lecture1423904331 (1).pdf

CONTROL SYSTEM ENGINEERING-I
Department of Electrical Engineering
VEER SURENDRA SAI UNIVERSITY OF TECHNOLOGY,
ODISHA, BURLA

1
Disclaimer
This document does not claim any originality and cannot be used as a substitute for
prescribed textbooks. The information presented here is merely a collection by the committee
members for their respective teaching assignments. Various sources as mentioned at the end
of the document as well as freely available material from internet were consulted for
preparing this document. The ownership of the information lies with the respective authors or
institutions. Further, this document is not intended to be used for commercial purpose and the
committee members are not accountable for any issues, legal or otherwise, arising out of use
of this document. The committee members make no representations or warranties with
respect to the accuracy or completeness of the contents of this document and specifically
disclaim any implied warranties of merchantability or fitness for a particular purpose. The
committee members shall be liable for any loss of profit or any other commercial damages,
including but not limited to special, incidental, consequential, or other damages.

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Syllabus
1.0 Introduction to Control system
1.1 Scope of Control System Engineer
1.2 Classification of Control System
1.3 Historical development of Control system
1.4 Analogues systems
1.5 Transfer function of Systems
1.6 Block diagram representation
1.7 Signal Flow Graph(SFG)
2.0 Feedback Characteristics of Control systems and sensitivity measures
2.1 The Concept of Feedback and Closed loop control
2.2 Merits of using Feedback control system
2.3 Regenerative Feedback
3.0Control System Components
3.1 Potentiometers
3.2 DC and AC Servomotors
3.3 Tachometers
3.4 Amplidyne
3.5 Hydralulic systems
3.6 Pneumatic systems
3.7 Stepper Motors
4.0 Time Domain Performance Analysis of Linear Control Systems
4.1 Standard Test Signals
4.2 Time response of 1st
order Systems
4.3 Unit step response of a prototype 2nd
order system
4.4 Unit Ramp response of a second order system
4.4 Performance Specification of Linear System in Time domain
4.5 The Steady State Errors and Error Constants
4.6 Effect of P, PI, PD and PID Controller
4.7 Effect of Adding a zero to a system
4.8 Performance Indices(ISE,ITSE,IAE, ITAE)
4.9 Approximations of Higher order Systems by Lower order Problems
5.0 The Stability of Linear Control Systems
5.1 The Concept of Stability
5.2 The Routh Hurwitz Stability Criterion
5.3 Relative stability analysis

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6.0 Root Locus Technique
6.1 Angle and Magnitude Criterion
6.2 Properties of Root Loci
6.3 Step by Step Procedure to Draw Root Locus Diagram
6.4 Closed Loop Transfer Function and Time Domain response
6.5 Determination of Damping ratio, Gain Margin and Phase Margin from Root Locus
6.6 Root Locus for System with transportation Lag.
6.7 Sensitivity of Roots of the Characteristic Equation.
7.0 Frequency Domain Analysis.
7.1 Correlation between Time and frequency response
7.2 Frequency Domain Specifications
7.3 Polar Plots and inverse Polar plots
7.4 Bode Diagrams
7.4.1 Principal factors of Transfer function
7.4.2 Procedure for manual plotting of Bode Diagram
7.4.3 Relative stability Analysis
7.4.4 Minimum Phase, Non-minimum phase and All pass systems
7.5 Log Magnitude vs Phase plots.
7.6 Nyquist Criterion
7.6.1 Mapping Contour and Principle of Argument
7.6.2 Nyquist path and Nyquist Plot
7.6.3 Nyquist stability criterion
7.6.4 Relative Stability: Gain Margin, and Phase Margin
7.7 Closed Loop Frequency Response
7.7.1 Gain Phase Plot
7.7.1.1 Constant Gain(M)-circles
7.7.1.2 Constant Phase (N) Circles
7.7.1.3 Nichols Chart
7.8 Sensitivity Analysis in Frequency Domain

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CHAPTER#1
1. Basic Concept of Control System
Control Engineering is concerned with techniques that are used to solve the following six
problems in the most efficient manner possible.
(a)The identification problem :to measure the variables and convert data for analysis.
(b)The representation problem:to describe a system by an analytical form or mathematical model
(c)The solution problem:to determine the above system model response.
(d)The stability problem:general qualitative analysis of the system
(e)The design problem: modification of an existing system or develop a new one
(f)The optimization problem: from a variety of design to choose the best.
The two basic approaches to solve these six problems are conventional and modern approach. The
electrical oriented conventional approach is based on complex function theory. The modern
approach has mechanical orientation and based on the state variable theory.
Therefore, control engineering is not limited to any engineering discipline but is equally
applicable to aeronautical, chemical, mechanical, environmental, civil and electrical engineering.
For example, a control system often includes electrical, mechanical and chemical components.
Furthermore, as the understanding of the dynamics of business, social and political systems
increases; the ability to control these systems will also increase.
1.1.Basic terminologies in control system
System: A combination or arrangement of a number of different physical components to form
a whole unit such that that combining unit performs to achieve a certain goal.
Control: The action to command, direct or regulate a system.
Plant or process: The part or component of a system that is required to be controlled.
Input: It is the signal or excitation supplied to a control system.
Output: It is the actual response obtained from the control system.
Controller: The part or component of a system that controls the plant.
Disturbances: The signal that has adverse effect on the performance of a control system.
Control system: A system that can command, direct or regulate itself or another system to
achieve a certain goal.
Automation: The control of a process by automatic means
Control System: An interconnection of components forming a system configuration that will
provide a desired response.
Actuator: It is the device that causes the process to provide the output. It is the device that
provides the motive power to the process.

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Design: The process of conceiving or inventing the forms, parts, and details of system to
achieve a specified purpose.
Simulation: A model of a system that is used to investigate the behavior of a system by
utilizing actual input signals.
Optimization: The adjustment of the parameters to achieve the most favorable or
advantageous design.
Feedback Signal: A measure of the output of the system used for feedback to control the
system.
Negative feedback: The output signal is feedback so that it subtracts from the input signal.
Block diagrams: Unidirectional, operational blocks that represent the transfer functions of
the elements of the system.
Signal Flow Graph (SFG): A diagram that consists of nodes connected by several directed
branches and that is a graphical representation of a set of linear relations.
Specifications: Statements that explicitly state what the device or product is to be and to do.
It is also defined as a set of prescribed performance criteria.
Open-loop control system: A system that utilizes a device to control the process without
using feedback. Thus the output has no effect upon the signal to the process.
Closed-loop feedback control system: A system that uses a measurement of the output and
compares it with the desired output.
Regulator: The control system where the desired values of the controlled outputs are more or
less fixed and the main problem is to reject disturbance effects.
Servo system: The control system where the outputs are mechanical quantities like
acceleration, velocity or position.
Stability: It is a notion that describes whether the system will be able to follow the input
command. In a non-rigorous sense, a system is said to be unstable if its output is out of
control or increases without bound.
Multivariable Control System: A system with more than one input variable or more than
one output variable.
Trade-off: The result of making a judgment about how much compromise must be made
between conflicting criteria.
1.2.Classification
1.2.1. Natural control system and Man-made control system:
Natural control system: It is a control system that is created by nature, i.e. solar
system, digestive system of any animal, etc.
Man-made control system: It is a control system that is created by humans, i.e.
automobile, power plants etc.
1.2.2. Automatic control system and Combinational control system:

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Automatic control system: It is a control system that is made by using basic theories
from mathematics and engineering. This system mainly has sensors, actuators and
responders.
Combinational control system: It is a control system that is a combination of natural
and man-made control systems, i.e. driving a car etc.
1.2.3. Time-variant control system and Time-invariant control system:
Time-variant control system: It is a control system where any one or more
parameters of the control system vary with time i.e. driving a vehicle.
Time-invariant control system: It is a control system where none of its parameters
vary with time i.e. control system made up of inductors, capacitors and resistors only.
1.2.4. Linear control system and Non-linear control system:
Linear control system: It is a control system that satisfies properties of homogeneity
and additive.
 Homogeneous property:      
f x y f x f y
  
 Additive property:    
f x f x
 

Non-linear control system: It is a control system that does not satisfy properties of
homogeneity and additive, i.e.   3
f x x

1.2.5. Continuous-Time control system and Discrete-Time control system:
Continuous-Time control system: It is a control system where performances of all
of its parameters are function of time, i.e. armature type speed control of motor.
Discrete -Time control system: It is a control system where performances of all of
its parameters are function of discrete time i.e. microprocessor type speed control of
motor.
1.2.6. Deterministic control system and Stochastic control system:
Deterministic control system: It is a control system where its output is predictable
or repetitive for certain input signal or disturbance signal.
Stochastic control system:It is a control system where its output is unpredictable or
non-repetitive for certain input signal or disturbance signal.
1.2.7. Lumped-parameter control system and Distributed-parameter control system:
Lumped-parameter control system: It is a control system where its mathematical
model is represented by ordinary differential equations.
Distributed-parameter control system:It is a control system where its mathematical
model is represented by an electrical network that is a combination of resistors,
inductors and capacitors.
1.2.8. Single-input-single-output (SISO) control system and Multi-input-multi-output
(MIMO) control system:
SISO control system: It is a control system that has only one input and one output.
MIMO control system:It is a control system that has only more than one input and
more than one output.
1.2.9. Open-loop control system and Closed-loop control system:
Open-loop control system: It is a control system where its control action only
depends on input signal and does not depend on its output response.

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Closed-loop control system:It is a control system where its control action depends
on both of its input signal and output response.
1.3.Open-loop control system and Closed-loop control system
1.3.1. Open-loop control system:
It is a control system where its control action only depends on input signal and does
not depend on its output response as shown in Fig.1.1.
Fig.1.1. An open-loop system
Examples: traffic signal, washing machine, bread toaster, etc.
Advantages:
 Simple design and easy to construct
 Economical
 Easy for maintenance
 Highly stable operation
Dis-advantages:
 Not accurate and reliable when input or system parameters are variable in
nature
 Recalibration of the parameters are required time to time
1.3.2. Closed-loop control system:
It is a control system where its control action depends on both of its input signal and
output response as shown in Fig.1.2.
Fig.1.2. A closed-loop system
Examples: automatic electric iron, missile launcher, speed control of DC motor, etc.
Advantages:
 More accurate operation than that of open-loop control system
 Can operate efficiently when input or system parameters are variable in
nature
 Less nonlinearity effect of these systems on output response
 High bandwidth of operation
 There is facility of automation
 Time to time recalibration of the parameters are not required
Dis-advantages:
 Complex design and difficult to construct

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 Expensive than that of open-loop control system
 Complicate for maintenance
 Less stable operation than that of open-loop control system
1.3.3. Comparison between Open-loop and Closed-loop control systems:
It is a control system where its control action depends on both of its input signal and
output response.
Sl.
No. Open-loop control systems Closed-loop control systems
1 No feedback is given to the control system A feedback is given to the control system
2 Cannot be intelligent Intelligent controlling action
3
There is no possibility of undesirable
system oscillation(hunting)
Closed loop control introduces the
possibility of undesirable system
oscillation(hunting)
4
The output will not very for a constant
input, provided the system parameters
remain unaltered
In the system the output may vary for a
constant input, depending upon the
feedback
5
System output variation due to variation in
parameters of the system is greater and the
output very in an uncontrolled way
System output variation due to variation in
parameters of the system is less.
6 Error detection is not present Error detection is present
7 Small bandwidth Large bandwidth
8 More stable Less stable or prone to instability
9 Affected by non-linearities Not affected by non-linearities
10 Very sensitive in nature Less sensitive to disturbances
11 Simple design Complex design
12 Cheap Costly

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1.4.Servomechanism
It is the feedback unit used in a control system. In this system, the control variable is
a mechanical signal such as position, velocity or acceleration. Here, the output signal
is directly fed to the comparator as the feedback signal, b(t) of the closed-loop control
system. This type of system is used where both the command and output signals are
mechanical in nature. A position control system as shown in Fig.1.3 is a simple
example of this type mechanism. The block diagram of the servomechanism of an
automatic steering system is shown in Fig.1.4.
Fig.1.3. Schematic diagram of a servomechanism
Fig.1.4. Block diagram of a servomechanism
Examples:
 Missile launcher
 Machine tool position control
 Power steering for an automobile
 Roll stabilization in ships, etc.
1.5.Regulators
It is also a feedback unit used in a control system like servomechanism. But, the
output is kept constant at its desired value. The schematic diagram of a regulating

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system is shown in Fig.1.5. Its corresponding simplified block diagram model is
shown in Fig.1.6.
Fig.1.5. Schematic diagram of a regulating system
Fig.1.6. Block diagram of a regulating system
Examples:
 Temperature regulator
 Speed governor
 Frequency regulators, etc.

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CHAPTER#2
2. Control System Dynamics
2.1.Definition: It is the study of characteristics behaviour of dynamic system, i.e.
(a) Differential equation
i. First-order systems
ii. Second-order systems
(b)System transfer function: Laplace transform
2.2.Laplace Transform: Laplace transforms convert differential equations into algebraic
equations. They are related to frequency response.
 
   
0
( ) st
x t e
x t X dt
s


  
L (2.1)
 
   
0
( ) st
x t e
x t X dt
s


  
L (2.2)
No. Function
Time-domain
x(t)=
ℒ-1
{X(s)}
Laplace domain
X(s)= ℒ{x(t)}
1 Delay δ(t-τ) e-τs
2 Unit impulse δ(t) 1
3 Unit step u(t)
s
1
4 Ramp t 2
1
s
5
Exponential
decay
e-αt


s
1
6
Exponential
approach  
t
e 


1 )
( 


s
s
7 Sine sin ωt
2
2



s
8 Cosine cos ωt
2
2


s
s
9 Hyperbolic
sine
sinh αt
2
2



s
10 Hyperbolic
cosine
cosh αt
2
2


s
s
11 Exponentiall
y decaying
sine wave
t
e t


sin

2
2
)
( 




s
12 Exponentiall
y decaying
cosine wave
t
e t


cos

2 2
( )
s
s

 

 
2.3.Solution of system dynamics in Laplace form: Laplace transforms can be solved using
partial fraction method.
A system is usually represented by following dynamic equation.
 
 
 
A s
N s
B s
 (2.3)
The factor of denominator, B(s) is represented by following forms,
i. Unrepeated factors

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ii. Repeated factors
iii. Unrepeated complex factors
(i) Unrepeated factors
( )
( )( )
( ) ( )
( )( )
N s A B
s a s b s a s b
A s b B s a
s a s b
 
   
  

 
(2.4)
By equating both sides, determine A and B.
Example 2.1:
Expand the following equation of Laplacetransform in terms of its partial fractionsand obtain
its time-domain response.
2
( )
( 1)( 2)
s
Y s
s s

 
Solution:
The following equation in Laplacetransform is expandedwith its partial fractions as follows.
2
( 1)( 2) ( 1) ( 2)
2 ( 2) ( 1)
( 1)( 2) ( 1)( 2)
s A B
s s s s
s A s B s
s s s s
 
   
  
 
   
By equating both sides, A and B are determined as 2, 4
A B
   . Therefore,
2 4
( )
( 1) ( 2)
Y s
s s
  
 
Taking Laplace inverse of above equation,
2
( ) 2 4
t t
y t e e
 
  
(ii) Unrepeated factors
2 2 2
( ) ( )
( )
( ) ( ) ( )
N s A B A B s a
s a
s a s a s a
 
  

  
(2.5)
Example 2.2:
2
2
( )
( 1) ( 2)
s
Y s
s s

 
Solution:
2 2
2
( 1) ( 2) ( 1) ( 1) ( 2)
s A B C
s s s s s
  
    
By equating both sides, A and B are determined as 2, 4
A B
   . Therefore,
2
2 4 4
( )
( 1) ( 1) ( 2)
Y s
s s s
   
  
2
( ) 2 4 4
t t t
y t te e e
  
   

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(iii)Complex factors: They contain conjugate pairs in the denominator.
2 2
( )
( )( ) ( )
N s As B
s a s a s  


   
(2.6)
Example 2.3:
2 1
( )
( 1 )( 1 )
s
Y s
s j s j


   
Solution:
2 2
2 1
( )
( 1) 1 ( 1) 1
s
Y s
s s
 
   
( ) 2 cos sin
t t
y t e t e t
 
 
2.4.Initial value theorem:
   
0
( ) lim ( )
lim
t
s
y t sY s


 (2.7)
Example 2.4:
Determine the initial value of the time-domain response of the following equation using the
initial-value theorem.
2 1
( )
( 1 )( 1 )
s
Y s
s j s j


   
Solution:
Solution of above equation,
( ) 2 cos sin
t t
y t e t e t
 
 
Applying initial value theorem,
(2 1)
2
( 1 )( 1 )
lim
s
s s
s j s j



   
2.5.Final value theorem:
   
0
( ) lim ( )
lim
t
s
y t sY s


 (2.8)
Example 2.5:
Determine the initial value of the time-domain response of the following equation using the
initial-value theorem.
2
2
( )
( 1) ( 2)
s
Y s
s s

 
Solution:
Solution of above equation,

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2
( ) 2 4 4
t t t
y t te e e
  
   
Applying final value theorem,
(2 1)
2
( 1 )( 1 )
lim
s
s s
s j s j



   

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CHAPTER#3
3. Transfer Function
3.1.Definition: It is the ratio of Laplace transform of output signal to Laplace transform of input
signal assuming all the initial conditions to be zero, i.e.
Let, there is a given system with input r(t) and output c(t) as shown in Fig.3.1 (a), then its
Laplace domain is shown in Fig.3.1 (b). Here, input and output are R(s) and C(s) respectively.
(a) (b)
(c)
Fig.3.1. (a) A system in time domain, (b) a system in frequency domainand (c) transfer function with differential
operator
G(s) is the transfer function of the system. It can be mathematically represented as follows.
 
 
  zero initial condition
C s
G s
R s
 Equation Section (Next)(3.1)
Example 3.1: Determine the transfer function of the system shown inFig.3.2.
Fig.3.2. a system in time domain
Solution:
Fig.3.1 is redrawn in frequency domain as shown in Fig.3.2.
Fig.3.2. a system in frequency domain

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Applying KVL to loop-1 of the Fig.3.2
   
1
i
V s R Ls I s
Cs
 
  
 
 
(3.2)
Applying KVL to loop-2 of the Fig.3.2
   
1
o
V s I s
Cs
 
  
 
(3.3)
From eq (2.12),
     
1
/
o o
I s V s CsV s
Cs
 
 
 
 
(3.4)
Now, using eq (2.13) in eq (2.10),
   
 
  2
1
1 1
1 1
i o
o
i
V s R Ls CsV s
Cs
V s
V s LCs RCs
R Ls Cs
Cs
 
  
 
 
  
   
 
 
 
(3.5)
Then transfer function of the given system is
  2
1
1
G s
LCs RCs

 
(3.6)
3.2.General Form of Transfer Function
 
    
    
 
 
1 2 1
1 2
1
...
...
m
i
m i
n
n
j
i
s z
K s z s z s z
G s K
s p s p s p
s z



  
 
  



(3.7)
Where, 1 2
, ... m
z z z are called zeros and 1 2
, ... n
p p p are called poles.
Number of poles n will always be greater than the number of zeros m
Example 3.2:
Obtain the pole-zero map of the following transfer function.
( 2)( 2 4)( 2 4)
( )
( 3)( 4)( 5)( 1 5)( 1 5)
s s j s j
G s
s s s s j s j
    

      
Solution:
Zeros Poles
s=2 s=3
s=-2-j4 s=4
s=-2+j4 s=5

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s=-1-j5
s=-1+j5
Fig.3.3. pole-zero map
3.3.Properties of Transfer function:
 Zero initial condition
 It is same as Laplace transform of its impulse response
 Replacing ‘s’ by
d
dt
in the transfer function, the differential equation can be obtained
 Poles and zeros can be obtained from the transfer function
 Stability can be known
 Can be applicable to linear system only
3.4.Advantages of Transfer function:
 It is a mathematical model and gain of the system
 Replacing ‘s’ by
d
dt
in the transfer function, the differential equation can be obtained
 Poles and zeros can be obtained from the transfer function
 Stability can be known
 Impulse response can be found
3.5.Disadvantages of Transfer function:
 Applicable only to linear system
 Not applicable if initial condition cannot be neglected
 It gives no information about the actual structure of a physical system

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CHAPTER#4
4. Description of physical system
4.1.Components of a mechanical system: Mechanical systems are of two types, i.e. (i)
translational mechanical system and (ii) rotational mechanical system.
4.1.1. Translational mechanical system
There are three basic elements in a translational mechanical system, i.e. (a) mass, (b)
spring and (c) damper.
(a) Mass: A mass is denoted by M. If a force f is applied on it and it displays
distance x, then
2
2
d x
f M
dt
 as shown in Fig.4.1.
Fig.4.1. Force applied on a mass with displacement in one direction
If a force f is applied on a massM and it displays distance x1in the direction of f and
distance x2 in the opposite direction, then
2 2
1 2
2 2
d x d x
f M
dt dt
 
 
 
 
as shown in Fig.4.2.
M
X1
f
X2
Fig.4.2. Force applied on a mass with displacement two directions
(b) Spring: A spring is denoted by K. If a force f is applied on it and it displays
distance x, then f Kx
Fig.4.3. Force applied on a spring with displacement in one direction
If a force f is applied on a springK and it displays distance x1in the direction of f and
distance x2 in the opposite direction, then  
1 2
f K x x
  as shown in Fig.4.4.

