DEPARTMENT OF ELECTRICAL AND ELECTRONICS ...Control system components can be mechanical, electrical, hydraulic, pneumatic, thermal, or it may be a computer program. It plays an important

M.I.E.T ENGG COLLEGE DEPT OF EEE

IV– CONTROL SYSTEMS 1

M.I.E.T. ENGINEERING COLLEGE

(Approved by AICTE and Affiliated to Anna University Chennai)

TRICHY – PUDUKKOTTAI ROAD, TIRUCHIRAPPALLI – 620 007

DEPARTMENT OF ELECTRICAL AND

ELECTRONICS ENGINEERING

COURSE MATERIAL

IC6501 CONTROL SYSTEMS ENGINEERING

III YEAR - V SEMESTER



SYLLABUS IC6501 CONTROL SYSTEMS ENGINEERING

L T P C 3 0 0 3 UNIT I: SYSTEMS AND THEIR REPRESENTATION [9 Hours] Basic Elements In Control Systems – Open And Closed Loop Systems – Electrical Analogy Of Mechanical And Thermal Systems – Transfer Function – Synchros – AC And DC Servomotors – Block Diagram Reduction Techniques – Signal Flow Graphs. UNIT II: TIME RESPONSE [9 Hours] Time Response – Time Domain Specifications – Types Of Test Input – I And II Order System Response – Error Coefficients – Generalized Error Series – Steady State Error – Root Locus Construction- Effects Of P, PI, PID Modes Of Feedback Control –Time Response Analysis. UNIT III : FREQUENCY RESPONSE [9 Hours] Frequency Response – Bode Plot – Polar Plot – Determination Of Closed Loop Response From Open Loop Response – Correlation Between Frequency Domain And Time Domain Specifications- Effect Of Lag, Lead And Lag-Lead Compensation On Frequency Response- Analysis. UNIT IV : STABILITY AND COMPENSATOR DESIGN [9 Hours] Characteristics Equation – Routh Hurwitz Criterion – Nyquist Stability Criterion- Performance Criteria – Lag, Lead And Lag-Lead Networks – Lag/Lead Compensator Design Using Bode Plots. UNIT V : STATE VARIABLE ANALYSIS [9 Hours] Concept Of State Variables – State Models For Linear And Time Invariant Systems – Solution Of State And Output Equation In Controllable Canonical Form – Concepts Of Controllability And Observability – Effect Of State Feedback. [TOTAL (L:45+T:15): 60 PERIODS]



UNIT-I

SYSTEMS AND THEIR REPRESENTATION

Introduction to Control Systems The goal of control engineering is to design and build real physical systems to perform given

tasks. An engineer is asked to design an install a heat exchanger to control the temperature and

humidity of a large building. He determines the required capacity of the exchanger and then

proceed to install the system. After installation the exchanger is found to be insufficiently

powerful to control the buildings environment. He has to replace the unit and place a powerful

one. This is an example of empirical method.

Consider again the task of sending astronauts in moon and bringing them back safely. It cannot

be carried out by empirical method. In this case analytical method is indispensable. This method

consists of the following steps:

- modeling

- setting up mathematical equations

- analysis and design

Empirical methods may be expensive and dangerous. Analytical methods are simulated in

computers to see the result. If the design is satisfactory, the system is implemented using

physical devices.

A control system is an interconnection of components or devices so that the output of the overall

system will follow as closely as possible a desired signal. The reasons of designing control

systems include:

- Automatic control (e.g., control of room temperature)

- Remote control (e.g., antenna position control)

- Power amplification(e.g., control system will generate sufficient power to turn the

heavy antennas)

Control system components can be mechanical, electrical, hydraulic, pneumatic, thermal, or it

may be a computer program. It plays an important role in the development of modern

civilization. We use heating and air-conditioning in domestic domain for comfortable living. It



has found application in quality control of manufacturing products, machine-tool control,

weapon systems, power systems, robotics and many other places.

Classification of Control Systems

Control systems are basically classified as –

Open-loop control system

Closed-loop control system

In open-loop system the control action is independent of output. In closed-loop system control

action is somehow dependent on output. Each system has at least two things in common, a

controller and an actuator (final control element). The input to the controller is called reference

input. This signal represents the desired system output.

Open-loop control system is used for very simple applications where inputs are known ahed of

time and there is no disturbance. Here the output is sensitive to the changes in disturbance

inputs. Disturbance inputs are undesirable inputs that tend to deflect the plant outputs from their

desired values. They must be calibrated and adjusted at regular intervals to ensure proper

operation.

Closed-loop systems are also called feedback control systems. Feedback is the property of the

closed-loop systems which permits the output to be compared with the input of the system so that

appropriate control action may be formed as a function of inputs and outputs. Feedback systems

has the following features:

- reduced effect of nonlinearities and distortion

- Increased accuracy

- Increased bandwidth

- Less sensitivity to variation of system parameters

- Tendency towards oscillations

- Reduced effects of external disturbances

The general block diagram of a control system is shown below.



Some Definitions

Reference input – It is the actual signal input to the control system.

Output (Controlled variable) – It is the actual response obtained from a control system.

Actuating error signal – It is the difference between the reference input and feedback signal.

Controller – It is a component required to generate control signal to drive the actuator.

Control signal – The signal obtained at the output of a controller is called control signal.

Actuator – It is a power device that produces input to the plant according to the control signal, so

that output signal approaches the reference input signal.

Plant – The combination of object to be controlled and the actuator is called the plant.

Feedback Element – It is the element that provides a mean for feeding back the output quantity

in order to compare it with the reference input.

Servomechanism – It is a feedback control system in which the output is mechanical position,

velocity, or acceleration.

Example of Control Systems Toilet tank filling system:

Figure: Toilet tank filling system Position control system: [antenna]



Figure: Position control system Velocity control system: [audio/ video recorder]

Figure: Velocity control system



Clothes Dryer:

Figure: Automatic dryer

Temperature control system: [oven, refrigerator, house]

Figure: Temperature control system Computer numerically controlled (CNC) machine tool:



(a)

(b)

Figure: CNC machine tool control system

Control System Design In order to design and implement a control system, we need knowledge about the following

things:

Knowledge of desired value, (performance specification)

Knowledge of the output value, (feedback sensor, its resolution and dynamic

response)

Knowledge of controlling device,

Knowledge of actuating device,

Knowledge of the plant.



With all of this knowledge and information available for the control system designer, he can start

the design steps shown below in the flow diagram.



Mathematical Models of Physical Systems

We use mathematical models of physical systems to design and analyze control systems.

Mathematical models are described by ordinary differential equations. If the coefficients of the

describing differential equations are function of time, then the mathematical model is linear time-

varying. On the other hand, if the coefficients describing differential equations are constants, the

model is linear time-invariant.

The differential equations describing a LTI system can be reshaped into different forms

for the convenience of analysis. For transient response or frequency response analysis of single-

input-single-output linear systems, the transfer function representation is convenient. On the

other hand, when the system has multiple inputs and outputs, the vector-matrix notation may be

more convenient.

Powerful mathematical tools like Fourier and Laplace transforms are available for linear

systems. Unfortunately no physical system in nature is perfectly linear. Certain assumptions must

always be made to get a linear model. In the presence of strong nonlinearity or in presence of

distributive effects it is not possible to obtain linear models.

A commonly adopted approach is to build a simplified linear model by ignoring certain

nonlinearities and other physical properties that may be present in a system and thereby get an

approximate idea of the dynamic response of the system. A more complete model is then built

for more complete analysis.

Transfer function

The transfer function of an LTI system is the ratio of Laplace transform of the output variable to

the Laplace transform of the input variable assuming zero initial conditions. Following are some

examples of how transfer functions can be determined for some dynamic system elements.

[See chapter 3 of Ogata for details]

2

( ) 1

( ) 1o

i

E s

E s LCs RCs



The insertion of an isolation amplifier between the two RC-circuits will produce no loading

effect.

A. Transfer Function of Armature Controlled DC Motor

In Laplace domain,

2

0

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

b b

a a a b

T a

E s K s s

L s R I s E s E s

Js f s s K I s

; 0

( ) ( )

( ) ( )( )T

a a T b

KsG s

E s s R sL Js f K K

Neglecting aL , 2

0

/ /( )

( / ) ( ) ( 1)T a T a m

T b a m

K R K R KG s

Js s f K K R s Js f s s ;

where, 0 /T b af f K K R and / ; / .m T a mK K R f J f

mK and m are called the motor gain and time constant respectively. These two parameters are

usually supplied by the manufacturer.The block diagram model is,

21 1 2 2 1 1 2 2 1 2

( ) 1

( ) ( ) 1o

i

E s

E s RC R C s RC R C RC s

1;f f M f f a T aK i T K K i i K i

; ab b a a a b

dide K L R i e e

dt dt

2

02 M T a

d dJ f T K i

dt dt



B. Transfer Function of a Field-controlled DC Motor

'1 1

2

2

TM a f f a f

ff f f

M T f

T K i K K i i K i

diL R i e

dt

d dJ f T K i

dt dt

; 2

( ) ( ) ( )

( ) ( ) ( ) ( )

f f f

M T f

L s R I s E s

Js fs s T s K I s

We obtain,

Mechanical System

Translational system and rotational system

( )( )

( ) ( )( )T

f f

KsG s

E s s L s R Js f

2

2

1 2 1 2 2 1( ); ( )

m

d s

du d xu ma m m

dt dtu k u u u k x x

2

2

1 1 2 2 1 2

;

( ); ( )

J

d s

d dT J J

dt dtT k T k



Mechanical Accelerometer

For a constant input acceleration y becomes constant.

Then, x ky . Taking Laplace transform of the previous equation,

2( ) ( ) ( )X s s c s k Y s ;

( ) ( )2

1X s Y s

s c s k

Rotary Potentiometer

2

1 22

( ) ( )( )

d y t dy tm k k y t u

dt dt

maxmax

max max

( )( ) ( )

or, ( ) ( )

o

o p

VtV t V t

V t k t



Example: Draw the block diagram of the following system.

Example: Control of flaps in airplane



Example: Figure below shows a reduction gearbox being driven by a motor that develops a torque( )mT t . It

has a gear reduction ratio of ‘n = b / a’. Find a differential equation relating the motor

torque ( )mT t and the output angular position( )o t .

2

2( ) ( ) m m

a m m m

d dT t T t I C

dt dt

; 2

02( ) o o

b o

d dT t I C

dt dt

; ( ) 1

( )a

b

T t a

T t b n ; m

o

bn

a

From above, 2 2

02 2( ) o o o o

m m m o

d d d dn T t nI nC I C

dt dt dt dt

; N.B. 2 2

2 2;m o m od d d d

n ndt dt dt dt

Gearbox parameters:

, , From above, 2 2( ) ( ) ( )o m o o m o mI n I C n C nT t

Inputting parameters, 0.0225 0.3 50 ( )o o mT t [Ans]



Example: Gear train and its equivalence

max

( ) ( ) ( )

/p r o

p batt

V s k s s

k V

1 1 1 1motor eq eqT J f

Tachometers

Error detector

( ) ( ) ( )t tV s k s k s s

max

( ) ( ) ( )

/p r o

p batt

V s k s s

k V



Example

Example

Sketch the analogous electrical circuit of the following mechanical system.

Example

Draw the analogous electric circuit of the system below using f-i analogy.

Error detector using op amp e = r - vw

Obtain the transfer function X(s)/E(s) for the electromechanical system shown left assuming

that the coil has a back emf 1b

dxe K

dt and the

coil current i2 produces a force 2 2cF k i on the

mass M.

24 3

21 2 1 2

( )

( ) ( ) (2

) ( 2 ) 2

KX s

E s RLCMs L M RCB s RC LK

k k s RB LK k k s RK



[Ans] Mechanical Systems: There are two Types

1. Mechanical Translation system. 2. Mechanical rotational system.

The basic elements used in Mechanical translation system and Mechanical Rotational Systems are

Mechanical Translational system

Mechanical Rotational system

1. The weight of mechanical translational system is represented by the element Mass(M)

1. The weight of the mechanical rotational system is represented by the moment of inertia of the mass (J)

2. The elastic deformation of the body can be represented by a Spring ( K)

2. The elastic deformation of the body can be represnted by a Torsional spring with stiffness (K).

3. The frictional existing in rotating mechanical system can be represented by Dash pot (B)

3. The friction existing in the mechanical rotational system can be represented by the Dash pot (B)



Symbols used in the mechanical system are:

Mechanical Translational Mechanical rotational

1. x = Displacement(m) 1. θ – Angular displacement (rad)

2. V = dx / dt (m/sec) 2. dθ / dt – Angular velocity.

3. a = dv / dt (m/s) 3. dω / dt – Angular acceleration

4. f = Applied force ( N) 4. T - Applied Torque (N- m)

5. fm = Opposing force offered by mass of the body (N)

5. Tj – Opposing torque due to moment of inertia.

6. fk = Opposing force offered by the elasticity of the body (N)

6. Tk – Opposing Torque due to elasticity.

7. fb = Opposing force offered by friction of the body (N)

7. Tb – Opposing torque due to friction.

8. M= Mass (Kg) 8. J = Moment of inertia (kg-m/rad)

9. K = Stiffness of Spring( N /m) 9. K - Stiffness of the spring (N-m /

rad)

10. B= Viscous friction coefficient 10. B - rotational frictional

coefficient(N-m/(rad/sec))

Balance equation of idealized elements:

Force Balance Equation Torque Balance Equation 1. An ideal mass element has

negligible friction and elasticity. A force applied on it. The opposing force offered by the mass is proportional to the acceleration.

2

2

dt

tdfm X

2

2

dt

XdMfm f

By Newton’s 2nd Law

1. An ideal mass element has negligible friction and elasticity. A force applied on it. The opposing force offered by the mass is proportional to the angular acceleration.

2

2

dt

dTj

2

2

dt

dJTj

T θ

By Newton’s 2nd Law 2. An ideal frictional element dashpot

which has negligible mass and elasticity. A force applied on it. The dash pot will offered the opposing

2. Negligible moment of inertia and elasticity.

dt

dTb

B

M J



force which is proportional to the velocity.

X f B

dt

dxfb

dt

dxBfb

By Newton’s 2nd Law

dt

dBTb

T θ

21 dt

dTb

21 dt

dBTb T1 θ1 θ2

3. Consider an Ideal elastic spring which has negligible mass and friction

xfk f

X1 X2

kxf k f

If the displacement at both the ends means 2121 XXKfXXf kk

3.

T θ k kT

kTk k

T θ1 θ2

21 kT

TkTk 21

Problems In Mechanical Translational System

1. Write the differential equation governing the mechanical system shown in the fig. And determine the transfer function.

X1 X k1 B

f(t) B1 k B2 Solution:

Transfer function = )(

)(

sF

sX

Step 1: Free Body diagram for M1

M1

M2



The opposing force acting on Mass M1 are fm1, fb, fk1, fb and fk

w.k.t. 2

12

11 dt

Xdmfm

dt

dXBfb

111

111 Xkf k

dt

XXdBfb

1

XXkf k 1

By Newton’s 2nd Law:

Applied force = Opposing Force

0111 kkbbm fffff - - - - - (1)

01

111

112

12

1 XXKdt

XXdBXK

dt

dXB

dt

XdM - - - - - (2)

Taking Laplace transform for equation (2)

0))()(()())()(()()( 11111112

1 sXsXKsXKsXsXBSsSXBsXM

0))()()())()()()( 11111112

1 sKXsKXsXKsBsXsBsXssXBsXsM

])[()]()()[( 112

11 KBssXKKsBBsMsX

)()(

)()()(

112

11 KKsBBsM

KBssXsX

- - - - - (3)

Step 2:

Free Body Diagram for M2

The opposing force acting on Mass M2 are fm2, fb, fk, fb2

w.k.t. 2

2

22 dt

XdMfm

)( 1XXdt

dBfb

)( 1XXKfk

dt

dXBfb 22

By Newton’s 2nd Law:



)(22 tfffff bkbm

)(2112

2

2 tfdt

dxBXXKXX

dt

dB

dt

XdM - - - - - (4)

Taking Laplace transform:

)()(())()(())()(()( 2112

2 sFsXsBsXsXKsXsXBssXsM

Subs X1(s) value in the above equation:

)()()(

)()(])()[(

112

1

2

22

2 sFKKsBBsM

KBssXKsBBsMsX

)()()(

)()()(][)([)(

112

1

112

122

2 sFKKsBBsM

KBsKKsBBsMKsBBsMsX

2

22

2112

1

112

1

)(])()][()([

)()(

)(

)(

kBsKsBBsMKKsBBsM

KKsBBsM

sF

sX

Result:

The differential equations governing the system are

1. 01

111

112

12

1 XXKdt

XXdBXK

dt

dXB

dt

XdM

2. )(2112

2

2 tfdt

dxBXXKXX

dt

dB

dt

XdM

Transfer function of the system is

1. 2

22

2112

1

112

1

)(])()][()([

)()(

)(

)(

kBsKsBBsMKKsBBsM

KKsBBsM

sF

sX

Block diagram reduction technique

Block diagram:

A block diagram is a short hand, pictorial representation of cause and effect

relationship between the input and the output of a physical system. It characterizes the functional

relationship amongst the components of a control system.



Lower case letters usually represent functions of time. Upper case letters denote

Laplace transformed quantities as a function of complex variable sor Fourier transformed

quantity i.e. frequency function as function of imaginary variablejωs .

Summing point: It represents an operation of addition and / or subtraction.

Negative feedback: Summing point is a subtractor.

Positive feedback: Summing point is an adder.

Stimulus: It is an externally introduced input signal affecting the controlled output.

Take off point: In order to employ the same signal or variable as an input to more than block or

summing point, take off point is used. This permits the signal to proceed unaltered along several

different paths to several destinations.

.



1.Find the transfer function of the closed-loop system below.

To simplify the inner feedback loop to obtain the following block diagram.

To combine the two series blocks into one.



To obtain the transfer function for the standard feedback system.

2.Find the transfer function of the closed-loop system below.

To simplify the inner feedback blocks.



To get the following block diagram.

To combine the two sets of series blocks.

Calculate the overall transfer function of the system

SIGNAL FLOW GRAPHS

An alternate to block diagram is the signal flow graph due to S. J. Mason. A signal flow graph is

a diagram that represents a set of simultaneous linear algebraic equations. Each signal flow graph

consists of a network in which nodes are connected by directed branches. Each node represents a

system variable, and each branch acts as a signal multiplier. The signal flows in the direction

indicated by the arrow.



Definitions:

1) Node: A node is a point representing a variable or signal.

2) Branch: A branch is a directed line segment joining two nodes.

3) Transmittance: It is the gain between two nodes.

4) Input node: A node that has only outgoing branche(s). It is also, called as source and

corresponds to independent variable.

5) Output node: A node that has only incoming branches. This is also called as sink and

corresponds to dependent variable.

6) Mixed node: A node that has incoming and out going branches.

7) Path: A path is a traversal of connected branches in the direction of branch arrow.

8) Loop: A loop is a closed path.

9) Self loop: It is a feedback loop consisting of single branch.

10) Loop gain: The loop gain is the product of branch transmittances of the loop.

11) Nontouching loops: Loops that do not posses a common node.

12) Forward path: A path from source to sink without traversing an node more than once.

13) Feedback path: A path which originates and terminates at the same node.

14) Forward path gain: Product of branch transmittances of a forward path.

M ASON ’S GAIN FORMULA: The relation between the i/p variable & the o/p variable of a signal flow graphs

is given by the net gain between the i/p & the o/p nodes and is known as Overall gain of the

system. Mason’s gain form ula for the determ ination of overall system gain is given by,

Properties of Signal Flow Graphs:

1) Signal flow applies only to linear systems.

2) The equations based on which a signal flow graph is drawn must be algebraic equations

in the form of effects as a function of causes.

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

following a succession of causes and effects through the system.

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

branches.



4) The branch directing from node Xk to Xj represents dependence of the variable Xj on Xk

but not the reverse.

5) The signal traveling along the branch Xk and Xj is multiplied by branch gain akj and

signal akjXk is delivered at node Xj.

Guidelines to Construct the Signal Flow Graphs: The signal flow graph of a system is constructed from its describing equations, or by direct

reference to block diagram of the system. Each variable of the block diagram becomes a node

and each block becomes a branch. The general procedure is

1) Arrange the input to output nodes from left to right.

