Control Systems
Engineering
S. K. Bhattacharya
Former Director
National Institute of Technical Teachers’ Training and Research
(Under Ministry of HRD, Government of India)
Kolkata and Chandigarh
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Dedicated lovingly to my dear grandson
Siddhant
who inspires me to remain focused
and never give up
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Contents
Preface
Roadmap to the Syllabus
xiii
xv
1.
IntroDUCtIon
1-1
1.1
Concept of Control System
1.1.1
More Examples of Control Systems
1.1.2
Basic Components of a Control System
1-1
1-2
1-5
1.2
Classification of Control Systems—Open-loop and Closed-loop
Control Systems and their Differences
1.2.1
Open-loop and Closed-loop Control Systems
1.2.2
Linear and Non-linear Control Systems
1.2.3
Time-invariant and Time-varying Control Systems
1.2.4
Continuous Time and Discrete Control Systems
1.2.5
Single-Input–Single-Output (SISO) and Multi-Input–
Multi-Output (MIMO) Control Systems
1.2.6
Lumped Parameter and Distributed Parameter
Control Systems
1.2.7
Deterministic and Stochastic Control Systems
1.2.8
Static and Dynamic Systems
1.2.9
Servomechanism, Regulator, Process Control and
Disturbance Signal
2.
1-6
1-6
1-7
1-8
1-8
1-8
1-8
1-9
1-9
1-9
1.3
Different Examples of Control Systems
1-10
1.4
Feedback Characteristics—Effect of Feedback
1.4.1
Effects of Feedback
1-12
1-13
Review Questions
1-13
MatHeMatICaL MoDeLS
2-1
2.1
Introduction
2-1
2.2
Impulse Response of a System
2-1
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vi
CONTENTS
2.3
Transfer
2.3.1
2.3.2
2.3.3
Function
Definition of Transfer Function
Poles and Zeros of a Transfer Function
Procedure for Determining the Transfer Function of a
Control System
2.4Differential Equations of Physical System and their Transfer
Function—Translations and Rotational Mechanical System
2.4.1
Electrical System
2.4.2
Mechanical System
2.4.3
Analogies of Mechanical and Electrical Systems
2.4.4
Hydraulic System
2.4.5
Pneumatic System
2.4.6
Thermal System
2.5
Concept
2.5.1
2.5.2
2.5.3
of State–Space Modelling
Concepts of State, State–Variables and State–Space
State–Space Equation
Concepts of Controllability and Observability
Review Questions
3.
Modelling a Control System—Block Diagram
Representation
3.1Block Diagram Representation of Systems Considering
Electrical System as Examples
3.1.1
Advantages of Block Diagram Representation
3.1.2
Block Diagram Representation of an Error Detector
3.1.3
Block Diagram of a Closed-loop Control System and Its
Transfer Function
3.1.4
Characteristic Equation of a Control System
3.1.5
Rules for Block Diagram Simplification
3.1.6
Block Diagram Representation of an Electrical Network
3.2
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Transfer Function of DC Synchro Motor
3.2.1
Block Diagram of a DC Motor Drive
3.2.2
Block Diagram of Gear Trains
3.2.3Block Diagram of a Servomechanism or a Position
Control System
2-3
2-4
2-5
2-5
2-6
2-6
2-8
2-12
2-15
2-16
2-17
2-31
2-32
2-33
2-34
2-35
3-1
3-1
3-1
3-2
3-2
3-4
3-4
3-29
3-31
3-32
3-37
3-41
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CONTENTS
3.3
Transfer Function of AC Synchro Motor
3.3.1
Transfer Function of a Two-phase AC Servo Motor
3.4Transfer Function of Synchro Transmitter and Synchro Receiver
Review Questions
4.
vii
3-44
3-44
3-46
3-46
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
4-1
4.1
Introduction
4-1
4.2
Construction of Signal Flow Graph
4.2.1
Steps Followed in Drawing SFG
4.2.2
Mason’s Gain Formula
4-2
4-2
4-4
4.3
SFG for Solution of Differential Equations
4-14
Review Questions
4-27
5.Time response analysis
5.1
Time Response of First Systems
5.1.1
Types of Input Signals
5.1.2
Classification of Control Systems
5.1.3
Time Response of First-order System
5.1.4
Response of First-order System to Step Input
5.1.5
Response of First-order System to Ramp Input
5.1.6
Response of First-order System to Impulse Input
5.2Characteristics Equation of Feedback Control System
5.2.1Transfer Function and Characteristic Equation in
Generalized Form
5.2.2Positional Servosystem as a Second-order
System and Its Analysis
5.2.3
Characteristic Equation and Position of Roots/Poles
5.2.4Classification of Time Response for Second-order
Control System
5.2.5Time Response of Second-order Control System
Subjected to Unit Step Input
5.3Transient Response of Second Order System—Time
Domain Specifications
5.3.1
Transient Response of a Second-order System
5.3.2
Dominant Closed-loop Poles of Higher Order Systems
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5-1
5-1
5-2
5-3
5-6
5-6
5-8
5-10
5-11
5-12
5-12
5-17
5-18
5-18
5-22
5-23
5-26
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viii
CONTENTS
5.3.3
5.3.4
Comments on Transient Response Specifications
Sensitivity of a Control System
5.4Steady State-Response-Steady State Error and Error Constants
5.4.1
Static Position Error Coefficient (K P)
5.4.2
Static Velocity Error Coefficient (K v)
5.4.3
Static Acceleration Error Coefficient (K a)
5.5Effects of Proportional, Derivative, and Proportional
Integral System
5.5.1
Proportional Control
5.5.2
Proportional Plus Derivative Control
5.5.3
Proportional Plus Integral Control (PI Control)
5.5.4Proportional Plus Integral Plus Derivative Control
(PID Control)
5.5.5
Derivative Feedback Control
5.6
6.
5-40
5-41
5-42
5-43
5-48
5-48
5-50
5-51
5-52
5-53
Transient Response Analysis Using MATLAB
5-66
Review Questions
5-70
Stability analysis in s-domain
6-1
6.1Concept of Stability
6.1.1
Pole-Zero Location and Conditions for Stability
6.1.2
Routh Stability Criterion and Its Application
6-1
6-1
6-2
6.2
7.
5-27
5-38
Limitations of Routh’s Stability
6-18
Review Questions
6-19
tHE rOOT lOCUS tECHNIQUE
7-1
7.1
The Root Locus Concept Construction of Root Loci
7.1.1
Root Locus Concept
7.1.2
Root Locus Construction Procedure
7.1.3
Root Locus Construction Rules
7.1.4Root Locus Construction Rules—Illustrated
Through Examples
7-1
7-2
7-4
7-7
7.2Effect of Adding Poles and Zeros to A(S) H(S) on the Root Loci
7-56
7.7
Root Locus Plot with MATLAB
7-57
Review Questions
7-60
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7-9
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CONTENTS
8.
ix
Frequency RESPONSE Analysis
8-1
8.1
8-1
8-2
Introduction
8.1.1
Polar Plot
8.2Frequency Domain Specifications— Bode Diagram, Determination
of Frequency Domain
8.2.1Correlation Between Time Response and
Frequency Response
8.2.2Correlation Between Time Domain and
Frequency Domain Parameters
8.2.3
Bandwidth
8.2.4
Relative and Absolute Stability
8.2.5
Presentation of Frequency Response in Graphical Form
8-5
8-5
8-8
8-9
8-10
8-10
8.5Specifications and Transfer Function for Bode Diagram
8.5.1
Method of Drawing Bode Plot
8.5.2
Initial Slope of Bode Plot
8.5.3
Bode Plot for Quadratic Form of Transfer Function
8.5.4Maximum Magnitude of the Second-order
Transfer Function
8.5.5Determination of Gain Margin and Phase Margin
for Stability Analysis
8.5.6
Plotting Bode Diagrams with MATLAB
8-12
8-13
8-18
8-21
8.6
Polar Plot and Nyquist Stability Criterion
8.6.1
Nyquist Path or Nyquist Contour
8-44
8-47
8.7
Drawing Nyquist Plots with MATLAB
8-63
8.8
Relative Stability
8-65
8.9Frequency Response of a Closed-loop System Using M -circle
and N -circle
8.9.1
Constant Magnitude Loci
8.9.2
Constant N -circles
8.9.3
Uses of M -circles and N -circles
Review Questions
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8-22
8-26
8-43
8-66
8-67
8-69
8-70
8-74
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x
9.
CONTENTS
STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-1
9.1
Polar Plots and Nyquist Stability Criterion
9.1.1
Nyquist Path or Nyquist Contour
9-1
9-4
9.2
Drawing Nyquist Plots with MATLAB
9-20
Review Questions
9-37
10. Compensation techniques
10-1
10.1
Necessity of Compensation
10.1.1 Effect of Adjustment of Gain
10.1.2 Compensation by Inserting a Network or Compensator
10.1.3 Lead Compensator
10.1.4 Lag Compensator
10.1.5 Lag–Lead Compensator
10.1.6 Design Procedure
10.1.7 Use of MATLAB Programming
10-1
10-2
10-2
10-3
10-4
10-6
10-8
10-30
10.2
PID Controllers
10.2.1 Proportional Controllers
10.2.2 Proportional Derivative Controllers
10.2.3 Proportional Integral Controllers
10.2.4 Basic Elements of a PID Controller
10.2.5 An Electronic PID Controller
10-40
10-40
10-41
10-43
10-43
10-46
Review Questions
10-46
11. STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-1
11.1
Introduction
11-1
11.2
Concepts of State, State Variables and State Model
11.2.1 State Model of Linear Systems
11.2.2State Model of Single-Input Single-Output Linear Systems
11-1
11-3
11-5
11.3
State Models of Linear Continuous Time Systems
11.3.1 State Space Representation Using Physical Variables
11.3.2 State Space Representation Using Phase Variables
11.3.3 State Space Representation Using Canonical Variables
11.4Co-relation Between State Model and Transfer
Function
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11-5
11-6
11-7
11-12
11-16
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CONTENTS
xi
11.5
Diagonalisation of State Matrix
11-17
11.6
Solution of State Equation
11.6.1 Computation of State Transition Matrix
11.6.2 Properties of State Transition Matrix
11-19
11-21
11-22
11.7
Concept
11.7.1
11.7.2
11.7.3
11-23
11-23
11-26
11-28
of Controllability and Observability
Controllability
Observability
Principle of Duality
Review Questions
Appendix 1: Laplace Transform
11-52
A-1
Appendix 2: MATLAB Fundamentals
A-10
Appendix 3: Key Control Terminologies
A-21
Model Question Papers
Index
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Q-1
I-1
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Preface
Control systems engineering is an exciting and challenging field. It is a multi-disciplinary
subject. This textbook is meant for the basic course in control systems for undergraduate students
of Electrical, Electronics, Instrumentation, Chemical, Mechanical, Production, Electronics and
Aerospace Engineering.
The purpose of writing this book has been to provide students with basics on control
systems engineering which will help them understand the subject easily, and also help them
prepare for university examinations.
The content of the book is based on the curriculum prescribed by various universities in
India. The classical methods of control systems engineering have been thoroughly covered.
Laplace transforms, transfer functions, root locus design, stability analysis, state-variable
method, frequency response methods, including Bode, Nyquist, steady-state error for standard
test signals and so on have been widely dealt with. This book includes a complete chapter on the
design of control systems, along with solved examples, and a chapter on state space analysis of
continuous system.
In addition to explaining the basic principles in a simplified manner, numerous examples
and exercises, design problems and MATLAB problems have been provided to help students
understand the subject better. The MATLAB exercises will help students practice computeraided design and re-work on some of the design examples.
This book also includes numerous illustrated examples to help the students develop
problem-solving skills. The review questions at the end of each chapter help in recollecting the
concepts learnt.
A good mathematical foundation is required for understanding the subject of control
system design. In this book an attempt has been made to provide the physical meaning of the
mathematical expressions so as to provide an insight into control system design.
Control systems engineering is basically taught in the parts—linear control systems and
non-linear digital control system. This basic book on control system has dealt extensively with
the linear control systems.
It is expected that this book will satisfy the requirements of students and teachers alike. I
thank Pearson Education for bringing out the book on time and in a presentable manner. I hope
the students will find this book easy to understand.
S. K. Bhattacharya
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About the Author
Dr S. K. Bhattacharya did his B.E. from Jadavpur University, his M.E. from Calcutta University
and his Ph.D. from BITS, Pilani. He is a fellow of the Institution of Engineers (India) and the
Institution of Electronics and Telecommunication Engineers (India).
Dr Bhattacharya is currently the Principal of SUS Women Engineering College, Mohali,
Punjab.
Formerly, he was the Principal of Technical Teachers Training Institute, Chandigarh;
Director of National Institute of Technical Teachers Training and Research (Under MHRD,
Government of India), Kolkata; Director of Hindustan Institute of Technology, Greater Noida;
and Principal of SUS College of Engineering and Technology, Tangori (Mohali), Punjab.
As a senior fellow of MHRD, Government of India, he attended the technical teacher’s
training program at the Bengal Engineering College, Shibpur. As a professor of electrical
engineering, Dr Bhattacharya taught all the major subjects of electrical engineering both
at the undergraduate and postgraduate levels. He has written numerous textbooks, which are
very popular, and has published over one hundred technical papers in journals and conference
proceedings.
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Roadmap to the Syllabus
Jawaharlal Nehru Technological University
Course: Control Systems
UnIt-I IntroDUCtIon
Concepts of control systems - Open loop and closed loop control systems and their differencesDifferent examples of control systems - Classification of control systems, Feed-back
Characteristics, Effects of feedback.
Mathematical models – Differential equations, Impulse response and transfer functions Translational and rotational mechanical systems
Refer Chapters 1 and 2
UnIt-II tranSFer FUnCtIon rePreSentatIon
Transfer function of DC servo motor - AC servo motor- Synchro transmitter and receiver, Block
diagram representation of systems considering electrical systems as examples -Block diagram
algebra – Representation by signal flow graph - Reduction using Mason’s gain formula.
Refer Chapters 3 and 4
UnIt-III tIMe reSPonSe anaLySIS
Standard test signals - Time response of first order systems – Characteristic equation of Feedback
control systems, Transient response of second order systems - Time domain specifications –
Steady state response - Steady state errors and error constants – Effects of proportional derivative,
proportional integral systems.
Refer Chapter 5
UnIt-IV StaBILIty anaLySIS In S-DoMaIn
The concept of stability – Routh’s stability criterion – qualitative stability and conditional stability
– Limitations of Routh’s stability
Root locus technique: The root locus concept—construction of root loci-effects of adding poles
and zeros to G(s)H(s) on the root loci.
Refer Chapters 6 and 7
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xvi
ROADMAP TO THE SYLLABUS
UNIT-V FREQUENCY RESPONSE ANALYSIS
Introduction, Frequency domain specifications-Bode diagrams - Determination of frequency
domain specifications and transfer function from the Bode diagram - Phase margin and gain
margin-stability analysis from Bode plots.
Refer Chapter 8
UNIT-VI STABILITY ANALYSIS IN FREQUENCY DOMAIN
Polar plots, Nyquist plots stability analysis.
Refer Chapter 9
UNIT-VII CLASSICAL CONTROL DESIGN TECHNIQUES
Compensation techniques – Lag, lead, lead-lag controllers design in frequency domain, PID
controllers.
Refer Chapter 10
UNIT-VIII STATE SPACE ANALYSIS OF CONTINUOUS SYSTEMS
Concepts of state, state variables and state model, derivation of state models from block diagrams,
diagonalization - solving the time invariant state equations - State transition matrix and it’s
properties – Concepts of controllability and observability.
Refer Chapter 11
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1
INTRODUCTION
1.1
CONCEPT OF CONTROL SYSTEM
A system is a combination of devices and components connected together by some form of
regular interactions to act together and perform a certain function. The system may be physical,
biological, economic and the like.
The term control means to regulate, to direct or to command. Thus, a control system may
be defined as a combination of devices and components connected or related so as to command,
direct or regulate itself or another system.
The part of the system which is to be controlled is given different names, for example,
plant, controlled system, process, etc. The control system designer develops a controller that will
control the plant or the controlled system.
Control systems are used in many applications, for example, the control of temperature,
liquid level, position, velocity, f low, pressure, acceleration, etc.
Let us consider the driving system of an automobile. Speed of the automobile is a function
of the position of its accelerator. The desired speed can be maintained by controlling the pressure
on the accelerator pedal. If the speed of the vehicle is to be increased, the driver has to increase
the pressure on the pedal. Figure 1.1 (a) shows the diagrammatic representation of a typical
driving system.
The input to the system is force, called the command signal, on the accelerator pedal. The
command signal is applied to the accelerator pedal which has various linkages and connections
with the carburetter. When the fuel is burnt, the engine causes the vehicle to increase its speed.
The output at variable speed is called controlled output because the signal has caused a control
on the speed of the vehicle.
Thus, the controlled output of the above driving system can be depicted as in Figure 1.1 (b).
Increase or decrease of the output, that is, the speed as the output, is achieved by increase
or decrease in the input, that is, the fuel input to the engine through the carburetter.
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1-2
CONTROL SYSTEMS ENGINEERING
Fig. 1.1 (a) Driving system of an automobile; (b) Control system of an automobile.
Change in the output signal requires a proportionate change in the input signal. This
proportionate change is compared with reference input by a comparator, which is also called an
error detector.
1.1.1
More Examples of Control Systems
Let us take an example of control of output of a d.c. generator. The d.c. generator is rotated at
constant speed with the help of a prime mover. Its output voltage can be controlled by changing
the field current, If . The field current gets changed when a variable resistance connected in series
with the field circuit is changed as shown in Figure 1.2. In manual control system, the human
operator will watch the output voltage using a voltmeter. He will compare the actual output
voltage with the desired output voltage. Let us assume that the output voltage should remain
constant at 220V when the load on the generator changes. As the load changes, the output voltage
changes. The operator has to keep a watch on the voltmeter reading. If the voltmeter reading is
less than the desired value of 220V, an error is created. Recognizing this error (negative error), the
operator will have to adjust the field regulating resistance so as to increase the field current. If the
field circuit resistance is decreased, field current, If , will increase. If the field current is increased,
the voltage output will increase. The field current should increase to such a value that the output
equals the desired value thereby reducing the error to zero. This is a manual control system.
This operation can be made automatic by using an automatic control system. A mechanism can
be developed to compare the actual value with the desired value and the error signal is fed to a
servomotor to control the field regular resistance. The error voltage, also called signal, has to be
sometimes amplified and fed to the servomotor. Such an automatic control system is also called
feedback control system.
Let us take another example of temperature control system. The heating element of the
heater can be kept on for heating a room using a timer. The heater will remain on for all the time
irrespective of the temperature of the room. To make the heater automatic, a thermostat system
has to be incorporated to measure the room temperature continuously. If the room temperature
attains the desired temperature, the heater will be automatically switched off. The control system
cannot be designed for fixed time based, that is, switching on and off of the heater cannot be done
after a fixed interval of time. This is because the room temperature will also depend upon the
outside temperature.
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INTRODUCTION
1-3
Fig. 1.2 Control of output voltage of a d.c. generator.
A control system can be represented using block diagram. The simplest form of representing
a system is as shown in Figure 1.3 (a). When the output is controlled by measuring the desired
output, the block diagram will be as shown in Figure 1.3 (b).
Fig. 1.3 (a) A non-feedback control system ; (b) A feedback control system.
Now let us draw the block diagrams of a few control systems.
Example 1.1 Represent a traffic control system in the form of a block diagram.
Fig. 1.4 Block diagram representation of a traffic control system.
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1-4
CONTROL SYSTEMS ENGINEERING
Solution Figure 1.4 shows the block diagram of a traffic light control system. Normally,
the timings of the time delay relays or the electronic circuits controlling the time of operation
of the traffic lights are fixed. This is irrespective of the volume of traffic on the road. Such a
system would have to wait for green signal on one road even if there is no vehicle moving on the
crossroad. The feedback sensor, if installed on the road crossings, will regulate the timings of the
signal lamps according to volume of traffic at a particular time.
Example 1.2
Represent the manually operated air conditioning system in an electrical train.
Solution The block diagram of a manually operated air conditioning system in an electric
train has been shown in Figure 1.5.
Fig. 1.5 Manually operated air conditioning system in a train.
Solution Once the thermostat setting is made for any desired temperature, say 24ºC, the air
conditioners will continue to work till the desired temperature is reached. During setting of the
time, the actual temperature is made equal to the desired temperature. But during night time,
the outside temperature falls hence less cooling time should be able to bring the temperature
inside to the desired temperature. But here, the setting of the thermostat being unchanged, the air
conditioner will over cool the compartments.
Example 1.3 Represent a room air conditioning system in the form of block diagram.
Fig. 1.6 Room air conditioning system.
Solution An air conditioner controls the temperature and humidity of a room to make it
comfortable. An air conditioner has a compressor and a thermostat. By measuring the actual
room temperature and comparing it with the desired temperature, say 24ºC, an error signal is
generated which either turns on the compressor or switches it off. The simple block diagram for
room air conditioning is shown in Figure 1.6.
After examining the above examples, we can develop a generalized block diagram of a
control system showing its components.
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INTRODUCTION
1.1.2
1-5
Basic Components of a Control System
The basic components of a control system are in the form of a control diagram as shown in
Figure 1.7.
We will consider room heating system using steam flowing through pipes fitted in the room.
The flow of steam through the pipes is regulated automatically by a control valve. The amount
of opening of the valve is regulated by a servomotor. The servomotor, controlling the opening
or closing of the valve, works as a controller for amount of heating of the room. A temperature
sensor will measure the room temperature and will provide a feedback signal for comparison
in the comparator. In case the output temperature is different from the desired temperature a
reference input is provided to the comparator, an error signal will be generated to control the hot
steam flow through the valve.
Fig. 1.7 Representation of a control system.
1.
2.
3.
4.
5.
6.
7.
Chapter 01.indd 5
The components shown in the diagram above are defined as follows:
Reference input: This provides input signal for the desired output.
Error detector: It is an element in which one system variable (feedback signal) is subtracted
from another variable (reference signal) to obtain a third variable (error signal). It is also
called comparator.
Feedback element: Feedback signal is a function of the controlled output which is
compared with the reference signal to obtain the actuating signal. Feedback element
measures the controlled output, converts or transforms to a suitable value for comparison
with the reference input.
Error signal: It is an algebraic sum of the reference input and the primary feedback.
Controller: The controller is an element that is required to generate the appropriate control
signal. The controller operates until the error between the controlled output and desired
output is reduced to zero.
Controlled system: It is a body, a plant, a process or a machine of which a particular
condition is to be controlled, for example, a spacecraft, reactor, boiler, CNC machine, etc.
Controlled output: Controlled output is produced by the actuating signal available as
input to the controller. Controlled output is made equal to the desired output with the help
of the feedback system.
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1-6
CONTROL SYSTEMS ENGINEERING
1.2CLASSIFICATION OF CONTROL SYSTEMS—OPEN-LOOP AND CLOSEDLOOP CONTROL SYSTEMS AND THEIR DIFFERENCES
1.2.1
Open-loop and Closed-loop Control Systems
The systems in which output has no effect on the control action are called open-loop control
systems. In other words, the output is neither measured nor fed back for comparison with
reference input as shown in Figure 1.8. For example, let us consider a washing machine in which
soaking, washing and rinsing in the washer operates on the time basis. The machine does not
measure the output signal, that is, cleanliness of clothes. Such open-loop control systems can be
used in practice, if the relationship between input and output is known and there are no external
or internal disturbances.
Fig. 1.8 Elements of an open-loop control system.
Feedback control systems are also referred to as closed-loop control systems. In a closed-loop,
the actuating error signal, which is the difference between the input signal and the feedback signal
(output signal), is fed to the controller so as to reduce the error and bring the output of the system to
the desired value. Such a system has already been shown in Figure 1.7. It is desirable that a closedloop system is not influenced by external disturbances and internal variations in system parameters.
The operation of a system may be controlled externally or automatically (by the system itself).
When the control action of a system is independent of the output, the system is said to be an openloop control system. However, if the control action is somehow dependent on the output, the system
is called a closed-loop or feedback control
system. If the feedback signal is equal
to the controlled output, the feedback
system is called unity feedback system.
Figure 1.9 shows an open-loop
system and a closed-loop system. In a
closed-loop system, the output is fed
back and compared with the input so
that any necessary corrective action
may be taken. The open-loop system,
however, relies for its action on the
settings of its components.
Fig. 1.9 Open-loop and closed-loop systems.
Some examples of open-loop
and closed-loop control systems are as
follows:
(1) Traffic-light systems vary in operation. Some systems are closed-loop and some are openloop. In the open-loop type, there is a timing mechanism which is set to switch the lights at
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INTRODUCTION
1-7
regular intervals, irrespective of the volume of traffic. In the closed-loop type, however, the
amount of traffic passing through the junctions is monitored (electronically or otherwise),
and the duration of the “red ” and “green” conditions are adjusted for each portion of the
roads of the junction accordingly.
(2) An example of the action of an open-loop system is a car-wash machine in which all cars
receive the same amount of washing irrespective of how dirty they are. Here, the output is
the cleanliness of the cars, which corresponds to a given setting of the machine (amount of
water and washing time). A human being, however, who washes cars, automatically makes
sure that the dirtier cars receive more attention than others. Car washing by a responsible
human being is, therefore, a closed-loop control system. An ordinary washing machine is
an open-loop system.
(3) A refrigerator is a closed-loop system. Its temperature is measured by a thermostat which
turns the motor ON when the temperature rises above the desired value and turns the motor
OFF when the temperature again reaches the desired value.
(4) The lighting of a room is an open-loop system. Once the light is turned on, it will stay on
(until it is switched off ) irrespective of whether the room is dark or light.
The differences between an open-loop system and a closed-loop system are shown in
Table 1.1.
Table 1.1 Differences between Open-loop and Closed-loop Systems.
Open-loop System
Closed-loop System
(a) No feedback used.
(a)
Feedback is there for comparison between
desired output and reference input.
(b)
(b)
Closed-loop system can become unstable under
certain conditions.
Open-loop system is generally stable.
(c) Their accuracy is determined by the calibration of (c) They are more complex. Complicated to construct
their elements. Simple to develop and cheap.
and costly.
(d) Affected by non-linearities in the system.
(d) Adjusts to the effects of non-linearities present in
the system.
Examples: Washing machine, fixed time traffic control
system, room heater, etc.
Examples: Servomotor control, generator output
voltage control system, refrigerator, biological system,
etc.
1.2.2
Linear and Non-linear Control Systems
A linear system is one which obeys the principle of superposition. The principle of superposition
states that the response produced by simultaneous application of two different forcing functions
is the sum of individual responses.
Non-linear systems do not obey the principle of superposition. Almost all practical systems
are non-linear to some extent. Non-linearities are introduced due to saturation effect of system
components, frictional forces, play between gear trains, mechanical linkages, non-linearities of
electronic components like power amplifiers, transistors etc. used.
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1-8
1.2.3
CONTROL SYSTEMS ENGINEERING
Time-invariant and Time-varying Control Systems
Time-invariant control system is one whose parameters do not vary with time. The response
of such a system is independent of the time at which input is applied. For example, resistance,
inductance and capacitance of an electrical network are independent of time.
A time-varying system is one in which one or more parameters vary with time. The
response depends on the time at which input is applied. A space vehicle control system where
mass decreases with time, as fuel it carries is consumed during flight, is an example of a timevarying system.
1.2.4
Continuous Time and Discrete Control Systems
In continuous control systems, all system parameters are functions of continuous time, t.
A discrete time control involves one or more variables that are known only at discrete instants
of time.
A continuous time or continuous data control system is one in which the signals at various
parts of the system are continuous functions of time. These signals are continuous time signals.
For example, when we consider speed control of a d.c. motor, we know that the output, that is,
the rotation in terms of radians per second is a function of voltage and current provided as input
to it on a continuous time basis. Such a system takes a continuous time input and provides a
continuous time output.
Sampled data control system is one in which the signals at one or more points of the system
are in the form of pulses or in digital code. Sampled data control system will have discrete time
input signal.
Discrete data control system is one in which a computer is used as one of the control circuit
element. The input and output are in binary numbers. Digital (A to D) converters also form a part
of the control system.
For example, in a generator excitation control system, the field current is adjusted
continuously on the basis of feedback from the output voltage generated so as to keep the
output voltage constant. Discrete signals are used in microprocessor and computer based control
systems.
1.2.5
Single-Input–Single-Output (SISO) and Multi-Input–Multi-Output
(MIMO) Control Systems
A system with one input and one output is called single-input–single-output control system. In
other words, there is only one command and one controlled output.
A system with multiple inputs and multiple outputs is called multi-input–multi-output
control system. For example, boiler drum level control, robot arm control, etc.
1.2.6
Lumped Parameter and Distributed Parameter Control Systems
Control systems that can be described by ordinary differential equations are lumped parameter
control systems whereas distributed parameter control systems are described by partial differential
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INTRODUCTION
1-9
equations. The parameters of a long transmission time, that is, the resistance, inductance and
capacitance, are distributed along the line but they may be considered as lumped parameters at
certain points.
1.2.7
Deterministic and Stochastic Control Systems
A control system is deterministic if the response is predictable and repeatable. If not, the control
system is a stochastic control system which involves random variable parameters.
1.2.8
Static and Dynamic Systems
A system is called dynamic or time dependent if its present output depends on past input, whereas,
a static system is the one whose current output depends only on current input.
To use/understand a control system efficiently, we must learn how to:
(1)
(2)
(3)
(4)
Develop mathematical system descriptions and reduce them to block diagram forms:
(Modelling/Mathematical Representation);
Manipulate and solve the resulting system equations;
Design system to satisfy general performance specifications and
Evaluate our results by analytical and simulation studies.
It is, thus, important that we become familiar with a variety of electrical and mechanical
control systems.
1.2.9
Servomechanism, Regulator, Process Control and
Disturbance Signal
Servomechanism is an automatic control system in which the controlled variable value is forced
to follow the variations of reference value, instead of regulating a variable value to “set point”.
For example, control of an industrial robot arm a position control system, etc. It is also called
tracking control system.
Regulator is a feedback control system in which controlled variable is maintained at a
constant value in spite of external load on the plant. The reference input or command signal,
although adjustable, is held constant for long period of time. The primary task is then to maintain
the output at the desired value in the presence of disturbances. Examples are regulation of
steam supply in steam engine by fly-ball governor, thermostat control of home heating systems,
regulation of the voltage of an alternator, frequency controller and speed controller, and so on.
Process control refers to control of such parameters as level, flow, pressure, temperature
and acidity of process variables. A particular parameter has only one desired value. The control
system is required to ensure that the process output is maintained at the desired level in spite of
external disturbances which affect the process.
Disturbance represents the undesired signals that tend to affect the controlled system.
Disturbance may be due to changes made in set point, amplifier noise, variation in load, wind
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CONTROL SYSTEMS ENGINEERING
powers disturbing outdoor installation, etc. Other disturbances that affect the performance of the
control system may be changes in parameters due to wear, ageing, environmental effects, high
frequency noise introduced by the measurement sensors, etc. Sometimes disturbance signals may
be too fast for the control system to take care of. Low pass filters may be used to take care of
high frequency disturbance signals thereby maintaining satisfactory performance of the control
system.
1.3
Different Examples of Control Systems
1. Temperature control system: The temperature of an electric furnace is measured by a
thermometer which is an analog device. The analog data (temperature) is converted into digital
data by A/D (Analog to Digital)
converter. The digital equivalent
data of temperature is fed to digital
controller where it is compared with
desired temperature and if there is
any error, the controller sends out a
signal to heater through an amplifier
and relay to bring the temperature of
the furnace to the desired value.
As shown in Figure 1.10, the
temperature of an electric furnace is
measured by a temperature sensor
(thermocouple) and the output of the
Fig. 1.10 Temperature control system.
transducer is amplified and applied
to an error detector (comparator). The output of the comparator drives a relay which makes the
heating system either ON or OFF. In this way, the temperature of the furnace is controlled as per
desired value (set point).
2. Robot hand grasping force
control system: Figure 1.11 shows
the schematic diagram of a grasping
force control system using force
sensing and slip sensing device.
If the grasping force is too
weak, the robot hand will drop the
mechanical object and if it is too
strong, the hand may crush and
Fig. 1.11 Robot hand control system.
damage the object. In the system
shown in Figure 1.11, the grasping
force is preset at moderate level before the hand touches the mechanical object. The hand picks
up and raises the object with preset grasping force. If there is slip in the raising motion, it will
be observed by the slip sensing device and a signal will be sent back to the controller which will
then increase the grasping force. In this way, a reasonable grasping force can be realized that can
prevent slipping but will not damage the mechanical object.
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INTRODUCTION
1-11
3. Speed control system: James Watt developed and used his fly-ball generator for
controlling the speed of a steam engine (see Fig. 1.12). It is recognized as the first automatic
feedback controller used in an industrial process.
A centrifugal Watts’ Governor uses the lift of the rotating ball as speedometer. The supply
of steam is automatically controlled as speed tends to increase or decrease beyond the set point.
In Figure 1.12, the position of the valve controls the opening of the steam supply to
the engine, thus regulating the speed. For a desired speed position, the valve is fixed to a set
point. An increase or decrease of speed of rotation will cause an increase or decrease in the
opening of the valve, thereby causing the variation in steam supply resulting in automatic control
of speed.
Fig. 1.12 Speed control system.
4. Pressure control system: Figure 1.13 shows a pressure control system of a furnace. Here
the pressure inside the furnace is automatically controlled by causing a change in the position of
the damper, which moves up and down.
Fig. 1.13 Pressure control system.
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1-12
CONTROL SYSTEMS ENGINEERING
The pressure inside the furnace is measured by pressure gauge. In case the pressure
increases or decreases beyond a desired value, the controller and the actuator will cause a change
in the position of the damper. The damper will increase or decrease the path of hot air to chimney
to maintain the desired pressure.
1.4
Feedback characteristics-Effect of Feedback
Feedback control is now a basic feature of modern industry. In present-day technological society,
in order to utilize natural resources optimally, some form of control is needed.
Control engineering is primarily concerned with controlling industrial processes and natural
resources, and forces of nature purposefully and for the benefit of mankind.
Early machine and equipment used for control were primarily, manually operated type,
requiring frequent adjustments so as to maintain and/or achieve the desired performance.
Advanced technology made revolutions in the procedure used for system analysis and
design. However, the fundamental theory to study system performance has not changed over the
years.
It is true that we design many complex systems using sophisticated computer control
concepts for process control, space vehicle guidance and similar applications. But such complex
system design requires routine design of control mechanism for items like temperature control,
speed control, voltage regulation and so on.
Let us consider some more examples of control systems which we commonly encounter.
A forced air gas furnace is used for heating the interior of a building. Depending on the
temperature level to be maintained in the building, the gas inlet to the furnace has to be turned on
or off with the help of a temperature sensor placed for control purpose. A blower fan is also used
to force the warm air to circulate inside the room.
When the room temperature drops below a preset reference level, measured by a temperature
sensor like a thermostat, a relay is actuated to switch on the furnace fire. Subsequently, a blower
fan is switched on to circulate the hot air. The blower fan runs till the heat is dissipated and then
turns off automatically.
Now, consider the control problems associated with learning to drive a motor car. At
the beginning, the driver must learn about the functions of various mechanisms provided for
controlling the vehicle and the way the car responds under all conditions.
One would learn that frequent and abrupt application of the brake or the accelerator on
slippery road should be avoided, since it results in the wheels sliding and loss of control of the
vehicle. Control inputs are provided by the accelerator, steering wheel and the brake system.
Inputs are obtained through the driver’s senses. The driver’s brain, his feet and his arms are
used to adjust the control mechanism by which these changes are made to influence the vehicle’s
motion. The fundamental controllers are the brakes, the steering mechanism and the accelerator.
The overall control system can become unstable as a result of skidding and hence total loss
of control due to excessive speed on icy/muddy roads.
It can be seen that we are surrounded by control system applications. All large-scale
chemical and industrial processes and automated production facilities depend heavily on control
technology.
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INTRODUCTION
1.4.1
1-13
Effects of Feedback
Use of feedback in control systems brings in significant changes in terms of improvement in
overall gain, improvement in system stability, reduction of sensitivity of the system to variations
in system parameters, and neutralizing or reducing the effect of disturbance signals. These will
be discussed in details in chapters that follow.
REVIEW QUESTIONS
1.1 Define a control system. Represent a control system by a block diagram.
1.2 Give examples of an open-loop control system and a closed-loop control system. State
their differences.
1.3 Represent in block diagram form, the control system used in automatic control of speed
of a steam turbine.
1.4 Discuss the advantages and disadvantages of an open-loop and closed-loop control
systems. Give one example of each control system.
1.5 What is a feedback control system? Give two examples to show the use of systems in our
everyday life.
1.6 State, giving reasons, whether the following are open-loop or closed-loop control systems:
(1) The room heater or a hot-air blower used to heat a room in winter.
(2) The refrigerator.
(3) The air conditioning system used in electric trains.
(4) The traffic control system used in road crossings in typical Indian cities.
(5) Maintenance of normal body temperature of a human being.
(6) Control of the speed of a motor vehicle by its driver depending upon road condition.
(7) The room air conditioner.
1.7 Human being is said to be the best example of an automatic control system. Explain.
1.8 Traffic at the road crossings are controlled using signal lamps (red, yellow, green).
Represent through block diagram a closed-loop traffic flow control system.
1.9 Define the following terms: System, control system, feedback, closed-loop system, openloop system, error detector, comparator, linear and non-linear control systems, controller
and servomechanism.
1.10 The following are examples of open-loop systems:
(1) A room heater.
(2) A traffic light control system.
(3) A domestic washing machine.
(4) A street light system.
(5) A water pump lifting water to an overhead tank.
How will you change each of them into closed-loop systems?
1.11 Explain the function of all the basic components of a feedback control system.
1.12 Explain how an error signal is generated in a feedback control system. What is a unity
feedback control system?
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2
MATHEMATICAL MODELS
2.1
INTRODuCTION
As we have seen in Chapter 1, a control system consists of a number of sub-systems. All the subsystems work in unison to achieve a desired output for a given input. Mathematical modelling of
sub-systems as well as of the whole system is required for carrying out performance studies. In
this chapter, modelling of a control system using transfer function approach has been dealt with.
2.2
IMPuLSE RESPONSE Of A SYSTEM
Impulse response of a linear time-invariant SISO system is defined as the output y(t) that results
from the application of the unit impulse input u(t) = δ(t) to a system which is initially at rest. The
unit impulse is shown in Figure 2.1(a).
u (t )
u (t )
1
1
u (t ) = δ(t )
a
o
ft
R (s)
t
t
C (s)
G (s)
Unit impulse
as a → o
(a)
Fig. 2.1
(b)
(a) Unit impulse; (b) a SISO system represented by its transfer function.
Consider the block diagram of an open-loop system as shown in Figure 2.1(b). R(s) and
C(s) are the Laplace transform of input and output respectively. One transfer function of the
system is expressed as
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2-2
CONTROL SYSTEMS ENGINEERING
G ( s) =
C ( s)
R( s )
Now suppose the system is subjected to a unit impulse. Then the output is
C(s) = G(s) u(s)
Since u(s) = ∝δ(t) = 1
C(s) = G(s)
Taking inverse transform,
e(t) = g(t) where g(t) is the response of it which is not possible to generate a unit impulse.
Example 2.1
The open-loop transfer function of a system is given as
G ( s) =
7 s 2 + 18s + 15
s3 + 5s 2 + 11s + 15
Show the impulse response of the system.
 7 s 2 + 18s + 15 
=
G
(
s
)
=
 3

