Fourier Transforms

−xfx dx f0 (3.1.1)or equivalently−x − afx dx fa (3.1.2)for any real a.No function can have this property in the ordinary sense but this is a very importantfunction and the usefulness of Fourier theory is dependent upon its existence.The function x can be considered to be a function that is extremely large in value atx 0 but is also only non-zero in a region very close to x 0. To gain an intuitiveunderstanding of what is happening and without attempting to be very precise or imposingstrict conditions, suppose that fx can be expanded as a Taylor series in the vicinity ofx 0 and that (3.1.1) holds, then−xfx dx xf0 xf ′ 0 ½x 2 f ′′ 0 ... dx− x f0 −∑ f0 −n1x dx ∑n1xnn! fn 0 dxf n 0 xxn! n dx−assuming that integration and summation can be interchanged. The first integral on theright-hand-side, together with (3.1.1), gives x dx 1,−which serves as a consistency condition for x; the remaining integrals can be evaluatedas xx n dx 0.−Suppose it was just given that xx n dx 0−for all strictly positive integers n and the question is then to determine what can be saidabout x apart from the obvious remark that x ≡ 0 is a solution. If x ≠ 0,then itsproperties are more difficult to determine but some elementary comments can be made. Ifx is even or odd then the integral is zero if n is odd or even respectively. For thepurpose of discussion it is enough to consider both x and n as being even and that2

x ≥ 0, resulting in an integrand that is always positive. As x n is positive and increaseswith |x|, then x must tend to zero very quickly with increasing x for the integral to give azero value and if x ≠ 0, then the only non-zero contribution can be in the vicinity ofx 0 and this must be of very small width if it is to produce a zero integral.A more precise approach is provided via the definition of a generalised function. Considerthe following sequence of good functions,g n x n exp−nx 2 ,in which the function defining the sequence becomes greater in value at x 0 and narrowerin width about x 0asn → . It can also be shown that gn xdx 1 for all n, sothat−the sequence is area preserving and in agreement with the condition x dx 1−established above. This serves as a possible sequence of good functions for the Dirac deltafunction, confirming that it is a generalised function.If x → 0 sufficiently quickly as |x| → then we can consider its derivative ′ x andobtain its properties using integration by parts,− ′ xfx dx − −xf ′ x dx − f ′ 0 (3.1.3)and more generally− n xfx dx −1 n f n 0 . (3.1.4)There are two other functions that are useful in the application of Fourier theories; these arethe Heaviside Unit Function and the Sign Function. Both can be considered to begeneralised functions.Heaviside Unit Function. ThisisdefinedbyHx 1 x 0 0 x ≤ 0 (3.1.5)and this is useful in stipulating the range of a functions. There is no universal definition ofHx and the lack of consistency centres on the value at x 0; all definitions agree whenx ≠ 0.ExampleConsider the function defined by fx 1for|x| a and fx 0for|x| ≥ a. This can bewritten in terms of a single definition on −, as3

fx Hx a 1 − Hx − aSignum (Sign) Function. Thisisdefinedbysgnx 1 x 0 0 x 0 − 1 x 0 (3.1.6)and simply identifies the sign of the argument.ExampleConsider the function fx |x 3 | which can also be written as fx x 3 sgnx.4

3.2 Fourier Transforms3.2.1 The basic conceptsLet fx be a function such that fx is of total bounded variation and absolutely integrable,i.e. the integral of |fx| exists and is finite, then the Fourier Transform (FT) of fx isdefined by f k −fx e ikx dx . (3.2.1)This is now a function of the real variable k and sometimes also called the Complex FourierTransform. It should be noted that this not the only definition and there are others incommon use. Note that fx can be real- or complex-valued and that f k will generally becomplex-valued. It can be shown that if fx is a good function, then f k is also a goodfunction.The Fourier Cosine and Sine Transforms are defined by f ck 0fx coskx dx , f sk 0fx sinkx dx . (3.2.2)and they play the same roles for odd and even functions as in Fourier Series.Example 1Determine the FT of the function fx exp−a|x| where a is a real constant and a 0.We havef̂k −e −a|x| e ikx dx ,0− 1a ike ax e ikx dx 0e −ax e ikx dx 1a − ik2aa 2 k 2 .Example 2Determine the FT of the function fx 1 when |x| a and fx 0 when |x| a. Fromthe previous section, we know that this function can also be represented in terms of theHeaviside Unit Function.We have5

