Digital Control Systems [MEE 4003] - Kckong.info

Digital Control Systems [MEE 4003] 

Kyoungchul Kong 

Assistant Professor 

Department of Mechanical Engineering 

Sogang University 

Draft date September 6, 2012

Preface 

For inspiration and motivation of my students. 

(Parts of this class note are copied and edited from articles available on the web.) 

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Chapter 1 

Introduction 

1.1 Problem Definition 

1.1.1 Control systems 

Control theory is an interdisciplinary branch of engineering and mathematics that deals 

with the behavior of dynamic systems. The desired output of a system is called the reference. 

When one or more output variables of a system need to follow a certain reference 

over time, a controller manipulates the inputs to a system to obtain the desired effect on 

the output of the system. 

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Figure 1.1: The concept of the feedback loop to control the dynamic behavior of 

the system: this is negative feedback, because the sensed value is subtracted from 

the desired value to create the error signal, which is amplified by the controller. 

1.1.2 Regulation and tracking control 

When the reference is constant, the control process is called regulation. On the other 

hand, when the reference is a time-varying quantity, the control process is called tracking 

control. Note that the control process is identical for both regulation and tracking control. 

2

1.1. PROBLEM DEFINITION 3 

(a) 

(b) 

(c) 

(d) 

Figure 1.2: Examples of controlled systems. (a): a robot arm, (b): a helicopter; 

(c): an active suspension, (d): an air conditioning system 

[Example 1-1] The systems in Fig. 1.2 are examples of controlled systems. The robot 

arm in (a) is a tracking control system, because the reference of the robot joint is 

time-varying in general. On the other hand, the remaining systems are all regulation 

systems. For example, once the desired height of a helicopter is set, the rotor speed is 

automatically controlled to maintain the height. For (c) and (d), find the reason why 

they are regulation systems. 

Disturbance 

Disturbance is an undesired input that affects the performance of the overall control system. 

Disturbance includes an environmental change, an external force, a change in system 

parameter, etc. 

1.1.3 System 

System is a set of interacting components forming an integrated whole. Every system has 

input(s) and output(s). Most systems share common characteristics, including: 

• Systems have structure, defined by components and their composition, 

Digital Control Systems, Sogang University 

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• Systems have behavior, which involves inputs, processing and outputs of material, 

energy, information, or data, 

• Systems have interconnectivity: the various parts of a system have functional as 

well as structural relationships to each other, 

• Systems may have some functions or groups of functions 

Plant 

A plant in control theory is the combination of process and actuator. In particular, a plant 

is the system to be controlled. 

1.1.4 Signal 

A signal is any time-varying or spatial-varying quantity. In the physical world, any quantity 

measurable through time or over space can be considered as a signal. More generally, 

any set of human information or machine data can also be considered as a signal. Such 

information or machine data must all be part of systems existing in the physical world. 

[Example 1-2] Examples of signals follow. 

• Motion: The motion of a particle through some space can be considered as a 

signal, or can be represented by a signal. In general the position signal is a 3- 

vector signal. If orientation is considered, it is a 6-vector signal. 

• Sound: Since a sound is a vibration of a medium (such as air), a sound signal 

is related to the pressure value of air. A microphone converts sound pressure at 

some place to a function of time, generating a voltage signal that is proportional 

to the sound signal. Sound signals can be sampled to a discrete set of time points. 

For example, compact discs (CDs) contain discrete signals representing sound, 

recorded at 44,100 samples per second. 

• Images: A picture or image consists of a brightness or color signal, a function of 

a two-dimensional location. A 2D image can have a continuous spatial domain, 

as in a traditional photograph or painting; or the image can be discretized in 

space, as in a raster scanned digital image. Color images are typically represented 

as a combination of images in three primary colors, so that the signal is vectorvalued 

with dimension three. 

• Videos: A video signal is a sequence of images. A point in a video is identified 

by its two-dimensional position and by the time at which it occurs, so a video 

signal has a three-dimensional domain. Analog video has one continuous domain 


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dimension (across a scan line) and two discrete dimensions (frame and line). 

1.1.5 Continuous-time and discrete-time signals 

If the quantities are defined only on a discrete set of times, we call it a discrete-time signal. 

The discrete-time signal can be indexed by an integer that represents the sequence of each 

data point. 

On the other hand, a continuous-time real signal is any real-valued function that is 

defined for all timetin an interval. 

[Example 1-3] An example of the continuous-time and discrete-time signals is shown 

in Fig. 1.3. In general, the discrete-time signal is a sampled version of its corresponding 

continuous-time signal. 

0.9 

0.9 

0.8 

0.8 

0.7 

0.7 

0.6 

0.6 

Amplitude 

0.5 

0.4 

Amplitude 

0.5 

0.4 

0.3 

0.3 

0.2 

0.2 

0.1 

0.1 

0 

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 

Time (sec.) 

0 

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 

Time (sec.) 

(a) 

(b) 

Figure 1.3: Continuous-time and discrete-time domain signals. 

(a): a continuous-time signal, (b): a discrete-time signal 

1.1.6 Digital control 

Digital control is a branch of control theory that uses digital computers to act as system 

controllers. Depending on the requirements, a digital control system can take the form 

of a microcontroller (costs less than 10,000 KRW) to a digital signal processor (DSP, 

costs more than 10,000,000 KRW). Since a digital computer has finite precision (i.e., 


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Figure 1.4: 4-channel analog-to-digital converter WM8775SEDS made by Wolfson 

Microelectronics placed on an X-Fi Pro sound card. 

quantization), extra care is needed to ensure that the errors in coefficients, A/D conversion, 

D/A conversion, etc. do not produce undesired or unplanned effects. 

The benefits of a digital control system include 

• Inexpensive: under $5 for many microcontrollers 

• Flexible: easy to configure and reconfigure through software 

• Scalable: programs can scale to the limits of the memory or storage space without 

extra cost 

• Adaptable: parameters of the program can change with time 

• Static operation: digital computers are not easily affected by environmental conditions 

than capacitors, inductors, etc. 

Analog-to-digital converter 

An analog-to-digital converter (ADC, A/D, or A to D) is a device that converts a continuous 

quantity to a discrete time digital representation. Typically, an ADC is an electronic 

device that converts an input analog voltage to a digital number proportional to the magnitude 

of the voltage. 

Digital-to-analog converter 

A digital-to-analog converter (DAC, D/A, or D-to-A) is a device that converts a digital 

code to an analog signal (voltage, current, or electric charge). 


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Figure 1.5: 8-channel digital-to-analog converter Cirrus Logic CS4382 as used 

in a soundcard. 

0.9 

0.8 

0.7 

0.6 

Amplitude 

0.5 

0.4 

0.3 

0.2 

T 

0.1 

0 

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 

Time (sec.) 

Figure 1.6: The definition of sampling period. 

1.1.7 Sampling Rate 

The sampling rate, sample rate, or sampling frequency defines the number of samples 

per unit of time (usually seconds) taken from a continuous signal to make a discrete 

signal. For time-domain signals, the unit for sampling rate is hertz[Hz] (inverse seconds, 

1/s, s −1 ), sometimes noted as Sa/s (samples per second). The inverse of the sampling 

frequency is the sampling period or sampling interval, which is the time between samples. 

T shown in Fig. 1.6 is the sampling period, and 1 is the sampling frequency. 

T 


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1.2. EXAMPLES 8 

1.2 Examples 

1.2.1 Vehicle Cruise Control 

Suppose you have a vehicle with a cruise control function. The cruise control unit (CCU) 

calculates the amount of acceleration from the measured speed error to maintain a set 

speed. The vehicle speed is measured by a tachometer. 

• What are the signals 

• What is the plant 

• What is the controller 

• What is the sensor 

• How is the block diagram represented 

• What are the expected disturbances 

1.2.2 Air Conditioner 

An air conditioner is installed in a room. The desired temperature is 24 ◦ C, and the current 

temperature is being measured by a thermocouple. The fan of the air conditioner is 

controlled such that its desired speed is proportional to the temperature error. The speed 

of the fan is controlled by an electric current flowing through the fan, where the electric 

current is controlled by a motor driver. 


• What are the systems 






1.2.3 Robot Arm Control 

A motor installed at a joint of an industrial robot is to be controlled. The desired angular 

position of the motor is a sine wave with an amplitude of1radian and a frequency of1Hz. 

It is known that the motor follows the equation of motionM¨Θ+C ˙Θ+KΘ = τ, whereΘ 

is the angular position of the motor andτ is the torque generated by the motor. The motor 

torque is proportional to the electric current flowing through the motor, where the electric 

current is regulated by a motor driver. The motor driver is connected to the computer via 

a D/A converter, and the torque command is transferred by an analog voltage signal. A 


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1.3. REFERENCES 9 

digital controller generates the torque command to be proportional to the angular position 

error. The angular position is measured by a potentiometer. 


• What are the systems 






• Is this system a continuous or digital control system 

1.3 References 

1. Wikipedia, available on-line: www.wikipedia.com 


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Chapter 2 

Review of Continuous-Time Domain 

Control Theory 

2.1 State Space Realization of Dynamic Systems 

2.1.1 Differential equations: state space realization 

Dynamics of systems can be described by equations of motions, which are in general 

described by differential equations. 

Modeling of complicated mass-spring-damper systems 

A mechanical system with multiple masses, springs, and dampers can be described as a 

mathematical model: 

Mẍ+Cẋ+Kx = u 

where 

M ∈ R n×n is the mass matrix, 

C ∈ R n×n is the damping coefficient matrix, 

K ∈⎡R n×n ⎤is the spring constant matrix, 

x 1 

x 2 

x = ⎢ ⎥ 

⎣ . ⎦ ∈ Rn is the position vector (x 1 , ...,x n are the position of each mass), 

x n 

⎡ 

u = ⎢ 

⎣ 

⎤ 

u 1 

u 2 

⎥ 

. 

u n 

⎦ ∈ Rn is the input vector (u 1 , ...,u n are the force applied to each mass). 

10

2.1. STATE SPACE REALIZATION OF DYNAMIC SYSTEMS 11 

k 

M 

x 

u 

x1 

2 

u1 

2 

m1 

u 

k 

x 

m 2 

(a) A mass-spring system 

(b) Double mass-spring system 

k1 

k2 

m1 

m2 

c 

c1 

2 

k 4 

m 3 

x 1 

x2 

x3 

(c) Complicated mass-spring-damper system 

k 3 

u 

c 

k 

x 1 

u 1 

(d) Creeping phenomenon of a steel material 

P 

x 

v 

(e) Buckling phenomenon of a rod 

Figure 2.1: Dynamic systems 

The matrices,M, C, and K, are defined as follows. 

• M ii is the mass value of thei th mass. 

• M ij,i≠j = 0. 

• C ii is the sum of damping coefficients of all dampers connected to thei th mass. 

• C ij,i≠j is the negative value of the sum of damping coefficients of all dampers connected 

between thei th mass and thej th mass. 

• K ii is the sum of spring constants of all springs connected to thei th mass. 

• K ij,i≠j is the negative value of the sum of spring constants of all springs connected 

between thei th mass and thej th mass. 


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[Example 2-1] Differential equations of the systems in Fig. 2.1 are as follows. 

(a) mẍ+kx = u 

[ ] 

x1 

(b) The output vector is x = ∈ R 

x 2 , and the input vector is u = 

2 

The entries of the matrices are defined as follows. 

• M 11 = m 1 and M 22 = m 2 . 

[ 

u1 

u 2 

] 

∈ R 2 . 

• M 12 = M 21 = 0. 

• Since there is no damper in the system,C = 0. 

• The spring k is the only spring connected to m 1 , i.e., K 11 = k. By the same 

reason, K 22 = k. 

• The spring k is the only spring connected between m 1 and m 2 . Since K ij,i≠j is 

the negative value of the sum of spring constants,K 12 = K 21 = −k. 

Therefore, the differential equation of the two mass-spring system is 

[ ] [ ] [ ] 

m1 0 0 0 k −k 

ẍ+ ẋ+ x = u 

0 m 2 0 0 −k k 

⎡ 

(c) The output vector is x = ⎣ 

⎤ 

x 1 

x 2 

x 3 

⎦ ∈ R 3 , and the input vector is u = ⎣ 

⎡ 

0 

0 

u 

⎤ 

⎦ ∈ R 3 . 

Note that no force is applied to m 1 and m 2 . The entries of the matrices are defined as 

follows. 

• M 11 = m 1 , M 22 = m 2 , and M 33 = m 3 . 

• M 12 = M 13 = M 21 = M 23 = M 31 = M 32 = 0. 

• The dampers connected to m 1 are c 1 and c 2 . Thus, C 11 = c 1 +c 2 . On the other 

hand, c 2 is the only damper connected to m 2 , i.e., C 22 = c 2 . Since no damper is 

connected tom 3 , C 33 = 0. 

• The damper(s) connected between m 1 and m 2 is c 2 . Thus, C 12 = C 21 = −c 2 . 

There is no damper between m 1 and m 3 , i.e., C 13 = C 31 = 0. Similarly, C 23 = 

C 32 = 0. 

• The springsk 1 ,k 2 , andk 4 are connected tom 1 , i.e.,K 11 = k 1 +k 2 +k 4 . Similarly, 

K 22 = k 2 +k 3 and K 33 = k 3 +k 4 . 


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c 

x 2 

k 

0 0 

x 1 

u 1 

Figure 2.2: An equivalent model of the creeping phenomenon. 

• The springk 2 is placed betweenm 1 andm 2 . Thus,K 12 = K 21 = −k 2 . Between 

m 2 and m 3 , the spring k 3 is placed, i.e., K 23 = K 32 = −k 3 . Similarly, K 13 = 

K 31 = −k 4 . 

⎡ 

⎣ 

Finally, the mathematical model of the three mass-spring-damper system is 

⎤ ⎡ ⎤ ⎡ 

⎤ 

m 1 0 0 c 1 +c 2 −c 2 0 k 1 +k 2 +k 4 −k 2 −k 4 

0 m 2 0 ⎦ẍ+ ⎣ −c 2 c 2 0 ⎦ẋ+ ⎣ −k 2 k 2 +k 3 −k 3 

⎦x 

0 0 m 3 0 0 0 −k 4 −k 3 k 3 +k 4 

(d) The modeling of creep phenomenon of materials is introduced in this example. In 

materials science, creep is the tendency of a solid material to slowly move or deform 

permanently under the influence of stresses. In order to find the mathematical model 

of the creep phenomenon, an equivalent model is introduced with fictitious masses 

(m = 0) as in Fig. 2.2. 

wherex = 

The mathematical model of the equivalent model is 

[ ] [ ] [ 

0 0 0 0 

ẍ+ ẋ+ 

0 0 0 c 

] [ ] 

∈ R 

x 2 u1 

and u = ∈ R 

2 0 

2 . 

[ 

x1 

k −k 

−k k 

] 

x = u 

(e) In the figure, x is the position from the top of the bar, andv is the distance from the 

vertical line to the center line of the deflected bar. The internal moment in the bar, M, 

is related to its deflected shape, v, by 

EI d2 v 

dx 2 = M 

The internal moment,M, is determined by the applied force, P , and the distance from 

the vertical line, v (i.e., M = −Pv). Therefore, the bar under a compressive load is 

modeled by 

EI d2 v 

dx 2 +Pv = 0 


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State space realization of linear systems 

Consider the equation of motion of a system: 

d n x 

a n 

dt +a d n−1 x 

n n−1 

dt +···+a dx 

n−1 1 

dt +a 0x = bu (2.1) 

where a 0 , a 1 , ..., a n , and b are constants all in R. The input is u ∈ R, and the output is 

x ∈ R. Note that the order of the equation of motion is n. 

The state space realization of the given equation of motion is obtained as follows. 

[Step 1] Find a variable that has the lowest order in the equation of motion and the output. 

In the example of (2.1), x is the variable with the lowest order. Let the variable bex 1 . 

[Step 2] Define the derivative ofx 1 as a new variable,x 2 , i.e., 

d 

dt x 1 = x 2 

[Step 3] Repeat defining new variables up to x n . For example, 

d 

dt x 2 = x 3 

d 

dt x 3 = x 4 

. 

d 

dt x n−1 = x n 

Note thatx 1 = x, x 2 = dx 

dt , x 3 = d2 x 

dt 2 , ...,x n = dn−1 x 

dt n−1 . 

[Step 4] Rearrange the equation of motion such that only the highest order term is remained 

on the left hand side, i.e., 

d n x 

dt = − 1 [ 

] 

d n−1 x 

a n n−1 

a n dt +···+a dx 

n−1 1 

dt +a 0x + b u 

a n 

The equation above can be rewritten using the new variables, i.e. 

d 

dt x n = − 1 [a n−1 x n +···+a 1 x 2 +a 0 x 1 ]+ b u 

a n a n 


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The n new first-order differential equations can be arranged into the following matrix 

form: 

⎡ ⎤ ⎡ 

⎤⎡ 

⎤ ⎡ ⎤ 

x 1 0 1 0 0 x 1 0 

d 

x 2 

0 0 1 ... 0 

x 2 

0 

x 3 

= 

0 0 0 0 

x 3 

+ 

u 

dt ⎢ ⎥ ⎢ 

⎣ . ⎦ ⎣ . 

.. ⎥⎢ 

⎥ ⎢ ⎥ 

. ⎦⎣ 

. ⎦ ⎣ 

0. ⎦ 

x n − a 0 

a n 

− a 1 

a n 

− a 2 

a n 

... − a n−1 

b 

a n 

x n a n 

where the output is 

⎡ 

y = [ 1 0 0 0 0 ] ⎢ 

⎣ 

⎤ 

x 1 

x 2 

x 3 

⎥ 

. ⎦ 

x n 

A state space realization has been obtained. If the original equation of motion consists of 

two or more differential equations, you may repeat the same process until the state space 

representation of the whole system is obtained. 

[Example 2-2] An equation of motion, mẍ = u, where the input is u ∈ R and the 

output isx ∈ R, is to be represented in a state space. 

The variable that has the lowest order in the equation of motion and the output is 

x, which is shown in the output. Let x be x 1 . Following the steps listed above, a new 

variable is defined, i.e., 

d 

dt x 1 = x 2 

Note thatx 2 = ẋ. The derivative ofx 2 is 

d 

dt x 2 = ẍ = 1 m u 

Arranging the two new first-order differential equations, 

[ ] [ ][ ] 

d x1 0 1 x1 0 

= +[ 

1 

dt x 2 0 0 x 2 m 

where the output is 

y = [ 1 0 ][ x 1 

x 2 

] 

] 

u 


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[Example 2-3] The governing equation of the two-mass-spring system 

[ ][ ] [ ][ ] [ ][ ] 

m1 0 ẍ1 0 0 ẋ1 k −k x1 

+ + = 

0 m 2 ẍ 2 0 0 ẋ 2 −k k x 2 

where the outputs are x 1 andx 2 , is to be represented in a state space. 

i.e., 

[ 

u1 

u 2 

] 

Note that the original equation of motion consists of two differential equations, 

ẍ 1 = 1 m 1 

[−kx 1 +kx 2 ]+ 1 m 1 

u 1 (2.2) 

ẍ 2 = 1 m 2 

[kx 1 −kx 2 ]+ 1 m 2 

u 2 (2.3) 

The variables with the lowest order in the equation of motion and the output arex 1 and 

x 2 . Let them be y 1 and z 1 . New variables, y 2 and z 2 , are defined as the derivatives of 

y 1 and z 1 , respectively, i.e. 

d 

dt y 1 = y 2 

d 

dt z 1 = z 2 

Differentiating once more, the original differential equations in (2.2)–(2.3) appear, i.e. 

d 

dt y 2 = 1 m 1 

[−ky 1 +kz 1 ]+ 1 m 1 

u 1 

d 

dt z 2 = 1 m 2 

[ky 1 −kz 1 ]+ 1 m 2 

u 2 

Arranging the new first-order differential equations, 

⎡ ⎤ ⎡ ⎤⎡ 

⎤ ⎡ 

y 1 0 1 0 0 y 1 

d 

⎢ y 2 

⎥ 

dt ⎣ z 1 

⎦ = −k k 

⎢ m 1 

0 

m 1 

0 

⎥⎢ 

y 2 

⎥ 

⎣ 0 0 0 1 ⎦⎣ 

z 1 

⎦ + ⎢ 

⎣ 

z 2 0 −k 

m 2 

0 z 2 

k 

m 2 

0 0 

1 

m 1 

0 

0 0 

0 

1 

⎤ 

[ ⎥ u1 

⎦ 

m 2 

u 2 

] 

Since the outputs of the system arey 1 = x 1 and z 1 = x 2 , 

⎡ ⎤ 

[ ] 

x 1 

1 0 0 0 

y = ⎢ x 2 

⎥ 

0 1 0 0 ⎣ x˙ 

1 

⎦ ∈ R2 

x˙ 

2 


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[Example 2-4] Find the state space realizations of the following dynamic models: 

(a) a 1 ẋ+a 2 x = u, where the input isu ∈ R and the output isx ∈ R. 

(b)ẍ+a 1 ẋ+a 2 x = b 1 u, where the input is u ∈ R and the output isẋ ∈ R. 

(c) a 1 ẍ+a 2 ẋ+a 3 x = b 1 u+b 2˙u, where the input isu ∈ R and the output is x ∈ R. 

[ ][ ] [ ][ ] [ ] 

m1 0 ẍ1 2k −k x1 u1 

(d) + = , 

0 m 2 

[ 

ẍ 2 

] 

−k 3k x 2 

[ 

u 2 

] 

u1 

where the input is ∈ R 

u 2 x1 + x˙ 

and the output is 1 

∈ R 

2 x 2 + x˙ 

2 . 

2 

[ ][ ] [ ][ ] [ ][ ] [ ] 

m1 0 ẍ1 c −c x1 ˙ k −k x1 u1 + u˙ 

(e) + + = 1 

, 

0 m 2 

[ 

ẍ 2 

] 

−c c x˙ 

2 

[ 

−k 

] 

k x 2 u 2 + u˙ 

2 

u1 

where the input is ∈ R 

u 2 x1 

and the output is ∈ R 

2 x 2 . 

2 

2.1.2 Solution to state space equations 

For a state space model, 

ẋ = Fx+Gu 

y = Hx 

whereF ∈ R n×n ,G ∈ R n×m , andH ∈ R p×n . The initial condition ofx(t) is given asx 0 . 

In order to find a solution (x(t)) for an arbitrary input (u(t)), e −Ft is multiplied to 

the both sides of the state space equation: 

e −Ft ẋ = e −Ft Fx+e −Ft Gu 

Rearranging the equation above and integrating the new equation, we get 

∫ t 

0 

( 

e 

−Fτẋ−e −Fτ Fx ) ∫ t 

dτ = e −Fτ Gu(τ)dτ fort ≥ 0 

Note that the integration variable has been replaced to τ to avoid confusion between the 

integration variable and the time index. The left hand side is the definite integration, and 

thus 

e −Ft x(t)−x(0) = 

∫ t 

0 

0 

e −Fτ Gu(τ)dτ fort ≥ 0 


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Figure 2.4: Matlab code for calculating the output signal from an arbitrary input. 

lsim.m function calculates the output from an arbitrarily defined input signal. You 

may also use step.m and impulse.m for obtaining step and impulse responses. 

By multiplying e Ft to the both sides of the equation above, and by using the initial condition 

(x(0) = x 0 ), the solution of the state space equation under an arbitrary input is 

obtained: 

∫ t 

x(t) = e Ft x 0 +e Ft e −Fτ Gu(τ)dτ fort ≥ 0 

Sincee Ft is not a function ofτ, the solution can be reduced to 

x(t) = e Ft x 0 + 

This process is called convolution. 

∫ t 

0 

0 

e F(t−τ) Gu(τ)dτ for t ≥ 0 (2.4) 

In order to find a complete solution of (2.4), it is necessary to solve e Ft . A simple 

method to the solution ofe Ft is the Taylor expansion. Namely, 

e Ft = I +Ft+ 1 2 F2 t 2 + 1 6 F3 t 3 +... = 

∞∑ 

i=0 

1 

i! Fi t i 


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[Example 2-5] If F = 

[ 

0 1 

0 0 

] 

, e Ft is 

e Ft = I +Ft+ 1 2 F2 t 2 + 1 6 F3 t 3 +... 

[ ] [ ] 

1 0 0 t 

= + 

0 1 0 0 

[ ] 

1 t 

= 

0 1 

[Example 2-6] Suppose a state space model 

[ 

d 0 1 

dt x = 0 0 

y = [ 1 0 ] x 

x(0) = 0 ∈ R 2 

] [ 0 

x+ 

1 

is under a unit step input, which is defined as 

{ 

1 fort ≥ 0 

u(t) = 

0 fort < 0 

] 

u 

By using the result of the previous example, the state is calculated as 

Then, the output is 

∫ t 

x(t) = e Ft x(0)+ e F(t−τ) Gu(τ)dτ 

0 

∫ t 

[ ][ ] 

1 t−τ 0 

= 

dτ 

0 

0 1 1 

[ 1 

] 

= 

2 t2 

fort > 0 

t 

y(t) = [ 1 0 ] x(t) = 1 2 t2 fort > 0 


Kyoungchul Kong


Solution of diagonalizable state space models 

For a diagonalizable matrix F = VΛV −1 ∈ R n×n , where Λ is a diagonal matrix with the 

eigenvalues ofF ,e Ft = e VΛV −1t is calculated by the Taylor expansion as follows. 

e (VΛV −1 )t 

= I +VΛV −1 t+ 1 2 VΛV −1 VΛV −1 t 2 + 1 6 VΛV −1 VΛV −1 VΛV −1 t 3 +... 