Solution 2
R( s )
 s + 5s + 11s + 15 
 3
2+ j
2− j 
+
C ( s) = 
+
 u( s )
2
s + 1− j 2 
 s + 3 s + 1− j
C ( s)
Since u(s) = ∝δ(t) = 1
 3
2+ j
2− j 
+
C ( s) = G ( s) 
+

2
s + 1− j 2 
 s + 3 s + 1− j
The impulse transform of the system is
2
2
C(t) = G(t) = 3 e–3t + (2 + j) e(–1 + j )t + (2 – j) e(–1 – j )
= 3 e–3t + e–t cos (2t + 0.464)
The response is shown in Figure 2.2.
8
7
6
C(t) 5
4
3
2
1
0
–1
–2
0.5
1.0
1.5
2.0
3.0
4.0
Time
5.0
–3
Fig. 2.2 Unit impulse response of a linear system of
Chapter 02.indd 2
G ( s) =
7 s 2 + 18s + 15
s3 + 5s 2 + 11s + 15
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MATHEMATICAL MODELS
2-3
The time-varying impulse response of the system will depend upon the location of the poles
of the system. A pole at the negative real axis contributes to exponentially decaying term to the
impulse response of the pole moves closer to zero, the time decay becomes slower. With complex
conjugate poles at s = – ∝ ± j β, the decay will be exponential sinusoidally. Such decay will be
slower when ∝ moves towards zero.
A pair of poles on the imaginary axis, i.e., s = ± jω will contribute sinusoidal term to the
impulse response whose amplitude neither grows nor decays. That means, the response will be
that of a sustained oscillation; such a response is often termed as marginal stability. Poles on
the right-hand side of the imaginary axis give rise to unstable operation of the system. Thus,
to sum up to state that a system with all its poles in the left-half plane of the imaginary axis is
characterized by exponentially decaying terms in the impulse response, and the system is called
a stable system as illustrated in Figure 2.3.
Im - axis
Pole
×
Pole
x
x
Pole
Stable half plane
Unstable half plane
×
Pole
×
– Real-axis
Pole
×
×
Pole
+ Real-axis
Fig. 2.4 Location of poles of the system transfer function and the corresponding impulse
response.
2.3
Transfer Function
In control theory, transfer functions are commonly used to characterize the input–output
relationship of components or systems that can be described by linear, time-invariant differential
equations. The relationship between input and output is represented by a diagram known as block
diagram as shown in Figure 2.4.
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2-4
CONTROL SYSTEMS ENGINEERING
Fig. 2.4 Block diagram representation of a system.
2.3.1
Definition of Transfer Function
The transfer function of a linear, time-invariant system is defined as the ratio of Laplace transform
of the output to the Laplace transform of the input, under the assumption that all initial conditions
are zero.
As shown in Figure 2.1, R(s) is the input and C(s) is the output. The output changes
according to the system transfer function, G(s).
Therefore,
C (s)
G(s) =
(2.1)
R( s )
Now let us consider a linear, time-invariant system defined by the following differential
equation:
a0 y n + a1 y n −1 + a2 y n − 2 +  + an −1 y + an = (b0 x m + b1 x m −1 +  + bm −1 x + bm )
where
(2.2)
n ≥ m,
y is the output of the system and
x is the input of the system.
Therefore, the transfer function of the system is obtained by taking Laplace transform of
both the sides under the assumption that the initial conditions are zero.
(a0 s n + a1s n −1 + a2 s n − 2 +  + an −1s + an ) Y ( s ) = (b0 s m + b1s m −1 +  + bm −1s + bm ) X ( s )
The transfer function is
G(s) =
=
L (output)
L (input) zero initial conditions
Y ( s)
X ( s)
=
b0 s m + b1 s m −1 +  + bm −1 s + bm
a0 s n + a1 s n −1 +  + an −1 s + an
(2.3)
Using the concept of transfer function it is possible to represent system dynamics and
at the same time, calculate the order of the system by knowing the highest power of s in the
denominator. The order of the system is the same as the highest power of s in the denominator.
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MATHEMATICAL MODELS
2.3.2
2-5
Poles and Zeros of a Transfer Function
The expression for transfer function can be represented, after factorization, as
G(s) =
k (s − z1 ) (s − z2 )  (s − zm )
(s − p1 ) (s − p2 )  (s − pn )
where k = bo/ao, called the gain factor of the transfer function.
If we substitute s equal to z1, z2, …, zn, the transfer function becomes zero. Hence z1, z2,
z3, … are called the zeros of the transfer function.
On the other hand, if we substitute s equal to p1, p2, …, pm, the transfer function
assumes infinite value. Hence p1, p2, p3, … are called the poles of the transfer function. The zeros
of G(s) are those values of s for which G(s) tends to zero. The poles of G(s) are those values of
s for which G(s) tends to infinity. The positions of poles and zeros in the s-plane are important
indicators of the performance of the system.
2.3.3
Procedure for Determining the Transfer Function of a
Control System
Transfer function of any system can be determined through the following steps:
Step 1: Formulate the mathematical equations for the system.
Step 2: Take the Laplace transform of the system equations assuming all the initial
conditions of the system as zero.
Step 3: Take the ratio of Laplace transform of the output to the Laplace transform of the
input.
Let us consider an example.
Let a system be described by the following differential equation:
Step 1: We write the equation representing the system
5
y + 3
y + 2 y + y = 4 
(2.4)
x + 2 x + x
where
y is the system output and
x is the system input. y, x are the derivatives.
Step 2: Take the Laplace transform of Equation (2.4). By assuming all the initial conditions
as zero, we get
(5s3 + 3s2 + 2s +1) Y(s) = (4s2 + 2s +1) X(s)
Step 3: Take the ratio of Laplace transform of the output to the Laplace transform of the
input. Transfer function is
G(s) =
Y ( s)
4s 2 + 2s + 1
(2.5)
5s 3 + 3s 2 + 2s + 1
After factorization of the numerator and denominator of the transfer function, the poles and
zeros can be determined.
Chapter 02.indd 5
X ( s)
=
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2-6
CONTROL SYSTEMS ENGINEERING
2.4Differential Equations of Physical System and Their
Transfer Function - Translations and Rotational mechanical
System
A physical system consists of a number of sub-systems connected together to serve a specific
purpose. If we consider a motor car as a mechanical system, it has a number of sub-systems like
ignition sub-system, pneumatic sub-system, power transmission sub-system and so on. Similarly,
we have electrical systems. In mechanical systems, we have rotational systems, translational systems
and so on. To have an insight and better understanding about the performance of a control system,
it is convenient to develop mathematical models of such systems and study and modify them for
getting better performance. It should be borne in mind that almost all physical systems are non-linear
to some extent and as such it may be difficult to write exact mathematical equations for all systems.
It may, therefore, be necessary to use the best possible linear approximation to analyse such systems.
There are a wide range of physical systems, therefore, it may be difficult to use the same
method of mathematical modelling and analysis in each case. In this section, we will discuss
briefly the electrical and mechanical systems and their modelling using transfer function
approach. We will assume that the systems are linear. The procedure will be same as listed in the
previous section.
2.4.1
Electrical System
A resistor, an inductor and a capacitor are the three basic elements of an electric circuit. The
circuit is analysed by the application of Kirchhoff ’s voltage and current laws.
The relationship that exists between the voltage and the current flowing through the circuit
elements may be expressed as
v = iR
For a resistive circuit,
Taking Laplace transform,
V(s) = RI(s)
v =
For a capacitive circuit,
Taking Laplace transform,
V(s) =
For an inductive circuit,
Taking Laplace transform,
1
C
1
t
∫ i dt
Cs
v =L
(Considering initial conditions to be zero)
o
I (s)
di
dt
V(s) = sLI(s)
When these basic elements form an electrical series circuit as shown in Figure 2.2,
mathematical formulation is made by using Kirchhoff’s laws.
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MATHEMATICAL MODELS
2-7
Fig. 2.5 An RLC circuit.
loop.
The RLC circuit of Figure 2.5 is analysed by Kirchhoff’s voltage law applied to the closed
The system equation is
e(t) = L
di
dt
+ Ri +
1
C
∫ i dt
(2.6)
Now taking Laplace transform on both sides, we get
1
E(s) = LsI (s) + RI (s) +
I (s)
Cs
(assuming all initial conditions to be zero)
1 

E(s) =  Ls + R +
 I ( s)
Cs 

 LCs 2 + RCs + 1 
E(s) = 
 I (s)
Cs


Let the output voltage vo(t) be taken across the capacitor, C. Then,
1
vo(t) = ∫ i dt
C
Taking Laplace transform on both sides of Equation (2.7), we get
1
Vo (s) =
I (s)
Cs
(assuming all initial conditions to be zero)
Therefore, the transfer function is given by
1
Vo (s)
I ( s)
Cs
G (s) =
=
Cs
2
E (s) ( LCs + RCs + 1)
I ( s)
V (s)
1
=
G (s) = o
2
E (s)
( LCs + RCs + 1)
(2.7)
The block diagram representation is shown in Figure 2.6.
Chapter 02.indd 7
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2-8
CONTROL SYSTEMS ENGINEERING
Fig. 2.6 Block diagram of the RLC network.
This is a second-order system since the highest power of s in the denominator is 2.
2.4.2
Mechanical System
In the analysis of mechanical systems, use is made of three idealized elements, namely, inertial
elements such as mass or moment of inertia, a spring and a damper. The inertial elements, that is,
mass and spring are capable of storing energy while the damper is capable of dissipating energy.
A damper is often referred to as a mechanical resistance. A mechanical system may have either
purely translational motion or purely rotational motion.
The equations of motions are generally formulated using Newton’s laws of motion. In
pure translational systems, motion is considered to take place in straight lines. Therefore, the
variables—force, displacement, velocity and acceleration—are aligned in a straight line. If a
mass m moves in a straight line with x as its position at any time with reference to a fixed
reference axis, then by applying Newton’s laws of motion, the force F acting upon it may be
expressed as
Force = Mass × Acceleration,
that is,
Chapter 02.indd 8
F =m
d2 x
= mx
dt 2
Again, if a mass m is moving with a velocity v, it will have a stored translational kinetic
energy of ½mv2. This provides us with an idea that mass is an energy-storing device.
A spring, when acted upon by a force, gets stressed. For a linear spring, the deformation
produced is directly proportional to the magnitude of the applied force. The equation describing
the relationship between F and x for a linear spring is F = Kx, where F is the force exerted on
the spring, x is the deformation of the spring and K is the stiffness coefficient of the spring.
A spring is usually represented by a metallic coil. A spring is compressed or elongated by an
amount x due to a force F.
An ideal spring will return the same amount of energy (in the form of work) when allowed
to come back to its original state.
Thus, we have seen that both mass and spring can store energy. Mass stores translational
kinetic energy which is due to its velocity while the spring stores translational potential energy
which is due to its position (elongated or compressed).
A damper is generally represented by a piston in a cylinder. A damper produces resistance
velocity which is often termed as friction. In a fluid medium, this friction is due to the viscosity.
When the force due to friction is proportional to relative velocity, the friction is known as linear
friction. The relationship of forces in a linear damper is given by
03/11/11 6:12 PM
MATHEMATICAL MODELS
f (t) = B
dx
dt
2-9
= Bx ,
where
f(t) is the damping (resisting) force due to relative velocity x between two movable parts and
B is the coefficient of viscous friction.
Rotation about a fixed axis takes place in a purely rotational system. The elements are
moment of inertia, that is, rotational mass, torsional spring and damping. Moment of inertia
is expressed as J. The governing equation relating angular velocity ω with applied torque T,
according to Newton’s second law of motion, is
T = Jω = J
d 2θ
= J 
θ
dt
Mechanical elements which experience deformation due to applied torque may be
considered rotational spring like a long shaft, helical spring and so on. The relationship of torque
T and angular displacement θ is expressed as
T = Kq
Rotational damping is the resistance to the rate of relative angular velocity between parts of
a rotating mechanical system. The relationship with torque T is
dθ
T =B
= Bθ
dt
Mechanical systems and devices can be modelled by three ideal translatory elements or
through three ideal rotational elements as shown in the following section.
2.4.2.1 Translational Mechanical System
The basic elements of any translatory mechanical system are the mass, the spring and the damper
as shown in Figure 2.7.
Fig. 2.7 Basic elements (mass, spring and damper) of a mechanical system.
The mass of the system is M (the unit of M is kilogram). The displacement of mass due
to applied force f (t) (the unit of force is newton) results in inertia force. This inertia force is
Chapter 02.indd 9
03/11/11 6:12 PM
2-10
CONTROL SYSTEMS ENGINEERING
the product of the mass and its acceleration. The spring deflection constant is K Newton/metre.
The restoring force fK for the spring is proportional to its displacement. The viscous damping in
the system is offered by a dashpot damper. The damping force varies in direct proportion to the
velocity. The coefficient of viscous damping is B Newton/rad/sec.
Application of force f (t) to the mass results in a displacement of x metre. The equation of
motion for the system is obtained by applying Newton’s second law of motion.
Inertia force = Mass × Acceleration = M
d2 x
dt 2
Applied force = Inertia force + Damping force + Restoring force by the spring.
Therefore,
f (t) = M
d2 x
d2 x
2
+B
dx
+ Kx
dt
dt
Rearranging Equation (2.8), the following equation is obtained
(2.8)
dx
(2.9)
+ Kx = f (t)
dt
dt
Taking Laplace transform on both sides and assuming all initial conditions to be zero,
Ms2X(s) + BsX(s) + KX(s) = F(s)
[Ms2 + Bs + K]X(s) = F(s)
X ( s)
1
G(s) =
(2.10)
=
2
F ( s)
Ms + Bs + K
This can be represented in the form of a block diagram as shown in Figure 2.8.
M
2
+B
Fig. 2.8 Block diagram representation of the mechanical system shown in Fig. 2.7.
This is a second-order system as the highest power of s in the denominator of the transfer
function is 2.
Figure 2.9 shows the equivalent circuit of the mechanical system of Figure 2.4.
Fig. 2.9 Equivalent circuit.
Chapter 02.indd 10
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MATHEMATICAL MODELS
2-11
How this circuit is analogous to the mechanical system will be understood from the force–
current analogy, which is described in Section 2.4.3.2.
2.4.2.2 Rotational Mechanical System
Mechanical systems involving rotation around a fixed axis are often seen in machineries such
as turbines, pumps, rotating discs, gears, generators, motors and so on. Figure 2.10 shows a
rotational mechanical system that consists of a rotating disc of moment of inertia J and a shaft of
stiffness K. The disc rotates in a viscous medium with a viscous friction coefficient B
Fig. 2.10 Rotational mechanical system.
Let T be the applied torque which tends to rotate the disc. The three basic components of
the rotational system are moment of inertia, viscous friction and spring action (torsional).
The system equation can be written using the relation
Applied torque = Inertia torque + Damping torque
+ Angular displacement (torsional) torque (2.11)
Inertia torque = Moment of inertia × Angular acceleration.
Thus,
d 2θ
dθ
T =J 2 +B
(2.12)
+ Kθ
dt
dt
The equivalent circuit diagram of the mechanical system is shown in Figure 2.11.
Fig. 2.11 Equivalent circuit of the rotational mechanical system shown in Fig. 2.10.
Chapter 02.indd 11
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2-12
CONTROL SYSTEMS ENGINEERING
Taking Laplace transform of Equation (2.12) and assuming all initial conditions as zero,
T (s) = Js 2 θ (s) + Bsθ (s) + K θ (s)
T (s) = ( Js 2 + Bs + K ) θ (s)
1
θ (s)
= 2
T (s)
Js + Bs + K
This is represented in the form of a block diagram as shown in Figure 2.12.
(2.13)
Fig. 2.12 Block diagram of a rotational mechanical system.
2.4.3
Analogies of Mechanical and Electrical Systems
Sometimes, mechanical and other systems are converted into electrical analogous systems for
the ease of design, modifications and analysis. Analogous systems have same type of differential
equations.
2.4.3.1Force–Voltage Analogy
Equation (2.9) which was written for a mechanical system represented by mass, spring and
damper is reproduced in the following as
M
d2 x
dt
2
+B
dx
dt
+ Kx = f (t)
Equation (2.9) reproduced.
The analogy of this equation can be established by the voltage equation of a RLC electrical
circuit is shown in Figure 2.13. The voltage equation for the circuit as established in Equation (2.6) is reproduced as follows:
di
1
L
+ Ri + ∫ idt = e (t)
C
dt
Fig. 2.13 Electrical circuit for force–voltage analogy.
Chapter 02.indd 12
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MATHEMATICAL MODELS
2-13
As the current is the rate of flow of electric charge, i = dq/dt. Thus, the equation becomes
L
d 2q
2
+R
dq
1
(2.14)
q = e (t)
dt C
dt
It is observed that Equation (2.9) is analogous to Equation (2.14). The equation for
the translational mechanical system represented by mass, spring and damper; the rotational
mechanical system represented by moment of inertia, viscous friction coefficient, stiffness; and
an electrical system of inductance, resistance, capacitance are rewritten together respectively as
+
M
J
dt
2
d 2θ
dt
d 2q
+B
+B
2
+R
dx
dt
dθ
dt
dq
+ Kx = f (t)
+ Kθ = T
1
q = e (t)
dt C
dt
From the above three equations, the analogies between the mechanical, translational as well
as rotational systems and electrical system are established as in Table 2.1.
In mechanical system, f(t) is the force applied whereas in electrical system, e(t) is the
voltage applied. This is the reason for which the analogy is called force–voltage analogy.
In rotational system, force is replaced by torque.
L
d2 x
+
Table 2.1 Analogous Quantities of Mechanical and Electrical Systems (Force–Voltage Analogy).
Mechanical Translational System
Mechanical Rotational System
Electrical System
Force, f
Torque, T
Voltage, e
Mass, M
Moment of inertia, J
Inductance, L
Viscous friction coefficient, B
Viscous friction coefficient, B
Resistance, R
Spring stiffness, K
Torsional spring stiffness, K
Reciprocal of capacitance, 1/C
Displacement, x
Angular displacement, q
Charge, q
Velocity, x
Angular velocity, θ
Current, i
2.4.3.2Force–Current Analogy
Applying Kirchhoff’s current law in the L–R–C parallel circuit shown in Figure 2.14, we can
write
iL + iC + iR = i(t)
(2.15)
we know that the emf induced in an inductor is
di
1
e = L L ; current through the inductor iL is therefore, iL = ∫ e dt
dt
L
and the charge on a capacitor is q = ∫ iC dt and q = Ce.
Chapter 02.indd 13
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2-14
CONTROL SYSTEMS ENGINEERING
p2
p1
Fig. 2.14 Electrical circuit for force–current analogy.
∫ iC dt
\
= Ce,
or current through the capacitor iC is, therefore,
iC = C
de
Also,
iR R = e;
iL =
dt
iR =
1
e dt;
L∫
1
R
e
iR =
1
R
e;
iC = C
de
dt
Substituting the values of iL, iR and iC in Equation (2.15), we get
1
de
e dt + C
L∫
dt
+
1
R
e = i(t)
(2.16)
In terms of flux linkage, ψ the above equation can be represented by substituting,
dψ
(2.17)
where y = Nf i.e. flux linkage
e =
dt
Equation (2.16) becomes
1
L
ψ+
1 dψ
R dt
+C
d 2ψ
dt 2
= i(t )
 e dt = ψ 
 ∫

Rearranging, we get
C
d 2ψ
dt
2
+
1 dψ
R dt
+
1
L
ψ = i (t)
(2.18)
This equation can be compared with the equation relating the mechanical translational
system of Equation (2.9) and equivalent rotational system of Equation (2.12) as in Table 2.2.
Chapter 02.indd 14
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MATHEMATICAL MODELS
2-15
Table 2.2 Analogous Quantities of Mechanical and Electrical Systems (Force–Current Analogy).
Mechanical Translational System
Mechanical Rotational System
Electrical System
Force, f
Torque, T
Current, i
Mass, M
Moment of inertia, J
Capacitance, C
Viscous friction coefficient, B
Viscous friction coefficient, B
Reciprocal of resistance, 1/R
Spring stiffness, K
Torsional spring stiffness, K
Reciprocal of inductance, 1/L
Displacement, x
Angular displacement, q
Flux linkage, y
Velocity, x
Angular velocity, θ
Voltage, e
On the basis of force–current analogy, we can derive the equations for the mechanical
system by developing an equivalent circuit diagram, as has been done in the case of force–voltage
analogy. We will now consider a few other systems and determine their transfer functions.
2.4.4
Hydraulic System
Let us determine the transfer function of a typical hydraulic system. A hydraulic system works
due to liquid pressure difference.
A dashpot (or damper) of a hydraulic system is shown in the Figure 2.15. Whenever a step
displacement x is applied to the piston, a corresponding displacement y becomes momentarily
equal to x. The applied force will make oil to flow through the restriction R and the cylinder will
return to its original position.
Assuming inertia force to be negligible, the forces balancing are represented as
A ( p1 – p2) = ky
(2.19)
where A is the piston area,
k is the spring constant and p1, p2 are the oil pressure existing on the right-hand side and
left-hand side of the piston.
Fig. 2.15 Damper of a hydraulic system.
The flow rate q through the restriction with resistance R is given by
p − p2
q = 1
R
Chapter 02.indd 15
(2.20)
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2-16
CONTROL SYSTEMS ENGINEERING
The oil is assumed to be incompressible (so oil density ρ = constant). As the mass of oil
flow through the restriction in time dt must balance with the change in mass to the left-hand side
of the piston, we have
q dt = Aρ (dx − dy)
Rearranging and using Equations (2.19) and (2.20), we obtain
q
p − p2
ky
dx dy
=
= 1
=
−
Aρ
RAρ
dt dt
RA2 ρ
dx
dy
ky
or
=
+
dt
dt RA2 ρ
Taking Laplace transform with zero initial conditions, we get
k
s × (s) = sY (s) +
Y (s)
RA2 ρ
So the transfer function of the system becomes
G (s) =
RA2 ρ
Let
k
Then,
2.4.5
Y ( s)
X ( s)
=
s


