control systems lecture notes

Control Systems

Lecture Notes for KJM597

Faculty of Mechanical Engineering, UiTM Shah Alam

2

KJM597 Control Systems Faculty of Mechanical Engineering

UiTM Shah Alam

CHAPTER 1.0 MATHEMATICAL REVIEW

1.1 COMPLEX VARIABLES AND COMPLEX FUNCTION

This chapter will outline an overview of basic mathematical formulation in solving control

systems problem that you will encounter throughout the course.

1.1.1 Complex Variables Concept

A complex variable denoted by s consists of two components: a real component x and an

imaginary axis component y. Graphically, the real component of s is represented by a x-axis in

the horizontal direction, and the imaginary component is measured along the vertical jy-axis.

Figure 1.0 illustrates the complex s-plane.

Figure 1.0: Complex s-plane

(source: http://mathworld.wolfram.com)

Using notation 𝑗 = −1, all numbers in engineering calculations can be re-written as

𝑧 = 𝑥 + 𝑗𝑦

Where z is called a complex number. Note that j is the only imaginary quantity in the expression.

The magnitude, |z| and angle, 𝜃 of z can be obtained mathematically,

Magnitude of z=|z|= 𝑥2 + 𝑦2, angle of z=𝜃 = tan−1 𝑦

𝑥

http://mathworld.wolfram.com/

3


UiTM Shah Alam

A complex number can be written in rectangular form or in polar form as follows:

i. Rectangular forms

𝑧 = 𝑥 + 𝑗𝑦

𝑧 = 𝑧 (cos 𝜃 + 𝑗 sin 𝜃)

ii. Polar forms

𝑧 = |𝑧|∠𝜃

𝑧 = |𝑧|𝑒𝑗𝜃

In converting complex numbers to polar form from rectangular, we use

𝑧 = 𝑥2 + 𝑦2, 𝜃 = tan−1 𝑦

𝑥

To convert complex number to rectangular form from polar, we employ

𝑥 = |𝑧| cos 𝜃, 𝑦 = |𝑧| sin 𝜃

(source: Ogata)

Notes:

4


UiTM Shah Alam

Notes:

5


UiTM Shah Alam

1.1.2 Complex Function Concept

a) Complex Function

A complex function F(s,) a function of s, has a real component and imaginary component, or

𝐹 𝑠 = 𝐹𝑥 + 𝐹𝑦

where 𝐹𝑥and 𝐹𝑦 are real quantities. The magnitude of 𝐹(𝑠)is 𝐹𝑥2 + 𝐹𝑦

2, and the angle 𝜃 of 𝐹(𝑠)

is tan−1 𝐹𝑦

𝐹𝑥. The angle is measured counterclockwise from the positive real axis.

b) Single-valued Function

In complex function analysis, we are interested in Single-Valued Function that can uniquely

determine the value of s. For instance, given the function

𝐹 𝑠 =1

𝑠(𝑠 + 1)

𝐹 𝑠 = ∞ is mapped onto two points, s=0 and s=-1, in the s-plane

c) Poles and zeros of a Function

Poles are the value of s that will make the function F(s) become infinity. In other words, poles

are the roots of the denominator of F(s). If the denominator of F(s) involves k-multiple factors

(𝑠 + 𝑝)𝑘 , then 𝑠 = −𝑝 is called a multiple poles and of order 𝑘or repeated pole of order 𝑘. If

𝑘 = 1, the pole is called a simple pole.

Zeros are the value of s that will make the function F(s) become zero. In other words, zeros are

the numerator of F(s).

As an illustrative example, consider the following complex function

𝐺 𝑠 = 𝑠 + 2 (𝑠 + 10)

𝑠 𝑠 + 1 𝑠 + 5 (𝑠 + 15)2

G(s) has zeros at 𝑠 = −2 and 𝑠 = −10, simple poles at 𝑠 = 0, 𝑠 = −1 and 𝑠 = −5, and a

double pole (multiple pole of order 2) at 𝑠 = −15. Note that G(s) becomes zero at 𝑠 = ∞.

G(s) is therefore has 2 zeros and 5 poles.

d) Singularities of a Function

The singularities of a function are the points in the s-plane at which the function or its

derivatives do not exist. A pole is the most common of singularities and plays a very important

role in studies of classical control theory. (source: ogata)

6


UiTM Shah Alam

1.2 REVIEW OF DIFFERENTIAL EQUATIONS, LINEAR SYSTEMS, IMPULSE RESPONSE AND LAPLACE

TRANSFORMATIONS. DEFINITION OF STABILITY. INTRODUCTION TO STATE EQUATIONS AND

TRANSFER FUNCTIONS.

1.2.1 Review of Differential Equations

Differential equations generally involve derivatives and integrals of the dependant variables

with respect to the independent variable. For instance, a shock absorber system of a car as in

figure 1.2 can be represented by the differential equation,

)()(1)(

)( tvdttiCdt

tdiLtRi

Figure 1.2: RLC Circuit

where R is the resistance, L the inductance, C the capacitance, i(t) the current and v(t) the

applied voltage. The dependent variable i(t) is determined by solving the equation.

In general, a differential equation of nth-order is written as

)()()()()(

011

1

1 tftyadt

tdya

dt

tyda

dt

tydn

n

nn

n

Which is also known as a linear ordinary differential equation if the coefficients a0, a

1, … ,a

n-1 are

not a function of y(t).

7


UiTM Shah Alam

1.2.2 Laplace Transforms

a) Laplace Transform

Laplace transform is used to convert from time domain to s-domain. Working with differential

equation is rather complicated. In analyzing and designing a control system it is easier to work in

s-domain. Laplace transform is defined as;

ℒ 𝑓 𝑡 = 𝐹 𝑠 = 𝑓 𝑡 𝑒−𝑠𝑡𝑑𝑡

∞

0

Where 𝑠 = 𝑥 + 𝑗𝑦, a complex variable.

Example 2.1: Let f(t) be a unit-step function that is defined as

0 ,0

0 ,1)(

t

ttu

The Laplace transform of f(t) is obtained as

se

sdtetusF stst 11

)()(0

0

Example 2.2: Consider the exponential function

0 ,)( tetf t

where α is real constant. The Laplace transform of f(t) is written as

00

)( 1)(

ss

edteesF

tsstt

8


UiTM Shah Alam

Table 2.1: Laplace Transform table for input responses

Notes:

9


UiTM Shah Alam

b) Laplace Transform Theorems

The laplace transform has a set of theorems to solve a complex mathematical equations. Table 2.2 summarizes the Laplace Transform theorems

Table 2.2: Laplace Transform Theorems

10


UiTM Shah Alam

c) Inverse Laplace Transformation Using Partial Fraction Method

Given the Laplace transform F(s), the operation of obtaining f(t) is termed the inverse Laplace Transformation and is denoted by:

𝑓 𝑡 = ℒ−1[𝐹 𝑠 ]

Inverse Laplace Transform is used when we want to convert from s-domain to time domain. The inverse Laplace transform of rational functions are normally carried out using partial-fraction expansion and the Laplace transform table. Consider a rational function

𝐺 𝑠 =𝑄(𝑠)

𝑃(𝑠)

where Q(s) and P(s) are polynomials of s. It is assume that the order of P(s) in s is greater than of Q(s). The polynomial P(s) may be written as

01

1

1)( asasassP n

n

n

where a

0, a

1, … ,a

n-1 are real coefficients. This method will be emphasized for the cases of

simple poles, multiple-order poles and complex poles. Case 1: Simple poles If all the poles of G(s) are simple and real, then G(s) can be written as


𝑃(𝑠)=

𝑄(𝑠)

𝑠 + 𝑠1 𝑠 + 𝑠2 … (𝑠 + 𝑠𝑛 ), 𝑤𝑕𝑒𝑟𝑒 𝑠1 ≠ 𝑠2 ≠ ⋯ 𝑠𝑛

Applying partial-fraction expansion, the equation can be written as

𝐺 𝑠 =𝐾−𝑠1

𝑠 + 𝑠1+

𝐾−𝑠2

𝑠 + 𝑠2+ ⋯ +

𝐾−𝑠𝑛

𝑠 + 𝑠𝑛

Where

isssP

sQssK isi

)(

)()(

The numerator of each fraction is called the residue. 𝐾−𝑠𝑖 is called the residue of G(s) for the pole 𝑠 = −𝑠𝑖 . The inverse transform is the written as

ts

s

ts

s

ts

sn

neKeKeKtg

2

2

1

1)(

11


UiTM Shah Alam

Example 2.3: Consider the function

)3)(2)(1(

35)(

sss

ssG

which is written in the partial-fraction expanded form:

321)( 321

s

K

s

K

s

KsG

The coefficients 𝐾−1 , 𝐾−2 , 𝐾−3 are determined as follows:

6)23)(13(

3)3(5

3)()3(

7)32)(12(

3)2(5

2)()2(

1)31)(21(

3)1(5

1)()1(

3

2

1

ssGsK

ssGsK

ssGsK

Thus,

3

6

2

7

1

1)(

ssssG

The inverse transform or time function is ttt eetg 32 67)(

Case 2: Multiple-order poles

If r of the n poles is identical, G(s) is written as


𝑃(𝑠)=

𝑄(𝑠)

𝑠 + 𝑠1 𝑠 + 𝑠2 … 𝑠 + 𝑠𝑛−𝑟 (𝑠 + 𝑠𝑖)𝑟

12


UiTM Shah Alam

Then G(s) can be expanded as

poles repeated of r terms

)()(

poles simple of r terms-n

)(2

21

21

)(21

r

i

r

iirn

sss

ss

A

ss

A

ss

A

ss

K

ss

K

ss

KsG

rn

The (n-r) coefficients K-s1, K-s2, … , K-s(n−r) which correspond to simple poles may be evaluated as explained before. The coefficients A

1 … Ar are evaluated as follows:

i

sssGss

ds

d

rA

iss

sGssds

dA

iss

sGssds

dA

iss

sGssA

r

ir

r

r

ir

r

ir

r

ir

)()()!1(

1

)()(!2

1

)()(

)()(

1

1

1

2

2

2

1

Example 2.4:

Consider the function

2)2)(1(

2)(

sssG

G(s) can be written as

2

211

)2()2()1()(

s

A

s

A

s

KsG

The coefficient corresponding to the simple pole is

2

1)2(

221

ss

K

13


UiTM Shah Alam

and those of second order-pole are

2

2)1(

2

2)1(

2

2

2)1(

2

21

2

sss

sds

dA

ss

A

The completed partial-fraction expansion is

2)2(

2

2

2

1

2)(

ssssG

The time function is

ttt teeetg 22 222)(

Case 3: Simple complex-conjugate poles

Suppose that G(s) contains a pair of complex poles:

j--s and js

The corresponding coefficients of these poles are

js

sGjsK j )()(

js

sGjsK j )()(

Example 2.5:

Considering transfer function G(s)

2121

)21)(21(

3

)52(

3)(

21210

2

js

K

js

K

s

K

jsjssssssG

jj

14


UiTM Shah Alam

5

3

052

320

sssK

21

12

21

12

20

353

)(

)12(20

3

21)21(

3

)12(20

3

21)21(

3

21

21

js

j

js

j

ssG

j

jsjss

K

j

jsjss

K

j

j

and the time function is given as

tte

j

eeeee

eejeee

ejejtg

t

tjtjtjtjt

tjtjtjtjt

tjtj

2sin2

12cos

5

3

5

3

22

24

20

3

5

3

)()22(20

3

5

3

)12()12(20

3

5

3)(

2222

2222

)21()21(

15


UiTM Shah Alam

TUTORIAL 1: MATHEMATICAL REVIEW

1. Derive equations for a unit step, ramp, impulse and sinusoidal response in time domain.

2. In a unit step response graph, what is the relationship between final value theorem and steady state error?

3. Find the Laplace transform of time function 𝑓 𝑡 = 5 + 3𝑒−2𝑡 .

4. Verify question (3) above by using MATLAB application. MATLAB hint >>syms s t; % Command to run MATLAB in s and t domains >>f=5+3*exp(-2*t) % Entering the function >>F=laplace(f,t,s) % Executing Laplace Transform command

5. Find the inverse Laplace Transform of a rational function and

𝐹 𝑠 =5

𝑠2 + 3𝑠 + 2

6. Find the inverse Laplace transform of a rational function

𝐹 𝑠 =2

𝑠 + 1 (𝑠 + 2)2

7. Verify the result in question (6) above using MATLAB application.

MATLAB hint >> syms st; >>F=2/((s+1)*(s+2)^2) >>f=ilaplace(F,s,t)

16


UiTM Shah Alam

CHAPTER 2.0 INTRODUCTION TO CONTROL SYSTEMS

Control systems can be placed into three broad functional groups:

Monitoring systems, such as Supervisory Control and Data Acquisition (SCADA) systems, which

provide information about the process state to the operator;

Sequencing systems, used where some process must follow a pre-defined sequence of discrete

events;

Closed-loop systems, which is widely taught in engineering course, are typically implemented to

give some process a set of desired performance characteristics

The history of feedback control system begun as early as in 1769 when James Watt’s steam engine and

governor are developed. The Watt stem engine often used to mark the beginning of the Industrial

Revolution in England. The revolution of automatic control system continues in which the first ever

autonomous rover vehicle, known as Sojourner was invented in 1997.

In summary below is the history of feedback control system

1769 - James Watt’s flyball governer

Figure 2.0: James Watt’s flyball governer

1868 - J. C. Maxwell’s model of governer

1927 - H. W. Bode’s feedback amplifiers

1932 - H. Nyquist’s stability theory

17


UiTM Shah Alam

1954 - George Devol’s robot design

1970 - State-variable models and optimal control theory

1980 - Robust control system design

1997 - First ever autonomous rover vehicle “Sojourner”

Figure 2.1: Sojourner

But before we go into further details, we have to know control systems’ terms and concepts. The

frequently used terms and concepts are as follow:

Automation - The control of a process by automatic means Control system - An interconnection of components forming a system

configuration that will provide a desired response Controlled variable

- Quantity or condition that is measured and controller. Normally it is the output of the system

Manipulated variable

- Quantity or condition that is varied by the controller so as to affect the value of the controlled variable

Plant - A plant is a piece of equipment, perhaps just a set of machine parts functioning together, the purpose of which to perform a particular operation. Any physical object to be controller (such as heating furnace, a chemical reactor etc) is called a plant

Processes - A process can be defined as a natural, progressively continuing operation or development marked by a

Info: The mobile Sojourner had a mass

of 10.5kg and 0.25 square meter solar

array

18


UiTM Shah Alam

series of gradual changes that succeed one another in a relatively fixed way and lead towards a particular result or end

Disturbances - A disturbance is a signal which tends to adversely affect the value of the output of the system. If a disturbance is generated within the system, it is called internal; which an external disturbance is generated outside the system.

Feedback control

- Feedback control is an operation which in the presence of disturbances, tends to reduce the difference between the output of a system and the reference input and which does so on the basis of the difference.

Feedforward - Feedforward has a reference signal which is act as an additional input.

Source: AAMI, Fac of Mech Eng., UiTM

Figure 2.2: Input-output configuration of control system (souce: AAMI)

19


UiTM Shah Alam

Figure 2.3: Input-output configuration of a closed-loop control system (source: AAMI)

2.1 OPEN LOOP AND CLOSED-LOOP SYSTEMS

2.1.1 Open Loop Control System

A system is said to be an open loop system when the system’s output has no effect on the control

action. In open loop system, the output is neither measured nor fed back for comparison with the input.

Figure 2.4: Open loop control system

An open loop control system utilizes an actuating device (or controller) to control the process directly

without using feedback as shown in Figure 2.4.

The advantages and the disadvantages of an open-loop control system is tabulated in table 2.1 below

ADVANTAGES DISADVANTAGES

Simple and ease of maintenance Disturbances and changes in calibration cause errors Less expensive

Stability is not a problem Output may be different from what is desired Convenient when output is hard to

measure

20


UiTM Shah Alam

2.1.2 Closed-loop control system

A system that maintains a prescribed relationship between the output and the reference input is called a

closed-loop system or a feedback control system. The system uses a measurement of the output and

feedback of the signal to compare it with the desired output.

Figure 2.5: Closed loop control system

In a closed-loop control system, the actuating error signal, which is the difference between the input

signal and the feedback signal, is fed to the controller so as to reduce the error and bring the output of

the system to a desired value.

2.1.3 Comparison between open loop and closed-loop control system.

The table below shows the comparison between the two systems:

OPEN LOOP CLOSED LOOP

System stability is not a major problem, therefore easier to build

The use of feedback makes the system response relatively insensitive to external disturbances and internal variations in system parameters

Use open loop only when the inputs are known ahead of time and there is no disturbances

System stability is a major problem because the system tends to overcorrect errors that can cause oscillations or changing amplitude.

2.2 TRANSFER FUNCTION

The transfer function of a linear system is defined as the ratio of the Laplace transform of the output

variable to the Laplace transform of the input variable, with all initial conditions assumed to be zero. The

Transfer function of a system (or element) represents the relationship describing the dynamics of the

system under consideration. A transfer function may be defined only for a linear, stationary (constant

parameter) system. A non-stationary system often called a time-varying system, has one or more time-

varying parameters, and the Laplace transformation may not be utilized. Furthermore, a transfer

function is an input-output description of the behavior of a system. Thus the transfer function

description does not include any information concerning the internal structure of the system and its

behavior.

21


UiTM Shah Alam

2.2.1 The Transfer function of linear systems

The transfer function of a LTI system is defined as the Laplace transform of the impulse response, with

all the initial conditions set to zero.

)]([)( tgLsG

The transfer function is related to the Laplace transform of the input and the output through the

following relation:

)(

)()(

sR

sYsG

where all the initial conditions set to zero, and )(sY and )(sR are the Laplace transform of )(ty and

)(tr respectively.

Although the transfer function of a linear system is defined in terms of the impulse response, in practice,

the input-output relation of a linear time-invariant system with continuous–data input is often described

by the differential equation, so it is more convenient to derive the transfer function directly from the

differential equation.

Let us consider that the input-output relation of a linear time-invariant system is described by the

following nth-order differential equation with constant real coefficients:

)()(

.....)()(

)()(

......)()(

011

1

1011

1

1 trbdt

tdrb

dt

trdb

dt

trdbtya

dt

tdya

dt

tyda

dt

tydm

m

mm

m

mn

n

nn

n

To obtain the transfer function of the linear system that is represented by Eq. (2.3), we simply take the

Laplace transform on both sides of the equation and assume zero initial conditions. The result is

R(s)bsbsbsbY(s)asasas m

m

m

m

n

n

n

01

1

101

1

1

The transfer function between )(tr and )(ty is given by:

01

1

1

01

......

..............

)(

)()(

asasas

bsbsb

sR

sYsG

n

n

n

m

m

22


UiTM Shah Alam

The transfer function is said to be strictly proper if nm . If nm then the transfer function is proper.

It is improper if nm .

Characteristic Equation: The characteristic equation of a LTI system is defined as the equation

obtained by setting the denominator polynomial of the transfer function to zero. Thus, the

characteristic equation of the system described by the Eq. (2.4) is

0a 01

1

1

assas n

n

n

Later, we shall show that the stability of a linear single-input single-output system is governed

completely by the roots of the characteristic equation.

2.2.2 Transfer function of multivariable system

The definition of a transfer function is easily extended to a system with multiple inputs and outputs. A

system of this type is often referred to as a multivariable system. Figure 2.6 shows a control system with

two inputs and two outputs.

Figure 2.6: General block representation of a two-input, two-output system

Since the principle of superposition is valid for linear systems, the total effect on any output due to all

the inputs acting simultaneously is obtained by adding up the outputs due to each input acting alone.

Thus, using transfer function relations we can write the simultaneous equations for the output variables

as

)()()()()(

)()()()()(

2221212

2121111

sRsGsRsGsY

sRsGsRsGsY

23


UiTM Shah Alam

where )(sG ij is the transfer function relating the ith output to the jth input variable. Thus

)(

)(

sR

sYG

j

iij

In general, for j inputs and i outputs, we can write the simultaneous equations for the output variables

as

)(

)(

)(

)()()(

)()()(

)()()(

)(

)(

)(

2

1

21

22221

11211

2

1

sR

sR

sR

sGsGsG

sGsGsG

sGsGsG

sY

sY

sY

jijii

j

j

i

It is convenient to express Eq. (2.7) in a matrix-vector form

G(s)R(s)Y(s)

where

)(

)(

)(

)(2

1

sY

sY

sY

sY

i

is the i 1 transformed output vector; whereas

)(

)(

)(

)(2

1

sR

sR

sR

sR

j

is the j 1 transformed input vector; and

)()()(

)()()(

)()()(

)(

21

22221

11211

sGsGsG

sGsGsG

sGsGsG

sG

ijii

j

j

is the i j transfer-function matrix.