2) Connect the nodes by appropriate branches.

3) If the desired output node has outgoing branches, add a dummy node and a unity gain

branch.

4) Rearrange the nodes and/or loops in the graph to achieve pictorial clarity.

Signal Flow Graph Algebra

Addtion rule

The value of the variable designated by a node is equal to the sum of all signals entering the

node.

Transmission rule

The value of the variable designated by a node is transmitted on every branch leaving the node.

Multiplication rule

A cascaded connection of n-1 branches with transmission functions can be replaced by a single

branch with new transmission function equal to the product of the old ones.

Masons Gain Formula



The relationship between an input variable and an output variable of a signal flow graph is given

by the net gain between input and output nodes and is known as overall gain of the system.

Masons gain formula is used to obtain the over all gain (transfer function) of signal flow graphs.

Gain P is given by

k

kkPP1

Where, Pk is gain of kth forward path,

∆ is determinant of graph

∆=1-(sum of all individual loop gains)+(sum of gain products of all possible combinations of

two nontouching loops – sum of gain products of all possible combination of three

nontouching loops) + ∙∙∙

∆k is cofactor of kth forward path determinant of graph with loops touching kth forward path. It is

obtained from ∆ by removing the loops touching the path Pk.

Example 1 Obtain the transfer function of C/R of the system whose signal flow graph is shown in Fig.1

Figure 1 Signal flow graph of example 1

There are two forward paths:

Gain of path 1 : P1=G1

R

G4

G2

-G3

G1

C 1 1



Gain of path 2 : P2=G2

There are four loops with loop gains:

L1=-G1G3, L2=G1G4, L3= -G2G3, L4= G2G4

There are no non-touching loops.

∆ = 1+G1G3-G1G4+G2G3-G2G4

Forward paths 1 and 2 touch all the loops. Therefore, ∆1= 1, ∆2= 1

The transfer function T = 42324131

212211

1 GGGGGGGG

GGPP

sR

sC

Example 2 Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in Fig.2.


There is one forward path, whose gain is: P1=G1G2G3

There are three loops with loop gains:

L1=-G1G2H1, L2=G2G3H2, L3= -G1G2G3

There are no non-touching loops.

∆ = 1-G1G2H1+G2G3H2+G1G2G3

Forward path 1 touches all the loops. Therefore, ∆1= 1.

R(s) C(s) 1 1 1 G1 G2 G3

H1

-1

-H2



The transfer function T = 321231121

32111

1 GGGHGGHGG

GGGP

sR

sC

Example 3 Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in Fig.3.


There are three forward paths.

The gain of the forward path are: P1=G1G2G3G4G5

P2=G1G6G4G5

P3= G1G2G7


L1=-G4H1, L2=-G2G7H2, L3= -G6G4G5H2 , L4=-G2G3G4G5H2

There is one combination of Loops L1 and L2 which are nontouching with loop gain product

L1L2=G2G7H2G4H1

∆ = 1+G4H1+G2G7H2+G6G4G5H2+G2G3G4G5H2+ G2G7H2G4H1

Forward path 1 and 2 touch all the four loops. Therefore ∆1= 1, ∆2= 1.

Forward path 3 is not in touch with loop1. Hence, ∆3= 1+G4H1.

The transfer function T =

2174225432254627214

14721654154321332211

1

1

HHGGGHGGGGHGGGHGGHG

HGGGGGGGGGGGGGPPP

sR

sC

G1 C(s) R(s)

G7 G6

-H1

G2 G3 G4 G5

-H2

X1 X2 X3 X4 X5

1



Example 4

Find the gains 1

3

2

5

1

6 ,,X

X

X

X

X

X for the signal flow graph shown in Fig.4.

Figure 4 Signal flow graph of MIMO system

Case 1: 1

6

X

X

There are two forward paths.

The gain of the forward path are: P1=acdef

P2=abef


L1=-cg, L2=-eh, L3= -cdei, L4=-bei


L1L2=cgeh

∆ = 1+cg+eh+cdei+bei+cgeh

Forward path 1 and 2 touch all the four loops. Therefore ∆1= 1, ∆2= 1.

The transfer function T = cgehbeicdeiehcg

abefcdefPP

X

X

1

2211

1

6

b

a d c f e

-h

-g

-i

X1 X6 X5

X4 X3 X2



Case 2: 2

5

X

X

The modified signal flow graph for case 2 is shown in Fig.5.

Figure 5 Signal flow graph of example 4 case 2

The transfer function can directly manipulated from case 1 as branches a and f are removed

which do not form the loops. Hence,

The transfer function T=cgehbeicdeiehcg

becdePP

X

X

1

2211

2

5

Case 3: 1

3

X

X

The signal flow graph is redrawn to obtain the clarity of the functional relation as shown in

Fig.6.

b

1 d c 1 e

-h

-g

-i

X2 X5 X5

X4 X3 X2



Figure 6 Signal flow graph of example 4 case 3

There are two forward paths.

The gain of the forward path are: P1=abcd

P2=ac

There are five loops with loop gains:

L1=-eh, L2=-cg, L3= -bei, L4=edf, L5=-befg


L1L2=ehcg

∆ = 1+eh+cg+bei+efd+befg+ehcg

Forward path 1 touches all the five loops. Therefore ∆1= 1.

Forward path 2 does not touch loop L1. Hence, ∆2= 1+ eh

The transfer function T =

ehcgbefgefdbeicgeh

ehacabefPP

X

X

1

12211

1

3

Example1 Draw the signal flow graph of the block diagram shown in Fig.1

G2 G1 G3

H2

H1

− −

R X1 X2 X3 X4 X5 X6 C

a e b f

-h

-g

-i

X1 X5

X4 X3

X2

c

d

X3 1



Figure 7 Multiple loop system Choose the nodes to represent the variables say X1 .. X6 as shown in the block diagram..

Connect the nodes with appropriate gain along the branch. The signal flow graph is shown in

Fig. 2

Figure 8 Signal flow graph of the system shown in Fig. 1

Example 2 Draw the signal flow graph of the block diagram shown in Fig.3.

R X1 X2 X3

X4 X5 X6

C G1

H1

-H2

G2 G3

-1

1 1 1 1

G1

G2

G3

G4

R

−

C

−

X1 X2

X3



Figure 9 Block diagram feedback system The nodal variables are X1, X2, X3. The signal flow graph is shown in Fig. 4.


Example 3 Draw the signal flow graph of the system of equations.

3332321313

223232221212

113132121111

XaXaXaX

ubXaXaXaX

ubXaXaXaX

The variables are X1, X2, X3, u1 and u2 choose five nodes representing the variables. Connect the various nodes choosing appropriate branch gain in accordance with the equations. The signal flow graph is shown in Fig. 5.


R

G4

G2

-G3

G1

C 1 1

X1

X2 X3

u1 b1

b2

a21

a12

a33

a31

X1

X2 a11

a13

a32

X3

u2

a23

a22



Synchros

Figure: Synchro error detector

( ) sinr r cv t V t ; 1

2

3

( ) sin cos( 120 )

( ) sin cos

( ) sin cos( 240 )

s r c

s r c

s r c

v t KV t

v t KV t

v t KV t

;

1 2

2 3

3 1

( ) 3 sin sin( 240 )

( ) 3 sin sin( 120 )

( ) 3 sin sin

s s r c

s s r c

s s r c

v t KV t

v t KV t

v t KV t

When 0 , 3 1( ) 0s sv t and maximum voltage is induced on S2 coil. This position of the rotor is

defined as the “electrical zero” of the transmitter and used as reference position of the rotor.

The output of the synchro transmitter is applied to the stator winding of a “synchro control

transformer”. Circulating current of the same phase but of different magnitude flows through the

two sets of stator coil. The pair acts as an error detector.

The voltage induced in the control transformer rotor is proportional to the cosine of the angle

between the two rotors and is given by,

( ) sin cos ; where, 90r ce t K V t .

( ) sin cos(90 ) sin sin( ) ( )sinr c r c r ce t K V t K V t K V t …(01)

The equation above holds for small angular displacement.

A synchro is an electromagnetic transducer that is used to convert angular shaft position into an electric signal. The basic element of a synchro is a synchro transmitter whose construction is very similar to that of the 3- alternator. An ac voltage is applied to the rotor winding through slip-rings. The schematic diagram of synchro synchro transmitter- control transformer pair is shown above.



Thus the synchro transmitter-control transformer pair acts as an error detector which gives a

voltage signal at the rotor terminal of the control transformer proportional to the angular

difference between the shaft positions.

Equation (01) is represented graphically in figure below for an arbitrary time variation of

( ).

It is seen that the output of the synchro error detector is a modulated signal, where the ac signal

applied to the rotor of synchro transmitter acts as carrier and the modulating signal is,

( ) ( ); m s s re t K K K V

A.C. Servomotor

For low power application a.c. motors are preferred, because of their light weight, ruggedness

and no brush contact. Two phase induction motors are mostly used in control system.

The motor has two stator windings displaced 90 electrical degree apart. The voltages applied to

the windings are not balanced. Under normal operating conditions a fixed voltage from a

constant voltage source is applied to one phase. The other phase, called control phase, is



energized by a voltage of variable magnitude which is 90 out of phase w.r.t. the voltage of fixed

phase.

The torque speed characteristic of the motor is different from conventional motor. X / R ratio is

low and the curve has negative slope for stabilization.



The torque-speed curve is not linear. But we assume it as linear for the derivation of transfer

function. The troque is afunction of both speed and the r.m.s. control voltage, ie., ( , )MT f E .

Using Tailor series expansion about the normal operating point0 0 0( , , )MT E we get,

0 0

0 0

0 0 0( ) ( )M MM M

E E E E

T TT T E E

E

For position control system, 0 0E , 0 0 , 0MT

Thus, the above equation may be simplified as,

0 ;MT kE m J f where, 0

0

M

E E

Tk

E

and 0

0

M

E E

Tm

.

Performing Laplace transform, 20( ) ( ) ( ) ( )kE s ms s Js s f s

Or, 2

0

( )( )

( ) ( ) ( 1)m

m

Ks kG s

E s Js f m s s s

; where,

0m

kK

f m and

0m

J

f m .

Since ‘m’ is negative the transient part is decaying as m is positive. If ‘m’ would positive and

0m f the transient part will increase with time and the system would be unstable. k and m are

the slope of the torque-voltage and torque-speed curve.



UNIT-II

TIME RESPONSE

Time Response:

The time response of the system is the output of the closed loop system as a function of

time. It is denoted by c(t).

It is obtained by solving the differential equation governing the system

Otherwise, it can be obtained from the transfer function of the system and input to the

system.

The closed loop transfer function,)()(1

)(

)(

)(

sHsG

sG

sR

sC

)()(

)(sM

sR

sC

)()()( sMsRsC

From the above equation the output (or) response in S-domain is obtained by product of

the transfer function and the input R(s).

Response in S – domain = C(s)

Response in t – domain = c(t)

Parts in Time Response:

It consists of two parts. They are

1. Transient Response

2. Steady State Response.

1. Transient Response:

It is the response of the system when the input changes from one state to another

2. Steady State Response:

It is the response as time t approaches infinity.

Test Signals:

Input Signals:

It is required to predict the response of a system.

Characteristics of Input Signals:



1. A sudden shock

2. A sudden Change

3. A constant velocity and

4. A constant acceleration

Need of Test Signals:

The test signals which resemble these characteristics are used as input signals to predict

the performance of the system.

Standard test signals are

1. i) Step Signal r(t) = 1; t ≥ 0

ii) Unit step Signal = 0; t < 0

r(t)

A

t

2. i) Ramp Signal r(t) = At; t ≥ 0

ii) Unit Ramp Signal = 0; t < 0

r(t)

A

1 2 t

3. i) Parabolic Signal r(t) = At2 / 2; t ≥ 0

ii) Unit Parabolic Signal = 0; t < 0

r(t)

2A

0.5

1 2 t

4. Impulse Signal δ (t) = 0; t ≠ 0

5. Sinusoidal Signal

Impulse Signal:

A signal of very large magnitude which is available for very short duration is called

impulse signal



It is denoted by δ (t) r(t) = δ (t) r(t) = δ (t)

δ (t) = ∞; t = 0 and Adtt)(

= 0; t ≠ 0

Note: 0 t 0 YΔ t

i) Impulse signal is derivative of a step signal

ii) Laplace transform of the Impulse function is Unity.

Weighting Function:

The response of the system, with input as impulse signal is called weighting function (or

impulse response ) of the system.

Impulse response, m(t) = L-1 {R(s), M(s)}

= L-1 {M(s)} where R(s) = 1

The response for any input can be obtained by convolution of input with impulse

response.

Name of the Signal Time domain r(t) Laplace Transform

Step A A / s

Unit Step 1 1 / s

Ramp At A / s2

Unit Ramp t 1 / s2

Parabolic At2 / 2 A / s3

Unit Parabolic t2 / 2 1 / s3

Impulse δ (t) 1

Note:

The zeroes and poles are critical value of s, at which the function T(s) attains extreme

values 0 or ∞.

1. When s takes the value of a zero, the function T(s) will be zero. [T(s)=0]

2. When s takes the value of a pole, the function T(s) will be ∞. [T(s) = ∞]



Response of First Order System:

Unit Step Input:

The closed loop order system with unity feedback is shown in below figure.

The closed loop transfer function of first order system, TssR

sC

1

1

)(

)(

If the input is Unit step r(t) = 1; R(s) = 1/s.

Ts

sRsC 1

1)()(

sT

sTTss 1

1

1

11

sT

s

TsC1

1

)(

Ts

B

s

A

Tss

TsC11

1

)(

BsTsAT

11

(i) Put s = 0: 001 T

AT

A = 1

(ii) Put S = -1 / T: TB

T1

B = -1

The response in s – domain is given by

TsssC

111

)(

The response in time domain is



TssLscLtc

111

)}({)( 11

Tt

etc1)(

The above equations is the response of the closed loop first order system for unit step

Input

1. For Closed Loop first order system,

Unit Step response = T

te1

Step Response = A( Tt

e1 )

When t = 0, c(t) = 1 – e0 = 1

When t = 1T, c(t) = 1 – e-1 = 0.632

When t = 2T, c(t) = 1 – e-2 = 0.865

When t = 3T, c(t) = 1 – e-3 = 0.95

When t = 4T, c(t) = 1 –e-4 = 0.9817

When t = 5T, c(t) = 1 – e-5 = 0.993

When t = ∞, c(t) = 1 – e-∞ = 1

T – is the time constant of the system

In a time of 5T, the system is assumed to have attained the steady state/

C(t)

r(t) 0.95

0.865

0.632

t=0 t 0 T 2T 3T t

Unit step Input

Second Order System:



ss

ss

ss

ss

ss

sR

sC

n

nn

n

n

n

n

n

n

2

2

2

21

2

)(

)(

2

22

2

2

2

2

2

2

22

2

2)(

)(

nn

n

sssR

sC

Where ωn = Undamped natural frequency in radians per second

ζ = Damping ratio

Damping Ratio:

Definition:

It is defined as the ratio of the actual damping to the critical damping

The response c(t) of second order system depends on the value of damping ratio.

The second order system can be classified into four types depending upon the value of

damping ratio

1. Undamped system ζ = 0

2. Under damped system 0 < ζ < 1

3. Critically damped system ζ = 1

4. Over damped system ζ > 1

The characteristics equation of the second order system is 22 2 nnss

It is a quadratic equation and the roots of this equation is given by

2

442,

222

21nnnss

2

)1(42 22 nn

)1( 2 nn



i) When ζ = 0, s1, s2 = ±jωn; ׀ roots are purely imaginary and the system

is under damped

ii) When ζ = 1, s1, s2 = -ωn ׀ roots are real and equal and the system is

critically damped

iii) When ζ > 1, s1, s2 = 12 nn roots are real and unequal and the system is over ׀

damped

iv) When 0 < ζ < 1, s1, s2 = 12 nn

)1(1 2 nn

21 nn j

dn j

Here ωd is called damping frequency of Oscillation

Response of Undamped Second Order system:

For Unit Step Input:

22

2

2)(

)(...

nn

n

sssR

sCtkw

For Undamped system ζ = 0

22

2

)(

)(

n

n

ssR

sC

When the input is unit step, r(t) = 1 and R(s) = 1/s

The response in s – domain 22

2

)()(n

n

ssRsC

22

21

n

n

ss



)( 22

2

n

n

ss

2222

2

)( nn

n

s

B

s

A

ss

BssA nn )( 222

i) Put s = 0

22nn A

A = 1

ii) Put s2 = - 2n s = j n nn Bj 2

Bj n

n 2

B = -jωn

B = -s

2222

)(1)(

nn s

s

ss

B

s

AsC

2222)(cos

1

nn s

stL

s

s

s

22

11 1)}({

ns

s

sLsCL

c(t) = 1 - cosωnt

r(t) c(t)

2

1 1

Input (Unit Step) output (Response) t



Every practical system has same amount of damping. Hence undamped system does not

exist in practice.

The response of the undamped closed loop second order system is

For Unit step Response = 1 - cosωnt

Step Response = A (1 - cosωnt)

The response is completely oscillatory.

Response of Under Damped Second Order System:

For Unit Step Input:

The standard form of closed loop transfer function of second order system is

22

2

2)(

)(

nn

n

sssR

sC

For under damped system, 0 < ζ < 1 and root of the denominator are complex conjugate.

The roots of the denominator are

12 nns

Since ζ < 1, ζ2 is also less than 1

Therefore, the damped frequency of oscillation, 21 nd

dn js

The response in s – domain, 22

2

2)()(

nn

n

sssRsC

For unit step input, r(t) = 1 and R(s) = 1/s

)2()(

22

2

nn

n

ssssC

)2()2()(

2222

2

nnnn

n

ss

CBs

s

A

ssssC



scBsssA nnn )()2( 222

1. Put s = 0

22nn A

A = 1

2. Equating the co-efficient of s2.

CsBsssA nnn 2222 )2(

0 = A + B

0 = 1 + B

B = -1

3. Equating co-efficient of s

0 = 2Aωnζ + C

= 2ωnζ + C

C = - 2ωnζ

22 2

21)(

nn

n

ss

s

ssC

Let us add and subtract 22n to denominator of the second term

222222 2

21)(

nnnn

n

ss

s

ssC

222222 2

21

nnnn

n

ss

s

s

2222)(

21

nnn

n

s

s

s

)1()(

21222

nn

n

s

s

s



22)(

21

dn

n

s

s

s

22 )()(

1

dn

n

dn

n

ss

s

s

Multiply and divide by ωd

2222 )()(

1)(

dn

d

d

n

dn

n

ss

s

ssC

The response in time domain is given by

222211

)()(

1)}({)(

dn

d

d

n

dn

n

ss

s

sLsCLtc

tete dt

d

nd

t nn sincos1

tte dd

nd

tn sincos1

tte d

n

nd

tn sin1

cos12

tte ddtn

sin1

cos12

tte

tc dd

tn

sincos11

1)( 2

2

Note:

On constructing on right angle triangle with ζ and 21 , we get



1

21

ζ

sin θ = 21

cos θ = ζ

2-1

tan

tt

etc dd

tn

sincoscossin1

1)(2

)sin(

11)(

2

t

etc d

tn

2

1 1tan

For closed loop under damped second order system:

Unit step response =

)sin(1 2

te

d

tn

Step Response =

)sin(1 2

te

A d

tn

r (t) c(t)

1 1

Input t Output Response t

θ



The response of the under damped second order system for unit step input sketched and

observed that the response oscillates before settling to a final value. The oscillation depends on

the value of damping ratio.

Response of Critically Damped second Order System for Step Input:

The standard form of closed loop transfer function of second order system is,

22

2

2)(

)(

nn

n

sssR

sC

For critically damping ζ = 1

22

2

2)(

)(

nn

n

sssR

sC

2

2

)( n

n

s

When input is unit step, r(t) = 1 and R(s) = 1/s

2

2

)()()(

n

n

ssRsC

2

2

)(

1)(

n

n

sssC

2

2

2 )()()()(

n

n

nn sss

C

s

B

s

AsC

ssCBssA nnn )()( 22

i) Put s = 0

022 nn A

ii) Put s = -ωn

))(()( 22nnnnnnn CBA



nn B 2

nB

iii) Equating the co-efficient of s2

0 = A + C

= 1 + C

C = -1

)(

1

)(

1

)()()(

22nn

n

nn ssss

C

s

B

s

AsC

The response in time domain,

nn

n

sssLscLtc

1

)(

1)}({)(

211

tt

nnn etetc 1)(

]1[1)( tetc ntn

For closed loop critically damped second order system,

Unit Step Response = ]1[1 te ntn

Step Response = ]1[1 teA ntn

r (t) c(t)

1

1

Input t Response t

Time response (Transient ) Specification (Time domain) Performance :-



The performance characteristics of a controlled system are specified in terms of the

transient response to a unit step i/p since it is easy to generate & is sufficiently drastic.