f̂k a−a1 e ikx dxa 1ike ikx −a2sinkakNote that the function has a discontinuity at x a but its FT is still defined.Example 3Suppose that fx is a real-valued even function, then determine its FT and show that it alsoeven.By definitionf̂k − −fx e ikx dxfx coskx isinkx dx −fx coskx dx i −fx sinkx dx 2−fx coskx dx 2f̂ckwhere the properties of coskx and sinxk being even and odd respectively has beenemployed together with the knowledge that fx is even. It is clear that f̂k is even, sincecoskx cos−kx and hence f̂−k f̂k.3.2.2 Derivatives and other useful identitiesFT of Derivative. The notation f̂k employed in the definition of the complex FTessentially defines a function of k. It is also possible to consider the definition as anoperator on fx, sothatFfx f k −fx e ikx dx .and F indicates the FT operator. This enables us to determine the FT of the derivatives offx,Ff ′ x −f ′ x e ikx dx fx e ikx −− ik −by definitionfx e ikx dxintegrating by parts Ff ′ x − ik Ffx − ik f̂k . (3.2.3)In this derivation , the property that fx → 0as|x| → if fx is absolutely integrable hasbeen employed.6

FT of Higher Derivatives. The higher derivatives can be obtained in a similar manner andit is straightforward to show thatFf ′′ x −ik 2 Ffx − k 2 f̂kFf n x −ik n Ffx −ik n f̂k (3.2.4)Derivative of FT. AstheFT f k of fx is a function of k, it can be differentiated in theusual way to gived f kdk d dk −−fx e ikx dx ix fx e ikx dx−ddk fx eikx dx d f kdk iFxfx or Fxfx − i d f kdk. (3.2.5)Thus if the FT of fx is known, then the FT of xfxcan be determined by differentiation.This can easily be extended to giveF x n fx −i n d n f kdk n .Both the result in (3.2.5) and the more general result require that x n fx is absolutelyintegrable for whatever value of n is being employed. This comes with a warning: if fx isabsolutely integrable, it should not be assumed that the same property holds for x n fx.Shift Operator. Consider the function fx e icx ,wherec is a real constant and the FT offx is known.F fx e icx −fx e icx e ikx dx −fx e ikcx dx f k c . (3.2.6)The appearance of c "shifts" the value of k at which the transform is evaluated.3.2.3 Convolution of two functionsAll of the previous discussion is solely concerned with a single function and sometimes it isnecessary to consider two or more functions. This can also arise when we considerfunctions that themselves can be considered to be the products of two or more functions.Suppose that we have two functions fx and gx, both of which are bounded andabsolutely integrable. The Convolution of fx and gx is a function, written f ∗ g, definedby7

fx fx −2 12−f̂k e −ikx dk ,analogous to the Dirichlet Conditions in Fourier Series. Usually we employ (3.2.9) but weknow that the above form is available if fx is discontinuous. This clearly requires f k tobe absolutely integrable as well.Tables of FTs and their inverses exist but beyond these standard forms, it becomesnecessary to evaluate (3.2.9) for the particular f̂k and this usually involves contourintegration. The integral in (3.2.9) is the same as the model form of Jordan’s Lemmapresented in (2.4.4) and the methods employed therein can be employed.All of this theory assumes that the integrals exist. With the stated conditions for fx, weknow that f̂k exists. The Riemann-Lebesque Lemma states if fx is also continuous thenf̂k → 0ask →, so this provides some theoretical support but not sufficient to ensureexistence in all cases.Combining (3.2.1) and (3.2.9) givesfx 12 fx 12−−f̂k e −ikx dk 12−−−fu e iku due −ikx dkfu e iku−x du dk (3.2.10)and this is the general Fourier Theorem that underpins the analysis.Return to (3.2.9) and replace x by −x to give2 f−x −f̂k e ikx dk ,and interchange x and k,2 f−k −f̂x e ikx dx .Thus the FT of f̂x is 2f−k and this identity can be useful in the manipulation anddetermination of transforms.9