[ 

= V I +Λt+ 1 2 Λ2 t 2 + 1 ] 

6 Λ3 t 3 +... V −1 

[ ∞ 

] 

∑ 1 

= V 

i! Λi t i V −1 

i=0 

= Ve Λt V −1 

Note thatVV −1 has been canceled in the equation above. Note thate Λt is 

⎡ ⎤ 

e λ 1t 

0 0 

e Λt 0 e λ 2t 

0 

= ⎢ 

⎣ 

. .. 

⎥ 

⎦ 

0 0 e λnt 

Therefore, the exponential of diagonalizable matrices can be solved as 

⎡ ⎤ 

e λ 1t 

0 0 

e Ft 0 e λ 2t 

0 

= V ⎢ 

⎣ 

. .. 

⎥ 

⎦ V −1 (2.5) 

0 0 e λnt 

From the result in (2.5), the solution to diagonalizable state space models can be 

obtained as follows. Suppose a state space model is given: 

ẋ = Fx+Gu 

y = Hx 

x(0) = x 0 

where F is a diagonalizable matrix in R n×n such that F can be eigendecomposed to 

F = VΛV −1 , whereV is a matrix consisting of the eigenvectors ofF . Then, a new state 

is defined as 

¯x = V −1 x ∈ R n 

orx = V ¯x. SubstitutingV ¯x forx, the state space model becomes 

V ˙¯x = FV ¯x+Gu 

y = HV ¯x 

¯x(0) = V −1 x 0 


Kyoungchul Kong


k 

c 

M 

y 

u 

Figure 2.5: A mass-spring-damper system. 

Multiplying V −1 to the both sides of the equation, a complete state space model is obtained 

as 

˙¯x = V −1 FV ¯x+V −1 Gu 

y = HV ¯x 

¯x(0) = V −1 x 0 

Note that the state space model has been transformed into a diagonal state space model, 

the solution of which is 

¯x(t) = e Λt¯x 0 + 

∫ t 

0 

e Λ(t−τ) Ḡu(τ)dτ fort ≥ 0 (2.6) 

y(t) = ¯H¯x(t) (2.7) 

whereΛ = V −1 FV , Ḡ = V −1 G, ¯H = HV . 

[Example 2-7] Suppose that a mass-damper-spring system shown in Fig. 2.5 is under 

a unit step input. In the figure, M = 1 is the mass, c = 3 is the damping coefficient, 

and k = 2 is the spring constant. The dynamic model of the system is 

and a state space model can be set as 

ÿ +3ẏ +2y = u 

d 

dt x = [ 0 1 

−2 −3 

y = [ 1 0 ] x 

x(0) = 0 ∈ R 2 

] [ 0 

x+ 

1 

where x(t) = [ y 

ẏ] 

∈ R 2 is the state. The initial condition was assumed to be zero for 

simplicity. The unit step input is defined as 

{ 

1 fort ≥ 0 

u(t) = 

0 fort < 0 

] 

u 

To solvee Ft , the state matrix F = [ 0 1 

−2 −3] is to be eigendecomposed. The char- 


Kyoungchul Kong


acteristic equation ofF is 

[ ] −λ 1 

det = λ 2 +3λ+2 = (λ+1)(λ+2) = 0 

−2 −3−λ 

Thus the eigenvalues ofF are λ 1 = −1 and λ 2 = −2. The eigenvalue equation is 

[ ] −λ 1 

v = 0 

−2 −3−λ 

and the associated eigenvectors (among many) are 

[ ] 1 

v 1 = 

and v 

−1 

2 = 

[ 1 

−2 

From the eigenvectors obtained, the transformation matrixV is set to 

[ ] 1 1 

V = 

−1 −2 

Finally, a new (diagonalized) state space model is set as 

˙¯x = Λ¯x+Ḡu 

y = ¯H¯x 

¯x(0) = 0 ∈ R 2 

] 

where 

[ ] −1 0 

Λ = V −1 FV = 

0 −2 

[ ][ ] [ ] 

2 1 0 1 

Ḡ = V −1 G = = 

−1 −1 1 −1 

¯H = HV = [ 1 0 ][ ] 

1 1 

= [ 1 1 ] 

−1 −2 

The state of the diagonalized state space matrix is 

∫ t 

[ ][ ] 

e 

−(t−τ) 

0 1 

¯x(t) = 

0 

0 e −2(t−τ) dτ fort ≥ 0 

−1 

[ ] 

1−e 

= 

−t 

0.5e −2t fort ≥ 0 

−0.5 

and the output isy(t) = ¯H¯x(t) = 0.5+0.5e −2t −e −t fort ≥ 0. 


Kyoungchul Kong


2.1.3 Solution of state space models in Jordan canonical form* 

This section is supplementary for students highly interested in Controls. 

Consider a state space model with a defective state matrix, i.e., some of the eigenvalues 

are repeated and there are fewer linearly independent eigenvectors than the number 

of the repeated eigenvalues. In order to form a transformation matrix, V , therefore, generalized 

eigenvectors should be obtained. 

For simplicity, suppose that a matrix F ∈ R n×n has only one distinct eigenvalue 

(λ) and one linearly independent eigenvector. The matrix F can be Jordan-decomposed 

to F = VJV −1 where 1 ⎡ 

λ 1 0 

0 λ . . . . 

J = ⎢ 

⎣ .. . 1 

0 0 λ 

By applying the Taylor expansion,e Ft = e VJV −1t is 

where 

⎤ 

e Ft = e VJV −1 t 

= Ve Jt V −1 

⎥ 

⎦ ∈ Rn×n 

1 

n tn 

⎡ 

1 

1 t 

2 t2 ··· 

1 

e Jt = e λt 0 1 t ··· 

⎢ 

⎣ 

. 

. . . .. . 

0 0 0 ··· 1 

n−1 tn−1 

The process to find a solution to the state space models with a defective state matrix is the 

same as in (2.7). Namely, for a given state space model with a defective state matrix 

ẋ = Fx+Gu 

y = Hx 

x(0) = x 0 ∈ R n 

⎤ 

⎥ 

⎦ 

the solution is 

¯x(t) = e Jt¯x 0 + 

∫ t 

0 

e J(t−τ) Ḡu(τ)dτ fort ≥ 0 (2.8) 

y(t) = ¯H¯x(t) (2.9) 

whereJ = V −1 FV , Ḡ = V −1 G, ¯H = HV . 

1 Note that the same notationJ is used for different matrix. 


Kyoungchul Kong


[Example 2-8] Consider the same mass-damper-spring system as in the previous example; 

but the properties are different so that M = 1, c = 2, and k = 1. The dynamic 

model of the system is 

ÿ +2ẏ +1y = u 

and a state space model can be set as 

d 

dt x = [ 

0 1 

−1 −2 

y = [ 1 0 ] x 

x(0) = 0 ∈ R 2 

] [ 

0 

x+ 

1 

where x(t) = [ y 

ẏ] 

∈ R 2 is the state. The initial condition was assumed to be zero for 

simplicity. The input is defined as 

{ 

1 fort ≥ 0 

u(t) = 

0 fort < 0 

] 

u 

To solve e Ft , the state matrix F = [ −1 0 −2] 1 is to be decomposed. The characteristic 

equation ofF is 

[ ] 

−λ 1 

det = λ 2 +2λ+1 = (λ+1)(λ+1) = 0 

−1 −2−λ 

Notice thatF has a repeated eigenvalue, i.e.,λ = −1. The eigenvalue equation is 

[ ] [ ] 

−λ 1 1 1 

v = v = 0 

−1 −2−λ −1 −1 

Since the nullity of[ −1 1 −1] 1 is1, there exists only one linearly independent eigenvector, 

which isv 1 = [ −1] 1 (among many). A generalized eigenvector is obtained from 

[ ] [ ] 

−λ 1 1 1 

v 

−1 −2−λ 2 = v 

−1 −1 2 = v 1 

and the generalized eigenvector isv 2 = [ 1 0] (among many). 

From the eigenvector and the generalized eigenvector obtained above, the transformation 

matrixV is set to [ ] 1 1 

V = 

−1 0 


Kyoungchul Kong


Finally, a new state space model in Jordan canonical form is obtained as 

where 

˙¯x = J¯x+Ḡu 

y = ¯H¯x 

¯x(0) = 0 ∈ R 2 

[ ] 

−1 1 

J = V −1 FV = 

0 −1 

[ ][ ] [ ] 

0 −1 0 −1 

Ḡ = V −1 G = = 

1 1 1 1 

¯H = HV = [ 1 0 ][ ] 

1 1 

= [ 1 1 ] 

−1 0 

The exponential ofJt is 

e Jt = e −t [ 1 t 

0 1 

] 

and the state is 

¯x(t) = 

= 

[ ][ 

e 

−(t−τ) 

(t−τ)e −(t−τ) −1 

0 e −2(t−τ) 1 

] 

−te −t 

0.5−0.5e −2t fort ≥ 0 

∫ t 

[ 

0 

] 

dτ fort ≥ 0 

In the calculation of the equation above, you may use the property of ∫ τe aτ dτ = 

e aτ 

a 2 (aτ −1). The output is 

y(t) = ¯H¯x(t) = 0.5−0.5e −2t −te −t fort ≥ 0 

2.1.4 Solution of state space models with complex eigenvalues* 

This section is supplementary for students highly interested in Controls. 

Suppose that a matrixF ∈ R 2×2 has complex eigenvalues(σ±jω) such thatF can 

be decomposed to 

[ ] σ ω 

F = VΣV −1 where Σ = 

−ω σ 


Kyoungchul Kong


Then the Taylor expansion ofe Ft gives 

e Ft = Ve σt [ cos(ωt) sin(ωt) 

−sin(ωt) cos(ωt) 

] 

V −1 

The remaining procedure to find a solution to the state space models is the same as 

above. 

[Example 2-9] Consider the same mass-damper-spring system as in the previous example; 

but the properties are different so that M = 1, c = 2, and k = 2. The system is 

under an impulse input. The state space model can be set as 

d 

dt x = [ 

0 1 

−2 −2 

y = [ 1 0 ] x 

x(0) = 0 ∈ R 2 

wherex(t) = [ y 

ẏ] 

∈ R 2 is the state. 

] 

x+ 

To solve e Ft , the state matrix F = [ −2 0 −2] 1 is to be decomposed. The characteristic 

equation ofF is 

[ ] 

−λ 1 

det = λ 2 +2λ+2 = (λ+1−j)(λ+1+j) = 0 

−2 −2−λ 

[ 

0 

1 

The eigenvalue equation forλ 1 = −1+j is 

[ ] [ −λ 1 1−j 1 

v = 

−2 −2−λ −2 −1−j 

] 

u 

] 

v = 0 

An eigenvector (among many) is v 1 = [ −1 

1−j] 

. Note that the remaining eigenvector is 

the complex conjugate of v 1 . Therefore, v 2 = [ ] 

−1 

1+j . Since vR and v I are v R = [ −1 

1 ] 

and v I = [ −1] 0 respectively, the state matrixAcan be decomposed to 

F = 

[ 

−1 0 

1 −1 

][ 

−1 1 

−1 −1 

][ 

−1 0 

1 −1 

Finally, a new state space model in oscillatory canonical form is obtained as 

˙¯x = Σ¯x+Ḡu 

y = ¯H¯x 

¯x(0) = 0 ∈ R 2 

] −1 


Kyoungchul Kong

2.2. LAPLACE TRANSFORMS AND TRANSFER FUNCTIONS 28 

where 

[ ] −1 1 

Σ = V −1 FV = 

−1 −1 

[ ][ ] [ ] 

−1 0 0 0 

Ḡ = V −1 G = = 

−1 −1 1 −1 

¯H = HV = [ 1 0 ][ ] 

−1 0 

= [ −1 0 ] 

1 −1 

The exponential ofΣt is 

e Σt = e −t [ 

cos(t) sin(t) 

−sin(t) cos(t) 

] 

and the state is 

∫ t 

[ cos(t−τ) sin(t−τ) 

¯x(t) = e −(t−τ) 

0 

−sin(t−τ) cos(t−τ) 

[ ] −sin(t) 

= e −t fort ≥ 0 

−cos(t) 

][ 0 

−1 

] 

δ(t)dτ fort ≥ 0 

The output is 

y(t) = e −t sin(t) fort ≥ 0 

2.2 Laplace Transforms and Transfer Functions 

In mathematics, the Laplace transform is a widely used integral transform. Denoted 

L{f(t)}, it is a linear operator of a function f(t) with a real argument t(t ≥ 0) that 

transforms it to a function F(s) with a complex argument s. The respective pairs of f(t) 

andF(s) are matched in tables. The Laplace transform has the useful property that many 

relationships and operations over the originals f(t) correspond to simpler relationships 

and operations over the images F(s). 

The Laplace transform is related to the Fourier transform, but whereas the Fourier 

transform resolves a function or signal into its modes of vibration, the Laplace transform 

resolves a function into its moments (i.e., poles and zeros). Like the Fourier transform, 

the Laplace transform is used for solving differential and integral equations. 


Kyoungchul Kong

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2.2.1 Formal definition 

The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the 

functionF(s), defined by: 

F(s) = L{f(t)} = 

where the parametersis a complex number. 

∫ ∞ 

0 

e −st f(t)dt 

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¬¢ ŸŸ¸»Ñ¾ œ ·¤°¬£¸ Ñ £° ¬¢ ŸŸ¸ £° ¬¥°°¤¹ 

œ ª¬»° Ÿ¾ »° Ó ¼ ¾ 

Figure 2.6: Matlab code for integrating a function defined with symbolic variables. 

You may utilize syms.m, int.m, and diff.m functions in various ways, e.g., 

Laplace transform, Fourier transform, Lagrangian mechanics, heat transfer, dynamics, 

etc. 

Note. Properties of Laplace transform 

• Linearity: L{af(t)+bg(t)} = aF(s)+bG(s), where a andbare any scalars. 

• Differentiation: L{f ′ (t)} = sF(s)−f(0) 

• Second differentiation: L{f ′′ (t)} = s 2 F(s)−sf(0)−f ′ (0) 

• Integration: L{ ∫ t 

0 f(τ)dτ} = 1 s F(s) 

• Convolution: L{(f ⋆g)(t)} = L{ ∫ t 

f(τ)g(t−τ)dτ} = F(s)G(s) 

0 

• Initial value theorem: f(0 + ) = lim s→∞ sF(s) 

• Final value theorem: f(∞) = lim s→0 sF(s), if the final value exists. 


Kyoungchul Kong


2.2.2 Output time histories 

Note. Laplace transform of selected signals 

All signals,f(t), are defined over t ≥ 0. 

• Unit impulse: L{δ(t)} = 1 

• Delayed impulse: L{δ(t−τ)} = e −τs 

• Unit step (integrate unit impulse): L{u(t)} = 1 s 

• Delayed unit step: L{u(t−τ)} = e−τs 

s 

• Ramp (integrate unit step): L{t} = 1 s 2 

• n th power for an integern: L{ tn 

n! } = 1 

s n+1 

• Exponential decay: L{e −αt } = 1 

s+α 

• Exponential approach: L{1−e −αt } = α 

s(s+α) 

• Sine: L{sin(ωt)} = ω 

s 2 +ω 2 

• Cosine: L{cos(ωt)} = s 

s 2 +ω 2 

• Hyperbolic sine: L{sinh(αt)} = α 

s 2 −α 2 

• Hyperbolic cosine: L{cosh(αt)} = s 

s 2 −α 2 

2.2.3 Transfer function 

Transfer functions are commonly used in the analysis of systems such as single-input 

single-output systems. 

In its simplest form for continuous-time input signalu(t) and output signaly(t), the 

transfer function is the linear mapping of the Laplace transform of the input,U(s), to the 

outputY(s), i.e. 

Y(s) = G(s)U(s) 

or 

G(s) = Y(s) 

U(s) = L{y(t)} 

L{u(t)} 

whereG(s) is the transfer function of a system. 


Kyoungchul Kong


[Example 2-10] Suppose that you adjusted the desired temperature of a room by−4 ◦ C 

at t = 0. By controlling an air-conditioner, the room temperature was changed by 

y(t) = −4 + 4e −3t . What is the transfer function of the room equipped with the airconditioner 

The transfer function is defined by the relationship between the input signal and 

the output signal, i.e. 

G(s) = Y(s) 

U(s) = L{−4+4e−3t } 

L{−4} 

= −4 s + 4 

s+3 

− 4 s 

= 3 

s+3 

whereG(s) is the transfer function of the room equipped with the air-conditioner. 

[Example 2-11] Suppose that a linear continuous system follows the equation of motion 

as 

ÿ +2ζω 0 ẏ +ω 2 0 y = K 0u(t) 

where the initial conditions are all zeros. 

The transfer function is obtained by taking the Laplace transform of the both 

sides of the equation, i.e. 

L{ÿ +2ζω 0 ẏ +ω 2 0y} = L{K 0 u(t)} 

s 2 Y(s)+2ζω 0 sY(s)+ω 2 0 Y(s) = K 0U(s) 

Thus, rearranging the equation above, the transfer function is obtained: 

G(s) = Y(s) 

U(s) = K 0 

s 2 +2ζω 0 s+ω 2 0 

2.2.4 Relationship between state space models and transfer functions 

Conversion from state space to transfer function 

Consider a state space model: 

ẋ = Fx+Gu 

y = Hx+Ju 


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(a) Actual system under an arbitrary input 

u(t) 

t 

0 

) 

0 

M & y 

+ ky = u y( t) 

= y ( t) 

+ ∫ g( 

t −τ 

) u( 

τ dτ 

(b) Time domain analysis: g(t) is an impulse response of the system 

u(t) 

x & = Fx + Gu 

y = Hx 

x = e 

Ft 

x 

t 

F ( t−τ 

) 

( 0) + ∫ e Gu( 

τ ) dτ 

0 

(c) State space analysis 

U (s) 

1 

2 

Ms + k 

(d) Laplace domain analysis 

1 

Y ( s) 

= U ( s) 

2 

Ms + k 

ω ω ω 

(e) Frequency domain analysis 

Figure 2.9: Various methods for analysis of an input-output relationship 


Kyoungchul Kong


where x ∈ R n , y ∈ R, u ∈ R, F ∈ R n×n , G ∈ R n×1 , H ∈ R 1×n , and J ∈ R. Note 

that the feedforward matrix,J, is included for the sake of generality. Since the state space 

model above consists only of linear functions 2 , the Laplace transform can be applied 3 . 

Taking the Laplace transform, we obtain 

Assuming thatx(0) = 0, 

L{ẋ} = L{Fx+Gu} 

L{y} = L{Hx+Ju} 

sX(s) = FX(s)+GU(s) 

Y(s) = HX(s)+JU(s) 

Rearranging the equation above, the transfer function fromU(s) toY(s) is obtained: 

Y(s) 

U(s) = H(sI −F)−1 G+J (2.10) 

[Example 2-12] A state space model with state matrices, F = [ 0 1 

−1 −2], G = [ 0 1 ], 

H = [ 1 0], J = 0, is converted into a transfer function as: 

G(s) = [ 1 0 ]( [ 

sI − 

0 1 

−1 −2 

]) −1 [ 0 

1 

] 

+0 = 

1 

s 2 +2s+1 

Recall that the dimensions of the state matrices are F ∈ R n×n , G ∈ R n×1 , H ∈ 

R 1×n , and J ∈ R, and a transfer function,G(s) is obtained uniquely via 

G(s) = H 

}{{} 

1×n 

(sI −F) −1 

} {{ } 

n×n 

}{{} G 

n×1 

+ }{{} J = b(s) 

a(s) 

1×1 

where b(s) is the numerator polynomial and a(s) is the denominator polynomial. Since 

(sI −F) −1 is reduced to 

(sI −F) −1 = 

1 

Adj(sI −F) 

det(sI −F) 

2 In other words, the state space equation is a linear system. 

3 This is not always true, if the state matrix includes nonlinear functions. For example, if ẋ = 

[ 

0 1 

−sin(t) −cos(t)] 

x+[ 

0 

1 ]u, the state space model cannot be transformed by the Laplace transform. Namely, 

the state space is capable of dealing with nonlinear or time-varying systems, while the Laplace transform 

can only be applied to linear systems. 


Kyoungchul Kong


where Adj(A) is the transpose of the cofactor matrix of A. 4 Therefore, the polynomials, 

a(s) and b(s), are 

a(s) = det(sI −F) 

= s n +a n−1 s n−1 +a n−2 s n−2 +...+a 0 

b(s) = H Adj(sI −F)G+J det(sI −F) 

= b m s m +b m−1 s m−1 +b m−2 s m−2 +...+b 0 

where n and m are the orders of a(s) and b(s), respectively. Note that m ≤ n (m = n 

only whenJ ≠ 0). Transfer functions withm ≤ n are called realizable systems. Transfer 

functions with m > n, i.e. unrealizable systems, cannot be realized in the reality. Most 

mechanical systems have transfer functions withm < n. 

Conversion from transfer function to state space 

Consider a differential equation of a mechanical system: 

... 

y +a 2 ÿ +a 1 ẏ +a 0 y = b 2 ü+b 1˙u+b 0 u (2.11) 

where the initial conditions are all 0. The transfer function of the differential equation 

above can be obtained by taking the Laplace transform, i.e. 

G(s) = 

b 2 s 2 +b 1 s+b 0 

s 3 +a 2 s 2 +a 1 s+a 0 

(2.12) 

Method 1: controllable canonical form (CCF) 

The differential equation in (2.11) can be reformulated into the following equations. 

x˙ 

1 = x 2 

x˙ 

2 = x 3 

x˙ 

3 = −a 0 x 1 −a 1 x 2 −a 2 x 3 +u 

y = b 0 x 1 +b 1 x 2 +b 2 x 3 

4 For example; if A = [ ] [ 

a b , then Adj(A) = d −b 

] 

. For the higher dimensional matrices, refer to 

c d 

Advanced Engineering Mathematics. 

−c a 


Kyoungchul Kong


In matrix form: 

d 

dt 

⎡ 

⎣ 

⎤ 

x 1 

x 2 

⎦ = 

x 3 

⎡ ⎤⎡ 

0 1 0 

⎣ 0 0 1 ⎦⎣ 

−a 0 −a 1 −a 2 

} {{ } 

F c 

⎡ 

y = [ ] 

b 0 b 1 b 2 

⎣ 

} {{ } 

H c 

Note thatH c (sI −F c ) −1 G c = G(s) in (2.12). 

⎤ 

x 1 

x 2 

⎦ 

x 3 

⎤ ⎡ 

x 1 

x 2 

⎦+ ⎣ 

x 3 

0 

0 

1 

⎤ 

} {{ } 

G c 

⎦u 

Method 2: observable canonical form (OCF) 

The differential equation in (2.11) can be reformulated into the following equations. 

d 

dt {ÿ +a 2ẏ −b 2˙u+a 1 y −b 1 u} = −a 

} {{ } 0 y +b 0 u 

:=x 3 

d 

dt {ẏ +a 2y −b 2 u} = x 

} {{ } 3 −a 1 y +b 1 u 

:=x 2 

d 

y 

dt }{{} 

:=x 1 

The equations above can be arranged in matrix form: 

⎡ ⎡ ⎤⎡ 

−a 

d 

2 1 0 

⎣ ⎣ −a 1 0 1 ⎦⎣ 

dt 

⎤ 

x 1 

x 2 

⎦ = 

x 3 

−a 0 0 0 

} {{ } 

F o 

⎡ 

y = [ 1 0 0 ] ⎣ 

} {{ } 

H o 

Note thatH o (sI −F o ) −1 G o = G(s) in (2.12). 

= x 2 −a 2 y +b 2 u 

⎤ 

x 1 

x 2 

⎦ 

x 3 

⎤ ⎡ 

x 1 

x 2 

⎦+ ⎣ 

x 3 

⎤ 

b 2 

b 1 

⎦ 

b 0 

} {{ } 

G o 

u 

Method 3-1: diagonal canonical form (DCF) 

When the transfer function in (2.12) has no repeated poles 5 , its partial fraction is 

G(s) = k 1 

s−p 1 

+ k 2 

s−p 2 

+ k 3 

s−p 3 

5 Poles of a transfer function are equivalent to the eigenvalues of the corresponding state matrix. 


Kyoungchul Kong


where p i ’s are the poles of G(s). Note that the new transfer function can be represented 

with the following differential equations. 

x˙ 

1 = p 1 x 1 +u 

x˙ 

2 = p 2 x 2 +u 

x˙ 

3 = p 3 x 3 +u 

y = k 1 x 1 +k 2 x 2 +k 3 x 3 


⎡ 

d 

⎣ 

dt 

⎤ 

x 1 

x 2 

⎦ = 

x 3 

⎡ ⎤⎡ 

p 1 0 0 

⎣ 0 p 2 0 ⎦⎣ 

0 0 p 3 

} {{ } 

F d 

⎡ 

y = [ ] 

k 1 k 2 k 3 

⎣ 

} {{ } 

H d 

⎤ ⎡ 

x 1 

x 2 

⎦+ ⎣ 

x 3 

⎤ 

x 1 

x 2 

⎦ 

x 3 

1 

1 

1 

⎤ 

} {{ } 

G d 

⎦u 

Note thatH d (sI −F d ) −1 G d = k 1 

s−p 1 

+ k 2 

s−p 2 

+ k 3 

s−p 3 

. 

