Controlengg Compiled Pillai(Session 1 7)

7/31/2019 Controlengg Compiled Pillai(Session 1 7)

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REFERENCE BOOKS :

1. Stefano III, J. J., Stubbered, A. R., Williams, I. J., Theory and Problems of

Feedback and Control Systems, Schaums Outline Series, 2001.2. Nagrath, I. J., and Gopal, M., Control Systems Engineering, Wiley Eastern Ltd.

3. Gopal, M., Control Systems Principles and Design, Tata McGraw Hill, 1997.

4. Sivanandom, S. N., Control Systems Engineering, Vikas Publishing House, 2001.5. Dorf, R. C., and Bishop, R. H., Modern Control Systems, Addison-Wesely, 8th

Editiuon, 1998.

6. Kuo, B. C., Automatic Control Systems, Perntice Hall (India), 1995.7. Eronini, E. U., System Dynamics and Control, Thomson Learning, 2002.

SUBJECT EXPERTS:

1. Pilli, S. C., Principal, K.L.E Societys College of Engineering and

Technology, Udyambag, Belgaum.2. Sridhar, B. K., Professor, Department of Mechanical Engineering,

National Institute of Engineering, Mysore.

3. Girish, D. V., Professor, Department of Mechanical Engineering,

Malanad College of engineering, Hassan.

4. Sreenidhi, R., Senior lecturer, Department of Mechanical

Engineering, S J C E, Mysore.

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CHAPTER 1

Introduction

A control system is an arrangement of physical components connected or related in such amanner as to command, direct, or regulate itself or another system, or is that means by

which any quantity of interest in a system is maintained or altered in accordance with a

desired manner.

Any control system consists of three essential components namely input, system

and out put. The input is the stimulus or excitation applied to a system from an external

energy source. A system is the arrangement of physical components and output is theactual response obtained from the system. The control system may be one of the following

type.

1) man made2) natural and / or biological and

3) hybrid consisting of man made and natural or biological.Examples:

1) An electric switch is man made control system, controlling flow of electricity.input : flipping the switch on/off

system : electric switch

output : flow or no flow of current2) Pointing a finger at an object is a biological control system.

input : direction of the object with respect to some direction

system : consists of eyes, arm, hand, finger and brain of a manoutput : actual pointed direction with respect to same direction

3) Man driving an automobile is a hybrid system.

input : direction or lanesystem : drivers hand, eyes, brain and vehicle

output : heading of the automobile

Classification of Control Systems

Control systems are classified into two general categories based upon the control action

which is responsible to activate the system to produce the output viz.1) Open loop control system in which the control action is independent of the out put.

2) Closed loop control system in which the control action is some how dependent

upon the output and are generally called as feedback control systems.

Open Loop System is a system in which control action is independent of output. To

each reference input there is a corresponding output which depends upon the system and itsoperating conditions. The accuracy of the system depends on the calibration of the system.

In the presence of noise or disturbances open loop control will not perform satisfactorily. A

closed loop control system is one in which the control action depends on the output. In

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Characteristics of open loop control system

1) Simple in construction and easy to maintain.

2) Less expensive than a corresponding closed loop system.3) There is no stability problem.

4) Convenient when output is hard to measure or economically not feasible.

5) Disturbances and changes in calibration cause errors.Characteristics of closed open loop control system

1) Feedback systems can faithfully reproduce the input.

2) Increased accuracy.

3) Reduced effects of nonlinearities and distortion.

4) Increased bandwidth over which the system will respond satisfactorily.

5) Closed loop control systems have tendency to oscillate or become unstable

Definitions:

Systems: A system is a combination of components that act together and perform a certainobjective. The system may be physical, biological, economical, etc.

Control system: It is an arrangement of physical components connected or related in a

manner to command, direct or regulate itself or another system.

Open loop: An open loop system control system is one in which the control action is

independent of the output.

Closed loop: A closed loop control system is one in which the control action is somehow

dependent on the output.Plants: A plant is equipment the purpose of which is to perform a particular operation. Any

physical object to be controlled is called a plant.

Processes: Processes is a natural or artificial or voluntary operation that consists of a series

of controlled actions, directed towards a result.Input: The input is the excitation applied to a control system from an external energy

source. The inputs are also known as actuating signals.

Output: The output is the response obtained from a control system or known as controlled

variable.

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 characterizesthe functional relationship amongst the components of a controlsystem.

Control elements: These are also called controller which are the components required to

generate the appropriate control signal applied to the plant.Plant: Plant is the control system body process or machine of which a particular quantity

or condition is to be controlled.

Feedback control: feedback control is an operation in which the difference between theoutput of the system and the reference input by comparing these using the difference as a

means of control.

Feedback elements: These are the components required to establish the functionalrelationship between primary feedback signal and the controlled output.

Actuating signal: also called the error or control action. It is the algebraic sum consisting

of reference input and primary feedback.

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Manipulated variable: it that quantity or condition which the control elements apply to

the controlled system.

Feedback signal: it is a signal which is function of controlled outputDisturbance: It is an undesired input signal which affects the output.

Forward path: It is a transmission path from the actuating signal to controlled output

Feedback path: The feed back path is the transmission path from the controlled output tothe primary feedback signal.

Servomechanism: Servomechanism is a feedback control system in which output is some

mechanical position, velocity or acceleration.

Regulator: Regulator is a feedback system in which the input is constant for long time.

Transducer: Transduceris a device which converts one energy form into other

Tachometer: Tachometeris a device whose output is directly proportional to time rate of

change of input.Synchros: Synchrosis an AC machine used for transmission of angular position synchro

motor- receiver, synchro generator- transmitter.

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 characterizesthe 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 s or Fourier transformed quantity

i.e. frequency function as function of imaginary variable js = .

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 thanblock or summing point, take off point is used. This permits the signal to proceed unaltered

along several different paths to several destinations.Time response: It is the output of a system as a function of time following the application

of a prescribed input under specified operating conditions.

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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. Thesignal flows in the direction indicated by the arrow.

Definitions:

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

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

Transmittance: It is the gain between two nodes.

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

corresponds to independent variable.

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

corresponds to dependent variable.

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

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

Loop: A loop is a closed path.

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

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

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

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

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

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

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 Xkbut not the reverse.

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

signal akjXkis delivered at node Xj.

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Example1

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

Figure 3 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 4 Signal flow graph of the system shown in Fig. 1

G2

G1

G3

H2

H1

+++ ++

R X1 X2 X3 X4 X5 X6 C

R X1 X2 X3

X4 X

5X

6

CG

1

H1

-H2

G2

G3

-1

11 1 1

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Example 2

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

Figure 5 Block diagram feedback system

The nodal variables are X1, X2, X3.

The signal flow graph is shown in Fig. 4.

Figure 6 Signal flow graph of example 2

Example 3Draw 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.

10

R

G4

G2

-G3

G1

C1 1

X1

X2

X3

G1

G2

G3

G4

R

C

X1

X2

X3


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u1 b1

b2

a21

a12

a33

a31

X1

X2

u2

a23

a22


Example 4

LRC net work is shown in Fig. 6. Draw its signal flow graph.