ExampleFrom Example 1 of §3.2.1, it is known that the FT of the function fx exp−a|x|, where2a2aa is a real constant and a 0isf̂k . What is the FT of gx ?a 2 k 2 a 2 x 2Using the terminology of the question, we have gx f̂x and the result above states thatthe FT of f̂x is 2f−k. Thus the FT of gx is 2f−k, i.e.as |−k| |k|.ĝk 2 f−k 2 exp−a|−k| 2 exp−a|k|3.2.5 Raleigh / Plancheral TheoremsThese results were originally due to Raleigh and rigorously proved by Plancherel; they arethe FT equivalent of Parseval’s theorem in Fourier Series. Let fx and gx be real- orcomplex-valued functions and both absolutely integrable. Consider the integral of theproduct of fx and the complex conjugate gx, employing (3.2.1) and (3.2.9) gives− −−fx gx dx fx 1 2 gk e −ikx dk dx− 12− fx gk e ikx dk dx− 1 2 gk− fx e ikx dx dk−fx gx dx 1 2 f k gk dk (3.2.11)−A special case occurs when f g and (3.2.11) becomes−|fx| 2 dx 1 22 f k dk (3.2.12)−This relationship is referred to as the Conservation of Energy, whereas (3.2.11) sometimescalled the Power Relation.ExampleThe FT of the real-valued function fx exp−a|x| has already been shown in one of theexamples above to be 2a/a 2 k 2 . Applying (3.2.12) gives−|e −a|x| | 2 dx 12−2aa 2 k 22dk.The first integral can be evaluated10

−|e −a|x| | 2 dx 2 0e −2ax dx 1 aOn substitution and re-arrangement−1 dk a 2 k 2 2 2a 33.2.6 Alternative notationsIt has been stated previously that there are a number of definitions of the Fourier Transformin common use. The attractiveness of one alternative definition can be seen clearly from acomparison of (3.2.1) and (3.2.9), which shows that the inverse FT has a 1/2multiplicative factor, whereas the FT does not. For this reason, some texts employ a 1/ 2factor in transform and inverse; occasionally the 1/2 factor is used in the transform but noton the inverse. There is no standard practice.However, all are of the form f k C2−fx e iAkx dx fx |A|C − f k e −iAkx dkfor constants A and C. Some of the common combinations are given below.A C Fourier Transform f k Inverse Transform fx1 2 fx e ikx 1dx −2 − f̂k e −ikx dk−1 2 fx e −ikx 1dx −2 − f̂k e ikx dk1−1 1211 22 fx e −ikx dx − − fx e ikx 1dx −2−2 2 fx e −2ikx dx − − −f̂k e ikx dkf̂k e −ikx dkf̂k e 2ikx dk11

3.3 Applications of Fourier TransformsThe principal uses of Fourier Transforms are the solution of partial differential equations,possessing a particular structure, and the processing and properties of time series. It is alsopossible to invoke transform methods to solve certain types of integral equations and toenable the summation of certain types of slowly-converging series. Some of theseproperties are shown in the following examples.3.3.1 Solution of Ordinary Differential EquationsAlthough not especially useful for the solution of ordinary differential equations (ODEs), itis instructive to show how transform methods work on ODEs before considering thebroader structure of partial differential equations (PDEs). This is best illustrated by anexample.Example 1Obtain a solution of the Ordinary Differential Equation (ODE)d 2 ydx 2 − 2 y gx ,which → 0asx →, assuming that gx is absolutely integrable.The general solution to a linear ODE is often written in the formyx y 1 x y 2 x ,where y 1 x and y 2 x are the Complimentary Function (CF) and Particular Integral (PI)respectively. The complementary function isy 1 x Ae x Be −x ,for arbitrary constants A and B; these solutions do not → 0asx →. Thus only the PI ispotentially capable of meeting this requirement. For convenience we write yx instead ofthe correct y 2 x.Note that it is assumed that ≠ 0. If 0 can occur, then it is necessary to consider thiscase separately and the CF solution is y 1 x Ax B and which is also incapable ofmeeting the large x requirement unless y 1 x ≡ 0.Taking the FT of the equation gives12

−−d 2 ydx 2d 2 y− 2 y e ikx dx gx e ikx dxdx 2 −eikx dx − 2 ye ikx dx − gx e ikx dx−and this step can be taken because the integration operator is linear. Thus it is necessary todetermine the FT of each term of the equation. The FT of the first term is given by (3.2.4)to be−d 2 ydx 2 e ikx dx − k 2 ŷ ,and the FT of each of remaining terms is simply the transform itself, so that the transformedequation is− k 2 2 ŷk ĝkThis must be inverted to obtain the solution yx 12 −21 ŷk − −−ŷk e −ikx dkĝkk 2 2 .ĝkk 2 2 e −ikx dk .The usual approach would then require some knowledge of gx and its transform ĝk forthe integral to be evaluated. However, in this case, another approach is possible since ŷkis the product of ĝk and 1/k 2 2 ; these two quantities are the FTs of gx andexp−|x|/2 respectively, with the latter given as an earlier example. Thus theconvolution (3.2.8) applies and the inversion can be written asyx − 12 −gu e −|x−u| du ,which satisfies all requirements. The modulus sign in the exponential is important andensures that the solution has the correct behaviour as x →. Note that it is often onlypossible to write the solution of a differential equation as an integral representation.3.3.2 Partial Differential EquationsOne of the principal uses of FTs is to obtain the solution of a certain class of PartialDifferential Equations (PDEs). The PDE should be linear and the boundary conditionsshould be of a certain form; this is a large subject and the details cannot be considered here,so we just present a solution of one particular equation.13