20
Fig.4.4. Force applied on a spring with displacement in two directions
(c) Damper: A damper is denoted by D. If a force f is applied on it and it displays
distance x, then
dx
f D
dt
Fig.4.5. Force applied on a damper with displacement in one direction
If a force f is applied on a damperD and it displays distance x1in the direction of f and
distance x2 in the opposite direction, then 1 2
dx dx
f D
dt dt
 
 
 
 
as shown in Fig.4.6.
Fig.4.6. Force applied on a damper with displacement in two directions
4.1.2. Rotational mechanical system
There are three basic elements in a Rotational mechanical system, i.e. (a) inertia, (b)
spring and (c) damper.
(a) Inertia: A body with aninertia is denoted by J. If a torqueT is applied on it and it
displays distanceӨ, then
2
2
d
T J
dt

 . If a torqueT is applied on a body with inertia
J and it displays distance Ө1 in the direction of T and distance Ө2 in the opposite
direction, then
2 2
1 2
2 2
d d
T J
dt dt
 
 
 
 
 
.
(b) Spring: A spring is denoted by K. If a torqueT is applied on it and it displays
distanceӨ, then T K
 . If a torqueT is applied on a body with inertia J and it
displays distance Ө1 in the direction of T and distance Ө2 in the opposite
direction, then  
1 2
T K  
  .
(c) Damper: A damper is denoted by D. If a torqueT is applied on it and it displays
distanceӨ, then
d
T D
dt

 . If a torqueT is applied on a body with inertia J and it

21
displays distance Ө1 in the direction of T and distance Ө2 in the opposite
direction, then 1 2
d d
T D
dt dt
 
 
 
 
 
.
4.2.Components of an electrical system: There are three basic elements in an electrical system,
i.e. (a) resistor (R), (b) inductor(L) and (c) capacitor (C). Electrical systems are of two types,
i.e. (i) voltage source electrical system and (ii) current source electrical system.
4.2.1. Voltage source electrical system: If i is the current through a resistor(Fig.4.7) and v
is the voltage drop in it, then v Ri
 .
If i is the current through an inductor (Fig.4.7) and v is the voltage developed in it,
then
di
v L
dt
 .
If i is the current through a capacitor(Fig.4.7) and v is the voltage developed in it,
then
1
v idt
C
  .
Fig.4.7. Current and voltage shown in resistor, inductor and capacitor
4.2.2. Current source electrical system:
If i is the current through a resistor and v is the voltage drop in it, then
v
i
R
 .
If i is the current through an inductor and v is the voltage developed in it, then
1
i vdt
L
  .
If i is the current through a capacitor and v is the voltage developed in it, then
dv
i C
dt
 .
4.2.3. Work out problems:
Q.4.1. Find system transfer function betweenvoltage drop across the capacitanceand
input voltage in the followingRC circuit as shown in Fig.4.8.
Fig.4.8.

22
Solution
Voltage across resistance, ( ) ( )
R
e t i t R

Voltage across capacitance,
1
( ) ( )
C
e t i t dt
C
 
Total voltage drop,
1
( ) ( )
i R C
e e e i t R i t dt
C
    
Laplace transform of above equation,
1
( ) ( )
i
E s I s R
Cs
 
 
 
 
System transfer function betweenvoltage drop across the capacitanceand input
voltage,
( ) 1 1
( ) 1 1
C
i
E s
E s RCs s

 
 
where, RC 
 is the time-constant
Q.4.2. Find system transfer function betweenfunction between the inductance
currentto the source currentin the followingRL circuit as shown in Fig.4.9.
Fig.4.9.
Voltage across the Resistance,
( )
( ) R R
e t
e t i R i
R
  
Voltage across the Inductance,
1
( ) ( )
L
L
di
e t L i e t dt
dt L
   
Total current,
( ) 1
( )
a R L
e t
i i i e t dt
R L
    
Laplace transform of the current source,
1 1
( ) ( )
a
I s E s
R Ls
 
 
 
 
and ( )
L
E
I s
Ls

Transfer function between the inductance current to the source current,
( ) 1 1
( ) 1
1
L
a
I s
L
I s s
s
R

 



23
where
L
R
  is the time-constant
Q.4.3. Find system transfer function betweenfunction between the capacitance
voltageto the source voltage in the followingRLC circuit as shown in Fig.4.10.
Fig.4.10.
Voltage across the Resistance, ( )
R
e t iR

Voltage across the Inductance, ( )
L
di
e t L
dt

Voltage across thecapacitance,
1
( )
C
e t idt
C
 
Total voltage,  
1
di
e t iR L idt
dt C
   
Laplace transform of the voltage source,
1
( ) ( )
E s I s R Ls
Cs
 
  
 
 
Transfer function between capacitance voltage and source voltage
 
2
2 2
( ) 1
1
( ) 2
C n
n n
E s
E s s s
Cs R Ls
Cs

 
 
   
 
 
 
where
1
n
LC
  and
2
R
L
C
 
Q.4.4.Find the transfer function of the following Spring-mass-damperas shown
in Fig.4.11.
Fig.4.11.

24
Solution
 
2 2 2
( ) 1 1
( ) 2 n n
X s
F s ms cs k m s s
 
 
   
4.3.Analogous system: Fig.4.12 shows a translational mechanical system, a rotational control
system and a voltage-source electrical system.
(a)
(b)
(c)
Fig.4.12. (a) a voltage-source electrical system,(b) a translational mechanical system and (c) a rotational control
system
From Fig4.12 (a), (b) and (c), we have
 
2
2
2
2
2
2
1
d q dq
L R q v t
dt C
dt
d d
J D K T
dt
dt
d x dx
M D Kx f
dt
dt
 

  
  
  
Equation Chapter 8 Section 0(4.1)
Where,

25
q idt
  (4.2)
The solutions for all the above three equations given by eq (4.2) are same. Therefore, the
above shown three figures are analogous to each other. There are two important types of
analogous systems, i.e. force-voltage (f-v) analogy and force-current analogy. From eq (4.2),
f-v analogy can be drawn as follows.
Translational Rotational Electrical
Force (f) Torque (T) Voltage (v)
Mass (M) Inertia (J) Inductance (L)
Damper (D) Damper (D) Resistance (R)
Spring (K) Spring (K) Elastance (1/C)
Displacement (x) Displacement (Ө) Charge (q)
Velocity (u) = x
 Velocity (u) = 
 Current (i) = q

Similarly, f-i analogy that can be obtainedfrom eq (4.1), can be drawn as follows.
Translational Rotational Electrical
Force (f) Torque (T) Current (i)
Mass (M) Inertia (J) Capacitance (C)
Damper (D) Damper (D) Conductance (1/R)
Spring (K) Spring (K) Reciprocal of Inductance (1/L)
Displacement (x) Displacement (Ө) Flux linkage (ψ)
Velocity (u) = x
 Velocity (u) = 
 Voltage (v) = 

4.4.Mathematical model of armature controlled DC motor: The armature control type speed
control system of a DC motor is shown in Fig.4.6. The following components are used in this
system.
Ra=resistance of armature
La=inductance of armature winding
ia=armature current
If=field current
Ea=applied armature voltage
Eb=back emf
Tm=torque developed by motor
Ө=angular displacement of motor shaft
J=equivalent moment of inertia and load referred to motor shaft
f=equivalent viscous friction coefficient of motor and load referred to motor shaft

26
J
,
f
Fig.4.6. Schematic diagram of armature control type speed control system of a DC motor
The air-gap flux  is proportional of the field current i.e.
f f
K I
  (4.3)
The torque Tm developed by the motor is proportional to the product of armature current and
air gap flux i.e.
1
=
m f f a
T k K I i (4.4)
In armature-controlled D.C. motor,the field current is kept constant,so that eq(4.4) can be
written as follows.
=
m t a
T K i (4.5)
The motor back emf being proportional to speed is given as follows.
=
b b
d
E K
dt

 
 
 
(4.6)
The differential equation of the armature circuit is
a
a a a b a
di
L R i E E
dt
 
  
 
 
(4.7)
The torque equation is
2
2 m t a
d d
J f T K I
dt
dt
 
   
  
   
   
 
(4.8)
Taking the Laplace transforms of equations (4.6), (4.7) and (4.8), assuming zero initial
conditions, we get
   
=
b b
E s sK s
 (4.9)
       
+
a a a a b
sL R I s E s E s
  (4.10)
2
( ) ( ) ( )
m t a
s J sf s T s K I

   (4.11)

27
From eq(4.9) to (4.11) the transfer function of the system is obtained as,
 
  
( )
( )
t
a a a t b
K
s
G s
E s s R sL sJ f K K


 


 
 
(4.12)
Eq(4.12) can be rewritten as
 
  
  
( ) 1
( )
1
t
a a
t b
a
a a
K
R sL sJ f
s
G s
K K
E s s
R sL sJ f

 
 
 
 

 

 
 



(4.13)
The block diagram that is constructed from eq (4.13) is shown in Fig.4.7.
1
sJ f

1
s
1
a a
sL R

 
s
s

Fig.4.7. Block diagram of armature control type speed control system of a DC motor
The armature circuit inductance La is usually negligible. Therefore, eq(4.13) can be
simplifiedas follows.
2
( )
( )
t t b
a a a
K K K
s
s J s f
E s R R
    
  
   
   
(4.14)
The term t b
a
K K
f
R
 

 
 
indicates that the back emf of the motor effectively increases the
viscous friction of the system. Let,
t b
a
K
f
K
f
R

  (4.15)
Where f  be the effective viscous friction coefficient. The transfer function given by eq(4.15)
may be written in the following form.
 
   
1
m
a
s K
E s s s




(4.16)
Here = t
m
a
K
K
R f
= motor gain constant, and
J
f
 

= motor time constant.Therefore, the motor
torque and back emf constant Kt, Kb are interrelated.
4.5.Mathematical model of field controlled DC motor: The field control type speed control
system of a DC motor is shown in Fig.4.8. The following components are used in this system.
Rf=Field winding resistance

28
Lf=inductance of field winding
If=field current
ef=field control voltage
Tm=torque developed by motor
Ө=angular displacement of motor shaft
J=equivalent moment of inertia and load referred to motor shaft
f=equivalent viscous friction coefficient of motor and load referred to motor shaft
Rf
If
Ia
(constant)
Lf
ef
M
Ө
J, f
Tm
Fig.4.8. Block diagram of field control type speed control system of a DC motor
In field control motor the armature current is fed from a constant current source.The air-gap
flux Φ is proportional of the field current i.e.
f f
K I
  (4.17)
The torque Tm developed by the motor is proportional to the product of armature current and
air gap flux i.e.
1
=
m f f a t f
T k K I I K I
 (4.18)
The equation for the field circuit is
f
f f f f
dI
L R I E
dt
  (4.19)
The torque equation is
2
2 m t f
d d
J f T K I
dt
dt
 
   (4.20)
Taking the Laplace transforms of equations (4.19) and (4.20) assuming zero initial conditions,
we get the following equations

29
     
f f f f
L s R I s E s
  (4.21)
and
       
2
m t f
Js fs s T s K I s

   (4.22)
From eq(4.21) and (4.22) the transfer function of the system is obtained as
 
 
    
t
f f f
s K
G s
E s s R sL Js f

 
 
(4.23)
The transfer function given by eq(4.23) may be written in the following form.
 
       
1 1
t m
a f f
s K K
E s s s s
s L s R Js f

 
 
 
 

(4.24)
Here t
m
f
K
K
R f
 = motor gain constant, and
f
f
L
R
  = time constant of field circuit and
J
f
  
= mechanical time constant.For small size motors field control is advantageous.The block
diagram that is constructed from eq (4.24) is shown in Fig.4.9.
Ef(s) Ө(s)
 
t
K
s sJ f

1
f f
sL R

Fig.4.9. Block diagram of field control type speed control system of a DC motor

30
CHAPTER#5
5. Block Diagram Algebra
5.1.Basic Definition in Block Diagram model:
Block diagram: It is the pictorial representation of the cause-and-response relationship
between input and output of a physical system.
(a) (b)
Fig.5.1. (a) A block diagram representation of a system and (b) A block diagram representation with gain of a
system
Output: The value of input multiplied by the gain of the system.
     
C s G s R s
 (5.1)
Summing point: It is the component of a block diagram model at which two or more signals
can be added or subtracted. In Fig.15, inputs R(s) and B(s) have been given to a summing
point and its output signal is E(s). Here,
     
E s R s B s
  (5.2)
Fig.5.2. A block diagram representation of a systemshowing its different components
Take-off point: It is the component of a block diagram model at which a signal can be taken
directly and supplied to one or more points as shown in Fig.5.2.
Forward path: It is the direction of signal flow from input towards output.
Feedback path: It is the direction of signal flow from output towards input.
5.2.Developing Block Diagram model from mathematical model:
Let’s discuss this concept with the following example.
Example: A system is described by following mathematical equations. Find its corresponding
block diagram model.
1 1 2 3
3 2 5
x x x x
  
 (5.3)
2 1 2 3
4 3
x x x x
  
 (5.4)

31
3 1 2 3
2
x x x x
  
 (5.5)
Example: Eq (5.3), (5.4) and (5.5) are combiningly results in the following block diagram
model.
++
++
++
5
3
2
1/s
1/s
1/s
4
2
x1(s)
x2(s)
x3(s)
+
+
+  
1
x s

 
2
x s

 
3
x s

x3(s)
x1(s)
x2(s)
3
x2(s)
x3(s)
x3(s)
x2(s)
x1(s)
Fig.5.3. A block diagram representation of the above example

32
5.3. Rules for reduction of Block Diagram model:
Sl.
No.
Rule
No.
Configuration Equivalent Name
1 Rule 1 Cascade
2 Rule 2 Parallel
3 Rule 3
   
( )
1
G s
G s H s

Loop
4 Rule 4
Associative
Law
5 Rule 5
Move take-
off point
after a block
6 Rule 6
Move take-
off point
before a
block
7 Rule 7
Move
summing-
point point
after a block
8 Rule 8
Move
summing-
point point
before a
block

33
9 Rule 9
Move take-
off point
after a
summing-
point
10 Rule 10
Move take-
off point
before a
summing-
point
Fig.5.4. Rules for reduction of Block Diagram model
5.4.Procedure for reduction of Block Diagram model:
Step 1: Reduce the cascade blocks.
Step 2: Reduce the parallel blocks.
Step 3: Reduce the internal feedback loops.
Step 4: Shift take-off points towards right and summing points towards left.
Step 5: Repeat step 1 to step 4 until the simple form is obtained.
Step 6: Find transfer function of whole system as
 
 
C s
R s
.
5.5.Procedure for finding output of Block Diagram model with multiple inputs:
Step 1: Consider one input taking rest of the inputs zero, find output using the procedure
described in section 4.3.
Step 2: Follow step 1 for each inputs of the given Block Diagram model and find their
corresponding outputs.
Step 3: Find the resultant output by adding all individual outputs.

34
CHAPTER#6
6. Signal Flow Graphs (SFGs)
It is a pictorial representation of a system that graphically displays the signal transmission in it.
6.1.Basic Definitions in SFGs:
Input or source node: It is a node that has only outgoing branches i.e. node ‘r’ in Fig.6.1.
Output or sink node: It is a node that has only incoming branches i.e. node ‘c’ in Fig.6.1.
Chain node: It is a node that has both incoming and outgoing branches i.e. nodes ‘x1’,
‘x2’,‘x3’,‘x4’,‘x5’and ‘x6’ in Fig.6.1.
Gain or transmittance: It is the relationship between variables denoted by two nodes or
value of a branch. In Fig.6.1, transmittances are ‘t1’, ‘t2’,‘t3’,‘t4’,‘t5’and ‘t6’.
Forward path: It is a path from input node to output node without repeating any of the nodes
in between them. In Fig.6.1, there are two forward paths, i.e. path-1:‘r-x1-x2-x3-x4-x5-x6-c’ and
path-2:‘r-x1-x3-x4-x5-x6-c’.
Feedback path: It is a path from output node or a node near output node to a node near input
node without repeating any of the nodes in between them (Fig.6.1).
Loop: It is a closed path that starts from one node and reaches the same node after trading
through other nodes. In Fig.6.1, there are four loops, i.e. loop-1:‘x2-x3-x4-x1’, loop-2:‘x5-x6-
x5’, loop-3:‘x1-x2-x3-x4-x5-x6-x1’ and loop-4:‘x1-x3-x4-x5-x6-x1’.
Self Loop: It is a loop that starts from one node and reaches the same node without trading
through other nodes i.e. loop in node ‘x4’ with transmittance ‘t55’ in Fig.6.1.
Path gain: It is the product of gains or transmittances of all branches of a forward path. In
Fig.6.1, the path gains are P1 = t1t2t3t4t5 (for path-1) and P2 = t9t3t4t5 (for path-2).
Loop gain: It is the product of gains or transmittances of all branches of a loop In Fig.6.1,
there are four loops, i.e. L1 = -t2t3t6, L2 = -t5t7, L3 = -t1t2t3t4t5t8, and L4 = -t9t3t4t5t8.
Dummy node: If the first node is not an input node and/or the last node is not an output node
than a node is connected before the existing first node and a node is connected after the
existing last node with unity transmittances. These nodes are called dummy nodes. In Fig.6.1,
‘r’ and ‘c’ are the dummy nodes.
Non-touching Loops: Two or more loops are non-touching loops if they don’t have any
common nodes between them. In Fig.6.1, L1 and L2 are non-touching loops
Example:
Fig.6.1. Example of a SFG model

35
6.2.Properties SFGs:
 Applied to linear system
 Arrow indicates signal flow
 Nodes represent variables, summing points and take-off points
 Algebraic sum of all incoming signals and outgoing nodes is zero
 SFG of a system is not unique
 Overall gain of an SFG can be determined by using Mason’s gain formula
6.3.SFG from block diagram model:
Let’s find the SFG of following block diagram model shown in Fig.6.2.
Ea(s)
Ө(s)
+
-
KT
Eb(s)
Kb
1
sJ f

1
s
1
a a
sL R

 
s
s

Fig.6.2. Armature type speed control of a DC motor
Step-1: All variables and signals are replaced by nodes.
Step-2: Connect all nodes according to their signal flow.
Step-3: Each ofgains is replaced by transmittances of the branches connected between two nodes
of the forward paths.
Step-4: Each ofgains is replaced by transmittances multiplied with (-1) of the branches connected
between two nodes of the forward paths.
1
sJ f

1
s
1
a a
sL R

 
s
s

(a)

36
1
a a
sL R

1
sJ f

1
s
(b)
Fig.6.3. Armature type speed control of a DC motor
6.4.Mason’s gain formula:
Transfer function of a system=
 
 
 
1
N
k k
k
P
C s
G s
R s


 


(6.1)
Where,
N= total number of forward paths
Pk= path gain of kth
forward path
∆= 1 - (∑loop gains of all individual loops) + (∑gain product of loop gains of all possible
two non-touching loops) - (∑gain product of loop gains of all possible three non-touching
loops) + …
∆k= value of ∆ after eliminating all loops that touches kth
forward path
Example:
Find the overall transfer function of the system given in Fig.6.1 using Mason’s gain
formula.
Solution:
In Fig.6.1,
No. of forward paths: 2
N 
Path gain of forward paths: 1 1 2 3 4 5
P t t t t t
 and 2 6 3 4 5
P t t t t

Loop gain of individual loops: 1 2 3 6
L t t t
  , 2 5 7
L t t
  , 3 1 2 3 4 5 8
L t t t t t t
  and 4 9 3 4 5 8
L t t t t t
 
No. of two non-touching loops = 2 i.e. L1 and L2
No. of more than two non-touching loops = 0

37
   
1 2 3 4 1 2 1 2 3 4 1 2
1 0 1
L L L L L L L L L L L L
             
1 1 0 1
    and 2 1 0 1
   
  1 1 2 2
P P
G s
  


 
     
1 2 3 4 5 6 3 4 5
2 3 6 5 7 1 2 3 4 5 8 9 3 4 5 8 2 3 5 6 7
1 1
1
t t t t t t t t t
G s
t t t t t t t t t t t t t t t t t t t t t

 
    
  1 2 3 4 5 6 3 4 5
2 3 6 5 7 1 2 3 4 5 8 9 3 4 5 8 2 3 5 6 7
1
t t t t t t t t t
G s
t t t t t t t t t t t t t t t t t t t t t

 
    

38
CHAPTER#7
7. Feedback Characteristics of Control System
7.1.Feedback and Non-feedback Control systems
Non-feedback control system: It is a control system that does not have any feedback paths.
It is also known as open-loop control system. It is shown in Fig.7.1 (a) and (b).
Feedback control system: It is a control system that has at least one feedback path. It is also
known as closed-loop control system. It is shown in Fig.7.2 (a) and (b).
(a)
(b)
Fig.7.1. (a) Block diagram of a non-feedback control system and (b) SFG of a non-feedback control system
(a)
(b)
Fig.7.2. (a) Block diagram of a feedback control system and (b) SFG of a feedback control system
7.2.Types of Feedback in a Control system
7.2.1. Degenerative feedback control system: It is a control system where the feedback
signal opposes the input signal. Here,
Error or actuating signal = (Input signal) – (Feedback signal).
Referring Fig.7.3,
     
E s R s B s
  (7.1)
and
 
 
   
1
1
G s
T s
G s H s


(7.2)
Fig.7.3. (a) Block diagram of a degenerative feedback control system

39
7.2.2. Regenerative feedback control system: It is a control system where the feedback
signal supports or adds the input signal. Here,
Error or actuating signal = (Input signal) + (Feedback signal).
Referring Fig.7.4,
     
E s R s B s
  (7.3)
and
 
 
   
2
1
G s
T s
G s H s


(7.4)
Fig.7.4. Block diagram of a regenerative feedback control system
7.3.Effect of parameter variation on overall gain of a degenerative Feedback Control system
The overall gain or transfer function of a degenerative feedback control system depends upon
these parameters i.e. (i) variation in parameters of plant, and (ii) variation in parameter of
feedback system and (ii) disturbance signals.
The term sensitivity is a measure of the effectiveness of feedback on reducing the influence of
any of the above described parameters. For an example, it is used to describe the relative
variations in the overall Transfer function of a system T(s) due to variation in G(s).
=
ℎ ( )
ℎ ( )
7.3.1. Effect of variation in G(s) on T(s) of a degenerative Feedback Control system
In an open-loop system,
     
C s G s R s

Let, due to parameter variation in plant G(s) changes to [G(s) + ∆G(s)] such that
|G(s)| >> |∆G(s)|. The output of the open-loop system then changes to
         
           
C s C s G s G s R s
C s C s G s R s G s R s
      
 
     
     
C s G s R s
    (7.5)
In an closed-loop system,

40
 
 
   
 
1
G s
C s R s
G s H s


Let, due to parameter variation in plant G(s) changes to [G(s) + ∆G(s)] such that
|G(s)| >> |∆G(s)|. The output of the open-loop system then changes to
   
   
     
 
   
   
       
 
1
1
G s G s
C s C s R s
G s G s H s
G s G s
C s C s R s
G s H s G s H s
   
 
  
    
 
 
   
  
Since, |G(s)| >> |∆G(s)|, then        
G s H s G s H s
 . Therefore,    
G s H s
 is
neglected. Now,
   
   
   
 
   
 
   
 
 
   
 
1
1 1
G s G s
C s C s R s
G s H s
G s G s
C s C s R s R s
G s H s G s H s
 
  


    
 
Or
 
 
   
 
1
G s
C s R s
G s H s

 

(7.6)
Comparing eq (42 and (43), it is clear that ∆ ( ) = (1 + ) ∆ ( )
This concept can be reproved using sensitivity. Sensitivity on T(s) due to variation in G(s) is
given by
T
G
T T T G
S
G G G T
 
  
 
(7.7)
For open-loop system,
1
T
G
T T G G
S
G G G G
 
   
 
(7.8)
For closed-loop system,
 
     
2
1 1
1 1
1
T
G
GH GH
T T G
S
G G G GH GH
GH
 

   
  

(7.9)
Therefore, it is proved that ( ) = (1 + ) ( ). Hence, the effect of
parameter variation in case of closed loop system is reduced by a factor of
( )
.

