16xy 2 sin(4xy),
8xcos(4xy) −16x 2 ysin(4xy)
(e) e x siny,e x cosy, −e x siny
(f) 9e 3x−y ,−3e 3x−y ,e 3x−y
(g) y 2 e xy ,e xy (1 +xy),x 2 e xy
∂ 2 z
∂x 2 = 3420(3x −2y)18 ,
∂ 2 z
∂x∂y = −2280(3x −2y)18 ,
5
∂ 2 z
= 1520(3x −2y)18
∂y2 (b) −(2x +5y) −3/2 ,
− 5 2 (2x +5y)−3/2 ,
− 25 (2x +5y)−3/2
4
(c) 2cos(x 2 +y 2 ) −4x 2 sin(x 2 +y 2 ),
−4xysin(x 2 +y 2 ),
2cos(x 2 +y 2 ) −4y 2 sin(x 2 +y 2 )
4
(d) −
(2x +5y) 2,
(e)
10
−
(2x +5y) 2,
25
−
(2x +5y) 2
18
(3x −2y) 3 , − 12
(3x −2y) 3 , 8
(3x −2y) 3
∂ 3 z
∂x 3 = 0, ∂ 3 z
∂x 2 ∂y = − 2
(y +1) 2,
∂ 3 z
∂x∂y 2 = 4x ∂3 z
(y +1) 3, ∂y 3 = − 6x2
(y +1) 4
6 (a)
∂ 2 f
(2,1) = 7.3891,
∂x2 ∂ 2 f
(2,1) = 7.3891,
∂x∂y
∂ 2 f
∂y 2 (2,1)=0
(b) −0.5645,0.2822, −0.1411
(c) 0.25,0, −1
25.5 PARTIALDIFFERENTIALEQUATIONS
Partial differential equations (p.d.e.s) occur in many areas of engineering. If a variable
dependsupontwoormoreindependentvariables,thenitislikelythisdependencecanbe
describedbyap.d.e.Theindependentvariablesareoftentime,t,andspacecoordinates
x,y,z.
One example is the wave equation. The displacement,u, of the wave depends upon
time and position. Under certain assumptions, the displacement of a wave travelling in
one direction satisfies
∂ 2 u ∂2 u
∂t 2 =c2 ∂x 2

25.5 Partial differential equations 833
wherecisthespeedofthewave.Thisp.d.e.iscalledtheone-dimensionalwaveequation.
Underprescribedconditionsthesubsequentdisplacementofthewavecanbecalculated
as a function of position and time.
Example25.8 Verifythat
u(x,t) = sin(x +2t)
isasolution ofthe one-dimensional wave equation
∂ 2 u
∂t = u
2 4∂2 ∂x 2
Solution The first partialderivatives arecalculated.
∂u
∂u
= cos(x +2t)
∂x
∂t =2cos(x+2t)
The second derivatives, ∂2 u
∂x and ∂2 u
, arenow found.
2 ∂t2 ∂ 2 u
∂x = −sin(x+2t) ∂ 2 u
= −4sin(x +2t)
2 ∂t2 Now
∂ 2 u
∂t = −4sin(x +2t) = 4[−sin(x +2t)] = u
2 4∂2 ∂x 2
Henceu(x,t) = sin(x +2t) isasolution ofthe given wave equation.
Another equally important p.d.e. is Laplace’s equation. This equation is used extensivelyinelectrostatics.Undercertainconditionstheelectrostaticpotentialinaregionis
describedby a function φ(x,y) which satisfiesLaplace’s equation intwo dimensions.
∂ 2 φ
∂x + ∂2 φ
2 ∂y = 0 2
Thisequationissoimportantthatawholeareaofappliedmathematics,calledpotential
theory, isdevoted tothe studyof its solution.
Example25.9 Verifythat
φ(x,y,z) =
1
√
x2 +y 2 +z 2
satisfies the three-dimensional Laplace’s equation
∂ 2 φ
∂x + ∂2 φ
2 ∂y + ∂2 φ
2 ∂z = 0 2
Solution We begin by calculating the first partialderivative, ∂φ . We aregiven
∂x
φ = (x 2 +y 2 +z 2 ) −1/2
and so
∂φ
∂x = −1 2 (x2 +y 2 +z 2 ) −3/2 (2x) = −x(x 2 +y 2 +z 2 ) −3/2

834 Chapter 25 Functions of several variables
We use the product ruletofind ∂2 φ
∂x . 2
∂ 2 φ
∂x = 2 −1(x2 +y 2 +z 2 ) −3/2 −x ( )
− 3 2 (x 2 +y 2 +z 2 ) −5/2 2x
= (x 2 +y 2 +z 2 ) −5/2 [−(x 2 +y 2 +z 2 ) +3x 2 ]
By a similaranalysis we have
∂ 2 φ
∂y 2 = (x2 +y 2 +z 2 ) −5/2 [−(x 2 +y 2 +z 2 ) +3y 2 ]
∂ 2 φ
∂z = 2 (x2 +y 2 +z 2 ) −5/2 [−(x 2 +y 2 +z 2 ) +3z 2 ]
So
∂ 2 φ
∂x + ∂2 φ
2 ∂y + ∂2 φ
2 ∂z = 2 (x2 +y 2 +z 2 ) −5/2 [−3(x 2 +y 2 +z 2 ) +3x 2 +3y 2 +3z 2 ]
= 0
1
Hence φ = √ isasolution of the three-dimensional Laplace’s equation.
x2 +y 2 +z2 Thetransmissionequationisanotherimportantp.d.e.Thepotential,u,inatransmission
cablewith leakagesatisfies a p.d.e. ofthe form
∂ 2 u u
∂x 2 =A∂2 ∂t +B∂u 2 ∂t +Cu
whereA,BandC areconstants relating tothe physical properties of the cable.
Theanalyticalandnumericalsolutionofp.d.e.sisanimportanttopicinengineering.
Coverage isbeyond the scope ofthis book.
EXERCISES25.5
1 Verify that
u(x,y) =x 2 +xy
is asolution ofthe p.d.e.
2 Verify that
∂u
∂x −2∂u ∂y =y
φ = sin(xy)
satisfiesthe equation
3 Verify that
∂ 2 φ
∂x 2 + ∂2 φ
∂y 2 +(x2 +y 2 )φ=0
u(x,y) =x 3 y +xy 3
isasolution ofthe equation
4 Verify that
xy ∂2 u
∂x∂y +x∂u ∂x +y∂u ∂y = 7u
u(x,y) =xy + x y
isasolution of
5 Verify that
satisfies
y ∂2 u
∂y 2 +2x ∂2 u
∂x∂y = 2x
φ(x,y)=xsiny+e x cosy
∂ 2 φ
∂x 2 + ∂2 φ
∂y 2 =−xsiny

25.6 Taylor polynomials and Taylor series in two variables 835
6 Given
φ =
(a) Show
√
x 2 +y 2
∂ 2 φ
∂x 2 =y2 (x 2 +y 2 ) −3/2
(b) Verify that φ is asolution of
∂ 2 φ
∂x 2 + ∂2 φ
∂y 2 = (x2 +y 2 ) −1/2
25.6 TAYLORPOLYNOMIALSANDTAYLORSERIES
INTWOVARIABLES
In Chapter 18 we introduced Taylor polynomials and Taylor series for functions of a
single variable. We now extend this to include functions of two variables. Recall the
mainideabehindTaylorpolynomialsandseriesforafunctionofonevariable.Knowing
thevaluesofafunction, f (x),anditsderivativesatx =awecanwritedowntheTaylor
polynomial generated by f aboutx = a. This polynomial approximates to the function
f. The values of the Taylor polynomial and the function are usually in close agreement
for values of x near to x = a. To put it another way, knowing the value of f and its
derivatives atx =aallows us toestimatethe value of f near tox =a.
The same idea holds when f is a function of two variables,xandy. If we know the
valueof f anditspartialderivativesatapointx =a,y =b,thentheTaylorpolynomials
allow ustoestimate f atpoints near to (a,b).
25.6.1 First-orderTaylorpolynomialintwovariables
Suppose f is a function of two independent variables, x and y, and that the values of
f, ∂f ∂f
and areknown atthe pointx =a,y =b, that isweknow
∂x ∂y
f(a,b)
∂f
∂x (a,b)
∂f
∂y (a,b)
The first-order Taylor polynomial, p 1
(x,y), generated by f about (a,b) isgiven by
p 1
(x,y)= f(a,b)+(x−a) ∂f
∂x (a,b)+(y−b)∂f ∂y (a,b)
We note the following properties of a first-order Taylor polynomial:
(1) ThevaluesoftheTaylorpolynomialandthefunctionareidenticalatthepoint (a,b).
(2) The values of the first partial derivatives of the Taylor polynomial and the function
are identical atthe point (a,b).
(3) Thehighestderivative needed tocalculatetheTaylor polynomialisthefirstderivative.
(4) The first-order Taylor polynomial contains only linear terms; that is, there are no
powers ofxoryhigher than 1.
The first-order Taylor polynomial represents a plane which is tangent to the surface
f(x,y)at (a,b).
We can use p 1
(x,y) toestimatethe value of f near to (a,b).

836 Chapter 25 Functions of several variables
Example25.10 Afunction, f, issuch that
f(3,1)=2
∂f
∂f
(3,1) = −1
∂x
∂y (3,1)=4
(a) State the first-order Taylor polynomial generated by f about (3, 1).
(b) Estimate the values of f(3.5,1.2) and f(3.2,0.7).
(c) Verifythatthe Taylor polynomial and the function have identical values at(3, 1).
(d) VerifythatthefirstpartialderivativesoftheTaylorpolynomialandthefunctionare
identical at(3, 1).
Solution (a) Inthisexamplea = 3 andb= 1.Hence
p 1
(x,y) = f(3,1)+(x−3) ∂f
∂x (3,1)+(y−1)∂f ∂y (3,1)
=2+(x−3)(−1)+(y−1)4
=4y−x+1
The first-order Taylor polynomial generated by f about (3, 1) is
p 1
(x,y)=4y−x+1
(b) We use p 1
(x,y) toestimate f(3.5,1.2) and f(3.2,0.7).
p 1
(3.5,1.2) = 4(1.2) −3.5 +1 = 2.3
p 1
(3.2,0.7) = 4(0.7) −3.2 +1 = 0.6
Hence 2.3 isan estimateof f(3.5,1.2) and 0.6 isanestimate of f(3.2,0.7).
(c) We aregiven f (3,1) = 2.Also
p 1
(3,1)=4(1)−3+1=2
andso f(3,1) = p 1
(3,1).
(d) We aregiven ∂f
∂f
(3,1) = −1and (3,1) =4.Now
∂x ∂y
p 1
(x,y)=4y−x+1
and so
∂p 1
∂x = −1 ∂p 1
∂y = 4
Hence
∂p 1
∂f
(3,1) =
∂x ∂x (3,1) = −1 ∂p 1
∂f
(3,1) =
∂y ∂y (3,1)=4
Example25.11 Afunction, f (x,y), isdefinedby
f(x,y)=x 2 +xy−y 3
(a) State the first-order Taylor polynomialgeneratedby f about(1, 2).
(b) Verify that the Taylor polynomial in (a) and the function f have identical values at
(1, 2).

25.6 Taylor polynomials and Taylor series in two variables 837
(c) VerifythatthefirstpartialderivativesoftheTaylorpolynomialand f haveidentical
values at(1, 2).
(d) Estimate f(1.1,1.9)usingtheTaylorpolynomial.Comparethiswiththetruevalue.
Solution (a) We aregiven f =x 2 +xy −y 3 and so
∂f
∂x =2x+y
∂f
∂y =x−3y2
Evaluating these atthe point(1, 2) gives
f(1,2)=−5
∂f
∂x (1,2)=4
∂f
(1,2) = −11
∂y
Puttinga = 1 andb = 2 in the formula for p 1
(x,y) we are able to write down the
Taylor polynomial:
p 1
(x,y) = f(1,2)+(x−1) ∂f
∂x (1,2)+(y−2)∂f ∂y (1,2)
= −5 + (x −1)4 + (y −2)(−11)
=4x−11y+13
The first-order Taylor polynomial is p 1
(x,y) = 4x −11y +13.
(b) We can evaluate the Taylor polynomial and the function at(1, 2).
p 1
(1,2)=4−22+13=−5
f(1,2)=−5
Hence p 1
(1,2) = f (1,2); that is, the Taylor polynomial and the function have
identical values at(1, 2).
(c) The first partialderivatives of p 1
(x,y) are found.
∂p 1
∂x = 4 ∂p 1
∂y = −11
∂f
∂x (1,2)=4
∂f
(1,2) = −11
∂y
Hence the first partialderivatives of p 1
and f are identical at(1, 2).
(d) p 1
(1.1,1.9) = 4(1.1) −11(1.9) +13 = −3.5
f(1.1,1.9) = (1.1) 2 + (1.1)(1.9) − (1.9) 3 = −3.559
ThevaluesoftheTaylorpolynomialandthefunctionareincloseagreementnearto
the point(1, 2).
25.6.2 Second-orderTaylorpolynomialintwovariables
Given a function, f, and its first and second partial derivatives at (a,b) we can write
down the second-order Taylor polynomial, p 2
(x,y).
p 2
(x,y) = f(a,b)+(x−a) ∂f
∂x (a,b)+(y−b)∂f ∂y (a,b)
+ 1 (
)
(x−a) 2∂2 f
2! ∂x (a,b)+2(x−a)(y−b) ∂2 f f
2 ∂x∂y (a,b)+(y−b)2∂2 ∂y (a,b) 2

838 Chapter 25 Functions of several variables
We note the following properties of the second-order Taylor polynomial:
(1) The values of the Taylor polynomial and the function areidentical at (a,b).
(2) The values of the first partial derivatives of the Taylor polynomial and the function
areidentical at (a,b).
(3) The values of the second partial derivatives of the Taylor polynomial and the function
areidentical at (a,b).
(4) Thesecond-orderTaylorpolynomialcontainsquadraticterms,thatistermsinvolvingx
2 ,y 2 andxy.
Example25.12 A function, f, and its first and second partial derivatives are evaluated at (2,−1). The
values are
f = 3
∂f
∂x = 4
∂f
∂y = −3
∂ 2 f
∂x 2 = 1
∂ 2 f
∂x∂y = 2
(a) State the second-order Taylor polynomial generated by f about (2,−1).
(b) Estimate f (1.8,−0.9).
(c) Verifythatthe Taylor polynomial and f have identical values at (2,−1).
∂ 2 f
∂y 2 = −1
(d) Verify that the first partial derivatives of the Taylor polynomial and f are identical
at (2,−1).
(e) VerifythatthesecondpartialderivativesoftheTaylorpolynomialand f areidentical
at (2,−1).
Solution (a) We puta = 2 andb=−1 inthe formulafor p 2
(x,y):
p 2
(x,y)=f +(x−2) ∂f
∂x +(y+1)∂f ∂y
+ 1 (
)
(x−2) 2∂2 f
2 ∂x +2(x −2)(y +1) ∂2 f f
2 ∂x∂y +(y+1)2∂2 ∂y 2
where f and its derivatives areevaluated at (2,−1). So
p 2
(x,y) =3+(x−2)4+(y+1)(−3)+ 1 2 {(x −2)2 +2(x −2)(y +1)2
+(y +1) 2 (−1)} = x2
2 − y2
2 +2xy+4x−8y−21 2
(b) The value of p 2
(1.8,−0.9) isan estimateof f (1.8,−0.9):
p 2
(1.8,−0.9) = (1.8)2
2
= 1.875
− (−0.9)2 +2(1.8)(−0.9)+4(1.8) −8(−0.9)− 21 2
2
(c) p 2
(2,−1) = 22
2 − (−1)2 +2(2)(−1) +4(2) −8(−1) − 21 2
2 = 3
f(2,−1)=3
Hence p 2
(2,−1) = f (2,−1); that is, the Taylor polynomial and the function have
identical values at (2,−1).
(d) The first partialderivatives of p 2
are
∂p 2
∂x =x+2y+4 ∂p 2
∂y =−y+2x−8

25.6 Taylor polynomials and Taylor series in two variables 839
so
∂p 2
∂x (2,−1)=2+2(−1)+4=4 ∂p 2
∂y (2,−1)=1+4−8=−3
Hence
∂p 2
∂f
(2,−1) =
∂x ∂x (2,−1) ∂p 2
∂f
(2,−1) =
∂y ∂y (2,−1)
that is, the first partial derivatives of the Taylor polynomial and the function have
identical values at (2,−1).
(e) The second partialderivatives of p 2
arefound:
∂ 2 p 2
∂x 2 = 1
Example25.13 Afunction f isgiven by
∂ 2 p 2
∂x∂y = 2 ∂ 2 p 2
∂y 2 = −1
These values areidentical tothe second partial derivatives of f at (2,−1).
f(x,y)=x 3 +x 2 y+y 4
(a) State the second-order Taylor polynomialgeneratedby f about(1, 1).
(b) Usethepolynomialtoestimate f(1.2,0.9).Comparethisvaluewiththetruevalue.
(c) Verify that the second partial derivatives of the function and the Taylor polynomial
are identical at(1,1).
Solution (a) Herea = 1 andb= 1.We aregiven that f =x 3 +x 2 y +y 4 and so
∂f
∂x = 3x2 +2xy
∂ 2 f
∂x∂y = 2x
∂f
∂y =x2 +4y 3
∂ 2 f
∂y 2 = 12y2
Evaluation of f and itsderivatives at(1, 1) yields
f = 3
∂f
∂x = 5
∂f
∂y = 5
∂ 2 f
∂x 2 = 8
The second-order Taylor polynomial is p 2
(x,y):
∂ 2 f
∂x 2 =6x+2y
∂ 2 f
∂x∂y = 2
p 2
(x,y)=f +(x−1) ∂f
∂x +(y−1)∂f ∂y
+ 1 (
)
(x−1) 2∂2 f
2 ∂x +2(x −1)(y −1) ∂2 f f
2 ∂x∂y +(y−1)2∂2 ∂y 2
∂ 2 f
∂y 2 = 12
=3+5(x−1)+5(y−1)+ 1 2 (8(x −1)2 +4(x −1)(y −1) +12(y −1) 2 )
=4x 2 +6y 2 +2xy−5x−9y+5
(b) The value of p 2
(1.2,0.9) isanestimate of f(1.2,0.9).
p 2
(1.2,0.9) = 4(1.2) 2 +6(0.9) 2 +2(1.2)(0.9) −5(1.2) −9(0.9) +5 = 3.68
The actual value is
f(1.2,0.9) = (1.2) 3 + (1.2) 2 (0.9) + (0.9) 4 = 3.6801

840 Chapter 25 Functions of several variables
(c) The partialderivatives of p 2
(x,y) arefound.
∂p 2
∂x =8x+2y−5 ∂p 2
∂y =12y+2x−9
∂ 2 p 2
∂x 2 = 8
∂ 2 p 2
∂x∂y = 2 ∂ 2 p 2
∂y 2 = 12
Thesecondpartialderivativesofp 2
areidenticaltothesecondpartialderivativesof
f evaluated at(1, 1).
25.6.3 Taylorseriesintwovariables
The third-order Taylor polynomial involves all the third-order partial derivatives of f,
the fourth-order Taylor polynomial involves all the fourth-order partial derivatives of f
and so on. Taylor polynomials approximate more and more closely to the generating
function as more and more terms are included. As more and more terms are included,
weobtainaninfiniteseriesknownasaTaylorseriesintwovariables.Thegeneralform
ofthis series isbeyond the scope ofthis book.
EXERCISES25.6
1 Useafirst-orderTaylor polynomial to estimate
f(2.1, 3.2) given
f(2,3)=4
∂f
∂y (2,3)=3
∂f
(2,3) = −2
∂x
2 Useafirst-orderTaylor polynomial to estimate
g(−1.1,0.2) given
g(−1,0) =6
∂g
(−1,0) = −1
∂y
∂g
(−1,0) =2
∂x
3 Useafirst-orderTaylor polynomial to estimate
h(−1.2, −0.7) given
h(−1.3, −0.6) = 4
∂h
(−1.3, −0.6) = −1
∂x
∂h
(−1.3, −0.6) = 1
∂y
4 Useasecond-order Taylor polynomial to estimate
f(3.1, 4.2) given
f(3,4)=1
∂f
∂x (3,4)=0
∂f
∂y (3,4)=2
∂ 2 f
∂x∂y (3,4)=3
∂ 2 f
(3,4) = −1
∂x2 ∂ 2 f
(3,4) =0.5
∂y2 5 Useasecond-order Taylor polynomial to estimate
g(−2.9,3.1) given
g(−3,3) =1
∂g
∂
(−3,3) =4
∂y
∂ 2 g
∂
(−3,3) = −2
∂x∂y
∂g
(−3,3) = −1
∂x
2 g
(−3,3) =3
∂x2 2 g
(−3,3) =2
∂y2 6 Useasecond-order Taylor polynomial to estimate
h(0.1,0.1) given
h(0,0) =4
∂h
∂
(0,0) = −3
∂y
∂ 2 h
∂x∂y (0,0)=2
∂h
(0,0) = −1
∂x
2 h
∂x 2 (0,0)=2
∂ 2 h
(0,0) = −1
∂y2

25.7 Maximum and minimum points of a function of two variables 841
7 Afunction, f (x,y),is defined by
f(x,y) =x 3 y+xy 3
(a) Calculate the first-orderTaylor polynomial
generated by f about (1,1).
(b) Calculate the second-order Taylor polynomial
generated by f about (1,1).
(c) Estimate f(1.2, 1.2) usingthe first-orderTaylor
polynomial.
(d) Estimate f(1.2, 1.2) usingthe second-order
Taylor polynomial.
(e) Compare your answers in (c) and (d)with the
true value of f(1.2, 1.2).
8 Afunctiong(x,y) is defined by
g(x,y)=xsiny+ x y
(a) Calculate the first-orderTaylor polynomial
generated bygabout (0,1).
(b) Calculate the second-order Taylor polynomial
generated bygabout (0,1).
(c) Estimateg(0.2, 0.9) usingthe polynomial in (a).
(d) Estimateg(0.2, 0.9) usingthe polynomial in (b).
(e) Compare your estimateswith the exact value of
g(0.2, 0.9).
9 A functionh(x,y) isdefined by
h(x,y) =e x y+x 2 e y
(a) Calculate the first-orderTaylor polynomial
generated byhabout (0,0).
(b) Calculate the second-order Taylor polynomial
generated byhabout (0,0).
(c) Estimateh(0.2, 0.15)usingthe polynomial
from (a).
(d) Estimateh(0.2, 0.15)usingthe polynomial
from (b).
(e) Compare youranswers in (c)and(d)with the
exact value ofh(0.2,0.15).
10 A function, f (x,y,z),is defined by
f(x,y,z)=x 2 +xyz+yz 2
(a) Writedown the first-orderTaylor polynomial
generated by f about (0,1,2).
(b) Use the polynomial from (a)to estimate
f(0.1,1.2, 1.9).
(c) Compare youranswerin (b)with the exact value
of f(0.1,1.2, 1.9).
Solutions
1 4.4
2 5.6
3 3.8
4 1.465
5 1.305
6 3.625
7 (a) 4x+4y−6
(b) 3x 2 +3y 2 +6xy−8x−8y+6
(c) 3.6 (d) 4.08 (e) 4.1472
8 (a) 1.8415x (b) 2.3012x −0.4597xy
(c) 0.3683 (d) 0.3775
(e) 0.3789
9 (a) y (b) x 2 +xy +y (c) 0.15
(d) 0.22 (e) 0.2297
10 (a) 2x+4y+4z−8 (b) 4.6
(c) 4.57
25.7 MAXIMUMANDMINIMUMPOINTSOFAFUNCTION
OFTWOVARIABLES
We saw inChapter12 thattofind the turningpoints ofy(x) wesolve
dy
dx = 0

842 Chapter 25 Functions of several variables
The sign of the second derivative, d2 y
dx2, is then used to distinguish between a maximum
pointandaminimumpoint.Theanalysisisverysimilarforafunctionoftwovariables.
Whenconsideringafunctionoftwovariables, f (x,y),weoftenseekmaximumpoints,
minimum points and saddle points. Collectively these are known as stationary points.
Figure 25.1 illustrates a maximum point at A, a minimum point at B and a saddle point
at C. When leaving a maximum point the value of the function decreases; when leaving
a minimum point the value of the function increases. When leaving a saddle point,
the function increases in one direction, axis D on the figure, and decreases in the other
direction, axis Eon the figure.
To locate stationary points we equate both first partial derivatives to zero, that is we
solve
∂f
∂x = 0
∂f
∂y = 0
Stationarypoints are locatedby solving
∂f
∂x = 0
∂f
∂y = 0
Example25.14 Locatethe stationarypoints of
(a) f(x,y)=2x 2 −xy−7y+y 2
(b) f(x,y)=x 2 −6x+4xy+y 2
Solution (a) Thefirst partialderivatives arefound:
∂f
∂x =4x−y
∂f
∂y =−x−7+2y
The stationary points are located by solving ∂f
∂x =0and∂f ∂y = 0 simultaneously,
thatis
4x−y=0
−x+2y−7=0
Solving these equations yields x = 1, y = 4. Hence the function f (x,y) has one
stationary pointand itislocated at(1, 4).
(b) The first partialderivatives arefound:
∂f
∂x =2x−6+4y
∂f
∂y =4x+2y
The first partialderivatives areequated tozero:
2x+4y−6=0
4x+2y=0
Solving the equations simultaneously yieldsx = −1,y = 2. Thus the function has
one stationary pointlocated at (−1,2).

25.7 Maximum and minimum points of a function of two variables 843
Example25.15 Locate the stationary points of
f(x,y)= x3
3 +3x2 +xy+ y2
2 +6y
Solution The first partialderivatives arefound:
∂f
∂x =x2 +6x+y
∂f
∂y =x+y+6
These derivatives are equated tozero:
x 2 +6x+y=0 (25.1)
x+y+6=0 (25.2)
Equations (25.1) and (25.2) are solved simultaneously. From (25.2)y = −x − 6, and
substituting this into Equation (25.1) yields x 2 + 5x − 6 = 0. Solving this quadratic
equationgivesx = 1, −6.Whenx = 1,y = −7,andwhenx = −6,y = 0.Thefunction
has stationary values at (1,−7) and (−6,0).
Equating the first partial derivatives to zero locates the stationary points, but does not
identifythemasmaximumpoints,minimumpointsorsaddlepoints.To distinguishbetween
these various points a test involving second partial derivatives must be made.
Note that this is similar to locating and identifying turning points of a function of one
variable.
To identifythe stationarypoints weconsider the expression
( )
∂ 2 f ∂ 2 f ∂ 2
∂x 2 ∂y − f 2
2 ∂x∂y
If the expression isnegative atastationary point,then thatpoint isasaddle point.
Iftheexpressionispositiveandinaddition ∂2 f
∂x ispositiveatastationarypoint,then
thatpoint isaminimum point.
2
Iftheexpressionispositiveandinaddition ∂2 f
∂x isnegativeatastationarypoint,then
thatpoint isamaximum point.
2
If the expression is zero then further tests are required. These are beyond the scope
of the book.
Insummary:
∂ 2 f ∂ 2 (
f ∂ 2 )
∂x 2 ∂y − f 2
< 0 Saddle point
2 ∂x∂y
( )
∂ 2 f ∂ 2 f ∂ 2
∂x 2 ∂y − f 2
>0 2 ∂x∂y
and
∂ 2 f ∂ 2 (
f ∂ 2 )
∂x 2 ∂y − f 2
>0 2 ∂x∂y
and
∂ 2 f
∂x 2 > 0
∂ 2 f
∂x 2 < 0
Minimum point
Maximum point

844 Chapter 25 Functions of several variables
Example25.16 Locate and identify the stationary values of
f(x,y)=x 2 +xy+y
Solution The first partialderivatives arefound:
∂f
∂x =2x+y
∂f
∂y =x+1
The first partialderivatives areequated tozero:
2x+y=0
x+1=0
This yieldsx = −1,y = 2.Thus thereisone stationary point,positioned at (−1,2).
The second partialderivatives arefound:
∂ 2 f
∂x 2 = 2
∂ 2 f
∂x∂y = 1
∂ 2 f
∂y 2 = 0
Now
∂ 2 f ∂ 2 (
f ∂ 2 )
∂x 2 ∂y − f 2
=2(0) − (1) 2 = −1
2 ∂x∂y
Since the expression isnegative, we conclude that (−1,2) isasaddle point.
Example25.17 Locateand identifythe stationarypoints of
f(x,y)=xy−x 2 −y 2
Solution Thefirst partialderivatives arefound:
∂f
∂x =y−2x
∂f
∂y =x−2y
Solving ∂f
∂x
= 0,
∂f
∂y =0yieldsx=0,y=0.
The second partialderivatives arefound:
∂ 2 f
∂x 2 = −2
∂ 2 f
∂x∂y = 1
∂ 2 f
∂y 2 = −2
The second derivative testtoidentify the stationary point isused. Now
∂ 2 f ∂ 2 (
f ∂ 2 )
∂x 2 ∂y − f 2
= (−2)(−2) − (1) 2 =3
2 ∂x∂y
( )
Since ∂2 f f ∂ 2 f ∂ 2
∂x 2 <0and∂2 ∂x 2 ∂y − f 2
> 0 then (0, 0) isamaximum point.
2 ∂x∂y

25.7 Maximum and minimum points of a function of two variables 845
Example25.18 Locate and identifythe stationarypoints of
f(x,y)= x3 y2
−x+
3 2 +2y
Solution The first partialderivatives arefound:
∂f
∂x =x2 −1
∂f
∂y =y+2
The first partialderivatives areequated tozero:
x 2 −1=0
y+2=0
These equations have two solutions:x = 1,y = −2 andx = −1,y = −2.
In order to identify the nature of each stationary point, the second derivatives are
found:
∂ 2 f
∂x 2 = 2x
∂ 2 f
∂x∂y = 0
∂ 2 f
∂y 2 = 1
Each stationary point isexamined inturn.
At (1,−2)
( )
∂ 2 f ∂ 2 f ∂ 2
∂x 2 ∂y − f 2
=2x(1)−0 2 =2x=2sincex=1
2 ∂x∂y
∂ 2 f
∂x 2 =2x=2
Since both expressions are positive the point (1,−2) isaminimum point.
At (−1,−2)
Here
∂ 2 f ∂ 2 (
f ∂ 2
∂x 2 ∂y − f
2 ∂x∂y
) 2
=2x=−2sincex=−1
Since the expression isnegative then (−1,−2) isasaddle point.
Figure 25.2 illustratesaplot of the surface defined by f (x,y).
z
1
0.5
0
–0.5
–1
–1.5
–2
–2.5
–3
0
–1
–2
y
–3
–4 –2
–1
0
1
2
x
Figure25.2
Thefunction
f(x,y)= x3
3
−x+
y2
2 +2y
has a minimumat (1, −2)
and asaddle point at
(−1, −2).

846 Chapter 25 Functions of several variables
EXERCISES25.7
1 Determine the position and nature ofthe stationary
pointsofthe following functions:
(a) f(x,y)=x+y+xy
(b) f(x,y)=x 2 +y 2 −2y
(c) f(x,y)=x 2 +xy−y
(d) f(x,y)=x 2 +y 2 −xy
(e) f(x,y)=x 2 +2y 2 −3xy+x
2 Determine the position and nature ofthe stationary
pointsofthe following functions:
(a) z(x,y) =x 2 +y 2 −3xy +2x
(b) z(x,y) =x 3 +xy +y 2
(c) z(x,y) = y3 3 −x2 −y
(d) z(x,y) = 1 x + 1 y − 1 xy
(e) z(x,y) = 4x 2 y −6xy
3 Locateandidentify the stationary pointsofthe
following:
(a) f(x,y)=x 2 y−x 2 +y 2
(b) f(x,y) = x y +x+y
(c) f(x,y) =x 4 +16xy +y 4
(d) f(x,y)=y−y 2 −e x x
(e) f(x,y) = ex
y 4 −xy
Solutions
1 (a) (−1, −1),saddle point
(b) (0,1),minimum
(c) (1, −2),saddle point
(d) (0,0),minimum
(e) (4,3),saddle point
(
2 (a) 45
, 5)
6 ,saddle point
)
,minimum
(
(b) (0,0),saddle point; 16
, −12
1
(c) (0,1),saddle point; (0, −1),maximum
(d) (1,1),saddle point
(e) (0,0), (1.5,0),saddlepoints
3 (a) (0,0),saddle point
(b) (1,−1),saddle point
(c) (0,0),saddle point; (2,−2),minimum; (−2,2),
minimum
(d) (−1,0.5),maximum
(e) (−4,0.4493), saddlepoint
REVIEWEXERCISES25
1 Find all firstpartialderivatives ofthe following
functions:
(a) z(x,y) = 9x 2 +2y 2
(b) z(x,y) = 3x 3 y 6
(c) z(x,y) = 4x 3 −5y 3
(d) z(x,y) =xy +x 2 y 2
(e) z(x,y) = sin(2xy)
(f) z(x,y) = 2e 3xy
2 Evaluate the firstpartialderivatives of f atx = −1,
y=2.
(a) f(x,y) =3x 3 −y 3 +x 2 y 2
(b) f(x,y) = 4x
y
(c) f(x,y)=sinx+2cosy
(d) f(x,y) = (xy) 3 +x
(e) f(x,y) = x +y
x −y
(f) f(x,y)=2e x e y
3 Findallsecond partialderivatives ofzwhere
(a) z(x,y) = 3x 4 −9y 4 +x 3 y 3
(b) z(x,y) = (x 2 +y 2 ) 2
(c) z(x,y) = 2e 4x−3y
(d) z(x,y) = 2
x +y
(e) z(x,y) = 2 √ x +y
(f) z(x,y) = (sinx)(cosy)

Review exercises 25 847
4 Find allfirstand second partialderivatives of
f(x,y) = (ax +by) n
wherea,bandnare constants.
5 Find the firstpartial derivatives of
f(x,y) = (ax 2 +by 2 +cxy) n
wherea,b,candnare constants.
6 Verifythat
z(x,y)=3x+2y+1
isasolution of
∂z
∂x − ∂z
∂y = 1
7 Verifythat
z(x,y)=2xy−x+y
isasolution of
∂z
∂x + ∂z
∂y =2(x+y)
8 Verifythat
f(x,y)=x 2 +y 2 −2xy
isasolution of
∂f
∂x + ∂f
∂y = 0
9 Verifythat
z(x,y) =sinx +cosy
isasolution of
∂ 2 z
∂x 2 + ∂2 z
∂y 2 +z=0
10 Verify that
z(x,y) =xye x
is asolution of
∂ 2 z
∂x 2 + ∂2 z
∂y 2 −y ∂2 z
∂x∂y =yex
11 (a) Writedown the second-order Taylor polynomial
generatedby f (x,y) aboutx = 2,y = 3 given
f(x,y) =3x 3 y−x 2 y 3
(b) Estimate f(2.1,2.9) usingyourpolynomial from
(a)andcompare thiswith the exact answer.
12 Writedown the second-order Taylor polynomial
generated byz(x,y) aboutx = 1,y = 1 given
z(x,y) = x +y
x
13 Calculate the second-order Taylor polynomial
generated by
√
f(x,y)= x 2 +y 2
aboutx=1,y=0.
14 Locateandidentify all the stationary pointsofthe
following functions:
(a) f(x,y)=x 2 +y 3 −3y
(b) f(x,y) =4xy−x 2 y
(c) f(x,y) =x 3 +2y 2 −12x
(d) f(x,y)=xy−y 2 −x 3
15 Locateandidentify the stationary pointsof
f(x,y)= y x −x2 +y 2
Solutions
1 (a)
∂z ∂z
= 18x,
∂x ∂y = 4y
(b) 9x 2 y 6 ,18x 3 y 5
(c) 12x 2 , −15y 2
(d) y+2xy 2 ,x+2x 2 y
(e) 2ycos(2xy),2xcos(2xy)
(f) 6ye 3xy ,6xe 3xy
∂f ∂f
2 (a) (−1,2) =1, (−1,2) = −8
∂x ∂y
(b) 2,1 (c) 0.5403, −1.8186
3 (a)
(d) 25, −12 (e) − 4 9 ,−2 9
(f) 5.4366,5.4366
∂ 2 z
∂x 2 = 36x2 +6xy 3 ,
∂ 2 z
∂x∂y =9x2 y 2 ,
∂ 2 z
∂y 2 = −108y2 +6x 3 y
(b) 12x 2 +4y 2 ,8xy,4x 2 +12y 2
(c) 32e 4x−3y , −24e 4x−3y ,18e 4x−3y

848 Chapter 25 Functions of several variables
4
4 4 4
∂f
(d)
(x +y) 3, (x +y) 3, (x +y) 3
5
∂x = (2ax +cy)n(ax2 +by 2 +cxy) n−1 ,
(e) − 1 2 (x +y)−3/2 ,
∂f
−2 1 (x ∂y = (2by +cx)n(ax2 +by 2 +cxy) n−1
+y)−3/2 ,
−2 1 11 (a)27x
(x 2 −36y 2 −72xy +108x +276y −432
+y)−3/2
(b)p
(f) −(sinx)(cosy),
2 (2.1,2.9) = −26.97,
f (2.1,2.9) = −26.985
−(cosx)(siny),
12 x 2 −xy−2x+2y+2
−(sinx)(cosy)
∂f
∂x =an(ax +by)n−1 ,
13 y2
2 +x
∂f
14 (a) (0,1)minimum; (0, −1)saddle point
∂y =bn(ax +by)n−1 ,
(b) (0,0),(4,0)saddlepoints
∂ 2 f
(c) (2,0)minimum; (−2,0)saddle point
∂x 2 =a2 n(n −1)(ax +by) n−2 ,
( )
(d) (0,0)saddlepoint; 16
, 12
1 maximum
∂ 2 f
( ) ( )
∂x∂y =abn(n −1)(ax +by)n−2 ,
1
15 √ , − 1
2
∂ 2 2 √ , − 1 1
√ ,
2 2 2 √ ;saddlepoints
2
f
∂y 2 =b2 n(n −1)(ax +by) n−2

26 Vectorcalculus
Contents 26.1 Introduction 849
26.2 Partialdifferentiationofvectors 849
26.3 Thegradientofascalarfield 851
26.4 Thedivergenceofavectorfield 856
26.5 Thecurlofavectorfield 859
26.6 Combiningtheoperatorsgrad,divandcurl 861
26.7 Vectorcalculusandelectromagnetism 864
Reviewexercises26 865
26.1 INTRODUCTION
This chapter draws together several threads from previous chapters. It builds upon differential
and integral calculus, functions of several variables and the study of vectors.
Thesetopicstogetherformabranchofengineeringmathematicsknownasvectorcalculus.
Vector calculus is used to model a vast range of engineering phenomena including
electrostaticcharges,electromagneticfields,airflowaroundaircraft,carsandothersolid
objects,fluidflowaroundshipsandheatflowinnuclearreactors.Thechapterstartsby
explainingwhatismeantbytheoperatorsgrad,divandcurl.Theseareusedtocarryout
various differentiation operations on the fields.
26.2 PARTIALDIFFERENTIATIONOFVECTORS
Consider the vector field v = v x
i + v y
j + v z
k, where each component v x
,v y
and v z
is a
function ofx,y andz. We can partiallydifferentiate the vector w.r.t.xas follows:
∂v
∂x = ∂v x
∂x i + ∂v y
∂x j + ∂v z
∂x k
This isanew vector with a magnitude and direction different fromthose ofv.

850 Chapter 26 Vector calculus
Partial differentiation w.r.t.yandzis defined in a similar way, as are higher derivatives.
For example,
∂ 2 v
∂x 2 = ∂2 v x
∂x 2 i + ∂2 v y
∂x 2 j + ∂2 v z
∂x 2 k
Example26.1 Ifv = 3x 2 yi +2xyzj −3x 4 y 2 k, find ∂v
∂x , ∂v
∂y , ∂v
∂z . Further, find ∂2 v
∂x 2 and ∂2 v
∂x∂z .
Solution We find
∂v
∂x = 6xyi +2yzj −12x3 y 2 k
∂v
∂y = 3x2 i +2xzj −6x 4 yk
∂v
∂z = 2xyj
∂ 2 v
∂x 2 = 6yi −36x2 y 2 k
∂ 2 v
∂x∂z = 2yj
EXERCISES26.2
1 Givenv = 2xi +3yzj +5xz 2 kfind
(a)
∂v
∂x
(b)
∂v
∂y
(c)
2 Iff =2i−xyzj+3x 2 zkfind
(a)
(d)
∂f
∂x
∂ 2 f
∂x 2
(b)
∂f
∂y
(e) ∂2 f
∂y 2
(c)
(f)
∂v
∂z
∂f
∂z
∂ 2 f
∂z 2
3 GivenE = (x 2 +y)i + (1 −z)j + (x +2z)kfind
(a)
(d)
∂E
∂x
∂ 2 E
∂x 2
(b)
(e)
∂E
∂y
∂ 2 E
∂y 2
(c)
(f)
∂E
∂z
∂ 2 E
∂z 2
4 Ifv=3xyzi+(x 2 −y 2 +z 2 )j+(x+y 2 )kfind
∂v
∂x , ∂v
∂y , ∂v
∂z , ∂2 v ∂2 v ∂2 v
∂x 2, ∂y 2,and ∂z 2.
5 Ifv = sin(xyz)i +ze xy j −2xykfind
∂v
∂x , ∂v
∂y , ∂v
∂z .
6 Ifv =xi+x 2 yj−3x 3 k,and φ =xyz,find
φv, ∂ ∂φ
(φv),
∂x ∂x , ∂v .Deduce that
∂x
∂
(φv) = φ∂v
∂x ∂x + ∂φ
∂x v
7 Ifv = ln(xy)i +2xycoszj −x 4 yzk,find
∂v
∂x , ∂v
∂y , ∂v
∂z , ∂2 v
∂x 2 , ∂2 v
∂y 2 , and ∂2 v
∂z 2.
Solutions
1 (a) 2i +5z 2 k (b) 3zj
(c) 3yj +10xzk
2 (a) −yzj +6xzk (b) −xzj
(c) −xyj +3x 2 k
(d) 6zk
(e) 0 (f) 0

26.3 The gradient of a scalar field 851
3 (a) 2xi+k (b) i
4
5
(c) −j +2k
(d) 2i
(e) 0 (f) 0
∂v
∂x =3yzi+2xj+k
∂v
= 3xzi −2yj +2yk
∂y
∂v
= 3xyi +2zj
∂z
∂ 2 v
∂x 2 = 2j
∂ 2 v
∂y 2 =−2j+2k
∂ 2 v
∂z 2 = 2j
∂v
∂x =yzcos(xyz)i +yzexy j −2yk
∂v
∂y =xzcos(xyz)i +xzexy j −2xk
∂v
∂z =xycos(xyz)i +exy j
6 φv =x 2 yzi+x 3 y 2 zj−3x 4 yzk
7
∂(φv)
= 2xyzi +3x 2 y 2 zj −12x 3 yzk
∂x
∂φ
∂x =yz
∂v
∂x =i+2xyj−9x2 k
∂v
∂x = i x +2ycoszj −4x3 yzk
∂v
∂y = i y +2xcoszj−x4 zk
∂v
∂z = −2xysinzj −x4 yk
∂ 2 v
∂x 2 = − i
x 2 −12x2 yzk
∂ 2 v
∂y 2 = − i
y 2
∂ 2 v
∂z 2 = −2xycoszj
26.3 THEGRADIENTOFASCALARFIELD
Given a scalar functionofx,y,z
φ = φ(x,y,z)
we can differentiate it partially w.r.t. each of its independent variables to find ∂φ
∂x , ∂φ
∂y
and ∂φ . If wedo this, the vector
∂z
∂φ
∂x i + ∂φ
∂y j + ∂φ
∂z k
turns out to be particularly important. We call this vector the gradient of φ and denote
itby
∇φ or grad φ
An alternative form of writing ∇φ isas threecomponents
( ∂φ
∂x , ∂φ
∂y , ∂φ )
∂z
gradφ=∇φ= ∂φ
∂x i + ∂φ
∂y j + ∂φ
∂z k
The process of forming a gradient applies only to a scalar field and the result is always
a vector field.