11

+

a11

a13

a32

X3


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Figure 8 LRC network

The governing differential equations are

( ) ( )

( ) ( )

( ) ( )3

2

11

tidt

deC

teeRidt

diL

or

teidtC

Ridt

diL

c

c

=

=++

=++

Taking Laplace transform of Eqn.1 and Eqn.2 and dividing Eqn.2 by L and Eqn.3 by C

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )510

411

0

sIC

essE

sEL

sEL

SIL

RissI

cc

c

=

=++

+

+

Eqn.4 and Eqn.5 are used to draw the signal flow graph shown in Fig.7.

Figure 9 Signal flow graph of LRC system

12

Cs

1( )L

RsL

1

+

( )L

RsL

1-

+

LRs

1

+s

1

i(0+)

Ec(s)

I(s)

E(s)

ec(0+)

R L

Ci(t)

+

ec(t)

e(t)


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SIGNAL FLOW GRAPHS

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.

Masons Gain Formula

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 threenontouching loops) +

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

Fig.1


There are two forward paths:

Gain of path 1 : P1=G1Gain of path 2 : P2=G2

There are four loops with loop gains:L1=-G1G3, L2=G1G4, L3= -G2G3, L4= G2G4There 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

13

R

G4

G2

-G3

G1

C1 1


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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=G1G2G3There are three loops with loop gains:L1=-G1G2H1, L2=G2G3H2, L3= -G1G2G3There are no non-touching loops.

= 1-G1G2H1+G2G3H2+G1G2G3Forward path 1 touches all the loops. Therefore, 1= 1.


( ) 321231121

32111

1 GGGHGGHGG

GGGP

sR

sC

++=

=

Example 3Obtain 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=G1G2G7There are four loops with loop gains:

L1=-G4H1, L2=-G2G7H2, L3= -G6G4G5H2 , L4=-G2G3G4G5H2There is one combination of Loops L1 and L2 which are nontouching with loop gain

product L1L2=G2G7H2G4H1 = 1+G4H1+G2G7H2+G6G4G5H2+G2G3G4G5H2+ G2G7H2G4H1Forward 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 =

( )

( )

( )

174225432254627214

14721654154321332211

1

1

HGGGHGGGGHGGGHGGHG

HGGGGGGGGGGGGGPPP

sR

sC

++++++++

=

++=

14

R(s) C(s)1 1 1G

1 G2 G3

H1

-1

-H2

G1

C(s)R(s)

G7

G6

-H1

G2 G3 G4 G5

-H2

X1

X2

X3

X4 X

5

1


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Example 4

Find the gains1

3

2

5

1

6 ,,X

X

X

X

X

Xfor the signal flow graph shown in Fig.4.

Figure 13 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=acdefP2=abef

There are four loops with loop gains:

L1=-cg, L2=-eh, L3= -cdei, L4=-beiThere is one combination of Loops L1 and L2 which are nontouching with loop gain

product L1L2=cgeh

= 1+cg+eh+cdei+bei+cgehForward 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

Case 2:2

5

X

X

15

b

a dc fe

-h

-g

-i

X1

X6

X5

X4X3X2


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

Figure 14 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 inFig.6.

Figure 15 Signal flow graph of example 4 case 3

There are two forward paths.

The gain of the forward path are: P1=abcdP2=ac

There are five loops with loop gains:

L1=-eh, L2=-cg, L3= -bei, L4=edf, L5=-befgThere is one combination of Loops L1 and L2 which are nontouching with loop gain

product 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


ehcgbefgefdbeicgeh

ehacabefPP

X

X

++++++++

=

+=

1

12211

1

3

Example 5

For the system represented by the following equations find the transfer function X(s)/U(s)

using signal flow graph technique.

+=

++=

+=

uXaX

uXXaX

uXX

1122

22111

31

Taking Laplace transform with zero initial conditions

16

b

1 dc 1e

-h

-g

-i

X2 X5X5

X4

X3

X2

a eb f

-h

-g

-i

X1

X5

X4 X

3

X2

c

d

X3

1


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( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )sUsXassX

sUsXsXassX

sUsXsX

1122

22111

31

+=++=

+=

Rearrange the above equation( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )sUs

sXs

asX

sUs

sXs

sXs

asX

sUsXsX

11

22

221

11

31

1

+

=

++

=

+=

The signal flow graph is shown in Fig.7.

Figure 16 Signal flow grapgh of example 5

There are three forward paths.

The gain of the forward path are: P1=3P2=1/ s2

P3=2/ s

There are two loops with loop gains:

2 22

11

s

aL

s

aL

=

=

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

There are no combination two Loops which are nontouching.

2

211s

a

s

a++=

Forward path 1 does not touch loops L1 and L2. Therefore

2

21

1 1 s

a

s

a

++=

Forward path 2 path 3 touch the two loops. Hence, 2= 1, 2= 1.

The transfer function T =(

21

2

1221

2

3332211

1

3

asas

sasasPPP

X

X

++++++

=

++=

STEADY STATE ERRORS IN UNITY FEEDBACK

17

U

X1

X2

3

s

a1s1

s

2

s

a2

s

1

1

X X


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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 giveninput 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 theseindividual 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

TK

sG

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 calledType 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 17 Unity feedback control system

The closed loop transfer function is

18

+

R(s)G(s)

E(s) C(s)


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( )( )

( )( )sG1

sG

sR

sC

+=

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

( )( )

( )( ) ( )sG1

1sRsC1

sRsE

+== , Which can be rearranged as ( )

( )( )sR

sG11sE

+=

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

( ) ( )( )

( )sG1

ssRlimssElimtelime

0s0stss +===

Three types of static error constants are1) Static position error constant Kp due to unit step input.

2) Static velocity error constant Kv due tounit 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

( ) ( )0GsGlimK0s

p==

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

by

p

ss K1

1e

+=

For a type 0 system,

K

1sp

T1s2

T1s1

T

1sm

T1sb

T1sa

TK

limK0s

p=

+

+

+

+

+

+

=

andK1

1e

ss +=

For a type 1 or higher system

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1N

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TK

limKN

0sp=

+

+

+

+

+

+

=

and =sse

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

020sss

=+

=s

The static velocity error constant Kv is defined by

( )ssGlimK0s

v =

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


0

1sp

T1s2

T1s1

T

1sm

T1sb

T1sa

TsK

limK0sv

=

+

+

+

+

+

+

=

and ==v

ss K

1e

For a type 1 system

K

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TsK

limK0s

v=

+

+

+

+

+

+

=

andK

1

K

1e

v

ss==

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2N

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TsK

limKN

0sv

=

+

+

+

+

+

+

=

and 0K

1e

v

ss==

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

( )sGslimK2

0sa

=

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

a

ss K

1e =


0

1sp

T1s2

T1s1

T

1sm

T1sb

T1sa

TKs

limK

2

0sa

=

+

+

+

+

+

+

=

and ==a

ss K

1e

For a type 1 system

0

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TKs

limK

2

0sa

=

+

+

+

+

+

+

=

and ==a

ss K

1e

For a type 2 system

21


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K

1spT1s

2T1s

1Ts

1sm

T1sbT1s

aTKs

limK2

2

0sa=

+

+

+

+

+

+

=

andK

1

K

1e

a

ss==


3N

1sp

T1s2

T1s1

Ts

1sm

T1sb

T1sa

TKs

limKN

2

0sa

=

+

+

+

+

+

+

=

and 0K

1e

a

ss==

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 ofsystem

Error constants Steady state error essKp 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 functionrespectively.