The transient response of a practical C.S often exhibits damped oscillations before

reaching steady state. In specifying the transient response characteristic of a C.S to unit step

i/p, it is common to specify the following terms.

1) Delay time (td)

2) Rise time (tr)

3) Peak time (tp)

4) Max over shoot (Mp)

5) Settling time (ts)

Response curve

1) Delay time :- (td)

It is the time required for the response to reach 50% of its final value for the 1st

time.



2) Rise time :- (tr)

It is the time required for the response to rise from 10% and 90% or 0% to 100% of

its final value. For under damped system, second order system the 0 to 100% rise time is

commonly used. For over damped system, the 10 to 90% rise time is commonly used.

3) Peak time :- (tp)

It is the time required for the response to reach the 1st of peak of the overshoot.

4) Maximum over shoot :- (MP)

It is the maximum peak value of the response curve measured from unity. The amount

of max over shoot directly indicates the relative stability of the system.

5) Settling time :- (ts)

It is the time required for the response curve to reach & stay with in a range about the

final value of size specified by absolute percentage of the final value (usually 5% to 2%).

The settling time is related to the largest time const., of C.S.

Transient response specifications of second order system :-

W. K.T. for the second order system,

T.F. = C(S) = Wn2 . ------------------------------(1)

R(S) S2+2 WnS+ Wn2

Assuming the system is to be under damped (< 1)

Rise time tr

W. K.T. C (tr) = 1- e-Wnt Cos Wdtr + . Sin wdtr

1-2

Let C(tr) = 1, i.e., substituting tr for t in the above Eq:



Then, C(tr) = 1 = 1- e-Wntr Cos wdtr + . sin wdtr

1-2

Cos wdtr + . sin wdtr = tan wdtr = - 1-2 = wd .

1-2

jW

Thus, the rise time tr is , jWd

tr = 1 . tan-1 - w d = - secs Wn 1-2 Wn

Wd wd

When must be in radians. -

Wn

Peak time :- (tp)

Peak time can be obtained by differentiating C(t) W.r.t. t and equating that

derivative to zero.

dc = O = Sin Wdtp Wn . e-Wntp

dt t = t p 1-2

Since the peak time corresponds to the 1st peak over shoot.

Wdtp = = tp = .

Wd

The peak time tp corresponds to one half cycle of the frequency of damped oscillation.

Maximum overshoot :- (MP)

The max over shoot occurs at the peak time.

S- Plane



i.e. At t = tp = .

Wd

Mp = e –(/ Wd) or e –( / 1-2)

Settling time :- (ts)

An approximate value of ts can be obtained for the system O < <1 by using the

envelope of the damped sinusoidal waveform.

Time constant of a system = T = 1 .

Wn

Setting time ts = 4x Time constant.

= 4x 1 . for a tolerance band of +/- 2% steady state.

Wn

Delay time :- (td)

The easier way to find the delay time is to plot Wn td VS . Then approximate

the curve for the range O<< 1 , then the Eq. becomes,

Wn td = 1+0.7

td = 1+0.7 Wn

PROBLEMS:

(1) Consider the 2nd order control system, where = 0.6 & Wn = 5 rad / sec, obtain the

rise time tr, peak time tp, max overshoot Mp and settling time ts When the system is subject

to a unit step i/p.,

Given :- = 0.6, Wn = 5rad /sec, tr = ?, tp = ?, Mp = ?, ts = ?

Wd = Wn 1- 2 = 5 1-(0.6) 2 = 4

= Wn = 0.6 x 5 = 3.

tr = -, = tan-1 Wd = tan-1 4 = 0.927 rad

Wd 3



tr = 3.14 – 0.927 = 0.55sec.

4

tp = = 3.14 = 0.785 sec.

Wd 4

. = e / Wd

MP = e 1- 2

MP = e (3/4) x 3.14 = 0.094 x 100 = 9.4%

ts :- For the 2% criteria.,

ts = 4 . = 4 . = 1.33 sec.

Wn 0.6x5

For the 5% criteria.,

ts = 3 = 3 = 1 sec

3

3) The open loop T.F. of a unity feed back system is given by G(S) = K . where,

S(1+ST)

T&K are constants having + Ve values.By what factor (1) the amplitude gain be reduced so

that (a) The peak overshoot of unity step response of the system is reduced from 75% to 25%

(b) The damping ratio increases from 0.1 to 0.6.

Solution: G(S) = K .

S(1+ST)

Let the value of damping ratio is, when peak overshoot is 75% & when peak

overshoot is 25%



Mp = .

e 1- 2

ln 0. 75 = . 0.0916 = . 1- 2 1- 2

1 = 0.091 (0.0084) (1-2) = 2

2 = 0.4037 (1.0084 2) = 0.0084

= 0.091

k .

S + S2T .

w.k.t. T.F. = G(S) = 1 + K . = K .

1+ G(S) . H(S) S + S2T S + S2T+K

T.F. = K / T .

S2 + S + K .

T T

Comparing with std Eq :-

Wn = K . , 2 Wn = 1 .

T T

Let the value of K = K1 When = 1 & K = K2 When = 2.

Since 2 Wn = 1 . , = 1 . = 1 .

T 2TWn 2 KT

1 .

1 . = 2 K1T = K2 .

2 1 K1

2 K2T



0.091 = K2 . K2 . = 0.0508

0.4037 K1 K1

K2 = 0.0508 K1

a) The amplitude K has to be reduced by a factor = 1 . = 20

0.0508

b) Let = 0.1 Where gain is K1 and

= 0.6 Where gain is K2

0.1 = K2 . K2 . = 0.027 K2 = 0.027 K1

0.6 K1 K1

The amplitude gain should be reduced by 1 . = 36

0.027

4) Find all the time domain specification for a unity feed back control system whose open loop

T.F. is given by

G(S) = 25 .

S(S+6)

Solution:

25 .

G(S) = 25 . G(S) . = S(S+6) .

S(S+6) 1 + G(S) .H(S) 1 + 25 .

S(S+6)

= 25 .

S2 + ( 6S+25 )

W2n = 25 , Wn = 5, 2 Wn = 6 = 6 . = 0.6

2 x 5



Wd = Wn 1- 2 = 5 1- (0.6)2 = 4

tr = - , = tan-1 Wd = Wn = 0.6 x 5 = 3

Wd

= tan-1 ( 4/3 ) = 0.927 rad.

tp = . = 3.14 = 0.785 sec.

Wd 4

MP = . = 0.6 . x3.4 = 9.5%

e 1- 2 e 1- 0.62

ts = 4 . for 2% = 4 . = 1.3 ………3sec.

Wn 0.6 x 5

5) The closed loop T.F. of a unity feed back control system is given by

C(S) = 5 .

R(S) S2 + 4S +5

Solution:

C(S) = 5 . , Wn2 = 5 Wn = 5 = 2.236

R(S) S2 + 4S +5

2Wn = 4 = 4 . = 0.894. Wd = 1.0018

2 x 2.236

MP = . = 0.894 . X 3.14 = 0.19%

e 1- 2 e 1-(0.894)2

Determine (1) Damping ratio (2) Natural undamped response frequency Wn. (3) Percent peak over shoot Mp (4) Expression for error resoponse.



W. K.T. C(t) = e-Wnt Cos Wdtr + . sin wdtr

1-2

= e-0.894x2.236t Cos 1.0018t + 0.894 . sin 1.0018t

1-(0.894)2

6) A servo mechanism is represent by the Eq:-

d2 + 10 d = 150E , E = R- is the actuating signal calculate the

dt2 dt value of damping ratio, undamped and damped

frequency of ascillation.

Soutions:- d2 + 10 d = 15 ( r - ) , = 150r – 150.

dt2 dt

Taking L.T., [S2 + 10S + 150] (S) = 150 R (S).

(S) = 150 .

R(S) S2 + 10S + 15O

Wn2 = 150 Wn = 12.25. ………………………….rad sec .1

2Wn = 10 = 10 . = 0.408.

2 x 12.25

Wd = Wn 1 - 2 = 12.25 1- (0.408)2 = 11.18. rad 1sec.

6) Fig shows a mechanical system and the response when 10N of force is applied to the

system. Determine the values of M, F, K,.



K x(t)inmt

f(t) 0.00193 The T.F. of the mechanical system is , 0.02 X(S) = 1 .

F(S) MS2 + FS = K

f(t) = Md2X + F dX + KX

F x dt2 dt

F(S) = (MS2 + FS + K) x (S)

1 2 3 4 5

Given :- F(S) = 10

S.

X(S) = 10 .

S(MS2 + FS + K)

SX (S) = 10 .

MS2 + FS + K

The steady state value of X is By applying final value theorem,

lt. SX(S) = 10 . = 10 = 0.02 ( Given from Fig.)

S O M(0) + F (0) + K K. ( K = 500.)

MP = 0.00193 = 0.0965 = 9.62%

0.02

MP = e . ln 0.0965 = . 1 - 2 1 - 2

= . 0.5539 = 2 . 0.744

1 - 2 1 - 2

M



0.5539 – 0.5539 2 = 2

= 0.597 = 0.6

tp = = .

Wn 1 – 2 Wd

3 = . Wn = 1.31…… rad / Sec.

Wn1 – (0.6)2

Sx(S) = 10/ M .

(S2 + F S + K )

M M

Comparing with the std. 2nd order Eq :-, then,

Wn2 = K Wn = K (1.31)2 = 500 . M = 291.36 kg.

M M M

F = 2Wn F = 2 x 0.6 x 291 x 1.31

M F = 458.7 N/M/ Sec.

8) Measurements conducted on sever me mechanism show the system response to be c(t) =

1+0.2e-60t – 1.2e-10t , When subjected to a unit step i/p. Obtain the expression for closed

loop T.F the damping ratio and undamped natural frequency of oscillation .

Solution:

C(t) = 1+0.2e-60t –1.2e-10t

Taking L.T., C(S) = 1 . + 0.2 . – 1.2 .



S S+60 S+10

C(S) . = 600 / S .

S2 + 70S + + 600

Given that :- Unit step i/p r(t) = 1 R(S) = 1 .

S

C(S) . = 600 / S .

R(S) S2 + 70S + + 600

Comparing, Wn2 = 600, 24.4 …..rad / Sec

2 Wn = 70, = 70 . = 1.428

2 x 24.4

Steady state Error :-

Steady state errors constitute an extremely important aspect of system

performance. The state error is a measure of system accuracy. These errors arise from the nature

of i/p’s type of system and from non-linearties of the system components. The steady state

performance of a stable control system are generally judged by its steady state error to step, ramp

and parabolic i/p.

Consider the system shown in the fig.

R(S) C(S)

G(S)

H(S)



C(S) = G(S) . …………………………(1)

R(S) 1+G(S) . H(S)

The closed loop T.F is given by (1). The T.F. b/w the actuating error signal e(t) and the

i/p signal r(t) is,

E(S) = R(S) – C(S) H(S) = 1 – C(S) . H(S)

R(S) R(S) R(S)

= 1 – G(S) . H(S) . = 1 + G(S) . H(S) – G(S)H(S)

1 + G(S) . H(S) 1+G(S) . H(S)

= 1 .

1 + G(S) . H(S)

Where e(t) = Difference b/w the i/p signal and the feed back signal

E(S) = 1 . .R(S) ……………………….(1)

1 + G(S) . H(S)

The steady state error ess may be found by the use of final value theorem

and is as follows;

ess = lt e(t) = lt SE(S)

t S O

Substituting (1), ess = lt S.R(S) . ……………….(2)

S O 1+G(S) . H(S)

Eq :- (2) Shows that the steady state error depends upon the i/p R(S) and the forward T.F.

G(S) and loop T.F G(S) . H(S).



The expression for steady state errors for various types of standard test signals are

derived below;

1) Steady state error due to step i/p or position error constant (Kp):-

The steady state error for the step i/p is

I/P r(t) = u(t). Taking L.T., R(S) = 1/S.

From Eq:- (2), ess = lt S. R(s) . = 1 .

S O 1 +G(S). H.S 1 + lt G(S). H(S)

S O

lt G(S) . H(S) = Kp

(S O )

Where Kp = proportional error constant or position error const.

ess = 1 .

1 + Kp

(1 + Kp) ess = 1 Kp = 1 - ess

ess

Note :- ess = R . for non-unit step i/p

1 + Kp

2) Steady state error due to ramp i/p or static velocity error co-efficient (Kv) :-

The ess of the system with a unit ramp i/p or unit velocity i/p is given by,

r ( t) = t. u(t) , Traking L -T, R(S) = 1/S2

Substituting this to ess Eq:-

ess = lt S . . 1 . = lt 1 .

S O 1 + G(S) . H(s) S2 S O S +S G(S) H(s)S

lt = SG(S) . H(S) = Kv = velocity co-efficient then

S O

ess = lt 1 . ess = 1 .



S O S + Kv Kv

Velocity error is not an error in velocity, but it is an error in position error due to a ramp

i/p

3) Steady state error due to parabolic i/p or static acceleration co-efficient (Ka) :-

The steady state actuating error of the system with a unit parabolic i/p (acceleration i/p)

which is defined by r(t) + 1 . t2 Taking L.T. R(S)= 1 .

2 S3

ess = lt S . 1 . lt 1 .

S O 1 + G(S) . H(S) S3 S O S2 + S2 G(S) . H(S)

lt S2 G(S) . H(S) = Ka.

S O

ess = lt 1 . = 1 .

S O S2 + Ka Ka

Note :- ess = R . for non unit parabolic.

Ka

Types of feed back control system :-

The open loop T.F. of a unity feed back system can be written in two std, forms;

1) Time constant form and 2) Pole Zero form,

G(S) = K (TaS +1) (TbS +1)…………………..

Sn(T1 S+1) (T2S + 1)……………….



Where K = open loop gain.

Above Eq:- involves the term Sn in denominator which corresponds to no, of

integrations in the system. A system is called Type O, Type1, Type2,……….. if n = 0, 1,

2, ………….. Respectively. The Type no., determines the value of error co-efficient. As

the type no., is increased, accuracy is improved; however increasing the type no.,

aggregates the stability error. A term in the denominator represents the poles at the origin

in complex S plane. Hence Index n denotes the multiplicity of the poles at the origin.

The steady state errors co-efficient for a given type has definite values. This is

illustration as follows.

1) Type – O system: - If, n = 0, the system is called type – 0, system. The

Steady state error is as follows;

Let, G(S) = K. [ .. . H(s) = 1]

S + 1

ess (Position) = 1 . = 1 . = 1 .

1 + G (O). H(O) 1 + K 1 + Kp

.. . Kp = lt G(S) . H(S) = lt K . = K

S O S O S + 1

ess (Velocity) = 1 . = 1 . =

Kv O

Kv = lt G(S) . H(S) = lt S K . = O.

S O S O S + 1

ess (acceleration) = 1 . = 1 . =

Ka O



Ka = lt S2 G(S) . H(S) = lt S2 K . = O.

S O S O S + 1

2) Type 1 –System :- If, n = 1, the ess to various std, i/p, G(S) = K .

S (S + 1)

ess (Position) = 1 . = O

1 +

Kp = lt G(S) . H(S) = lt K . =

S O S O S( S + 1)

Kv = lt S K . = K

S O S(S+1)

ess (Velocity) = 1 .

K

ess (acceleration) = 1 . = 1 . =

O O

Ka = lt S2 K . = O.

S O S (S + 1)

3) Type 2 –System :- If, n = 2, the ess to various std, i/p, are , G(S) = K .

S2 (S + 1)

Kp = lt K . =

S O S2 (S + 1)

. . . ess (Position) = 1 . = O



Kv = lt S K . =

S O S2 (S + 1)

. . . ess (Velocity) = 1 . = O

Ka = lt S2 K . = K.

S O S2 (S + 1)

. . . ess (acceleration) = 1 .

K

STEADY STATE ERRORS IN UNITY FEEDBACK CONTROL SYSTEMS

Errors in control systems may be attributed to many factors. Changes in the reference input will

cause unavoidable errors during transient periods, and also cause steady state errors. Any

physical control system inherently suffers from steady state error in response to certain types of

inputs. Whether a given system will exhibit steady state error for a given input depends on the

type of the open loop transfer function of the system.

Classification of Control Systems

Control systems may be classified according to their ability to follow step inputs, ramp inputs,

parabolic inputs etc. The magnitudes of the steady state errors due to these individual inputs are

indicative of the goodness of the system. Consider the unity feedback control system with the

following open loop transfer function G(s):

1s

pT1s

2T1s

1TNs

1sm

T1sb

T1sa

TKsG

It involves the term sN in the denominator, representing the pole of multiplicity N at the origin

which also indicates the number of integrations in the open loop. A system is called Type 0,

Type 1, Type 2 …. If N = 0, 1, 2, … respectively. As the type number is increased accuracy is

improved; however, increasing the type number aggravates the stability problem.



Steady State Errors

A unity feedback closed loop control system is shown in Fig.1.

Figure 12 Unity feedback control system

The closed loop transfer function is

sG1

sG

sR

sC

The transfer function between the error signal e(t) and input signal r(t) is

sG1

1

sR

sC1

sR

sE

, Which can be rearranged as sRsG1

1sE

The final value theorem provides a convenient way to find the steady state error

sG1

ssRlimssElimtelime

0s0stss

Three types of static error constants are

1) Static position error constant Kp due to unit step input.

2) Static velocity error constant Kv due to unit ramp input.

3) Static acceleration error constant Ka due to unit parabolic input

Which indicate the figures of merit of control systems?

Static Position Error Constant

The steady state error of the system for a unit step input is

0G1

1

s

1

sG1

slime

0sss

The static position error constant Kp is defined by

0GsGlimK0sp

−

R(s) G(s)

E(s) C(s)



Thus, the steady state error in terms of the static position error constant Kp is given by

pss K1

1e

For a type 0 system,

K

1sp

T1s2

T1s1

T

1sm

T1sb

T1sa

TK

limK0sp

and

K1

1e

ss

For a type 1 or higher system

1N

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TK

limKN0sp

and ss

e

Response of a feedback control system to a step input involves a steady state error if there

is no integration in the feed forward path. If a zero steady state error for a step input is desired,

the type of the system must be one or higher.

Static Velocity Error Constant

The steady state error of the system with a unit ramp input is given by

ssG

1lim

s

1

sG1

slime

02

0sss

s

The static velocity error constant Kv is defined by

ssGlimK0sv

Thus, the steady state error in terms of the static velocity error constant Kv is given by

vss K

1e

The term velocity error is used here to express the steady state error for a ramp input. The

velocity error is an error in position due to ramp input.




0

1sp

T1s2

T1s1

T

1sm

T1sb

T1sa

TsK

limK0sv

and

vss K

1e

For a type 1 system

K

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TsK

limK0sv

and

K

1

K

1e

vss

For a type 2 or

higher system

2N

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TsK

limKN0sv

and 0K

1e

vss

Type 0 system is incapable of following a ramp input in the steady state. Type 1 system

with unity feedback can follow the ramp input with a finite error. Type 2 and higher order

systems can follow a ramp input with zero steady state error.

Static Acceleration Error Constant

The steady state error of the system with a unit parabolic input is given by

sGs

1lim

s

1

sG1

slime

20

30s

ss

s

The static acceleration error constant Ka is defined by

sGslimK 2

0sa

Thus, the steady state error in terms of the static acceleration error constant Ka is given by

ass K

1e




0

1sp

T1s2

T1s1

T

1sm

T1sb

T1sa

TKs

limK

2

0sa

and

ass K

1e

For a type 1 system

0

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TKs

limK

2

0sa

and

ass K

1e

For a type 2 system

K

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TKs

limK2

2

0sa

and K

1

K

1e

ass

For a type 3 or higher system

3N

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TKs

limKN

2

0sa

and 0K

1e

ass

Type 0 and type 1 systems are incapable of following a parabolic in the steady state. Type 2

system with unity feedback can follow the parabolic input with a finite error. Type 3 and higher

order systems can follow a parabolic input with zero steady state error.