41
7.3.2. Effect of variation in H(s) on T(s) of a degenerative Feedback Control system
This concept can be reproved using sensitivity. Sensitivity on T(s) due to variation in H(s) is
given by
T
H
T T T H
S
H H H T
 
  
 
(7.10)
For closed-loop system,
     
2
1 1
1
T
H
T H G H GH
S G
H T G GH GH
GH
 
  
     
  
 

 
(7.11)
For higher value of GH, sensitivity approaches unity. Therefore, change in H affects
directly the system output.
Equation Chapter (Next) Section 1

43
CHAPTER#8
8. Time Domain Analysis of Control Systems
8.1.Time response
Time response c(t)is the variation of output with respect to time. The part of time response
that goes to zero after large interval of time is called transient response ctr(t). The part of time
response that remains after transient response is called steady-state response css(t).
Fig.7.1. Time response of a system
8.2.System dynamics
System dynamics is the study of characteristic and behaviour of dynamic systems
i.e.
i. Differential equations: First-order systems and Second-order systems,
ii. Laplace transforms,
iii. System transfer function,
iv. Transient response: Unit impulse, Step and Ramp
Laplace transforms convert differential equations into algebraic equations. They are related to
frequency response
 
   
0
( ) st
x t e dt


  
x t X s
L (8.1)

44
No. Function Time-domain
x(t)= ℒ-1
{X(s)}
Laplace domain
X(s)= ℒ{x(t)}
1 Delay δ(t-τ) e-τs
2 Unit impulse δ(t) 1
3 Unit step u(t)
s
1
4 Ramp t
2
1
s
5 Exponential decay e-αt


s
1
6 Exponential approach  
t
e 


1
)
( 


s
s
7 Sine sin ωt
2
2



s
8 Cosine cos ωt
2
2


s
s
9 Hyperbolic sine sinh αt
2
2



s
10 Hyperbolic cosine cosh αt
2
2


s
s
11 Exponentially decaying sine
wave
t
e t


sin

2
2
)
( 




s
12 Exponentially decaying cosine
wave
t
e t


cos

2 2
( )
s
s

 

 
8.3.Forced response
1 2
1 2
( )( ) ( )
( ) ( ) ( ) ( )
( )( ) ( )
m
n
K s z s z s z
C s G s R s R s
s p s p s p
  
 
  


(8.2)
R(s) input excitation
8.4.Standard test signals
8.4.1. Impulse Signal: An impulse signal δ(t) is mathematically defined as follows.
 
; 0
0 ;t 0
undefined t
t

 
 
 
(8.3)
Laplace transform of impulse signal is

45
  1
s
  (8.4)
Fig.7.2. Impulse signal
Dirac delta function
( ) ( )
i
x t x t a

  (8.5)
Integral property of Dirac delta function
( ) ( ) ( )
o o
t t t dt t
  


 
 (8.6)
Laplace transform of an impulse input
0
( ) ( )
st sa
i i
X s e x t a dt x e


 
  
 (8.7)
8.4.2. Step Signal: A step signal u(t) is mathematically defined as follows.
 
0 ; 0
;t 0
t
u t
K
 
 
 
(8.8)
Laplace transform of step signal is
 
K
U s
s
 (8.9)

46
Fig.7.2. Step signal
8.4.3. Ramp Signal: A step signal r(t) is mathematically defined as follows.
 
0 ; 0
;t 0
t
r t
Kt
 
 
 
(8.10)
Laplace transform of ramp signal is
  2
K
R s
s
 (8.11)
Fig.7.3. Ramp signal
8.4.4. Parabolic Signal A step signal a(t) is mathematically defined as follows.
  2
0
; 0
;t 0
2
t
a t Kt

 
 
 

(8.12)
Laplace transform of parabolic signal is
  3
K
A s
s
 (8.13)
Fig.7.4. Parabolic signal
8.4.5. Sinusoidal Signal A sinusoidal x(t) is mathematically defined as follows.

47
  sin
x t t

 (8.14)
Laplace transform of sinusoidal signal is
  2 2
0
sin
st
X s e t dt
s





 

 (8.15)
Fig.7.4. Sinusoidal signal
8.5.Steady-state error:
A simple closed-loop control system with negative feedback is shown as follows.
Fig.7.5. A simple closed-loop control system with negative feedback
Here,
     
E s R s B s
  (8.16)
     
B s C s H s
 (8.17)
     
C s E s G s
 (8.18)
Applying (1) in (9),
       
E s R s C s H s
  (8.19)
Using (11) in (12),
         
E s R s E s G s H s
  (8.20)
       
1 G s H s E s R s
    
  (8.21)

48
 
 
   
1
R s
E s
G s H s
 

(8.22)
Steady-state error,
   
0
lim lim
ss
t s
e e t sE s
 
  (8.23)
Using (15) in (16),
 
 
   
0 0
lim lim
1
ss
s s
sR s
e sE s
G s H s
 
 

(8.24)
Therefore, steady-state error depends on two factors, i.e.
(a) type and magnitude of R(s)
(b) open-loop transfer function G(s)H(s)
8.6.Types of input and Steady-state error:
8.6.1. Step Input
 
A
R s
s
 (8.25)
Using (18) in (17),
       
0 0
lim lim
1 1
ss
s s
A
s
A
s
e
G s H s G s H s
 
 
 
 
 
 
(8.26)
   
0
1 lim 1
ss
P
s
A A
e
G s H s K

  
 
(8.27)
Where,
   
0
lim
P
s
K G s H s

 (8.28)
8.6.2. Ramp Input
  2
A
R s
s
 (8.29)
Using (18) in (17),
       
   
   
2
0 0
0
0
lim lim
1 1
lim
lim
ss
s s
ss
s
ss
V
s
A
s
A
s
e
G s H s s G s H s
A
e
s sG s H s
A A
e
sG s H s K
 


 
 
 
 
   
 
 

  
(8.30)
Where,

49
   
0
lim
V
s
K sG s H s

 (8.31)
8.6.3. Parabolic Input
  3
A
R s
s
 (8.32)
Using (18) in (17),
       
   
   
3
2
0 0
2 2
0
2
0
lim lim
1 1
lim
lim
ss
s s
ss
s
ss
A
s
A
s
A
s
e
G s H s s G s H s
A
e
s s G s H s
A A
e
K
s G s H s
 


 
 
 
 
   
 
 

  
(8.33)
Where,
   
2
0
lim
A
s
K s G s H s

 (8.34)
Types of input and steady-state error are summarized as follows.
Error Constant Equation Steady-state error (ess)
Position Error Constant (KP)    
0
lim
P
s
K G s H s


1
ss
P
A
e
K


Velocity Error Constant (KV)    
0
lim
V
s
K sG s H s


ss
V
A
e
K

Acceleration Error Constant (KA)
   
2
0
lim
A
s
K s G s H s


ss
A
A
e
K

8.7.Types of open-loop transfer function G(s)H(s)and Steady-state error:
8.7.1. Static Error coefficient Method
The general form of G(s)H(s) is
   
    
    
1 2
1 1 ... 1
1 1 ... 1
n
j
a b m
K T s T s T s
s T s
G
T s s
s
T
s H
  

  
(8.35)
Here, j = no. of poles at origin (s = 0)
or, type of the system given by eq (28) is j.
8.7.1.1. Type 0
   
    
    
1 2
1 1 ... 1
1 1 ... 1
n
a b m
K T s T s T s
T s T s
G s
s
s
T
H
  

  
(8.36)
Here,

50
   
0
lim
P
s
K G s H s K

  (8.37)
Therefore,
1
ss
A
e
K


(8.38)
8.7.1.2. Type 1
   
    
    
1 2
1 1 ... 1
1 1 ... 1
n
a b m
K T s T s T
G s
s
s T s T s T s
H s
  

  
(8.39)
Here,
   
0
lim
V
s
K sG s H s K

  (8.40)
Therefore,
ss
A
e
K
 (8.41)
8.7.1.3. Type 2
   
    
    
1 2
2
1 1 ... 1
1 1 ... 1
n
a b m
K T s T s T s
s T s T s T s
G s H s
  

  
(8.42)
Here,
   
2
0
lim
A
s
K s G s H s K

  (8.43)
Therefore,
ss
A
e
K
 (8.44)
Steady-state error and error constant for different types of input are summarized as follows.
Type
Step input Ramp input Parabolic input
KP ess KV ess KA ess
Type 0 K
1
A
K

0  0 
Type 1  0 K
A
K
0 
Type 2  0  0 K
A
K
The static error coefficient method has following advantages:
 Can provide time variation of error
 Simple calculation

51
But, the static error coefficient method has following demerits:
 Applicable only to stable system
 Applicable only to three standard input signals
 Cannot give exact value of error. It gives only mathematical value i.e. 0 or ∞

52
8.7.2. Generalized Error coefficient Method
From eq (15),
 
   
 
1
1
E s R s
G s H s
 
  

 
 
So,
     
1 2
E s F s F s
 (8.45)
Where,
   
1
1
1
F
G s H s


and    
2
F s R s

Using convolution integral to eq (38)
         
1 2 1
0 0
t t
e t f f t d f r t d
     
   
  (8.46)
Using Taylor’s series of expansion to  
r t 
 ,
         
2 3
...
2! 3!
r t r t r t r t r t
 
    
      (8.47)
Now, applying eq (40) in eq (39),
                 
2 3
1 1 1 1
0 0 0 0
...
2! 3!
t t t t
e t f r t d r t f d r t f d r t f d
 
        
  
    
    (8.48)
Now, steady-state error, ess is
 
lim
ss
t
e e t

 (8.49)
Therefore,
                 
               
2 3
1 1 1 1
0 0 0 0
2 3
1 1 1 1
0 0 0 0
lim lim ...
2! 3!
...
2! 3!
t t t t
ss
t t
ss
e e t f r t d r t f d r t f d r t f d
e f r t d r t f d r t f d r t f d
 
        
 
        
 
   
 
  
     
 
 
 
  
     
   
   
(8.50)
Eq (44) can be rewritten as
       
3
2
0 1 ...
2! 3!
ss
C
C
e C r t C r t r t r t
  
     (8.51)
Where, C0, C1, C2, C3, etc. are dynamic error coefficients. These are given as

53
   
 
 
 
 
 
 
0 1 1
0
0
1
1 1
0
0
2
2
1
2 1 2
0
0
3
3
1
3 1 3
0
0
lim
lim
lim
2!
lim
3!
s
s
s
s
C f d F s
dF s
C f d
ds
d F s
C f d
ds
d F s
C f d
ds
 
  

 

 








 
  
 
   




, and so on… (8.52)
8.8.First-order system:
A Governing differential equation is given by
( )
y y Kx t

 
 (8.53)
Where, Time constant, sec =  ,
Static sensitivity (units depend on the input and output variables) = K ,
y(t) is response of the system and
x(t) is input excitation
The System transfer function is
( )
( )
( ) (1 )
Y s K
G s
X s s

 

(8.54)
Pole-zero map of a first-order system
Normalized response
In this type of response

54
 Static components are taken out leaving only the dynamic component
 The dynamic components converge to the same value for different physical systems of
the same type or order
 Helps in recognizing typical factors of a system
8.8.1. Impulse input to a first-order system
Governing differential equation
( )
i
y y Kx t
 
 
 (8.55)
Laplacian of the response
1
( )
1
(1 )
i i
Kx Kx
Y s
s s
 

 
 
   
  

 
(8.56)
Time-domain response
( )
t
i
Kx
y t e 


 (8.57)
Impulse response function of a first-order system
( )
t
K
h t e 


 (8.58)
By putting x
i
=1 in the response
Response of a first-order system to any force excitation
0
( ) ( )
t t
K
y t e F t d

 


 
 (8.59)
The above equation is called Duhamel’s integral. Normalized response of a first-order system to
impulse input is shown below.
8.8.2. Step input to a first-order system
( )
i
y y Kx u t

 
 (8.60)
( )
i
y t
Kx

/
t 

55
( )
1
(1 )
i i i
Kx Kx Kx
Y s
s s s s


  
 
(8.61)
( ) 1
t
i
y t Kx e 

 
 
 
 
 
(8.62)
Normalized response of a first-order system to impulse input is shown below.
8.8.3. Ramp input to a first-order system
y y Kt

 
 (8.63)
2 2
1
( )
1
(1 )
K
Y s
s s s s s
 


   
 
(8.64)
( )
t
y t
t e
K

 

   (8.65)
( )
i
y t
Kx

/
t 

56
8.8.4. Sinusoidal input to a first-order system
sin
y y KA t
 
 
 (8.66)
 
2
2 2 2 2 2 2
1
( )
(1 ) 1/
1
K A s
Y s
s s s s s
   
    

   
   
 
 
    
   

(8.67)
 
/
2
( ) 1
cos sin
1
t
y t
e t t
KA


   



 
  
 
 

(8.68)
8.9.Second-order system
A Governing differential equation is given by
( )
i
y t
Kx

/
t 

57
( )
my cy ky Kx t
  
  (8.69)
Where,  = Time constant, sec,
K = Static sensitivity (units depend on the input and output variables),
m = Mass (kg),
c = Damping coefficient (N-s/m),
k = Stiffness (N/m),
y(t) is response of the system and
x(t) is input excitation
The System transfer function is
 
2 2
( )
( ) 2 n n
Y s K
X s m s s
 

 
(8.70)
Pole-zero map
(a) ζ>1 over damped
Poles are:
 
2
1,2 1
n
s   
    (8.71)
Graphically, the poles of an over damped system is shown as follows.
(b) ζ =1 critically damped
Poles are:
1,2 n
s 
  (8.72)
Graphically, the poles of an critically damped system is shown as follows.

58
(c) ζ<1 under damped
Poles are:
 
2
1,2
1,2
j 1
n
n d
s
s j
  
 
   
   
(8.73)
Where, d
  Damped natural frequency
2
1
d n
  
  (8.74)
Graphically, the poles of an critically damped system is shown as follows.
Here,
2
tan
1





(d) ζ = 0 un-damped
Poles are:
1,2 j n
s 
   (8.75)

59
Solved problems:
1. A single degree of freedom spring-mass-damper system has the following data: spring stiffness 20
kN/m; mass 0.05 kg; damping coefficient 20 N-s/m. Determine
(a) undamped natural frequency in rad/s and Hz
(b) damping factor
(c) damped natural frequency n rad/s and Hz.
If the above system is given an initial displacement of 0.1 m, trace the phasor of the system for three
cycles of free vibration.
Solution:
3
20 10
632.46 rad/s
0.05
n
k
m


  
632.46
100.66Hz
2 2
n
n
f

 
  
3
20
0.32
2 2 20 10 0.05
c
km
   
 
2 2
1 632.46 1 0.32 600rad/s
d n
  
    
600
95.37 Hz
2 2
d
d
f

 
  
0.32 632.46
( ) 0.1
nt t
y t Ae e

  
 
2. A second-order system has a damping factor of 0.3 (underdamped system) and an un-damped
natural frequency of 10 rad/s. Keeping the damping factor the same, if the un-damped natural
frequency is changed to 20 rad/s, locate the new poles of the system? What can you say about the
response of the new system?
Solution:
Given, 1 10 rad/s
n
  and 2 20 rad/s
n
 
1 1
2 2
1 10 1 0.3 9.54rad/s
d n
  
    
2 2
2 2
1 20 1 0.3 19.08rad/s
d n
  
    
1 1
1,2 3 9.54
n d
p j j
 
     

60
2 2
3,4 6 19.08
n d
p j j
 
     
2 2
0.3
tan 17.45
1 1 0.3
o



  
 
8.9.1. Second-order Time Response Specifications with Impulse input
(a) Over damped case (ζ>1)
General equation
2
2 ( )
i
n n
Kx
y y y t
m
  
  
  (8.76)
Laplacian of the output
2 2
2 2 2
1
( )
2
1 1
2 1 ( 1) ( 1
i
n n
i
n n n n n
Kx
Y s
m s s
Kx
m s s
 
       
 
  
 
 
 
 
 
 
      
 
 
(8.77)
 
2
2
( ) sinh 1
1
nt
i
n
n
Kx
y t e t
m

 
 

 
 
 
 

 
(8.78)
(b) Critically damped case (ζ=1)
General equation
2
( )
i
n
Kx
y y t
m
 
 
 (8.79)
2 2
1
( ) i
n
Kx
Y s
m s 
 
  

 
(8.80)
( ) nt
i
n
n
Kx
y t te
m




 
  
 
(8.81)
(c) Under damped case (ζ<1)

61
Poles are: 1,2 n d
s j
 
  
General equation
2
2 ( )
i
n n
Kx
y y y t
m
  
  
  (8.82)
1
( )
( )( )
i
n d n d
Kx
Y s
m s j s j
   
 
  
   
 
(8.83)
( ) sin
nt
i
d
d
Kx
y t e t
m




 
  
 
(8.84)
Normalized impulse-response of a second-order system with different damping factors are shown
graphically as follows.
Solved problems:
3. A second-order system has an un-damped natural frequency of 100 rad/s and a damping factor of
0.3. The value of the coefficient of the second time derivative (that is m) is 5. If the static
sensitivity is 10, write down the response (do not solve) for a force excitation shown in the figure
in terms of the Duhamel’s integral for the following periods of time: 0<t<t1, t1<t<t2 and t>t2.
Solution:
Given, Undamped natural frequency ωn=100 rad/s
Damping factor  =0.3
Coefficient of the second time derivative m=5

62
Static sensitivity K=10
2 2
1 100 1 0.3 95.39rad/s
d n
  
    
Here,
1
1
( ) ;0
t
F t F t t
t
  
 
2 1 2
2 1
( ) ;
F
F t t t t t t
t t
   

 
0
( ) ( ) sin
n
t
d
d
K
y t F t e d
m
 
   


 

 
 
0.3 100
1 0
1
30
1 0
10
( ) sin 95.39 ( )
5 95.39
;0
0.057
sin 95.39 ( )
t
t
F
y t e t d
t
t t
F
e t d
t


  
  
 

  

 
 


,
 
 
1
1
30
1 0
1 2
30
2
2 1
0.057
( ) sin 95.39 ( )
;
0.057
sin 95.39 ( )
t
t
t
F
y t e t d
t
t t t
F
e t t d
t t


  
  


  
 
  



and
 
 
1
2
1
30
1 0
2
30
2
2 1
0.057
( ) sin 95.39 ( )
;
0.057
sin 95.39 ( )
t
t
t
F
y t e t d
t
t t
F
e t t d
t t


  
  


  

  



8.9.2. Second-order Time Response Specifications with step input
2 2
1
( )
( 1)( 1)
i
n n n n
Kx
Y s
m s s s
     
 
 
  
     
 
 
(8.85)
   
2 2
2 2
( ) 1 cosh 1 sinh 1
1
nt
i
n n
n
Kx
y t e t t
m
 
   
 

 
 
 
    
 
 
 

 
 
 
(8.86)
1
( )
( )( )
i
n d n d
Kx
Y s
m s s j s j
   
 
  
   
 
(8.87)
2 2
( ) 1 cos sin
1
nt
i
d d
n
Kx
y t e t t
m
 
 
 

 
 
 
  
 
 
 

 
 
 
(8.88)

63
8.10. Time Response Specifications with step-input for under-damped case
For under-damped case, the step-response of a second-order is shown as follows
2 2
( ) 1 sin( )
1
nt
i
d
n
Kx e
y t t
m

 
 

 
 
  
 

 
 
(8.89)
2
1 1
tan



 
 (8.90)
For this case, different time-domain specifications are described below.
(i) Delay time, td

64
(ii) Rise time, tr
(iii) Peak time, tp
(iv) Peak overshoot, Mp
(v) Settling time
For unity step input,
(i)Delay time, td: It is the time required to reach 50% of output.
  2
1
1 sin( )
2 1
n d
t
d d d
e
y t t

 


   

1 0.7
d
n
t
w


  (8.91)
(ii) Rise time, tr:The time required by the system response to reach from 10% to 90% of the
final value for over-damped case, from 0% to 100% of the final value for under-damped case
and from 5% to 95% of the critically value for over-damped case.
.
  2
1 1 sin( )
1
n r
t
r d r
e
y t t

 


   

2
sin( ) 0
1
n r
t
d r
e
t

 


  

d r
t
  
  
r
d
t
w
 

  (8.92)
(iii) Peak time, tp:The time required by the system response to reach the first maximum value.
  0
p
dy t
dt

2
1 sin( )
1
0
n p
t
d p
e
d t
dt

 


 
 
 
 

 
 
2
sin( )
1
0
n p
t
d p
e
d t
dt

 


 
 
 
 

 
 
2
1 1
tan
d p
w t n

  

 
     ; where 1,2,3,...
n 
For n=1,
d p
w t n
 
p
d
n
t
w

  (8.93)
(iv) Peak overshoot, Mp: It is the time required to reach 50% of output.
 