s+ k 2 
 RA ρ 
= t i.e. time constant
G(s) =
s
s+
1
τ
=
τs
1 + τs
Pneumatic System
A pneumatic system works due to the pressure difference of air or any other gas. We shall consider
a simple pneumatic system with an actuating value as shown in Figure 2.16.
Fig. 2.16 Schematic diagram of a simple pneumatic system
complete arrangement not shown.
Chapter 02.indd 16
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MATHEMATICAL MODELS
2-17
As shown in Figure 2.13, air at a pressure, pi is injected through the input manifold. The
plunger arrangement has a mass M, B and K being the coefficients of viscous friction and spring
constant, respectively. If A is the area of diaphragm, then the force exerted on the system will
be Api. The output is the displacement of the plunger shown as y.
Force
F = Pressure × Area
F = pi × A = APi
that is,
Again,
d2 y
dy
F =M 2 +B
+ Ky
dt
dt
\
M
d2 y
dt
2
+B
dy
dt
+ Ky = APi
(2.21)
Taking Laplace transform of both sides and assuming initial conditions to be zero, we get
Ms2Y(s) + BsY(s) + KY(s) = APi(s)
The transfer function is
2.4.6
G (s) =
Y ( s)
Pi (s)
=
2
A
Ms + Bs + K
(2.22)
Thermal System
Figure 2.17 shows a water heating system (a thermal system) where input water temperature is
θi and output water temperature is θo .
The following assumptions are made to analyse the thermal system and determine its
transfer function.
(1) The temperature of the medium is uniform
(2) The tank is insulated from the surrounding atmosphere.
Fig. 2.17 A simple water heating system.
Let the steady-state temperature of inflowing water is θi and that of outflowing water is θo .
The steady-state heat input rate from the heater is Q.
Chapter 02.indd 17
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2-18
CONTROL SYSTEMS ENGINEERING
Let the water flow rate be constant. Any increase in heat input rate, ∆H, will be balanced
by increase in heat outflow rate, ∆H1 , and heat storage rate in the tank, ∆H 2 .
Using heat balance equation, we can write
∆H = ∆H1 + ∆H 2
(2.23)
∆H1 = QS ∆θo
where
Q is water flow rate and S is the specific heat.
∆θo
∆H1 =
R
where R = 1/QS is called the thermal resistance.
Heat storage rate in the tank,
∆H 2 = MS
or,
∆H 2 = C
d
d
dt
dt
(2.24)
(∆θo )
(∆θo )
(2.25)
where
M is the mass of water in the tank,
C = MS, is the thermal capacitance of water and
d ( ∆θo )
is the rate of rise of temperature of water in the tank.
dt
Now from Equations (2.23), (2.24) and (2.25), we can write,
∆ θo
d
( ∆ θo )
R
dt
In Laplace transform form, the quantities can be expressed as
∆H =
H (s) =
1
R
+C
θo (s) + Csθo (s)
 RCs + 1 
H (s) = 
 θ ( s)
 R  o
The transfer function is
G (s) =
θo ( s )
H ( s)
=
R
RCs + 1
Example 2.2 Write the equation relating the various forces acting on a mechanical system
shown in Figure 2.18 (a). Show that the electrical circuit of Figure 2.18 (b) is its analogous
system.
Chapter 02.indd 18
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MATHEMATICAL MODELS
2-19
Fig. 2.18 Mechanical and electrical analogous systems.
Solution The forces acting on the system of Figure 2.18 (a) are external applied force f(t)
and the internal resisting forces, which are inertia force, M d 2x/dt 2; damping force, B dx/dt; and
spring force, kx.
Equating the applied force with the resisting forces, we get
d2 x
f (t) = M
dt
dx
+B
2
dt
+ kx
(2.26)
The voltage equation for the R–L–C circuit shown in Figure 2.18 (b) is written as
e (t) = L
di
dt
+ Ri +
1
C
∫ i dt
(2.27)
since i = dq/dt and q = ∫ i dt, the above equation is represented as
e (t) = L
d 2q
dt
2
+R
dq
dt
+
1
C
q
(2.28)
By observing Equations (2.26) and (2.28), we conclude that the two systems shown in
Figure 2.18 are analogous systems.
Example 2.3 Show that circuit of Figure 2.19 (b) is the analogous electrical network for the
system shown in Figure 2.19 (a) using force–voltage analogy.
Solution The forces acting on mass M1 and M2 are represented using free body diagrams as
shown in Figure 2.20. The balancing equations are then written from Figures 2.20 (a) and (b) as
Chapter 02.indd 19
f (t) = M1
d 2 x1
dt
2
+ B1
d
dt
( x1 − x2 )
(2.29)
03/11/11 6:12 PM
2-20
CONTROL SYSTEMS ENGINEERING
Fig. 2.19 (a) A mass-spring-damper system; (b) Electrical equivalent circuit using
force-voltage analogy.
Fig. 2.20 Free-body diagrams of the system of Fig. 2.19 (a).
d
( x1 − x2 ) = M2
d 2 x2
B1
e (t) = L1
and
d
d 2 q1
dt
2
d 2 q2
+ R1
Chapter 02.indd 20
R1
2
+ R2
d
dt
(q1 − q2 )
dq2
1
q
dt
C 2
dt
dt
since q = ∫ i dt or dq/dt = i, the equation can also be written as
di
e (t) = L1 1 + R1 (i1 − i2 )
dt
(q1 − q2 ) = L2
2
+ B2
dx2
(2.30)
+ kx2
dt
dt
dt
We will now apply force-voltage analogy for which we shall refer to Table 2.1.
Replacing f (t) by e (t), M1 by L1, M2 by L2, x1 by q1, x2 by q2, B1 by R1, B2 by R2 and k by
1/C, these equations are written as
and,
+
03/11/11 6:12 PM
MATHEMATICAL MODELS
2-21
and
R1 (i1 – i2) = L2
di2
+ R2 i2 +
1
i dt
c∫ 2
dt
These equations are the equations for the analogous electrical circuit shown in Figure 2.19 (b).
The student can write the Kirchhoff’s voltage equations for the two loops of the electrical circuit
shown in Figure 2.19 (b).
Example 2.4 Obtain the transfer function of the mechanical system shown in Figure 2.21
and draw its analogous circuit.
Fig. 2.21 A mechanical translational system.
Solution Figure 2.22 shows the free body diagram of the various forces acting on mass M1
and M2 respectively.
Fig. 2.22 Free body diagrams of the system shown in Fig. 2.21.
The equations of the balancing forces on M2 and M1, respectively, are
f (t) = M 2 
x + B2 x + B ( x − x1 ) + K ( x − x1 )
Using Laplace transform,
(M2s2 + B2s + Bs + K) X(s) – (Bs + K) X1(s) = F(s)
(2.31)
Again,
M1 
x1 + B1 x1 + K1 x1 = B ( x − x1 ) + K ( x − x1 )
or
( M1 s 2 − B1 s + Bs + K + K1 ) X1 (s) = ( Bs + K ) X (s)
Substituting the value of X1(s) from Equation (2.32) in Equation (2.31), we get
( M 2 s 2 + B2 s + Bs + K ) X ( s ) −
Chapter 02.indd 21
( Bs + K ) ( Bs + K ) X ( s )
( M1s 2 + B1s + Bs + K + K1 )
(2.32)
= F (s)
03/11/11 6:13 PM
2-22
or
CONTROL SYSTEMS ENGINEERING
{( M2 s2 + B2 s + Bs + K ) ( M1s2 + B1s + Bs + K + K1 ) − (Bs + K )2 } X (s)
= ( M1 s 2 + B1 s + Bs + K + K1 ) F (s)
The transfer function is
X ( s)
F ( s)
=
M1 s 2 + B1 s + Bs + K + K1
( M2 s 2 + B2 s + Bs + K ) ( M1 s 2 + B1 s + Bs + K + K1 ) − ( Bs + K )2
Converting the nodal equations [Equations (2.31) and (2.32)] into comparable electrical
analogous equations, we get
di
1
+ R2 i + R (i − i1 ) + ∫ (i − i1 ) dt = e (t)
(2.33)
L2
C
dt
di
1
1
i1 dt = R (i − i1 ) + ∫ (i − i1 ) dt
L1 1 + R1i1 +
(2.34)
∫
dt
C1
C
Using Equations (2.33) and (2.34), the electrical analogous circuit based on force–voltage
analogy is drawn as shown in the below diagram. Equation (2.33) is for loop ABCDA and
Equation (2.34) is for loop AEFDA.
Fig. 2.23 Electrical equivalent of the mechanical system shown in Fig. 2.21.
Here, L2 = M2, R2 = B2, C = K, R1 = B1, L1 = M1, C1 = K1, i = x, i1 = x1, e(t) = f (t) and R = B.
Example 2.5 Obtain the transfer function of the mechanical system shown in Figure 2.24.
Fig. 2.24 A mechanical rotational system.
Solution The circuit diagram for the above system has been shown in the Figure 2.25.
Chapter 02.indd 22
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MATHEMATICAL MODELS
2-23
Fig. 2.25 The mechanical system of Fig. 2.24 has been represented
in the form of a network diagram.
The nodal equations at Nodes θ1 and θ are
At Node θ1 ,
J1
θ1 + K (θ1 − θ) = T (t)
At Node θ,
J 2 
θ + B θ = K (θ1 − θ)
or
(2.35)
( J1 s 2 + K ) θ1 (s) − K θ(s) = T (s)
and
(2.36)
( J 2 s 2 + Bs + K ) θ(s) = K θ1 (s)
The equations representing the mechanical system with moment of J1 and J2 respectively
are written as
T(t) = J1
θ1 + K (θ1 − θ) and J 2 
θ + Bθ = K (θ1 − θ2 )
Substituting value of θ1 (s) from Equation (2.36) in Equation (2.35), we get
 ( J1 s 2 + K ) ( J 2 s 2 + Bs + K )

− K  θ(s) = T (s)

K


Therefore, the transfer function is
θ(s)
T (s)
=
=
2
K
2
( J1 s + K ) ( J 2 s + Bs + K ) − K 2
K
J1 J 2 s 4 + J1 Bs 3 + (KJ1 + KJ 2 ) s 2 + KBs
Example 2.6 Draw the network diagram for the system in Figure 2.26 and also draw its
analogous circuit.
Solution The network diagram for the system in Figure 2.26 is shown in Figure 2.27.
Node x1,
f (t) = M1 
x1 + B1 x1 + K1 x1 + K3 x1 − K3 x2
Chapter 02.indd 23
03/11/11 6:13 PM
2-24
CONTROL SYSTEMS ENGINEERING
Fig. 2.26 A mechanical translational system.
Fig. 2.27
Node x2,
Circuit diagram for the system of Example 2.5.
K2 x1 – K2 x2 = M 2 
x2 + B2 x2 + K3 x2 + K 4 x2 − K 4 x3
or
M 2 
x2 + B2 x2 + (K 2 + K3 + K 4 ) x2 − K 2 x1 − K 4 x3 = 0
Node x3,
M3 
x3 = K 4 x 2 − K 4 x3
or
M3 
x3 − K 4 x 2 + K 4 x3 = 0
Using force–current analogy (replacing force by current, mass by capacitance, velocity x
by voltage v, damping coefficient B by 1/R and stiffness k by 1/L):
i (t) = C1
and
C2
Chapter 02.indd 24
dv2
dt
+
v2
dv1
dt
+
v1
R1
+
1
1
1
v dt +
v dt −
v dt
L1 ∫ 1
L2 ∫ 1
L2 ∫ 2
1
1 
1
 1
+
+
+
 ∫ v2 dt −
R2  L2 L3 L4 
L4
1
∫ v3 dt − L2 ∫ v1 dt = 0
03/11/11 6:13 PM
MATHEMATICAL MODELS
and
dv3
2-25
 dx


= v, x = ∫ v dt 

dt
 dt

Using the above three equations, the analogous electrical circuit is drawn as shown in
Figure 2.28.
C3
+
1
1
v dt −
v dt
L4 ∫ 3
L4 ∫ 2
=0
Fig. 2.28 Electrical equivalent circuit of the mechanical system of Fig. 2.26 has been
drawn using force-current analogy.
Example 2.7 Write the nodal equations for the mechanical system shown in Figure 2.29 and
draw its analogous electrical network.
Fig. 2.29 A mechanical system represented by mass-spring-damper.
Solution The mechanical network diagram for the given system is shown in Figure 2.30.
The following are the nodal equations:
For Node x1,
f (t) = M1 
x1 + Bx1 + Bx1 + K1 x1 + K 2 ( x1 − x2 )
In Laplace transform form,
Chapter 02.indd 25
F (s) = (s 2 M1 + sB + K1 + K 2 ) X1 (s) − K 2 X 2 (s)
03/11/11 6:13 PM
2-26
CONTROL SYSTEMS ENGINEERING
Fig. 2.30 Network diagram of the mechanical system shown in Fig. 2.29.
For Node x2,
K2 (x1 – x2) = M2 
x2
Taking Laplace transform,
(s2M2 + K2) X2(s) – K2 X1(s) = 0
Electrical analog circuit based on force–voltage analogy is drawn as shown in Figure 2.28.
Fig. 2.31 Electrical equivalent circuit of the mechanical system represented as in Fig. 2.29.
The differential equations for the electrical analog are
di
1
1
v (t) = Ri1 + L1 1 +
i1 dt +
∫
dt C1
C2
1
∫ i1 dt − C2 ∫ i2 dt
or
or
or
Chapter 02.indd 26
v (t) = R
dq1
dt
+ L1
d 2 q1
dt
2
+
q1
C1
+
q1
C2
−
q2
C2
1
1 
1

V (s) =  sR + s 2 L1 +
+
Q (s)
 Q1 (s) −
C1 C2 
C2 2

1
1 
Q2 (s)

+
V (s) =  s 2 L1 + sR +
 Q1 (s) −
C1 C2 
C2

03/11/11 6:13 PM
MATHEMATICAL MODELS
2-27
and
L2
di2
dt
+
1
1
i dt −
i dt
C2 ∫ 2
C2 ∫ 1
=0
or
L2
d 2 q2
dt
2
+
q2
C2
−
q1
C2
=0
1 
Q1 (s)
 2
=0
 s L2 +
 Q2 (s) −
C2 
C2

These equations are similar to the nodal equations formed earlier.
Example 2.8 Draw the network diagram for the system shown in Figure 2.32 and write the
equations of performance.
Fig. 2.32
Solution Equating force × distance on both sides of the hinge,
or
f (t ) × R1 = f ′(t ) × R2
f ′(t ) = f (t ) ×
R1
R2
Equating forces acting on mass M2,
f ′(t ) = M2 
x2 + K ( x2 − x1 )
(a)
Equating forces acting on mass M1,
K (x2 – x1) = M1 
x1
(b)
Using the Equations (a) and (b), the network diagram is drawn as shown in Figure 2.33.
Chapter 02.indd 27
03/11/11 6:13 PM
2-28
CONTROL SYSTEMS ENGINEERING
Fig. 2.33 Network diagram of the mechanical system of Fig. 2.32.
Example 2.9 A hydraulic actuator with incompressible oil inside the cylinder is shown in
Figure 2.34. The mass of piston and leakage are assumed negligible. Obtain the transfer function
relating displacement x of the valve piston and displacement y of the main piston.
Solution Figure 2.34 shows partial diagram of a hydraulic system where linear motion of a
small piston valve (not shown in the figure) in a cylinder can cause movement of large piston in
the main cylinder.
Fig. 2.34 A simplified hydraulic pump.
On application of a small force, piston valve in the control cylinder moves from left to right
by a distance x. The two ports A and B of the main cylinder opens. Oil at high pressure (actuated
by a pressure pump) enters through port A and develops a pressure P1. Oil at the other side of
the piston will have a comparatively lower pressure. The low pressure oil will return to pump
through the outlet valve. Let us call the pressure difference on the two sides of the main piston
as PD.
The quantity of oil flow is q which is a function of displacement x and PD.
Thus we can write,
q = ∫ ( x, PD )
(2.37)
By partial differentiation, we get
Chapter 02.indd 28
∂q =
∂q
∂x
dx +
∂q
∂PD
dPD
(2.38)
03/11/11 6:13 PM
MATHEMATICAL MODELS
2-29
with zero initial conditions for all the variables. On integration, we get
∂q
 ∂q 
q =  x +
P
∂PD D
 ∂x 
or
q = K1 x – K2 PD
(2.39)
If the displacement of the main piston is dy in time dt (see Fig. 2.34) and assuming oil to be
incompressible and neglecting leakage, the rate of flow of oil into the main cylinder with piston
area A is given by
dy
(2.40)
Quantity of oil flow q = A
dt
From Equations (2.39) and (2.40),
dy
= K1 x − K 2 PD
A
dt
or
1 
dy 
PD =
 K1 x − A 
K2 
dt 
Since pressure is force per unit area, the force (F) developed by the piston is APD,
F =
\
A 
dy 
 K1 x − A 
K2 
dt 
(2.41)
The equation relating force F applied to a mass M and having coefficient of viscous friction
B at the load (see Fig. 2.34), is written as
F =M
d2 y
2
+B
dy
(2.42)
dt
dt
Equating Equations (2.41) and (2.42) we get,
M
d2 y
dt
2
+B
dy
dt
=
A 
dy 
 K1 x − A 
K2 
dt 
Taking Laplace transform of both sides and assuming initial conditions as zero, we obtain
A
Ms 2Y (s) + BsY (s) =
[ K X (s) − AsY (s)]
K2 1
or

A2 s 
AK1
Y (s)  Ms 2 + Bs +
 =
X ( s)
K 2 
K2

The transfer function is
Chapter 02.indd 29
G (s) =
Y (s)
X ( s)
=
AK1
K2
Ms 2 + Bs + ( A2 s/ K 2 )
03/11/11 6:13 PM
2-30
CONTROL SYSTEMS ENGINEERING
A
=
The transfer function is
where
G (s) =
K =
t =
AK1
K2
K 2 BK 2 + A2
K1
K2
=
MK 2




A2 
s
s 1 +
s   B +
 + Ms 
2 
BK 2 + A 
K2 


 
[Dividing numerator and denominator by (BK + A2)/K2]
K
(2.43)
s[1 + sτ]
AK1
( BK 2 + A2 )
MK 2
and
( BK 2 + A2 )
Example 2.10 Derive the transfer function of a simple mercury glass thermometer and
draw its electrical analogous electrical circuit.
Solution The cross-section of mercury in a simple glass thermometer is shown in Figure 2.35.
The thermometer is assumed to have thermal capacitance C which stores heat and thermal
resistance R which limits that flow.
Fig. 2.35 Cross-sectional view of a mercury glass thermometer.
The thermometer responds to temperature changes because of the heat absorbed or lost by
the mercury. Heat input from the surrounding is hA(θi − θo ).
dθ
Heat stored = C o
dt
Balancing heat absorbed with heat stored,
dθ
hA(θi − θo ) = C o
dt
C d θo
= θi − θo
or
hA dt
dθ
or,
(2.44)
τ o = θi − θo
dt
Chapter 02.indd 30
03/11/11 6:13 PM
MATHEMATICAL MODELS
2-31
where
τ = C hA is the time constant of the thermometer,
C is the thermal capacitance,
A is the surface area and
h is the film coefficient of solid–liquid interface.
It may be noted that the heat capacity of glass has been neglected as it is insignificant as
compared to that of mercury. The time constant can be written as t = RC, where R = 1 hA is the
thermal resistance for convection.
Taking Laplace transform of Equation (2.44) with zero initial conditions, we get
θo ( s )
1
(a)
=
θi (s)
τs + 1
Referring to Equation (2.44), an electrical analog for the thermometer system is given by
the circuit shown in Figure 2.36. The voltage equations are:
1
1
ei = RC + ∫ i dt and eo = ∫ i dt
C
C
Taking Laplace transform,
1
ei(s) = RI (s) +
I (s)
Cs
1
eo(s) =
I (s)
Cs
1
eo (s)
= G( s ) =
1 + RCs
ei (s)
1
=
(b)
1 + τs
Equations (a) and (b) are analogous.
Fig. 2.36 Electrical analogous circuit of a thermometer.
2.5
Concept of State-Space Modelling
For the analysis and design of any feedback control system using any technique, the system is
first modelled in the form of transfer function. Although the transfer function method of system
modelling is useful, the method suffers from certain drawbacks.
Chapter 02.indd 31
03/11/11 6:13 PM
2-32
CONTROL SYSTEMS ENGINEERING
The following are the drawbacks of transfer function method of system modelling:
(i) The transfer function is defined only under zero initial conditions.
(ii) It is only applicable to linear time-invariant systems and generally restricted to single-input
single-output (SISO) systems.
(iii) Although it provides output for certain input, it provides no information about the internal
state of the system.
To overcome these drawbacks of the transfer function modelling approach of system analysis
and design, a more generalized and powerful technique of state-space modelling technique in the
time domain was developed.
The state-variable method of modelling, analysis and design is applicable to linear and nonlinear, time-invariant and time-varying, multi-input multi-output (MIMO) systems. It is a direct
time domain approach and provides a basis for modern control theory. However, the classical
approach of control system design and analysis, described in this book, provides a deep insight
into the system and greatly assist in the preliminary design by approximating a complex system
into a simple system.
2.5.1
Concepts of State, State-Variables and State-Space
Some basic concepts like state, state-variables and state-space are provided below before taking
examples of state-space modelling.
State: The state or status of a system represents the minimum amount of information that
is required to be known about the system at time t = t0 so that the behaviour of the system can be
determined for t ≥ t0 without referring to input before time t0. For example, consider a network
shown in Figure 2.37.
Fig. 2.37
If the initial current through the conductor and the initial voltage across the capacitor is
known, then for any input voltage, the behaviour of the network can be determined. For this
network, we may consider inductor current, iL (0) capacitor voltage, vC (0) at t = 0 as the state of
the network.
State-variables: The state-variables of a dynamic system are the smallest set of variables
that completely determines the state of the system.
State-vector: This is a vector consisting of n number of state-variables that completely
determines the behaviour of a given system. These n state-variables will be taken as n components
of a vector, x. Such a vector will be called a state-vector.
We can now define a state-vector as a vector which determines the system state x (t) for time
t ≥ 0 if the state at t = t0 is known and the input u(t) for t ≥ t0 is also known.
Chapter 02.indd 32
03/11/11 6:13 PM
MATHEMATICAL MODELS
2-33
State-space: The state-space is an x-dimensional space whose coordinate axes are n-number
of state-variables, x1, x2, x3, …, xn. Any state can be represented by a point in the state-space. If
x1, x2, x3, …, xn are the n state-variables of a dynamic system, then its state-vector is represented
by a matrix as
 x1 
x 
 2
x =  x3 
 
 
x 
 n
where x1, x2, x3, …, xn are all functions of time t.
2.5.2
State-Space Equation
The input variables, the output variables and the state-variables are the three types of variables
used in state-space modelling of dynamic systems. Let us assume that there are r inputs u1(t),
u2(t), …, un(t), in outputs y1(t), y2(t), …, yn(t); and n state-variables x1(t), x2(t), …, xn(t) that
completely describe a multiple-input multiple-output system. Then the state equations of a timeinvariant system will be represented by n differential equations of first-order. The general form
of state equations will be
x (t ) = Ax (t ) + Bu(t )
y(t) = Cx(t) + Du(t)
where
y = n-dimensional output vector,
u = r-dimensional input vector,
A = n × n system matrix,
B = n × r input matrix,
C = m × n output matrix and
D = m × r transmission matrix
The matrices A, B, C, D are defined as follows.
Chapter 02.indd 33
 a11
a
 21
A =  a31

 
a
 n1
 c11
c
C =  21
 

 cm1
a12
an 2
c12
cm 2
... a1n 





... ann 
 b11
b
B =  21
 

 bn1
... c1n 




... cmn 
 d11
d
D =  12
 

 dm1
b12
...
d12
...
... b1r 




... bnr 
d1r 




... dmr 
...
03/11/11 6:13 PM
2-34
CONTROL SYSTEMS ENGINEERING
The block diagram of the above time-invariant MIMO (r-input, m-output) system can be
drawn as shown in Figure 2.38.
Fig. 2.38
Block diagram of a linear time-invariant continuous time MIMO control system
in state-space.
Example 2.10 The differential equation describing a system is given as
d3y
dt
3
+5
d2y
dt
2
+ 10
dy
dt
+ 9 y = u (t )
where y is the output and u is the input to the system. Represent the system in state-space.
Solution Let the state-variables be
x1 = y,
x2 = y,
x3 = 
y
The given equation is written as
x3 = u(t ) − 9 x1 − 10 x2 − 5 x3
1 0   x1 
 0
 x1 
 x  =  0
0
1  x2  +
 2

 x3 
 −9 −10 −5   x3 
0 
 0  u(t )
 
 1 
The general form of state equation is
x (t ) = Ax (t ) + Bu(t )
Thus, here
2.5.3
1 0
 0

0
1
A = 0
 −9 −10 −5 
0 
B =  0 
 1 
and
 x1 
x (t ) =  x2 
 x3 
Concepts of Controllability and Observability
The concepts of controllability and observability were introduced by R. Kalman and play an
important role in the design of modern control system in state-space. There are four possible
states for a control system: (i) the system is controllable and observable, (ii) the system is
controllable and unobservable, (iii) the system is observable and uncontrollable and (iv) the
system is uncontrollable and observable.
Chapter 02.indd 34
03/11/11 6:14 PM
MATHEMATICAL MODELS
2-35
A system is said to be completely controllable if it is possible to find an input u(t) that will
transfer the system from any initial state x(t0) to any given final state x(t1) over a time interval
t1 – t0.
A system is said to be completely observable if its state can be determined knowing the
input u(t) and output y(t) over a finite interval of time.
Kalman’s test tells us whether it is possible to control all the states of the system completely
by a suitable choice of an input or whether it is possible to reconstruct the states of a system from
its input and output.
The concepts of controllability and observability help us apply state-variable feedback and
reconstruct system states from the system input and output. System performance can be improved
by using state-variable feedback to place the poles of a closed-loop system at suitable locations.
The students will be introduced to the subject in more detail while dealing with non-linear
control system design and analysis.
REVIEW QUESTIONS
2.1
Draw the corresponding network diagram for the simple mass-spring-damper mechanical
system as shown in the Figure 2.39 and write its system equations.
Fig. 2.39
2.2
Draw the corresponding network diagram for the damped mechanical system as shown in
the Figure 2.40 and write its system equations.
Fig. 2.40
2.3
Chapter 02.indd 35
Write the force or torque equations for the mechanical systems as shown in the Figure 2.41.
Draw their equivalent mechanical circuits.
03/11/11 6:14 PM
2-36
CONTROL SYSTEMS ENGINEERING
Fig. 2.41
2.4
Determine a mathematical model for the operational amplifier circuit as shown in the
Figure 2.42.
Fig. 2.42
2.5
Obtain the transfer function of the mechanical system as shown in the Figure 2.43 and draw
its analogous circuit.
Fig. 2.43
Chapter 02.indd 36
03/11/11 6:14 PM
MATHEMATICAL MODELS
2.6
2-37
A pneumatic bellow (Fig. 2.44) made from corrugated copper, when employed with
a variable resistance (turbulent) RT , is useful in pneumatic controllers as a feedback
element to sense and control the pressure pi. The displacement x is transmitted by
a linkage or beam balance to a pneumatic valve regulating the air supply. The elasticity
of the bellow is modelled as a spring with constant K. Develop a model for the displacement x.
Fig. 2.44
2.7
Considering small deviations of variables from their steady-state values, obtain a model for
outlet temperature θo of the air heating system as shown in the Figure 2.45.
Fig. 2.45
2.8
θi is the temperature of the inlet air and h is the heat input.
Draw the mechanical network for the system as shown in the Figure 2.46 and draw its
analogous circuit.
Fig. 2.46
Chapter 02.indd 37
03/11/11 6:14 PM
2-38
2.9
CONTROL SYSTEMS ENGINEERING
Using force–voltage analogy, show that the mechanical system as shown in the Figure 2.47
is analogous to the electrical system.
Fig. 2.47
2.10 With the help of suitable example of any one physical system, illustrate the procedure for
determining the transfer function.
2.11 Tabulate the analogous quantities of mechanical and electrical systems using
(a) force–voltage analogy; and
(b) force–current analogy.
2.12 Derive the system equations for the system as shown in the Figure 2.48 and find the value
of x2(s)/F(s).
Fig. 2.48
Chapter 02.indd 38
03/11/11 6:14 PM
MATHEMATICAL MODELS
2-39
2.13 Show that the systems as shown in the Figure 2.49 are analogous to each other.
Fig. 2.49
2.14 Show that the two systems as shown in the Figure 2.50 are analogous.
Fig. 2.50
2.15 Using force–voltage analogy of the mechanical system as shown in the Figure 2.51, draw
its analogous electrical circuit.
Fig. 2.51
Chapter 02.indd 39
03/11/11 6:14 PM
2-40
CONTROL SYSTEMS ENGINEERING
2.16 Using force–voltage analogy, draw the analogous electrical circuit of the system as shown
in the Figure 2.52.
Fig. 2.52
2.17 Using force–voltage analogy and force–current analogy, draw the equivalent analogous
system for the following system below.
Fig. 2.53
Chapter 02.indd 40
03/11/11 6:14 PM
3
MODELLING A CONTROL
SYSTEM—BLOCK DIAGRAM
REPRESENTATION
3.1
BLOCK DIAGRAM REPRESENTATION OF SYSTEMS CONSIDERING
ELECTRICAL SYSTEM AS EXAMPLES
Any control system will have a number of control components. A control system can be
represented in the form of a block diagram. For making a block diagram, first the transfer function
of the system components is determined and then showed in respective blocks. These blocks are
connected by arrows indicating the direction of flow of signals in the control system whose block
diagram is being represented. The signals can pass only in the direction of arrows represented in
the diagram. The arrow head pointing towards a particular block indicates the input to the system
component and the arrow head leading away from the block indicates the output. All the arrows
in a block diagram are referred to as signals.
For complicated systems, the block diagram interconnecting all the sub-systems becomes a
complex one. Reduction of such complex block diagram becomes necessary to determine the overall
transfer function of the whole system and then apply various techniques to analyse the system.
3.1.1
Advantages of Block Diagram Representation
The following are the main advantages of block diagram representation of control systems:
(i) The overall block diagram of a system can be easily drawn by connecting the blocks
according to the signal flow. It is also possible to evaluate the contribution of each of the
components towards the overall performance of the control system.
(ii) Block diagram helps in understanding the functional operation of the system more readily
than examination of the actual control system physically.
Chapter 03.indd 1
05/11/11 8:39 AM
3-2
CONTROL SYSTEMS ENGINEERING
It may be noted that a block diagram drawn for a system is not unique, that is, there may be
alternative ways of representing a system in block diagram form.
3.1.2
Block Diagram Representation of an Error Detector
In a feedback control system, there is one reference input signal which corresponds to the desired
output and one feedback signal. The feedback signal is received from the output (as a fraction
of the output) if there is any deviation of the output from the desired output. Error detector is a
summing device which produces a resultant signal. This error signal is the difference between
the reference input and the feedback signal of the control system. As shown in Figure 3.1, R(s)
is the reference input corresponding to the desired output, C(s) is the desired output and B(s) is
the feedback signal.
Fig. 3.1 Error detector.
3.1.3
Block Diagram of a Closed-loop Control System and Its
Transfer Function
The basic block diagram representation of a unity feedback control system has been shown in
Figure 3.2.
Fig. 3.2 A closed-loop system in block diagram form.
When the output is fed back to the summing point for comparison with the input to create
an error signal, it is necessary to convert the form of the output signal to that of the input signal.
The quantities being added or subtracted at the summing point should have the same dimensions
and, therefore, should have the same units. The conversion of the fraction of output signal form
to input signal form on feedback path is done by the feedback element whose transfer function
is H(s). The feedback element modifies the output before it is compared with the input signal.
Chapter 03.indd 2
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-3
Figure 3.3 shows the standard form of representation of a feedback control system in block
diagram form. Here a fraction of output B(s) = C(s) H(s) is brought to the summing point for
producing an error signal. This block diagram is also called the canonical form of representation
of a control system. G(s) is the system transfer function.
Fig. 3.3 Standard way or canonical form of block diagram representation of a
system with negative feedback.
Referring to Figure 3.3, we can write
E(s) = R(s) – B(s)
B(s) = H(s) C (s)
and
C(s) = E(s) G (s)
\
E(s) = R(s) – H(s) C(s)
or,
R(s) = E(s) + H(s) C(s)
= E(s) + H(s) E(s) G(s)
Therefore, the closed-loop transfer function is
C (s)
E (s) G (s)
G (s)
=
=
R( s) E ( s) + H ( s) G ( s) E ( s) 1 + G ( s) H ( s)
(3.1)
(3.2)
(3.3)
The transfer function is
C (s)
G ( s)
=
R( s)
1 + G ( s) H ( s)
The above transfer function is for the closed-loop system with negative feedback. The
transfer function for a closed-loop system with positive feedback is given by
C (s)
G ( s)
(3.4)
=
R( s)
1 − G ( s) H ( s)
This is because, the error signal for positive feedback will be, E(s) = R(s) + B(s).
For a unity feedback control system, H(s) = 1. The transfer function for a unity feedback
system is
C (s)
G (s)
=
R( s)
1  G (s)
Chapter 03.indd 3
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3-4
CONTROL SYSTEMS ENGINEERING
where plus sign in the denominator stands for negative feedback. The block diagram of the unity
feedback control system is the same as shown in Figure 3.2.
3.1.4
Characteristic Equation of a Control System
Consider the transfer function of the general closed-loop system described earlier.
The transfer function (TF) is written as
TF =
C (s)
R( s)
=
G (s)
1 + G (s) H (s)
The denominator of the transfer function when equated to zero is called the characteristic
equation of the control system.
Thus, 1 + G(s) H(s) = 0 is the characteristic equation of the system.
The roots of the characteristic equation are called the characteristic roots or closed-loop
poles.
The location of the poles in the s-plane indicates the condition of stability of the control
system. Use of characteristic equation in stability analysis will be dealt with in subsequent
chapters.
3.1.5
Rules for Block Diagram Simplification
Six rules are given so as to help simplify a block diagram of a control system and represent in
canonical form. Their standard form as has been shown in Figure 3.3.
The rules are illustrated as under.
(1) When two or more blocks are connected in series, we are to multiply the transfer functions
of each block and put as one block as shown in the Figure 3.4.
Fig. 3.4
(2)
When two blocks are connected in parallel, the transfer functions are to be added as shown
in Figure 3.5.
Fig. 3.5
Chapter 03.indd 4
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
(3)
3-5
When shifting the summing point ahead of, that is before, a block.
Fig. 3.6
Note that, in this shifting, the original input, output and feedback quantities must not
change. Figure 3.6 shows the shifting of the summing point before block G with the help of
a dotted line.
A block 1/G has to be introduced in the feedback path to maintain the input, output, and
feedback signals unaltered.
(4)
Shifting the summing point beyond, that is after, a block.
Fig. 3.7
As shown in Figure 3.7, to keep the output signals unaltered, a block G has been placed in
the feedback path.
(5)
Moving a take-off point ahead of, that is before, a block.
Fig. 3.8
As shown in Figure 3.8, a block G has to be introduced in the feedback path to keep
feedback signal as AG unaltered.
Chapter 03.indd 5
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3-6
(6)
CONTROL SYSTEMS ENGINEERING
Moving a take-off point beyond, that is after, a block.
Fig. 3.9
A block of 1/G has to be introduced in the feedback path to maintain the equivalence as
shown in Figure 3.9.
Now let us consider a few examples of block diagram reduction.
Example 3.1 Reduce the block diagram as shown in Figure 3.10 into canonical form and
determine its transfer function.
Fig. 3.10 Block diagram representation of a system.
Solution Combining elements G1 and G4 in series and elements G2 and G3 in parallel, the
block diagram (applying rules 1 and 2) gets reduced to as shown in Figure 3.11 (a)
G1G4 and H1 form the feedback loop. Note that feedback is positive and therefore the TF
will be of the form
Chapter 03.indd 6
G( s)
(1 − G ( s ) H ( s )).
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-7
G1G4 (1 − H1G1G4 ) ,(G2 + G3 ) are in series. They together form one block, and H2 form the
feedback loop. The simplified block diagram has been shown in Figure 3.11 (c).
Fig. 3.11 Block diagram of Fig. 3.10 gets reduced to canonical form through
successive simplification.
G1G4 (G2 + G3 )
G1G4 (G2 + G3 )
C (s)
1 − H1G1G4
TF =
=
=
(
)
G
G
G
+
G
1 − H1G1G4 + H 2G1G4 (G2 + G3 )
R( s)
3
1+ 1 4 2
H2
1 − H1G1G4
Example 3.2 Reduce the block diagram of a system represented in Figure 3.12 and determine
the net output response.
Fig. 3.12 Block diagram representation of a system.
Solution There are two input signals, R(s) and U(s), and one output signal, C(s). We apply
the superposition theorem to reduce the block diagram, considering the effect of one input at a
time when other is set to zero.
Chapter 03.indd 7
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3-8
(1)
CONTROL SYSTEMS ENGINEERING
To determine the total response, the individual responses will be added.
Let us consider input R(s) first, keeping U(s) = 0. The block diagram of Figure 3.12 will be
as shown in Figure 3.13 (a). Later on, we will consider the input U(s), keeping R(s) = 0.
Combining G1 and G2 in series
Fig. 3.13
Block diagram representation using superposition theorem with only one input R (s).
G1G2 and H2 form the feedback loop.
The transfer function is
TF1 =
C1 ( s )
R( s)
=
G1G2
1 − H 2G2G1
(since the feedback is positive, we have used minus sign in the denominator)
or output,
(2)
Chapter 03.indd 8
G1G2