Table 1 shows the summary of error constants and steady state errors for unit step, unit

ramp and unit parabolic inputs to a unity feedback loop.

Table 1: Summary of steady state errors for unity feedback systems

Type of

system

Error constants Steady state error ess

Kp Kv Ka Unit step

input

Unit ramp

input

Unit parabolic

input

0 K 0 0 1/(1+K)

1 K 0 0 1/K

2 K 0 0 1/K

3 0 0 0



Note:

1) The step, ramp, parabolic error constants are significant for the error analysis only

when the input signal is a step, ramp and parabolic function respectively.

2) Since the error constants are defined with respect to forward path transfer function

G(s), the method is applicable to a unity feedback system only.

3) Since error analysis relies on use of final value theorem, it is important first to

check to see if sE(s) has any poles on the j axis or in the right half of s-plane.

4) Principle of superposition can be used if combination of the three basic inputs is

present.

5) If the configuration differs, we can either simplify to unity feedback system or

establish error signal and apply final value theorem.

Examples:

For a negative unity feedback system determine the steady state error due to unit step, unit ramp

and unit parabolic input of the following systems.

102sss

4s12s1KsGiii)

2004sss

KsGii)

2s10.1s1

50sGi)

22

2

Solution



ass

2

0sv

vss0sv

pss0sp

K

1e G(s)slimK

K

1e sG(s)limK

K1

1e G(s)limK

Table 2 shows the results of error constants and steady state errors.

Table 2 Results of example

Problem Error constants Steady state error ess

Kp Kv Ka Unit step

input

Unit ramp

input

Unit parabolic

input

i 50 0 0 1/51

ii K/200 0 0 200/K

iii K/10 0 0 10/K

CONTROL ACTIONS

An automatic controller compares the actual value of the system output with the reference input

(desired value), determines the deviation, and produces a control signal that will reduce the

deviation to zero or a small value. The manner in which the automatic controller produces the

control signal is called the control action. The controllers may be classified according to their

control actions as

1) Two position or on-off controllers

2) Proportional controllers

3) Integral controllers

4) Proportional-plus- integral controllers

5) Proportional-plus-derivative controllers

6) Proportional-plus-integral-plus-derivative controllers



A two position controller has two fixed positions usually on or off.

A proportional control system is a feedback control system in which the output forcing function

is directly proportional to error.

A integral control system is a feedback control system in which the output forcing function is

directly proportional to the first time integral of error.

A proportional-plus-integral control system is a feedback control system in which the output

forcing function is a linear combination of the error and its first time integral.

A proportional-plus-derivative control system is a feedback control system in which the output

forcing function is a linear combination of the error and its first time derivative.

A proportional-plus-derivative-plus-integral control system is a feedback control system in

which the output forcing function is a linear combination of the error, its first time derivative

and its first time integral.

Many industrial controllers are electric, hydraulic, pneumatic, electronic or their

combinations. The choice of the controller is based on the nature of plant and operating

conditions. Controllers may also be classified according to the power employed in the operation

as

1) Electric controllers

2) Hydraulic controllers

3) Pneumatic controllers

4) Electronic controllers.

The block diagram of a typical controller is shown in Fig.1. It consists of an automatic

controller, an actuator, a plant and a sensor. The controller detects the actuating error signal and

amplifies it. The output of a controller is fed to the actuator that produces the input to the plant

according to the control signal. The sensor is a device that converts the output variable into



another suitable variable to compare the output to reference input signal. Sensor is a feedback

element of the closed loop control system.

Figure 13 Block diagram of control system

Two Position Control Actions

In a two position control action system, the actuating element has only two positions which are

generally on and off. Generally these are electric devices. These are widely used are they are

simple and inexpensive. The output of the controller is given by Eqn.1.

0

0

2

1

teU

teUtu ….. (1)

Where, U1 and U2 are constants (U2= -U1 or zero)

The block diagram of on-off controller is shown in Fig. 2

−

Actuating Error signal e(t)

Reference Input r(t)

Error Detector

Feed back signal b(t)

Controller Output u(t)

−

Output c(t)

automatic controller

actuating error signal e(t)

reference input r(t)

error detector

Amplifier Actuator Plant

Sensor feed back signal b(t)

Controller output u(t)



Figure 14 Block diagram of on off controller

Proportional Control Action

The proportional controller is essentially an amplifier with an adjustable gain. For a controller

with proportional control action the relationship between output of the controller u(t) and the

actuating error signal e(t) is

teKtup

..... (2)

Where, Kp is the proportional gain.

Or in Laplace transformed quantities

pK

sE

sU ….. (3)

Whatever the actual mechanism may be the proportional controller is essentially an

amplifier with an adjustable gain. The block diagram of proportional controller is shown in

Fig.3.

Figure 15 Block diagram of a proportional controller

Integral Control Action

The value of the controller output u(t) is changed at a rate proportional to the actuating error

signal e(t) given by Eqn.4

t

0ii

dtteKtuor teKdt

tdu …..(4)

−



Error Detector



Kp



Where, Ki is an adjustable constant.

The transfer function of integral controller is

s

K

sE

sU i ….(5)

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

value of u(t) remains stationary. The integral control action is also called reset control. Fig.4

shows the block diagram of the integral controller.

Figure 16 Block diagram of an integral controller

Proportional-plus-Integral Control Action

The control action of a proportional-plus-integral controller is defined by

t

0i

p

pdtte

T

KteKtu …..(6)

The transfer function of the controller is

sT

11K

sE

sU

ip

.....(7)

Where, Kp is the proportional gain, Ti is the integral time which are adjustable.

The integral time adjusts the integral control action, while change in proportional gain

affects both the proportional and integral action. The inverse of the integral time is called reset

rate. The reset rate is the number of times per minute that a proportional part of the control action

is duplicated. Fig.5 shows the block diagram of the proportional-plus-integral controller. For an

actuating error of unit step input, the controller output is shown in Fig.6.

−



Error Detector



s

Ki



Figure 17 Block diagram of a proportional-plus-integral control system

Figure 18 Response of PI controller to unit actuating error signal

Proportional-plus-Derivative Control Action

The control action of proportional-plus-derivative controller is defined by

dt

tdeTKteKtu

dpp …..(8)

The transfer function is

sT1KsE

sUdp

…..(9)

Where, Kp is the proportional gain and Td is a derivative time constant.

−



Error Detecto

r



sT

sT1K

i

ip

t 0

Kp

u(t)

2Kp

proportional only

P-I control action

Ti



Both, Kp and Td are adjustable. The derivative control action is also called rate control. In

rate controller the output is proportional to the rate of change of actuating error signal. The

derivative time Td is the time interval by which the rate action advances the effect of the

proportional control action. The derivative controller is anticipatory in nature and amplifies the

noise effect. Fig.7 shows the block diagram of the proportional-plus-derivative. For an actuating

error of unit ramp input, the controller output is shown in Fig.8

Figure 19 Block diagram of a proportional-plus-derivative controller

Figure 20 Response of PD controller to unit actuating error signal

Proportional-plus-Integral-plus-Derivative Control Action

It is a combination of proportional control action, integral control action and derivative control

action. The equation of the controller is

dt

tdeTKdtte

T

KteKtu

dp

t

oi

p

p …..(10)

or the transfer function is

t 0

u(t)

proportional only

P-D control action

Td

−



error detector

feed back signal b(t)


ST1KDP



sT

sT

11K

sE

sUd

ip

…..(11)

Where, Kp is the proportional gain, Ti is the integral time, and Td is the derivative time. The

block diagram of PID controller is shown in Fig.9. For an actuating error of unit ramp input, the

controller output is shown in Fig.10.

Figure 21 Block diagram of a proportional-plus-integral-plus-derivative controller

Figure 22 Response of PID controller to unit actuating error signal

t 0

u(t)

proportional only

PD control action PID control action

−



error detector

feed back signal b(t)


sT

sTTsTK

d

diip

21



UNIT-III

FREQUENCY RESPONSE ANALYSIS

FREQUENCY RESPONSE

The Frequency response is the steady state response (output) of a system when the input

to the system is a sinusoidal signal.The frequency response of a system is normally obtained by

varying the frequency of the input signal by keeping the magnitude of the input signal at a

constant value.

In a system transfer function T(s),if s is replaced by jω then the resulting transfer

function T(jω) is called sinusoidal transfer function. The frequency response of the system can

be directly obtained from the sinusoidal transfer function T(jω) of the system.

Open loop transfer function, G(jω) => | G(jω)| G(jω) G(jω)

Loop transfer function, G(jω)H(jω) =>| G(jω) H(jω) | G(jω) H(jω)

Closed loop transfer function,C(jω)/R(jω)=> M((jω)= | M((jω)| M((jω)|

ADVANTAGES OF FREQUENCY RESPONSE ANALYSIS

1. The absolute and relative stability of the closed loop system can be estimated from the

knowledge of the open loop frequency response

2. The practical testing of system can be easily carried with available sinusoidal signal

generators and precise measurements equipments

3. The transfer function of complicated function can be determined experimentally by

frequency response test

4. The design and parameter adjustment can be carried more easily

5. The corrective measure for noise disturbance and parameter variation can be easily

carried

6. It can be extended to certain non-linear systems.

FREQUENCY DOMAIN SPECIFICATIONS



The performance and characteristics of a system in frequency domain are measured in

terms of frequency domain specifications. The frequency domain specifications are

1. Resonant peak,Mr

2. Resonant frequency, ωr

3. Bandwidth

4. cut-off rate

5. phase margin,Kp

Peak response Mp :

The peak response Mp is defined as the maximum value of M() that is given in

Eq.(1.4). In general, the magnitude of Mp gives an indication of the relative stability of a feed

back control system. Normally, a large Mp corresponds to a large peak overshoot in the step

response. For most design problems it is generally accepted that an optimum value Mp of should

be somewhere between 1.1 & 1.5.

Resonant frequency p :

The resonant frequency p is defined as the frequency at which the peak resonance Mp occurs.

Bandwidth :

The bandwidth , BW, is defined as the frequency at which the magnitude of M(j), M(),

drops at 70.7 percent of its zero-frequency level, or 3 dB down from the zero-frequency gain. In

general, the bandwidth of a control system indicates the noise-filtering characteristics of the

system. Also, bandwidth gives a measure of the transient response properties, in that a large

bandwidth corresponds to a faster rise time, since higher-frequency signals are passed on to the

outputs. Conversely, if the bandwidth is small, only signals of relatively low frequencies are

passed, & the time response will generally be slow & sluggish.

Cutoff rate:

Often, bandwidth alone is inadequate in the indication of the characteristics of the system in

distinguishing signals from noise. Sometimes it may be necessary to specify the cutoff rate of the

frequency response at the higher frequencies. However, in general, steep cutoff characteristics

may be accompanied by a large Mp, which corresponds to a system with a low stability margin.

The performance criteria defined above for the frequency-domain analysis are illustrated on

the closed-loop frequency response, as shown in Fig. 1.3.



There are other criteria defined that may be used to specify the relative stability &

performance of a feedback control system. These are defined in the ensuring sections of this

chapter.

M() Mp 1 Bandwidth 0 p BW Fig.1.3. Typical magnification curve of a feedback control system. Mp, , p & the bandwidth of a second-order system: For a second-order feedback control system, the peak resonance Mp, the resonant frequency p,

& the bandwidth are all uniquely related to the damping ratio & the natural undamped

frequency n of the system. Consider the second-order sinusoidal steady-state transfer function

of a closed-loop system,

C(j) 2n

M(j) = = R(j) (j)2 + 2 n (j) + 2

n

1 = 1 + j 2 (/n) - (/n)2

We may simplify the last expression by letter u = / n. The Eq. (1.6) becomes 1 M(ju) = 1 + j2 u - u2

The magnitude & phase of M (j) are 1 M(ju) = M(u) = [( 1 – u2 )2 + ( 2 u)2] ½



And 2 u M(ju) = m(u) = - tan -1 1 – u2

The resonant frequency is determined first by taking the derivative of M(u) with respect to u &

setting it equal to zero. Thus

dM(u) 1 = - [( 1 – u2 )2 + ( 2 u)2] –3/2 ( 4 u3 – 4u + 8u2) = 0 du 2 from which 4u3 – 4u + 8u2 = 0 The root of Eq. (1.11) are up = 0 and up = 1 - 22 The solution Eq. (1.12) merely indicates that the slope of the M() versus curve is zero at = 0; it is not true maximum. The solution of eq. (1.13) gives the resonant frequency, p = n 1 - 22 (1.14) Since frequency is a real quantity, Eq. (1.14) is valid only for 1 22 or 0.707. This means

simply that for all values of greater than 0.707, the solution of p = 0 becomes the valid one, &

Mp = 1.

Substituting Eq. (1.13) into Eq. (1.8) & simplifying, we get 1 Mp = 2 1 - 2 It is important to note that Mp is a function of only, whereas p is a function of & n



5 4 3 Mp 2 1 0 0.5 0.707 1.0 1.5 2.0 Damping ratio 1 Fig.1.4 Mp versus-damping ratio for a second – order system, Mp = 2 1 - 2 1.0 0.8 up = p /n 0.6 0.4 0.2 0 0.5 0.707 1.0 Damping ratio Fig 1.5. Normalized resonant frequency- versus-damping ratio for a second order system,



Up = 1 - 2 . Fig.1.4 & 1.5 illustrate the relationship between Mp & , & u = p / n & , respectively. Bandwidth: Bandwidth BW of a system is a frequency at which M() drops to 70.7% of its zero

frequency level or 3 dB down from the zero frequency gain. Equating the Eq.

1 M(u) = = 0.707 [( 1 –u2)2 + ( 2u)2] ½

1 = [( 1 –u2)2 + ( 2u)2] ½ = 0.707 = 2. Squaring both sides ( 1 –u2)2 + 42u2 = 2 1 + u4 – 2u2 + 42u2 = 2

Let u2 = x 1 + x2 – 2x + 42 x = 2 x2 – 2x + 42 x = 1

x2 – x ( 2 - 42 ) = 1 x2 – x ( 2 - 42 ) – 1 = 0 a = 1, b = - ( 2 - 42 ) , c = -1 -b b2 – 4ac x = 2a (2 – 42 ) (2 – 42 )2 + 4 = 2 (2 – 42 ) (4 + 164 – 16 2 + 4 =



2 (2 – 42 ) 16 4 – 16 2 + 8 = 2 2 (1 – 22 ) 4 + 164 – 16 2 + 4 = 2 2 (1 – 22 ) 22 + 44 – 42 = 2 2 (1 – 22 ) 22 + 44 – 42 = 2 2 [(1 – 22 ) 2 + 44 – 42] = 2 u2 = x = (1 – 22 ) 2 + 44 – 4 2 / n = u = [(1 – 22 ) (2 + 44 – 4 2 ] 1/2 BW = n [ ( 1 – 22 ) + 44 - 42 +2]1/2

For the second order system under consideration, we easily establish some simple relationship

between the time – domain response & the frequency-domain response of the system.

1. The maximum over shoot of the unit step response in the time domain depends upon

only.

2. The response peak of the closed - loop frequency response Mp depends upon only.

3. The rise time increases with , & the bandwidth decreases with the increase of , for a

fixed n, therefore, bandwidth & rise time are inversely proportional to each other.

4. Bandwidth is directly proportional to n.

5. Higher bandwidth corresponds to larger Mp.



Determination of closed loop response from open loop response

• The steady-state response of a linear system to sinusoidal excitation may be determined

from the system transfer function T(s) by replacing s by j

• The resulting frequency response function T(j) has the polar form :

• For any value of , T(j) reduces to a complex number whose amplitude, A(j) gives

, and () gives .

\ Steady-state vs. general response • In control work. the response of the system to general input functions, e.g. step input,

impulse input, are of most interest (rather than steady state sinusoidal response)

• However, with Fourier analysis, any time function can be represented as a series sum of

sinusoids :

Therefore it is possible to predict the nature of a response, y(t) to any given x(t) from a

knowledge of how the gain, A, and phase shift, , vary with angular frequency

• In particular, the closed-loop behaviour of a feedback system may be usefully predicted

from the open-loop gain and phase-shift characteristics

Open loop response: Bode plots

T(s) X(s ) Y(s)

)()()( jeAjT

)(

)(

sX

sY

)(

)(arg

sX

sY

22

2)(

2 ss

sT

......)2()()( 22110 tSintSintx



• Amplitude is plotted in dB

• Phase is plotted in degrees

• Used to identify system order, corner frequencies, etc

Open loop response: Polar plots • Polar plots are more useful in control system design work

– open-loop polar plot indicates whether the closed-loop system is stable

– also gives a measure of how stable the system is

• Represent T(jw) on an Argand diagram :

• Polar plot is traced out by the trajectory of T(jw) as w increases progressively from zero

Re Im

T(jw)

A(jw)

)

fai(jw)

T(s)-plane



• A polar plot may be constructed from experimental data or from a system transfer

function

• If values of w are marked along the contour, a polar plot has the same information as a

Bode plot .Usually, the shape of a polar plot is of most interest.

FREQUENCY RESPONSE PLOTS Frequency response analysis of control system can be carried either analytically (or)

graphically. The various graphical techniques available for frequency response analysis are

1. Bode plot

2. Polar plot

3. Nichols plot

4. M & N Circles

5. Nichols chart

Among the above five, Bode plot, Polar plot and Nichols plot are used to get the frequency

response of closed loop system

BODE PLOT It is one of the plot used to obtain the frequency response for the open loop or closed loop

system it consist of

1. Log magnitude plot

2. Phase magnitude plot

The log magnitude plot can be drawn with the help of magnitude in decibel and varying

frequencies (in angular)

The phase angle plot ca be drawn with the help of phase angle and varying frequencies (in

angular)

ADVANTAGE OF BODE PLOT The main advantage of the bode plot is that multiplication of magnitudes can be

converted into addition. also simple method for sketching an approximate log=magnitude curve

is available



THE FACTORS USED TO DRAW THE BODE PLOT ARE GIVEN BELOW

1. Constant gain factor, K

2. Integral factor, K / jω or K / (jω)n

3. Derivative factor, K(jω) or K (jω) n

4. First order factor found in dr

5. First order factor found in nr

6. Second order factor found in dr

7. Second order factor found in nr

CONSTANT GAIN FACTOR, K Let G(S) = K

G(jω)= K=K 0˚

A=|G(jω)| in db

=20 log K

Φ= G(jω)|= 0˚

When K>1,20 log K is Positive

When 0<K<1,20 log K is Negative

When K=1,20 log K is Zero



INREGRAL FACTOR Let G(S) = K/s

G(jω)= K/(jω)=K /ω -90˚

A=|G(jω)| in db

=20 log K/ω

Φ= G(jω)|= -90˚

When ω =0.1K, A=20 log K is 20db

When ω =K , A=20 log K is Zero db

When ω =10K , A=log K is -20db

Let G(S) = K/sn

G(jω)= K/(jω) n=K /ω -90n˚

A=|G(jω)| in db

=20 log K/ωn=20 log [K1/n/ω]n

Φ= G(jω)|= -90n˚

Now the magnitude plot of integral factor is a straight line with a slope of -20ndb/dec and

passing through zero db when ω=KI/n.the phase plot is a straight line at -90 n˚



DERIVATIVE FACTOR Let G(S) = Ks

G(jω)= K(jω) =Kω -90

A=|G(jω)| in db

=20 log Kω=20 log Kω

Φ= G(jω)|= +90˚

When ω =0.1/K, A=20 log K is -20db

When ω =1/K , A=20 log K is Zero db

When ω =10/K , A=log K is +20db

Let G(S) = Ksn

G(jω)= K(jω)n =Kωn -90˚

A=|G(jω)| in db

=20 log Kωn=20 log Kω

Φ= G(jω)|= +90n˚



FIRST ORDER FACTOR IN DENOMINATOR G(S) =1/(1+ST)

G(jω) =1/(1+jωT)=1/√(1+ω2T2) -tan-1ωT

A=|G(jω)|in db=20log1/√(1+ω2T2)= -20 log√(1+ω2T2)

At very low frequencies ωT<<1

A=-20log√(1+ω2T2)≈-20log1=0

At very high frequencies ωT>>1

A=-20 log√(1+ω2T2)≈-20log√ ω2T2=-20log√ ω2T2=-20logωT

At ω=1/T,A=-20log1=0

At ω=10/T,A=-20log1=-20db

The actual magnitude at corner frequency, ωc=1/T is A=-20log√1+1=-3db

Hence by approximation the loss in db at corner frequency is 3db

The phase angle is obtained by calculating the phase angle of G(jω) for various values of ω

Phase angle, Φ= G(jω)=-tan-1 ωT

At corresponding, ω=ωc=1/T, Φ= -tan-1 ωT=-tan-11=-45˚

At ω0, Φ0 and At ω∞, Φ=-90˚



G(S)=1/(1+ST)m=1/(√(1+ ω2T2)m m tan-1 ωT

A=|G(j ω)| in db=20log 1/√(1+ ω2T2)m=-20 m log1/√(1+ ω2T2) Φ= G(jω)= -m tan-1 ωT

FIRST ORDER FACTOR IN THE NUMERATOR G(S)=1+ST

G(jω)=1+ jωT=/√(1+ ω2T2) tan-1 Ωt

A= |G(jω)| in db=20 log√(1+ ω2T2)



Φ= G(jω)= -tan-1 ωT

QUADRATIC FACTOR IN DENOMINATOR G(S)=1/(1+2ζs/ωn+(s/ωn)2)

G(jω)=1/(1+j2ζω/ωn+(jω/ωn)2=1/(1-(ω/ωn)2+j2 ζω/ωn)=1/√(1- ω2/ωn2)2+4ζ2 ω2/ωn

2

-tan-1(2ζ ω/ωn)/(1- ω2/ωn2)

A=|G(jω)| in db

A=20log 1//√(1- ω2/ωn2)2+4ζ2 ω2/ωn

2=- -20log√(1- ω2/ωn2(2-4ζ2)+ ω4/ωn

4)

At low frequencies when ω<< ωn

A= -20 log √(1- ω2/ωn2(2-4ζ2)+ ω4/ωn

4)≈-20log 1=0

At very high frequencies, when ω>> ωn

A=-20 log √(1- ω2/ωn2(2-4ζ2)+ ω4/ωn

4≈-20log√ ω4/ωn4=-20log ω2/ωn

2

A =-40 log ω/ωn

At ω= ωn,,A=-40log 1=0db

At ω= 10 ωn=-40log10=-40db



Φ= G(jω)= tan-1 2 ζ ω/ωn At ω= ωn Φ=-90˚

1-ω2/ωn2 At ω0 Φ0˚

ω ∞ Φ180˚

QUADRATURE FACTOR IN THE NUMERATOR

G(S)=1+2ζs/ωn+(s/ωn)2

G(jω)= √(1- ω2/ωn2)2+4ζ2 ω2/ωn

2 -tan-1(2ζ ω/ωn)/(1- ω2/ωn2)

ζ=0.1

ζ=0.3

ζ=0..5.