24


UiTM Shah Alam

2.3 DEFINITION OF STABILTY

A stable system is defined as a system which gives a bounded output in response to a bounded input.

The concept of stability can be illustrated by considering a circular cone placed on a horizontal surface,

as shown in Fig. 2.7 and Fig. 2.8.

Figure 2.7: The stability of a cone.

----------------------------------------------------------------------------------------------------

Figure 2.8: Stability in the s-plane.

The stability of a dynamic system is defined in a similar manner. Let u(t), y(t), and g(t) be the input,

output, and impulse response of a linear time-invariant system, respectively. The output of the system is

given by the convolution between the input and the system's impulse response. Then

0

)()()( dgtuty

25


UiTM Shah Alam

This response is bounded (stable system) if and only if the absolute value of the impulse response, g(t),

integrated over an infinite range, is finite. That is

0

)( dg

Mathematically, Eq. (4.24) is satisfied when the roots of the characteristic equation, or the poles of G(s),

are all located in the left-half of the s-plane.

A system is said to be unstable if any of the characteristic equation roots is located in the right-half of

the s-plane. When the characteristic equation has simple roots on the j-axis and none in the right-half

plane, we refer to the system as marginally stable.

The following table illustrates the stability conditions of a linear continuous system with reference to the

locations of the roots of the characteristic equation.

STABILITY CONDITION LOCATION OF THE ROOTS

Stable All the roots are in the left-half s-plane

Marginally stable of marginally unstable At least one simple root and no multiple

roots on the j-axis; and no roots in the right-half s-plane.

Unstable At least one simple root in the right-half s-plane or at least one multiple-order root

on the j-axis.

The following examples illustrate the stability conditions of systems with reference to the poles of the

closed-loop transfer function M(s).

321

20)(

ssssM

Stable

)22)(1(

)1(20)(

2

sss

ssM

Unstable due to the pole at s = 1

)4)(2(

)1(20)(

2

ss

ssM

Marginally stable or marginally unstable due to s =

j2.

26


UiTM Shah Alam

)10()4(

10)(

22

sssM

Unstable due to the multiple-order pole at s = j2.

2.3.1 Open loop and Closed loop stability

A system is open-loop stable if the poles of the loop transfer function G(s)H(s) are all in the left hand side of s-plane.

Figure 2.9: A typical closed-loop system

A system is closed0loop stable (or simply stable) if the poles of the closed-loop transfer function (or zeros of 1+G(s)H(s) are all in the left hand side of s-plane

2.4 BASIC CONTROL ACTIONS

The following six basic control actions are very common among industrial automatic controllers:

1. Two-position or on-off controller 2. Proportional controller 3. Integral controller 4. Proportional-plus-integral controller 5. Proportional-plus-derivative controller 6. Proportional-plus-derivative-plus-integral controller

2.4.1 Two-position of on-off control action

In a two-position control system, the actuating element has only two fixed positions which are, in many cases, simply on and off. Two-position or on-off control is relatively simple and inexpensive and, for this reason, is very widely used in both industrial and domestic control systems.

Let the output signal from the controller be m(t) and the actuating error signal be e(t). In two position control, the signal m(t) remains at either a maximum or minimum value, depending on whether the actuating error signal is positive or negative, so that

𝑚 𝑡 = 𝑀1 𝑓𝑜𝑟 𝑒(𝑡) > 0

= 𝑀2 𝑓𝑜𝑟 𝑒(𝑡) < 0

+- e(s)

yysp H(s)

PlantController

G(s)

27


UiTM Shah Alam

Where 𝑀1and 𝑀2, are constants. The minimum value 𝑀2, is usually either zero or −𝑀1. Two-position controllers are generally electrical devices, and an electric, solenoid-operated valve is widely used in such controller. Pneumatic proportional controller with very high gain act as two-position controller and are sometimes called pneumatic two-position controller.

Figure 2.10 show the block diagrams for two-position controller. The range through which the actuating error signal must move before the switching occurs is called the differential gap.

Figure 2.10: Two-position controller

2.4.2 Proportional controller

For a controller with proportional control action, the relationship between the output of the controller m(t) and the actuating error signal e(t) is

𝑚 𝑡 = 𝐾𝑝𝑒(𝑡)

or, in Laplace Transform

𝑀(𝑠)

𝐸(𝑠)= 𝐾𝑝

Where 𝐾𝑝 , is termed the proportional sensitivity or the gain.

28


UiTM Shah Alam

Whatever the actual mechanism may be and whatever the form of the operating power, the proportional controller is essentially an amplifier with and adjustable gain.

The proportional action has the following two properties:

1. Reduce rise time 2. Does not eliminate steady state error

Example 2.1:

Given a system consist of mass-spring and damper

a) The second order PDE is: b) Taking the LT c) The TF is therefore: d) Let M=1kg, b=10N.s/m, k=20 N/m & F(s)=1, therefore X(s) / F(s): e) From the Transfer Function, the DC gain is: f) Corresponding to the steady state error of: g) The settling time is:

b

M

x

F

k

Open Loop Response

Time (sec)

Dis

pla

ce

me

nt

(m)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

29


UiTM Shah Alam

P control (K) reduces the rise time, increases the overshoot and reduces the steady state error.

h) The closed-loop transfer function of the system with P controller is X(s)/F(s)=G/(1+G): i) Let the P gain (K) equal 300

Rise time and ss error reduced, slightly reduced settling time but increased overshoot.

2.4.3 Integral controller

In a controller with integral control action, the value of the controller output m(t) is changed at a rate proportional to, the actuating error signal e(t). That is

𝑑𝑚(𝑡)

𝑑𝑡= 𝐾𝑖𝑒(𝑡)

Therefore; 𝑚 𝑡 = 𝐾𝑖 𝑒 𝑡 𝑑𝑡𝑡

0

Where 𝐾𝑖 is an adjustable constant. The transfer function of the integral controller is

𝑀(𝑠)

𝐸(𝑠)=

𝐾𝑖

𝑠

If the value of e(t) is doubled, then the value of m(t) varies twice as fast. For zero actuating error, the value of m(t) remains stationary.

Closed Loop Step : K = 300

Time (sec)

Dis

pla

ce

me

nt

(m)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

30


UiTM Shah Alam

The integral controller has the following properties:

1. Proportional controllers often give a steady-state error. Integral controller arose from trying to add a “reset” term to the control signal to eliminate steady state error. In other words, the integral controller “resets” the bias error from the P controller.

2. Gives large gain at low frequencies resulting in “beating down” load disturbances. 3. May make the transient response worse. 4. Controller phase starts out at -90° and increases to 0° at the break frequency. This phase lag can

be compensated by derivative action.

The integral controller act as “automatic reset” as shown in figure 2.11

Figure 2.11: Automatic reset action

Almost always used in conjunction with P control.

Figure 2.12: PI control

The integral term may be expressed in (i) 𝑇𝑖 and (ii) 𝑘𝑖

The integral term 𝑇𝑖 is known as the integral time constant. 𝑇𝑖 = ∞ corresponds to pure (proportional) gain.

The integral term 𝑘𝑖 is known as integral gain (e.g: in MATLAB)

The relationship between 𝑇𝑖 and 𝑘𝑖 is as follows:

𝑘𝑖

𝑠=

𝐾

𝑇𝑖𝑠

+-

ysp yplantK

load disturbance

1sTi

u

e

+-

ysp yplant

Kload disturbance

1sTi

u

e K

31


UiTM Shah Alam

Example 2.2:

a) I control reduces the rise time, increases both settling time and overshoot, and eliminates the steady-state error

b) The closed-loop transfer function of the system with a PI controller is: X(s)/F(s) = ______________ .

c) Let k = 30 and ki = 70. P gain (k) was reduced because the I controller also reduces the rise time and increases the overshoot as does the P controller (double effect).

2.4.4 Derivative controller

Introducing a derivative controller will add damping and in doing so:

1. increases system stability (add phase lead)

2. reduces overshoot

3. generally improves transient response

A derivative controller may able to provide anticipative action but derivative action can make the system become noisy.

Almost always used in conjunction with P control.

Figure 2.12: PD control

The integral term may be expressed in (i) 𝑇𝑑 and (ii) 𝑘𝑑

The integral term 𝑇𝑑 is known as the derivative time constant.

Closed Loop Step : K = 30, Ki = 70

Time (sec)

Dis

pla

ce

me

nt

(m)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

+-

ysp yplantc

load disturbance

KTd s

sTd /N1+

32


UiTM Shah Alam

The integral term 𝑘𝑑 is known as derivative gain (e.g: in MATLAB)

The relationship between 𝑇𝑑 and 𝑘𝑑 is as follows:

𝑘𝑑𝑠 = 𝐾𝑇𝑑𝑠

Example 2.3:

a) D control reduces both settling time and overshoot. b) The closed-loop transfer function of the system with a PD controller is:

X(s)/F(s)=______________ c) Let k = 300 and kd = 10.

d) Reduced overshoot and settling time, small effect on rise time and ss error

Closed Loop Step : K = 300, Kd = 10

Time (sec)

Dis

pla

ce

me

nt

(m)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

33


UiTM Shah Alam

2.4.5 PID controller

In some system the commonly implemented controller consist of the P, I and D control action. We call

this type of controller as PID controller.

Figure 2.13: PID control

The standard form of PID controller according to ISA (Instrument Society of America) is as follows:

𝐺𝑐 𝑠 = 𝐾(1 +1

𝑠𝑇𝑖+ 𝑇𝑑𝑠)

Or 𝐺𝑐 𝑠 = 𝐾 +𝑘𝑖

𝑠+ 𝑘𝑑𝑠

Closed Loop Step : K = 350, Ki = 300, Kd = 50

Time (sec)

Dis

pla

ce

me

nt

(m)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

K

1/( )Tsi

ysp

T sd

G s( )u

-

+ e y

34


UiTM Shah Alam

Example 2.4:

a) The closed-loop transfer function of the system with a PID controller is: X(s)/F(s) = (kd s

2 +ks+ki )/(s3 + (10+kd)s2 + (20+k)s + ki ) b) Let k = 350, ki = 300 and kd = 50.

c) No overshoot, fast rise and settling time and no steady-state error

2.4.6 PID tuning

Introducing the P, I and D controller has certainly proven to contribute some effect to our system’s response. These effects are summarized as in table below.

CLOSED LOOP RESPONSE

RISE TIME OVERSHOOT SETTLING TIME

SS ERROR

K Decrease Increase Small change Decrease

𝑘𝑖 =𝐾

𝑇𝑖

Decrease Increase Increase Eliminate

𝑘𝑑 = 𝐾𝑇𝑑 Small change Decrease Decrease Small change

When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.

1. Obtain an open-loop response and determine what needs to be improved 2. Add a proportional control to improve the rise time 3. Add a derivative control to improve the overshoot 4. Add an integral control to eliminate the steady-state error 5. Adjust each of K, Ki, and Kd until you obtain a desired overall response referring to the table

shown previously to find out which controller controls what characteristics.

Closed Loop Step : K = 350, Ki = 300, Kd = 50

Time (sec)

Dis

pla

ce

me

nt

(m)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

35


UiTM Shah Alam

6. It is not necessary to implement all three controllers (P, I & D) into a single system. For example, if a PI controller gives a good enough response, then you don't need to add D control to the system. Simple is better.

2.5 BLOCK DIAGRAM & REDUCTION METHODS

A block diagram is used to describe the composition and interconnection of a system, or it can be used

together with the transfer functions to describe the cause-and-effect relationships throughout the

system. For instance, Figure 2.14 (a) shows a dc motor wiring diagram, (b) sketch, and (c) shows the

block diagram with transfer function.

Figure 2.14: A dc motor: (a) wiring diagram (b) sketch

Figure 2.14 (c): A dc motor: Block diagram with transfer functions

36


UiTM Shah Alam

2.5.1 Block diagram reduction method

We shall now define the block diagram elements used frequently in linear control systems and the

related algebra. All component parts of a block diagram for linear time-invariant systems are shown in

Figure 2.15.

The characteristic of the summing junction as shown in Figure 2.15 (c) is that the output signal, )(sC , is

the algebraic sum of the input signals. The figure shows three inputs, but any number can be presented.

A pickoff point, as shown in Figure 2.15 (d), distributes the input signal, )(sR , undiminished, to several

output points.

Figure 2.15: Components of a block diagram for LTI systems

Figure 2.16 shows the block diagram of a linear feedback control system. The following terminology is

defined with reference to the diagram.

Figure 2.16: Feedback control system

37


UiTM Shah Alam

)(),( trsR = reference input (command)

)(),( tcsC or )(),( tysY = output (controlled variable)

)(),( tbsB = feedback signal

)(),( tesE = actuating signal = error signal

)(sH = feedback transfer function

)()( sHsG = )(sL = loop transfer function

)(sG = forward-path transfer function

)()()( sRsCsM or )()( sRsY = closed-loop transfer function or system transfer function.

)(sM can be expressed as a function of )(sG and )(sH . From Figure 2.16, we write

)()()(

)()()(

sYsHsB

sEsGsY

The actuating signal is written as

)()()( sBsRsE

Thus,

)()(1

)(

)(

)()(

)()()()()(

sHsG

sG

sR

sYsM

sBsGsRsGsY

The block diagram representation of a given system often can be reduced by block diagram reduction

techniques to a simplified block diagram with fewer blocks than the original diagram. Table below shows

some of the block diagram reduction techniques.

The block diagram reduction technique is based on the utilization of rule 6 in which eliminates feedback

loops. Therefore, the other transformations are used to transform the diagram to a form ready for

eliminating feedback loops.

38


UiTM Shah Alam

For parallel subsystems as shown below in (a), the reduction technique is shown in (b).

39


UiTM Shah Alam

Example 2.5: Block Diagram Reduction.

A block diagram of a multiple-loop feedback control system is shown in Figure 2-5. It is interesting to

note that the feedback signal H1(s)Y(s) is a positive feedback signal, and the loop G3(s)G4(s)H1(s) is called

a positive feedback loop. First, to eliminate the loop G3G4H1, we move H2 behind block G4 by using rule

4, and therefore obtain Figure 2-6 (a).

Multiple-loop feedback control system

40


UiTM Shah Alam

Eliminating the loop G3G4H1 by using rule 6, we obtain Figure 2-6 (b). Then, eliminating the inner loop

containing H2/G4, we obtain Figure 2-6 (c). Finally, by reducing the loop containing H3, we obtain the

closed-loop system transfer function as shown in Figure 2-6 (d).

Block diagram reduction of the system

41


UiTM Shah Alam

Example 2.6: Reduce the system shown to a single transfer function.

Block diagram for Example 2.6

42


UiTM Shah Alam

Steps in the block diagram reduction for Example 2.6

The block diagram representation of feedback control systems is a valuable and widely used approach.

The block diagram provides the analyst with a graphical representation of the interrelationships of

controlled and input variables. Furthermore, the designer can readily visualize the possibilities for

adding blocks to the existing system block diagram to alter and improve the system performance. The

transition from the block diagram method to a method utilizing a line path representation instead of a

block representation is readily accomplished and is presented in the following section.

43


UiTM Shah Alam

2.6 SIGNAL FLOW DIAGRAM & REDUCTION METHODS

Block diagrams are adequate for the representation of the interrelationships of controlled and input

variables. However, for a system with reasonably complex interrelationships, the block diagram

reduction technique is cumbersome and often quite difficult to complete. An alternative method for

determining the relationship between system variables has been developed by Mason and is based on a

representation of the linear system by line segments called Signal-Flow Graph (SFG).

The advantage of the SFG method is the availability of a flow graph gain formula, which provides the

relation between system variables without requiring any reduction procedure or manipulation of the

flow graph.

2.6.1 Basic elements of SFG

When constructing a SFG, junction points or nodes are used to represent variables. The nodes are

connected by line segments, called branches. A signal can transmit through a branch only in the

direction of the arrow.

For instance, consider that a linear system is represented by a simple algebraic equation

1122 yay

where y1 is the input, y2 the output, and a12 the gain between two variables. The SFG is shown in Figure

2-9.

Figure 2.17: Signal-flow graph of 1122 yay

Example 2.7: Consider the following set of algebraic equations:

4452255

4443342244

4432233

3321122

yayay

yayayay

yayay

yayay

Y1 Y2

a12

44


UiTM Shah Alam

The SFG for these equations is constructed, step by step, as shown:

Step-by-step construction of the SFG of Example 2.7

45


UiTM Shah Alam

2.6.2 Summary of the basic properties of SFG

The important properties of the SGF are summarized as follows:

1. SFG applies only to linear systems.

2. Nodes are used to represent variables. Normally, the nodes are arranged from left to right,

from input to output.

3. Signals travel along branches only in the direction described by the arrows of the branches.

2.6.3 Definitions of the SFG terms

Input Node (source)

- An input node is a node that has only outgoing branches.

Output Node (sink)

- An output node is a node that has only incoming branches. In general, we can make any non input node an output node, simply by connecting a branch with unity gain from the existing node to a new node with the same name (Example: node y2 in Figure 2.18(b)). If we attempt to convert y2 into input node, by using the same unity gain branch (Figure 2.18 (c)), then y2 output will differ from the original (y2 = y2 + a12y1 + a32y3).

Path - A path is any collection of a continuous succession of branches traversed in the same direction.

Forward Path - A forward path is a path that starts at an input node and ends at an output node, and along which no node is traversed more than once.

Loop - A loop is a path that originates and terminates on the same node and along which no other node is encountered more than once. For example, there are four loops in the SFG of Example 2.7. These are shown in Figure 2.19

Path Gain - The product of the branch gains encountered in traversing a path is called the path gain

Loop Gain - The loop gain is the path gain of a loop Non-touching Loops

- Two parts of a SFG are non-touching if they do not share a common node. For example, the loop y2-y3-y2 and y4-y4 of the SFG in Figure (d) of Example 2.7 are non-touching loops.

46


UiTM Shah Alam

Figure 2.18 (a & b): Modification of SFG so that y2 become output node

(c): Erroneous way to make node y2 an input node

47


UiTM Shah Alam

Figure 2.19: Four loops in the signal-flow graph of Example 2.7

2.6.4 SFG Algebra

Based on the properties of the SFG, we can outline the following manipulation rules and algebra of SFG.

1. The value of the variable represented by a node is equal to the sum of all the signals

entering the node. For the SFG of Figure 2.20 (a),

5514413312211 yayayayay

2. The value of the variable represented by a node is transmitted through all branches

leaving the node. In Figure 2.20 (a), we have

1188

1717

1166

yay

yay

yay

48


UiTM Shah Alam

Figure 2.20

3. Parallel branches in the same direction connecting two nodes can be replaced by a

single branch with the gain equal to the sum of gains of the parallel branches. Example:

Figure 2.20 (b).

4. A series connection of unidirectional branches can be replaced by one branch with gain

equal to the product of branch gains.

49


UiTM Shah Alam

2.6.5 Gain formula for SFG (Mason’s Rule)

The overall gain between the input node yin and output node yout of a SFG with N forward paths and L

loops is given by

Nk

P

y

yM k

kk

in

out ,2,1 ,

where

yin = input-node variable

yout = output-node variable

M = gain between yin and yout

N = total number of forward paths between yin and yout

Pk = kth forward-path gain

= 1 – (sum of all individual loop gains)

+ (sum of all gain products of two non-touching loops)

– (sum of all gain products of three non-touching loops) + …

k = , which is evaluated by eliminating all loops that touch kth forward-path

Procedures to solve SFG by using Mason’s rule:

1. Identify the no. of forward paths and determine the forward-path gains.

2. Identify the no. of loops and determine the loop gains.

3. Identify the non-touching loops taken two at a time, three at a time and so on. Determine the

product of the non-touching loop gains.

4. Determine and k.

5. Substitute all of the above information into the gain formula:

Nk

P

y

yM k

kk

in

out ,2,1 ,

50


UiTM Shah Alam

Care must be taken when applying the gain formula to ensure that it is applied between an input node

and an output node.

Example 2.8: Consider the SFG of a closed loop control system as given in Figure below. By using the

gain formula, find the transfer function )()( sRsY .

SFG of a feedback control system

1. There is only one forward path between )(sR and )(sY , and the forward-path gain is

P1 = )(sG .

2. There is only one loop; the loop gain is L1 = )()( sHsG .

3. There are no non-touching loops.

4. = 1 - L1 = )()(1 sHsG and 1 = 1.

5. Thus,

)()(1

)(

)(

)( 11

sHsG

sGP

sR

sY

51


UiTM Shah Alam

Example 2.9: For the system shown in Figure below, determine the gain between y1 and y5.