C1(s) = 
 R( s)
 1 − G1G2 H 2 
Considering only the second input U(s), and keeping R(s) = 0, the network becomes as
shown in Figure 3.14 (a).
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-9
Combining H2, G1 in series, and drawing the network in canonical form,
Fig. 3.14 Canonical form of block diagram representation using superposition theorem
with only the second input U(s).
G2 and G1 H2 form the feedback loop.
The transfer function is
TF2 =
C2 ( s )
U (s)
=
G2
1 − G1G2 H 2
Therefore,


G2
C2(s) = 
 U (s)
 1 − G1G2 H 2 
Combining the responses in cases (1) and (2), the net output response can be calculated.
The net output response is

G1G2
C(s) = C1(s) + C2(s) = 
 1 − G2 G1 H 2


G2
 R( s ) + 

 1 − G1G2 H 2

 U (s)

Example 3.3 Determine the net response of the system shown in Figure 3.15 using block
diagram reduction technique.
Fig. 3.15 Block diagram of a system with three inputs.
Solution We will follow the same procedure as in the previous example. Here we have three
inputs. We will take one input at a time, keeping other two inputs equal to zero.
Chapter 03.indd 9
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3-10
(1)
CONTROL SYSTEMS ENGINEERING
Considering only input R(s),
U1(s) = U2 (s) = 0, Output = C1 (s)
G1, G2 and H1, H2 are in series. There the circuit gets reduced to
Fig. 3.16 Response of the system of Fig. 3.15 with only one input.
G1G2 and H1 H2 form the feedback loop.
C1 ( s )
R( s)
=
G1G2
1 − G1G2 H1 H 2
(minus sign in the denominator is due to positive feedback) (3.5)
(2)
Considering input U1 only
Chapter 03.indd 10
U2(s) = R(s) = 0, Output = C2(s)
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-11
H1, H2 and G1 are in series.
Fig. 3.17 Response of the system of Fig. 3.15 with the second input.
G1H1H2 and G2 form the feedback loop.
C2 ( s )
U1
=
G2
1 − G1G2 H1 H 2
(3.6)
(3) Considering input U2 only,
U1(s) = R(s) = 0, Output = C3(s)
H1, G1 and G2 in the series
H1, G1 G2 and H2 form the feedback loop.
G1G2 H1
C3 ( s )
=
1 − G1G2 H1 H 2
U 2 (s)
(3.7)
By adding Equations (3.5), (3.6) and (3.7), we get the net response as
C(s) = C1(s) +C2(s) +C3(s)
Chapter 03.indd 11
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3-12
CONTROL SYSTEMS ENGINEERING
Fig. 3.18 Response of the system of Fig. 3.15 with the third input.
or,
Combined output,
G1G2

C(s) = 
 1 − G1G2 H1 H 2
G2


 R( s) + 

 1 − G1G2 H1 H 2
G1G2 H1


 U1 + 

 1 − G1G2 H1 H 2

 U2

Example 3.4 Reduce the block diagram of Figure 3.19 to canonical form and derive the
overall transfer function (Note that this problem has also been solved by Signal Flow Graph
Method as in Example 4.3 in Chapter 4).
Fig. 3.19 Block diagram of a system.
Solution Shifting the take-off point after G2, beyond the block G3 (Figure 3.20).
G2, G3 and H1,
Chapter 03.indd 12
1
G3
are in series, respectively in their individual circuits.
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-13
H2 and G2G3 form a feedback loop.
G1 and
G2G3
(1+ G2G3 H 2 )
G1G2 G3
1 + G2 H 2 H3
are in a series.
and H1/G3 form a feedback system whose TF is calculated as shown here.
The simplified block diagram is shown in Figure 3.20.
Fig. 3.20
The block diagram of Fig. 3.19 gets reduced to a canonical form
through successive steps.
This is a unity feedback system, where H(s) = 1
Chapter 03.indd 13
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3-14
CONTROL SYSTEMS ENGINEERING
So, the overall transfer function is given by
G1G2 G3
G1G2 G3
C (s)
1 + G2 G3 H 2 + G1G2 H1
=
=
G1G2 G3
1 + G2 G3 H 2 + G1G2 H1 + G1G2G3
R( s)
1+
1 + G2 G3 H 2 + G1G2 H1
Example 3.5 Reduce the block diagram of Figure 3.21 to canonical form.
Solution There are two inputs, R1 and R2, and there are two outputs, C1 and C2.
(1) First of all, we shall find the output C1, when C2 = 0
Now, we shall consider the inputs independently and then add the two responses C11 and
C12 to find the overall response C1.
C11 is the output with only R1 as the input.
C12 is the output with only R2 as the input.
C1 = C11 + C12
C11 and C12 will be calculated with one input.
Fig. 3.21 Block diagram of a system with two inputs and two outputs.
(a) To find C11
Putting R2 = C2 = 0 in the network, G2, G3, G4 are in series, G2G3G4 and G1 form
the feedback loop and we get
G1
C11 ( s )
=
R1 ( s ) 1 − G1G2G3G4
Fig. 3.22 Calculation of output of the system of Fig. 3.21 at
output point C 1 with input R 1 i.e. C 11 .
Chapter 03.indd 14
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
(b)
3-15
To find C12,
Putting R1 = C2 = 0 in the network, G1, G2, G4 are in series, G3 and G1G2G4 form the
feedback loop and we get
Fig. 3.23 Output of the system of Fig. 3.21 at C 1 with input R 2 i.e. C 12 .
C12 ( s )
R2 ( s )
∴
=
−G1G2G4
1 − G1G2G3G4
C1(s) = C11(s) + C12(s)


 −G1G2G4 
G1
= 
 R1 ( s ) + 
 R2 ( s )
 1 − G1G2G3G4 
 1 − G1G2G3G4 
(2)
Now, we shall find C2, when C1 = 0
(a) For finding C21, consider input R1 only, G1, G3, G4 are in series and G1G3G4 and G2
form the feedback loop.
Fig. 3.24 Output of the system of Fig. 3.21 at C 2 with input R 1 i.e. C 21 .
Putting R2 = C1 = 0,
(b)
Chapter 03.indd 15
we get
C21 ( s )
R1 ( s )
=
−G1G3G4
1 − G1G2G3G4
For finding C22, consider input R2 only and R1 = C1 = 0. In the network, G1, G2, G3 are
in series, and G4 and G1G2G3 form the feedback loop.
05/11/11 8:39 AM
3-16
CONTROL SYSTEMS ENGINEERING
Fig. 3.25 Output of the system of Fig. 3.21 at C 2 with input R 2 i.e. C 22 .
\
\
Thus,
C22 ( s )
R2 ( s )
=
G4
1 − G1G2G3G4
C2(s) = C21(s) + C22(s)

G4
C2(s) = 
 1 − G1G2 G3G4

 −G1G3G4
 R2 (s) + 

 1 − G1G2 G3G4

 R1 (s)

Example 3.6 Obtain the transfer function for the block diagram shown in Figure 3.26.
Fig. 3.26 Block diagram of a system.
Solution Moving the take-off point (Point A) between blocks G2 and G3 ahead of G2 (Point
B) and inserting block G2 with H1 and G4
Chapter 03.indd 16
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-17
G2G4 and G2G3 are in parallel,
Moving take-off point 1 beyond block (G2G4 + G2G3) at point 2,
Eliminating the feedback loops,
Fig. 3.27 Reduction of the block diagram of the system of Fig. 3.26
through successive steps.
Chapter 03.indd 17
G1G2 (G4 + G3 )
1
+
G
H
C
2 2 (G4 + G3 ) + H1G1G2
=
G1G2 (G4 + G3 )
R
1+
1 + G2 H 2 (G4 + G3 ) + H1G1G2
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3-18
CONTROL SYSTEMS ENGINEERING
or,
\
C
R
C
R
=
=
G1G2 (G4 + G3 )
1 + G2 H2 (G4 + G3 ) + H1G1G2 + G1 (G4 + G3 )G2
G1G2 (G4 + G3 )
1 + G2 (G3 + G4 ) + ( H 2 + G1 ) + H1G1G2
Example 3.7 Obtain the transfer function for the block diagram shown in Figure 3.28.
Fig. 3.28 Block diagram of a system.
Solution Moving take-off point 4 beyond block G3 (at point 5).
Eliminating feedback with transfer function H2,
Chapter 03.indd 18
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-19
Eliminating feedback loop,
Fig. 3.29 Determination of transfer function of the system of Fig. 3.28
through successive reduction technique.
The two blocks being parallel,
C
R
= G4 +
G1G2G3
1 + H 2 G2G3 + G2 H1 (1 − G1 )
Example 3.8 Evaluate C R1 and C R2 for a system, the block diagram representation of which
is shown in Figure 3.30. R1 and R2 are the inputs at the summing points.
Fig. 3.30 Block diagram of a system with two inputs.
Solution First we will find C R1 with R2 = 0 and then find C R2 with R1 = 0.
(1) Evaluation of C R1 . Assume R2 = 0. Therefore, summing point 5 can be removed. Shift takeoff point 4 beyond block G3.
Chapter 03.indd 19
05/11/11 8:39 AM
3-20
CONTROL SYSTEMS ENGINEERING
Eliminate the feedback loop between points 3 and 4 and then combining with G2 in series.
Eliminating the feedback loop again, we get the block diagram as shown in the below
diagram.
With G1 and
G2G3
(1+ G3 H 2 + G2 H 3 )
C
R1
=
in series, we get
G1G2G3
1 + G3 H 2 + H 3G2 + G1G2G3 H1
Fig. 3.31 Output at C with one input R 1 of the system shown in Fig. 3.30.
Chapter 03.indd 20
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
(2)
Evaluation of
C
R2
3-21
. Assume R1 = 0. Thus, summing point 1 is removed.
Shifting the summing point 2 beyond G2 and rearranging,
Rearranging again, we get
Rearranging another time and eliminating the feedback loop,
Chapter 03.indd 21
05/11/11 8:39 AM
3-22
CONTROL SYSTEMS ENGINEERING
Rearranging further,
Fig. 3.32 Output at C with another input R 2 of the system of Fig. 3.30.
So, we get
C
=
G3 (1 + G2 H3 )
1 + G2 H3 + G3 H 2 + G1G2 G3 H1
R2
Example 3.9 Find the closed-loop transfer function of the system shown in Figure 3.33.
Fig. 3.33 Block diagram of a control system.
Solution Redrawing the diagram having input R to the extreme light and C taken to the
extreme right as output and then joining each block step by step will help solve the problem with
ease. Accordingly, the block diagram has been redrawn as shown in Figure 3.34.
Fig. 3.34 Figure 3.33 redrawn showing input and output terminals.
Chapter 03.indd 22
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-23
Shifting the take-off points at 1 and 2 beyond the block G1, we get
Eliminating summing point at 4, we get
Eliminating inner feedback loop and then combining G3 in series, we get
Fig. 3.35 Successive steps of reduction of block diagram of Fig. 3.34.
Since this is unity feedback system, we calculate the transfer function as
Chapter 03.indd 23
C
R
=
G1G2 G3G5G8
1 + G6 G7 G8 + G2 G5G8 (G1G3 + G4 G6 )
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3-24
CONTROL SYSTEMS ENGINEERING
Example 3.10 Find the closed-loop transfer function of the system shown in Figure 3.36.
Fig. 3.36 Block diagram representation of a control system.
Solution Moving the summing point beyond the block H2 and inserting G4 H2 block with H1,
H2 in series, we get
Further simplification gives,
●
●
●
G4 H2 and 1 are in parallel;
G2 and G3 are in parallel;
G1 and (G2 + G3) are in series.
Fig. 3.37 Reduction of the block diagram of Fig. 3.36 through successive steps.
Chapter 03.indd 24
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-25
Therefore, the transfer function is found as
(1 + G4 H 2 ) (G1G2 + G1G3 )
C
=
1 + G1G2 H1 H 2 + G1G3 H1 H 2
R
Example 3.11 Determine
reduction technique.
C
R
of the system shown in Figure 3.38 by the block diagram
Fig. 3.38 Block diagram of a system.
Solution Shifting the take-off point ahead of G2 to a position beyond G2 and thereafter
combining blocks G2 and H2 in feedback loop with G1 in series,
Eliminating the negative feedback loop,
Fig. 3.39 Reduction of block diagram of Fig. 3.38 through successive steps.
Eliminating the positive feedback loop, we get the transfer function as
G1G2
C
=
1 + G1G2 H1 + G2 H1 − G1
R
Example 3.12 Using the block diagram reduction technique, reduce the following block
diagram shown in Figure 3.40 and determine the overall transfer function.
Fig. 3.40 Block diagram of a control system.
Chapter 03.indd 25
05/11/11 8:39 AM
3-26
CONTROL SYSTEMS ENGINEERING
Solution
Shift the take off point 1 to point 3 i.e. beyond G3. The feedback element becomes
H1/G3. Take point 2 to point 3. TF of G3 with unity feedback becomes G3/1 + G3. This in series
with G2 becomes G2G3/1 + G3.
With blocks
G2G3
(1+ G3 )
and
H1
G3
in feedback loop, we get
Fig. 3.41 Simplification of block diagram of Fig. 3.40 by
block diagram reduction technique.
With
G1G2G3
(1+ G3 + G2 H1 )
and H2 blocks in feedback loop, we get
C (s)
R( s)
=
G1G2G3
1 + G3 + G2 H1 + G1G2G3 H 2
Example 3.13 Using the block diagram reduction technique, find the transfer function for
the control system shown in Figure 3.42.
Fig. 3.42 Block diagram of a control system.
Solution
Removing feedback loop with blocks G2 and H1 and then shifting the lower
summing point beyond block G5.
Chapter 03.indd 26
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-27
Fig. 3.43 Simplification of the block diagram of Fig. 3.42.
The first loop is a parallel circuit with a negative sign on one element, and the second
loop is a positive feedback loop. The diagram can further be reduced by calculating the transfer
function of the two loops and then combining the two loops.
Example 3.14 Using the block diagram reduction technique, reduce the system shown in
Figure 3.44 to the simplest possible form and find the transfer function.
Fig. 3.44 Block diagram of a control system.
Solution Shifting take-off point at 2 to point 3, we get
Chapter 03.indd 27
05/11/11 8:39 AM
3-28
CONTROL SYSTEMS ENGINEERING
Removing the feedback loop with G2, G3 and H1 blocks, we get
Shifting take-off point at 3 to point 1, we get
Eliminating parallel blocks, we get
Fig. 3.45 Determination of transfer function of the system shown in Fig. 3.44
through block diagram reduction technique.
C (s)
R( s)
=
G1G2G3G4 + G1G2G5
1 + G2G3 H1 + G1G2G3G4 H 2 + G1G2G5 H 2
Example 3.15 Determine the ratio
C (s)
R(s)
for the multiple-loop system shown in Figure 3.46.
Fig. 3.46 Block diagram of a control system.
Chapter 03.indd 28
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-29
Solution Shifting take-off point of feedback loop after G2 to the point ahead of G2 and
eliminating feedback loop with G4 and H1, we get
Eliminating parallel blocks and the feedback loop, we get
Fig. 3.47 Block diagram reduction technique applied to the system of Fig. 3.46.
Simplifying, the transfer function is determined as
G1G2G4 + G1G3G4
C (s)
=
1 + G1G2 H 2 + G4 H1 + G1G2G4 H1 H 2 + G1G2G4 + G1G3G4
R( s)
3.1.6
Block Diagram Representation of an Electrical Network
A system may be electrical, mechanical or a combination of both having a number of elements or
components interconnected to perform a specific task. For drawing the block diagram of a system,
we have to first write the equations describing the dynamic behaviour of each component. Then,
we have to take the Laplace transform of these equations, assuming zero initial conditions, and
represent each Laplace transformed equation individually in the form of block diagram. Finally,
we have to assemble the elements into a complete block diagram. Let us consider a simple
example of an electrical network.
Example 3.16 Draw the block diagram for the electrical network shown in Figure 3.48.
This is an integrating circuit used in many practical applications.
Solution
(1) The following are the equations describing the system:
Chapter 03.indd 29
ei = Ri +
1
C
∫ i dt
05/11/11 8:39 AM
3-30
CONTROL SYSTEMS ENGINEERING
and
eo =
1
C
∫ i dt
Fig. 3.48 Determination of input-output relation of an R-C circuit.
(2)
Taking Laplace transform of above equations, we get
Ei(s) = RI ( s ) +
and
Eo(s) =
From Equation (3.8),
Cs
Cs
I (s)
I (s)
(3.8)
(3.9)
1 

Ei(s) =  R +
 I (s)
Cs 

or
I(s) =
or
1
1
I(s) =
Ei (s)
 1 
R+

 Cs 
CsEi ( s )
1 + RCs
Fig. 3.49 Block diagram representation of the electrical circuit shown in Fig. 3.47.
Substituting the value of I(s), from Equation (3.9), we get
1 CsEi ( s )
Ei ( s )
Eo(s) =
=
Cs (1 + RCs ) 1 + RCs
Chapter 03.indd 30
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-31
Therefore,
Eo ( s )
Ei ( s )
=
1
1 + RCs
We have the standard form of representation of a system as shown in Figure 3.49 as
Eo ( s )
Ei ( s )
=
G(s)
1 + G(s) H (s)
where H(s) = 1 (see Figure 3.49).
The transfer function is written as
1
1
1
= RCs =
=
TF =
1
1 + RCs 1 + τs
1 + G(s)
1+
RCs
G(s)
or
where t is the line constant of the circuit.
or
1
G(s) =
RCs
Therefore, the standard form is represented as shown in Figure 3.50.
Fig. 3.50 Representation of the R-C circuit in canonical form.
From Equations (3.8) and (3.9) we have [Ei(s) – Eo(s)]1/R = I(s) and Eo(s) = (1/Cs) I(s)
Therefore, the control diagram of Figure 3.50 can be represented as shown in the Figure 3.51.
Fig. 3.51 Representation of the electrical circuit of Fig. 3.48
in the form of control circuit diagram.
3.2
Transfer function of DC synchro motor
A servomechanism is a feedback control system and consists of a mechanism in which the output
of the system may be some mechanical position, velocity or acceleration. Servomechanism and
position control systems are synonymous.
Chapter 03.indd 31
05/11/11 8:39 AM
3-32
CONTROL SYSTEMS ENGINEERING
In almost all servomechanism applications d.c. motor drives and gear mechanism are used.
Block diagram representation and transfer function of these are discussed in the following subsections.
3.2.1
Block Diagram of a DC Motor Drive
In servo applications, a d.c. motor is required to produce rapid acceleration from stand still
(position of rest). Therefore, the physical requirements of such a motor are low inertia and high
starting torque. Low inertia is attained with reduced armature diameter.
In control systems, d.c. motors are used in two different control modes:
(1)
Armature control mode with constant field current
(2)
Field control mode with fixed armature current.
These are discussed as follows
1.
Armature controlled d.c. motor: In d.c. motors, we know that emf (Eb) is directly
proportional to armature speed (N) and field flux (φ). Since,
Eb =
φ ZNP
60 A
or
ZP
Also,
where
Thus,
Eb = kN φ where k = 60 A
Eb = V – Ia Ra
V = Terminal voltage
Ia = Armature current
Ra = Armature resistance
Eb =
Again,
φZNP
60 A
= KφN
Eb = V – Ia Ra
Therefore,
N =
V − I a Ra
Kφ
(3.10)
From Equation (3.10), it can be seen that the speed of motor, N, can be varied by changing
V, that is, the terminal voltage, or by changing the field flux, φ, or by changing the armature
circuit resistance.
In armature control method, speed is varied by changing the armature current keeping the
field flux constant. Figure 3.52 shows an armature controlled d.c. motor. The armature resistance
and inductance have been shown separately.
Chapter 03.indd 32
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-33
Fig. 3.52 Armature controlled d.c. motor.
The motor has a rotating armature with load on its shaft having a moment of inertia J and
a viscous friction coefficient B0. Torque developed by the motor is Tm which causes rotation. The
angular velocity is ω and angular acceleration is α. All these constants and variables of the motor
drive are defined as
V = Applied armature voltage
Tm = Torque developed by the motor
θ = Angular displacement of the motor shaft
ω = d θ dt = Angular velocity of the motor shaft
ω = 2 πN 60 , where N is the number of revolutions per minute
ω = 2πN, if N is expressed in rps.
J = Equivalent moment of inertia of the rotor of the motor and that of the
load referred to the motor shaft
2
α = d θ dt 2 = Angular acceleration
Torque of the motor is due to armature current Ia and the field flux φ. Again field flux is
proportional to the field current If.
Therefore,
Tm = K1φIa
= K1Kf If Ia
If If is constant,
Tm = KT Ia, where KT = K1Kf If
(3.11)
Back emf,
K φω KK1 I f d θ
dθ
Eb = KφN =
(3.12)
=
= Kb
2π
2π dt
dt
Inertia torque = Moment of inertia × Angular acceleration
Therefore,
d 2θ
(3.13)
Ti = J 2
dt
Torque due to bearing friction, windage friction and any other drag opposing rotation has
been accounted for an equivalent viscous friction coefficient B0.
Chapter 03.indd 33
05/11/11 8:39 AM
3-34
CONTROL SYSTEMS ENGINEERING
Frictional torque = Coefficient of viscous friction × Angular velocity
Therefore,
dθ
Tf = B0 ω = B0
(3.14)
dt
Torque developed by the motor is opposed by inertia torque and frictional torque. Thus,
from Newton’s second law of rotational motion,
Ti + Tf = Tm
or
d 2θ
dθ
= Tm = KT Ia
(3.15)
J 2 + B0
dt
dt
The differential equation involving quantities of the armature circuit can be written as
dI a
(3.16)
+ Ra I a + Eb = V
dt
Taking Laplace transform of Equations (3.12), (3.15), and (3.16) respectively, we get
La
Eb(s) = Kbs θ(s)
2
(Js + B0s) θ(s) = KT Ia(s)
(Las + Ra) Ia(s) = V(s) – Eb(s)
From Equations (3.18) and (3.19),
V ( s ) − Eb ( s )
[Js2 + B0s] θ(s) = KT
La s + Ra
Using Equation (3.12) in the above,
(3.17)
(3.18)
(3.19)
[Js2 + B0s] [Las + Ra] θ(s) = KT V(s) – KT Kbs θ(s)
or
[(Js2 + B0s) (Las + Ra)] θ(s) + KT Kbs θ(s) = KT V(s)
The transfer function is
θ( s )
V (s)
=
2
KT
( Js + B0 s ) ( La s + Ra ) + KT Kb s
KT
=
s [( Js + B0 ) ( La s + Ra ) + KT Kb ]
Block diagram representation:
From Equation (3.19), we have
I a (s)
V ( s ) − Eb ( s )
=
1
La s + Ra
(3.20)
From Equation (3.18), we have
Chapter 03.indd 34
θ( s )
I a (s)
=
KT
s ( Js + B0 )
(3.21)
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-35
Thus, from Equation (3.21) we have the following block diagram
and from Equation (3.20), we have
as
With V(s) as the input signal and Eb(s) as the feedback signal, the comparator is represented
Again, from Equation (3.17),
Eb(s) = Kb sθ (s)
Thus, combining all the above blocks, the complete block diagram is represented in
Figure 3.53 (a).
Alternatively, the block diagram can be represented as in Figure 3.53 (b).
Fig. 3.53 Block diagram representation of an armature controlled d.c. motor.
Chapter 03.indd 35
05/11/11 8:39 AM
3-36
2.
CONTROL SYSTEMS ENGINEERING
Field controlled d.c. motor
Fig. 3.54 Field controlled d.c. motor.
As shown in Figure 3.54,
ef = Field voltage (control input)
ia = Armature current (constant)
\
Tm = K1φia = K1Kf if ia = K if
where
K = K1Kf ia
Also, the field circuit equation is
di f
ef = L f
dt
The torque equation is
d 2θ
+ Rf if
dθ
= Tm = K if
dt
dt
Taking Laplace transform of the above equations
(Lf s + Rf ) If (s) = Ef (s)
or,
(Js2 + Bos) θ(s) = K I f (s)
or,
E f (s)
(Js2 + Bos) θ(s) = K
[from Equation (3.22)]
Lf s + Rf
J
2
+B
TF =
From Equation (3.22),
I f (s)
E f (s)
From Equation (3.23),
Chapter 03.indd 36
θ(s)
I f ( s)
=
=
θ(s)
E f ( s)
=
(3.23)
K
s( Js + Bo )( L f s + R f )
1
(L f s + R f )
K
Js + B0 s
2
(3.22)
(3.24)
(3.25)
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-37
Block diagram representation: Using Equations (3.24) and (3.25), the block diagrams are
developed as shown in Figure 3.55.
Combining the above two blocks, we get
Fig. 3.55 Block diagram representation of a field controlled d.c. motor.
3.2.2
Block Diagram of Gear Trains
Gear trains are referred to as a number of gears of different number of teeth connected together
for power transmission. Power is a product of torque and speed. If speed is reduced, torque
increases and vice versa, which we experience while driving motor vehicles.
Gear trains are often used in servomechanism to reduce speed, to increase torque or to
obtain the most efficient power transfer by matching the driving member to the driven load.
Thus, in mechanical systems, gear trains act as matching devices like transformers in electrical
systems. Figure 3.56 shows a motor, driving a load through a gear train which consists of two
gears meshed together. The gear with N1 teeth is called the primary gear (analogous to primary
winding of a transformer) and the gear with N2 teeth is called the secondary gear (analogous to
secondary winding of a transformer).
T1 = Load torque on gear 1 due to the rest of the gear train
Fig. 3.56 A motor driving a load through a gear train.
Chapter 03.indd 37
05/11/11 8:39 AM
3-38
CONTROL SYSTEMS ENGINEERING
J1 and J2 are the moment of inertia
B1 and B2 are viscous friction of the systems as shown in Figure 3.56.
T2 = Torque transmitted to gear 2
The torque equations are
J1
θ1 + B1θ 1 + T1 = Tm
J 2
θ2 + B2 θ 2 + TL = T2
(3.26)
(3.27)
Since the linear distance travelled along the surface of each gear is the same,
\
θ1r1 = θ2r2
where r1 and r2 are the radius of gear 1 and gear 2, respectively
The number of teeth on the gear surface being proportional to the gear radius,
N
θ
r1
= 2 = 1
N2
θ1
r2
In an ideal case, i.e. when there is no slip, work done by gear 1 = work done by gear 2, that is,
T1θ1 = T2θ2
T1
T2
=
θ2
θ1
=
N1
N2
If the ratio of N1 and N2 is less than 1, then the gear train reduces the speed but increases
the torque.
N
\
T1 = T2 1
N2
Substituting this value of T1 in Equation (3.26),
N
J1
θ1 + B1θ 1 + T2 1 = Tm
N2
(3.26A)
Substituting the value of T2 from Equation (3.27) in Equation (3.26A),
N
J1
θ1 + B1θ 1 + 1 ( J 2
θ2 + B2 θ 2 + TL ) = Tm
N2
We had
θ2
θ1
=
(3.28)
N1
N2
Differentiating, we have
Therefore,

θ2
θ
N
= 2 = 1


N2
θ1
θ1
2

 N1   
 J1
θ1 + J 2 
 θ +
N2  1 




Chapter 03.indd 38
2

 N 
N 
 B1θ 1 + B2  1  θ 1  +  1  TL = Tm
N


 N2 

  2
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-39
or,
2
2

 N1    
 N1    N1 




θ1 J1 + J 2 
 + θ1 B1 + B2 
 +
 TL = Tm
N
N



 2 
 2    N 2 


(3.29)
or
N
J1eq 
θ1 + B1eq θ 1 +  1
 N2

 TL = Tm

(3.30)
where the equivalent moment of inertia and viscous friction of the gear train referred to shaft-1
are respectively,
2
J1eq
N 
= J1  1  J 2
 N2 
B1eq
N 
= B1 +  1  B2
 N2 
2
and
Similarly, we can express
J 2eq 
θ2 + B2eq θ 2 + TL = Tm
where
2
N 
J2eq = J 2  2  J1
 N1 
N 
= B2  2  B1
 N1 
2
B2eq
Taking Laplace transform of Equation (3.30)
N
J1eq s 2 θ1 ( s ) + B1sθ1 ( s ) = Tm ( s ) −  1
 N2