  1
% 100
1
p
p
y t
M

 

65
  2
% 100 1 sin( ) 1
1
n r
t
p d r
e
M t

 


 
 
     
 

 
  2 2
% 100 sin( ) 100 sin( )
1 1
n
n p d
t
p d p d p
e e
M t t


 
   
 


 
   
 
        
 
 
 
 
 
 
 
2 2
1 1
2 2 2
% 100 sin( ) 100 sin( )
1 1 1
p d
e e
M
 
 

   
  
 
 
   
   
        
   
   
  
   
   
 
2 2
1 1
2
2 2
% 100 sin 100 1
1 1
p
e e
M
 
 
 
 
 
 
   
   
     
   
   
 
   
   
 
2
1
% 100
p
M e




   (8.94)
(iv) Settling time, ts: It is the time taken by the system response to settle down and stay with in 2%

or 5%
 its final value.
For 2%
 error band,
4
s
n
t
w

 (8.95)
For 5%
 error band,
3
s
n
t
w

 (8.96)
Sl. No.
Time Specifications
Type Formula
1 Delay time
1 0.7
d
n
t
w



2 Rise time r
d
t
w
 


3 Peak time p
d
t
w


4 Maximum overshoot
 
2
1
% 100
p
M e




 
5 Settling time
4
s
n
t
w



66
Solved Problems:
1. Consider the system shown in Figure 1. To improve the performance of the system a feedback is
added to this system, which results in Figure 2. Determine the value of K so that the damping
ratio of the new system is 0.4. Compare the overshoot, rise time, peak time and settling time and
the nominal value of the systems shown in Figures 1 and 2.
Figure 1 Figure 2
Solution:
For Figure 1,
 
 
 
 
2
20
1
( ) 20
20
1 ( ) 20
1
1
c s s s
G s
R s G s s s
s s

  
  


Here, 2
20
n
  and 2 1
n
 
20
n
  rad/s and
1 1
0.112
2 2 20
n


  

For Figure 2,
 
   
2
20
( ) 20
( 1 20 )
20
1 ( ) 1 20 20
1
( 1 20 )
c s G s s s K
R s G s s K s
s s K
 
  
   

 
Here, 2
20
n
  and 2 1 20K
n
  
20
n
  rad/s
But, given that
1 20 1 20
0.4
2 2 20
n
K K


 
  
0.128
K
 
Transient characteristics of Figures 1 and 2
CharacteristicS Figure 1 Figure 2
Overshoot, Mp 70% 25%
Rise time, tr, sec 0.38 0.48
Peak time, tp, sec 0.71 0.77
Settling time (2%), sec 8 2.24
Steady-state value, c∞ 1.0 1.0

67
1.1.Transient Response using MATLAB
Program 1: Find the step response for the following system
 
  2
3 20
5 36
C s s
R s s s


 
Solution:
>> num=[3 20]
num=
3 20
>> den=[1 5 36]
den=
1 5 36
>>sys=tf(num,den)
Transfer function:
3s+20
--------------------
s^2+5s+36
>>step(sys)
Program 2: Find the step response for the following system
 
  2
20
4 25
C s
R s s s

 
Solution:
>> num=[20]
num=
20
>> den=[1 425]
den=
1 4 25
>>sys=tf(num,den)
Transfer function:
20
--------------------
s^2+4s+25
>>step(sys)

68
2. Stability
2.1.Concept of stability
Stability is a very important characteristic of the transient performance of a system. Any working
system is designed considering its stability. Therefore, all instruments are stable with in a boundary of
parameter variations.
A linear time invariant (LTI) system is stable if the following two conditions are satisfied.
(i) Notion-1: When the system is excited by a bounded input, output is also bounded.
Proof:
A SISO system is given by
 
 
 
1
0 1
1
0 1
...
...
m m
m
n n
n
C s b s b s b
G s
R s a s a s a


  
 
  
(9.1)
So,
     
1
c t G s R s
 
  
  (9.2)
Using convolution integral method
     
0
c t g r t d
  

 
 (9.3)
   
1
g G s
 
 = impulse response of the system
Taking absolute value in both sides,
     
0
c t g r t d
  

 
 (9.4)
Since, the absolute value of integral is not greater than the integral of absolute value of the integrand
     
     
     
0
0
0
c t g r t d
c t g r t d
c t g r t d
  
  
  



 
  
  



(9.5)
Let, r(t) and c(t) are bounded as follows.
 
 
1
2
r t M
c t M
  
  
(9.6)
Then,

69
   
1 2
0
c t M g d M
 

 
 (9.7)
Hence, first notion of stability is satisfied if  
0
g d
 

 is finite or integrable.
(ii) Notion-2: In the absence of the input, the output tends towards zero irrespective of initial
conditions. This type of stability is called asymptotic stability.
2.2.Effect of location of poles on stability
Pole-zero map Normalized response
Over-damped close-loop poles
Critically damped close-loop poles
Under-dampedclose-loop poles

70
Un-dampedclose-loop poles
Negative Under-dampedclose-loop poles
Negative Over-dampedclose-loop poles

71
2.3.Closed-loop poles on the imaginary axis
Closed-loop can be located by replace the denominator of the close-loop response with s=jω.
Example:
1. Determine the close-loop poles on the imaginary axis of a system given below.
Solution:
Characteristics equation, 2
( ) 0
B s s s K
   
Replacing s jw

2
( ) ( ) ( ) 0
B j j j K
  
   
2
( ) 0
K j
 
   
Comparing real and imaginary terms of L.H.S. with real and imaginary terms of R.H.S., we get
K
  and 0
 
Therefore, Closed-loop poles do not cross the imaginary axis.
2. Determinetheclose the imaginary axis of a system given below.
3 2
( ) 6 8 0
B s s s s K
     .
Solution:
Characteristics equation,
3 2
( ) ( ) 6( ) 8 0
B j j j j K
   
    
2 3
( 6 ) (8 ) 0
K j
  
    
Comparing real and imaginary terms of L.H.S. with real and imaginary terms of R.H.S., we get
8
   rad/s and 2
6 48
K 
 
Therefore, Close-loop poles cross the imaginary axis for K>48.
( )
( 1)
K
G s
s s



72
2.4.Routh-Hurwitz’s Stability Criterion
General form of characteristics equation,
1
1 1 0
( ) 0
n n
n n
B s a s a s a s a


    

1 2
( )( ) ( ) 0
n
s r s r s r
    

Where, i
r  Roots of the characteristics equation
2.4.1. Necessary condition of stability:
Coefficients of the characteristic polynomial must be positive.
Example:
3. Consider a third order polynomial 3 2
( ) 3 16 130
B s s s s
    . Although the coefficients of the
above polynomial are positive, determine the roots and hence prove that the rule about
coefficients being positive is only a necessary condition for the roots to be in the left s-plane.
Solution:
Characteristics equation, 3 2
( ) 3 16 130 0
B s s s s
    
By using Newton-Raphson’s method 1 5
r   and 2,3 1 5
r j
 
Therefore, from the above example, the condition that coefficients of a polynomial should be positive
for all its roots to be in the left s-plane is only a necessary condition.
2.4.2. Sufficient condition of stability:
2.4.2.1.Method I (using determinants)
The coefficients of the characteristics equation are represented by determinant form
as follows.
1 3 5
2 4
1 3
0
n n n
n n n
n
n n
a a a
a a a
a a
  
 
 
 



(9.8)
Here, the determinant decreases by two along the row by one down the
column. For stability, the following conditions must satisfy.
1 3 5
1 3
1 1 2 3 2 4
2
1 3
0, 0, 0
0
n n n
n n
n n n n
n n
n n
a a a
a a
a a a a
a a
a a
  
 
  

 
          (9.9)

73
2.4.2.2.Method II (using arrays)
The coefficients of the characteristics equation are represented by array form as
follows.
2 4
1
1 3 5
2
1 3 5
3
1 3 5
n
n n n
n
n n n
n
n n n
n
n n n
a a a
s
a a a
s
b b b
s
c c c
s
 

  

  

  

(9.10)
Where,
1 2 3
1
1
1 4 5
3
1
1 3 1 3
1
1
( )( ) ( )
( )( ) ( )
( )( ) ( )
n n n n
n
n
n n n n
n
n
n n n n
n
n
a a a a
b
a
a a a a
b
a
b a a b
c
b
  


  


   








(9.11)
For stability, the following conditions must satisfy.
The number of roots of B(s) with positive real parts is equal to the number of sign
changes an, an-1, bn-1, cn-1, etc.
Example:
4. Find stability of the following system given by ( )
( 1)
K
G s
s s


and ( ) 1
H s  using Routh-
Hurwitz stability criterion.
Solution:
In the system,   2
( ) ( 1)
1 ( ) ( ) 1
( 1)
K
G s K
s s
T s
K
G s H s s s K
s s

  
  


Method-I,
Characteristics equation,   2
0
B s s s K
   
Here,
1
2
1
1 0
1
K
K
 
  
For stability, 1
2
0
0
 
 
The system is always stable for K>0.
Method-II,

74
Characteristics equation,   2
0
B s s s K
   
Here, Routh array is
2
1
0
1
1 0
s K
s
K
s
There are no sign changes in first column elements of this array.Therefore, the system is always stable
for K>0.
5. Find stability of the following system given by ( )
( 2)( 4)
K
G s
s s s

 
and ( ) 1
H s  using
Routh-Hurwitz stability criterion.
Solution:
In the system,
 
 
    3 2
( ) ( 2)( 4)
1 6 8
1
( 2)( 4)
K
G s
C s K
s s s
K
R s G s H s s s s K
s s s
 
  
   

 
Method-I,
General form of characteristics equation,   3 2
3 2 1 0 0
B s a s a s a s a
    
And in this system, characteristics equation is   3 2
6 8 0
B s s s s K
    
Here,sufficient condition of stability suggests
 
 
1 2
3
6
8 0, 48 0,
1 8
6 0
1 8 0 48 0
0 6
K
K
K
K K
K
       
    
Therefore, the system is always stable for 48
K  .
Method-II,
Characteristics equation is   3 2
6 8 0
B s s s s K
    
andRouth’s array
3
2
1
0
1 8
6
48
0
6
s
K
s
K
s
s K

There are no sign changes in first column elements of this array if 48
K  . Therefore, the system is
always stable for 0 48
K
  .

75
6. Find stability of the following system given by   3 2
5 10 3
B s s s s
    using Routh-Hurwitz
stability criterion.
Solution:
In this problem, given Characteristics equation is   3 2
5 10 3 0
B s s s s
     , andRouth’s array is
3
2
1
0
1 10
5 3
9.4 0
3
s
s
s
s
There are no sign changes in first column elements of this array. Therefore, the system is always
stable.
7. Find stability of the following system given by   3 2
2 3 10
B s s s s
    using Routh-Hurwitz
Solution:
In this problem, given characteristics equation is
  3 2
2 3 10 0
B s s s s
     and
Routh’s array is
3
2
1
0
1 3
2 10
2 0
10
s
s
s
s

There are two sign changes in first column elements of this array. Therefore, the system is unstable.
8. Examine stability of the following system given by 5 4 3 2
2 4 8 3 1
s s s s s
     using Routh-Hurwitz
Solution:
In this problem, Routh’s array is
5
4
3
2
1
0
1 4 3
2 8 1
0 2.5
s
s
s
s
s
s

Here, the criterion fails. To remove the above difficulty, the following two methods can be used.
Method-1
(i) Replace 0 by ε(very small number) and complete the array with ε.
(ii) Examine the sign change by taking 0
 
Now, Routh’s array becomes

76
5
4
3
2
1
0
1 4 3
2 8 1
2.5 0
5 8
1 0
5 8
2.5
5 8
1
s
s
s
s
s
s










 

 
 

Now putting 0
  , Routh’s array becomes
5
4
3
2
1
0
1 4 3
2 8 1
2.5 0
5 8
1 0
5 8
2.5
5 8
1
s
s
s
s
s
s










 

 
 

Method-2
Replace s by . The system characteristic equation 5 4 3 2
2 4 8 3 1 0
s s s s s
      becomes
5 4 3 2
1 2 4 8 3
1 0
Z Z Z Z Z
     
5 4 3 2
3 8 4 2 1 0
Z Z Z Z Z
      
Now, Routh’s array becomes
5
4
3
2
1
0
1 8 2
3 4 1
6.67 1.67 0
3.25 1 0
0.385 0 0
1 0 0
s
s
s
s
s
s

9. Examine stability of the following system given by 5 4 3 2
2 2 4 4 8
s s s s s
     using Routh-Hurwitz
Solution:
1
Z

77
5
4
3
2
1
0
1 2 4
2 4 8
0 0 0
s
s
s
s
s
s
Here, the criterion fails. To remove the above difficulty, the following two methods can be used.
The auxillary equation is
  4 2
2 4 8
A s s s
  
  3
8 8
dA s
s s
ds
  
Now, the array is rewritten as follows.
5
4
3
2
1
0
1 2 4
2 4 8
8 8 0
2 8 0
24 0
8
s
s
s
s
s
s

There are two sign changes in first column elements of this array. Therefore, the system is
unstable.
10. Examine stability of the following system given by 4 3 2
5 2 3 1 0
s s s s
     using Routh-Hurwitz
stability criterion. Find the number of roots in the right half of the s-plane.
Solution:
4
3
2
1
0
1 2 2
5 3 0
1.4 2
4.14 0
2
s
s
s
s
s

There are two poles in the right half of the s-plane.
2.4.3. Advantages of Routh-Hurwitz stability
(i) Stability can be judged without solving the characteristic equation
(ii) Less calculation time
(iii)The number of roots in RHP can be found in case of unstable condition
(iv) Range of value of K for system stability can be calculated
(v) Intersection point with the jw-axis can be calculated
(vi) Frequency of oscillation at steady-state is calculated

78
2.4.4. Advantages of Routh-Hurwitz stability
(i) It is valid for only real coefficient of the characteristic equation
(ii) Unable to give exact locations of closed-loop poles
(iii)Does not suggest methods for stabilizing an unstable system
(iv) Applicable only to the linear system

79
Equation Chapter 1 Section 1
MODULE#3

80
CHAPTER#10
10. Root locus
10.1. Definition:
The locus of all the closed-loop poles for various values of the open-loop gain K is called root locus.
The root-locus method is developed by W.R. Evans in 1954. It helps to visualize the various
possibilities of transient response of stable systems.
Closed-loop response function
( ) ( )
( ) 1 ( ) ( )
C s G s
R s G s H s


(10.1)
Characteristic equation
1 2
1 2
( )( )...( )
1 ( ) ( ) 1 0
( )( )...( )
m
n
K s z s z s z
G s H s
s p s p s p
  
   
  
(10.2)
Vector from open-loop pole to the root-locus
Vector from open-loop zero to the root-locus
Behaviors of closed-loop poles
Closed-loop poles negative
and real
Exponential decay Stable
Closed-loop poles complex
with negative real parts
Decaying and oscillatory Stable
Closed-loop poles positive and
real
Exponential increase Unstable
with positive real parts
Exponential and oscillatory
increase
Unstable
10.2. BASIS for CONSTRUCTION

81
10.2.1. Construction steps
1. Determine the number of open-loop poles and zeros
2. Mark open-loop poles and zeros on the s-plane
3. Determine parts of the root-locus on the real axis
4. Determine breakaway and break-in points
5. Draw asymptotes to the root-locus
6. Determine angles of departure
7. Determine angles of arrival
8. Determine points on the root-locus crossing imaginary axis
9. Obtain additional points and complete the root-locus
10.2.2. Starting points
Characteristics equation of a closed-loop system
1 2
1 2
( )( )...( )
1 ( ) ( ) 1 0
( )( )...( )
m
n
K s z s z s z
G s H s
s p s p s p
  
   
  
(10.3)
For K=0,
1 2 1 2
1 2
( )( )...( ) ( )( )...( )
0
( )( )...( )
n m
n
s p s p s p K s z s z s z
s p s p s p
      
 
  
1 2
( )( )...( ) 0
n
s p s p s p
     (10.4)
Open-loop poles are also closed-loop poles for K=0. A root-locus starts from every open-loop pole.
10.2.3. Ending points
1 2
1 2
( )( )...( )
1 ( ) ( ) 1 0
( )( )...( )
m
n
K s z s z s z
G s H s
s p s p s p
  
   
  
(10.5)
For K=∞,
1 2
1 2
( )( )...( )
1
( )( )...( )
m
n
K s z s z s z
s p s p s p
  

  
1 2
( )( )...( ) 0
m
s z s z s z
     (10.6)
Root-locus ends at an open-loop zero or at infinity.
10.2.4. Magnitude and angle criterion
1 ( ) ( ) 1 ( ) ( ) (cos sin ) 0
G s H s G s H s j
 
     (10.7)
Angle criterion:
0
1 1
180 360
n m
i j
i j
k
   
 
  
  (10.8)
Where, i
  angle in case of ith
pole and j
  angle in case of jth
zero
Magnitude criterion:

82
( ) ( ) 1
G s H s  (10.9)
10.2.5. Determining gain at a root-locus point
Using the magnitude of vectors drawn from open-loop poles and zeros to the root-locus point, we get
1 1 2
1 2
1
( )
| ( ) ||( ) |...| ( ) |
| ( ) || ) |...| ( ) |
( )
n
i
i n
m
m
j
j
s p
s p s p s p
K
s z s z s z
s z



  
 
  



(10.10)
Gain at a root-locuspoint is determinedusing synthetic division.
Example:
Determine K of the characteristic equation for the root s=-0.85.
Solution:
3 2
6 8 0
S s s K
    (10.11)
1 6 8 K
-0.85 -4.378 -3.079
1 5.15 3.622 K-3.079=0
10.2.6. Determine parts of the root-locus on the real axis
1. Start from open-loop poles on the real axis, extend on the real axis for increasing
values of the gain and end at an open-loop zero on the real axis.
values of the gain and end at an infinite value on the real axis.
3. Start from a pair of open-loop poles on the real axis, extend on the real axis for
increasing values of gain, meet at a point and then leave the real axis and end at a
complex open-loop zero or infinity.
increasing values of gain, meet at a point and then leave the real axis. They may once
again enter the real axis and end at open-loop zeros or at a large value on the real axis.
5. Start from a pair of complex open-loop poles, enter the real axis and end at an open-
loop zero or an infinite value on the real axis. They could leave the real axis again and
end at a complex open-loop zero or infinity.
10.2.7. Angle contributions from complex poles
Complex poles and zeros do not contribute to the angle criterion on the real axis

83
10.2.8. Determine breakaway and break-in points of the root-locus
( )
1 ( ) ( ) 1 0
( )
A s
G s H s K
B s
    (10.12)
( ) ( ) ( ) 0
f s B s KA s
   (10.13)
( )
)
B s
K
As
  (10.14)
1 2 1
( ) ( ) ( )...( ) 0
r
n r
f s s s s s s s  
     (10.15)
1
1 2 1 1 3 1
( )
( ) ( )...( ) ( ) .( )..( ) ...
r r
n r n r
df s
r s s s s s s s s s s s s
ds

   
         (10.16)
1
( )
0
s s
df s
ds 
 (10.17)
' ' '
( ) ( ) ( ) 0
f s B s KA s
   (10.18)
'
'
( )
( )
B s
K
A s
   (10.19)
Therefore,
' '
( ) ( ) ( ) ( ) 0
B s A s B s A s
  (10.20)
At breakaway and break-in points of the root-locus,

84
' '
2
( ) ( ) ( ) ( )
0
( )
dK B s A s B s A s
ds A s

   (10.21)
10.2.9. Draw asymptotes to the root-locus
Angle of asymptotes
0
180 360
( )
c
k
n m




where, k=0, 1, 2, 3..
Location of asymptotes
1 2
1 2
( )( )...( )
( )( )...( )
n
m
s p s p s p
K
s z s z s z
  
 
  
(10.22)
1
1 2
1
1 2
( ... )
( ... ) ...
n n
n
m m
m
s p p p s
K
s z z z s


   
 
   

(10.23)
1
1 2 1 2
[( ... ) ( ... )] ...
n m n m
n m
K s p p p z z z s
  
         (10.24)
i c
s p s 
   (10.25)
( )
i c
s z s 
   (10.26)
1
( )
( ) ...
( )
n
n m n m
c
c
m
c
s
K s n m s
s



  

     

(10.27)
1 2 1 2
( ... ) ( ... )
( )
n m
c
p p p z z z
n m

    


(10.28)
Angle of departure
1 2 1
180 ( )
d
   
    (10.29)

85
d=180
o
- angles of vectors to the complex open-loop pole in question from other open - loop poles
+ angles of vectors to the complex open-loop pole in question from all open-loop zeros
Angle of arrival
1 3 1 2 3
180 ( ) ( )
a
     
      (10.30)
a=180
o
- angles of vectors to the complex open-loop zero in question from other open- loop zeros
+ angles of vectors to the complex open-loop zero in question from all open-loop poles
Determine points on the root-locus crossing imaginary axis
Re [1 ( ) ( )] 0
al G j H j
 
  (10.31)
[1 ( ) ( )] 0
imaginary G j H j
 
  (10.32)
Example
Problem-1: Draw the root-locus of the feedback system whose open-loop transfer function is given
by ( ) ( )
( 1)
K
G s H s
s s


Solution:
Step 1: Determine the number of open-loop poles and zeros
Number of open-loop poles n=2
Number of open-loop zeros m=0
Open-loop poles: s=0 and s=-1

86
Step 2: Mark open-loop poles and zeros on the s-plane
Step 3: Determine parts of the root-locus on the real axis
Test points on the positive real axis
Test points in between the open-lop poles
Step 4: Determine breakaway and break-in point
Characteristic equation, ( 1)
K s s
  
2 1 0
dK
s
ds
   
breakaway point as σ
b
=-0.5

87
Gain at the breakaway point
| 0.5 0|| 0.5 ( 1) | 0.25
b
K       
Step 5: Draw asymptotes of the root-locus
Angle of asymptotes:
0
0
0
180 360 180 360
( ) 2
90 0
270 1
c c
c
c
k k
n m
k
k
 


 
  

 
 
Centroid of asymptotes
1 2 1 2
( ... ) ( ... ) 0 1
0.5
( ) 2
n m
c
p p p z z z
n m

     
   

Steps 6 & 7: Since there are no complex open-loop poles or zeros, angle of departure and arrival need
not be computed
Step 8: Determine points on the root-locus crossing imaginary axis
2
1 1 0
( 1)
K
GH s s K
s s
      

2 2
( ) ( ) ( ) ( )
B j j j K K j
    
     
2
0 0
K j
 
   
The root-locus does not cross the imaginary axis for any value of K>0

88
Here,
1 1 4
2
K
s
  

by ( ) ( )
( 2)( 4)
K
G s H s
s s s

 
Solution:
Open-loop poles: s=0, s=-2 and s=-4

89
Characteristic equation, ( 2)( 4)
K s s s
   
( 2)( 4) ( 4) ( 2) 0
dK
s s s s s s
ds
        
Breakaway point as σb=-0.85 and –3.15
σb = –3.15 is not on the root-locus and therefore not a breakaway or break-in point

90
| 0.85 0|| 0.855 ( 2) || 0.85 ( 4) | 3.079
b
K          
1 6 8 K
-0.85 -4.378 -3.079
1 5.15 3.622 K-3.079=0
0
0
0
0
180 360 180 360
( ) 3
60 0
180 1
300 2
c
c
c
c
k k
n m
k
k
k




 
 

 
 
 
1 2 1 2
( ... ) ( ... ) 0 2 4
2
( ) 3
n m
c
p p p z z z
n m

      
   

not be computed
3 2
1 1 6 8 0
( 2)( 4)
K
GH s s s K
s s s
       
 
3 2 2 3
( ) ( ) 6( ) 8 ( 6 ) (8 ) 0
B j j j j K K j
      
        

91
When imaginary-part is zero, then 8 j 8
s
      and when real-part is zero,
then 2
6 48
K 
  .
The root-locus does not cross the imaginary axis for any value of K>48.
1 6 8 48
+j2.828 -8+j16.97 -48
1 6+j2.828 J16.97 0
1 6+j2.828 J16.97
-j2.828 -j16.97
1 6 0
Therefore, closed-loop pole on the real axis for K=48 at 6
s  
No.
Closed-loop pole
on the real axis
K
Second and third
closed- loop poles
Remarks
1 -4.309 3.07 -0.85,-0.85 Already computed
2 -4.50 5.625 -0.75j0.829
3 -5.00 15 -0.5j1.6583
4 -5.50 28.875 -0.25j2.2776
5 -6.00 48 j2.8284 Already computed
6 -6.5 73.125 0.25j3.448
Determine the gain corresponding to s=-4.5
K=|-4.5-(-4)||-4.5-(-2)||-4.5-0|= 5.625
3 2
6 8 0
s s s K
   
1 6 8 K
-4.5 -6.75 -5.625
1 1.5 1.25 K-5.625=0
2
( 1.5 1.25) 0
s s
  
2,3 0.75 0.829
s j
  

92
by 2
( ) ( )
( 1)
K
G s H s
s s


Solution:
Open-loop poles: s=0, s=0 and s=-1

93
Characteristic equation, 2
( 1)
K s s
  
 
0
2 ( 1) 0
2 3 0
dK
ds
s s s
s s

    
   
Breakaway point as σb= -2/3and 0
σb = -2/3is not on the root-locus and therefore not a breakaway or break-in point.
Therefore σb= 0 and the two loci start from the origin and breakaway at the origin itself.
0
0
0
0
180 360 180 360
( ) 3
60 0
180 1
300 2
c
c
c
c
k k
n m
k
k
k




 
 

 
 
 
1 2 1 2
( ... ) ( ... ) 0 1 1
( ) 3 3
n m
c
p p p z z z
n m

     
   

not be computed.