ζ=1.0

ζ=0.1 ζ=0.3

ζ=0..5

ζ=1.

0

0˚

-90˚

-180˚

+db

-db

A

Log ω



PROBLEMS 1. Sketch bode plot for the following transfer function and determine the system gain K for the

gain cross over frequency to be 5rad/sec

G(S) = Ks2/(1+0.2s)(1+0.02s)

Solution:

The sinusoidal transfer function G(j ω) is obtained by replacing S by j ω in the given S-domain

transfer function

G(j ω)=K(j ω)2/(1+0.2 j ω)(1+0.002 j ω)

Let K=1, G(j ω)= (j ω)2/(1+0.2 j ω)(1+0.002 j ω)

MAGNITUDE PLOT The corner frequencies are ωc1=1/0.2=5rad/sec and ωc2=1/0.02=50rad/sec

The various values of G(j ω) are below tabulation , in the increasing order of their corner

frequencies. also slope contributed by each term and the change in slope at the corner frequency

ζ=0.5

ζ=1.0

ζ=0.3

ζ=0.1

ζ=1.0

ζ=0.3

ζ=0.5

ζ=0.1

A in db

Log ω



TREM CORNER FREQUENCY

SLOPE IN db/dec CHANGE IN SLOPE db/dec

(jω)2 ----- +40 1/(1+j0.2) ωc1=1/0.2=5 -20 40-20=20 ωc2=1/0.02=50 -20 20-20 Choose a low frequency ωl such that ωl< ωc1& Choose a high frequency ωh such that

ωh> ωc2.Let ωl=0.5 rad/sec and ωh=100 rad/sec

Let A=|G(jω)| in db

Let us calculate A at ωl, , ωc1, ωc2, ωh

At ω= ωl, A=20 log|(jω)2| =20 log(ω2)=20 log(0.5)2= -12db

At ω= ωc1, A=20 log|(jω)2|=20log(5)2=28db

At ω= ωc2, A=[slope from ωc1 to ωc2×log ωc2/ωc1]+A (at ω= ωc1)

=20log50/5+28=48db

At ω= ωh,A=[slope from ωc2 to ωh×log ωh/ωc2]+A (at ω= ωc2)

=0×log (100/50)+48=48db

Let points a,b,c,d be the points corresponding to frequencies ωl, , ωc1, ωc2, ωh respectively on the

magnitude plot.

PHASE PLOT The phase angle G(jω) as a function of ω is given by Φ= G(jω)= 180˚-tan-10.2ω-tan-10.02 ω The phase angle of G(jω) are calculated for various values of ω ω rad/sec Tan-1(0.2 ω) deg Tan-1(0.02 ω) deg Φ= G(jω)

0.5 5.7 0.6 173.7≈174 1 11.3 1.1 167.6≈168 5 45 5.7 129.3≈130 10 63.4 11.3 105.3≈106 50 84.3 45 50.7≈50 100 87.1 63.4 29.5≈30 CALCULATION OF K Given that the gain crossover frequency is 5rad/sec

At ω=5,gain= 28db

At ω=5,frequency the db gain should be zero



20log K=-28db

Log K=-28/20

K=10-28/20=0.0398

NOTE: The frequency ω rad/sec is a corner frequency .hence in the exact plot the db gain at

ω=5rad/sec will be 3db less than the approximate plot.therefore for exact plot the 20logK=-25

LogK=-25/20=>K=10-25//20=0.0562

2. Given that G(S)=Ke-0.2s/s(s+2)(s+8). Find K so that the system is stable with

A) Gain margin equal to 6db

b) Phase margin equal to 45˚

Solution:

G(S)= Ke-0.2s/s(s+2)(s+8)= Ke-0.2s/s×2(1+s/2)×8(1×s/8)

Let K=1; G(j ω)= e-0.2jω/j ω(j ω+2)( j ω+8)

NOTE:

|0.0625 e-0.2jω|=0.0625 and e-0.2jω= -0.2 ω rad/sec

The various values of G(j ω) are below tabulation , in the increasing order of their corner

frequencies.also slope contributed by each term and the change in slope at the corner frequency

TREM CORNER FREQUENCY

SLOPE IN db/dec CHANGE IN SLOPE db/dec

0.0625/jω ----- -20 1/(1+j0.2) ωc1=1/0.2=5 -20 -20-20=-40 ωc2=1/0.02=50 -20 -40-20=-60 Choose a low frequency ωl such that ωl< ωc1& Choose a high frequency ωh such that

ωh> ωc2.Let ωl=0.5 rad/sec and ωh= 50 rad/sec

Let A=|G(jω)| in db

Let us calculate A at ωl, , ωc1, ωc2, ωh

At ω= ωl, A=20 log|0.0625/jω|=20log0.0625/0.5=-18db

At ω= ωc1, A=20 log|0.0625/jω|=20log 0.0625/2=-30db

At ω= ωc2, A=[slope from ωc1 to ωc2 × log ωc2/ ωc1]+A at ω= ωc1

=-40×log8/2+(-30)=-54db

At ω= ωh, A=[slope from ωc2 to ωh × log ωh/ ωc2]+A at ω= ωc2

=-60log50/8+(-54)=-102db



Example

Problem : Construct the polar plot for the (critically damped system)

defined by :

Solution :

Limiting conditions :

(i)

(ii)

(iii)

0j1)j(T:0 o

2

180j

1

e0)j(T.e.i

)j(T:

2j)j(T:1

Re-0.2 0 0.2 0.4 0.6 0.8 1

-0.6

-0.4

-0.2

Im

2)1s(

1)s(T

2)j1(

1)j(T

w 0 0.25 0.5 0.75 1.0 1.5 2.0 3.0 5.0

T(jw) 1 +j0

0.83 -j0.44

0.48 -j0.64

0.18 -j0.61

0 -j0.5

-0.12 -j0.28

-0.12 -j0.16

-0.08 -j0.06

-0.04 -j0.01



ssT

1)(

Re

Im

1

22

2

2)(

nn

n

sssT

Re

Im



UNIT –IV

STABILITY OF CONTROL SYSTEM

Introduction

The stability of a linear closed loop system can be determined from the locations of closed loop poles in the S-plane. If the system has closed loop T.F. C(s) 10 R(s) (S+2) (S+4) 1 Output response for unit step input R(s) = S C(s) 10 A B C S(S+2) (S+4) S S+2 S+4 Find out partial fractions 1 1 1 C(s) = 8 4 8 10 S S+2 S+4

= 1.25 2.5 1.25

S S+2 S+4 C(s) = 1.25 – 2.5e-2t+1.25e-4t =Css + Ct (t) If the closed loop poles are located in left half of s-plane, Output response contains exponential

terms with negative indices will approach zero & output will be the steady state output.

i.e. Ct (t) = 0 t Transient output = 0 Such system is called absolutely stable systems. Now let us have a system with one closed loop pole located in right half of s- plane C(s) 10 R(s) S(S-2)(s+4) A + B + C 10 = S S-2 S + 4 C(t) = - 1.25 + 0.833e 2t + 0.416e – 4 t Here there is one exponential term with positive in transient output



Therefore Css = - 1.25

t C(t) 0 0 1 + 4.91 2 + 44.23 4 +2481.88

From the above table, it is clear that output response instead of approaching to steady state value

as t due to exponential term with positive index, transients go on increasing in amplitude.

So such system is said to be unstable. In such system output is uncontrollable & unbounded one.

Output response of such system is as shown in fig(4).

C(t) C(t) OR Steady state output t (a) fig(4) t (b) For such unstable systems, if input is removed output may not return to zero. And if the input

power is turned on, output tends to . If no saturation takes place in system & no mechanical

stop is provided then system may get damaged.

If all the closed loop poles or roots of the characteristic equation lie in left of s-plane, then in the

output response contains steady state terms & transient terms. Such transient terms approach to

zero as time advances eventually output reaches to equilibrium & attains steady state value.

Transient terms in such system may give oscillation but the amplitude of such oscillation will be

decreasing with time & finally will vanish. So output response of such system is shown in fig5

(a) & (b).



C(t) C(t) Damped oscillations Steady state ----------------------- --------------------------- OR Steady state output t t (a) fig 5 (b) BIBO Stability: This is bounded input bounded output stability. Definition of stable system:

A linear time invariant system is said to be stable if following conditions are satisfied.

1. When system is excited by a bounded input, output is also bounded & controllable.

2. In the absence of input, output must tend to zero irrespective of the initial conditions.

Unstable system:

A linear time invariant system is said to be unstable if,

1. for a bounded input it produces unbounded output.

2. In the absence of input, output may not be returning to zero. It shows certain output

without input.

Besides these two cases, if one or more pairs simple non repeated roots are located on the

imaginary axis of the s-plane, but there are no roots in the right half of s-plane, the output

response will be undamped sinusoidal oscillations of constant frequency & amplitude. Such

systems are said to be critically or marginally stable systems.



3. Complex

conjugate with negative real part

J x J 1

-a1 x -J 2

C(t) t Damped oscillation

Absolutely stable.

4. Complex conjugate with positive real part

J1 x -J 1 -x

Ct

t

oscillations with increasing amplitude

Unstable.

Sl.No

Nature of closed loop poles.

Location of closed loop poles in s-plane

Step response Stability condition

1. Real negative i.e in LHS of splane

J x x -02 -01

C(t) --------------------- t

Absolutely stable

2.

Real positive in RHS of s-plane

J x a1

C(t) ------------------------ t increasing towards

Unstable



5. Non repeated pair on imaginary axis

J x J1 x -J2

OR J X J2

x J1 x –J1 x – J2

Two non repeated pairs on imaginary axis.

C(t) t C(t) t Sustained oscillations with two frequencies 1 & 2

Marginally or critically stable Marginally or critically stable

6. Repeated pair on imaginary axis

J x x J1

x x -J1

C(t) ------------------------ t oscillation of increasing amplitude

Unstable



Routh – Hurwitz Criterion:

This represents a method of determining the location of poles of a characteristics

equation with the respect to the left half & right half of the s-plane without actually solving the

equation.

The T.F.of any linear closed loop system can be represented as,

C(s) b0 sm + b1 sm-1 +….+ bm = R(s) a0 sn + a1 sn-1 + …. + an

Where ‘a’ & ‘b’ are constants.

To find the closed loop poles we equate F(s) =0. This equation is called as Characteristic

Equation of the system.

F(s) = a0 sn + a1 sn-1 + a2 sn-2 + ….. + an = 0.

Thus the roots of the characteristic equation are the closed loop poles of the system which decide

the stability of the system.

Necessary Condition to have all closed loop poles in L.H.S. of s-plane. In order that the above characteristic equation has no root in right of s-plane, it is necessary

but not sufficient that,

1. All the coefficients off the polynomial have the same sign.

2. Non of the coefficient vanishes i.e. all powers of ‘s’ must be present in descending order

from ‘n’ to zero.

These conditions are not sufficient.

Hurwitz’s Criterion :

The sufficient condition for having all roots of characteristics equation in left half of s-plane is

given by Hurwitz. It is referred as Hurwitz criterion. It states that:



The necessary & sufficient condition to have all roots of characteristic equation in left half of s-

plane is that the sub-determinants DK, K = 1, 2,………n obtained from Hurwitz determinant ‘H’

must all be positive.

Method of forming Hurwitz determinant:

a1 a3 a5 …….. a2n-1 a0 a2 a4 ..…… a2n-2 0 a1 a3 .……. a2n-3 0 a0 a2 …….. a2n-4 H = 0 0 a1 ……... a2n-5 - - ……... - - - - ……... - 0 - - ……. an

The order is n*n where n = order of characteristic equation. In Hurwitz determinant all

coefficients with suffices greater than ‘n’ or negative suffices must all be replaced by zeros.

From Hurwitz determinant subdeterminants, DK, K= 1, 2, ….n must be formed as follows:

a1 a3 a5 D1 = a1 D2 = a1 a3 D3 = a 0 a2 a4 DK = H a0 a2 0 a1 a3 For the system to be stable, all above determinants must be positive. Determine the stability of the given characteristics equation by Hurwitz,s method.



Ex 1: F(s)= s3 + s2 + s1 + 4 = 0 is characteristic equation. a0 = 1, a1 = 1, a2 = 1, a3 = 4, n = 3 a1 a3 a5 1 4 0 H = a 0 a2 a4 = 1 1 0 0 a1 a3 0 1 4 D1 = 1 = 1 1 4 D2 = 1 1 = -3 1 4 0 D3 = 1 1 0 = 4 –16 = -12. 0 1 4 As D2 & D3 are negative, given system is unstable. Disadvantages of Hurwitz’s method:

1. For higher order system, to solve the determinants of higher order is very complicated &

time consuming.

2. Number of roots located in right half of s-plane for unstable system cannot be judged by

this method.

3. Difficult to predict marginal stability of the system.

Due to these limitations, a new method is suggested by the scientist Routh called Routh’s

method. It is also called Routh-Hurwitz method.

Routh’s Stability Criterion:

It is also called Routh’s array method or Routh-Hurwitz’s method

Routh suggested a method of tabulating the coefficients of characteristic equation

in a particular way. Tabulation of coefficients gives an array called Routh’s array.



Consider the general characteristic equation as,

F(s) = a0 sn + a1 sn-1 + a2 sn-2 + ….. + an = 0. Method of forming an array :

Sn a0 a2 a4 a6 ………. Sn-1 a1

a3 a5 a7 Sn-2 b1 b2 b3 Sn-3 c1 c2 c3 - - - - - - - - S0 an

Coefficients of first two rows are written directly from characteristics equation.

From these two rows next rows can be obtained as follows.

a1 a2 – a0 a3 a1 a4 – a0 a5 a1 a6 – a0 a7 b1 = , b2 = , b3 =

a1 a1 a1 From 2nd & 3rd row , 4th row can be obtained as b1 a3 – a1 b2 b1 a5 – a1 b3 C1 = , C2 = b1 b1

This process is to be continued till the coefficient for s0 is obtained which will be an. From this

array stability of system can be predicted.



Routh’s criterion : The necessary & sufficient condition for system to be stable is “All the terms in the first

column of Routh’s array must have same sign. There should not be any sign change in first

column of Routh’s array”.

If there do sign changes exist then?

1. System is unstable.

2. The number of sign changes equals the number of roots lying in the right half of the

S-plane.

Examine the stability of given equation using Routh’s method: Ex.2: s3+6s2 + 11s + 6 =0

Sol: a0 = 1, a1 = 6, a2 =11, a3 = 6, n = 3

S3 1 11 S2 6 6 S1 11 * 6 – 6 =10 0 6 S0 6 As there is no sign change in the first column, system is stable. Ex. 3 s3 + 4s2 + s + 16 = 0 Sol: a0 =1, a1 = 4, a2 = 1, a3 = 16 S3 1 1 S2 +4 16 S1 4 - 16 = -3 0 4 S0 +16 As there are two sign changes, system is unstable.

Number of roots located in the right half of s-plane = number of sign changes = 2.



Special Cases of Routh’s criterion: Special case 1 :

First element of any of the rows of Routh’s array is zero & same remaining rows contains at

least one non-zero element.

Effect : The terms in the new row become infinite & Routh’s test fails.

e.g. : s5 + 2s4 + 3s3 + 6s2 + 2s + 1 = 0 S5 1 3 2 S4 2 6 1 Special case 1 Routh’s array failed S3 0 1.5 0 S2 …. … Following two methods are used to remove above said difficulty. First method: Substitute a small positive number ‘’ in place of a zero occurred as a first element in the row. Complete the array with this number ‘’. Then examine lim Sign change by taking . Consider above Example. 0 S5 1 3 2 S4 2 6 1 S3 1.5 0 S2 6 - 3 1 0 S1 1.5(6 - 3) 0 - (6 - 3) S0 1



To examine sign change, Lim Lim = 6 - 3 = 6 - 3 0 0 = 6 - = - sign is negative. Lim 1.5(6 – 3) - 2 = Lim 9 - 4.5 - 2 0 6 -3 0 6 - 3 = 0 – 4.5 – 0 0 -3 = + 1.5 sign is positive Routh’s array is, S5 1 3 2 S4 2 6 1 S3 + 1.5 0 S2 - 1 0 S1 +1.5 0 0 S0 1 0 0 As there are two sign changes, system is unstable.

Second method:

To solve the above difficulty one more method can be used. In this, replace‘s’ by ‘1/Z’ in

original equation. Taking L.C.M. rearrange characteristic equation in descending powers of ‘Z’.

Then complete the Routh’s array with this new equation in ‘Z’ & examine the stability with this

array.

Consider F(s) = s5 + 2s4 + 3s3 + 6s2 + 2s + 1 = 0 Put s = 1 / Z



1 + 2 + 3 + 6 + 2 + 1 = 0 Z5 Z4 Z3 Z2 Z Z5 + 2Z4+ 6Z3+3Z2+2Z+ 1 = 0 Z5 1 6 2 Z4 2 3 1 Z3 4.5 1.5 0 Z2 2.33 1 0 Z1 - 0.429 0 Z0 1 As there are two sign changes, system is unstable. Special case 2 : All the elements of a row in a Routh’s array are zero.

Effect : The terms of the next row can not be determined & Routh’s test fails.

S5 a b c

S4 d e f

S3 0 0 0 Row of zeros, special case 2

This indicates no availability of coefficient in that row.

Procedure to eliminate this difficulty :

1. Form an equation by using the coefficients of row which is just above the row of zeros.

Such an equation is called an Auxillary equation denoted as A(s). For above case such

an equation is,

A(s) = ds4 + es2 + f



Note that the coefficients of any row are corresponding to alternate powers of ‘s’ starting

from the power indicated against it.

So ‘d’ is coefficient corresponding to s4 so first term is ds4 of A(s).