 TL ( s )

or,
J1eq s 2 θ1 ( s ) + B1eq sθ1 ( s ) = Tm(s) – nTL(s)
where n =
or
∴
Chapter 03.indd 39
N1
N2
( sJ1eq + B1eq s ) θ1 ( s ) = Tm(s) – nTL (s)
θ1 ( s ) =
Tm (s) − nTL (s)
s(sJ1eq + B1eq )
05/11/11 8:39 AM
3-40
CONTROL SYSTEMS ENGINEERING
Block diagram representation: Block diagram representation is shown in Figure 3.57.
Fig. 3.57 Block diagram representation of a gear train.
Example 3.17 A d.c. motor has been shown schematically in Figure 3.58. L and R represent
the inductance and resistance of the motor’s armature circuit, and the voltage Eb represents the
generated back emf which is proportional to the shaft velocity d θ dt . The torque T developed by
the motor is proportional to the armature current i. The inertia J represents the combined inertia
of the motor armature and the load, and B is the total viscous friction acting on the output shaft.
Determine the transfer function between the input voltage V and the angular position θ of the
output shaft.
Fig. 3.58 A d.c. motor during a load.
Solution The differential equations of the motor armature circuit and the inertia load are
written as follows:
di
(3.31)
V = iR + L + Eb
dt
i.e.,
di
dθ
= V – Eb = V – K
iR + L
dt
dt
and
d 2θ
dθ
= KT i
dt
dt
Taking the Laplace transform of the above equations,
(R + sL) I(s) = V(s) – Ks θ(s)
and
(Js2 + Bs) θ(s) = KT I(s)
From the above two equations,
[V ( s ) − Ks θ( s )]
(JS 2 + BS) θ(s) = KT
( Ls + R)
Chapter 03.indd 40
J
2
+B
(3.32)
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3.2.3
3-41
θ( s )  ( Js 2 + Bs )( Ls + R) KKT s  = KT V(s)
TF =
θ( s )
V (s)
=
2
Kr
( Js + Bs)( Ls + R) + KKT s
Block Diagram of a Servomechanism or a Position Control System
The servomechanism shown in Figure 3.59 is used to control the position of the mechanical load
in accordance with the reference position. A pair of potentiometers forms an error-measuring
device which produces an error voltage of V = kp (r – c) with respect to input and output angular
positions of r and c, respectively, of potentiometers’ arms, where kp is a constant of proportionality
and e = (r – c) is the error signal. An amplifier of gain kA with high input impedance is used to
feed an amplified voltage of kAV to the armature circuit of a d.c. servomotor, whose field is
excited by a fixed voltage source. The armature inductance and low output impedance have been
neglected as they are usually small. If an error exists, the motor develops a torque to rotate the
output load in such a way as to reduce the error to zero.
Fig. 3.59 Positional servomechanism.
The complete block diagram for the system can be represented block by block using transfer
functions for each block as has been shown in Figure 3.60.
The transfer function of the above block diagram is given by
K
C (s)
= 2
R( s)
Js + Bs + K
Fig. 3.60 Block diagram representation of a position control system.
Chapter 03.indd 41
05/11/11 8:39 AM
3-42
CONTROL SYSTEMS ENGINEERING
B = B0 +
and
K =
K b KT
Ra
K p K A KT n
Ra
where
K is a constant and
J and B0 are the moment of inertia and viscous friction coefficient, respectively, referred to
motor shaft for the combination of motor, load and gear train with gear ratio n.
The quantities J, B and K when multiplied by 1/n2 produces inertia, viscous friction
coefficient and constant referred to as output shaft.
C (s)
can also be written as
R( s )
K
2
C (s)
= Js + Bs
K
R( s)
1+ 2
Js + Bs
The block diagram of Figure 3.60 can, therefore, be simplified as shown in the Figure 3.61.
Fig. 3.61 Simplified block diagram of the system shown in Fig. 3.60.
Example 3.18 For the positional servomechanism shown in Figure 3.62, the input to the
system is the reference shaft position and the system output is the output shaft position. The
numerical values for the system constants are as follows:
r = Angular displacement of the reference input shaft, radians
c = Angular displacement of the output shaft, radians
ω = Angular displacement of the motor shaft, radians
K1 = Gain of the potentiometer error detector = 24 π V/rad
Kp = Amplifier gain = 10 V
Rf = Field-winding resistance = 2 Ω
Lf = Field-winding inductance = 0.1 H
If = Field-winding current, amperes
ef = Applied field voltage, volts
Chapter 03.indd 42
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-43
Fig. 3.62 Positional servomechanism.
K = Motor torque constant = 0.05 N-m/A
n = Gear ratio = 1/10
J = Equivalent moment of inertia of the motor and load referred to the motor
B = Equivalent viscous-friction coefficient of the motor and load referred to the
motor shaft = 0.02 N-m/rad/sec.
Draw a block diagram of the system showing the transfer function of each block. Finally,
simplify the block diagram.
Solution The equations describing the system dynamics are as follows:
For the potentiometric error detector
24
[ R ( s ) − C ( s )]
E(s) = K1[ R( s ) − C ( s )] =
π
(3.33)
For the amplifier
Ef (s) = Kp E(s) = 10E(s)
(3.34)
For field controlled d.c. motor
θ( s )
E f (s)
θ( s )
E f (s)
=
=
K
s ( L f s + R f ) ( Js + B)
Km
s (T f s + 1) (Tm s + 1)
where,
Chapter 03.indd 43
Km =
K
Rf B
=
0.05
(2) (0.02)
05/11/11 8:39 AM
3-44
CONTROL SYSTEMS ENGINEERING
= 1.25 rad/V-sec.
Tf =
Tm =
θ( s )
∴
E f (s)
=
Lf
=
Rf
J
B
=
0.1
2
0.02
0.02
= 0.05 sec.
= 1 sec.
1.25
(3.35)
s (0.05s + 1) ( s + 1)
The block diagram of the system is drawn using Equations (3.33) through (3.35) and is
shown in Figure 3.63 (a).
The simplified block diagram of the above system is shown in the general form as shown
in Figure 3.63 (b).
Fig. 3.63 Block diagram representation of the system shown in Fig. 3.61.
3.3
3.3.1
Transfer function of AC synchro motor
Transfer Function of a Two-phase AC Servo Motor
AC servo motors are two-phase induction motors.
VR
TM (t )
Reference
Winding
T4
load
T2
Ic
From
Servo
amplifier
ec (t )
Co ntrol
Winding
w
J, fo
VC 4
VC 3
VC 2
VC 1
T3
T1
VC 2
VC 3
VC 2
VC 1
Rotor
W1
(a)
W2
W3
W4
WM (t )
(b)
Fig. 3.64 (a) A two-phase servo motor (b) Torque–speed characteristic
Chapter 03.indd 44
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-45
The two stator fields are placed at right angles to each other. One winding is called the
reference winding and the other winding is called the control winding. A fixed reference voltage,
VR, is applied across the reference winding. The control winding is supplied from the output of
a servo amplifier.
The speed of angular rotation, w, is determined by the control current, IC. A two-phase
servo motor is shown in Figure 3(a) and its torque–speed characteristics are shown in Figure 3(b).
Let ec(t) be the sinusoidal control voltage. The rms value of the control voltage is E. TM is
the torque developed by the motor. The angular speed w = θ .
Then, TM = f (θ, E)
The torque developed is a function of both speed θ and runs value of control voltage E. At
normal operating point with
T = T , E = E and θ = θ
M
M0
0
0
the equation TM = f ( θ , E) is expanded using Taylor’s series. Neglecting terms of higher order
derivatives, we get
TM = TM +
∂TM
0
∂E
( E − E0 ) +
∂TM  
(θ − θ0 )
∂θ
or, TM − TM 0 = K1 ( E − E0 ) + K 2 (θ − θ0 )
or, ∆TM = K1 ∆E + K2∆ θ
(i)
The load side consists of inertiate and viscon friction, J and B, respectively. The equation
relating torque, speed, inertial and viscon friction in incremental notation is
∆T = J ∆θ + B∆θ
(ii)
M
Equating (i) and (ii)
J ∆θ + B∆θ = K1∆E + K 2 ∆θ
Taking Laplace transform,
[Js2 + (B + K2)s] ∆E(s) = K1∆E(s)
Transfer function,
∆θ( s)
∆E ( s)
=
=
where K M =
K1
K1
Js + ( B + K 2 )s
2
KM
s(τ M s + 1)
and τ M =
J
B + K2
B + K2
KM is the motor gain constant and
τM is the motor time constant.
When the motor is used in a position control system, the operating point corresponds to
E0 = 0 and θ = 0 so that ∆θ = θ and ∆E = E.
So the transfer function of the servo motor is
Chapter 03.indd 45
05/11/11 8:39 AM
3-46
CONTROL SYSTEMS ENGINEERING
G ( s) =
∂( s)
=
KM
E ( s) s(τ M s + 1)
The value of motor constants can be found by conducting no-load test and blocked rotor test.
3.4Transfer Function of Synchro transmitter and Synchro
Receiver
These have been discussed in the chapter on control circuit components.
REVIEW QUESTIONS
3.1 Determine the transfer function C(s)/R(s) for the block diagram in Figure 3.64.
Fig. 3.64
3.2 Obtain the overall transfer function relating C(s) and R(s) for the block diagram shown in
the Figure 3.65.
Fig. 3.65
3.3 Simplify the block diagram shown in Figure 3.66 and find the transfer function
C(s)/R(s).
Fig. 3.66
Chapter 03.indd 46
05/11/11 8:39 AM
MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-47
3.4 Obtain the closed-loop transfer function C(s)/R(s) after simplifying the following block
diagram shown in Figure 3.67.
Fig. 3.67
3.5 Obtain the transfer function relating V2(s) and V1(s) and represent in the form of a block
diagram for the differentiating circuit shown in Figure 3.68.
Fig. 3.68
3.6 Figure 3.69 shows the block diagram representation of a control system. You are to do the
following:
Fig. 3.69
(1) Represent the system, by block diagram reduction technique, in the form shown in
Figure 3.70 and find G(s) and H(s).
Chapter 03.indd 47
05/11/11 8:39 AM
3-48
CONTROL SYSTEMS ENGINEERING
Fig. 3.70
(2) For the same system, find Geq(s) for an equivalent unity-feedback system of the
form shown in Figure 3.71.
Fig. 3.71
3.7 Find the closed-loop transfer function of the system shown in Figure 3.72 by block
diagram reduction technique.
Fig. 3.72
3.8 Find the transfer function of the system represented by block diagram as shown in
Figure 3.73.
Fig. 3.73
Chapter 03.indd 48
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MODELLING A CONTROL SYSTEM—BLOCK DIAGRAM REPRESENTATION
3-49
3.9 Draw the block diagram and derive the transfer function of an armature controlled
d.c. motor shown in Figure 3.74.
Fig. 3.74
3.10 Determine the transfer function C(s)/R(s) for the system shown in Figure 3.75.
Fig. 3.75
3.11 Simplify the following block diagram and determine the transfer function.
Fig. 3.76
Chapter 03.indd 49
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3-50
CONTROL SYSTEMS ENGINEERING
3.12 Determine the transfer function of the system represented in the form of a block diagram
as shown in Figure 3.77.
Fig. 3.77
Chapter 03.indd 50
05/11/11 8:39 AM
4
MODELLING A CONTROL
SYSTEM—SIGNAL FLOW GRAPH
4.1
INTRODuCTION
Block diagrams are quite convenient in representing control systems. However, for very
complicated systems, the block diagram reduction approach for arriving at the transfer function
relating the input and output variables is tedious and time consuming. An alternative approach
is that of the signal flow graph (SFG) developed by S. J. Mason. A SFG does not require any
reduction process because of the availability of a flow graph gain formula which relates the input
and output system variables.
Definition: A signal flow graph is a graphical representation of the relationship between
the variables of a set of linear algebraic equations. It consists of a network in which nodes
representing each of the system variables are connected by directed branches.
The meanings of the following terms need to be understood before proceeding further.
Node: It represents a system variable which is equal to the sum of all incoming signals
at the node. Outgoing signals from the node do not affect the value of the node variable.
Figure 4.1 (b) shows an SFG. Points designated by R, E and C are the nodes of this SFG.
Fig. 4.1
Chapter 04.indd 1
Representation of a control system (a) Block diagram form and;
(b) As equivalent signal flow graph.
03/11/11 6:22 PM
4-2
CONTROL SYSTEMS ENGINEERING
Branch: A signal travels along a branch from one node to another in the direction indicated
by the branch arrow and, in the process, gets multiplied by the gain or the transmittance. For
example, the signal reaching the node C from the node E is given by GE, where G is the branch
transmittance and the branch is directed from node E to node C (see Figure 4.1).
Input Node or Source: It is a node with only outgoing branches; for example, R in
Figure 4.1 is an input node.
Output Node or Sink: It is a node with only incoming branches. However, this condition
is not always met. An additional branch with unit gain may be introduced in order to meet the
specified condition; for example, node C in Figure 4.1 has one outgoing branch. However,
after introducing an additional branch with unit transmittance as shown in Figure 4.2, the node
becomes an output node.
Fig. 4.2 Input node and output node illustrated.
Path: It is the traversal of connected branches in the direction of the branch arrows such
that no node is traversed more than once.
Forward Path: It is a path from the input node to the output node, when no node is
encountered twice. R–E–C is a forward path.
Forward Path Gain: It is the product of branch gains in the forward path; for example, the
forward path gain of the path R–E–C in Figure 4.2 is G.
Loop: It is a path which originates and terminates at the same node; for example, E–C–B–E
is a loop.
Loop Gain: It is the product of the branch gains encountered in traversing the loop, for
example, the loop gain of the loop E–C–B–E in Figure 4.2 is GH.
Non-touching Loops: Loops are said to be non-touching, if they do not possess any
common node.
4.2
4.2.1
Construction of Signal Flow Graph
Steps Followed in Drawing SFG
The equations, describing a system, are used to draw the signal flow graph. Consider a system
described by the following set of equations. The input node is x1 and the output node is x2.
x2 = a12x1 + a32 x3 + a42 x4 + a52 x5
(4.1)
x3 = a23x2
(4.2)
x4 = a34 x3 + a44 x4
(4.3)
x5 = a35x3 + a45x4 + a55x5
(4.4)
Here, the input variable is x1 and output variable is x5.
Chapter 04.indd 2
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MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
4-3
First, locate the nodes of the system, that is, x1, x2, x3, x4, x5 as in (i) of Figure 4.3 below and
then draw SFG of the given equations as in (ii) through (v).
Fig. 4.3 Development of signal flow graph for a set of equations describing a system.
The overall SFG can be drawn by adding the individual flow graphs of Equations (4.1)
through (4.4) as shown in Figure 4.4.
Fig. 4.4 Complete signal flow graph of a system.
The overall gain from input to output may be obtained by using Mason’s gain formula.
Chapter 04.indd 3
03/11/11 6:22 PM
4-4
CONTROL SYSTEMS ENGINEERING
4.2.2
Mason’s Gain Formula
According to Mason’s gain formula, the overall gain T is expressed as
T =
1
∆
∑ Pk ∆ k
k
where
Pk = Path gain of the kth forward path.
∆ = Determinant of the path
= 1 – (sum of loop gains of all individual loops)
+ (sum of gain products of all possible combinations of two non-touching loops)
– (sum of gain products of all possible combinations of three
non-touching loops) + …
D = 1 − ∑ Pm1 + ∑ Pm 2 − ∑ Pm 3 + ......
m
m
m
∆ k = Value of for that part of the graph non-touching the kth forward path
T = Overall gain of the system
Pmr = Gain product of mth possible combination of r non-touching loops
Example 4.1
(1)
Obtain the transfer function of the system shown in Figure 4.5 by block diagram reduction
technique and also using SFG.
Fig. 4.5 Block diagram of a control system.
(2) Draw the SFG for the following set of linear equations:
3y1 + y2 + 5y3 = 0
y1 + 2y2 – 4y3 = 0
–y2 – y3 = 0
Solution
(1)
Chapter 04.indd 4
Using block diagram reduction technique, the transfer function of system shown is
determined as follows.
03/11/11 6:22 PM
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
4-5
Fig. 4.6 Determination of transfer function of the system of Fig. 4.5 using
block diagram reduction technique.
Transfer function =
G1G2
(1 + G1 H1 ) (1 + G2 H 2 ) + G1G2
The SFG of the system is shown in Figure 4.7 below.
Fig. 4.7 SFG of the system shown in Fig. 4.5.
(i)
The forward path is only one
The gain is P1 = G1G2
(ii)
There are three individual loops; the loop gains are
P11 = –G1H1; P12 = –G2H2; P13 = –G1G2
(iii) There are two non-touching loops with gain
Chapter 04.indd 5
P12 = (–G1H1) (–G2H2) = G1G2H1H2
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4-6
CONTROL SYSTEMS ENGINEERING
(iv) There is one forward path and all the loops touch the forward path
Therefore, D1 = 1
(v)
D = 1 – (sum of loop gains of all individual loops) + (sum of gain products of all possible
combinations of two non-touching loops) – …
= 1 − (−G1 H1 − G2 H 2 − G1G2 ) + G1 H1G2 H 2
Transfer function is
G1G2
P1∆1
=
1 + G1 H1 + G2 H 2 + G1G2 + G1G2 H1 H 2
∆
(2)
To draw the SFG for the given set of linear equations, first rearrange these equations as
1
5
y1 = − y2 − y3
3
3
y2 = 2 y3 −
y3 = –y2
1
2
y1
Since there are three variables, we consider three nodes of the SFG as shown. The SFG is
drawn such that the graphs satisfy the three equations as shown in Figure 4.8 below.
Fig. 4.8 SFG for a set of linear equations.
Example 4.2 Find the gain of the system represented by the following equations:
x2 = t12x1 + t32x3
x3 = t23x2 + t43x4
x4 = t24x2 + t34x3 + t44x4
x5 = t25x2 + t45x4
(4.5)
(4.6)
(4.7)
(4.8)
Here, the input node is x1 and output node is x5.
Solution First of all, let us locate the nodes of the system.
Now, draw the SFGs of the four equations as shown in (ii) through (v).
Chapter 04.indd 6
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MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
4-7
The overall SFG is obtained by adding the graphs of the individual equations, as shown in
Figure 4.9.
Fig. 4.9 Signal flow graph for a set of equations of Example 4.2.
Procedure for finding gain
(1) There are three forward paths as shown in (i), (ii) and (iii).
∴ Gain, P1 = t12t23t34t45
Chapter 04.indd 7
03/11/11 6:22 PM
4-8
CONTROL SYSTEMS ENGINEERING
∴ Gain, P2 = t12t24t45
∴ Gain, P3 = t12t25
(2)
There are four individual feedback loops as shown in (i), (ii), (iii) and (iv).
∴ Loop gain P11 = t23t32
∴ Loop gain P21 = t34 t43
∴ Loop gain P31 = t44
∴ Loop gain P41 = t32t43t24
(3)
There is only one possible combination of two non-touching loops as shown
Chapter 04.indd 8
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MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
(4)
4-9
∴ Gain, P12 = t23 t32 t44
(i)
The first forward path touches all the loops, so no individual loops are formed.
∴ D1 = 1
(ii) Also, when the second forward path is eliminated, no individual loop is formed.
∴ D2 = 1
(iii) When the third forward path is eliminated, two loops are formed.
∴ D3 = 1 − t34 t43 − t44
Now,
TF
=
=
x5
x1
=
P1∆1 + P2 ∆ 2 + P3 ∆3
∆
(t12 t 23t 34 t 45 ) + t12 t 24 t 45 + t12 t 25 (1 − t 34 t 43 − t 44 )
1 − (t 23t 32 + t 34 t 43 + t 44 + t 32 t 43t 24 ) + t 23t 32 t 44
Example 4.3 Find the gain of the control system represented in block diagram form as in
Figure 4.10, using Mason’s gain formula.
Fig. 4.10 Block diagram of a control system.
Solution It was mentioned at the beginning of the chapter that for complicated systems, the
SFG method is preferred over the block diagram reduction technique for finding the closed-loop
transfer function. We have already solved this problem by the block diagram reduction technique
in Example 3.4. Now, we shall solve it by the SFG method. The signal flow graph of the system
of Figure 4.10 has been drawn and shown in Figure 4.11.
Chapter 04.indd 9
03/11/11 6:22 PM
4-10
CONTROL SYSTEMS ENGINEERING
Fig. 4.11 Signal flow graph of the system of Fig. 4.10.
(1)
There is only one forward path
∴ Gain P1 = G1G2G3
(2)
There are three individual loops
(3)
∆1 = 1 − L1
∴
Chapter 04.indd 10
D = 1 − (−G1G2 H1 − G2 G3 H 2 − G1G2 G3 )
D = 1 + (G1G2 H1 + G2 G3 H 2 − G1G2 G3 )
TF =
1
∆
( P1 ∆)
03/11/11 6:22 PM
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
C (s)
R( s )
=
4-11
G1G2 G3 (1 + G1G2 H1 )
1 + G1G2 H1 + G2 G3 H 2 + G1G2 G3
The result is the same as obtained in Example 3.4. This shows that we can use either block
diagram reduction technique or SFG method for determining the transfer function.
Example 4.4 Find the outputs of the system, the SFG of which is as shown in Figure 4.12.
Fig. 4.12 Representation of a system through SFG.
Solution Consider output C1 with only one input R1. The system has one forward path with
path gain P1 = G1.
It has two individual loops with loop gains as
P11 = G3H2 and P21 = G4H1
The sum of the product of two non-touching loops will be
= G3G2H1H2
∆1 = 1, as there is no non-touching loop with the forward path.
Chapter 04.indd 11
03/11/11 6:22 PM
4-12
CONTROL SYSTEMS ENGINEERING
Hence,
Similarly,
C1
=
C1
=
R1
G1
1 − (G3 H 2 + G4 H1 ) + (G3G4 H1 H 2 )
G4
R2
1 − G4 H1
Therefore, the output C1 is given by
C1 =
1 − (G3 H 2 + G4 H1 ) + G3 H1 H 2 G4
Similarly, the output C2 can be calculated as
C2 =
G1 R1
G2 R2
+
1 − (G3 H 2 + G4 H1 ) + (G3G4 H1 H 2 )
G4 R2
1 − G4 H1
+
G3 R1
1 − G3 H 2
Example 4.5 Find the transfer function of the system with SFG as shown in Figure 4.13.
Fig. 4.13 A system represented by signal flow graph.
Solution It has two forward paths with path gains as shown in (i) and (ii) below.
Chapter 04.indd 12
03/11/11 6:22 PM
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
P1 = G1G2G3G4
P2 = G1G2G3G4
4-13
There are four loops in the SFG shown in Figure 4.13.
The loop gains are:
P11 = –G1 P21 = – G2G3
P31 = –G1
P41 = –G2G3
The sum of product of two non-touching loops will be
= (−G1 )(−G1 ) + (−G1 )(−G2 G3 ) + (−G1 )(−G2 G3 ) + (−G2 G3 )(−G2 G3 )
= G12 + 2G1G2 G3 + G22 G32
Hence,
D = 1 + 2(G1 + G2 G3 ) + (G12 + 2G1G2 G3 + G22 G32 )
D1 = 1 − (−G1 ) − (−G2 G3 ) = ∆ 2
∴ System transfer function by application of Mason’s gain formula will be
C
R
=
2G1G2 G3G4 (1 + G1 + G2 G3 )
1 + 2(G1 + G2 G3 ) + (G12 + 2G1G2 G3 + G22 G32 )
Example 4.6 Consider a node having three input signals from source nodes R1, R2 and R3.
The transfer function from each source node is g10, g20 and g30, respectively. There are two sink
nodes C1 and C2 having transfer functions t01 and t02 through which signal is transmitted. This is
shown in Figure 4.14. Simplify the circuit to arrive at the transfer function.
Fig. 4.14 SFG of a system with three input nodes and two output nodes.
Solution From Figure 4.14, the output at sink nodes will be
C1 = g10t01R1 + g20t01R2 + g30t01R3
Chapter 04.indd 13
03/11/11 6:22 PM
4-14
CONTROL SYSTEMS ENGINEERING
C2 = g10t02R1 + g20t02R2 + g30t02R3
This state can also be shown by the SFG shown in Figure 4.15.
Fig. 4.15 Representation of the SFG of Fig. 4.14.
This indicates that the series paths or cascaded nodes in any SFG can be replaced by a
single path or pair of nodes having transfer function equal to the product of the individual paths.
Similarly, the parallel paths can be combined by adding the transfer functions.
(a)
(b)
(c)
(d)
Fig. 4.16 Node elimination method.
The mixed nodes can be eliminated by using series, parallel and Mason’s gain formula as
shown in Figure 4.15 for SFG of a closed-loop control system.
4.3
SFG for Solution of Differential Equations
SFG technique can also be used to simplify the solution of differential equations. The procedure
is explained with the help of the following example.
Example 4.7 Draw a SFG of a control system with differential equation for its response as

x + 2 x + 2 x = y
The system has all initial conditions as equal to zero.
Chapter 04.indd 14
03/11/11 6:22 PM
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
4-15
Solution
x = y − 2 x − 2 x.
Step 1: Write the equation for the highest order derivative as 
Step 2: Consider the highest derivative as a dependent variable and all the R.H.S. terms as
independent variables. Prepare the SFG as shown in Figure 4.17.
Fig. 4.17 SFG of a differential equation x + 2 x + 2 x = y .
Step 3: Insert branches which represent the auxiliary equations. Thereby, connect the nodes
of the highest order derivative with the one whose order is lower than this and the node with
the lower order than the highest and so on. The flow of the signal must be from a higher order
derivative to lower order derivative and insert the transfer function as 1/s. The resultant graph is
shown below.
Step 4: Reverse the signs of the branches which indicate integration. While applying this
step observe the following rule.
Rule: To reverse the sign of a branch connecting the mth node to nth node (entering nth node)
of a SFG, without disturbing the TF of the graph, it is necessary to reverse the signs of all the
remaining branches entering as well as leaving the nth node as shown below.
x to x branch, the sign reversal of x to
It is to be noted that when the sign is reversed for 
x branch is already achieved and therefore, further reversal is not necessary.
Chapter 04.indd 15
03/11/11 6:22 PM
4-16
CONTROL SYSTEMS ENGINEERING
Step 5: Redraw the SFG to indicate direct flow from independent variable to dependent
variable as shown in Figure 4.18.
Fig. 4.18 Resultant SFG of the control system described by a differential equation.
Example 4.8 Draw signal flow diagrams for the differential equations, as stated in (a), (b)
and (c),
(a)
y2 = a1
(b)
y3 =
(c)
d2 y
dx
2
dy1
dt
d 2 y2
dt
+
2
+
2 dy
3 dx
dy1
dt
+
− y1
11
2
y=x
Solution The signal flow graphs for the differential equations have been drawn as in (a), (b),
and (c) below respectively.
Chapter 04.indd 16
03/11/11 6:22 PM
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
4-17
Alternatively,
d2 y
dx
2
+
2 dy
3 dx
+
11
2
y =x
or
s 2Y ( s ) +
2s
3
Y ( s) +
11
2
Y (s) = X(s)
1
1
2
1
s2
s
=
=
=
2
11
2
11
X ( s)
s2 + s +
1+
+ 2 1 −  − 2 − 11 


3
2
3s 2 s
 3s 2 s 2 
Y ( s)
Comparing with Mason’s gain formula, there is
(1)
one forward path with gain of 1/s2 and
(2)
two feedback loops with gain of –2/3s, –11/3s2.
The SFG is shown in Figure 4.19.
Fig. 4.19 An alternative SFG for the differential equation
Chapter 04.indd 17
d2y
dx 2
+
2 dy 11
+
y = x.
3 dx
2
03/11/11 6:22 PM
4-18
CONTROL SYSTEMS ENGINEERING
Example 4.9 Find by SFG technique, the transfer function of the control system, the block
diagram repretsentation of which is shown in Figure 4.20.
Fig. 4.20 Block diagram of a control system.
Solution The SFG for the given block diagram representation is shown in Figure 4.21.
Fig. 4.21 SFG of the control system where block diagram has been shown in Fig. 4.20.
Forward path gains are
Loop gains are
P1 = G1G3
P2 = G1G2
L1 = –G1G3H1H2
L2 = –G1G2H1H2
D = 1 − ( L1 + L2 ) = 1 + G1G2 H1 H 2 + G1G3 H1 H 2
TF =
=
=
Chapter 04.indd 18
∴ D1 = 1
∴ D2 = 1
P1∆1 + P2 ∆ 2
∆
G1G3 + G1G2
1 + G1G3 H1 H 2 + G1G2 H1 H 2
G1 (G3 + G2 )
1 + G1 H1 H 2 (G3 + G2 )
03/11/11 6:22 PM
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
Example 4.10
4-19
Find the transfer function of the system shown in Figure 4.22.
Fig. 4.22 A control system represented by signal flow graph.
Solution Forward path gains are calculated as
P1 = G2G4G6
Loop gains are
P2 = G3G5G7
P3 = G2G1G7
P4 = G3G8G6
P5 = –G2G1H2G8G6
P6 = –G3G8H1G1G7
L1 = –G4H1
L2 = –G5H2
L3 = G1H2G8H1
Non-touching loops: There is one pair having gain product = G4H1G5H2
D = 1 + G4 H1 + G5 H 2 − G1 H 2 G8 H1 + G4 H1G5 H 2
D1 = 1 + G5H2
D2 = 1 + G4H1
D3 = ∆ 4 = ∆ 5 = ∆ 6 = 1
TF =
=
P1∆1 + P2 ∆ 2 + P3 ∆3 + P4 ∆ 4 + P5 ∆ 5 + P6 ∆ 6
∆
G2 G4 G6 (1 + G5 H 2 ) + G3G5G7 (1 + G4 H1 ) + G2 G1G7 + G3G8G6 − G2 G6 G8G1 H 2 − G3G7 G8G1 H1
Chapter 04.indd 19
1 + G4 H1 + G5 H 2 + G4 G5 H1 H 2 − G1G8 H1 H 2
03/11/11 6:22 PM
4-20
CONTROL SYSTEMS ENGINEERING
Example 4.11 Are the two systems shown in Figure 4.23 (a) and (b) equivalent? Otherwise,
prove that they are not equivalent.
Fig. 4.23 Systems represented by SFGs.
Solution Two systems can be considered equivalent to each other if their transfer functions
are the same. The systems are not equivalent as the transfer function for Figure 4.23 (a) is
abc
TF =
1 − (ad + be + cf ) + (adbe + adcf + becf ) − abcdef
and the transfer function for Figure 4.23 (b) is
TF =
abc
1 − (ad + be + cf ) + (adcf )
Example 4.12 Find the transfer function for the system shown in Figure 4.24 by Mason’s
gain formula.
Fig. 4.24 Block diagram of a control system.
Chapter 04.indd 20
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MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
4-21
Solution The SFG for the system is shown in Figure 4.25.
Fig. 4.25 A control system shown in the form of a signal flow graph.
Forward path gains are
Loop gains are
P1 = G1G2G3
P2 = G4
L1 = –G1G2H1
L2 = –G1G2G3
L3 = –G2G3H2
L4 = –G4
L5 = G4H2G2H1
D = 1 + G1G2 H1 + G1G2 G3 + G2 G3 H 2 + G4 − G2 G4 H1 H2
D1 = ∆ 2 = 1
TF =
G1G2 G3 + G4
1 + G1G2 H1 + G1G2 G3 + G2 G3 H 2 + G4 G2 G4 H1 H 2
Example 4.13 Find C/R for the control system shown in Figure 4.26.
Solution The SFG has been drawn as in Figure 4.27.
Forward path gains are
5(1 + k2 s)
and ∆1 = 1
(1) P1 =
s
(2)
P2 =
(3)
P3 =
Chapter 04.indd 21
50(1 + k2 s)(1 + k1 s)
s2
10(1 + k1s)
s
and ∆ 2 = 1
and ∆3 = 1
03/11/11 6:22 PM
4-22
CONTROL SYSTEMS ENGINEERING
Fig. 4.26 Block diagram of a control system.
Fig. 4.27 SFG of the control system of Fig. 4.26.
(4)
P4 =
−50(1 + k1s)(1 + k2 s)
s2
and ∆ 4 = 1
Loop gains are
(1)
P11 =
(2)
P21 =
(3)
P31 =
(4)
P41 =
Chapter 04.indd 22
−5(1 + k2 s)
s
−10(1 + k1s)
s
−50(1 + k2 s)(1 + k1 s)
s2
50(1 + k1s)(1 + k2 s)
s2
03/11/11 6:23 PM
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
(5)
P51 =
−50(1 + k1s)(1 + k2 s)
s2
D =1−
∴
TF =
4-23
C
R
=
=1+
5
∑ Pm1
m =1
5(1 + k2 s)
s
+
10(1 + k1s)
s
+
50(1 + k1s)(1 + k2 s)
s2
 5(1 + k2 s) 10(1 + k1s) 50(1 + k2 s)(1 + k1 s) 50(1 + k1s)(1 + k2 s) 
+
+
−


s
s
s2
s2


5(1 + k2 s) 10(1 + k1 s) 50(1 + k2 s)(1 + k1 s) 

+
+
1 +

s
s
s2


TF =
C
R
=
2
5s(1 + k2 s) + 10 s(1 + k1 s)
s + 5s(1 + k2 s) + 10 s(1 + k1s) + 50(1 + k2 s)(11 + k1s)
Example 4.14 Draw a block diagram of the SFG shown in Figure 4.28. Find the overall
transfer function by block diagram reduction technique. Verify the result by applying Mason’s
gain formula.
Fig. 4.28 A control system represented through a signal flow graph.
Solution Mason’s gain formula
Forward paths are
(1) G1G2G3G4
(2)
G1G2G5G4
(1)
(2)
(3)
(4)
Loops are
–G1G2G3G4
–G1G2G5G4
–G2H1
–G4H2
Chapter 04.indd 23
03/11/11 6:23 PM
4-24
CONTROL SYSTEMS ENGINEERING
Non-touching loops are –G2H1 and –G4H2
C
R
=
G1G2 G3G4 + G1G2 G4 G5
1 + G1G2 G3G4 + G1G2 G4 G5 + G2 H1 + G4 H2 + G2 G4 H1 H2
The block diagram of the given SFG is shown in Figure 4.29.
Fig. 4.29 Block diagram representation of the system whose SFG was shown in Fig. 4.28.
By the block diagram reduction technique we get Figure 4.30.
Fig. 4.30 Block diagram reduction technique applied to the system of Fig. 4.29.
 G1G2 (G3 + G5 )G4 
 (1 + G H )(1 + G H ) 