852 Chapter 26 Vector calculus
Itisoften useful towrite ∇φ inthe form
( ∂
∂x , ∂ ∂y , ∂ ∂z)
φ
where the quantity in brackets is called a vector operator and is regarded as operating
on the scalar φ. Thus the vector operator, ∇, isgiven by
( ∂
∂x , ∂ ∂y , ∂ ∂z)
Example26.2 If φ = φ(x,y,z) = 4x 3 ysinz,find ∇φ.
Solution
φ(x,y,z) = 4x 3 ysinz
so thatby partialdifferentiation weobtain
Therefore
∂φ
∂x = 12x2 ysinz
∂φ
∂y = 4x3 sinz
∂φ
∂z =4x3 ycosz
∇φ = 12x 2 ysinzi +4x 3 sinzj +4x 3 ycoszk
We often need to evaluate ∇φ at a particular point, say, for example, at (4, 2, 3). We
write ∇φ| (4,2,3)
todenotethe value of ∇φ atthe point(4,2,3).
Example26.3 If φ =x 3 y +xy 2 +3yfind
(a) ∇φ
(b) ∇φ| (0,0,0)
(c) |∇φ|at(1, 1,1)
Solution (a) If φ =x 3 y +xy 2 +3y then
so that
∂φ
∂x =3x2 y+y 2
∂φ
∂y =x3 +2xy+3
∂φ
∂z = 0
∇φ = (3x 2 y+y 2 )i+ (x 3 +2xy+3)j+0k

(b) At (0,0,0), ∇φ =0i+3j+0k =3j.
26.3 The gradient of a scalar field 853
(c) At(1,1,1),∇φ=(3×1 2 ×1+1)i+(1 3 +2×1×1+3)j+0k=4i+6j+0k
so that |∇φ| at (1,1,1) isequal to √ 4 2 +6 2 = √ 52.
So far we have been given φ and have calculated ∇φ. Sometimes we will be given ∇φ
and will needtofind φ. ConsiderExample 26.4.
Example26.4 IfF = ∇φ find φ whenF = (3x 2 +y 2 )i + (2xy +5)j.
Solution Note that in this example F has only two components. Consequently ∇φ will have two
components,thatisF = ∇φ = ∂φ
∂x i + ∂φ j. Therefore
∂y
(3x 2 +y 2 )i + (2xy +5)j = ∂φ
∂x i + ∂φ
∂y j
Equating theicomponents we have
∂φ
∂x =3x2 +y 2 (26.1)
Equating thejcomponents we have
∂φ
∂y
=2xy+5 (26.2)
Integrating Equation (26.1) w.r.t.xand treatingyas a constant we find
φ=x 3 +xy 2 +f(y) (26.3)
where f (y) is an arbitrary function of y which plays the same role as the constant of
integration does when there is only one independent variable. Note in particular that
∂
∂φ
f (y) = 0.Check by partialdifferentiation that
∂x ∂x =3x2 +y 2 .
Integrating Equation (26.2) w.r.t.yand treatingxas a constant we find
φ=xy 2 +5y+g(x) (26.4)
∂
whereg(x) is an arbitrary function ofx. Note that g(x) = 0. Check by partial differentiationthat
∂φ = 2xy+5.Comparingbothformsfor φ giveninEquations(26.3)and
∂y
∂y
(26.4) wesee thatby choosingg(x) =x 3 and f (y) = 5y wehave
φ=x 3 +xy 2 +5y
Check that F is indeed equal to ∇φ. Also check that by adding any constant to φ the
same property holds, thatisFisstillequal to ∇φ.

854 Chapter 26 Vector calculus
26.3.1 Physicalinterpretationof ∇φ
Suppose wethinkofthescalarfield φ(x,y,z)asdescribingthetemperaturethroughout
a region. This temperature will vary from point to point. At a particular point it can be
shown that ∇φ is a vector pointing in the direction in which the rate of temperature increaseisgreatest.
|∇φ|isthemagnitudeoftherateofincreaseinthatdirection.Similarly,
the rate of temperature decrease is greatest in the direction of −∇φ. Analogous interpretationsarepossibleforotherscalarfieldssuchaspressureandelectrostaticpotential.
Engineeringapplication26.1
Electrostaticpotential
Engineers working on the design of equipment such as cathode ray tubes and electricalvalves,whicharealsocommonlyknownbytheirgenerictermvacuumtubes,
needtocalculatetheelectrostaticpotentialthatresultsfromanaccumulationofstatic
chargesatvariouspointsinaregionofspace.Complicatedexamplesrequiretheuse
of a computer. Let us consider a simple example.
The electrostatic potential,V, inaregion isgiven by
y
V =
(x 2 +y 2 +z 2 ) 3/2
Suppose a unit charge is located in the region at the point with coordinates (2, 1, 0).
Find the direction atthis point,inwhich the rate ofdecrease inpotential isgreatest.
Solution
The rate of decrease isgreatest inthe direction of −∇V.
We first calculate the first partialderivatives ofV. Writing
we find
∂V
V =y(x 2 +y 2 +z 2 ) −3/2
∂x = (− 3 2
)
y(x 2 +y 2 +z 2 ) −5/2 (2x) = −3xy(x 2 +y 2 +z 2 ) −5/2
∂V
(−
∂y = 3 )
y(x 2 +y 2 +z 2 ) −5/2 (2y) + (x 2 +y 2 +z 2 ) −3/2
2
= −3y 2 (x 2 +y 2 +z 2 ) −5/2 + (x 2 +y 2 +z 2 ) −3/2
∂V
(−
∂z = 3 )
y(x 2 +y 2 +z 2 ) −5/2 (2z) = −3yz(x 2 +y 2 +z 2 ) −5/2
2
These partialderivatives can each be evaluated atthe point(2, 1,0). Thatis,
∂V
∂V
∂x ∣ = −0.107
∂V
(2,1,0)
∂y ∣ = 0.036
(2,1,0)
∂z ∣ = 0
(2,1,0)
and so
Finally,
∇V| (2,1,0)
= −0.107i +0.036j +0k
−∇V| (2,1,0)
= 0.107i −0.036j −0k

26.3 The gradient of a scalar field 855
Atthepoint (2,1,0)thisvectorpointsinthedirectionofgreatestrateofdecreasein
potential. This is also the direction of the electrostatic force experienced by the unit
charge atthatpoint.
EXERCISES26.3
1 If φ =x 2 −y 2 −3xyz,(a)find ∇φ,(b)evaluate ∇φ at
the point (0,0,0).
2 If φ =x 2 yz 3 ,find(a) ∇φ,(b) ∇φat (1,2,1),(c) |∇φ|
at (1,2,1).
3 Ifv = ∇φ,find φwhenv = (2x−4y 2 )i−8xyj.
4 Given φ =xyzfind(a) ∇φ,(b) −∇φ,(c) ∇φ
evaluated at (3,0, −1).
5 Find ∇V when
(a)V =x 2 +y 2 +z
( )
2
(b)V =zsin −1 y
x
(c)V = e x+y+z .
6 An electrostatic potential is given byV =xe −√y .
Find ∇V and deduce the direction in which the
decrease ofpotential isgreatest atthe point with
coordinates (1,1,1).
7 Inthe theoryoffluid mechanics the scalar field φ is
known asthevelocity potential ofafluidflow. The
fluid velocityvector, v,atapoint canbe foundfrom
the equationv = ∇φ.Foraparticulartypeofflow
φ =Ux,whereU is aconstant. Show thatthe
corresponding fluidmotion is entirely parallelto thex
axis, andatany point the fluidspeed isU.Finda
velocitypotential forasimilar flow whichisentirely
parallelto theyaxis.
8 Is −∇φ the same as ∇(−φ)?Explain youranswer.
9 Given φ = 3x 2 y +xz,
(a) determine ∇φ
(b) determine ∇(7φ)
(c) determine7∇φ
(d) Is ∇(7φ)thesameas7∇φ?
10 Foranyscalarfield φ andanyconstantk,is ∇(kφ)the
sameask∇φ?
Solutions
1 (a) (2x −3yz)i − (2y +3xz)j −3xyk
(b) 0
2 (a) 2xyz 3 i +x 2 z 3 j +3x 2 yz 2 k
(b) 4i+j+6k
√
(c) 53
3 φ=x 2 −4xy 2 +c
4 (a) yzi +xzj +xyk
(b) −yzi −xzj −xyk
(c) −3j
5 (a) 2xi +2yj +2zk
−zy
(b)
x 2√ 1−y 2 /x 2i
( )
z
+
x √ y
1−y 2 /x 2j +sin−1 k
x
(c) e x+y+z (i +j+k)
6 e −√y i − xe−√ y
2 √ j,−0.368i +0.184j
y
7 φ=Uy
8 yes
9 (a) (6xy +z)i +3x 2 j +xk
(b) (42xy +7z)i +21x 2 j +7xk
(c) (42xy +7z)i +21x 2 j +7xk
(d) yes
10 yes

856 Chapter 26 Vector calculus
26.4 THEDIVERGENCEOFAVECTORFIELD
Given a vector field v = v(x,y,z) let us consider what happens when we differentiate
its individual components. If
v=v x
i+v y
j+v z
k
wecantakeeachcomponentinturnanddifferentiateitpartiallyw.r.t.x,yandz,respectively;
thatis, we can evaluate
∂v x
∂x
∂v y
∂y
∂v z
∂z
Ifweaddthecalculatedquantitiestheresultturnsouttobeaveryusefulscalarquantity
known asthedivergenceof v, thatis
divergence ofv = ∂v x
∂x + ∂v y
∂y + ∂v z
∂z
Thisisusuallyabbreviatedtodivv.Alternatively,thenotation ∇ ·visoftenused.Ifwe
use the vector operator notation introduced inthe previous section wehave
( ∂
∇·v=
∂x , ∂ ∂y ∂z)
, ∂ ·v
=
( ∂
∂x , ∂ ∂y , ∂ ∂z)
·(v x
,v y
,v z
)
Interpreting the · as a scalar product wefind
∇·v= ∂v x
∂x + ∂v y
∂y + ∂v z
∂z
( ∂
asbefore,althoughthisisnotascalarproductintheusualsensebecause
∂x , ∂ ∂y ∂z)
, ∂
is a vector operator. We note that the process of finding the divergence is always performed
on a vector field and the resultisalways a scalar field:
divv=∇·v= ∂v x
∂x + ∂v y
∂y + ∂v z
∂z
Example26.5 Ifv =x 2 zi +2y 3 z 2 j +xyz 2 kfind divv.
Solution Partially differentiating the first component ofvw.r.t.xwefind
∂v x
∂x = 2xz
Similarly,
∂v y
∂y = 6y2 z 2
and
Adding these results wefind
∂v z
∂z = 2xyz
divv=∇·v=2xz+6y 2 z 2 +2xyz

26.4.1 Physicalinterpretationof ∇ · v
26.4 The divergence of a vector field 857
If the vector field v represents a fluid velocity field, then, loosely speaking, the divergence
of v evaluated at a point represents the rate at which fluid is flowing away from
or towards that point. If fluid is flowing away from a point then either the fluid density
must be decreasing there or there must be some source providing a supply of new
fluid.
If the divergence of a flow is zero at all points then outflow from any point must be
matchedbyanequalinflowtobalancethis.Suchavectorfieldissaidtobesolenoidal.
Example26.6 Show thatthe vector field
v=xsinyi+ysinxj−z(sinx+siny)k
issolenoidal.
Solution We have
v x
=xsiny so that
Also,
v y
=ysinx so that
Finally,
v z
= −z(sinx +siny)
Therefore,
∂v x
∂x =siny
∂v y
∂y =sinx
sothat
∂v z
∂z
= −(sinx +siny)
∇·v=siny+sinx−(sinx+siny)=0
and hencevissolenoidal.
Engineeringapplication26.2
ElectricfluxandGauss’slaw
WesawinEngineeringapplication7.5thatelectricchargesproduceanelectricfield,
E, around them which can be visualized by drawing lines of force. Suppose we surroundaregioncontainingchargeswithasurfaceS.Ifasmallportionofthissurface,
δS,ischosenwecandrawthefieldlineswhichpassthroughthisportionasshownin
Figure 26.1.
The flux of E through δS is a measure of the number of lines of force passing
through δS. Gauss’s law states that the total flux out of any closed surfaceSis proportional
to the total charge enclosed. It is possible to show that this law can be
expressed mathematically as
∇·E= ρ ε 0
➔

858 Chapter 26 Vector calculus
E
dS
Figure26.1
Theflux ofEthrough δS isa
measure ofthe number oflines
offorce passingthrough δS.
where ρ is the charge density and ε 0
is a constant called the permittivity of free
space. Note that in a charge-free region, ρ = 0, and so ∇ ·E = 0. This means there
isno net flux ofE.
EXERCISES26.4
1 Avector fieldvisgiven by
Find
v =3x 2 yi+2y 3 zj+xz 3 k
(a)v x , v y , v z
(b) ∂v x
∂x , ∂v y
∂y , ∂v z
∂z
(c)∇·v
2 Avector fieldFis defined by
F=(x+y 2 )i+(y 2 −z)j+(y+z 2 )k
(a) Find ∇ ·F.
(b) Calculate ∇ ·F atthe point (3,2, −1).
3 IfA = 3yzi +2xyj +xyzkfind ∇ ·A.
4 Find the divergence ofeachofthe following vector
fields:
(a) v =x 2 i+y 2 j+z 2 k
(b) v = e xy i +2zsin(xy)j +x 3 zk
(c) v=xyi−2yzj+k
(d) v =x 2 y 2 i−y 2 j−xyzk
5 IfE=xi+z 2 j−yzkfind
(a) E·i
(b) E·j
(c) E·k
(d) ∇·E
6 Avector field is given by
F = (3x 2 −z)i + (2x +y)j
+(x +3yz)k
Find ∇ ·F atthe point (1,2,3).
7 Given the scalar field φ =x 2 +y 2 −2z 2 ,find ∇φ and
showthat ∇ ·(∇φ) =0.
8 Forany vectorfield F is ∇ · (−F)the same as
−(∇ ·F)?
9 ThevectorfieldFis given by
F =xz 2 i+2xy 2 zj+xz 3 k
(a) Find ∇ ·F.
(b) State4F.
(c) Find ∇ · (4F).
(d) Is4(∇ ·F)thesameas ∇ · (4F)?
10 Forany vectorfield F andany scalar constantkis
∇ · (kF)thesameask∇ ·F?
11 Give an example ofavectorfield
v = v x i+v y j+v z ksuchthat ∂v x
∂x ≠ 0, ∂v y
∂y ≠ 0,
∂v z
∂z ≠0,but∇·v=0.

26.5 The curl of a vector field 859
Solutions
1 (a) 3x 2 y,2y 3 z,xz 3
(b) 6xy,6y 2 z,3xz 2
(c) 6xy +6y 2 z +3xz 2
2 (a) 1+2y+2z (b) 3
3 ∇·A=2x+xy
4 (a) 2x+2y+2z
(b) ye xy +2xzcos(xy) +x 3
(c)y−2z
(d)2xy 2 −2y −xy
5 (a) x (b) z 2 (c) −yz (d) 1−y
6 13
7 ∇φ=2xi+2yj−4zk
8 yes
9 (a)z 2 +4xyz +3xz 2
(b)4xz 2 i +8xy 2 zj +4xz 3 k
(c)4z 2 +16xyz +12xz 2
(d)yes
10 yes
11 v = 2xyi +y 2 j −4yzk forexample
26.5 THECURLOFAVECTORFIELD
Athird differential operator isknown ascurl. Itisdefined rather like a vector product.
curlv=∇×v
( ∂
=
∂x , ∂ ∂y ∂z)
, ∂ × (v x
,v y
,v z
)
∣ i jk ∣∣∣∣∣∣∣∣
=
∂ ∂ ∂
∂x ∂y ∂z
∣
v x
v y
v z
This determinant is evaluated in the usual way except that we must regard ∂ ∂x , ∂ ∂y and
∂
as operators, notmultipliers. Thus, forexample,
∂z
∣ ∂ ∂ ∣∣∣∣∣
∂x ∂y
∣ v x
v y
Explicitly wehave
curlv =
means
∂v y
∂x − ∂v x
∂y
( ∂vz
∂y − ∂v ) (
y ∂vx
i +
∂z ∂z − ∂v ) ( ∂vy
z
j +
∂x ∂x − ∂v )
x
k
∂y

860 Chapter 26 Vector calculus
Example26.7 Ifv =x 2 yzi −2xyj +yzk find ∇ ×v.
i j k
Solution ∇ ×v=
∂ ∂ ∂
∂x ∂y ∂z
∣
x 2 yz −2xy yz
∣
[ ∂(yz)
= − ∂(−2xy) ] [ ] [ ]
∂(yz)
i − − ∂(x2 yz) ∂(−2xy)
j + − ∂(x2 yz)
k
∂y ∂z ∂x ∂z ∂x ∂y
=zi+x 2 yj− (2y+x 2 z)k
Notethatthecurloperationisonlyperformedonavectorfieldandtheresultisanother
vector field.
Adetaileddiscussionofthephysicalinterpretationofthecurlofavectorfieldisbeyondthescopeofthisbook.However,ifthevectorfieldvunderconsiderationrepresents
a fluid flow then it may be shown that curl v is a vector which measures the extent to
whichindividualparticlesofthefluidarespinningorrotating.Forthisreason,avector
field whosecurliszero forallvalues ofx,y andzissaidtobe irrotational.
Example26.8 Show thatthe vectorfield
F=ye xy i+xe xy j+0k
isirrotational.
i j k
Solution ∇ ×F =
∂ ∂ ∂
∂x ∂y ∂z
∣
ye xy xe xy 0
∣
( ∂
=
∂y 0 − ∂ ) ( ∂
∂z xexy i −
∂x 0 − ∂ ) ( ∂
∂z yexy j +
∂x xexy − ∂ )
∂y yexy k
=0i +0j + ((xye xy +e xy ) − (yxe xy +e xy ))k
=0 forallx,yandz
The field istherefore irrotational.
EXERCISES26.5
1 Find the curlofthe vectorfield
v =xi −3xyj +4zk.
2 Ifv =3xi−2y 2 zj+3xyzkfind ∇ ×v.
3 SupposeF =P(x,y)i +Q(x,y)jisa
two-dimensionalvector field. Show thatFis
irrotationalif ∂P
∂y = ∂Q
∂x .
4 Findthe divergence andcurlofthe vector field
E = cosxi +sinxj.

26.6 Combining the operators grad, div and curl 861
5 Find the curlofeach ofthe following vector fields:
(a) E =x 2 yi +7xyzj +3x 2 k
(b) v =y 2 xi+4xzj+y 2 xk
(c) F = sinxi +cosxj +3xyzk
6 Avector fieldFisgiven by
F=x 3 yi+2y 2 j+(x+z 2 )k
(a) Find ∇ ×F.
(b) State 3F.
(c) Find ∇ × (3F).
(d) Is3(∇ ×F)thesameas ∇ × (3F)?
7 F is avector field andkis aconstant. Is ∇ × (kF)the
sameask(∇ ×F)?
Solutions
1 −3yk
2 (3xz +2y 2 )i −3yzj
4 −sinx,cosxk
5 (a) −7xyi −6xj + (7yz −x 2 )k
(b) x(2y −4)i −y 2 j + (4z −2xy)k
(c) 3xzi −3yzj −sinxk
6 (a) −j−x 3 k
(b) 3x 3 yi +6y 2 j +3(x +z 2 )k
(c) −3j −3x 3 k
(d) yes
7 yes
26.6 COMBININGTHEOPERATORSGRAD,DIVANDCURL
We have now metthreevector operators; these aresummarized inTable 26.1.
Itisimportanttobeabletocombinethethreeoperatorsgrad,divandcurlinsensible
ways.Forinstance,becausethegradientofascalarisavectorwecanconsiderevaluating
its divergence, thatis( ) ∂φ
∇·(∇φ)=∇·
=
∂x , ∂φ
∂y , ∂φ
∂z
( ∂
∂x , ∂ ∂y ∂z)
, ∂ ·
= ∂2 φ
∂x + ∂2 φ
2 ∂y + ∂2 φ
2 ∂z 2
( ∂φ
∂x , ∂φ
∂y , ∂φ
∂z
This lastexpression isvery important and isoften abbreviated tosimply
∇ 2 φ
Table26.1
Thethree vectoroperators.
Operator Acts on Resultisa Definition
grad scalar field vector field ∇φ = grad φ = ∂φ
∂x i + ∂φ
∂y j + ∂φ
∂z k
div vector field scalar field ∇ ·v = divv = ∂v x
∂x + ∂v y
∂y + ∂v z
∂z
∣ i jk ∣∣∣∣∣∣∣
∂ ∂ ∂
curl vector field vector field ∇ ×v = curlv =
∂x ∂y ∂z
∣ v x v y v z
)

862 Chapter 26 Vector calculus
pronounced‘del-squared φ’,andoccursinLaplace’sequation ∇ 2 φ = 0andotherpartial
differential equations.
Example26.9 If φ = 2x 2 −y 2 −z 2 ,find ∇φ, ∇ · (∇φ)anddeducethat φ satisfiesLaplace’sequation.
Solution ∇φ = 4xi −2yj −2zk
∇·(∇φ)=4−2−2=0
that is,
∇ 2 φ = 0
Hence φ satisfies Laplace’s equation.
Example26.10 If φ(x,y,z)isanarbitrarydifferentiablescalarfield,showthatcurl(grad φ) = ∇×(∇φ)
isalways zero.
Solution Given φ = φ(x,y,z) we have, by definition,
Then
∇φ = ∂φ
∂x i + ∂φ
∂y j + ∂φ
∂z k
curl(grad φ) = ∇ × (∇φ)
i j k
∂ ∂ ∂
=
∂x ∂y ∂z
∂φ ∂φ ∂φ
∣
∂x ∂y ∂z
∣
( ( ∂ ∂φ
=
∂y ∂z
( ( ∂ ∂φ
+
∂x ∂y
)
withsimilarresultsfortheothermixedpartialderiva-
Now,since ∂ ∂x
tives, itfollows that
∇×(∇φ)=0
( ) ∂φ
= ∂ ∂y ∂y
( ∂φ
∂x
for any scalar field φ whatsoever.
)
− ∂ ∂z
( ∂φ
∂y
)
− ∂ ∂y
( ∂φ
∂x
)) ( ∂
i −
∂x
))
k
( ) ∂φ
− ∂ ∂z ∂z
( )) ∂φ
j
∂x
Foranarbitrarydifferentiable scalar field φ
∇×(∇φ)=0

26.6 Combining the operators grad, div and curl 863
Engineeringapplication26.3
Poisson’sequation
Recall from Engineering application 26.1 that engineers sometimes need to solve
complex electrostatic problems when designing electrical equipment. For example,
this need arises when calculating the electrical field strength in a high-voltage electrical
distribution station to ensure there is no danger of electrical discharge across
the air gap between components that have different voltages. Poisson’s equation is
an equation thatcan beusefulwhen carrying outthis work.Consider the following.
If E is an electric field andV an electrostatic potential, then the two fields are
related by
E=−∇V
FromExample26.2weknowGauss’slaw: ∇ · E = ρ ε 0
.Combiningthesetwoequations
we can write
that is
∇·(−∇V)= ρ ε 0
∇ 2 V = − ρ ε 0
ThispartialdifferentialequationisknownasPoisson’sequationandbysolvingitwe
coulddeterminetheelectrostaticpotentialinaregionoccupiedbycharges.Notethat
inacharge-freeregion, ρ = 0andPoisson’sequationreducestoLaplace’sequation
∇ 2 V = 0.
EXERCISES26.6
1 Ascalar field φ is given by
φ = 3x +y−y 2 z 2 .Show that φ satisfies
∇ 2 φ = −2(y 2 +z 2 ).
2 If φ =2x 2 y−xz 3 showthat ∇ 2 φ =4y−6xz.
3 Ifv =xyi −yzj + (y +2z)k findcurl(curl(v)).
4 If φ =xyzandv =3x 2 i+2y 3 j+xykfind ∇φ, ∇·v,
and ∇ · (φv).Show that
∇·(φv)=(∇φ)·v+φ∇·v.
5 Verifythat φ =x 2 y +y 2 z +z 2 x satisfies
∇·(∇φ)=2(x+y+z).
6 IfAis an arbitrarydifferentiablevectorfield show
thatthe divergence ofthe curlofAis always 0.
7 Expresseachofthe following in operator notation
using‘∇’, ‘∇ ·’ and‘∇×’:
(a) grad (divF)
(b) curl(grad φ)
(c) curl(curlF)
(d) div (curlF)
(e) div (grad φ)
8 Scalar fields φ 1 and φ 2 are givenby
φ 1 =2xy+y 2 z φ 2 =x 2 z
(a) Find ∇φ 1 .
(b) Find ∇φ 2 .
(c) State φ 1 φ 2 .
(d) Find ∇(φ 1 φ 2 ).
(e) Find φ 1 ∇φ 2 +φ 2 ∇φ 1 .
(f) Whatdo you conclude from(d)and(e)?

864 Chapter 26 Vector calculus
9 Ascalar field φ and avector fieldFare given by
φ=xyz 2 F=x 2 i+2j+zk
(a) Find ∇φ.
(b) Find ∇ ·F.
(c) Calculate φ(∇ ·F) +F · (∇φ).[Hint:recall the
dot product oftwo vectors.]
(d) State φF.
(e) Calculate ∇ · (φF).
(f) What do you conclude from(c)and (e)?
Solutions
3 j−k
(d) (6x 2 yz +2xy 2 z 2 )i + (2x 3 z +2yx 2 z 2 )j
+(2x 3 y +2y 2 x 2 z)k
4 ∇φ=yzi+xzj+xyk
∇·v=6x+6y 2
(e) same as(d)
∇·(φv) =9x 2 yz+8xy 3 z+x 2 y 2
(f) ∇(φ 1 φ 2 ) = φ 1 ∇φ 2 +φ 2 ∇φ 1
9 (a) yz 2 i +xz 2 j +2xyzk
7 (a) ∇(∇·F) (b) ∇ ×(∇φ)
(b) 2x+1
(c) ∇×(∇×F) (d) ∇·(∇×F)
(e) ∇·(∇φ)
(c) 3x 2 yz 2 +2xz 2 +3xyz 2
8 (a) 2yi + (2x +2yz)j +y 2 (d) x 3 yz 2 i +2xyz 2 j +xyz 3 k
k
(b) 2xzi +x 2 k
(e) same as(c)
(c) 2x 3 yz +y 2 x 2 z 2 (f) ∇·(φF)=φ(∇·F)+F·(∇φ)
26.7 VECTORCALCULUSANDELECTROMAGNETISM
Vector calculus provides a useful mechanism for expressing the fundamental laws of
electromagnetisminaconcisemanner.Theselawscanbesummarizedbymeansoffour
equations,knownasMaxwell’sequations.Muchofelectromagnetismisconcernedwith
solving Maxwell’s equations for different boundary conditions.
Equation1
divD=∇·D=ρ
where D = electric flux density, and ρ = charge density. This equation is a general
formofGauss’stheoremwhichstatesthatthetotalelectricfluxflowingoutofaclosed
surface isproportional tothe electriccharge enclosed by thatsurface.
Equation2
divB=∇·B=0
whereBisthe magnetic fluxdensity.
This equation arises from the observation that all magnetic poles occur in pairs and
therefore magnetic field lines are continuous; that is, there are no isolated magnetic
poles.Incontrast,electricfieldlinesoriginateonpositivechargesandterminateonnegative
charges and so a net positive charge in a region leads to an outflow of electric
flux.

Equation3
Review exercises 26 865
curlE=∇×E=− ∂B
∂t
where E is the electric field strength. This equation is a statement of Faraday’s law. A
time-varying magnetic field produces a space-varying electric field.
Equation4
curlH=∇×H=J+ ∂D
∂t
where H is the magnetic field strength and J is the free current density. This equation
statesthatatime-varying electric field gives risetoaspace-varying magnetic field.
The derivation of these equations is beyond the scope of this text but can be found in
manybooksonelectromagnetism.Thepoweroftheseequationsliesintheirgenerality.
Thebrevitywithwhichthemainlawsofelectromagnetismcanbeexpressedisatribute
tothe utility ofvector calculus.
REVIEWEXERCISES26
1 Find ∇φif
(a) φ = 3xyz
(b) φ =x 2 yz+xy 2 z+xyz 2
(c) φ =xy 2 z 2 +x 2 yz 2 +x 2 y 2 z
2 Ifφ=1/ √ x 2 +y 2 +z 2 ,showthat
∂ 2 φ
∂x 2 + ∂2 φ
∂y 2 + ∂2 φ
∂z 2 = 0
3 Functions satisfyingLaplace’sequationare called
harmonicfunctions.Showthat the following
functionsare harmonic:
(a) z =x 4 −6x 2 y 2 +y 4
(b) z = 4x 3 y −4xy 3
4 IfA=xi+yj+zkand
B = cosxi −sinxj,find
(a) A×B
(b) ∇·(A×B)
(c) ∇×A
(d) ∇×B
Verifythat
∇·(A×B)=B·(∇×A)−A·(∇×B).
5 Forarbitrary differentiablescalar fields φ and ψ show
that∇(φψ) = ψ∇φ +φ∇ψ.
6 Ifψ=x 2 yanda=xi+yj+zk,find∇ψ,
∇ ×a,∇ × (ψa).Showthat
∇×(ψa)=ψ∇×a+(∇ψ)×a.
7 Ascalarfield φ isafunctionofx,zandt only.Vectors
E and Hare defined by
( )
E = 1 ∂φ
ε ∂z i − ∂φ
∂x k H = − ∂φ
∂t j
where ε isaconstant.
(a) Show that ∇ ·E = 0.
(b) Show that ∇ ·H = 0.
Given that ∇ ×E = −µ ∂H ,where µ isaconstant,
∂t
show that φ satisfiesthe partial differential equation
∂ 2 φ
∂x 2 + ∂2 φ
∂z 2 = φ
µε∂2 ∂t 2
8 Ifv = (2x 2 y+3x 5 )i+e xy j+xyzkfind ∂v
∂x and ∂2 v
∂x 2.
9 An electrostatic potential isgiven byV = 5xyz.Find
(a) the associatedelectric field E,
(b) |E| atthe point (1,1,1).
10 An electrostatic field is given by
E = 3(x+y)i+2xyj.Findthedirectionofthisfieldat
(a) the point (2,2)
(b) the point (3,4).
11 Find the curlofA =yi +2xyj +3zk.
12 Find the curlofthe vector field
F = (x 2 −y)i + (xy −4y 2 )j.
13 If φ = 5e x+2y coszfind ∇φ atthepoint (0,0,0).

866 Chapter 26 Vector calculus
14 Thevector fieldsAand B are given by
A =x 2 yi+2xzj+xk
B=yzi+x 2 yj+3k
(a) Calculate the vector productA ×B.
(b) Calculate ∇ · (A ×B).
(c) CalculateB·(∇ ×A) −A·(∇ ×B).
(d) What doyou conclude from (b)and(c)?
(e) Can you prove thisresult istrueforany two
vector fields?
15 Given
F = (3x,2y,4z)
φ =x 2 yz
G = (1,−1,3)
find
(a) ∇·F
(b) ∇φ
(c) ∇×F
(d) ∇ ·(φF)
(e) ∇ × (φG)
(f) ∇(F·G)
(g) ∇×(F×G)
(h) ∇·(F+G)
(i) ∇ × (3F +4G)
Solutions
1 (a) 3yzi +3xzj +3xyk
(b) (2xyz +y 2 z +yz 2 )i
+ (x 2 z +2xyz +xz 2 )j
+ (x 2 y +xy 2 +2xyz)k
(c) (y 2 z 2 +2xyz 2 +2xy 2 z)i
+ (2xyz 2 +x 2 z 2 +2x 2 yz)j
+ (2xy 2 z +2x 2 yz +x 2 y 2 )k
4 (a) zsinxi+zcosxj+(−ycosx−xsinx)k
(b) zcosx (c) 0 (d) −cosxk
6 ∇ψ =2xyi+x 2 j
∇×a=0
∇ × (ψa) =x 2 zi −2xyzj + (2xy 2 −x 3 )k
8 (4xy +15x 4 )i +ye xy j +yzk,
(4y +60x 3 )i +y 2 e xy j
9 (a) E = −∇V = −5yzi −5xzj −5xyk
√
(b) 75
10 (a) 12i +8j
(b) 21i +24j
11 (2y −1)k
12 (y +1)k
13 5i +10j
14 (a) (6xz −x 3 y)i + (xyz −3x 2 y)j
+(x 4 y 2 −2xyz 2 )k
(b) −3x 2 (y +1) −xz(4y −1) +6z
(c) −3x 2 (y +1) −xz(4y −1) +6z
(d) ∇·(A×B)=B·(∇×A)−A·(∇×B)
15 (a) 9
(b) 2xyzi +x 2 zj +x 2 yk
(c) 0
(d) 21x 2 yz
(e) x 2 (y +3z)i + (x 2 y −6xyz)j
− (x 2 z +2xyz)k
(f) 3i −2j +12k
(g) −6i +7j −15k
(h) 9
(i) 0

27 Lineintegralsand
multipleintegrals
Contents 27.1 Introduction 867
27.2 Lineintegrals 867
27.3 Evaluationoflineintegralsintwodimensions 871
27.4 Evaluationoflineintegralsinthreedimensions 873
27.5 Conservativefieldsandpotentialfunctions 875
27.6 Doubleandtripleintegrals 880
27.7 Somesimplevolumeandsurfaceintegrals 889
27.8 ThedivergencetheoremandStokes’theorem 895
27.9 Maxwell’sequationsinintegralform 899
Reviewexercises27 901
27.1 INTRODUCTION
In this chapter a number of new sorts of integral are introduced. These are intimately
connected with the developments of the previous chapter on differential vector calculus.
The chapter starts by explaining the physical significance of line integrals and how
theseareevaluated.Thisleadsnaturallyintothetopicsofconservativevectorfieldsand
potentialfunctions.Theseareimportantinthestudyofelectrostatics.Doubleandtriple
integrals are then introduced; these generalize the earlier work on integration to integrands
which contain two and threeindependent variables.
Finallysomesimplevolumeandsurfaceintegralsareintroduced,togetherwiththedivergence
theorem and Stokes’ theorem. These enable Maxwell’s equations to be
expressed inintegral form.
27.2 LINEINTEGRALS
Consider an object of massmplaced in a gravitational field. Because the force of gravity
is a vector the gravitational field is an example of a vector field. The gravitational
force on the mass is known as its weight and is given by mg where g is a constant

868 Chapter 27 Line integrals and multiple integrals
A
s
m
M
N
ds
B
Figure27.1
An object ofmassmfallsfrom A to B.
vector called the acceleration due to gravity. Suppose we release the mass and allow
it to fall from point A in Figure 27.1. The vertical displacement measured downwards
from Aiss.
Workisbeingdonebythegravitationalforceinordertomakethemassaccelerate.We
wish to calculate the work done by the field in moving the mass from A to B. Suppose
we consider the amount of work done as the mass moves from point M to point N, a
distance δs.Elementaryphysicstellsusthattheworkdoneisequaltotheproductofthe
magnitudeoftheforceandthedistancemovedinthedirectionoftheforce.Inthiscase
themagnitudeoftheforceismg,andsothesmallamountofworkdone, δW,inmoving
from M toN,is
δW =mgδs
from which we have δW δs
=mg. As δs → 0 weobtain
δW
lim
δs→0 δs = dW ds =mg
To find the total work done as the mass falls from A to B we must add up, or integrate,
the contributions over the whole interval of interest,thatis
total work done =W =
∫ B
A
mgds
This is an elementary example of a line integral, so called because we are integrating
alongthelinefromAtoB.Inthiscaseitisstraightforwardtoevaluate.Sincebothgand
mare constants the integral becomes
W=mg
∫ B
A
ds
which equalsmg× (distancefrom AtoB).
Engineeringapplication27.1
Theworkdonebythegravitationalfield
AnobjectofmassmfallsverticallyfromAtoB.IfAisthepointwheres = 0andB
is the point wheres = 10 find the total work done by gravity as the mass falls from
AtoB.

27.2 Line integrals 869
Solution
The work done by gravity isfound by evaluating the line integral
W=mg
∫ s=10
s=0
ds =mg[s] 10
0 = 10mg
Note thatthisis alsothe potential energy lostbyminfallingadistance of 10 units.
Inthe previous example the path along which weintegrated was a straightline, butthis
need not always bethe case.Consider the following example.
Engineeringapplication27.2
Theworkdonebyanelectricfield
Figure 27.2 shows a unit charge moving along a curveC from point A to point B in
an electric fieldE.
At the particular point of interest, M, we have resolved the electric field vector
into two components. Resolving a vector into perpendicular components has been
described in Example 7.3. One component is tangential toC, namelyE t
, and one is
normaltoC,namelyE n
.AsaunitchargemovesfromMtoN,adistance δsalongthe
curve, the work done by the electric field isE t
δs. The componentE n
does no work
sincethereisnomotionperpendiculartothecurveC.Tofindthetotalworkdonewe
mustadd up all contributions, resulting inthe integral
total work done =
∫ B
A
E t
ds
Thisisasecondexampleofalineintegral,thelinebeingthecurve,C,joiningAand
B. Itisusual todenote thisby
∫
total work done = E t
ds
∫
C
where the symbol tells us tointegrate along the curveC.
C
C
B
M
E t
ds
N
E n
A
E
Figure27.2
Asacharge movesfrom Mto N,
the fieldEdoesworkE t δs.

870 Chapter 27 Line integrals and multiple integrals
If the coordinates of the end points of the curveC are known, say (x 1
,y 1
) and (x 2
,y 2
),
we often write ∫ (x 2
,y 2
)
(x 1
,y 1
)
to show this, but care must then be taken to define the intended
route from AtoB.
Wenowexplainhowtointegrateafunctionalongacurve.Consideravectorfield,F,
through which runs a curve,C, as shown inFigure 27.3.
F(x,y)
F
F n
M
u
F t
N
C
MN 5 ds
Figure27.3∫
Theintegral F · dsis equal to the work
C
done by Fasthe particle movesalongC.
Suppose we restrict ourselves to two-dimensional situations. In the general case the
vector field will vary asxandyvary, that is F = F(x,y). Consider the small element
ofC joining points M and N. Let θ be the angle between the tangent to the curve at M
andthedirectionofthefieldthere.WeshalldenotethevectorjoiningMandN(i.e. −→ MN)
by δs. Consider the quantity
F·δs
where · represents the scalar product. When F represents a gravitational force field,
F·δsrepresentsthesmallamountofworkdonebythefieldinmovingaparticleofunit
mass from M to N. The appropriate integral along the whole curve represents the total
work done. From the definition of the scalar product wenote that
F·δs = |F‖δs|cosθ
Writingthe modulus ofFas simplyF, and the modulus of δs, as δs, we have
F·δs=Fδscosθ
= (Fcosθ)δs
=F t
δs
whereF t
is the component of F tangential toC. This result is of the same form as the
expressions for work done obtained previously. We are therefore interested in integrals
of the form
∫
F(x,y)·ds
C

27.3 Evaluation of line integrals in two dimensions 871
ds
dxi
Figure27.4
Thevector dscan be
written asdxi +dyj.
dyj
Since F is a vector function of x and y it will have Cartesian components P(x,y) and
Q(x,y) and so wecan writeFinthe form
F(x,y) =P(x,y)i +Q(x,y)j
Similarly,referringtoFigure27.4, wecanwritethevectordsasds = dxi +dyj.Hence
∫ ∫
F(x,y)· ds = (P(x,y)i +Q(x,y)j)· (dxi +dyj)
C
Taking the scalar product gives
∫
P(x,y)dx +Q(x,y)dy
C
and thisisthe lineintegral inCartesian form.
C
IfF(x,y) =P(x,y)i +Q(x,y)jthen
∫ ∫
F(x,y)· ds = (P(x,y)i +Q(x,y)j)· (dxi +dyj)
C
C
∫
= P(x,y)dx +Q(x,y)dy
C
In order to proceed further we now need to explore how such integrals are evaluated
inpractice. This isthe topic of the next section.
27.3 EVALUATIONOFLINEINTEGRALSINTWODIMENSIONS
Example27.1 Evaluate
∫
F·ds
C
whereF = 5y 2 i +2xyj andC isthe straightline joiningthe originand the point (1,1).
Solution We compare the integrand with the standard form and recognize that in this case
P(x,y) = 5y 2 andQ(x,y) = 2xy. The integralbecomes
∫
5y 2 dx +2xydy
C
The curve of interest, in this case a straight line, is shown in Figure 27.5. Along this
curve it is clear that y = x at all points. We use this fact to simplify the integral by
writing everything in terms of x. We could equally well choose to write everything in
terms of y. If y = x then dy = 1, that is dy = dx. As we move along the curveC, x
dx
ranges from0to1,and the integral reduces to
∫
∫ ∫ x=1
[ 7x
5x 2 dx +2x 2 dx = 7x 2 dx = 7x 2 3
dx = = 7 3 3
C
C
x=0
] 1
0

872 Chapter 27 Line integrals and multiple integrals
y
1 (1, 1)
y
1
(1, 1)
C
C 2
C 1
C
0 1 x
Figure27.5
Thepath ofintegration joins
(0,0)and(1,1)and has equation
y=x.
0 1 x
Figure27.6
ThepathC is made upoftwo distinct
parts.Integrationis performed
separately oneach part.
Let us now see what happens when we choose to evaluate the integral of Example 27.1
along a different path joining (0,0) and (1,1).
Example27.2 Evaluate the integral ∫ C 5y2 dx +2xydy along the curve,C, consisting of thexaxis betweenx
= 0 andx = 1 and the linex = 1 asshown inFigure 27.6.
Solution WeevaluatethisintegralintwopartsbecausethecurveCismadeupoftwopieces.The
firstpieceC 1
ishorizontal,andthesecondC 2
isvertical.Therequiredintegralisthesum
of the two separate ones. Along thexaxis,y = 0 and dy = 0. This means that both the
terms5y 2 dx and 2xydy arezero,and so the integral reducesto
∫ x=1
x=0
0dx=0
andsothereiszerocontributiontothefinalanswerfromthispartofthecurveC.Along
the linex = 1, the quantity dx equals zero. Hence 5y 2 dx also equals zero, and 2xydy
equals2ydy. Becauseyrangesfrom0to1the integral becomes
∫ y=1
y=0
2ydy = [y 2 ] 1 0 = 1
Note thatthis isadifferent answerfrom thatobtained inExample 27.1.
As we have seen from Examples 27.1 and 27.2 the value of a line integral can depend
upon the pathtaken. Itisthereforeessential tospecifythispathclearly.
Example27.3 Evaluatetheintegral ∫ C F(x,y)·ds,whereF(x,y) = (3x2 +y)i+(5x−y)jandCisthe
portionofthe curvey = 2x 2 between A(2,8)and B(3, 18).
∫ ∫
Solution F(x,y)·ds = (3x 2 +y)dx + (5x −y)dy
C
C
The curveC has equationy = 2x 2 for 2 x3. Along this curve we can replaceyby
2x 2 . Note also that dy = 4x and so we can replace dy with 4xdx. This will produce an
dx

27.4 Evaluation of line integrals in three dimensions 873
integral entirely in terms of the variablexwhich is then integrated betweenx = 2 and
x = 3.Thus
∫
C
(3x 2 +y)dx+(5x−y)dy=
=
∫ 3
2
∫ 3
2
(3x 2 +2x 2 )dx + (5x −2x 2 )(4xdx)
25x 2 −8x 3 dx
[ ] 25x
3 3
= −2x 4
3
2
= 28.333
EXERCISES27.3
1 Evaluate ∫ C3ydx +2xdyalong the straightlineC
between(1,1)and(3,3).
2 Evaluate ∫ C 2yxdx +x2 dyalong the straightline
y = 4xfrom (0,0)to (3,12).
3 Evaluate ∫ C (7x +3y)dx +2ydy alongthe curve
y =x 2 between (0,0)and (2,4).
4 IfE = (x +2y)i + (x −3y)j,Aisthepoint (0,0)and
Bis the point (3,2),evaluate
∫ B
E·ds
A
(a)along the straightline joining AandB,
(b)horizontally along thexaxisfromx = 0 tox = 3
andthenverticallyfromy = 0 toy = 2.
5 IfF
∫
= (2xy −y 4 +3)i + (x 2 −4xy 3 )jevaluate
CF · dswhereC is the straightlinejoining A(1,3)
andB(2,5).
6 Evaluate the integral
∫
(3x 2 +2y)dx + (7x +y 2 )dy
C
from A(0,1)to B(2, 5)along the curveC defined by
y=2x+1.
Solutions
1 20
2 108
3 38
4 (a)
15
2
5 −1149
6
268
3
(b)
9
2
27.4 EVALUATIONOFLINEINTEGRALSINTHREEDIMENSIONS
Evaluation of line integrals along curves lying in three-dimensional space is performed
inasimilar way. It isoften helpful inthis case toexpress the independentvariablesx,y
andzintermsofasingleparameter. Consider the following example.

874 Chapter 27 Line integrals and multiple integrals
Example27.4 Acurve,C, isdefined parametrically by
x=4 y=t 3 z=5+t
and islocated within a vector fieldF =yi +x 2 j + (z +x)k.
(a) Find the coordinates of the point P on the curve where the parameter t takes the
value 1.
(b) Find the coordinates of the point Qwhere the parametert takes the value 3.
(c) By expressing the line integral ∫ F·ds entirely in terms oft find the value of the
C
lineintegral from PtoQ.Note thatds = dxi +dyj +dzk.
Solution (a) Whent = 1,x = 4,y = 1 andz = 6,and so Phas coordinates (4, 1,6).
(b) Whent = 3,x = 4,y = 27 andz = 8,and so Qhas coordinates (4, 27, 8).
(c) To express the line integral entirely in terms of t we note that if x = 4,
dx
= 0 so that dx is also zero. If y = t 3 then dy = 3t 2 so that dy = 3t 2 dt.
dt
dt
Similarlysincez = 5 +t, dz = 1 so thatdz = dt. The line integralbecomes
dt
∫ ∫
F·ds= (yi +x 2 j + (z +x)k)·(dxi +dyj +dzk)
C
C
∫
= ydx+x 2 dy+(z+x)dz
=
=
C
∫ t=3
t=1
∫ 3
1
0 +16(3t 2 )dt+(9+t)dt
48t 2 +9+tdt
[ ] 48t
3 3
=
3 +9t+t2 2
1
( 48(3 3 )
=
3
= 438
+ (9)(3) + 32
2
) ( 48
−
3 +9+ 1 2)
EXERCISES27.4
1 You are requiredto evaluate the lineintegral
∫
CF ·ds where F is the vector field
F = (2y +3x)i + (yz +x)j +3xykand
ds = dxi +dyj +dzk.The curveC is defined
parametrically byx =t 2 ,y = 3t andz = 2t for
values oft between0and1.
(a) Findthe coordinatesofthe point A, wheret = 0.
(b) Findthe coordinatesofthe point B,wheret = 1.
(c) Byexpressing the line integralentirely in terms
oft,evaluate the lineintegral from Ato B along
the curveC.