94
3 2
( )
B s s s K
  
3 2 2 3
( ) ( ) ( ) ( )
B j j j K K j
    
     
When imaginary-part is zero, then 0 0
s
    and when real-part is zero,
then 2
0
K 
  .
Additional closed-loop poles
No. Closed-loop pole on the
real axis
K Second and third
closed- loop poles
1 -1.5 1.125 0.25±j0.82
2 -2.0 4 0.50±j1.32
3 -2.5 9.375 0.75±j1.78
4 -3.0 18 1.00±j2.23
K=|-1.5-(-1)||-1.5-(0)||-1.5-0|= 1.125
3 2
1.125 0
s s
  
1 1 0 1.125
-1.5 0.75 -1.125
1 -0.5 0.75 0
2
( 1.5 1.25) 0
s s
  
2,3 0.25 0.82
s j
  

95
by 4 3 2
( ) ( )
5 8 6
K
G s H s
s s s s

  
Solution:
4 3 2 2
5 8 6 ( 2 2)( 3) ( 1 )( 1 )( 3)
s s s s s s s s s j s j s s
            
Open-loop poles: s=0 and s=-3, s=-1+j and s=-1-j

96
Characteristic equation, 4 3 2
( 5 8 6 )
K s s s s
    
3 2
3 2
0
4 15 16 6 0
3.75 4 1.5 0
dK
ds
s s s
s s s

    
    
' 2
( ) 3 7.5 4
f s s s
  
This equation is solved using Newton-Raphson’s method
1 '
( )
( )
n
n n
n
f s
s s
f s
  
No.
n
s ( )
n
f s '
( )
n
f s 1
n
s 
1 -3.75 -13.5 18.0625 -3.0026
2 -3.0026 -3.7721 8.5273 -2.5602
3 -2.5602 -0.9421 4.4624 -2.3491
4 -2.3491 -0.1658 2.9364 -2.2926
5 -2.2926 -0.0103 2.5737 -2.2886
6 -2.2886
-5.03x10
-5
Breakaway point as σb= -2.3
Gain at the breakaway point, | 2.3 ( 3) || 2.3 0 || 2.3 ( 1 ) || 2.3 ( 1 ) | 4.33
K j j
              
1 5 8 6 K
-2.2886 -6.2053 -4.1073 -4.3316

97
1 2.7114 1.7947 1.8926 0
Other closed-loop poles for K=4.3
1 2.7114 1.7947 1.893
-2.2886 -0.9676 -1.893
1 0.4228 0.8270 0
s
3,4
=-0.2114±j0.8814
0
0
0
0
0
180 360 180 360
( ) 4
45 0
135 1
225 2
315 3
c
c
c
c
c
k k
n m
k
k
k
k





 
 

 
 
 
 
1 2 1 2
( ... ) ( ... ) 0 3 1 1
1.25
( ) 4
n m
c
p p p z z z j j
n m

         
   


98
Steps 6:Determine angles of departure
0 0 0 0 0 0
180 (135 26.56 90 ) 71.56 288.44
d
       

99
Step 7: As there are no complex open-loop zeros, angle of arrival need not be computed.
4 3 2
( ) 5 8 6
B s s s s s K
    
4 3 2 4 2 3
( ) ( ) 5( ) 8( ) 6 ( 8 ) (6 5 )
B j j j j j K K j
        
         
When imaginary-part is zero, then
6 6
5 5
s j
then
2
6 6
8 8.16
5 5
K
   
   
   
   
.
There are two closed-loop poles on the imaginary axis for any value of K>0.
No. S1 S2 S3,4 K
1 -0.25 -2.9217 -0.9142±0.7969j 1.0742
2 -0.50 -2.8804 -0.8098±0.655j 1.5625
3 -0.75 -2.8593 -0.6953±0.5938j 1.7930
4 -1.0 -2.8393 -0.5804±0.6063j 2.0000
5 -1.25 -2.8055 -0.4722±0.6631j 2.3242
6 -1.75 -2.6562 -0.3763±0.7354j 2.8125
7 -2.0 -2.5214 -0.2393±0.8579j 4.0

100
Additional Information from Root-Locus Plot
1. Gain Margin
2
1
20log
K
GM
K
 (10.33)
K1 is the gain of a feedback system at some point on the root-locus
K2 is the gain at which the system becomes unstable
2. Transient Characteristics
Where,
2
1 1
tan



 

3. Percentage overshoot
/tan
p
M e  

 (10.34)
4. Settling time
4
s
n
t

 (10.35)
5. Steady-state error is also related to K.
Example
by
   
2
4 3 2
10 100
( ) ( ) , 1
20 100 500 1500
K s s
G s H s H s
s s s s
 
 
   
(a) Determine the value of gain at which the system will be stable and as well have a maximum
overshoot of 5%.
(b) What is the gain margin at this point?
(c) What is the steady-state error for a unit step excitation at the above point?
Solution:

101
(a)
0
tan 1.0487
ln
46
p
M




 
 
2
1
0.690
1 tan


 

(10.36)
(b)
192.2
20log 2.65
261
GM dB
  
(c) Position error
2
4 3 2
0
( 10 100) 100
20 100 500 1500 1500
lim
s
s
K s s K
K
s s s s

 
 
   
Steady-state error,
1 1 1500
( )
1 1 100 /1500 1500 100
e
s
S
K K K
   
  
1500
( ) 5.4%
1500 100 261
e
S   
 

102
Root locus
The locus of all the closed-loop poles for various values of the open-loop gain K is called root locus.
The root-locus method is developed by W.R. Evans in 1954. It helps to visualize the various
possibilities of transient response of stable systems.
Closed-loop response function
( ) ( )
( ) 1 ( ) ( )
C s G s
R s G s H s


(10.37)
Characteristic equation
1 2
1 2
( )( )...( )
1 ( ) ( ) 1 0
( )( )...( )
m
n
K s z s z s z
G s H s
s p s p s p
  
   
  
(10.38)
Vector from open-loop pole to the root-locus
Vector from open-loop zero to the root-locus
Behaviors of closed-loop poles
Closed-loop poles negative
and real
Exponential decay Stable
with negative real parts
Decaying and oscillatory Stable
Closed-loop poles positive and
real
Exponential increase Unstable
with positive real parts
Exponential and oscillatory
increase
Unstable
BASIS for CONSTRUCTION
Construction steps
10. Determine the number of open-loop poles and zeros

103
11. Mark open-loop poles and zeros on the s-plane
12. Determine parts of the root-locus on the real axis
13. Determine breakaway and break-in points
14. Draw asymptotes to the root-locus
15. Determine angles of departure
16. Determine angles of arrival
17. Determine points on the root-locus crossing imaginary axis
18. Obtain additional points and complete the root-locus
Starting points
1 2
1 2
( )( )...( )
1 ( ) ( ) 1 0
( )( )...( )
m
n
K s z s z s z
G s H s
s p s p s p
  
   
  
(10.39)
For K=0,
1 2 1 2
1 2
( )( )...( ) ( )( )...( )
0
( )( )...( )
n m
n
s p s p s p
      
 
  
1 2
( )( )...( ) 0
n
s p s p s p
     (10.40)
Open-loop poles are also closed-loop poles for K=0. A root-locus starts from every open-loop pole.
Ending points
1 2
1 2
( )( )...( )
1 ( ) ( ) 1 0
( )( )...( )
m
n
K s z s z s z
G s H s
s p s p s p
  
   
  
(10.41)
For K=∞,
1 2
1 2
( )( )...( )
1
( )( )...( )
m
n
K s z s z s z
s p s p s p
  

  
1 2
( )( )...( ) 0
m
s z s z s z
     (10.42)
Root-locus ends at an open-loop zero or at infinity.
Magnitude and angle criterion
1 ( ) ( ) 1 ( ) ( ) (cos sin ) 0
G s H s G s H s j
 
     (10.43)
Angle criterion:
0
1 1
180 360
n m
i j
i j
k
   
 
  
  (10.44)
Where, i
  angle in case of ith
pole and j
  angle in case of jth
zero
Magnitude criterion:
( ) ( ) 1
G s H s  (10.45)
Determining gain at a root-locus point

104
Using the magnitude of vectors drawn from open-loop poles and zeros to the root-locus point, we get
1 1 2
1 2
1
( )
| ( ) ||( ) |...| ( ) |
| ( ) || ) |...| ( ) |
( )
n
i
i n
m
m
j
j
s p
s p s p s p
K
s z s z s z
s z



  
 
  



(10.46)
Gain at a root-locus point is determined using synthetic division.
Example:
Determine K of the characteristic equation for the root s=-0.85.
Solution:
3 2
6 8 0
S s s K
    (10.47)
1 6 8 K
-0.85 -4.378 -3.079
1 5.15 3.622 K-3.079=0
Determine parts of the root-locus on the real axis
values of the gain and end at an open-loop zero on the real axis.
values of the gain and end at an infinite value on the real axis.
increasing values of gain, meet at a point and then leave the real axis and end at a
complex open-loop zero or infinity.
increasing values of gain, meet at a point and then leave the real axis. They may once
again enter the real axis and end at open-loop zeros or at a large value on the real axis.
10. Start from a pair of complex open-loop poles, enter the real axis and end at an open-
loop zero or an infinite value on the real axis. They could leave the real axis again and
end at a complex open-loop zero or infinity.
Angle contributions from complex poles
Complex poles and zeros do not contribute to the angle criterion on the real axis

105
Determine breakaway and break-in points of the root-locus
( )
1 ( ) ( ) 1 0
( )
A s
G s H s K
B s
    (10.48)
( ) ( ) ( ) 0
f s B s KA s
   (10.49)
( )
)
B s
K
As
  (10.50)
1 2 1
( ) ( ) ( )...( ) 0
r
n r
f s s s s s s s  
     (10.51)
1
1 2 1 1 3 1
( )
( ) ( )...( ) ( ) .( )..( ) ...
r r
n r n r
df s
r s s s s s s s s s s s s
ds

   
         (10.52)
1
( )
0
s s
df s
ds 
 (10.53)
' ' '
( ) ( ) ( ) 0
f s B s KA s
   (10.54)
'
'
( )
( )
B s
K
A s
   (10.55)
Therefore,
' '
( ) ( ) ( ) ( ) 0
B s A s B s A s
  (10.56)
At breakaway and break-in points of the root-locus,
' '
2
( ) ( ) ( ) ( )
0
( )
dK B s A s B s A s
ds A s

   (10.57)
Draw asymptotes to the root-locus

106
Angle of asymptotes
0
180 360
( )
c
k
n m




where, k=0, 1, 2, 3..
Location of asymptotes
1 2
1 2
( )( )...( )
( )( )...( )
n
m
s p s p s p
K
s z s z s z
  
 
  
(10.58)
1
1 2
1
1 2
( ... )
( ... ) ...
n n
n
m m
m
s p p p s
K
s z z z s


   
 
   

(10.59)
1
1 2 1 2
[( ... ) ( ... )] ...
n m n m
n m
K s p p p z z z s
  
         (10.60)
i c
s p s 
   (10.61)
( )
i c
s z s 
   (10.62)
1
( )
( ) ...
( )
n
n m n m
c
c
m
c
s
K s n m s
s



  

     

(10.63)
1 2 1 2
( ... ) ( ... )
( )
n m
c
p p p z z z
n m

    


(10.64)
Angle of departure
1 2 1
180 ( )
d
   
    (10.65)

107
d=180
o
- angles of vectors to the complex open-loop pole in question from other open - loop poles
+ angles of vectors to the complex open-loop pole in question from all open-loop zeros
Angle of arrival
1 3 1 2 3
180 ( ) ( )
a
     
      (10.66)
a=180
o
- angles of vectors to the complex open-loop zero in question from other open- loop zeros
+ angles of vectors to the complex open-loop zero in question from all open-loop poles
Determine points on the root-locus crossing imaginary axis
Re [1 ( ) ( )] 0
al G j H j
 
  (10.67)
[1 ( ) ( )] 0
imaginary G j H j
 
  (10.68)
Example
by ( ) ( )
( 1)
K
G s H s
s s


Solution:
Open-loop poles: s=0 and s=-1

108
Characteristic equation, ( 1)
K s s
  
2 1 0
dK
s
ds
   
breakaway point as σ
b
=-0.5

109
| 0.5 0|| 0.5 ( 1) | 0.25
b
K       
0
0
0
180 360 180 360
( ) 2
90 0
270 1
c c
c
c
k k
n m
k
k
 


 
  

 
 
1 2 1 2
( ... ) ( ... ) 0 1
0.5
( ) 2
n m
c
p p p z z z
n m

     
   

not be computed
2
1 1 0
( 1)
K
GH s s K
s s
      

2 2
( ) ( ) ( ) ( )
B j j j K K j
    
     
2
0 0
K j
 
   
The root-locus does not cross the imaginary axis for any value of K>0

110
Here,
1 1 4
2
K
s
  

by ( ) ( )
( 2)( 4)
K
G s H s
s s s

 
Solution:
Open-loop poles: s=0, s=-2 and s=-4

111
Characteristic equation, ( 2)( 4)
K s s s
   
( 2)( 4) ( 4) ( 2) 0
dK
s s s s s s
ds
        
Breakaway point as σb=-0.85 and –3.15
σb = –3.15 is not on the root-locus and therefore not a breakaway or break-in point

112
| 0.85 0|| 0.855 ( 2) || 0.85 ( 4) | 3.079
b
K          
1 6 8 K
-0.85 -4.378 -3.079
1 5.15 3.622 K-3.079=0
0
0
0
0
180 360 180 360
( ) 3
60 0
180 1
300 2
c
c
c
c
k k
n m
k
k
k




 
 

 
 
 
1 2 1 2
( ... ) ( ... ) 0 2 4
2
( ) 3
n m
c
p p p z z z
n m

      
   

not be computed
3 2
1 1 6 8 0
( 2)( 4)
K
GH s s s K
s s s
       
 
3 2 2 3
( ) ( ) 6( ) 8 ( 6 ) (8 ) 0
B j j j j K K j
      
        

113
When imaginary-part is zero, then 8 j 8
s
then 2
6 48
K 
  .
1 6 8 48
+j2.828 -8+j16.97 -48
1 6+j2.828 J16.97 0
1 6+j2.828 J16.97
-j2.828 -j16.97
1 6 0
Therefore, closed-loop pole on the real axis for K=48 at 6
s  
No.
Closed-loop pole
on the real axis
K
Second and third
closed- loop poles
Remarks
1 -4.309 3.07 -0.85,-0.85 Already computed
2 -4.50 5.625 -0.75j0.829
3 -5.00 15 -0.5j1.6583
4 -5.50 28.875 -0.25j2.2776
5 -6.00 48 j2.8284 Already computed
6 -6.5 73.125 0.25j3.448
K=|-4.5-(-4)||-4.5-(-2)||-4.5-0|= 5.625
3 2
6 8 0
s s s K
   
1 6 8 K
-4.5 -6.75 -5.625
1 1.5 1.25 K-5.625=0
2
( 1.5 1.25) 0
s s
  
2,3 0.75 0.829
s j
  

114
by 2
( ) ( )
( 1)
K
G s H s
s s


Solution:
Open-loop poles: s=0, s=0 and s=-1

115
Characteristic equation, 2
( 1)
K s s
  
 
0
2 ( 1) 0
2 3 0
dK
ds
s s s
s s

    
   
Breakaway point as σb= -2/3and 0
σb = -2/3is not on the root-locus and therefore not a breakaway or break-in point.
Therefore σb = 0 and the two loci start from the origin and breakaway at the origin itself.
0
0
0
0
180 360 180 360
( ) 3
60 0
180 1
300 2
c
c
c
c
k k
n m
k
k
k




 
 

 
 
 
1 2 1 2
( ... ) ( ... ) 0 1 1
( ) 3 3
n m
c
p p p z z z
n m

     
   

not be computed.

116
3 2
( )
B s s s K
  
3 2 2 3
( ) ( ) ( ) ( )
B j j j K K j
    
     
When imaginary-part is zero, then 0 0
s
    and when real-part is zero,
then 2
0
K 
  .
No. Closed-loop pole on the
real axis
K Second and third
closed- loop poles
1 -1.5 1.125 0.25±j0.82
2 -2.0 4 0.50±j1.32
3 -2.5 9.375 0.75±j1.78
4 -3.0 18 1.00±j2.23
K=|-1.5-(-1)||-1.5-(0)||-1.5-0|= 1.125
3 2
1.125 0
s s
  
1 1 0 1.125
-1.5 0.75 -1.125
1 -0.5 0.75 0
2
( 1.5 1.25) 0
s s
  
2,3 0.25 0.82
s j
  

117
by 4 3 2
( ) ( )
5 8 6
K
G s H s
s s s s

  
Solution:
4 3 2 2
5 8 6 ( 2 2)( 3) ( 1 )( 1 )( 3)
s s s s s s s s s j s j s s
            
Open-loop poles: s=0 and s=-3, s=-1+j and s=-1-j

118
Characteristic equation, 4 3 2
( 5 8 6 )
K s s s s
    
3 2
3 2
0
4 15 16 6 0
3.75 4 1.5 0
dK
ds
s s s
s s s

    
    
' 2
( ) 3 7.5 4
f s s s
  
This equation is solved using Newton-Raphson’s method
1 '
( )
( )
n
n n
n
f s
s s
f s
  
No. n
s ( )
n
f s '
( )
n
f s 1
n
s 
1 -3.75 -13.5 18.0625 -3.0026
2 -3.0026 -3.7721 8.5273 -2.5602
3 -2.5602 -0.9421 4.4624 -2.3491
4 -2.3491 -0.1658 2.9364 -2.2926
5 -2.2926 -0.0103 2.5737 -2.2886
6 -2.2886
-5.03x10
-5
Breakaway point as σb= -2.3
Gain at the breakaway point, | 2.3 ( 3) || 2.3 0|| 2.3 ( 1 ) || 2.3 ( 1 ) | 4.33
K j j
              
1 5 8 6 K
-2.2886 -6.2053 -4.1073 -4.3316

119
1 2.7114 1.7947 1.8926 0
Other closed-loop poles for K=4.3
1 2.7114 1.7947 1.893
-2.2886 -0.9676 -1.893
1 0.4228 0.8270 0
s
3,4
=-0.2114±j0.8814
0
0
0
0
0
180 360 180 360
( ) 4
45 0
135 1
225 2
315 3
c
c
c
c
c
k k
n m
k
k
k
k





 
 

 
 
 
 
1 2 1 2
( ... ) ( ... ) 0 3 1 1
1.25
( ) 4
n m
c
p p p z z z j j
n m

         
   


120
Steps 6: Determine angles of departure
0 0 0 0 0 0
180 (135 26.56 90 ) 71.56 288.44
d
       

121
Step 7: As there are no complex open-loop zeros, angle of arrival need not be computed.
4 3 2
( ) 5 8 6
B s s s s s K
    
4 3 2 4 2 3
( ) ( ) 5( ) 8( ) 6 ( 8 ) (6 5 )
B j j j j j K K j
        
         
When imaginary-part is zero, then
6 6
5 5
s j
then
2
6 6
8 8.16
5 5
K
   
   
   
   
.
There are two closed-loop poles on the imaginary axis for any value of K>0.
No. S1 S2 S3,4 K
1 -0.25 -2.9217 -0.9142±0.7969j 1.0742
2 -0.50 -2.8804 -0.8098±0.655j 1.5625
3 -0.75 -2.8593 -0.6953±0.5938j 1.7930
4 -1.0 -2.8393 -0.5804±0.6063j 2.0000
5 -1.25 -2.8055 -0.4722±0.6631j 2.3242
6 -1.75 -2.6562 -0.3763±0.7354j 2.8125
7 -2.0 -2.5214 -0.2393±0.8579j 4.0

122
Additional Information from Root-Locus Plot
6. Gain Margin
2
1
20log
K
GM
K
 (10.69)
K1 is the gain of a feedback system at some point on the root-locus
K2 is the gain at which the system becomes unstable
7. Transient Characteristics
Where,
2
1 1
tan



 

8. Percentage overshoot
/tan
p
M e  

 (10.70)
9. Settling time
4
s
n
t

 (10.71)
10. Steady-state error is also related to K.
Example
by
   
2
4 3 2
10 100
( ) ( ) , 1
20 100 500 1500
K s s
G s H s H s
s s s s
 
 
   
(a) Determine the value of gain at which the system will be stable and as well have a maximum
overshoot of 5%.
(b) What is the gain margin at this point?
(c) What is the steady-state error for a unit step excitation at the above point?
Solution:

123
(b)
0
tan 1.0487
ln
46
p
M




 
 
2
1
0.690
1 tan


 

(10.72)
(b)
192.2
20log 2.65
261
GM dB
  
(c) Position error
2
4 3 2
0
( 10 100) 100
20 100 500 1500 1500
lim
s
s
K s s K
K
s s s s

 
 
   
Steady-state error,
1 1 1500
( )
1 1 100 /1500 1500 100
e
s
S
K K K
   
  
1500
( ) 5.4%
1500 100 261
e
S   
 
a. Root Locus using MATLAB
Program 1:Draw the root locus for the following system
 
      
1 2 4 5
C s K
R s s s s s

   
Solution:
>> num=[01]

124
num=
0 1
>>q1=[1 1];
>> q2=[1 2];
>> q3=[1 3];
>> q4=[1 4];
>>den=conv(q1,q2);
>> den=conv(den,q3);
den=
1 12 49 78 40
>>sys=tf(num,den)
Transfer function:
1
--------------------
s^4+12s^3+49s^2+78s+40
>>rlocus(sys)

125
11. Frequency Response Analysis
11.1. Frequency Response
This is defined as the steady-state response of a system due to a sinusoidal input.
Here,
 
 
 
 
   ...
C s N s
G s
R s s a s b s c
 
  
(11.1)
 
   
   ...
N s R s
C s
s a s b s c
 
  
(11.2)
Let,   sin
r t A t

 , then
  2 2
A
R s
s




(11.3)
Using eq (3) in eq (2),
 
 
   
 
2 2
1 1 1 1 2
...
...
N s A
C s
s a s b s c s
A A A B B
C s
s a s b s c s j s j


 
 