Consider the PDE∂ 2 u∂x 2 ∂2 u∂y 2 0on the domain defined by − x and y ≥ 0. This equation is called thetwo-dimensional Laplace Equation and the domain is often referred to as the upper halfplane. The solution is required to satisfy ux,y gx on y 0,for some given functiongx, andux,y → 0asy → . The latter condition known as a Far-field or Boundednesscondition.The strategy is to take the FT of the PDE and the boundary condition. This gives an ODEthat we attempt to solve subject to the transformed boundary condition. It is hoped that thefinal solution can be obtained by inversion but from previous experience we know that anintegral equation may arise.As the boundary conditions are given in terms of y, wetaketheFTinx. Define uk,y −ux,y e ikx dx gk −gx e ikx dx uk,0 .The FT of the two terms in the PDE are−−∂ 2 u∂x 2 e ikx dx − k 2 uk,y using (3.2.4)∂ 2 u∂y 2 e ikx dx ∂2∂y 2−ux,y e ikx dx ∂2 u∂y 2 .We regard the resulting equation as one in which y is the variable and and k is a constant,so partial differentiation in y can be replaced by full differentiation. Thus we need to solvethe equationd 2 u − k 2 u 0,dy 2subject to uk,0 gk and uk,y → 0asy → .It may seem that the solution of this ODE is straightforward to obtain, since equations ofthis type are associated with expontentially increasing or decreasing solutions. However,the parameter k lies within the range − k and so there are three cases: k 0,k 0and k 0. The first and third can be combined by using k 2 |k| 2 and then the questionarises as to how k 0 should be accommodated.For k ≠ 0, the solution of this ODE is easily shown to be14

uk,y Ak e|k|y Bk e −|k|y .Now apply the boundary and far-field conditions. The far-field condition can only besatisfied if Ak 0 and the condition on y 0givesBk gk, thus uk,y gk e−|k|yandbyinversionux,y 12 gk exp−ikx |k|y dk .−In fact the transform uk,y is of convolution form and results obtained previously can beemployed to determine the functions involved. The details are not presented here and thefinal form isux,y y −gkx − k 2 y 2 dk .and this is known as Poisson’s formula.3.3.3 Integral EquationsIntegral equations occur when the unknown function occurs as part of an integrand ratherthan in a differential equation. If the integral is over a fixed domain - such as the finitedomain a,b or an infinite one −,, then the integral equation is called a FredholmEquation. If the domain varies with the unknown x, then the equation is known as aVolterra Equation. Only equations of Fredholm type are considered here.Fredholm equations are divided into two main categories, known as the First Type and theSecond Type; there is a Third Type but this can be re-written as one of second type and isnot considered here. The general form of these equations types, for unknown fx, isType 1 hx abkx,t ft dtType 2fx hx abkx,t ft dtwhere is a constant. The function kx,t is called the Kernel and hx is called the FreeTerm. Ifhx 0 in type 2, then the equation is sometimes said to be homogeneous.Both of these definitions apply only to linear equations. Thus if f 1 x and f 2 x are bothsolutions of either a type 1 or type 2 equation, then fx Af 1 x Bf 2 x is also a solutionfor constants A and B. This comment is made despite the fact that we do not yet know how15