27.5 Conservative fields and potential functions 875
Solutions
1 (a) (0,0, 0)
(b) (1,3, 2)
(c) 17
27.5 CONSERVATIVEFIELDSANDPOTENTIALFUNCTIONS
We have seen that, in general, the value of a line integral depends upon the particular
pathchosen.However,inspecialcases,theintegralofagivenvectorfieldFturnsoutto
be the same on any path with the same end points; that is, it is independent of the path
chosen.InthesecasesthevectorfieldFissaidtobeconservative.Thereisasimpletest
which tells us whether or notFis conservative:
Fisaconservative vector field if
curlF = 0
everywhere
Ifwearedealingwithatwo-dimensionalvectorfieldF(x,y) =P(x,y)i+Q(x,y)jthen
thistestbecomes:Fisconservative if ∂P
∂y = ∂Q . (SeeQuestion 7 inExercises 27.5.)
∂x
Example27.5 Show thatthe vector field
F =ze x sinyi +ze x cosyj +e x sinyk
isaconservative field.
Solution We find curlF:
i j k
∇×F=
∂ ∂ ∂
∂x ∂y ∂z
∣
ze x sinyze x cosy e x siny
∣
= (e x cosy −e x cosy)i − (e x siny −e x siny)j + (ze x cosy −ze x cosy)k
= 0
Wehaveshownthat ∇ ×F = 0andsothefieldisconservative.NotefromSection26.5
thatsuch a field isalsosaidtobe irrotational.
Therearetwoimportantpropertiesofconservativevectorfieldswhichwenowdiscuss.
27.5.1 Thepotentialfunction
Itcanbeshownthatanyconservativevectorfield,F,canbeexpressedasthegradientof
some scalar field φ. That is, we can find a scalar field φ such that F = ∇φ. The scalar
field φ issaidtobeapotentialfunctionandFissaidtobederivablefromapotential.

876 Chapter 27 Line integrals and multiple integrals
Inthree dimensions, if
F =P(x,y,z)i +Q(x,y,z)j +R(x,y,z)k
isconservative, wecan write
So
F=∇φ
= ∂φ
∂x i + ∂φ
∂y j + ∂φ
∂z k
P = ∂φ
∂x
Q = ∂φ
∂y
R = ∂φ
∂z
You should refer back toExample 26.4 where thisidea was introduced.
Example27.6 Thevector field visderivable from the potential φ = 2xy +zx. Find v.
Solution If v isderivable from the potential φ, then v = ∇φ and so
v=∇φ=(2y+z)i+2xj+xk
This vector field isconservative asiseasilyverified by findingcurl v. Infact,
i jk
∇×v=
∂ ∂ ∂
∂x ∂y ∂z
∣
2y+z2x x
∣
=0i−(1−1)j+(2−2)k
= 0
Indeed, recall from Example 26.10 thatcurl (grad φ) isidentically zero forany φ.
When the vector field F is conservative there is an alternative method of evaluating the
line integral ∫ F·ds which involves the use of the potential function φ. Consider the
C
following example.
Example27.7 Thetwo-dimensional vector fieldF = i +2j isconservative.
(a) Findasuitablepotential function φ suchthat F = ∇φ = ∂φ
∂x i + ∂φ
∂y j.
(b) Evaluate ∫ (3,2)
F·ds along any convenient path.
(0,0)
(c) Find the value of φ at B(3, 2), and the value of φ at A(0, 0), and show that the
differencebetweentheseisequaltothevalueofthelineintegralobtainedinpart(b).
Solution (a) We are given that ∂φ = 1 so that φ = x + f (y), where f is an arbitrary function
∂x
ofy. We are also given that ∂φ = 2 so that φ = 2y +g(x), wheregis an arbitrary
∂y
function ofx. Itiseasy toverify that φ =x +2yisasuitablepotential function.

27.5 Conservative fields and potential functions 877
(b) Tofind thelineintegralweshallchoosethestraightlinepath,C,joining (0,0) and
(3,2). This has equationy = 2 x. The lineintegral becomes
3
∫ (3,2)
(0,0)
F·ds=
=
=
=
∫ (3,2)
(0,0)
∫ (3,2)
(0,0)
∫ (3,2)
(0,0)
∫ x=3
x=0
[ ] 7 3
=
3 x 0
= 7
(i +2j)·(dxi +dyj)
dx+2dy
dx + 4 3 dx
7
3 dx
since dy = 2 3 dxonC
(c) The value of φ at B(3, 2) is 3 + 2(2) = 7. The value of φ at A(0, 0) is 0. Clearly
the difference between these values is 7, the same as the value of the line integral
obtained inpart(b).
Theresultobtained inthe previousexample isimportant:
IfFisaconservative vectorfield suchthat F = ∇φ then, forany points Aand B,
∫ B
A
F· ds = φ(B)−φ(A)
27.5.2 Thelineintegralaroundaclosedloop
AsecondimportantpropertyofconservativefieldsariseswhenthecurveCofintegration
forms a closed loop. Suppose we evaluate a line integral from A to B firstly along the
curveC 1
and secondly alongC 2
asshown inFigure 27.7.
If the field,F, isconservative bothintegrals will yield the same answer, thatis
∫ ∫
F·ds= F·ds
C 1
C 2
But
∫ ∫
F·dsfromAtoB =−
F·dsfromBtoA
Consequently,thelineintegralaroundtheclosedcurveAtoBandbacktoAmustequal
zero.When the pathofintegration isaclosedcurve weuse the symbol
∮

878 Chapter 27 Line integrals and multiple integrals
B
y
1
B
C 2
C 1
C 2
C 1
A
Figure27.7 ∫
Theline integral
F·dscanbe
evaluated along different paths
between Aand B.
A
0 1
Figure27.8
Thepath ofintegration is the curveC 1
followed bythe lineC 2 .
x
torepresent the integral.So, foraconservative field wehave the result
∮
F·ds=0
C
for any closed curveC.
Insummarywe have the following results:
Foraconservative field, F,all the following statements areequivalent:
∇×F=0
∮
F·ds=0
∫ B
A
F·ds isindependentofthe pathbetween Aand B
Fisderivable from a scalar potential,thatisF = ∇φ
Example27.8 Evaluate
∫
F·ds
C
whereFisthe vectorfieldy 2 i +2xyj and where:
(a) C =C 1
isthe curvey =x 2 goingfrom A(0,0)toB(1, 1).
(b) C =C 2
isthe straightlinegoingfromB(1, 1)toA(0,0).
(c) Deduce that ∮ F·ds = 0 where the closed line integral is taken around the pathC 1
and thenC 2
.
Solution Thesituation isdepictedinFigure 27.8.
(a) C 1
hasequationy =x 2 andsody = 2xdx.NotethatweareintegratingfromAtoB.
Therefore
∫
F·ds=
C 1
∫
(y 2 i +2xyj)·(dxi +dyj) =
C 1
∫
y 2 dx +2xydy
C 1

Substitutingy =x 2 and dy = 2xdx we have
∫ ∫
F·ds= x 4 dx + (2x)x 2 (2xdx)
C 1
C 1
=
∫ x=1
x=0
27.5 Conservative fields and potential functions 879
5x 4 dx = [x 5 ] 1 0 = 1
(b) C 2
hasequationy =xandsody = dx.NotethatthistimeweareintegratingfromB
toA.Therefore
∫
F·ds=
C 2
∫
(y 2 i +2xyj)·(dxi +dyj) =
C 2
∫
y 2 dx +2xydy
C 2
Substitutingy =xand dy = dx we have
∫ ∫ x=0
F·ds= x 2 dx+2x 2 dx=
C 2
x=1
∫ 0
(c) Now
∮ ∫ ∫
F·ds= F·ds+ F·ds=1−1=0
C 1
C 2
1
3x 2 dx = [x 3 ] 0 1 = −1
The line integral around the closed path is zero because the field is conservative.
You should check that ∇ ×F = 0and thatasuitablepotential function is φ =y 2 x.
EXERCISES27.5
1 Verifythat the vector field
F = (4x 3 +y)i +xj is conservative.
2 Forthe vector field
F =ycosxyi +xcosxyj +2zkfind ∇ ×Fandverify
that the field isconservative.
3 Considerthe fieldF =yi +xj.
(a) ShowthatFis aconservative field.
(b) Find afunction φ such that ∂φ
∂x =yand∂φ ∂y =x.
(c) Find asuitable potential function φ for F.
(d) IfAisthe point (2,1)and Bis the point
(5,8)evaluate ∫ B
A F·ds.
(e) Find φ at Band φ atAandshow that the value of
the line integralcalculated in part (d)is equal to
thedifferencebetweenthevaluesof φ atBandA.
4 Showthat the vectorfield,
F = (2xy −y 4 +3)i + (x 2 −4xy 3 )j,ofQuestion5in
Exercises27.3, isconservative andfindasuitable
potential function φ from which Fcan be derived.
Showthat the difference between φ evaluated at
B(2, 1)andatA(1,0)is equalto the value ofthe line
integral ∫ B
A F·ds.
5 Show that
∫ (1,1)
I = 3x 2 y 2 dx +2x 3 ydy
(−1,−1)
is independentofthe pathofintegration,andevaluate
the integral.
6 The function φ = 4xy is apotential functionfor
whichF = ∇φ.
(a) Find F.
(b) Evaluate ∫ F ·dsalong the curvey =x 3 from
A(0, 0)to B(2, 8).
(c) Evaluate φ at BandatAandshowthat the
difference between these values is equalto the
value ofthe lineintegral obtainedin part (b).
7 IfF =P(x,y)i
(
+Q(x,y)j
)
+0k,showthat
∂Q
∇×F=
∂x − ∂P k.Deduce that ifFis
∂y
conservative then ∂Q
∂x = ∂P
∂y .

880 Chapter 27 Line integrals and multiple integrals
Solutions
3 (b) φ=xy+c (c) xy+c (d) 38
4 φ=x 2 y−y 4 x+3x+c,5
5 2
6 (a) F = 4yi +4xj (b) 64 (c) 64
27.6 DOUBLEANDTRIPLEINTEGRALS
27.6.1 Doubleintegrals
Expressions such as
∫ y=y2
x=x 1
y=y 1
∫ x=x2
f(x,y)dxdy
are known asdouble integrals. What ismeant by the above is
∫ (
y=y2 ∫ )
x=x2
f(x,y)dx dy
y=y 1
x=x 1
wherefirstlytheinnerintegralisperformedbyintegrating f withrespecttox,treatingy
asifitwereaconstant.Thelimitsofintegrationareinsertedasusualandthenthewhole
expression isintegrated with respect toybetween the limitsy 1
andy 2
.
It is important that you can distinguish between double integrals and line integrals.
Physically they mean quite different things, and they are evaluated in different ways.
Whilstbothtypesofintegralcontaintermsinvolvingx,y,dxanddy,inadoubleintegral
dx and dy always occur as a product, that is as dxdy. In a line integral they occur separately.
When evaluating a line integral we integrate along a curve. When evaluating a
double integral we integrate over a two-dimensional region.
Double integrals may alsobe writteninthe form
∫ ∫
f(x,y)dxdy
R
whereRiscalledtheregionofintegration.TheregionRmustbedescribedmathematically.
Inthe integral
∫ y=y2
x=x 1
y=y 1
∫ x=x2
f(x,y)dxdy
R is the rectangular region defined by x 1
x x 2
, y 1
y y 2
. Non-rectangular
regions are alsocommon.
Example27.9 Sketch the regionRdefined by 1 x4 and 0 y2.
Solution TheregionisshowninFigure27.9.Weseethatxliesbetween1and4,andyliesbetween
0 and 2.Consequently the region isrectangular.

27.6 Double and triple integrals 881
y
4
y
4
3
R
3
R
2
2
1
dy
1
dx
0
1 2
3 4 x
Figure27.9
Therectangular region over which the
integration in Example 27.10 is
performed.
0
1 2
3 4 x
Figure27.10
Theintegral could have been foundbyfirst
integrating over a vertical stripsuchasthat
shown.
Example27.10 Evaluate
∫ y=2 ∫ x=4
y=0 x=1
x+2ydxdy
over the region,R, given inExample 27.9 and defined by 1 x4 and 0 y2.
Solution The inner integral is found first by integrating with respect tox, keepingyfixed, that is
constant.
∫ x=4
[ ] x
2 x=4
x+2ydx=
2 +2xy =8+8y− 1 2 −2y=15 2 +6y
x=1
x=1
Then the outer integral isfound by integrating the resultwith respecttoy:
∫ y=2
y=0
[ ]
15 15 2
2 +6ydy= 2 y+3y2 =15+12=27
0
Now consideragain Example 27.10. Byperformingthe innerintegral first,
∫ x=4
x=1
x+2ydx
weareeffectivelysinglingoutahorizontalstrip,suchasthatshowninFigure27.9,and
integrating along it fromx = 1 tox = 4. When we then integrate the result fromy = 0
toy = 2 weare effectively adding upcontributions fromall suchhorizontalstrips.
Theintegraloverthesameregioncouldalsohavebeenperformedbyintegratingfirst
over vertical stripsfor0 y2(Figure 27.10)yielding
∫ y=2
y=0
x+2ydy =[xy+y 2 ] y=2
y=0 =2x+4

882 Chapter 27 Line integrals and multiple integrals
The result is then integrated fromx = 1 tox = 4 effectively adding up contributions
from each vertical strip:
∫ x=4
x=1
2x+4dx=[x 2 +4x] 4 1 =32−5=27
as expected. We note that
∫ y=2 ∫ x=4
y=0
x=1
x+2ydxdy=
∫ x=4 ∫ y=2
x=1
y=0
x +2ydydx (27.1)
In particular, note the order of dx and dy in Equation (27.1). The simple interchange of
limits isonly possible because inthis example the regionRisarectangle.
WhenevaluatingadoubleintegraloverarectangularregionR,theintegrationmay
becarried outinany order, thatis
∫ ∫
∫ ∫
f(x,y)dxdy= f(x,y)dydx
R R
The double integral in Example 27.10 could have the following interpretation. Consider
the surfacez = x +2y, for 0 y 2, 1 x 4, as shown in Figure 27.11. In
this example the surface is a plane but the theory applies to more general surfaces. The
variable z represents the height of the surface above the point (x,y) in the x--y plane.
For example, consider a point in thex--y plane, say (2, 3), that isx = 2,y = 3. Then at
(2,3),z =x +2y = 2 +2(3) = 8. Thus the point on the surface directly above (2,3)
is8unitsfrom (2,3).
Let us now consider dxdy. We call dxdy an element of area. It is a rectangle with
dimensionsdxanddyandareadxdy.Hencezdxdyrepresentsthevolumeofarectangular
column of base dxdy. As we integrate with respect toxfor 1 x4and integrate
with respect toyfor 0 y2weareeffectively summing all such volumes. Thus
∫ y=2 ∫ x=4
y=0
x=1
x+2ydxdy
represents the volume bounded by the surfacez = x + 2y and the regionRin thex--y
plane.
Example27.11 Sketch the regionRover which wewould evaluate the integral
∫ y=1 ∫ x=2−2y
y=0
x=0
f(x,y)dxdy
Solution Firstconsidertheouterintegral.Therestrictiononymeansthatinterestcanbeconfined
to the horizontal strip 0 y 1. Then examine the inner integral. The lower limit on
x means that we need only consider values ofxgreater than or equal to 0. The upperx
limit depends upon the value ofy. Ify = 0 this upper limit isx = 2 −2y = 2. Ify = 1
theupperlimitisx = 2 −2y = 0.Atanyotherintermediatevalueofywecancalculate
thecorrespondingupperxlimit.Thisupperlimitwilllieonthestraightlinex = 2−2y.
With this information the region of integration can be sketched. The region is shown in
Figure 27.12 and isseen tobe triangular.

27.6 Double and triple integrals 883
z
8
7
6
5
4
3
2
1
0 0
1
2
3
4
x
dx dy
Figure27.11
Thesurfacez=x+2y,0y2,1x4;the
quantityzdxdyis the volume ofthe rectangular column
with base area dxdy.
z
y
y
1
dy
2–x
x = 2 – 2y, y = ––— 2
0 2
Figure27.12
Theregion over which the integration in Example 27.12
isperformed.
x
Example27.12 Evaluate
∫ y=1 ∫ x=2−2y
y=0 x=0
4x+5dxdy
over the region described inExample 27.11.
Solution We first perform the inner integral
∫ x=2−2y
x=0
4x+5dx
integrating with respect tox. This gives
[2x 2 +5x] x=2−2y
x=0
=2(2−2y) 2 +5(2−2y)−0
=8y 2 −26y+18
Next weperform the outer integral
∫ y=1
y=0
[ 8y
8y 2 3
−26y+18dy=
3 − 26y2
2
= 8 3 − 26 2 +18
= 7.667
] 1
+18y
0
The region of integration is shown in Figure 27.12. The first integral, with respect tox,
corresponds to integrating along the horizontal strip fromx = 0 tox = 2 − 2y. Then
asyvaries from 0 to 1 in the second integral, the horizontal strips will cover the entire
region.

884 Chapter 27 Line integrals and multiple integrals
If we wish to change the order in which the integration is carried out, care must be
takenwiththelimitsofanon-rectangularregion.ConsideragainFigure27.12.Theline
x = 2 − 2y can be written asy = 2 −x . We can describe the regionRby restricting
2
attention to the vertical strip 0 x 2, and then letting y vary from 0 up to 2 −x
2 ,
that is
∫ x=2 ∫ y=(2−x)/2
x=0
y=0
4x+5dydx
You should check by evaluating this integral that the same result is obtained as in
Example 27.12.
Example27.13 Evaluate the double integral of f (x,y) = x 2 + 3xy over the regionRindicated in Figure
27.13by
(a) integrating first with respecttox, and thenwith respecttoy,
(b) integrating first with respecttoy, and thenwith respecttox.
Solution (a) If we integrate first with respect toxwe must select an arbitrary horizontal strip as
shown in Figure 27.13 and integrate in the x direction. On OB, y = x so that the
lower limit of thexintegration isx =y. On AB,x = 1 so the upper limit isx = 1.
Therefore
∫ x=1
[ ] x
x 2 3
+3xydx=
x=y 3 + 3x2 y 1
2
y
( 1
=
3 + 3y ) ( ) y
3
−
2 3 + 3y3
2
= 1 3 + 3y
2 − 11y3
6
Asyvaries from 0 to 1 the horizontal strips will cover the entire region. Hence the
limitsof integration of theyintegral are 0 and 1.So
∫ y=1
y=0
1
3 + 3y
2 − 11 6 y3 dy=
[ 1
3 y + 3 ] 1
4 y2 − 11y4
24
0
= 1 3 + 3 4 − 11
24
= 15
24
= 5 8
thatis,
∫ y=1 ∫ x=1
y=0
x=y
x 2 +3xydxdy= 5 8

27.6 Double and triple integrals 885
y
y
1
B
1
B
x = y
y = x
y = x
R
dy
x = 1
R
O 1
A
x
0 y = 0 1
A
x
Figure27.13
Theintegral with respect toxhas
limitsx =yandx = 1.
Figure27.14
We integrate along a vertical strip
fromy=0toy=x.
(b) If we choose to integrate with respect toyfirst we must select an arbitrary vertical
stripasshowninFigure27.14.Atthelowerendofthestripy = 0.Attheupperend
y=x.
To integrate along the stripweevaluate
∫ y=x
y=0
] x
x 2 +3xydy=
[x 2 y + 3xy2 =x 3 + 3x3
2
0
2 = 5x3
2
Weaddcontributionsofallsuchverticalstripsbyintegratingwithrespecttoxfrom
x=0tox=1:
∫ 1
0
5x 3
[ 5x
4
2 dx = 8
] 1
0
= 5 8
We see that the prudent selection of the order of integration can yield substantial
savings inthe effort required.
27.6.2 Tripleintegrals
The techniques we have used for evaluating double integrals can be generalized naturally
to triple integrals. Whereas double integrals are evaluated over two-dimensional
regions, tripleintegrals areevaluated over volumes.
Example27.14 Evaluate
I =
∫ x=1 ∫ y=1 ∫ z=1
x=0 y=0 z=0
x+y+zdzdydx
Solution What ismeantby thisexpression is
I =
∫ x=1
(∫ y=1
(∫ z=1
x=0
y=0
z=0
) )
x+y+zdz dy dx

886 Chapter 27 Line integrals and multiple integrals
where,asbefore,theinnerintegralisperformedfirst,integratingwithrespecttoz,with
x andybeing treated asconstants. So
∫ (
x=1 ∫ y=1
] ) 1
I =
[xz +yz+ z2
dy dx
x=0 y=0 2
0
∫ x=1
(∫ y=1
= x+y+ 1 )
2 dy dx
=
=
x=0
∫ x=1
x=0
∫ x=1
x=0
y=0
[xy + y2
[ ] x
2 1
=
2 +x 0
x + 1 2 + 1 2 dx
2 + 1 ] 1
2 y dx
0
= 3 2
Considerationofthelimitsofintegrationshowsthattheintegralisevaluatedoveraunit
cube.
27.6.3 Green’stheorem
There is an important relationship between line and double integrals expressed in
Green’s theorem inthe plane:
If the functions P(x,y) and Q(x,y) are finite and continuous in a region of the
x--y plane, R, and on its boundary, the closed curveC, provided the relevant partialderivatives
exist and arecontinuousinand onC, then
∮
∫ ∫ ( ∂Q
Pdx+Qdy=
C
R
∂x − ∂P
∂y
)
dxdy
where the direction of integration alongC is such that the region R is always to
the left.
Theconditionsgiveninthetheoremarepresentformathematicalcompleteness.Most
of the functions that engineers deal with satisfy these conditions, and so we will not
consider these further. The important thing to note is that this relationship states that a
line integral around a closed curve can be expressed in terms of a double integral over
the region,R, enclosed byC.
Example27.15 (a) Evaluate ∮ C xydx + x2 dy around the sides of the square with vertices D(0, 0),
E(1,0),F(1,1)and G(0,1).
(b) Convert the lineintegral toadouble integral and verify Green’s theorem.

y
1
0
G
D
R
27.6 Double and triple integrals 887
Solution (a) The path of integration is shown in Figure 27.15. To apply Green’s theorem the
path of integration should be followed in such a way that the region of integration
isalways toits left.We musttherefore travel anticlockwise aroundC.
1
F
E
Figure27.15
ThepathC ischosen so
that the regionRis
always to itsleft.
x
OnDE,y=0,dy=0and0x1.
OnEF,x=1,dx=0and0y1.
On FG,y = 1, dy = 0 andxdecreases from1to0.
On GD,x = 0, dx = 0 andydecreases from1to0.
The integral around the curveC can thenbe written as
∮
C
=
∫ E
D
+
∫ F
E
+
∫ G
Therefore
∮
xydx+x 2 dy=0+
C
F
∫ D
+
G
∫ 1
0
= [y] 1 0 + [ x
2
1dy+
2
] 0
1
∫ 0
1
xdx+0
= 1 − 1 2
= 1 2
(b) Applying Green’s theorem withP(x,y) = xy andQ(x,y) = x 2 we can convert the
line integral into a double integral. Note that ∂Q ∂P
= 2x and = x. Clearly the
∂x ∂y
region of integration isthe squareR. We find
∮
∫ ∫
xydx+x 2 dy= (2x −x)dxdy
R
C
=
=
=
∫ 1 ∫ 1
0
∫ 1
0
∫ 1
0
0
[ x
2
2
1
2 dy
[ ] 1 1
=
2 y 0
xdxdy
] 1
dy
0
= 1 2
We see that the same result as that in part (a) is obtained and so Green’s theorem
has been verified.

888 Chapter 27 Line integrals and multiple integrals
EXERCISES27.6
1 Evaluate
∫ 1 ∫ 3
(a) z 2 dxdz
0 0
∫ 2 ∫ 3
(c) x 2 y+1dxdy
1 2
∫ 4 ∫ 2
(b) xdydz
0 0
∫ 1 ∫ 3 y
(d)
−1 2 x dxdy
2 R isthe shaded region shown in Figure27.16.
Evaluate ∫∫ R x + √ ydxdy
(a) performing the integration with respect toxfirst,
(b) performing the integration with respect toyfirst.
y
4
2
0 1
y = x 2
2 x
Figure27.16
Theregion ofintegration forQuestion2.
3 Evaluate ∫∫ R (5x2 +2y 2 )dxdywhereRis the interior
ofthe triangularregion boundedby
A(1,1),B(2, 0)andC(2, 2).
4 (a) Sketch the region ofintegration ofthe double
integral
∫ 3 ∫ √ 4−y
ydxdy
1 1
(b) Evaluate the integral byfirst reversingthe order
ofintegration.
5 Evaluate
∫ 1 ∫ 5
(x 2 +y 2 )dxdy
−1 1
6 Evaluate
∫ ∫
(x 2 +y 2 )dxdy
R
over the triangularregionRwith vertices at
(0,0),(2,0)and(1,1).
7 Evaluate
∫
(x +2y) −1/2 dxdy
R
over the regionRgiven byx −2y 1 and
xy 2 +1.
8 Evaluate ∮ C (sinx +cosy)dx +4ex dywhereC isthe
boundary ofthe triangle formedby the points(1,0),
(3,0)and(3,2).Byconverting thisline integralinto a
doubleintegral, verifyGreen’stheorem.
9 Evaluate the line integral
∮
(x+y)dx+3xydy
C
whereC is the boundary ofthe triangle formedbythe
points(0,0),(2,0)and(0,5).Expressthelineintegral
in terms ofan appropriatedoubleintegral and
evaluate this.Verify Green’stheorem.
Solutions
1 (a) 1 (b) 8x (c) 10.5 (d) 0
2 (a)
3
33
2
20
3
(b)
20
3
7
2
.TheregionRis shown in FigureS.26.
3
y
2
x = y 2 + 1
y = –
x 1
– –
1
2 2
4 (b) 1.35435
5
6
256
3
4
3
0 1 2 3 4 5
FigureS.26
8 92.306
9 20
x

27.7 Some simple volume and surface integrals 889
27.7 SOMESIMPLEVOLUMEANDSURFACEINTEGRALS
Volume and surface integrals arise frequently in electromagnetism and fluid mechanics.
We will illustrate these concepts through some simple examples. For a thorough
treatment ofmore general cases you will need torefer toamore advanced text.
Engineeringapplication27.3
Themassofasolidobject
Consider a solid object such as that shown in Figure 27.17. The density, ρ, of this
objectmayvaryfrompointtopoint.So,atanypointPwithcoordinates (x,y,z),the
density is a function of position, that is ρ = ρ(x,y,z). Since density is a scalar, this
isan example ofascalar field, like those discussed inSection 7.4.
Supposeweselectaverysmallpieceofthisobjecthavingvolume δV andlocated
atP(x,y,z). Recall fromelementary physics that
density = mass
volume
Then the mass of thissmallpiece, δm, isgiven by
δm=ρδV
If we wish to calculate the total mass,M, of the object we must sum all such contributionsfromtheentirevolume.Thisisfoundbyintegratingthroughout
thevolume.
We write thisas ∫
total mass,M = ρdV
V
This is an example of a volume integral, so called because the integration is performed
throughout the volume. It will usually take the form of a triple integral such
as those discussed in Section 27.6.2. Technically, there are three integral signs, but
forbrevitythesehavebeenreplacedbythesingle ∫ V whereitistobeunderstoodthat
the integral is to be performed over a volume. In any specific problem care must be
takentoensurethattheentirevolumeisincludedwhentheintegrationisperformed.
For example, consider the case of a solid cube with sides of length 1 unit. Let
one corner be positioned at the origin and let the edges coincide with the positive
x, y and z axes. Suppose the density of the cube varies from point to point, and is
∫givenby ρ(x,y,z) =x+y+z.Thentheintegralwhichgivesthemassofthecubeis
(x + y + z)dV, where the volume V is the region occupied by the cube. This
V
integral has been evaluated in Example 27.16 and found to be 3 , representing the
2
mass of the cube.
volume V
volume dV
densityr(x, y,z)
P (x, y,z)
Figure27.17
Thetotal massofabodyis foundby
summing,orintegrating,throughout
the wholevolume.

890 Chapter 27 Line integrals and multiple integrals
Engineeringapplication27.4
Theelectricchargeenclosedinaregion
Suppose electric charge is distributed throughout a volume,V, and at any point has
charge density ρ(x,y,z). Charge density isascalar field.
Supposeweselectaverysmallportionhavingvolume δV andlocatedatP(x,y,z)
as shown inFigure 27.18. The charge inthisportion, δQ, isgiven by
δQ=ρδV
Ifwewishtocalculatethetotalchargeenclosedwithinthevolume,wemustsumall
such contributions from the entire volume. This is found by integrating throughout
the volume. We write thisas
∫
total charge,Q = ρdV
This isanother example of a volume integral.
volume V
V
volume dV
charge density
r(x, y,z)
P (x, y,z)
Figure27.18
The total charge enclosed is foundby
summing, orintegrating, throughout
the whole volume.
Engineeringapplication27.5
Fluidflowacrossasurface
Figure27.19(a)representsthemotionofabodyoffluidthroughoutaregion.Atany
pointP(x,y,z) fluid will be moving with a certain speed in a certain direction. So,
each small fluid element has a particular velocityv, which varies with position, that
isv = v(x,y,z). This isan example of a vector field, as discussedinSection 7.4.
surface S
v = v(x, y,z)
v = v(x, y,z)
P(x, y,z)
dS
small portion of surface dS
(a)
(b)
Figure27.19
Thevector fieldvrepresentsfluid velocity andSisan imaginary surface through whichthe
fluidflows.

27.7 Some simple volume and surface integrals 891
In Figure 27.19(b) we have placed an imaginary surface,S, within the flow and
it is reasonable to ask ‘what is the volume of fluid which crosses this surface in
any given time?’ Suppose we select a very small portion of the surface having area
δS. Note that δS is a scalar. We can also define an associated vector δS, which has
magnitude δS and whose direction is normal to this portion of the surface. The
component of fluid velocity in the direction of δS is given by the scalar product
v· ̂δS where ̂δS is a unit vector in the direction of δS. The volume of fluid crossing
this portion each second is v·δS. This is also known as the flux of v. If we
wishtocalculatethetotalfluxthroughthesurfaceSwemustsumallsuchcontributions
over the entire surface. This is found by integrating over the surface. We write
this as
∫
volume flowper second =total flux = v·dS
This is an example of a surface integral, so called because we must integrate over
a surface. It will usually take the form of a double integral such as those discussed
inSection27.6.1.Technically,therearetwointegralsigns,butforbrevitythesehave
been replaced by the single ∫ where it is to be understood that the integral is to be
S
performed over the surfaceS. In any specific problem care must be taken to ensure
that the entire surface is included when the integration is performed. The double
integralsevaluatedinSection27.6arespecialcasesofsurfaceintegrals,inwhichthe
surface isaplane region (thex--y plane).
Forexample,supposethevelocityfieldisgivenbyv = (x+2y)k.Thisrepresents
a flow in the z direction whose magnitude varies with x and y. Suppose S is the
plane surface defined by 1 x 4, 0 y 2,z = 0. This surface is shown in
Figure 27.20. Note that the normal to this surface is parallel to thezaxis and so we
can writedS = dxdyk. Then
v·dS=(x+2y)k·dxdyk
=(x+2y)dxdy
S
y
S
y= 2
dS = dxdyk
v= (x+2y)k
x
y= 0
0
x = 1
z
x = 4
Figure27.20
The vectorfieldvrepresents
fluid flowing acrossthe
surfaceS.Theintegral over
the surface ofv ·dSgives the
volume flow persecond
acrossS.
➔

892 Chapter 27 Line integrals and multiple integrals
The surfaceintegral which gives the flux ofvthroughSis
∫ y=2 ∫ x=4
y=0
x=1
(x +2y)dxdy
Note how the limits of integration are chosen so that contributions from the whole
surfaceareincluded.ThisintegralhasalreadybeenevaluatedinExample27.10and
found tobe 27. This represents the volume flow per second throughS.
Engineeringapplication27.6
Electriccurrentdensity
Electric current is the flow of electric charges. Figure 27.21 represents an electric
current flowing across a surfaceS.
If ρ(x,y,z)isthechargedensityatapoint,thatisthechargeperunitvolume,and
v(x,y,z) isthe velocity of the charges, thenthe quantityJdefined as
J=ρv
is called the current density and has units of amperes per square metre. Suppose
weselectaverysmallportionofthesurfacehavingarea δS.Notethat δS isascalar.
Supposewedefineavector, δS,whichhasmagnitude δSandwhichisnormaltothis
portion of surface. The component of current density in the direction of δS is given
by the scalar productJ· ̂δS. The current crossing this portion isJ·δS. If we wish to
calculate the total current,I, through the surfaceSwe must sum all such contributions
over the entire surface. This is found by integrating over the surface. We write
thisas
∫
current,I = J·dS
S
This isan example of asurface integral.
surface S
J = J(x, y,z)
dS
small portion of surface dS
Figure27.21
Theelectric current through asurface is
obtained usingasurface integral.
Note that this surface integral involves integrating a current density with units of
amperes per square metre over an area with units of square metres. Therefore the
result is a quantity with units of amperes and so the units balance on both sides of
the equation.

27.7 Some simple volume and surface integrals 893
Engineeringapplication27.7
ElectricfluxandGauss’slaw
Electric charges produce an electric field, E, which can be visualized by drawing
lines of force. Suppose we surround a region containing charges with a surface S.
ThefluxofEthroughSisameasureofthenumberoflinesofforcepassingthrough
it. The flux isgiven by
∫
flux = E·dS
S
Gauss’slawstatesthatthetotalfluxoutofanyclosedsurfaceisproportionaltothe
totalchargeenclosed.Soconsideraclosedsurfaceinfreespace,enclosingavolume
V. The total charge enclosed
∫
inV isgiven by the volume integral
total charge = ρdV
Gauss’s lawthen states:
∮
E·dS= 1 ∫
ρdV
ε 0
S
V
V
where ε 0
isthepermittivityoffreespace.Thenotation ∮ indicatesthatthesurfaceis
a closed surface. Note how a surface integral can be expressed as a volume integral.
This relationship isgeneralized inthe divergence theorem inthe following section.
Example27.16 Avectorfield isgiven by v =x 2 i + 1 2 y2 j+ 1 2 z2 k.
(a) Find divv.
(b) Evaluate ∫ V divvdVwhereVistheunitcube0x1,0y1,0z1.
Solution (a) divv = 2x +y+z.
(b) We seek ∫ (2x +y+z)dV whereV is the given unit cube. The small element of
V
volume dV is equal to dzdydx. With appropriate limits of integration the integral
becomes
∫ x=1 ∫ y=1 ∫ z=1
x=0
y=0
z=0
(2x+y+z)dzdydx
This isatripleintegral of the kind evaluated inSection 27.6.2:
∫ x=1 ∫ y=1 ∫ z=1
∫ x=1 ∫ y=1
] 1
(2x+y+z)dzdydx =
[2xz+yz+ z2
dydx
x=0 y=0 z=0
x=0 y=0 2
0
∫ x=1 ∫ y=1
(
= 2x+y+ 1 )
dydx
2
=
x=0
∫ x=1
x=0
y=0
[2xy + y2
2 2] + y 1
dx
0

894 Chapter 27 Line integrals and multiple integrals
=
∫ x=1
x=0
= [x 2 +x] 1 0
= 2
2x+1dx
Example27.17 Evaluate ∮ S v·dSwherev =x2 i+ 1 2 y2 j + 1 2 z2 k and whereSis the surface of the unit
cube 0 x 1, 0 y 1, 0 z 1. The vector dS should be drawn on each of the
sixfaces inan outward sense.
Solution ThecubeisshowninFigure27.22.Weevaluatethesurfaceintegralovereachofthesix
faces separately and then add the results.
OnsurfaceA,x = 1and0 y1,0 z1.dSisavectornormaltothissurface,
drawn inanoutward sense,and so wecan write itas dydzi. Then
v·dS= ( x 2 i + 1 2 y2 j+ 1 2 z2 k ) ·dydzi
=x 2 dydz
=dydz
sinceonA,x=1
The required surfaceintegral overAisthen
∫ z=1 ∫ y=1
z=0
y=0
1dydz=
= 1
∫ z=1
z=0
[y] 1 0 dz
OnsurfaceB,x = 0and0 y1,0 z1.dSisavectornormaltothissurfaceand
sowecanwriteitas −dydzi.Thenv·dSbecomes −x 2 dydz = 0sincex = 0.Overthis
surface, the integral iszero.
You should verify in a similar manner that over each ofC and E the integral is 1 2 ,
whilst integrals overDandF arezero. The total surfaceintegral isthen 2.
on E, dS = dxdy k
z
z = 1
E
dx
dy
D
B
dz
dx
on C, dS = dxdz j
x
x = 1
A
y= 1
C
dy
F
dz
on A, dS = dydz i
y
Figure27.22
Theintegral isevaluated over
allsixsurfaces.

27.8 The divergence theorem and Stokes’ theorem 895
EXERCISES27.7
1 Given F = −yi −xk evaluate ∮ SF·dSwhereSis the
surface ofaunitcube 0 x1, 0 y1,
0z1.
2 Given F = 6xy 2 z 2 k evaluate ∫ SF ·dS whereSisthe
plane surfacez = 2, 0 x2, 0 y2. Take the
direction ofthe vector element ofarea to bek.
3 Evaluate the volume integral ∫ V 6xdV whereV isthe
parallelepiped 0 z2,0 x1, 0 y3.
4 Evaluate ∫ V 1 +zdV whereV isaunitcube
0x1,0y1,0z1.
Solutions
1 0
3 18
2 128
4
3
2
27.8 THEDIVERGENCETHEOREMANDSTOKES’THEOREM
Thereareanumberoftheoremsinvectorcalculuswhichallowline,surfaceandvolume
integrals to be expressed in alternative forms. One of these, Green’s theorem, has been
describedinSection27.6.3.Thisallowsalineintegraltobewrittenintermsofadouble
integral.Now wegive details ofthe divergence theoremand Stokes’ theorem.
27.8.1 Thedivergencetheorem
The divergence theorem relates the integral over a volume,V, to an integral over the
closedsurface,S, whichsurroundsthatvolume, asillustratedinFigure 27.23.
When calculating such surface integrals vectors drawn normal to the surface should
always be drawn in an outward sense, that is away from the enclosed volume. Recall
thatwhen a surfaceisclosedthe symbolforasurfaceintegralis ∮ S .
Thedivergence theorem:
∮ ∫
v·dS= divvdV
S V
Example27.18 Verify the divergence theorem for the vector field v = x 2 i + 1 2 y2 j + 1 2 z2 k over the unit
cube0x1,0y1,0z1.
Solution Firstlyweneedtoevaluate ∮ S v·dSwhereSisthesurfaceofthecube.Thisintegralhas
been calculated inExample 27.17 and shown tobe2.
Secondlyweneedtocalculate ∫ divvdV overthevolumeofthecube.Thishasbeen
V
done inExample 27.16 and again the resultis2.
We have verified the divergence theorem that ∮ S v·dS=∫ V divvdV.

896 Chapter 27 Line integrals and multiple integrals
surface S
open surface S
dS
volume V
outwardly drawn normal
dS
Figure27.23
The divergence theoremrelates avolume integral
to asurface integral.
bounding curve C
Figure27.24
An open surfaceSwith bounding
curveC.
27.8.2 Stokes’theorem
Stokes’ theorem relates the integral over an open surface,S, to a line integral around a
closed curve,C, which bounds thatsurfacesuch as thatshown inFigure 27.24.
When a surface is open we adopt the following convention when drawing vectors
normal to the surface. When the direction of the normal vector has been specified, use
theright-handscrewruletoobtainasenseofturningaroundthenormalvector,asshown
inFigure27.24.Imaginenowmovingthecirclewhichsurroundsthenormalvectoralong
the surface until it just meets the curveC. Transfer its sense of turning to the curveC.
WhencalculatingthelineintegralaroundthecurveC,itshouldbetraversedinthesame
sense.
SpecificallyconsiderthetwoopensurfacesshowninFigure27.25.Inbothcaseswe
are considering cubes. In the first case the cube has no top face. In the second case the
cubehasnobottomface.Drawingvectordxdzjnotethatthesenseofturningrequiredby
theright-handscrewruleisthatshown.ThecurveC mustbetraversedinthedirections
shown.
Recall thatwhen a curve isclosed the symbol foralineintegral is ∮ C .
z
z
dS = dx dy k
C
dS=–dxdzj
dS=dxdzj
dS=–dxdzj
dS=dxdzj
y
y
C
x
x
dS= –dxdyk
Figure27.25
The right-handscrew rulegivesthe directionin whichC must be traversed.

27.8 The divergence theorem and Stokes’ theorem 897
Stokes’ theorem:
∮ ∫
v·dr= curlv·dS
C S
HeredrisanelementoflengthalongthecurveC.Recallthatdr = dxi +dyj +dzk.
WehaveusedthesymboldrratherthandsaswedidinSection27.2toavoidconfusion
with the element ofarea dS.
Example27.19 Acubeofside2unitsisconstructedwithfivesolidfacesandoneopenface.Itislocated
in the region defined by 0 x 2, 0 y 2, 0 z 2 and its open face is its top
face, bounded by the curveC, lying inthe planez = 2,asshown inFigure 27.26.
Throughout thisregion a vector field isgiven by
v=(x+y)i+(y+z)j+(x+z)k
(a) Evaluate ∮ C v·dr.
(b) Evaluate curlv.
(c) Evaluate ∫ curlvdS, and verifyStokes’ theorem.
S
Solution (a) The open face is highlighted in Figure 27.26. It is bounded by the curveC around
whichthelineintegral ∮ C v·drmustbeperformedinthesenseshown.Inthisplane
z = 2 and dz = 0,and hence
v·dr=(x+y)dx+(y+2)dy
We perform the lineintegral aroundC infour stages.
OnI,x = 0,dx = 0andhencev·dr = (y +2)dy.Notingthatyvariesfrom0to
2,the contribution tothe lineintegral is
∫ 2
[ ] y
2 2
y+2dy=
0 2 +2y = 6
0
z
E
z = 2 I bounding curve C
dS =
dxdzj
D
IV
III
II
B
dS= dxdzj
x
x = 2
A
F
y= 2
C
y
dS =
dxdyk
Figure27.26
Acubicalbox with open topbounded bycurveC.

898 Chapter 27 Line integrals and multiple integrals
On II,y = 2, dy = 0 and hence v·dr = (x +2)dx. Noting thatxvaries from 0
to2,the contribution tothe lineintegral is
∫ 2
[ ] x
2 2
x+2dx=
0 2 +2x = 6
0
On III,x = 2, dx = 0 and hence v·dr = (y +2)dy. Hereyvaries from 2 to 0.
This contribution tothe lineintegral is
∫ 0
[ ] y
2 0
y+2dy=
2 2 +2y = −6
2
On IV,y = 0, dy = 0 and hence v·dr = xdx. Herexvaries from 2 to 0. The
contribution tothe lineintegral is
∫ 0
[ x
2
xdx= = −2
2
2
] 0
2
Putting all these results together we find
∮
v·dr=6+6−6−2=4
C
i j k
(b) curlv =
∂ ∂ ∂
∂x ∂y ∂z
=−i−j−k
∣
x+yy+zx+z
∣
(c) Now we calculate the surface integral which must be performed over the five solid
surfaces separately. Refer to Figure 27.26. On surfaceA, the front face lying in the
planex = 2,dS = dydzi. Hence curlv·dS = −dydz. Then
∫ ∫ z=2 ∫ y=2
curlv·dS= −dydz
A
=
=
z=0
∫ z=2
z=0
∫ z=2
z=0
= [−2z] 2 0
= −4
y=0
[−y] 2 0 dz
−2dz
On B, the back face lying in the plane x = 0, dS = −dydzi. It follows that
curlv·dS = dydz. The required integral overBis
∫ z=2 ∫ y=2
z=0
y=0
dydz=4
OnC, the right-hand face, dS = dxdzj. Hence curlv·dS = −dxdz. The required
integral overC is
∫ z=2 ∫ x=2
z=0
x=0
−dxdz = −4

27.9 Maxwell’s equations in integral form 899
Similarly, onD, the left-hand face, dS = −dxdzj. Hence curlv·dS = dxdz. The
required integral overDis
∫ z=2 ∫ x=2
z=0
x=0
dxdz=4
On surfaceF, the base,z = 0 and dS = −dxdyk. Hence curlv·dS = dxdy. The
required integral overF is
∫ y=2 ∫ x=2
y=0
x=0
dxdy=4
RecallthatthetopsurfaceE isopen,andsowehavecompletedthesurfaceintegrals.
Finally, putting these results together,
∫
curlv·dS=−4+4−4+4+4=4
S
Notefrompart(a)thatthisequals ∮ v·drandsowehaveverifiedStokes’theorem.
EXERCISES27.8
1 VerifyStokes’theorem forthe field v =xyi +yzj
whereSisthe surface ofthe cube 0 x1,
0y1,0z1andthefacez=0isopen.
2 Ifv = 3x 2 i +xyzj −5zkverify Stokes’theorem
whereSisthe surface ofthe cube 0 x1,
0y1,0z1andthefacez=0isopen.
3 Supposev =x 3 i −3x 2 y 2 z 2 j + (7x +z)kandSisthe
surface ofacube ofside2units lying in the region
0x2,0y2,0z2withanopentopin
the planez = 2. Verify Stokes’theorem forthisfield.
4 Consideracube given by 0 x1, 0 y1,
1 z2,above the surfacez = 1. Suppose the
surfacez = 1 is the only open face. LetSbe the
surface ofthiscube. Verify Stokes’theorem forthe
field v =yi + (x −2xz)j −xyk. TakeC asthe square
withcorners (0,0,1), (1,0,1), (1,1,1), (0,1,1)in
the planez = 1.
5 Considerthat part ofthe positiveoctant, that iswhere
x,yandzare allpositive, bounded bythe planes
x=0,y=0,z=0andx+y+2z=2.Assumethat
the tetrahedron soformedhas three solidfaces, and
one open face onthe planex+y+2z = 2.Verify
Stokes’theorem forthe field v =x 2 i −2xzk.TakeC
asthe triangle (2,0,0), (0,2,0)and (0,0,1).
Solutions
When calculating the relevant line and surface integrals
the signofthe result dependsupon the orientation of
the curveC.
1 ∮ v·dr=± 1 2
2 ∮ v·dr=0
3 ∮ v·dr=±128
4 ±2
5 ± 2 3
27.9 MAXWELL’SEQUATIONSININTEGRALFORM
InSection26.7Maxwell’sequationswerestatedasexamplesoftheapplicationofvector
calculus.Infact,alternativeintegralformsoftheseequationsareoftenmoreusefuland
theseare given here forcompleteness.