Next coefficient ‘e’ is corresponding to alternate power of ‘s’ from 4 i.e. s2 Hence the term es 2 & so on.

2. Take the derivative of an auxillary equation with respect to ‘s’.

i.e. dA(s)

= 4d s3 + 2e s

ds

3. Replace row of zeros by the coefficients of dA(s) ds

S5 a b c

S4 d e f S3 4d 2e 0

4. Complete the array of zeros by the coefficients. Importance of auxiliary equation: Auxiliary equation is always the part of original characteristic equation. This means the roots of

the auxiliary equation are some of the roots of original characteristics equation. Not only is this

but roots of auxillary equation the most dominant roots of the original characteristic equation,

from the stability point of view. The stability can be predicted from the roots of A(s)=0 rather

than the roots of characteristic equation as the roots of A(s) = 0 are the most dominant from the

stability point of view. The remaining roots of the characteristic equation are always in the left

half & they do not play any significant role in the stability analysis.

e.g. Let F(s) = 0 is the original characteristic equation of say order n = 5.



Let A(s) = 0 be the auxiliary equation for the system due to occurrence of special case 2 of

the order m = 2.

Then out of 5 roots of F(s) = 0, the 2 roots which are most dominant (dominant means very

close to imaginary axis or on the imaginary axis or in the right half of s-plane) from the stability

point of view are the 2roots of A(s) = 0. The remaining 5 –2 = 3 roots are not significant from

stability point of view as they will be far away from the imaginary axis in the left half of s-plane.

The roots of auxillary equation may be,

1. A pair of real roots of opposite sign i.e.as shown in the fig. 8.10 (a).

j j x x x x Fig. 8. 10 (b) Fig 8. 10(a)

2. A pair of roots located on the imaginary axis as shown in the fig. 8.10(b).

3. The non-repeated pairs of roots located on the imaginary axis as shown in the

fig.8.10 (c).

j j x xx x x

x xx

Fig. 8.10(c) Fig. 8.10(d).



4. The repeated pairs of roots located on the imaginary axis as shown in the Fig.8.10

(d).Hence total stability can be determined from the roots of A(s) = 0, which can be out

of four types shown above.

Change in criterion of stability in special case 2 :

After replacing a row of zeros by the coefficients of dA(s) , complete the Routh’s array.

ds

But now, the criterion that, no sign in 1st column of array for stability, no longer remains

sufficient but becomes a necessary. This is because though A(s) is a part of original

characteristic equation, dA(s) is not, which is in fact used to complete the array.

ds

So if sign change occurs in first column, system is unstable with number of sign changes

equal to number of roots of characteristics equation located in right half of

s-plane.

But there is no sign changes, system cannot be predicted as stable. And in such case

stability is to be determined by actually solving A(s) = 0 for its roots. And from the location

of roots of A(s) = 0 in the s-plane the system stability must be determined. Because roots

A(s) = 0 are always dominant roots of characteristic equation.

Application of Routh’s of criterion: Relative stability analysis: If it is required to find relative stability of system about a line s = - . i.e. how many roots

are located in right half of this line s = - , the Routh’s method can be used effectively.

To determine this from Routh’s array, shift the axis of s – plane & then apply Routh’s

array i.e. substitute s = s 1 - , ( = constant) in characteristic equation. Write polynomial in

terms of s1. Complete array from this new equation. The number of sign changes in first

column is equal to number of roots those are located to right of the vertical line

s = - .



Imaginary j - 0 Determining range of values of K :

In practical system, an amplifier of variable gain K is introduced.

The closed loop transfer function is

C(s) KG(s) = R(s) 1+ KG(s) H(s)

Hence the characteristic equation is

F(s) = 1+ KG(s) H(s) = 0 So the roots of above equation are dependent on the proper selection of value of ‘K’.

So unknown ‘K’ appears in the characteristic equation. In such case Routh’s array is to be

constructed in terms of K & then the range of values of K can be obtained in such away that it

will not produce any sign change in first column of the Routh’s array. Hence it is possible to

obtain the range of values of K for absolute stability of the system using Routh’s criterion. Such

a system where stability depends on the condition of parameter K, is called conditionally stable

system.

Advantages of Routh’s criterion: Advantages of routh’s array method are:

1. Stability of the system can be judged without actually solving the characteristic equation.

2. No evaluation of determinants, which saves calculation time.

3. For unstable system it gives number of roots of characteristic equation having

Positive real part.



4. Relative stability of the system can be easily judged.

5. By using the criterion, critical value of system gain can be determined hence

Frequency of sustained oscillations can be determined.

6. It helps in finding out range of values of K for system stability.

7. It helps in finding out intersection points of roots locus with imaginary axis.

Limitation of Routh’s criterion:

1. It is valid only for real coefficients of the characteristic equation.

2. It does not provide exact locations of the closed loop poles in left or right half of s-plane.

3. It does not suggest methods of stabilizing an unstable system.

4. Applicable only to linear system.

Ex.1. s6 + 4s5 +3s4 – 16s2- 64s – 48 = 0 Find the number of roots of this equation with positive

real part, zero real part & negative real part

Sol: S6 1 3 -16 -48 S5 4 0 -64 0

S4 3 0 -48 0 S3 0 0 0 dA A(s) = 3S4 – 48 = 0 = 12s3

ds

S6 1 3 -16 -48 S5 4 0 -64 0

S4 3 0 -48 0 S3 12 0 0 0 S2 ( )0 -48 0 0 S1 576 0 0 0 S0 -48



Lim 576

0 + Therefore One sign change & system is unstable. Thus there is one root in R.H.S of the s – plane

i.e. with positive real part. Now

solve A(s) = 0 for the dominant roots

A(s) = 3s4 – 48 =0 Put S2 = Y 3Y2 = 48 Y2 =16, Y = 16 = 4 S2 = + 4 S2 = -4 S = 2 S = 2j So S = 2j are the two parts on imaginary axis i.e. with zero real part. Root in R.H.S. indicated

by a sign change is S = 2 as obtained by solving A(s) = 0. Total there are 6 roots as n = 6.

Roots with Positive real part = 1

Roots with zero real part = 2

Roots with negative real part = 6 –2 – 1 = 3

Ex.2 : For unity feed back system, k G(s) = , Find range of values of K, marginal value of K S(1 + 0.4s) ( 1 + 0.25 s) & frequency of sustained oscillations. Sol : Characteristic equation, 1 + G (s) H (s) = 0 & H(s) = 1 K 1 + = 0 s(1 + 0.4s) ( 1 + 0.25s) s [ 1 + 0.65s + 0.1s2} + K = 0



0.1s3 + 0.65s2 +s + K = 0

S3 0.1 1 From s0, K > 0 S2 0.65 K from s1, S1 0.65 – 0.1K 0 0.65 – 0.1K > 0 0.65 0.65 > 0.1 K S0 K 6.5 > K Range of values of K, 0 < K < 6.5

Now marginal value of ‘K’ is that value of ‘K’ for which system becomes marginally stable. For

a marginal stable system there must be row of zeros occurring in Routh’s array. So value of ‘K’

which makes any row of Routh array as row of zeros is called marginal value of ‘K’. Now K = 0

makes row of s0 as row of zeros but K = 0 can not be marginal value because for K = 0, constant

term in characteristic equation becomes zeros ie one coefficient for s0 vanishes which makes

system unstable instead of marginally stable.

Hence marginal value of ‘K’ is a value which makes any row other than s0 as row of zeros.

0.65 – 0.1 K mar = 0

K mar = 6.5

To find frequency, find out roots of auxiliary equation at marginal value of ‘K’

A(s) = 0.65 s2 + K = 0 ;

0.65 s2 + 6.5 = 0 Because K = 6.5

s2 = -10

s = j 3.162

Comparing with s = j

= frequency of oscillations = 3.162 rad/ sec.



Ex : 3 For a system with characteristic equation

F(s) = s5 + s4 + 2s3 + 2s2 + 3s +15 = , examine the stability

Solution : S5 1 2 3

S4 1 2 15 S3 0 -12 0 S2 S1

S0

S5 1 2 3 S4 1 2 15 S3 -12 0 S2 (2 + 12) 15 0 S1 (2 + 12)( -12 ) – 15 0 2 + 12 S0 15 Lim 2 + 12 12 0 = 2 + = 2 + = +



Lim (2 + 12)( -12 ) – 15 Lim -24 - 144 – 15 2

0 = 0 2 + 12 2 + 12 0 – 144 - 0 = = - 12 0 + 12

S5 1 2 3 S4 1 2 15 S3 -12 0 There are two sign changes, so system is unstable. S2 + 15 0 S1 - 12 0 S0 15 Ex : 4 Using Routh Criterion, investigate the stability of a unity feedback system whose open loop transfer function is e -sT G(s) = s ( s + 1 ) Sol : The characteristic equation is 1 + G(s) H(s) = 0 e -sT 1 + = 0 s ( s + 1 ) s2 + s + e –sT = 0 Now e – sT can be Expressed in the series form as s2 T2 e –sT = 1 – sT + + …… 2!



Truncating the series & considering only first two terms we get esT = 1 – sT s2 + s + 1 – sT = 0 s2 + s ( 1- T ) + 1 = 0

So routh’s array is S2 1 1 S 1-T 0 S0 1 1 – T > 0 for stability T < 1 This is the required condition for stability of the system.

Ex: 5 Determine the location of roots with respect to s = -2 given that F(s) = s4 + 10 s3 + 36s2 + 70s + 75 Sol: shift the origin with respect to s = -2 s = s1 – 2 (s – 2) 4 + 10 (s – 2)3 + 36(s – 2)2 + 70 (s –2) + 75 = 0 S4 + 2s3 + 0s2 + 14s + 15 = 0 S4 1 0 15 S3 2 14 0 S2 -7 15 0 S1 18.28 0 0 S 0 15 Two sign change, there are two roots to the right of s = -2 & remaining ‘2’ are to the left of

the line s = -2. Hence the system is unstable.



NYQUIST STABILITY CRITERIA CAN BE STATED AS FOLLOWS “ If The G(S) H(S) Contour In The G(S)H(S) Plane Corresponding To Nyquist Contour In The

S-Plane Encircles The Point -1+J0 In The Anticlockwise Direction As Many Times As The

Number Of Right Half S-Plane Poles Of G(S)H(S). Then the Closed Loop System Is Stable”

1. No of encirclement of -1+j0 point: This implies that the system is stable if there are no

poles of G(S)H(S) in the right half S-Plane. If there are poles on right half s-plane then

the system is unstable

2. Anticlockwise encirements of -1+j0 point: In this case the system is stable if the number

of anticlockwise encirclement is same as the number of poles of G(S)H(S) in the right

half of S-Plane. If the number of anticlockwise encirclement is not equal to the number of

poles on right half of S-plane then the system is unstable.

3. clockwise encirclement of the -1+j0 point: In this case the system is always unstable.

Also this incase if n poles of G(S)H(S) in right half of S-Plane, then the number of

clockwise encirclement is equal to no of poles of closed loop system on right half S-Plane

NOTE: If the Nyquist contour must not endure the poles (or) zeros lying on imaginary axis.

1+G(S)H(S)

PROBLEMS

(-1,0)

G(S) Contour



SOLUTION: Given that G(S) H(S) = K/S(S+2) (S+10) = K/(Sx2(S/2+1)x10(S/10+1))

= 0.05 K/S(1+0.5S)(1+0.1S)

The open loop transfer function has a pole at origin. Hence choose the Nyquist contour on S-

plane enclosing the entire right half plane except the origin

The Nyquist contour has four sections C1, C2, C3, C4 the mapping of each section is

performed separately and the overall Nyquist plot is obtained by combining the individual

sections

Mapping of section C1 In section c1,ω varies from 0 to ∞. The mapping of section C1 is given by the locus

G(jω)H(jω ) as ω is varied from 0 to ∞. This polar plot of G(jω)H(jω)

G(S)H(S)= 0.05K/S(1+0.5S)(1+0.1S)

Let S=jω

G(jω)H(jω S)= 0.05K/ jωS(1+0.5 jω )(1+0.1 jω )

= 0.05K/-0.6ω2+jω(1-0.05ω2)

when the locus of G(jω) H(jω ) crosses real axis the imaginary term will be zero and the

corresponding frequency is the phase crossover frequency ‘



ωPC.

At ω=ωpc ωpc (1-0.05 ωpc2)=0 1-0.05 ωpc

2=0

At ω=ωpc = 4.427 rad/sec

G(jω)H(jω) = 0.05K/-0.6ω2 = -0.00417 K

The open loop system is type -1 and third order system. Also it is a minimum phase system with

all poles. Hence the polar plot of G(jω)H(jω) starts at -90˚axis at infinity crosses real axis at -

0.00417 K and ends at origin in second quadrant. The section C1 and its mapping show below

diagram A and B

R∞

C1

C3

C4

σ

ω=+∞

ω=0

ω=-∞

ω=∞

ω=0

σ

ω=∞

-0.00417K

jv

Mapping of section c1 in G(S)H(S)



Mapping of section C2 The mapping of section C2 from S-plane to G(S)H(S) –Plane is obtained by letting

Lt

S=R∞ Rejθ in G(S)H(S) and varying θ from +∏/2 to -∏/2. Since S Rejθ and R, the

G(S)H(S) can be approximately as shown below [ ie (1+ST)≈ ST ]

G(S)H(S) = 0.05K/S(1+0.5S)(1+0.1S) ≈ 0.05K/S x 0.5S x 0.1S = K/S3

Lt

Let S=R∞ Rejθ

G(S)H(S)| S=R∞ Rejθ K/S3| = = 0e-j3θ

when θ=+∏/2 , G(S)H(S) = 0e-j3∏/2-----------------1a

when θ=-∏/2 , G(S)H(S) = 0e+j3∏/2--------------------------1b

Section c1 in s-plane

K Lt R∞ (Rejθ)3



From the equation 1a and 1b we can study that section c2 in S-plane fig1a

is mapped as circular are of zero radius around origin in G(S)H(S) plane with argument (phase

varying from -3∏/2 and +3∏/2 as shown in fig 1b

Mapping of section C3

In Section C3, ω varies from -∞ to 0. The mapping of section C3 is given by the locus of

G(jω)H(jω) as a ω is varied from -∞ to 0. This locus is the inverse polar plot of G(jω)H(jω)

The inverse polar plot is given by the mirror image of polar plot with respect to real axis.

The section C3 in S-Plane and its corresponding contour in G(S)H(S) plane shown in fig 2a and

fig 2b

R∞ ∞

C2 S=Plane jω

Section c2 in S-Plane

jv

u

G(S)H(S) plane



Mapping of section C4

The mapping of section c4 from s-plane to G(S)H(S) plane is obtained by letting

Lt

S=R∞ Rejθ in G(S)H(S) and varying θ from +∏/2 to -∏/2. Since S Rejθ and R, the

G(S)H(S) can be approximately as shown below [ ie (1+ST)≈ 1 ]

0.05K

G(S)H(S) = ----------------------- = 0.05K/S

S(1+0.5S)(1+0.1S)

Lt

Let S= R0 Rejθ

G(S)H(S) = αe-jθ

when θ=+∏/2 , G(S)H(S) = αe-j3∏/2-----------------2a

when θ=∏/2 , G(S)H(S) = αe+j3∏/2--------------------------2b

From equation above we can say that sec c4 in S-Plane fig 3a is mapped as a circular are of

infinity radius with argument(phase) varying from +∏/2 to-∏/2 shown in figure

σ

jω

ω=0

ω=-∞

S-Plane

jv

ω=0

ω=-∞ -0.00417k



Complete Nyquist plot The entire Nyquist plot in G(S)H(S) plane can be obtained by combining the mappings of individual sections shown

R->∞

jv

R->0

C4

S-Plane

G(S)H(S) plane

Mapping of section c4

R0

-0.00417K

-1+j0 for K<240 -1+j0K>240

G(S) H(S) Contour

K G(S)H(S) = ---------------- S(S+2)(S+10)

Section C4 in s-plane



STABILITY ANALYSIS When -0.00417K =-1, the contour passes through (-1+j0) point and corresponding value of K

is the limiting value for K for stability

Limiting value of K = 1/0.00417 = 240

When K<240 When K is less than 240, the contour crosses real axis at appoint between 0 & -1+j0 . On

traveling through Nyquist plot along the indicated direction if it found that the point -1+j0 is not

encircled. Also the open loop transfer function has no poles on the right half of S-Plane.

Therefore the closed loop system is stable.

When K>240 When K is graeter than 240, the contour crosses real axis at a point B/W -1+j0 and -∞. On

traveling through Nyquist plot along the indicated direction it is found that the point -1+j0 is

encircled in clockwise direction two time. [ since there are two clockwise encirclement and no

right half open loop poles, the closed loop system has 2 poles on right half of S-Plane. Therefore

the closed loop system is unstable

RESULT

The value of K for stability is 0<K<240

RELATIVE STABIITY The relative stability indicates the closeness of the system to stable region. It is an indication

of strength or degree of stability

in time domain, the relative stability may be measured by relative settling times of each root

or pair of roots. The settling time is inversely proportional to the location of roots of

characteristic equation. If the root is located for away from the imaginary axis, then the transients

dies out faster and so the relative stability of system will improve. the transient response and so

the relative stability for various of roots in S-Plane is shown below fig



ROOT LOCUS INTRODUCTION The characteristics of the transient response of a closed loop control system is related to location

of the closed loop poles. If the system has a variable loop gain, then the location of the closed

loop poles depends on the value of the loop gain chosen. It is important, that the designer knows

how the closed loop poles move in the s-plane as the loop gain is varied. W. R. Evans introduced

a graphical method for finding the roots of the characteristic equation known as root locus

method. The root locus is used to study the location of the poles of the closed loop transfer

function of a given linear system as a function of its parameters, usually a loop gain, given its

open loop transfer function. The roots corresponding to a particular value of the system

parameter can then be located on the locus or the value of the parameter for a desired root

location can be determined from the locus. It is a powerful technique, as an approximate root

locus sketch can be made quickly and the designer can visualize the effects of varying system

parameters on root locations or vice versa. It is applicable for single loop as well as multiple loop

system.

t t

t t

jω

σ



ROOT LOCUS CONCEPT

To understand the concepts underlying the root locus technique, consider the second

order system shown in Fig. 1.

Fig. 23 Second order control system

The open loop transfer function of this system is

)1(a)s(s

KG(s)

Where, K and a are constants. The open loop transfer function has two poles one at origin s = 0

and the other at s = -a. The closed loop transfer function of the system shown in Fig.1 is

(2)Kass

K

G(s)H(s)1

G(s)

R(s)

C(s)2

The characteristic equation for the closed loop system is obtained by setting the

denominator of the right hand side of Eqn.(2) equal to zero. That is,

(3)0 KassG(s)H(s)1 2

The second order system under consideration is always stable for positive values of a and

K but its dynamic behavior is controlled by the roots of Eqn.(3) and hence, in turn by the

magnitudes of a and K, since the roots are given by

(4)K2

a

2

a

2a

4K)(a

2

as,s

22

21

R(s) E(s)

− a)s(s

K

C(s)



From Eqn.(4), it is seen that as the system parameters a or K varies, the roots change.

Consider a to be constant and gain K to be variable. As K is varied from zero to infinity, the two

roots s1 and s2 describe loci in the s-plane. Root locations for various ranges of K are:

1) K= 0, the two roots are real and coincide with open loop poles of the system s1 =

0, s2 = -a.

2) 0 K < a2/4, the roots are real and distinct.

3) K= a2/4, roots are real and equal.

4) a2/4 < K < , the rots are complex conjugates.

The root locus plot is shown in Fig.2

Fig. 24 Root loci of s2+as+K as a function of K

Figure 2 has been drawn by the direct solution of the characteristic equation. This

procedure becomes tedious. Evans graphical procedure helps in sketching the root locus quickly.

The characteristic equation of any system is given by

(5)0Δ(s)

Where, (s) is the determinant of the signal flow graph of the system given by Eqn.(5). ∆=1-(sum of all individual loop gains)+(sum of gain products of all possible combinations of

two no touching loops – sum of gain products of all possible combination of three no

touching loops) + ∙∙∙

Or



(6)PPP1Δ(s)m m3m m2m m1

Where, Pmr is gain product of mth possible combination of r no touching loops of the graph.