Method 3-2: Jordan canonical form (JCF) 

If the transfer function in (2.12) has a repeated pole (p 2 = p 3 = p m ), it can be reduced to 

G(s) = k 1 k 2 

+ 

s−p 1 (s−p m ) + k 3 

2 s−p m 

The new transfer function can be represented with the following differential equations. 

x˙ 

1 = p 1 x 1 +u 

x˙ 

2 = p 2 x 2 +x 3 

x˙ 

3 = p 3 x 3 +u 

y = k 1 x 1 +k 2 x 2 +k 3 x 3 


⎡ 

d 

⎣ 

dt 

⎤ 

x 1 

x 2 

⎦ = 

x 3 

⎡ ⎤⎡ 

p 1 0 0 

⎣ 0 p 2 1 ⎦⎣ 

0 0 p 3 

} {{ } 

F j 

⎡ 

y = [ ] 

k 1 k 2 k 3 

⎣ 

} {{ } 

H j 

⎤ ⎡ 

x 1 

x 2 

⎦+ ⎣ 

x 3 

⎤ 

x 1 

x 2 

⎦ 

x 3 

1 

0 

1 

⎤ 

} {{ } 

G j 

⎦u 

Note thatH j (sI −F j ) −1 G j = k 1 

s−p 1 

+ k 2 

(s−p m) 2 + k 3 

s−p m 

. 


Kyoungchul Kong


[Summary] A transfer function 

G(s) = 

can be converted into a state space model 

b 2 s 2 +b 1 s+b 0 

s 3 +a 2 s 2 +a 1 s+a 0 

= k 1 

s−p 1 

+ k 2 

s−p 2 

+ k 3 

s−p 3 

ẋ = Fx+Gu 

y = Hx 

where the state matrices are 

Controllable Canonical Form 

Observable Canonical Form 

Diagonal Canonical Form 

⎡ 

⎣ 

F 

⎤ ⎡ 

G 

⎤ 

H 

0 1 0 0 

0 0 1 ⎦ ⎣ 0 ⎦ [ ] 

b 0 b 1 b 2 

−a 

⎡ 0 −a 1 −a 

⎤ 2 

⎡ 

1 

⎤ 

−a 2 1 0 b 2 [ ] 

⎣ −a 1 0 1 ⎦ ⎣ b 1 

⎦ 1 0 0 

⎡ 

−a 0 0 0 

⎤ ⎡ 

b 0 

⎤ 

p 1 0 0 1 

⎣ 0 p 2 0 ⎦ ⎣ 1 ⎦ [ ] 

k 1 k 2 k 3 

0 0 p 3 1 

When the transfer function has a repeated pole such that 

G(s) = 

b 2 s 2 +b 1 s+b 0 

= k 1 k 2 

+ 

s 3 +a 2 s 2 +a 1 s+a 0 s−p 1 (s−p m ) + k 3 

2 s−p m 

the corresponding state space model is 

ẋ = Fx+Gu 

y = Hx 




Jordan Canonical Form 

⎡ 

F 

⎤ ⎡ 

G 

⎤ 

H 

0 1 0 0 

⎣ 0 0 1 ⎦ ⎣ 0 ⎦ [ ] 

b 0 b 1 b 2 

−a 

⎡ 0 −a 1 −a 

⎤ 2 

⎡ 

1 

⎤ 

−a 2 1 0 b 2 [ ] 

⎣ −a 1 0 1 ⎦ ⎣ b 1 

⎦ 1 0 0 

⎡ 

−a 0 0 0 

⎤ ⎡ 

b 0 

⎤ 

p 1 0 0 1 

⎣ 0 p m 1 ⎦ ⎣ 0 ⎦ [ ] 

k 1 k 2 k 3 

0 0 p m 1 


Kyoungchul Kong


2.2.5 Response versus pole locations 

Modes in an impulse response 

Given the transfer function of a linear system, 

G(s) = Y(s) 

U(s) = b(s) 

a(s) 

the roots of a(s) = 0, called poles, make G(s) infinity, and those of b(s) = 0, called 

zeros, makeG(s) zero. 

Each pole location in thes-plane can be identified with a particular type of response. 

When the poles of a transfer function are not repeated (i.e., all the roots of a(s) = 0 are 

distinct), it can be expanded by a partial fraction expansion as 

G(s) = 

n∑ 

i=1 

k i 

s−p i 

where k i ’s are scalars, p i ’s are the poles of G(s), and n is the order of a(s). Suppose that 

the system is under an impulse input, i.e.,U(s) = 1. Then, the output is 

y(t) = L −1 {G(s)U(s)} = L −1 {G(s)} 

n∑ 

= L −1 k i 

{ } 

s−p i 

= 

i=1 

n∑ 

k i e p it 

i=1 

Note that the impulse response is a linear combination of e p it , which is called a mode. 

When all the poles have strictly negative real parts 6 , every mode converges to zero as 

t → ∞, and thusy(∞) → 0. 

[Example 2-13] Consider a transfer function: 

G(s) = 

8s+4 

s 3 +4s 2 +s−6 

Using the partial fraction expansion,G(s) can be expanded to 

6 That is, R{p i } < 0 for all i’s 

G(s) = −5 

s+3 + 4 

s+2 + 1 

s−1 


Kyoungchul Kong


f g hi jkl 

m g hn j n opkl 

q g rstfumvl 

wxyz{|}tqvl 

~€ rr‚ƒ„ mƒ ‚…†‡€ˆ‰ Š‰ˆ† ƒˆ‰ 

hŠu †u ‹k g Œ}|w z}tfu mvl 

~mŠr‚m€ sŠmŽr‚ ƒ ‰ †mƒˆ‚ ƒ 

Figure 2.10: Partial fraction expansion and plotting an impulse response 

The impulse response (i.e., U(s) = 1) ofG(s) is 

y(t) = −5e −3t +4e −2t +e t 

Sincee t diverges as t → ∞, the output does not converges to0. 

Complex poles 

Complex poles can be described in terms of their real and imaginary parts as 

p = −σ ±jω d 

Since complex poles always come in complex conjugate pairs for real polynomials, the 

denominator corresponding to a complex pair is 

a(s) = (s+σ −jω d )(s+σ +jω d ) = (s+σ) 2 +ω 2 d 

When finding the transfer function from differential equations, we typically 7 write the 

result in the polynomial form 

G(s) = 

b 0 

s 2 +2ζω n s+ω 2 n 

(2.13) 

where b 0 is any scalar, σ = ζω n , and ω d = ω n 

√ 

1−ζ2 . The parameter ζ is called the 

damping ratio, and ω n is called the natural frequency. 


G(s) = 

3 

s 2 +s+1 

7 That is, it is a common option, but is not the only option. 


Kyoungchul Kong


‘ ’“”• 

– ‘ ’— — —”• 

˜ ‘ š› œ– • 

žŸ ¡›˜ • ¢£¤¥¦¤¥§ ¥–¨©–ª š©«¬¨«¥¤«® –¥¦ ¦–¯°¤¥§ ©–¤±® 

²¤§«¥³–ª¨« ´–¯°¤¥§ £©«¬µ ›©–¦® 

·¸µ¹¹«·¹¹— º »µ¼¼«·¹¹—¤ ¸µ¹¹«·¹¹— —µ¹¹«º¹¹¹ 

·¸µ¹¹«·¹¹— · »µ¼¼«·¹¹—¤ ¸µ¹¹«·¹¹— —µ¹¹«º¹¹¹ 

Figure 2.11: Natural frequency and damping ratio of a transfer function 

G(s) can be expressed in the following form: 

G(s) = 

3 

s 2 +2ζω n s+ω 2 n 

where ω n = 1, and ζ = 0.5. One can say that the transfer function has the natural 

frequency of1and the damping ratio of0.5. 

When a transfer function has more than two poles, it is not clear how the damping 

ratio and the natural frequency are defined. Originally the transfer function in (2.13) is 

obtained from a mass-spring-damper system, which follows a second-order linear differential 

equation 8 . Therefore, in a strict sense the damping ratio and the natural frequency 

can be defined only when the transfer function has two poles. In a general sense, ζ and 

ω n are defined for each pair of complex poles. 

Repeated poles 

Suppose that a transfer function,G(s), hasmdistinct poles andn−m repeated poles. By 

the partial fraction expansion, it can be expanded as 

m∑ 

n−m 

k i 

∑ κ k 

G(s) = + 

s−p 

i=1 i (s−p m ) k+1 

k=1 

where k i ’s and κ k ’s are scalars, p i ’s are the distinct poles and p m is the repeated pole. If 

the system is under an impulse input, i.e.,U(s) = 1, the output is 

8 mÿ +cẏ +ky = u 

y(t) = L −1 { 

= 

m∑ 

i=1 

m∑ 

k i e pit + 

i=1 

n−m 

k i 

∑ κ k 

+ 

s−p i (s−p m ) k+1} 

k=1 

n−m 

∑ 

k=1 

α k t k e pmt 


Kyoungchul Kong


Im 

Asymptotically 

stable 

Unstable 

Re 

Stable in the sense of Lyapunov, 

if not repeated on the imaginary axis 

Figure 2.12: Stability and the location of poles 

where α k ’s are scalars. Note that the last term includes t k e pmt . It satisfies the following 

conditions: 

f(t) = t k e σmt > 0 for all k andt > 0 

f(0) = 0 

{ 

f(t) ˙ = (k +σ m t)t k−1 e σmt > 0 for all k and0 < t < −kσm 

−1 

< 0 for all k and −kσm −1 < t 

where σ m is the real part of p m . Therefore, f(t) converges to 0 as t → ∞, if and only if 

Re(p m ) < 0. f(t) does not converge to zero if Re(p m ) = 0, unless k = 0, which means 

that no pole is repeated. 

In summary, a transfer function is asymptotically stable (or, simply stable) if all 

the poles have strictly negative real parts. It is unstable if any of the poles has positive 

real part. It is marginally stable (or, stable in the sense of Lyapunov) if all the poles 

have non-positive real parts and no pole is repeated on the imaginary axis. This theorem 

is depicted in Fig. 2.12. 

2.2.6 Poles and eigenvalues 

Recall that 

G(s) = b(s) 

a(s) = H(sI −F)−1 G+J 

where F , G, H, and J are the state matrices of a state space equation, and G(s) is the 

corresponding transfer function. The poles are the roots of 

a(s) = 0 


Kyoungchul Kong


which is called the characteristic equation. Also remind that a(s) = det{sI −F} and 

the eigenvalues of the matrixF are calculated from 

det{λI −F} = 0 

Therefore, the poles ofG(s) are equivalent to the eigenvalues of the state matrix F . 

A system is asymptotically stable if 

• all the poles have strictly negative real parts (in the Laplace domain), or 

• all the eigenvalues have strictly negative real parts (in the state space). 


Kyoungchul Kong

2.3. FREQUENCY RESPONSE ANALYSIS 44 

2.3 Frequency Response Analysis 

2.3.1 Fourier transform versus Laplace transform 

There are several common conventions for defining the Fourier transform F(jω) of an 

integrable functionf(t). A common expression is 

F(jω) = 

∫ ∞ 

0 

f(t)e −jωt dt 

where ω ∈ R + is a positive scalar. Notice that the Fourier transform is related to the 

Laplace transform 

F(s) = 

∫ ∞ 

0 

f(t)e −st dt 

byjω = s. Therefore, Fourier-transformed functions can be obtained by replacingjω for 

s. 

[Example 2-15] The following functions (signals) are equivalent: 

Description Time-domain Laplace-domain Frequency-domain 

(t ∈ R + ) (s ∈ C) (ω ∈ R + ) 

Unit impulse δ(t) 1 1 

Delayed impulse δ(t−τ) e −τs e −jωτ 

Unit step u(t) 

1 

s 

e 

Delayed unit step u(t−τ) −τs 

s 

1 

Ramp t 

n th power for an integern 

t n 

n! 

Exponential decay e −αt 1 

Sine sin(ω 0 t) 

1 

jω 

e −jωτ 

jω 

− 1 

s 2 ω 2 

1 

1 

s n+1 

s+α 

ω 0 

s 2 +ω0 

2 

(jω) n+1 

1 

jω+α 

ω 0 

ω0 2−ω2 

Notice that the unit impulse signal includes all frequency components. In the 

Laplace domain, all the coefficients are real numbers but the Laplace operator, s, is 

a complex number. In the frequency domain, however, the coefficients are complex 

numbers while the frequency term,ω, is a positive real number. 

As the input and output signals can be transformed into the frequency domain by 

taking the Fourier transform, its relationship can also be represented in the frequency 

domain. The frequency-domain input-output relationship, which is usually represented 

by a Bode plot, is obtained by replacing jω forsof a transfer function. 


Kyoungchul Kong


2.3.2 Drawing bode plots by hand and by Matlab 

Consider a transfer function: 

∏ (s−zi ) 

G(s) = k∏ (s−pj ) 

where k is a gain, z i ’s are the zeros of G(s), and p j ’s are the poles of G(s). In the 

frequency domain,G(jω) is 

∏ (jω −zi ) 

G(jω) = k∏ (jω −pj ) 

The Bode plot can be drawn by following steps: 

[Step 1] Sort z i ’s and p j ’s by their absolute values. 

[Step 2] From ω = 0 to ω → ∞, make sections by |Re{z i }|’s and |Re{p j }|’s. 

[Step 3] Calculate |G(0)|. If |G(0)| does not exist, calculate |G(jω)| for any selected 

ω ∈ R + . This value will be used as the starting point of the magnitude plot. 

[Step 4] For every section, find an asymptote, and draw it on a magnitude graph (log(ω)- 

dB) graph 9 and a phase graph. The slope of a magnitude plot is 20(m ω −n ω ) where m ω 

and n ω are the numbers of (jω)’s in the numerator and denominator of each asymptote, 

respectively. Similarly, the phase is90(m ω +2m − −n ω −2n − ) in degrees, wherem − and 

n − are the numbers of minus signs in the numerator and denominator of each asymptote, 

respectively. Tip: do not replace (jω) 2 with−ω 2 for easier calculation. 

[Step 5] Smoothly connect the asymptotes. When any pole or zero is complex, you may 

calculate the exact value at the section boundary. 

[Example 2-16] The Bode plot of a transfer function 

G(s) = 

G(jω) = 

2s+10 

s 2 +9s−10 = 

2(jω +5) 

(jω −1)(jω +10) 

2(s+5) 

(s−1)(s+10) 

is obtained by: 

[Step 1] Since z 1 = −5, p 1 = 1, and p 2 = −10, they are sorted by their absolute 

9 dB is20log(x) 


Kyoungchul Kong


0 

−10 

dB 

−20 

degree 

−30 

−40 

10 −1 10 0 10 1 10 2 

0 

−50 

−100 

−150 

Frequency (rad/sec) 

−200 

10 −1 10 0 10 1 10 2 


Figure 2.13: Asymptotes of a transfer function in the frequency domain. 

values as 1, 5,10. 

[Step 2-3-4] There are four sections: 0 < ω < 1,1 < ω < 5,5 < ω < 10, and10 < ω. 

For each section, the asymptote is 

2(5) 

• at ω = 0: |G(j0)| = ∣ ∣ = 1 (i.e.,0dB) 

(−1)(10) 

• 0 < ω < 1: G(jω) ≈ 2(5) (slope: 0, phase: -180) 

(−1)(10) 

• 1 < ω < 5: G(jω) ≈ 2(5) (slope: -20, phase: -90) 

(jω)(10) 

• 5 < ω < 10: G(jω) ≈ 2(jω) (slope: 0, phase: 0) 

(jω)(10) 

• 10 < ω: G(jω) ≈ 2(jω) (slope: -20, phase: -90) 

(jω)(jω) 

They are plotted on a magnitude graph and a phase graph as in Fig. 2.13. 

[Step 5] Finally, the Bode plot is obtained by smoothly connecting the asymptotes as 

in Fig. 2.14. 


Kyoungchul Kong


0 

−10 

dB 

−20 

degree 

−30 

−40 

10 −1 10 0 10 1 10 2 

0 

−50 

−100 

−150 


−200 

10 −1 10 0 10 1 10 2 


Figure 2.14: Smoothly connected asymptotes. 

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ØÙÚÛÁ½Ê 

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Ì àÎÓ×Ô ×À ½ÁÔÊ 

áÙâÛÁ½ÊË 

Ì ÍÓÒã Ò ä×åÎ æÝ×¿ 

Figure 2.15: Matlab code for drawing a Bode plot 


Kyoungchul Kong


[Example 2-17] The Bode plot of a transfer function 

G(s) = 

G(jω) = 

4 

s 2 +s+4 

4 

(jω) 2 +jω +4 

is obtained by: 

√ 

15 

[Step 1] Since p 1,2 = −0.5 ± j , they are sorted by the absolute values (i.e., 2). 

4 

There is only one section boundary at 2. 

[Step 2-4] There are two sections: 0 < ω < 2, 2 < ω. For each section, the asymptote 

is 

• at ω = 0, |G(j0)| = | 4 4 | = 1 (i.e.,0dB) 

• 0 < ω < 2: G(jω) ≈ 4 4 

(slope: 0, phase: 0) 

• 2 < ω: G(jω) ≈ 4 

(jω) 2 (slope: -40, phase: -180) 

The exact value of|G(jω)| at the section boundary is 

∣ |G(2j)| = 

4 

∣∣∣ ∣(2j) 2 +2j +4∣ = 4 

2j∣ = 2 

[Step 5] Finally, the Bode plot is obtained by smoothly connecting the asymptotes as 

in Fig. 2.16, where the line must pass through the exact value at the section boundary. 

Graphical representation ofG(s) in the Laplace-domain andG(jω) in the frequency- 

Domain 

Remind that the Laplace domain and the frequency domain are related by s = jω. Since 

s is a complex number, which can be defined as 

s = σ +jω 

where σ = Re{s} ∈ R and ω = Im{s} ∈ R, the Bode plot is the line of intersection 

between G(s) and a plane of σ = 0. Namely, the Bode plot is a part of G(s) plotted on 

the s-plane, and thus information on the locations of poles and zeros of G(s) is resolved 

in the Bode plot. 


Kyoungchul Kong


10 

0 

at 2 (rad/sec) 

dB 

−10 

−20 

−30 

10 0 10 1 

degree 

0 

−50 

−100 

−150 

−200 

10 0 10 1 


Figure 2.16: Smoothly connected asymptotes. 


Kyoungchul Kong


Magnitude plot 

Figure 2.17: Magnitude plot ofG(s) = (s+2)/(s 2 +6s+5) on thes-plane and 

the bode plot ofG(jω) 

Phase plot 

Figure 2.18: Phase plot ofG(s) = (s+2)/(s 2 +6s+5) on thes-plane and that 

ofG(jω) 


Kyoungchul Kong


2.3.3 Nyquist plot 

A Nyquist plot is a parametric plot of a transfer function, G(jω). The most common use 

of Nyquist plots is for assessing the stability of a system with feedback. In Cartesian 

coordinates, the real part of G(jω) is plotted on the x-axis, and the imaginary part is 

plotted on they-axis. 

ç è éêëìí îïðìí íñ ò ñïóô 

õö÷øùúéëçóô 

û üýþééùõÿ ö÷øùúé ýþé 

ÿù þõô 

û ùúýöùõÿ ÿùú 

Figure 2.19: Matlab code for drawing a Nyquist plot 

r 

+ − 

C 

u 

G 

y 

Figure 2.20: A typical feedback control system 

Nyquist stability criterion 

Suppose a plantG(s) is under feedback control by C(s) as shown in Fig. 2.20. 

To assess stability of the closed loop system using Nyquist theorem, the Nyquist 

contour should be constructed first. The Nyquist contour consists of: 

• G(jω)C(jω) whereω ∈ (−∞,∞), and 

• a semicircular arc that travels clock-wise, with radius of ∞. This semicircular line 

is necessary only whenG(jω)C(jω) → ∞ for someω ∈ R. 

Given a Nyquist contour, let 

• P be the number of unstable poles ofGC, and 

• Z be the number of unstable poles of the closed loop system 

Z is desired to be zero. 

GC . In most cases, 

1+GC 

The resultant contour should encircle counterclockwise the point −1 +0j N times such 

thatN = P −Z. 

Notice that ifGC is stable 10 ,P = 0 and thus the Nyquist contour must not encircle 

the critical point−1+0j. 

10 That is, all the poles ofGC have strictly negative real parts. 


Kyoungchul Kong


[Summary] Given: 

• GC(s) is an open loop transfer function withP unstable poles, 

• T(s) = GC(s) is a closed loop transfer function withZ unstable poles, 

1+GC(s) 

• N is the number of counterclockwise encirclement of (−1 + 0j) point by the 

locus ofG(jω) for−∞ < ω < ∞, and 

• n is the degree of the characteristic polynomial ofGC(s). 

Nyquist stability criterion states that: 

N = P −Z 

and Z = 0 implies stability of the closed loop system. 

This implies that for a system to be stable by feedback control, the number of 

encirclement of (−1 + 0j) point by the locus of GC(jω) for −∞ < ω < ∞ in the 

counterclockwise direction is equal to the number of unstable open loop poles. 

Proof of nyquist stability criterion 

Let GC(s) = N(s) and F(s) be 

D(s) 

F(s) = 1+GC(s) = 1+ N(s) 

D(s) = D(s)+N(s) 

D(s) 

:= D C(s) 

D(s) 

Note that the roots of D C (s) = 0 are the closed loop poles 11 , and those of D(s) = 0 are 

the open loop poles. F(s) can be organized as 

∏ n 

i=1 

F(s) = ∏ (s−p c,i) 

n 

k=1 (s−p o,k) 

wherep c,i ’s are the closed loop poles (i.e., the roots ofD C (s) = 0), andp o,k ’s are the open 

loop poles. In the frequency domain, 

∏ n 

i=1 

F(jω) = ∏ (jω −p c,i) 

n 

k=1 (jω −p o,k) 

From the equation above, the phase ofF(jω) is calculated as 

11 GC 

1+GC = N/D 

1+N/D = N 

D+N = N D C 

ϕ F (ω) = 

n∑ 

ϕ c,i (ω)− 

i=1 

n∑ 

ϕ o,k (ω) 

k=1 


Kyoungchul Kong


then, 

∆ϕ F := ϕ F (∞)−ϕ F (−∞) 

n∑ 

= (ϕ c,i (∞)−ϕ c,i (−∞))− 

i=1 

n∑ 

(ϕ o,k (∞)−ϕ o,k (−∞)) 

where ϕ(∞)−ϕ(−∞) for each pole is either+π or−π as shown in Fig. 2.21. Namely, 

ϕ i (∞) − ϕ i (−∞) is +π if p i is located on the left half plane (LHP), and is −π if p i is 

located on the right half plane (RHP). 

k=1 

Im 

∞j 

Im 

∞j 

p 

c , i 

π 

−π 

p 

c , i 

Re 

Re 

Recall that 

− ∞j 

− ∞j 

(a) ϕ ( ∞) 

−ϕ( 

−∞) 

for a zero in LHP (b) ϕ( ∞) 

−ϕ( 

−∞) 

for a zero in RHP 

Figure 2.21: ϕ(∞)−ϕ(−∞) for a stable and unstable pole 

• the open loop transfer function,GC(s), has P unstable poles, and 

• the closed loop transfer function,T(s) = GC(s) , has Z unstable poles. 

1+GC(s) 

Therefore, 

where 

∆ϕ F = 

n∑ 

(ϕ c,i (∞)−ϕ c,i (−∞))− 

i=1 

n∑ 

(ϕ o,k (∞)−ϕ o,k (−∞)) 

k=1 

= [(π)(n−Z)+(−π)(Z)]−[(π)(n−P)+(−π)(P)] 

= 2π(P −Z) = 2πN 

N = counterclockwise encirclement of origin in (1+GC(jω)) plane 

= counterclockwise encirclement of(−1+0j) in GC(jω) plane 


Kyoungchul Kong


Im 

−1 

GM 

−1 

PM 

Re 

Nyquist plot 

of GC 

Figure 2.22: Graphical Representation of Phase Margin and Gain Margin 

Gain margin and phase margin 

Recall that if the open loop system, GC, is stable, the corresponding Nyquist contour 

must not encircle the critical point, −1+0j. If GC crosses the critical point 12 such that 

G(jω 0 )C(jω 0 ) = −1 + 0j for some ω 0 ∈ R, the closed loop system (GC/(1 + GC)) 

has an infinitely large gain at the frequency ω 0 , which means the system amplifies the 

frequency component ofω 0 infinitely 13 . In general,GC must encircle(−1+0j) the right 

number of times according to the number of unstable open-loop poles. 

There are two measures of the stability margin, the gain margin and phase margin, 

that represent how far the Nyquist plot ofGC is from the critical point. 

Suppose that an open loop system, GC, has the gain margin of k and the phase 

margin ofθ. Then, you can expect that 

• the closed loop system becomes unstable if the open loop system is amplified byk, 

or 

• the closed loop system becomes unstable if the open loop system has an additional 

phase delay ofθ. 