2 1
4 2 
C (s)
=
G1G2 (G3 + G5 )G4
R( s )
1+
(1 + G2 H1 )(1 + G4 H 2 )
=
G1G2 G3G4 + G1G2 G4 G5
1 + G1G2 G3G4 + G1G2 G4 G5 + G2 H1 + G4 H 2 + G2 G4 H1 H2
This shows that the transfer function calculated by both the methods is the same.
Example 4.15 Using SFG and Mason’s gain formula, obtain the overall gain of the system
depicted in Figure 4.31. Also verify by block diagram reduction technique.
Chapter 04.indd 24
03/11/11 6:23 PM
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
4-25
Fig. 4.31 Block diagram of a control system.
Solution The SFG of the given block diagram is shown in Figure 4.32.
Fig. 4.32 SFG of the system shown in block diagram form in Fig. 4.31.
Forward path gains are, G1G2G3 and G1G2
Loop gains are
(1)
–G1G2H1H2
(2)
–G1G2H2
(3)
–G2
(4)
–G1G2G3H1H2
Overall gain, that is, the transfer function is
TF =
G1G2 + G1G2 G3
1 + G2 + G1G2 H 2 + G1G2 H1 H 2 + G1G2 G3 H1 H 2
Block diagram reduction technique
Chapter 04.indd 25
G1G2 (1 + G3 )
1 + G2 + G1G2 H 2
C (s)
=
G1G2 (1 + G3 )
R( s )
1+
HH
1 + G2 + G1G2 H 2 1 2
03/11/11 6:23 PM
4-26
CONTROL SYSTEMS ENGINEERING
=
G1G2 + G1G2 G3
1 + G2 + G1G2 H 2 + G1G2 H1 H 2 + G1G2 G3 H1 H 2
The results are the same in both the cases.
Fig. 4.33 Transfer function of the system shown in block diagram form in Fig. 4.31.
Example 4.16 Use Mason’s gain formula for determining the overall transfer function of the
system shown in Figure 4.34.
Fig. 4.34 A system shown in the form of a signal flow graph.
Solution Forward path gains are
P1 = G1G2G3
P2 = G4
Chapter 04.indd 26
03/11/11 6:23 PM
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
4-27
Loop gains are
P11 = –H1
P21 = –H2
P31 = –G3H4
P41 = –G2G3H3
Non-touching loops are
P12 = H1H2
P22 = H1H4G3
∴
D = 1 + H1 + H 2 + G3 H 4 + G2 G3 H3 + H1 H 2 + H1 H 4 G3
D1 = 1
and
D2 = 1 + H1 + H2 + H1H2
P ∆ + P2 ∆ 2
G1G2 G3 + G4 (1 + H1 + H 2 + H1 H 2 )
∴
TF = 1 1
=
∆
1 + H1 + H 2 + G3G4 + G2 G3 H3 + H1 H 2 + G3 H1 H 4
REVIEW QUESTIONS
4.1 Draw the SFG for the system whose block diagram is shown in the Figure 4.35 and find
its overall transfer function.
Fig. 4.35
4.2 Determine the transfer function relating to C and R for the system represented in
Figure 4.36.
Fig. 4.36
Chapter 04.indd 27
03/11/11 6:23 PM
4-28
CONTROL SYSTEMS ENGINEERING
4.3 Apply Mason’s gain formula to determine the overall transfer function of a control system
represented through block diagram as shown in Figure 4.37.
Fig. 4.37
4.4 Draw the SFG for the system represented by the block diagram as shown in Figure 4.38
and determine the C/R ratio.
Fig. 4.38
4.5 Obtain the transfer functions G6, G7 and G8 so that the following three systems of
Figure 4.39 (a), (b) and (c) become equivalent. In the SFG of Figure 4.39 (a), G1, G2, G3,
G4 and G5 are given.
Chapter 04.indd 28
03/11/11 6:23 PM
MODELLING A CONTROL SYSTEM—SIGNAL FLOW GRAPH
4-29
Fig. 4.39
4.6 Derive C(s)/R(s) by the SFG technique for the system represented by a block diagram as
shown in Figure 4.40.
Fig. 4.40
4.7 Find the transfer function of the system shown in the Figure 4.41 by Mason’s gain formula.
Fig. 4.41
Chapter 04.indd 29
03/11/11 6:23 PM
4-30
CONTROL SYSTEMS ENGINEERING
4.8 Draw a block diagram of the SFG given in Figure 4.42. Find the overall transfer function
by block diagram reduction and verify the result by Mason’s gain formula.
Fig. 4.42
4.9 The SFG of two input two output system is shown in Figure 4.43. Find the outputs of the
system.
Fig. 4.43
4.10 Draw the SFG of the system whose block diagram is given in Figure 4.44. Find the outputs
using Mason’s gain formula.
Fig. 4.44
Chapter 04.indd 30
03/11/11 6:23 PM
5
TIME RESPONSE ANALYSIS
5.1
TIME RESPONSE OF FIRST SYSTEMS
Total time response c(t) of a control system consists of transient response (dynamic response)
ct(t) and steady-state response css(t).
c(t) = ct(t) + css(t)
where
c(t) = Total time response
ct(t) = Transient response
css(t) = Steady-state response
A transient state of the system remains for short time while steady state is that stage of the
system as time t approaches infinity. A feedback control system has the inherent capability that its
parameters can be adjusted to alter both its transient and steady-state response or behaviour. In order
to analyse the transient and steady-state behaviour of control systems, we obtain a mathematical
model of the system. For any specific input signal, a complete time response expression can
then be obtained through the Laplace transform inversion. This expression yields the steady-state
behaviour of the system with time tending to infinity. In case of simple deterministic signals,
steady-state response expression can be calculated by the use of the final value theorem.
Before proceeding with the time response analysis of a control system, it is necessary to test
the stability of the system through indirect tests without actually obtaining the transient response.
In case, the system happens to be unstable, hence of no practical use, we need not proceed with
its transient response analysis.
The following are the typical test signals, that is, signals which can be generated in the
laboratory:
(1) Unit step
(2) Unit ramp
(3) Parabolic
(4) Sinusoidal
Chapter 05.indd 1
03/11/11 7:03 PM
5-2
CONTROL SYSTEMS ENGINEERING
The nature of the transient response of a system depends only upon the system poles and not
on the type of input. Therefore, we shall analyse the transient response to one of the standard test
signals. A step signal is generally used for this purpose. The time-consuming transient analysis
need not be carried out for all the test signals.
5.1.1
Types of Input Signals
In analysing or designing a particular control system, we must have a basis of comparison of
performance of various control systems. This basis may be set by specifying particular test input
signals and by comparing the responses of various systems to these input signals. Usually the
input signals to control systems are not fully known. However, from experience, it has been
observed that the actual input signals which severely strain a control system are a sudden shock,
a sudden change, a constantly increasing change or a constantly accelerating change. Therefore,
system dynamic behaviour for analysis and design can be studied and compared under application
of standard test signals such as an impulse signal (sudden shock), a step signal (sudden change),
a ramp signal (constant velocity) or a parabolic signal (constant acceleration). Sinusoidal signal
is also another important test signal. With these test signals, mathematical and experimental
analysis of control systems can be carried out easily since these signals are very simple functions
of time. The following are the standard test signals.
Step signal: A step signal gives an instantaneous change in the value of the reference r(t)
1.
as shown in Figure 5.1 (a).
Fig. 5.1 Various types of test signals used for the performance analysis of control system.
A step signal can be expressed mathematically as r(t) = 0; t < 0 and r(t) = A; t > 0. The
Laplace transform is R(s) = A/s
Chapter 05.indd 2
03/11/11 7:03 PM
TIME RESPONSE ANALYSIS
2.
5-3
Ramp signal: A ramp signal gives a constant change the value of the reference variable r(t)
with respect to time, as shown in Figure 5.1 (b). It is also the integral of a step signal.
r(t) = 0; t < 0 and r(t) = At; t > 0. The Laplace transform is R(s) = A/s2
Parabolic signal: A parabolic signal gives an accelerating change in the value of the
reference variable r(t). This is the integral of ramp signal. It is shown in Figure 5.1 (c).
3.
r(t) = 0; t < 0 and r(t) =
At 2
2
; t>0
The Laplace transform is R(s) = A/s3.
Impulse signal: The unit-impulse signal gives an infinite magnitude to the value of the
reference variable at t = 0 and a zero value everywhere except at t = 0. Since a perfect
impulse cannot be achieved in practice, it is approximated as a pulse of small width and of
unit area shown in Figure 5.1 (d). Mathematically, the impulse function r(t) can be written
as r(t) = u(t), where u(t) = 0 for t < 0 and u(t) = 1 for t > 0, that is, u(t) is a unit step function.
4.
r(t) = 0; t ≠ 0 and r (t) = 1; t = 0
The Laplace transform is R(s) = 1.
Whether or not a given system will exhibit steady-state error for a particular type of input
will depend upon the type of open-loop transfer function of the system.
5.1.2
Classification of Control Systems
Control systems may be of different orders, viz. zero-order system, first-order system, secondorder system and so on. The generalized relation between a particular input, say qi , and the
corresponding output, say q0, with proper simplified assumptions, can be written as
d n q0
d n −1q0
dq0
d m qi
d m −1qi
dqi
+ b0 qi (5.1)
dt
dt
dt
dt
dt
dt
where
q0 is the output quantity
qi is the input quantity
t is the time
a’s and b’s are combination of system parameters which are assumed to be constants.
Taking Laplace transform and assuming all initial conditions to be zero, we get
an
n
+ an −1
n −1
+  + a1
+ a0 q0 = bm
m
+ bm −1
m −1
+  + b1
[an s n + an −1s n −1 +  + a0 ]Q0 ( s ) = [bm s m + bm −1s m −1 +  + b0 ]Qi ( s )
Transfer function is
Q0 ( s ) bm s m + bm −1s m −1 +  + b0
=
G(s) =
Qi ( s )
an s n + an −1s n −1 +  + a0
The TF can also be written as
k (τ s + 1)(τb + 1) 
G(s) = N a
s (τ1s + 1)(τ2 s + 1) 
(5.1a)
The power of sN (N = 0, 1, 2) determines the type of the system.
Chapter 05.indd 3
03/11/11 7:03 PM
5-4
CONTROL SYSTEMS ENGINEERING
Zero-order System
In the generalized system equation of Equation (5.1), if all the a’s and b’s are made zero
except a0 and b0, we will get
b
b
(5.2)
q0 = 0 qi = kqi
where k = 0
a0
a0
(1)
Therefore, q0 = kqi represents a zero-order system.
A simple example of a zero-order system is a potentiometer where the output voltage is a
fraction of the input voltage, that is, e0 = kei.
First-order System
If in the generalized system equation of Equation (5.1), all a’s and b’s other than a1, a0 and
b0 are taken as zero, we will get
dq
a1 0 + a0 q0 = b0qi
dt
or
b
a1 dq0
+ q0 = 0 qi
a0
a0 dt
or
dq
(5.3)
τ 0 + q0 + kqi dt
where
τ = a1/a0 is called the time constant
k = b0/a0 is called the static sensitivity
Any system that follows the above relation of Equation (5.3) is called a first-order system.
Taking Laplace transform,
τsQ0 (s) + Q0 (s) = kQi(s)
(2)
TF, G(s) =
Q0 ( s )
Qi ( s )
=
k
τs + 1
We have seen that a simple R–C network having an input ei and output across C as e0 will
have a similar transfer function and, therefore, can be called a first-order system.
(3)
Second-order System
A second-order system is one which follows the following equation
a2
d 2 q0
dt
2
+ a1
dq0
dt
+ a0 q0 = b0qi
or
a2 d 2 q0
Chapter 05.indd 4
a0 dt
2
+
a1 dq0
a0 dt
+ q0 =
b0
a0
qi
03/11/11 7:03 PM
TIME RESPONSE ANALYSIS
or
1 d 2 q0
2ζ dq0
+ q0 = kqi
ωn dt
dt
where
ω2n
2
a0
ωn =
k =
b0
a0
a2
+
5-5
(5.4)
, is called the undamped natural frequency
;
ζ =
a1
2 a0 a2
, is called the damping ratio.
 a0
a a
2ζ
aω
a1 
=
, ∴ζ = 1 n = 1 0 =


2a0
2a0 a2
2 a0 a2 
 a0 ωn
Taking Laplace transform of Equation (5.4), we get
 s 2 2ζ s

+ 1 Q0 ( s ) = kQi(s)
 2 +
ωn
 ωn

TF is
G(s) =
Q0 (s)
Qi (s)
=
s2
ω2n
k
+
2ζ s
+1
ωn
=
k ω2n
s 2 + 2ζωn s + ω2n
A mass-spring-damper system is an example of second-order system. By referring to
Figure 2.4 and Equation (2.9), we can rewrite the equation of forces as
M
d2x
dt
+B
dx
dt
+ kx = f(t)
Taking Laplace transform,
1
K1
1
X (s)
M
=
=
= 2
2
B
F (s)
s + 2ζωn s + ω2n
Ms + Bs + K
s +1
s2 +
M
where
ωn =
and
K
M
, is the undamped natural frequency in rad/sec.
B
, is the damping ratio.
2 KM
It will be seen later that as the system order number increases, the accuracy improves but
the system tends to become unstable.
Chapter 05.indd 5
z =
03/11/11 7:03 PM
5-6
CONTROL SYSTEMS ENGINEERING
Before going to error and stability analysis, we will discuss steady-state and dynamic error
coefficients.
5.1.3
Time Response of First-order System
We will examine the time response of a first-order system to different types of input signals.
5.1.4
Response of First-order System to Step Input
Let us consider a simple RC circuit as shown in Figure 5.2.
Fig. 5.2 A simple first-order system.
We have
ei = Ri +
and
eo =
1
C
1
C
∫ idt
∫ idt
Taking Laplace transform of these two equations
Ei(s) = RI ( s ) +
or
1
Cs
I (s)
1 

Ei(s) =  R +
 I (s)
Cs 

and
Eo(s) =
TF =
1
Cs
I (s)
Eo ( s )
1


=

Ei ( s )  1 + RCs 
where τ = RC = Time constant of RC circuit
Chapter 05.indd 6
03/11/11 7:03 PM
TIME RESPONSE ANALYSIS
∴
TF =
5-7
1
1 + τs
Note that the power of s in the denominator of TF determines the order of the system.
The block diagram of the system is shown in Figure 5.3.
Fig. 5.3 Block diagram representation of a first-order system.
C(s) = R(s)
1
sτ + 1
Time response to the unit step input R(s) = 1/s is calculated as
1
C(s) =
s (τs + 1)
or
∴
C(s) =
1
s
−
τ
τs + 1
c(t) = [1 − e−t/τ ]
The unit step response of the above system is shown in Figure 5.4. It is seen that as time
tends to infinity, the error e(t) goes on reducing and finally becomes zero. The steady-state error
becomes zero.
Fig. 5.4 Response of a first-order system to unit step input.
The time constant is indicative of how fast the system tends to reach the final value.
The speed of the response can be quantitatively defined as the time for the output to become a
particular percentage of its final value. A large time constant corresponds to a sluggish response
and a small time constant corresponds to a fast response as shown in Figure 5.5.
Chapter 05.indd 7
03/11/11 7:03 PM
5-8
CONTROL SYSTEMS ENGINEERING
Error is
or
or
e(t) = r(t) – c(t)
e(t) = 1 − (1 − et/τ )
e(t) = e−t/τ
Fig. 5.5 Effect of time constant on system response—response shown
in b is slower than that of a.
Steady-state error is given by
ess = lim e(t ) = lim e−t/τ = 0
t →∞
t →∞
The error response is shown in Figure 5.4.
5.1.5
Response of First-order System to Ramp Input
C (s)
R( s )
=
R(s) =
or
C(s) =
1
τs + 1
1
s2
R( s )
τs + 1
or
Chapter 05.indd 8
C(s) =
(for ramp input)
2
1
s (τs + 1)
03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
or
∴
C(s) =
1
s2
−
τ
s
+
5-9
τ2
τs + 1
c(t) = t − τ(1 − e−t/τ )
Error is given by
e(t) = r(t) – c(t)
or
e(t) = t − [t − τ (1 − e−t/τ )]
e(t) = τ (1 − e−t/τ )
Steady-state error is given by
ess = lim e(t ) = τ
t →∞
Therefore, reducing the system time constant improves the speed of response of the system
as well as reduces its steady-state error to a ramp input. We, therefore, need to examine only the
steady-state error to ramp input which can also be obtained by applying the final value theorem
as follows:
ess = lim e(t )
t →∞
= lim sE (s)
s →0
= lim s [ R( s ) − C ( s )]
s →0
1
1

= lim s  2 − 2

s →0
s (τs + 1) 
s
= lim
s 
1 
1−
2 
(τs + 1) 
s 
= lim
1  τs + 1 − 1 
s  τs + 1 
= lim
1  τs 


s  τs + 1 
s →0
s →0
s →0
 τ 
= lim 

s → 0  τs + 1 
∴
e =t
ss
Thus, there is no need of taking inverse Laplace transform, if we use the final value theorem.
Chapter 05.indd 9
03/11/11 7:04 PM
5-10
CONTROL SYSTEMS ENGINEERING
Fig. 5.6 Response of a first-order system to (a) unit ramp input; (b) unit impulse input.
5.1.6
Response of First-order System to Impulse Input
For unit impulse input R(s) =1.
C (s)
∴
R( s )
=
C(s) =
or
Steady-state error
c(t) =
1
τs + 1
1
τs + 1
1
τ
R(s) =
1
τs + 1
=
1
s+
1
τ
e −t/τ
ess = lim sE ( s )
s →0
= lim s[ R( s ) − C ( s )]
s →0
1 

= lim s 1 −
τs + 1 
s →0 
 τs + 1 − 1 
= lim s 

s → 0  τs + 1 
 τs 
= lim s 

s → 0  τs + 1 
Chapter 05.indd 10
ess = 0
03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
5-11
To summarize, the response of a first-order system having transfer function 1/ts + 1 to
different input signals has been tabulated in Table 5.1.
Table 5.1 Response of a First-order System to Different Types of Input Signals.
Signal
Input r(t)
Output c(t)
Steady-state Error ess(t)
(i) Ramp
t
t − τ(1 − e −t/τ )
t
(ii) Unit step
1
1 − e −t/τ
0
(iii) Impulse
u (t ) = δ(t )
1 −t/τ
e
τ
0
It may be seen from Table 5.1 that input at (ii) is derivative of input at (i); also output at
(ii) is derivative of output at (i). This rule can also be generalized for higher order systems.
We have found the step response of the second-order system. We can find the impulse
response of the second-order system by differentiating the step response found earlier. Similarly,
the ramp response can be found by integrating the step response. Since initial conditions are
already zero, we can find the constants of integration easily.
5.2
CHARACTERISTICS EQUATION OF FEEDBACK CONTROL SYSTEM
A second-order system is represented by the general equation
a2
d 2 q0
dt
2
+ a1
dq0
dt
+ a0 q0 = b0qi
Fig. 5.7 Block diagram representation of a second-order system with unity feedback.
The block diagram of a second-order system with unity feedback is shown in Figure 5.7
As
Chapter 05.indd 11
C (s)
R( s )
=
G(s)
1 + G(s) H (s)
H(s) = 1,
C (s)
R( s )
=
G(s)
1 + G(s)
=
k ω2n / s ( s + 2ζωn )
1+
ω2n
s ( s + 2ζωn )
03/11/11 7:04 PM
5-12
CONTROL SYSTEMS ENGINEERING
5.2.1
Transfer Function and Characteristic Equation in Generalized Form
The transfer function of a second-order system is given as
C (s)
where
R( s )
=
ωn =
k ω2n
s 2 + 2ζωn s + ωn2
a0
a2
and
ζ=
a1
2 a0 a2
ωn is the undamped natural frequency and z is the damping ratio. The characteristic
equation is written by equating the denominator of the transfer function to zero.
The characteristic equation of the second-order system is
s 2 + 2ζωn s + ω2n = 0
The roots of the characteristic equation are
s1, s2 = −ζωn ± ωn ζ 2 − 1
= σ ± jωd
where
and
s = zwn is called the attenuation
ωd = ωn 1 − ζ 2 is called the frequency of damped oscillation
The time domain response of the system will depend upon the roots of the characteristic
equation. For over-damped system, z > 1 and for an under-damped system, z < 1. For a critically
damped system, z = 1.
5.2.2
Positional Servosystem as a Second-order System and Its Analysis
As an example, let us consider a second-order system. A positional servomechanism system has
been shown in Figure 5.8. Let us examine the positional servomechanism.
Let us consider,
Ra = Resistance of armature
ia = Armature current
if = Field current
va = Applied armature voltage
eb = Back emf
Tm = Torque developed by motor
q = Angular displacement of motor shaft
J = Equivalent moment of inertia of motor and load referred to the motor shaft
Chapter 05.indd 12
03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
5-13
Fig. 5.8 Positional servomechanism.
Bo = Equivalent viscous friction coefficient of motor and load referred to the motor
shaft
f = Air-gap flux
KA = Amplifier gain
Kp = Potentiometer sensitivity
Ve = Voltage across the amplifier
r = Reference shaft position
c = Output shaft position
n = Gear train ratio of load shaft speed (c) to motor shaft speed (θ )
The servomechanism controls the position of the mechanical load in accordance with the
position of the reference shaft. The error signal (ve) appears at the leads of the potentiometer
wiper arms. From Figure 5.8 we may write the following:
ve = Kp(r – c)
(5.5)
va = KAve = KAKp(r – c)
eb = kb
dθ
dt
(5.6)
(5.7)
where kb = back emf constant. Neglecting the small armature inductance, the equation of armature
circuit is
Raia + eh = va
or
Raia + kb
dθ
dt
= va
(5.8)
In servo application, a dc motor is used in the linear zone of magnetization curve and
if = constant for armature-controlled dc motor.
Chapter 05.indd 13
03/11/11 7:04 PM
5-14
CONTROL SYSTEMS ENGINEERING
∴
φ ∝ i f and Tm ∝ φia
Tm = KT ia
where KT = motor torque constant.
The torque equation is given by
J
d 2θ
dt
2
+ Bo
dθ
dt
(5.9)
= Tm
or
(5.10)
Ve(s) = KP[R(s) −C(s)]
(5.12)
Va(s) = KAVe(s) = KAKP[R(s) − C(s)]
E (s) = K θ (s)
(5.13)
or
or
Also,
J
d θ
+ Bo θ = KTia
dt
For the gear train, we have
C = nθ
Taking Laplace transforms of Equations (5.5) through (5.11), we have
b
b
(5.11)
(5.14)
Ra I a (s) + K b θ (s) = Va(s)
Tm(s) = KTIa(s)
Jsθ (s) + B θ (s) = K I (s)
(5.15)
(5.16)
( Js + Bo )θ (s) = KTIa(s)
(5.17)
o
T a
sC(s) = nsθ(s)
C(s) = nθ(s)
dθ
dt
= θ and its Laplace transform is sθ(s) = θ (s)
(5.18)
(5.19)
Hence, using Equations (5.12) through (5.19), we may draw the block diagram as shown
in Figure 5.9.
Fig. 5.9 Block diagram of servomechanism shown in Fig. 5.5.
Chapter 05.indd 14
03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
5-15
Using Equations (5.13), (5.15) and (5.17) through (5.19), we get
K K [R(s) – C(s)] = R I (s) + K θ (s)
P
a a
A
b
= Ra I a (s) + K b sθ(s)
=
=
Ra
KT
RA
KT
( Js + Bo )θ ( s ) + Kb sθ( s )
( Js + Bo ) sθ( s ) + Kb sθ( s )
R
 C (s)
=  a ( Js + Bo ) s + Kb s 
K
 n
 T
or
Ra
K s

KP KA R(s) =  K P K A + a ( Js + Bo ) + b  C ( s )
nKT
n 

or
C (s)
R( s )
=
=
Substituting
Ra
and
K P K A KT n
Bo +
K b KT
Ra
KP K A
Rs
K s
K P K A + a ( Js + Bo ) + b
nKT
n
K P K A nKT / Ra J
K P K A KT n  Bo K b KT 
2
+
+
 s+s
Ra J
J
R
J
a


=K
= B,
we get
K
J
=
K B
R( s )
+ s + s2
J
J
C (s)
=
Chapter 05.indd 15
ω2n
ω2n + 2ζωn s + s 2
03/11/11 7:04 PM
5-16
CONTROL SYSTEMS ENGINEERING
or
C (s)
R( s )
ω2n
where
ω2n =
or
1+
2ζωn =
or
=
ωn2
G(s)
1 + G(s)
(5.20)
s 2 + 2ζωn s
K
J
K
ωn =
and
s 2 + 2ζωn s
=
J
B
J
B
J
ζ =
Bc = 2 JK
=
B
J 2 K
2 JK
Here, ωn is the undamped natural frequency; ζ is the damping ratio and B is the viscous
friction coefficient referred to the motor shaft. B is the actual damping and for ζ = 1, the critical
damping is
∴
ζ =
B
Bc
=
Actual damping value
Critical damping value
The damping ratio actually refers to the damping of oscillations.
From the expression for the derived transfer function, the simplified block diagram of the
servomechanism is shown in Figure 5.10.
Fig. 5.10 Simplified block diagram of Fig. 5.8.
A plot of c(t) against ωn t for all values of damping has been shown in Figure 5.12. It can
be observed that the way c(t) reaches its final value from initial zero value will depend on the
damping condition, that is, on the value of z. For an undamped system when z = 0, the system
output is (1 − cos ωn t ). The system will continue to oscillate around its final value of 1, with peak
values changing from 0 and 2. As the damping is increased, z increases towards unity and system
oscillation gets reduced.
Chapter 05.indd 16
03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
5.2.3
5-17
Characteristic Equation and Position of Roots/Poles
As mentioned earlier, characteristic equation is the equation obtained by equating the denominator
of the overall closed-loop transfer function to zero. The characteristic equation of the closed-loop
second-order system is given by
s 2 + 2ζωn s + ω2n = 0
If ζ ≥ 1, the roots of the equation are real and the system response will be over-damped.
On the other hand, if ζ < 1, the roots of the equation are complex conjugate and the system
response will be oscillatory in nature. The condition of stability of the system is that the value of
ζ should be positive. So for ζ ≥ 1, the real roots the of equation are
s 1, s 2 =
−2ζωn ±
4ζ 2 ω2n − 4ω2n
2
= −ζωn ± ωn ζ 2 − 1
For ζ < 1, the complex conjugate roots of the equation are
s1, s2 = −ζωn ± jωn 1 − ζ 2
= −σ ± jωd
where ζ is the damping ratio, ωn is the undampd natural frequency or nature frequency of
oscillation, σ = ζωn is called the attenuation and ωd = ωn 1 − ζ 2 is called the frequency of
damped oscillation. Characteristic roots are also the poles of the system. So the poles of secondorder system for different values of damping ratio ζ are depicted in Figure 5.11.
Fig. 5.11
Chapter 05.indd 17
Location of poles of second-order system for different values of damping ratio ζ.
03/11/11 7:04 PM
5-18
CONTROL SYSTEMS ENGINEERING
5.2.4
Classification of Time Response for Second-order Control System
According to different values of damping ratio, ζ, the time response of a second-order control
system can be classified as follows:
(1) Under-damped: For 0 < z < 1, the transient response will be oscillatory in nature, but
decays exponentially to give a stable response.
(2) Critically damped: For ζ = 1, the response just becomes non-oscillatory and gives a stable
response after transient disappears.
(3) Over-damped: For ζ > 1, the response is non-oscillatory and gives a stable response after
transient disappears.
(4) Undamped: For ζ = 0, the transient does not disappear and the response gives a sustained
oscillation.
The time response c(t) of a second-order control system for different values of damping
ratio z are shown in Figure 5.12.
Fig. 5.12 Time response for different values of damping ratio.
5.2.5
(1)
Time Response of Second-order Control System Subjected to Unit
Step Input
Under-damped case (when 0 < ζ < 1)
Here R(s) = 1/s and so C(s) can be written as
C(s) =
=
ω2n
s ( s 2 + 2ζωn s + ωn2 )
1
s
−
s + ζωn
2
(s + ζωn ) +
ω2d
−
ζωn
ωn
ωd (s + ζωn )2 ω2d
[where ωd = ωn 1 − ζ 2 = damped natural frequency]
Chapter 05.indd 18
03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
5-19
Taking inverse Laplace transform, we get
ζωn


sin ωd t 
c(t) = 1 − e−ζωnt  cos ωd t +
ωd


e−ζωn t 
=1−
1 − ζ 2 cos ωd t + ζ sin ωd t 
2 
−ζ
=1−
e− ζωn t
1 − ζ2
[where cos φ = ζ ]
e−ζωnt
=1−
1 − ζ2
= 1−
(sin φ cos ωd t + cos φ sin ωd t )
sin (ωd t + φ)


1 − ζ2 
sin  ω n 1 − ζ 2 t + tan −1
ζ 

1 − ζ2
(
e−ζω t
n
)
As lim c(t ) = 1, the time response reaches its steady-state value of unity and hence the
t →∞
steady-state error (ess) becomes zero as deduced in the following:
ess = lim [r (t ) − c(t )]
t →∞
= lim s[ R( s ) − C ( s )]
s →0
1

ω2n
= lim s  −

2
2
s →0  s
s ( s + 2ζωn s + ωn ) 



ω2n
= lim 1 − 2

2
s →0 
s + 2ζωn s + ωn 

= 0
The under-damped response is shown in Figure 5.13.
(2)
Critically damped case (when ζ = 1)
As R(s) = 1/s, so C(s) can be written as
C(s) =
=
Chapter 05.indd 19
ω2n
s(s 2 + 2ωn s + ω2n )
[Putting ζ = 1]
(s + ωn )2 − s(s + ωn ) − sωn
s(s + ωn )2
03/11/11 7:04 PM
5-20
CONTROL SYSTEMS ENGINEERING
=
1
s
−
1
s + ωn
−
ωn
(s + ωn )2
Taking inverse Laplace transform on both sides, we get
c(t) = 1 − e−ωnt − ωn te−ωnt = 1 − e−ωnt (1 + ωn t )
Fig. 5.13 Comparison of unit step input response of a second-order system for
different values of damping ratio ζ.
The unit step response for this case also approaches unity as lim c(t ) = 1 and also steadyt →∞
state error ess = 0 as before. The response curve is shown in Figure 5.13.
(3)
Over-damped case (when ζ > 1)
Here also, R(s) = 1/s, and C(s) may be written as
C(s) =
=
Chapter 05.indd 20
ω2n
s  ( s + ζωn ) 2 − ωn2 (ζ 2 − 1) 
ω2n
s  s + ωn (ζ +
ζ 2 − 1   s + ωn (ζ −
ζ 2 − 1) 
1
1
= +
s 2 ζ 2 − 1 ζ + ζ 2 − 1 s + ω ζ + ζ 2 − 1 


n
(
)
(
)
03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
=−
1


2 ζ 2 − 1 ζ − ζ 2 − 1 s + ω n ζ − ζ 2 − 1 
(
)
(
Taking inverse Laplace transform on both sides, we have
5-21
c( t ) = 1 +
e
(
)
− ζ+ ζ 2 −1 ω nt
(
2 ζ2 − 1 ζ + ζ2 − 1
)
−
)
e
(
)
− ζ− ζ 2 −1 ω nt
(
2 ζ2 − 1 ζ − ζ2 − 1
)
The time response curve is shown in Figure 5.13.
If ζ becomes comparatively larger than 1, then the second term having a smaller time
constant 1 / ζ + ζ 2 − 1 ω n  decays more quickly than the third term with the larger time
(
constant 1 / ( ζ +
)

− 1) ω 
So after the time of the smaller time constant has elapsed, the
ζ2
n
response is similar to that of a first-order system and C(s) may be approximated as
C(s) =
(
1/ s
)
1 + 1 / ζ − ζ 2 − 1 ω n  s
( ζ − ζ − 1) ω
=
s s + ( ζ − ζ − 1) ω 
2
n
2
n
Taking inverse Laplace transform, we get
−
c(t ) = 1 − e (ζ−
)
ζ 2 −1 ω nt
for ζ >> 1
Here also, lim c(t ) = 1 and ess = 0
t →∞
(4)
Undamped case (when ζ = 0)
And R(s) = 1/s, z = 0, we may write C(s) as given here.
C(s) =
=
ω2n
s(s 2 + ω2n )
(s 2 + ω2n ) − s 2
s(s 2 + ω2n )
1
s
= − 2
s s + ωn2
Taking inverse Laplace transform on both sides, we have
c(t) = 1 − cos ωn t
In this case, the time response oscillates with the undamped natural frequency of ωn and so
it does not follow the step input. The response curve is shown in Figure 5.13.
Chapter 05.indd 21
03/11/11 7:04 PM
5-22
CONTROL SYSTEMS ENGINEERING
5.3Transient Response OF SECOND ORDER SYSTEM-TIME DOMAIN
Specifications
Since the unit step response is easy to generate and mathematically the response to any input can
be derived if the response to a step input is known, the performance characteristics of a control
system are described in terms of transient response to a unit step input with standard initial
conditions of output and all time derivatives being zero when the system is at rest. The time
response of second and higher order control systems to a unit step input is generally damped
oscillatory in nature before reaching the steady state. The transient response specifications of a
control system to a unit step input are illustrated in Figure 5.14.
Fig. 5.14 Transient response specifications illustrated for a control system
Rise time (tr): It is the time required for the response to rise from 10% to 90% (for overdamped or critically damped systems) and 0 to 100% (for under-damped systems) of its
final value. The 10–90% and 0–100% rise time are commonly used for over-damped and
under-damped second-order systems, respectively.
(2) Peak time (tp): The peak time is the time required for the response to reach the first peak of
the overshoot.
(3) Maximum overshoot and maximum percentage overshoot: The maximum overshoot
(Mp) is the maximum peak value of the response measured from unity. So Mp is given by
Mp = c(tp) – 1.
If the steady-state value is not unity, then the maximum per cent overshoot as defined below
is used.
Maximum per cent overshoot
c(t p ) − c(∞)
=
× 100
c(∞)
(1)
Chapter 05.indd 22
03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
(4)
(5)
5-23
Settling time (ts): It is the time required for the response curve to reach and stay within a
specified tolerance band (either 2% or 5%) of final value.
Delay time (td): It is the time required by the response to reach half of its final value at the
first attempt.
5.3.1
Transient Response of a Second-order System
Assuming the system to be under-damped second-order, we shall obtain the rise time (tr), peak
time (tp), maximum overshoot (Mp) and settling time (ts) in terms of ζ and ωn. We have deduced
earlier that time response c(t) under unit step input is given by
(1)
c( t ) = 1 −

2
sin  1 − ζ 2 ω t + tan −1 1 − ζ
n
ζ

1 − ζ2
e
− ζω nt
(
)



Rise time (tr): Rise time is obtained from c(tr) = 1
Substituting c(t) = 1 in the above equation, we get
1−

1 − ζ2
sin  ω n 1 − ζ 2 tr + tan −1
ζ

1 − ζ2
e
(
− ζω ntr
)

 = 1

or

1 − ζ2
sin  ω n 1 − ζ 2 tr + tan −1
ζ

(
)

 = 0 = sin π

Now, as
−1
0 < tan
∴
(ω
n
)
1 − ζ 2 tr + tan −1
1 − ζ2
ζ
<
π
2
for 0 < ζ < 1
1 − ζ2
=π
ζ
or
(2)
1 − ζ2
ζ
ωn 1 − ζ
2
= The response time
Peak time (tp): Peak time is obtained by differentiating c(t) with respect to t and equating
to zero. At maxima, the torque is zero. Therefore, peak time is obtained from
Chapter 05.indd 23
tr =
π − tan −1
dc(t )
dt
t =t p
=0
03/11/11 7:04 PM
5-24
CONTROL SYSTEMS ENGINEERING
or
ζω n e
− ζω nt p
1 − ζ2


1 − ζ2 
sin  ω n 1 − ζ 2 t p + tan −1
ζ 

(
)

2
−ω n e− ζωnt p cos  ω n 1 − ζ 2 t p + tan −1 1 − ζ
ζ

(
)