2) Since the error constants are defined with respect to forward path transferfunction 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 firstto 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 are present.

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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 erroe due to unit step, unit

ramp and unit parabolic input of the following systems.

( )( ) ( )

( ) ( )

( )( ) ( )

( )102sss4s12s1K

sGiii)

2004sss

KsGii)

2s10.1s1

50sGi)

22

2

++++

=

++=

++=

Solution

a

ss

2

0sv

v

ss0sv

p

ss0s

p

K

1eG(s)slimK

K

1esG(s)limK

K1

1eG(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 essKp 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

23


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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 controllerproduces 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 controllers4) Proportional-plus- integral controllers

5) Proportional-plus-derivative controllers6) 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 theoperation as

1) Electric controllers

2) Hydraulic controllers3) Pneumatic controllers

24


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4) Electronic controllers.

The block diagram of a typical controller is shown in Fig.1. It consists of anautomatic 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 convertsthe 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 18 Block diagram of control system

Two Position Control Action

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

Figure 19 Block diagram of on off controller

++

Output c(t)

automatic controller

actuating

error signal e(t)

reference

input r(t)

errordetector

Amplifier Actuator Plant

Sensorfeed backsignal b(t)

Controller

output u(t)

++

actuating

error signal e(t)

referenceinput r(t)

error

detector

feed back

signal b(t)

Controller

output u(t)

25


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Proportional Control Action

The proportional controller is essentially an amplifier with an adjustable gain. For acontroller with proportional control action the relationship between output of the controller

u(t) and the actuating error signal e(t) is

( ) ( )teKtu p= .....(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 20 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 actuatingerror signal e(t) given by Eqn.4

( )( ) ( ) ( )==

t

0ii

dtteKtuorteKdt

tdu..(4)

Where, Ki is an adjustable constant.

The transfer function of integral controller is

( )( ) s

K

sE

sU i= .(5)

++

actuating

error signal e(t)

referenceinput r(t)

error

detector

feed back

signal b(t)

Controller

output u(t)

Kp

26


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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 21 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

i

p.....(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 ofthe 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.

++

actuating

error signal e(t)

reference

input r(t)

errordetector

feed back

signal b(t)

Controller

output u(t)

s

Ki

++

actuating

error signal e(t)

reference

input r(t)

error

detector

feed back

signal b(t)

Controller

output u(t)( )sT

sT1K

i

ip+

27


28/114

Figure 22 Block diagram of a proportional-plus-integral control system

Figure 23 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.

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 theeffect 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 24 Block diagram of a proportional-plus-derivative controller

t0

Kp

u(t)

2Kp

proportional only

P-I control action

Ti

++

actuatingerror signal e(t)

reference

input r(t)

error

detector

feed backsignal b(t)

Controlleroutput u(t)

( )ST1K DP +

28


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Figure 25Response 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 derivativecontrol action. The equation of the controller is

( ) ( ) ( )( )

dt

tdeTKdtte

T

KteKtu

dp

t

oi

p

p++= ..(10)

or the transfer function is

( )( )

++= sT

sT

11K

sE

sUd

i

p..(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 26Block diagram of a proportional-plus-integral-plus-derivative controller

t0

u(t)

proportional only

P-D control action

Td

++

actuating

error signal e(t)

reference

input r(t)

errordetector

feed backsignal b(t)

Controller

output u(t)

( )sT

sTTsTK

d

diip

21 ++

29


30/114

Figure 27 Response of PID controller to unit actuating error signal

t0

u(t)

proportional only

PD control actionPID control action

30


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CHAPTER II

MATHEMATICAL MODELING

2.1 Introduction: In chapter I we have learnt the basic concepts of control systems such as

open loop and feed back control systems, different types of Control systems like regulator

systems, follow-up systems and servo mechanisms. We have also discussed a few simple

applications. In this chapter we learn the concepts of modeling.

The requirements demanded by every control system are many and depend on the system

under consideration. Major requirements are 1) Stability 2) Accuracy and 3) Speed of

Response. An ideal control system would be stable, would provide absolute accuracy

(maintain zero error despite disturbances) and would respond instantaneously to a change

in the reference variable. Such a system cannot, of course, be produced. However, study of

automatic control system theory would provide the insight necessary to make the most

effective compromises so that the engineer can design the best possible system. One of the

important steps in the study of control systems is modeling. Following section presents

modeling aspects of various systems that find application in control engineering.

2.2 Modeling of Control Systems: The first step in the design and the analysis of control

system is to build physical and mathematical models. A control system being a collection

of several physical systems (sub systems) which may be of mechanical, electrical

electronic, thermal, hydraulic or pneumatic type. No physical system can be represented in

its full intricacies. Idealizing assumptions are always made for the purpose of analysis and

synthesis.An idealized representation of physical system is called a Physical Model.

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Control systems being dynamic systems in nature require a quantitative mathematical

description of the system for analysis. This process of obtaining the desired

mathematical description of the system is called Mathematical Modeling.

The basic models of dynamic physical systems are differential equations obtained by the

application of appropriate laws of nature. Having obtained the differential equations and

where possible the numerical values of parameters, one can proceed with the analysis.

When the mathematical model of a physical system is solved for various input conditions,

the results represent the dynamic response of the system. The mathematical model of a

system is linear, if it obeys the principle ofsuperposition and homogeneity.

A mathematical model is linear, if the differential equation describing it has coefficients,

which are either functions of the independent variable or are constants. If the coefficients

of the describing differential equations are functions of time (the independent variable),

then the mathematical model is linear time-varying. On the other hand, if the coefficients

of the describing differential equations are constants, the model is linear time-invariant.

Powerful mathematical tools like the Fourier and Laplace transformations are available for

use in linear systems. Unfortunately no physical system in nature is perfectly linear.

Therefore certain assumptions must always be made to get a linear model.

Usually control systems are complex. As a first approximation a simplified model is built

to get a general feeling for the solution. However, improved model which can give better

accuracy can then be obtained for a complete analysis. Compromise has to be made

between simplicity of the model and accuracy. It is difficult to consider all the details for

mathematical analysis. Only most important features are considered to predict behaviour of

32


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the system under specified conditions. A more complete model may be then built for

complete analysis.