900 Chapter 27 Line integrals and multiple integrals
Equation1
∮ ∫
D·dS=
S
V
ρdV
whereD =electricfluxdensity,and ρ iselectricchargedensity.Notethatther.h.s.isthe
totalchargeenclosedbythevolumeV.Thisequationstatesthatthetotalfluxcrossinga
closedsurfaceSwhichenclosesavolumeV isequaltothetotalchargeenclosedbythe
surface.ThisisalsotheintegralformofGauss’slaw.(SeeEngineeringapplication27.7
which isobtained by lettingD = ε 0
E.)
Equation2
∮
B·dS=0
S
where B = magnetic flux density. This law states that the net magnetic flux crossing
any closed surface is zero. So whilst charges can be thought of as sources or sinks of
electric flux,there areno equivalent sources or sinks ofmagnetic flux.
Equation3
∮
CE·dr=− ∂ ∂t
∫
S
B·dS
Note that in the theory of electrostatics, differentiating
∮
partially with respect to time
always yields zero, and so this equation reduces to E·dr = 0. This is a condition
discussed inSection 27.5 and confirms thatan electrostatic field isaconservative field.
Equation4
∮
CH·dr= ∂ ∂t
∫
S
∫
D·dS+ J·dS
S
This isthe integral formof Ampère’s circuitallaw. The closed curveC bounds an open
surfaceS. Acurrentwith density ∂D +Jflows through the surfaceS.
∂t
C
EXERCISES27.9
1 Startingwith Maxwell’sequation ∇ ·D = ρ,by
integrating both sides over anarbitraryvolumeV and
usingthe divergence theoremobtain Equation1
above.
2 Startingwith Maxwell’sequation ∇ ·B = 0, by
integrating both sides over anarbitraryvolumeV and
usingthe divergence theoremobtain Equation2
above.
3 Startingwith the equation ∇ ×E = 0forstatic
electric fields, use Stokes’theoremto showthat
∮
E·dr=0.
4 Startingfrom ∇ ×H = ∂D +J andusingStokes’
∂t
theorem obtain Equation 4.
( )
∂
5 Inmagnetostatics
∂t = 0 ,Ampère’slaw (orthe
magnetic circuit law) states:
∮
H·dr=I
C
whereI is the currentenclosed bythe closedpathC.
Obtain
∫
thislaw from Equation 4 andusing
S J·dS=I.

Review exercises 27 901
REVIEWEXERCISES27
1 Find ∫ C (2x −y)dx +xydy along the straightline
joining (−1, −1)and (1,1).
2 Find ∫ C (2x+y)dx+y3 dy
(a)along the curvey =x 3 between (1,1)and (2,8),
and
(b)along the straightline joining these points.
3 IfF = 3xyi +2e x yjfind ∫ CF ·ds whereC is the
straightliney = 2x +1 between (1,3)and (4,9).
4 ShowthatthefieldF = (2x +1)yi + (x 2 +x+1)jis
conservative and findasuitable potential function φ.
5 Find ∮ C (x2 i +4xyj) ·ds whereC is aclosed pathin
the form ofatriangle with vertices at
(0,0),(3,0)and(3,5).
6 Find ∫ y=3
y=0
7 Find ∫ y=2
y=0
∫ x=2
x=1 (3x −5y)dxdy.
∫ x=1+y
x=0 (2y +5x)dxdy.
8 Find
∫ x=2 ∫ y=1 ∫ z=3
x=0 y=0 z=0 (x2 +y 2 +z 2 )dzdydx.
9 Evaluate ∫∫ R (4y +x2 )dxdy whereRisthe interior of
the squarewith verticesat (0,0),
(4,4),(0,4)and(4,0).
10 Evaluate ∮ C ey dx +e x dywhereC isthe boundary
ofthe triangle formedbythe linesy =x,y = 5
andx = 0. Byconverting thisline integral intoa
doubleintegral verifyGreen’s theoremin the
plane.
11 TheregionRis bounded bytheyaxisand the lines
y=xandy=3−2x.
(a) Sketch the regionR.
(b) Find the volume between the surfacez =xy +1
and the regionR.
(a) Evaluate ∮ CF ·dswhereC is the curve enclosing
the regionR.
(b) Verify Green’stheorem in the plane.
(c) Thesurface,z(x,y),isgivenbyz = 1 +x+xy.
Calculatethevolumeunderthesurfaceandabove
the regionR.
13 The regionRisbounded bythexaxisand the curve
y=8+2x−x 2 .
(a) SketchR.
(b) Evaluate ∫∫ R 3x+2ydxdy
14 Evaluate
∫ 0 ∫ 4
(a) −1 3
∫ 3xydxdy
3 ∫ 2
(b) 2 0
∫ 4+x−ydydx
1 ∫ 4
(c) 0 2y x2 +y 2 dxdy
∫ −1 ∫ x 3
(d) −2 0 1dydx
15 Sketch the regionsofintegration ofthe double
integrals in Question14.
16 Ifa,b,candd are constantsshow that
∫ d ∫ b
f(x)g(y)dxdy
c a
is identical to
[∫ b
f(x)dx
] [∫ d ]
g(y)dy
a c
y
9
C
R
y = 9 – x 2
12 TheregionRis shown in Figure 27.27.
Thevector fieldFisgiven by
Solutions
F =y 2 i+3xyj
0
3
Figure27.27
The regionRforQuestion12.
x
1
2
3
2 (a) 1030.5 (b) 1031.25
3 1666.37
4 φ=x 2 y+yx+y+c

902 Chapter 27 Line integrals and multiple integrals
10 −452.239
5 50
9
640
3 14 (a) − 21 4 (b) 11 (c) 21.5
6 −9
7 31
17
11 (b) 8
12 (a) 64.8 (c) 99
8 28
13 (b) 367.2
(d) − 15 4

28 Probability
Contents 28.1 Introduction 903
28.2 Introducingprobability 904
28.3 Mutuallyexclusiveevents:theadditionlawofprobability 909
28.4 Complementaryevents 913
28.5 Conceptsfromcommunicationtheory 915
28.6 Conditionalprobability:themultiplicationlaw 919
28.7 Independentevents 925
Reviewexercises28 930
28.1 INTRODUCTION
Probability theory is applicable to several areas of engineering. One example is reliabilityengineeringwhichisconcernedwithanalysingthelikelihoodthatanengineering
system will fail. For most systems calculating the exact time of failure is not feasible
but it is often possible to obtain a good estimate of whether or not a system will fail in
a certain time interval. This is useful information to have for any engineering system,
butitisvitalifthefailureofthesystemresultsinthepossibilityofinjuryorlossoflife.
Examplesincludethefailureofhigh-voltageswitchgearsothatthecasingbecomeslive,
or electricalequipment producing a spark whilebeing usedunderground inamine.
Probability theory is also used extensively in production engineering, particularly in
the field of quality control. No manufacturing process produces components of exactly
the same quality each time. There is always some variation in quality and probability
theory allows this variation to be quantified. This enables some predictability to be introduced
into the activity of manufacturing and gives an engineer the confidence to say
components of a certainquality can be supplied toacustomer.
Afinalexampleisinthefieldofcommunicationengineering.Communicationchannelsaresubjecttonoisewhichisrandominnatureandsoismostsuccessfullymodelled
usingprobabilitytheory. We will examine some ofthese concepts inmore detailinthis
chapter.

904 Chapter 28 Probability
E
A
28.2 INTRODUCINGPROBABILITY
B = A
Figure28.1
A:acomponent meets
the specification.B =A:
acomponent fails to
meet the specification.
Consideramachinewhichmanufactureselectroniccomponents.Thesemustmeetacertain
specification. The quality control department regularly samples the components.
Suppose, on average, 92 out of 100 components meet the specification. Imagine that a
component is selected at random and letAbe the outcome that a component meets the
specification; let B be the outcome that a component does not meet the specification.
Then we say the probability of A occurring is 92 = 0.92 and the probability of B
100
8
occurring is = 0.08. The probability is thus a measure of the likelihood of the
100
occurrence of a particular outcome. We write
P(A) = probability ofAoccurring = 0.92
P(B) = probability ofBoccurring = 0.08
We note that the sum of the probabilities of all possible outcomes is 1. The process of
selecting a component is called a trial. The possible outcomes are also called events.
In this example there are only two possible events,AandB. We can depict this situation
using a Venn diagram as shown in Figure 28.1. Recall from Section 5.2 that Venn
diagrams areusedtodepictsets.InthisdiagramwearedepictingtheeventsAandBas
sets. The set of all possible outcomes is called the sample space and is represented by
the universal set E. The setArepresents the event that a component meets the specification.
The setBrepresents the event that a component fails to meet the specification.
In this instance, when a trial takes place there are only two possible outcomes, either a
component meets the specification or it does not; that is, either eventAoccurs or event
Boccurs. An alternative notation istowrite
B =A
whereAissaidtobethecomplementofA. We could alsowrite
A =B
When a bar appears over a set then wesay, for example, ‘notA’, or‘notB’.
Wenowseektodefineprobabilityinamoreformalway.LetE beanevent.Wewish
toobtaintheprobabilitythatE willoccur,thatisP(E),whenatrialtakesplace.Inorder
to do so we repeat the trial a large number of times,n. We count the number of times
that eventE occurs,denoted bym. We then conclude that
P(E)= m n
(28.1)
Thelargerthenumberoftrialsthattakeplace,themoreconfidentweareofourestimate
of the probability of E occurring. For example, consider the trial of tossing a coin. If
we wish to calculate the probability of a head occurring then measuring the results of
1000 tosses of the coin is likely to yield a more accurate estimate than measuring the
results of 10 tosses of the coin. Various consequences flow from Equation (28.1). The
numberoftimesE occursmustbenon-negativeandlessthanorequaltothenumberof
trials,thatis0 mn. So,
0P(E)1

28.2 Introducing probability 905
Ifm = 0,corresponding toeventE never occurring inntrials,then
P(E)= 0 n = 0
We conclude thatE isan impossible event.
Ifm =n, corresponding toeventE always occuring inntrials,then
P(E)= n n = 1
We deduce thatE isacertainevent.
IfP(E) > 0.5 then weconclude thatE ismore likely tooccur thannot.
Theapproachtodefiningprobabilitythatwehaveadoptedsofarisessentiallyexperimental.Wecarryoutaseriesoftrialsandmeasuretheprobabilityofaneventoccurring.
Sometimes it is possible to deduce the probability of an event purely from theoretical
considerations.Consideragainthetrialoftossingacoin.Ifthecoinisfairthendefining
H tobe the event thatthe coin lands with the head facing up we can easilydeduce that
P(H) = 0.5
because we intuitively recognize that a head has the same likelihood of occurring as a
tail. Similarly, if we roll a fair die -- die is the singular of dice -- andE is the event that
a 6 isobtained, thenP(E) = 1 6 .
Suppose there aretwo possible outcomes,E 1
andE 2
, of a trial. Then
P(E 1
)+P(E 2
)=1
If a trialhasnpossible outcomes,E 1
,E 2
,...,E n
, then
P(E 1
)+P(E 2
)+···+P(E n
)=1
The sum of the probabilities of all possible outcomes is 1, representing the total
probability.
28.2.1 Compoundevent
Consider the situation of rolling a fair die. It is possible to define a variety of events
or outcomes associated with this trial. We choose to define two events, E 1
and E 2
, as
follows:
So,
E 1
: a 1,2,3 or 4 isobtained
E 2
: an even number isobtained
E 1
= {1,2,3,4}
E 2
= {2,4,6}
Now the universal set is
E={1,2,3,4,5,6}
which embraces all possible outcomes.
Suppose we now define the eventE 3
as
E 3
:E 1
occurs andE 2
occurs

906 Chapter 28 Probability
TheeventE 3
isknownasacompoundeventandonlyoccurswhenbothE 1
andE 2
occur
atthe same time.Usingset notation we can write
E 3
=E 1
∩E 2
= {1,2,3,4} ∩ {2,4,6}
= {2,4}
E
SoE 3
occurswhena2ora4isrolled.HenceE 1
∩E 2
canoccurintwooutofsixequally
likely ways.Therefore,
E 1
E 2
1
3
2
4
5
6
P(E 3
)=P(E 1
∩E 2
)= 2 6 = 1 3
Figure 28.2 shows a Venn diagram forthis example.
We could alsodefine the compound eventE 4
as
Figure28.2
E 1 ∩E 2 corresponds to
the compound eventE 1
occursandE 2 occurs.
E 4
:E 1
occurs orE 2
occurs
The eventE 4
occurs when eitherE 1
occurs orE 2
occurs,or both. We can write
So,
E 4
=E 1
∪E 2
= {1,2,3,4} ∪ {2,4,6}
= {1,2,3,4,6}
P(E 4
)=P(E 1
∪E 2
)= 5 6
If eventsAandBboth occur thenthiscompound event isdenotedA ∩B.
If either eventAoreventBoccursthenthiscompound event isdenotedA ∪B.
Engineeringapplication28.1
Powersupplytoacomputer
Computersusedinthecontroloflife-criticalsystemsoftenhavetwoseparatepower
supplies.Ifonepowersupplyfailsthentheothertakesover.LetE 1
betheeventthat
powersupply1fails,andletE 2
betheeventthatpowersupply2fails.Forthepower
supplytothecomputertofailcompletelythecompoundeventE 1
andE 2
mustoccur,
thatisE 1
∩E 2
occurs.
A system such as this is said to have dual modular redundancy. One of the
difficultieswithsuchasystemisindeterminingwhichsystemisoperatingcorrectly
andwhichunitisswitchedtobeingtheactiveone.Insomesituationsitisnecessary
to have three systems operating in parallel and then using a voting system to decide
on the correct output. This way the decision is taken based on the majority of the
threesystems, thatis, when atleasttwo outofthe threeagree.

28.2 Introducing probability 907
Engineeringapplication28.2
Qualitycontrolonafactoryproductionline
Quality control is an important aspect of factory production. So much so that some
engineersspendtheirwholecareersinthisfieldandareknownasqualitycontrolengineers.Thefundamentalgoalistoensurethatmanufacturedgoodsareproducedto
aspecifiedstandardinspiteofvariationsinthefinishedproductthatareaninevitable
partof the manufacturing process.Consider the following problem.
Machines A and B make components, which are then placed on a conveyor belt.
OfthosemadebymachineA,93%areacceptable.OfthosemadebymachineB,95%
areacceptable.MachineAmakes60%ofthecomponentsandmachineBmakesthe
rest. Find the probability that a component selected at random from the conveyor
belt is
(a) made by machine A
(b) made by machine Aand acceptable
(c) made by machine Band acceptable
(d) made by machine Band unacceptable
Solution
(a) We are given that 60% of the components are made by machine A. Converting
thispercentage toadecimal number gives
P(component ismade by machine A) = 0.6
(b) Weknowthat60%ofthecomponentsaremadebymachineAand93%ofthese
areacceptable. Converting these percentages todecimal numbers wehave
P(component ismade by machine Aand is acceptable) = 60
100 × 93
100
= 0.60 ×0.93 = 0.558
(c) Weknowthat40%ofthecomponentsaremadebymachineBand95%ofthese
areacceptable. So,
P(component ismade by machine Band isacceptable) = 40
100 × 95
100
= 0.40 ×0.95 = 0.38
(d) We know that 40% of the components are made by machine B and 5% of these
areunacceptable. So,
P(component ismade by machine Band isunacceptable) = 40
100 × 5
100
= 0.40 ×0.05 = 0.02
An alternative way to solve this problem is by constructing a tree diagram. To
do so, consider the case when 1000 components are picked from the conveyor
belt. We know that machine A makes 60% of the components and so 600 of these
components, on average, will be made by machine A and the other 400 must
therefore be made by machine B. This isshown inFigure 28.3.
➔

908 Chapter 28 Probability
acceptable
558
1000
machine A
machine B
600
1000
machine A
machine B
600
400
not acceptable
acceptable
not acceptable
42
380
400
Figure28.3
Partial tree diagram forEngineering
application 28.2.
Figure28.4
Full tree diagram forEngineering
application 28.2.
Of the 600 components made by machine A, we know that 93% of these are acceptable,
thatis93% of 600 = 558.So 600 −558 = 42 are unacceptable.
Of the 400 components made by machine B, we know that 95% of these are acceptable,
thatis95% of 400 = 380. So 400 −380 = 20 are unacceptable.
We can now complete a full tree diagram for this problem, which is shown in
Figure 28.4.
Usingthetreediagramitisstraightforwardtocalculatetherequiredprobabilities.
(a) Theprobabilityacomponent ismadebymachineAisfoundbynotingthat600
of the 1000 components aremade by machine A,that is
P(component is made by machine A) = 600
1000 = 0.6
(b) TheprobabilityacomponentismadebymachineAandisacceptableisfoundby
notingthat558ofthe1000componentsaremadebymachineAandacceptable,
that is
P(component is made by machine Aand acceptable) = 558
1000 = 0.558
(c) P(component is made by machine Band acceptable) = 380
1000 = 0.38
(d) P(component is made by machine Band unacceptable) = 20
1000 = 0.02
20
EXERCISES28.2
1 Afair die isrolled. TheeventsE 1 , . . . ,E 5 are defined
asfollows:
E 1 :aneven number is obtained
E 2 :anodd numberis obtained
E 3 :ascore ofless than2isobtained
E 4 :a3is obtained
E 5 :ascore ofmore than3is obtained
Find
(a)P(E 1 ),P(E 2 ),P(E 3 ),P(E 4 ),P(E 5 )
(b)P(E 1 ∩E 3 )
(c)P(E 2 ∩E 5 )
(d)P(E 2 ∩E 3 )
(e)P(E 3 ∩E 5 )
2 Atrialcan have threeoutcomes,E 1 ,E 2 andE 3 .E 1
andE 2 are equally likely to occur.E 3 is threetimes
more likely to occur thanE 1 .FindP(E 1 ),P(E 2 ) and
P(E 3 ).
3 Acomponent ismadeby machines AandB.Machine
Amakes 70%ofthe components andmachine B
makesthe rest. Forboth machines, the proportion of

28.3 Mutually exclusive events: the addition law of probability 909
acceptable components is90%. Find the probability
that acomponent selected atrandom is
(a) unacceptable
(b) acceptable and ismade bymachine A
(c) unacceptable and ismade by machine B
4 MachinesA, Band Cmake components. Machine A
makes20% ofthe components, machine B makes
30% ofthe components and machine C makesthe
rest.Theprobabilitythatacomponentisfaultyis0.07
when madeby machine A, 0.06 when made by
machine B and0.05 when made bymachine C.A
component is picked atrandom. Calculate the
probability that the component is
(a) made bymachine C
(b) made bymachine A andis faulty
(c) made bymachine B and isnotfaulty
(d) made bymachine C and isfaulty
(e) made bymachine A andis notfaulty
(f) faultyand isnotmade by machine B
5 Circuit boards are made bymachines A, B,CandD.
Machine A makes15%ofthe components, machine
B makes30%, machine C makes35%and machine D
makes the remainder. Theprobability that aboard is
acceptable is0.93 when made bymachine A, 0.96
when made by machine B,0.95 when made by
machine Cand 0.93 when made by machine D. A
board ispicked atrandom. Calculate the probability
that itis
(a) made bymachine D
(b) made bymachine A andis acceptable
(c) made bymachine B and isunacceptable
(d) made bymachine C and isacceptable
(e) made bymachine D andis unacceptable
(f) unacceptable and made bymachine C
(g) Inabatch of1000 boards,how many would be
expected to be acceptable andmade by
machine D?
Solutions
1
1 (a)
2 , 1 2 , 1 6 , 1 6 , 1 2
1
(d)
6
2 0.2, 0.2, 0.6
(b) 0
(e) 0
(c)
1
6
3 (a) 0.1 (b) 0.63 (c) 0.03
4 (a) 0.5 (b) 0.014 (c) 0.282
(d) 0.025 (e) 0.186 (f) 0.039
5 (a) 0.2 (b) 0.1395 (c) 0.012
(d) 0.3325 (e) 0.014 (f) 0.0175
(g) 186
Table28.1
Theprobability ofacar
componentfallinginto
oneoffourcategories.
Category
28.3 MUTUALLYEXCLUSIVEEVENTS:THEADDITION
LAWOFPROBABILITY
Probability
topquality 0.18
standard 0.65
substandard 0.12
reject 0.05
Consideramachinewhichmanufacturescarcomponents.Supposeeachcomponentfalls
into one offourcategories:
topquality
standard
substandard
reject
After many samples have been taken and tested, it is found that under certain specific
conditions the probability that a component falls into a category is as shown in Table
28.1. The four categories cover all possibilities and so the probabilities must sum
to 1. If 100 samples are taken, then on average 18 will be top quality, 65 of standard
quality, 12 substandardand 5 will berejected.

910 Chapter 28 Probability
Example28.1 Using the data in Table 28.1 calculate the probability that a component selected at random
iseither standard or topquality.
Solution On average 18 out of 100 components are top quality and 65 out of 100 are standard
quality.So83outof100areeithertopqualityorstandardquality.Hencetheprobability
that a component is either top quality or standard quality is 0.83. The solution may be
expressed more formally as follows. LetAbe the event that a component is top quality.
LetBbe the event thatacomponent isstandard quality.
Then,
P(A) = 0.18 P(B) = 0.65
P(A ∪B) = 0.18 +0.65 = 0.83
Note thatinthisexample
P(A ∪B) =P(A) +P(B)
In Example 28.1 the eventsAandBcould not possibly occur together. A component is
either top quality or standard quality but cannot be both. We sayAandBare mutually
exclusivebecausetheoccurrenceofoneexcludestheoccurrenceoftheother.Theresult
applies more generally.
E
E i
E j
If the occurrence of either of eventsE i
orE j
excludes the occurrence of the other,
thenE i
andE j
aresaidtobemutuallyexclusive events.
IfE i
andE j
aremutuallyexclusive wedenotethis by
Figure28.5
E i andE j are mutually
exclusive eventsand so
are depicted asdisjoint
sets.
E i
∩E j
= ◦/
Weuse◦/todenotetheemptyset,thatisasetwithnoelements.Ineffectwearestating
that the compound eventE i
∩E j
is an impossible event and so will never occur. On a
Venn diagramE i
andE j
are shown as disjoint sets (see Figure 28.5). Suppose thatE 1
,
E 2
,...,E n
arenevents and that in a single trial only one of these events can occur. The
occurrenceofanyevent,E i
,excludestheoccurrenceofallotherevents.Sucheventsare
mutuallyexclusive.
Formutuallyexclusive events the addition lawofprobabilityapplies:
P(E 1
orE 2
or ... orE n
)=P(E 1
∪E 2
∪···∪E n
)
=P(E 1
)+P(E 2
)+···+P(E n
)

28.3 Mutually exclusive events: the addition law of probability 911
Engineeringapplication28.3
Electricalcomponentreliability
Electrical components fail after a certain length of time in use. Their lifespan is not
fixed but subject to variation. It is important to be able to quantify the reliability of
these components. Consider the following problem.
Thelifespansof5000electricalcomponentsaremeasuredtoassesstheirreliability.
The lifespan (L) is recorded and the results are shown in Table 28.2. Find the
probability thatarandomly selected component will last
(a) more than 3 years
(b) between 3 and 5 years
(c) lessthan 4 years
Table28.2
Thelifespansof5000electricalcomponents.
Lifespanofcomponent(years)
Number
L>5 500
4<L5 2250
3<L4 1850
L3 400
Solution
We define eventsA,B,C andD:
A: the component lasts more than 5 years
B: the component lasts between 4 and 5 years
C: the component lasts between 3 and 4 years
D: the component lasts 3 years or less
P(A) = 500 2250 1850
= 0.1, P(B) = = 0.45, P(C) =
5000 5000 5000 = 0.37,
P(D) = 400
5000 = 0.08.
The eventsA,B,C andDare clearly mutually exclusive and so the addition law
may be applied.
(a) P(component lasts more than 3 years) =P(A ∪B∪C)
=P(A) +P(B) +P(C)
=0.1 +0.45 +0.37
=0.92
There isa92% chance a component will lastformore than 3 years.
(b) P(component lasts between 3 and 5 years) =P(B ∪C)
=P(B) +P(C)
=0.45 +0.37
=0.82 ➔

912 Chapter 28 Probability
(c) P(component lasts less than 4 years) =P(C ∪D)
=P(C) +P(D)
=0.37 +0.08
=0.45
Engineeringapplication28.4
Calculatingtheprobabilityacomponentisacceptableon
afactoryproductionlinewithseveralmachines
Sometimesafactoryproductionlinemaybefedbyseveralmachinesthatmakecomponents.
It is important to be able to calculate the overall quality of the product that
emerges given knowledge of the performance of the individual machines. Consider
the following problem.
MachinesAandBmakecomponents.MachineAmakes60%ofthecomponents.
The probability that a component is acceptable is 0.93 when made by machine A
and0.95whenmadebymachineB.Acomponentispickedatrandom.Calculatethe
probability thatitis
(a) made by machine Aand isacceptable
(b) made by machine Band isacceptable
(c) acceptable
Solution
WehavealreadylookedatthisprobleminEngineeringapplication28.2.Figure28.4
shows the tree diagram for the problem.
(a) P (component ismade by machine Aand isacceptable) = 558
1000 = 0.558.
(b) P (component ismade by machine Band isacceptable) = 380
1000 = 0.38.
(c) Note that the events described in (a) and (b) are mutually exclusive and so the
addition lawcan beapplied.
P(component is acceptable) =P(component ismade by machine Aand isacceptable)
+P(component ismade by
machine Band isacceptable)
= 0.558 +0.38
= 0.938
We can now obtain this probability directly from the tree diagram. We see that
558 +380 = 938 components areacceptable and so
P(component isacceptable) = 938
1000 = 0.938

28.4 Complementary events 913
EXERCISES28.3
1 Acomponent is classifiedasone oftopquality,
standard quality orsubstandard, with respective
probabilities of0.07,0.85 and 0.08.Findthe
probability that acomponent is
(a) either topquality orstandard quality
(b) not topquality
2 Thelifespan (L)ofeach of2000 valvesis measured
andgiven in Table 28.3.
Table28.3
Thelifespans of2000valves.
Lifespan(hours)
Number
L 1000 119
800 L<1000 520
600 L<800 931
400 L<600 230
L < 400 200
Calculate the probability that the lifespanofavalve is
(a) more than800 hours
(b) less than600 hours
(c) between 400 and 800 hours
3 Chipsare manufactured bymachines A andB.
Machine Amakes65% ofthe chips and machine B
makesthe remainder. Theprobability that a chipis
faulty is0.03 when made bymachine A and 0.05
when made by machine B.Achip isselected at
random. Calculate the probability that itis
(a) faulty andmade by machine A
(b) faulty ormade bymachine A
(c) faulty ormade bymachine B
(d) faulty andmade by machine B
(e) faulty
4 Componentsare made bymachines A, B,Cand D.
Machine A makes25%ofthe components, machine
B makes16% ofthe components, machine Cmakes
21% ofthe components and machine Dmakes the
remainder. The probability that acomponent is
acceptable is0.92 when made bymachine A, 0.90
when made by machine B,0.96 when made by
machine Cand 0.93 when made by machine D. A
component ispicked at random. Calculate the
probability that itis
(a) made by machine Aormachine C
(b) made by machines A, B orD
(c) acceptable andmade by machine B
(d) acceptable andmade by machine C
(e) acceptable andmade by machine B oracceptable
and made bymachine C
(f) acceptable ormade bymachine A
(g) acceptable
Solutions
1 (a) 0.92 (b) 0.93
2 (a) 0.3195 (b) 0.215 (c) 0.5805
3 (a) 0.0195 (b) 0.6675 (c) 0.3695
(d) 0.0175 (e) 0.037
4 (a) 0.46 (b) 0.79 (c) 0.144
(d) 0.2016 (e) 0.3456 (f) 0.949
(g) 0.929
28.4 COMPLEMENTARYEVENTS
Consider an electronic circuit. Clearly, either the circuit works correctly or it does not
workcorrectly.LetAbetheeventthatthecircuitworkscorrectly,andletBbetheevent
that the circuit does not work correctly. EitherAorBmust happen when the circuit is
tested, and one excludes the other. The eventsAandBare said to be complementary.

914 Chapter 28 Probability
TheVenndiagramcorrespondingtothissituationisidenticaltothatinFigure28.1where
the setAcorresponds tothe eventA, and the eventBisrepresented byA.
Two events, A and B, are complementary if they are mutually exclusive and in a
singletrialeitherAorBmusthappen.
Hence,ifAandBarecomplementary then
and
P(A) +P(B) =1
P(A∩B)=0
Itis usualtodenotecomplementary events asAandA. For example,
A : the component istop quality
A : the component isnot topquality
B:n>6
B:n6
C : the circuithas failed tomeet the specification
C : the circuithas met the specification
Recall from Engineering application 28.3 that D was the event the component lasts
3 years or less, and thishad probabilityP(D) = 0.08. We could say
D : the component lasts 3 years orless
D : the component lasts more than3years
DandDarecomplementary events and so
P(D) +P(D)=1
P(D)=1−P(D)=1−0.08=0.92
Engineeringapplication28.5
Testingmultipleelectroniccomponents
Three transistors are tested. The probability that none of them works is 0.03. What
isthe probability thatatleastone transistor works?
Solution
We definethe eventsAandA:
A : all threetransistors fail
A : not all three transistorsfail
AandAare complementary events and so
P(A) = 0.03 P(A) =1−P(A) =0.97

28.5 Concepts from communication theory 915
To say that not all three transistors fail is the same as saying that one or more transistorswork.
So we could say
A : atleastone transistor works
Hence the probability thatatleastone transistor works is0.97.
EXERCISES28.4
1 Which pairsofevents are complementary?
A:the componentis reliable
B:there is only onecomponent
C:there are less thantwo components
D:more thantwo components are
reliable
E:the componentis unreliable
F:thereismore thanonecomponent
G:most ofthe components are
unreliable
2 EventsA,B,C andDare defined.
A:the lifespanis90 daysorless
B:the machine is reliable
C:allcomponents have been tested
D: atleastthree components from
the sample are unreliable
State the eventsA,B,C andD.
Solutions
1 AandE;CandF
2 A:the lifespanis more than 90days
B:the machine is unreliable
C:some components have not been tested
D:two orfewer components are unreliable
28.5 CONCEPTSFROMCOMMUNICATIONTHEORY
Communicationengineersfinditusefultoquantifyinformationforthepurposesofanalysis.Inordertodosoaveryrestrictedviewofinformationisused.Informationisseenin
termsofknowledgeofaneventoccurring.Ahighlyimprobableeventoccurringconstitutes
more information than an almost certain event occurring. This correlates to some
extent with human experience as people tend to be much more interested in hearing
aboutunlikely events. The information,I, associatedwith anevent, isdefined by
I=−logp
0<p1
where p = probability of an event occurring. Notice that p = 0 is excluded from the
domain of the function as the logarithm is not defined at 0. In practice this is not a
problem because an event with zero probability never occurs. Often logarithms to the
base 2 are used when calculating information as in many cases information arrives in
the form of a ‘bit stream’ consisting of a series of binary numbers. For this caseI has
unitsofbits and isgiven by
I=−log 2
p
Aformulaforevaluating logarithmstothe base2isgiven inChapter2.

916 Chapter 28 Probability
Engineeringapplication28.6
Informationcontentofabinarydatastream
Suppose that a computer generates a binary stream of data and that 1s and 0s occur
with equal probability, that is P(0) = P(1) = 0.5. Calculate the information per
binary digitgenerated.
Solution
Here, p = 0.5 whether the binary digitis0or 1,so
I = −log 2
(0.5) = −log 10 (0.5)
log 10
2
=1bit
Engineeringapplication28.7
Informationcontentofanalphabeticcharacterstream
Supposethatasystemgeneratesastreamofuppercasealphabeticcharactersandthat
the probability ofacharacteroccurring isthe same forall characters. Calculate
(a) the informationassociatedwith the character Goccurring
(b) the informationassociatedwith any singlecharacter.
Solution
(a) P(Goccurring) = 1
( ) 1
26 , I=−log 2
26
(b) Allcharacters areequally likely tooccur and so
( ) 1
I=−log 2
= −log 10 (1/26) = 4.70 bits
26 log 10
2
= −log 10 (1/26)
log 10
2
= 4.70 bits
Often a series of events may occur that do not have the same probability. For example,
ifastreamofalphabeticcharactersisbeinggeneratedthenitislikelythatsomecharacters
will occur more frequently than others and so have a higher probability associated
with them. For this situation it becomes convenient to introduce the concept of average
information. Given a sourceproducingasetofevents
E 1
,E 2
,E 3
,...,E n
with probabilities
p 1
,p 2
,p 3
,...,p n
thenforalong series ofevents theaverage informationper event isgiven by
∑i=n
H = − p i
log 2
p i
bits
i=1
H isalsotermedtheentropy.

28.5 Concepts from communication theory 917
Engineeringapplication28.8
Entropyofasignalconsistingofthreecharacters
Asourceproducesmessagesconsistingofthreecharacters,A,BandC.Theprobabilities
of each of these characters occurring isP(A) = 0.2,P(B) = 0.5,P(C) = 0.3.
Calculate the entropy of the signal.
Solution
H = −0.2log 2
(0.2) −0.5log 2
(0.5) −0.3log 2
(0.3) = 1.49 bits
Engineeringapplication28.9
Entropyofabinarydatastream
Asourcegeneratesbinarydigits0,1,withprobabilitiesP(0) = 0.3andP(1) = 0.7.
Calculate the entropy of the signal.
Solution
H = −0.3log 2
(0.3) −0.7log 2
(0.7) = 0.881 bits
Note that in Engineering application 28.9, on average, each binary digit only carries
0.881bitsofinformation.Infactthemaximumaverageamountofinformationthatcan
be carried by a binary digit occurs whenP(0) = P(1) = 0.5, as seen in Engineering
application 28.6. For this caseH = 1.
Foranydatastream,themaximumaverageamountofinformationthatcanbecarried
by a digit occurs when all digits have equal probability, that isH is maximized when
p 1
=p 2
=p 3
=···=p n
.
H is maximized when p 1
= p 2
= p 3
= ··· = p n
. The maximum value of H is
denotedH max
.
When the probabilities are not the same then one way of viewing the reduction in
H is to think of the likely event being given too much of the signalling time given its
lower information content. It is interesting to explore the two limiting cases, that is (a)
P(0) = 0,P(1) = 1;(b)P(0) = 1,P(1) = 0.InbothcasesitcanbeshownthatH = 0.
However,onexaminationthisisreasonablebecauseacontinuousstreamof1sdoesnot
relayanyusefulinformationtotherecipientandneitherdoesacontinuousstreamof0s.
The fact that some streams of symbols do not contain as much information as other
streams of the same symbols leads to the concept of redundancy. This allows the efficiency
with which information isbeing sent tobe quantified and isdefined as
redundancy =
maximum entropy -- actual entropy
maximum entropy
Alowvalue of redundancy corresponds toefficient transmission ofinformation.

918 Chapter 28 Probability
Engineeringapplication28.10
Redundancyofabinarydatastream
Consider the source of binary digits examined in Engineering application 28.6 and
28.9. The maximum entropy for a binary stream is 1 bit per binary digit. Calculate
the redundancy ineach case.
Solution
For Engineering application 28.6
redundancy = 1 −1 = 0
1
For Engineering application 28.9
redundancy = 1 −0.881
1
= 0.119
Engineeringapplication28.11
Redundancyofacharacterdatastream
A stream of data consists of four characters A, B, C, D with probabilities 0.1, 0.3,
0.2,0.4,respectively. Calculate the redundancy.
Solution
It can be shown that the maximum entropy, H max
, corresponds to the situation in
whichthe probabilityofeachsymbolisthe same, thatis0.25.
H max
= 4 × (−0.25log 2
(0.25)) = 2 bits
Theactual entropy,H act
, isgiven by
H act
= −(0.1log 2
(0.1) +0.3log 2
(0.3) +0.2log 2
(0.2) +0.4log 2
(0.4))
= 1.846bits
redundancy = 2 −1.846
2
= 0.0770
Intheexampleswehaveexaminedsofarwehaveusedthebitastheunitofinformation
because the most common form of digital signalling uses binary digits. When there are
only two possible events it is possible to represent an event by a single binary digit.
However, if there is a larger number of possible events then several binary digits are
neededtorepresentasingleevent.Whencalculatingvaluesforinformationandentropy
in these examples an assumption was made that each event was represented by binary
sequences or codes of the same length. It is only possible to do this efficiently if the
numberofeventsisapowerof2,thatis2,4,8,16, ....Inpracticethisproblemdoesnot
arise because it is more common to produce codes that have a small number of binary
digitsforlikelyeventsandalongnumberofbinarydigitsforunlikelyevents.Thisallows

28.6 Conditional probability: the multiplication law 919
the redundancy of a data stream to be reduced. The design of such codes is known as
coding theory. One complication is that most streams of data are not transmitted with
100%accuracyasaresultofthepresenceofnoisewithinthecommunicationchannel.It
isoftennecessarytobuildextraredundancyintoacodeinordertorecovertheseerrors.
EXERCISES28.5
1 Asourcegenerates sixcharacters, A, B,C,D, E, F,
with respective probabilities 0.05, 0.1, 0.25, 0.3, 0.15,
0.15.Calculate the average information percharacter
andthe redundancy.
2 Avisual displayunit has aresolutionof600 rowsby
800 columns.Ten different grey levelsare associated
with each pixel andtheir probabilitiesare 0.05,0.07,
0.09,0.10,0.11,0.13,0.12,0.12,0.11,0.10.Calculate
theaverageinformationcontentineachpictureframe.
3 Asourcegenerates five characters A, B,C,DandE
with respective probabilities of0.1, 0.15,0.2, 0.25
and0.3.
(a)Calculate the informationassociated with the
character B.
(b)Calculate the entropy.
(c)Calculate the redundancy.
4 Adata stream comprisesthe characters A, B,C,D
andEwithrespectiveprobabilitiesof0.23,0.16,0.11,
0.37 and 0.13.
(a) Calculate the informationassociated with the
character B.
(b) Calculate the informationassociated with the
character D.
(c) Calculate the entropy.
(d) Calculate the redundancy.
5 A data stream comprisesthe characters A, B,C,D
andEwithrespectiveprobabilitiesof0.12,0.21,0.07,
0.31 and 0.29.
(a) Which character carries the greatest information
content?
(b) Which character carries the leastinformation
content?
(c) Calculate the informationassociated with the
letter D.
(d) Calculate the entropy.
(e) Calculate the redundancy.
Solutions
1 2.3905,0.0752
2 3.2790
3 (a) 2.7370 (b) 2.2282 (c) 0.0404
4 (a) 2.6439 (b) 1.4344 (c) 2.1743
(d) 0.0636
5 (a) C (b) D (c) 1.6897
(d) 2.1501 (e) 0.0740
28.6 CONDITIONALPROBABILITY:THEMULTIPLICATIONLAW
Suppose two machines, M and N, both manufacture components. Of the components
made by machine M, 92% are of an acceptable standard and 8% are rejected. For machine
N, only 80% are of an acceptable standard and 20% are rejected. Consider now
the eventE:
E: a component isof an acceptable standard
If all the components are manufactured by machine M then P(E) = 0.92. However,
if all the components are manufactured by machine N then P(E) = 0.8. If half the
componentsaremanufacturedbymachineMandhalfbymachineNthenP(E) = 0.86.
Toseewhythisissoconsider1000components.OfthehalfmadebymachineM,92%×

920 Chapter 28 Probability
500 = 460 will be of an acceptable standard. Of the half made by machine N, 80% ×
500 = 400 will be acceptable. Hence 860 of the 1000 components will be acceptable
andsoP(E) = 860
1000 = 0.86.Clearly,therearedistinctprobabilitiesofthesameevent;
the probability changes as the conditions change. This is intuitive and leads to the idea
of conditional probability.
We introduce a notation forconditional probability. Define eventsAandBby
A: the component ismanufactured by machine M
B: the component ismanufactured by machine N
Thentheprobabilitythatacomponentisofanacceptablestandard,givenitismanufactured
by machine M, is written asP(E|A). We read this as the conditional probability
ofE givenA. SimilarlyP(E|B) is the probability ofE happening, givenBhas already
happened.
P(E|A) = 0.92 P(E|B) = 0.8
To be pedantic, all probabilities are conditional since the conditions surrounding any
event can change. However, for many situations there is tacitly assumed a definite set
of conditions which is always satisfied. The probability of an event calculated under
onlytheseconditionsisknownastheunconditionalprobability.Iffurtherwell-defined
conditions areattached, the probability isconditional.
Engineeringapplication28.12
Productionlineproductfedbytwomachines
MachinesMandNmanufactureacomponent.Theprobabilitythatthecomponentis
of an acceptable standard is 0.95 when manufactured by machine M and 0.83 when
manufactured by machine N. Machine M supplies 65% of components; machine N
supplies 35%. Acomponent ispicked atrandom.
(a) What isthe probability thatthe component isofanacceptable standard?
(b) Whatistheprobabilitythatacomponentisofanacceptablestandardandismade
by machineM?
(c) What is the probability that the component is of an acceptable standard given it
is made by machineM?
(d) What isthe probability thatthe component wasmade bymachine M?
(e) What is the probability that the component was made by machine M given it is
ofanacceptable standard?
(f) The component is not of an acceptable standard. What is the probability that it
wasmade by machineN?
Solution
We definethe events:
A: the component ismanufactured by machineM
B: the component ismanufactured by machineN
C: the component isofanacceptable standard

28.6 Conditional probability: the multiplication law 921
(a) Consider 1000 components. Then 650 are manufactured by machine M, 350 by
machineN.Ofthe650manufacturedbymachineM,95%willbeacceptable,that
is650× 95
100 = 617.5.Ofthe350manufacturedbymachineN,83%willbeacceptable,thatis350×
83
100 = 290.5.Soonaveragein1000components,617.5+
290.5 = 908 will be acceptable, thatisP(C) = 0.908.
(b) From part (a) we know that out of 1000 components 617.5 will be made by
machine M and be of an acceptable standard. Hence P(A ∩C) = 617.5
1000 =
0.6175.
(c) We requireP(C|A). The probability a component is acceptable given it is manufactured
by machine M is
P(C|A) = 0.95
(d) Machine M makes 65%of components, thatisP(A) = 0.65.
(e) We requireP(A|C). Consider again the 1000 components. On average 908 are
acceptable.Ofthese908acceptablecomponents,617.5aremanufacturedbymachine
M and 290.5 by machine N. We are told the component is acceptable and
sowemustrestrictattentiontothe908acceptablecomponents.So,outof908acceptablecomponents,617.5aremadebymachineM,thatisP(A|C)
= 617.5
908 =
0.68.
(f) We require P(B|C). Consider 1000 components. Machine M manufactures
650 components of which 617.5 are acceptable and hence 32.5 are unacceptable.
Machine N manufactures 350 components of which 290.5 are acceptable
and59.5areunacceptable.Thereare92unacceptablecomponentsofwhich59.5
weremadebymachineN.Theprobabilityofthecomponentbeingmadebymachine
Ngiven itisunacceptable is
P(B|C) = 59.5
92 = 0.647
that is, almost 65% of unacceptable components are manufactured by
machine N.
As an alternative method of solution, we can represent the information via a tree
diagram. This isshown inFigure 28.6.
617.5 (acceptable)
M
650
32.5 (unacceptable)
1000
N
350
290.5 (acceptable)
59.5 (unacceptable)
Figure28.6
Tree diagram forEngineeringapplication
28.12.
➔

922 Chapter 28 Probability
(a) There are617.5 +290.5 = 908 acceptable components, so
P(component isacceptable) = 0.908
(b) There are617.5 components which are acceptable and made by machine M. So
P(component isacceptable and made by machine M)
=P(C ∩A) = 0.6175
(c) Of the 650 components made by machine M, 617.5 areacceptable, and so
P(component isacceptable given itismade by machine M) =P(C|A)
= 617.5
650 = 0.95
(d) Of the 1000 components, 650 aremade by machine M, so
P(component ismade by machine M) =P(A) = 0.65
(e) Thereare908acceptablecomponents.Ofthese,617.5aremadebymachineM.
Hence
P(component ismade by machine M given itisacceptable) =P(A|C)
= 617.5
908 = 0.68
(f) There are 32.5 + 59.5 = 92 unacceptable components. Of these 92, 59.5 are
made by machine N.Thus
P(component ismade by machine Ngiven itisnot acceptable) =P(B|C)
= 59.5
92 = 0.647
28.6.1 Themultiplicationlaw
Consider events A and B for which A ∩B ≠ ◦/ as shown in Figure 28.7. Suppose we
knowthateventAhasoccurredandweseektheprobabilitythatBoccurs,thatisP(B|A).
Knowing event A has occurred we can restrict our attention to the set A. Event B will
occur ifany outcome is inA ∩B. Hence,
E
P(B|A) =
P(A ∩B)
P(A)
A
B
Themultiplication lawofprobabilitystates:
P(A ∩B) =P(A)P(B|A)
A > B
Figure28.7
AandBare not
mutually exclusive.
Since the compound event ‘Aand B’ isidentical to‘B and A’ wemay alsosay
P(A ∩B) =P(B ∩A) =P(B)P(A|B)
Consider Engineering application 28.12(e). We require the probability that the component
was manufactured by machine M, given itisacceptable. This isP(A|C). Now

P(A ∩C) =P(C ∩A) =P(C)P(A|C)
P(A|C) = P(A∩C)
P(C)
28.6 Conditional probability: the multiplication law 923
NowP(A∩C)istheprobabilitythatthecomponentismanufacturedbymachineMand
isacceptable.Thisisknowntobe0.6175,usingEngineeringapplication28.12(b).Also
fromEngineering application 28.12(a) we seeP(C) = 0.908. Hence,
P(A|C) = 0.6175
0.908 = 0.68
Engineeringapplication28.13
Reliabilityofmanufacturedcomponents
A manufacturer studies the reliability of a certain component so that suitable guarantees
can be given: 83% of components remain reliable for at least 5 years; 92%
remain reliable for at least 3 years. What is the probability that a component which
has remainedreliable for3yearswill remainreliablefor5years?
Solution
We definethe events:
A: a component remainsreliableforatleast3years
B: a component remainsreliableforatleast5years
ThenP(A) = 0.92,P(B) = 0.83.Notethattheseareunconditionalprobabilities.We
requireP(B|A), a conditionalprobability.
P(A ∩B) =P(A)P(B|A)
A ∩Bisthecompoundeventacomponentremainsreliableforatleast3yearsandit
remainsreliableforatleast5years. Clearlythis isthe same asthe eventB. So
Hence
P(A ∩B) =P(B)
P(B) =P(A)P(B|A)
P(B|A) = P(B)
P(A) = 0.83
0.92 = 0.90
thatis,90%ofthecomponentswhichremainreliablefor3yearswillremainreliable
for atleast5years.
Alternatively, a tree diagram is shown in Figure 28.8. Starting with 100 components,92remainreliable3yearslater.Ofthese92,2yearslater,83arestillreliable.
So
P(component isreliable after 5 years, given itisreliable after 3 years) = 83
92
= 0.90 ➔

924 Chapter 28 Probability
3 years 5 years
100
92 (reliable)
83 (reliable)
}
9 (unreliable)
17 (unreliable)
8 (unreliable) 8 (unreliable)
Figure28.8
Tree diagram forEngineering application 28.13.
EXERCISES28.6
1 Acomponent is manufactured bymachines 1 and2.
Machine 1 manufactures 72% oftotal production of
the component. Thepercentage ofcomponents which
are acceptable varies, depending uponwhich machine
is used.Formachine 1, 97% ofcomponents are
acceptable and formachine 2, 92% are acceptable. A
component ispicked at random.
(a) What is the probability itwasmanufactured by
machine 1?
(b) What is the probability itisnot acceptable?
(c) What is the probability that itis acceptable and
made bymachine 2?
(d) If the component isacceptable what is the
probability itwasmanufactured bymachine 2?
(e) If the component isnot acceptable what isthe
probability itwasmanufactured bymachine 1?
2 Themeasured lifespans (L)of1500components are
recorded in Table 28.4.
Table28.4
Lifespansof1500components.
Lifespan(hours)
Numberofcomponents
L 1000 210
900 L<1000 820
800 L<900 240
700 L<800 200
L<700 30
(a) Whatistheprobabilitythatacomponentwhichis
still workingafter800 hours willlast foratleast
900 hours?
(b) Whatistheprobabilitythatacomponentwhichis
still workingafter900 hours willcontinue to last
foratleast 1000 hours?
3 Thelifespan,L,of1000 components is measuredand
detailed in Table 28.5.
Table28.5
Lifespansof1000 components.
Lifespan(hours)
Numberofcomponents
L 1750 70
1500 L<1750 110
1250 L<1500 150
1000 L<1250 200
750 L<1000 317
500L<750 96
250L<500 42
L<250 15
(a) Calculate the probability that the lifespanofa
component is more than 1000hours.
(b) Calculate the probability that the lifespanofa
component is lessthan 750 hours.
(c) Calculate the probability that a component which
isstill workingafter 500 hours willcontinue to
last forat least1500hours.
4 MachinesA, Band C manufacture components.
Machine Amakes 50%ofthe components, machine
Bmakes 30%ofthe components andmachine C
makesthe rest. Theprobability that acomponent is
reliable is0.93 when made bymachine A, 0.90 when
made bymachine Band 0.95 when made by machine
C.Acomponent is picked atrandom.
(a) Calculate the probability that itis reliable.
(b) Calculate the probability that itis made by
machine B given itisunreliable.