many solutions to expect for such equations.Fredholm equations with a kernel of convolution form.Consider the particular formAfx ft gx − t dt hx ,−where A is a real constant and the functions gx and hx are known; the kernel g appearsas a convolution and the range of integration is infinite. Note that by a change of variablein the integral, the equation can also be written asAfx fx − t gt dt hx ,−and so the two equations can be considered as equivalent. This equation is of type 2 ifA ≠ 0 and of type 1 if A 0.Take the complex Fourier Transform of the equation, assuming that f,g and h all possessthe required properties for the transform to exist,and this can be rearranged to giveAf̂k f̂k ĝk ĥk A ĝk f̂k ĥkf̂k ĥkA ĝk .Thus the solution is given by the inverse Fourier Transformfx 12 f̂k e −ikx dk 1−2−ĥkA ĝk e−ikx dkand it is assumed that this can be integrated, if only numerically. If A ≠ 0 and both ĥkand ĝk are absolutely integrable, then the integral exists.The nonlinear equation: special case fx gxIf fx gx, then the equation becomesAfx ft fx − t dt hx .−This is no longer of Fredholm type, since the Fredholm equation is linear whereas this oneis not.Fourier techniques do not work generally on nonlinear equations but will work in this case16

ecause of the convolution structure. The transformed equation becomesf̂k2 Af̂k − ĥk 0and this is a quadratic equation that can be solved to givef̂k − A2 A22 ĥk .Both roots are important. It is sometimes possible to deduce what the solution should be bylooking at the structure. Write the two solutions asf̂1k − A 2 A22 ĥk , f̂2k − A 2 − A22 ĥk .If ĥk is absolutely integrable, then ĥk → 0ask → and sof̂1k → 0,f̂2k → −A.Thus if we require f̂k to be absolutely integrable, then only f̂1k is acceptable. If A 0,then both solutions can be important and an example has been previously given.ExampleUse the FT to solve the integral equation−fx − t ft dt exp − x22and check your answer by direct substitution.Outline of solution onlyStep 1: From (3.2.7), the integral is the convolution f ∗ fx. Take the FT of the equationand use (3.2.8) to obtainf̂k f̂k F exp − x 22 f̂k F exp − x22 gk sayStep 2: Invert transform using (3.2.9)fx 12−gk e −ikx dkStep 3: Determine fx, using contour integration if necessary. The solution should be17

substituted into the original equation to check that it is correct.The final result is found to be of the formfx A exp−ax 2 where the constants A and a are determined as part of the solution process (determinethem!). If we had been given this at the start, then A and a could have been determined bysubstituting directly into the integral equation without having to employ FTs.Note: Step1 requires the use of contour integration to determine gk and the result−e −x2 dx .3.3.4 Signal ProcessingSignal processing usually involves the time t and frequency as the two variables, givingrise to time-domain and frequency domain modelling. These quantities are related via theFT, i.e.Fft f ft eitdt .−If the signal ft is such that ft 0 for all |t| ≥ T, then the signal is called Time-Limitedand corresponds to a signal of finite length. If the signal is such that f 0 for all|| ≥ L, then the signal is called Band Limited and is restricted to a range of frequencies. Aband limited signal can be obtained by passing a full signal through a low-pass filter. It canbe shown, though we do not provide a proof, that a signal cannot be both time- andband-limited.The Shannon Sampling Theorem demonstrates the link between FT and FS for aband-limited signal. Let ft be the measured time signal and f be its band-limitedtransform. Using the inverse FT givesft 12 f e−itd 1−2L−L f e−itdand at the point t n/L, we obtainf nL 12L−L f exp −i nLd .Let H be a periodic complex-valued function of period 2L, defined to be equal to the18

complex FS of f on −L ≤ ≤ L. This function can be written asH ∑n−c nexp i nL~ f .From the earlier theory on Fourier series, we know that the series H converges to f at points of continuity and that the coefficients c n are given byLc n 12L−L H L ∑ n− f exp −i nLd L f nLf nLexp i nL.The FT f is band-limited and we can relate this to H by introducing the functionG such thatto giveG 1 when || L 0 when || L. f ~ H G .Thus f is the product of two functions and can be considered to be the product of twotransforms, so that ft could be written in convolution form. It is slightly easier, andequivalent, to employ the direct inversion formula to obtainft ~ 12 H G e−itd .−Now substitute for the function G andthenforH, ftft ~ 12 12 12L~ ∑n−L−L∑n− L−LH e−itdL∑n−f nL−Lf nLsinn − Ltn − Ltf nLexp i nLLexp i nL− t de −it dand which can be made exact by careful justification. This enables an approximation to be19

made to the signal form by sampling at a finite number of chosen points and using atruncated series.20