SFG for Example 2.9

1. There are three forward paths

Path 1: y1 – y2 – y3 – y4 – y5 P1 = a12 a23 a34 a45

Path 2: y1 – y2 – y4 – y5 P2 = a12 a24 a45

Path 3: y1 – y2 – y5 P3 = a12 a25

2. There are four loops

Loop 1: y2 – y3 – y2 L1 = a23 a32

Loop 2: y3 – y4 – y3 L2 = a34 a43

Loop 3: y2 – y4 – y3 – y2 L3 = a24 a43 a32

Loop 4: y4 – y4 L4 = a44

3. Non-touching loops: y2 – y3 – y2 and y4 – y4

Thus the product of the gains of the two non-touching loops:

L1L4 = a23 a32 a44

4. = 1 – (L1 + L2 + L3 + L4) + L1L4

= 1 – (a23 a32 + a34 a43 + a24 a43 a32 + a44) + a23 a32 a44

52


UiTM Shah Alam

All the loops are in touch with forward path P1, thus 1 = 1.


Two loops (y3 – y4 – y3 and y4 – y4) are not touching with forward path P3.

Thus, 3 = 1 - a34a43 – a44.

5. Thus,

4432234443322443343223

444334251245241245342312

332211

1

5

)(1

)1)(()()(

aaaaaaaaaaa

aaaaaaaaaaaa

PPP

y

yM

Example 2.10: Consider the SFG as shown in the figure. The following input-output relation is obtained

by use of the gain formula:

)1( 235143212211

1

7 HGGGGGGGPP

y

y

where

4213143321423411

2131433212311

1

HHHGGHHGGGHHGHHG

HHGGHHGGGHGHG

SFG for Example 2.10

53


UiTM Shah Alam

2.7 CONVERSION FROM BLOCK DIAGRAMS TO SFG

An equivalent SFG for a block diagram can be drawn by performing the following steps:

1. Identify the input/output signals, summing junctions & pickoff points → they are replaced with

nodes.

2. Interconnect the nodes & indicate the directions of signal flow by using arrows.

3. Identify the blocks - they are replaced with branches.

For each negative sum, a negative sign is included with the branch.

4. Add unity branches as needed for clarity or to make connections.

5. Simplify the SFG → eliminate redundant nodes/branches (only if the node is connected to branches

of a single flow in & a single flow out with unity gain).

6. Label the input/output signals and the branches accordingly.

Example 2.11: Convert the block diagram in the figure to a signal flow graph and determine the transfer

function using Mason’s gain formula.

54


UiTM Shah Alam

The equivalent SFG:

1. There are two forward paths; the forward-path gains are:

P1 = G1G2G3

P2 = G1G4

2. There are five individual loops; the loop gains are:

L1 = −G1G2H1

L2 = −G2G3H2

L3 = −G1G2G3

L4 = −G1G4

L5 = −G4H2

3. There are no non-touching loops.

4. ∆ = 1 – (L1 + L2 + L3 + L4 + L5)

= 1 + G1G2H1 + G2G3H2 + G1G2G3 + G1G4 + G4H2


55


UiTM Shah Alam

All the loops are in touch with forward path P2, thus 2 = 1

5. Thus,

2441321232121

41321

2211

1 HGGGGGGHGGHGG

GGGGG

PP

R

Y

2.8 STATE SPACE EQUATIONS

State space approach is an alternative method for representing physical system. In order to use this

approach, we have to limit our approach to linear, time-invariant systems or system that can be

linearized by the methods we have covered previously.

In state space method, the models are constructed in the time domain. This means we can work directly

with the governing differential equations to model, analyze and design a wide range of system. In

contrast, classical control design practices looking at the frequency domain output to interpret system’s

physical dynamics. With the arrival of space exploration, requirements for control systems increased in

scope. Hence the use of classical control design seems inadequate.

Many systems do not have just a single input. Multiple-input, multiple-output systems can be compactly

represented in state space with a model similar in form and complexity to that used for single-input,

single-output systems. To address the multiple input and output system a convenient matrix based is

used in representing the state space. In addition, the state space approach is also attractive because of

the availability of numerous state-space software packages for the personal computer

56


UiTM Shah Alam

The table below outlines the advantages and disadvantages of state space models

ADVANTAGES DISADVANTAGES

Multiple input / output models are now possible

Difficult to examine robustness (stability margins)

Possible to minimize “error critera” (optimal control)

More work than classical control for “simple” problems

Possible to examine stability in more depth

“optimal” systems require “optimal” error criteria Ideally suited to computer-based design

and analysis

2.8.1 Definition of state space terms

State of a system

- A set of quantities which completely determine the evolution of the response of a system (in the absence of external inputs)

State Variables - Set of variables that define the state. These variables are not unique. For example x1, x2,….

State Vector - The (column) vector of the nth state variables: 𝑥 𝑡 = [𝑥1 𝑡 𝑥2 𝑡 … 𝑥𝑛 𝑡 ] Note: system is of order n (i.e it is described by an nth order D.E.

State Space - The n-dimensional space in which the components of the state vector are the co-ordinate axes.

State trajectory - The path in state space produced by the state vector as it changes with time.

Note: The selection of state variables is not unique. In the first instance, it is often reasonable to

choose something with “physical meaning”, often something associated with system “energy”

57


UiTM Shah Alam

2.8.2 State space model

We begin our state space equation with a state equation. A state equation consist of the state equation

and output equation as follows:

𝑥 = 𝐴𝑥 + 𝐵𝑢 State Equation

𝑦 = 𝐶𝑥 + 𝐷𝑢 Output Equation

Now A, B, C and D are all matrices involved in a state space equation

A = (n x n) state matrix that describes “internal (homogenous) motion

B = (n x r) input matrix that describes how r inputs affect n states

C = (m x n) output matrix that describes how n states contribute to m outputs

D = (m x r) direct transmission matrix that describes how r inputs are fed through to m outputs.

LECTURER’S NOTES:

58


UiTM Shah Alam

LECTURER’S NOTES:

59


UiTM Shah Alam

CHAPTER 3.0 SYSTEM PERFORMANCE ANALYSIS

The ability to adjust the transient and steady-state response of a control system is a beneficial outcome

of the design of feedback systems. Since time is used as an independent variable in most of control

systems, it is usually of interest to evaluate the state and output responses with respect to time, or

simply the time response.

In the analysis problem, we will use selected input signals to test the response of control systems. This

response will be characterized by a selected set of response measures. In this chapter, we will strive to

delineate a set of quantitative performance measures that adequately represent the performance of the

control systems.

3.1 Time Response and Test Signals

The time response of a control system is usually divided into two parts: the transient response and the

steady-state response. Let y(t) denote the time response of a continuous-data system; then, in general,

it can be written as

y(t) = yt(t) + yss(t) (3.1)

where yt(t) denotes the transient response and yss(t) denotes the steady-state response.

In control systems, the transient response is defined as the part of the time response that goes to zero

as time becomes very large. Thus yt(t) has the property

0)(lim

tytt

(3.2)

The steady-state response is simply the part of the total response that remains after the transient has

died out. All real stable systems exhibit transient phenomena to some extent before the steady state is

reached.

In the design problem, specifications are usually given in terms of the transient and steady-state

performance, and controllers are designed so that the specifications are all met by the design system.

60


UiTM Shah Alam

Since it is difficult to design a control system that will perform satisfactorily for all possible forms of

input signals, it is necessary, for the purpose of analysis and design, to assume some basic types of test

signals properly for the prediction of the system's performance to other more complex inputs.

3.1.2 Step-Input Function

The step-function input represents an instantaneous change in the reference input. The mathematical

representation of a step function of magnitude A is

0 0

0 )(

t

tAtr

Mathematically, r(t) = Aus(t), where us(t) is the unit-step function. The step function is shown in Fig.

3.1(a).

3.1.3 Ramp-Input Function

The ramp function is a signal that changes constantly with time. Mathematically, a ramp function is

represented by

)()( tAtutr s

where A is a real constant. The ramp function is shown in Fig. 3.1(b).

3.1.4 Parabolic-Input Function

The parabolic function represents a signal that is one order faster than the ramp function.

Mathematically, it is represented by

)(2

)(2

tuAt

tr s

The parabolic function is shown in Fig. 3.1(c).

61


UiTM Shah Alam

Fig. 3.1 shows the three time-domain test signals.

Figure 3.1: Test input signals: (a) Step, (b) Ramp, (c) Parabolic.

3.2 First & Second Order System: Transient & Steady State Response

For linear control systems, the time response is characterized by using the unit-step input. The response

of the control system to the unit-step input is called the unit-step response. Fig. 3.2 illustrates a typical

unit-step response of a linear control system. With reference to the unit-step response, the following

performance criteria (parameters) are defined:

1. Maximum overshoot: Let ymax denotes the maximum value of y(t) and yss be the steady-state value of

y(t) and ymax yss. The maximum overshoot of y(t) is defined as,

Maximum overshoot = ymax − yss

%100overshoot maximum

overshoot maximum of Percentage ssy

(3.3)

2. Delay time: The delay time, td is defined as, the time required for the step response to reach 50% of

its final value.

3. Rise time: The rise time, tr is defined as, the time required for the step response to rise from 10 to 90

percent of its final value.

4. Settling time: The settling time, ts is defined as, the time required for the step response to reach and

stay within a specified percentage (5%) of its final value.

62


UiTM Shah Alam

Figure 3.2: Step response of a control system.

Analytically, these quantities are difficult to establish, except for simple systems that are lower than the

third order.

3.2.1 Transient Response of a Prototype of Second-Order Systems

Although it is true that second-order control systems are rare in practice, their analysis generally helps

to form a basis for the understanding of analysis and design of higher-order systems, especially the ones

that can be approximated by second-order systems.

Consider that a second-order control system with unity feedback is represented by the block diagram

shown in Fig. 3.3. The open-loop transfer function of the system is

n

n

sssG

2)(

2

(3.5)

63


UiTM Shah Alam

where ζ and n are real constants. The closed-loop transfer function of the system is

22

2

2)(

)(

nn

n

sssR

sY

(3.6)

The characteristic equation of the prototype of the second-order system is obtained by setting the

denominator of Eq. (3.6) to zero

02)( 22 nnsss (3.7)

The system is stable (Bounded output for bounded input) if the roots of the characteristic equation is

located on the left half of s-plane, and marginally stable (Oscillation for a bounded input) if the

characteristic equation has simple roots on the imaginary axis with all other roots on the left half of s-

plane. For an unstable (Unbounded output for any bounded input) system, the characteristic equation

has at least one root on the right half of the s-plane or it has a repeated j roots.

Figure 3.3: A prototype of a second-order control system.

For a unit-step input, R(s) = 1/s, the output response is given as

)2()(

22

2

nn

n

ssssY

(3.8)

By taking inverse Laplace transform, we obtain the unit step response of the control system

0 cos1sin1

1)( 12

2

tte

ty n

tn

(3.9)

64


UiTM Shah Alam

Fig. 3.4 shows the unit-step response of the second-order system for various values of . It may be

noted that the response becomes more oscillatory with larger overshoot as decreases.

Figure 3.4: Unit-step response of a second-order system with various ζ values.

3.2.2 Damping Ratio and Damping Factor

The effects of the system parameters ζ and n on the step response y(t) can be studied by referring to

the roots of the characteristic equation in Eq. (4.7). The roots can be expressed as

j

jss nn

1, 2

21 (3.10)

where

= ζn (3.11)

and

65


UiTM Shah Alam

21 n (3.12)

The physical significance of ζ and is now investigated. As seen from Eq. (4.9), the factor = n

appears as a constant multiplied by t in the exponential term of the response y(t). Therefore, controls

the rate of rise or decay of the unit-step response y(t). In other words, controls the "damping" of the

system and is called the damping factor.

The inverse of , 1/ is proportional to the time constant of the system. When = 1, the oscillations

disappear and the system is said to be critically damped. Under this condition, = n. Thus, we can

regard as

dampingcriticaltheatfactordamping

factordampingactual

n

ratio, Damping (3.13)

When < 1, the system is under-damped and when > 1, the system is over-damped.

3.2.3 Natural Undamped Frequency

The parameter n is defined as the natural undamped frequency. As seen from Eq. (3.10), when = 0,

the roots of the characteristic equation are imaginary. Thus, the unit-step response of the system

becomes purely oscillatory with angular frequency of n. For 0 < < 1, the imaginary parts of the roots

have the magnitude of the actual (damped) frequency of oscillation. Thus

21 n

Fig. 3.5 illustrates the relationships between the location of the roots of the characteristic equation and

, ζ, and n.

66


UiTM Shah Alam

Figure 3.5: The relationships between the location of the roots of the characteristic equation and , ζ,

and n.

The effect of the roots of the characteristic equation on the damping of the second-order system is

illustrated in Fig. 3.6.

67


UiTM Shah Alam

Figure 3.6: Step-response comparison for various locations of the roots of the characteristic equation in

the s-plane.

68


UiTM Shah Alam

3.2.4 Analytical Expression for Maximum Overshoot

By taking the derivative of Eq. (3.9) with respect to time t and setting the result to zero, we get

tedt

tdyn

tn n .1 .sin1

)( 2

2

(3.14)

,...2,1,01 2 nntn

From which we get

,...2,1,0 1 2

nn

t

n

(3.15)

For the unit-step responses shown in Fig. 3.4, the first overshoot is the maximum overshoot. This

corresponds to n = 1 in Eq. (4.15). Thus, the time at which the maximum overshoot occurs is

2max

1

n

t (3.16)

With reference to Fig. 3.4, the overshoots occur at odd values of n, that is, n =1, 3, 5, …, and

undershoots occur at even values of n.

The magnitude of the overshoots and undershoots can be determined by subistituting Eq. (3.14) into Eq.

(3.9). This results in y(t)max or min . Therefore

21/

max 1overshoot maximum

ey (3.17)

and the percentage of maximum overshoot is

2-1/-100eovershoot maximum of percentage

(3.18)

The relationship between the percent maximum overshoot and the damping ratio, as given in Eq. (3.18),

is plotted in Fig. 3.7.

69


UiTM Shah Alam

Figure 3.7: The relationship between the percent maximum overshoot and the damping ratio.

3.2.5 Delay Time and Rise Time

It is more difficult to determine the exact analytical expressions of the delay time td, rise time tr, and

settling time ts. However, we can utilize the linear approximation

1.00 7.01

n

dt (3.19)

The plot of ntr versus ζ is shown in Fig. 3.8. This relation can be approximated by a straight line over a

limited range of ζ.

10 16.260.0

n

rt (3.20)

70


UiTM Shah Alam

Figure 3.8: Normalized rise time versus ζ for the prototype second-order system.

From this discussion, the following conclusions can be made:

1. tr and td are proportional to ζ and inversely proportional to n.

2. Increasing (decreasing) the natural undamped frequency n will reduce (increase) tr and td.

The settling time ts can be approximated as

n

st

3 (3.21)

We can summarize the relationships between ts and the system parameters as follows:

1. For ζ < 0.69, the settling time is inversely proportional to ζ and n. A practical way of reducing

the settling time is to increase n while holding ζ constant.

2. For ζ > 0.69, the settling time is proportional to ζ and inversely proportional to n. Again, ts can

be reduced by increasing n.

71


UiTM Shah Alam

3.3 STABILITY & PERFORMANCE SPECIFICATIONS – ROUTH-HURWITZ STABILITY TEST

The discussions in the preceding sections lead to the conclusion that the stability of a linear time-

invariant system can be determined by checking on the location of the roots of the characteristic

equation. When the system parameters are all known, the roots of the characteristic equation can be

solved by means of a root-finding computer program.

For design purposes, there will be unknown or variable parameter embedded in the characteristic

equation, and it will be feasible to use the root-finding programs. The method outlined below is well

known for the determination of stability of a LTI system without involving root solving.

3.3.1 Routh-Hurwitz Criterion

The Routh-Hurwitz criterion represents a method of determining the location of zeros of a polynomial

with constant real coefficients with respect to the left and right half of the s-plane, without actually

solving for the zeros.

Consider that the characteristic equation of a linear time-invariant SISO system is of the form

0)( 01

1

1

asasasasF n

n

n

n (3.25)

where all the coefficients are real. In order that Eq. (3.25) does not have roots in the right half of s-

plane, it is necessary and insufficient that the following conditions hold:

1. All the coefficients of the equation have the same sign

2. None of the coefficients vanishes

However, these conditions are not sufficient, for it is quite possible that an equation with all its

coefficients nonzero and of the same sign still will not have all the roots in the left half of the s-plane.

The first step in the Routh-Hurwitz criterion is to arrange the coefficients of the Eq. (3.25) as follows:

72


UiTM Shah Alam

531

42

1

nnn

nnn

n

n

aaa

aaa

s

s

Further rows of the schedule are then completed as follows:

1

531

531

531

42

0

3

2

1

n

nnn

nnn

nnn

nnn

n

n

n

n

h

ccc

bbb

aaa

aaa

s

s

s

s

s

where

31

31

1

1

51

4

1

3

31

2

1

1

1

1

1

nn

nn

n

n

nn

nn

n

n

nn

nn

n

n

bb

aa

bc

aa

aa

ab

aa

aa

ab

and so on.

Once the Routh's tabulation has been completed, we investigate the signs of the coefficients in the first

column of the tabulation. The roots of the equation are all in the left half of the s-plane if all the

elements of the first column of the Routh's tabulation are of the same sign. The number of changes of

signs in the elements of the first column equal the number of roots with positive real parts or in the right-

half s-plane.

73


UiTM Shah Alam

Example 3.1: Consider the equation

064312 23 ssssss

This equation has one negative coefficient. Thus, we know without applying Routh's test that not all the

roots of the equation are in the left-half s-plane. In fact, from the factored form of the equation, we

know that there are two roots in the right-half s-plane, at s = 2 and s = 3. For the purpose of illustrating,

the Routh's tabulation is made as follows:

06

05.2

64

11

0

1

2

3

s

s

s

s

Since there are two sign changes in the first column of the tabulation, the equation has two roots

located in the right-half s-plane.

Example 3.2: Consider the equation

010532 234 ssss

Since this equation has no missing terms and the coefficients are all of the same sign, it satisfies the

necessary conditions for not having roots in the right half or on the imaginary axis of the s-plane.

However, since these conditions are necessary but not sufficient, we have to check the Routh's

tabulation.

0010

0043.6

0107

051

1032

0

1

2

3

4

s

s

s

s

s

Since there are two changes in the first column of the tabulation, the equation has two roots in the right

half of the s-plane.

74


UiTM Shah Alam

Special Cases When Routh's Tabulation Terminates Prematurely

Depending on the coefficients of the equation, the following difficulties may occur that prevent the

Routh's tabulation from completing properly:

1. The first element in any one row of Routh's tabulation is zero, but the others are not.

2. The elements in one row of Routh's tabulation are all zero.

In the first case, we replace the zero element in the first column by an arbitrary small positive number ,

and then proceed with Routh's tabulation. This is illustrated by the following example:

Example 3.3: Consider the characteristic equation of a linear system:

0322 234 ssss

Since all the coefficients are nonzero and of the same sign, we need to apply the Routh-Hurwitz

criterion. Routh's tabulation is carried out as follows:

30

021

321

2

3

4

s

s

s

Since the first element of the s2 row is zero, the element in the s1 row would all be infinite. To overcome

this difficulty, we replace the zero in the s2 row by a small positive number and then proceed with the

tabulation.

03

03

3

0

1

2

s

s

s

Since there are two sign changes in the first column of Routh's tabulation, the equation has two roots in

the right-half s-plane.

75


UiTM Shah Alam

In the second special case, when all the elements in one row of Routh's tabulation are zeros before the

tabulation is properly terminated, it indicates that one or more of the following conditions may exist.

1. The equation has at least one pair of real roots with equal magnitude but opposite signs.

2. The equation has one or more pairs of imaginary roots.

3. The equation has pairs of complex-conjugate roots forming symmetry about the origin of the s-

plane (e.g. s = -1 j1, s = 1 j1).

The situation with the entire row of zeros can be remedied by using the auxiliary equation A(s) = 0,

which is formed from the coefficients of the row just above the row of zeros in Routh's tabulation. The

roots of the auxiliary equation also satisfy the original equation. To continue with Routh's tabulation

when a row of zeros appears, we conduct the following steps:

1. Form the auxiliary equation A(s) = 0 by use of the coefficients from the row just preceding the

row of zeros.

2. Take the derivative of the auxiliary equation with respect to s; this gives dA(s)/ds = 0.

3. Replace the row of zeros with the coefficients of dA(s)/ds = 0.

4. Continue with Routh's tabulation in the usual manner.

Example 3.4: Consider the following characteristic equation of a linear control system:

047884 2345 sssss

The Routh's tabulation is

00

44

066

484

781

1

2

3

4

5

s

s

s

s

s

76


UiTM Shah Alam

A(s) = 4s2 + 4 = 0

The derivative of A(s) with respect to s is

dA(s)/ds = 8s = 0

From which the remaining portion of the Routh's tabulation is

4

080

1

s

s

Since there are no sign changes in the first column, the system is stable. Solving the auxiliary equation

A(s) = 0, we get the two roots at s = j and s = -j, which are also two of the roots of the characteristic

equation. Thus, the equation has two roots on the j-axis, and the system is marginally stable. These

imaginary roots caused the tabulation to have an entire row of zeros in the s1 row.