  
   
 
      
    
(11.4)
In time domain, eq (5) becomes
  1 2 3 1 2
...
at bt ct j t j t
c t A e A e A e B e B e
 
   
      (11.5)
The term with i
A terms are decaying components. So, they tend to zero as time tends to infinity.
Then, eq (5) becomes
  1 2
j t j t
ss
C t B e B e
 

  (11.6)
Where,
 
   
 
   
1
2
2
2
j G j
s j
j G j
s j
A G s A
B G j e
s j j
A G s A
B G j e
s j j










 




  

 

(11.7)
Since,    
G j G j
 
  and    
G j G j
  
    

126
     
   
2 2
j t j t
A A
c t G j e G j e
j j
   
 
  

  (11.8)
   
2
j j
j t e e
c t A G j e
j
 



  

    
 
(11.9)
     
sin
c t A G j t
  
   (11.10)
     
sin
c t B t
  
   (11.11)
Where,    
B A G j
 

Therefore, the steady-state response of the system for a sinusoidal input of magnitude A and
frequency  is a sinusoidal output with a magnitude  
B  , frequency  and phase shift  .
The following plots are used in frequency response.
 Polar plot
 Bode plot
 Magnitude versus phase angle plot
11.2. Definition of frequency domain specifications
(i) Resonant peak  
r
M : Maximum value of  
M j when  is varied from 0 to ∞.
(ii) Resonant frequency  
r
 : The frequency at which r
M occurs
(iii)Cut-off frequency  
c
 : The frequency at which  
M j has a value
1
2
. It is the frequency
at which the magnitude is 3dB below its zero frequency value
(iv) Band-width  
b
 : It is the range of frequencies in which the magnitude of a closed-loop
system is
1
2
times of r
M

127
(v) Phase cross-over frequency: The frequency at which phase plot crosses -1800
(vi) Gain margin (GM): It is the increase in open-loop gain in dB required to drive the closed-loop
system to the verge of instability
(vii) Gain cross-over frequency: The frequency at which gain or magnitude plot crosses 0dB line
(viii) Phase margin (PM): It is the increase in open-loop phase shift in degree required to
drive the closed-loop system to the verge of instability
11.3. Correlation between time and frequency response
For a second order system
 
 
2
2 2
2
n
n n
C s
R s s s

 

 
(11.12)
Putting s j

 
 
 
 
2
2 2
2
2
2
1
1 2
n
n n
n
n
C j
R j j
C j
R j
j
 
    

  




 
 
   
 
   
 
 
(11.13)
Let,
n
u


 , then
 
   
2
1
1 2
C j
R j u j u

 

 
(11.14)
Now,
     
M j M j M j
  
  (11.15)
Where,
 
   
2 2
2
1
2
1
1 2
2
tan
1
M j
u u
u
u



 

 
 
   

 
(11.16)
Now,
2
1
2 1
r
M
 


(11.17)
2
1 2
r n
  
  (11.18)

128
2 4 2
1 2 4 4 2
b n
    
     (11.19)
0
180
PM 
   (11.20)
Where, 1
2 2
2
tan
4 1 2


 


 
11.4. Advantages
 Good accuracy
 Possible to test in lab
 Can be used to obtain transfer function that is not possible with analytical techniques
 Easy to design open-loop transfer function from closed-loop performance in frequency
domain
 It is very easy to visualize the effect of disturbance and parameter variations.
11.5. Disadvantages
 Applied only to linear systems
 Frequency response for existing system is possible to obtain if the time constant is up to few
minutes
 Time consuming procedure
 Old and back dated method
Equation Chapter 12 Section 1

129
12. Bode Plots
12.1. Magnitude plot and phase plot on a semi-log paper
Magnitude plot on a semi-log paper
20log | ( ) ( ) |
M G j H j
 
 dB
Phase plot on a semi-log paper

130
12.2. Magnitude versus phase Bode plot Nichols plot
Table 12.1 Basic frequency response factors
No Laplace term Frequency response Type of factor
1 K K Constant
2 s j Derivative factor
3 1/s 1/ j Integral factor
4 s+1 (1+ j) First order derivative factor
5 1/(s+1) 1/(1+ j) First order integral factor
6 2 2
2 n n
s s
 
  2 2
2
n n
j
   
  Second order derivative factor
7 2 2
1
2 n n
s s
 
  2 2
1
2
n n
j
   
 
Second order integral factor

131
12.3. Derivative factor: magnitude
20log 20log
M j 
  dB (12.1)
0
90
j
  (12.2)
2
2 1
1
20log 20log 20log
M

 

    dB/decade (12.3)
20log10 20
M
   dB/decade (12.4)
20log 2 6
M
   dB/octave (12.5)
Table 12.2 Magnitude variation of a derivative factor for various multiples of the initial
frequency
2
1


1 2 3 4 5 6 7 8 9 10
M
 dB 0 6 10 12 14 16 17 18 19 20

132
12.4. Derivative Factor: (phase)
Table 15.3Derivative factor
Frequency, rad/s
0.1 1 10 30 100
Magnitude, dB -20 0 20 30 40
Phase, degrees 90 90 90 90 90
12.5. Integral factor: magnitude
1
20log 20log
M
j


   dB (12.6)
0
270
j
  (12.7)
2
2 1
1
20log 20log 20log
M

 

      dB/decade (12.8)
20log10 20
M
     dB/decade (12.9)
20log2 6
M
    dB/octave (12.10)

133
Table 12.4Magnitude variation of an integral factor for various multiples of the initial frequency
2
1


1 2 3 4 5 6 7 8 9 10
M
 , dB 0 -6 -10 -12 -14 -16 -17 -18 -19 -20
12.6. Integral factor: phase

134
Table 12.5Bode magnitude and phase of an integral factor
Frequency, rad/s
0.1 1 10 20 100
Magnitude, dB 20 0 -20 -26 -40
Phase, degrees 270 270 270 270 270
12.7. First-order derivative factor: magnitude
 
2
20log 1 20log( 1
M j 
    dB (12.11)
For <<
c
, M≈0 dB
For >>
c
,
20log
c
M


 dB (12.12)
Here, 
c
=1/ = corner frequency
For >
c
2
2 1
1
20log 20log 20log
M

 

    (12.13)
20log10 20
M
   dB/decade (12.14)
20log 2 6
M
   dB/octave (12.15)
Table 12.6Magnitude variation of a first-order derivative factor for various multiples of the corner
frequency
c


1 2 3 4 5 6 7 8 9 10
M
 ,
dB
0 6 10 12 14 16 17 18 19 20
12.8. First-order derivative factor: phase
 
1 arctan
j
  
    (12.16)
0
45 1 log
0 ;
10
1
;
;
0
10
90 10
c
c
c
c
c
w
w
w
w w
w w





 
 

 
 
 

 
(12.17)

135
Table 12.7Phase angles of a first-order derivative factor around the corner frequency
c


1 2 3 4 5 6 7 8 9 10
 , deg 45 59 66 72 76 80 83 86 88 90
c


0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
 , deg 0 2 4 7 10 14 18 24 31 45
12.9. First-order derivative factor
For 1
 
Table 15.8Bode magnitude and phase
Frequency, rad/s
0.1 1 5 10 20 100
Magnitude, dB 0 3 14 20 26 40
Phase, degrees 0 45 76 90 90 90
First-order derivative factor: magnitude (3 dB correction at the corner frequency)

136
First-order derivative Factor: phase
12.10. First-order integral factor: magnitude
 
2
1 1
20log 20log
1 1
M
j 
 
 
 
 
  

 
dB (12.18)
20l g
0,
,
o
c
c
c
M w w
M dB w w





 

(12.19)
2
2 1
1
20log 20log 20log
M

 

      dB/decade (12.20)
20log 2 6
M
     dB/octave (12.21)
Table 12.9Magnitude variation of a first-order integral factor for various multiples of the corner
frequency
c


1 2 3 4 5 6 7 8 9 10
M
 ,
dB
0 -6 -10 -12 -14 -16 -17 -18 -19 -20

137
Table 12.10Phase angles of a first-order integral factor around the corner frequency
c


1 2 3 4 5 6 7 8 9 10
 deg 315 301 294 288 284 280 277 274 272 270
c


0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
 deg 360 358 356 353 350 346 342 336 329 315
First-order integral factor: phase
=360, <c /10
0
360 45 1 log
c



 
  
 
 
, c/10<<10 c
0
360 45 1 log
c



 
  
 
 
=270, >10 c
Table 12.11Bode magnitude and phase of a first-order integral factor
Frequency, rad/s
0.01 0.1 0.7 1 7 10 20 100
Magnitude, dB 0 0 -2 -3 -17 -20 -26 -40
Phase, degrees 360 360 322 315 277 270 270 270
First-order integral factor: magnitude

138
First-order integral factor: phase
12.12. Second-order derivative factor: magnitude
2 2
2 2
2
2
2
20log | 2 |
20log 1 2
n n
n
n n
M j
  
 
 
 
  
 
   
 
  
   
 
   
 
(12.22)
,
,
( )
,
 
 
 
n n
2
n n
n
M 40logw w w
M 20log 2zw w w
M 40 logw w w
(12.23)
For n
w w


139
2
2 1
1
40log 40log 40log
M

 

    dB/decade (12.24)
40log10 40
M
   dB/decade (12.25)
40log 2 12
M
   dB/octave (12.26)
Magnitude variation of a second-order derivative factor for various multiples of the resonant
frequency
n


1 2 3 4 5 6 7 8 9 10
M
 dB 0 12 20 24 28 32 34 36 38 40
Second-order derivative factor: phase
2 2
2
2
2
| 2 | arctan
1
n
n n
n
j



   


 
 
 
    
 

 
 
(12.27)
0
0
0
0 ,
10
90 ,
180 , 10
n
n
n
w
w
w w
w w



 
 
 
(12.28)
Bode magnitude and phase
ωn=1 rad/s, ζ=0.3
Frequency, rad/s 0.01 0.1 0.7 1 3 10 100
Magnitude, dB 0 0 -4 -4 18 40 80
Phase, degrees 0 0 39 90 167 180 180

140
Second-order integral factor
2 2 2 2
2
2
2
1 1
20log 20log
2
1 2
n n
n
n n
M dB dB
j
    
 
 
 
 
 
   
   
   
 
 
   
 
   
 
(12.29)
M≈ - 40log n, <<n
M=-20log (2n
2
), = n
M= - 40 log , >>n
2
2 1
1
40log 40log 40log /
M dB decade

 

      (12.30)

141
40log10 40
M dB dB
     (12.31)
Magnitude variation of a second-order integral factor for various multiples of the resonant frequency
n


1 2 3 4 5 6 7 8 9 10
M

, dB
0 -12 -20 -24 -28 -32 -34 -36 -38 -40
2
2 2
2
2
1
360 arctan
2
1
n
n n
n
j





  

 
 
 
   
 
 

 
 
(12.32)
=0, <n
=2700
, =n
=1800
, >n
Bode magnitude and phase
Frequency, rad/s 0.01 0.1 0.7 1 3 10 100
Magnitude, dB 0 0 4 4 -18 -40 -80
Phase, degrees 360 360 321 270 193 180 180
Magnitude plot

142
Phase plot
Example 14.1
Draw the Bode magnitude and phase plot of the following open-loop transfer function and determine
gain margin, phase margin and absolute stability?
1
( ) ( )
( 1)
G s H s
s s


Solution
Applying s j
 ,

143
1
( ) ( )
( 1)
G j H j
j j
 
 


The above frequency response function has two factors: (1) Integral factor and (2) First order integral
factor with a corner frequency of 1 rad/s
Bode magnitude of the transfer function
Frequency, radians/s
0.01 0.1 1 10 100
1
20log
j
dB
40 20 0 -20 -40
1
20log
1
j 
dB
0 0 -3 -20 -40
Magnitude, dB 40 20 -3 -40 -80
p
= 100 rad/s
Frequency, rad/s
0.01 0.1 1 10 100
1
j
 degrees 270 270 270 270 270
1
1
j


degrees 360 360 315 270 270
Bode phase, degrees 270 270 225 180 180

145
Example 14.2
 
1
( ) ( )
( 2) 4)
G s H s
s s s

 
Solution
1
( ) ( )
8 1 1
2 4
G j H j
j j
j
 
 


  
 
  
  
The corner frequencies corresponding to first order integral factors are 2 rad/s and 4 rad/s. Minimum
frequency is chosen as 0.01 rad/s and maximum frequency 100 rad/s.
Table 14.1 Computation of Bode magnitude using asymptotic properties of the integral first-order
term
1
2
 
x1 x2 x1 x10 x2 x1 x1 x2 x1 x10
Frequency, rad/s 2 4 2 20 20 10 20 40 10 100
Magnitude, dB 0 -6 0 -20 -20 -14 -20 -26 -14 -34
Table 14.2 Computation of Bode magnitude using asymptotic properties of the integral first-order
term
1
4
 
x1 x10 x2 x1 x2 x1 x1 x10
Frequency, rad/s 4 40 40 20 20 10 10 100
Magnitude, dB 0 -20 -20 -14 -14 -8 -8 -28
Table 12.3 Bode magnitude
Frequency, rad/s
Factor 0.01 0.1 0.2 0.4 1 2 4 10 20 40 100
1
20log
8
-18 -18 -18 -18 -18 -18 -18 -18 -18 -18 -18
40 20 14 8 0 -6 -12 -20 -26 -32 -40
1
2
1
log
20


j
0 0 0 0 -1 -3 -6 -14 -20 -26 -34
1
20log
1
4
j

0 0 0 0 0 -1 -3 -8 -14 -20 -28
Bode
magnitude,
22 2 -4 -10 -18 -28 -39 -60 -78 -96 -120

j
1
log
20

146
dB
Bode magnitude
Bode phase
Frequency, rad/s
Factor 0.01 0.1 0.2 0.4 1 2 4 10 20 40 100
1
8

0 0 0 0 0 0 0 0 0 0 0
1
j

270 270 270 270 270 270 270 270 270 270 270
1
1
2
j


360 360 360 346 328 315 301 284 270 270 270
1
1
4
j


360 360 360 360 342 326 315 297 285 270 270
Phase
degrees
270 270 270 256 220 191 166 131 105 90 90
Phase plot

148
Bode plot
Example 12.1
2
1
( ) ( )
( 1)
G s H s
s s


Solution
1
( ) ( )
( )( )( 1)
G j H j
j j j
 
  


There are two integral factors and an integral first-order term with a corner frequency of 1 rad/s
Bode magnitude
Frequency, rad/s
0.01 0.1 1 10 100
1
20log
j
dB 40 20 0 -20 -40
1
20log
j
dB 40 20 0 -20 -40

149
1
20log
1
j 
dB 0 0 -3 -20 -40
Bode magnitude, dB 80 40 -3 -60 -120
Example 12.2
4 3 2
1
( ) ( )
5 8 6
G s H s
s s s s

  
Solution
2
1
( ) ( )
( 2 2)( 3)
G s H s
s s s s

  
 
2
1
( ) ( )
( ) 2( ) 2 (( ) 3)
G j H j
j j j j
 
   

  
 
2
1
3
( ) ( )
(2 ) 2 ) 2 ( 1)
3
G j H j
j j j
 

  

   
Comparing the second order term with a standard second order term,
2 2
2
n n
j
  
 
2
n
  and
1
2
  .
For the first order integral factor, c=3 rad/s
For ζ> 0.5, the response at resonance is less than theresponse at frequencies less than the resonant
frequencies
Table Computation of Bode magnitude using asymptotic properties of the integral second-order term
x1 x10 x1 x2 x3 x1 x1 x10 x3 x1
Frequency, rad/s 1.4 14 14 30 30 10 10 100 30 3
Magnitude, dB -6 -46 -46 -58
-
58
-
38
-38 -78 -58 -18
Table Computation of Bode magnitude using asymptotic properties of the integral first-order term
x1 x3 x2 x1 x3 x1 x1 x10
Frequency, rad/s 3 30 30 14 30 10 10 100
Magnitude, dB 0 -20 -20 -14 -20 -10 -10 -30

150
Bode magnitude
Frequency, rad/s
n c
0.01 0.1 0.14 0.3 1 3 10 14 30 100
1
20log
3
-10 -10 -10 -10 -10 -10 -10 -10 -10 -10 -10
1
20log
j
40 20 17 10 0 -3 -10 -20 -23 -30 -40
 
2
1
20log
(2 ) (2 )
j
 
 
-6 -6 -6 -6 -6 -9 -18 -38 -46 -58 -78
1
20log
1
3
j


0 0 0 0 0 -1 -3 -10 -14 -20 -30
Bode magnitude, dB 24 4 1 -6 -16 -23 -41 -78 -93 -118 -158
2

151
Bode phase
Frequency, rad/s
n c
0.01 0.1 0.14 0.3 1 3 10 14 30 100
1
3

0 0 0 0 0 0 0 0 0 0 0
1
j
 degrees
270 270 270 270 270 270 270 270 270 270 270
 
2
1
(2 ) (2 )
j
 

 
degrees
360 360 360 343 297 270 221 192 180 180 180
1
1
3
j



, degrees
360 360 360 360 336 330 315 291 285 270 270
Bode phase, degrees 270 270 250 253 183 150 86 33 15 0 0
2

152
Nichols plot
13. Polar Plots
It is a graphical method of determining stability of feedback control systems by using the polar plot of
their open-loop transfer functions.
Example 13.1
Draw a polar plot of the open-loop transfer functionfor
( ) ( )
( 1)
K
G s H s
s s


(14.33)
Frequency response
( ) ( )
( 1)
K
G j H j
j j
 
 


(14.34)
Magnitude
2
( ) ( )
1
K
G j H
 
 


(14.35)
Angle
1
( ) ( ) tan
2
G j H j

  

    (14.36)
0 0
270 ( ) ( ) 180
G j H j
 
   (14.37)
Magnitude and phaseof the open-loopfrequency transfer function
No.
Frequency,
rad/s
Magnitude
Phase,
degrees
1 0 ∞ 270

153
2 0.2 4.9029 259
3 0.4 2.3212 248
4 0.8 0.9761 231
5 1 0.7071 225
6 4 0.0606 194
7 10 0.01 186
8 50 0.0004 181
9 100 0.0001 181
10 200 ≈0 ≈180
Polar plot of the transfer function
 
1
K
s s 
and K=1
Example 14.2
Draw a polar plot of the open-loop transfer functionfor K=1, 10, 25, 55
( 2)( 4)
K
GH
s s s

 
Solution
Frequency response
( ) ( )
( 2)( 4)
K
G j H j
j j j
 
  

 
ω=0.2
ω=∞

154
Magnitude
2 2
( ) ( )
4 16
K
G j H j
 
  

 
Angle
1 1
( ) ( ) tan tan
2 2 4
G j H j
  
   
    
The lies in II and III quadrants as 0 0
90 ( ) ( ) 270
G j H j
 
  
Magnitude and phase of the open-loop frequencytransfer function (K=1)
No. Frequency,
rad/s
Magnitude Phase,
degrees
1 0.1 1.2481 266
2 0.2 0.6211 261
4 0.4 0.3049 253
5 0.8 0.1423 237
6 1 0.1085 229
7 4 0.0099 162
8 10 0.0009 123
9 50 0 97
( 2)( 4)
K
GH
s s s

 
for K=1, 10, 25, 55
Example 14.3
Draw a polar plot of the open-loop transfer function 2
( ) ( )
( 1)
K
G s H s
s s



155
Solution
Frequency response
2
( ) ( )
( ) ( 1)
K
G j H j
j j
 
 


Magnitude
2 2
( ) ( )
1
K
G j H j
 
 


Angle
0 1
( ) ( ) 180 tan
G j H j
  

   
The lies in II quadrant only as 0 0
90 ( ) ( ) 180
G j H j
 
  
Magnitude and phase of the open-loop frequency transfer function (K=1)
No.
Frequency,
rad/s
Magnitude
Phase,
degrees
1 0.4 5.803 158
2 0.5 3.5777 153
4 0.8 1.2201 141
5 1 0.7071 135
6 2 0.1118 117
7 3 0.0351 108
8 4 0.0152 104
9 5 0.0078 101
( 2)( 4)
K
GH
s s s

 
for K=1, 10, 25, 55

156
Bode plot using MATLAB
Program 1: Sketch the bode plot for the open loop transfer function
 
    
40
1 4
C s
R s s s s

 
.
Determine the gain margin, phase margin, gain cross over frequency and phase cross over frequency.
Solution:
>> num=[0 40]
num=
0 40
>> q1=[1 0];
>> q2=[1 1];
>> q3=[1 4];
>> den=conv(q1,q2);
den=
1 9 24 16 0
>>sys=tf(num,den)
Transfer function:
1
--------------------
s^4+9s^3+24s^2+16s
>>bode(sys)
>>margin(sys)

157
14. Nyquist plot
14.1.Definition
Nyquist criterion is a graphical method of determining stability of feedback control systems
by using the Nyquist plot of their open-loop transfer functions.
14.2.Theory
Feedback transfer function
( ) ( )
( ) 1 ( ) ( )
C s G s
R s G s H s


(14.1)
Poles and zeros of theopen-loop transfer function
1 2
1 2
( )( )...( )
( ) ( )
( )( )...( )
m
n
K s z s z s z
G s H s
s p s p s p
  

  
(14.2)
1 2 1 2
1 2
( )( )...( ) ( )( )...( )
1 ( ) ( )
( )( )...( )
n m
n
G s H s
s p s p s p
      
 
  
(14.3)
Number of closed-loop poles - Number of zeros of 1+GH = N umber of open-loop poles
1 2
1 2
( )( )( )
1 ( ) ( )
( )( )...( )
n
c c c
n
s z s z s z
G s H s
s p s p s p
  
 
  
(14.4)
1 2
, ...
n
c c c
z z z = zeros of 1+G(s)H(s)
These are also poles of the close-loop transfer function
Magnitude
1 2
1 2
...
1 ( ) ( )
( ) ( ) ... ( ) .
n
c c c
n
s z s z s z
G s H s
s p s p s p
  
 
  
(14.5)
Angle
1 2
1 2
1 ( ) ( )
( ) ( ) ( )
n
c c c
n
s z s z s z
G s H s
s p s p s p
     
  
     
(14.6)
The s-plane to 1+GH plane mappingphase angle of the 1+G(s)H(s) vector, corresponding to a point on
the s-plane is the difference between the sum of the phase of all vectors drawn from zeros of
1+GH(close loop poles) and open loops on the s plane. If this point s is moved along a closed contour
enclosing any or all of theabove zeros and poles, only the phase of the vector of each of the enclosed
zeros or open-loop poles will change by 3600
. The directionwill be in the same sense of the contour
enclosing zeros and in the opposite sense for the contour enclosing open-loop poles.