2.3 Modeling of Mechanical Systems: Mechanical systems can be idealized as spring-

mass-damper systems and the governing differential equations can be obtained on the basis

of Newtons second law of motion, which states that

F = ma: for rectilinear motion: where F: Force, m: mass and a: acceleration (with

consistent units)

T = I : orJ for rotary motion: where T: Torque, I or J: moment of inertia and

: angular acceleration (with consistent units)

Mass / inertia and the springs are the energy storage elements where in energy can be

stored and retrieved without loss and hence referred as conservative elements. Damper

represents the energy loss (energy absorption) in the system and hence is referred as

dissipative element. Depending upon the choice of variables and the coordinate system, a

given physical model may lead to different mathematical models. The minimum number of

independent coordinates required to determine completely the positions of all parts of a

system at any instant of time defines the degrees of freedom (DOF) of the system. A large

number of practical systems can be described using a finite number of degrees of freedom

and are referred as discrete or lumped parameter systems. Some systems, especially those

involving continuous elastic members, have an infinite number of degrees of freedom and

are referred as continuous or distributed systems. Most of the time, continuous systems

are approximated as discrete systems, and solutions are obtained in a simpler manner.

Although treatment of a system as continuous gives exact results, the analysis methods

available for dealing with continuous systems are limited to a narrow selection of

33


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problems. Hence most of the practical systems are studied by treating them as finite

lumped masses, springs and dampers. In general, more accurate results are obtained by

increasing the number of masses, springs and dampers-that is, by increasing the number of

degrees of freedom.

Mechanical systems can be of two types: 1) Translation Systems 2) Rotational Systems.

The variables that described the motion are displacement, velocity and acceleration.

2.3.1 Characteristics of Elementary parts of Mechanical Systems

Translational Systems

Mass: Property or means Kinetic energy is stored

Figure 2.1:

F= Mass * Acceleration

= m x

Linear Spring

Elasticity (Spring): Property or means Potential energy is stored. Spring force is

proportional to displacement

m

x

F

34

..


35/114

Figure 2.2 (a) and (b):

Spring Force = Stiffness * displacement

Fs = Kx

Damping

Exists in all system and opposes motion. The types which are generally considered are:

Static Friction

Coulomb Friction

Viscous Friction

Viscous Friction (Damping)

Viscous Damping: Means by which energy is absorbed. Damping Force is proportional to

velocity

xF k

Linear

Non Linear

Fs

x

35


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Figure 2.3 (a) and (b):

Damping Force =

Damping Coefficient * Velocity

Fd = C x

Rotational Systems

Inertia Element

Figure 2.4:

Torque = Inertia * Angular Acceleration

T = J

Torsional Spring

Fd

x.

Oil

Piston Cylinder

Fd

Fd

= C x.

T

J

36

.

..


37/114

Figure 2.5:

Torque = Torsional Stiffness * Angular displacement

T = kt *

Friction (Viscous Damping)

Figure 2.6:

Damping Torque = Damping Coefficient * Angular velocity

Td = B *

T

kt

T

B

37

.


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CHAPTER II

MATHEMATICAL MODELING (Continued)(Continued from session III: 30.08.2006)

2.3.2 Models for Translational Systems:

Illustration 1: Figure 2.7 shows a spring mass system that represents the simplest possible

mechanical system. It is a single DOF system since one coordinate (x) is sufficient to

specify the position of mass at any time.

Figure 2.7: Spring Mass System Figure 2.7 a: Free Body Diagram

Using Newtons second law of motion, we will consider the derivation of the equation of

motion in this section. The procedure we will use can be summarized as follows:

1) Select a suitable coordinate (x) to describe the position of the mass or rigid body in

the system. Use a linear coordinate to describe the linear motion of a point mass or the

centroid of a rigid body, and an angular coordinate () to describe the angular motion

of a rigid body.

2) Determine the static equilibrium configuration of the system and measure the

displacement of the mass or rigid body from its static equilibrium position.

38

mmx

KKx (Spring force)

Static equilibrium position


39/114

3) Draw the free-body diagram of the mass or rigid body when a positive

displacement and velocity are given to it. Indicate all the active and reactive forces

acting on the mass or rigid body.

4) Apply Newtons second law of motion to the mass or rigid body shown by the free-

body diagram. Newtons second law of motion can be stated as follows:

Note:

1. Spring Force: Stiffness*displacement = K*x

2. Sign Convention: Forces and torques in the direction of motion are positive. Therefore

in the above example spring force (Kx) is to be taken negative as it is in the direction

opposite to that of displacement (motion).

Illustration 2: Here the mass m slides on a frictionless surface. Therefore there is no

damping

Figure 2.8:

From NSL

Kx

m

x

Friction less surfacem

K

39

F = ma ---- Newtons Second Law of Motion

m x = - Kx

mx + Kx = 0

..

..


40/114

F = ma

m x = - Kx

mx + Kx = 0

Points to Note:

The displacement of the mass is to be considered zero with the system at equilibrium i.e.,

the spring is neither stretched nor compressed

Sign Convention: Adopted sign convention must not be changed during the course of the

problem

All the forces / torque in the direction of the displacement are considered positive,

otherwise, negative

Behaviour of the system is independent of the sign convention

Illustration 3: The weight of the mass was not factor in the system just analyzed.

However, weight is a factor in the spring mass system shown in figure 2.9

Figure 2.9:

From NSL

F = ma

m x = k ( x) - mg

k ( x)

x

k

m mg

x

40

..

..

Position with sprinrelaxed

Reference position

(x = 0)

..


41/114

= k kx mg

= (mg / k) = Static deflection

mx = - kx mx + kx = 0

The weight of the mass has no effect on the differential equation when the reference is at

the equilibrium position. Hence, the differential equation is same as that of the previous

system in which weight was not a factor.

Illustration 4: Figure 2.10 (a) shows a spring mass damper system and a corresponding

free body diagram is shown in figure 2.10 (b) which assumes positive (downward)

displacement and velocity. If the reference x = 0 is chosen at the equilibrium position, the

mass can be considered weight less. Both the spring force and the damping force act

vertically upwards.

Figure 2.10 (b)

Figure 2.10 (a)

From NSL

F = ma

mx = - c x kx

41

xm

k

c

x

k x (Spring Force)

c x (damping Force).

.

mx + c x + kx = 0.. .

.. ..

..


42/114

Illustration 5: A variation of the previous system is shown in figure 2.11 (a) where a

provision is made for input displacement x at the top of spring. Figure 2.11 (b) shows the

free body diagram of the mass responding to a positive input at x. It is assumed that x is

displaced upward and that the mass is responding with a positive displacement y (less than

x) and with a positive velocity. With this assumption the spring force acts upward with a

magnitude k (x-y) and the damping force is acts downwards as shown in the free body

diagram.

Figure 2.11 (b)

Figure 2.11 (a)

--- GDE

42

m

X (Input)

y (Response)

m

k (x-y)

c y.

m

kx

c y.

ky

From NSLmy = + k (x-y) - c y.. .

my + c y + ky = kx.. .

c

k

x > y


43/114

2.3.3 Models for Rotational or Torsional Systems: The principles presented previously

can be extended to obtain the mathematical model for torsional or rotational systems.