For example, if an open loop system has an infinitely large gain margin, the closed loop 

system will not become unstable by increasing its gain. If the open loop system has an 

infinitely large phase margin, the closed loop system will always be stable even in the 

presence of unknown phase delay. 

12 −1+0j; it is also called “dangerous point.” 

13 This is the response of unstable systems 


Kyoungchul Kong


[Example 2-18] Suppose that a plantG(s) is under feedback control, where 

( ) 

1 

C(s) = k P 1+T d s+T i 

s 

and it has found thatGC has the gain margin of5. Then, the maximum gain ofk P that 

stabilizesG(s) is 5k P . 


Kyoungchul Kong

2.4. CONTROLLER DESIGN 56 

2.4 Controller Design 

2.4.1 Objectives of feedback control 

Suppose that a plant, G, is under feedback control with a controller, C. There are three 

main objectives of feedback control: stability, performance, and robustness. 

Stability 

The feedback controller must stabilize the plant such that 

• In the time-domain, an output, y(t), is bounded for an bounded reference input, 

r(t). When an impulse signal is exerted as a reference input,y(t) converges to zero 

as t → ∞. 

• In the Laplace-domain, the following four transfer functions have poles with strictly 

negative real parts: 

GC 

1+GC 

C 

1+GC 

G 

1+GC 

1 

1+GC 

• In the state space, the state matrix has eigenvalues with strictly negative real parts. 

• In the frequency-domain, the Nyquist contour encircles(−1+0j) counterclockwise 

N times such that N = P , where P is the number of unstable poles of the open 

loop transfer function,GC(s). 

The four statements above are all equivalent. You may select the most appropriate domain 

according to the characteristics of your plant. 

Performance 

Once the plant is stabilized by feedback control, it is desired to achieve the desired performance. 

The feedback-controlled system should show the following characteristics: 

• In the time-domain, an error,e(t) = r(t)−y(t), should converge to zero ast → ∞ 

for any constant reference input,r(t) = r 0 < ∞. 

Moreover, it is desired fore(t) to converge to zero as quick as possible. 

In addition, the output should not be affected by a constant disturbance, d(t) = 

d 0 < ∞. In other words, the error, e(t) = r(t)−y(t), should converge to zero as 

t → ∞ for any constant disturbance. 

• In the Laplace-domain, all the closed-loop poles should be placed as far as possible 

from the imaginary axis. 


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• In the state space, all the eigenvalues should be placed as far as possible from the 

imaginary axis. 

GC 

• In the frequency-domain, (jω) ≈ 1 and | G 

(jω)| ≈ 0 for a sufficiently 

1+GC 1+GC 

large frequency range 14 . In other words, the magnitudes of GC(jω) and C(jω) 

should be large enough for a sufficiently large frequency range. 

Moreover,|GC(0j)| and |C(0j)| should be infinitely large. 

Robustness 

Most of the controller design techniques are based on a mathematical model of a plant. 

In practice, however, the exact model parameters are difficult to identify. Moreover, in 

many cases the governing equation is not clear, so that an approximated model is used. 

Therefore, a controller designed for a particular set of parameters is required to be robust 

such that it works well under a different set of assumptions. 

This subject will be discussed more thoroughly in the Robust Control Systems, 

a graduate level course. You may notice that we have already learned two criteria for 

stability robustness in the frequency-domain: 

• The open loop transfer function, GC(jω), should have a large phase margin to 

account for the phase uncertainty. 

• The open loop transfer function, GC(jω), should have a large gain margin to account 

for the gain uncertainty. 

2.4.2 Controller design in time-domain 

PID control law 

Currently, more than half of the controllers used in industry are proportional-integralderivative 

(PID) controllers. When a mathematical model of a system is available, the 

parameters of the controller can be explicitly determined. However, when a mathematical 

model is unavailable, the parameters must be determined experimentally. Controller 

tuning is the process of determining the controller parameters which produce the desired 

output. 

The equation below is the PID control law represented in the time-domain: 

u(t) = K p e(t)+K i 

∫ t 

0 

e(τ)dτ +K d 

de(t) 

dt 

14 Also, it can be explained as “the closed-loop system should have a large frequency bandwidth.” 


Kyoungchul Kong


where, 

u(t) is the control signal, 

e(t) is the difference between the current measurement,y(t), and the reference input,r(t), 

K p is the gain for a proportional controller, 

K i is the gain for an integral controller, and 

K d is the gain for a derivative controller. 

Trial and error method 

The trial and error tuning method is based on guess-and-check. In this method, the proportional 

action is the main control, while the integral and derivative actions refine it. The 

procedure is as follows. 

1. SettingK i and K d to zero, increaseK p from zero until the output oscillates. 

2. HoldingK p , increaseK d until the oscillation disappears. 

3. HoldingK p andK d , increaseK i such that the error converges to zero for a constant 

reference input. 

4. Repeat 1.-3. until the desired performance is obtained. In each step, you should 

check the magnitude and oscillation of the control input, u(t). If u(t) reaches the 

saturation range too often or includes a high frequency noise, the K p , K d , and K i 

should not be increased further. 

Ziegler-Nichols method 

The Ziegler-Nichols tuning method is a heuristic method of tuning a PID controller. It 

was developed by John G. Ziegler and Nathaniel B. Nichols. The procedure is as follows. 

1. Setting K i and K d to zero, increase K p from zero until the closed-loop system is 

marginally stable. The marginally stable gainK p is called the ultimate gain,K u . 

2. K u and the oscillation periodT u are used to set theK p ,K i , andK d gains depending 

on the type of controller used: 

Control Type K p K i K d 

1 

P K 2 u - - 

1 

PI K 2.2 u 1.2 Kp 

T u 

- 

classic PID 0.6K u 2 Kp 1 

K T u 8 pT u 

0.15K p T u 

Pessen Integral Rule 0.7K u 2.5 Kp 

T u 

some overshoot 0.33K u 2 Kp 

T u 

1 

3 K pT u 

no overshoot 0.2K u 2 Kp 

T u 

1 

3 K pT u 


Kyoungchul Kong


2.4.3 Controller design in Laplace-domain 

Pole placement 

Pole placement is a method employed to place the closed-loop poles of a plant at predetermined 

locations in the s-plane. Placing poles is desirable because the location of the 

poles corresponds directly to the characteristics of the response of the system. 

Consider a plant: 

G(s) = b(s) 

a(s) 

wherea(s) andb(s) are the denominator and numerator polynomials, respectively. Without 

loss of generality, it is assumed that a(s) and b(s) are coprime, such that there is no 

pole-zero cancelation. The orders of a(s) and b(s) are assumed to be n and m ≤ n, 

respectively. 

Suppose thatG(s) is under feedback control with a controller: 

C(s) = β 1s n−1 +β 2 s n−2 +...+β n−1 s+β n 

s m +α 1 s m−1 +...+α m−1 s+α m 

= β(s) 

α(s) 

Note that the orders ofα(s) andβ(t) aremandn−1, respectively. Then, the closed-loop 

characteristic polynomial 15 is 

a(s)α(s)+b(s)β(s) = d 0 (s r +d 1 s r−1 +...+d r−1 s+d r ) 

} {{ } 

:=D c(s) 

(2.14) 

wherer = n+m. Note thatD c (s) hasr free variables (i.e., controller gains,α 1,2,...,m and 

β 1,2,...,n ), which are to be determined. 

Suppose that you haver desired poles, such that 

D d c(s) = 

r∏ 

(s−p i ) 

i=1 

= s r +d d 1 sr−1 +...+d d r−1 s+dd r (2.15) 

where d d i ’s are the coefficients of the desired closed-loop characteristic equation. The superscriptddenotes 

“desired.” Then, the controller gains can be determined by comparing 

(2.14) and (2.15). 

15 That is obtained from GC 

1+GC . 


Kyoungchul Kong


[Example 2-19] Consider a plant: 

G(s) = s+1 

s 2 +s+1 

where the desired closed-loop poles are: −10,−15, and−20. Sincen = 2 andm = 1, 

a possible feedback controller is: 

C(s) = β 1s+β 2 

s+α 1 

The closed-loop characteristic polynomial is 

D c (s) = s 3 +(1+α 1 +β 1 )s 2 +(1+α 1 +β 1 +β 2 )s+(α 1 +β 2 ) 

and the desired closed-loop characteristic polynomial is 

D d c(s) = (s+10)(s+15)(s+20) = s 3 +45s 2 +650s+3000 

Thus, the controller gains should be selected to 

α 1 = 2395 

β 1 = −2351 

β 2 = 605 

Pole placement for plants without RHP zeros 

Suppose that a plant 

G(s) = b(s) 

a(s) 

does not have zeros on the closed right-half plane 16 , i.e., all the zeros have negative real 

parts. The orders of a(s) and b(s) are n and m ≤ n, respectively. Then a controller can 

be designed as 

C(s) = β 1s n−1 +β 2 s n−2 +...+β n−1 s+β n 

b(s) 

= β(s) 

b(s) 

With this feedback controller, the closed-loop transfer function from the reference 

16 The closed RHP includes the imaginary axis. 


Kyoungchul Kong


input,r(t), to the output,y(t), is 

b(s) 

GC(s) 

1+GC(s) = 

β(s) 

a(s) b(s) 

1+ b(s) β(s) 

a(s) b(s) 

= 

β(s) 

a(s)+β(s) 

Notice that the closed-loop characteristic polynomial has been reduced toa(s)+β(s), the 

order of which is n. Since β(s) includes n design parameters (i.e., β 1 , β 2 , ..., β n ), the n 

closed- loop poles can be arbitrarily assigned. 

[Example 2-20] Consider the plant in [Example 2-19]. Since the zero ofG(s) is placed 

on the open left half plane, the two closed-loop poles can be arbitrarily assigned by a 

feedback controller 

C(s) = β 1s+β 2 

s+1 

The closed-loop characteristic polynomial is 

D c (s) = s 2 +(1+β 1 )s+(1+β 2 ) 

Suppose that the desired closed-loop poles are −p 1 and−p 2 , i.e. 

Dc(s) d = (s+p 1 )(s+p 2 ) = s 2 +(p 1 +p 2 )s+p 1 p 2 

Thus, the controller gains should be selected to 

β 1 = p 1 +p 2 −1 

β 2 = p 1 p 2 −1 

2.4.4 Controller design in state space 

Full state feedback 

In the state space, pole placement can be achieved in a more convenient way. Remind 

that the poles in the Laplace-domain are equivalent to the eigenvalues of the state matrix 

in the state space. Therefore, in the state space a feedback controller is designed such 

that the eigenvalues of the state matrix can be assigned at pre-determined locations. If a 

system is controllable, there always exists a state feedback gain such that the closed-loop 

eigenvalues can be arbitrarily placed. 


Kyoungchul Kong

£ ¤ ¥ ¦§ ¡ 

¨§ 

¨© 

£ ¤¥§ 

¦§ 


£ ¤¨¢ ¨¦¥§ ! "#$ 

% £ &'()*+¡,, - ¡$## !./ 

Figure 2.23: Matlab code for the design of the state feedback gain,K 

When an open loop transfer function is represented in the state space 

ẋ = Fx+Gu 

y = Hx 

wherex ∈ R n is the state. The eigenvalues of the state matrix are the roots of the characteristic 

equation given by 

det[sI −F] = 0 

Full state feedback is realized by manipulating the input. u(t). Consider a feedback 

controller: 

u = −Kx+v 

whereK ∈ R 1×n is the state feedback gain,v is an auxiliary input. Substitutingu = −Kx 

into the state space equations above, 

ẋ = [F −GK]x+Gv 

y = Hx 

The eigenvalues of the FSF system are given by the characteristic equation,det[sI−(F− 

GK)] = 0. Comparing the terms of this equation with those of the desired characteristic 

equation yields the values of the feedback matrix which force the closed-loop eigenvalues 

to the locations specified by the desired characteristic equation. 

[Example 2-21] Consider a control system given by the following state space equation 

[ ] [ ] 0 1 0 

ẋ = x+ u 

−2 −3 1 

The open loop system has eigenvalues at s = −1 and s = −2. Suppose, for considerations 

of the response, we wish the closed-loop system eigenvalues to be located at 

s = −1 and s = −5; i.e., the desired characteristic equation is then s 2 +6s+5 = 0. 


Kyoungchul Kong


Following the procedure given above, K = [ k 1 k 2] ∈ R 2×1 , and the FSFcontrolled 

system characteristic polynomial is 

[ ] 

s −1 

det[sI −(F −GK)] = det 

= s 2 +(3+k 

2+k 1 s+3+k 2 )s+(2+k 1 ) 

2 

Upon setting this characteristic equation equal to the desired characteristic equation, 

we find 

K = [ 3 3 ] 

Therefore, setting u = −Kx forces the closed-loop eigenvalues to the desired locations, 

affecting the response as desired. 

Controllability* 

Controllability is an important property of a dynamic system, and the controllability property 

plays a crucial role in many control problems. Roughly, the concept of controllability 

denotes the ability to move the state of a system to an arbitrary desired state in a finite 

time using certain admissible inputs. 

Consider a state space model 

ẋ = Fx+Gu 

y = Hx 

whereF ∈ R n×n , G ∈ R n , and H ∈ R 1×n . 

Recall that the solution of the state space model is 

x(t) = e Ft x(0)+ 

∫ t 

0 

e F(t−τ) Gu(τ)dτ 

Let us assume zero initial conditions (i.e., x(0) = 0 ∈ R n ) for simplicity. Now, since 

the controllability means that the system can reach any x(t), for controllable system the 

integral converges to x(t) with someu(t). Using the Taylor expansion of exponential, 

∫ t 

] 

x(t) = 

[I +F(t−τ)+ F2 

2 (t−τ)2 + F3 

6 (t−τ)3 +... Gu(τ)dτ 

0 

The constant terms can be taken out such that 

⎡ 

x(t) = [ G FG F 2 G ... ] ⎢ 

⎣ 


∫ t 

∫ t 

0 

∫ t 

0 u(τ)dτ 

(t−τ)u(τ)dτ 0 

1 

2 (t−τ)2 u(τ)dτ 

. 

⎤ 

⎥ 

⎦ ≡ C ∞U 

Kyoungchul Kong


where the matrix part is denoted as C ∞ . For controllable systems, it should be able to 

obtain anyx(t) ∈ R n , and therefore C ∞ must be full rank (i.e., rankn). From the Cayley- 

Hamilton theorem, we know thatF n is linearly dependent onF n−1 ,F n−2 , ...,F ,I. 17 This 

means that no more information about the rank of C ∞ is added after F n−1 G, because the 

remaining terms are linear combinations of the first n terms. Therefore the system is 

controllable if 

rank(C ∞ ) = rank(C) = rank ([ G FG F 2 G ··· F n−1 G ]) 

has full rank. 

If a system is controllable, there always exists a state feedback gain,K, such that the 

closed-loop eigenvalues ofF −GK can be arbitrarily assigned. Therefore, the following 

statements are all equivalent: 

• A system is controllable. 

• The state of a system, x(t), can reach any point in R n with an appropriate input, 

u(t). 

• [ G FG F 2 G ··· F n−1 G ] has full rank. 

• There existsK such that the eigenvalues ofF −GK can be arbitrarily determined. 

State observer 

A state observer is a computer algorithm that simulates a real system in order to provide 

an estimate of its internal state, given measurements of the input and output of the real 

system. While the state is necessary to implement the state feedback control, the 

physical state of the system cannot be determined by direct observation. Instead, the state 

is indirectly observed (i.e., estimated) from the system output. A simple example is that 

of vehicles in a tunnel: the rates and velocities at which vehicles enter and leave the tunnel 

can be observed directly, but the exact velocity inside the tunnel can only be “estimated.” 

If a system is observable, it is possible to fully estimate the system state from its output 

measurements using the state observer. 

For a system represented in the state space, 

ẋ = Fx+Gu (2.16) 

y = Hx 

17 For more information on the Cayley-Hamilton theorem, refer to the wikipedia, or the classnote of 

Mechanical Systems Analysis. 


Kyoungchul Kong


Accelerometer 

Encoder 

Figure 2.24: A car model. 

the state x is to be estimated. For simulation of the model, a new state space equation 

with the same model parameters is constructed: 

˙ˆx = Fˆx+Gu 

ŷ = Hˆx 

where ˆx(t) is the simulated (i.e., estimated) state. ˆx(t) may be different from x(t) because 

of the initial condition, disturbance, sensor noise, model mismatch, etc. Therefore, 

feedback is necessary to force ˆx(t) to converge to x(t) as t → ∞. Since the output is 

measurable, it is utilized for correcting the state, i.e. 

˙ˆx = Fˆx+Gu+L(y −ŷ) 

ŷ = Hˆx 

where L ∈ R n is the observer gain that corrects the estimation error. Note that y(t) − 

ŷ(t) = 0 when ˆx(t) = x(t). Sincey = Hx andŷ = Hˆx, the state observer equation is: 

˙ˆx = [F −LH]ˆx+Gu+Ly (2.17) 

Recall that it is desired for the estimation error, ˜x(t) := ˆx(t) − x(t), to converge 

to zero as t → ∞. The dynamics of the estimation error can be obtained by subtracting 

(2.16) from (2.18), i.e. 

˙˜x = ˙ˆx−ẋ 

= ([F −LH]ˆx+Gu+Ly)−(Fx+Gu) 

= [F −LH](ˆx(t)−x(t)) 

= [F −LH]˜x 

Therefore, the observer gain,L, must be determined such that[F −LH] has eigenvalues 

with strictly negative real parts. 


Kyoungchul Kong


[Example 2-22] The example in Fig. 2.24 is related to sensor-fusion. 

Suppose that the velocity of a car is required in a control algorithm. Since there is 

no direct method to measure the velocity, an accelerometer and an encoder a are utilized 

to estimate the velocity. Notice that there exists a kinematic relationship between the 

acceleration and the position, i.e. 

d 2 

dt2y(t) = u(t) 

whereu(t) is the acceleration measurement, andy(t) is the position measurement. The 

above equation, however, does not hold in the reality because of sensor noise, parameter 

uncertainty, etc. Therefore, the velocity can be directly calculated from neitheru(t) 

nory(t). In order to obtain more precise information by fusing two different measurements, 

a state observer can be utilized. In the state space, the differential equation can 

be formulated into: 

ẋ = 

[ 0 1 

0 0 

y = [ 1 0 ] x 

] [ 0 

x+ 

1 

[ ] ẏ 

where x = is the state to be estimated. Note that the velocity is included in the 

y 

state. 

] 

u 

To estimate the state, a state observer is constructed: 

[ ] [ ] [ ] 

0 1 0 l1 

˙ˆx = ˆx+ u(t)+ (y(t)− [ 1 0 ]ˆx) 

0 0 1 l 2 

The L = [ l 1 

] 

l2 

matrix should be selected such that the eigenvalues of [F − LH] have 

strictly negative real parts. The corresponding characteristic equation is 

det 

[ 

sI − 

([ 0 1 

0 0 

] 

− 

[ 

l1 0 

l 2 0 

])] 

= s 2 +l 1 s+l 2 = 0 

A possible solution set tol 1 and l 2 (among many) isl 1 = 2 andl 2 = 1, which yield the 

eigenvalues of−1 and −1. Note that the velocity can be estimated by 

ˆv = [ 0 1 ]ˆx 

a An encoder is often used to measure the travel distance of a vehicle. 


Kyoungchul Kong


Observability 

A state space model without an input 

ẋ = Fx 

y = Hx 

x(0) = x 0 

is said to be observable, if ‖y(t)‖ ≠ 0 for all t > 0 and for all nonzero initial conditions 

x 0 . Note that the outputy(t) is 

y(t) = He Ft x 0 

Taking the Taylor expansion ofe Ft , 

[ 

y(t) = H I +Ft+ 1 ] 

2 F2 t 2 +... 

⎡ 

H 

= [ 1 

1 t 

2 t2 ... ] HF 

⎢ 

⎣ HF 2 

. 

= T O ∞ x 0 

x 0 

⎤ 

⎥ 

⎦ x 0 

where T and O ∞ are ∞-dimensional. Notice that the nullity of O ∞ must be zero for 

‖y(t)‖ to be nonzero for all nonzero x 0 (or, equivalently O ∞ must have full rank). From 

the Cayley-Hamilton theorem, we know thatF n is linearly dependent onF n−1 , F n−2 , ..., 

F , I. Therefore, 

⎛⎡ 

⎤⎞ 

H 

HF 

rank(O ∞ ) = rank(O) = rank 

HF 2 

⎜⎢ 

⎥⎟ 

⎝⎣ 

. ⎦⎠ 

HF n−1 

Therefore, the system is observable if the observability matrix O has full rank. 

If a system is observable, there always exists an observer gain, L, such that the 

closed-loop eigenvalues ofF −LH can be arbitrarily assigned. Therefore, the following 


• A system is observable. 

• The output is nonzero (i.e., ‖y(t)‖ ≠ 0) for all t > 0 and for all nonzero initial 

conditionsx 0 . 


Kyoungchul Kong


− 

u 

x& 

= Fx + Gu 

y = Hx 

y 

L 

+ 

− 

G 

+ 

x 0ˆ xˆ 

+ + 

∫ 

H 

ŷ 

F 

K 

Computer algorithm 

Figure 2.25: Block diagram of observer-based state feedback-controlled system: 

the thick lines represent vector quantities. 

• The initial condition,x 0 , can be calculated from the output,y(t), in a finite time. 

⎡ ⎤ 

H 

HF 

• 

HF 2 


⎢ ⎥ 

⎣ . ⎦ 

HF n−1 

• There existsLsuch that the eigenvalues ofF −LH can be arbitrarily determined. 

• There existsLsuch that the state estimate, ˆx(t), converges to the actual state, x(t), 

as t → ∞. 

Observer-based state feedback 

In the full state feedback control system, the information on the state, which is not measurable 

in general, was required. Therefore, for implementation the full state feedback 

always accompanies the state observer, which estimates the state. Recall that the full state 

feedback control law is 

u = −Kx 

Since x is not measurable, the estimate of x is alternatively used in the observer-based 

state feedback control: 

u = −Kˆx 

Remind that the state feedback gain, K, was designed such that the eigenvalues of 

[F −GK] have strictly negative real parts. In the state observer system, the observer gain, 

L, was designed such that the eigenvalues of [F −LH] have strictly negative real parts. 

When both the state feedback and the state observer are used at the same time, however, 


Kyoungchul Kong


it is not clear if the overall system (i.e., the closed-loop system with observer-based state 

feedback) is stable. To assess the stability of the observer-based state feedback-controlled 

system, the state space equations are arranged as: 

Rearranging the equations above, 

ẋ = Fx+Gu 

y = Hx 

˙ˆx = [F −LH]ˆx+Gu+Ly 

u = −Kˆx 

ẋ = Fx−GKˆx 

˙ˆx = [F −LH −GK]ˆx+LHx 

The two equations above can be represented in a single state space equation, i.e. 

[ ] F −GK 

ẋ = 

x (2.18) 

LH F −LH −GK 

[ ] 

wherex = ∈ R xˆx 

2n . Define another state, ¯x, by 

¯x = 

[ I 0 

I −I 

] 

x 

Then (2.18) is converted into 

˙¯x = 

[ 

F −GK GK 

0 F −LH 

] 

¯x 

Remind that the eigenvalues of [ ] 

F−GK GK 

0 F−LH are equal to those ofF −GK andF −LH 

due to the property of determinant of upper diagonal block matrix. Therefore, the closedloop 

eigenvalues of observer-based feedback-controlled system are the eigenvalues of 

F −GK and F −LH, which implies thatK and L can be separately designed such that 

F −GK andF −LH respectively have eigenvalues with strictly negative real parts. 

Linear quadratic (LQ) optimal control 

We have found that the state feedback allows us to assign any closed-loop system eigenvalues 

if the system is controllable. The desired eigenvalues must be selected by a designer. 

Another way to determine feedback control gains is to solve linear quadratic (LQ) 

optimal control problem. The LQ problem is one of the most frequently appearing optimal 

control problems. 


Kyoungchul Kong


Suppose, a system under the full state feedback control is described by 

ẋ = Fx+Gu 

y = Hx 

x(0) = x 0 

The optimal control is sought to minimize the quadratic performance index, 

or in more general form, 

J = 

J = 

∫ ∞ 

0 

∫ ∞ 

0 

[ 

y T y +u T Ru ] dt 

[ 

x T Qx+u T Ru ] dt (2.19) 

whereQ = H T H is a positive semidefinite matrix, and R is positive definite. Notice that 

the first term penalizes the deviation of x from the origin, while the second penalizes the 

control energy. 

The LQ problem as formulated above is concerned with the regulation of the system 

around the origin of the state space. The resulting system is called the Linear Quadratic 

Regulator (LQR). 