158
14.3.Principle of argument
When a closed contour in the s-plane encloses a certain number of poles and zeros of 1+G(s)H(s) in
the clockwise direction, the number ofencirclements of the origin by the corresponding contour in the
G(s)H(s)plane will encircle the point (-1,0) a number of times given by thedifference between the
number of its zeros and poles of 1+G(s)H(s) it enclosed on the s-plane.

159
Modified contour on the s-plane forchecking the existence of closed-looppoles
j
s e 


Magnitude of GH remains the same alongthe contourPhase of β changes from 270 to 90 degrees
14.4.Gain Margin and Phase Margin
Phase crossover frequency p
 is the frequency at which the open-loop transfer function has a phase of
1800
. The gain crossover frequency g
 is the frequency at whichthe open-loop transfer function has a
unit gain
Gain margin
20log ( ) ( )
p p
M G j H j
 
  (14.7)
Phase margin
( ) ( ) 180o
g g
G j H j
  
   (14.8)
GH from the polar plot
GH from the mirror image of the
polar plot
Magnitude zero since n >m

160
14.5.Procedure
(1) Locate open-loop poles on the s-plane
(2) Draw the closed contour and avoid open-loop poles on the imaginary axis
(3) Count the number of open-loop poles enclosed in the above contour of step 2, say P
(4) Plot G(j)H(j) and its reflection on the GH plane and map part of the small semi-circle
detour on the s-plane around poles (if any) on the imaginary axis.
(5) Once the entire s-plane contour is mapped on to the GH plane, count the number of
encirclements of the point (-1,0) and its direction. Clockwise encirclement is considered
positive, say N.
(6) The number of closed-loop poles in the right-half s-plane is given by Z=N+P. if Z >0, the
system is unstable.
(7) Determine gain margin, phase margin, and critical value of open-loop gain.

161
Example 14.1
Using Nyquist criterion, determine the stability of a feedback systemwhose open-loop transfer
function is given by
( ) ( )
( 1)
K
G s H s
s s


Solution
Step 1Locate open-loop poles on the s-plane. Open-loop poles are at s=0 and –1. Let K=1
Step 2 Draw the closed contour on the s-plane to check the existenceof closed-loop poles in the right-
half s-plane.
Open-loop poles and s-plane contour
2
1
( ) ( )
1
G j H
 
 


1
( ) ( ) tan
2
G j H j

  

   
No. Frequency,
rad/s
Magnitude Phase,
degrees , s-plane, deg
, GH plane,
deg
1 0.2 Positive
frequencies
4.9029 259 270 101
2 0.4 2.3212 248 280 91
3 0.8 0.9761 231 290 80
4 1 0.7071 225 300 69
5 4 0.0606 194 310 58
6 10 0.01 186 320 46
7 50 0.0004 181 330 35
8 100 0.0001 181 340 23

162
9 200 0 180 350 12
10 -200 Negative
frequencies
0 180 0 0
11 -100 0.0001 179 10 348
12 -50 0.0004 179 20 337
13 -10 0.01 174 30 325
14 -4 0.0606 166 40 314
15 -1 0.7071 135 50 302
16 -0.8 0.9761 129 60 291
17 -0.4 2.3212 112 70 280
18 -0.2 4.9029 101 80 269
The above system is stable. Here, phase crossover frequency is very large (infinity) and gain
crossover frequency 0.786 rad/s. Phase angle corresponding to gain crossover frequency= 2320
and
Phase margin is 52
o
Example 14.2.
Using Nyquist criterion, determine the stability of a feedback systemwhose open-loop transfer
function is given by
55
( ) ( )
( 2)( 4)
G s H s
s s s

 
Solution
Step 1Locate open-loop poles on the s-plane. Open-loop poles are at s=0, -2 and –4. Let K=1
Step 2 Draw the closed contour on the s-plane to check the existenceof closed-loop poles in the right-
half s-plane.
Open-loop poles and s-plane contour

163
The number of open-loop pole enclosed, P is zero
2 2
( ) ( )
4 16
K
G j H j
 
  

 
1 1
( ) ( ) tan tan
2 2 4
G j H j
  
   
    
No.
Frequency Magnitude
Phase,
degrees
, s-plane,
deg
1 1.5 Positive
frequencies
3.4332 213 270
2 2 2.1741 198 280
3 2.5 1.4568 187 290
4 2.83 1.1446 180 300
5 3 1.017 177 310
6 3.5 0.7334 169 320
7 4.5 0.4122 156 330
8 5 0.319 150 340
9 5.5 0.2513 146 350
10 6 0.201 142 0
11 7 0.1339 136 10
12 8 0.0932 131 20
13 9 0.0673 126 30
14 -9 Negative
frequencies
0.0673 234 40
15 -8 0.0932 229 50
16 -7 0.1339 224 60

164
17 -6 0.201 218 70
18 -5.5 0.2513 214 80
19 -5 0.319 210 90
20 -4.5 0.4122 204 0
21 -3.5 0.7334 191 343
22 -3 1.017 183 326
23 -2.83 1.1446 180 309
24 -2.5 1.4568 173 292
25 -2 2.1741 162 276
26 -1.5 3.4332 147 259
Here, Z=N+P=2.
Hence, the above system is unstable.
Again,
Phase crossover frequency 2.83 rad/s
The gain at which the system becomes marginally stable, *
55 /1.1446 48
K  
Gain margin
20log ( ) ( )
20log 1.1446 1.17dB
p p
M G j H j
 
 
   
Gain crossover frequency =3 rad/s and the corresponding angle of GH=177
o
Phase margin=177-180=-3
o

165
Nyquist plot using MATLAB
Program 1:Sketch the nyquist plot for the open loop transfer function
 
    
40
1 4
C s
R s s s s

 
.
Solution:
>> num=[040]
num=
0 40
>> q1=[1 0];
>> q2=[1 1];
>> q3=[1 4];
>> den=conv(q1,q2);
den=
1 9 24 16 0
>>sys=tf(num,den)
Transfer function:
1
--------------------
s^4+9s^3+24s^2+16s
>>nyquist(sys)

166
MODULE#4

167
15. CLOSED LOOP FREQUENCY RESPONSE
15.1. Peak Magnitude
( )
20log dB
( )
r
C j
M
R j


 (16.1)
3 dB is considered good
15.2. Constant M-circles for unity feedback systems
( )
( )
1 ( )
G j
M j
G j





(16.2)
( )
G j x jy
   (16.3)
2 2
2 2
( )
(1 )
x y
M j
x y



 
(16.4)
2 2 2 2 2 2
(1 )
M x M y x y
    (16.5)
2 2 2 2 2 2
(1 ) (1 ) 2
x M M y M x M
     (16.6)
2 2
2 2
2 2
2
1 1
M M
x y x
M M
  
 
(16.7)
Adding
2
2
2
1
M
M
 
 

 
in both sides, we get
2 2
2
2
2 2
1 1
M M
x y
M M
   
  
   
 
 
 
(16.8)
The above equation represents a family of circles with its center at
2
2
,0
1
M
M
 
 

 
and radius 2
1
M
M

.
15.3. Family of M-circles
Family of M-circles corresponding to the closeloop magnitudes (M) of aunit feedback system
Constant M-circles for unityfeedback systems

168
15.4. Constant N-circles
( )
1 ( )
G j
M
G j




  
 
(16.9)
1 1
tan tan
1
y y
x x
  
 

(16.10)
1 1
tan tan tan
1
y y
N
x x
 
 
 
 

 
(16.11)
tan tan
tan( )
1 tan tan
A B
A B
A B

 

(16.12)
Here, tan()=N
2 2
y
N
x x y

 
(16.13)
2 2 2
1 1 1 1
2 2 4 2
x y
N N
     
    
     
     
(16.14)
The above equation represents a family of circles with its center at (
2
1
,
N
2
1
) and radius
2
1 1
4 2N
 
  
 

169
Example 15.1.
Determine the closed-loop magnitude ratio and bandwidth of the feedback system whose forward
transfer function is given by
10
( )
( 2)( 4)
G s
s s s

 
and H(s)=1, by (1) direct computation and (2)
using M and N circles.
Solution

170
Closed-loopmagnitude and phase values
Frequency,
rad/s
Closed-loop
Magnitude
ratio
Closed-loop
Phase angle, deg
0.1 1.0 355
0.5 1.1 335
0.8 1.2 316
0.9 1.2 308
1.0 1.2 300
1.1 1.3 290
1.2 1.3 280
1.3 1.2 269
1.4 1.2 258
1.5 1.1 248
1.6 1.0 238
1.7 0.9 230
1.8 0.8 222
1.9 0.7 216
2.0 0.6 210

171
15.5. Nichols chart for closed-loop response
( ) ( )
( ) 1 ( ) ( )
C j G j
R j G j H j
 
  


(16.15)
( )
20log dB
( )
r
C j
M
R j


 (16.16)
( )
( )
C j
R j



  (16.17)

172
Closed-loopconstant magnitude (dB)and phase angle(degrees)lines of a unity feedbackfunction
Example 15.2.
Determine the closed-loop magnitude ratio and bandwidth of the feedback system whose forward
transfer function is given by
10
( )
( 2)( 4)
G s
s s s

 
and H(s)=1, by (1) direct computation and (2)
using Nichols chart.
Solution

173
Closed-loop magnitude and phase values
Frequency, rad/s
Closed-loop Magnitude
ratio, dB
Closed-loop Phase
angle, deg
0.1 0.02 355
0.2 0.10 351
0.3 0.22 346
0.4 0.38 341
0.5 0.59 335
0.6 0.84 330
0.7 1.11 323
0.8 1.39 316
0.9 1.66 308
1.0 1.87 300
1.1 1.99 290
1.2 1.95 280

174
1.3 1.72 269
1.4 1.27 258
1.5 0.62 248
1.6 -0.19 238
1.7 -1.12 230
1.8 -2.11 222
1.9 -3.13 216
2.0 -4.15 210
Open-loop magnitude and phase values
Frequency,
rad/s
open-loop
magnitude, dB
open-loop phase
angle, deg
0.1 22 266
0.2 16 261
0.3 12 257
0.4 10 253
0.5 8 249
0.6 6 245
0.7 4 241
0.8 3 237
0.9 2 233
1.0 1 229
1.1 0 226
1.2 -1 222
1.3 -2 219
1.4 -3 216
1.5 -4 213
1.6 -5 210

175
1.7 -6 207
1.8 -7 204
1.9 -7 201
2.0 -8 198
2.1 -9 196
2.2 -10 193
2.3 -10 191
2.4 -11 189

176
16. Controllers
16.1. Basic Control Action and response of Control systems
An automatic controller compares the actual value of the plant output with the reference input
(desired value), determines the deviation, and produces a control signal that will reduce the
deviation to zero or to a small value. The manner in which the automatic controller produces the
control signal is called the control action. Fig.1 is a block diagram of an industrial control
system, which consists of an automatic controller, an actuator, a plant and a sensor (measuring
element). The controller detects the actuating error signal, which is usually at a low power level,
and amplifies it to a sufficiently high level. The output of the controller is fed to an actuator
such as pneumatic motor or valve, hydraulic motor or electric motor. The actuator is the device
that produces the input to the plant according to the control signal so that the output signal will
approach the reference input signal.
The sensor or measuring element is device that converts the output variable into another
suitable variable such as a displacement, pressure or voltage that can be used to compare the
output to the reference input signal. This element is in the feedback path of the closed-loop
system. The set point of the controller must be converted to a reference input with the same
units as feedback signal from sensor.
Fig.16.1. Basic Control Action and response of Control systems
16.2. APPLICATION OF CONTROL THEORY TO NON-ENGINEERING FIELDS
Engineering is concerned with understanding and controlling the materials and forces of
nature for the benefit of human kind. Control system engineers are concerned with
understanding and controlling segments of their environment often termed SYSTEMS to
provide useful economic products for society. There has been considerable interest recently in
applying the feedback control concepts to processes prevalent in the social, economic and
political spheres. Some of the examples with block diagram models are discussed.
16.3. ECONOMIC INFLATION PROBLEM
A model of the vicious price-wage inflationary cycle, assuming simple relationship between
wages, product costs and cost living is shown in Fig.1. The economic system depicted in this
figure is found to be a positive feedback system.
Ref I/P Amplifier Actuator Plant
Sensor
Error Detector
Output

177
Fig.16.2. ECONOMIC INFLATION DYNAMICS
16.4. POLLUTION CONTROL IN AUTO ENGINE
To meet the emission standards for automobiles, Hydrocarbons (HC), Carbon monoxide (CO), and
Nitrogen oxides (NOx) emissions can be controlled by employing a three way catalyst in conjunction
with a closed loop engine control system as shown in Fig.2. The exhaust gas sensor gives an
indication of a rich or lean exhaust and compares it to a reference. The difference signal is processed
by the controller, and the output of the controller modulates the vacuum level in the carburetor to
achieve the best air-fuel ratio for proper operation of the catalytic converter.
Fig.16.3
16.5. CONTROL OF BLOOD PRESSURE WITH ANESTHESIA
Anesthesia is used in surgery to induce unconsciousness. One problem with drug-induced
unconsciousness is large differences in patient responsiveness. Furthermore the patient
response changes during an operation. A model of drug induced Anesthesia Control is shown
in Fig.3. The proxy for unconsciousness is arterial blood pressure.
Fig.16.4
16.6. Types of Controllers
Sensor
2
2
s 
Body dynamics
2 sT
e
s

Controller
2(s+5)
Desired
blood
pressure
Actual
blood
pressu
+
_ Exhaust
Reference
Sensor
Engine
Controller
Carburetor
Three
way
catalytic
converter
Dissatisfactionf
actor
K2
Industry
K1
Present
wages
Initial
wages
Cost of
living
Wages
increment
Product
cost

178
(i) P-controller
(ii) PI-controller
(iii)PD-controller
(iv) PID-controller
P-controller
(a)
(b)
Fig.16.5
Control system with P-controller with inertia load
Fig.16.6
For this system, closed-loop response is

179
2
2 2 2
2
( )
( ) ( )
1
p
p p
p p n
K
K K
C s Js
K
R s Js K J s
Js

  
 

(16.1)
2 2
( ) ( )
( )
p
n
K
C s R s
J s 
 

(16.2)
For step-input,  
1
R s
s

Step response becomes
( ) 1 cos n
c t t

  (16.3)
Where,
p
n
K
J
 
Fig.16.7
Solved problem
1. Consider the unity feedback system of Fig. 16.8. Let Kp=20 and J=50. Determine the equation of
response for a unit step input and determine the steady-state error.

180
Fig.16.8
Solution
2
2
2
( )
( )
1
P
p
p p
K
K
C s Js
K
R s Js K
Js
 


 
2 2
( )
( )
p
n
K
C s
R s J s 


2
rad/s
5
p
n
K
J
  
2
( ) (1 cos )
p
n
n
K
c t t
J


 
2
( ) 1 cos
5
c t t
 
 
 
 
 
2 2
( ) 1 1 cos cos
5 5
e
s t t t
 
   
 
 
 
2. Find the step response of the system shown inFig.16.9.
Fig.16.9
Solution
1
1
p
K
G
s



181
1
1
( ) 1 1
( ) 1 ( ) 2
E s s
R s G s s

 
 
1
1
1 1
( ) ( ) ( )
1 ( ) 2
s
E s R s R s
G s s

  
 
Step response is
 
1
1
( )
2
s
E s
s s

 

1
( 1)
( )
( ) ( 2)
p
K s
U s
R s s



1( ) 1 1
( ) 1 2
p
C s
R s s K s
 
  
 
1
1 2
( ) 1
2
t
e t e
 
 
1
1 2
( ) 1
2
t
c t e
 
Fig.16.10
I-controller

182
(a)
+
-
ei
eo
R1
C2
(b)
Fig.16.11
2
( 1)
i
K
G
s s


2
2
2
( ) 1 1 ( 1)
( ) 1 ( ) 1
1
( 1)
i
E s s s
K
R s G s s s
s s

  
  


2
2
( 1)
( )
( ) 1
i
K s
U s
R s s s


 
2
2 2
( ) 1 1
( ) 1
i
C s
R s s s K s s
 
   
0.5
2
3 1 3
( ) cos sin
2 2
3
t
e t e t t

 
 
 
 
 
0.5
2
1 3 3
( ) 1 sin cos
2 2
3
t
c t e t t

 
 
  
 
 
 

183
Fig.16.12
PD-controller
(a)
C
R
+
-
ei
eo
R1
R2
+
-
R
(b)
Fig.16.13
Control system with P-controller with inertia load
Fig.16.14

184
2
2
2
2
(1 )
(1 ) (1 )
( )
(1 )
( ) (1 )
1
p d
p d p d
p d p d p
p d
K T s
K T s K T s
C s Js
K T s K T K
R s Js K T s
J s s
Js J J

 
  
    
  
 
 
(16.4)
2
(1 )
( ) ( )
p d
p d p
K T s
C s R s
K T K
J s s
J J

 
 
 
 
 
(16.5)
For step-input,  
1
R s
s

Step response becomes
 
2 2
2
2
( ) 1 cos sin
1
1
sin 1
1
n
n
p t
d d
n
p d t
n
n
K
c t e t t
J
K T
e t
J



 
 
 
 


 
 
 
  
 
 
 

 
 
 
 
 
 
 

 
 
(16.6)
Where,
p
n
K
J
 
Solved problem
3. Consider the unity feedback system of Figure 3. Let Kp=20 and J=50. Determine the equation of
response for a unit step input and determine the steady-state error. Here, Kp
=20, Td
=1 and J=50.
Fig.16.14
Solution
2 2 2
( ) 20( 1) 20(1 )
( ) 50 20 20 50( 2 )
n n
C s s s
R s s s s s
 
 
 
   

185
 
2 2
2
2
( ) 1 cos sin
1
1
sin 1
1
n
n
p t
d d
n
p d t
n
n
K
c t e t t
J
K T
e t
J



 
 
 
 


 
 
 
  
 
 
 

 
 
 
 
 
  
 

 
 
2
( ) 1
p
n
K
c
J
  
Transient characteristic
Only PD control
No system damping
Maximum overshoot, %
35.09
unsatisfactory
Rise time tr, sec 3.15
Peak time tp, sec 5.24
Settling time ts, sec (5% criterion) 15
PI-controller
(a)
(b)
Fig.16.15
Control system with PI-controller with inertia load

186
Fig.16.16
1
1 ( 1)
p i
i
p
sK K
K
G K
s s s s

  
  
 
   
 
 
(16.7)
2
1
( ) 1 ( 1)
( ) 1 ( ) (1 )
p i
E s s s
R s G s s s K K

 
   
(16.8)
2
( 1)( )
( )
1
( ) (1 )
p i
p i
s sK K
U s
R s s s K K
 
 
  
(16.9)
2
( )
( ) 1
( ) (1 ) ( 1)
p i
p i
sK K
C s
R s s s K K s

 
   
(16.10)
Step response
( ) 1
u t 
( ) t
e t e

 
( ) 1 t
c t e
 
Fig.16.17

187
Fig.16.18
Fig.16.19
PID-controller
(a)
(b)
Fig.16.20

188
Here, transfer function of PID-controller,
  i
c p d
K
G s K K s
s
   (16.11)
  1 i
c p d
T
G s K T s
s
 
   
 
 
(16.12)
Where,
i
i
p
d
d
p
K
T
K
K
T
K


(16.13)
Tuning of PID-controller
A. First Method (Ziegler and Nichols)
The Setup for obtaining system parameters for PID tuning
Fig.16.21
( )
( ) 1
Ls
C s Ke
U s Ts



(16.14)
( ) ( )
1
Ls
Ke
C s U s
Ts



(16.15)

189
Fig.16.22
Ziegler-Nichols tuning rules based on step response
Type of controller Kp Ki Kd
P T
L
0 0
PI 0.9T
L 0.3
L 0
PID
1.2
T
L
2L 0.5L
1
( ) 1
1.2 1
1 0.5
2
c p d
i
G s K T s
T s
T
Ls
L Ls
 
  
 
 
 
  
 
 
2
1
( ) 0.6
c
s
L
G s T
s
 

 
 

Ziegler-Nichols tuning rule based on critical gain Kcr and critical period Pcr.
Type of controller Kp Ki Kd
P 0.5 Kcr 0 0
PI 0.45 Kcr 1/1.2 Pcr 0
PID 0.6 Kcr 1/0.5Pcr 0.125 Pcr

190
Where, Kcr
proportional constant of a switched-off integral and derivative controls at which sustained
oscillations of period Pcr occur.
Second Method
( ) 1
1
0.6 1 0.125
0.5
i
c p d
cr cr
cr
T
G s K T s
s
K P s
P s
 
  
 
 
 
  
 
 
(16.16)
2
4
( ) 0.075 cr
c cr cr
s
P
G s K P
s
 

 
 
 (16.17)

191
17. Components
17.1. AC SERVOMOTORS
A two phase servomotor (Induction Motor) (A few watts to hundred watts) is commonly used
in feedback control systems. In servo applications, an induction motor is required to produce
rapid accelerations from standstill.
Schematic Diagram
Constructional features
 Squirrel Cage rotor with Cu or Al conductor
 High Rotor resistance
 Small diameter to length ratio to minimize inertia
 Two stator windings in space quadrature(One called reference winding and the other Control
winding)
 The two voltages to stator windings must derived from same source(Or they must be in
synchronism)
Principle of Operation
(i) The two applied AC voltage to stators with a phase difference produce a rotating flux.
(ii) As this moving flux sweeps over the rotor conductors, small emf is induced in rotor. Rotor being
short circuited, currents will flow and this current interacts with rotating flux to produce a torque
in the rotor. This torque causes the rotor to turn so that it chases the rotating magnetic flux.
Torque-Speed Characteristics of AC Servomotor
For induction motor in high power applications, rotor resistance is low in order to obtain maximum
torque. Positive slope part of the characteristics is not suitable to control applications as this results
instability.

192
In AC servomotor high resistance rotor results in negative torque-speed characteristics.This
characteristic is needed for positive damping and good stability. The rotor has a small diameter –to-
length ratio to minimize the moment of inertia and to give a good accelerating characteristic.
However, more rotor resistance results more loss and less efficiency.
Transfer Function
The torque developed is a function of shaft angular position (Ө) and control voltage Ec.
 