The characteristic equation can be written in the form

(7)0B(s)

KA(s)1

0P(s)1

For single loop system shown in Fig.3

(8)G(s)H(s)P(s)

Where, G(s)H(s) is open loop transfer function in block diagram terminology or transmittance in

signal flow graph terminology.

Fig. 25 Single loop feedback system

From Eqn.(7) it can be seen that the roots of the characteristic equation (closed loop

poles)occur only for those values of s where

(9)1P(s)

Since, s is a complex variable, Eqn.(9) can be converted into the two Evans conditions

given below.

)10(1)( sP

)11(2,1,0);12(180)( qqsP

Roots of 1+P(s) = 0 are those values of s at which the magnitude and angle condition

given by Eqn.(10) and Eqn.(11). A plot of points in the complex plane satisfying the angle

−

R(s) E(s) G(s)

C(s)

H(s)



criterion is the root locus. The value of gain corresponding to a root can be determined from the

magnitude criterion.

To make the root locus sketching certain rules have been developed which helps in

visualizing the effects of variation of system gain K ( K > 0 corresponds to the negative feed

back and K < 0 corresponds to positive feedback control system) and the effects of shifting

pole-zero locations and adding in anew set of poles and zeros.

GENERAL RULES FOR CONSTRUCTING ROOT LOCUS

1) The root locus is symmetrical about real axis. The roots of the characteristic equation are

either real or complex conjugate or combination of both. Therefore their locus must be

symmetrical about the real axis.

2) As K increases from zero to infinity, each branch of the root locus originates from an

open loop pole (n nos.) with K= 0 and terminates either on an open loop zero (m nos.)

with K = along the asymptotes or on infinity (zero at ). The number of branches

terminating on infinity is equal to (n – m).

3) Determine the root locus on the real axis. Root loci on the real axis are determined by

open loop poles and zeros lying on it. In constructing the root loci on the real axis choose

a test point on it. If the total number of real poles and real zeros to the right of this point

is odd, then the point lies on root locus. The complex conjugate poles and zeros of the

open loop transfer function have no effect on the location of the root loci on the real axis.



4) Determine the asymptotes of root loci. The root loci for very large values of s must be

asymptotic to straight lines whose angles are given by

)12(1-mn0,1,2,;mn

1)(2q180asymptotesofAngle

A

q

5) All the asymptotes intersect on the real axis. It is denoted by a , given by

)13(mn

)zz(z)pp(pmn

zerosofsumpolesofsumσ

m21n21

a

6) Find breakaway and break-in points. The breakaway and break-in points either lie on the

real axis or occur in complex conjugate pairs. On real axis, breakaway points exist

between two adjacent poles and break-in in points exist between two adjacent zeros. To

calculate this polynomial 0ds

dK must be solved. The resulting roots are the breakaway /

break-in points. The characteristic equation given by Eqn.(7), can be rearranged as

)z(s)z)(szK(sA(s)

and )p(s)p)(sp(s B(s) where,

(14)0KA(s)B(s)

m21

n21

The breakaway and break-in points are given by

)15(0Bds

dABA

ds

d

ds

dK



Note that the breakaway points and break-in points must be the roots of Eqn.(15), but

not all roots of Eqn.(15) are breakaway or break-in points. If the root is not on the root

locus portion of the real axis, then this root neither corresponds to breakaway or break-

in point. If the roots of Eqn.(15) are complex conjugate pair, to ascertain that they lie on

root loci, check the corresponding K value. If K is positive, then root is a breakaway or

break-in point.

7) Determine the angle of departure of the root locus from a complex pole

)16()zerosotherfromquestioninpolecomplexatovectorsofanglesof(sum

poles)otherfromquestioninpolecomplexatovectorsofanglesof(sum

180pcomplexafromdepartureofAngle

8) Determine the angle of arrival of the root locus at a complex zero

(17)poles)otherfromquestioninzerocomplexatovectorsofanglesof(sum

zeros)otherfromquestioninzerocomplexatovectorsofanglesof(sum

180zerocomplexatarrivalofAngle

9) Find the points where the root loci may cross the imaginary axis. The points where the

root loci intersect the j axis can be found by

a) use of Routh’s stability criterion or

b) letting s = j in the characteristic equation , equating both the real part and

imaginary part to zero, and solving for and K. The values of thus found give

the frequencies at which root loci cross the imaginary axis. The corresponding K

value is the gain at each crossing frequency.

10) The value of K corresponding to any point s on a root locus can be obtained using the

magnitude condition, or



)18(zerostopointsbetweenlengthofproduct

polestopointsbetweenlengthsofproductK

PHASE MARGIN AND GAIN MARGIN OF ROOT LOCUS Gain Margin

It is a factor by which the design value of the gain can be multiplied before the closed

loop system becomes unstable.

(19)KofvalueDesign

overcrossimaginaryatKofValueMarginGain

The Phase Margin

Find the point j1 on the imaginary axis for which 1jHjG for the design value

of K i.e. design

Kj/AjB .

The phase margin is (20))H(jωjωargG180φ11

Problem No 1 Sketch the root locus of a unity negative feedback system whose forward path transfer function

is s

KG(s) .

Solution:

1) Root locus is symmetrical about real axis.

2) There are no open loop zeros(m = 0). Open loop pole is at s = 0 (n = 1). One branch of

root locus starts from the open loop pole when K = 0 and goes to asymptotically when

K .

3) Root locus lies on the entire negative real axis as there is one pole towards right of any

point on the negative real axis.



4) The asymptote angle is A = .01,)12(180

mnq

mn

q

Angle of asymptote is A = 180.

5) Centroid of the asymptote is mn

zeros)of(sumpoles)of(sumσA

0.01

0

6) The root locus does not branch. Hence, there is no need to calculate the break points. 7) The root locus departs at an angle of -180 from the open loop pole at s = 0.

8) The root locus does not cross the imaginary axis. Hence there is no imaginary axis cross

over. The root locus plot is shown in Fig.1

Figure 26 Root locus plot of K/s Comments on stability: The system is stable for all the values of K > 0. Th system is over damped. Problem No 2

The open loop transfer function is 21)(s

2)K(sG(s)

. Sketch the root locus plot

Solution:



1) Root locus is symmetrical about real axis. 2) There is one open loop zero at s=-2.0(m=1). There are two open loop poles at

s=-1, -1(n=2). Two branches of root loci start from the open loop pole when K= 0. One branch goes to open loop zero at s =-2.0 when K and other goes to (open loop zero) asymptotically when K .

3) Root locus lies on negative real axis for s ≤ -2.0 as the number of open loop poles plus

number of open loop zeros to the right of s=-0.2 are odd in number.

4) The asymptote angle is A = .01,)12(180

mnq

mn

q

Angle of asymptote is A = 180.



0.01

)2()11(

6) The root locus has break points.

The root loci brakes out at the open loop poles at s=-1, when K =0 and breaks in onto the real axis at s=-3, when K=4. One branch goes to open loop zero at s=-2 and other goes to along the asymptotically.

7) The branches of the root locus at s=-1, -1 break at K=0 and are tangential to a line s=-

1+j0 hence depart at 90. 8) The locus arrives at open loop zero at 180. 9) The root locus does not cross the imaginary axis, hence there is no need to find the

imaginary axis cross over. The root locus plot is shown in Fig.2.

4K3,s0;K1,s

02)(s

1)(s2)1)(s2(s

0ds

dKbygivenispointBreak

2)(s

1)(sK

21

2

2

2



Figure 27 Root locus plot of K(s+2)/(s+1)2

Comments on stability:

System is stable for all values of K > 0. The system is over damped for K > 4. It is critically damped at K = 0, 4.

Problem No 3

The open loop transfer function is 2)s(s

4)K(sG(s)

. Sketch the root locus.

Solution:

1) Root locus is symmetrical about real axis. 2) There are is one open loop zero at s=-4(m=1). There are two open loop poles at s=0, -

2(n=2). Two branches of root loci start from the open loop poles when K= 0. One branch goes to open loop zero when K and other goes to infinity asymptotically when K .

3) Entire negative real axis except the segment between s=-4 to s=-2 lies on the root locus.

4) The asymptote angle is A = .01,1,0,)12(180

mnq

mn

q

Angle of asymptote are A = 180.



0.21

)4()2(



6) The brake points are given by dK/ds =0. 7) Angle of departure from open loop pole at s =0 is 180. Angle of departure from pole at

s=-2.0 is 0. 8) The angle of arrival at open loop zero at s=-4 is 180 9) The root locus does not cross the imaginary axis. Hence there is no imaginary cross over.

The root locus plot is shown in fig.3.

Figure 3 Root locus plot of K(s+4)/s(s+2) Comments on stability: System is stable for all values of K. 0 > K > 0.343 : > 1 over damped K = 0.343 : = 1 critically damped 0.343 > K > 11.7 : < 1 under damped K = 11.7 : = 1 critically damped

11.7K6.828,s

0.343;K1.172,s

04)(s

2s)(s4)2)(s(2s

ds

dK

4)(s

2)s(sK

2

1

2

2



K > 11.7 : >1 over damped. Problem No 4

The open loop transfer function is 3.6)(ss

0.2)K(sG(s)

2 . Sketch the root locus.

Solution:

1) Root locus is symmetrical about real axis. 2) There is one open loop zero at s = -0.2(m=1). There are three open loop poles at

s = 0, 0, -3.6(n=3). Three branches of root loci start from the three open loop poles when

K= 0 and one branch goes to open loop zero at s = -0.2 when K and other two go to

asymptotically when K .

3) Root locus lies on negative real axis between -3.6 to -0.2 as the number of open loop

poles plus open zeros to the right of any point on the real axis in this range is odd.

4) The asymptote angle is A = 1,01,)12(180

mnq

mn

q

Angle of asymptote are A = 90, 270.



7.12

)2.0()6.3(

6) The root locus does branch out, which are given by dK/ds =0.

The root loci brakeout at the open loop poles at s = 0, when K =0 and breakin onto the

real axis at s=-0.432, when K=2.55 One branch goes to open loop zero at s=-0.2 and

other goes breaksout with the another locus starting from open loop ploe at s= -3.6. The

break point is at s=-1.67 with K=3.66. The loci go to infinity in the complex plane with

constant real part s= -1.67.

ly.respective 3.662.55,0,Kand1.670.432,0,s

01.44s4.8s2s

0.2)(s

)3.6s(s0.2)7.2s)(s(3s

ds

dK

0.2s

)3.6s(s-K

23

2

232

23



7) The branches of the root locus at s=0,0 break at K=0 and are tangential to imaginary axis

or depart at 90. The locus departs from open loop pole at s=-3.6 at 0. 8) The locus arrives at open loop zero at s=-0.2 at 180. 9) The root locus does not cross the imaginary axis, hence there is no imaginary axis cross

over. The root locus plot is shown in Fig.4.

Figure 4 Root locus plot of K(s+0.2)/s2(s+3.6)

Comments on stability:

System is stable for all values of K. System is critically damped at K= 2.55, 3.66. It is under

damped for 2.55 > K > 0 and K >3.66. It is over damped for 3.66 > K >2.55.

Problem No 5

The open loop transfer function is 25)6ss(s

KG(s) . Sketch the root locus.

Solution:

1) Root locus is symmetrical about real axis. 2) There are no open loop zeros (m=0). There are three open loop poles at s=-0,



-3j4(n=3). Three branches of root loci start from the open loop poles when K= 0 and all the three branches go asymptotically when K .

3) Entire negative real axis lies on the root locus as there is a single pole at s=0 on the real

axis.

4) The asymptote angle is A = .2,1,01,1,0,)12(180

mnq

mn

q

Angle of asymptote are A = 60, 180, 300.



0.23

)33(

6) The brake points are given by dK/ds =0.

j18.0434K

j2.0817and2s

02512s3sds

dK

25s)6s(s25)6ss(sK

1,2

1,2

2

232

For a point to be break point, the corresponding value of K is a real number greater than or equal to zero. Hence, S1,2 are not break points.

7) Angle of departure from the open loop pole at s=0 is 180. Angle of departure from

complex pole s= -3+j4 is

zeros) from inquestion polecomplex a to vectorsof angles theof (sum

poles)other fromquestion in polecomplex a to vectorsof angles theof (sum

180p

87.36)90

3

4tan180(180 1

p

Similarly, Angle of departure from complex pole s= -3-j4 is 36.87or323.13)270(233.13180φp

8) The root locus does cross the imaginary axis. The cross over point and the gain at the

cross over can be obtained by Rouths criterion The characteristic equation is 0K25s6ss 23 . The Routh’s array is

6s6

K150s

K6s

251s

0

1

2

3



For the system to be stable K < 150. At K=150 the auxillary equation is 6s2+150=0. s = ±j5. or substitute s= j in the characteristic equation. Equate real and imaginary parts to zero. Solve for and K.

The plot of root locus is shown in Fig.5.

Figure 5 Root locus plot of K/s(s2+6s+25) Comments on stability: System is stable for all values of 150 > K > 0. At K=150, it has sustained oscillation of 5rad/sec. The system is unstable for K >150. Problem No 1 Sketch the root locus of a unity negative feedback system whose forward path transfer function

is j)3j)(s31)(s(s

2)K(sG(s)H(s)

. Comment on the stability of the system.

1500,Kj50,ω

025ωjωK)6ω(

0Kjω25jω6jω

0K25s6ss

22

23

23



Solution:

9) Root locus is symmetrical about real axis. 10) There is one open loop zero at s = -2 (m = 1). There are three open loop poles at

s = -1, -3 ± j (n=3). All the three branches of root locus start from the open loop poles when K = 0. One locus starting from s = -1 goes to zero at s = -2 when K , and other two branches go to asymptotically (zeros at ) when K .

11) Root locus lies on the negative real axis in the range s=-1 to s= -2 as there is one pole to the right of any point s on the real axis in this range.

12) The asymptote angle is A = .0,11mnq,mn

1)(2q180

Angle of asymptote is A = 90, 270.



2.51

2)(3)31(

14) The root locus does not branch. Hence, there is no need to calculate break points. 15) The angle of departure at real pole at s=-1 is 180. The angle of departure at the complex

pole at s=-3+j is 71.57.



180p

The angle of departure at the complex pole at s=-3-j is -71.57. 16) The root locus does not cross the imaginary axis. Hence there is no imaginary axis cross

over. The root locus plot is shown in Fig.1

57.71135)90(153.43180

900

2tanθ,135or -45

1-

1tan

153.43)atan2(-2,1θ

153.43or 57.262-

1tanθ

p

1

3

1

1

1

1

57.71522)270(206.57180p



Figure 1 Root locus plot of K(s+2)/(s+1)(s+3+j)(s+3-j)

Comments on stability: The system is stable for all the values of K > 0.

Problem No 2

The open loop transfer function is10)0.6s0.5)(ss(s

KG(s)H(s)

2 Sketch the root locus

plot. Comment on the stability of the system. . Solution:

10) Root locus is symmetrical about real axis. 11) There are no open loop zeros (m=0). There are four open loop poles (n=4) at s=0,

-0.5, -0.3 ± j3.1480. Four branches of root loci start from the four open loop poles when K= 0 and go to (open loop zero at infinity) asymptotically when K .

12) Root locus lies on negative real axis between s = 0 to s = -0.5 as there is one pole to the right of any point s on the real axis in this range.

13) The asymptote angle is A = .3,2,1,01,)12(180

mnq

mn

q

Angle of asymptote is A = 45, 135, 225, ±315.





275.04

)3.03.05.0(

The value of K at s=-0.275 is 0.6137. 15) The root locus has break points.

K = -s(s+0.5)(s2+0.6s+10) = -(s4+1.1s3+10.3s2+5s) Break points are given by dK/ds = 0

0520.6s3.3s4sds

dK 23

s= -0.2497, -0.2877 j 2.2189 There is only one break point at -0.2497. Value of K at s = -0.2497 is 0.6195.

16) The angle of departure at real pole at s=0 is 180 and at s=-0.5 is 0. The angle of

departure at the complex pole at s = -0.3 + j3.148 is -91.8



180p

The angle of departure at the complex pole at s = -0.3 - j3.148 is 91.8 17) The root locus does cross the imaginary axis, The cross over frequency and gain is

obtained from Routh’s criterion.

The characteristic equation is s(s+0.5)(s2+0.6s+10)+K =0 or s4+1.1s3+10.3s2+5s+K=0 The Routh’s array is

Ks5.75

1.1K-28.75s

K5.75s

51.1s

K10.31s

0

1

2

3

4

8.91)9086.4(95.4180

900

6.296tanθ, 4.68

0.2

3.148tan

4.95or6.840.3-

3.148tanθ

p

1

3

1

2

1

1

91.8 )270273.6(264.6180p



The system is stable if 0 < K < 26.13 The auxiliary equation at K 26.13 is 5.75s2+26.13 = 0 which gives s = ± j2.13 at imaginary axis crossover. The root locus plot is shown in Fig.2.

Figure 28 Root locus plot of K/s(s+0.5)(s2+0.6s+10)

Comments on stability: System is stable for all values of 26.13 >K > 0. The system has sustained oscillation at = 2.13 rad/sec at K=26.13. The system is unstable for K > 26.13.

Problem No 3

The open loop transfer function is 20)4s)(s4s(s

KG(s)

2 . Sketch the root locus.

Solution:

10) Root locus is symmetrical about real axis.



11) There are no open loop zeros (m=0). There are three open loop poles (n=3) at s = -0, -4, -2 j4. Three branches of root loci start from the three open loop poles when K= 0 and to infinity asymptotically when K .

12) Root locus lies on negative real axis between s = 0 to s = -4.0 as there is one pole to the

right of any point s on the real axis in this range.

13) The asymptote angle is A = 3,2,1,01,)12(180

mnq

mn

q

Angle of asymptote are A = 45, 135, 225, 315.



0.24

)0.40.20.2(

15) The root locus does branch out, which are given by dK/ds =0.

The root loci brakeout at the open loop poles at s = -2.0, when K = 64 and breakin and breakout at s=-2+j2.45, when K=100

16) The angle of departure at real pole at s=0 is 180 and at s=-4 is 0. The angle of departure at the complex pole at s = -2 + j4 is -90.



180p

100Kj2.45,2.0s

64;K2.0,s

40)16s2)(4s(s

08040s32s16s8s4s

08072s24s4s

0ds

dKbygivenispointBreak

80s)36s8s(s

20)4s4)(ss(sK

2

1

2

223

23

234

2

90)90.463(116.6-180

900

8tanθ,4.36

2

4tanθ

116.6)atan2(4,-2θ

116.6or63.42-

4tanθ

p

1

3

1

2

1

1

1



The angle of departure at the complex pole at s = -2 – j4 is 90

17) The root locus does cross the imaginary axis, The cross over point and gain at cross over

is obtained by either Routh’s array or substitute s= j in the characteristic equation and

solve for and gain K by equating the real and imaginary parts to zero.

Routh’s array

The characteristic equation is 0K80s36s8ss 234

For the system to be stable K > 0 and 2080-8K > 0. The imaginary crossover is given by 2080-8K=0 or K = 260. At K = 260, the auxiliary equation is 26s2+260 = 0. The imaginary cross over occurs at s= j10.

or

Ks26

8K2080s

K26s

808s

K361s

isarrayRouthsThe

0

1

2

3

4

260K0K36ωω

10js;10j0,ω080ω8ω

zerotopartsimaginaryandrealEquate

080ω8ωjK36ωω0Kjω80jω36jω8jω

jωsput

0K80s36s8ss

24

3

324

234

234

90270

)270296.6(243.4-180p



The root locus plot is shown in Fig.3.

Figure 29 Root locus plot of K/s(s+4)(s2+4s+20)

Comments on stability: For 260 > K > 0 system is stable

K = 260 system has stained oscillations of 10 rad/sec.

K > 260 system is unstable.

UNIT-V

COMPENSATOR DESIGN Basic ideas of compensator design

In the following some analytical design methods will be discussed, which will lead directly to the

design solution in a strongly systematic way. In contrary to these direct design methods, the

methods hitherto discussed, e.g. the design method using frequency-domain characteristics or the

root-locus method, are indirect methods based more on systematic trial and error techniques

iterating through some design steps. The success depends strongly on the experience and skill of

the designer. The starting point was always the open loop, which was modified iteratively by



adding lead and lag elements until the closed loop shows the desired behavior. Whereas in the

direct design methods one will always start from the behavior of the closed loop. Mostly a

desired transfer Function

is given. In general this follows from the specification of some performance indices, which, for

example are required for the step response . For a series of appropriate transfer

functions a table of numerator and denominator polynomials of the associated transfer function

and its distribution of zeros and poles to yield a specific response are given. Then for

known plant behavior the required controller can be directly calculated.