Example 3.5: Consider that a third-order control system has the characteristic equation

0105.11012043.3408 7323 ksss

Determine the crucial value of k for stability.

70

371

72

33

105.1

03408

1012043408105.1105.13.3408

1012041

s

ks

ks

s

For the system to be stable, all the coefficients in the first column must have the same sign. This lead to

the following conditions:

03408

1036.410105.1 77

k

Therefore, the condition of k for the system to be stable is

57.2730 k

77


UiTM Shah Alam

If we let k = 273.57, the characteristic equation will have two roots on the j-axis. To find these roots,

we substitute k = 273.57 in the auxiliary equation, as follows:

0101036.43.3408)( 92 ssA

which has roots at s = j1097.27 and s = -j1097.27.

Thus if the system operate with k = 273.57, the system response will be an undamped sinusoid with a

frequency of 1097.27 rad/sec.

3.4 STEADY STATE RESPONSE – STEADY STATE ERROR

One of the objectives of most control systems is that the system output response follows a specific

reference signal accurately in the steady state. Steady-state error is the difference between the output

and the reference in the steady state. Steady-state errors in control systems are almost unavoidable and

generally derive from the imperfections, frictions, and the natural composition of the system. In the

design problem, one of the objectives is to keep the steady-state error below a certain tolerable value.

3.4.1 Definition of the Steady-State Error with respect to System Configuration

Let us refer to the closed-loop system shown in Fig. 3.11, where r(t) is the input, e(t) the actuating signal,

and y(t) is the output. The error of the system may be defined as:

Figure 3.11: Closed-Loop Control System.

78


UiTM Shah Alam

)(signal reference)( tyte (3.26)

where the reference signal is the signal that the output is to track. When the system has unity feedback

(i.e. H(s) = 1), the error is simply

)()()( tytrte

The steady-state error is defined as

)(1

)(lim

)(lim)(lim

0s

0

sG

ssR

ssEteest

ss

(3.27)

Clearly, ess depends on the characteristics of G(s). More specifically, ess depends on the number of poles

that G(s) has at s = 0. This number is known as the system type. Fig. 3.12 shows steady state errors for

different input functions.

79


UiTM Shah Alam

Figure 3.12: Steady-state errors (a) step input, (b) ramp input

Now let us investigate the effects of the types of inputs on the steady-state error.

80


UiTM Shah Alam

3.4.2 Steady-State Error of System with a Step-Input Function

When the input r(t) to a control system with unity-feedback is a step function with magnitude A, then

R(s) = A/s and the steady-state error is written from Eq. (4.27),

)(lim1)(1lim

)(1

)(lim

0

00 sG

A

sG

A

sG

ssRe

s

ssss

(3.28)

For convenience, we define

)(lim0

sGks

p

as the step-error constant. Then Eq. (4.28) becomes

p

ssk

Ae

1 (3.29)

We can summarize the steady-state error due to a step-function input as follows:

Type 0 system: p

ssk

Ae

1= constant

Type 1 or higher system: ess = 0

3.4.3 Steady-State Error of System with a Ramp-Input Function

When the input to the unity-feedback control system is a ramp function with amplitude A,

)()( tAtutr s

where A is a real constant, the Laplace transform of r(t) is

2)(

s

AsR

The steady-state error is written using Eq. (4.27) as follows:

81


UiTM Shah Alam

)(lim)(lim

0

0 ssG

A

ssGs

Ae

s

sss

(3.30)

We define the ramp-error constant as

)(lim0

ssGks

v

Then Eq. (3.30) becomes

v

ssk

Ae (3.31)

The following conclusions may be stated with regard to the steady-state error of a system with ramp

input:

Type 0 system: ess =

Type 1 system: ess = A/kv = constant


3.4.4 Steady-State Error of System with a Parabolic Input

When the input is described by the standard parabolic form

)(2

)(2

tuAt

tr s

The Laplace transform of r(t) is

3)(

s

AsR

The steady-state error of the system is

)(lim 2

0sGs

Ae

s

ss

(3.32)

82


UiTM Shah Alam

Defining the parabolic-error constant as

)(lim 2

0sGsk

sa

(3.33)

the steady-state error becomes

a

ssk

Ae (3.34)

The following conclusions are made with regard to the steady-state error of a system with parabolic

input:



Type 2 system: ess = A/ka = constant


Example 3.5: Find the steady state errors of the following system

1 )5.0)(5.1(

)15.3()( H(s)

sss

sksG

It is clear that this system is a type 1 system. The steady-state errors are:

Step input Step-error constant, kp = ess = A/1+kp = 0

Ramp input Ramp-error constant, kv = 4.2k ess= A/kv = A/(4.2k)

Parabolic input Parabolic-error constant, ka = 0 ess = A/ka =

3.4.5 Steady-State Error for Non-unity Feedback System

For non-unity feedback control, we usually find the equivalent unity-feedback system, as shown in Fig.

3.13.

83


UiTM Shah Alam

Figure 3.13: Forming an equivalent unity feedback for nonunity feedback system.

We have to take into consideration, that the above steps require that input and output of the same

units. The following example summarizes the concepts of steady-state error, system type, and the

steady state errors.

Example 3.6: For the system shown in Fig. 4.14, find the system type and the steady state error for the

unit step function. Assume input and output units are the same.

84


UiTM Shah Alam

Figure 3.14: Nonunity feedback control system for Example 3.6.

The first step in solving the problem is to convert the system of Fig. 3.14 into an equivalent unity

feedback system. Using the equivalent forward transfer function of Fig. 3.13(e) along with

)10(

100)(

sssG

and

5

1)(

ssH

we find

4005015

)5(100

)()()(1

)()(

23

sss

s

sGsHsG

sGsGe

Thus, the system is type 0, and

4

5

400

5100)(lim

0

sGk e

sp

The steady-state error is

41

1

p

ssk

e

85


UiTM Shah Alam

3.5 FREQUENCY RESPONSE ANALYSIS

In practice, the performance of a control system is measured more realistically by its time-domain

characteristics. The reason is that the performance of most control systems is judged based on the time

response due to certain test signals.

In design problems, there are no unified methods of arriving at a designated system that meets time-

domain performance specifications. On the other hand, in frequency domain, a wealth of graphical and

other techniques are available that are useful for system analysis and design, irrespective of the order of

the system.

It is important to realize that there are correlating relations between the frequency- and time-domain

performances in linear system so that time-domain properties of the system can be predicted based on

the frequency–domain characteristics. With these in mind, we shall study the frequency response

analysis of control systems.

3.5.1 Frequency Response of a System

It is well known from linear system theory that, when the input to a linear time invariant system is

sinusoidal with amplitude R and frequency o, i.e.,

tωRr(t) osin

the steady-state output of the system, y(t), will be a sinusoid with the same frequency o, but possibly

with different amplitude and phase; i.e.

φ)t(ωYy(t) o sin

where Y is the amplitude of the output sine wave and is the phase shift.

Let the transfer function of a SISO system be M(s); the output Y(s) and the input R(s) are related through

M(s)R(s)Y(s)

86


UiTM Shah Alam

For sinusoidal steady-state analysis, we replace s by j, and equation (6.3) becomes

)) jRjΜ)Y(j

By writing Y(j) and M(j) as (similar expression for )jR also):

)()()( jYjYjY

)()()( jMjMjM

)()()( jRjMjY

and the phase relation:

)()()( jRjMjY

Thus, for the input and output signals described by equations (6.1) and (6.2),

RjMY o )(

)( ojM

Thus, by knowing the transfer function M(s), the frequency response of the system can be obtained.

The frequency response of the loop transfer function G(s)H(s) [G(s) if H(s) is unity] can be plotted in

several ways. The two commonly used representations are:

a. Bode diagram, or Logarithmic plot.

b. Polar plot, or Nyquist plot.

87


UiTM Shah Alam

3.5.2 Frequency Response – Bode Diagram

A Bode diagram consists of two graphs. One is a plot of the logarithm of the magnitude of a sinusoidal

transfer function; the other is a plot of the phase angle; both are plotted against the frequency on a

logarithmic scale.

The standard representation of the logarithmic magnitude of G(j) is 20 log |G(j)|, where the base of

the logarithm is 10. The unit used in this representation is the decibel (dB). The curves are drawn on a

semilog paper, using the log scale for frequency and linear scale for either magnitude (in dB) or phase

angle (degrees).

The main advantage of Bode diagrams is that the multiplication of magnitudes can be converted into

addition. Furthermore, a simple asymptotic method is available for sketching the approximate curve.

Should the exact curve be desired, corrections could be made easily to these basic asymptotic plots.

In Bode diagrams, the frequency ratios are expressed in terms of octaves or decades. An octave is a

frequency band from 1 to 21, where 1 is any frequency. A decade is a frequency band from 1 to

101, where 1 is any frequency.

Basic Factors of G(j)H(j):

The basic factors that very frequently occur in an arbitrary open-loop transfer function G(j)H(j) are:

a. Constant gain, K.

b. Zeros and poles at the origin, nj

.

c. Simple zeros and poles, 11

Tj .

d. Quadratic factors, 12//21

nn jj .

a. Real Constant: G(s)H(s) = K

G(jω)H(jω) = K

Magnitude: |G(jω)H(jω)| (dB) = 20 log10 |K| (dB).

Phase angle: G(jω)H(jω) = 0°.

88


UiTM Shah Alam

The Bode plot for any value of K is shown in Fig. 6.1.

Figure 6.1: Bode plot for gain K.

b. Poles and zeros at the origin: G(s)H(s) = s±n

For sn:

G(jω)H(jω)=(jω)n

Magnitude: |G(jω)H(jω)| (dB) = 20n log10 |jω| (dB) = 20n log10 ω (dB) (6.11)

Phase angle: G(jω)H(jω) = 90n° (a constant).

For s-n:

G(jω)H(jω)=(jω)-n

Magnitude: |G(jω)H(jω)| (dB) = −20n log10 |jω| (dB) = −20n log10 ω (dB) (6.12)

Phase angle: G(jω)H(jω) = −90n° (a constant).

The Bode magnitude plots are a straight line in semi log coordinate. The slope of the line is ±20n

dB/decade i.e. the magnitude change by ±20n dB for the frequency change of 10 times. The straight line

passes through 0 dB at = 1. The phase angle () of ±j is constant and equal to ±900.The Bode plots

are shown in Fig. 6.2.

89


UiTM Shah Alam

Figure 6.2: Bode diagrams for (a)G(j) = 1/j (b)G(j) = j.

c. Simple zeros and poles: G(s)H(s) = (1+sT)±1

G(jω)H(jω)= (1+ jωT)±1

Magnitude:

|G(jω)H(jω)| (dB) = ±20 log10 |1 + jωT| (dB)

= ±20 log10 √*1 + ω2T2] (dB) (6.11)

90


UiTM Shah Alam

To obtain asymptotic approximation we consider both very large and very small values of . For low

frequencies, such that T << 1, the log magnitude may be approximated by

dBT 01log 201log 20 22

For high frequencies, such that T >> 1,

dBTT log 201log 20 22

At =1/T, log magnitude = 0 dB while at =10/T, log magnitude = ±20 dB. Thus, the value of

Tlog20 increases/decreases with 20 dB/decade. Hence, the magnitude plot can be approximated

by two straight-line asymptotes, one a straight line at 0 dB for the frequency range 0 < < 1/T and the

other a straight line with slope ±20 dB/decade for the frequency range 1/T < < . The frequency,

=1/T, at which the two asymptotes meet is called the corner frequency or break frequency.

Phase angle: G(jω)H(jω) = T1tan .

At corner frequency, G(jω)H(jω) = ±45. The phase plot can be approximated by a straight line passing

through 0 at one decade below corner frequency and ±90 at one decade above corner frequency. The

Bode plots are shown in Fig. 6.3 and Fig. 6.4.

An advantage of the Bode diagram is that for reciprocal factors, for example the factor 1/(1+jT), the

log-magnitude and phase angle curves need only be changed in sign, since

TjTj

1log201

1log20 and phase angle of 1/(1+jT) = T1tan = (phase angle of

(1+jT).

91


UiTM Shah Alam

Figure 6.3: Bode plot for (1+jT).

92


UiTM Shah Alam

Figure 6.4: Log-magnitude curve (with asymptotes) for 1/(1+jT).

d. Quadratic factors: 12//21)()(

nn sssHsG

12//21)()(

nn jjjHjG

Magnitude:

|G(jω)H(jω)| (dB) = ±20 log10 |1 + 2ζ(jω/ωn)+ (jω/ωn)2| (dB)

= ±20 log10

22

2

2

21

nn

(dB) (6.12)

93


UiTM Shah Alam

If ζ > 1, this quadratic factor can be expressed as a product of two first-order factors with real

zeros/poles. If 0 < ζ < 1, this quadratic factor is the product of two complex-conjugate zeros/poles. The

asymptotic frequency response curves can be obtained as follows.

For low frequencies such that /n << 1, the log magnitude becomes ±20 log 1 = 0 dB. The low

frequency asymptote is thus a horizontal line at 0 dB. For high frequencies such that /n >> 1, the log-

magnitude becomes nn

log40log20

2

2

dB. The equation for the high frequency asymptote is a

straight line with a slope of ±40 dB/decade.

The frequency n is the corner frequency. The two asymptotes just derived are independent of the value

of ζ. Fig. 6.5 shows exact curves with the straight-line asymptotes and the exact phase angle curves.

Phase angle of the quadratic factor is:

2

1

1

2

tan

n

n

(6.13)

94


UiTM Shah Alam

Figure 6.5: Bode plot for Eqn. (6.12) and (6.13).

95


UiTM Shah Alam

Example 6.1: Sketch the Bode plot for the following function: 2

1)(

ssG

Solution: 2

1)(

ssG ,

21

1

2

1

2

1)(

jjjG

Magnitude: 2

120log-2

1log20)(log20

2

jG

2tan)( 1 jG

Figure 6.6: Bode plots for the system in Example 6.1.

Frequency (rad/sec)

Phase (

deg);

Magnitude (

dB

)

-25

-20

-15

-10

-5

10-1

100

101

-80

-60

-40

-20

0

To: Y

(1)

96


UiTM Shah Alam

General Procedure for Plotting the Bode Diagrams:

Rewrite the sinusoidal transfer function as a product of the basic factors discussed above.

Identify the corner frequencies associated with these basic factors.

Draw the asymptotic log-magnitude curves with proper slopes between the corner frequencies

considering all the basic factors together. The exact curve, which lies very close to the

asymptotic curve, can be obtained by adding contributions from all the factors and proper

corrections.

Phase-angle curve can be drawn by adding the phase-angle curves of individual factors.

3.5.3 Polar Plot (Nyquist Plot)

The polar plot of a sinusoidal transfer function G(j) is a plot of the magnitude of G(j) versus the phase

angle of G(j) on polar coordinates as is varied from zero to infinity. Note that, in polar plots, a

positive (negative) phase angle is measured counterclockwise (clockwise) from the positive real axis. The

polar plot is very often called the Nyquist plot in control system engineering. An example of such a plot

is shown in Fig. 6.7. Each point on the polar plot of G(j) represents the terminal point of a vector at a

particular value of . In the polar plot, it is important to show the frequency graduation of the locus.

97


UiTM Shah Alam

Example of a Polar Plot.

a. Poles and zeros at the origin: G(s)H(s) = s±n

1)()(

jjHjG

The polar plot of G(j)H(j) = 1/j is the negative imaginary axis since

09011

)(

j

jjHjG (6.14)

The polar plot of G(j)H(j) = j is the positive imaginary axis.

98


UiTM Shah Alam

b. Simple zeros and poles: G(s)H(s) = (1+sT)±1

11)()(

TjjHjG

For the sinusoidal transfer function

TTTj

jHjG

1

22tan

1

1

1

1)(

(6.15)

the values of G(j)H(j) at = 0 and at = 1/T are, respectively,

001)0(0 jHjG and 0452

111

TjH

TjG

If approaches infinity, the magnitude approaches 0 and the phase angle approaches –900. The polar

plot of this transfer function is a semicircle as the frequency is varied from 0 to . It is shown in Fig. 6.8.

The center is located at 0.5 in the real axis and the radius is equal to 0.5. The lower semicircle

corresponds to 0 , and the upper semicircle corresponds to 0 .

Figure 6.8: (a) Polar plot of 1/(1+jT) ; (b)Same plot in X-Y plane.

99


UiTM Shah Alam

The polar plot of the transfer function 1+jT is simply the upper half of the straight line passing through

the point (1, 0) in the complex plane and parallel to the imaginary axis as shown in Fig. 6.7.

Figure 6.7: Polar plot of 1+jT.

c. Quadratic factors: 12//21)()(

nn sssHsG

12//21)()(

nn jjjHjG

The low and high frequency portions of the polar plot of the following transfer function

2//21

1)(

nn jjjHjG

are given, respectively, by

0

001)(lim

jHjG and 01800)(lim

jHjG

Thus, the high frequency portion is tangent to the negative real axis. The polar plots are shown in Fig.

6.8.

*Phase angle of the quadratic factor is the same as Eqn. (6.13):

100


UiTM Shah Alam

2

1

1

2

tan

n

n

Figure 6.8: Polar plots of 2//21

1

nn jj for ζ > 0.

Next, consider the following transfer function:

nn

nn

j

jjjHjG

21

//21)(

2

2

2

The low-frequency portion of the curve is:

0

001)(lim

jHjG

and the high-frequency portion of the curve is:

101


UiTM Shah Alam

0180)(lim

jHjG

The general shape of the polar plot is shown in Fig. 6.9.

Figure 6.9: Polar plot of 2//21 nn jj for ζ > 0.

Example 6.2: Draw polar plot of 2

1)(

ssG

Solution: First substitute s = j in G(s).

2

1)(

jjG

2tan)(

4

1)( 1

2

jGjG

102


UiTM Shah Alam

Figure 6.10: Polar plot of G(s) in Example 6.2.

Example 6.3: Draw polar plot for the system with )2)(1(

10)(

ssssG

Solution: )2)(1(

10)(

jjjjG

)2()1(

10)(

22

jG

11 tan2

tan90)( jG

103


UiTM Shah Alam

The plot is shown in Fig. 6.11.

Figure 6.11: Polar plot for the system in Example 6.3.

NYQUIST STABILITY TEST – THE CAUCHY CRITERION

The Cauchy criterion (from complex analysis) states that when taking a closed contour in the complex

plane, and mapping it through a complex function G(s), the number of times, N, that the plot of G(s)

encircles the origin is equal to the number of zeros, Z, of G(s) enclosed by the frequency contour minus

the number of poles, P, of G(s) enclosed by the frequency contour.

N = Z – P

Encirclements of the origin are counted as positive if they are in the same direction as the original closed

contour or negative if they are in the opposite direction.

When studying feedback control, we are not as interested in G(s)H(s) as in the closed-loop transfer

function G(s)H(s)/[1+G(s)H(s)]

If 1+G(s)H(s) encircles the origin, then G(s)H(s) will enclose the point -1. Since we are interested in the

closed-loop stability, we want to know if there are any closed-loop poles (zeros of 1+G(s)H(s)) in the

right-half plane.

104


UiTM Shah Alam

The Nyquist Stability Criterion usually written as Z = P + N, where

> Z is the number of right hand plane poles for the closed loop system (or zeros of

1+G(s)H(s))

> P is the number of open-loop poles (in the RH side of the s-plane) of G(s)H(s) (or poles of

1+G(s)H(s)), and

> N is the number of clockwise encirclements of (-1,0)

“A feedback control system is stable if and only if the number of counter-clockwise encirclements of the

critical point (-1,0) by the GH polar plot is equal to the number of poles of GH with positive real parts.”

(Nyquist Stability Criterion Definition)

Example:

• Consider the unity feedback applied to the following system

G(s)=K/[s(s+3)(s+5)]

• The loop transfer function is

G(j)H(j)=K/[s(s+3)(s+5)]|K=1,s= j

• The number of open-loop poles in the RH side of the s-plane, P = __

105


UiTM Shah Alam

• For stability Z = 0, therefore N must also be __

• From Nyquist diagram it can be seen that K can be increased by ____________ before the

Nyquist diagram encircles -1.

• For marginal stability, K = _____

RELATIVE STABILITY – GAIN AND PHASE MARGIN

• K is a variable (constant) gain

• G(s) is the plant under consideration

Gain margin is defined as the change in open loop gain required to make the system unstable. Systems

with greater gain margins can withstand greater changes in system parameters before becoming

unstable in closed loop. Phase margin is defined as the change in open loop phase shift required to

make a closed loop system unstable.