Illustration 6: Consider a system with rotating inertia J with torsional spring of stiffness kt

and a rotary damper with damping coefficient b as shown in figure 2.12 (a). If J is

displaced by the corresponding free body diagram is as shown in figure 2.12 (b).

Figure 2.12 (a) Figure 2.12 (b)

From NSL for rotating system

= TJ bkJ t =

0=++ tkbJ

Illustration 7: A variation of the previous system is shown in figure 2.13 (a) where a

provision is made for input displacement i at the end of torsional spring. Figure 2.13 (b)

shows the free body diagram of the inertia responding to a positive input i. It is assumed

that i is displaced clockwise (looking from left) and that the inertia is responding with a

clockwise displacement 0 (less than i) and a positive velocity. With this assumption the

kt

bJ

Jkt. b.

43

.. .

.. .


44/114

spring torque is acting clockwise with a magnitude kt (i -0) and the damping torque is

acting counter clockwise as shown in the free body diagram.

Figure 2.13 (a)

Figure 2.13 (b)

Let i: Input

And 0: ResponseLet (i > 0)


= TJ

Illustration 8: A torsional system with torque T as input and angular displacement asoutput.

kt

bJ

0

i

J bKti

Kt(

i-

0) J

b

kt

0

.

.

44

J0 = kt (i - 0) - b

J 0 + b0 + kt0 = kt i --- Model

.. .

.. . .


45/114

Figure 2.14 (a)

Figure 2.14 (b)


= TJ

J = T - b

J + b = T --- Model

2.3.4 System with more than one mass: System involving more than one mass is

discussed below.

Illustration 9: For a two DOF spring mass damper system obtain the mathematical model

where F is the input x1 and x2 are responses.

b

T

J

b

T

.J

45

.. .

.. .


46/114

Figure 2.15 (a)

Figure 2.15 (b)

From NSL F= ma

For mass m1

m1x1 = F - b1 (x1-x2) - k1 (x1-x2) --- (a)

For mass m2

m2 x2 = b1 (x2-x1) + k1 (x2-x1) - b2 x2 - k2 x2 --- (b)

m2

m1

k2

b2

(Damper)

x2 (Response)

x1

(Response)

k1

F

b1

m2

m1

F

k2

x2

b2

x2

k1

x2

k1

x1

b1

x1

b1

x2

x2

k1

x2

k1

x1

b1

x1 b1 x2

.

. .

.

m2

m1

F

k2

x2

b2

x2

k1

(x1-x

2)

.

.b

1(x

1-x

2)

.

x1

.

46

.. . .

.. . . .

Draw the free body diagram for mass m1and m2 separately as shown in figure 2.15

(b)

Apply NSL for both the masses separatelyand get equations as given in (a) and (b)


47/114

Illustration 10: For the system shown in figure 2.16 (a) obtain the mathematical model if

x1 and x2 are initial displacements.

Let an initial displacement x1 be given to mass m1 and x2 to mass m2.

Figure 2.16 (a)

K1

K2

K3

m2

m1

X1

X2

47


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Figure 2.16 (b)

Based on Newtons second law of motion: F = maFor mass m1

m1x1 = - K1x1 + K2 (x2-x1)

m1x1 + K1x1 K2 x2 + K2x1 = 0

m1x1 + x1 (K1 + K2) = K2x2 ----- (1)

For mass m2

m2x2 = - K3x2 K2 (x2 x1)

m2x2 + K3 x2 + K2 x2 K2 x1

m2 x2 + x2 (K2 + K3) = K2x1 ----- (2)

Mathematical models are:

m1x1 + x1 (K1 + K2) = K2x2 ----- (1)

m2x2 + x2 (K2 + K3) = K2x1 ----- (2)

K1

X1

K2

X2

K2

X2

K2

X1

K2

X1

K3

X2 K

3X

2

K1

X1

X1 X

1

m1

m1

m2 m

2X

2 X2

K2

(X2

X1)

K2

(X2

X1)

48

..

..

..

..

..

..

..

..


49/114

CHAPTER VI

FREQUENCY RESPONSE

6.1 Introduction

We have discussed about the system response to step and ramp input in time domain earlier

in Chapter 3. When input signals are frequency dependent, frequency response assume

greater importance. In such systems the time domain analysis is difficult from the design

point of view.

Frequency response of a control system refers to the steady state response of a system

subject to sinusoidal input of fixed (constant) amplitude but frequency varying over a

specific range, usually from 0 to . For linear systems the frequency of input and output

signal remains the same, while the ratio of magnitude of output signal to the input signal

and phase between two signals may change. Frequency response analysis is a

complimentary method to time domain analysis (step and ramp input analysis). It deals

with only steady state and measurements are taken when transients have disappeared.

Hence frequency response tests are not generally carried out for systems with large time

constants.

The frequency response information can be obtained either by analytical methods or by

experimental methods, if the system exits. The concept and procedure is illustrated in

Figure 6.1 (a) in which a linear system is subjected to a sinusoidal input. I(t) = a Sin t and

the corresponding output is O(t) = b Sin (t + ) as shown in Figure 6.1 (b).

49


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The following quantities are very important in frequency response analysis.

M () = b/a = ratio of amplitudes = Magnitude ratio or Magnification factor or gain.

() = = phase shift or phase angle

These factors when plotted in polar co-ordinates give polar plot, or when plotted in

rectangular co-ordinates give rectangular plot which depict the frequency response

characteristics of a system over entire frequency range in a single plot.

6.2 Frequency Response Data

The following procedure can be adopted in obtaining data analytically for frequency

response analysis.

1. Obtain the transfer function of the system

)()()(

SISOSF = , Where F (S) is transfer function, O(S) and I(S) are the Laplace

transforms of the output and input respectively.

2. Replace S by (j) (As S is a complex number)

)(

)()(

jI

jOjF =

)()()()(

jBAjIjO +== (another complex number)

3. For various values of, ranging from 0 to determine M () and .

jBAjBAjI

jOM +=+== )()(

)(

)()(

50


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22 BAM +=

jBAjI

jO+==

)(

)(

A

B1tan =

4. Plot the results from step 3 in polar co-ordinates or rectangular co-ordinates. These

plots are not only convenient means for presenting frequency response data but are also

serve as a basis for analytical and design methods.

6.3 Comparison between Time Domain and Frequency Domain Analysis

An interesting and revealing comparison of frequency and time domain approaches is based

on the relative stability studies of feedback systems. The Rouths criterion is a time domain

approach which establishes with relative ease the stability of a system, but its adoption to

determine the relative stability is involved and requires repeated application of the criterion.

The Root Locus method is a very powerful time domain approach as it reveals not only

stability but also the actual time response of the system. On the other hand, the Nyquist

criterion (discussed later in this Chapter) is a powerful frequency domain method of

extracting the information regarding stability as well as relative stability of a system

without the need to evaluate roots of the characteristic equation.

6.4 Graphical Methods to Represent Frequency Response Data

Two graphical techniques are used to represent the frequency response data. They are: 1)

Polar plots 2) Rectangular plots.