28.7 Independent events 925
5 Componentsare made by machines A, B,Cand D.
Machine Amakes17% ofthe components, machine
Bmakes 21% ofthe components, machine Cmakes
20% ofthe components and machine D makesthe
remainder. Formachine A, 96% ofthe components
are reliable, formachine B,89% are reliable, for
machine C,92% are reliable and formachine D, 97%
are reliable. Acomponent ispicked atrandom.
Calculate the probability that itis
(a) reliable
(b) notreliable
(c) reliable, given itis madeby machine B
(d) notreliable, given itis made bymachine D
(e) made bymachine Agiven itis reliable
(f) made bymachine Cgiven itis unreliable
Solutions
1 (a) 0.72 (b) 0.044 (c) 0.2576
(d) 0.2695 (e) 0.4909
2 (a) 0.8110 (b) 0.2039
4 (a) 0.925 (b) 0.4
5 (a) 0.9415 (b) 0.0585 (c) 0.89
(d) 0.03 (e) 0.1733 (f) 0.2735
3 (a) 0.53 (b) 0.153 (c) 0.1909
28.7 INDEPENDENTEVENTS
Twoeventsareindependentiftheoccurrenceofeithereventdoesnotinfluencethe
probability of the other event occurring.
Engineeringapplication28.14
Acceptabilityofmanufacturedelectronicchips
Machine 1 manufactures an electronic chip, A, of which 90% are acceptable. Machine2manufacturesanelectronicchip,B,ofwhich83%areacceptable.Twochips
are picked at random, one of each kind. Find the probability that they are both
acceptable.
Solution
The eventsE 1
andE 2
aredefined:
E 1
: chip Aisacceptable
E 2
: chip Bis acceptable
P(E 1
) = 0.9 P(E 2
) = 0.83
A single trial consists of choosing two chips at random. We require the probability
that the compound event,E 1
∩E 2
, istrue.Usingthe multiplication lawwe have
P(E 1
∩E 2
) =P(E 1
)P(E 2
|E 1
) = 0.9P(E 2
|E 1
)
➔

926 Chapter 28 Probability
P(E 2
|E 1
) is the probability ofE 2
happening givenE 1
has happened. However, machine1andmachine2areindependent,sotheprobabilityofchipBbeingacceptable
is in no way influenced by the acceptability of chip A. The events E 1
and E 2
are
independent.
Therefore
P(E 2
|E 1
) =P(E 2
) = 0.83
P(E 1
∩E 2
) =P(E 1
)P(E 2
) = (0.9)(0.83) = 0.75
For independent eventsE 1
andE 2
:
(1)P(E 1
|E 2
) =P(E 1
), P(E 2
|E 1
) =P(E 2
)
(2)P(E 1
∩E 2
) =P(E 1
)P(E 2
)
The concept of independence may be applied to more than two events. Three or more
events are independent if every pair of events is independent. If E 1
,E 2
,...,E n
are n
independent events then
P(E i
|E j
)=P(E i
)
foranyiandj,i≠j
and
P(E i
∩E j
) =P(E i
)P(E j
)
i≠j
Acompoundeventmaycompriseseveralindependentevents.Themultiplicationlawis
extended inanobvious way:
P(E 1
∩E 2
∩E 3
) =P(E 1
)P(E 2
)P(E 3
)
P(E 1
∩E 2
∩E 3
∩E 4
) =P(E 1
)P(E 2
)P(E 3
)P(E 4
)
and soon.
Engineeringapplication28.15
Probabilityoffaultycomponents
The probability that a component is faulty is 0.04. Two components are picked at
random. Calculate the probability that
(a) both components arefaulty
(b) both components arenot faulty
(c) one ofthe components isfaulty
(d) one ofthe components isnotfaulty
(e) atleastone of the components isfaulty
(f) atleastone of the components isnot faulty

28.7 Independent events 927
Solution
We define eventsF 1
andF 2
tobe
F 1
: the first component isfaulty
F 2
: the second component isfaulty
Then eventsF 1
andF 2
are
Then
F 1
: the first component isnotfaulty
F 2
: the second component isnotfaulty
P(F 1
) =P(F 2
) = 0.04
P(F 1
)=P(F 2
)=1−0.04=0.96
(a) We requireboth components tobefaulty:
P(F 1
∩F 2
) =P(F 1
)P(F 2
)
= (0.04) 2
= 0.0016
since events areindependent
(b) We requireboth components tobenot faulty:
P(F 1
∩F 2
)=P(F 1
)P(F 2
)
= (0.96) 2
= 0.9216
(c) Consider the two components. Then either both components are faulty or one
component isfaulty or neither of the components isfaulty. So,
P(one component isfaulty) = 1 −P(both components are faulty)
−P(neither of the components isfaulty)
= 1 −0.0016 −0.9216
= 0.0768
Analternativeapproachisasfollows.Eitherthefirstcomponentisfaultyandthe
second one is not faulty, or the first component is not faulty and the second one
isfaulty. These two cases arerepresented byF 1
∩F 2
andF 1
∩F 2
.So,
P(one component isfaulty) =P(F 1
∩F 2
)+P(F 1
∩F 2
)
=P(F 1
)P(F 2
)+P(F 1
)P(F 2
)
= (0.04)(0.96) + (0.96)(0.04)
= 0.0768
(d) Werequireonecomponenttobenotfaulty.Sincetherearetwocomponentsthen
requiring one to be not faulty is equivalent to requiring one to be faulty. Hence
the calculation isthe same as thatin(c):
P(one component isnot faulty) = 0.0768
(e) At least one of the components is faulty means that one or both of the components
isfaulty:
➔

928 Chapter 28 Probability
P(atleastone component isfaulty) =P(exactly one component isfaulty)
+P(both components arefaulty)
= 0.0768 +0.0016
= 0.0784
Asanalternativewecannotethatthecomplementof‘atleastoneofthecomponentsisfaulty’is‘noneofthecomponentsisfaulty’.Theprobabilitythatneither
of the components is faulty isgiven in(b). Hence
P(atleastone component isfaulty)
= 1 −P(none of the components isfaulty)
= 1 −0.9216
= 0.0784
(f) At least one of the components is not faulty means that one or both of the components
isnotfaulty. So
P(atleastone component isnotfaulty)
=P(exactly one component isnot faulty)
+P(both components are notfaulty)
= 0.0768 +0.9216
= 0.9984
Asanalternativewenotethatthecomplementof‘atleastoneofthecomponents
is not faulty’ is ‘none of the components are not faulty’. The last statement is
equivalent to‘both the components arefaulty’.Hence
P(atleastone component isnot faulty) = 1−P(both components arefaulty)
= 1 −0.0016
= 0.9984
Engineeringapplication28.16
Probabilityofacceptablecomponents
Machines 1, 2 and 3 manufacture resistors A, B and C, respectively. The probabilities
of their respective acceptabilities are 0.9, 0.93 and 0.81. One of each resistor is
selected atrandom.
(a) Find the probability thatthey are all acceptable.
(b) Find the probability thatatleastone resistor isacceptable.
Solution
Define eventsE 1
,E 2
andE 3
by
E 1
: resistor Ais acceptable P(E 1
) = 0.9
E 2
: resistor Bisacceptable P(E 2
) = 0.93
E 3
: resistor Cisacceptable P(E 3
) = 0.81

28.7 Independent events 929
E 1
,E 2
andE 3
areindependent events.
(a) P(E 1
∩E 2
∩E 3
) =P(E 1
)P(E 2
)P(E 3
) = (0.9)(0.93)(0.81) = 0.68
(b) LetE 4
andE 5
be the events
E 4
: atleastone resistor isacceptable
E 5
: no resistorisacceptable
E 4
andE 5
arecomplementary events and so
P(E 4
)+P(E 5
)=1
E 5
may be expressed as
E 5
: resistor Aisnotacceptable and resistor Bisnot acceptable and
resistor Cisnotacceptable
thatis,E 1
∩E 2
∩E 3
P(E 5
) =P(E 1
∩E 2
∩E 3
)=P(E 1
)P(E 2
)P(E 3
)
= (1 −0.9)(1 −0.93)(1 −0.81)
= (0.1)(0.07)(0.19)
= 0.00133
P(E 4
) = 1 −P(E 5
) = 1 −0.00133 = 0.99867
EXERCISES28.7
1 AandBare two independenteventswithP(A) = 0.7
andP(B) = 0.4. The compound eventAoccurs, then
Aoccurs, thenBoccurs, is denotedAAB, andother
compound eventsare denotedin a similarway.
Calculatethe probability ofthe following compound
events:
(a) AAB
(b) BAB
(c) AAAA
2 Theprobabilityacomponentis faultyis 0.07.Two
components are picked at random.Calculate the
probabilitythat
(a) both are faulty
(b) both are not faulty
(c) the firstonepicked is faultyandthe second one
picked isnot faulty
(d) atleast oneisnot faulty
3 Theprobabilityacomponentis acceptable is 0.92.
Three components are picked atrandom. Calculate
the probability that
(a) allthree are acceptable
(b) none are acceptable
(c) exactlytwo are acceptable
(d) atleast two are acceptable
4 Componentsare madebymachines A, BandC.
Machine A makes30%ofthe components,machine
B makes25% ofthe components andmachine C
makes the rest. Two components are picked at
random. Calculate the probability that
(a) they are both madebymachine C
(b) oneismadeby machine Aandoneismadeby
machine B
(c) exactlyoneofthe components ismadeby
machine B
(d) atleast oneofthe components ismadeby
machine B
(e) both components are madebythe same machine
5 Capacitorsaremanufacturedbyfourmachines,1,2,3
and4.Theprobability a capacitor is manufactured
acceptablyvaries according to the machine.The

930 Chapter 28 Probability
probabilitiesare 0.94, 0.91,0.97 and0.94,
respectively, formachines 1, 2, 3 and 4.
(a) Acapacitor istaken from each machine. What is
the probability all fourcapacitors are acceptable?
(b) Twocapacitorsaretakenfrommachine1andtwo
from machine 2. What isthe probability all four
capacitors are acceptable?
(c) Acapacitor istaken from each machine.
Calculate the probability that atleastthree
capacitors are acceptable.
(d) Acapacitor istaken from each machine. From
thissampleoffour capacitors, one istaken at
random.
(i) What is the probability itis acceptable and
made by machine 1?
(ii) What is the probability itis acceptable and
made by machine 2?
(e) Acapacitor istaken from each machine. From
thissampleoffour capacitors, one istaken at
random. What isthe probability itis acceptable?
[Hint: use the resultsin (d).]
Solutions
1 (a) 0.196 (b) 0.112 (c) 0.2401
2 (a) 0.0049 (b) 0.8649 (c) 0.0651
(d) 0.9951
3 (a) 0.7787 (b) 5.12 ×10 −4
(c) 0.2031 (d) 0.9818
4 (a) 0.2025 (b) 0.15 (c) 0.375
(d) 0.4375 (e) 0.355
5 (a) 0.7800 (b) 0.7317 (c) 0.9808
(d) (i) 0.235 (ii) 0.2275 (e) 0.94
REVIEWEXERCISES28
1 Componentsare made bymachines A, B and C.
Machine Amakes 30% ofthe components, machine
B makes50% ofthe components and machine C
makes the rest. Theprobability that a component is
acceptable when made by machine Ais 0.96, when
made bymachine Bthe probability is0.91 andwhen
made bymachine Cthe probability is0.93.
(a) Acomponent is picked atrandom. Calculate the
probability that itis made bymachine C.
(b) Acomponent is picked atrandom. Calculate the
probability that itis made byeither machine Aor
machine C.
(c) Acomponent is picked atrandom. Calculate the
probability that itis made bymachine Band is
acceptable.
(d) Acomponent is picked atrandom. Calculate the
probability that itis notacceptable.
(e) Acomponent picked atrandom is notacceptable.
Calculate the probability that itismade by
machine A.
(f) Acomponent picked atrandom is acceptable.
Calculate the probability that itismade by either
machine Aormachine B.
(g) Two components are picked atrandom. Calculate
the probability that they are both acceptable.
(h) Two components are picked atrandom. Calculate
the probability that oneis acceptable and oneis
unacceptable.
2 Three machines, A, Band C,manufacture a
component. Machine A manufactures 35%ofthe
components, machine B manufactures 40%ofthe
components and machine C makesthe rest.A
component is eitheracceptable ornotacceptable: 7%
ofcomponents made bymachine Aare not
acceptable, 12%ofcomponents made by machine B
are not acceptable and 2% ofthosemade bymachine
Care alsonotacceptable.
(a) Findthe probability that acomponent is made by
eithermachine Aormachine B.
(b) Two components are picked atrandom. What is
the probability that they are both made by
machine B?
(c) Three components are picked atrandom. What is
the probability they are each made by adifferent
machine?

Review exercises 28 931
(d) A component is picked atrandom. What is the
probability itis notacceptable?
(e) A component is picked atrandom. Itisnot
acceptable. What isthe probability itwas made
by machine B?
(f) A component is picked atrandom and is
acceptable. What isthe probability itwas made
by eithermachine Aormachine B?
3 MachinesMandNmanufacture components. The
probability that acomponent isofan acceptable
standard is 0.93 when manufactured bymachine M
and0.86 when manufactured by machine N. Machine
Msupplies70% ofthe components andmachine N
suppliesthe rest.
(a) Calculate the probability that acomponent picked
atrandom is ofan acceptable standard.
(b) Acomponent isnot ofan acceptable standard.
Calculate the probability that itismade by
machine N.
(c) Two components are picked at random. Calculate
the probability that they are made bydifferent
machines.
4 E 1 is the event the component is reliable.E 2 is the
event the component ismade bymachine A. State
(a) E 1
(b) E 2
(c) E 1 ∩E 2
(d) E 1 ∪E 2
(e) E 2 ∩E 1
5 Thelifespans,L,of2500components were measured
andthe resultswere recorded in Table 28.6.
Table28.6
Lifespan(hours)
Numberofcomponents
0L<200 76
200 L<300 293
300 L<400 574
400 L<500 1211
500 L<600 346
(a) Calculate the probability thatacomponenthas a
lifespanofover 400 hours.
(b) Calculate the probability thatacomponenthas a
lifespanofless than300 hours.
(c) Calculate the probability thatacomponentthat
lasts for300 hours willcontinueto lastforat
least 500 hours.
6 Given the eventsE 1 ,E 2 andE 3 are
E 1 :the circuitisfunctioning
E 2 :the circuitismadeby machine A
E 3 :the circuitwilllast foratleast600 hours state
(a) E 2
(b) E 3
(c) E 1 ∪E 2
(d) E 1 ∩E 3
(e) E 2 ∩E 3
7 A data stream consistsofthe characters A, B,C,D
andE, with respective probabilities of0.16,0.23,
0.31, 0.12 and0.18.
(a) Calculate the informationassociated with the
character B.
(b) Calculate the entropy.
(c) Calculate the redundancy.
8 The probabilityacomponentis reliableis0.93.Three
components are picked atrandom. Calculatethe
probability that
(a) all components are unreliable
(b) exactly onecomponent isreliable
(c) at leastonecomponentis reliable
(d) exactly two components are reliable
9 A circuitis asshown in Figure 28.9.Thecircuitis
operational ifcurrent can flowfrom Pto Q alongany
route. Theprobability thatresistor typeAis faultyis
0.09, andforresistor typeBthe probability ofafault
is 0.14.Calculate the probability thatthe circuitis
operational.
P
Figure28.9
A
A
10 The probabilityacomponentis reliableis0.96.Five
components are picked atrandom. Calculatethe
probability that
(a) allfive components are reliable
(b) none are reliable
(c) atleast oneis reliable
B
A
B
Q

932 Chapter 28 Probability
Solutions
1 (a) 0.2 (b) 0.5 (c) 0.455
(d) 0.071 (e) 0.1690 (f) 0.7998
(g) 0.8630 (h) 0.1319
2 (a) 0.75 (b) 0.16 (c) 0.21
(d) 0.0775 (e) 0.619 (f) 0.734
3 (a) 0.909 (b) 0.4615 (c) 0.42
4 (a) Thecomponent isnot reliable.
(b) Thecomponent isnot made bymachine A.
(c) Thecomponent isreliable and made by
machine A.
(d) Thecomponent isreliable oritis made by
machine A.
(e) Thecomponent ismade by machine A anditis
notreliable.
5 (a) 0.6228 (b) 0.1476 (c) 0.1624
6 (a) Thecircuit isnot made bymachine A.
(b) Thecircuit willlast forlessthan 600 hours.
(c) Thecircuit isfunctioning oritis made by
machine A.
(d) Thecircuit isfunctioning and willlast foratleast
600 hours.
(e) Thecircuit ismade by machine Aand willlast
foratleast600 hours.
7 (a) 2.1203 (b) 2.2469 (c) 0.0323
8 (a) 3.43 ×10 −4 (b) 0.0137 (c) 0.9997
(d) 0.1816
9 0.9948
10 (a) 0.8154 (b) 1.024 ×10 −7
(c) 0.9999999

29 Statisticsandprobability
distributions
Contents 29.1 Introduction 933
29.2 Randomvariables 934
29.3 Probabilitydistributions--discretevariable 935
29.4 Probabilitydensityfunctions--continuousvariable 936
29.5 Meanvalue 938
29.6 Standarddeviation 941
29.7 Expectedvalueofarandomvariable 943
29.8 Standarddeviationofarandomvariable 946
29.9 Permutationsandcombinations 948
29.10 Thebinomialdistribution 953
29.11 ThePoissondistribution 957
29.12 Theuniformdistribution 961
29.13 Theexponentialdistribution 962
29.14 Thenormaldistribution 963
29.15 Reliabilityengineering 970
Reviewexercises29 977
29.1 INTRODUCTION
Engineersoftenneedtomeasuremanydifferentvariables,forexampletheoutputvoltage
of a system, the strength of a beam or the cost of a project. Variables are classified into
one of two types,discrete orcontinuous. These arediscussed inSection 29.2.
It is of interest to know the probability that a variable has of falling within a given
range. There is a high probability that a given variable will fall within some ranges,
and a small probability for other ranges of values. The way in which the probability is

934 Chapter 29 Statistics and probability distributions
distributedacrossvariousrangesofvaluesgivesrisetotheideaofaprobabilitydistribution.Thestudyofprobabilitydistributionsformsthemajorpartofthischapter.Themost
important distributions, the binomial, the Poisson, the uniform, the exponential and the
normal,areallincluded. Thechapter concludes withastudyofreliabilityengineering.
29.2 RANDOMVARIABLES
Engineeringquantitieswhosevariationcontainsanelementofchancearecalledrandom
variables. Some examples are listedbelow:
(1) the diameter of a motor shaft of nominal size0.2 m;
(2) the weight of a steelbox used tocontain anelectronic circuitboard;
(3) thenumberofcomponentspassingapointonafactoryproductionlinein1minute;
(4) the nominal resistance value of resistors;
(5) the length oftime a machine works withoutfailing.
All of these quantities vary. In (1), (2) and (5) the quantities vary continuously; that
is, they can assume any value in some range. For example, a motor shaft may have any
diameterbetween0.197mand0.203m;asteelboxcouldhaveanyweightbetween,say,
0.345 kg and 0.352 kg. These are examples of continuous variables. The value itself
will only be recorded to a certain accuracy, which depends on the measuring device
and the use to which the data will be put. For example, the shaft diameter may only
be measured to the nearest tenth of a millimetre. Although such a measurement is only
integerinmultiplesofatenthofamillimetre,thevariablebeingmeasurediscontinuous.
In(3)and(4)thevariablescanassumeonlyalimitednumberofvalues.Thenumberof
componentspassingapointinaminutewillbeanon-negativeinteger0,1,2,3....The
nominal resistance value of resistors has a limited number of values which is specified
by manufacturers in their catalogues. Variables such as these, which can assume only a
limitedset of values, arecalleddiscrete variables.
EXERCISES29.2
1 Isthe length oftime amachine workswithout failing
a continuous oradiscrete variable?
2 State whether the following variables are continuous
ordiscrete:
(a)the length ofabridge
(b)the number ofelectrical sockets in ahouse
(c)the length ofcable used to wire ahouse
(d)the weight ofsolderused to build acircuit board
3 Statewhether the following variablesare discreteor
continuous:
(a) the force requiredto stretch a spring bya
specifiedlength
(b) the output voltage ofasystem
(c) the height ofacolumn ofliquid
(d) the number ofresistorsin acircuit
(e) the number ofbitsofmemory ofacomputer
Solutions
1 continuous
2 (a) continuous (b)discrete
(c) continuous (d)continuous
3 (a)continuous (b)continuous (c)continuous
(d)discrete (e)discrete

29.3 Probability distributions -- discrete variable 935
29.3 PROBABILITYDISTRIBUTIONS--DISCRETEVARIABLE
The range of values that a variable can take does not give sufficient information about
the variable. We need to know which values are likely to occur often and which values
willoccuronlyinfrequently.Forexample,supposexisadiscreterandomvariablewhich
cantakevalues0,1,2,3,4,5and6.Wemayaskquestionssuchas‘Whichvalueismost
likelytooccur?’,‘Isa6morelikelytooccurthana5?’,andsoon.Weneedinformation
on the probability of each value occurring. Suppose that information is provided and is
given in Table 29.1. Ifxis sampled 100 times then on average 0 will occur 10 times, 1
will occur 10 times, 2 will occur 15 times and so on. Table 29.1 is called aprobability
distribution for the random variable x. Note that the probabilities sum to 1; the table
tells us how the total probability isdistributed among the various possible values of the
random variable. Table 29.1 may be represented ingraphical form (see Figure 29.1).
P(x)
0.3
0.2
Table29.1
The probability of a discrete
value occurring.
x 0 1 2 3 4 5 6
P(x) 0.1 0.1 0.15 0.3 0.2 0.1 0.05
0.1
0 1 2 3 4 5 6
Figure29.1
Plotted data ofTable 29.1.
x
EXERCISES29.3
1 Theprobabilitydistribution forthe randomvariable,
x,is
x 2 2.5 3.0 3.5 4.0 4.5
P(x) 0.07 0.36 0.21 0.19 0.10 0.07
(a) StateP(x = 3.5)
(b) CalculateP(x 3.0)
(c) CalculateP(x < 4.0)
(d) CalculateP(x > 3.5)
(e) CalculateP(x 3.9)
(f) The variable,x,is sampled50000 times. How
many timeswould you expectxto have a value
of2.5?
2 The probabilitydistribution ofthe randomvariable,y,
is given as
y −3 −2 −1 0 1 2 3
P(y) 0.63 0.20 0.09 0.04 0.02 0.01 0.01
Calculate
(a) P(y 0) (b) P(y 1)
(c) P(|y| 1) (d) P(y 2 > 3)
(e) P(y 2 < 6)
Solutions
1 (a) 0.19 (b) 0.57 (c) 0.83 (d) 0.17
(e) 0.83 (f) 18000
2 (a) 0.08 (b) 0.98 (c) 0.15 (d) 0.85
(e) 0.36

936 Chapter 29 Statistics and probability distributions
29.4 PROBABILITYDENSITYFUNCTIONS--CONTINUOUS
VARIABLE
Suppose x is a continuous random variable which can take any value on [0, 1]. It is
impossible to list all possible values because of the continuous nature of the variable.
There are infinitely many values on [0, 1] so the probability of any one particular value
occurring is zero. It is meaningful, however, to ask ‘What is the probability ofxfalling
inasub-interval,[a,b]?’Dividing[0,1]intosub-intervalsandattachingprobabilitiesto
each sub-interval will result in a probability distribution. Table 29.2 gives an example.
The probability thatxwill lie between 0.4 and 0.6 is 0.35, that isP(0.4 x < 0.6) =
0.35. Similarly,
P(0.2 x < 0.4) = 0.25
Figure 29.2 shows the table in a graphical form. By making the sub-intervals smaller a
more refined distributionisobtained. Table 29.3 and Figure 29.3 illustratethis.
The probability thatxlies in a particular interval is given by the sum of the heights
of the rectangles on that interval. For example, the probability thatxlies in [0.5, 0.8] is
0.2+0.1+0.1 = 0.4;thatis,thereisaprobabilityof0.4thatxliessomewherebetween
0.5 and 0.8.Note thatthe sum ofall the heights is1,representing total probability.
Considerthesub-interval[a,b].Werequiretheprobabilitythatxliesinthisinterval.
The way this is answered is by means of a probability density function (p.d.f.), f (x).
Suchap.d.f.isshowninFigure29.4,whereitistheareaunderthegraphbetweenaand
Table29.2
Probability thatxlies in agiven sub-interval.
x [0,0.2) [0.2, 0.4) [0.4,0.6) [0.6,0.8) [0.8,1.0]
P(x) 0.1 0.25 0.35 0.2 0.1
Table29.3
Refining the sub-intervals in Table 29.2.
x [0,0.1) [0.1, 0.2) [0.2,0.3) [0.3,0.4) [0.4, 0.5)
P(x) 0.03 0.07 0.1 0.15 0.15
x [0.5, 0.6) [0.6, 0.7) [0.7,0.8) [0.8,0.9) [0.9, 1.0]
P(x) 0.2 0.1 0.1 0.07 0.03
P(x)
0.4
0.3
0.2
P(x)
0.2
0.1
0.1
0
0.2 0.4 0.6 0.8 1.0
x
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x
Figure29.2
Plotted data ofTable 29.2.
Figure29.3
Plotted data ofTable 29.3.

29.4 Probability density functions -- continuous variable 937
f(x)
0 a b
x
Figure29.4
Shaded area representsP(a xb).
bthatgivesP(a xb).
P(a xb) = areaabove[a,b] =
The total area under a p.d.f. isalways 1.
∫ b
a
f(x)dx
Example29.1 Supposexisacontinuousrandomvariabletakinganyvalueon[1,4].Itsp.d.f., f (x),is
given by
f(x)= 1
2 √ x
1x4
(a) Check that f (x)isasuitable function forap.d.f.
(b) What isthe probability that (i)xlieson [2, 3.5],(ii)x 2,(iii)x < 3?
Solution (a) x can have any value on [1, 4]. For f (x) to be a p.d.f., then the total area under it
should equal 1,thatis
∫ 4
1
∫ 4
1
f(x)dx=1
f(x)dx=
∫ 4
1
1
2 √ x dx=[√ x] 4 1 = 1
Hence f (x) isasuitablefunction forap.d.f.
(b) (i) P(2 x 3.5) =
(ii) P(x 2) =
(iii) P(x < 3) =
∫ 4
2
∫ 3
1
∫ 3.5
2
f(x)dx=[ √ x] 3.5
2
= 0.457
f(x)dx=[ √ x] 4 2 = 0.586
f(x)dx=[ √ x] 3 1 = 0.732
Example29.2 Arandomvariable,z, has a p.d.f. f (z)where
f(z)=e −z
0z<∞
Calculate the probabilitythat
(a) 0z2
(b) z ismorethan1
(c) z islessthan0.5

938 Chapter 29 Statistics and probability distributions
Solution Note that ∫ ∞
0
e −z dz = 1 sothat f (z) = e −z issuitableas a p.d.f.
(a)P(0z2)= ∫ 2
0 e−z dz = [−e −z ] 2 0 = 0.865
(b) P(z >1) = ∫ ∞
1
e −z dz = [−e −z ] ∞ 1
= 0.368
(c) P(z < 0.5) = [−e −z ] 0.5
0
= 0.393
EXERCISES29.4
1 f(x) =kx 2 ,k constant, −1 x1. f(x)isap.d.f.
(a) What is the value ofk?
(b) Calculate the probability thatx > 0.5.
(c) IfP(x >c) = 0.6 then what isthe value ofc?
2 f (x)is ap.d.f. forthe random variablex,which can
varyfrom0to10.ItisillustratedinFigure29.5.What
is the probability thatxlies in [2,4]?
f(x)
0 10 x
Figure29.5
Probability densityfunctionforQuestion 2.
3 Ap.d.f. isgiven by
f(z)=2e −2z
0z<∞
(a) If200 measurements ofzare made, howmany,
onaverage, willbe greaterthan1?
(b) If50% ofmeasurementsare less thank,findk.
4 Ap.d.f.,h(x),is defined by
Calculate
h(x) = 3 4 (1−x2 )
(a) P(0 x 0.5)
(b) P(−0.3 x0.7)
(c) P(|x| < 0.5)
(d) P(x > 0.5)
(e) P(x 0.7)
5 (a) Verify that
f(t)=λe −λt t 0
issuitable asap.d.f.
(b) CalculateP(t 2) if λ = 3.
−1x1
Solutions
1 (a) 1.5 (b) 0.4375 (c) −0.5848
2 0.28
3 (a) 27 (b) 0.3466
4 (a) 0.3438 (b) 0.6575 (c) 0.6875
(d) 0.1563 (e) 0.9393
5 (b) 2.479 ×10 −3
29.5 MEANVALUE
If {x 1
,x 2
,x 3
,...,x n
} is a set ofnnumbers, then the mean value of these numbers, denotedbyx,is
x =
sum ofthe numbers
n
=
∑
xi
x issometimes called thearithmetic mean.
n

29.5 Mean value 939
Example29.3 Find the mean of −2.3, 0,1,0.7.
Solution
x = −2.3+0+1+0.7
4
= −0.6
4
= −0.15
The mean value is a single number which characterizes the set of numbers. It is useful
inhelpingtomake comparisons.
Engineeringapplication29.1
Comparisonoftwoproductionmachines
A component is made by two machines, A and B. The lifespans of six components
made byeachmachineare recordedinTable 29.4. Whichisthe preferred machine?
Table29.4
Thelifespans ofsixcomponents madeby machines AandB.
Lifespan(hours)
Machine A 92 86 61 70 58 65
Machine B 64 75 84 80 63 70
Solution
For componentsmanufacturedby machineA
average lifespan = 432
6 = 72
For components manufactured by machine B
average lifespan = 436
6 = 72.7
Machine B produces components with a higher average lifespan and so is the preferredmachine.
Example29.4 A variable, x, can have values 2, 3, 4, 5 and 6. Many observations of x are made and
denotedbyx i
. They have correspondingfrequencies, f i
. Theresults are asfollows:
value(x i ) 2 3 4 5 6
frequency (f i ) 6 9 3 7 4
Calculate the mean ofx.
Solution Thesumofallthemeasurementsmustbefound.Thevalue2occurssixtimes,contributing12
tothe total.Similarly, 3 occursninetimescontributing 27 tothe total.Thus
total =2(6) +3(9) +4(3) +5(7) +6(4) =110

940 Chapter 29 Statistics and probability distributions
The number of measurements made is6 +9+3+7+4 = 29. So,
mean =x= 110
29 = 3.79
Example 29.4 illustratesageneralprinciple.
If valuesx 1
,x 2
,...,x n
occur with frequencies f 1
, f 2
,..., f n
then
∑
xi f
x = ∑ i
fi
EXERCISES29.5
1 Engineersinadesigndepartmentareassessedbytheir
leader. A ‘0’is ‘Terrible’ anda‘5’is‘Outstanding’.
The29 members ofthe department are evaluated and
their scores recorded asfollows:
score 0 1 2 3 4 5
number ofstaff 2 5 6 9 4 3
Calculate the meanoutputvoltage.
4 Thecurrent, in amps,through aresistor is measured
140 times. The resultsare:
current(amps) 2.25 2.50 2.75 3.00 3.25
number of 32 27 39 22 20
measurements
Whatis the meanscore forthe wholedepartment?
2 Two samples ofnailsare taken.Thefirst sample has
12nailswith ameanlength of2.7cm, the second
sample has 20nailswith a meanlength of2.61 cm.
Whatis the meanlength ofall32nails?
3 Theoutput,in volts, from asystem is measured40
times. Theresults are recorded asfollows:
voltage(volts) 9.5 10.0 10.5 11.0 11.5
number of 6 14 8 7 5
measurements
Calculate the meancurrent through the resistor.
5 Inacommunication network, packets ofinformation
travel alonglines. The numberoflinesusedby each
packet variesaccording to the following table:
number oflinesused 1 2 3 4 5
number ofpackets 17 54 32 6 1
Calculate the meannumber oflinesusedperpacket.
Solutions
1 2.586
2 2.64
4 2.70
5 2.27
3 10.39

29.6 Standard deviation 941
29.6 STANDARDDEVIATION
Althoughthemeanindicateswherethecentreofasetofnumberslies,itgivesnomeasure
ofthespreadofthenumbers.Forexample, −1,0,1and −10,0,10bothhaveameanof
0 but clearly the numbers in the second set are much more widely dispersed than those
inthe first. Acommonly used measure of dispersion isthestandard deviation.
Let x 1
,x 2
,...,x n
be n measurements with a meanx. Then x i
−x is the amount by
whichx i
differs from the mean. The quantityx i
−x is called the deviation ofx i
from
the mean. Some of these deviations will be positive, some negative. The mean of these
deviations is always zero (see Question 3 in Exercises 29.6) and so this is not helpful
in measuring the dispersion of the numbers. To avoid positive and negative deviations
summing to zero the squared deviation is taken, (x i
−x) 2 . The variance is the mean of
the squared deviations:
variance =
∑ (xi −x) 2
n
and
standarddeviation = √ variance
Standard deviation has the same units as thex i
.
Example29.5 Calculate the standarddeviation of
(a) −1,0,1
(b) −10, 0,10
Solution (a) x 1
= −1,x 2
= 0,x 3
= 1.Clearlyx=0.
x 1
−x=−1 x 2
−x=0 x 3
−x = 1
variance = (−1)2 +0 2 +1 2
3
= 2 3
standard deviation =
√
2
3 = 0.816
(b) x 1
= −10,x 2
= 0,x 3
= 10. Againx=0andsox i
−x=x i
,fori=1,2,3.
variance = (−10)2 +0 2 +10 2
= 200
3 3
√
200
standard deviation =
3 = 8.165
As expected, the second set has a much higher standard deviation than the first.

942 Chapter 29 Statistics and probability distributions
Example29.6 Find the standard deviation of −2,7.2,6.9, −10.4, 5.3.
Solution x 1
= −2,x 2
= 7.2,x 3
= 6.9,x 4
= −10.4,x 5
= 5.3,x=1.4
x 1
−x = −3.4
x 2
−x=5.8
x 3
−x=5.5
x 4
−x = −11.8
x 5
−x=3.9
variance = (−3.4)2 + (5.8) 2 + (5.5) 2 + (−11.8) 2 + (3.9) 2
√ 5
229.9
standard deviation = = 6.78
5
= 229.9
5
Calculatingx i
−x,i = 1,2,...,n,istediousforlargenandsoamoretractableformof
the standard deviation is sought. Firstly observe that ∑ x i
=nx and ∑ x 2 =nx 2 , since
x 2 isaconstant. Now
∑
(xi −x) 2 = ∑ x 2 i
− ∑ 2x i
x + ∑ x 2
= ∑ x 2 i
−2x ∑ x i
+nx 2
= ∑ x 2 i
−2x(nx)+nx 2
= ∑ x 2 i
−nx 2
Hence,
and
variance =
∑ x
2
i
−nx 2
n
√ ∑x
2
i
−nx 2
standard deviation =
n
Using these formulae itisnot necessary tocalculatex i
−x.
Example29.7 RepeatExample 29.6 usingthe newly derived formulae.
Solution ∑ xi 2 = (−2) 2 + (7.2) 2 + (6.9) 2 + (−10.4) 2 + (5.3) 2 = 239.7
√
239.7 −5(1.4)
2
standard deviation =
= 6.78
5

29.7 Expected value of a random variable 943
EXERCISES29.6
1 Calculate the means andstandard deviationsof:
(a) 1, 2, 3, 4, 5
(b) 2.1, 2.3, 2.7, 2.6
(c) 37, 26,19, 21,19,25, 17
(d) 6, 6, 6, 6, 6, 6
(e) −1, 2, −3, 4, −5, 6
2 Asetofmeasurements {x 1 ,x 2 ,x 3 , . . . ,x n }hasamean
ofxandastandard deviation ofs.What are the mean
and standard deviation ofthe set
{kx 1 ,kx 2 ,kx 3 ,...,kx n }wherek isaconstant?
3 The meanofthe numbers {x 1 ,x 2 ,x 3 , . . . ,x n }isx.
Show that the sum ofthe deviations about the mean
is 0; that is,show ∑ n
i=1 (x i −x)=0.
Solutions
1 (a) mean = 3; st.dev. = 1.414
(b) 2.425,0.238
(c) 23.4,6.321
(d) 6,0
(e) 0.5,3.862
2 mean=kx,st.dev. =ks
29.7 EXPECTEDVALUEOFARANDOMVARIABLE
InSections 29.5and29.6 weshowedhowtocalculatethe mean andstandarddeviation
ofagivensetofnumbers.Noreferencewasmadetoprobabilitydistributionsorp.d.f.s.
Supposenowthatwehaveknowledgeoftheprobabilitydistributionofadiscreterandom
variable or the p.d.f. of a continuous random variable. The mean value of the random
variable can still be found. Under these circumstances the mean value is known as the
expectedvalueor expectation.
29.7.1 Expectedvalueofadiscreterandomvariable
Supposefordefinitenessthatxisadiscreterandomvariablewithprobabilitydistribution
asgiven inTable29.5.
Table29.5
Probabilitydistribution foradiscrete randomvariablex.
x 0 1 2 3 4
P(x) 0.1 0.2 0.4 0.15 0.15
In100trialsxwillhaveavalueofzero10timesonaverage,avalueofone20timeson
average and soon. Themean value, thatisthe expected value,istherefore
expected value =
0(10) +1(20) +2(40) +3(15) +4(15)
100
We could have arranged the calculation as follows:
expected value = 0 ( ) (
10
100 +1 20
) (
100 +2 40
100
)
+3
( 15
100
= 2.05
) (
+4
15
)
100
= 0(0.1) +1(0.2) +2(0.4) +3(0.15) +4(0.15)

944 Chapter 29 Statistics and probability distributions
Each termisof the form (value) ×(probability). Thus,
∑i=5
expected value = x i
P(x i
)
i=1
The symbol, µ, isused todenote the expected value of a random variable.
If a discrete random variable can take values
x 1
,x 2
,...,x n
with probabilitiesP(x 1
),P(x 2
),...,P(x n
), then
∑i=n
expected value ofx = µ = x i
P(x i
)
i=1
Example29.8 Arandom variable,y, has a known probability distributiongiven by
y 2 4 6 8 10
P(y) 0.17 0.23 0.2 0.3 0.1
Find the expected value ofy.
Solution We have
µ = expectedvalue = 2(0.17) +4(0.23) +6(0.2) +8(0.3) +10(0.1) = 5.86
29.7.2 Expectedvalueofacontinuousrandomvariable
Suppose a continuous random variable,x, has p.d.f. f (x),a x b. The probability
thatxliesinavery smallinterval, [x,x + δx], is
∫ x+δx
x
f(t)dt
Since the interval is very small, f will vary only slightly across the interval. Hence the
probability is approximately f (x)δx: see Figure 29.6. The contribution to the expected
value as a resultofthis interval is
(value) ×(probability)
that is,xf(x)δx. Summing all such termsyields
expected value = µ =
∫ b
a
xf(x)dx

29.7 Expected value of a random variable 945
f(x)
Area ≈ f(x).dx
x
x + dx
x
Figure29.6
The shaded area represents the probability thatxlies
in the smallinterval [x,x + δx].
Example29.9 Arandom variable has p.d.f. given by
f(x)= 1
2 √ x
1x4
Calculate the expected value ofx.
Solution µ =
∫ 4
1
[ x
3/2
=
3
x 1
2 √ x dx = ∫ 4
] 4
1
= 7 3
1
√ x
2 dx
So, if several values ofxare measured, the mean of these values will be near to 7 3 . As
more and more values are measured the mean will get nearer and nearer to 7 3 .
EXERCISES29.7
1 Calculatethe expected value ofthe discrete random
variable,h,whose probabilitydistribution is
h 1 1.5 1.7 2.1 3.2
P(h) 0.32 0.24 0.17 0.15 0.12
2 Calculatethe expected value ofthe random variable,
x,whose probabilitydistribution is
x 2 2.5 3.0 3.5 4.0 4.5
P(x) 0.07 0.36 0.21 0.19 0.10 0.07
4 A randomvariable,x,has p.d.f. f (x) given by
f(x)= 5
4x 2
1x5
(a) Calculate the expected value ofx.
(b) Ten values ofxare measured. They are
1.9, 2.9, 2.8, 2.1, 3.2, 3.4, 2.7, 2.3, 2.8, 2.7
Calculate the mean ofthe observations and
comment on yourfindings.
5 A p.d.f.h(x)is defined by
h(x) = 3 4 (1−x2 )
−1x1
Calculate the expected value ofx.
3 Arandom variable,z,has p.d.f. f (z) = e −z ,
0 z<∞. Calculatethe expected value ofz.
6 Is the expected value ofadiscrete random variable
necessarily oneofits possiblevalues?