106


UiTM Shah Alam

Stability Analysis with Bode Plot

Bode plot is a very useful graphical tool for the analysis and design of linear control systems.

Advantages of the Bode Plot over Nyquist plot:

1. Gain crossover, phase crossover, gain margin, and phase margin are more easily determined on the

Bode plot.

2. For design purposes, the effects of adding controllers and their parameters are more easily visualized

on the Bode plot.

Identifying Marginal Values from Bode Plot

The gain margin is the difference between the magnitude curve and 0dB at the point corresponding to

the frequency that gives us a phase of −180° (the phase cross over frequency, ωp).

The phase margin is the difference in phase between the phase curve and −180° at the point

corresponding to the frequency that gives us a gain of 0dB (the gain cross over frequency, ωg).

107


UiTM Shah Alam

Figure 6.12: Determination of GM and PM from the Bode plot.

Example 6.4: Consider the loop transfer function given as

)50)(5(

2500)(

ssssL

From the provided Bode diagram, find the GM and PM and corresponding frequencies.

108


UiTM Shah Alam

Solution:

Figure 6.13: Bode diagram for Example 6.4.

Fig. 6.14 illustrates PM and GM of a stable and unstable system in Bode diagrams.

109


UiTM Shah Alam

Figure 6.14: Phase and Gain margins for stable and unstable systems.

Stability Analysis with Nyquist Plot

OPEN LOOP CLOSED LOOP

Gain Margin, GM

The change in open-loop gain, expressed in dB, required at -180° to make the closed-loop system unstable. A good range is 2<GM<5 (equivalent to 6dB < GM < 14dB).

Phase Margin, M

The change in the open-loop phase shift required at unity gain to make the closed-loop unstable. Good range 30 < PM < 60 degrees.

Gain margin and phase margin

quantitative measures of stability.

systems with large gain and phase margins

can withstand greater changes in system

parameter before becoming unstable.

related to root locus, in that systems with

poles farther from the imaginary axis have

a greater degree of stability.

110


UiTM Shah Alam

A gain of a will move the system response to the critical point

If a phase shift of degrees occurs, then the system will become unstable

111


UiTM Shah Alam

CHAPTER 4.0 PID CONTROLLER

There are three broad categories of PID tuning techniques:

i) Feature-based techniques

ii) Techniques that require an analytical model

iii) Optimisation (minimisation of an error criterion)

But before looking at PID tuning, we need to look at modeling of simple process dynamics. There are

two common approaches:

transient response methods, which look at the time domain characteristics of the system

response to a step or impulse

frequency response methods, which look at the response to an impulse, white noise or one

or more sinusoids

Stick to transient response models and very simple frequency response for the moment.

112


UiTM Shah Alam

4.1 Transient Response Method of Modelling

Step response modeling probably the most common approach.

Step Response

i) Peak Overshoot

(peak - final value)/final value*100%

Measure of maximum value of response

Indication of the largest error between input and output

Increases as damping decreased

Well designed systems generally have overshoot less than 30%

ii) Rise Time

Measure of the speed of response

Time necessary for the response to rise from 10% to 90% of its final steady state

error

iii) Time Delay

Time for system to show any response

Time

Am

plitu

de

00

0.5

1

1.5

Settling Time

Peak Overshoot

Unit Step Input

Time Delay

1.05

0.950.9

0.1

Rise Time

113


UiTM Shah Alam

iv) Delay Time

Time necessary for the step response to reach some value (often 50%) of the steady

state value. Not to be confused with Dead time = Time Delay

v) Settling Time

The time taken for the step response to decrease and stay within a specified range

of the final value.

Often 1%, 2% or 5%

vi) Decay Ratio

Defined as the ratio between two consecutive maxima of the error for a step change

in the set-point

The value d=1/4, which is called quarter amplitude damping, is used traditionally

but is often too high

For a second order system given by

The decay ration is given by

2

22

2

)/(/21

1

2

)(1

)()(

nn

nn

n

ss

ss

sG

sGsCL

21/2 ed

114


UiTM Shah Alam

4.1.1 Step Response Model 1 – 2 parameter model

Model of Integrator with time delay (first order response)

Response rises linearly over time

characterised by two parameters

L - essentially the “dead time”

a - where a/L is the slope.

2 4 6 8 10 12 14 16 18 20

2

4

6

8

10

12

14

16

18

step response for1

e-2ss

aL time

am

plit

ude

G(s) = e-sLaLs

115


UiTM Shah Alam

4.1.2 Step Response Model 2 – 3 parameter model (first order)

Response rises smoothly and is stable:

characterised by three parameters, gain k, time constant T and time delay L.

Most common model for PID tuning

Problem:

Tangent to step response must be drawn at the location of the largest slope

Need some alternatives that are more robust

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

time

am

plitu

de

step response for e-2s21+s

L

T

k

G(s) = e-sLk1+sT

116


UiTM Shah Alam

4.1.3 Step Response Model 3 – Alternative 3 parameter model (first order)

Response is 63% of final value at t = T

63% =1-e-1

Still some sensitivity to high frequency noise

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

time

am

plit

ude

step response for e-2s21+s

L

T

k

G(s) = e-sLk1+sT

.63k

117


UiTM Shah Alam

4.1.4 Step Response Model 4 – Another alternative 3 parameter model (first order)

3 Parameter Model Alternative:

Effective at removing noise

Let (L+T) = A1/k

Can show that T = (A2/k)e1

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

time

am

plit

ude

L

L+T

k

G(s) = e-sLk1+sT

A1

A2

118


UiTM Shah Alam

4.1.5 Step Response Model 5 – Second order response

Second order response generally has the following features:

Oscillatory

3 parameters - need k, w, z

Time delay can be added to this model and determined as done previously in the tutorial.

time

am

plit

ude

0 2 4 6 8 10 12 14 16 18 200

0.5

1

1.5

2

2.5

3

3.5

k

Tp

d = e or =

-2

(1- )

2 1/2 1

(1 +

(2 /

log

d) )

2 1/2

G(s) = k 2

s2 + 2 s + 2

Tp = or = 2

(1- )

2 1/2

2

Tp(1- )

2 1/2

d = e / 2 e1e -e =o(1-d)1 2

119


UiTM Shah Alam

4.2 Frequency Response Modelling

Some tuning formulas are based upon the frequency response of the plant

Parameters of interest are the ultimate gain Ku and the ultimate period Tu

Find these by first closing the loop and then disabling the integral and derivative parts of the controller

(Td=0, Ti=very large), and increasing the proportional gain until the system begins to oscillate. Gain at

this point = Ku and period of oscillation = Tu

Problems with looking for ultimate gain in this way

Often time consuming in practice, requiring several trials

Process can be detrimental to plant equipment and product quality

Can easily mistake other responses for the ultimate gain

a. small amplitude “limit cycles” due to valve friction or hysteresis

b. large amplitude oscillations due to actuator saturation

Alternative method for finding ultimate gain and period:

for the closed-loop system, find controller gain that produces a 1/4 decay ratio

(overshoot of one peak is 25% of the peak before it). Let this be K25%

Ku = 2 K25%

Period of oscillation, T25%, will be approximately Tu (a little longer in practice, but close

enough)

+- e(s)

yysp G(s)K

PlantController

120


UiTM Shah Alam

4.2.1 Nyquist Plot Frequency Reponse Modelling

On a Nyquist plot:

Look at open-loop response

static gain Kp = point on plot where w=0

Ku = -1 divided by ultimate point

Tu = 2p divided by w at ultimate point

-1+

=0

ultimate pointIm G(j )

Re G(j )

121


UiTM Shah Alam

4.3 Simple Tuning Law

There are several tuning law can be implemented in order to obtain the desired response. These

tuning law are as follows:

i) Ziegler – Nichols

Step Response

Ultimate Gain method

Generalised ZN

ii) Chien, Hrones and Reswick Method

iii) Cohen Coon Method

4.3.1 Ziegler – Nichols Method

Ziegler-Nichols rule was first presented in 1942. This tuning law was developed empirically

based on large number of cases. It can be said as a standard starting point.

There are some drawbacks of using this rule mainly because it needs additional manual tuning

and not particularly robust.

Now we will employ this method for both step response and frequency response modeling.

122


UiTM Shah Alam

4.3.1.1 Z-N method for Step Response modeling

Uses 2 parameter model “a” and “L”

CONTROLLER TYPE

K Ti Td Tp

P 1/a 4L

PI 0.9/a 3L 5.7L

PID 1.2/a 2L L/2 3.4L

Tp is the estimate of the period of the closed loop system

Example:

Consider the plant G(s) = (s+1)-3

From step response

a = _____

L = _____

PID controller

a= 0.218L= 0.806

123


UiTM Shah Alam

K = ____, Ti = ____ and Td = ____

Overshoot in setpoint response is too large

124


UiTM Shah Alam

4.3.1.2 Z-N Method for Frequency Response Modelling

PID settings based upon ultimate gain, Ku, and ultimate period, Tu

Aims to achieve effective disturbance rejection, and acceptable set point following

CONTROLLER TYPE

K Ti Td Tp

P 0.5Ku Tu

PI 0.4Ku 0.8Tu 1.4Tu

PID 0.6Ku 0.5Tu 0.125Tu 0.85Tu

Example:

Consider the plant G(s) = (s+1)-3

125


UiTM Shah Alam

From the Nyquist diagram

Ultimate Gain, Ku = _____

Ultimate Period, Tu = _____ = _____

PID controller

K = ____ Ti = ____ and Td = ____

4.3.2 The Chien, Hrones and Reswick Method – Improved Z-N step response modeling

The CHN method is a modified Z-N step response rules which can gives better damped closed

loop response.

The tuning method gives you two options depending on the desired response and they can be

either:

quickest response without overshoot, or

quickest response with 20% overshoot

126


UiTM Shah Alam

CONTROLLER TYPE

K Ti Td

P 0.3/a

PI 0.6/a 4L

PID 0.95/a 2.4L 0.42L

Option for 0% overshoot

CONTROLLER TYPE

K Ti Td

P 0.7/a

PI 0.7/a 2.3L

PID 1.2/a 2L 0.42L

Option for 20% overshoot

4.3.3 Cohen-Coon Method

Based on plant model

Attempts to position dominant poles that give quarter amplitude decay ratio by employing:

Method same as the ZN rules

This minimises the SS error due to load disturbances.

For PID control, 3 poles are assigned, two complex conjugate poles and the third real pole is

positioned at the same distance from the origin as the other 2 poles.

CONTROLLER TYPE

K Ti Td

P (1/a)*[1+0.35/(1-)]

PI (0.9/a)*[1+0.92/(1-)] [(3.3-3)/(1+1.2)]*L

PD (1.24/a)*[1+0.13/(1-)] [(0.27-0.36) / (1-

0.87)]*L

PID (1.35/a)*[1+0.18/(1-)] [(2.5-2)/(1-0.39)]*L [(0.37-0.37)/(1-

0.81)]*L

Where a = K0L/T and = L / (L+T) and K0 is the Open Loop DC Gain.

sLesT

KsG

1)( 0

127


UiTM Shah Alam

4.4 PID Tuning Rule of Thumb

PARAMETER SPEED STABILITY

K increases Increases Decreases

Ti increases Decreases Increases

Td increases Does not really change

to much (Increase)

Increases

4.5 PID Tuning – A Summary

CONTROLLER TYPE

Z-N Step Z-N Nyquist

P K=1/a 0.5Ku

PI K=0.9/a Ti=3L K=0.4Ku Ti=0.8Pu

PID K=1.2/a Ti=2L Td=L/2 K=0.6ku Ti=0.5Pu Td=0.125Pu

a

L

time

am

pli

tud

e

open loop stepresponse

128


UiTM Shah Alam

4.5.1 Interpretation of ZN Ultimate Gain Approach

The ZN ultimate gain approach can be interpreted as shifting a point in the Nyquist curve. The

technique is based around finding the “ultimate point”, where the Nyquist curve intercepts the real axis.

P moves in direction of G(jw), or radially out.

I moves in direction of G(jw)/jw, or at -90 degrees to P

D moves in direction of jwG(jw), or at 90 degrees to P

time

am

plitu

de

step response with closed loop proportionalcontrol with gain Ku Pu

-1+

=0

ultimate point= -1/Ku

Im G(j )

Re G(j )

D

I

P

A point on the Nyquist curve canbe moved to an arbitrary positionusing PI, PD or PID control.

129


UiTM Shah Alam

CHAPTER 5.0 ANALYSIS OF CONTROL SYSTEM STATE SPACE

The introduction of the state space representation has been discussed earlier in chapter 2.0. Please

refer to the chapter for basic overview of state space representation.

In this chapter, we will cover InsyaAllah the extension of state space representation for example the

conversion between the transfer function and state space equation. Also in this chapter we will gonna

look at how to solve the time invariant state equation, controllability and observability.

5.1 State Space Representation Extended

The transfer function of any system can be converted to state space equation and vice versa. Consider a

transfer function given by:

𝑌(𝑠)

𝑈(𝑠)= 𝐺(𝑠)

This system may also be represented in state space as:

𝐱 = 𝐀𝐱 + 𝐁𝑢

𝑦 = 𝐂𝐱 + 𝐷𝑢

Where x is the state vector, u is the input and y is the output. The Laplace Transform of the equations:

𝑠𝐗 𝑠 − 𝐱 0 = 𝐀𝐗 𝑠 + 𝐁𝑈(𝑠)

𝑌 𝑠 = 𝐂𝐗 𝑠 + 𝐷𝑈(𝑠)

Assuming the initial conditions are zero

𝑠𝐗 𝑠 − 𝐀𝐗 𝑠 = 𝐁𝑈(𝑠)

Or

𝑆𝐈 − 𝐀 𝐗 𝑠 = 𝐁𝑈(𝑠)

By premultiplying (𝑠𝐈 − 𝐀)−1 to both sides of this equation, we obtain

𝑋 𝑠 = 𝑠𝐈 − 𝐀 −1𝐁𝑈(𝑠)

Substitute into output equation;

𝑌 𝑠 = 𝐂 𝑠𝐈 − 𝐀 −1𝐁 + 𝐷 𝑈(𝑠)

130


UiTM Shah Alam

Comparing with the above equation, we see that

𝐺 𝑠 = 𝐂(𝑠𝐈 − 𝐀)−1𝐁 + 𝐷

5.1.1 Transfer Matrix

For MIMO system that the r inputs 𝑢1 , 𝑢2 , … … . , 𝑢𝑟 and m outputs 𝑦1 , 𝑦2 , … … . , 𝑦𝑚 define as:

𝑦 =

𝑦1

𝑦2

.

.𝑦𝑚

𝑢 =

𝑢1

𝑢2

.

.𝑢𝑟

The transfer matrix G(s) relates the output Y(s) to the input U(s), or

𝐘 𝑠 = 𝐆 𝑠 𝐔(𝑠)

Since the input vector u is r dimensional and the ouput vector y is m dimensional, the transfer matrix is

an m x r matrix.

5.2 Converting State Space to transfer function

A modern complex system may have many inputs and outputs. Let say we have a state space

representations of the following:

𝑦𝑛 + 𝑎1𝑦𝑛−1 + ⋯ + 𝑎𝑛−1𝑦 + 𝑎𝑛𝑦 = 𝑏0𝑢𝑛 + 𝑏1𝑢𝑛−1 + ⋯ + 𝑏𝑛−1𝑢 + 𝑏𝑛𝑢

Controllable canonical form

𝑥1 𝑥2 ..

𝑥 𝑛−1

𝑥𝑛

=

0 1 0 … 00 0 1 … 0. . . … .. . . … .0 0 0 … 1

−𝑎𝑛 −𝑎𝑛−1 −𝑎𝑛−2 … −𝑎1

𝑥1

𝑥2

.

.𝑥𝑛−1

𝑥𝑛

+

00..01

𝑢

𝑦 = 𝑏𝑛 − 𝑎1𝑏0 𝑏𝑛−1 − 𝑎𝑛−1𝑏0 … . 𝑏1 − 𝑎1𝑏0

𝑥1

𝑥2

.

.𝑥𝑛−1

𝑥𝑛

+ 𝑏0𝑢

131


UiTM Shah Alam

Observable canonical form

𝑥1 𝑥2 ..

𝑥 𝑛−1

𝑥𝑛

=

0 0 … 0 −𝑎𝑛

1 0 … 0 −𝑎𝑛−1

. . … . −𝑎𝑛−2

. . … . .0 0 … 0 .0 0 … 1 −𝑎1

𝑥1

𝑥2

.

.𝑥𝑛−1

𝑥𝑛

+

𝑏𝑛 − 𝑎1𝑏0

𝑏𝑛−1 − 𝑎𝑛−1𝑏0

.

.

.𝑏1 − 𝑎1𝑏0

𝑢

𝑦 = 0 0 … 0 1

𝑥1

𝑥2

.

.𝑥𝑛−1

𝑥𝑛

+ 𝑏0𝑢

Diagonal Canonical Form

Consider the transfer function system defined by equation below. In this case the denominator

polynomial involves only distinct roots only.

𝑌(𝑠)

𝑈(𝑠)=

𝑏0𝑠𝑛 + 𝑏1𝑠𝑛−1 + ⋯ + 𝑏𝑛−1𝑠 + 𝑏𝑛

𝑠 + 𝑝1 𝑠 + 𝑝2 … (𝑠 + 𝑝𝑛 )

= 𝑏0 +𝑐1

𝑠+𝑝1+

𝑐2

𝑠+𝑝2+ ⋯ +

𝑐𝑛

𝑠+𝑝𝑛

𝑥 1𝑥 2..

𝑥 𝑛−1

𝑥 𝑛

=

−𝑝1 0 … 0 0

0 −𝑝2 … 0 0. . … . .. . … . .0 0 … 0 .0 0 … 0 −𝑝𝑛

𝑥1

𝑥2

.

.𝑥𝑛−1

𝑥𝑛

+

11...1

𝑢

𝑦 = 𝑐1 𝑐2 … 𝑐𝑛−1 𝑐𝑛

𝑥1

𝑥2

.

.𝑥𝑛−1

𝑥𝑛

+ 𝑏0𝑢

132


UiTM Shah Alam

Jordan Canonical Form

Consider the case where the denominator polynomial involves multiple roots.

𝑌(𝑢)

𝑈(𝑠)=

𝑏0𝑠𝑛 + 𝑏1𝑠𝑛−1 + ⋯ + 𝑏𝑛−1𝑠 + 𝑏𝑛

𝑠 + 𝑝1 3 𝑠 + 𝑝4 𝑠 + 𝑝5 … (𝑠 + 𝑝𝑛 )

The partial fraction expansion becomes

𝑌(𝑠)

𝑈(𝑠)= 𝑏0 +

𝑐1

(𝑠 + 𝑝1)3+

𝑐2

(𝑠 + 𝑝1)2+

𝑐3

(𝑠 + 𝑝1)+

𝑐4

(𝑠 + 𝑝4)+ ⋯ +

𝑐𝑛

(𝑠 + 𝑝𝑛 )

A state space representation of this system in the Jordan canonical form is given by:

𝑥 1𝑥 2𝑥 3𝑥 4...

𝑥 𝑛

=

−𝑝1 1 0 0 … 0

0 −𝑝1 1 0 … 00 0 −𝑝1 0 … 00 0 0 −𝑝4 … 0

. . .

. . .

. . .0 0 0 0 … −𝑝𝑛

𝑥1

𝑥2

𝑥3

𝑥4

.

.

.𝑥𝑛

+

0011...1

𝑢

𝑦 = 𝑐1 𝑐2 … 𝑐𝑛−1 𝑐𝑛

𝑥1

𝑥2

⋮𝑥𝑛−1

𝑥𝑛

+𝑏0𝑢

133


UiTM Shah Alam

LECTURER’S NOTE

134


UiTM Shah Alam

5.3 Controllability and Observability

5.3.1 Controllability

Controllability is a test of the ability of the actuators. A system is controllable if it is possible to transfer

any state with any set of initial conditions to any final state in some finite time period. Alternatively, a

system is only controllable if every mode (or state) is connected to the control input.

A system is referred to as “stabilizable” so long as we can state control all unstable modes. This might

mean that there are some stable uncontrollable states. Strictly speaking the dynamical system described

by the pair (A;B) is said to be (state-feedback) stabilizable if there exists a state feedback u=-Kx such that

A+BK is stable.

In order to test the controllability of a LTI system, the “Controllability Matrix” must be of full rank. The

Controllability matrix, 𝐂𝐨 = 𝐁 𝐀𝐁 𝐀𝟐𝐁… 𝐀𝐧−𝟏𝐁 i.e the controllability matrix must be invertible. Note

the difference between rank and determinant. Often in uncontrollable systems, part of the system is

unconnected from input.