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6.5Polar Plot

The frequency response data namely magnitude ratio M() and phase angle () when

represented in polar co-ordinates polar plots are obtained. The plot is plotted in complex

plane shown in Figure 6.2. It is also called Nyquist plot. As is varied the magnitude and

phase angle change and if the magnitude ratio M is plotted for varying phase angles, the

locus obtained gives the polar plot. It is easier to construct a polar plot and ready

information of magnitude ratio and phase angle can be obtained.

Figure 6.2: Complex Plane Representation

A typical polar plot is shown in Figure 6.3 in which the magnitude ratio M and phase angle

at a given value of can be readily obtained.

52

M

Real

Img

0, + 360, -360

+90, -270

+180, -180

+270, -90

Positive angles

Negative angles


53/114

Figure 6.3: A Typical Polar Plot

6.6 Rectangular Plot

The frequency response data namely magnitude ratio M() and phase angle () can also

be presented in rectangular co-ordinates and then the plots are referred as Bode plots which

will be discussed in Chapter 7.

6.7 Illustrations on Polar Plots: Following examples illustrate the procedure followed in

obtaining the polar plots.

Illustration 1: A first order mechanical system is subjected to a input x(t). Obtain the polar

plot, if the time constant of the system is 0.1 sec.

53

x (t) (input)

y (t) (output)

C

K


54/114

Transfer function F(S) = 11)( )( += SSX SY

Given = 0.1 sec

SSSX

SY

1.01

1

11.0

1

)(

)(

+=

+=

To obtain the polar plot (i.e., frequency response data) replace S by j.

)1.0(11

11.0

1

)(

)(

jjjX

jY

+=+=

Magnification Factor M =)1.0(1

)0(1

)1.0(1

1

)(

)(

j

j

jjX

jY

++

=+

=

2)1.0(1

1

+=M

Phase angle = )1.01()0(1)()(

)(

++=== jXjjYjXjY

)1.0(tan1

0tan 11

=

)1.0(tan 1 =

54

Governing Differential Equation:

KxKydt

dyC =+. by K

xydtdy

KC =+.

xydt

dy=+. Take Laplace transform

SY(S) + Y(S) = X(S)

(S+1) Y(S) = X(S)


55/114

Now obtain the values of M and for different values of ranging from 0 to as givenin Table 6.1.

Table 6.1 Frequency Response Data

2

)1.0(1

1

+=M )1.0(tan

1 =

0 1.00 0

2 0.98 -11.13

4 0.928 -21.8

5 0.89 -26.6

6 0.86 -30.9

10 0.707 -45

20 0.45 -63.4

40 0.24 -7650 0.196 -78.69

100 0.099 -84.29

0 -90

The data from Table 6.1 when plotted on the complex plane with as a parameter polar

plot is obtained as given below.

55

= 6

= 50

= 20


56/114

Illustration 2: A second order system has a natural frequency of 10 rad/sec and a damping

ratio of 0.5. Sketch the polar plot for the system.

The transfer function of the system is given by

22

2

2)(

)(

nn

n

SSSX

SY

++= Given n = 10 rad/sec and = 0.5

10010

100

)(

)(2 ++= SSSX

SY

Replace S by j

10010

)0(100

10010)(

100

)(

)(22 ++

+=

++==

j

j

jjjX

jYas 1=j

222

2

2)10()100(

100

)10()100(

)0(100

)(

)(

+=

++

==j

j

jX

jYM

Magnification Factor = 222 )10()100(

100

+=M

Phase angle = )10()100()0(100)10()100(

)0(100 22

jjj

j++=

++

=

56

m

x (Input)

y (Response)

C

K


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=

2

11

100

10tan

100

0tan

Now obtain the values of M and for various value of ranging from 0 to as given inthe following Table 6.2.

Table 6.2 Frequency Response Data: Illustration 2

M( )

0 1.00 0.0

2 1.02 -11.8

5 1.11 -33.7

8 1.14 -65.8

10 1.00 -90.0

12 0.78 -110.1

15 0.51 -129.8

20 0.28 -146.3

40 0.06 -165.1

70 0.02 -171.7

0.00 -180.0

The data from Table 6.2 when plotted on the complex plane with as a parameter polar

plot is obtained as given below.

Note: The polar plot intersects the imaginary axis at a frequency equal to the natural

frequency of the system = n = 10 rad/sec.

57


58/114

Illustration 3: Obtain the polar plot for the transfer function

)1(

10)(

+=

S

SF Replace S by j

1

10)(

+=

jjF

1

10)()(

2 +==

jFM

() = F (j) = 10 - (j+1)

1tan 1 =

Table 6.3 Frequency Response Data: Illustration 3

M( )

0 1.00 0

0.2 9.8 -11

0.4 9.3 -21

0.6 8.6 -31

0.8 7.8 -392.0 4.5 -63

3.0 3.2 -72

4.0 2.5 -76

5.0 1.9 -79

10 0.99 -84

0.00 -90

58


59/114

Polar plot for Illustration 3

6.8 Guidelines to Sketch Polar Plots

Polar plots for some typical transfer function can be sketched on the following guidelines.

I )()(

))((SFjBA

jfe

jdcjba=+=

+

++--- (Transfer function)

Magnitude Ratio =22

2222 *

fe

dcba

jfe

jdcjbaM

+

++=

+++

=

Phase angle:

( ) ( )( )

)()()( jfejdcjbajfe

jdcjba++++=

+

++=

e

f

c

d

a

b 111 tantantan +=

59

= M() = 0


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II Values of tan functions

IQ: tan-1 (b/a) Positive:

IIQ: tan-1 (b/-a) Negative: 180 -

IIIQ: tan-1 (-b/-a) Positive: 180 +

IVQ: tan-1 (-b/a) Negative: 360 -

tan-1 (0) = 00 tan-1 (-0) =1800

tan-1 (1) = 450 tan-1 (-1) = 1350

tan-1 () = 900 tan-1 (-) = 2700

III Let K = Constant

= K + j(0)

Therefore KKK =+= 22 0 Applicable for both K>0 and K0

= tan-1 (-0) = 1800, if K


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IV: Sn = (j)n = (0+j)n

= (0+j) (0+j) . . . . . . . . . . . . . n times

a. Magnitude

)0()0()( jjjS nn ++== . . . . . . . . . . . . . n times

22 0()0( ++= . . . . . . . . . . . . . n times

Therefore S

n

=

n

b. Angle

++++== )0()0()( jjjS nn . . . . . . . . . . . . n times

= tan-1 (/0) + tan-1 (/0)+ . . . . . . . . . . . . n times

= 900 + 900 + . . . . . . . . . . . . n times

Sn = n * 900

Illustration 4: Sketch the polar plot for the system represented by the following open loop

transfer function.

)2)(10(

10)()(

++=

SSSSHSG , obtain M and for different values of

i) As 0,0 S jS =

61


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==== 0

5.05.0

2*10*

10)()( 0

SSSHSG S

S= 5.0

0900 =

090=

ii) As S,

010))((

10)()(3===

SSSSSHSG S

310 S=

027090*30 ==

62


63/114

Illustration 5: Sketch the polar plots for the system represented by the following open loop

transfer function.