Let P ∈ R n×n be a positive definite matrix. Then 

x T (∞)Px(∞)−x T 0Px 0 = 

= 

= 

∫ ∞ 

0 

∫ ∞ 

0 

∫ ∞ 

0 

∫ ∞ 

d [ 

x T Px ] dt 

dt 

[ (dx 

Since (2.20) is true for any P , selectP such that 

From (2.20) and (2.21) 

0 = x T 0 Px 0 + 

∫ ∞ 

0 

= 

0 

) ] 

T 

Px+x T P dx dt 

dt 

dt 

[ 

] 

(Fx+Gu) T Px+x T P (Fx+Gu) dt 

[ ( x 

T 

F T P +PF ) x+u T G T Px+x T PGu ] (2.20) dt 

F T P +PF = PGR −1 G T P −Q (2.21) 

[ 

x 

T ( PGR −1 G T P −Q ) x+u T G T Px+x T PGu ] dt (2.22) 


Kyoungchul Kong


where x T (∞)Px(∞) in (2.20) was neglected because x(t) → 0 as t → ∞ if the closedloop 

system is stable. Adding (2.22) to the cost function in (2.19), 

J = x T 0 Px 0 + 

= x T 0 Px 0 + 

∫ ∞ 

0 

∫ ∞ 

0 

[ 

x T PGR −1 G T Px+u T G T Px+x T PGu+u T Ru ] dt 

[ 

(R −1 G T Px+u) T R(R −1 G T Px+u) ] dt 

SinceRis positive definite, in order forJ to be minimum, 

u = −R −1 G T Px 

and the minimum value ofJ is 

J o = x T 0 Px 0 

[Summary] For a controllable system 

ẋ = Fx+Gu 

y = Hx 

x(0) = x 0 

the state feedback controller that minimizes 

is 

J = 

∫ ∞ 

whereP is the positive definite solution of 

0 

[ 

x T Qx+u T Ru ] dt, Q ≥ 0,R > 0 

u = −R −1 G T Px 

F T P +PF −PGR −1 G T P +Q = 0 

which is called Riccati equation. The Riccati equation has a proper solution, as long as 

the state space equation is controllable. 

The linear quadratic regulator 18 is very useful in practice, because (1) the LQcontrolled 

systems are always stable, and (2) the number of parameters to be tuned is 

reduced to one 19 . 

18 It is called “regulator” often, because the LQ-gains are optimal for regulation problems (i.e.,r(t) = 0). 

Nevertheless, you can use the LQ-gains for tracking control problems. 

19 R is the only parameter to be tuned. 


Kyoungchul Kong


Moreover, performance can easily be expected according to the choice ofR. Namely, 

the smaller R, the shorter settling time, vice versa. Fig. 2.26 shows the experimental results 

of the LQR for a servo motor. Note in the experimental results that the settling time 

reduced, as R was decreased. In practice, however, R cannot be too small, because the 

magnitude of control signal is limited by hardware issues. In the experiment, the control 

signal was in the range of [−10,10], and thus the performance was not improved for 

R < 0.01. 

[Example 2-23] For a state space equation, 

[ 0 1 

ẋ = 

0 0 

y = [ 1 0 ] x 

[ ] 

1 

x(0) = 

0 

the full state feedback gain that minimizes 

J = 

∫ ∞ 

wherer > 0, is calculated as follows. 

0 

] [ 0 

x+ 

1 

[ 

y 2 (t)+ru 2 (t) ] dt 

The positive definite solution of the Riccati equation 

[ ] [ ] [ ] 

0 0 0 1 0 

P +P −P r −1[ 0 1 ] P + [ 1 0 ][ ] 

1 

= 0 

1 0 0 0 1 

0 

] 

u 

is 

[ √ 

2r 

0.25 

r 

P = √ 0.5 

r 0.5 2r 

0.75 

Thus, the optimal state feedback law is 

] 

u = − [ r −0.5 √ 

2r 

−0.25 ] x 

and the minimal cost function is J o = x T 0 Px 0 = √ 2r 0.25 . Notice that J o is decreased, 

as r decreases. 


Kyoungchul Kong


1143 

(a) Parker BE series motor; its transfer function is G( s) 

= 

s 

2 + 1.714s 

1 

1 

Position (Rev.) 

0.5 


0.5 

0 

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 

Time (sec.) 

0 

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 

Time (sec.) 

Controller Output (V) 

0 

-0.5 

-1 

-1.5 

-2 

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 

Time (sec.) 

(b) LQ Regulation (R=1, five experiments) 


5 

0 

-5 

-10 

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 

Time (sec.) 

(c) LQ Regulation (R=0.1, five experiments) 

1 

1 


0.5 

0 

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 

Time (sec.) 


0.5 

0 

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 

Time (sec.) 


5 

0 

-5 

-10 

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 

Time (sec.) 

(d) LQ Regulation (R=0.05, five experiments) 


10 

5 

0 

-5 

-10 

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 

Time (sec.) 

(e) LQ Regulation (R=0.01, three experiments) 

Figure 2.26: Experimental results of the linear quadratic regulator for a Parker 

BE series servo motor. Notice that the only parameter to be tuned is R, and the 

LQ algorithm finds the optimal controller gains based for the selected R. 


Kyoungchul Kong

2 34 564 476 

1 

2 346576 8 

2 ;6 < : =>@AB 

: 

2 9DE96 

< C =>@AB 

C 

F 2 AN@OP>Q R@>@S =SSTU>V W>OXG 

< 

H 2 RAQY@OAX @A @ZS :O>@O S[\ 

< 


9 2 35 476 

3FGH7 2 IJKL1G8GCG:M 

Figure 2.27: Matlab code for calculating the linear quadratic optimal state feedback 

gain 

2.4.5 Controller design in frequency-domain 

Requirements of controller design 

Remind that the information about the performance of the closed-loop system, obtained 

from the open-loop frequency response, are: 

• Low frequency region indicates the steady-state behavior. 

• Medium frequency region (around−1+0j in the Nyquist plot, or around gain and 

phase crossover frequencies 20 in the Bode plot) is related to stability. 

• High frequency region indicates the transient behavior. 

The requirements on open-loop frequency response are: 

• The gain at low frequency should be large enough to give a high value for error 

constants 21 . 

• At medium frequencies, the phase and gain margins should be large enough. 

• At high frequencies, the gain should be small enough as rapidly as possible to minimize 

noise effects. 

Lead compensator 

A lead compensator improves the transient response by increasing the gains at high frequencies. 

Since it increases the high frequency gain, the system bandwidth is increased 

20 The gain crossover frequency is ω Cg , where |GC(ω Cg )| = 1. The phase crossover frequency is ω Cp , 

where∠GC(ω Cp ) = −180. 

21 The error constant isK e , where steady state error(%) = 100 1 

1+K e 


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as well. In addition, addition of phase lead near the gain crossover frequency improves 

the phase margin. On the other hand, the disadvantage is that it amplifies high-frequency 

noise, which is not desired in practice. 

A lead compensator is 

C(s) = K c a Ts+1 

aTs+1 = K c 

s+ 1 T 

s+ 1 

aT 

where T > 0 and 0 < a < 1. Note that the lead compensator has a pole at − 1 and a aT 

zero at − 1 . The maximum phase-lead angle ϕ T m occurs at ω m , where ω m is the middle 

point between 1 T and 1 

aT 

in the log scale, i.e. 

logω m = 1 [ 

log 1 2 T +log 1 

aT 

1 

ω m = √ aT 

] 

and thus 

[ 

ϕ m = ∠C(jω m ) = ∠ K c a Tjω ] 

m +1 

aTjω m +1 

[ ] 

T √ 

1 

aT 

j +1 

= ∠ 

aT √ 1 

aT 

j +1 

[√ ] aj +a 

= ∠ √ aj +1 

[ ] 1−a 

= sin −1 1+a 

Moreover, note that 

|C(jω m )| = K c 

√ a 

Using these properties, a lead compensator is designed as follows. 

1. Determine the compensator gainK c a as the performance enhancement ratio 22 . 

2. Find the gain margin and phase margin of the gain-adjusted open loop system, i.e., 

K c aG(s). 

3. Determine the additional phase leadϕ m required for the desired gain margin+10% ∼ 

15%. 

22 If the desired steady state error is E(%) is given instead of the performance enhancement ratio, K c a 

should be selected such thatE = 100 

1 

1+K caG(0) . 


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14 

12 

Bode Diagram 

Magnitude (dB) 

10 

8 

6 

4 

Lead compensator 

Lag compensator 

2 

0 

45 

Phase (deg) 

0 

Phase lead 

Phase lag 

−45 

10 −2 10 −1 10 0 10 1 10 2 


Figure 2.28: Bode plots of a lead compensator and a lag compensator 

4. Obtainafrom sinϕ m = 1−a 

1+a . 

5. Determineω m such thatω m is the gain crossover frequency, i.e., 

|GC(jω m )| = 1 ⇔ |G(jω m )| = 1 

K c 

√ a 

6. Find T from ω m and the transfer function ofC(s): 

1 

T = √ aωm 

C(s) = K c a Ts+1 

aTs+1 


G(s) = 

4 

s(s+2) 

(2.23) 

with the following performance requirements. 

• performance enhancement ratio =10 

• phase margin> 50 ◦ 

Following the procedure, a lead compensator is designed: 


Kyoungchul Kong


50 

G(s): Gm = Inf dB (at Inf rad/sec) , Pm = 51.8 deg (at 1.57 rad/sec) 

GC(s): Gm = Inf dB (at Inf rad/sec) , Pm = 50.5 deg (at 8.9 rad/sec) 


0 

−50 

−100 

G(s) 

GC(s) 

−150 

−90 

Phase (deg) 

−135 

−180 

10 −1 10 0 10 1 10 2 10 3 


1. K c a = 10. 

Figure 2.29: Bode plots ofG(s) in (2.23) and GC(s) 

2. The phase margin of10G(s) is about 17 ◦ . 

3. The additional phase lead isϕ m = 0.15(50 ◦ −17 ◦ ) ≈ 38 ◦ . 

4. a = 1−sinϕm 

1+sinϕ m 

≈ 0.24. 

5. ω m is determined such that 

4 

∣jω m (jω m +2) ∣ = 1 √ 

K c a 

By simple calculation,ω m ≈ 9(rad/sec). 

6. T = 1 √ aωm 

= 0.227. Finally, 

C(s) = 10 0.227s+1 

0.0545s+1 = 41.65s+4.41 s+18.4 


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1.4 

1.2 

Step Response 

Amplitude 

1 

0.8 

0.6 

0.4 

0.2 

G(s)/(1+G(s)) 

GC(s)/(1+GC(s)) 

0 

0 1 2 3 4 5 6 

Time (sec) 

Figure 2.30: Feedback step responses ofG(s) in (2.23) and GC(s) 

Lag compensator 

In contrast to the lead compensator, a lag compensator decreases gain at high frequencies 

and moves the gain crossover frequency lower to obtain the desired phase margin. 

A lag compensator is 

C(s) = K c b Ts+1 

bTs+1 = K c 

s+ 1 T 

s+ 1 

bT 

whereT > 0 and b > 1. The design procedure is as follows: 

1. Let the desired phase margin beϕ d . 

2. Determine the compensator gainK c b to be the performance enhancement ratio. 

3. Find the gain margin and phase margin ofK c bG(s). 

4. Find the frequency point where the phase ofK c bG(s) is equal to(−180 ◦ +ϕ d +5 ◦ ∼ 

12 ◦ ). This will be the new gain crossover frequency,ω C . 

5. DetermineT such thatT ∈ ( 5 

ω C 

, 10 

ω C 

). 

6. Determinebsuch that 

Approximately, 

|GC(jω C )| = 1 

−K c b|G(jω C )| = −20logb 


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G(s) = 

1 

s(s+1)(0.5s+1) 

(2.24) 

with the following performance requirements. 

• performance enhancement ratio =5 

• phase margin> 40 ◦ 

Following the procedure, a lag compensator is designed: 

1. ϕ d = 40. 

2. K c b = 5. 

3. The phase margin of 5G(s) is about −13 ◦ . Thus the closed-loop system is unstable 

for K c b = 5. From the Bode plot of 5G(jω), you may note that at about 

0.5rad/sec, the phase is −130 ◦ = −180 ◦ +ϕ d +10 ◦ . 

Therefore, the new gain crossover frequency will beω C = 0.5rad/sec. 

4. T = 5 

ω C 

= 10. 

5. b = 10 is roughly selected from 

−K c b|G(jω C )| = −20logb 

6. Finally, 

C(s) = 5 10s+1 

100s+1 

= 0.5 

s+0.1 

s+0.01 


Kyoungchul Kong


G(s): Gm = 14.3 dB (at 1.32 rad/sec) , Pm = 41.6 deg (at 0.454 rad/sec) 

GC(s): Gm = 9.54 dB (at 1.41 rad/sec) , Pm = 32.6 deg (at 0.749 rad/sec) 

100 

50 

G(s) 

GC(s) 


0 

−50 

−100 

−150 

−90 

Phase (deg) 

−135 

−180 

−225 

−270 

10 −4 10 −3 10 −2 10 −1 10 0 10 1 10 2 


Figure 2.31: Bode plots ofG(s) in (2.24) and GC(s) 

Step Response 

Amplitude 

1.4 

1.2 

1 

0.8 

0.6 

0.4 

0.2 

G(s)/(1+G(s)) 

GC(s)/(1+GC(s)) 

0 

0 5 10 15 20 25 30 35 

Time (sec) 

Figure 2.32: Feedback step responses ofG(s) in (2.24) and GC(s) 


Kyoungchul Kong

Chapter 3 

Basics of LabVIEW for 

Control Systems Implementation 

LabVIEW is a graphical programming language that uses icons instead of lines of text to 

create applications. In contrast to text-based programming languages, where instructions 

determine program execution, LabVIEW uses dataflow programming, where the flow of 

data determines execution. 

Project 

In order to control a system, multiple functions and programs are required. The Project 

in LabVIEW defines a group of custom-made programs that will interact with each other. 

An empty project can be made by clicking New → Empty Project on the main window. 

Once a project is made, the main window disappears. 

Virtual instruments 

LabVIEW programs are called virtual instruments, or VIs, because their appearance and 

operation imitate physical instruments, such as oscilloscopes and multimeters. Every VI 

uses functions that manipulate input from the user interface or other sources and display 

that information or move it to other files or other computers. A VI can be created by 

clicking the right button on my computer of the project explorer, New, and VI, as shown 

in Fig. 3.1. 

A VI contains the following components: 

• Front panel serves as the user interface. 

• Block diagram contains the graphical source code that defines the functionality of 

the VI. 

81

82 

Figure 3.1: Creating a Virtual Instrument on an empty project 

Figure 3.2: The front panel (left) and block diagram (right) of a VI 

You may press Ctrl + e, or Ctrl + t to switch the window from one to another. In 

order to run a VI, press Ctrl + r, or click the⇒button on the top left. 

Controls palette 

The Controls palette contains the controls and indicators you use to create the front panel. 

The Controls palette is available only on the front panel. The controls and indicators are 

located on subpalettes based on the types of controls and indicators. In order to place an 

indicator on the front panel, select Window → Show Controls Palette or rightclick 

the front panel workspace to display the Controls palette. You can place the Controls 

palette anywhere on the screen. LabVIEW retains the Controls palette position and size 

so when you restart LabVIEW, the palette appears in the same position and has the same 


Kyoungchul Kong

83 

]^_à bc^bde 

Figure 3.3: Controls palette 

size. 

Functions palette 

The Functions palette is available only on the block diagram. The Functions palette contains 

the VIs and functions you use to build the block diagram. The VIs and functions 

are located on subpalettes based on the types of VIs and functions. To select a function 

icon, select Window→ Show Functions Palette or right-click the block diagram 

workspace to display the Functions palette. 

Tools palette 

A tool is a special operating mode of the mouse cursor. The cursor corresponds to the 

icon of the tool selected in the palette. Use the tools to operate and modify front panel 

and block diagram objects. 

Data types 

Fig. 3.6 shows the symbols for the different types of control and indicator terminals. 

The color and symbol of each terminal indicate the data type of the control or indicator. 

Control terminals have a thicker border than indicator terminals. Also, arrows appear on 

front panel terminals to indicate whether the terminal is a control or an indicator. An 

arrow appears on the right if the terminal is a control, and an arrow appears on the left if 

the terminal is an indicator. 


Kyoungchul Kong

84 

Figure 3.4: Functions palette 

Figure 3.5: Tools palette 

Control 

(Input) 

Indicator 

(Output) 

Double-precision floating-point numeric 

16-bit signed integer numeric 

32-bit signed integer numeric 

16-bit unsigned integer numeric 

Boolean 

Figure 3.6: Control and indicator data types 


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85 

For loop While loop Timed while loop 

Figure 3.7: Fundamental three loops 

Loops and structures 

Structures are graphical representations of the loops and case statements of text-based 

programming languages. Use structures on the block diagram to repeat blocks of code 

and to execute code conditionally or in a specific order. 

Like other nodes, structures have terminals that connect them to other block diagram 

nodes, execute automatically when input data are available, and supply data to 

output wires when execution completes. Each structure has a distinctive, resizable border 

to enclose the section of the block diagram that executes according to the rules of the 

structure. The section of the block diagram inside the structure border is called a subdiagram. 

The terminals that feed data into and out of structures are called tunnels. A tunnel 

is a connection point on a structure border. 

A For loop executes a subdiagram a set number of times. The count terminal, 

shown as [N], indicates how many times to repeat the subdiagram. Set the count explicitly 

by wiring a value from outside the loop to the left or top side of the count terminal. 

The iteration terminal, shown as [i], contains the number of completed iterations. The 

iteration count always starts at zero. 

A While loop executes a subdiagram until a condition is met. The While loop 

executes the subdiagram until the conditional terminal, an input terminal, receives a specific 

Boolean value. 

A Timed While loop is similar to the While loop, but executes a subdiagram 

at a constant calculation period. 

Shift registers 

Use shift registers when you want to pass values from previous iterations through the 

loop. A shift register appears as a pair of terminals directly opposite each other on the 

vertical sides of the loop border. The right terminal contains an up arrow and stores data 

on the completion of an iteration. LabVIEW transfers the data connected to the right side 

of the register to the next iteration. Create a shift register by right-clicking the left or right 

border of a loop and selecting Add Shift Register from the shortcut menu. 


Kyoungchul Kong

86 

Figure 3.8: Creating a shift register 

Formula node 

The Formula Node is a convenient text-based node you can use to perform mathematical 

operations on the block diagram. You do not have to access any external code or 

applications, and you do not have to wire low-level arithmetic functions to create equations. 

Formula Nodes are useful for equations that have many variables or are otherwise 

complicated and for using existing text-based code. You can copy and paste the existing 

text-based code into a Formula Node rather than recreating it graphically. 

When you work with variables, remember the following points: 

• There is no limit to the number of variables or equations in a Formula Node. 

• No two inputs and no two outputs can have the same name, but an output can have 

the same name as an input. 

• Declare an input variable by right-clicking the Formula Node border and selecting 

Add Input from the shortcut menu. You cannot declare input variables inside the 

Formula Node. 

• Declare an output variable by right-clicking the Formula Node border and selecting 

Add Output from the shortcut menu. The output variable name must match either 

an input variable name or the name of a variable you declare inside the Formula 

Node. 

• You can change whether a variable is an input or an output by right-clicking it 

and selecting Change to Input or Change to Output from the shortcut 

menu. 

• You can declare and use a variable inside the Formula Node without relating it to 

an input or output wire. 

• You must wire all input terminals. 


Kyoungchul Kong

87 

The indicator on the front panel 

is connected to this icon. 

Formula node is set to calculate 

y ( k) 

= y( 

k −1) 

+ k 

for k = 10. 

The shift register is initialized 

such that y(0) = 0. 

A subdiagram repeats 10 times. 

Figure 3.9: An example using a Formula Node and a For loop 

File input and output 

Use the high-level File I/O VIs to perform common I/O operations, such as writing 

to or reading from the following types of data: 

• Characters to or from text files. 

• Lines from text files. 

• 1D or 2D arrays of single-precision numerics to or from spreadsheet text files. 

• 1D or 2D arrays of single-precision numerics or 16-bit signed integers to or from 

binary files. 

[Example 3-1] A bank account simulator is to be designed with the following functions: 

• The calculation is updated every 1 second, which correspond to one month in 

reality. 

• The monthly interest is0.5%. 

• The user can specify the amount of money to be saved for every month. 

Using a Timed While loop, the algorithm can be realized as shown in Fig. 3.11. 


Kyoungchul Kong

‚ƒ} ~€ 

pvŠˆ‰||} ‹Œ…†zz € zŽƒ xyzy † zy ƒx ‘y ’“”• 

„…†zvpv‡ˆ‰|ˆ 

88 

(Setting significant digits) 

Two 1D arrays are stacked to a 2D array. 

The data is stacked to an array by indexing. 

g h i j k l m n o 

f 

g i l gf gk hg hn il jk 

f 

Saved data (opened with Excel) 

p q rstuvwxyzy{|} 

Matlab code for plotting the data obtained by LabVIEW 

Figure 3.10: Saving data generated by a For loop and plotting with Matlab 

Program runs every 1 sec. 

Program runs until the stop button is clicked. 

The chart is updated at every iteration. 

Figure 3.11: A bank account simulator in [Example 3-1] 


Kyoungchul Kong

89 

[Example 3-2] Suppose a plant described in state space: 

[ ] [ 0 1 0 

ẋ = x+ 

−1 −2 1 

y = [ 1 0 ] x 

[ ] 

1 

x(0) = 

0 

] 

u 

Since the differential equation cannot be realized by a computer algorithm, the equations 

above are approximated by 

ẋ ≈ 

x(kT +T)−x(kT) 

T 

where T is the calculation period of the computer algorithm, and k is the time index. 

Note that t is replaced by kT in the discretization process. Substituting the equations 

above, we get 

x(kT +T)−x(kT) 

T 

≈ 

[ 

0 1 

−1 −2 

] [ 

0 

x(kT)+ 

1 

] 

u(kT) 

OmittingT in the index for simplicity, 

[ 1 T 

x(k +1) ≈ 

−T 1−2T 

] [ 0 

x(k)+ 

T 

] 

u(k) 

Assuming T = 1ms and u = 0, the approximated plant model is realized as shown in 

Fig. 3.12. In the program, x(k) = [ x1 

x2] and x(k +1) = [ x1f 

x2f]. 

[Example 3-3] For the system in [Example 3-2], an appropriate full state feedback 

controller is 

u = − [ 1 1 ] x 

The full state feedback controller is implemented as shown in Fig. 3.13. 


Kyoungchul Kong

› œ ž Ÿ 

–—˜š 

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œ Ÿ£¢Ÿ ¤ ›£¢¥ 

¢Ÿ– 

œ ¦›£¢Ÿ ¤ §Ÿ¦¥£›¨£¢¥ ¤ ›£¡ 

¢¥– 

¯ ° ±²±±³´ 

ª«¬® 

° ³·¸³ ¹ ³·¸º´ 

µ 

° ³·¸³ » ¯·¸º´ 

¸³ª 

° ¯·¸³ » ¼³º·¯½·¸º » ¯·µ´ 

¸ºª 

90 

© œ Ÿ£¢Ÿ ¤ £¢¥ 

Figure 3.12: An approximated state space equation in [Example 3-2] 

¾ ° ³·¸³ » ±·¸º´ 

Figure 3.13: An approximated state space equation with full state feedback 


Kyoungchul Kong

Chapter 4 

Discrete-Time Domain Analysis 

4.1 Discrete-Time Domain Signals 

In computer-based control systems, the output signal is measured once in a loop, and the 

measured signal is regarded as the “current signal” until the next loop is executed. This 

process is called “sampling,” and the measured signal is in the “discrete-time domain.” 

Computer 

¿ÀÁÂÃÄ 

ÅÂÆÄ 

Ç 

Figure 4.1: Sampling of a signal. 

[Example 4-1] Suppose that a signal 

y(t) = sin(t) 

is measured by a computer at a sampling frequency of10Hz, as shown in Fig. 4.1. 

Since the sampling period is 

T = 1 10 = 0.1 

the measured signal is 

y(k) = sin(0.1k) 

wherek = 0,1,2... is the time index. 

91

4.2. Z-TRANSFORM AND TRANSFER FUNCTIONS 92 

Fourier Transform 

y(t) 

Continuous-time domain signal 

Laplace Transform 

Inverse Laplace 

Transform 

 

È É ÊË 

 

Sampling 

Ì É Í ÎÏ 

y(k) 

z-Transform 

Inverse z-Transform 

 

Ì É Í ÐÑÏ 

ÒÓÔ 

Discrete-time domain signal 

Discrete-Time Fourier Transform 

Figure 4.2: Various domains for analysis of signals 

4.2 z-Transform and Transfer Functions 

The z-transform converts a discrete-time domain signal, which is a sequence of real 

or complex numbers, into a complex frequency-domain representation. The name “ztransform” 

may have been derived from the idea of the letter “z” being a sampled/digitized 

version of the letter “s” used in Laplace transforms. This seemed appropriate since the 

z-transform can be viewed as a sampled version of the Laplace transform. 

Definition 

For a signal,x(k) wherek ∈ Z + 1 , thez-transform is defined as 

X(z) = Z{x(k)} = 

∞∑ 

x(k)z −k 

k=0 

Properties 

Thez-transform has the following properties: 

1 Z represents an integer, andZ + means a positive integer. 


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• Linearity 2 : Z{a 1 x 1 (k) + a 2 x 2 (k)} = a 1 X 1 (z) + a 2 X 2 (z), where a 1 and a 2 are 

scalars. 

• Time shifting 3 : Z{x(k −n)} = z −n X(z), wheren ∈ Z. 