 
     
2
1
m m
c m
s K K K
G s
E s s Js D s T s
Js Ds

   
 

Where, m
K
K
D
 = motor gain constant, m
J
T
D
 = motor time constant
Merits of AC Servomotors
(i) Lower cost,
(ii) less weight and inertia,
(iii)higher efficiency and
(iv) fewer maintenance requirements(since no commutator or brush)
Demerits of AC Servomotors
(i) Nonlinear characteristics,
(ii) Used for low power applications(e.g. instrument servo),
(iii)Difficult for speed control and positioning
17.2. Synchros
It is also known as selsyn. It is a self-synchronizing device widely used in servomechanisms as
a position indicator.
Important synchro systems are
 Synchro system with transmitter and control transformer
 Synchro system with synchro transmitter and motor
 Synchro system with transmitter, differential and motor

193
General Constructional features of Synchro
(i) The construction of synchro transmiiter, motor and transformer are almost same.
(ii) Stator laminated silicon steel, slotted to house distributed 3-ϕ,Y-connected
windings with axes 1200
apart.
(iii)Stator not directly connected to supply
(iv) Rotor is 2-pole (dumb-bell shaped for synchro transmitter and cylindrical shape for
control transformer) with single winding connected to AC source. The magnetic
field in excited rotor induces voltages in stator coils. The magnitude of voltage
induced in any stator coil depends on the angular position of coil’s axis with respect
to rotor axis.
(v) Synchro control transformer has cylindrical shape rotor so that air gap flux is
uniformly distributed around the rotor.
Constructional features
(a) Constructional features
(b) Electrical Circuit

194
(c) Schematic Symbol
Fig. Synchro Transmitter
Synchro transmitter
It is not a three phase machine. It is a single phase machine. Here, input is angular position of
its rotor shaft.Output is a set of three stator coil-to-coil voltages. Common connection
between the stator coils is not accessible.
Synchro system with transmitter and control transformer
 A synchro error detector system may consist of synchro transmitter and synchro control
transformer.
 It compares two angular displacements and the output voltage is approximately linear
with angular difference or misalignment between shafts of transmitter & Control
transformer.
 Usedas error detector in feedback systems.
Synchro system with synchro transmitter and synchro motor
The rotors of both the synchro devices are connected to same AC source. Figure (b) shows a
circuit configuration, using two synchros, for maintaining synchronism between two shafts.
When rotor windings are excited, emfs are induced by transformer action in the stator windings
of transmitter and motor. If the two shafts are in similar positions (relative to that of the stator
windings), then there are two emfs of equal value are induced in the two stator windings.Also
no circulating current exists and hence no torque is produced. If the two shaft positions do not
match, the emfs are unequal and result circulating current to flow. The circulating current in
conjunction with air gap magnetic field produce torque which tend to align the shafts.
Synchro system with transmitter, differential and motor
The function of this system is to permit the rotation of a shaft to be a function of sum or
difference of the rotations of two other shafts. The differential has 3-phase distributed windings
on both stator and rotor. The voltages impressed on its stator windings induce corresponding
voltages in its rotor windings.
r=Displacement of receiver shaft

195
s=Displacement of transmitter shaft
d=Displacement of differential shaft
Then, r t d t
  
 
If the phase sequence of stator and rotor windings of differential are reversed then
r t d t
  
 
17.3. TACHOGENERATOR OR TACHOMETER
In many control systems, it is necessary to feedback a voltage proportional to speed of shaft.
Tachogenerator serves the purpose.
17.3.1. DC Tachometer
It is a permanent magnet DC generator. It resembles a small DC machine having a PM
stator, rotating armature, brush and commutator assembly. The rotor is connected to the
shaft to be measured. The output voltage is proportional to the angular velocity of the
shaft. The direction of rotation decides the polarity of the voltage.
DC tachogenerator suffers from the drawback of output in ripple, commutator &
brush problem.
A DC tachometer can be used in AC servomechanism by converting the DC output
voltage to an AC voltage by using a rectifier circuit.

196
17.3.2. AC Tachometer
 Used in AC servomechanism. It resembles 2-phase AC induction motor.
 It comprises two stator windings arranged in space quadrature and a rotor which is
not conductively connected to external circuit.
 One stator phase winding is excited by a suitable AC voltage of constant
magnitude and frequency. A voltage of the same frequency is generated across the
other winding known as control winding.
 It is necessary that the voltage developed across the control winding is linearly
proportional to shaft speed and the phase of this voltage be fixed with respect to
voltage applied to reference winding.
 The output voltage is connected to high impedance circuit of amplifier so that the
winding is considered open circuit.
 An AC tacogenerator should have low inertia when rapid speed variations are
encountered. The drag cup construction gives low inertia and is used many times
17.4. HYDRAULIC OPERATED DEVICES IN FEEDBACK CONTROL SYSTEM
Hydraulic systems must be stable and satisfactory under all operating conditions. Hydraulic
output devices are generally of two types.
A. Hydraulic linear actuator(Produce linear motion)
B. Hydraulic motors(rotary motion)

197
17.5.1 Hydraulic Linear Actuator(Hydraulic Servomotor)
Hydraulic linear actuator consists of pilot valve and a power cylinder. The piston inside
the power cylinder divides the cylinder into two chambers. The pilot valve is known as
spool valve because of its shape control the flow rate of the hydraulic fluid to the power
cylinder. It is a four port valves. It is connected to fluid supply at constant pressure. The
two ports connected to each chamber of power cylinder. One drain port is connected to
reservoir.
Principle of Operation
If input x moves the pilot valve to the right, port II is uncovered, and so high pressure oil
enters the right side of the power piston. Since port I is connected to the drain port, the
oil in the left side of the power piston is returned to the drain. The oil flowing into the
power cylinder is at high pressure; the oil flowing out from the power cylinder into the
drain is at low pressure. The resulting difference in pressure on both sides of the power
piston will cause it to move to the left.
Transfer Function
Rate of flow of fluid Q(kg/sec) time dt(sec) is equal to the power piston displacement
dy(m) times the piston area A(sq.m) times the density of fluid P (kg/m3
).Fluid flow rate
is proportional to pilot valve displacement x. So,
Q x

Q Kx
 
where K is a positive constant.
dy
Q AP
dt

dy
AP Kx
dt
 

198
   
APsY s KX s

 
 
Y s K
X s APs

Advantages
(i) Hydraulic fluid acts as a lubricant and heat transfer medium
(ii) Comparatively small size hydraulic actuators can develop large forces or torques
(iii)Fast start, stop, and speed reversals(Faster response)
(iv) Hydraulic actuators can be operated under any type of load (i.e. continuous,
intermittent, reversing or stalled loads)
(v) Availability of linear and rotary actuators(e.g. motors)
(vi) Better speed regulation.
Disadvantages
(i) Hydraulic fluid acts as a lubricant and heat transfer medium
(ii) Like electric power, hydraulic source not readily available
(iii)Presence of dirt contaminate the hydraulic fluid
(iv) Fire and explosion hazards exist
(v) For a similar function, cost of hydraulic system may be higher compared to
electrical system.
17.5.2 DC and AC Motors in Control Systems to Position an Inertia Load
i. The load may be massive (e.g. radar antenna) or light weight precision
instrument.
ii. The actuator should have
 Desired dynamic response
 Desired cost, size, and weight
iii. Electric power is readily available, cleaner and quieter and easier to transmit.
So electric motors is mostly preferable compared to hydraulic and pneumatic actuation)
Merits of DC motor
(i) Linear characteristics,
(ii) Used for large power applications,
(iii)Easier control
Demerits of DC motor
(i) Lower torque to volume and
(ii) Lower torque to inertia ratio.
Future developments
(i) Development of rare earth magnet results in DC motor high torque to volume ratio.
(ii) Advances in brush commutator technology make trouble free maintenance.
(iii)Development of brushless DC motors.

199
SUMMARY
CONTROL SYSTEM ENGINEERING
3.0 Introduction to Control system
3.1 Scope of Control System Engineer
3.2 Classification of Control System
3.3 Historical development of Control system
3.4 Analogus systems
3.5 Transfer function of Systems
3.6 Block diagram representation
3.7 Signal Flow Graph(SFG)
4.0 Feedback Characteristics of Control systems and sensitivity measures
4.1 The Concept of Feedback and Closed loop control
4.2 Merits of using Feedback control system
4.3 Regenerative Feedback
3.0Control System Components
3.1 Potentiometers
3.2 DC and AC Servomotors
3.3 Tachometers
3.4 Amplidyne
3.5 Hydralulic systems
3.6 Pneumatic systems
3.7 Stepper Motors
4.0 Time Domain Performance Analysis of Linear Control Systems
4.1 Standard Test Signals
4.2 Time response of 1st
order Systems
4.3 Unit step response of a prototype 2nd
order system
4.4 Unit Ramp response of a second order system
4.4 Performance Specification of Linear System in Time domain
4.5 The Steady State Errors and Error Constants
4.6 Effect of P, PI, PD and PID Controller
4.7 Effect of Adding a zero to a system
4.8 Performance Indices(ISE,ITSE,IAE, ITAE)
4.9 Approximations of Higher order Systems by Lower order Problems
5.0 The Stability of Linear Control Systems

200
5.1 The Concept of Stability
5.2 The Routh Hurwitz Stability Criterion
5.3 Relative stability analysis
6.0 Root Locus Technique
6.1 Angle and Magnitude Criterion
6.2 Properties of Root Loci
6.3 Step by Step Procedure to Draw Root Locus Diagram
6.4 Closed Loop Transfer Fuction and Time Doamin response
6.5 Determination of Damping ratio, Gain Margin and Phase Margin from Root Locus
6.6 Root Locus for System with transportation Lag.
6.7 Sensitivity of Roots of the Characteristic Equation.
7.0 Frequency Domain Analysis.
7.1 Correlation between Time and frequency response
7.2 Frequency Domain Specifications
7.3 Polar Plots and inverse Polar plots
7.4 Bode Diagrams
7.4.1 Principal factors of Transfer function
7.4.2 Procedure for manual plotting of Bode Diagram
7.4.3 Relative stability Analysis
7.4.4 Minimum Phase, Nonminimum phase and All pass systems
7.5 Log Magnitude vs Phase plots.
7.6 Nyquist Criterion
7.6.1 Mapping Contour and Principle of Argument
7.6.2 Nyquist path and nyquist Plot
7.6.3 Nyquist stability criterion
7.6.4 Relative Stability: Gain Margin, and Phase Margin
7.7 Closed Loop Frequency Response
7.7.1 Gain Phase Plot

201
7.7.1.1 Constant Gain(M)-circles
7.7.1.2 Constant Phase (N) Circles
7.7.1.3 Nichols Chart
7.8 Sensitivity Analysis in Frequency Domain

202
Reference Books
1. K. Ogata, “Modern Control Engineering”, 4th
Edition, PHI.
2. I. J. Nagrath and M. Gopal, “Control System Engineering”, 4th
Edition, New Age.
3. J. J. Distefano, III, A. R. Stubberud and I. J. Williams, “Feedback and Control Systems”,
2nd
Edition, TMH, Schaums Outlines.
4. G. F. Franklin, J. D. Powell, A. Emami, Naini, “Feedback Control of Dynamic Systems”,
4th
Edition, Pearson Education.
5. B. C. Kuo and F. Golnaraghi, “Automatic Control Systems”, 8th
Edition, John Wiley and
Sons.
6. S. Ghosh, “Control Systems: Theory and Applications”, 2nd
Edition, Pearson.
7. D. RoyChaudhury, “Modern Control Engineering”, 4th
Edition, PHI.

203
Sampled Question Sets
SET-I
CONTROL SYSTEM ENGINEERING-I
Time:-3Hrs Full Marks:70
Answer Question no.1 and any five questions from the rest.
Answer all parts of question at one place only
The figures in right hand margin indicate marks.
(Semi log graph papers are allowed)
1 Answer all the following questions briefly (Compulsory) [2x10]
(a)Distinguish between regulator and servo-control problem in control system study.
(b)Sketch the underdamped time response of a typical second order feedback control system subjected
to a unit step input. State the time domain performance indices.
(c)Prove that a Type-1 system has no steady state error for step input while the steady state error for
ramp input decreases for increase of Velocity error Constant(Kv).
(d)Give the equation of intersect of asymptotes in root locus plot.
(e) What is system type number? Explain the practical significance of this number.
(f)Show that the Phase Margin=tan-1
where ξ is the damping ratio of the standard second
order system.
(g)List the advantages and disadvantages of carrying frequency analysis with Nyquist plot.
(h) State the Zeigler-Nichols tuning Rules for PID Controller.
(i) Give all the properties of a minimum phase transfer function.
(j)Explain with sketch the use of drag cup rotor in servo application.
2(a) The Block diagram of a feedback control system is given below. The output
Y(s)=C(s)R(s)+D(s)W(s). Find the transfer functions C(s) and D(s). [5]

204
(b)Describe the construction and working of a two phase motor suitable for use in AC servo systems.
[5]
3(a)Show that high loop gain in feedback control system results in
(i)good steady state tracking accuracy (ii)low sensitivity to process parameter variations
(iii)good disturbance signal rejection (iv)good relative stability
What are the factors limiting the gain? [6]
(b)Explain drawing a neat diagram, the principle of operation of a position servo using a synchro
system as error transducer. [4]
4.The peak overshoot (%Mp) in a unit feedback control system is specified to be within 20% to 40%
range.
(a)Sketch the area in the s-plane in which dominant roots of the systems characteristic equation must
lie. This system has a settling time ts=0.85 sec. [4]
(b)Determine the smallest value of third root such that dominance of the complex roots corresponding
to part (a) is preserved. Further, Determine the open loop transfer function of the system if Mp =50%
[6]
5.(a)State the merits and demerits of using static error coefficients. The open loop transfer function in
a unity feedback control system, is given by G(s)=
( )
( )
; Find the steady state error of the system
using generalized error constants when subjected to an input signal given by r(t)=1+4t+3t2
. [5]
(b) )In a unity feedback control system, the open loop transfer function is given by
G(s)= ( )( )
; Using Routh Hurwitz Criterion, determine the range of K for which the given
system is stable. [5]
6.(a) The Open loop transfer function of a control system is given as G(s)H(s)=
( )
( )( )
;
Sketch the Root Locus. Determine the value of K such that damping ratio(ξ) is 0.4. [7]
(b) State the use of Nichol’s Chart. [3]
7.Using Bode Plot, determine gain crossover frequency, phase crossover frequency, gain margin and
phase margin in a unity feedback control system, where, the open loop transfer function is given by
G(s)=
( . ) .
( . )
. [10]
8. (a)A unity feedback system has open loop transfer function G(s) =
( )
( )
;

205
Use Nyquist criterion to determine if the system is stable in the closed loop configuration [7]
(b) State the merits and demerits of PI Controller. [3]

206
CONTROL SYSTEM ENGINEERING(IC323)
Time:-3Hrs Full
Marks:70
Answer all parts of question at one place only
1. Answer all the following questions briefly (Compulsory) [2x10]
(a)What are the constraints in developing the transfer function of a device a part of larger system?
(b)The transfer function of a control system is T(s)=K/[S2
+4S+K]; Find K if the system is critically
damped.
(c) What are the steady state errors of a Type-3 unity feedback system subjected to step input, ramp
input and parabolic input?
(d)Explain what do you mean by Root Contours.
(e) The magnitude of frequency response of a second order system is 5 at 0 rad/sec and peaks to
3
10
at 5 2 rad / sec. Determine the transfer function of this underdamped system.
(f)Show that the bandwidth(ωb)=ωn [(1 − 2ξ ) + 4ξ − 4ξ + 2 ] where ξ is the damping ratio
and ωn is the natural frequency of the standard second order system.
(g) Sketch the constant gain loci for the unity feedback system whose feed forward transfer function is
G(s)=
)
1
( 
S
S
K
(h) Show that high loop gain in feedback control system results in good steady-state tracking accuracy
(i) State the use of Nichol’s Chart.
(j) State the merits and demerits of PI Controller
2(a) Obtain the signal flow graph representation for a system represented by a block diagram as
shown below and determine the overall gain G(s)= ; [5]

207
(b)Explain giving a schematic diagram how a synchro pair would be embodied in an AC position
control system. [5]
3(a)In a negative feedback control system, calculate separately, the sensitivity of the system transfer
function at s= jω=j1.6 rad/sec with respect to
(i)the forward path transfer function G(s) where G(s)=
( )
(ii)feedback path transfer function H(s) where H(s)=0.8 [5]
(b)Describe in detail along with a schematic diagram, a typical position control system employing an
armature controlled DC Motor with a fixed field separately excited system. Derive the transfer
function. [5]
4.(a)What is system type number? Explain the practical significance of this number. [2]
(b) In a unit feedback control system, the open loop transfer function is given by G(s)=
( )
By what factor should the amplifier gain k be multiplied so that the damping ratio(ξ) is enhanced
from 0.35 to 0.95. [8]
5(a)In a unit feedback control system, the open loop transfer function is given by
G(s)= ( )( )
; Find the static error coefficients (Kp, Kv, and Ka) and the steady state error of
the system when subjected to an input signal given by r(t)=10+20t+30t2
. [5]
(b) )In a unit feedback control system, the open loop transfer function is given by
G(s)= ( )( )
; Using Routh Hurwitz Criterion, determine the range of K for which the given
system is stable. [5]
6. (a)The Open loop transfer function of a control system is given as G(s)H(s)=
( )
;
Sketch the Root Locus. [6]

208
(b)Given the open loop frequency response G(jω) = U+ jV;
Obtain the radii and center locations of constant M and N circles [4]
7(a)Define minimum phase, non-minimum phase and All pass system. [2]
(b)Draw the Bode Plot of the open loop transfer function of a feedback system given by
G(s)H(s)=
( )
( )( )
; Also determine the system Stability. [8]
8(a)A unity feedback system has open loop transfer function G(s) =
( )
( )( )
;
Use Nyquist criterion to determine if the system is stable in the closed loop configuration.[7]
(b)Describe two tuning methods, one based on ultimate gain and the other based on process reaction
curve. [3]
SET-III
Time:-3Hrs Full Marks:70
1.Answer all the following questions briefly (Compulsory) [2x10]
(a)Sketch the underdamped time response of a typical second order feedback control system
subjected to a unit step input. State the time domain performance indices.
(b)The transfer function of a control system is T(s)=K/[S2
+4S+K]; Find K if the system is critically
damped.
(c)Prove that a Type-1 system has no steady state error for step input while the steady state error
for ramp input decreases for increase of Velocity Error Constant (Kv).
(d)Give the equation of intersect of asymptotes in root locus plot.
(e)Explain what do you mean by Root Contours.
(f)Show that the Phase Margin=tan-1 ξ
ξ ξ
where ξ is the damping ratio of the standard
second order system.

209
(g)List the advantages and disadvantages of carrying frequency analysis with Nyquist plot.
(h) State the Zeigler-Nichols tuning Rules for PID Controller.
(i) Give all the properties of a minimum phase transfer function.
(j)Explain with sketch the use of drag cup rotor in servo application.
2(a)Obtain the signal flow graph representation for a system represented by a block diagram as
shown below and determine the overall gain G(s)= ; [6]
(b)Describe the construction and working of a two phase motor suitable for use in AC servo systems.
[4]
3(a)Show that high loop gain in feedback control system results in
(i)good steady state tracking accuracy (ii)low sensitivity to process parameter variations
(iii)good disturbance signal rejection (iv)good relative stability
What are the factors limiting the gain? [5]
(b)Explain drawing a neat diagram, the principle of operation of a position servo using a synchro
system as error transducer. [5]
4.The peak overshoot (%Mp) in a unit feedback control system is specified to be within 20% to 40%
range.
(a)Sketch the area in the s-plane in which dominant roots of the systems characteristic equation
must lie. This system has a settling time ts=0.85 sec. [4]
(b)Determine the smallest value of 3rd
root such that dominance of the complex roots
corresponding to part (a) is preserved. Further, Determine the open loop transfer function of the
system if Mp=50% [6]
5.(a)State the merits and demerits of using static error coefficients. The open loop transfer function
in a unity feedback control system, is given by G(s)=
( )
( )
; Find the steady state error of the
system using generalized error constants when subjected to an input signal given by r(t)=1+4t+3t2
.
[5]

210
(b)In a unity feedback control system, the open loop transfer function is given by G(s)= ( )( )
;
Using Routh Hurwitz Criterion, determine the range of K for which the given system is stable.
[5]
6.(a) The Open loop transfer function of a control system is given as G(s)H(s)=
( )
( )( )
;
Sketch the Root Locus. Determine the value of K such that damping ratio(ξ) is 0.4. [8]
(b) State the use of Nichol’s Chart. [2]
7.Using Bode Plot, determine gain crossover frequency, phase crossover frequency, gain margin and
phase margin in a unity feedback control system, where, the open loop transfer function is given by
G(s)=
( . ) .
( . )
. [10]
8. A unity feedback system has open loop transfer function G(s) =
( )
( )
; Use Nyquist criterion to
determine if the system is stable in the closed loop configuration. [10]
Set-IV
Sub: Control System Engineering (3:1:0)
Time Duration : Two Hours Date 8.10.2013 Maximum 20 Marks
A.The figures in the right hand margin indicate marks Answer any four including Question No.1
B.The symbols carry usual meaning
1. Answer the following questions (Compulsory) [5x1]
(a) If
2
3
3
)
(
)
(
2




s
s
s
s
U
s
Y
obtain the SFG representation of this transfer function.
(b) Experimental measurements yield a plot of the magnitude of the frequency response function with a
resonance peak 1.35 at a frequency of 10rad/sec.
(a)Estimate  and n of the dominating system poles
(c)Draw the schematic diagram of a DC closed loop position control system consisting of
(I) a pair of Potentiometers (II)Amplifier (III)Armature controlled DC Servomotor
(IV)Gear Train as major component and explain the operation of this system
(d) .The open loop transfer function of a unity feedback control system is given by
G(s)=K/[(s+2)(s+4)(s2
+6s+25)]; By applying Routh-Hurwitz criterion determine the range of K for
which the closed loop system will be stable:
(e)Draw the polar plot for the transfer function of the system G(s)H(s)=10/[s(s+1)2
].

211
2(a)Consider a -ve unity feedback system with following OLTF. Obtain peak overshoot, damped
frequency of oscillation, settling time on 2% tolerance band and response of the system to unit
step input.
( ) =
.
( . )
(b)The OLTF of a system is G(s)H(s)=100/[s(s+100)]. (i)Obtain Static and Dynamic error Constants.
(ii)If the input is r(t)=A+Bt+Ct2
, obtain the steady state error and the dynamic error.
[3+2]
3.Sketch the complete Bode plot of the unity feedback system whose open loop frequency function
)
1
05
.
0
)(
1
1
.
0
(
10

 s
s
s
; Determine the GM, PM and open loop gain for a GM of 20db. [5]
4.Given the G(s)=
)
1
( 
s
s
K
, and H(s)=(s+4)-1
. Sketch the root locus of the system.
(i)Determine the value of K for which the system is at the verge of instability
(ii)For the damping ratio 0.34, determine the value of K and the GM. [5]
5.(a) Show that the bandwidth of a linear standard second order control system = ωn
[ 1 − 2ξ + 4ξ − 4ξ + 2 ] where ξ is the damping ratio and ωn is the natural frequency of
system. What will be the resonant peak for the system whose transfer function is 5/(s2
+2s+5)
(b)Determine the critical value of K for stability of a unity feedback system with loop transfer
function(S)=K/(S-1) using Nyquist stability criterion. [3+2]

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