Additional tests are to show that the controllability Gramian P is positive definite, where P may be found

by the solution to the Lyapunov equation: 𝐀𝐏 + 𝐏𝐀𝐓 = −𝐁𝐁𝐓.

Alternatively;

𝐏 ≡ 𝑒𝐀𝑡

∞

0

𝐁𝐁𝑇𝑒𝐀𝑇𝑡𝑑𝑡

135


UiTM Shah Alam

Example 1:

Investigate the controllability of: 𝐀 = −1 01 −2

, 𝐁 = 12

Finding the Controllability Matrix,

𝐂0= 𝐁 𝐀𝐁

𝐂0= 1 −12 −3

Look at the determinant,

𝐂𝟎 = −1 ≠ 0

The rank of this matrix is 2, hence the system is controllable.

Example 2:

Investigate the controllability of: 𝐀 = −1 00 −1

, 𝐁 = 13

Example 3:

Investigate the controllability of:

𝑌(𝑠)

𝑈(𝑠)= 𝑇 𝑠 =

1

𝑆3 + 𝑎2𝑠2 + 𝑎1𝑠 + 𝑎0

136


UiTM Shah Alam

5.3.2 Observability

Observability is a test of the ability of the sensors. A system is observable if every initial state x(0) can be

determined by observing the system output over some finite time period. A system is referred to as

“detectable” if all unstable modes are state observable. This may mean the system has unobservable

states which are stable. Strictly speaking the pair (C;A) is said to be detectable if there exists a matrix L

such that A+LC is stable.

In order to test the observability of an LTI system, the “Observability Matrix” must be of full rank. The

observability matrix, 𝐎𝐛 =

𝐂𝐂𝐀𝐂𝐀𝟐

⋮𝐂𝐀𝐧−𝟏

i.e the observability matrix must be invertible.

Additional tests are to show that the controllability Gramian Q is positive definite, where Q may be

found by the solution to the Lyapunov equation: 𝐀𝐓𝐐 + 𝐐𝐀 = −𝐂𝐓𝐂.

Alternatively;

𝐐 ≡ 𝑒𝐀𝑇𝑡

∞

0

𝐂𝑇𝐂𝑒𝐀𝑡𝑑𝑡

Example 1:

Investigate the observability of: 𝐀 = −1 01 −2

, 𝐂 = 1 0 and 𝐂 = 0 1

Finding the observability matrix,

𝐂 = 1 0

𝐂𝐀 = −1 0

𝐎𝐛 = 1 0

−1 0

137


UiTM Shah Alam

Look at the determinant,

𝐂𝟎 = 0

The rank of this matrix is 1, hence the system is non observable.

What about the other case?

Example 2:

Investigate the observability of: 𝐀 = −1 01 −1

, 𝐂 = 1 3

Example 3:

Investigate the observability of:

𝑌(𝑠)

𝑈(𝑠)= 𝑇 𝑠 =

1

𝑆3 + 𝑎2𝑠2 + 𝑎1𝑠 + 𝑎0

138


UiTM Shah Alam

5.4 Solving the time invariant State equation

The homogeneous State Equations is 𝐱 = 𝐀𝐱 which gives:

𝑥 𝑡 = 𝑒𝐀𝑡𝑥(0)

Matrix 𝑒𝐀𝑡 is a matrix exponential. This matrix is also known as the state transition matrix. It is

sometimes labeled as 𝜙 𝑡 = 𝑒𝐀𝑡 .

The state transition matrix is difficult to calculate. Hence there are two common ways of expressing it:

1. Expansion:

𝜙 𝑡 = 𝑒𝐴𝑡 = 𝐼 + 𝐴𝑡 +𝐴2𝑡2

2!+ ⋯ +

𝐴𝑘𝑡𝑘

𝑘!

2. Inversion:

𝜙 𝑡 = 𝐿−1 Φ 𝑠 = 𝐿−1 𝑠𝐈 − 𝐀 −1

The forced response (in-homogeneous solution) is given by (assuming 𝑡0 = 0):

𝒙 𝑡 = 𝑒𝐀𝑡𝐱 0 + 𝑒𝐀 𝑡−𝜏 𝐁𝐮 𝜏 𝑑𝜏

𝑡

0

The output is therefore given by:

𝐲 𝑡 = 𝐂𝐱 𝑡 + 𝐃𝐮(𝑡)

Example 1:

Obtain the state transition matrix Φ(𝑡) of the following system.

𝑥1 𝑥 2

= 0 1

−2 −3

𝑥1

𝑥2

Obtain also the inverse of the state transition matrix, Φ−1 𝑡

For this sytem

𝐴 = 0 1

−2 −3

The state transition matrix is given by

139


UiTM Shah Alam

Φ 𝑡 = 𝑒𝐀𝑡 = 𝐿−1[ 𝐬𝐼 − 𝐀 −1]

Since 𝑠𝐼 − 𝐴 = 𝑠 00 𝑠

− 0 1

−2 −3 =

𝑠 −1−2 𝑠 + 3

(sI − A)−1 =1

s + 1 (s + 2) s + 3 −1−2 s

𝑠 + 3

𝑠 + 1 (𝑠 + 2)

1

𝑠 + 1 (𝑠 + 2)−2

𝑠 + 1 (𝑠 + 2)

𝑠

𝑠 + 1 (𝑠 + 2)

Hence Φ 𝑡 = 𝑒𝐀𝑡 = 𝐋−1 𝑠𝐼 − 𝐴 −1 = 2𝑒−𝑡 − 𝑒−2𝑡 𝑒−𝑡 − 𝑒−2𝑡

−2𝑒−𝑡 + 2𝑒−2𝑡 −𝑒−𝑡 + 2𝑒−2𝑡

Noting that Φ−1 𝑡 = Φ(𝑡)

Φ−1 𝑡 = 𝑒−𝐀𝑡 = 2𝑒−𝑡 − 𝑒−2𝑡 𝑒−𝑡 − 𝑒−2𝑡

−2𝑒−𝑡 + 2𝑒−2𝑡 −𝑒−𝑡 + 2𝑒−2𝑡

Example 2: EM/APR 2008/KJM597/MEC522

a) An electro-hydraulic car suspension system can be modeled by the following state matrix

equation

𝑥 1𝑥 2

= −1

𝑘 − 1

𝑘

0 −1

𝑘

𝑥1

𝑥2 +

𝑥 1𝑥 2

𝑢

𝑦 = 𝑐1 𝑐2 𝑥1

𝑥2

Where 𝑥1 and 𝑥2 are suspension displacements, u is an electrical actuating signal and k is the

suspension stiffness.

i. Determine the condition for the system to be controllable

ii. If y is a single output displacement, (given as 𝑐1 𝑐2 where 𝑐1 and 𝑐2 are constants),

establish the conditions which must be avoided if the system is to remain observable.

b) If the values of k=2N/m, 𝑐1=1 and 𝑐2=0, determine

i. The eigenvalue of the system

ii. The state transition matrix

iii. The variation of x(t) response to a step change in u(t) at time t=0 from u=0 to u=1N for

initial conditions 𝑥1(𝑡) and 𝑥2(𝑡) equal to zero.

140


UiTM Shah Alam

CHAPTER 6.0 CONTROL SYSTEM DESIGN

We have discussed so far the importance of the closed-loop system poles on the dynamic performance

of the system. The transient-response specifications can be translated into desired locations for

dominant closed-loop poles. The roots of the characteristic equation, which are the poles of the closed-

loop system, determine the absolute and relative stability of the system. Therefore, an important study

in linear control systems is the investigation of the trajectories of the roots of the characteristic

equation, or simply, the root loci when a certain system parameter varies. The basic properties and

construction of root loci are first due to W.R. Evans (1948).

In this chapter, we will discuss the construction of root loci using simple rules. For plotting the root loci

accurately, one can always use standard computer program packages like MATLAB. The basics of root

loci should be thoroughly understood so that the engineers may be able to interpret the data provided

by root loci for system analysis and design.

6.1 Root locus technique

Consider the second-order system shown in Fig. 6.1, which represents a typical position control system.

The plant consists of a servomotor and load, driven by power amplifier with gain K. The open-loop

transfer function of the system is

)2()(

ss

KsG (6.1)

141


UiTM Shah Alam

Figure 6.1: A position control system

Figure 6.2: Root locus for Eq. (6.3)

The open-loop poles, marked in Fig. 5.2, are at s = 0 and s = -2. The closed-loop transfer function of the

system is

Kss

K

sG

sG

sR

sY

2)(1

)(

)(

)(2

(6.2)

142


UiTM Shah Alam

The characteristic equation is

02)( 2 Ksss (6.3)

This second order system is always stable for positive values of K. The relative stability of the system

depends upon the location of the closed-loop poles

Ks 112,1 (6.4)

and hence on the choice of the parameter K.

As K is varied from zero to infinity, the closed-loop poles move in the s-plane as shown in Fig. 6.2. At K =

0, the root s1 is equal to the open-loop pole at s = 0, and root s2 is equal to the open-loop pole at s = −2.

As K increases, the roots move toward each other. The two roots meet at s = −1 for K = 1. As K is

increased further, the roots breakaway from the real axis, become complex conjugate, and since the real

part of both roots remains fixed at s = −1, the roots move along the line = −1.

A root locus of a system is a plot of the roots of the system characteristic equation (poles of the closed-

loop transfer function) as some parameters of the system are varied.

The two branches A-C-E and B-C-D of the plot of Fig. 6.2 are thus two root loci of the system of Fig. 6.1.

Each root locus starts at an open-loop pole with K = 0 and terminates at infinity as K . Each root

locus gives one characteristic root (closed-loop pole) for a specific value of K.

The root locus plot gives us considerable information about the transient behavior of the system as gain

K is varied. From Fig. 6.2:

For 0 < K < 1, the roots are real and distinct and the system is overdamped.

For K = 1, the roots are real and repeated. Thus, the system is critically damped.

For K > 1, the roots are complex conjugate and the system is underdamped with the value of

decreasing as K increases.

143


UiTM Shah Alam

Thus, by choosing appropriate value of K, we can cause a characteristic root at any point on root locus.

For example, the dashed lines in Fig. 5.2 correspond to = 0.707. The points where the root loci cross

the dashed lines have been marked . These points corresponds to the closed loop poles for K = 2.

6.1.1 Basic properties of root loci

Figure 6.3: Typical control system

Consider the control system shown in Fig. 6.3. The closed-loop transfer function is

)()(1

)(

)(

)(

sHsKG

sKG

sR

sY

(6.5)

Let K be a positive quantity. The roots of the characteristics equation must satisfy the expression

1 + KG(s)H(s) = 0 (6.6)

or, G(s)H(s) = -1/K (6.7)

KG(s)

KG(S)

144


UiTM Shah Alam

Thus, any point s is a closed-loop pole or a root of the characteristic equation, if it satisfies the following

conditions (K > 0):

Magnitude condition: |G(s)H(s)| = 1/K (6.8)

Angle condition: G(s)H(s) = (2q+1) = (2q+1) 180o

where q = 0, ±1, ±2, … (6.9)

The angle condition is used to determine the trajectory of the loci in the s-plane.

Once the root loci are drawn, the values of K on the loci are determined by using the magnitude

condition.

Graphical Interpretation

Let )())((

)())(()()(

21

21

n

m

pspsps

zszszsKsHsKG

(6.10)

The magnitude condition becomes

Kps

zs

sHsGn

j

j

m

i

i1

)()(

1

1

(6.11)

The angle condition becomes

o

j

n

j

i

m

i

qpszssHsG 180)12()()()()(11

where q = 0, 1, 2, … (6.12)

145


UiTM Shah Alam

Figure 6.4: Zero (z1) and pole (p1) on a complex plane

In Fig 6.4, let us assume a complex pole and real zero: s+p1 and s+z1 represent the respective vectors in

the complex plane. A and B are magnitudes of vectors (s+z1) and (s+p1) and 2 and 1 are angles of (s+z1)

and (s+p1), respectively.

The graphical interpretation is:

The difference between the sums of the angles of the vectors drawn from the zeros and those from

the poles of G(s)H(s) to s is an odd multiple of 180.

Once the root loci are constructed, the values of K along the loci can be determined. Thus, the

construction of root loci involves:

1. A search for all the points in the s-plane.

2. Find the magnitude of K on the root loci.

146


UiTM Shah Alam

6.1.2 Properties and construction of root loci

The purpose of root locus is to show in graphical form the general trend of the roots of the characteristic

equation

(s) = 1+KG(s)H(s) = 1+F(s) = 0 (6.13)

where

nm

ps

zs

sHsGn

j

j

m

i

i

;)()(

1

1 (6.14)

as the parameter K is varied from zero to infinity. Every point s = + j in the complex plane that

satisfies the angle criterion

o

j

n

j

i

m

i

qpszssHsG 180)12()()()()(11

; q = 0, 1, 2, ….

is on the root locus. The value of the parameter K corresponding to a point on the root locus can be

obtained from the magnitude criterion

Kps

zs

sHsGn

j

j

m

i

i1

)()(

1

1

In principle, the root locus for a given F(s) can be sketched by measuring F(s) at all the points of the

complex plane and marking down those places where we find F(s) equal to an odd multiple of 1800.

However, this trial-and-error method would be a very tedious task. Therefore, certain rules have been

developed for making a quick approximate sketch of the root locus. This approximate sketch provides a

guide for the selection of trial points such that a more accurate root locus can be obtained by a few

trials. Further, the approximate root locus sketch is very useful in visualizing the effects of variation of

the parameter K, the effects of shifting of pole-zero locations and of bringing a new set of poles and

zeros.

147


UiTM Shah Alam

Rules for Construction of Root Loci

The root locus for a given F(s) is to be sketched. F(s) has m zeros at s = -zi and n poles at s = -pj (refer to

Eq. (6.14)) where m n. These m zeros and n poles of F(s) are referred to as open-loop zeros and open-

loop poles, respectively.

Rule 1: Number of Root Loci (Branches)

The root locus plot consists of n root loci (branches) as K varies from 0 to . The loci are symmetric with

respect to the real axis.

The characteristic equation can be written as:

0)()()(11

m

i

i

n

j

j zsKpss (6.15)

This equation has degree n. Thus, for each real K, there are n roots. As the roots are continuous function

of the coefficients of equation, the n roots form n continuous loci as K varies from 0 to . Since the

complex roots occur in complex conjugate pairs, the root loci must be symmetrical about the real axis.

Rule 2: Starting and Ending Points of Root Loci

As K increases from 0 to , each root locus starts from an open-loop pole with K = 0 and ends on an

open-loop zero or on with K = . The number of root loci ending at equals the number of open-loop

poles minus zeros.

Refer to Eq. (6.15). When K = 0, the equation has roots at -pj (j = 1,, n), which are open-loop poles.

Thus, the root loci start at open-loop poles.

Eq. (6.15) can be rearranged as

0)()(1

11

m

i

i

n

j

j zspsK

When K = , the equation has roots at –zi (i = 1, , m), which are open-loop zeros. Therefore, m root

loci end on the open-loop zeros.

148


UiTM Shah Alam

In case m < n, the open-loop transfer function has (n - m) zeros at infinity. From the magnitude criterion,

Kps

zs

n

j

j

m

i

i1

1

1

, we find that this is satisfied by s ej as

K . Thus, (n - m) root loci end on infinity.

Rule 3: Asymptotes to Root Loci (Behavior at Infinity)

The (n − m) root loci which tend to do so along straight line asymptotes radiating out from a single

point s= −a on the real axis (called the centroid) where

mn

zerosloopopenofpartrealpolesloopopenofpartreala

)()(

These (n − m) asymptotes have angles

)1(,,1,0;180)12( 0

mnq

mn

qa

This rule will be justified by referring to a pole-zero patterns shown in Fig. 6.5. For a point far away from

the origin, the poles and zeros can be considered to cluster at the same point, say −a, as shown in Fig.

6.5. Thus, Eq. (6.15) can be approximated as

0

)(11

1

1

mn

a

n

j

j

m

i

i

s

K

ps

zsK

(6.16)

This means that all m zeros are cancelled by poles, and only (n - m) poles are left at -a.

149


UiTM Shah Alam

Figure 6.5: Asymptotes to root loci

From Eq. (6.16),

mn

am

i

i

n

j

j

s

zs

ps

)(

1

1 (6.17)

By simplifying this, we get,

11

11

mn

a

mnmnm

i

i

n

j

j

mn smnsszps

Thus, by comparison of coefficients, we get

mn

zpn

j

m

i

ij

a

1 1

)()(

(6.18)

Moreover, for the point s0 to be on the root locus,

,1,0;180)12()( 0 qqmn

Thus, )1(,,1,0;180)12( 0

mnq

mn

qa

150


UiTM Shah Alam

The (n - m) angles given by the above equation divide 3600 equally and are symmetric with respect to

real axis. The (n - m) root loci tend to along (n - m) asymptotes radiating out from s = -a at angles a.

Example 6.1: The pole-zero map of Fig. 6.5 corresponds to

))(s)(sj)(sj(s

)K(sF(s)

434141

2

(6.19)

The root loci has four branches, each starting from an open-loop pole with K = 0. One root locus will

terminate on open-loop zero with K = . The other three loci will terminate on as K along the

asymptotes radiating out from s = -a

where 3

7

14

)2(4311

a

at angles 600, 1800, and 3000, respectively. Fig. 6.6 shows the asymptotes.

Figure 6.6: Asymptotes for Eq. (5.19)

151


UiTM Shah Alam

Rule 4: On-Locus Segments on the Real Axis

A point on the real axis lies on the locus if the number of open-loop poles plus zeros on the real axis to

the right of this point is odd.

For the system of Eq. (6.19), the open-loop pole-zeros are shown in Fig. 6.7(a). Take a point s0 on the

real axis. Join this point to all the open-loop poles and zeros. It is seen that (i) poles and zeros on the real

axis to the right of this point contribute an angle of 1800 each, (ii) poles and zeros to the left of this point

contribute angle of 00 each, and (iii) the net angle contribution of a complex conjugate pole or zero pair

is always zero.

Thus, F(s)=(mr – nr)1800 = (2q+1) 1800 , q = 0, 1, 2, …

where mr = number of open-loop zeros on the real axis to the right of s0 and nr = number of open-loop

poles on the real axis to the right of s0. Thus, the angle criterion is satisfied if (nr – mr) or (nr + mr) is odd

and hence the rule. Thus, the real axis can be divided into segments on-locus and not-on-locus; the

dividing points being the real open-loop poles and zeros. The on-locus segments of the real axis

alternate as shown in Fig. 6.7(b).

Figure 6.7: On-locus segments of the real axis

152


UiTM Shah Alam

Rule 5: On-Locus Points of the Imaginary Axis

The intersections (if any) of root loci with the imaginary axis can be determined by use of the Routh

criterion.

Segments of root loci can exist in the right half of s-plane. This signifies instability. The points at which

the root loci cross the imaginary axis define the stability limits. The Routh Table determines the gains at

the stability limit. By using this gain in the auxiliary equation, the value s = j0 at the stability limit is

computed.

Example 6.2: The characteristic equation of system in Eq. (5.19) is

02204)143(439 234 KsKsss (6.20)

The corresponding Routh Table is shown below.

s4 1 43 204+2K

s3 9 143+K

s2 (244 – K)/9 204+2K

s1 (18368 – 61K – K2)/(244 – K)

s0 204+2K

For stability, 244 – K > 0, 18368 – 61K – K2 > 0, and 204 + 2K > 0. It can be seen that these conditions are

satisfied if K < 108.4. For K = 108.4, all the coefficients in s1 row are zero. Thus, the auxiliary equation is

formed from the coefficients of s2 row and is given by

0)2204(9

244 2

KsK

For K = 108.4, the roots of the above equation lie on the j axis and are given by s = j5.28. Thus, the

root loci intersect the imaginary axis at s = j5.28 and the corresponding value of K is 108.4.

153


UiTM Shah Alam

Rule 6: Angle of Departure from Complex Poles

The angle of departure, p, of a locus from a complex open-loop pole is given by p = 1800+ where is

the net angle contribution at this pole of all other open-loop poles and zeros.

Example 6.3: For the system of Eq. (6.19), the characteristic equation is

0434141

211

))(s)(sj)(sj(s

)K(sF(s) (6.21)

The pole zero-map is shown in Fig. 6.8.

Figure 6.8: Angle of departure from complex poles

Let s0 be an arbitrary point on the root locus starting from s = -1+j4. The phase from this pole to s0 is p.

The net angle contribution of all other open-loop poles and zeros at s0 is

)( 4312

Thus, the total phase of F(s) at s0 is - p. For s0 to be on the root locus, the total phase must be 1800.

So, p = 1800 +. This is the angle of departure from the complex open-loop pole.