• Scaling bya k 4 : Z{a k x(k)} = X(a −1 z), wherea ∈ C. 

• Time reversal: Z{x(−k)} = X(z −1 ). 

• First difference (differentiation): Z{x(k)−x(k −1)} = (1−z −1 )X(z). 

• Accumulation (integration): Z{ ∑ k 

i=0 x(i)} = 1 

1−z −1 X(z). 

• Convolution 5 : Z{ ∑ ∞ 

l=0 x 1(l)x 2 (k −l)} = X 1 (z)X 2 (z). 

2 Proof: 

3 Proof: 

4 Proof: 

X(z) = 

∞∑ 

(a 1 x 1 (k)+a 2 x 2 (k))z −k 

k=0 

∑ ∞ 

∑ ∞ 

= a 1 x 1 (k)z −k +a 2 x 2 (k)z −k = a 1 X 1 (z)+a 2 X 2 (z) 

k=0 

Z{x(k −n)} = 

Z{a k x(k)} = 

k=0 

∞∑ 

x(k −n)z −k = 

k=0 

∞∑ 

i=−n 

x(i)z −(i+n) 

∑ ∞ 

= z −n x(i)z −i note: x(i) = 0 if i < 0, 

i=−n 

∑ 

∞ 

= z −n x(i)z −i = z −n X(z) 

i=0 

∞∑ ∞∑ 

a k x(k)z −k = x(k)(a −1 z) −k 

k=0 

= X(a −1 z) 

k=0 

5 Proof: 

∞∑ 

Z{ x 1 (l)x 2 (k −l)} = 

l=0 

= 

= 

( 

∞∑ ∑ ∞ 

) 

x 1 (l)x 2 (k −l) 

k=0 

l=0 

z −k 

( 

∑ ∞ 

) 

∑ ∞ 

x 1 (l) x 2 (k −l)z −k 

l=0 

k=0 

∞∑ 

x 1 (l)z −l X 2 (z) = X 1 (z)X 2 (z) 

l=0 


Kyoungchul Kong


Initial value theorem 

Recall that 

∞∑ 

X(z) = x(k)z −k 

k=0 

= x(0)+x(1)z −1 +x(2)z −2 +x(3)z −3 ... 

Ifz → ∞, the terms ofz −1 , z −2 , z −3 ... become all zero. Therefore, 

x(0) = lim 

z→∞ 

X(z) 

which is called the initial value theorem. The initial value theorem holds only if x(0) 

exists. 

Final value theorem 

For a signal,x(k) wherek ≥ 0, define the difference∆x(k) such that 

∆x(k) := x(k)−x(k −1) 

where∆x(0) = x(0). The summation of∆x(k) results in 

∞∑ 

∆x(k) = (x(0))+(x(1)−x(0))+(x(2)−x(1))+... 

k=0 

= x(∞) 

Sincelim z→1 z −k = 1 for all k, it can be multiplied to the equation above, i.e. 

x(∞) = lim 

z→1 

∞ 

∑ 

k=0 

∆x(k)z −k = lim 

z→1 

Z{∆x(k)} 

Note thatZ{∆x(k)} = Z{x(k)−x(k −1)} = X(z)−z −1 X(z). Therefore, we get 

x(∞) = lim 

z→1 

(1−z −1 )X(z) 

which is called the final value theorem. The final value theorem holds only for signals 

that converge to a certain value. 

Table of commonz-transform pairs 

Signals in the discrete-time domain, defined for k ≥ 0, 6 are converted into thez-domain 

as follows. 

6 It is assumed thatx(k) = 0 fork < 0. 


Kyoungchul Kong


• Impulse signal: Z{δ(k)} = 1 

• Delayed impulse signal 7 : Z{δ(k −k 0 )} = z −k 0 

• Step signal 8 : Z{u(k)} = 1 

1−z −1 

• Delayed step signal: Z{u(k−k 0 )} = z−k 0 

1−z −1 

• Exponential decay: Z{e −ak } = 

1 

1−e −a z −1 

• Ramp signal: Z{k} = z−1 

(1−z −1 ) 2 

• 2 nd -order polynomial signal: Z{k 2 } = z−1 (1+z −1 ) 

(1−z −1 ) 3 

• k-th power: Z{a k } = 1 

1−az −1 

• k-th power times ramp: Z{ka k } = kz−1 

(1−az −1 ) 2 

• Cosine signal: Z{cos(ω 0 k)} = 1−z−1 cos(ω 0 ) 

1−2z −1 cos(ω 0 )+z −2 

• Sine signal: Z{sin(ω 0 k)} = 

z −1 sin(ω 0 ) 

1−2z −1 cos(ω 0 )+z −2 

• Decaying cosine signal: Z{a k cos(ω 0 k)} = 

• Decaying sine signal: Z{a k sin(ω 0 k)} = 

1−az −1 cos(ω 0 ) 

1−2az −1 cos(ω 0 )+a 2 z −2 

az −1 sin(ω 0 ) 

1−2az −1 cos(ω 0 )+a 2 z −2 

Transfer functions inz-domain 

The convenience in convolution makes z-transform useful to analyze input-output relationships, 

i.e., transfer functions. 

[Example 4-2] Consider a difference equation: 

y(k+2)+3y(k+1)−y(k)+2y(k−1) = 2u(k)+u(k−1) 

7 Proof: 

Z{δ(k −k 0 )} = 

∞∑ 

δ(k −k 0 )z −k = 0+0+...+0+z −k0 +0+... = z −k0 

k=0 

8 Note that a step signal is the integrated impulse signal. 


Kyoungchul Kong

ßæàäëèð 

õäåæëèð 


Õ Ö ×Ø××ÙÚ 

Û ÜÝÞßàáâã ßäåáæç 

è Ö éêëìÙ Ùíî ìÙ Ù Ùíî ïðÚ 

Û ñäêáâáâã Ý éåÝâòêäå êóâôéáæâ 

áÞßóàòäëèðÚ 

Û ñåÝöáâã Ýâ áÞßóàòä åäòßæâòä æê èëõð 

òéäßëèðÚ 

Û ñåÝöáâã Ýâ áÞßóàòä åäòßæâòä æê èëõð 

÷æçäëèðÚ 

Û ñåÝöáâã éøä ùæçä ßàæé æê èëõð 

âúûóáòéëèðÚ 

Û ñåÝöáâã éøä üúûóáòé ßàæé æê èëõð 

Figure 4.3: Matlab code for defining a transfer function in the discrete-time domain 

Taking thez-transform, the equation is converted into 

Z{y(k+2)+3y(k+1)−y(k)+2y(k−1)} = Z{2u(k)+u(k −1)} 

thus the transfer function fromu(k) toy(k) is 

(z 2 +3z −1+2z −1 )Y(z) = (2+z −1 )U(z) 

Y(z) 

U(z) = 

2+z −1 

z 2 +3z −1+2z −1 

It is common to express a transfer function such that the highest order of the denominator 

polynomial isz 0 = 1, i.e. 

Y(z) 

U(z) = 

2z −2 +z −3 

1+3z −1 −z −2 +2z −3 

The physical meaning of a transfer function in the discrete-time domain is the same 

as one in the continuous-time domain. For discrete-time input signal u(k) and output 

signal y(k), the transfer function is the linear mapping of the z-transform of the input, 

U(z), to the output,Y(z), i.e. 

Y(z) = G(z)U(z) 

or 

G(z) = Y(z) 

U(z) = Z{y(k)} 

Z{u(k)} 

whereG(z) is the transfer function of the system. 


Kyoungchul Kong


Response versus pole locations 

Given the transfer function of a linear system in the discrete-time domain, 

G(z) = Y(z) 

U(z) = b(z) 

a(z) 

the roots of a(z) = 0, called poles, make G(z) infinity, and those of b(z) = 0, called 

zeros, makeG(z) zero. 

Each pole location in thez-plane can be identified with a particular type of response. 

When the poles of a transfer function are not repeated (i.e., all the roots of a(z) = 0 are 

distinct), it can be expanded by a partial fraction expansion as 

G(z) = 

n∑ 

i=1 

k i 

z −p i 

where k i ’s are scalars, p i ’s are the poles ofG(z), and n is the order ofa(z). Suppose that 

the system is under an impulse input, i.e.,U(z) = 1. Then, the output is 

y(k) = Z −1 {G(z)U(z)} = Z −1 {G(z)} 

n∑ 

= Z −1 k i 

{ } 

z −p i 

= Z −1 { 

= 

i=1 

n∑ 

i=1 

n∑ 

k i p k−1 i = 

i=1 

k i z −1 

1−p i z −1} 

n∑ 

i=1 

k i p −1 

i p i 

k 

Note that the impulse response is a linear combination of p k i , which is called a mode. 

When all the poles are located in the unit circle 9 , every mode converges to zero ask → ∞, 

and thus y(∞) → 0. If any single pole is located on the unit circle (i.e., |p i | = 1), 

the associated mode maintains its magnitude for all k. Therefore, for a system in the 

discrete-time domain to be asymptotically stable, all the poles must be located in the 

unit circle. If any pole on the unit circle is repeated or any pole is located outside of the 

unit circle, the system in unstable. 

9 That is, |p i | < 1 for all i’s 


Kyoungchul Kong



G(z) = 

−1.55z +0.24 

(z +1.2)(z −0.2)(z −0.3) 

Using the partial fraction expansion,G(z) can be expanded to 

G(z) = 

= 

1 

z +1.2 + 0.5 

z −0.2 − 1.5 

z −0.3 

z −1 0.5z−1 1.5z−1 

+ − 

1+1.2z−1 1−0.2z−1 1−0.3z −1 

The impulse response (i.e., U(z) = 1) ofG(z) is 

y(k) = (−1.2) k−1 +0.5(0.2) k−1 −1.5(0.3) k−1 

Since(−1.2) k−1 diverges as k → ∞, the output does not converges to0. 


G(z) = 

z 3 −z 2 +0.5z 

(z −0.5)(z 2 −2z +1) 

Notice thatG(z) has a repeated pole on the unit circle (p 1 = 0.5,p 2 = p 3 = 1). Using 

the partial fraction expansion,G(z) can be expanded to 

G(z) = 

The impulse response ofG(z) is 

= 

Note thaty(k) diverges as k → ∞. 

z 

z 2 −2z +1 + z 

z −0.5 

z −1 

(1−z −1 ) + 1 

2 1−0.5z −1 

y(k) = k +(0.5) k 


Kyoungchul Kong


Im 

1.0 

0.5 

0.5 

1.0 

Re 

Asymptotically stable 

Marginally stable, if not repeated. 

Figure 4.4: Time sequences associated with pole locations in thez-plane 


Kyoungchul Kong


4.2.1 Frequency responses in discrete-time domain 

The z-transform is a generalization of the discrete-time Fourier transform (DTFT). The 

DTFT can be found by replacinge jωT forz ofX(z). In order for the DTFT of a signal to 

exist, the signal must not diverge. 

Nyquist frequency 

Recall that the frequency response of a transfer function in the discrete-time domain is a 

function ofe jωT = cosωT+jsinωT . Therefore, the computer system cannot measure or 

process signals in the frequency range higher thanω = 2π . T Moreover,ejωT forω ∈ [ π, 2π] 

T T 

is the complex conjugate of e jωT for ω ∈ [0, π ], and thus the effective frequency range is 

T 

onlyω ∈ [0, π ]. T 

The maximum frequency that a computer system can measure or process is called 

Nyquist frequency and is half the sampling frequency, i.e. π T . 


G(z) = 0.9 

z −0.1 

where the sampling period T = 1ms. The Nyquist frequency is 

ω N = π T = 1000π 

Thus, G(z) can accept an input signal or generate an output signal in the frequency 

range of0 ∼ 500Hz. 

The frequency response ofG(z) is obtained by replacinge jωT forz, i.e. 

G(e jωT ) = 

= 

0.9 

e jωT −0.1 

0.9 

(cos(ωT)+jsin(ωT))−0.1 


Kyoungchul Kong

4.3. STATE SPACE IN DISCRETE-TIME DOMAIN 101 

The magnitude and phase ofG(e jωT ) are 

|G(e jωT 0.9 

)| = √ 

1.01−0.2cos(ωT) 

( ) sin(ωT) 

φ G(e jωT ) = −tan −1 cos(ωT)−0.1 

Notice that the magnitude is changed from 1 to 0.818 as ω changes from 0 to π T , and 

the phase is changed from0 ◦ to −180 ◦ . 

4.3 State Space in Discrete-Time Domain 

4.3.1 Definition of state space in discrete-time domain 

As in the continuous-time domain, realizable difference equations can be represented in 

the state space. The general form of a single-input single-output 10 state space equation in 

the discrete-time domain is 

x(k +1) = Φx(k)+Γu(k) 

y(k) = Hx(k)+Ju(k) (4.1) 

whereΦ ∈ R n×n ,Γ ∈ R n ,H ∈ R 1×n , andJ ∈ R. Similar to the continuous-time case,J 

is 0 for most of the mechanical systems. 

Solution ofx(k) in discrete-time domain 

Assume that the initial condition of (4.1) is x(0) = x 0 ∈ R n . Then, following (4.1), x(1) 

is 

x(1) = Φx(0)+Γu(0) 

Likewise,x(2) is 

x(2) = Φx(1)+Γu(1) 

= Φ(Φx(0)+Γu(0))+Γu(1) 

1∑ 

= Φ 2 x(0)+ Φ 1−i Γu(i) 

10 Single-input single-output (SISO) means u ∈ R1 and y ∈ R 1 . When u ∈ R m and y ∈ R q , where 

m,q > 1, the system is called multi-input multi-output (MIMO). In this class, we deal with SISO systems 

only. 

i=0 


Kyoungchul Kong


Figure 4.5: Implementation of a state space equation, where Φ = [0 1;−0.5 − 

0.2], Γ = [0 1] T , and H = [0.1 0.9]. 

Repeating this calculation, the solution of a state space equation in the discrete-time domain 

is obtained: 

∑k−1 

x(k) = Φ k x 0 + Φ k−1−i Γu(i) 

It is important to solve Φ k . Suppose that V is a matrix in R n×n that consists of the 

eigenvectors ofΦ, i.e. 

ΦV = VΛ 

where Λ is a diagonal matrix with the eigenvalues ofΦ. Since V is always invertible 11 , it 

satisfies 

Φ = VΛV −1 

ThusΦ k is 

i=0 

Φ k = (VΛV −1 )(VΛV −1 )...(VΛV −1 ) = VΛ k V −1 

11 When Φ does not have n independent eigenvectors (i.e., Φ is defective), generalized eigenvectors can 

be utilized to makeV invertible, which results in Jordan form. 


Kyoungchul Kong


Notice that 

Λ k = 

J k = 

where 

⎡ 

⎢ 

⎣ 

⎡ 

⎢ 

⎣ 

λ k 1 0 0 

0 λ k 2 0 

. .. 

⎤ 

⎥ 

⎦ 

⎡ 

if Λ = ⎢ 

⎣ 

0 0 λ k n 

⎤ 

λ k ρ 12 λ (k−1) ρ 1n λ (k−n+1) 

0 λ k ρ 2n λ (k−n+2) 

.. 

⎥ 

. ⎦ 

0 0 λ k 

ρ ij = 

j−i−1 

1 ∏ 

(k −p) 

(j −i)! 

p=0 

⎤ 

λ 1 0 0 

0 λ 2 0 

. .. 

⎥ 

⎦ , and 

0 0 λ n 

⎡ 

λ 1 0 

0 λ . . . . 

if J = ⎢ 

⎣ . .. 1 

0 0 λ 

⎤ 

⎥ 

⎦ 

4.3.2 Relationship between state space equations and transfer functions 

in discrete-time domain 

Conversion from state space to transfer function 

Takingz-transform, 

Arranging the equation above, 

SinceY(z) = HX(z), 

Z{x(k +1)} = Z{Φx(k)+Γu(k)} 

zX(z) = ΦX(z)+ΓU(z) 

X(z) = (zI −Φ) −1 ΓU(z) 

G(z) = Y(z) 

U(z) = H(zI −Φ)−1 Γ 

Conversion from transfer function to state space 

A transfer function 

G(z) = 

can be converted into a state space model 


b 2 z 2 +b 1 z +b 0 

z 3 +a 2 z 2 +a 1 z +a 0 

= k 1 

z −p 1 

+ k 2 

z −p 2 

+ k 3 

z −p 3 

x(k +1) = Φx(k)+Γu(k) 

y(k) = Hx(k) 


Kyoungchul Kong




Diagonal Canonical Form 

⎡ 

⎣ 

Φ 

⎤ ⎡ 

Γ 

⎤ 

H 

0 1 0 0 

0 0 1 ⎦ ⎣ 0 ⎦ [ ] 

b 0 b 1 b 2 

−a 

⎡ 0 −a 1 −a 

⎤ 2 

⎡ 

1 

⎤ 

−a 2 1 0 b 2 [ ] 

⎣ −a 1 0 1 ⎦ ⎣ b 1 

⎦ 1 0 0 

⎡ 

−a 0 0 0 

⎤ ⎡ 

b 0 

⎤ 

p 1 0 0 1 

⎣ 0 p 2 0 ⎦ ⎣ 1 ⎦ [ ] 

k 1 k 2 k 3 

0 0 p 3 1 

When the transfer function has a repeated pole such that 

G(z) = 

the corresponding state space model is 


b 2 z 2 +b 1 z +b 0 

= k 1 k 2 

+ 

z 3 +a 2 z 2 +a 1 z +a 0 z −p 1 (z −p m ) + k 3 

2 z −p m 

x(k +1) = Φx(k)+Γu(k) 

y(k) = Hx(k) 



Jordan Canonical Form 

⎡ 

Φ 

⎤ ⎡ 

Γ 

⎤ 

H 

0 1 0 0 

⎣ 0 0 1 ⎦ ⎣ 0 ⎦ [ ] 

b 0 b 1 b 2 

−a 

⎡ 0 −a 1 −a 

⎤ 2 

⎡ 

1 

⎤ 

−a 2 1 0 b 2 [ ] 

⎣ −a 1 0 1 ⎦ ⎣ b 1 

⎦ 1 0 0 

⎡ 

−a 0 0 0 

⎤ ⎡ 

b 0 

⎤ 

p 1 0 0 1 

⎣ 0 p m 1 ⎦ ⎣ 0 ⎦ [ ] 

k 1 k 2 k 3 

0 0 p m 1 

4.3.3 Controllability and observability 

Controllability 

Remind that the concept of controllability is the ability to move the state of a system to 

an arbitrary desired state in a finite time using a certain input sequence. 

Consider a state space model in the discrete-time domain: 

x(k +1) = Φx(k)+Γu(k) 

y(k) = Hx(k) 

x(0) = 0 


Kyoungchul Kong


whereΦ ∈ R n×n ,Γ ∈ R n , andH ∈ R 1×n . The initial condition is assumed to be zero for 

simplicity. The solution of the state space model is 

x(k) = 

∑k−1 

Φ k−1−i Γu(i) 

i=0 

⎡ 

= [ Γ ΦΓ Φ 2 Γ ··· Φ k−1 Γ ] ⎢ 

⎣ 

:= CU 

u(k −1) 

u(k −2) 

u(k −3) 

. 

u(0) 

Since Φ ∈ R n×n , C ∈ R n×k becomes n by n matrix if k = n. Therefore, x(n) can reach 

any point ifC n has full rank, where 

C n = [ Γ ΦΓ Φ 2 Γ ··· Φ n−1 Γ ] ∈ R n×n 

Moreover, the corresponding input sequence is 

U = C n −1 x(n) 

If a system is controllable, there always exists a state feedback gain,K, such that the 

closed loop eigenvalues of Φ−ΓK can be arbitrarily assigned. Therefore, the following 


• A system is controllable. 

• The state of a system,x(k), can reach any point inR n atk = n with an appropriate 

input sequence, U = [ u(n−1) u(n−2) ··· u(0) ] T 

. 

• [ Γ ΦΓ Φ 2 Γ ··· Φ n−1 Γ ] has full rank. 

• There existsK such that the eigenvalues ofΦ−ΓK can be arbitrarily determined. 

⎤ 

⎥ 

⎦ 

Observability 

A state space model without an input 

x(k +1) = Φx(k) 

y(k) = Hx(k) 

x(0) = x 0 


Kyoungchul Kong


is said to be observable, if x 0 can be observed by the sequence of y(0), y(1), ..., y(k). 

Note that 

⎡ ⎤ ⎡ ⎤ 

y(0) H 

y(1) 

⎢ ⎥ 

⎣ . ⎦ = HΦ 

⎢ ⎥ 

⎣ . ⎦ x 0 ∈ R k 

y(k−1) HΦ k−1 

:= Ox 0 

The dimension of an observability matrix, O ∈ R k×n , becomes n by n if k = n. Therefore, 

the system is observable ifO n has full rank, where 

⎡ ⎤ 

H 

HΦ 

O n = ⎢ ⎥ 

⎣ . ⎦ ∈ Rn×n 

HΦ k−1 

Moreover,x 0 can be observed by 

⎡ 

−1 

x 0 = O n ⎢ 

⎣ 

y(0) 

y(1) 

. 

y(n−1) 

If a system is observable, there always exists an observer gain, L, such that the 

closed loop eigenvalues of Φ−LH can be arbitrarily assigned. Therefore, the following 


• A system is observable. 

• The output is nonzero (i.e., ‖y(k)‖ ≠ 0) for all k > 0 and for all nonzero initial 

conditionsx 0 . 

• The initial condition,x 0 , can be calculated from the output sequence,y(0),y(1), ..., 

y(n−1). 

⎡ ⎤ 

H 

HΦ 

• 

HΦ 2 


⎢ ⎥ 

⎣ . ⎦ 

HΦ n−1 

• There existsLsuch that the eigenvalues ofΦ−LH can be arbitrarily determined. 

• There existsLsuch that the state estimate, ˆx(k), converges to the actual state,x(k), 

as k → ∞. 

⎤ 

⎥ 

⎦ 


Kyoungchul Kong

4.4. DISCRETIZATION 107 

4.4 Discretization 

4.4.1 Discretization of transfer functions by approximation 

Exact mapping ofz-plane to s-plane. 

Recall that z −1 represents one step delay in the discrete-time domain, and e −sT is the 

time delay of T in the continuous-time domain. Since they are equivalent in physical 

phenomenon, 

z −1 = e −sT 

where T is the sampling period. Therefore, the exact mapping of the z-plane to the s- 

plane is z = e sT . This exact mapping, however, cannot be utilized to convert from one 

domain to another, becausez = e sT = 1+sT + 1 2 s2 T 2 + 1 6 s3 T 3 +... results in the infinite 

order ofs. 

Forward approximation 

The forward approximation is a first-order approximation ofz = e sT , i.e. 

z = e sT 

≈ 1+sT (4.2) 

whereT is the sampling period. The inverse of (4.2) is 

s ≈ z −1 

(4.3) 

T 

Using this relationship, a transfer function in the discrete-time domain can be obtained 

from one in the continuous-time domain. 

Backward approximation 

Notice that the difference equation transformed by the forward approximation often requires 

future information of the input, i.e., it often results in an unrealizable transfer function. 

Therefore, the backward approximation, which is a first order approximation of 

z = (e −sT ) −1 , is alternatively used: 

The inverse of (4.4) is 

z = e sT = 1 

≈ 

1 

1−sT 

e −sT 

s ≈ 1−z−1 

T 

The backward approximation is also called Euler method. 

(4.4) 

(4.5) 


Kyoungchul Kong


Bilinear approximation 

The bilinear approximation, which is a first-order approximation ofz = e sT = e 0.5sT (e −0.5sT ) −1 , 

enables more precise conversion between the discrete-time and continuous-time domains. 

z = e sT = esT/2 

e −sT/2 

≈ 1+sT/2 

1−sT/2 

The inverse of this mapping is 

s ≈ 2 z −1 

T z +1 

This method is also called Tustin’s method or trapezoid rule. 

(4.6) 

[Example 4-6] Suppose you have designed a PD controller in thes-domain: 

C(s) = U(s) 

E(s) = k P +k D s (4.7) 

For implementation, it should be converted into thez-domain as follows. 

1. Forward Approximation: s = z−1 

T 

z −1 

C(z) = k P +k D 

T 

= k ( 

D 

T z + k P − k ) 

D 

T 

The implementable difference equation is 

u(k) = k ( 

D 

T e(k +1)+ k P − k ) 

D 

e(k) 

T 

Notice that the difference equation above is not realizable, i.e., it requires future 

information to generate the current signal. Therefore, this controller cannot be 

implemented. 

2. Backward Approximation: s = 1−z−1 

T 

C(z) = k P +k D 

1−z −1 

= 

( 

k P + k D 

T 

) T 

− k D 

T z−1 


Kyoungchul Kong



( 

u(k) = k P + k ) 

D 

e(k)− k D 

e(k −1) 

T T 

3. Bilinear Approximation: s = 2 T 

z−1 

z+1 

2 

C(z) = k P +k D 

T 


( 

u(k) = −u(k −1)+ k P + 2k D 

T 

= 

= 

z −1 

z +1 

( ) ( ) 

kP + 2k D 

T 

z + kP − 2k D 

T 

z +1 

( 

kP + 2k D 

T 

) 

+ 

( 

kP − 2k D 

T 

) 

z 

−1 

1+z −1 

) 

e(k)+ 

( 

k P − 2k ) 

D 

e(k −1) 

T 

Figure 4.6 shows the frequency responses of a PD controller, and its discretized 

controllers by the three approximation methods. Notice that their magnitudes and 

phases are similar but not the same. In particular, the phases of C(z) obtained by 

the forward and backward approximation methods show large deviation from that of 

C(s), which may cause an instability problem. 