946 Chapter 29 Statistics and probability distributions
Solutions
1 1.668
2 3.05
3 1
4 (a) 2.012 (b) 2.68
5 0
6 no
29.8 STANDARDDEVIATIONOFARANDOMVARIABLE
29.8.1 Standarddeviationofadiscreterandomvariable
RecallfromSection29.6thatthestandarddeviationofasetofnumbers, {x 1
,x 2
,...,x n
},
isgiven by
√∑ (xi −x)
standard deviation =
2
n
Nowsupposethatxisadiscreterandomvariablewhichcanhavevaluesx 1
,x 2
,x 3
,...,x n
with respective probabilities of p 1
,p 2
,p 3
,...,p n
; thatis, we have
x x 1 x 2 x 3 ... x n
P(x) p 1 p 2 p 3 ... p n
Let the expected value of x be µ. Then the square of the deviation from the expected
value has anidentical probabilitydistribution:
value (x 1 − µ) 2 (x 2 − µ) 2 ... (x n − µ) 2
probability p 1 p 2 . . . p n
Theexpectedvalueofthemeansquareddeviationisthevariance.Thesymbol σ 2 isused
todenotethe varianceofarandom variable:
variance = σ 2 =
n∑
p i
(x i
− µ) 2
1
As before the standarddeviation isthe squarerootofthe variance:
standard deviation = σ =
√ ∑pi
(x i
− µ) 2
Example29.10 Adiscrete randomvariable has probabilitydistribution
x 1 2 3 4 5
P(x) 0.12 0.15 0.23 0.3 0.2

29.8 Standard deviation of a random variable 947
Calculate
(a) the expected value
(b) the standard deviation
Solution (a) µ = ∑ x i
p i
= 1(0.12) +2(0.15) +3(0.23) +4(0.3) +5(0.2) = 3.31
(b) σ 2 = ∑ p i
(x i
− µ) 2
=0.12(1 −3.31) 2 +0.15(2 −3.31) 2 +0.23(3 −3.31) 2
+0.3(4 −3.31) 2 +0.2(5 −3.31) 2
=1.6339
standard deviation = σ = √ 1.6339 = 1.278
29.8.2 Standarddeviationofacontinuousrandomvariable
Wesimplystatetheformulaforthestandarddeviationofacontinuousrandomvariable.
It is analogous to the formula for the standard deviation of a discrete variable. Letxbe
a continuousrandom variablewith p.d.f. f (x),a xb. Then
√ ∫ b
σ =
a
(x − µ) 2 f(x)dx
Example29.11 Arandomvariable,x, has p.d.f. f (x)given by
f(x)=1
0x1
Calculate the standarddeviation ofx.
Solution Theexpected value, µ,is found:
∫ 1
[ x
2
µ = xf(x)dx= = 1 2 2
0
] 1
0
The variance can now befound:
∫ 1
(
variance = σ 2 = x − 1 2
1dx
2)
Hence
=
σ =
√
1
12 = 0.29
0
∫ 1
0
x 2 −x+ 1 4 dx
The standard deviation ofxis0.29.
[ x
3
=
3 − x2
2 4] + x 1
= 1
0
3 − 1 2 + 1 4 = 1
12

948 Chapter 29 Statistics and probability distributions
EXERCISES29.8
1 Ap.d.f.h(x)ofthe random variablexisdefined by
h(x) = 3 4 (1−x2 )
−1x1
(a) Calculate the expected value ofx.
(b) Calculate the standard deviation ofx.
2 Arandom variable,t,has p.d.f.H(t)given by
H(t)=3e −3t
t0
(a) Calculate the expected value oft.
(b) Calculate the standard deviation oft.
3 Adiscrete random variable, w,has a known
probability distribution
w −1 −0.5 0 0.5 1
P(w) 0.1 0.17 0.4 0.21 0.12
Calculate the standard deviation of w.
4 Adiscrete randomvariable,y,has a probability
distribution
y 6 7 8 9 10 11
P(y) 0.13 0.26 0.14 0.09 0.11 0.27
Calculate the standard deviation ofy.
5 Acontinuous random variablehas p.d.f.
{ x+1 −1x0
f(x)=
−x+1 0<x1
(a) Calculate the expected value ofx.
(b) Calculate the standard deviation ofx.
Solutions
1 (a) 0 (b) 0.4472
4 1.84
2 (a)
3 0.5598
1
3 (b)
1
3
5 (a) 0 (b) 0.4082
29.9 PERMUTATIONSANDCOMBINATIONS
Bothpermutationsandcombinationsareusedextensivelyinthecalculationofprobabilities.
29.9.1 Permutations
Thefollowing problem introducespermutations.
Engineeringapplication29.2
Linkingprocesscontrolcomputers
A primary and a secondary route must be chosen from available routes A, B and C,
for the linking of two process control computers in order to provide redundancy in
case one fails.Inhow many ways can the choicesbemade?

29.9 Permutations and combinations 949
Solution
The various possibilitiesare listed.
Primary route
A
B
A
C
B
C
Secondary route
B
A
C
A
C
B
There are six ways in which the choices can be made. Alternatively we could argue
as follows. Suppose the primary route is chosen first. There are three choices: any
oneofA,BorC.Thesecondaryrouteisthenchosenfromthetworemainingroutes,
givingtwopossiblechoices.Togetherthereare3×2 = 6waysofchoosingaprimary
route and a secondary route.
InEngineeringapplication29.2thechoiceABisdistinctfromthechoiceBA,thatisthe
order is important. Choosing two routes from three and arranging them in order is an
example of apermutation. More generally:
Apermutationofndistinctobjectstakenr atatimeisanarrangementofr ofthen
objects.
Informingpermutations,theorderoftheobjectsisimportant.Ifthreelettersarechosen
fromthealphabetthepermutationXYZisdistinctfromthepermutationZXY.Wepose
the question ‘How many permutations are there of n objects taken r at a time?’ The
following example will helptoestablish a formulafor the number ofpermutations.
Example29.12 Calculate the number of permutations thereare of
(a) four distinctobjects taken two atatime
(b) five distinctobjects taken threeatatime
(c) seven distinctobjects taken four atatime
Solution (a) Listing all possible permutations is not feasible when the numbers involved are
large. In choosing the first object, four choices are possible. In choosing the second
object, three choices are possible. There are thus 4 × 3 = 12 permutations of
four objects taken two atatime.Note that12 may be writtenas
12=4×3= 4!
2! = 4!
(4 −2)!
(b) There are five objects available for the first choice, four for the second choice and
threeforthethirdchoice.Hencethereare5×4×3 = 60permutations.Againnote
that
60 = 5!
2! = 5!
(5 −3)!

950 Chapter 29 Statistics and probability distributions
(c) Therearesevenobjectsavailableforthefirstchoice,sixforthesecond,fiveforthe
third and four for the fourth. The number of permutations is 7 ×6×5×4 = 840.
7!
This may bewritten as
(7 −4)! .
Theexample illustratesthe following generalrule.
Thenumberofpermutationsofndistinctobjectstakenratatime,writtenP(n,r),is
P(n,r) =
n!
(n−r)!
Example29.13 Findthe numberofpermutations of
(a) 10 distinctobjects takensixatatime
(b) 15 distinctobjects takentwo atatime
(c) sixdistinctobjects takensixatatime
Solution (a) P(10,6) = 10!
(10 −6)! = 10!
4! = 151200
15!
(b) P(15,2) =
(15 −2)! = 15!
13! = 210
6!
(c) P(6,6) =
(6 −6)! = 6! = 720 (Note that0! = 1)
0!
P(6,6) issimplythe number of ways of arranging all sixof the objects.
Note that
P(n,n) =n!
This isthe numberofways ofarrangingngiven objects.
29.9.2 Combinations
Closelyrelatedto,but nevertheless distinctfrom, permutations arecombinations.
Acombination isaselection ofrdistinctobjectsfromnobjects.
Inmakingaselectiontheorderisunimportant.Forexample,giventhelettersA,BandC,
ABandBAarethesamecombinationbutdifferentpermutations.Aswithpermutations
wedevelopanexpressionforthenumberofcombinationsofnobjectstakenratatime.
Examples29.14and 29.15helpwith thisdevelopment.

29.9 Permutations and combinations 951
Example29.14 There are three routes, A, B and C, joining two computers. In how many ways can two
routes be chosen from A,Band C?
Solution The possible combinations (selections) can be listedas
AB,BC, AC
thatis,therearethreepossiblewaysofmakingtheselection.Wecanalsouseourknowledge
of permutations to calculate the number of combinations. There areP(3,2) ways
of arranging the threeroutes taken two atatime.
P(3,2) =
3!
(3 −2)! = 6
EachcombinationoftworoutescanbearrangedinP(2,2)ways;thatis,eachcombinationgivesrisetotwopermutations.Forexample,thecombinationABcouldbearranged
as AB or BA, giving two permutations. Thus, the number of combinations is half the
number of permutations. There are 6/2 = 3 combinations.
Example29.15 Calculate the numberofcombinationsof
(a) sixdistinctobjectstaken fouratatime
(b) 10 distinctobjects takensixatatime
Solution (a) Consider one combination of four objects. These four objects can be arranged in
4! ways; that is, each combination gives rise to 4! permutations. The number of
permutations ofsixobjects takenfouratatime is
P(6,4) = 6!
2!
Hence the number of combinations is
P(6,4)
4!
= 6!
2!4! = 15
There are15 combinations of sixobjects taken four atatime.
(b) Eachcombination,comprisingsixobjects,givesriseto6!permutations.Thenumber
of permutations of 10 objects taken sixatatime is
P(10,6) = 10!
4!
Hence the number of combinations = P(10,6)
6!
= 10!
4!6! = 210.
( n
We write to denote the number of combinations ofnobjects takenrat a time. A
r)
( n
formulafor isnow developed.
r)

952 Chapter 29 Statistics and probability distributions
Each combination ofr objects gives rise tor! permutations, but
P(n,r) =
n!
(n−r)!
Thenumberofcombinationsofndistinctobjects takenratatime is
( n
=
r)
P(n,r) n!
=
r! (n −r)!r!
Example29.16 Calculate the numberofcombinationsof
(a) sixdistinctobjects takenfive atatime
(b) ninedistinctobjectstaken nineatatime
(c) 25 distinctobjects takenfive atatime
Solution (a)
( 6
5)
= 6! ( ) 9
1!5! =6 (b) = 9! ( ) 25
9 0!9! =1 (c) = 25!
5 5!20! = 53130
( n
We can generalize the resultofExample 29.16(b)and statethat = 1.
n)
Example29.17 There are k identical objects and n compartments (n k). Each compartment can
hold only one object. In how many different ways can thekobjects be placed in then
compartments?
Solution The order in which the objects are placed is unimportant since all the objects are identical.
Placing the k objects is identical to selecting k of the n compartments (see
(
Figure29.7).Butthenumberofwaysofselectingkcompartmentsfromnisprecisely
. n
(
k)
n
Hence thekobjects can beplacedinthencompartmentsin different ways.
k)
n compartments
k objects
Figure29.7
Placingkobjectsinncompartments.

29.10 The binomial distribution 953
EXERCISES29.9
1 Evaluate
(a) P(8,6)
( ) 12
(c)
9
(b) P(11,7)
( ) 15
(d)
12
2 Writeoutexplicitly
( ( n n
(a) (b)
0)
1)
(c)
(e)
( ) 15
3
( n
2)
3 Theexpansionof (a+b) n wherenisapositiveinteger
maybe written with the help ofcombination notation.
(
(a+b) n =a n n
+ a
1)
n−1 b
( n
+ a
2)
n−2 b 2 +···
( ) ( n
+ ab n−1 n
+ b
n −1 n)
n
n∑
( n
= a
k)
n−k b k
k=0
Expand
(a) (a +b) 4 (b) (1 +x) 6 (c) (p+q) 5
4 Primary and secondary routesconnecting two
computers need to be chosen. Two primaryroutesare
needed from eightwhich are suitable and three
secondary routesmust be chosen from four available.
Inhow many ways can the routesbe chosen?
5 A combination lock can be opened by diallingthree
correct lettersfollowed bythree correct digits. How
many different possibilitiesare there forarranging the
lettersand digits? Isthismore secure than alock
which has seven digits? Isthe word ‘combination’
being usedcorrectly?
6 A nuclear power station isto be builton oneof20
possiblesites.Ateamofengineersiscommissionedto
examine the sites andrank the three most favourable
in order. Inhow many ways can thisbe done?
Solutions
1 (a) 20160 (b) 1663200 (c) 220 (c)p 5 +5p 4 q+10p 3 q 2 +10p 2 q 3 +5pq 4 +q 5
(d) 455 (e) 455
4 112
n(n −1)
2 (a) 1 (b) n (c)
2 5 17576000. Yes,itis more secure. Combination is not
3 (a)a 4 +4a 3 b+6a 2 b 2 +4ab 3 +b 4
being usedcorrectly:permutation lockwould be
better.
(b)1 +6x +15x 2 +20x 3 +15x 4 +6x 5 +x 6 6 6840
29.10 THEBINOMIALDISTRIBUTION
In a single trial or experiment a particular result may or may not be obtained. For example,
in an examination, a student may pass or fail; when testing a component it may
workornotwork.Theimportantpointisthatthetwooutcomesarecomplementary.Assumingthattheprobabilityofanoutcomeisfixed,suchatrialiscalledaBernoullitrial
inhonour ofthe mathematician, J.Bernoulli.
Weaddressthefollowingproblem.InasingletrialtheoutcomeiseitherAorB,thatis
A.WerefertoAasasuccessandBasafailure.IfweknowP(A) = pthenP(B) = 1−p.
Ifnindependenttrialsareobserved,whatistheprobabilitythatAoccursktimes,andB
occursn −k times? The number of successful trials innsuch experiments is a discrete
randomvariable;suchavariableissaidtohaveabinomialdistribution.Letusconsider
a particular problem.

954 Chapter 29 Statistics and probability distributions
Engineeringapplication29.3
Samplingcomponentsmadebyamachine
Amachinemakescomponents.Theprobabilitythatacomponentisacceptableis0.9.
(a) If three components are sampled find the probability that the first is acceptable,
the second isacceptable and the thirdisnot acceptable.
(b) If three components are sampled what is the probability that exactly two are
acceptable?
Solution
Let the eventsAandBbe defined thus:
A:thecomponentis acceptable,P(A) = 0.9
B:thecomponentis notacceptable,P(B) = 0.1
(a) We denote byAAB the compound event that the first is acceptable, the second
is acceptable, the third is not acceptable. Since the three events are independent
the multiplication lawinSection 28.7 gives
P(AAB) =P(A)P(A)P(B) = (0.9)(0.9)(0.1) = (0.9) 2 (0.1) = 0.081
(b) Weareinterestedinthecompoundeventinwhichtwocomponentsareacceptable
and one is not acceptable. We denote byAAB the compound event that the first
is acceptable, the second is acceptable, the third is not acceptable. Compound
eventsABA andBAA have obvious interpretations.
If exactly two components are acceptable then either AAB or ABA or BAA
occurs.Thesecompoundeventsaremutuallyexclusiveandwecanthereforeuse
the addition law (see Section 28.3). Hence
P(exactly two acceptable components) =P(AAB) +P(ABA) +P(BAA)
From (a) P(AAB) = 0.081 and by similar reasoning P(ABA) = P(BAA) =
0.081. Hence,
P(exactly two acceptable components) = 3(0.081) = 0.243
29.10.1 Probabilityofksuccessesfromntrials
Let us now return to the general problem posed earlier. We define the compound
eventC:
C:Aoccursktimes andBoccursn −k times
( n
The k occurrences of event A can be distributed amongst the n trials in different
k)
ways (see Example 29.17). The probability of a particular distribution
(
ofkoccurrences
n
ofAandn−koccurrencesofBisp k (1−p) n−k .Sincethereare distinctdistributions
k)
possible then
P(C) =
( n
k)
p k (1−p) n−k
k=0,1,2,...,n

29.10 The binomial distribution 955
Example29.18 The probability that a component is acceptable is 0.93. Ten components are picked at
random. What is the probability that
(a) atleastnine areacceptable
(b) atmostthree areacceptable?
Solution (a) P(exactly 9 components are acceptable) =
Hence,
P(exactly 10 components areacceptable) =
( 10
9
)
(0.93) 9 (0.07) = 0.364
( ) 10
(0.93) 10 = 0.484
10
P(atleast9components areacceptable) = 0.364 +0.484 = 0.848
(b) We require the probability that none are acceptable, one is acceptable, two are acceptable
and threeare acceptable:
( ) 10
P(0are acceptable) = (0.93) 0 (0.07) 10 = 2.825 ×10 −12
0
( ) 10
P(1is acceptable) = (0.93) 1 (0.07) 9 = 3.753 ×10 −10
1
( ) 10
P(2are acceptable) = (0.93) 2 (0.07) 8 = 2.244 ×10 −8
2
( ) 10
P(3are acceptable) = (0.93) 3 (0.07) 7 = 7.949 ×10 −7
3
Hence,
P(atmost3are acceptable) = 2.825 ×10 −12 +3.753 ×10 −10
+2.244 ×10 −8 +7.949 ×10 −7
= 8.18 ×10 −7
that is, the probability that at most three components are acceptable is almost zero;
itisvirtually impossible.
29.10.2 Meanandstandarddeviationofabinomialdistribution
Let the probability of success in a single trial be p and let the number of trials be n.
The number of successes in n trials is a discrete random variable, x, with a binomial
distribution. Thenxcan have any value from {0,1,2,3,...,n}, although clearly some
valuesaremorelikelytooccurthanothers.Theexpectedvalueofxcanbeshowntobe
np. Thus,ifmany values ofxarerecorded, the mean of these will approachnp.
Expected value of the binomial distribution =np
The standard deviation of the binomial distributioncan alsobefound. This isgiven by
standard deviation of the binomial distribution = √ np(1−p)

956 Chapter 29 Statistics and probability distributions
29.10.3 Mostlikelynumberofsuccesses
When conducting a series of trials it is sometimes desirable to know the most likely
outcome. For example, what is the most likely number of acceptable components in a
sample of five tested?
Example29.19 Theprobabilityacomponentisacceptableis0.8.Fivecomponentsarepickedatrandom.
What isthe mostlikely number of acceptable components?
( 5
Solution P(no acceptable components) = (0.8)
0)
0 (0.2) 5 = 3.2 ×10 −4
( 5
P(1 acceptable component) = (0.8)
1)
1 (0.2) 4 = 6.4 ×10 −3
( 5
P(2 acceptable components) = (0.8)
2)
2 (0.2) 3 = 0.0512
( 5
P(3 acceptable components) = (0.8)
3)
3 (0.2) 2 = 0.2048
( 5
P(4 acceptable components) = (0.8)
4)
4 (0.2) 1 = 0.4096
( 5
P(5 acceptable components) = (0.8)
5)
5 (0.2) 0 = 0.3277
The mostlikely number of acceptable components isfour.
Example 29.19 illustrates an important general result. Suppose we conductnBernoulli
trialsandwishtofindthemostlikelynumberofsuccesses.Ifp =probabilityofsuccess
on a single trial,andi =mostlikely number ofsuccesses inntrials,then
p(n+1)−1<i<p(n+1)
InExample 29.19, p = 0.8,n = 5 and so
(0.8)(6) −1 <i < (0.8)(6)
3.8<i<4.8
Sinceiisan integer, theni = 4.
EXERCISES29.10
1 Theprobabilityacomponentisacceptableis0.8.Four
components are sampled. What is the probability that
(a) exactly one is acceptable
(b) exactly two are acceptable?
2 Amachine requires allseven ofitsmicro-chips to
operate correctly in order to be acceptable. The
probability amicro-chip isoperating correctly is 0.99.
(a) What isthe probability the machine is
acceptable?
(b) What isthe probability that sixofthe seven chips
are operating correctly?
(c) Themachine is redesigned sothat the original
seven chips are replaced by four new chips.The
probability a new chipoperates correctly is0.98.

29.11 The Poisson distribution 957
Isthe new designmore orlessreliable than the
original?
3 Theprobability a machine has a lifespanofmore than
5 years is0.8. Ten machines are chosen atrandom.
What isthe probability that
(a) eightmachines have a lifespanofmore than 5
years
(b) allmachines have alifespan ofmore than 5 years
(c) atleast eight machines have alifespanofmore
than5years
(d) nomore than two machines have alifespanof
lessthan 5 years?
4 Theprobability a valve remainsreliable formore than
10years is 0.75.Eightvalves are sampled. What is
the most likely number ofvalvesto remain reliable
formore than 10years?
5 The probability a chipis manufactured to an
acceptable standard is 0.87.Asampleofsixchipsis
picked at random from a large batch.
(a) Calculate the probability all sixchipsare
acceptable.
(b) Calculate the probability none ofthe chipsis
acceptable.
(c) Calculate the probability that fewer than five
chips in the sampleare acceptable.
(d) Calculate the mostlikely number ofacceptable
chips in the sample.
(e) Calculate the probability that more than two
chips are unacceptable.
Solutions
1 (a) 0.0256 (b) 0.1536
2 (a) 0.9321 (b) 0.0659 (c) 0.9224.
New designis lessreliable
3 (a) 0.3020 (b) 0.1074 (c) 0.678 (d) 0.678
4 6
5 (a) 0.4336 (b) 4.826 ×10 −6 (c) 0.1776 (d) 6
(e) 0.0324
29.11 THEPOISSONDISTRIBUTION
ThePoissondistributionmodelsthenumberofoccurrencesofaneventinagiveninterval.
Consider the number of emergency calls received by a service engineer in one day.
Wemayknowfromexperiencethatthenumberofcallsisusuallythreeorfourperday,
butoccasionallyitwillbeonlyoneortwo,orevennone,andonsomedaysitmaybesix
or seven, or even more. This example suggests a need for assigning a probability to the
numberofoccurrencesofaneventduringagiventimeperiod.ThePoissondistribution
serves this purpose.
Thenumberofoccurrencesofanevent,E,inagiventimeperiodisadiscreterandom
variable which we denote by X. We wish to find the probability that X = 0, X = 1,
X = 2, X = 3, and so on. Suppose the occurrence of E in any time interval is not
affectedbyitsoccurrenceinanyprecedingtimeinterval.Forexample,acarisnotmore,
or less, likely to pass a given spot in the next 10 seconds because a car passed (or did
notpass) the spot inthe previous 10 seconds, thatisthe occurrences areindependent.
Let λ be the expected (mean) value ofX, the number of occurrences during the time
period. IfX is measured for many time periods the average value ofX will be λ. Under
thegivenconditionsX followsaPoissondistribution.TheprobabilitythatX hasavalue
r isgiven by
P(X=r)= e−λ λ r
r!
r=0,1,2,...

958 Chapter 29 Statistics and probability distributions
The expected value and variance of the Poisson distributionareboth equal to λ.
Engineeringapplication29.4
Emergencycallstoaserviceengineer
Recordsshowthatonaveragethreeemergencycallsperdayarereceivedbyaservice
engineer. What isthe probability thaton a particular day
(a) three (b) two (c) four
calls will be received?
Solution
Thenumber ofcallsreceived follows aPoissondistribution.Theaverage number of
calls isthreeper day, thatis λ = 3.
(a) P(X =3) = e−3 3 3
= 0.224 (b) P(X = 2) = e−3 3 2
= 0.224
3!
2!
(c) P(X =4) = e−3 3 4
= 0.168
4!
The engineer will receive three calls on approximately 22 days in 100, two calls on
approximately 22 days in100 and four calls on approximately 17 days in100.
Engineeringapplication29.5
Machinebreakdownsinaworkshop
A workshop has several machines. During a typical month two machines will break
down. What arethe probabilitiesthatinamonth
(a) none (b) one (c) morethantwo
will break down?
Solution
λ = average number ofmachinesthat break down = 2
X = numberofmachines brokendown
(a) P(X =0) = e−2 2 0
= 0.135
0!
(b) P(X =1) = e−2 2 1
= 0.271
1!
(c) P(X >2)=1−P(X =0)−P(X =1)−P(X =2)
=1−e −2 −2e −2 −2e −2 =0.323

Table29.6
Theprobabilitiesforbinomial and Poissondistributions.
29.11 The Poisson distribution 959
Binomial (n,p) Poisson (λ)
P(X =r);n =15,p=0.05 P(X =r); λ =0.75
r = 0 0.46339 0.47237
r = 1 0.36576 0.35427
r = 2 0.13475 0.13285
r = 3 0.03073 0.03321
r = 4 0.00485 0.00623
r = 5 0.00056 0.00090
r = 6 0.00005 0.00007
r = 7 0.00000 0.00001
29.11.1 Poissonapproximationtothebinomial
ThePoissonandbinomialdistributionsarerelated.Consider abinomialdistribution,in
whichntrialstakeplaceandtheprobabilityofsuccessisp.Ifnincreasesandpdecreases
such that np is constant, the resulting binomial distribution can be approximated by a
Poisson distribution with λ = np. Recall thatnp is the expected value of the binomial
distribution,and λisthe expected value of the Poisson distribution.
Toillustratetheabovepoint,Table29.6liststheprobabilitiesforbinomialandPoissondistributionswithn
= 15,p= 0.05andhence λ = 15(0.05) = 0.75.Theremaining
probabilities are all almost 0.
Asnincreases and p decreases withnp remaining constant, agreement between the
two distributions becomes closer.
Engineeringapplication29.6
Workforceabsentees
A workforce comprises 250 people. The probability a person is absent on any one
day is0.02. Find the probability thaton a day
(a) three
people are absent.
(b) seven
Solution
This problem may be treated either as a sequence of Bernoulli trials or as a Poisson
process.
Bernoullitrials
The probabilities followabinomial distribution.
E:apersonis absent
n =number oftrials = 250
p = probability thatE occurs inasingletrial =0.02
X = numberof occurrencesof eventE
➔

960 Chapter 29 Statistics and probability distributions
( 250
(a) P(X =3) =
3
( 250
(b) P(X =7) =
7
Poissonprocess
)
(0.02) 3 (0.98) 247 = 0.140
)
(0.02) 7 (0.98) 243 = 0.105
Sincenislargeand pissmallthePoissondistributionwillbeagoodapproximation
tothe binomial distribution
λ=np=5
(a) P(X =3) = e−5 (5) 3
= 0.140 (b) P(X = 7) = e−5 (5) 7
= 0.104
3!
7!
EXERCISES29.11
1 Acomputer network has several hundredcomputers.
During an 8 hour period,thereare on averageseven
computers notfunctioning. Find the probabilitythat
duringan 8 hour period
(a) nine (b) five
donot function.
2 Aworkforce has on averagetwo people absent
through illnessonany given day. Find the probability
that onatypicalday
(a) two
(b) atleast three
(c) less thanfour
people are absent.
3 Amachine manufactures 300 micro-chips perhour.
Theprobability an individual chipis faultyis 0.01.
Calculate the probabilitythat
(a) two
(b) four
(c) more thanthree
faultychipsaremanufacturedinaparticularhour.Use
both the binomial andPoisson approximations and
comparethe resulting probabilities.
4 Theprobability ofadisk drive failurein any weekis
0.007. Acomputer servicecompany maintains900
disk drives.Usethe Poisson distribution to calculate
the probability of
(a) seven (b) more thanseven
disk drivefailures in a week.
5 Theprobability an employee fails to cometo work is
0.017. Alarge engineeringfirm employs650 people.
Whatisthe probability thatonaparticularday
(a) nine (b) 10
people are away from work?
6 Amachine manufactures electrical components for
the car industryatthe rate of750 per hour.The
probability acomponentis faultyis 0.013. Useboth
the binomialdistribution andthe corresponding
Poisson approximation to findthe probabilitythat in a
sampleof200 components
(a) none are faulty
(b) oneisfaulty
(c) two are faulty
(d) three are faulty
(e) more thanthree are faulty
Solutions
1 (a) 0.1014 (b) 0.1277
2 (a) 0.2707 (b) 0.3233
(c) 0.8571
3 (a) Binomial0.2244; Poisson 0.2240
(b) 0.1689,0.1680
(c) 0.353,0.353
4 (a) 0.1435 (b) 0.2983

29.12 The uniform distribution 961
5 (a) 0.1075 (b) 0.1188
6 (a) Binomial0.0730; Poisson 0.0743
(b) 0.1923,0.1931
(c) 0.2521, 0.2510
(d) 0.2191, 0.2176
(e) 0.2634, 0.2640
29.12 THEUNIFORMDISTRIBUTION
Wenowconsideracontinuousdistribution--theuniformdistribution.Supposetheprobability
of an event occurring remains constant across a given time interval. The p.d.f.,
f (t),of such a distributiontakes the form shown inFigure 29.8.
f(t)
1 –
T
0
Area = 1
T
t
Figure29.8
Theuniform p.d.f.
The area under f (t) must equal 1 and so if the interval is of lengthT, the height of
the rectangle is 1 T .
Thep.d.f. forthe uniformdistributionisgiven by
⎧
⎨ 1
0<t<T
f(t)= T
⎩
0 otherwise
EXERCISES29.12
The probability an event occurs in an interval [a,b] is ∫ b
a f(t)dt.If0abT
thisprobabilityissimply b −a
T .WeshallmakeuseofthisdistributioninSection29.15
when we deal with reliabilityengineering.
1 Arandom variable,x,has a uniform p.d.f.with
T = 10.Calculate the probability that
(a) 1x3 (b) 1.6x9.3
(c) x 2.9 (d) x < 7.2
(e) −1<x<2
(f) 9.1<x<12.3
2 A randomvariablet has a uniform p.d.f.with
T = 1.5. Calculate the probability that
(a) 0.7t1.3 (b) 1<t<2
(c) |t| <0.5 (d) |t| >1
Solutions
1 (a) 0.2 (b) 0.77 (c) 0.71 (d) 0.72
(e) 0.2 (f) 0.09
2 (a) 0.4 (b) 0.3333 (c) 0.3333
(d) 0.3333

962 Chapter 29 Statistics and probability distributions
29.13 THEEXPONENTIALDISTRIBUTION
SupposearandomvariablehasaPoissondistribution;forexample,therandomvariable
could be the number of customers arriving at a service point, the number of telephone
callsreceivedataswitchboard,thenumberofmachinesbreakingdowninaweek.Then
the time between events happening is a random variable which follows an exponential
distribution. Note that whereas the number of events is a discrete variable, the time
between events isacontinuous variable.
Lett be the time between events happening.
The exponential p.d.f. f (t)isgiven by
{
αe
−αt
t 0
f(t)=
0 otherwise
where α > 0.
The probability of an event occurring in a time interval, T, is given by ∫ T
f(t)dt.
0
Figure29.9shows f (t)forvariousvaluesof α.Theexpectedvalueofthedistributionis
given, by definition, as
expected value = µ =
For example, if
∫ ∞
0
tαe −αt dt = 1 α
f(t) =3e −3t tinseconds t 0
then the mean time between events is 1 3
happening per second.
s; that is, on average there are three events
f(t)
a increasing
t
Figure29.9
Theexponential p.d.f.forvariousvalues of α.
Engineeringapplication29.7
Timebetweenbreakdownsofamachine
Thetimebetweenbreakdownsofaparticularmachinefollowsanexponentialdistribution,withameanof17days.Calculatetheprobabilitythatamachinebreaksdown
ina15 day period.

29.14 The normal distribution 963
Solution
Themeantimebetweenbreakdowns = 17 = 1 α ,soα= 1 .Thusthep.d.f., f (t),is
17
given by
f(t)= 1 17 e−t/17 t 0
We require the probability thatthe machine breaks down ina15 day period:
∫ 15
P(0t15)= f(t)dt
0
∫ 15
1
=
0 17 e−t/17 dt
] 15
=
[−e −t/17
0
= −e −15/17 +1
= 0.5862
There isa58.62% chance thatthe machine will break down ina15 day period.
EXERCISES29.13
1 Aservice engineer receives onaveragean emergency
callevery 3 hours. If the time betweencalls follows
an exponential distribution, calculatethe probability
thatthe time from oneemergency callto the next is
(a) greaterthan3hours
(b) less than4.5hours
2 Themeantimebetweenbreakdownsforacertaintype
ofmachineis400hours.Calculatetheprobabilitythat
the time between breakdowns foraparticular
machine is
(a) greater than450 hours
(b) less than350 hours
3 The meantime taken byan engineerto repair an
electrical faultin asystem is 2.7hours.Calculate the
probability thatthe engineerwillrepair afaultin less
thanthe meantime.
Solutions
1 (a) 0.3679 (b) 0.7769
3 0.6321
2 (a) 0.3247 (b) 0.5831
29.14 THENORMALDISTRIBUTION
The normal probability density function, commonly called the normal distribution, is
one of the most important and widely used. It is used to calculate the probable values
of continuous variables, for example weight, length, density, error measurement. Probabilities
calculated using the normal distribution have been shown to reflect accurately
thosewhichwould befound usingactual data.

964 Chapter 29 Statistics and probability distributions
N(x)
N(x)
N(x) (a) (b)
(a)
m
x
(b)
m
x
m 1 m 2 x
Figure29.10
Two typical normal curves.
Figure29.11
Two normalcurveswith the same
standarddeviationbutdifferentmeans.
Letxbeacontinuous random variable with a normal distribution,N(x). Then
N(x) = 1
σ √ 2π e−(x−µ)2 /2σ 2
−∞<x<∞
where µ =expected(mean)valueofx, σ =standarddeviationofx.Figure29.10shows
two typical normal curves. All normal distributions are bell shaped and symmetrical
about µ.
In Figure 29.10(a) the values ofxare grouped very closely to the mean. Such a distribution
has a low standard deviation. Conversely, in Figure 29.10(b) the values of the
variable are spread widely about the mean and so the distribution has a high standard
deviation.
Figure29.11showstwonormaldistributions.Theyhavethesamestandarddeviation
butdifferentmeans.ThemeanofthedistributioninFigure29.11(a)is µ 1
whilethemean
of that in Figure 29.11(b) is µ 2
. Note that the domain ofN(x) is (−∞,∞); that is, the
domain is all real numbers. As for all distribution curves the total area under the curve
is1.
29.14.1 Thestandardnormal
A normal distribution is determined uniquely by specifying the mean and standard deviation.
The probability thatxliesinthe interval [a,b] is
P(axb)=
∫ b
a
N(x)dx
Themathematicalformofthenormaldistributionmakesanalyticintegrationimpossible,
so all probabilities must be computed numerically. As these numerical values would
change every time the value of µ or σ was altered some standardization is required. To
this end we introduce the standard normal. The standard normal has a mean of 0 and
a standard deviation of 1.
Considertheprobabilitythattherandomvariable,x,hasavaluelessthanz.Forconvenience
wecall thisA(z).
A(z) =P(x <z) =
∫ z
−∞
N(x)dx
Figure29.12illustratesA(z).ValuesofA(z)havebeencomputednumericallyandtabulated.TheyaregiveninTable29.7.Usingthetableandthesymmetricalpropertyofthe
distribution,probabilities can becalculated.

29.14 The normal distribution 965
N(x)
A(z)
0
z
x
Figure29.12
A(z) =P(x <z) = ∫ z
−∞ N(x)dx.
Example29.20 Thecontinuousrandomvariablexhasastandardnormaldistribution.Calculatetheprobability
that
(a) x < 1.2 (b) x > 1.2 (c) x > −1.2 (d) x < −1.2
Solution (a) From Table 29.7
P(x < 1.2) = 0.8849
This isdepicted inFigure 29.13.
(b) P(x > 1.2) = 1 −0.8849 = 0.1151
This isshown inFigure 29.14.
(c) By symmetryP(x > −1.2) isidentical toP(x < 1.2) (see Figure 29.15). So,
P(x > −1.2) = 0.8849
(d) Using part(c)we find
P(x < −1.2) = 1 −P(x > −1.2) = 0.1151
(see Figure 29.16).
N(x)
N(x)
Figure29.13
P(x < 1.2) = 0.8849.
0 1.2 x
Figure29.14
P(x > 1.2) = 0.1151.
0 1.2 x
N(x)
N(x)
–1.2 0
x
Figure29.15
P(x > −1.2) = 0.8849.
–1.2 0
x
Figure29.16
P(x < −1.2) = 0.1151.

966 Chapter 29 Statistics and probability distributions
Example29.21 Thecontinuousrandomvariable vhasastandardnormaldistribution.Calculatetheprobability
that
(a)0<v<1 (b) −1<v<1 (c) −0.5v2
Solution (a) Figure 29.17 shows the area (probability) required.
P(v < 1) = 0.8413 using Table 29.7
P(v < 0) = 0.5
using symmetry
P(0 < v < 1) = 0.8413 −0.5 = 0.3413
(b) Figure 29.18 shows the area (probability) required.
P(−1 < v <1) = 2 ×P(0 < v <1)
= 2 ×0.3413 = 0.6826
usingsymmetry
This tells us that 68.3% of the values of v are within one standard deviation of the
mean.
(c) Figure 29.19 shows the area (probability) required.
P(v 2) = 0.9772
P(v −0.5) =P(v >0.5) =1−P(v <0.5)
= 1 −0.6915 = 0.3085
P(−0.5 v 2) = 0.9772 −0.3085 = 0.6687
Note that whether or not inequalities defining v are strict is of no consequence in
calculating the probabilities.
N(v)
N(y)
N(y)
Figure29.17
P(0 < v < 1) = 0.3413.
0 1 v
–1 0 1 y
Figure29.18
P(−1 < v < 1) = 0.6826.
–0.5 0 2 y
Figure29.19
P(−0.5 v 2) = 0.6687.
29.14.2 Non-standardnormal
Table 29.7 allows us to calculate probabilities for a random variable with a standard
normal distribution. This section show us how to use the same table when the variable
has a non-standard distribution. A non-standard normal has a mean value other than 0
and/or a standard deviation other than 1. The non-standard normal is changed into a
standard normal by application of a simple rule. Suppose the non-standard distribution
hasamean µandastandarddeviation σ.Thenallnon-standardvaluesaretransformed
tostandard values using
non-standard → standard
X → X − µ
σ

29.14 The normal distribution 967
Table29.7
Cumulative normal probabilities.
z A(z) z A(z) z A(z) z A(z) z A(z) z A(z)
0.00 0.5000000 0.40 0.6554217 0.80 0.7881446 1.20 0.8849303 1.60 0.9452007 2.00 0.9772499
0.01 0.5039894 0.41 0.6590970 0.81 0.7910299 1.21 0.8868606 1.61 0.9463011 2.01 0.9777844
0.02 0.5079783 0.42 0.6627573 0.82 0.7938919 1.22 0.8887676 1.62 0.9473839 2.02 0.9783083
0.03 0.5119665 0.43 0.6664022 0.83 0.7967306 1.23 0.8906514 1.63 0.9484493 2.03 0.9788217
0.04 0.5159534 0.44 0.6700314 0.84 0.7995458 1.24 0.8925123 1.64 0.9494974 2.04 0.9793248
0.05 0.5199388 0.45 0.6736448 0.85 0.8023375 1.25 0.8943502 1.65 0.9505285 2.05 0.9798178
0.06 0.5239222 0.46 0.6772419 0.86 0.8051055 1.26 0.8961653 1.66 0.9515428 2.06 0.9803007
0.07 0.5279032 0.47 0.6808225 0.87 0.8078498 1.27 0.8979577 1.67 0.9525403 2.07 0.9807738
0.08 0.5318814 0.48 0.6843863 0.88 0.8105703 1.28 0.8997274 1.68 0.9535213 2.08 0.9812372
0.09 0.5358564 0.49 0.6879331 0.89 0.8132671 1.29 0.9014747 1.69 0.9544860 2.09 0.9816911
0.10 0.5398278 0.50 0.6914625 0.90 0.8159399 1.30 0.9031995 1.70 0.9554345 2.10 0.9821356
0.11 0.5437953 0.51 0.6949743 0.91 0.8185887 1.31 0.9049021 1.71 0.9563671 2.11 0.9825708
0.12 0.5477584 0.52 0.6984682 0.92 0.8212136 1.32 0.9065825 1.72 0.9572838 2.12 0.9829970
0.13 0.5517168 0.53 0.7019440 0.93 0.8238145 1.33 0.9082409 1.73 0.9581849 2.13 0.9834142
0.14 0.5556700 0.54 0.7054015 0.94 0.8263912 1.34 0.9098773 1.74 0.9590705 2.14 0.9838226
0.15 0.5596177 0.55 0.7088403 0.95 0.8289439 1.35 0.9114920 1.75 0.9599408 2.15 0.9842224
0.16 0.5635595 0.56 0.7122603 0.96 0.8314724 1.36 0.9130850 1.76 0.9607961 2.16 0.9846137
0.17 0.5674949 0.57 0.7156612 0.97 0.8339768 1.37 0.9146565 1.77 0.9616364 2.17 0.9849966
0.18 0.5714237 0.58 0.7190427 0.98 0.8364569 1.38 0.9162067 1.78 0.9624620 2.18 0.9853713
0.19 0.5753454 0.59 0.7224047 0.99 0.8389129 1.39 0.9177356 1.79 0.9632730 2.19 0.9857379
0.20 0.5792597 0.60 0.7257469 1.00 0.8413447 1.40 0.9192433 1.80 0.9640697 2.20 0.9860966
0.21 0.5831662 0.61 0.7290691 1.01 0.8437524 1.41 0.9207302 1.81 0.9648521 2.21 0.9864474
0.22 0.5870604 0.62 0.7323711 1.02 0.8461358 1.42 0.9221962 1.82 0.9656205 2.22 0.9867906
0.23 0.5909541 0.63 0.7356527 1.03 0.8484950 1.43 0.9236415 1.83 0.9663750 2.23 0.9871263
0.24 0.5948349 0.64 0.7389137 1.04 0.8508300 1.44 0.9250663 1.84 0.9671159 2.24 0.9874545
0.25 0.5987063 0.65 0.7421539 1.05 0.8531409 1.45 0.9264707 1.85 0.9678432 2.25 0.9877755
0.26 0.6025681 0.66 0.7453731 1.06 0.8554277 1.46 0.9278550 1.86 0.9685572 2.26 0.9880894
0.27 0.6064199 0.67 0.7485711 1.07 0.8576903 1.47 0.9292191 1.87 0.9692581 2.27 0.9883962
0.28 0.6102612 0.68 0.7517478 1.08 0.8599289 1.48 0.9305634 1.88 0.9699460 2.28 0.9886962
0.29 0.6140919 0.69 0.7549029 1.09 0.8621434 1.49 0.9318879 1.89 0.9706210 2.29 0.9889893
0.30 0.6179114 0.70 0.7580363 1.10 0.8643339 1.50 0.9331928 1.90 0.9712834 2.30 0.9892759
0.31 0.6217195 0.71 0.7611479 1.11 0.8665005 1.51 0.9344783 1.91 0.9719334 2.31 0.9895559
0.32 0.6255158 0.72 0.7642375 1.12 0.8686431 1.52 0.9357445 1.92 0.9725711 2.32 0.9898296
0.33 0.6293000 0.73 0.7673049 1.13 0.8707619 1.53 0.9369916 1.93 0.9731966 2.33 0.9900969
0.34 0.6330717 0.74 0.7703500 1.14 0.8728568 1.54 0.9382198 1.94 0.9738102 2.34 0.9903581
0.35 0.6368307 0.75 0.7733726 1.15 0.8749281 1.55 0.9394292 1.95 0.9744119 2.35 0.9906133
0.36 0.6405764 0.76 0.7763727 1.16 0.8769756 1.56 0.9406201 1.96 0.9750021 2.36 0.9908625
0.37 0.6443088 0.77 0.7793501 1.17 0.8789995 1.57 0.9417924 1.97 0.9755808 2.37 0.9911060
0.38 0.6480273 0.78 0.7823046 1.18 0.8809999 1.58 0.9429466 1.98 0.9761482 2.38 0.9913437
0.39 0.6517317 0.79 0.7852361 1.19 0.8829768 1.59 0.9440826 1.99 0.9767045 2.39 0.9915758
(continued)

968 Chapter 29 Statistics and probability distributions
Table29.7
Cumulativenormal probabilities (continued).
z A(z) z A(z) z A(z) z A(z) z A(z) z A(z)
2.40 0.9918025 2.45 0.9928572 2.50 0.9937903 2.55 0.9946139 2.60 0.9953383 3.20 0.9993129
2.41 0.9920237 2.46 0.9930531 2.51 0.9939634 2.56 0.9947664 2.70 0.9965330 3.40 0.9996631
2.42 0.9922397 2.47 0.9932443 2.52 0.9941323 2.57 0.9949151 2.80 0.9974449 3.60 0.9998409
2.43 0.9924506 2.48 0.9934309 2.53 0.9942966 2.58 0.9950600 2.90 0.9981342 3.80 0.9999277
2.44 0.9926564 2.49 0.9936128 2.54 0.9944574 2.59 0.9952012 3.00 0.9986501 4.00 0.9999683
4.50 0.9999966
5.00 0.9999997
5.50 0.9999999
This table is condensed from Table 1 of theBiometrikaTablesforStatisticians, Vol. 1 (1st ed.), edited by E. S. Pearson
andH. O. Hartley. Reproduced with the kind permissionofE. S.Pearsonand the trusteesofBiometrika.
Source:StatisticsVol.1ProbabilityInferenceandDecisionbyHays,W.L.andWinkler,R.L.(HoltRinehart&Winston,
New York, 1970).
Example29.22 A random variable,h, has a normal distribution with mean 7 and standard deviation 2.
Calculate the probability that
(a)h>9 (b)h<6 (c)5<h<8
Solution (a) Applying the transformation gives
9 → 9 −7 = 1
2
Soh > 9 has the same probability asx > 1, wherexis a random variable with a
standard normal distribution:
P(h >9) =P(x >1) =1−P(x <1) =1−0.8413 =0.1587
(b) Applying the transformation gives
6 → 6 −7 = −0.5
2
Soh < 6 has the same probability asx < −0.5:
P(x < −0.5) =P(x > 0.5) = 1 −P(x < 0.5) = 0.3085
(c) Applying the transformation to5and 8 gives
5 → 5 −7 =−1 8→ 8 −7 = 0.5
2 2
and so werequireP(−1 <x<0.5). Therefore
and then
P(x < 0.5) = 0.6915 P(x < −1) = 0.1587
P(−1 <x<0.5) = 0.6915 −0.1587 = 0.5328

29.14 The normal distribution 969
EXERCISES29.14
1 Arandom variable,x,has a standard normal
distribution. Calculate the probability thatxlies in the
followingintervals:
(a) (0.25, 0.75)
(b) (−0.3,0.1)
(c) within 1.5standard deviationsofthe mean
(d) more thantwo standard deviations from the mean
(e) (−1.7, −0.2)
2 Arandom variable,x,has a normal distribution with
mean 4 andstandard deviation 0.8. Calculate the
probability that
(a) 3.0 x4.4
(b) 2.5 <x<3.9
(c) x > 4.6
(d) x < 4.2
(e) x is within 0.6ofthe mean
3 Arandom variable,t,has anormal distribution with
mean 1 andstandard deviation 2.5. Calculate the
probability that
(a) −1t2 (b)t>0
(c) |t| 0.9 (d) |t| > 1.6
4 Thescores fromIQ tests have amean of100 anda
standard deviation of15. What should aperson score
in order to be described asin the top10% ofthe
population?
5 Amachine produces car pistons.Thediameter ofthe
pistonsfollows anormal distribution, mean 6.04 cm
with astandard deviation of0.02 cm. Thepistonis
acceptable ifitsdiameter is in the range 6.010 cm to
6.055cm. What percentage ofpistonsis
acceptable?
6 The random variable,x, has anormal distribution.
How many standard deviations above the mean must
the pointPbe placed ifthe tail-end isto represent
(a) 10% (b) 5% (c) 1%
ofthe total area? (See Figure29.20.)
N(x)
? P
Figure29.20
Graph forQuestion6.
7 ConsiderFigure29.21. Thetwo tail-endshave equal
area. How many standard deviations from the mean
mustAandBbe placed ifthe tail-ends are
(a) 10% (b) 5% (c) 1%
ofthe total area?
A
N(x)
Figure29.21
Graph forQuestion7.
B
x
x
Solutions
1 (a) 0.1747 (b) 0.1577 (c) 0.8664
(d) 0.0455 (e) 0.3762
2 (a) 0.5858 (b) 0.4199 (c) 0.2266
(d) 0.5987 (e) 0.5467
3 (a) 0.4436 (b) 0.6554 (c) 0.2604
(d) 0.5544
4 119
5 71%
6 (a)1.28 (b)1.645 (c)2.33
7 (a)1.64 (b)1.96 (c)2.57

970 Chapter 29 Statistics and probability distributions
29.15 RELIABILITYENGINEERING
Reliability engineering is an important area of study. Unreliable products lead to human
frustration, financial loss and in the case of life-critical systems can lead to death.
As the complexity of engineering systems has increased, mathematical methods of assessing
reliability have grown in importance. Probability theory forms a central part of
thedesignofhighlyreliablesystems.Formostitemsthefailureratechangeswithtime.
A common pattern is exhibited by the appropriately named ‘bath tub’ curve, illustrated
inFigure 29.22.
Carpartfailuresarequitewellmodelledbythisdistribution.Forexample,acrankshaft
may fail quite quickly as a result of a manufacturing defect. If it does not, then there is
usuallyalongperiodduringwhichthelikelihoodoffailureislow.Aftermanyyearsthe
probability offailure increases.
ConsidertheprobabilityofanitemfailingoveratotaltimeperiodT.Thisperiod,T,
isthetimeduringwhichtheitemisfunctioning.Thetimetakentorepairtheitemisnot
consideredinthiscalculationbutisconsideredalittlelateron.Supposetheprobabilityof
failureisevenlydistributedoverthisperiod;thatis,theprobabilityoffailureismodelled
bytheuniformdistribution(seeSection29.12).Itisimportanttonotethatthisisafairly
simplistic assumption. IfN = number of failures of the item over a time periodT, it is
possible todefine a mean failure rate,k, by
k = N T
For example, if an item fails 10 times in a period of 5 years we define the mean failure
ratetobek = 10/5 = 2,thatistwofailuresperyear.Becauseoftheuniformdistribution
ofthefailuresacrossthetimeperiod,thequantitykisconstant.Toillustratethisconsider
the previous example with a period of 10 years. During this time the item will fail 20
times and so
k = 20 = 2 failures per year
10
as found earlier.
Another usefultermisthemean timebetween failures(MTBF), which isgiven by
MTBF = 1 k
The termMTBF isonly used foritems thatarerepairable.
Let the interval T be divided into n small sub-intervals each of length δT, that is
nδT = T. Suppose each sub-interval is so small that only one failure can occur during
it. Note that since repair time has been neglected it is always possible to have a failure
in a sub-interval. So theN failures which occur during timeT occur inN distinct sub-
Failure rate
Time
Figure29.22
The‘bath tub’curve.