)5()()(

2 +=

SS

KSHSG

M( ) ( )

0 -900

0 -2700

63

Real

Img

0

-90

-180

-270

= , M = 0

= 0, M =


64/114

i) As 0,0 S jS =

, S is far lesser than 5 and can be neglected

22

5/

5 S

K

S

K==

===n

K

S

KM

2

5/)(

2

5/)(

S

K

= 2)5/( SK = = 0 2*90 = 0 180 = - 1800

ii) As S,

32 )5()()(

S

K

SS

KSHSG S =+

=

0)( 33 === K

S

K

M

3)( SK =

= 0 - 3*90 = - 2700

64

)5()()(

200

+=

SS

KSHSG

S


65/114

M( ) ( )

0 -1800

0 -2700

65

Real

Img

0

-90

-180

-270

= , M = 0 = 0, M =


66/114


transfer function.

)8)(5(10)()( 2 ++= SSSSHSG

i) As 0,0 S jS =

2200

4/1

8.5.

10)()(

SSSHSG S ==

====0

4/14/14/1)(

22

SM

24/1)( S=

= 0 - 2*90

= - 2*90

0

= 180

0

ii) As S,

42

10

..

10)()(

SSSSSHSG S ==

44

1010

)( == SM

as , M () = 0

410)( S=

66


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= 0 4*90 = - 3600


transfer function.

)5)(4(

)2(10)()(

3 +++

=SSS

SSHSG

i) As 0,0 S jS =

)5)(4(

)2(10)()(

300 ++

+==

SSS

SSHSG S

33

1

5.4.

2.10

SS==

M( ) ( )

0 -1800

0 -3600

67

Real

Img

0, -360

-90

-180

-270

= , M = 0 = 0, M =


68/114

===33

11)(

SM

31)( S= = 0 3*90 = - 2700

ii) As S,

43

10

)..

.10)()(

SSSS

SSHSG S ===

01010

)(44===

SM

410)( S= = 0 4*90 = -3600

68

Img-270


69/114

M( ) ( )

0 -2700

0 -3600

69

Real

0, -360

-90

-180

= , M = 0

= 0, M =


70/114


transfer function.

)4)(2(

10)()(

+=

SSSSHSG

i) As 0,0 S jS =

SSSHSGS

8/10

4).2(

10

)()(00

===

=

=

=

8/108/10

)(S

M

S= 8/10)( = 1800 - 900 = 900

ii) As S,

310)()( SSHSG S ==

01010

)(33===

SM

310)( S= = 0 3*900 = - 2700

70

-270

+90


71/114

Illustration 9: Sketch the polar plot for the system represented by the following transfer

function.

M( ) ( )

0 900

0 -2700

71

Real

Img

0, -360

+270-90

+180-180

= , M = 0

= 0, M =


72/114

10010

100

)(

)(2 +

=SSSX

SY

i) As 0,0 S jS =

11)( ==M , for both K positive and negative

01801)( ==

ii) As S,

2

100

S= 0

100)(

2== S

SM ,

2

100)(

S

= = 0 2* = - 900*2 = -1800

6.9 Experimental determination of Frequency Response

Many a times the transfer function of a physical system may not be available in such

circumstances it is necessary to obtain frequency response information experimentally.Such data may then be used to establish the transfer function. This method requires the

actual system.

6.10 System Analysis using Polar Plots: Nyquist Criterion: Continued in Session 21 on

17.10.2006

CHAPTER VI

FREQUENCY RESPONSE

(Continued from Session 20)

M( ) ( )

0 1 1800

0 -1800

72

Real

Img

0, -360

+270-90

+180

-180

-270

+90

= 0, M = 1

= , M = 0


73/114

SYSTEM ANALYSIS USING POLAR PLOTS: NYQUIST

CRITERION

6.10 System Analysis using Polar Plots: Nyquist Criterion

Polar plots can be used to predict feed back control system stability by the application of

Nyquist Criterion, and therefore are also referred as Nyquist Plots. It is a labor saving

technique in the analysis of dynamic behaviour of control systems in which the need for

finding roots of characteristic equation of the system is eliminated.

Consider a typical closed loop control system which may be represented by the simplified

block diagram as shown in Figure 6.4

Figure 6.4 Simplified System Block Diagram

The closed-loop transfer function or the relationship between the output and input of the

system is given by

)()(1

)(

)(

)(

SHSG

SG

SR

SC

+=

The open-loop transfer function is G(S) H(S) (the transfer function with the feedback loop

broken at the summing point).

73

H(S)

G(S)C(S)

R(S) +

-


74/114

1+ G(S) H(S) is called Characteristic Function which when equated to zero gives the

Characteristic Equation of the system.

1 + G(S) H(S) = 0 Characteristic Equation

The characteristic function F(S) = 1 + G(S) H(S) can be expressed as the ratio of two

factored polynomials.

Let).().........)()((

)....().........)(()()(1)(

321

21

n

k

n

ZSPSPSPSS

ZSZSZSKSHSGSF

+++++++

=+=

The Characteristic equation in general can be represented as

F(S) = K (S+Z1) (S+Z2) (S+Z3) . (S+Zn) = 0

Then:

Z1, -Z2, -Z3 . Zn are the roots of the characteristic equation

at S= -Z1, S= -Z2, S= -Z3, 1+ G(S) H(S) becomes zero.

These values of S are termed as Zeros of F(S)

Similarly:

at S= -P1, S= -P2, S= -P3 . Etc. 1+ G (S) H (S) becomes infinity.

These values are called Poles of F (S).

6.10.1 Condition for Stability

For stable operation of control system all the roots of characteristic equation must be

negative real numbers or complex numbers with negative real parts. Therefore, for a

system to be stable all the Zeros of characteristic equation (function) should be either

negative real numbers or complex numbers with negative real parts. These roots can be

plotted on a complex-plane or S-plane in which the imaginary axis divides the complex

74


75/114

plane in to two parts: right half plane and left half plane. Negative real numbers or complex

numbers with negative real parts lie on the left of S-plane as shown Figure 6.5.

Figure 6.5 Two halves of Complex Plane

Therefore the roots which are positive real numbers or complex numbers with positive real

parts lie on the right-half of S-plane.

In view of this, the condition for stability can be stated as For a system to be stable all

the zeros of characteristic equation should lie on the left half of S-plane.

Therefore, the procedure for investigating system stability is to search for Zeros on the

right half of S-plane, which would lead the system to instability, if present. However, it is

impracticable to investigate every point on S-plane as to which half of S-plane it belongs to

and so it is necessary to have a short-cut method. Such a procedure for searching the right

half of S-plane for the presence of Zeros and interpretation of this procedure on the Polar

plot is given by the Nyquist Criterion.