154


UiTM Shah Alam

If s0 is very close to the pole -1+j4, then the vectors drawn from all other poles and zeros to s0 can be

approximated by the vector drawn to the pole at -1+j4, i.e., we consider s0 to be -1+j4 for measurement

of angles 1, 2, 3, and 4. With this approximation, for this example, 1 = 900, 2 = 760, 3 = 630, and 4 =

530. So, 0

4312 130)( and p = 1800 + = 500. A rough sketch of the root locus for this

system is shown in Fig. 5.9.

Figure 6.9: Root locus plot for Eq. (6.21)

There are four open-loop poles, so there are four loci. One locus departs from real pole at –3 and ends

on the zero at –2 along the real axis. The second locus departs from real pole at –4 and moves along the

asymptote on the negative real axis. The third locus departs from the complex pole at –1+j4 with a

departure angle of p = 500 and moves toward the asymptote radiating from the centroid at –7/3 at an

angle of +600; it crosses the imaginary axis at j5.28. Using the symmetry property, the fourth locus is

obtained immediately by reflection about the real axis.

155


UiTM Shah Alam

Rule 7: Angle of Arrival at Complex Zeros

The angle of arrival, z , of a locus at a complex zero is given by z = 1800 – , where is the net angle

contribution at this zero of all other open-loop poles and zeros.

Example 6.4: Let us consider the characteristic equation

0)2(

)1(1)(1

2

ss

sKsF (6.22)

The pole-zero map of this F(s) is shown in Fig. 6.10. Open loop poles: s = 0, –2. Open-loop zeros: s = j1.

Let s0 be an arbitrary point on the root locus terminating on the zero at s = j1. Let the phase from this

zero to s0 = z. If the point s0 is very close to the zero at j1, then the vectors drawn from the other zero at

–j1 and poles at 0 and –2 to s0 can be approximated by vectors to the zero at j1. Under this

approximation, the net angle contribution at s0 is given by

= 900 – 900 – 26.50 = – 26.50.

For s0 to be on the root locus, the total phase must be 1800. Thus, z = 1800 – = 206.50. The complete

root locus plot is shown in Fig. 6.10.

Figure 6.10: Angle of arrival at complex zero

156


UiTM Shah Alam

Rule 8: Location of Multiple Roots

Points at which multiple roots of the characteristic equation occur (breakaway points of root loci) are the

solutions of

0ds

dK (6.23)

where

m

i

i

n

j

j

zs

ps

K

1

1 (6.24)

Let us assume that the characteristic equation has a multiple root at s = s0 of multiplicity r. Then,

)()()(1 0 sMsssF r , r 2 (6.25)

where M(s) does not contain the factor (s - s0). Thus, by differentiating Eq. (6.25), we have

)()()()( '

0

1

0 sMsssrMssds

dF r (5.26)

At s = s0, the RHS of Eq. (6.26) is zero. Thus, at s = s0,

0ds

dF

In pole-zero form, the characteristic equation is:

0

)(

)(11)(1

1

1

sA

sKB

ps

zsK

sFn

j

j

m

i

i

(6.27)

Thus,

0)(

)()()()(2

sA

sBsAsBsAK

ds

dF (6.28)

Therefore, the breakaway points are the roots of

0)()()()( '' sBsAsBsA (6.29)

157


UiTM Shah Alam

This equation can be equivalently represented as

0ds

dK , where

m

i

i

n

j

j

zs

ps

B(s)

A(s)K

1

1 (6.30)

Example 6.5: Consider the characteristic equation

0)(

)(1

)1(

)3)(2(1

sA

sKB

ss

ssK (6.31)

Fig. 6.11 shows the open-loop poles and zeros on the complex plane. Root loci segments exist on the

negative real axis between 0 and –1 and between –2 and –3. At K = 0, the roots are at s = 0 and s = –1.

As K increases, the two roots move away from poles at 0 and –1 toward each other inside the segment [-

1,0]. At some K, the two real roots will become repeated real roots and then break away from the real

axis into two complex conjugate roots. Such a point is called a breakaway point.


Similarly as K approaches , one root will approach zero at s = –2 along the negative real axis and

another will approach zero at s = –3. As the root loci are continuous, the two complex conjugate roots

will approach the real axis somewhere inside the segment [–3, –2] and then depart in opposite

directions along the real axis. This point is also another breakaway point. Sometimes, such a point is also

called as break-in point.

Applying Eq. (6.30) to this case, we get the solutions of dK/ds = 0 as s = –0.634 and s = –2.366. Thus, the

root locus has two breakaway points.

158


UiTM Shah Alam

It is important to note that the condition for the breakaway point (as derived above) is necessary but

not sufficient. In other words, all breakaway points on root locus must satisfy Eq. (6.30), but not all

points that satisfy Eq. (6.30) are breakaway points.

SUMMARY OF RULES FOR ROOT LOCUS PLOTTING

The characteristic equation of the system is

0;;

01)(1)()(1

1

1

Knm

ps

zsK

sFsHsKGn

j

j

m

i

i

1. The root locus plot consists of n root loci (branches) as K varies from 0 to . The loci are symmetric

with respect to real axis.

2. As K increases from 0 to , each root locus starts from an open-loop pole with K = 0 and ends on an

open-loop zero or on with K = . The number of root loci ending at equals the number of open-

loop poles minus zeros.

3. The (n - m) root loci which tend to do so along straight line asymptotes radiating out from a single

point s= -a on the real axis (called the centroid) where

mn

zerosloopopenofpartrealpolesloopopenofpartreala

)()(

These (n - m) asymptotes have angles

)1(,,1,0;180)12( 0

mnq

mn

qa

4. A point on the real axis lies on the locus if the number of open-loop poles plus zeros on the real axis

to the right of this point is odd. By use of this fact, the real axis can be divided into segments on-

locus and not-on-locus; the dividing points being the real open-loop poles and zeros.

5. The intersections (if any) of root loci with the imaginary axis can be determined by use of Routh

criterion.

6. The angle of departure p of a locus from a complex open-loop pole is given by p = 1800 + , where

is the net angle contribution at this pole of all other open-loop poles and zeros.

159


UiTM Shah Alam

7. The angle of arrival z of a locus at a complex zero is given by z = 1800 - , where is the net angle

contribution at this zero of all other open-loop poles and zeros.

8. Points at which multiple roots of the characteristic equation occur (breakaway points of root loci)

are the solutions of 0ds

dK where

m

i

i

n

j

j

zs

ps

K

1

1

6.1.3 A complete example

Question: Consider a feedback system with the characteristic equation

0;0)2)(1(

1

Ksss

K (6.32)

Plot the root locus for this system.

Solution:

The open-loop poles are located at s = 0, −1, −2. There are no finite open-loop zeros. The pole-zero

configuration is shown in Fig. 5.12.

Rule 1 tells that the root locus plot consists of three root loci as K varies from 0 to .

Rule 2 tells that the three root loci originate from the three open loop poles with K = 0 and terminate on

with K = .

Rule 3 tells that the three root loci tend to along asymptotes radiating out from

103

12

)()(

zerosofnumberpolesofnumber

zerosofpartsrealpolesofpartsreals a

with angles

000

0

0

300,180,60

2,1,0;3

180)12(

,2,1,0;180)12(

qq

qzerosofnumberpolesofnumber

qa

160


UiTM Shah Alam

The asymptotes are shown by dotted lines.


Rule 4 tells that the segments of real axis between 0 and –1, and between –2 and - lie on the root

locus. On-locus segments are shown by thick lines in the Figure.

From Fig. 6.12, it is seen that out of the three loci, one is a real-root locus originating from s = −2 and

terminating on −. The other two loci originate from s = 0 and s = −1, and move on the real axis towards

each other as K increases. Their meeting point corresponds to a double root. As K increases further, the

root loci breakaway from the real axis to give complex conjugate pair of roots.

Rule 5 is used to calculate the intersection points on the imaginary axis by Routh Table. The

characteristic equation can be written as

023 23 Ksss

The Routh Table is given below.

s3 1 2

s2 3 K

s1 (6-K)/3

s0 K

161


UiTM Shah Alam

For all roots to lie on the left half of the s-plane, the following conditions must be satisfied.

K > 0, and (6 − K)/3 > 0

Therefore, the critical value of K, which corresponds to the roots on the imaginary axis, is 6. K = 6 makes

all the coefficients on s1 row to be zero. The auxiliary equation is formed from the coefficients of the s2

row as:

0633 22 sKs

The roots of this equation lie on the j axis and are given by 2js which are also the points where

the two root loci intersect the imaginary axis and the intersection points correspond to K = 6.

Rule 6 and Rule 7 are not necessary in this case since there are no open-loop complex poles or zeros.

Rule 8 is used to determine the breakaway points. From the characteristic equation of the system,

K = −(s3 + 3s2 + 2s).

Thus, by differentiating K and equate it to zero,

0)263( 2 ssds

dK

The solutions of this equation are:

s = −0.4226 and s = −1.5774

Thus, s = −0.4226 is the breakaway point and, since the other point s = −1.5774 is not on the root locus,

it is not a breakaway point.

If two loci breakaway from a breakaway point, their tangents will be 1800 apart. In general, if r loci

breakaway from a breakaway point, then their tangents will be 3600/r apart, i.e., the tangents will

equally divide 3600.

The complete root loci are shown in Fig. 6.12. For K > 6, the system has two closed-loop poles in the

right half s-plane.

A closed-loop pole with = 0.5 lies on a line passing through the origin and making an angle cos-1 = 600

with the negative real axis. From Fig. 6.12, the points of intersection are s = −0.33 j0.58 which are the

dominant closed-loop poles. From the magnitude criterion, the corresponding K can be found.

04.12158.033.0

jssssK

162


UiTM Shah Alam

Additional Example

Question: Sketch the root lo

cus of a unity feedback system with forward path transfer function G(s) given as follows:

)ss(s

KG(s)

542

Solution:

The open-loop poles are located at s = 0, −2+j, −2−j. There are no finite open-loop zeros.

Rule 1 tells that the root locus plot consists of three root loci as K varies from 0 to .

Rule 2 tells that the three root loci originate from the three open loop poles with K = 0 and terminate on

with K = .

Rule 3 tells that the three root loci tend to along asymptotes radiating out from

3/403

22

)()(

zerosofnumberpolesofnumber

zerosofpartsrealpolesofpartsreals a

with angles

000

0

0

300,180,60

2,1,0;3

180)12(

,2,1,0;180)12(

qq

qzerosofnumberpolesofnumber

qa

Rule 4 tells that the segments of real axis between 0 and – lie on the root locus.

Rule 5 is used to calculate the intersection points on the imaginary axis by Routh Table. The

characteristic equation can be written as

054 23 Ksss

163


UiTM Shah Alam

The Routh Table is given below.

s3 1 5

s2 4 K

s1 (20-K)/4

s0 K

For all roots to lie on the left half of the s-plane, the following conditions must be satisfied.

K > 0, and (20−K)/4 > 0

Therefore, the critical value of K, which corresponds to the roots on the imaginary axis, is 20. K = 20

makes all the coefficients on s1 row to be zero. The auxiliary equation is formed from the coefficients of

the s2 row as:

02044 22 sKs

The roots of this equation lie on the j axis and are given by 5js which are also the points where

the two root loci intersect the imaginary axis and the intersection points correspond to K = 20.

Rule 6 tells the angle of departure for complex poles.

For pole −2+j,

9043153 .φ and p = 1800 + = -63.430

For pole −2−j,

p = 63.430

Rule 8 is used to determine the breakaway points. From the characteristic equation of the system, K =

−(s3 + 4s2 + 5s).

Thus, by differentiating K and equate it to zero,

0583 2 )ss(ds

dK

The solutions of this equation are:

s = −1 and s = −1.667

Since the complete negative real axis is on the root loci, both are valid breakaway or break-in points.

21 sK , 85216671 .K .s

164


UiTM Shah Alam

165


UiTM Shah Alam

6.1.4 Effects of addition of poles and zeros to G(s)H(s)

The controller design in control systems may be treated as an investigation of the effects to root loci

when poles and zeros are added to the loop transfer function KG(s)H(s).

Addition of Poles to G(s)H(s)

Adding a pole to G(s)H(s) has the effect of pushing the root loci toward the right-half s-plane.

Example 6.6: Consider the loop transfer function )2(

)()(

ss

KsHsKG

The root loci are shown in Fig. 6.13(a). It is noted that the system is stable for all K. Let us introduce a

pole at s = −b (b > 2). The loop transfer function G(s)H(s) becomes, with b = 3,

)3)(2())(2()()(

sss

K

bsss

KsHsKG

The root loci are shown in Fig. 6.13(b) where the root loci bend towards the right-half s-plane. The

asymptote angles and centroid are changed from 90 to 60 and –1 to –(2+b)/3, respectively. The

addition of a pole may make the system unstable if K exceeds the stability limit.

Figure 6.13(a): Root loci for )2( ss

K

166


UiTM Shah Alam

Figure 6.13(b): Root loci for )3)(2( sss

K

Addition of Zeros to G(s)H(s)

Adding left-half plane zeros to the function G(s)H(s) generally has the effect of moving and bending the

root loci toward the left-half s-plane.

Fig. 6.14 shows the root loci of G(s)H(s) with a zero added at s = −3. The complex conjugate parts of root

loci of the original system are bent towards the left and form a circle. Thus, the relative stability is

improved by the addition of the zero.

167


UiTM Shah Alam

Figure 6.14: Root locus for )2(

)3(

ss

sK.

6.1.5 Compensator design via root locus

The preceding chapters have shown that it is often possible to adjust the system parameters in order to

provide the desired system response. However, we often find that it is not sufficient to reconsider the

structure of the system and redesign the system in order to obtain a suitable one. That is, we must

examine the scheme or plan of the system and obtain a new design or plan that results in a suitable

system. Thus the design of a control system is concerned with the arrangement, or the plan, of the

system structure and the selection of suitable components performance is called compensation.

Compensation is the adjustment of a system in order to make up for deficiencies or inadequacies. In

redesigning a control system to alter the system response, an additional component is inserted within

the structure of the feedback system. It is this additional component or device that equalizes or

compensates for the performance deficiency.

168


UiTM Shah Alam

The compensating device may be electric, mechanical, hydraulic, pneumatic, or some other type of

device or network and is often called a compensator. Commonly an electric circuit serves as a

compensator in many control systems. A compensator is an additional component or circuit that is

inserted into a control system to compensate for a deficient performance. The transfer function of a

compensator is designated as 𝐺𝑐 𝑠 = 𝐸0(𝑠)/𝐸𝑖𝑛 (𝑠), and the compensator can be placed in a suitable

location within the structure of the system. Several types of compensation are shown in Figure 6.15 for

a simple, single-loop feedback control system. The compensator placed in the feedforward path is called

a cascade, or series, compensator (6.15a)

Figure 6.15: Types of compensation (a) Cascaded compensation. (b) Feedback compensation. (c) Output,

or load compensation. (d) Input compensation

The objectives of introducing compensator can be categorized as follows:

i. Use PI and phase-lag compensators to improve steady-state error.

ii. Use PD and phase-lead compensators to improve transient response.

169


UiTM Shah Alam

Ideal PI compensator design

The PI compensator’s transfer function is given by:

𝐺𝑐 𝑠 = 𝐾1 +𝐾2

𝑠=

𝐾1(𝑠 +𝐾2𝐾1

)

𝑠

The ideal PI compensator’s transfer function is given by;

𝐺𝑐 𝑠 =𝑠 + 𝑎

𝑠

170


UiTM Shah Alam

Example 6.7:

Given a uncompensated system operating with a damping ration of 0.174. Find out the steady state

error for a unit step input. Design an ideal PI compensator to reduce the steady-state error to zero

without appreciably affecting transient response.

If the original OLTF is 𝐺 𝑠 =1

𝑠+1 𝑠+2 (𝑠+10) operating with a damping ratio of 0.174, then design a PI

compensator to reduce the steady-state error to zero for a step input without appreciably affecting

transient response. The compensator has a zero at -0.1, close to the compensator pole.

To achieve these requirements, the compensated system should have a dominant closed-loop pole at

𝑠1 = −0.694 + 𝑗3.926, 𝐾 = 164.6

171


UiTM Shah Alam

Thus, the steady-state error is

𝑒𝑠𝑠 ,𝑠𝑡𝑒𝑝 =1

1 + 𝐾𝑝=

1

1 + 164.6/20= 0.108

The dominant pole of the compensated system and the gain are approximately the same as for the

uncompensated system

𝑒𝑠𝑠 ,𝑠𝑡𝑒𝑝 =1

1 + 𝐾𝑝=

1

1 + lim𝑠→0

𝐺𝑐𝐺(𝑠)= 0

172


UiTM Shah Alam

General first-order compensators

Consider the first-order compensator with the transfer function

𝐺𝑐 =𝐾(𝑠 − 𝑧0)

𝑠 − 𝑝0

When |𝑧0|<|𝑝0|, the compensator is called a phase-lead compensator, because this results in a

contribution to the angle criterion of the root locus that is always positive.

∠𝐺𝑐 𝑠 = ∠𝑠 − 𝑧0 − ∠𝑠 − 𝑝0 = 𝜃𝑧 − 𝜃𝑝 > 0

When |𝑧0|>|𝑝0|, the compensator is called a phase-lag compensator, because this results in a

contribution to the angle criterion of the root locus that is always positive.

∠𝐺𝑐 𝑠 = ∠𝑠 − 𝑧0 − ∠𝑠 − 𝑝0 = 𝜃𝑧 − 𝜃𝑝 < 0

173


UiTM Shah Alam

Phase-lead compensator design

The compensator transfer function is given by:

𝐺𝑐 𝑠 =𝐾𝑐(𝑠 − 𝑧0)

𝑠 − 𝑝0

and it can be re-written in a form as:

𝐺𝑐 𝑠 =𝑎1𝑠 + 𝑎0

𝑏1𝑠 + 1

Where,

𝐾𝑐 =𝑎1

𝑏1, 𝑧0 = −

𝑎0

𝑎1, 𝑝0 = −

1

𝑏1

The compensator DC gain is 𝐷𝐶𝑔𝑎𝑖𝑛 = lim𝑠→0 𝐺𝑐 𝑠 = 𝑎0

Assume that the parameter 𝑎0 is either known or can be determined. The design problem is to find

𝑎1and 𝑏1 such that the compensated system will have a closed-loop pole at 𝑠 = 𝑠1.

First we express 𝑠1 and G(s)H(s) as

𝑠1= 𝑠1 𝑒𝑗𝛽 𝐺 𝑠1 𝐻 𝑠1 = 𝐺 𝑠1 𝐻(𝑠1) 𝑒𝑗𝜓

From the characteristics equation, we get 1 +𝑎1𝑠+𝑎0

𝑏1𝑠+1𝐺 𝑠 𝐻 𝑠 = 0

Equating magnitudes and angles, we can rewrite as

𝑎1𝑠1 + 𝑎0

𝑏1𝑠1 + 1 𝐺 𝑠1 𝐻(𝑠1) = 1

∠𝑎1𝑠1 + 𝑎0

𝑏1𝑠1 + 1+

∠𝐺 𝑠1 𝐻(𝑠1)

𝜓= 180°

Where

𝑎1 =sin 𝛽 + 𝑎0 𝐺 𝑠1 𝐻(𝑠1) sin(𝛽 − 𝜓)

𝑠1 𝐺 𝑠1 𝐻(𝑠1) sin 𝜓

174


UiTM Shah Alam

𝑏1 =sin(𝛽 + 𝜓) + 𝑎0 𝐺 𝑠1 𝐻(𝑠1) sin 𝛽

− 𝑠1 sin 𝜓

Example 6.8:

Design a phase-lead compensator such that the closed-loop compensated system has a settling time

around 4 sec. and a percent overshoot around 4.32%. The compensator has a DC gain as 0.15.


𝑠1= − 1 + 𝑖. Because the DC gain for 𝐺𝑐(𝑠) is 0.15, so we have 𝑎0=0.15.

At 𝑠1= − 1 + 𝑖, we have 21

𝑠 𝑠+1 (𝑠+3) 𝑠1=−1+𝑖

= −2.1 + 6.3𝑗 = 6.64∠108.43°

Also, we have 𝑠1= − 1 + 𝑖 = 2∠135°

𝑎1 =sin 𝛽 + 𝑎0 𝐺 𝑠1 𝐻(𝑠1) sin(𝛽 − 𝜓)

𝑠1 𝐺 𝑠1 𝐻(𝑠1) sin 𝜓= 0.1924

𝑏1 =sin(𝛽 + 𝜓) + 𝑎0 𝐺 𝑠1 𝐻(𝑠1) sin 𝛽

− 𝑠1 sin 𝜓= 0.1417

175


UiTM Shah Alam

Phase-lag compensator design

The negative angle contributed by the phase-lag compensator will tend to shift the root locus to the

right in the s-plane, i.e., towards the unstable region. Thus, in general, the angle contribution of the

phase-lag compensator must be small, which is assured by placing the pole and the zero of the

compensator very close to each other.