29.15 Reliability engineering 971
intervals. The probability of failure occurring in a particular sub-interval, P f
, is then
given by
P f
=
number of sub-intervals inwhich failure occurs
total number ofsub-intervals
= N n
Ineach sub-interval, δT, the probability ofan item notfailing,P nf
, is
P nf
=1−P f
=1− N n = n −N
n
Astheitemprogressesthrougheachofthesuccessivesub-intervals δT itcanbethought
ofasundergoingaseriesoftrials,theresultofwhichisfailureornon-failure.Therefore,
the probability of the item not failing as it passes through all of thensub-intervals can
be obtained by multiplying each of the sub-interval probabilities of notfailing, thatis
( ) n −N n
probability of notfailingintimeT =
n
As the sub-interval δT becomes small, that is δT → 0, the number of sub-intervals,n,
becomes large, thatisn → ∞. Hence weneed toconsider the quantity
( ) n −N n
lim
n→∞
n
This limitcan be shown tobe e −N (see Appendix IV). Hence
probability of notfailingintimeT = e −N
Finally,theprobabilityofanitemfailingoneormoretimes--itemscanberepairedand
fail again -- inthe time intervalT isgiven by
P(T)=1−e −N =1−e −kT
P(T ) is the probability of at least one failure in the time period T. As a consequence
of the constancy of k, this formula can be used to calculate the probability of an item
failingone or more times inan arbitrarytimeperiod,t, inwhich case
P(t) =1−e −kt (29.1)
Itisimportanttostressthatthisformulaonlyappliesiftheprobabilityoffailureisevenly
distributed.
Engineeringapplication29.8
Breakdownsonafactoryprocessline
A factory process line makes use of 12 controllers to maintain process variables at
their correct values; all 12 controllers need to be working in order for the process
line to be operational. Records show that each controller fails, on average, once
every six months. Calculate the probability of the process line being stopped as a
resultofacontrollerfailure within a time periodofone month.
➔

972 Chapter 29 Statistics and probability distributions
Solution
The mean failure rate of each controller is 1/6 breakdowns per month. Given that
thereare12controllerstheoverallfailurerateforcontrollersis12/6 = 2breakdowns
per month. Using Equation (29.1) withk = 2 andt = 1 gives the probability,P(t),
of the process line being stopped because of a controller failure within a period of
one month as
P(t) =1−e −kt =1−e −2 =0.865
So far we have ignored the ‘downtime’ associated with an item waiting to be repaired
after it has failed. This can be a significant factor with many engineering systems. The
simplestpossibilityisthatanitemisoutofactionforafixedperiodoftimeT r
whileitis
being repaired. If there areN failures during a periodT then the total downtime isNT r
.
The time the item isavailable,T a
, isgiven by
T a
=T−NT r
Ausefulquantityisthefractionaldeadtime,D,whichistheratioofthemeantimethe
item isinthe dead statetothe total time.Inthiscase
D = NT r
(29.2)
T
Anotherusefulquantityistheavailability,A,whichistheratioofthemeantimeinthe
working statetothe total time. For the present model
A = T−NT r
=1−D
T
Thefailureratemodeldevelopedpreviouslywasbasedonthetimetheitemwasworking
rather thanthe total time.When the repair time isincludedkbecomes
and so
k = N T a
N =kT a
=k(T −NT r
)
N+kNT r
=kT
and therefore
N =
kT
1+kT r
So using Equation (29.2)
Also
D =
kT
1+kT r
T r
T =
A=1−D=1−
kT r
1+kT r
kT r
1
=
(29.3)
1+kT r
1+kT r
For more complicated repair characteristics the equations forDandAare correspondingly
more complicated.

29.15 Reliability engineering 973
Engineeringapplication29.9
Availabilityofanelectricalsupplytoalargefactory
Theelectricalsupplytoalargefactoryhasameantimebetweenfailuresof350hours.
When the supply fails it takes 3 hours to repair the failure and restore the supply.
Calculate the average availability of the electricalsupply tothe factory.
Solution
Failure rate of the supply is k where k = 1/350 failures per hour. Using Equation
(29.3) withT r
= 3 hours gives
A =
1
= 0.992 that is,99.2%
1 +3/350
The supply isup and running for99.2% of the time.
So far we have only examined systems in which failure was caused by one or more
componentseachwiththesamefailurerateorMTBF.Amorecommonsituationisonein
whichthedifferentcomponentsofasystemhavedifferentdegreesofreliability.Itisstill
useful to be able to calculate the overall reliability of the system although the analysis
ismorecomplicated.Inordertodosoitisnecessarytodefinethetermreliability.From
Equation(29.1)weknowthatP(t)definestheprobabilityofoneormorefailuresduring
atimeperiod,t.Thereforetheprobabilityofnofailuresisgivenby1−P(t).Thequantity
1−P(t)iscalledthereliabilityofthesystemduringatimeperiodt,andisdenotedR(t),
thatis
R(t) =1−P(t) =e −kt (29.4)
R(t) can be interpreted as the probability a component works properly during a period
t. We now examine the reliabilityoftwo simple systemconfigurations.
29.15.1 Seriessystem
A series system is one in which all the components of a system must operate satisfactorily
if the system is to function correctly. Consider a system consisting of three
components, shown in Figure 29.23. The reliability of the system is the product of the
reliabilitiesofthe individual components,that is
R =R A
R B
R C
(29.5)
This formula is a direct consequence of the fact that the failure of any one of the components
is an independent event. So the probability of the system not failing, that is its
reliability,istheproductoftheprobabilitiesofeachofthecomponentsnotfailing.(See
Section28.7 forindependentevents.)
Input A B C Output Figure29.23
Seriessystem.

974 Chapter 29 Statistics and probability distributions
Engineeringapplication29.10
Reliabilityofaradiosystem
A radio system consists of a power supply, a transmitter/amplifier and an antenna.
During a 1000 hour period the reliabilityofthe various components is asfollows:
R ps
= 0.95
R ta
= 0.93
R a
= 0.99
Calculate the overall reliabilityof the radio system.
Solution
R =R ps
R ta
R a
= 0.95 ×0.93 ×0.99 = 0.87
thatis,thereisan87%chancethattheradiowillnotfailduringa1000hourperiod.
Using Equations (29.4) and (29.5) it is possible to generate a formula for the reliability
of a systemconsisting ofnseries components:
R = e −k 1 t e −k 2 t ...e −k n t = e −(k 1 +k 2 +···+k n )t
wherek 1
,k 2
,...,k n
arethe mean failure ratesof thencomponents.
Engineeringapplication29.11
Reliabilityofasatellitelink
AsatellitelinkisbeingusedtotransmittelevisionpicturesfromAmericatoEngland.
ThecomponentsofthissystemarethetelevisionstudioinNewYork,thetransmitter
groundstation,thesatelliteandthereceivergroundstationinLondon.TheMTBF of
each ofthe components isasfollows:
MTBF ts
= 1000 hours
MTBF tgs
= 2000 hours
MTBF s
= 500000 hours
MTBF rgs
= 5000 hours
Calculate the overall reliability of the system during a 28 day period and a yearly
period of 365 days.
Solution
Firstwecalculate the mean failure rate of each of the components:
k ts
= 1
1000 = 1 ×10−3 failuresper hour
k tgs
= 1
2000 = 5 ×10−4 failures per hour

29.15 Reliability engineering 975
k s
=
1
500000 = 2 ×10−6 failuresper hour
k rgs
= 1 = 2 ×10 −4 failuresper hour
5000
So the overall reliabilityof the systemis
R = e −(k ts +k tgs +k s +k rgs )t = e −(1×10−3 +5×10 −4 +2×10 −6 +2×10 −4 )t
= e −1.702×10−3 t
Fort = 28 ×24 hours,R = 0.319.
Fort = 365 ×24hours,R = 3.35 ×10 −7 .Clearlyduringayearlyperiodthesystem
isalmost certain tofail atleastonce.
29.15.2 Parallelsystem
A parallel system is one in which several components are in parallel and all of them
mustfailforthesystemtofail.ThecaseofthreecomponentsisshowninFigure29.24.
The probability of all three components failing in a time period,t, is the product of the
individual probabilities of each component failing, that is (1 −R A
)(1 −R B
)(1 −R C
).
So the overall systemreliabilityis
R=1−(1−R A
)(1−R B
)(1−R C
)
This formulacan begeneralizedtothe case ofncomponentsinparallelquite easily.
A
Input
B
Output
C
Figure29.24
Parallel system.
Engineeringapplication29.12
Reliabilityofprocesscontrolcomputerpowersupplies
A process control computer is supplied by two identical power supplies. If one fails
then the other takes over. TheMTBF of the power supplies is 1000 hours. Calculate
thereliabilityofthepowersupplytothecomputerduringa28dayperiod.Compare
this with the reliabilityifthereisno standbypower supply.
Solution
k = 1 = 0.001 failures per hour
1000
R ps
= e −kt = e −0.001×24×28 = 0.511
➔

976 Chapter 29 Statistics and probability distributions
Therefore the overall reliabilityof the power supply tothe computer is
R = 1 − (1 −0.511)(1 −0.511) = 0.761
Clearly the existence of a standby power supply does improve the reliability of the
power supply to the computer. In practice, the figure would be much higher than
this because once a power supply failed the standby would take over and maintenance
engineers would quickly repair the failed power supply. Therefore the period
of time in which the computer was relying on one power supply would be very
small.
We have only examined the simplest possible reliability models. In practice, reliability
engineering can often be very complicated. Some models can cater for maintenance
strategiesofthesortwetouchedoninthepreviousexample.Theeffectofnon-uniform
failureratescanalsobecateredfor.Also,manysystemsconsistofamixtureofseriesand
parallel subsystems of the type we have discussed as well as more complicated failure
modes than we have examined.
EXERCISES29.15
1 Acomputer contains 20circuit boards each with an
MTBF of3months. Calculate the probability that the
computer willsufferacircuit boardfailure duringa
monthly period.
2 Threepumpsarerequiredtofeedwatertoaboilerina
power station in order foritto be fullyoperational.
TheMTBF ofeach ofthe pumps is 10days.Calculate
the probability that the boiler willnotbe fully
operational during amonthly periodifthere are only
three pumps available.
3 Aprocesscontrol computer has apower supply with
anMTBF of600 hours. If the power supply fails then
ittakes 2 hoursto replace it.Calculate the average
availability ofthe computer power supply.
4 Aradar station has anMTBF of1000hours. The
average repair time is 10hours. Calculate the average
availability ofthe radar station.
5 Aprocessline to manufacture bread consistsoffour
main stages:mixing ofingredients, cooking,
separation andfinishing, packaging. Each ofthe
stageshas anMTBF of
MTBF m = 30 hours
MTBF c = 10 hours
MTBF sf = 20 hours
MTBF p = 15 hours
Calculate the probability that the processline will be
stopped during an 8 hour shift.
6 Aremote water pumpingstation hastwo pumps.Only
one pumpis needed in normal operation; the other
actsasastandby. Access to thisstation isdifficult and
soifapumpfails itis noteasy to repair it
immediately. Given that theMTBF ofthe pumps is
3000hours, calculate the overall reliability ofthe
pumping station duringa28 dayperiod in which the
maintenance engineers are unable to repair apump
which fails.Calculate the improvement in reliability
that would be obtained ifasecond standby pump was
installed in the pumping station.
7 Repeat the calculation carried out in Question6with
the following information.The pumps,which are
different from those in Question 6,can be considered
to consist oftwo components: a motor andthe pump
itself.TheMTBF ofthe motors is 4000hours and that
ofthe pumps is 6000 hours.

Review exercises 29 977
Solutions
1 0.999
2 0.9998
3 0.9967
5 0.865
6 0.960, 0.992
7 0.940, 0.985
4 0.9901
REVIEWEXERCISES29
1 Find the mean and standard deviation of
8,6.9,7.2,8.4,9.6,10.3,7.4,9.0
2 Adiscrete random variable, v,has a probability
distributiongiven by
v −2 −1.5 −1 −0.5 0 0.5 1.0
P(v) 0.23 0.17 0.06 0.03 0.17 0.31 0.03
Calculate
(a)P(v = −1) (b)P(v = −2or v =0)
(c)P(v 0) (d)P(v < −0.5)
3 Acontinuous random variable,x, has ap.d.f., f (x),
given by
f(x)=3x 2
(a) FindP(0 x0.5).
(b) FindP(x > 0.3).
(c) FindP(x < 0.6).
0x1
(d) Find the expected value ofx.
(e) Find the standard deviation ofx.
4 Adiscrete random variable,y, has aprobability
distributiongiven by
y −0.25 0.25 0.75 1.25 1.75
P(y) 0.25 0.20 0.10 0.15 0.30
(a) Find the expected value ofy.
(b) Find the standard deviation ofy.
5 Therandom variable,t,has ap.d.f.,H(t),given by
H(t)=λe −λt
t0
(a) Calculate the expected value oft.
(b) Calculate the standard deviation oft.
6 Calculate the number ofpermutations of
(a) five distinct objects,taken two atatime
(b) eight distinct objects,taken four at atime
7 Calculate the number ofcombinations of
(a) seven distinct objects,taken four at atime
(b) 200 distinct objects,taken 198 atatime
8 The probability a component is manufactured to an
acceptable standard is 0.92.Twelve components are
picked at random. Calculate the probability that
(a) sixare acceptable
(b) 10are acceptable
(c) more than10are acceptable
(d) atleast two are notacceptable
9 The number ofbreakdowns ofacomputer systemin a
month isarandom variable with aPoisson
distribution. The mean number ofbreakdowns per
month istwo. Calculate the probability that the
number ofbreakdowns ofthe system in a monthis
(a) two
(b) three
(c) one
(d) more thantwo
10 The probability ofacomputer failingin asystem in a
given week is 0.03. Inasystem thereare 150
computers.
(a) Usethe binomial distribution to calculate the
probability that in a given week fivecomputers
fail.
(b) Usethe binomial distribution to calculate the
probability that in a given week more than two
computers fail.
(c) Usethe Poissonapproximation to calculate the
probability that in a given week four computers
fail.

978 Chapter 29 Statistics and probability distributions
(d) Usethe Poissonapproximation to calculate the
probability that in a given week more than three
computersfail.
11 Arandom variable,t,has a uniform distribution, f (t),
given by
{ 14
0<t<4
f(t)=
0 otherwise
Calculate the probability that
(a) 1t3
(b) 0.2t2.7
(c) |t| >1 (d) |t| 1.5
12 Thetime,t,betweenfailures foraparticular typeof
electrical component follows an exponential
distribution with a meanvalue of24weeks. Calculate
the probabilitythat
(a)t>24 (b)t<22
13 Theresistance ofresistorsis arandom variablewith a
normal distributionwith ameanof3ohmsanda
standard deviation of0.1ohm. Calculate the
probability thataresistor has aresistance of
(a) between 2.9and3.05 ohms
(b) more than2.95 ohms
(c) less than2.83 ohms
14 Thecapacitance ofcapacitors isarandom variable
with anormal distribution, with ameanof12farads
andastandard deviation of0.6farads.Inabatchof
300 capacitors how many would you expectto have
with capacitance
(a) between 11and12.3 farads
(b) more than11.6 farads
(c) less than12.8 farads?
15 Acomputer fails,onaverage, onceevery 4 months.
An engineering systemusesninecomputers, all of
whichneedto be workingforthe system to operate.
Calculate the probability thatthe system willfail
sometimeduring
(a) a1monthperiod
(b) a2weekperiod(i.e.half amonth)
16 TheMTBF foramotor is270 hours.When the motor
fails ittakes7hours to repair. Calculate the
availabilityofthe motor.
17 Asystem comprisesfour components in series.The
MTBF foreachcomponentis 170 hours,200 hours,
250 hours and210 hours. Calculate the reliabilityof
the system over a100 hour period.
18 Asystem comprisestwo components in parallel.The
components have anMTBF of150 hours and
170 hours. Calculate the probabilitythe systemwill
failin a 200 hourperiod.
Solutions
1 mean = 8.35;st.dev. = 1.1314
2 (a) 0.06 (b) 0.4 (c) 0.51 (d) 0.46
3 (a) 0.125 (b) 0.973 (c) 0.216
(d) 0.75 (e) 0.194
4 (a) 0.775 (b) 0.799
5 (a)
1
λ
(b)
1
λ
6 (a) 20 (b) 1680
7 (a) 35 (b) 19900
8 (a) 1.4687 ×10 −4 (b) 0.1835
(c) 0.7513 (d) 0.2487
9 (a) 0.2707 (b) 0.1804 (c) 0.2707
(d) 0.3233
10 (a) 0.1736 (b) 0.8307 (c) 0.1898
(d) 0.6577
11 (a) 0.5 (b) 0.625 (c) 0.75
(d) 0.375
12 (a) 0.3679 (b) 0.6002
13 (a) 0.5328 (b) 0.6915 (c) 0.0446
14 (a) 193 (b) 225 (c) 272
15 (a) 0.8946 (b) 0.6753
16 0.9747
17 0.1402
18 0.5093

Appendices
Contents I Representingacontinuousfunctionandasequenceasa
sumofweightedimpulses 979
II TheGreekalphabet 981
III SIunitsandprefixes 982
IV Thebinomialexpansionof ( )
n−N n
n
982
AppendixI REPRESENTINGACONTINUOUSFUNCTIONANDA
SEQUENCEASASUMOFWEIGHTEDIMPULSES
Inthisappendixweshowhowacontinuousfunction, f (t),oradiscretesequence, f[k],
obtained by sampling this function, can be represented as a sum of weighted impulses.
SucharepresentationisimportantinthestudyoftheztransformandthediscreteFourier
transform.
Thedeltafunction
ThedeltafunctionisintroducedinChapter2.Foreaseofreferencewerestateitsdevelopmenthere.Thedeltafunction
δ(t),orunitimpulsefunction,isthelimitofarectangle
function bounding an area of 1 unit, and located at the origin, as its width approaches
zero,anditsheightincreasesaccordinglytoensurethatthearearemains1.Theenclosed
area is known as the strength, or the weight, of the impulse. This is illustrated in Figure
AI.1. Note that the impulse is represented by an arrow and the height of the arrow
gives the strength of the impulse. If the impulse occurs at the pointt = d, then this is
written δ(t −d).Further,bymultiplyingthedeltafunctionbyanumberA,togiveAδ(t),
we obtain the limit of a rectangle function bounding an area ofA. This is an impulse of
strengthA.
Samplingacontinuousfunction
Now consider a function f (t) defined fort 0 as shown in Figure AI.2. Suppose this
function is sampled at timest = 0,T,2T,3T,...,kT,..., to generate a sequence of
values f(0), f(T), f(2T), f(3T),..., f(kT),..., which we shall write as f[0], f[1],
f[2],f[3],...,f[k],....
Note from the discussion above that the quantity f[0]δ(t) is the limit of a rectangle
functionlocated atthe originand boundinganarea equal to f[0].

980 Appendices
d(t)
as h 0
1
1–
h
0 t
0
h
t
FigureAI.1
The limitofthe rectangle function is the delta function.
f(t)
f(3T)
f(kT)
f(0)
f(2T)
f(T)
0 t 0 T 2T 3T....... kT t
FigureAI.2
Sampling a continuous function f (t)gives the sequence f[k].
Similarly, f[1]δ(t −T )isthelimitofarectanglefunctionboundinganareaequalto
f[1]and located att =T.
In general f[k]δ(t −kT) is the limit of a rectangle function bounding an area equal
to f[k]andlocatedatt =kT.
The quantity
∞∑
f[k]δ(t −kT)
k=0
isaseries of delta functions. The area bounded by these isgiven by
∞∑
area=f[0]+f[1]+f[2]+···+f[k]+···,
thatis
f[k]
(AI.1)
k=0
Approximatingtheareaunderacurve
Now consider approximating the area under the continuous function f (t) by a series of
rectangular areas as shown in Figure AI.3. The first rectangle has widthT and height
f (0) and so its area isTf (0). Using our sampling notation this may be writtenTf[0].
Similarly, the second rectangle has widthT and height f (T) and so its area isTf (T).
Using the sampling notation this is written Tf[1]. The third rectangle has area Tf[2]
and soon. The total area isthus given by
area=Tf[0]+Tf[1]+Tf[2]+···+Tf[k]+···=T
∞∑
f[k]
k=0
(AI.2)

II The Greek alphabet 981
f(t)
T
0 t 0 T 2T kT t
FigureAI.3
Thearea under f (t)can be approximated byaseriesofrectangular areas.
Now compare Equations (AI.1) and (AI.2). If we multiply the series of delta functions
byT we see that
∞∑
T f[k]δ(t −kT)
k=0
will bound the same area as our approximation to the area under the graph of f (t). In
thissense wecan regard
∞∑
T f[k]δ(t −kT)
k=0
as anapproximation tothe function f (t).We will denote thisapproximation by ˜f(t).
In summary, a continuous function f (t) can be represented as a sum of weighted
impulses each ofstrength f[k]occurring att =kT:
∑ ∞
˜f(t)=T f[k]δ(t −kT)
k=0
Thisrepresentationcanalsobethoughtofasawayofexpressingadiscretesequenceof
values, f[k],asacontinuousfunction ˜f (t).Thisisusefulwhenstudyingtheztransform
andthediscreteFouriertransform.Sometimesitwillbeconvenienttoworkwithoutthe
factorT, inwhich casewedefine
∞∑
f ∗ (t) = f[k]δ(t −kT)
k=0
It should be remembered that when using this form, f ∗ needs to be multiplied by the
factorT inordertoapproximatethe function f (t).
AppendixII THEGREEKALPHABET
A α alpha I ι iota P ρ rho
B β beta K κ kappa σ sigma
Ŵ γ gamma λ lambda T τ tau
δ delta M µ mu Y υ upsilon
E ε epsilon N ν nu φ phi
Z ζ zeta ξ xi X χ chi
H η eta O o omicron ψ psi
θ theta π pi ω omega

982 Appendices
AppendixIII SIUNITSANDPREFIXES
ThroughoutthisbookSIunitshavebeenused.Belowisalistoftheseunitstogetherwith
their symbols.
Quantity SIunit Symbol
Prefix
Symbol
length metre m
mass kilogram kg
time second s
frequency hertz Hz
electric current ampere A
temperature kelvin K
energy joule J
force newton N
power watt W
electric charge coulomb C
potential difference volt V
resistance ohm
capacitance farad F
inductance henry H
10 15 peta P
10 12 tera T
10 9 giga G
10 6 mega M
10 3 kilo k
10 2 hecto h
10 1 deca da
10 −1 deci d
10 −2 centi c
10 −3 milli m
10 −6 micro
10 −9 nano n
10 −12 pico p
10 −15 femto f
AppendixIV THEBINOMIALEXPANSIONOF ( n−N
( ) n −N n
Consider the binomial expansion of .
( )
n
n −N
n (
= 1 − N ) n
n n
(
=1+n − N )
+
n
n(n −1)
2!
n
(
− N ) 2
n
(
n(n −1)(n −2)
+ − N ) 3
+···
3! n
( ) ( )( ) n −1 N
2 n −1 n −2 N
3
=1−N+
n 2! − n n 3! +···
(
=1−N+ 1 − 1 ) N
2
(1
n 2! − − 1 )(
1 − 2 ) N
3
n n 3! +···
Suppose now weletn → ∞. We find
( n −N
lim
n→∞
n
) n
=1−N+ N2
2! − N3
3! +···
But thisisthe power series expansion of e −N (see Section 6.5).We conclude that
( ) n −N n
= e −N
lim
n→∞
n
Inparticular, note thatifN = −1
( ) n +1 n
= e
lim
n→∞
n
) n

Index
absolute quantity88
absorption laws 179, 185, 186
acceleration 690, 868
addition:
algebraic fraction 30--1
complex numbers329
law ofprobability 909--13, 954
matrix 259--60
vectors 225--6,336--7
adjoint matrix281--2
algebraseeBoolean algebra; matrixalgebra
algebraic equations 627, 649
algebraic fractions 26--33
addition 30--1
division29--30
equivalent fractions 27--8
expressing fraction in simplestform28--9
multiplication 29--30
proper andimproper fractions 27
resistorsin parallel 32
subtraction 30--1
algebraic techniques 1--53
algebraic fractionsseealgebraic fractions
indices, laws of2--11
dividing expressions 4--5
firstlaw 2
fractional indices 8--11
multiple7
multiplying expressions 2--4
negative indices 5, 6
positiveindices 6
power density ofsignal transmitted byradio antenna 6
power dissipationin resistor3--4
radar scattering7--8
second law 4
skindepth, in radial conductor 9--10
thirdlaw, 8 7
inequalities, solution of33--8
number basesseebase, number
partial fractionsseepartial fractions
polynomial equations 20--6
current used by electric vehicle 21--3
ofhigher degree 24--5
quadratic equations 20--4
summation notation 46--50
Kirchhoff’s currentlaw 47--8
Kirchhoff’s voltage law 48--9
alphabetic character stream,informationcontent of916
alternating current:
circuits 324
waveforms 134
Ampère’s circuital law 900
amplitude 581, 724, 744, 751--3
Fourier transform766
modulation 767--8
trigonometric functions133, 134, 135, 136
ofwave 131
analog-to-digital converter 690
analogue simulation 603--5
analytical integration 457
analytical solutions534, 618--19,621--2
AND gate 184, 187, 188
angular acceleration 613
angular frequency 133, 134, 135, 136, 340
Fourier series723--4,753
fundamental 723--4
ofwave 131
antennae 169--70
approximate solution 615
arbitraryconstant 538, 589, 608
arbitraryfunction 876
arbitraryscalingconstant 304
Argand diagram332, 334, 337, 345--6, 351
argument:
ofangle 333
offunction 57--8
arithmetic mean (mean value) 938--40
arithmetic progressions204
arithmetic series,finite 209--10
armaturecurrent/voltage 612--13
arrays1
arrows(signals)661
associative laws179, 186
associative matrixaddition 260
associative matrixmultiplication 264
asymptote 76--7, 81
Bode plotoftransfer function with real poles andzeros
671--5
ofgraph75
oblique 76
983

984 Index
attenuation 87,88, 578
in step-index opticalfibre 89--90
augmented matrix 309
autocorrelation 814, 815--16, 817
automated vehicles, routing226--7
auxiliary equation 562--4,572
particular integral574
availability, in reliability 972
average information perevent 916
average value:
offunction471--5
ofperiodic waveform 477--8
back substitution287--90
backward wave 578--80
base, number 2, 11--20, 80
binarycoded decimal 11,17--20
binarysystem 11--14
decimal system 11
hexadecimal system 11,14--17
seven-segment displays18--19
‘bath tub’curve 970
Bernoulli, J.953
Bernoulli trials953, 956, 959
Besselfunctions584, 586, 587--601
firstkind oforder one591
firstkind oforder zero589
in frequency modulation 598--601
and generating function595--7
ofhigherorder 592--3
Jacobi-Anger identities 597--8
modesofcylindrical microwave cavity 593--5
second kind oforder one 591
second kind oforder zero 589
tabulated values of601
Bessel’sequations 584, 587--601
ofhigherorder 592--3
oforder one 589--91
oforder zero 587--9
power seriessolution of587--8, 590--1
biasedcircuit 379
binarycoded decimal 11, 17--20
binarydata stream:
entropy of917
informationcontent of916
redundancy of918
binarydigits (bits)11,17
binaryfull-adder circuit 190--2
binarysystem 11--14
binaryto decimal conversion 12
decimal to binary conversion 12--14
binarywords190--1
binomial distribution 953--7
mean andstandard deviation 955
most likely number ofsuccesses 956
probability ofksuccesses fromntrials954--5
binomial expansion 982
binomial theorem 214--18, 700, 717--18
Biot-Savart law 250
bit stream 915
bits(binarydigits)11,17
block 627--8, 661
basic 661
delay 688--9
diagram661, 663--4,665, 688--94
Bode plot oflinearcircuit 96--7
Bode plot oftransferfunction with real polesand zeros
671--5
Boolean algebra 175, 183, 185--97
expressions185, 187--8
lawsof185--7
logicalequivalence 187--93
variables 185
bounded sequence 205
branch current 307--10
breakpoints 673, 674
byte 11,17
capacitance 575, 637
ofcoaxial cable 448--50
mutual 102
between two parallel wires102--3
capacitive reactance 435
capacitors 577, 637
with capacitance 435
currentthrough 368--9
discharge of82, 655
energystoredin 487--8
phasors341--3
in series484--5
voltage across435
cardinal sinefunction 123--4
carrier 599
carrier signal 767
carry-out value 190
Cartesian components 232--40
line and multipleintegrals 871
Cartesian coordinate system160--1, 162, 166--7, 171
complex numbers333--4
functionsofmore than onevariable 823--5
threedimensions 157--8
two dimensions 154--7
vectors 240
Cartesian equations 169
Cartesian form 334
complex numbers342
discrete Fourier transform793
cathode raytube, electrical potential inside 824
chain rule389--91,395, 463
character data stream, redundancy of918
characteristic equation 299, 300--2
characteristic impedance 348
ofcoaxial cable 450--1
charged particle,movement in electric field 244--5

Index 985
chord 357--8
circuital law (Ampère)900
circuits:
biased 379
binary full-adder 190--2
digital electronic 175
integrated 428
inverting 437
LC 572--3
LCR560--1
linear 97
RC 535--6, 554--7
RL 544--5,553--4, 651--2
RLC 109, 573--5
thyristorfiring473--4
truthtable for187--8
circular (periodic) convolution 803--7
circular convolution theorem 807--12
circular correlation theorem816--20
circular cross-correlation815--16, 818--20
closed form(binomial theorem)216
closed loop system with negative feedback 665
CMOS(complementary metal-oxide semiconductor) logic
192--3
co-domain 180--1
coaxial cable 576--8
capacitance of448--50
characteristic impedance of450--1
codes (communication theory)918
coding theory919
coefficient:
damping 610
difference equations and the z transform685--6
differential equations, linearconstant 649--59
equating 40--1, 585, 586
Fourier 734, 739, 741--2, 748--50, 766
matrix 285, 289
polynomial 20
reflection 581--2
cofactor ofelement 279, 282
column vector 233, 258--9, 261--2
n-component 612
p-component 612
r-component 612
combinations 950--2
common derivatives 372--4
common difference 204
common ratio 204--5, 211, 213
communication engineering 903
communication theory915--19
commutative laws 179, 186, 191
commutative numbers259--60
complement:
laws 179, 186, 189--90, 191
in probability 904
in set theory 178--9
complementary events913--15
complementary function558, 561--9,572
inhomogeneous term appears in 575--83
particular integral574--5
complementary metal-oxide semiconductor (CMOS)logic
192--3
completeness andperiodic functions733
completing the square 23,37--8
complex frequency:
function 751
variable 636
complex notation andFourier series749--51
complex numbers97, 324--55
addition 329
complex conjugate pairs 327
De Moivre’stheorem 344--51
division330
in polar form 335
exponential form337--8
Fourier series751
graphical representation 332--3
imaginary part325--6, 328--30, 334
and inverse Laplace transform643--6
loci and regionsofcomplex plane 351--3
modulus333
multiplication 329--30
in polar form 335
operationswith 328--30
phasors 340--4
polar form 333--6
Poynting vector343--4
real part 326, 328--30, 334
subtraction329
vectors and 336--7
complex plane 332
complex poles 668, 669
complex roots24,564, 567
complex translation theorem 714
component laws,differential equations 572, 577
composition, offunctions61--2
compound events 905--8,954
compression, image and audio 795
computer software packages 73
Fourier transform816--17
graph plotting 723
Laplace transform675
seealso MATLAB ®
computer solutionsformatrixalgebra 319--21
concave down function402--3, 415--17
concave upfunction 402--3,415--17
conditional probability 919--25
conduction current 369
conjunction (AND gate) 184
conservative fields875--80
constant689
arbitrary538, 589, 608
arbitraryscaling 304
coefficient differential equations, linear627, 649--59

986 Index
constant(Continued)
coefficient equations 560--84
exponential 81
ofintegration 429
phase581
time 82, 545
constantpolynomial 71
continuous function 64--5,361
andsequencerepresentationassumofweightedimpulses
979--81
approximating area under acurve 980--1
delta function 979
sampling a continuous function979--80
continuous integrand 491
continuous random variable 965--6
expected value 944--5
standard deviation 947
continuous signal sampling 702--4
continuous spectra 766
continuous variable 962, 963
control engineering 319
control systems628
controllersignal 693
converged series211
convergence, radiusof218, 698
convergence criteria 212
converges sequence 205
convolution 647
circular (periodic) 803--7
linear801--3, 809--11
theorem 647--9,676, 772--8
circular 807--12
seealso discrete convolution and correlation
coordinate frame 268
coordinate system 154--74
CartesianseeCartesian coordinate system
electrode 156
polarsee polar coordinates
polar curves 163--6
corner(derivatives) 371
correlation:
theorem 780--2
circular 816--20
seealso discrete convolution and correlation
cosecant (cosec)120, 434
hyperbolic 100
cosine(cos):
complex notation 749
complex numbers333--4, 338--9
De Moivre’stheorem 344--50
definite andindefinite integrals 445
differentialequations 539
differentiation 391, 394, 420
discrete Fourier transform795--7
first-orderlinear differentialequations 547--9,555
Fourier series734--43
odd andeven functions 740--3
functions120--3
half-range series746--8
hyperbolic (cosh)100--1,102--3
improper integrals 486, 488
integration, elementary 429--33, 438--9
integrationby parts 458, 460
integrationby substitution464--5
Jacobi-Anger identities 598
Maclaurin series525--6
modellingwaves using131--44
odd and even functions726--7,730--1
orthogonalfunctions 481--3
orthogonality relations 732--1
partialdifferential equations 825, 827
polar coordinates 160, 171
polar curves164--5
second-order linear differentialequations 541, 559--60,
564, 566--7,571--3
Taylor polynomials 522
Taylor series526--7
trigonometricequations 144, 146, 147, 148
trigonometricidentities 125--30
trigonometricratios 116--20
vectors 244
waves734, 757
cosine (cos)periodic waveforms 723--4
cotangent (cot)120, 434
hyperbolic 100
coupled difference equations 694
coupled equations 606--7
coupled first-orderequations 607--8
coupled-tanks system 614--15
Cramer’s rule280, 658
critical resistance 412
critically damped response 667
cross-correlation780
circular 815--16, 818--20
linear812--15
cubic equation 543
cubic polynomial 71
cumulative normalprobabilities 967--8
current 577--8
alternating 134, 324
branch307--10
conduction 369
density892
direct510
displacement 369
eddy 827
lawsee Kirchhoff’scurrent law
leakage 576--7
mesh 254, 307, 308--10
node 47--8, 310--12, 317--19
sinusoidalsignals 136--7
through capacitor 368--9
usedbyelectric vehicle 21--3
cusp(derivatives) 371

Index 987
cycle ofsint 131
cycles oflogarithmic scale94--5
cylindrical microwave cavities, modesof593--5
cylindrical polar coordinates166--70
decaying exponential 670
fluid flow along apipe 167
helical antennae 169--70
damped sinusoidalsignal 387--8
damping coefficient 610
damping ratio 411--12
DCT-II795
DCT-III795
De Moivre’stheorem 344--51
De Morgan’s laws 179, 186, 191
decaying exponential 578
decibels (dB)87
decimal system 11
deconvolution 810
definite integrals 442--51
degree 116, 117
ofequation 20
ofpolynomial 71
delta function111--12,979, 980, 981
Dirac 480
discrete Fourier transform787
Fourier transform770--1
integral properties489--91
Laplace transform675--6
sifting property of490
demodulation 817--18
demodulator 598
denominator 27--8, 29--30, 39,75, 465
dependent variable 58,59, 366
difference equations 682, 683, 684--5, 686--7
differential equations 535--7,543, 550--2
derivable from a potential 875
derivatives 365, 370--2
common 372--4
higherseehigher order derivatives
and Laplace transform635--8
partialseepartial derivatives
secondseesecond-derivative
table 356
determinants 275--6,278--81
to evaluate vector products248--9
and vector products279--80
deviation:
from mean 941
seealsostandard deviation
diagonal matrix 271
diagonally dominant, matrix ofcoefficients 317
dielectric 82
difference, common 204
difference equations 175, 203, 681--98
block diagram representation 688--93
coefficient 685--6
coupled 694
dependent andindependent variables 682
discrete-time controllerdesign693--5
discrete-time filter689--90
homogeneous and inhomogeneous equations 684--5
linear andnon-linear equations 683
low-pass filter696--7
non-recursive 684
numerical solution 695--8
order684
recursive684
rewriting686--8
second-order 719
signal processingusingamicroprocessor684--5
solution ofdifference equation 683
andztransform718--20
differentialcalculus 357
differentialequations 534--626, 638
analogue simulation 603--5
condition 538
coupled-tanks system 614--15
Euler’smethod 616--19
first-orderseefirst-orderequations
first-orderlinearseefirst-orderlinear differential
equations
higherorder equations 606--9,625
Laplace transform635, 659--61, 663
LC circuit with sinusoidalinput 572--3
linear 536--7
linear constantcoefficient 627, 649--59
numerical methods 615--16
order536
oscillatingmass-springsystem 565--6
parallelRLC circuit573--5
RC charging circuit 535--6
RC circuit: zero-input response and zero-state response
554--8
RL circuitwith ramp input 553--4
RL circuitwith stepinput 544--5
Runge-Kutta method offourth-order623--5
second-order 538--9,605, 610
second-order linearseesecond-order linearordinary
differential equation
seriessolution of584--6
solution of537--9
standing waves on transmissionlines579--82
state-space models609--15
voltage reflection coefficient 578--80
differentiation356--85,429, 457--8
applicationsseedifferentiation applications
common derivatives 372--4
currentthrough capacitor 368--9
defined 366
derivatives, existence of370--2
dynamic resistance ofsemiconductor diode 378--9
explicit 394
from firstprinciples 366, 372

988 Index
differentiation(Continued)
fluid flow intotank376--7
graphical approach to 357--8
implicit 394--7
limitsand continuity 358--61
linearmodel forsimple fluid system 380--3
aslinearoperator 375--83
linearity430
logarithmic 397--8
parametric 393--4
rate ofchange
atgeneral point 364--9
atspecific point 362--4
techniquesseedifferentiation techniques
voltage across an inductor 368
differentiationapplications 406--27
inflexion, pointsof415--18
maximum pointsandminimum pointssee
maximum/minimum points
maximum power transfer413--14
Newton-Raphson method 406, 418--23
risetime forsecond-order electrical system411--13
seriesdiode-resistor circuit421--3
vectors 423--6
differentiationtechniques 386--405
chain rule389--91, 395
damped sinusoidalsignal 387--8
higherderivatives 400--4
logarithmic differentiation397--8
parametric differentiation 393--4
product rule 386--8,397--8
quotient rule388--9,401
differentiator691
digital computers 615
digital controller 694
digital electronic circuits 175
digital filtering692
digital signal processing200, 692
diode equation 83--4
Dirac delta function 480
directcurrent electrical network 510
directtransmissionmatrix 612
Dirichlet conditions 735, 740
discontinuous function64--5, 361
discrete,usage ofterm 175
discrete convolution and correlation 801--20
circular (periodic) convolution 803--7
circular convolution theorem 807--12
circular correlation theorem 816--20
circular cross-correlation815--16,818--20
convolution reverb 810--12
linearconvolution 801--3,809--11
linearcross-correlation812--15
radar,use ofcorrelation in 817--20
discrete cosine transform795--801
definition andits inverse 795--801
truncation ofset ofsamples 798--9
two-dimensional andimage compression799--801
discrete Fourier transform783--7,979, 981
definition 784--6
derivation 787--90
to estimateFourier transform790--1
inverse 786--7
matrixrepresentation 792--3
properties793--5
linearity793
Parseval’s theorem 794
periodicity 793
Rayleigh’s theorem 794
discrete independent variable 202
discrete mathematicsseeBoolean algebra; settheory
discrete random variable 953, 957
expected value 943--4
standard deviation 946--7
discrete spectra766
discrete-time:
controllerdesign693--5
data 688
filter689--90, 696
interval702
discrete variable 962
disjointsets 178
disjunction connective 183
disjunctive normal form188
displacement:
current369
vector226, 227, 423
dissipation, power, in resistor3--4
functionto model 59
distance travelled by aparticle 434
distance travelled by rocket498
distributive laws179, 186, 189, 191
divergence theorem 893, 895
divergent sequence 205
diverging integral 486--7
division:
algebraic fractions 29--30
complex numbers330
indices4--5
in polar form 335
domain (sets)180--1
domain, function 58
doping 83
dot product 241
double integrals 880--5,886--7,891, 895
double the input rule56--7, 61
dual modular redundancy 906
dummy variable, and integration 451
duty cycle 743--4
dynamic resistance, ofsemiconductor diode 378--9
dynamic systems535
echelon form 290--1
eddy current losses827