6.10.2 Nyquist Criterion: Cauchys Principle of Argument:

75

2+j1

2-j1

-3-j2

-3+j2

Real

Img Right half of

S Plane

Left half of S

Plane


76/114

In order to investigate stability on the Polar plot, it is first necessary to correlate the region

of instability on the S-plane with identification of instability on the polar plot, or 1+GH

plane. The 1+GH plane is frequently the name given to the plane where 1+G(S) H(S) is

plotted in complex coordinates with S replaced by j. Likewise, the plot of G(S) H(S) with

S replaced by j is often termed as GH plane. This terminology is adopted in the remainder

of this discussion.

The Nyquist Criterion is based on the Cauchys principle of argument of complex variable

theory. Consider [F(S) = 1+G(S) H(S)] be a single valued rational function which is

analytic everywhere in a specified region except at a finite number of points in S-plane. (A

function F(S) is said to be analytic if the function and all its derivatives exist). The points

where the function and its derivatives does not exist are called singular points. The poles of

a point are singular points.

Let CS be a closed path chosen in S-plane as shown Figure 6.6 (a) such that the function

F(S) is analytic at all points on it. For each point on CS represented on S-plane there is a

corresponding mapping point in F(S) plane. Thus when mapping is made on F(S) plane, the

curve CG mapped by the function F(S) plane is also a closed path as shown in Figure 6.6

(b). The direction of traverse of CG in F(S) plane may be clockwise or counter clockwise,

depending upon the particular function F(S).

Then the Cauchy principle of argument states that: The mapping made on F(S) plane will

encircle its origin as many number of times as the difference between the number of

Zeros and Poles of F(S) enclosed by the S-plane locus CS in the S-plane.

76


77/114


Figure 6.6 Mapping on S-plane and F(S) plane

Thus N = Z P

N0+j0 = Z P

77

S-plane+j

-j

S5

S1

S2

S3S4

CS-

+j

-j

-

GS1

GS2

GS3

GS4

GS5

(0+j )

CG

F(S) = 1+G(S) H(S) plane


78/114

Where N0+j0: Number of encirclements made by F(S) plane plot (CG) about its

origin. Z and P: Number of Zeros and Poles of F(S) respectively enclosed by the locus

CS in the S-plane.

Illustration: Consider a function F(S)

)25)(25)(5)(3(

)22)(22)(1()(

jSjSSSS

jSjSSKSF

+++++++++

=

Zeros: -1, (-2-j2), (-2+j2) indicated by O (dots) in the S-plane

Poles: 0, -3, -5, (-5 j2), (-5 +j2) indicated by X (Cross) in S-plane: As shown in Figure 6.6

(c)

78


79/114

Figure 6.6 (c) Figure 6.6 (d)

Now consider path CS1 (CCW) on S-plane for which:

Z: 2, P =1

79

S-plane+j

-j

CS2

-

CS1

O: ZEROS

X: POLES

+j

-j

CG2

-

CG1

CG2

CG2

(0+j0)

F(S) = 1+G(S) H(S) plane

CS1

CS2


80/114

Consider another path CS2 (CCW) in the same S plane for which:

Z = 1, P = 4

CG1 and CG2 are the corresponding paths on F(S) plane [Figure 6.6 (d)].

Considering CG1 [plot corresponding to CS1 on F (S) plane]

N0+j0 = Z P = 2-1 = +1

CG1 will encircle the origin once in the same direction of CS1 (CCW)

Similarly for the path CG2

N0+j0 = Z P = 1 4 = - 3

CG2 will encircle the origin 3 times in the opposite direction of CS2 (CCW)

Note: The mapping on F(S) plane will encircle its origin as many number of times as the

difference between the number of Zeros and Poles of F(S) enclosed by the S-plane locus.

From the above it can be observed that

In the expression

N= Z - P,

N can be positive when: Z>P

N = 0 when: Z = P

N can be negative when: Z


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N is negative the map CG encircles the origin N times in the opposite direction as

that of CS

6.10.3 Nyquist Path and Nyquist Plot

The above Cauchys principle of argument can be used to investigate the stability of

control systems. We have seen that if the Zeros of characteristic function lie on the right

half of S-plane it will lead to system instability. Now, to encircle the entire right half of S-

plane, select a closed path as shown in Figure 6.6 (e) such that all the Zeros lying on the

right-half of S-plane will lie inside this path. This path in S-plane is known as Nyquist

path. Nyquist path is generally taken in CCW direction. This path consists of the

imaginary axis of the S-plane (S = 0+j, - < < ) and a closing semicircle of infinite

radius. If the system being tested has poles of F(S) on the imaginary axis, it is customary to

modify the contour as shown Figure 6.6 (f) excluding these poles from the path.

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Figure 6.6 (e): Nyquist Path Figure 6.6 (f)

82

S-plane

+j

-j

-

-j

+j

0+j00-j0

r =

+j

-j

-

S=+j0

S= -j0

r =

S= -j

r 0


83/114

Corresponding to the Nyquist path a plot can be mapped on F(S) = 1+G(S) H(S) plane as

shown in Figure 6.6 (g) and the number of encirclements made by this F(S) plot about its

origin can be counted.

Figure 6.6 (g)

Now from the principle of argument

N0+j0 = Z-P

N0+j0 = number of encirclements made by F(S) plane plot

Z, P: Zeros and Poles lying on right half of S-plane

For the system to be stable: Z = 0

N0+j0 = - P Condition for Stability

Apart from this, the Nyquist path can also be mapped on G(S) H(S) plane (Open-loop

transfer function plane) as shown in Figure 6.6 (h).

Now consider

F(S) = 1+ G(S) H(S) for which the origin is (0+j0) as shown in Figure 6.6 (g).

Therefore G(S) H(S) = F(S) 1

= (0+j0) 1 = (-1+j0) Coordinates for origin on G(S) H(S) plane as shown

in Figure 6.6(h)

83

ImgG(S) H(S) plane

Real

Img

Real

Img

0+j0

1+ G(S) H(S) plane


84/114

Figure 6.6 (h)

Thus a path on 1+ G(S) H(S) plane can be easily converted to a path on G(S) H(S) plane or

open loop transfer function plane. This path will be identical to that of 1+G(S) H(S) path

except that the origin is now shifted to the left by one as shown in Figure 6.6 (h).

This concept can be made use of by making the plot in G(S) H(S) plane instead of 1+ G(S)

H(S) plane. The plot made on G(S) H(S) plane is termed as the Niquist Plot and its net

encirclements about (-1+j0) (known as critical point) will be the same as the number of net

encirclements made by F(S) plot in the F(S) = 1+G(S) H(S) plane about the origin.

Now, the principle of argument now can be re-written as

N-1+j0 = Z-P

Where N-1+j0 = Number ofnet encirclements made by the G(S) H(S) plot (Nyquist Plot) in

the G(S) H(S) plane about -1+j0

For a system to be stable Z = 0

N-1+j0 = -P

Thus the Nyquist Criterion for a stable system can be stated as The number of net

encirclementsmade by the Nyquist plot in the G(S) H(S) plane about the critical point

(-1+j0) is equal to the number of poles of F(S) lying in right half of S-plane.

84

0+j0

(-1+j0)

Origin of the plot forG(S) H(S)


85/114

[Encirclem

Controlengg Compiled Pillai(Session 1 7)

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