The controllers designed in this way are not always optimal. They guarantee the compliance of

the desired specification, e.g. maximum overshoot and settling time. A drawback of theses

methods is that they cannot be applied directly to systems with dead time.

Design by specifying the closed-loop transfer function

The desired transfer function of the closed loop is given by

(10.1)

where and are polynomials in . In the following design methods, the distribution of

poles and zeros of will be chosen, such that the performance indices for the step response

are fulfilled. Often a detailed investigation of the distribution of poles and zeros of the

desired transfer function is not necessary, especially when the transfer function does not contain

zeros and when - because of the requirement - just yields and in the simplest

case

(10.2)



For a closed loop with the transfer function according to Eq. (10.2) different possibilities exist,

the so called standard forms, which can be used by a table lookup for the step response ,

distribution of poles of and the coefficients of the denominator polynomial .

A first possibility is a distribution of poles with a real multiple pole at . Here and in the

following sections, the term is a relative frequency, not the natural frequency. Thus one

obtains for the step response of the desired behavior

(10.3)

This is a series connection of elements with the same time constant . This

representation is also called a binomial form. The standard polynomials of different order

are given in Table A.2. As this table further shows, the normalized step response will

become slower with increasing order. A design using this binomial form is only considered

when the step response is required to have no overshoot.

A further possibility of a standard form for of Eq. (10.2) is the Butterworth form. In this

form, all poles of are equally distributed on a semicircle with radius in the left-half

plane and centered at the origin. Table A.2 contains the standard polynomials and the

associated normalized step responses .

Numerous further possibilities for the development of standard forms of Eq. (10.2) can be

derived from the integral criteria given in Table 7.2. For example, the minimum performance

index is the basis of a standard form that is also shown in Table A.2. Further, often the

http://virtual.cvut.cz/dynlabmods/syscontrol/node72.html#eq:8.4.2

http://virtual.cvut.cz/dynlabmods/syscontrol/node154.html#tab:A8.4.1




http://virtual.cvut.cz/dynlabmods/syscontrol/node58.html#tab:8.2.1




minimum settling time is used as the criterion. This table contains for the

corresponding standard form.

Furthermore, for pole assignment the Weber method can be used. This specifies the desired closed-loop transfer function

(10.4)

by a real pole with multiplicity and a pair of complex poles. Table A.1 contains for

different values of and the normalized step responses . By a proper choice of , and

a closed-loop transfer function can be found that fulfils in many instances the desired performance.

The method of Truxal and Guillemin

For the closed loop shown in Figure 10.1

Figure 10.1: Block diagram of the closed loop to be designed

the behavior is described by the transfer function

(10.5)

Where the numerator and denominator polynomials and must have no common roots.

Furthermore, is normalized to and must be valid.

It is assumed that is stable and minimum phase. For the controller to be designed, the

transfer function


http://virtual.cvut.cz/dynlabmods/syscontrol/node73.html#fig:8.4.1



(10.6)

is chosen and normalized to . Because of the reliability of the controller the relation

must be valid. Now, the controller must be designed such that

the closed loop behaves like a given transfer function for Eq. (10.1), whereby should be

freely chosen under the condition of the reliability of the controller. From the closed-loop

transfer function

(10.7)

one obtains the controller transfer function

(10.8)

or with the numerator and denominator polynomials given above

(10.9)

The condition of reliability for the controller is

or

(10.10)

The pole excess ( ) of the desired closed-loop transfer function must be larger than or

equal to the pole excess ( ) of the plant. Within these constraints the order of is free.

According to Eq. (10.8) the controller contains the inverse plant transfer function . This

is a total compensation of the plant as shown in the block diagram of Figure 10.2. For the

realization






Figure 10.2: Compensation of the plant

of the controller Eq. (10.9) is used, not the controller structure as shown in this figure with the

plant inverse . As the controller implicitly contains the plant inverse, i.e. the plant zeros

are in the set of the controller poles and the plant poles are in the set of the controller zeros, the

plant must be stable and minimum phase as mentioned at the beginning. Otherwise, the

manipulated variable and/or the controlled variable will show unstable behavior.

Example 10.3.1 The plant transfer function is given as

(10.11)

The pole excess of the plant is . According to (10.10) the pole excess of the desired

closed-loop transfer function must be

The coefficients of the transfer function that obeys the reliability condition (10.10) are

subjected to practical constraints, like the maximum range of the manipulated variable, plant

parameter errors and measurement noise in the controlled variable, which is disturbing the

controller output. The procedure for the design of will be demonstrated by the following

example.

Example 10.3.2 For a plant with the transfer function

(10.12)






a controller should be designed such that the closed loop shows optimal behavior in the sense of

the performance index and has a rise time of .

First, it follows from the reliability condition Eq. (10.10) and from that the

pole excess of the desired transfer function is

Inspecting Table A.2 one obtains from the form for and the standard

polynomial

(10.13)

From the associated step response it follows from Table A.2 that the normalized rise

time

and with this value from the specified rise time the relative frequency is . Eq. (10.13) is now

As for the chosen standard form for the numerator polynomial is , so it follows

from Eq. (10.9) that the compensator transfer function is

or

This controller contains an integrator. The time responses are shown in Figure 10.3.









Figure 10.3: Closed-loop behavior for the example 10.3.2: step response of the controlled

variable on step in the set point, step response of the associated controlled variable,

step response of the uncontrolled plant

If as a further example the plant given by Eq. (10.11) instead of Eq. (10.12) is taken, then for the

same the controller is

For these two very different plants the same closed-loop behavior for the controlled variable can

be achieved.

In the considerations of this section it has been hitherto assumed that is stable and

minimum phase. For plants that do not have these properties this design method cannot be

applied in this form. The method must be extended to the following:

A direct compensation of the plant poles and zeros by the controller must be avoided, otherwise

stability problems would arise. In these cases, the closed-loop transfer function cannot be

http://virtual.cvut.cz/dynlabmods/syscontrol/node73.html#8.4.2





arbitrary. For a stable non-minimum phase plant the transfer function must be given such

that the zeros of contain the right-half-plane zeros of . Whereas for an unstable

plant the zeros of the transfer function must contain the right-half-plane poles of

. Of course, this restricts the choice of as the following examples demonstrate.

Example 10.3.3 For an all-pass plant with the transfer function

a controller is to be designed such that the closed loop has the desired transfer function

Using Eq. (10.9) one gets for the controller transfer function

This controller gives a direct compensation (cancellation) of the plant zero. This is undesirable as

already discussed above, and must be selected as

With Eq. (10.9) one obtains the controller transfer function as

Because of this choice of , the closed loop shows also all-pass behavior. This effect is

more intense the smaller the time constant. Figure 10.4 shows the time responses of this

control system.






Figure 10.4: Closed-loop behaviour of the example 10.3.3: step response of the controlled

variable on step in the set point, step response of the associated controlled variable,

step response of the uncontrolled plant ( ; )

Example 10.3.4 The transfer function of the unstable plant

is given and a controller is required for which fulfills the reliability condition

and for which the zeros of must contain the plant pole . This is expressed by the approach

Whereby is chosen such that




is valid. In the present case should be chosen such that . From this

follows . In order to have a stable one can take

and obtain

Observing the reliability condition, it follows that

Comparing the coefficients and taking one obtains

and

and finally

and

The parameters and are still free and may now be chosen such that taking an acceptable

behavior of the manipulated variable into account, a given damping ratio and natural frequency

for can be reached. Without going into details,

and

will be chosen for the present case and from this it follows that

and

The desired closed-loop transfer function will be



and

The conditions for the design are fulfilled and for the controller transfer function one obtains

from Eq. (10.8) or Eq. (10.9)

This design obviously produces a PI controller. The time responses of this control system are

shown in Figure 10.5 for . The relatively large maximum overshoot cannot be avoided

with an acceptable behavior of the manipulated variable.

Figure 10.5: Closed-loop behavior of the example 10.3.4: step response of the controlled

variable on step in the set point, step response of the associated controlled variable Generalized compensator design method

The basic idea

With the following method a control system according to Figure 10.1 using the controller given

by Eq. (10.6) will be designed for a plant described by Eq. (10.5) such that the closed loop










behaves like the desired transfer function Eq. (10.1). Hereby the orders of the controller

numerator and denominator polynomials are equal, i.e. . The

closed-loop poles are the roots of the characteristic equation, which one obtains from

With respect to the polynomials defined in Eqs. (10.5) and (10.6) this gives

(10.14)

On the other hand it follows from Eq. (10.1) that

(10.15)

This polynomial has order , the coefficients depend linearly on the plant and controller parameters. Comparing both equations, the first coefficient is

(10.16)

and the last because of and

(10.17)

A general representation is given by

(10.18)

whereby

for and

for and

and . The coefficients are obtained from the poles. For the first, second last and last one gets







(10.19)

(10.20)

(10.21)

While the coefficients according to Eqs. (10.19), (10.20) and (10.21) are directly given by the

closed-loop poles, the coefficients of Eq. (10.16), (10.17) and (10.18) contain the required

controller parameters. Comparing both sides of the latter equation one obtains the synthesis

equation, which is a system of linear equations for unknown controller

coefficients . The number of equations is . A unique solution

exists if .

A detailed analysis shows, however, that a controller obtained in this way does not usually

achieve the desired goals. Because of its small gain a finite steady-state error may occur. This

must be taken into consideration during the design. For plants with integral behavior an order

for the controller is sufficient; for proportional behavior or when disturbances at the

input of an integral plant are taken into consideration, the gain must be influenced so that an

integral behavior of the controller can be obtained. This happens if the order of the controller is

increased by one, i.e. , such that the system of equations is of lower rank. This gives an

additional degree of freedom and allows one to choose the controller gain , which is usually

introduced as a reciprocal gain factor:

(10.22)

Indeed, the order of the closed loop will be increased; it is now double that of the plant order.

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Zeros of the closed loop

In the method presented above, the zeros of the closed loop transfer function for command

changes

(10.23)

are obtained automatically. In fact, the zeros of the plant, i.e. the roots of , can be

considered during the choice of the pole distribution and may be compensated, but the

polynomial arises not in the design and must possibly be compensated after this step. This

can be done by introducing a pre-filter in the feed-forward path according to 10.6a with a

transfer function

Figure 10.6: Compensation of the plant zeros (a) with a controller in the feed-forward path and (b) in the feedback path

The zeros of the controller and plant are compensated in this way. For stability reasons, this is

only possible for left-half-plane zeros. If and are polynomials with only left-half-




plane zeros and and the corresponding polynomials with only right-half-plane zeros

including the imaginary axis, the polynomials of and can be factorized as

(10.24)

(10.25)

with

(10.26)

(10.27)

and

(10.28)

(10.29)

For the case that and , and and do not have common divisors, i.e. the controller does not compensate plant poles and zeros, the denominator polynomial of the pre-filter can be determined as

(10.30)

The transfer function for a command input is then

(10.31)



If both, the controller and the plant, show minimum phase behaviour and their transfer functions

do not have zeros on the imaginary axis, all zeros of the closed loop can be compensated, such

that one obtains instead of Eq. (10.31)

(10.32)

If the closed-loop transfer function also contains given zeros, the transfer function should

have a corresponding numerator polynomial. The coefficient in the numerator is used to make

the gain of the closed-loop transfer function equal to 1. From Eq. (10.31) it therefore

follows that

(10.33)

The expression in the denominator is the first coefficient of the characteristic polynomial

, and therefore with Eq. (10.33)

(10.34)

For a controller with integral action the coefficient is zero and according to Eq. (10.22)

. From Eqs. (10.33) and (10.24) to (10.28) and (10.29) it follows directly that

(10.35)

When the controller is inserted into the feedback path according to Figure 10.6b the inherent

closed-loop dynamics will not be changed compared with the configuration according to

Figure 10.6a, because the denominator polynomial of the transfer function, and therefore the

characteristic equation of the closed loop, are preserved. Indeed, the zeros of the controller

transfer function do no longer arise, but their poles as zeros in the closed-loop transfer function.

Analogous considerations for lead to

(10.36)

Whereby the polynomial contains the poles of the controller and the plant zeros in

the left-half plane. The transfer function













(10.37)

is the same as for the case of a stable controller and a minimum-phase plant according to Eq. (10.32).

The constant for a proportional controller is

(10.38)

For an integral controller in the feedback loop a feed-forward path is not realizable.

The synthesis equations

The system of equations described by Eq. (10.18) can be rewritten in matrix notation. Thus the

required controller parameters are combined into one parameter vector. The matrix of the plant

parameters applies for both the cases, controller order and .

For integral plants

( ) with controller order

and normalized

the system is:

(10.39)

and

(10.40)





For proportional plants or in the case of disturbances at the input of integral plants, where the

order of the controller is increased by one to , with Eqs. (10.22) and (10.16) it follows that

(10.41)

(10.42)

Here the system is:

(10.43)

and

(10.44)

The matrices on the left side of Eqs. (10.39) and (10.43) are equal for . This matrix is always regular and therefore the solution is always unique.

Application of the method Example 10.4.1 An integral plant has the transfer function

As we do not consider a disturbance at the input of the plant, the controller coefficients can be calculated by Eqs. (10.39) and (10.40).

According to section 10.4.1 one obtains for this second-order plant the order of the

controller. The closed loop has therefore order . The step response of this loop







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should comply with the binomial form given in Table A.2, and should be met. This

corresponds to a value of . The associated characteristic polynomial is

and Eqs. (10.39) and (10.40) provide the synthesis equation

with

from which the controller coefficients follow as

The transfer function of the controller is

It can be seen that this design leads to an unstable controller, which is in addition non-minimum

phase. The closed-loop transfer function is

and it contains in the numerator polynomial, besides the zero of the plant, the right-half-plane

zero of the controller. According to the considerations in section 10.4.2 this zero cannot be

compensated by a pre-filter for stability reasons. Also this would not be necessary, as the plant

zero is dominant and has a stronger influence on the closed-loop step response (see Figure 10.7).

The denominator polynomial




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Figure 10.7: Step responses of the controlled variable for the case: (a) without pre-filter

, (b) with pre-filter , and the associated manipulated variable and , and the

step response of the given closed-loop transfer function .

for the transfer function of the pre-filter is determined as

and the numerator is

The closed-loop transfer function including the pre-filter is thus

This in fact still contains in the numerator polynomial the controller zero, but as can be seen

from Figure 10.7, the corresponding step response does not show a large deviation from

the step response of the given transfer function




Conspicuous is the fact that the step response of the closed loop without pre-filter has a

large overshoot and does not show any similarity with the other step responses, though all three

other systems have the same inherent behavior. Here, the dominant behavior of the plant zero

has a noticeable effect.

Example 10.4.2 Given the third-order plant transfer function

The step response of this plant is shown in Figure 10.8. A controller should be designed

such that the step response of the closed loop has a desired standard form chosen from

Table A.1. In the current case of a plant the order of the controller must be chosen equal

to the order of the plant that is

Thus one obtains a sixth-order closed-loop transfer function . The desired transfer

function according to Eq. (10.4), which is the basis for Table A.1, has then exactly the total order

, if

is chosen. For this case the step response with

(see Table A.1) will be desired. The last unspecified parameter of the transfer function is

the relative frequency . All step responses in Table A.1 are normalised by this value such that

the time scale can still be chosen. Consequently, by a proper choice of the normalised step

response from Table A.1 can be scaled to the desired time scale. E.g., for a value of a

rise time would follow, for this would be . If a large value is










taken for to obtain a small rise time, this would result in controller coefficients of very

different order such that this controller would not be realisable from a numerical point of view.

Therefore here

is a good choice. For the determined values of and one obtains for the desired closed-loop

transfer function from Eq. (10.4)

The Eqs. (10.41) and (10.42) deliver at first the absolute coefficients of the controller transfer

function

and

The remaining coefficients are obtained from Eq. (10.43) by solving

The solution is







If the reciprocal gain is chosen according to Eq. (10.22) as

for

then the controller coefficients have the same order of magnitude and the controller transfer

function is

From Eq. (10.34) one obtains for the coefficient of the pre-filter

The transfer function of the pre-filter is





Figure 10.8: (a) Step responses of the controlled variable for the plant in the uncontrolled

case and in the controlled case for the designed compensator and for the optimal PI

controller ; (b) Step responses of the associated manipulated variables and

The step response of the controlled variable of the closed loop is shown in Figure 10.8.

For comparison this figure also shows the corresponding step response of the closed loop

using a PI controller, which is optimal in the sense of the performance index according to

section 7.3.3. Both, the maximum overshoot and the rise time of this control system with a PI

controller are clearly worse than for the controller designed here. From the behaviour of the

manipulated variables and , respectively, it can be seen that in general a smaller rise

time must be bought by a larger amplitude of the controller output. Because of the always


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existing limitation on the value of the manipulated variable, too demanding specifications for the

transfer function cannot be realised.

DESIGN OF PHASE-LEAD COMPENSATOR Phase-lead compensator, which corresponds to PD controller, is appropriate, when speed is

required, because it will speed up the original response. Typical applications are in

servos.Transfer function of phase-lead compensators is given by

Gain at high frequencies

and at low frequencies

Upper cut-off frequency

and lower cut-off frequency

Maximum phase shift f max of phase-lead compensator

occurs at frequency

The value of gain at that frequency is

A typical Bode diagram of a phase-lead controller is given in Figure 4.1.



WLead(s) is leading ahead the phase of the input signal. As can be seen from Figure 4.1 the gain

increases at higher frequencies. The gain at higher frequencies is a2. Since a2 > 1, this means

that WLead ( ) dB (a2 )dB is positive, which is clear from Figure 4.1. Let us study more

carefully how much phase margin can be added for various values of a2 .

Steps in design of phase-lead compensator STEP 1.

Choose gain K to satisfy steady-state requirements.

STEP 2.

Draw Bode-diagram of KG(s).

STEP 3.



Determine the new crossover frequency wc, i.e.., the frequency at which the uncompensated

system has phase (-180o+PM’), where PM’ = desired phase margin by choosing a few values of

a2 starting with a2 =2 and going up to 10-20. Construct a table showing how much this choice

will add to phase margin and also to magnitude. In this way a2 and

Can be determined.This procedure also reveals if one phase lead circuit is enough.Remember

also that the bigger a2 is, the more susceptible it is for noise.

STEP 4.

The new crossover frequency

This completes the design of the phase-lead compensator.

STEP 5.

Check the design by simulating step response. Draw also the open-loop Bode diagram of the

compensated system for comparison purposes.

STEP 6.

If requirements are met, stop. Otherwise go back to STEP 1.

EXAMPLE

Open-loop transfer function of a unity feedback system of Fig.4.5 is given by

Fig.4. 5. The block diagram of the overall system. Here WLead(s) represents the phase-lag

controller or compensator and G(s) the open-loop system transfer function. The system has



Unity feedback, H(s) = 1.

Specifications for the system are:

• Accuracy for a unit ramp input < 2%, i.e. steady-state error < 0.02.

• Maximum percent overshoot = PO < 20%.

Design a phase-lead compensator that satisfies the requirements.

SOLUTION: STEP 1. (is the same as in phase lag)

Compute the steady-state for unit ramp input ess

This implies that

K > 1/0.02 = 50, choose K = 50. (4- 12)

Gain is now sufficient. The controller is at this point a P (proportional) controller.

STEP 2. (As in phase lag)

From Fig.2.5 percent overshoot PO <20% corresponds to > 48° phase margin, PM. This

holds for second order systems and therefore we will make the dominant pole assumption.

Draw the Bode diagram of the transfer function KG(s) = 50 / s(1 + 0.2s) as before. It is

redrawn in Fig.4.6.

» kg=zpk([] ,[0 -5],250)

Zero/pole/gain:



STEP 3

Recall that maximum phase shift





Let us compare the Bode diagrams of the uncompensated and compensated systems by

Drawing them in the same figure, Figures 4.8.



The step responses are plotted in the same figure, Fig.4.10:



The zero is not very close to the pole p = -17.4439 but seems to cancel most of its

effect, but not quite enough. This is probably the reason why the PO requirement was not

satisfied right away.

Let us finally draw the Bode diagram of the phase-lead compensator.



UNIT-V

STATE SPACE ANALYSIS























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