Index 989
eigenvalues andeigenvectors 294--307
eigenvalues 297--303, 305
eigenvectors 303--6
linear homogeneous equations 294--7
electric charge density 900
electric displacement, electric field strength and
240--1
electric field:
ofelectrostatic dipole 529--31
movement ofcharged particle in 244--5
strength, and electric displacement 240--1
work done by 869
electric flux857--8, 893, 900
electrical components:
multiple, testing 914--15
reliability 911--12
electrical networks, analysis of307--12
electrical potential inside acathode ray tube824
electrode coordinates 156
electromagnetism andvector calculus 864--6
electronic chips, manufactured, acceptability of
925--6
electronic circuits, digital 175
electronic thermometer measuring oven temperature
656--7
electrostatic dipole, electric field of529--31
electrostatic potential 241, 854--5,863
electrostatics theory900
elements 176, 259
ofarea 882
cofactor of279, 282
minor of278--9
ofsystem 661
empty set 178, 910
energy 798
storedin capacitor 487--8
used by electric motor 447
engineering functions 54--114
argument offunction 57--8
composition of61--2
continuous function 64--5
discontinuous function64--5
exponential functions80--4
graph offunction58--9
hyperbolic functionsseehyperbolic functions
inverse offunction62--4
logarithm functionsseelogarithm functions
many-to-one functions60--1
numbersand intervals 55--6
one-to-many function 60
one-to-one functions60--1
parametric definition 61
periodic function 65--6
piecewise continuous functions 64--5
polynomial functions70--4
power dissipatedin resistor59
rational functions75--9
saw-tooth waveform 66--7
squarewaveform 67
triangularwaveform 66
entropy 916--19
ofbinarydata stream 917
ofsignal consistingofthree characters 917
equal sets177
equating coefficients 40--1,585, 586
equations:
algebraic 627, 649
auxiliary 562--4,572, 574
Bessel’sseeBessel’sequations
Cartesian 169
characteristic 299, 300--2
circle, centre onthe origin 163, 164
coupled 606--7
coupled difference 694
coupled first-order607--8
cubic 543
degree of20
differenceseedifference equations
differentialseedifferentialequations
diode 83--4
exact 547--9
familyof587
first-orderseefirst-orderequations
higherorder 606--9,625
homogeneous 558--60, 684--6
inconsistent288
inhomogeneous 558--60,572--3
Laplace 833--4, 862, 863
linear 163, 164, 220, 536--7, 683, 685
non-linear 219--22, 536--7, 683, 685
non-recursive 691
output614
partialdifferential 832--5, 862
Poisson’s863
quadratic 20--4,103, 325--6
second-order 684, 685, 696
separable 541
simple 540--1
simultaneoussee simultaneous equations
state609, 613
state-space 615
third-order684, 685
transmission834
trigonometric 144--50
wave 833
zero order684
equipotential surface 245
equivalent fractions 27--8
equivalent resistance 77--8
error-correcting code 291
errorsignal 693
errorterm 521
Euler’smethod 616--19, 623
improved 620--2

990 Index
Euler’srelations339, 749
differentialequations 564, 566
discrete Fourier transform787
Fourier transform760, 771--2
even functions726--32
exact equations 547--9
exact solution 616, 617--18,620--1,624
exclusive OR gate189
expansion along the firstrow278
expected value 943--5, 946--7,955, 959, 962
experiment and binomial distribution 953
explicitdifferentiation 394
exponentseepower
exponent, defined 80
exponential constant 81
exponential decay 81
exponential distribution 962--3
exponential expressions 80
exponential form,ofcomplex number 337--8
exponential functions80--4, 728
diode equation 83--4
discharge ofcapacitor 82
seealso logarithm functions
exponential growth 81
exponential term 669, 670, 671
factorialnotation 55
factorization 20--1,28--9,39
factory production line, quality control on907--8
failurein asingle trial953
Faraday, M. 240
Faraday’s law 368, 865
feedback:
loop, negative, elimination of662--8
negative 436
path689
Fibonacci sequence 203
field:
irrotational860
line and multiple integrals 876
solenoidal 857
filters722
digital 692
discrete-time 689--90,696
high pass689
lowpass689, 696--7
final value theorem 634, 657
finite sequences 801, 809, 813
finite series209
arithmetic 209--10
geometric 211
firstcondition (differentialequations) 539
first-derivative417
Laplace transform635
Taylor polynomials 508--9
test 407, 411
firstlaw ofindices 3
first-orderequations 540--5, 573, 606--8, 685
coupled 607--8
difference equations 696
Laplace transform656
separation ofvariables 541--5
simplest540--1
first-orderlinear differential equations 547--57
exact equations 547--9
integratingfactor method 550--7
separation ofvariables 549
first-orderTaylor polynomial 509--13
directcurrent electrical network 510
gravityfeed water supply 510--11
linearity509--13
power dissipationin resistor511--12
first-orderTaylor polynomials, in two variables 835--7
firstshifttheorem 632--3,719, 768
Fourier transform762--4
ztransform711--12
fluidflow:
acrossasurface 890--2
along a pipe167
coupled-tanks system614--15
intotank376--7
flux 891
folding (signalprocessing)775
forward wave 578--80
Fourier analysis 482, 745
Fourier coefficients 734, 739, 741--2,748--50, 766
Fourier components 752, 754--5
Fourier series722--56, 723--4,728, 731--42, 746, 749--50
complex notation 749--51
frequency response oflinearsystem 751--5
half-range series745--8
low-passfilter752--5
odd and even functions726--32, 740--4
Parseval’s theorem 748--9
periodicwaveforms 723--6
spectrum ofpulse width:
modulation controlled solar charger743--4
Fourier synthesis 734
Fourier transform749, 757--822
amplitude modulation 767--8
convolution and convolution theorem 774--82
definitions 758--61
ofdelta function 770--1
discrete cosinetransform795--81
fast785
integration480
inverse 758--9,774
andLaplace transform772--4
pair758
periodicfunctions 771--2
properties761--6
firstshifttheorem 762--4
linearity761--2
second shift theorem 764--5

Index 991
spectra 766--8
t-w duality principle 768--70, 771
seealsodiscrete convolution and correlation; discrete
Fourier transform
fourth-orderTaylor polynomials 518
fractional dead time 972
fractional indices 8--11
fractions:
algebraicseealgebraic fractions
equivalent 27--8
expressing,in simplestform 28--9
improper 27,43--4
negative 35--6
partialseepartial fractions
positive 35--7
proper 27
freevariables 288--9
freevectors 225
frequency 132, 724
angularseeangular frequency
component, zero736
domain 722, 766
Fourier series723
natural 411
resonant 343, 411, 594
response ofasystem636--7
response oflinear system751--5
sinusoidalsteady-state 636--7
spectrum 744
variable, generalized 636
frequency modulation (FM),Besselfunctionsin 598--601
full-wave rectifier 106--7
function:
average value of471--5
orthogonal 480--3
rootmean square value of475--9
function handle 61
functionsofseveral variables 823--48
dependent variable 823--4
eddy current losses827
electrical potential inside acathode raytube 824
first-orderTaylor polynomial in two variables 835--7
functionsofmore thanone variable 823--5
higher order derivatives 829--32
independent variables 823--4,825
maximum and minimumpointsofafunctionoftwo
variables 841--6
partial derivatives 825--9
partial differentialequations 832--5
power dissipatedin a variable resistor824
second-order Taylor polynomial in two variables 837--40
Taylor seriesin two variables 840
gain/attenuation 87,88, 578
Gauss-Seidel method 314--17
Gaussian elimination 286--94,306, 307, 309
augmented matrix287--91, 292
back substitution 287--90
rowoperations 287--90
Gauss’stheorem 448, 857--8,864, 893, 900
general solution 537
complementary function558--9
difference equations 683
differential equation 584--5
first-orderlinear differentialequations 547
higherorder equations 607--8
particular integral578
second-order differentialequation 538, 561--4,566--8,
569--70
generalized frequency variable 636
generating function 509
Besselfunctionsand 595--7
geometric progressions204--5
geometric series:
finite 211
infinite 212--13
gradient 620
ofchord357--8
ofsolution 616
oftangent 366--7,508
graph,offunction 58--9
graphics calculators 73,723
gravitational field, work done by868--9
gravityfeed water supply 510--11
Greek alphabet 981
Green’s theorem 895
half-range cosine series746--8
half-range Fourier series745--8
half-range sineseries746
half-wave dipole, radiation patternsof172--3
half-wave rectifier 107
Halleffect, in semiconductor 251--2
harmonics723
components 724
even 744
fundamental orfirst723
nth 744
odd 736, 744
second 723
waves 725, 734
helical antennae 169--70
hexadecimal system 11,14--17
binaryto hexadecimal conversion 16--17
conversion to decimal 14--15
decimal to hexadecimal conversion15--16
high passfilter689
higherorderderivatives 400--4,829--32
fifth402
first400--2
fourth402, 503--4,831
mixed 831
second 400--2,831
third402, 831

992 Index
higherorder equations 606--9,625
homogeneous equations 558--60, 572, 684--6
linear294--7
horizontalaxis turbines 73
horizontalshift 138--44
horizontalstrips881--4
hyperbolic functions100--4
capacitance between two parallelwires102--3
cosecant (cosech) 100
cosine(cosh) 100--1
cotangent (coth) 100
identities 101
inverse 102
sine(sinh)100--1
iandjcomponents ofr233
identity
hyperbolic function 101
Jacobi-Anger 597--8
laws 179, 186, 189--90, 191
matrix271--2, 274, 284, 292--3
trigonometric 125--30
imaginaryaxis 670, 709
imaginaryaxis (yaxis) 332
imaginaryparts 578, 669
complex numbers325--6, 334, 348
implicit differentiation 394--7
impossibleevent 910
improper fraction 27, 43--4
improper integrals 480, 483--9
impulseresponse 112, 676, 810--12
impulsetrain 111--12
inclusiveOR gate 189
inconsistentequations 288
increment (rateofchange at aspecific point) 363
indefinite integrals 442--51
independent events 925--30
independent solution 568
independent variable 58,59,366, 374
difference equations 682, 683, 684--5,686, 691
differentialequations 535--7, 540--1,543, 550--2
discrete 202, 682
integration 429, 431--2
indices,laws ofseeunder algebraic techniques
inductor577
phasors341
inequalities, solution of33--8
inertia,moment of612, 613
infinite integrand 483, 486
infinite limitsofintegration 483
infinite series209, 211--12
geometric, sum of212--13
inflexion, pointsof415--18
informationcontent:
ofalphabetic character stream 916
ofbinarydata stream 916
informationperevent 915
inhomogeneous equations 294, 558--60, 569--70, 572--3
difference equations 684--6
inhomogeneous term,in complementary function
575--83
initial conditions539, 542, 556--7, 566
analogue simulation 604--5
difference equations 686, 696
differentialequations 616
Laplace transform659
particularintegral 578
zero660, 663
initial values 696
input:
matrix612
sequence 685, 689, 691, 697
signal 661, 669
to system659--61
transferfunctions663, 665
vector612
voltage 676
integers 55,345
non-negative 682
positive55,214--16
integrals 438, 460, 463, 480, 490
differentialequations 541--2
diverging 486--7
double880--5,886--7,891, 895
improper 480, 483--9
indefinite/definite 442--51
infinite limits483
inner 881--3,886
Laplace transform628, 635--8
limits443, 446, 488
lower limits443--5
odd and even functions730--1
outer 881--3
periodicfunctions 677--8
scalar 493
second-order linear ordinary differential equation
569--75
Simpson’srule502
surface 889--95, 896, 898
trapezium rule499--500
triple 885--6,889, 893
upperlimits443--5
volume 889--95
seealso line integrals and multiple integrals
integrands 458, 465, 480
continuous 491
infinite 483, 488
integrating factor 551, 552
integrating factor method550--7
integration 428--56, 480--95
analytical 457
applications ofsee integration applications
capacitance ofcoaxial cable 448--50
capacitors in series484--5

Index 993
characteristic impedance ofcoaxial cable 450--1
constant of429
delta function, integralproperties of489--91
differential equations 542
distance travelled by aparticle 434
elementary 429--41
electronic integrators 435--7
energy storedin capacitor 487--8
energy used by electric motor 447
improprerintegrals 483--9
integrals, definiteand indefinite 442--51
limits491
aslinear operator432--4
linearity 438
numericalsee numerical integration
orthogonal functions480--3
ofpiecewise continuous functions491--2
region of880
techniquessee integration techniques
oftrigonometric functions437--9
vectors 493
voltage across a capacitor 435
integration applications 471--9
average value and r.m.s.value ofperiodic waveform
477--8
average value offunction471--5
load resistor476--7
rootmean square value offunction475--9
saw-tooth waveform 472--3
thyristorfiringcircuit473--4
integration techniques 457--70
by parts457--63
by substitution463--6
usingpartial fractions 466--8
integrator 604--5
circuits 428
intersection (settheory) 178
intervals 56
closed 56
ofconvergence 218
open 56
semi-open 56
inverse:
offunction 62--4
ofmatrix 285
usingrow operations292--3
of3×3matrix281--2
trigonometric functions121--3
of2×2matrix274--7
finding 275--6
orthogonalmatrix 276
inverse square law 6
inverters 188
inverting circuit437
irrational number 55
irrotationalvector field 860
isotropic antenna 6
iteration420
ofnon-linear equations,sequences arisingfrom 219--22
simple 220
Jacobi-Anger identities 597--8
Jacobi’s method 313--14, 315--16
Kirchhoff’scurrent law 310
algebraic techniques 47--8
complex numbers342
Kirchhoff’svoltage law72, 308
algebraic techniques 48--9
complex numbers342
differential equations 535, 544, 560, 572, 577
differentiation 413
Fourier series752
integration 435, 485
Laplace transform655
Kraus,J.169
Kronecker delta sequence 202, 203, 698
Laplace equation 862, 863
three-dimensional 833--4
Laplace transform411, 627--80, 681--2,706, 709, 759, 774
asymptotic Bodeplotoftransfer functionwith real poles
and zeros671--5
ofcommon functions(look-uptable) 629--30
convolution theorem 647--9
definition 628
delta function675--6
ofderivatives and integrals 635--8
discharge ofacapacitor 655
electronic thermometer measuring oven temperature
656--7
and Fourier transform772--4
frequency response ofasystem 636--7
integration 480
inverse 627, 638--41
usingcomplex numbers 643--6
usingpartial fractions 641--3
linear constantcoefficient differentialequations 649--59
linearity 631--2
partialdifferential equations 833
periodic functions677--8
poles,zerosandsplane 668--75
rule1670
rule2670
rule3670--5
position controlsystem 665--8
properties631--5
final value theorem 634
firstshift theorem 632--3
linearity 631--2
second shifttheorem 633--4
RL circuitwith ramp input 651--2
rule1: combining two transfer functionsin series
661--2

994 Index
Laplace transform(Continued)
rule2: eliminating negative feedback loop 662--8
transferfunctions 659--68, 752
transport lag 664--6, 664--5
voltage across acapacitor 636
andztransform, relationshipbetween 704--9
LCRcircuit560--1
leakage, current 576--7
left-handlimits360
lifespan, electrical components 911
limitsequence 205
limit5sequence 206
limits463
differentiation 358--61
ofintegrals 443, 446, 486
integration 491
left-hand360
right-hand360
line integrals and multiple integrals 867--902
conservative fields875--80
line integralaround closedloop 877--9
potential function 875--7
divergence theorem 895
double integrals 880--5,886--7, 891
electric charge enclosed in a region 890
electric current density 892
electric field, work done by869
electric fluxand Gauss’slaw 893
evaluation ofline integrals in three dimensions
873--5
evaluation ofline integrals in two dimensions 871--3
fluid flow acrossasurface 890
gravitational field, work done by868--9
Green’s theorem 886--7
line integrals 867--71
massofasolid object 889
Maxwell’s equations in integral form 899--900
simple volume and surface integrals 889--95
Stokes’theorem 896--9
three-dimensional line integrals 873--5
triple integrals 885--6, 889
two-dimensional line integrals 871--3
line spectra 766
linearcircuit, Bode plotof96--7
linearcombination 237
ofsinusoids723
linearconstant coefficient:
differentialequations 627, 649--59
linearconvolution 801--3,809--11
linearcross-correlation812--15
linearequations 163, 164, 220, 536--7, 683, 685
linearfactors 39--41
equating coefficients 40--1
evaluation usingspecific value ofx40
Laplace transform651, 668
repeated 39, 41--2
linearhomogeneous equations 294--7
linear modelsforsimple fluidsystem 380--3
linear operator:
differentiationas375--83
integrationas432--4
Laplace transform638
linear polynomial 71
linear scale94
linear system:
frequency response 751--5
Taylor polynomials 511
linear time-invariant system 560--1
linearity:
differentialequations 536--7
ofdifferentiation 430
discrete Fourier transform793
first-orderTaylor polynomial 509--13
Fourier transform761--2
integration438
Laplace transform631--2
ztransformanddifference equations 710, 718--19
linearly dependent vectors 238
linearly independent functions558
linearly independent vectors 238
linesofforce 240
linking processcontrolcomputers 948--9
load:
line 422
resistor476--7,655
voltage waveform 473--4
log-linear graph92
log-linear scales 92--4
log-log graph94--6
log-log plot92--3
log-log scales 92--4
log scale92
logarithm functions85--100, 187--93
attenuation in a step-index optical fibre89--90
decibelsin radio frequency engineering 88--9
delta function 111--12
exponential functionsand 91--2
logarithms85--7
modulusfunction104--8
ramp function 108
reference levelsseereference levels
signal ratiosand decibels 87--8
unitimpulsefunction 111--12
unitstepfunction108--10
logarithmic differentiation 397--8
logarithmic frequency scale 672
logarithmic scale 94
logarithms85--7
laws86,398, 433, 672
natural 85, 89
logical equivalence:
AND gate184, 187, 188
binaryfull-adder circuit design190--2
connective 183

Index 995
exclusive OR gate 189
NAND gate 185, 192--3
NOR gate 184--5
NOT gate 184
OR gateseeOR gate
realization oflogicalgates 198--9
losslessline 578
lossycompression 799
low passfilter689, 692
analogue 752--5
digital 696--7
lower bound 56, 499, 504
lowest common denominator (l.c.d.)30
m-files 61
Maclaurin series524--31
magnetic field:
due to moving charge 250
intensity249--50
strength 249--50
magnetic flux 900
magnitude:
ofphasor 340
ofvector 225
many-to-one functions60--1, 122
maps 180
mark time 743
mass610--11
mass-springsystem, oscillating565--6
mathematical model 54
MATLAB ® 1--2,58, 320--1,615, 785, 790, 808--9
toolboxes 67
matrices 1
matrixalgebra 257--328
addition 259--60
adjoint matrix281--2
augmented matrix309
computer solutionsfor319--21
definitions 258--9
determinants 275--6, 278--81
Cramer’srule280
andvector products279--80
diagonal matrix 271
direct transmission612
eigenvalues and eigenvectorsseeeigenvalues and
eigenvectors
electrical networks,analysis of307--12
Gaussian eliminationseeGaussian elimination
identity matrix271--2, 292--3
input 612
inverse of2×2matrix 274--7
finding 275--6
orthogonalmatrix 276
inverse of3×3matrix 281--2
multiplication 261--4, 285
non-singularmatrix 276
output 612
representation 792--3
robot coordinate frames268
scalar multiplication 260--1
simultaneous equations and 283--6
simultaneous equations, iterative techniques forsolution
of312--19
singular matrix276, 282
skewsymmetric matrix272--3
squarematrix 271, 272--3, 275, 300
state612
subtraction259--60
symmetric matrix 272
translation and rotation ofvectors 267--71
transpose ofmatrix272
Vandermonde matrix 291--2
maximum/minimum points406--15, 824--5
first-derivativetest 407, 411
ofafunction oftwo variables 841--6
local maximum 406--7
local minimum406--7
second-derivative test 410, 412
turningpoints407--9, 412, 414
Maxwell’s equations 864--6
in integral form 899--900
mean 964
deviation and binomial distribution 955
time between failures (MTBF)970, 973, 975
value (arithmetic mean) 938--40
mesh current307, 308--10
vector 254
microwave cavities 593--5
minimization laws179, 185, 186
minimumpointsseemaximum/minimum points
minor ofelement 278--9
mixed partial derivatives 862
modelling systems566
modulation index 599
modulation signal 767
modulus:
ofcomplex number 333
function 104--8
ofvector 225
moment ofinertia 612, 613
moving average filter692
multipath-induced fading 142
multipleintegralsseeline integrals andmultiple integrals
multiplication:
algebraic fractions 29--30
complex numbers329--30
indices2--4
law 919--25, 926, 954
matrix261--4
in polar form335
scalar 260--1
mutual capacitance 102
mutuallyexclusive events:addition law ofprobability
909--13

996 Index
n-component column vector 612
n-typemetal-oxide semiconductor (NMOS) transistors
192--3
n-typesemiconductor 83, 251
NAND gate 185, 192--3
natural frequency 411
natural logarithms85, 89
natural numbers55
negative feedback 436
negative number 325
neper (Np)89
Newton-Raphson method221, 406, 418--23
Newton’s Second Law 610, 613
nibble 17
NMOS(n-typemetal-oxide semiconductor) transistors
192--3
node 47--8
reference 310
voltages/currents47--8, 310--12, 317--19
non-linear equations 219--22, 536--7,683, 685
non-linear system, in Taylor polynomials 511
non-negative number 333
non-periodic function757--8
non-rectangular region 880, 884
non-recursiveequations 691
non-singularmatrix276
non-standard normal 966--9
non-trivial solution 294--6,298--300
non-uniformfailurerates976
non-zeroinitial value ofstate vector 660
NOR gate 184--5
normal distribution 963--9
non-standard normal 966--9
standard normal 964--6
normalized sincfunction123, 124--5
norms254
NOTgate 184
nth-order Taylor polynomial 517--21
numbers55--6
commutative 259--60
complexseecomplex numbers
numerator 27--8, 29--30, 75,465
numerical integration 457, 496--506
distance travelled byrocket 498
energy dissipationin resistor502--3
Simpson’srule 500--5
trapezium rule496--500, 502
numerical methods,and differentialequations 615--16
numerical solution 220, 534, 617--18, 695--8
obliqueasymptote 76
odd and even functions726--32, 740--4
integral properties730--1
Ohm’s law3, 55
differentialequations 535, 544
differentiation 413
fluid equivalent 614
Fourier series752
integration436, 487
Laplace transform655
phasor form341
polynomial functions71--2
Taylor polynomials 510
one-to-many function 60
one-to-one functions60--1, 122
operating point 379--80
OR gate 183, 184--5,187, 188, 192
exclusive 189
inclusive189
truthtable 183
order258, 263, 587
difference equations and theztransform684
differentialequations 536
ofobjects 949
ofsystem 609
ordinary differentialequationsseedifferential equations
origin (coordinate systems)154
orthogonal functions480--3
orthogonal matrix276
orthogonal vectors 232, 242
oscillate sequence 205
oscillating mass-springsystem565--6
oscillatory term669
output 663, 665
equation 614
matrix612
reduction 436
sequence 685, 689, 691, 697
signal 661, 669
to system659--61
variables 614
vector612
voltage 676
overdamped response 666--7
overshooting 666
p-component columnvector 612
p-n junction 83
p-type metal-oxide semiconductor (PMOS)transistor
192--3
p-type semiconductor 83,251
padding signals with zeros 809--11
parallel system 975--7
parallel vectors 242
parallelogram 336--7
parameters 393, 587
functions61
parametric differentiation 393--4
Parseval’s theorem 748--9, 794
partial derivatives 825--9,835
first-order825, 829, 833, 835--7, 842--5
fourth-order840
line and multipleintegrals 886
mixed 862

Index 997
second-order 829--30, 833, 838--40, 842--5
third-order840
vector calculus 854
partial differentialequations 832--5,862
partial fractions 39--45
improper fractions 43--4
and integration 466--8
linear factors 39--41
quadratic factors 42--3
repeated linearfactors 39, 41--2
partial fractions:
inverse Laplace transform641--3,644--5,651, 653--4,
656
Laplace transform666
z transform716--17,720
partial sumssequence 212
particular integralfunction 558
particular solution 538, 556, 563--4,575
parts,integration by457--63
Pascal’s triangle 214--15
period 131
offunction 65
periodic convolutionsee circular (periodic) convolution
periodic extension 746--7
periodic functions65--6,131--2, 725, 733
Fourier transform766, 771--2
Laplace transform677--8
periodic waveforms 477--8, 723--6,752
periodicity and discrete Fourier
transform793
permeability, ofconductor 10
permittivity offree space 102, 858
permutations 948--50, 951--2
phase 136, 724, 751--3
angle 132, 133, 138--9, 723
constant 581
Fourier transform766
phasors 340--4
capacitor 341--3
inductor 341
magnitude of340
reference 340
resistor341
pick andplace robot 159--60
piecewise continuous functions64--5, 648, 780
integration of491--2
place sign 279
PMOS(p-typemetal-oxidesemiconductor)transistor192--3
Poissondistribution 957--61,962
Poissonprocess959
Poisson’sequation 863
polar coordinates 159--63, 333--4
cylindricalsee cylindrical polar coordinates
pick and place robot 159--60
spherical 170--3
polar curves 163--6
equation ofcircle, centre on the origin 163, 164
equation ofline 163, 164
and radiationpatterns fromantennas 164--5
polar form 333--6,338, 345--6
poles 706--9
complex 668, 669
dominant 670
offunction 76
Laplace transform668--75
atthe origin 673
positions667
real 668, 669
stablecomplex 671
system 668
polynomial:
coefficients 20
constant 71
cubic 71
ofdegree 2 300
ofdegreen--d 43--4
degree of71
equationsseeunder algebraic techniques
expression 70
functions70--4
non-ideal voltage source 72--3
Ohm’s law 71--2
wind power turbines 73
linear 71
quadratic 71
quartic 71
Taylorsee Taylor polynomials
position controlsystem (servo-system)108, 665--8
position vector423--4
positiveintegers 55
positivenumber 325
postmultiplying263
potential function875--7
potential theory833
potentiometer 604--5
power:
dissipationin resistor3--4,511--12, 824
function to model 59
gain 87
series218--19, 337--8
supply to a computer 906
power density6
power seriessolution, ofBessel’sequation 587--8,
590--1
Poyntingvector 343--4
premultiplying 263, 272
primaryroute (permutations)948--9
probability 903--32
acceptability ofmanufactured electronic chips925--6
ofacceptable components 928--9
addition law909--13,954
alphabetic character stream,information content
of916
binarydata stream,information content of916

998 Index
probability (Continued)
breakdowns onfactory processline 971--2
character data stream,redundancy of918
communication theory915--19
complementary events913--15
component, calculating 912
compound event 905--8
conditional 919--25
cumulative normal 967--8
electrical component reliability911--12
electrical supply to a large factory,availability of973
emergency calls to service engineer 958
entropy ofbinary data stream 917
entropy ofsignal consistingofthree characters 917
events 904
offaultycomponents 926--8
independent events925--30
machine breakdowns in workshop 958
mutually exclusive events909--13
power supply to acomputer 906
production line product fedbytwo machines 920--2
quality control onfactory production line 907--8
redundancy ofbinary data stream 918
reliabilityofmanufactured components 923--4
testing multiple electronic components 914--15
trial904
unconditional 920, 923
workforce absentees 959--60
seealso statisticsand probability
product rule543
differentialequation 548, 551
differentiation 386--8,397--8, 412
partialdifferential equations 827--8,833
productionengineering 903
projection(cylindrical polar coordinates) 166
properfraction 27
proportional/integral/derivative (p.i.d.) controller 694
Pythagoras’s theorem106, 141, 161, 233, 235--6, 333
quadratic approximation to diode characteristic
515--16
quadratic equations 20--4, 103, 325--6
quadratic expression 37
quadratic factors 39,42--3
quadratic polynomial 71
quality control 903, 907--8
quantity, absolute88
quartic polynomial 71
quotientrule 388--9,401, 414
r-component column vector 612
radar:
and correlation817--20
defined 8
scattering7--8
radar cross-section (RCS)8
radians116, 117
radiation patterns:
ofhalf-wave dipole 172--3
polar curvesand 164--5
radio antenna, power densityofsignal transmitted by6
radio-frequency (RF)engineering 579
decibelsin 88--9
radiusofconvergence 218, 698
ramp function108
random variables 934, 962, 968
continuous 944--5, 947, 965--6
discreteseediscrete random variable
range, sets andfunctions180
rate ofchange 357--8, 362--4
atgeneral point 364--9
negative 365
positive365
atspecific point 362--4
rational functions75--9
rational numbers55, 176, 345
Rayleigh’s theorem 794
RC charging circuit535--6
real axis(x axis) 332
real line 55--6
real numbers55,176, 181
real parts 578
complex numbers326, 334, 348
negative 669, 670
real roots24
real system670
rectangle function979--80
rectangular region 880--3
rectangular wave 743
rectifier:
defined 106
full-wave106--7
half-wave 107
recurrence relations203, 220
recursive difference equations 684
recursive filter689
recursive sequence 203
reduction formula460--1
redundancy 917--19
Reed-Solomon codes291
reference levels 88,90--7
Bode plotoflinearcircuit 96--7
exponential and logarithmfunctions, connection between
91--2
log-linearand log-loggraph paper 94--6
log-logandlog-linear scales 92--4
logarithmfunctions90--1
reference node 310
reference phasor 340
relation (setsand functions)180--1
reliability engineering 903, 970--7
parallelsystem 975--7
seriessystem 973--5
remainder ofordern522

Index 999
remainder term,Taylor’s formulaand 521--4
repeated linear factors 39,41--2
resistors577
load 476--7
in parallel 32
phasors 341
power dissipationin 3--4,511--12
functionto model 59
resolvingforce intotwo perpendicular directions
227--8
resonant frequencies 343, 411, 594
resultantoftwo forcesacting onabody 227
reverberation 810--12
reverse saturated diode 83
right-hand limits360
right-handed screwrule 896
risetime 411--13
RL circuitwith ramp input 651--2
RL circuitwith stepinput 544--5
RLC circuit 109
parallel 573--5
robot coordinate frames268
robot positions237
rootmean square(r.m.s.)340, 471
value offunction 475--9
value ofperiodic waveform 477--8
roots20,219--21, 563--4,568, 668
complex 24,564, 567
formula 21
real 24
routing, ofautomated vehicle 226--7
row258--9, 261--2
operations 287--90, 292--3
vector 233
Runge-Kutta method offourth-order623--5
s domain661, 665
s plane 668--75
mapping ofto the zplane 706--9
saddle points825, 842--5
sample space 904
sampled signal 690
sampling 202, 688
saw-tooth waveform 66--7, 472--3,733--4, 742--3
scalar 224
field 240, 851--5, 856, 875, 889
integrals 493
multiple 230--1,260--1
product 241--5,870--1, 891
secant (sech)120, 434
hyperbolic 100
second-derivative 416--17
difference equations 691
differential equations 536
Laplace transform635
numerical integration 499
test 410, 412
second law ofindices 4
second-order difference equations 719
second-order differential equations 538--9,604, 610
second-order electrical systems,risetime for411--13
second-order equations 573, 606--8,684, 685, 696
second-order linear ordinary differential equation 558--84
complementary function561--9
constant coefficient equations 560--84
integrals 569--75
property 1 558
property 2 558--60
transmissionlines576--8
second-order system 610
second-order Taylor polynomials 513--17
second shift theorem 633--4,665, 712--14, 716, 764--5
secondary route (permutations)948--9
semiconductors192--3
diode:
dynamic resistance of378--9
ideal, operating point for379--80
Hall effect in 251--2
n-type83, 251
p-type83, 251
separable equations 541
separation ofvariables 541--5,549
sequences 200--9,683, 688--9,976
arithmetic progressions204
binomial theorem 214--18
bounded sequence 205
converges sequence 205
divergent sequence 205
Fibonacci sequence 203
finite sequence 801, 809, 813
geometric progressions204--5
input sequence 685, 689, 691, 697
from iterative solution ofnon-linear equations 219--22
Kronecker delta sequence 202, 203
limitsequence 205
limit5sequence 206
oscillate sequence 205
outputsequence 685, 689, 691, 697
partialsums sequence 212
recursivesequence 203
unbounded sequence 205
unitramp sequence 699--700
unitstepsequence 202, 699
seealso series
series200, 209--14
binomial theorem 214--18
converged 211
diode-resistorcircuit 421--3
finite 209
arithmetic, sum of209--10
geometric, sum of211
inductance series576
infinite 209, 211--12
geometric, sum of212--13

1000 Index
series(Continued)
power 218--19
sequence ofpartial sums212
system series973--5
seealso sequences
servo-systems665
settheory 55,175--83
absorption laws 179, 185, 186
associative laws 179, 186
commutative laws 179, 186, 191
complement 178--9, 186, 189--90
De Morgan’slaws 179, 186, 191
disjointsets178
distributivelaws 179, 186, 189, 191
empty set178
empty sets910
equal sets177
identitylaws 179, 186, 189--90, 191
intersection 178
logic183--5
minimization laws 179, 185, 186
setsand functions179--81
subsets178
union178
universalset 177
Venn diagrams 177, 178--9
seven-segment displays18--19
shrinking intervalmethod 363, 364, 372
SIunits and prefixes982
sidebands601
siftingproperty, ofdelta function490, 787--8
sigma notation 209
signals133, 688
damped sinusoidal387--8
input 661, 669, 697
output 661, 669
processing200, 201
digital 200, 692
usingamicroprocessor 684--5
ratios, anddecibels 87--8
sampling 690, 702--4
sequence 697
sinusoidal751--2
sinusoidalcurrent 136--7
sinusoidalmodulating 599
sinusoidalvoltage 134--5
ofsystem661
telescope drive 108
simple iteration220
Simpson’srule500--5
simultaneous equations:
algebraic 658
directmethods 312
and iterativetechniques 312--19
Gauss-Seidel method 314--17
Jacobi’s method 313--14, 315--16
and matrixalgebra 283--6
sincfunction 123--5,744, 760
normalized 123, 124--5
zerocrossingpoints124--5
sine(sin)115
complex numbers333--4,338--9
DeMoivre’s theorem 344--50
definiteand indefinite integrals 444--5
differentialequations 534, 537, 539
differentiation387--9,391, 394, 412, 420
elementary integration 432--3,434, 438--9
first-orderlineardifferential equations 547--8
Fourier transform763, 766
functions120--3
hyperbolic 100--1
improper integrals 486, 488
integration, elementary 429--33
integrationby parts 458
integrationby substitution464
inverse function 122
Jacobi-Anger identities 598
Maclaurin series525--6
modellingwaves using131--44
orthogonalfunctions 481--3
partialdifferential equations 826, 827
polar coordinates 160, 171
polar curves164--5
rootmean squarevalue offunction475--6
second-order linear differentialequations 541, 559--60,
564, 566--7,571--3
series,half-range 746
Taylor polynomials 522--3
Taylor series526--7
trigonometricequations 144, 145, 148, 149
trigonometricidentities 125--30
trigonometricratios 116--20
vectors 246--7,857
waves723, 734, 757
single loop industrial controlsystem 693
single trial953, 956
singular matrix 276, 282
sinusoidalfunctions121
sinusoidalmodulating signal 599
sinusoidalsignals 751--2
current136--7
damped 387--8
voltage 134--5
sinusoidalsteady-state frequency 636--7
sinusoidalterm670--1
sinusoidalwaveform 340
sinusoids723
linearcombination 723
skew symmetric matrix272--3
skin depth, in radial conductor 9--10
slope (differentiation) 357
slotted line apparatus 582
small-angle approximations 219
smoothing the input signal 696

Index 1001
solenoidal vectorfield 857
solution:
analytical 534, 618--19,621--2
differential equations 537--9
exact 616, 617--18, 620--1,624
generalsee general solution
independent 568
non-trivial 294--6,298--300
numerical 220, 534, 617--18,695--8
particular 538, 556, 563--4, 575
specific 683
trivial 294--6, 298
true 615
solve, defined 219
sonar817, 820
specific solution 683
spectra:
continuous 766
discrete 766
Fourier transform766--8
line 766
ofpulsewidthmodulationcontrolledsolarcharger743--4
spherical polar coordinates 170--3
squarematrix 271, 272--3, 275, 300
squarethe input rule58--9, 61,64
squarewave input 753--4
squarewaveform 67
standard deviation 941--3,964
binomial distribution 955
ofacontinuous random variable 947
ofadiscrete random variable 946--7
standard formsandLaplace transform639--40
standard normal 964--6
standing waves on transmissionlines580--2
state equations 609, 613
state matrix 612
state-space equations 615
state-space models603, 609--15,660
forcoupled-tanks system 614--15
forspring-mass-dampersystem610--11
state variables 609--11, 613--14
state vector 254, 612
stationary points842--5
stationary values 407
statisticsand probability
binomial distribution 953--7
combinations 950--2
continuous random variable 944--5
continuous variable 933, 934, 936--8
discrete random variable 943--4
discrete variable 933, 934, 935
distributions933--53
exponential distribution 962--3
linking processcontrol computers 948--9
mean value (arithmetic mean) 938--40
normal distribution 963--9
permutations 948--50, 951--2
Poissondistribution 957--61
probability densityfunctions936--8
random variables 934
reliabilityengineering 970--7
standard deviation 941--3
ofacontinuous random variable 947
ofadiscrete random variable 946--7
uniform distribution 961
seealso probability
steady-stateerror 657
steady-stateresponse 556
step-index optical fibre,attenuation in 89--90
stepinput 411
stepsize615, 616--18
Stokes’theorem896--9
straightline approximation 620
sub-intervals936, 971
subscripts46
subsets178
substitution:
back 287--90
integration by463--6
subsystems,parallel 976
subtraction:
algebraic fraction 30--1
complex numbers329
matrix259--60
vectors 228--30, 337
successin asingle trial953
sum to infinity213
summationnotation 46--50
Kirchhoff’s currentlaw 47--8
Kirchhoff’s voltage law 48--9
summer 604--5,689
summing890
point 661, 662
superposition principle509
suppressedcarrier amplitude modulation 768
surface integrals 889--95, 896, 898
symmetric matrix272
t-w duality principle 768--70, 771
take-off point 661
‘takeplus orminusthe squarerootofthe input’ 60, 64
‘takethe positivesquare rootofthe input’ rule60
tangent (tan)115, 358
complex numbers333--4
differential equations 543
functions120--3
gradient of366--7
hyperbolic 100
integration 430, 434, 464
line approximationseestraightline approximation
polar coordinates 161--2
trigonometric equations 144, 147, 148
trigonometric identities 125--8
trigonometric ratios116--20

1002 Index
Taylor polynomials:
first-orderseefirst-orderTaylor polynomial
fourth-order518, 840
non-linear system511
ofnth order 517--21
second-order 513--17,837--40
third-order518--19, 840
Taylor series524--31
Taylor seriesexpansion 617, 620, 623, 751
truncation 620
Taylor seriesin two variables 840
Taylor’sformula, andremainder term 521--4
telescope drivesignal 108
thirdlawofindices 7, 8
third-orderequation 684, 685
third-orderTaylor polynomials 518--19
thyristorfiringcircuit473--4
time:
constant82, 545, 753
delayseetransport lag
dependency 340
displacement 132, 133, 723
domain 340, 628, 661, 665
interval 689
lineartime-invariant system560--1
seealso discrete-time
toolboxes67
‘top-hat’ function777
torque613
transferfunctions609, 627--8,637, 659--68, 669
Laplace transform674--6
rule1: combining two transferfunctions in series661--2
rule2: eliminating a negative feedback loop 662--8
system 663
transform,discrete 795--801
transients556, 669--71
response 659
transistors192--3
transmissionequation 834
transmissionlines576--8
transportlag 664--5
transverse wave 343
trapezium rule496--500, 502
tree diagram 907, 912, 921, 923--4
triangle law 225--6,228--30,233, 336, 351
triangularwaveform 66
trigonometricfunctions115--53,349
cosinefunction 120--3
degrees 116
equations, trigonometric144--50
integration 431
integration of437--9
inverse 121--3
radians116
ratios, trigonometric116--20
sincfunction123--5
sinefunction120--3
tangent function120--3
seealso trigonometricidentity
trigonometric identity 345, 348
Fourier series724, 732
integration481--3
trivial solution 294--6,298
true solution 615
truthtable 186--7
exclusive OR gate189
full-adder circuit191
AND gate184
NAND gate 185
NOR gate 184
NOTgate 184
ORgate 183
turbines, wind 73
turning points407--9,412, 414
two raypropagation model 140--2
unbounded sequence 205
unconditional probability 920, 923
underdamped response 667
underdamped system 411
uniform distribution 961
union (settheory) 178
unit impulsefunctionsee delta function
unit ramp sequence 699--700
unit step:
function108--10
functions712
input 669
sequence 202, 699
unit vectors 231
universal set177
upper bound 56,499--500, 503--4
upper limits484
ofintegrals 443--5
Vandermonde matrix291--2
variables 58
Boolean185
complex frequency 636
continuous 962, 963
dependentsee dependent variable
discrete 962
dummy451
free288--9
generalized frequency 636
independentsee independent variable
Laplace transform637
output614
randomseerandom variables
separation of541--5, 549
state613--14
variance 941--2,946--7, 958
vectors 1,224--55
addition225--6,336--7

Index 1003
calculus 849--66
curl859--61, 861--4
divergence 856--9, 861--4
electric fluxand Gauss’slaw 857--8
andelectromagnetism 864--6
electrostatic potential 854--5
gradient861--4
partialdifferentiation ofvectors 849--51, 852--3
Poisson’sequation 863
scalar field 861--2
scalar field gradient 851--5
vectorfield 861
Cartesian components 232--40
column 233, 258--9
and complex numbers336--7
differentiation 423--6
displacement 226, 227, 423
electric field strength andelectric displacement 240--1
electrostatic potential 241
equal 225
field 240, 244, 867
line and multipleintegrals 890, 893
free 225
Hall effect in semiconductor 251--2
input 612, 615
integration of493
linearly dependent 238
linearly independent 238
magnetic field due to moving charge 250
magnitude 225
mesh current 254
modulus225
ofndimensions 253--4
negative 225
operator 852
orthogonal 232, 242
output 612
parallel 242
partial differentiation 849--51, 852--3
position 423--4
Poynting 343--4
product 246--53
applications of249--53
determinantsand 248--9,279--80
magnetic fluxdensityand magnetic field strength
249--50
resolvingforce intotwo perpendicular directions
227--8
resultantoftwo forcesacting onabody227
robot positions237
routing ofautomated vehicle 226--7
row 233
scalar multiple230--1
scalar product 241--5
state 254, 612
subtraction 228--30, 337
translation androtation of267--71
unit231
work done by force 244
zero 236
Venn diagram 904, 906, 910, 914
Venn diagrams 177, 178--9
vertical strips881--2,884--5
very-large-scale integration (VLSI)192--3
VHDL design languages 193
virtualearth 436
visual artefacts800--1
voltage 577--8,580--2
across acapacitor 636, 655
across an inductor 368
across capacitor 435
gain 88,90
input 676
lawseeKirchhoff’s voltage law
node 47--8,310--12, 317--19
non-ideal source72--3
output676
reflection coefficient 578--80
saw-tooth waveform 66--7
sinusoidalsignals 134--5
voltage controlled oscillator(VCO) 599
wave:
amplitude of131
backward 578--80
combining 134--8
destructive interfering 142
equation 833
forward578--80
input, square 753--4
modelling, usingsineand cosine131--47
number 138--44
rectangular 743
transverse 343
two-raypropagation model 140--2
waveforms 133
alternating current 134
periodic 477--8,677--8, 723--6,752
saw-tooth 66--7, 472--3
sawtooth 733--4,742--3
sinusoidal340
square67
time dependency of340
triangular66
waveguides 593--5
wavelength 138--44
weighted impulses112, 703--4,979--81
whole numbers176
windpower turbines 73
work done byforce 244
worldcoordinate frame 268
x axis58, 157, 159, 168, 171
sequences and series220

1004 Index
x coordinate 154--5
x-yplane 166, 169, 170, 171, 233, 336, 351
yaxis58,157
matrixalgebra 270--1
ycoordinate 154--5
zaxis157, 167, 169, 171
ztransforms175, 200, 979, 981
continuous signal sampling 702--4
definition 698--702
and difference equations 718--20
and Laplace transform,relationship between 704--9
mapping thesplane to thezplane 706--9
propertiesof709--15
complex translation theorem714
directinversion 717--18
firstshifttheorem 711--12
inversion715--18
linearity710
second shift theorem 712--14
zero frequency component 736
zero initialconditions 660, 663
zero-input response 554--7,573, 575
zero orderequations 684
zero real part 670
zero-state response 554--7,573, 575
zero vector 236
zeros668, 706, 709
Laplace transform668--75
padding signalswith 809--11