 = 0

or

1 − ζ2
tan  ωn 1 − ζ 2 t p + tan −1
ζ



 =

1 − ζ2
(
ζ
)
The general solution of the above equation is ω n 1 − ζ 2 t p = nπ
where n = 0, 1, 2, 3, … and so on.
As tp is the time required for the response to reach the first peak of the overshoot, so tp is
obtained for n = 1. (By putting n = 2 we can get tp for the second peak and so on).
∴
or
(ω
n
)
1 − ζ2 t p = π
tp =
π
ωn 1 − ζ 2
= Peak for the first maxima
Here
n = 2 first undershoot
n = 3 second overshoot etc.
(3) Maximum overshoot (Mp): Maximum overshoot Mp is found by substituting the value of
tp in the expression for c(t) and subtracting the steady-state response value from it.
Thus,
π


MP = c(t p ) − 1 = c 
 −1
 ω 1 − ζ2 
 n

=1−
e
− ζωn
π
ωn 1− ζ 2
1 − ζ2

π
+ tan −1
sin  ωn 1 − ζ 2
2

ωn 1 − ζ

1 − ζ2 
 −1

ζ

−ζ π
=1−
e
1−ζ 2

sin  π + tan −1

1 − ζ2
1 − ζ2 


ζ

−ζπ
=+
Chapter 05.indd 24
e

1 − ζ2
sin  tan −1
ζ

1 − ζ2
1−ζ 2




03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
5-25
Let
tan −1
1 − ζ2
ζ
=θ
∴
tan θ =
∴
sin θ =
1 − ζ2
ζ
1
cosec θ
1 + cot 2 θ
1
=
1+
1
=
ζ2
1 − ζ2
2
= 1 − ζ
− ζπ
∴
Mp =
e
1− ζ 2
1−ζ
−
1 − ζ2
2
−ζπ
2
MP = e 1−ζ
Maximum per cent overshoot
c(t p ) − c(∞)
=
× 100
c(∞)
∴
−ζπ
=
1−ζ 2
1+ e
−1
1
−
× 100
−ζπ
2
= 100e 1−ζ
(4) Settling time (ts): The transient response curve c(t) always lies in between its two envelope
(
)
curves 1 ± e− ζωnt / 1 − ζ 2 whose time constant is T = (1/ ζωn ). So, the settling time (ts)
for the tolerance band of x per cent may be obtained from the given equation as follows:
x
1 − 1 ± e− ζωnt / 1 − ζ 2
=
100
or
x
1 − ζ2
e−ζωnt =
100
{ (
Chapter 05.indd 25
)}
03/11/11 7:04 PM
5-26
CONTROL SYSTEMS ENGINEERING
For 0 < ζ < 0.9 and for convenience in comparing the responses of systems, we commonly
define the settling time to be
4
(for 2% tolerance band)
ts = 4T =
ζωn
or
3
ts = 3T =
(for 5% tolerance band)
ζωn
The settling time reaches a minimum value a value of ζ around 0.76 (for 2% tolerance
band) or ζ = 0.68 (for 5% tolerance band) and then increases almost linearly for large values of ζ.
5.3.2
Dominant Closed-loop Poles of Higher Order Systems
The closed-loop poles that have dominant effects on the transient response behaviour are called
dominant closed-loop poles. The dominant closed-loop poles often occur in the form of a
complex conjugate pair. The results obtained in the second-order system could easily be extended
to higher order systems with a dominant pair of complex poles. That is, as long as the real part of
these poles is much less than those of the other poles, the time response of higher order systems
can be approximated by their having only dominant poles.
For example, let us consider a third-order system with transfer function,
G (s) =
=
where
p =
C (s)
R( s)
=
ω2n
(1 + sT )( s 2 + 2ζωn s + ω2n )
pω2n
( s + p)( s 2 + 2ζωn s + ω2n )
1
τ
The unit step response by Laplace transform can be obtained as
(
)
c(t ) = 1 − K1e− pt + K 2 e− ζωnt sin ω n 1 − ζ 2 t
where
K1 =
K2 =
ω2n
ω2n − 2ζωn p + p2
1
(
)
1 − ζ 2 1 − 2ζω / p + ω2 / p 2 
n
n


and
Chapter 05.indd 26
φ = tan −1
ωn 1 − ζ 2
p − ζωn
+ tan −1
1 − ζ2
−ζ
03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
5-27
Now two cases arise.
Case I: If p << ζωn , then it can be shown that lim K1 = 1 and lim K2 = 0 so the pole
p→0
p→0
at s = –p being closer to the origin dominates in the response than the two conjugate poles at
s = −ζωn ± jωn 1 − ζ 2 .
Case II: If p >> ζωn , then it can be shown that lim K1 = 0 and lim K2 =
1
so
1 − ζ2
the poles at s = −ζωn ± jωn 1 − ζ 2 being closer to the origin dominate in the response than
the pole at s = –p. It is observed that for p = 6ζωn , the response of the third-order system can be
approximated as the response of the second-order system with transfer function:
p→∞
C (s)
R( s )
=
p→∞
ω2n
s 2 + 2ζωn s + ω2n
The relative locations of the dominant poles of the third-order system are shown in
Figure 5.15.
Fig. 5.15 Location of dominant poles of a third-order system.
5.3.3
Comments on Transient Response Specifications
It is always desirable that the transient response should be sufficiently fast and well damped.
The output should reach the final value in a small interval of time, that is, the system response
is quick. We have already seen from the response curves of the second-order system that for
values of ζ (0.5–0.8), steady-state conditions are obtained earlier but for such an under-damped
system (ζ = 0.5–0.8), there are oscillations at the equilibrium position. For large values of ζ, the
maximum overshoot is more but settling time (Ts = 1/ζωn ) for higher values of ζ is less. Thus,
both the maximum overshoot and the rise time cannot be made smaller simultaneously. If one of
them is made smaller, the other unnecessarily becomes larger.
Chapter 05.indd 27
03/11/11 7:04 PM
5-28
CONTROL SYSTEMS ENGINEERING
Example 5.1 For a second-order system, ζ = 0.6, ωn = 5 rad/sec. Find the values of tr , tp, Mp, ts.
Solution
2
ωd = ωn 1 − ζ
= 5 1 − (0.6)2
ωd = 4 rad/sec
σ = ζωn
= 0.6 × 5
σs = 3
π−β
(1)
tr =
tr =
tr =
tr = 0.55 sec
π
tp =
ωd
(2)
=
ωd
3.14 − β
4
,
β = tan −1
ωd
σ
= tan −1
4
3
= 0.93 rad
3.14 − 0.93
4
3.14
4
tp = 0.785 sec
(3)
Mp =
 σ 
−
π
ω
e  d
= e
3
−  × 3.14
4
Mp = 0.095
(4)
ts =
ts =
4
σ
3
σ
=
=
4
3
3
3
= 1.33 sec (for 2% criterion)
= 1 sec (for 5% criterion)
Example 5.2 Consider the servomechanism shown in Figure 5.16. Determine the values
of k and k1 so that the maximum overshoot in unit step response is 25% and the peak time
is 2 sec.
Chapter 05.indd 28
03/11/11 7:04 PM
TIME RESPONSE ANALYSIS
5-29
Fig 5.16 A feedback control system in block diagram form.
Solution
Mp = e( −ζ /
and
∴
∴
tp =
ζ
1 − ζ2
π
ωd
1−ζ 2 ) π
= 0.25
= 2 given
π = 1.39 and ωd = 1.57
ζ = 0.4
and
ωn =
=
ωd
1 − ζ2
1.57
1 − (0.4)2
= 1.71
Comparing,
∴
C (s)
R( s )
=
k
s 2 + kk1s + k
=
ω2n
s 2 + 2ζωn s + ω2n
k = ω2n = (1.71)2 = 2.93 and kk1 = 2ζωn
k1 =
2(0.4)(1.71)
2.93
= 0.47
Example 5.3 The block diagram of a servo system is shown in Figure 5.17. Determine the
characteristic equation of the system. Also, calculate the following when the unit step input is
given:
(1) Undamped frequency of oscillation
(2) Damped frequency of oscillation
(3) Damping ratio and damping factor
(4) Maximum overshoot
(5) Time interval after which maximum and minimum output will occur
Chapter 05.indd 29
03/11/11 7:04 PM
5-30
CONTROL SYSTEMS ENGINEERING
Fig. 5.17 Block diagram of a servo system.
(6)
(7)
Settling time
Number of cycles completed before the output is settled within 2%, 5% of final value.
Solution The problem indicates a second-order controlled system of the type having output
function for unit step input as
C(s) =
ω2n
2
(s + 2ζωn s +
c(t) = 1 −
ω2n )
1
s
ωn
ω 

e−σt sin  ωd t + tan −1 d 
ωd
σ 

where
ωd = ωn 1 − ζ 2
σ = ωn ζ
Now, for the given problem, the open-loop transfer function is
10 × 2.4
G(s) =
s(s + 1)(0.2s + 1)
G(s) =
and the feedback is
H(s) =
s(s + 1)(0.2s + 1)
s
6
24
C (s)
s ( s + 1)(0.2s + 1)
=
=
24
1
(
)
(
)
+
G
s
H
s

s
R( s )
1 + s ( s + 1)(0.2s + 1)  6


Chapter 05.indd 30
24
G(s)
=
=
2
24
s(0.2s + 1.2s + 1) + 4s
2
=
2
24
s(0.2s + 1.2s + 5)
120
s(s + 6s + 25)
03/11/11 7:05 PM
TIME RESPONSE ANALYSIS
5-31
This gives the characteristic equation as (s2 + 6s + 25) = 0. Comparing it with the generalized
equation s 2 + 2ζωn s + ω2n = 0, we get
ω2n = 25 or
ωn = 5 rad/sec
2ζωn = 6 or
ζ = 0.6
∴ Damping factor= Damping ratio × Natural frequency
= ζωn = 0.6 × 5 = 3
(1)
(2)
(3)
= ωn 1 − ζ 2
Damping frequency
= 5 1 − (0.6)2
= 4 rad/sec
(4)
Maximum overshoot
Mp = e− πζ
1− ζ 2
2
= e−0.6 π (1−0.6)
= 0.095
(5) Time interval after which maximum and minimum output will occur is given by
2π
2π
2tp =
=
= 1.57 sec
2
4
ωn 1 − ζ
(6)
Settling time (2%)
ts =
Settling time (5%)
(7)
ts =
4
ζωn
3
ζωn
= 1.33 sec
= 1 sec
Number of cycles completed before settling within 2% is
2 1 − ζ2
2 × 0.8
= 0.85 cycle
πζ
π × 0.6
Setting time changes from 1.33 sec to 1 sec for a change of tolerance from 2% to 5%.
Hence, the number cycles completed before reaching 5% tolerance will be reduced linearly
to
0.85
= 0.64
1.33
=
Example 5.4 A flywheel is driven by a motor which develops a torque of 1600 N m/rad. The
moment of inertia of the system is 100 kg m2. The flywheel is controlled by a handwheel. If the
handwheel is suddenly given a movement through π/4 rad from rest, calculate the response or
output of the flywheel. The coefficient of viscous friction is 400 N m/rad/sec.
Solution Let Qi be the reference position of the handwheel and Qo be the output of the
flywheel.
Chapter 05.indd 31
03/11/11 7:05 PM
5-32
CONTROL SYSTEMS ENGINEERING
Torque developed = 1600 [θi (s) − θo (s)]
Also, torque developed is equated to
J
d 2 θo
2
+B
d θo
dt
dt
Taking Laplace transform, equating and substituting the values,
or
1600 [θi (s) − θo (s)] = [100 s 2 + 400 s ] θo (s)
=
θo ( s )
θi (s)
1600
2
100 s + 400 s + 1600
=
2
16
s + 4s + 16
When a unit step input of π/4 rad is applied, we express
π 1
θi (s) =
4 s
Thus,
π
16
θo ( s ) =
2
4s (s + 4 s + 16)
=
π
16
2
4s s ( s + 4s + 16)
=
π 1
s+4

−

2
4  s ( s + 2) + 12 
The output can be determined by taking inverse Laplace transform as
π
θo (t ) = 1 − e−2t (cos 3.46t + 0.578 sin 3.46t ) 

4
Example 5.5 A unity feedback system has an open-loop transfer function
G(s) =
25
s(s + 8)
Determine its damping ratio, peak overshoot and time required to reach the peak output.
Now a derivative component having transfer function of s/5 is introduced in the system. Discuss
its effect on the above-obtained values.
Solution Without derivative feedback,
G(s) =
C (s)
R( s )
=
25
s(s + 8)
,
G(s)
1 + G(s) H (s)
25
s(s + 8)
=
25
1+
1
s(s + 8)
Chapter 05.indd 32
03/11/11 7:05 PM
TIME RESPONSE ANALYSIS
=
2
5-33
25
s + 8s + 25
Its characteristic polynomial can be written as
s 2 + 2ζωn s + ω2n = s2 + 8s + 25
Comparing,
and
∴ Damping ratio
ωn = 5 rad/sec
2ζωn = 8
ζ =
8
2×5
= 0.8
Percentage peak overshoot
Mp = 100e−( πζ
1− ζ 2 )
= 100e− π×0.8 1−(.8)
= 100 × 0.0152
= 1.52%
π
tp =
ωn 1 − ζ 2
2
π
=
2
5 1 − 0.8
tp = 1.047 sec
When the derivative control is introduced in the feedback,
s
H(s) = 1 +
5
25
s(s + 8)
M(s) =
=
25 (s + 5)
R( s )
1+
s(s + 8) 5
C (s)
=
2
25
s + 8s + 5s + 25
25
= 2
s + 13s + 25
So the characteristic polynomial can be written as
Chapter 05.indd 33
s 2 + 2ζωn s + ω2n = s2 + 13s + 25
03/11/11 7:05 PM
5-34
CONTROL SYSTEMS ENGINEERING
Comparing
ω2n = 25,
ωn = 5
13
ζ =
ζ = 1.3
2×5
There is no change in ωn whereas the value of ζ is now more than unity. The value of
ζ has changed from 0.8 to 1.3. An unde-damped system has now become an over-damped one.
Since the system becomes over-damped, the peak overshoot will not be attained.
Thus, the effect of derivative control in a unity feedback system is summarized as follows:
(1) There is no change in natural frequency, but the damped frequency will change as per the
damping ratio.
(2) There is an increase in the value of the damping ratio.
(3) The time to reach the peak overshoot is increased.
(4) The value of percentage peak overshoot is reduced.
Example 5.6 A feedback system having derivative feedback control is shown in Figure 5.18.
Fig. 5.18 A feedback control system with derivative feedback.
(1)
If k0 = 0, determine the damping ratio and natural frequency of the system. What is the
steady-state error to a unit ramp input?
(2)
Determine the value of k0 for which the damping ratio of the system will be 0.6. Find the
steady-state error to a unit ramp input for this value of k0.
(3)
How can the steady-state error and damping ratio of the system with derivative feedback be
kept to 0.2 and 0.6, respectively, by applying unit ramp input?
Solution
(1) When k0 = 0, then the transfer function of the system is
M(s) =
C (s)
R( s )
=
2
20
s + 2s + 20
Its characteristic equation will be
Chapter 05.indd 34
s 2 + 2ζωn s + ω2n = s2 + 2s +10
03/11/11 7:05 PM
TIME RESPONSE ANALYSIS
5-35
Comparing to get natural frequency,
ωn =
Damping ratio
ζ =
Steady-state error
R(s) =
=
2ωn
s →0
Now for unit ramp input,
2
ess = lim
20 rad/sec,
2
2 20
2ζωn = 2
=
1
20
sR(s)
1 + G( s )
1
s2
ess = lim
s →0
= lim
s →0
s
s2
20
1+
s(s + 2)
2
(s + 2)
s + 2s + 20
= 0.1
(2) With derivative feedback, the forward path transfer function of the system will be
1


 s(s + 2) 

G(s) = 20 
sk0 

 1 + (s + 2) 


20
= 2
s + 2(s + k0 )
So the characteristic polynomial of the system can be written as
s 2 + 2ζωn s + ω2n = s2 + (2 + k0)s + 20
Comparing the coefficients, we get
or
ωn =
ζ =
ζ =
Chapter 05.indd 35
20 and 2ζωn = 2 + k0
2 + k0
2ωn
2 + k0
2 20
03/11/11 7:05 PM
5-36
CONTROL SYSTEMS ENGINEERING
To have its value as 0.6,
or
2 + k0 = ζ + 2 20
k0 = (0.6 × 2 × 20) − 2
= 5.4 – 2 = 3.4
Steady-state error
s
sR( s )
s + 2 + k0
s2
= lim
= 2
ess = lim
20
s →0 1 + G ( s )
s →0
s + 2 s + k0 s + 2 0
1+ 2
s + 2 s + k0 s
= lim =
s →0
Substituting the value of K = 3.4,
ess =
(3)
s + 2 + k0
s 2 + (2 + k0 )s + 20
2 + 3.4
20
=
2 + k0
20
= 0.27
To keep ess = 0.2 and ζ = 0.6 let the amplifier gain be adjusted to a higher value of K. Hence
the characteristic polynomial can be written as
s2 + (2 + k0)s + k = s 2 + 2ζωn s + ω2n
Comparing, we get
∴
or
2ζωn = 2 + k0 and ωn =
2 + k0 = 2(0.6) K
2 + k0 = 1.2 K
Again, steady-state error
2 + k0
ess =
= 0.2
K
or
K = 5(2 + k0)
or
or,
∴
K
K = 5(1 .2) k
(5.21)
[from Equation (5.21)]
K = 6.0
K = 36.0
Substituting the value of K in Equation (5.21),
∴
Chapter 05.indd 36
2 + k0 = (1.2)(6.0)
k0 = 5.2
03/11/11 7:05 PM
TIME RESPONSE ANALYSIS
5-37
Example 5.7
Block diagram representation of a control system has been shown in Figure
5.19. Determine the value of gain K and velocity feedback constant Kh, so that the maximum
overshoot in unit step response is 25% and the peak time is 2 sec.
Fig. 5.19 Block diagram of a feedback control system.
Solution The system transfer function C(s)/R(s) is
C (s)
R( s )
K
=
s2
K
(1 + K h s )
s2
The characteristic equation is obtained by putting the denominator of the transfer function
to 3000
Thus,
K
1 + 2 (1 + K h s) = 0
s
or
s2 + KKhs + K = 0
By comparing this equation with the basic equation
1+
s 2 + 2ζωn s + ω2n = 0
We find,
∴
ω2n = K or ωn =
2ζωn = KKh
ζ =
Maximum overshoot
KK h
2ωn
=
= 0.25 =
Peak time,
z = 0.404
tp =
Chapter 05.indd 37
K
KK h
2 K
−πζ
e 1 − ζ2
π
ωn 1 − ζ
2
=2=
π
ωn 1 − (0.404)2
03/11/11 7:05 PM
5-38
CONTROL SYSTEMS ENGINEERING
from which,
ωn = 1.72
K = ω n2 = 2.96
and
5.3.4
Kh =
2ζωn
K
=
2ζ K
K
=
2ζ
K
=
2 × 0.404
1.72
= 0.47
Sensitivity of a Control System
Consider the case of an open-loop system as shown in Figure 5.20, where the overall transfer
function is
C (s)
= G( s )
M(s) =
R( s )
Input
Fig. 5.20 Block diagram of an open-loop control system.
A very small change DG(s) in the transfer function G(s) changes it to G(s) + DG(s). The
changed output of the system is
C(s) + DC(s) = [G(s) + DG(s)]R(s)
or
C(s) + DC(s) = G(s)R(s) + DG(s)R(s)
or
C(s) + DC(s) = C(s) + [DG(s)]R(s)
or
DC(s) = [DG(s)]R(s)
(5.22)
Now take the case of a closed-loop system the diagram of which has been shown in
Figure 5.21.
Fig. 5.21 Block diagram of a closed-loop control system.
The overall transfer function is
M(s) =
Chapter 05.indd 38
C (s)
R( s)
=
G(s)
1 + G(s) H (s)
03/11/11 7:05 PM
TIME RESPONSE ANALYSIS
5-39
A very small change DG(s) in the forward path transfer function changes G(s) to G(s) + DG(s).
C(s) + DC(s) =
or
C(s) + DC(s) =
or
C(s) + DC(s) =
G( s ) + ∆ G( s )
1 + [G(s) + ∆G(s)]H (s)
G( s ) + ∆ G( s )
1 + G( s ) H ( s )
G( s )
1 + G( s ) H ( s )
or
C(s) + DC(s) = C (s) +
or
∆ C (s) ≅
R( s )
R( s ) +
∆ G( s )
1 + G( s ) H ( s )
∆G(s)
1 + G( s ) H ( s )
R( s )
∆ G( s )
1 + G( s ) H ( s )
R( s )
R( s )
R( s )
(5.23)
From Equations (5.22) and (5.23) it is observed that feedback reduces the variation of
output by a factor [1 + G(s)H(s)]. Since [1 + G(s)H(s)] >> 1, output variation is more sensitive
(to variation of G(s)) in case of open-loop systems as compared to closed-loop systems.
Sensitivity of a control system is expressed as
C (s)
Percentage change in overall transfer function H ( s ), i.e.
R( s)
Sensitivity =
Percentage change in G ( s )
The sensitivity function for the overall transfer function M(s) with respect to variation in
G(s) is written as
G( s ) ∂ M ( s )
∂M ( s) / M ( s)
SGM =
; or sGM =
M ( s ) ∂ G( s )
∂ M ( s ) / G( s )
For the closed-loop control system of Figure 5.20, the sensitivity SGM is given by
SGM =
=
G( s )
[1 + G(s) H (s)] − G(s) H (s)
G( s )
[1 + G(s) H (s)]2
1 + G( s ) H ( s )
1
1 + G(s) H (s)
From the above analysis, the following conclusions are drawn:
(1) Use of feedback reduces sensitivity of the system to parameter variations.
(2) For open-loop systems to be more accurate, components of G(s) must be selected to strictly
meet the desired specifications.
Chapter 05.indd 39
03/11/11 7:05 PM
5-40
CONTROL SYSTEMS ENGINEERING
(3)
For closed-loop systems, the components of G(s) may not be so tightly specified.
(4)
Improvement of sensitivity by use of feedback is off-set by the need for reduction of system
gain.
5.4Steady state-RESPONSE-STEADY STATE ERROR AND ERROR
CONSTANTS
Steady-state error is a measure of the accuracy of a control system. The steady-state error of a
control system is judged by the steady-state error due to step, ramp or acceleration input. We shall
investigate a type of steady-state error which is caused by the incapability of a system to follow
particular types of inputs. Any physical control system inherently suffers steady-state error in
response to certain types of inputs. A system may have no steady-state error to a step input but the
same system may exhibit non-zero steady-state error to a ramp input. We can, however, eliminate
this error by modifying the system structure. Generally, steady-state error should be as low as
possible. Figure 5.22 shows the block-diagram of a closed-loop system.
Fig. 5.22 Block diagram of a closed-loop system.
Here,
C (s)
R( s )
=
G(s)
1 + G(s) H (s)
and
E(s) = R(s) – B(s)
or
E(s) = R(s) – C(s)H(s)
Dividing both sides by R(s),
E (s)
R( s )
Chapter 05.indd 40
=
R(s) − C (s) H (s)
=1−
R(s)
= 1−
C (s)
R(s)
H (s)
G(s) H (s)
1 + G(s) H (s)
03/11/11 7:05 PM
TIME RESPONSE ANALYSIS
or
E (s)
R( s )
∴ Error,
=
E(s) =
5-41
1
1 + G(s) H (s)
R( s)
1 + G(s) H (s)
The steady-state error is
lim f (t ) = lim sF ( s )
t →∞
[from final value theorem]
s →∞
ess = lim e(t )
t →∞
So, steady-state error,
ess = lim e(t ) lim sE ( s ) = lim
t →∞
sR( s )
s →∞ 1 +
s →∞
G(s) H (s)
5.4.1 Static Position Error Coefficient ( K P )
The steady-state error of the system for a unit step input is
s
ess = lim
s →∞ 1 +
1
G(s) H (s) s
1


 R(s) = for unit step input 
s


or
1
ess = lim
where,
Kp = lim G ( s ) H ( s );
s →0 1 +
G(s) H (s)
s →0
=
1
1 + KP
K p is called the static position error coefficient.
Now, we shall find the value of Kp for different types of systems, that is, type 0, type 1 and
type 2.
(1) For a type 0 system,
K (T s + 1)(Tb s + 1)
[from Equation (5.1a)]
G(s)H(s) = 0 a
s (T1s + 1)(T2 s + 1) 
(2)
Kp = lim G ( s ) H ( s ) =
s →0
(0 + 1)(0 + 1) 
=K
For a type 1 or higher system,
G(s)H(s) =
K (Ta s + 1)(Tb s + 1) 
s N (T1s + 1)(T2 s + 1) 
KP = lim G ( s ) H ( s ) =
KP = ∞
Chapter 05.indd 41
K (0 + 1)(0 + 1) 
s →0
K (0 + 1)(0 + 1) 
0(0 + 1)(0 + 1) 
(for N ≥ 1)
03/11/11 7:05 PM
5-42
CONTROL SYSTEMS ENGINEERING
Now, we shall find the static position error for different systems at steady state.
For type 0 system,
1
1
[ K P = K ]
ess =
=
1 + KP 1 + K
(1)
(2)
For type 1 or higher systems,
ess =
1
1 + KP
=
1
1+ ∞
[ K P = ∞]
=0
Thus, for a unit step input, steady-state error for different types of systems is finite. Hence,
every type of system is capable of following step input. Though type 0 system shows some error,
higher type of systems can respond to step input very accurately.
5.4.2 Static Velocity Error Coefficient ( K v )
Static velocity error coefficient is associated with ess for unit ramp input. The steady-state
actuating error of the system with a unit ramp input (unit velocity input) is given by
sR( s )
ess = lim
s →0 1 + G ( s ) H ( s )
1
s
1

= lim
 R(s) = 2 
s →0 1 + G ( s ) H ( s ) s 2
s 

1
1
= lim
=
s →0 s + sG ( s ) H ( s )
Kv
where,
Kv = lim sG ( s ) H ( s );
s →0
K v is called the static velocity error coefficient.
Now, we shall find the static velocity error coefficient for different types of systems.
(1) For a type 0 system,
K (T s + 1)(Tb s + 1) 
G(s)H(s) = 0 a
s (T1s + 1)(T2 s + 1) 
∴
K = lim sG ( s ) H ( s )
v
s →0
= lim
(2) For a type 1 system,
(3)
s →0
sK (Ta s + 1)(Tb s + 1) 
(T1s + 1)(T2 s + 1) 
Kv = lim sG ( s ) H ( s ) = lim s
s →0
s →0
=0
K (Ta s + 1)(Tb s + 1) 
s (T1s + 1)(T2 s + 1) 
=K
For a type 2 or higher system,
Kv = lim sG ( s ) H ( s ) = lim s
K (Ta s + 1)(Tb s + 1) 
s N −1 (T1s + 1)(T2 s + 1) 
K (T s + 1)(Tb s + 1) 
[ N ≥ 2]
= lim N −1 a
=∞
s →0 s
(T1s + 1)(T2 s + 1) 
Chapter 05.indd 42
s →0
s →0
03/11/11 7:05 PM
TIME RESPONSE ANALYSIS
5-43
Now, we shall find static velocity error for different types of systems at steady state.
For type 0 system,
1
1
[ K v = 0]
ess =
= =∞
Kv
0
(1)
(2)
For type 1 system,
(3)
ess =
For a type 2 or higher systems,
ess =
1
Kv
1
Kv
=
=
1
[ K v = K ]
K
1
0
[ K v = ∞]
=∞=0
For ramp input, steady-state error for type 0 system is infinite. Hence, a type 0 system is not
capable of following ramp input. The static velocity error for type 1 system is finite. But static
velocity error for type 2 system or higher system is zero. So type 2 or higher systems are capable
of following a ramp input very accurately. We see from this analysis that as we move from type
0 to type 2 or higher systems, the static velocity error goes on decreasing. But with increase in
the type number, the number of poles on the jw-axis also goes on increasing, which will make the
system response more oscillatory.
Hence, we have to make a compromise between stability and accuracy.
5.4.3 Static Acceleration Error Coefficient ( K a )
Static acceleration error coefficient is associated with ess for unit parabolic input. The steady-state
actuating error of the system with unit-parabolic input (acceleration) is given by
sR( s )
ess = lim
s →0 1 + G ( s ) H ( s )
s
1
1