Pole-zero matching method 

The three discretization methods above are based on the approximation of z = e sT . Another 

idea is to convert the transfer function in the continuous-time domain such that 

the poles and zeros are matched in both domains. Consider a transfer function in the 

continuous-time domain, 

C(s) = k c 

∏ m 

j=1 (s−z i) 

∏ n 

i=1 (s−p i) 

wherez i ’s andp i ’s are zeros and poles ofC(s), respectively. Since the exact mapping between 

the continuous-time domain and the discrete-time domain isz = e sT , the equivalent 

poles and zeros in the discrete-time domain aree z iT ande p iT , respectively. Therefore, the 

corresponding discrete-time domain transfer function is 

C(z) = k d 

∏ m 

j=1 (z −ez iT ) 

∏ n 

i=1 (z −ep iT 

) 


Kyoungchul Kong


Bode Plots 


20 

15 

10 

5 

C(s) 

C(z) by forward approximation 

C(z) by backward approximation 

C(z) by bilinear approximation 

0 

180 

135 

Phase (deg) 

90 

45 

0 

10 −2 10 −1 10 0 


Figure 4.6: Bode plots of C(s) and three approximated C(z)’s for k P = 1, 

k D = 1, andT = 1. 

wherek d is a constant to match the magnitude ofC(s = jω) andC(z = e jωT ) at a certain 

frequency 12 . This conversion method is called pole-zero matching method. Notice that 

in spite of the theoretical background, the pole-zero matching method is still not an exact 

conversion between the two domains 13 . Remind that the exact conversion between the 

continuous-time domain and the discrete-time domain transfer functions is impossible, 

becausez = e sT = 1+sT + 1 2 s2 T 2 +... has an infinitely large order. 

Fig. 4.8 shows the step responses of a transfer function in the continuous-time domain 

and its discrete-time models discretized by the forward approximation, the backward 

approximation, the bilinear approximation, and the pole-zero matching method. Although 

all of the discretized models show similar responses, notice that none of them follows the 

response of the continuous-time model exactly. 

[Example 4-7] A controller designed in the continuous-time domain 

C(s) = 

0.5s+1 

(s+1)(s+3) 

12 In general cases,k d is designed to match the magnitudes at zero frequency. 

13 The exact mapping of the locations of poles and zeros does not necessarily mean that the responses of 

two systems are the same. 


Kyoungchul Kong


0.5 

0.4 

Output y(t=kT) 

0.3 

0.2 

0.1 

C(s) 

C Forward 

(z) 

C Backward 

(z) 

C Bilinear 

(z) 

C Matched 

(z) 

0 

0 1 2 3 4 5 6 7 8 

Time (t=kT) [sec] 

Figure 4.7: Step responses ofC(s) = 1/(s 2 +3s+2) and its discretized models. 

is to be discretized by the pole-zero matching method with the sampling period of 

T = 1ms. The poles and zeros ofC(s) are 

z 1 = −2 

p 1 = −1 

p 2 = −3 

The corresponding poles and zeros in the discrete-time domain are 

z 1d = e −0.002 

p 1d = e −0.001 

p 2d = e −0.003 

Therefore, C(z) obtained by the pole-zero matching method is 

C(z) = k d 

z −e −0.002 

(z −e −0.001 )(z −e −0.003 ) 

where k d is determined such that|C(s = jω)| = |C(z = e jωT )| at a certain frequency. 

By settingω = 0, 

C(s = j0) = 1 3 

Therefore, k d = (1−e−0.001 )(1−e −0.003 ) 

3(1−e −0.002 ) 

. 

C(z = e j0T ) = k d 

1−e −0.002 

(1−e −0.001 )(1−e −0.003 ) 


Kyoungchul Kong

þ ÿ¥¦§¨© § 

ý 

þ ¨©¨¨ 


ý þ ¡¢£¦ý ÿÿ 

!"ÿ 

ý þ ¡¢£¦ý "ÿ#$% & ' "ÿ#$( "ÿ$ % 

Figure 4.8: Matlab code for discretization by approximation. 

4.4.2 Stability mapping of approximation-based discretization methods 

Since the discretization methods are based on approximation, the stability of the discretized 

transfer functions is not guaranteed. Recall the discretization methods 

[Forward approximation] z = 1+sT 

[Backward approximation] 

1 

z = 

1−sT 

[Bilinear approximation] z = 1+sT/2 

1−sT/2 

If we let s = jω in these equations, we obtain the boundaries of the regions in the z- 

plane which originate from the stable portion of the s-plane. The shaded areas sketched 

in the z-plane in Fig. 4.9 are these regions 14 for each case. To show that the backward 

approximation results in a circle, 1 is added to and subtracted from the right hand side, 

2 

i.e. 

z = 1 ( 1 

2 + 1−sT − 1 ) 

2 

= 1 2 − 1 1+sT 

21−sT 

Now it is easy to see that withs = jω, the magnitude ofz− 1 2 ∣ z − 1 2∣ = 1 1+jωT 

2 ∣1−jωT∣ = 1 2 

is constant, i.e. 

and the curve is thus a circle as drawn in Fig. 4.9(b). Because the unit circle is the stability 

boundary in thez-plane, it is apparent from Fig. 4.9 that the forward approximation 

method can cause a stable continuous transfer function to be mapped into an unstable 

digital filter. It is especially interesting to notice that the bilinear approximation 

method maps the stable region of thes-plane exactly into the stable region of thez-plane. 

14 That is, the regions that correspond to the stable region in the s-plane. 


Kyoungchul Kong


Re{ s} 

< 

0 

Im 

Im 

Re 

Re 

s = 

z −1 

T 

2 

s = 

T 

z 

z 

−1 

+ 1 

(d) Pole-zero matching 

Im 

1− 

z 

s = 

T 

−1 

Im 

Im 

Re 

Re 

Re 

(a) Forward approx. (b) Backward approx. (c) Bilinear approx. 

Figure 4.9: Maps of the left-half of the s-plane by the approximation methods 

onto thez-plane. Stables-plane poles map onto the shaded regions of thez-plane. 

The unit circle is shown for reference. 


Kyoungchul Kong


u(k) 

D/A 

u(t) 

u(t) 

Computer 

Discrete-time domain signal 

Continuous-time domain signal 

Figure 4.10: Zero-order-hold effect of common digital-to-analog converters: 

they hold the output (analog signal) until the input (digital signal) is updated. 

4.4.3 Discretization of plant models with D/A 

The zero-order-hold (ZOH) is a mathematical model of the practical signal reconstruction 

done by a conventional digital-to-analog converter (D/A). That is, it describes the effect 

of converting a discrete-time signal to a continuous-time signal by holding each sample 

value for one sample interval, as shown in Fig. 4.10. In fact, the control signal generated 

by a computer through a D/A is a combination of step functions delayed bykT . 

as 

Remind that the transfer function of a plant is described in the discrete-time domain 

G(z) = Y(z) 

U(z) 

whereU(z) is an input to the plant, andY(z) is an output from the plant. Since an impulse 

signal is1in thez-domain, the transfer function is equivalent to thez-transformed impulse 

response, i.e. 

G(z) = Z{y(k)} ifu(k) is δ(k). 

As shown in Fig. 4.11, suppose a computer is generating an impulse signal, i.e. 

{ 

1 fork = 0 

u(k) = 

(4.8) 

0 fork > 0 

The continuous-time domain signal generated through a D/A is 

{ 

1 for0 ≤ t < T 

u(t) = 

0 fort ≥ T 

The signal above can be expressed as a combination of two step functions, i.e. 

u(t) = u 1 (t)−u 2 (t) 

where u 1 (t) is a unit step function, and u 2 (t) is another unit step function delayed by T . 

Taking the Laplace transform ofu(t), we get 

U(s) = U 1 (s)−U 2 (s) = 1 s − e−sT 

s 


= (1−e −sT ) 1 s 

Kyoungchul Kong


u(k) 

1 

D/A 

u(t) 

1 

0 

u(t) 

0 

T 

Computer 

Discrete-time impulse signal 

Output from D/A 

Figure 4.11: The output of a D/A whenU(z) = 1 

Notice that the output of the plant 15 is 

or in the time domain 

Y(s) = G(s)U(s) = (1−e −sT ) G(s) 

s 

y(t) = L −1 {G(s)U(s)} = L −1 {(1−e −sT ) G(s) 

s } 

Sampling the signal above and taking thez-transform, 

Y(z) = Z{L −1 {(1−e −sT ) G(s) }} (4.9) 

s 

Notice that Z{L −1 {1 − e −sT }} is (1 − z −1 ), because e −sT is an one-step delay, i.e., 

z −1 . Since the input signal was an impulse function (i.e., δ(k)) as in (4.8), the output in 

(4.9) is equivalent to the transfer function of the plant. Therefore, the plant model in the 

discrete-time domain is obtained by 

G(z) = (1−z −1 )Z{L −1 { G(s) 

s }} 

which is called the zero-order-hold (ZOH) equivalent. Note that this discretization 

method is not an approximation method, but is an exact mapping method. However, the 

ZOH equivalent can be applied only when the signal passes through a D/A. For example, 

the ZOH equivalent cannot be utilized to obtain a digital controller from one designed in 

the continuous-time domain. 

[Example 4-8] A plant 

G(s) = a 

s+a 

is connected to a computer via a D/A and an A/D. The calculation period is T . Then, 

the plant model including the D/A and A/D in the discrete-time domain is 

G(z) = (1−z −1 )Z{L −1 a 

{ 

s(s+a) }} 

15 The physical plants are always in the continuous-time domain. 


Kyoungchul Kong


ZOH 

D/A 

+¤)* 

G(s) 

¤)* 

/ 

A/D 

¤,* 

C(z) 

+¤,* -¤,* 

Computer 

− 

+ 

.¤,* 

(a) Structure of a typical digital feedback control system 

ZOH 

G(s) 

¤)* 

+¤)* 

¤,* 

+¤,* 

/ 

D/A 

A/D 

(b) Actual plant dynamics with a D/A and an A/D 

+¤,* 

G ¤,* 

(z) 

(c) Discretized plant dynamics 

Figure 4.12: Block diagram of a typical digital control system with a D/A and 

an A/D.G(z) must be obtained by the zero-order-hold equivalent, since the plant 

dynamics include a D/A. 

0 1 234567856 97:; < =>@2 ABCD> 42E@F3DE 3G@H2IB@: 

J 1 KLKK6; 

0M 1 NOP408 J8 QRSTQ:; 

< UDEBVBECDEVWB>CDE DXGIY>D@2 

Z 1 5K 6;V[ V97; 

\ 1 5K; 67; 

] 1 56 K7; 

^ 1 K; 

0F 1 FF4Z8\8]8^:; 

< =>@2 ABCD> 4F22D F_HD: 

0FM 1 H9C40F8 J8 QRSTQ:; 

< UDEBVBECDEVWB>CDE DXGIY>D@2 

Figure 4.13: Matlab code to obtain the zero-order-hold equivalent of a plant 

model. 


Kyoungchul Kong


The partial fraction expansion of 

a 

s(s+a) is 

Taking the inverse Laplace transform, 

a 

s(s+a) = 1 s − 1 

s+a 

L −1 { 1 s − 1 

s+a } = 1−e−at 

Sampling the signal and taking thez-transform, G(z) is obtained, i.e. 

G(z) = (1−z −1 )Z{1−e −akT } 

( ) 1 

= (1−z −1 ) 

1−z − 1 

−1 1−e −aT z −1 

= (1−z −1 z −1 −e −aT z −1 

) 

(1−z −1 )(1−e −aT z −1 ) 

= z−1 −e −aT z −1 

(1−e −aT z −1 ) 

= 1−e−aT 

(z −e −aT ) 

4.4.4 Discretization of state space models by 

zero-order-hold equivalent 

Recall that a state space equation of a plant in the continuous-time domain is 

and its solution is 

ẋ = Fx+Gu (4.10) 

y = Hx 

x(t) = e F(t−t 0) x(t 0 )+ 

wheret > t 0 , and t 0 is the initial time 16 . 

∫ t 

t 0 

e F(t−τ) Gu(τ)dτ (4.11) 

When the plant is accompanied with a D/A and an A/D, the input signal, u(t), is 

constant during one step, i.e. 

u(t) = u(k) 

forkT ≤ t < (k +1)T 

which represents the zero-order-hold effect. Settingt = (k +1)T and t 0 = kT in (4.11), 

x((k +1)T) = e FT x(kT)+ 

16 t 0 has been regarded as 0 in the previous chapters. 

∫ (k+1)T 

kT 

e F((k+1)T−τ) Gu(τ)dτ 


Kyoungchul Kong


Sinceu(τ) = u(k) is constant in one calculation period, 

[ ∫ ] 

(k+1)T 

x((k +1)T) = e FT x(kT)+ e F((k+1)T−τ) Gdτ u(k) 

Setting t = τ −kT in the integration term and omittingT in the index ofx(kT), 

[∫ T 

] 

x(k +1) = e FT x(k)+ e F(T−t) Gdt u(k) 

Notice that the new state space equation is in the discrete-time domain. Therefore, the 

zero-order-hold equivalent of the state space equation in (4.14) is 

where 

kT 

x(k +1) = Φx(k)+Γu(k) 

0 

y(k) = Hx(k) 

Φ = e FT 

Γ = 

∫ T 

0 

e F(T−t) Gdt (4.12) 

[Example 4-9] Consider the equation of motion of a mass-damper-spring system: 

mÿ +cẏ +ky = u 

where the output is the position of the mass, i.e., y. It is often desirable to construct 

a state space equation with the state variables of the continuous-time domain position 

and velocity in the discrete-time domain, i.e. 

[ ] 

y(k) 

x(k) = 

(4.13) 

ẏ(k) 

Remind that if the state space equation is obtained in the discrete-time domain 

as in Section 4.3.2, it is possible ] to have the state in the form of x(k) = 

[ 

y(k) y(k +1) ··· y(k+n−1) , but it is not possible to set the state as in 

(4.13). For this purpose, the method in (4.12) is applicable. First, the continuous-time 

domain state space equation is obtained, for example: 

[ 

0 1 

ẋ = 

− k − c m m 

y = [ 1 0 ] x 

] 

x+[ 0 

1 

m 

] 

u 


Kyoungchul Kong


where the state is 

x(t) = 

[ 

y(t) 

ẏ(t) 

This state can be simply sampled as in (4.13), if the state space equation is discretized 

as 

[ ([ 

] )[ 

x(k +1) = exp 

(T −t) 

y(k) = [ 1 0 ] x(k) 

0 1 

− k m − c m 

] )] [∫ T 

([ 

T x(k)+ exp 

0 

] 

0 1 

− k m − c m 


Kyoungchul Kong

Chapter 5 

Controller Design in Discrete Time 

Domain 

5.1 Controller Design Process 

All the plants are operated in the continuous-time domain, while controllers are usually 

running in the discrete-time domain. The controller in the discrete-time domain, C(z), 

can be directly designed from the discretized plant model,G(z), or obtained by converting 

C(s) with an appropriate approximation method. 

In general, designing a controller in the continuous-time domain is more preferred 

compared to the direct design of C(z) because of various useful controller design methods, 

such as PID control, lead-lag compensation, Nyquist plot analysis, and so on. However, 

there are many useful controller design methods available only in the discrete-time 

domain, such as the zero phase error tracking control, repetitive control, learning control, 

adaptive control, parameter adaptation, and so on. Therefore, it is important to know the 

usages of controllers in both the continuous-time domain and the discrete-time domain. 

The general controller design processes are depicted in Fig. 5.1. 

5.2 System Identification by Least Squares 

System identification 1 is a process to identify the model parameters of a plant based on 

the input and output signals obtained from experiments. The most fundamental method 

is to minimize an error between the actual (i.e., measured) output and a simulated output. 

The Least Squares method is utilized for this purpose. 

1 System identification is called “SysID” in short. 

120

5.2. SYSTEM IDENTIFICATION BY LEAST SQUARES 121 

 

D/A 

A/D 

Fourier transform of input/output signals 

` 

Time-domain analysis 

` 

` Parameter identification 

G (s) 

G(z) 

` C.C.F. 

` O.C.F. 

PID control 

` 

Lead-lag 

` 

compensator 

Pole ` 

assignment 

F , G, 

H Φ,Γ, H 

` D.C.F. 

` Zero-order-hold equivalent 

K 

Zero-order-hold 

` 

equivalent 

Eigenvalue 

` 

assignment 

LQ ` 

` C.C.F. 

` O.C.F. 

` D.C.F. 

K 

Eigenvalue 

` 

assignment 

LQ ` 

` Pole assignment 

C(s) 

Discretization 

` 

by forward, 

backward, or 

bilinear 

approximation 

Observer design 

` 

by eigenvalue 

assignment 

u( t) 

= −Kxˆ( 

t) 

Observer 

` 

discretization by 

forward, backward, or 

bilinear approximation 

Observer design 

` 

by eigenvalue 

assignment 

C(z) 

u( k) 

= −Kxˆ( 

k) 

u( k) 

= −Kxˆ( 

k) 

C(z) 

` Implementation 

Figure 5.1: Controller design processes. 


Kyoungchul Kong


Consider a single-input (u(k)) single-output (y(k)) system described by 

G(z) = Y(z) 

U(z) = B(z) 

A(z) 

where the order of A(z) is n and that of B(z) is m. Assuming that m < n, G(z) can be 

written as 

G(z) = z−1 B(z −1 ) 

A(z −1 ) 

where 

A(z −1 ) = 1+a 1 z −1 +a 2 z −2 +...+a n z −n 

B(z −1 ) = b 0 +b 1 z −1 +b 2 z −2 +...+b m z −m 

Taking the inversez-transform, the outputy(k) is obtained: 

Z −1 {A(z −1 )Y(z)} = Z −1 {z −1 B(z −1 )U(z)} 

y(k)+a 1 y(k−1)+...+a n y(k−n) = b 0 u(k −1)+...+b m u(k−m−1) 

Moving they(k−1), ..., y(k−n) terms to the right hand side, 

y(k) = −a 1 y(k −1)−...−a n y(k−n)+b 0 u(k −1)+...+b m u(k−m−1) 

The equation above can be expressed in vector form, i.e. 

where 

y(k) = θ T φ(k) 

θ = [ a 1 ... a n b 0 ... b m 

] T 

∈ R 

n+m+1 

φ(k) = [ −y(k−1) ... −y(k −n) u(k−1) ... u(k −m−1) ] T 

∈ R 

n+m+1 

In general, θ is called a parameter vector, and φ(k) is called a regressor. 

Suppose that we have a data set[u(0),u(1),...,u(N)] and [y(0),y(1),...,y(N)] but 

do not know the model parameters, a 1 , ..., a n , b 0 , ..., b m . One way to identify the model 

parameters is to minimize 

N∑ [ 

J = y(k)−θ T φ(k) ] 2 

k=0 

which requires solving a Least Squares problem. The solution to the Least Squares problem 

can be found by setting the partial derivative ofJ with respect toθ to zero, i.e. 

N 

∂J 

∂θ = −2 ∑ 

k=0 

[ 

y(k)−θ T φ(k) ] φ(k) 


Kyoungchul Kong


Sinceθ T φ(k) ∈ R 1 ,(θ T φ(k))φ(k) = (φ T (k)θ)φ(k) = φ(k)(φ T (k)θ). Therefore 

N 

∂J 

∂θ = −2 ∑ 

k=0 

[ 

φ(k)y(k)−φ(k)φ T (k)θ ] 

In order to make ∂J 

∂θ 

= 0, θ should be 

[ N 

] −1 [ 

∑ 

N 

] 

∑ 

θ = φ(k)φ T (k) φ(k)y(k) 

k=0 

k=0 

(5.1) 

Using (5.1), the model parameters can be identified. Notice that the calculated θ may be 

different from the actual values, if sensor noise or disturbance exists. Moreover, when the 

assumed system orders (i.e., m and n) do not match those of actual plant, the calculated 

θ may not be the actual value. 

[Example 5-1] Consider a plant in the discrete-time domain: 

G(z) = 

z +0.5 

z 2 +0.2z −0.8 

Suppose that the model parameters are unknown. To identify the parameters, an openloop 

control experiment has been carried out with an input signal 

The measured output signal is 

u(0) = 1.0000 

u(1) = 0.5000 

u(2) = −1.0000 

u(3) = −0.8000 

u(4) = 0.6000 

u(5) = 0.0000 

y(0) = 0.0000 

y(1) = 1.0000 

y(2) = 0.8000 

y(3) = −0.1100 

y(4) = −0.6380 

y(5) = 0.2396 


Kyoungchul Kong

5.3. FEEDFORWARD CONTROLLER DESIGN 124 

From the data above, φ(k) = [ −y(k −1) −y(k −2) u(k −1) u(k −2) ] T 

can 

be constructed as 

φ(0) = [ 0.0000 0.0000 0.0000 0.0000 ] T 

φ(1) = [ 0.0000 0.0000 1.0000 0.0000 ] T 

φ(2) = [ −1.0000 0.0000 0.5000 1.0000 ] T 

φ(3) = [ −0.8000 −1.0000 −1.0000 0.5000 ] T 

φ(4) = [ 0.1100 −0.8000 −0.8000 −1.0000 ] T 

φ(5) = [ 0.6380 0.1100 0.6000 −0.8000 ] T 

The model parametersθ = [ ] T 

a 1 a 2 b 0 b 1 can be identified by the Least Squares, 

i.e. [ 5∑ 

] −1 [ 5∑ 

] 

θ = φ(k)φ T (k) φ(k)y(k) 

(5.2) 

k=0 

k=0 

The calculated result is θ = [ 0.2000 −0.8000 1.0000 0.5000 ] T 

, which is the 

same as the actual parameters. 

5.3 Feedforward Controller Design 

5.3.1 Perfect tracking control 

Suppose that a plant,G(z), is under feedback control withC(z). The closed-loop transfer 

function is 

G C (z) = G(z)C(z) 

1+G(z)C(z) = B C(z) 

A C (z) = b 0z q +b 1 z q−1 +b 2 z q−2 +...+b q 

z p +a 1 z p−1 +a 2 z p−2 +...+a p 

where the orders of A C (z) and B C (z) are p and q ≤ p, respectively. Multiplyingb −1 

0 z −p 

to both the numerator and denominator,G C (z) becomes 

G C (z) = z−d β C (z −1 ) 

α C (z −1 ) 

wheredis the relative order, i.e.,d = p−q. α C (z −1 ) andβ C (z −1 ) are 

α C (z −1 ) = α 0 +α 1 z −1 +α 2 z −2 +...+α p z −p 

β C (z −1 ) = 1+β 1 z −1 +β 2 z −2 +...+β q z −q 

Notice that the first term ofβ(z −1 ) is set to 1. 2 

(5.3) 

2 Note thatβ 1 = b1 

b 0 

, ...,β q = bq 

b 0 

, α 0 = 1 b 0 

, ...,α p = ap 

b 0 

. 


Kyoungchul Kong


d 

G C 

− 

1 ( z) 

C (z) G(z) 

y 

y r 

+ 

− 

) 

G( 

z) 

C( 

z) 

G C 

( z) 

= 

1+ 

G( 

z) 

C( 

z 

Figure 5.2: Block diagram of a feedforward and feedback control system 

In addition to the feedback controller, suppose that the reference input, r(k), is 

obtained by filtering the desired output,y d . The most intuitive filter for this purpose is the 

inverse of the closed-loop transfer function,G −1 

C 

(z), as shown in Fig. 5.2. 

and 

Suppose that G C (z) has no zero outside of the unit circle. Then, G −1 

C 

(z) is stable 

G −1 

C (z) = zd α C (z −1 ) 

β C (z −1 ) 

Since the input toG −1 

C (z) isy d(k), the reference input,r(k), is obtained by 

r(k) = Z −1 {G −1 

C (z)Y d(z)} 

Notice that the overall transfer function from y d (k) toy(k) is 

Y(z) 

Y d (z) = Y(z) R(z) 

R(z) Y d (z) = G C(z)G −1 

C (z) = 1 

(5.4) 

which results in y(k) = y d (k), i.e., perfect tracking control. For this scheme to work, at 

least the following three conditions must be satisfied: 

1. the desired output must be known at leastdsteps ahead 3 . 

2. the mathematical inverse of the closed-loop system is asymptotically stable, which 

implies that the closed-loop system does not possess any unstable zeros, and 

3. G C (z) must be an accurate representation of the closed-loop system. 

In most of the mechanical control problems such as robot control and machining, the 

desired output 4 is usually predetermined or known in advance, and thus the first condition 

is satisfied. The second condition requires that the open plant model does not possess 

any zero on the right half plane and that a feedback controller does not introduce unstable 

zeros. The third condition requires accurate identification of system parameters. 

3 When the future information is not available, the z d term in G −1 

C 

(z) can be ignored. In this case, the 

output will bedstep delayed such that y(k) = y d (k −d), which is still good performance. 

