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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
Figure 7 Signal flow graph of example 2
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
Figure 10 Signal flow graph of example 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.
Figure 11 Signal flow graph of example 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.
The transfer function T =( )
( ) 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.
Figure 12 Signal flow graph of example 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
The transfer function T =( )
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.
For a type 0 system,
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==
20
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For a type 2 or higher system
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 =
For a type 0 system,
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
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K
1spT1s
2T1s
1Ts
1sm
T1sbT1s
aTKs
limK2
2
0sa=
+
+
+
+
+
+
=
andK
1
K
1e
a
ss==
For a type 3 or higher system
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
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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
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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)
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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
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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+
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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 +
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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 ++
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Figure 27 Response of PID controller to unit actuating error signal
t0
u(t)
proportional only
PD control actionPID control action
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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
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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
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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
..
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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
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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
.
..
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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
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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
..
..
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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)
..
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= 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.. .
.. ..
..
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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
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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
.. .
.. .
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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)
From NSL for rotating system
= 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
.. .
.. . .
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Figure 2.14 (a)
Figure 2.14 (b)
From NSL for rotating system
= 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
.. .
.. .
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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)
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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
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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
..
..
..
..
..
..
..
..
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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).
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Figure 6.1 (a) Figure 6.1 (b)
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 +=+== )()(
)(
)()(
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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
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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
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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)
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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.
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= 6
= 50
= 20
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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.
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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
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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 =
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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 ==
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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 =
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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
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M( ) ( )
0 -1800
0 -2700
65
Real
Img
0
-90
-180
-270
= , M = 0 = 0, M =
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Illustration 6: Sketch the polar plots for the system represented by the following open loop
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=
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= 0 4*90 = - 3600
Illustration 7: Sketch the polar plots for the system represented by the following open loop
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 =
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===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
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M( ) ( )
0 -2700
0 -3600
69
Real
0, -360
-90
-180
= , M = 0
= 0, M =
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Illustration 8: Sketch the polar plots for the system represented by the following open loop
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
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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 =
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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
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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).
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H(S)
G(S)C(S)
R(S) +
-
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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
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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
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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.
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Figure 6.6 (a) Figure 6.6 (b)
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
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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)
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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
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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
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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)
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ImgG(S) H(S) plane
Real
Img
Real
Img
0+j0
1+ G(S) H(S) plane
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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.
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0+j0
(-1+j0)
Origin of the plot forG(S) H(S)
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[Encirclem