For convenience in the design, we assume that the compensator has a unit DC gain, i.e.,

𝐺𝑐(𝑠) 𝑠=0 =𝐾𝑐𝑧0

𝑝0= 1

176


UiTM Shah Alam

𝐾𝑐 =𝑃0

𝑧0< 1

Suppose that the root locus of the point of the uncompensated system passes through the point 𝑠1for

𝐾0.

1 + 𝐾0𝐺 𝑠1 𝐻 𝑠1 = 0

𝐾0 =−1

𝐺 𝑠1 (𝐻(𝑠1)

As we choose the value of 𝑧0 and 𝑝0 to be approximately equal, and the magnitudes of 𝑧0 and 𝑝0 to be

small compared to 𝑠1, so

𝐺𝑐 𝑠1 =𝐾𝑐(𝑠1 − 𝑧0)

𝑠1 − 𝑝0≈ 𝐾𝑐

Now the gain required to place a root of the locus at approximately 𝑠1 for the uncompensated system is

given by

𝐾 =−1

𝐺𝑐 𝑠1 𝐺(𝑠1)=

−1

𝐾𝑐𝐺𝑐 𝑠1 𝐺 𝑠1 =

𝐾0

𝐾𝑐

Since 𝐾𝑐 < 1, so 𝐾 > 𝐾0. The compensator has been chosen to have a unity DC gain; thus the open-loop

DC gain has been increased, but the transient response appears to remain unaffected.

The steady-state error 𝑒𝑠𝑠 , when H(s)=1, is

𝑒𝑠𝑠 = lim𝑡→∞

𝑒 𝑡 = lim𝑠→0

𝑠𝐸 𝑠 = lim𝑠→0

𝑠𝑅(𝑠)

1 + 𝐺(𝑠)

177


UiTM Shah Alam

The sready-state error 𝑒𝑠𝑠 has been improved, and this is the principal use of the phase-lag

compensator.

Example 6.9: Phase-lag compensator design

Design a radar tracking system of the uncompensated OLTF given by 𝐾𝐺 𝑠 𝐻 𝑠 =𝐾

𝑠(𝑠+2). Suppose that

the design requirements are such that a time constant of 1 second and damping coefficient of 0.707 are

satisfactory and the compensator has a DC gain as 1.

So 𝑠1 = −1 + 𝑗 𝑤𝑕𝑒𝑛 𝐾0 = 2 , is acceptable. Suppose that the system is required to track aircraft that

have essentially constant velocity, which will appear to the control system as a ramp input, i.e. the

antenna must rotate at a constant velocity to remain pointed directly at the aircraft. Also, it is required

that the 𝑒𝑠𝑠 of 0.2° with a unit ramp input.

178


UiTM Shah Alam

179


UiTM Shah Alam

Ideal PD compensator

The PD compensator’s transfer function is given by:

𝐺𝑐 𝑠 = 𝐾1 + 𝐾2 = 𝐾2(𝑠 +𝐾1

𝐾2)

The ideal PD compensator’s transfer function is given by;

𝐺𝑐 𝑠 = 𝑠 + 𝑧𝑐

If the original open-loop transfer function is

𝐺 𝑠 =1

𝑠 + 1 𝑠 + 2 (𝑠 + 5)

Then design the PD compensator zero at -2.

𝐺𝑠 𝑠 = 𝑠 + 2

180


UiTM Shah Alam

181


UiTM Shah Alam

Example 6.9

Given the system below, design an ideal derivative compensator such that the closed-loop compensated

system has a threefold reduction in settling time and a 16% percent overshoot.


𝑠1 = −1.205 + 𝑗2.064𝐾 = 43.45. Thus, the uncompensated system’s settling time is 𝑇𝑠 =4

𝜁𝜔𝑛=

4

1.205= 3.302.

The desired real part of the closed-loop pole is 𝜁𝜔𝑛 =4

𝑇𝑠=

4

1.107= 3.613. The desired imaginary part of

the closed-loop pole is 𝜔𝑑 = 3.613 tan 180 − 120.26 = 6.193.

182


UiTM Shah Alam

183


UiTM Shah Alam

6.2 State Space Applications

An alternative method yet powerful tools of control system design by using a State Space

representation. The concept of using state space is by placing a pole at a desired location. We call this as

pole placement method.

Similar in concept to classical control system design where we have to firstly formulate desired pole

locations to satisfy some performance criteria, then formulate control gains to make this happen.

In designing the control system by using state space application, we have to assume that all states can

be measured and used in control implementation; this is called full state feedback.

6.2.1 Controller Pole Placement Method

The controller pole placement method mainly concerns with the controllability matrix. It takes

measurement and/or estimates of the state variables, multiplies them by the control gains, and produce

the control signal. This can be designed by pole placement or optimal control.

Let say we have a control input, u in the form of,

u = -kx

where k = vector (or matrix) of proportional control gains applied to each state given by:

k = [𝒌𝟏 𝒌𝟐 … 𝒌𝒏]

Essentially, we are trying to modify the underlying differential equations.

Now, consider the state equation:

𝐱 = 𝐀𝐱 + 𝐁𝐮 = 𝐀𝐱 − 𝐁𝐤𝐱 = 𝐀 − 𝐁𝐤 𝐱

Taking the Laplace,

𝑠𝐈 − 𝐀 + 𝐁𝐤 𝐱 𝐬 = 0

184


UiTM Shah Alam

Poles defined by:

det 𝑠𝐈 − 𝐀 + 𝐁𝐤 = 0

Suppose we have some desired pole locations:

𝑝1 , 𝑝2 , 𝑝3 , ….

Then the desired characteristics equation is:

𝑠 − 𝑝1 𝑠 − 𝑝2 … 𝑠 − 𝑝𝑛 = 0

This can be expanded to (desired characteristics equation):

𝑠𝑛 + 𝛼1𝑠𝑛−1 + ⋯ + 𝛼𝑛−1𝑠 + 𝛼𝑛 = 0

Now suppose that the state space equations are in control canonical form:

𝐀 =

0 1 0 … 00 0 1 … 0. . . … .. . . … .0 0 0 … 1

−𝑎𝑛 −𝑎𝑛−1 −𝑎𝑛−2 … −𝑎1

𝐁 =

00..01

𝐂 = 𝑏𝑛 − 𝑎1𝑏0 𝑏𝑛−1 − 𝑎𝑛−1𝑏0 … . 𝑏1 − 𝑎1𝑏0

The poles of the feedback system are defined by the expression:


Poles are therefore given by

det 𝑠𝐈 − 𝐀 + 𝐁𝐤 = 𝑠𝑛 + (𝛼1 + 𝑘1)𝑠𝑛−1 + ⋯ + (𝛼2 + 𝑘2)𝑠𝑛−2 + 𝛼𝑛 + 𝑘𝑛 = 0

Compare this to the “desired” characteristics equation:

𝑠𝑛 + 𝛼1𝑠𝑛−1 + 𝛼2𝑠𝑛−2 … + 𝛼𝑛−1𝑠 + 𝛼𝑛 = 0

The conclusion is that when the system is in control canonical form, then control gains can be calculated

by simple comparison of coefficients:

𝑘𝑖 = 𝛼𝑖 − 𝑎𝑖

185


UiTM Shah Alam

Example 6.10

Consider an undamped oscillator with frequency 𝜔𝑛 and a SS model given by

𝑥 1𝑥 2

= 0 1

−𝜔𝑛2 0

𝑥1

𝑥2 +

01

Find the controller that places the both CL poles of the system at −2𝜔𝑛 . In other words, you want to

double the natural frequency and increase the damping ratio from 0 to 1.

Solution:

Assumption: Must be a full state feedback. We need to prove the system has a full state feedback by

assessing its controllability matrix.

We have an open loop poles, 𝑠 = 𝑗𝜔𝑛 and closed loop poles, 𝑠 = −2𝜔𝑛 .

The open loop characteristics equation is: 𝑠 + 𝑗𝜔𝑛 𝑠 + 𝑗𝜔𝑛 = 𝑠2 + 𝜔𝑛2

And the desired characteristics equation:

(𝑠 + 2𝜔𝑛 )2 = 𝑠2 + 4𝜔𝑛𝑠 + 4𝜔𝑛2

Now we have to determine the poles of closed-loop system;


det 𝑠 00 𝑠

− 0 1

−𝜔𝑛2 0

+ 01 𝑘1 𝑘2

𝑠2 + 𝑘2𝑠 + 𝜔𝑛2 + 𝑘1 = 0

Comparing with the desired characteristics equation gives:

𝑘2 = 4𝜔𝑛

𝑘1 + 𝜔𝑛2 = 4𝜔𝑛

2

We have now,

𝑘1 = 3𝜔𝑛2

𝑘2 = 4𝜔𝑛

Hence,

𝐤 = 𝑘1 𝑘2

𝐤 = 3𝜔𝑛2 4𝜔𝑛

186


UiTM Shah Alam

Generalised Strategy

Step 1: Transform state equations into control canonical form.

𝐱′ = 𝐓−𝟏𝐱 (𝐱′ = transformed state vector)

𝐀′ = 𝐓−𝟏𝐀𝐓, 𝐁′ = 𝐓−𝟏𝐁 (A’=transformed state matrix)

Where T is the transformation matrix: 𝐓 = 𝐂𝟎𝐍

𝐂𝟎 = controllability matrix = 𝐁 𝐀𝐁 … 𝐀𝐧−𝟏𝐁

𝐍 = the following Toeplitz matrix:

1 0 … 0⋮

𝑎𝑛−2

𝑎𝑛−1

⋱ ⋮ ⋮0 1 0… 𝑎1 1

Where 𝑎1 , 𝑎2 , … are from the characteristics equation of the uncontrolled system:

𝑠𝑛 + 𝛼1𝑠𝑛−1 + 𝛼2𝑠𝑛−2 … + 𝛼𝑛−1𝑠 + 𝛼𝑛 = 0

Step 2: Calculate control gains by comparison with the desired characteristics equation

Step 3: Transform back to original state

Transform has the form:

𝐤𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑠𝑡𝑎𝑡𝑒 = 𝐤𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑐𝑎𝑛𝑜𝑛𝑖𝑐𝑎𝑙 𝐓−𝟏

Controller Pole Placement using Ackermann’s Formula

For SISO systems the control gains using Ackermann’s Formula are

𝐤 = 0 0 … 0 1 𝐂0−1𝛾(𝐀)

Where 𝐂0=controllability matrix (note inversion again), and

𝛾 𝐀 = 𝐀𝑛 + 𝛼1𝐀𝑛−1 + 𝛼2𝐀𝑛−2 + ⋯ + 𝛼𝑛 𝐈

A = state matrix,

Where 𝑎1 , 𝑎2 , … = coefficient of the desired characteristics equation.

187


UiTM Shah Alam

Example 6.11:

Consider an undamped oscillator with frequency 𝜔𝑛 and a SS model given by

𝑥 1𝑥 2

= 0 1

−𝜔𝑛2 0

𝑥1

𝑥2 +

01

Find the controller that places the both CL poles of the system at −2𝜔𝑛 . In other words, you want to

double the natural frequency and increase the damping ratio from 0 to 1.

Solution:

Our objective is to get the vector k. According to Ackermann’s Formula:

𝐤 = 0 0 … 0 1 𝐂0−1𝛾(𝐀)

The SS model must be controllable before we can proceed. Checking the controllability matrix,

𝐂𝟎 = 0 11 0

Where

𝛾 𝐀 = 𝐀𝑛 + 𝛼1𝐀𝑛−1 + 𝛼2𝐀𝑛−2 + ⋯ + 𝛼𝑛 𝐈

From the desired characteristics equation we have,

(𝑠 + 2𝜔𝑛 )2 = 𝑠2 + 4𝜔𝑛𝑠 + 4𝜔𝑛2

So we have,

𝛾 𝐀 = 𝐀2 + 4𝜔𝑛𝐀 + 4𝜔𝑛2𝐈

Matrices 𝐀2 , 4𝜔𝑛𝐀 and 4𝜔𝑛2𝐈 are given by,

𝐀2 = −𝜔𝑛

2 0

0 −𝜔𝑛2

4𝜔𝑛𝐀 = 0 4𝜔𝑛

−4𝜔𝑛3 0

4𝜔𝑛2𝐈 =

4𝜔𝑛2 0

0 4𝜔𝑛2

Therefore matrix 𝛾 𝐀 is given by,

𝛾 𝐀 = 3𝜔𝑛

2 4𝜔𝑛

−4𝜔𝑛3 3𝜔𝑛

2

188


UiTM Shah Alam

Hence,

𝐤 = 0 1 0 11 0

3𝜔𝑛

2 4𝜔𝑛

−4𝜔𝑛3 3𝜔𝑛

2

𝐤 = 3𝜔𝑛2 4𝜔𝑛

Example 6.12:

Consider the SS system

𝑥 1𝑥 2

= −1 01 −2

𝑥1

𝑥2 +

12 𝑢

Design a control system to move the system poles to 𝑠 = −1 ± 𝑗

189


UiTM Shah Alam

6.2.2 Observer Pole Placement Method

The observer pole placement mainly concerns with Observability matrix. The objective of designing the

observer is to estimate some or all of the states of the system. This can be achieved by linear observers

(pole placement) or optimal observers (Kalman filters).

Let’s say a system is described by:

𝐱 = 𝐀𝐱 + 𝐁𝐮, 𝐲 = 𝐂𝐱

Assume that we know 𝐀, 𝐁, 𝐂, 𝐮.

We need to extract 𝐱 by constructing a second linear system (a model of the target system), using the

known parameters (𝐀, 𝐁, 𝐂, 𝐮) of the target system, which predicts the (measurable) target system

output. If the predicted output is acceptably close to the actual output, then we can use the estimated

states in place of the actual states.

In other words, we want to minimize the difference between the actual and predicted states. This

difference or error in estimate of state is given by,

𝐱𝑒 = 𝐱 − 𝐱

Which converges to zero if 𝐀 is stable.

The open loop dynamics is then:

𝐱 𝑒 = 𝐱 − 𝐱 = 𝐀𝐱 + 𝐁𝐮 − 𝐀 𝐱 − 𝐱𝑒 − 𝐁𝐮 = 𝐀𝐱𝑒

And the characteristics equation,

det 𝑠𝐈 − 𝐀 = 0

The observer (model) system response is now:

𝐱 = 𝐀𝐱 + 𝐁𝐮 + 𝐋𝐂(𝐱 − 𝐱 )

190


UiTM Shah Alam

where L = vector (or matrix) of estimator (observer) gains applied to each state given by:

𝑳 =

ℓ1

ℓ2

⋮ℓ𝑛

The dynamics of the error become:

𝐱 𝑒 = 𝐀𝐱𝑒 − 𝐋𝐂𝐱𝑒

Therefore, the characteristics equation = det 𝑠𝐈 − (𝐀 − 𝐋𝐂) = 0 i.e we can change convergence speed

by adding feedback.

Suppose we have some desired pole locations:

𝑝1 , 𝑝2 , 𝑝3 , ….

Then the desired characteristics equation is:

𝑠 − 𝑝1 𝑠 − 𝑝2 … 𝑠 − 𝑝𝑛 = 0

This can be expanded to (desired characteristics equation):

𝑠𝑛 + 𝛼1𝑠𝑛−1 + ⋯ + 𝛼𝑛−1𝑠 + 𝛼𝑛 = 0

Now suppose that the state space equations are in observer canonical form:

𝐴 =

0 0 … 0 −𝑎𝑛

1 0 … 0 −𝑎𝑛−1

. . … . −𝑎𝑛−2

. . … . .0 0 … 0 .0 0 … 1 −𝑎1

𝐵 =

𝑏𝑛 − 𝑎1𝑏0

𝑏𝑛−1 − 𝑎𝑛−1𝑏0

.

.

.𝑏1 − 𝑎1𝑏0

𝐶 = 0 0 … 0 1

With the observer feedback, the poles are defined by the expression: det 𝑠𝐈 − (𝐀 − 𝐋𝐂) = 0

191


UiTM Shah Alam

𝑠𝐈 − 𝐀 − 𝐋𝐂 =

𝑠 0 … 0 𝑎𝑛 + ℓ𝑛

−1 𝑠 … 0 𝑎𝑛−1 + ℓ𝑛−1

. . … . .

. . … . .0 0 … 𝑠 𝑎2 + ℓ2

0 0 … −1 𝑠 + 𝑎1 + ℓ1

Poles are therefore given by

det 𝑠𝐈 − 𝐀 + 𝐋𝐂 = 𝑠𝑛 + (𝛼1 + ℓ1)𝑠𝑛−1 + (𝛼2 + ℓ2)𝑠𝑛−2 + ⋯ + 𝛼𝑛 + ℓ𝑛 = 0

Compare this to the “desired” characteristics equation:

𝑠𝑛 + 𝛼1𝑠𝑛−1 + 𝛼2𝑠𝑛−2 … + 𝛼𝑛−1𝑠 + 𝛼𝑛 = 0

The conclusion is that when the system is in control canonical form, then control gains can be calculated

by simple comparison of coefficients:

ℓ𝑖 = 𝛼𝑖 − 𝑎𝑖

Example 6.13:

Compute the estimator (observer) gain matrix which will place both estimator poles at −10𝜔𝑛 , given

𝑥 1𝑥 2

= 0 1

−𝜔𝑛2 0

𝑥1

𝑥2 +

01

𝑦 = 1 0 𝑥1

𝑥2

Solution:

The desired characteristics equation is given by;

(𝑠 + 10𝜔𝑛 )2 = 𝑠2 + 20𝜔𝑛𝑠 + 100𝜔𝑛2

Now we have to determine the poles of closed-loop system;

det 𝑠𝐈 − 𝐀 + 𝐋𝐂 = 0

det 𝑠 00 𝑠

− 0 1

−𝜔𝑛2 0

+ ℓ1 0ℓ2 0

192


UiTM Shah Alam

det 𝑠 + ℓ1 −1

𝜔𝑛2+ℓ2 𝑠

det 𝑠2 + ℓ1𝑠 + 𝜔𝑛2 + ℓ2 = 0

Comparing with the desired characteristics equation gives:

ℓ1 = 20𝜔𝑛

ℓ2 + 𝜔𝑛2 = 100𝜔𝑛

2

We have now,

ℓ1 = 20𝜔𝑛

ℓ2 = 99𝜔𝑛2

Hence,

𝐤 = ℓ1

ℓ2

𝐤 = 20𝜔𝑛

99𝜔𝑛2

Generalised Strategy

Step 1: Transform state equations into control canonical form.

𝐱′ = 𝐓−𝟏𝐱 (𝐱′ = transformed state vector)

𝐀′ = 𝐓−𝟏𝐀𝐓, 𝐂′ = 𝐓−𝟏𝐂 (A’=transformed state matrix)

Where T is the transformation matrix: 𝐓 = (𝐍𝐎𝐛)−𝟏 and

𝐎𝐛 = observability matrix =

𝐂𝐂𝐀𝐂𝐀𝟐

⋮𝐂𝐀𝐧−𝟏

𝐍 = the following Toeplitz matrix:

1 0 … 0⋮

𝑎𝑛−2

𝑎𝑛−1

⋱ ⋮ ⋮0 1 0… 𝑎1 1

Where 𝑎1 , 𝑎2 , … are from the characteristics equation of the uncontrolled system:

𝑠𝑛 + 𝛼1𝑠𝑛−1 + 𝛼2𝑠𝑛−2 … + 𝛼𝑛−1𝑠 + 𝛼𝑛 = 0

193


UiTM Shah Alam

Step 2: Calculate control gains by comparison with the desired characteristics equation

Step 3: Transform back to original state

Transform has the form:

𝐋𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑠𝑡𝑎𝑡𝑒 = 𝐓 ∗ 𝐋𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑟 𝑐𝑎𝑛𝑜𝑛𝑖𝑐𝑎𝑙

Observer Pole Placement using Ackermann’s Formula

For SISO systems the observer gains using Ackermann’s Formula are

𝐋 = 𝛾(𝐀)𝐎b−1 0 0 … 0 1 T

Where 𝐂0=controllability matrix (note inversion again), and

𝛾 𝐀 = 𝐀𝑛 + 𝛼1𝐀𝑛−1 + 𝛼2𝐀𝑛−2 + ⋯ + 𝛼𝑛 𝐈

A = state matrix,

Where 𝑎1 , 𝑎2 , … = coefficient of the desired characteristics equation.

Example 6.14:

Compute the estimator (observer) gain matrix which will place both estimator poles at −10𝜔𝑛 , given

𝑥 1𝑥 2

= 0 1

−𝜔𝑛2 0

𝑥1

𝑥2 +

01

𝑦 = 1 0 𝑥1

𝑥2

control systems lecture notes

Documents