4 That is, the desired trajectory 


Kyoungchul Kong


In addition to the necessary conditions stated above, there exists a recommended 

condition in the design of desired output, y d (k). Since the magnitude of G C (e jωT ) is 

small at a high frequency 5 ,G −1 

C 

(z) amplifies the high frequency component in the desired 

output. Thus, the desired output should be designed such that y d (k) is continuous and 

smooth for all k. Notice that a step function is not a good choice for the desired output. 

[Example 5-2] Suppose a plant is under feedback control, where 

G(z) = 

The closed-loop transfer function is 

1 

z 2 +0.2z +0.5 

C(z) = z +0.5 

z 

G C (z) = 

z +0.5 

z 3 +0.2z 2 +1.5z +0.5 = z −2 (1+0.5z −1 ) 

1+0.2z −1 +1.5z −2 +0.5z −3 

Since G C (z) does not have any zero outside of the unit circle, the perfect tracking 

control is applicable. The inverse of the closed-loop transfer function is 

G −1 

C (z) = z2 (1+0.2z −1 +1.5z −2 +0.5z −3 ) 

1+0.5z −1 

Therefore, the reference input is obtained from the actually desired output by 

R(z) 

Y d (z) 

= z2 (1+0.2z −1 +1.5z −2 +0.5z −3 ) 

1+0.5z −1 

(1+0.5z −1 )R(z) = z 2 (1+0.2z −1 +1.5z −2 +0.5z −3 )Y d (z) 

Taking the inversez-transform, we get 

r(k) = −0.5r(k −1)+y d (k +2)+0.2y d (k +1)+1.5y d (k)+0.5y d (k −1) 

5.3.2 Zero phase error tracking control 

Remind that the perfect tracking control in (5.4) is available only when the closed loop 

system does not possess any zero outside of the unit circle. The undesired zero, however, 

often appears through the discretization process or is inherited from the physical model. 

For example, see the following example. 

5 Remind in the continuous-time domain that |G(jω)| → 0 as ω → ∞, if G(s) has n poles and m < n 

zeros. Most mechanical systems satisfy this condition. 


Kyoungchul Kong


a b cdefghifg jhkl 

m nopqc rstuo 

v b cdefg ghifg whkl 

m nx ysqczso 

{ b w|wwgl 

a} b yjteai {kl m ~uzssztuz€sotuz u ‚ƒ„pouqc 

v} b yjtevi {i …{‚†cƒq…kl m ‡ƒoƒqupz pˆˆzs‰ƒrpcƒsq 

ay b duutŠpy‹ea}Œv}i gkl m vos†utossˆ czpq†duz d‚qycƒsq 

n{v b Ž ‘’“eay”gk m rƒqzupoek †ƒrôƒdƒu† c€u 

m ysuddƒyƒuqc† †‚y€ c€pc • – ƒ† g 

{zpq†duz d‚qycƒsq— 

gwwg }”j g˜˜ } š ˜˜›|œ 

 

} w|˜˜˜ 

prôƒqž cƒru— w|wwg 

Figure 5.3: Matlab code to obtain the feedforward controller transfer function. 

[Example 5-3] A pure-mass system 

is controlled by a proportional controller 

Then the closed loop transfer function is 

G(s) = 1 

ms 2 

C(s) = k D s+k P 

G C (s) = 

k D s+k P 

ms 2 +k D s+k P 

which does not possess any zero on the right half plane, ifk P > 0 and k D > 0. 

Suppose that the control system above is implemented into a computer with the 

calculation period of T . The plant is accompanied by a D/A and an A/D such that the 

plant model is discretized by the zero-order-hold equivalent, i.e. 

G(z) = T2 z −1 (1+z −1 ) 

2m(1−z −1 ) 2 


Kyoungchul Kong


and the controller is discretized by the backward approximation, i.e. 

( 

C(z) = k P + k ) ( 

D 

+ − k ) 

D 

z −1 := k 1 +k 2 z −1 

T T 

The resulting closed loop transfer function in the discrete-time domain is 

G C (z) = 

T 2 z −1 (1+z −1 )(k 1 +k 2 z −1 ) 

2m+(k 1 T 2 −4m)z −1 +(k 1 T 2 +k 2 T 2 +2m)z −2 +k 2 T 2 z −3 

Notice that G C (z) possesses a zero at −1, which is on the stability boundary. 

Therefore, the inverse of G C (z) will possess a marginally stable pole. Moreover, the 

mode of −1 causes a large oscillation a , and thus the perfect tracking control is not 

recommended to this system. 

a Recall that y(k) = (−1) k ifY(z) = 1 

1+z −1 . 

Suppose that a closed loop transfer function is 

G C (z) = z−d β C (z −1 ) 

α C (z −1 ) 

In order to develop a feedforward controller for systems with uncancellable 6 zeros,β C (z −1 ) 

is factorized into two parts, i.e. 

β C (z −1 ) = β s C (z−1 )β u C (z−1 ) 

whereβ s C (z−1 ) contains stable zeros, andβ u C (z−1 ) contains uncancellable zeros. Since the 

inverse ofβ u C (z−1 ) cannot be realized, an alternative feedforward controller is introduced: 

G ZPET (z) = zd α C (z −1 )β u C (z) 

β s C (z−1 )β u C (1)2 (5.5) 

where β u C (z) is obtained by replacing z−1 in β u C (z−1 ) by z. β u C (1)2 is introduced to make 

the magnitude ofG ZPET (z)G C (z) one at the zero frequency. 

Using the feedforward controller in (5.5), the reference input (r(k)) is obtained from 

the desired output (y d (k)), i.e. 

R(z) = G ZPET (z)Y d (z) 

6 That is, unstable zeros and marginally stable zeros. 


Kyoungchul Kong


The overall transfer function fromy d (k) to y(k) is 

Y(z) 

Y d (z) 

= G ZPET (z)G C (z) 

= zd α C (z −1 )βC u(z) 

z −d βC s(z−1 )βC u(z−1 ) 

βC s(z−1 )βC u(1)2 α C (z −1 ) 

= βu C (z)βu C (z−1 ) 

βC u(1)2 

Notice that in the frequency domain, βC u(ejωT ) is the complex conjugate of βC u(e−jωT ). 

Therefore the phase of βu C (ejωT )βC u(e−jωT ) 

is zero for the entire frequency range. Moreover, 

βC u(1)2 

at the zero frequency (i.e., at z = 1), βu C (1)βu C (1−1 ) 

= 1. Since this feedforward control 

βC u(1)2 

method does not introduce any phase error, it is called zero phase error tracking (ZPET) 

control. 

[Example 5-4] Consider the closed loop transfer function in [Example 2]. For simplicity, 

assume that T = 0.1,m = 1,k P = 1 andk D = 0.1. 

G C (z) = 0.02z−1 (1+z −1 )(1−0.5z −1 ) 

2−3.98z −1 +2.01z −2 −0.01z := z−d βC s(z−1 )βC u(z−1 ) 

−3 α C (z −1 ) 

G C (z) can be factorized into 

α C (z −1 ) = 0.02 −1 (2−3.98z −1 +2.01z −2 −0.01z −3 ) 

β s C (z−1 ) = 1−0.5z −1 

β u C(z −1 ) = 1+z −1 

d = 1 

To obtain a feedforward filter by the zero phase error tracking control, BC u (z) and 

BC u(1) are necessary, i.e. βC u (z) = 1+z 

βC(1) u = 2 


Kyoungchul Kong

5.4. CONTROLLER DESIGN IN DISCRETE-TIME STATE SPACE 130 

ThusG ZPET (z) is 

G ZPET (z) = zd α C (z −1 )β u C (z) 

β s C (z−1 )β u C (1)2 

The implementable control law is 

= z1 0.02 −1 (2+−3.98z −1 +2.01z −2 −0.01z −3 )(1+z) 

(1−0.5z −1 )2 2 

= 25z2 −24.75z 1 −24.625+25z −1 −0.125z −2 

1−0.5z −1 

r(k) = 0.5r(k−1)+25y d (k+2)−24.75y d (k+1)−24.625y d (k)+25y d (k−1)−0.125y d (k− 

Although the relative order d was 1, the two-step advanced information is required in 

the ZPET control. Using this feedforward control method, the output is 

y(k) = Z −1 {G C (z)G ZPET (z)Y d (z)} 

= Z −1 { βu C (z−1 )βC u(z) 

Y 

βC u d (z)} 

(1)2 

= Z −1 { (1+z−1 )(1+z) 

Y d (z)} 

4 

= 0.25y d (k +1)+0.5y d (k)+0.25y d (k −1) 

Notice that ify d is constant,y(k) = y d . 

5.4 Controller Design in Discrete-Time State Space 

5.4.1 Discrete-time full state observer 

For an n-dimensional discrete-time system described by 

x(k +1) = Φx(k)+Γu(k) 

y(k) = Hx(k) 

x(0) = x 0 

a full state observer is 

ˆx(k +1) = Φˆx(k)+Γu(k)+L[y(k)−Hˆx(k)], ˆx(0) = 0 (5.6) 

The estimation error equation is 

e(k +1) = [Φ−LH]e(k) 


Kyoungchul Kong


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Figure 5.4: Matlab code to obtain the feedforward controller transfer function 

by zero phase error tracking control. 

where 

e(k) = x(k)− ˆx(k), e(0) = x 0 

As in the continuous-time case, the eigenvalues of [Φ−LH] can be arbitrarily assigned 

if the system is observable. Therefore, the observer gain matrix, L, should be designed 

such that the eigenvalues of[Φ−LH] are all inside of the unit circle. 

[Example 5-5] Consider a state space equation 

[ 

1 1 

x(k +1) = 

0 1 

] 

x(k)+ 

y(k) = [ 1 0 ] x(k) 

[ 

0.5 

1 

] 

u(k) 

A full state observer is constructed as 

[ ] [ 1 1 0.5 

ˆx(k +1) = ˆx(k)+ 

0 1 1 

] 

u(k)+L[y(k)− [ 1 0 ]ˆx(k)] 

The observer gain matrix, L ∈ R 2 , is designed such that the eigenvalues of [Φ−LH] 


Kyoungchul Kong


are all placed in the unit circle. The eigenvalue equation is 

[ [ ] [ ] 1 1 l1 [ ] ] 

det zI − + 1 0 = z 2 +(l 

0 1 l 1 −2)z +(1−l 1 +l 2 ) = 0 

2 

Suppose that the desired eigenvalues are all zeros, then 

l 1 = 2 l 2 = 1 

5.4.2 Discrete-time full state observer with predictor 

In the discrete-time case, we note that the observer in (5.6) does not make use ofy(k) for 

the estimation of the state at timek in spite that the current measurementy(k) is strongly 

correlated with the current state x(k). Therefore, we consider a new observer, which 

makes use ofy(k) for the estimation ofx(k). The new observer involves two stages: prediction 

and correction. 

The new observer is constructed as 

[Corrector] ˆx(k) = ˆx p (k)+L[y(k)−Hˆx p (k)], ˆx p (0) = 0 

[Predictor] 

ˆx p (k +1) = Φˆx(k)+Γu(k) 

If the predicted state, ˆx p (k), is eliminated from the equations: 

ˆx(k +1) = [I −LH]Φˆx(k)+[I −LH]Γu(k)+Ly(k +1) 

Then, the estimation error equation becomes 

where 

e p (k +1) = [I −LH]Φe p (k) 

e p (k) = x(k)− ˆx(k), e p (0) = x 0 

The estimation error is governed by [I −LH]Φ, instead of [Φ−LH]. Therefore, in the 

full state observer with a predictor the observer gain, L, should be designed such that the 

eigenvalues of[I −LH]Φ are all inside of the unit circle. 

[Example 5-6] Consider the same state space equation as in [Example 5-5]. A full 

state observer with predictor is constructed as 

[Corrector] ˆx(k) = ˆx p (k)+L[y(k)− [ 1 0 ]ˆx p (k)], ˆx p (0) = 0 

[ ] [ ] 1 1 0.5 

[Predictor] ˆx p (k +1) = ˆx(k)+ u(k) 

0 1 1 


Kyoungchul Kong


The observer gain matrix, L ∈ R 2 , is designed such that the eigenvalues of [I − 

LH]Φ are all placed in the unit circle. The eigenvalue equation is 

[ [ ] [ ] 1 1 l1 [ ] [ ]] 

1 1 

det zI − + 1 0 = z 2 +(l 

0 1 l 2 0 1 1 +l 2 −2)z+(1−l 1 ) = 0 

Suppose that the desired eigenvalues are all zeros, then l 1 = 1 l 2 = 1. 

[Example 5-7] Consider the full state observers in [Example 5-5] and [Example 5-6]. 

Suppose the initial state is [ ] 

a0 

x(0) = ∈ R 2 

a 1 

wherea 0 and a 1 are any scalars. 

Recall that the estimation error equations are 

e(k +1) = [Φ−LH]e(k) 

e p (k +1) = [I −LH]Φe p (k) 

where e p (k) and e(k) are the estimation errors (i.e., x(k) − ˆx(k)) with and without a 

predictor, respectively. 

Using the observer gain matrices obtained in [Example 1] and [Example 2], 

which assign the eigenvalues at zero, the estimation error equations become 

[ ] 

−1 1 

e(k +1) = e(k) 

−1 1 

[ ] 

0 0 

e p (k +1) = e 

−1 0 p (k) 

At k = 0, the estimation errors are 

e(0) = x(0) = 

At k = 1, they become 

[ ][ ] 

−1 1 a0 

e(1) = = 

−1 1 a 1 

[ 

a0 

a 1 

] 

[ ] 

a1 −a 0 

a 1 −a 0 

e p (0) = x(0) = 

e p (1) = 

[ 

a0 

[ 

0 0 

−1 0 

a 1 

] 

][ 

a0 

] 

= 

a 1 

[ 

0 

−a 0 

] 


Kyoungchul Kong


At k = 2, 

[ ][ ] [ ] [ 

−1 1 a1 −a 

e(2) = 

0 0 0 0 

= e 

−1 1 a 1 −a 0 0 p (1) = 

−1 0 

Notice that the estimation errors become a zero within two steps. 

][ ] [ 

0 0 

= 

−a 0 0 

a It does not only “converge” to zero but also “become” exactly zero. This is another advantage of 

using the discrete-time domain. 

] 

5.4.3 Discrete-time linear quadratic optimal control 

Consider a plant under the full state feedback control: 

x(k +1) = Φx(k)+Γu(k) (5.7) 

y(k) = Hx(k) 

u(k) = −Kx(k) (5.8) 

x(0) = x 0 

The optimal control is sought to minimize a quadratic performance index, 

J = 

∞∑ 

y T (k)y(k)+u T (k)Ru(k) 

k=0 

or in more general form, 

J = 

∞∑ 

x T (k)Qx(k)+u T (k)Ru(k) (5.9) 

k=0 

whereQ = H T H is a positive semidefinite matrix, and R is positive definite. Notice that 

the first term penalizes the deviation of x(k) from the origin, while the second penalizes 

the control energy. 

In order to find a control law that minimizes (5.9), let P ∈ R n×n be a positive 

definite matrix. Note that 

∞∑ [ 

x(k +1) T Px(k +1)−x(k) T Px(k) ] = x(∞) T Px(∞)−x(0) T Px(0) 

k=0 

Assuming that the system is asymptotically stable (i.e., x(k) → 0 as k → ∞), 

x(0) T Px(0)+ 

∞∑ [ 

x T (k +1)Px(k+1)−x T (k)Px(k) ] = 0 

k=0 


Kyoungchul Kong


Plugging (5.7) into the equation above, we get 

∞∑ [ 

x(0) T Px(0)+ [Φx(k)+Γu(k)] T P[Φx(k)+Γu(k)]−x T (k)Px(k) ] = 

k=0 

x(0) T Px(0)+ 

∞∑ 

k=0 

k=0 

[ x T (k) [ Φ T PΦ−P ] x(k)+u T (k)Γ T PΦx(k) 

+x T (k)Φ T PΓu(k)+u T (k)Γ T PΓu(k) 

Since (5.10) is zero, it can be added to the cost function in (5.9), i.e. 

∞∑ 

[ x 

J = x(0) T Px(0)+ 

T (k) [ Φ T PΦ−P +Q ] ] 

x(k)+u T (k)Γ T PΦx(k) 

+x T (k)Φ T PΓu(k)+u T (k) [ R+Γ T PΓ ] u(k) 

] 

= (5.10) 0 

The equation above can be arranged into 

∞∑ 

J = x(0) T Px(0)+ [u+K LQ x(k)] [ T R+Γ T PΓ ] [u+K LQ x(k)] (5.11) 

where 

k=0 

K LQ = [ R+Γ T PΓ ] −1 

Γ T PΦ (5.12) 

Φ T PΦ−P +Q = Φ T PΓ [ R+Γ T PΓ ] −1 

Γ T PΦ (5.13) 

where (5.12) is the full state feedback gain obtained by the LQ method, and (5.13) is 

called the discrete-time algebraic Riccati equation (DARE). The cost function in (5.11) is 

minimized if 

u(k) = −K LQ x(k) 

and the minimal cost is 

J o = x(0) T Px(0) 

[Summary] For a controllable system 

x(k +1) = Φx(k)+Γu(k) 

y(k) = Hx(k) 

x(0) = x 0 

the state feedback controller that minimizes 

∞∑ 

J = x T (k)Qx(k)+u T (k)Ru(k), Q ≥ 0,R > 0 

k=0 


Kyoungchul Kong

- ì ïð 

, ì ó'.óð 

/ ì ¡¢£¤õ)(âö òéææéö ,ö -÷ Ý +, ßùáâæéê %éâà 

5.5. DISTURBANCE OBSERVER 136 

ÝÝ Þßàáâàãßãäåáâæç èßæéâà æßèçê 

ë ì íî ïð î îñð 

ò ì íîð ïñð 

ó ì íï îñð 

ô ì îð 

ò ì ääõëöòöóöô÷ð 

Ý øáéáç äùéúç æßèçê 

ÝÝ ûâäúüçáâýéáâßà 

þ ì îÿîîïð 

Ý øéæùêâà% ùçüâßè 

òè ì ú&èõòöþö'ýß('÷ð 

)(â ì òèÿéð 

Ý ûâäúüçáâýçè äáéáç æéáüâúçä 

òéææé ì òèÿ*ð 

ó ì òèÿúð 

ô ì òèÿèð 

ÝÝ +, Þßàáüßê òéâà 

Figure 5.5: Matlab code for discrete-time LQ optimal control. 

is 

u(k) = −K LQ x(k) 

where 

K LQ = [ R+Γ T PΓ ] −1 

Γ T PΦ 

and P is the positive definite solution of 

Φ T PΦ−P +Q−Φ T PΓ [ R+Γ T PΓ ] −1 

Γ T PΦ = 0 

which is called the discrete-time algebraic Riccati equation. The Riccati equation has 

a proper solution, as long as the state space equation is controllable. 

5.5 Disturbance Observer 

Various useful control methods have been introduced in this lecture. The performance 

of the control methods, however, is highly dependent on the accuracy of a plant model, 

which is nearly impossible to obtain in practice. Most of the mechanical systems exhibit 


Kyoungchul Kong


nonlinear behavior due to the Coulomb friction, the input/output limitation 7 , the actuator 

nonlinearity, and so on. Moreover, in many cases the high-frequency plant dynamics is 

neglected for the sake of simplicity in the design of control algorithms. Another practical 

challenge in control systems is a disturbance. Mechanical systems interact with environments, 

and thus they cannot be free from environmental disturbances. 

d + 

u 

+ 

~ G ( z ) 

y 

Figure 5.6: An open-loop system with model discrepancy and disturbance. 

Figure 5.6 shows a plant with a model discrepancy and a disturbance. The actual 

plant dynamics, ˜G(z), is not necessarily the same asG(z), which is a mathematical (called 

nominal) model identified by a system identification process. Moreover, a disturbance, 

d, is exerted to the plant so that the output is affected by the disturbance. Note that the 

output is 

y = ˜G(z)[u+d] (5.14) 

d + 

u 

+ 

~ G ( z ) 

y 

G(z) 

+ 

− 

ŷ 

− 

Q( 

z) 

G 

1 ( z) 

dˆ 

Computer program 

Figure 5.7: A disturbance observer system. 

In order to reject the disturbance and the model discrepancy, they should be estimated. 

Suppose that the nominal model is being simulated simultaneously as in Fig. 5.7. 

Then the simulated and actual outputs should be matched if d = 0 and ˜G(z) = G(z). 

Subtracting the simulated output, ŷ = G(z)u, from the actual output, y, the effect of the 

disturbance and the model discrepancy can be estimated, i.e. 

ˆd = Q(z)G −1 (z)[y −G(z)u] (5.15) 

where Q(z) is a filter introduced to make (5.15) realizable 8 . If ˜G(z) = G(z), (5.15) is 

reduced to ˆd = Q(z)d. Therefore,Q(z) should be designed such thatQ(z) ≈ 1 for a large 

frequency range. Since (5.15) estimates the disturbance, it is often called a disturbance 

observer (DOB). 

The estimated disturbance is fed back into the system in order to reject the actual 

disturbance, as shown in Fig. 5.10. Such control method is called a DOB-based control 

7 It is often called “saturation.” 

8 Remind that G −1 (z) is not realizable. 


Kyoungchul Kong


u C 

d 

u 

+ 

~ 

+ + 

G ( z ) 

y 

− 

Q (z) − + 

− 

Q( 

z) 

G 

1 ( z) 

dˆ 

Figure 5.8: A disturbance-observer-based control system. 

system. The closed-loop transfer functions from u C and d toy are 

where 

y = G u (z)u C +G d (z)d (5.16) 

G u (z) = 

G d (z) = 

˜G(z)G(z) 

G(z)+[˜G(z)−G(z)]Q(z) 

˜G(z)G(z)[1−Q(z)] 

G(z)+[˜G(z)−G(z)]Q(z) 

(5.17) 

(5.18) 

Note that G u (z) = G(z) and G d (z) = 0 if Q(z) = 1. In other words, if Q(z) = 1 

the disturbance is perfectly rejected and the system follows the nominal plant model. 

This makes the DOB-based control system attractive in practice, because it solves two 

major problems, the disturbance and the model discrepancy, all at once. 

The Q(z) filter, however, cannot be 1, because Q(z)G −1 (z) must be realizable. 

Alternatively, suppose that Q(z) is a low pass filter with the DC-gain of 1, then G u (z) ≈ 

G(z) and G d (z) ≈ 0 at frequencies where Q(e jωT ) ≈ 1 + 0j. Among many choices, a 

typicalQ(z) is obtained by converting 

( ) p ωb 

Q(s) = 

s+ω b 

by the pole-zero matching method. ω b determines the bandwidth of the filter, and p can 

be selected such that Q(z)G −1 (z) is realizable. 

if 

Moreover, note from (5.15) and (5.17) that the DOB-based control system is stable 

1. G(z) and G −1 (z) are all stable. 

2. G(z)+[˜G(z)−G(z)]Q(z) = 0 does not possess any roots outside of the unit circle. 

For the first condition, the nominal plant must be selected such that it possesses neither 

unstable pole nor unstable zero. 


Kyoungchul Kong


In order to satisfy the second condition, the Nyquist criterion can be applied. Namely, 

˜G(e 

the Q(z) filter should be designed such that 

jωT )−G(e jωT ) 

Q(e jωT ) does not encircle 

G(e jωT ) 

−1+0j. In practice, however, checking the Nyquist plot is not convenient, and the bandwidth 

ofQ(z), ω b , is often adjusted by trials and errors such thatω b is as large as possible 

and the overall DOB-controlled system is stable. 

[Example 5-8] Consider a plant 

˜G(z) = 0.4 

z −0.7 

with a sampling period of 1ms. A nominal model has been identified through the 

system identification process, i.e. 

G(z) = 0.3 

z −0.6 

In order to compensate for the model discrepancy, a disturbance observer is designed 

with aQ(z) filter with the bandwidth of20rad/sec, i.e. 

Then the disturbance observer is 

Q(z) = 0.06 

z −0.94 

ˆd = Q(z)G −1 (z)[y −G(z)u] 

0.06 z −0.6 

= 

z −0.94 0.3 

= 

[ 

y − 0.3 

z −0.6 u ] 

1 [ 

(0.2−0.12z −1 )y −0.06z −1 u ] 

1−0.94z −1 

The implementable form of the disturbance observer is 

ˆd(k) = 0.94ˆd(k −1)+0.2y(k)−0.12y(k−1)−0.06u(k−1) 

The estimated disturbance is subtracted from the control signal,u C (k), calculated by a 

feedback controller,C(z), i.e. 

u(k) = u C (k)− ˆd(k) 


Kyoungchul Kong


Frequency responses with and withput DOB 

0 


Phase (deg) 

−5 

−10 

−15 

45 

0 

−45 

−90 

Actual plant dynamics 

Nominal plant 

DOB−controlled system 

−135 

−180 

10 0 10 1 10 2 10 3 


Figure 5.9: G u (z) with and without disturbance observer. 

Frequency responses with and withput DOB 

0 


Phase (deg) 

−10 

−20 

−30 

−40 

90 

45 

0 

−45 

−90 

Actual plant dynamics 

DOB−controlled system 

−135 

−180 

10 0 10 1 10 2 10 3 


Figure 5.10: G d (z) with and without disturbance observer. 


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Kyoungchul Kong

Digital Control Systems [MEE 4003] - Kckong.info

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