3.4 Fourier Transforms and Generalised Functions3.4.1 The Dirac Delta FunctionWe know from (3.1.1.) that the Dirac Delta Function is defined by−xgx dx g0 ,for any piecewise continuous function gx. Ifwetakegx expikx, whichiscontinuous, then−x e ikx dx 1 k . (3.4.1)Thus the FT of x is given by k 1 provided that the definition of the FT in (3.2.1)remains applicable.The two conditions for the existence of the FT is that fx is of total bounded variation andabsolutely integrable. The function x satisfies the appropriate conditions as |x| → butmay not be bounded. Conversely, k 1 is a bounded function but not absolutelyintegrable.Suppose that the Fourier Theorem (3.2.10) holds for x, thenx 12 x 12−−−u e iku−x due −ikx dk 12−dke ikx dk (3.4.2)and this is a real-valued integral, since the imaginary part is zero. If this representation canbe justified, which the requires the use of generalised functions, it also means that x canbe represented as an improper integral whose integrand is bounded but is not absolutelyintegrable.By interchanging k and x in (3.4.2), we obtaink 12−e ikx dx , (3.4.3)and the resulting integral is real-valued and an even function of k, as it is easily shown thatits imaginary part is zero. A consequence of (3.4.3) is that k can be considered to be theFT of the function fx 1/2.The same integral appears in (3.4.2) and (3.4.3) and does not exist in a conventional sense.Denote this integral by Ik and adopt a slightly different perspective. Define two functionsI 1 k and I 2 k by21

I 1 k limX→X−Xe ikx dx , I 2 k lim n→n k−nke ikx dx ,where X and n are a real variable and positive integer respectively. If Ik exists in theconventional sense, then Ik I 1 k I 2 k.For k ≠ 0, we obtainI 1 k limX→sinkXk, I 2 k lim n→sinnn .The function I 1 k does not exist for a fixed value of k, since its value varies continuouslywith X in an oscillatory manner and within the range −1/k,1/k. However, I 2 k does existand furthermore I 2 k 0. This is because the limit is attained via a sequence in whichevery value is zero.Thus the interpretation of Ik via I 2 k is consistent with the sequence approach for thedelta function, provided that k ≠ 0. If k 0, then integrals in (3.4.1) and (3.4.2) are bothinfinite and this matches our intuitive interpretation of the delta function. This is not aformal justification of the FT of x, nor is intended to be, but simply shows the type ofapproach necessary to employ generalised functions.3.4.2 ApplicationsConsider the equation of simple harmonic motiond 2 ydx 2 2 y 0,for which the solution is known to be given by yx Ccosx Ssinx, for realconstants C and S. An equivalent form is yx Aexpix Bexp−ix, for complexconstants A and B. Both forms of the solutions are bounded but neither is absolutelyintegrable and so standard FT theory may not seem to apply. However, this is a commonODE and often provides the LHS of a more complicated equation. It also arises quitecommonly in PDE systems after the FT has been taken and is associated especially withwavelike systems.Nevertheless, take the FT of the equation and assume that this can be justified,− k 2 − 2 yk 0 k k − yk 0.This tells us that yk 0 when k ≠ but what happens when k ?There is a relevant result in the theory of generalised functions: if gk is a generalised22

function and kgk 0, then gk is a constant multiple of k. Hence gk Qk, forsome constant Q. This means that yk is of the form yk 2A k 2B k − for constants A and B and where the 2 factor has been included for convenience.Inverting this givesyx 12 2 A k B k − e −ikx dk− Ae ix Be −ix ,in agreement with the earlier form.This illustrates clearly why generalised functions are required for the deployment of theFourier theory. The fundamental solutions of the equations cannot be obtained withoutthem.3.4.3 Fourier Transforms and Fourier SeriesThe complex FS for a real or complex-valued function fx, defined on −,, is given in§1.7 to befx ~ ∑n−c n e inx , where c n 12 fxe −inx dx−and generally the coefficients c n will be complex. We assume that the series converges.Suppose we wish to obtain the FT of this FS, how do we proceed? The function fxdescribes a function that is 2-periodic and if we attempt to integrate over the range−, we meet the same problems as encountered previously. However, if we deal withintervals that are multiples of 2, as with the quantity I 2 k above, then the limiting processand the integral are well-defined. Furthermore, the sequence is composed of goodfunctions.Assuming that such issues can be settled, take the FT of fx in the usual way23

f k −fx e ikx dx−∑n−c n∑n−c n e inx−e ikx dxe iknx dx f k 2 ∑ c n k n . (3.4.4)n−Thus the FS is a sum of delta functions and each is associated with a particular frequency(or harmonic) in the series.A series like (3.3.4) is sometimes called a row of delta functions. Furthermore the terms in(3.3.4) are spaced a unit distance apart, so may be considered as a periodic function.24