= lim
 R( s ) = 3 

3
s →0 1 + G ( s ) H ( s ) s
s 

1
1
= lim 2
=
2
s →0 s + s G ( s ) H ( s )
Ka
where
(1)
Ka = lim s 2 G ( s ) H ( s ) = lim
s →0
s →0
For a type 1 system,
Chapter 05.indd 43
s →0
Now, we shall find the static acceleration error coefficient for different types of systems.
For type 0 system,
(2)
Ka = lim s 2 G ( s ) H ( s )
Ka = lim s 2 G ( s ) H ( s ) = lim
s →0
s →0
s 2 K (Ta s + 1)(Tb s + 1) 
s 0 (T1s + 1)(T2 s + 1) 
=0
s 2 K (Ta s + 1)(Tb s + 1) 
s (T1s + 1)(T2 s + 1) 
03/11/11 7:05 PM
5-44
CONTROL SYSTEMS ENGINEERING
= lim
(3) For a type 2 system
s →0
s →0
s →0
ess =
For a type 1 system,
(3)
ess =
For a type 2 system,
(4)
s 2 K (Ta s + 1)(Tb s + 1) 
s N (T1s + 1)(T2 s + 1) 
s →0
K (Ta s + 1)(Tb s + 1) 
[ N ≥ 3]
=∞
s (T1s + 1)(T2 s + 1) 
Now we shall find the static acceleration error for different types of systems at steady-state.
For a type 0 system,
= lim
(2)
=K
s 2 (T1s + 1)(T2 s + 1) 
s →0
Ka = lim s 2 G ( s ) H ( s ) = lim
=0
s 2 K (Ta s + 1)(Tb s + 1) 
For a type 3 or higher system,
(1)
(T1s + 1)(T2 s + 1) 
Ka = lim s 2 G ( s ) H ( s ) = lim
(4)
sK (Ta s + 1)(Tb s + 1) 
s →0
ess =
For a type 3 or higher system,
ess =
1
Ka
1
Ka
1
Ka
1
Ka
N −2
=∞
[ K a = 0]
=∞
[ K a = 0]
=
=
1
[ K a = K ]
K
1
∞
[ K a = ∞]
=0
Thus, type 0 and type 1 systems are capable of following a parabolic input. For a type 2
system, the error is finite, but for type 3 or higher systems the error is zero. Again, as we increase
the type numbers, the error goes on reducing.
The steady-state errors in terms of forward path gain (K) of type 0, type 1 and type 2 closedloop systems have been summarized in Table 5.2.
Table 5.2 Steady-state Errors of Closed-loop Systems of Different Types.
System
Step Input
r(t) = 1
Ramp Input
r(t) = t
Acceleration Input
r(t) = 1/2t 2
Type 0
1/1 + K
∞
∞
Type 1
0
1/K
∞
Type 2
0
0
1/K
Chapter 05.indd 44
03/11/11 7:05 PM
TIME RESPONSE ANALYSIS
5-45
The terms “position error”, “velocity error”, “acceleration error” mean steady-state
deviations in the output position. A finite velocity error implies that after transients have died
out, the input and output move at the same velocity but have a finite position difference.
The error coefficient Kp, Kv and Ka describe the ability of a system to reduce or eliminate
steady-state error. It is desirable to increase the error coefficients while maintaining the transient
response within an acceptable range. From Table 5.2, it can be seen that a type 0 system gives
error for all the three types of inputs. A type 2 system gives error due to one type of input only,
which is finite. So a type 2 system is better than a type 0 system or a type 1 system from the
viewpoint of steady-state error. Higher types of systems are better from the steady-state error
point of view but are less stable.
Example 5.8 Consider the system shown in Figure 5.23 (a). The steady-state error to a unit
ramp input is ess = 2z/wn. Show that the steady-state error for following ramp input may be
eliminated if the input is introduced to the system through a proportional plus derivative element
in Figure 5.23 (b).
Fig. 5.23 (a) A unity feedback control system; (b) A unity feedback control system with a
proportional plus derivative input element.
Solution From Figure 5.15 (b), we get
C (s)
Again
Chapter 05.indd 45
R( s )
=
(1 + Ks )ω2n
s 2 + 2ζωn s + ωn2
E(s) = R( s ) − C ( s ) = R( s ) −
(1 + Ks )ω2n
s 2 + 2ζωn s + ωn2
R( s)
03/11/11 7:05 PM
5-46
CONTROL SYSTEMS ENGINEERING
∴
E(s) =
s 2 + 2ζωn s − ω2n Ks
s 2 + 2ζωn s + ω2n
R( s)
So the steady-state error (ess) is given by
ess = lim sE (s) = lim s
s →0
s →0
s 2 + 2ζωn s − ω2n Ks 1
2
s + 2ζωn s +
= lim
s →0
ω2n
s
1

 for unit ramp input R(s) = 2 
s 

2
s + 2ζωn − ωn2 K
2
s + 2ζωn s +
ωn2
=
2ζωn − ωn2 K
ωn2
=
2ζ
−K
ωn
Thus, the value of K, for which steady-state error will be eliminated, is given by
ess = 0
or
2ζ
−K =0
ωn
∴
K =
2ζ
ωn
Example 5.9 Consider a unity feedback control system whose open-loop TF is
K
G(s) =
s( Js + F )
Discuss the effect of variation of gain K and friction coefficient F on the steady-state error
in unit ramp response. Sketch typical unit ramp response curves for a small, medium and large
values of K.
Solution
or
Again,
or
∴
Chapter 05.indd 46
C (s)
R( s )
C (s)
R( s )
E (s)
R( s )
=
=
=
E(s) =
G( s )
1 + G( s )
2
K
Js + Fs + K
R( s) − C ( s)
R( s)
Js 2 + Fs
=1−
Js 2 + Fs + K
C (s)
R( s)
=1−
K
2
Js + Fs + K
R( s )
 Js 2 + Fs  1
ess = lim sE ( s ) = lim s  2

s →0
s →0  Js + Fs + K  s 2
1

 R(s) = 2 
s 

03/11/11 7:05 PM
TIME RESPONSE ANALYSIS
5-47
Fig. 5.24 Effect of variation of gain k and friction coefficient of response curve.
∴
ess =
F
K
From the result, we see that steady-state error can be reduced by either increasing the gain
K or reducing the viscous friction coefficient F. However, this causes damping to decrease and
the transient response becomes more oscillatory (see Fig. 5.24).
Example 5.10
For a unity feedback system having an open-loop transfer function as
K ( s + 2)
,
G(s) =
3
s ( s + 7 s 2 + 12s )
determine the (1) type of system, (2) error constant Kp, Kv and Ka and (3) steady-state error for
parabolic input.
Solution
G(s) =
2
K (s + 2)
s (s 2 + 7s + 12)
(1) The open-loop transfer function G(s) has poles of order two at the origin. So the type of
system is two.
(2)
K ( s + 2)
=∞
Kp = lim G ( s ) = lim 2 2
s →0
s →0 s ( s + 7 s + 12)
K ( s + 2)
Kv = lim sG ( s ) = lim
Ka = lim s 2 G ( s ) = lim
s →0
s →0
s →0
2
s ( s + 7 s + 12)
s →0
K ( s + 2)
2
s + 7 s + 12
=
=∞
K
6
(3) Let the parabolic input be
r(t) = t2
Chapter 05.indd 47
03/11/11 7:05 PM
5-48
CONTROL SYSTEMS ENGINEERING
∴
∴
2
R(s) =
s3
ess = lim sE ( s ) = lim
s →0 1 +
s →0
= lim
s →0
s R( s )
= lim
s →0
G(s) H (s)
[ H (s) = 1]
1 + G( s )
s
1+
2
s3
K ( s + 2)
= lim
s →0
sR( s )
s 2 ( s 2 + 7 s + 12)
2( s 2 + 7 s + 12)
2
2
s ( s + 7 s + 12) + K (d + 2)
=
24
2K
=
12
K
5.5effects of proportional, derivative and proportional
Integral system
Control actions involves steps to be taken to maintain the system output within the desired
limit. For automatic control, the error detector detects any variation in the output and generates
an actuating signal to be fed to a controller. Various types of control actions, viz. proportional
control, derivative control, integral control, etc., are discussed as follows.
5.5.1 Proportional Control
After obtaining the error signal, a proportional actuating signal is obtained by introducing a
proportional gain controller having a gain Kp and its output is fed to the system as shown in
Figure 5.25. A proportional controller is basically an amplifier with controllable gain.
Fig. 5.25 Second-order unity feedback control system using proportional control.
The actuating signal is given by
Ea(t) = KpE(t)
and its Laplace transform is given by
Chapter 05.indd 48
Ea(s) = KpE(s)
03/11/11 7:05 PM
TIME RESPONSE ANALYSIS
For the system shown in Figure 5.17,
E(s) = R(s) – C(s) = R(s) – E(s) =
or
E (s)
R( s )
=
2
5-49
K p ω2n
s(s + 2ζωn )
s ( s + 2δωn )
s + 2ζωn s + K p ω2n
For a unit ramp input, the steady-state error is given by
ess = lim sE (s)
s →0
= lim s
s →0
=
2ζ
1
2
2
s ( s + 2δωn )
s s + 2ζωn s + K p ω2n
K p ωn
As ess is inversely proportional to Kp, steady-state error will decrease as the proportional
gain Kp is increased.
To study the effect of change of gain Kp on the transient response of the system, we write
the closed-loop transfer function, which is given as
C (s)
R( s )
K p ω2n
=
s 2 + 2ζωn s + K p ω2n
The roots of the characteristic equation
is given as
s 2 + 2ζωn s + K p ω2n = 0
−ζωn ± jωn K p − ζ 2
The damped natural frequency, ωd = ωn K p − ζ 2 , will, thus, increase as Kp is increased,
indicating that the system output will oscillate at a higher frequency. The system, however,
remains stable as the two roots lie in the left half of s-plane. Due to increase in ωd , the rise time
will decrease, making the system respond faster. However, minimum overshoot will increase.
This follows from the expression for rise time and Mp is as follows.
π−β
ω
, where β = tan −1 d
tr =
ωd
ζωn
and
Mp =
 ζω 
− n  π
ω
e  d 
Thus, a proportional controller reduces the steady-state error and makes the system respond
faster but at the expense of increased maximum overshoot and oscillatory response.
For third and higher order systems, the system becomes unstable beyond a particular value
of Kp as one or more roots shifts to the right-hand side of the s-plane as the value of Kp increases.
Chapter 05.indd 49
03/11/11 7:06 PM
5-50
CONTROL SYSTEMS ENGINEERING
5.5.2 Proportional Plus Derivative Control
Basically, derivative control involves using an actuating signal which is proportional to the rate
of change of the error signal. However, for achieving proportional plus derivative control action,
we should have an actuating signal consisting of proportional error signal plus the derivative of
the error signal. The actuating signal, thus, will be
de(t )
ea(t) = e(t ) + K d
, where Kd is a constant and Kp = 1
dt
Taking Laplace transform, we get
Ea(s) = E(s) + sKdE(s)
or
Ea(s) = (1 + sKd)E(s)
Derivative control action when added to a proportional controller increases the sensitivity
of the controller and tends to increase the stability of the system by producing early corrective
action before the magnitude of the error signal e(t) becomes too large.
Fig. 5.26 Second-order unity feedback control system using derivative control.
The overall transfer function of the closed-loop second-order control system using
derivative control (Fig. 5.26) is given by
C (s)
(1 + sK d )ωn2
=
R( s )
s 2 + (2ζωn + ω2n K d ) s + ω2n
So the characteristic equation is
s 2 + (2ζωn + ω2n K d )s + ω2n = 0
The standard form is
s 2 + 2ζωn s + ω2n = 0
From the co-efficient of s, the effective damping with derivative control and considering ζ
as ζn is given by
New damping ratio
ζn =
2ζωn + ω2n K d
2ωn
=ζ+
ωn K d
2
Thus, derivative control increases the damping of the second-order system. Since natural
frequency ωn remains unchanged, the maximum overshoot, rise time and settling time are
reduced.
Chapter 05.indd 50
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
5-51
For a closed-loop system, we can write
C (s)
or
R( s )
E (s)
R( s )
=1−
=1−
E (s)
R( s )
C (s)
R( s )
From Figure 5.18, we get
E (s)
R( s )
=
2
s(s + 2ζωn )
s + 2ζωn + ω2nTkd )s + ω2n
For unit ramp input, the steady-state error is given by
ess = lim sE ( s )
s →0
= lim s
=
s →0
1
2
2
s(s + 2ζωn )
s s + 2ζωn + ω2n K d )s + ω2n
2ζ
ωn
Kd = 0 will also give the same result of ess = 2ζ / ωn . Thus, steady-state error is not affected
by derivative control. While derivative control action has an advantage of being anticipatory
thereby improving transient response, it has the disadvantages that it amplifies noise signals and
may cause a saturation effect in the actuator. Derivative control action can never be used alone
because this control action is effective only during transient period.
5.5.3 Proportional Plus Integral Control (PI Control)
In integral control, use is made of an actuating signal which is proportional to the integral of
error signal. In PI control, we use actuating signal which is proportional error signal plus the
integral of the error signal. The actuating signal for integral control action can, therefore, be
written as
ea(t) = e(t ) + ki ∫ e(t ) dt , where ki is a constant and assuming Kp = 1
Fig. 5.27 Second-order unity feedback system using proportional plus integral control.
Chapter 05.indd 51
03/11/11 7:06 PM
5-52
CONTROL SYSTEMS ENGINEERING
Taking Laplace transform, we get
E (s)
Ea(s) = E ( s ) + ki
s
The overall transfer function of the closed-loop second-order control system (Fig. 5.27) is
given by
C (s)
R( s )
=
( s + Ki )ωn2
s 2 + 2ζωn s 2 + ωn2 s + Ki ω2n
So the characteristic equation is obtained by equating denominator of the TF = 0 as
s 2 + 2ζωn s 2 + ω2n s + Ki ω2n = 0
E (s)
or
R( s )
E (s)
R( s )
=1−
=
C (s)
R( s )
=1−
(s + Ki )ω2n
s 3 + 2ζωn s 2 + ω2n s + Ki ω2n
s 2 ( s + 2ζωn )
s 3 + 2ζωn s 2 + ωn2 s + Ki ω2n
or
E(s) = R( s )
s 2 ( s + 2ζωn )
s 3 + 2ζωn s 2 + ωn2 s + Ki ω2n
For unit ramp input where R(s) = 1/s2, steady-state error is given by
ess = lim sE ( s )
s →0
or
ess = lim s
s →0
1
s 2 (s + 2ζωn )
s 2 s 3 + 2ζωn s 2 + ω2n s + Ki ω2n
=0
Ki = 0 will give the steady-state error ess = 2ζ / ωn . So without integral control action, a
steady-state error or offset ess = 2z/wn exists in a second-order closed-loop control system with
unit ramp input. This offset or steady-state error can be reduced to zero by using integral control
action. The integral control action, while improving steady-state accuracy by removing offset,
may lead to oscillatory response of a slowly decreasing amplitude or, even increasing amplitude,
both of which are usually undesirable.
5.5.4
Proportional Plus Integral Plus Derivative Control (PID Control)
Proportional plus integral control (PI control) was introduced to meet the high accuracy
requirement and improve steady-state performance. Introduction of PI control changed a secondorder system into a third-order system. Third-order system introduces a distinct possibility of
Chapter 05.indd 52
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
5-53
system instability. In PID control, we take advantage of PI control with derivative control. PID
controller is, therefore, a three action controller which combines proportional, integral and
derivative control. The actuating signal is given by
ea (t) = e(t ) + K d
d
dt
e(t ) + K i ∫ e(t ) dt
In Laplace transform form,
Ea(s) = E (s) + K d sE (s) +
or
Ki
s
E (s)
K 

Ea(s) = E ( s ) 1 + sK d + i 
s 

PID controller is represented in the block diagram as shown in Figure 5.28.
Fig. 5.28 PID control contains the advantages of proportional and
derivative integral control actions.
5.5.5 Derivative Feedback Control
For achieving derivative feedback control action, the actuating signal is obtained as the difference
between the proportional error signal and derivative of the output signal. The actuating signal
can, therefore, be expressed as
dc(t )
ea(t) = e(t ) − Kt
, where Kt is a constant
dt
Taking Laplace transform, we get
Ea(s) = E(s) − sKtC(s)
Fig. 5.29
Chapter 05.indd 53
Derivative (or rate) feedback control.
03/11/11 7:06 PM
5-54
CONTROL SYSTEMS ENGINEERING
The overall transfer function for the unity feedback second-order control system (Fig. 5.29)
is given by
C (s)
ω2n
=
R( s )
s 2 + (2ζωn + ω2n Kt )s + ω2n
The characteristics equation is
s 2 + (2ζωn + ω2n Kt )s + ω2n = 0
So, the effective damping is given by
ζ′ =
2ζωn + ω2n Kt
2ωn
=ζ+
ωn Kt
2
Therefore, like derivative control, the damping ratio is increased by using derivative feedback
control. The maximum overshoot is thereby decreased and rise time, however, is increased. Since
natural frequency ωn is unchanged, this controller decreases the response settling time.
From Figure 5.21 we get,
E (s)
=
s 2 + (2ζωn + ω2n Kt ) s
R( s )
s 2 + (2ζωn + ω2n Kt ) s + ω2n
For unit ramp input, steady-state error is given by
ess = lim sE ( s )
s →0
= lim s
s →0
1
2
2
s 2 + (2ζωn + ω2n K t )s
s s + (2ζωn +
ω2n K t )s
+
ω2n
=
2ζ
ωn
+ Kt
So for a ramp input, the steady-state error is increased by using derivative feedback control action.
Example 5.11 The control valve of a hydraulic servomotor is shown in Figure 5.30 (a).
Assume that the hydraulic fluid is incompressible and the oil flow rate is proportional to the
control valve displacement (x). The power piston and load are of negligibly small mass.
Fig. 5.30 (a) Hydraulic servomotor; (b) Hydraulic proportional controller.
Chapter 05.indd 54
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
5-55
Show that
(1)
The servomotor of Figure 5.30 (a) is an integral controller.
(2)
The servomotor of Figure 5.30 (a) can be modified to a proportional controller by means of
a floating feedback link ABC as shown in Figure 5.30 (b).
Solution
(1)
If the input x moves the pilot valve to the right, high pressure oil enters the right side of the
power piston through port II and low pressure oil in the left side of the power piston returns
to the drain through port I.
The rate (q) of flow of oil times dt is equal to the power piston displacement (dy) times the
piston area (A) times the density (ρ) of oil.
That is,
Aρdy = qdt
Again from the given assumption,
q = K1x
where K1 is a positive constant.
Combining the above two equations, we get
dy
= K1x
dt
Taking Laplace transform with zero initial conditions,
Aρ
Y (s)
X ( s)
=
K1
Aρs
=
K
s
(5.24)
So the hydraulic servomotor shown in Figure 5.30 (a) acts as an integral controller.
Fig. 5.30
(2)
Chapter 05.indd 55
In the Figure 5.30 (b), if input e moves the control valve to the right, the power piston will
move to the left with feedback link ABC, thereby moving the pilot valve to the left until it
returns to its original position.
03/11/11 7:06 PM
5-56
CONTROL SYSTEMS ENGINEERING
The displacement x is given by
x =
be
a+b
−
ay
[see Fig. 5.30 (c)]
a+b
Taking Laplace transform with zero initial condition,
X(s) =
b
a+b
E (s) −
a
a+b
Y (s)
(5.25)
Fig. 5.31 Block diagram of Fig. 5.30 (b).
Again from Equation (5.24),
Y(s) =
K
s
X ( s)
(5.26)
Using Equations (5.25) and (5.26), we can determine the ratio of
 b  K 
K
b

 
Y ( s)
a
b
s
+
s
a
(
+ b)
  =
= 
K
E (s)
 K  a 
1+  
 1 + a s ( a + b)
 s  a + b 
Under normal operation, we have K a [ s(a + b)] >> 1. So the above transfer function is
simplified as
K
Y ( s)
s(a + b) b
=
= = Kp
K
E (s)
a
a
s(a + b)
b
Thus, the hydraulic servomotor with a feedback link works as a proportional controller of
gain Kp.
Example 5.12
controller.
Chapter 05.indd 56
Show that the electronic controller shown in Figure 5.32 works like a PD
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
Fig. 5.32
5-57
Electronic controller.
Solution
Output,
eo = (ei – ef )K
In Laplace transform form,
Eo(s) = K[Ei(s) – Ef(s)]
Again,
1
1
eo = Ri + ∫ i dt , Eo ( s ) = RI ( s ) +
I (s)
c
cs
and
1
1
Ef = ∫ i dt , E f ( s ) =
I (s)
c
cs
1
I (s)
E f ( s)
1
cs
=
=
1 
1 + Rcs
Eo ( s )

 R +  I (s)
cs 

∴
Ef(s) =
We had,
1
1 + Rcs
Eo ( s )
Eo(s) = K [ Ei (s) − E f (s)]
Substituting the value of Ef (s), we have
Eo(s) = KEi ( s ) −
K
1 + Rcs
Eo ( s )
or
or
Chapter 05.indd 57
K 

Eo ( s ) 1 +
 = KEi(s)
 1 + Rcs 
Eo ( s )
Ei (s)
=
K
1+
K
1 + Rcs
03/11/11 7:06 PM
5-58
CONTROL SYSTEMS ENGINEERING
Assuming
K
1 + Rcs
>> 1,
Eo ( s )
Ei ( s )
=
K
= 1 + Rcs
K
1 + Rcs
= (1 + Kd s)
where
Kd = Rc
This (1 + Kds) is equivalent to a PD controller and is represented as in Figure 5.33.
Fig. 5.33 PD controller.
Example 5.13 A unity feedback control system with proportional plus derivative control is
shown in Figure 5.34. Calculate the value of Kd so that the system is critically damped. Also for
a ramp input, calculate the values of steady-state error, maximum overshoot, and settling time for
the system without and with control actions.
Fig. 5.34
Use of PD controller.
Solution With control action,
2(1 + K d s )
2(1 + K d s )
s ( s + 2)
=
=
2(1 + K d s ) s ( s + 2) + 2(1 + K d s )
R( s )
1+
s ( s + 2)
C (s)
The characteristic equation is
s(s + 2) + 2(1 + Kds) = 0
or
s2 + (2 + 2Kd)s + 2 = 0
Putting in the standard form and comparing
Chapter 05.indd 58
s 2 + 2ζωn s + ω2n = 0
2ζωn = 2 + 2K d = 2(1 + K d )
ζωn = 1 + Kd
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
5-59
and
ω2n = 2,
i.e. ωn =
2
Substituting the value of ωn and taking ζ = 1 for cortically damped system,
1 × 2 = 1 + Kd
Kd = 0.414
Without control action
2
C (s)
G(s)
2
s ( s + 2)
=
=
= 2
2
R( s )
1 + G ( s )1 + ( s )
s + 2s + 2
1+
s ( s + 2)
as 1 + (s) = 1
The characteristic equation is
s2 + 2s + 2 = 0
Comparing with the standard equation s 2 + 2ζωn s + ω2n = 0,
ω2n = 2
2ζωn = 2
Steady-state error is
ess =
Setting time is
ts =
or ωn =
ωn
=
4
ζωn
=
1.414
4
1
2
−πζ
Mp = e
1−ζ 2
2ωn
2 × 0.707
Maximum overshoot, Mp is calculated as
2
or ζ =
2ζ
2 = 1.414
=e
=
2
2 2
= 0.707
= 1 rad
= 4 sec
2
−3.14×0.707
1− (.707)2
= e3.11 = 0.12 = 12%
With control action
The characteristic equation is
s(s + 2) + 2(1 + Kds) = 0
Kd was calculated as 0.414
Therefore, the equation becomes
s2 + 2s + 2 + 2 × 0.414s = 0
or
s2 + 2.828s + 2 = 0
Chapter 05.indd 59
03/11/11 7:06 PM
5-60
CONTROL SYSTEMS ENGINEERING
Comparing with the standard form s 2 + 2ζωn s + ω2n = 0,
ω2n = 2
ess =
ts =
or ωn =
2ζ
ωn
=
4
ζωn
2 ×1
1.414
=
2 = 1.414
ζ =1
= 1.414 rad
4
1× 2
= 2 2 = 2.828 sec
Maximum overshoot
−πζ
Mp = e
1−ζ 2
=e
−π×1
1−1
= e−∞ = 0
Example 5.14 A feedback control system is shown in Figure 5.35. Determine the value of K
so that the system will have a damping ratio of 0.6. Also for this value of K, calculate maximum
overshoot, setting time and steady-state error for a unit ramp input.
Fig. 5.35 Proportional control action.
Solution
K
C (s)
G(s)
K
s ( s + 5)
=
=
=
K × 0.2 s ( s + 5) + 0.2 K
R( s )
1 + G(s) H (s)
1+
s ( s + 5)
The characteristic equation is
s2 + 5s + 0.2K
Comparing with the standard equation s 2 + 2ζωn s + ω2n ,
or
Chapter 05.indd 60
ω2n = 0.2K or ωn =
2ζωn = 5
0.2K
or 2 × 0.5ωn = 5
5 = ωn =
or ωn = 5
0.2K
0.2K = 52 = 25
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
or
K =
25
5-61
= 125
0.2
The characteristic equation with this value of K is
s2 + 5s + 25 = 0
ω2n = 25
Settling time
Steady-state error
ts =
ess =
or ωn = 5 and ζ = 0.6
4
ζωn
2ζ
ωn
=
=
4
0.6 × 5
2 × 0.6
4
−πζ
Maximum overshoot
Mp = e
1−ζ 2
= 1.33 sec
= 0.3 rad
−3.14×0.6
=e
1−.062
= e−2.35 = 0.15
Example 5.15 A feedback control system with proportional gain controller and having deriderivative feedback is shown in Figure 5.36. Determine the damping ratio and steady-state error
for ramp input without the derivative feedback. Consider the gain of the amplifier KA = 9. Now
calculate with derivative control, the values of KA and Kt so that the damping ratio is increased to
0.6 without any change in the steady-state error.
Fig. 5.36 Feedback control system.
Solution When derivative feedback loop is absent, the system is represented as shown in
Figure 5.37.
Fig. 5.37 Control system of Figure 5.36 without the derivative feedback.
Chapter 05.indd 61
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5-62
CONTROL SYSTEMS ENGINEERING
With amplifier gain KA = 9, the overall transfer function is
KA
KA
9
4.5
s(1 + 2s)
=
=
= 2
= 2
KA
R( s )
s(1 + 2s) + K A 2s + s + 9 s + 0.5s + 4.5
1+
s(1 + 2s)
The characteristic equation is
s2 + 0.5s + 4.5 = 0
Comparing with equation
C (s)
s 2 + 2ζωn s + ω2n = 0
ω2n = 4.5 or ωn =
4.5 = 2.14 rad/sec
and
2ζωn = 0.5 or ζ =
ess =
Steady-state error
2ζ
ωn
=
0.5
2ωn
2 × 0.117
2.14
=
0.5
2 × 2.14
= 0.117
= 0.1
When the derivative feedback loop is present,
1
C (s)
1
1
s (1 + 2s )
=
=
= 2
sKt
Ea ( s )
s (1 + 2s ) + 2 K t
2s + s (1 + K t )
1+
s (1 + 2s )
C (s)
E (s)
=
KA
2
2s + s(1 + Kt )
KA
KA
0.5K A
2s + s(1 + Kt )
= 2
=
KA
(1 + Kt )s K A
2 s + s(1 + Kt ) + K A
R( s )
+
1+ 2
s2 +
2
2
2 s + s(1 + Kt )
The characteristic equation is
C (s)
s2 +
1 + Kt
2
2
=
s+
KA
2
=0
Comparing with
Chapter 05.indd 62
s 2 + 2ζωn s + ω2n = 0,
ω2n =
KA
2
or ωn =
2ζωn = 0.5(1 + Kt)
KA
2
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
2 × 0.6 ×
KA
2
= 0.5(1 + Kt)
1.7 K A = 1 + Kt
or
E(s) = R(s) – C(s)
E (s)
R( s )
=1−
(5.27)
C (s)
R( s )
C (s) 
KA



E(s) = R(s) 1 −
 = R(s) 1 − 2

R( s ) 
2 s + s(1 + Kt ) + K A 


= R(s)
5-63
ess = lim sE (s) = s
s →0
=
1 + Kt
s[2s + (1 + Kt )]
2
2 s + s(1 + Kt ) + K A
1
2
s[2s + (1 + Kt )]
2
s 2s + s(1 + Kt ) + K A
KA
This new ess must be equal to the earlier calculated ess as 0.1
Therefore,
1 + Kt
0.1 =
KA
=
[2 s + (1 + Kt )]
2
2s + s(1 + K t ) + K A
(5.28)
From Equations (5.27) and (5.28),
or
0.1KA = 1 + Kt = 1.7 K A
and
K A = 17, i.e. KA = 289
Kt = 0.1KA – 1 = 28.9 – 1 = 27.9.
Example 5.16 Determine the resistance values to obtain the following electronic PI controller
with proportional and integral gain of 2 and 0.04, respectively. Use a 1μF capacitor (Fig. 5.38).
Fig. 5.38 Electronic PI controller.
Chapter 05.indd 63
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5-64
CONTROL SYSTEMS ENGINEERING
Solution The transfer function of the above controller with zero initial conditions is given by
Eo ( s )
Ei (s)
=−
RCs + 1
R1Cs
K 

= −  Kp + I 
s 

where Kp = R/R1 = proportional gain and KI = 1/R1C = integral gain.
Given, Kp = 2 and KI = 0.04
R
∴
=2
R1
or
R = 2R1
and
1
= 0.04
R1C
or
R1C = 25
Given that C = 1µF
From Equation (5.30)
From Equation (5.29)
R1 =
(5.29)
(5.30)
25
1 × 10−6
Ω = 25 M Ω
R = (2 × 25) M Ω = 50 M Ω
Example 5.17 Determine the resistance values to obtain the following electronic PID
controller with proportional, integral and derivative gain of 5, 0.7 and 2, respectively. The circuit
should limit frequencies above 250 rad/sec. Take one capacitance to be 1 μF (Fig. 5.39).
Fig. 5.39 Electronic PID controller.
Solution The transfer function of the above controller with zero initial conditions is given by
Eo ( s )
Chapter 05.indd 64
Ei (s)
=−
( RCs + 1)( R2C1s + 1)
Cs( R1 + R2 + R1 R2C1s)
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
5-65
1
β
 RC + R2C1


= −
+
+ RC1s  

R2C
R2Cs
β R1C1s + 1 



where
K
1


= −  K p + I + KD s 
s

 (β R1C1s + 1)
R2
β =
Kp = β
KI =
KD = β RC1
R1 + R2
RC + R2C1
R2C
β
R2C
(5.31)
(5.32)
(5.33)
(5.34)
The denominator term (β R1C1s + 1) limits the effect of frequencies above ω = 1/ β R1C1 .
As ω = 250 rad/sec (given),
1
∴
= 250
(5.35)
β R1C1
Eliminating β from Equations (5.34) and (5.35)
R
R1
= 250KD = 500
[KD = 2 given]
or
(5.36)
R = 500R1
Eliminating β from Equations (5.32) and (5.33)
Kp
5
[Kp = 5 and KI = 0.7 given]
RC + R2C1 =
=
KI
0.7
or
50
RC + R2C1 =
7
or
50
(5.37)
500R1C + R2C1 =
7
Substituting the value of β from Equation (5.31) into Equation (5.33), we get
10
(R1 + R2)C =
7
or
10
C =
(5.38)
7( R1 + R2 )
Chapter 05.indd 65
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5-66
CONTROL SYSTEMS ENGINEERING
Substituting the value of β from Equation (5.31) into Equation (5.35), we get
C1 =
R1 + R2
(5.39)
250 R1 R2
Substituting the values of C and C1 into Equation (5.37), we get
5000 R1
or
7( R1 + R2 )
+
R1 + R2
250 R1
=
50
7
2
 R + R2 
 R1 + R2 
7 1
 − 12, 500 
 + 1, 250, 000 = 0
 R1 
 R1 
R1 + R2
∴
= 106.33 (chosen)
R1
or
R2 = 105.33R1
From Equations (5.38) and (5.40), we get
10
106.33R1 =
7C
Taking C = 1 μF, we have
and
R1 =
(5.40)
107
Ω = 13.435 KΩ
106.33 × 7
R2 = 105.33R1 = 1.415 MΩ
R = 500R1 = 6.717 MΩ
C1 =
R1 + R2
250 R1 R2
= 0 . 3 µF
5.12 Transient Response Analysis Using MATLAB
MATLAB (the abbreviated form of MATrix LABoratory) is a matrix-based system for
mathematical and engineering calculations. It has many commands and predefined matrix
functions that can be called by the user through MATLAB programs to solve different types of
problems on control systems. In this section, transient responses will be illustrated by solving a
few problems through MATLAB.
Example 5.18 Using MATLAB, obtain the unit impulse response of the system the transfer
function of which is as follows.
1
C (s)
G(s) =
= 2
R(s) s + 0.2s + 1
Assume zero initial conditions and use ‘step (num, den)’ command in place of ‘impulse
(num, den)’.
Chapter 05.indd 66
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TIME RESPONSE ANALYSIS
5-67
Solution For unit impulse input, R(s) = 1. Hence
C (s)
R( s )
= G( s ) =
2
s
1
s + 0.2s + 1 s
Thus, as can be seen from the above transfer function, with zero initial conditions, the unit
step response of sG(s) is the same as the unit impulse response of G(s). Here we shall use
arrays for numerator and denominator polynomials with coefficients in descending powers of s
as follows.
num
= [0
1
0]
den
= [1
0.2
1]
The MATLAB program for unit step response of sG(s) is given below.
MATLAB PROGRAM 5.1
% - - - - Unit-step response of sG(s) ————
% *** Enter the numerator and denominator of the transfer function***
num = [0 1 0];
den = [1 0.2 1];
% *** Enter the following step-response command***
step (num, den);
% *** Enter grid and title of the plot ***
grid;
title (‘Unit-step Response of sG(s) = s/(s^2+0.2s+1)’)
Program lines beginning with ‘%’ are comments or remarks which are not executed. A
semicolon at the last character of a statement will display no result, though the command is still
executed. The last two statements put the grid lines and title to the graph. The ‘step (num, den)’
command will generate the unit step response (Fig. 5.40) with automatic labels of ‘Time (sec)’
and ‘Amplitude’ on x-axis and y-axis, respectively.
Fig. 5.40 Unit impulse response curve of G(s) or unit step response curve of SG(s).
Chapter 05.indd 67
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5-68
CONTROL SYSTEMS ENGINEERING
Example 5.19 Using MATLAB, obtain the unit ramp response of the closed-loop system
given by
C (s)
1
= 2
R( s )
s + s +1
Assume zero initial conditions.
Solution For a unit ramp input,
1
R(s) = 2
s
Hence,
1
1
C(s) = 2
(s + s + 1)s s
As there is no ramp command in MATLAB, we have to use step command on the transfer
function G(s)/s to obtain unit ramp response through the following program.
MATLAB PROGRAM 5.2
% - - - - Unit-ramp response ----------------% *** Enter the numerator and denominator of G(s)/s ***
num = [0 0
0
1];
den = [1
1
1
0 ];
% *** Assume the computing time points to be t = 0:0.1:7 and then
% enter step-response command as stated below ***
t = 0:0.1:7;
c = step (num, den, t);
% *** The left-hand argument in the step command will not show plot on
% the screen. The following plot command will show input curve of solid(-)
% line type and output curve of circular (0) point type ***
plot (t, c, ‘0’, t, t, ‘-‘)
% *** Add grid, title of response, x-axis label and y-axis label ***
grid;
title (‘Unit-Ramp Response of G(s) = 1/(s^2+s+1)’)
x label (‘t sec’)
y label (‘Input and Output’)
Fig. 5.41 Unit ramp response.
Chapter 05.indd 68
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
5-69
Note that the step command with left-hand argument is used here to place user-defined
labels on the x-axis and y-axis. The above program, when executed, gives us the following
response curve. Unit ramp response for G(s) = 1/(s^2 + s + 1) (Fig. 5.41)
Example 5.20 A second-order system is given by
1
C (s)
= 2
s + 2ζ s + 1
R( s )
Using MATLAB, obtain the unit impulse response curves for each zeta given as ζ = 0.1,
0.3, 0.5, 0.7 and 1.0. Assume zero initial conditions.
Solution For a unit impulse input, R(s) = 1, the MATLAB program for drawing all the
response curves in one diagram is as follows
MATLAB PROGRAM 5.3
% - - - - Unit-impulse response ——————
% *** This is obtained as step response of sG(s) ***
% *** Enter numerator and denominator of sG(s) for zeta = 0.1 ***
num = [0 1
0];
1];
den1 = [1 0.2
% *** Let computing time point be t = 0:0.1:10. The following step and
% text commands given below***
t = 0:0.1:10;
step (num, den1, t);
text (2.2, 0.89, ‘zeta = 0.1’
% ***Hold this plot to add other response curves***
hold
current plot held
% ***Enter denominators of sG(s) for other zeta values ***
den2 = [1 0.6
1];
den3 = [1 1
1];
den4 = [1 1.4
1];
1];
den5 = [1 2
% ***Superimpose other response curves to the first curve by giving
% step and text command as before***
step (num, den2, t);
text (1.32, 0.73, ‘0.3’)
step (num, den3, t);
text (1.14, 0.59, ‘0.5’)
step (num, den4, t);
text (1.1, 0.47, ‘0.7’)
step (num, den5, t);
text (0.8, 0.29, ‘1.0’)
% ***Add grid and title of the plot grid***
title (‘Impulse-Response Curves for G(s) = 1[(s^2+2(zeta)s+1]’)
% ***Clear hold command on graphics***
hold
current plot released
Chapter 05.indd 69
03/11/11 7:06 PM
5-70
CONTROL SYSTEMS ENGINEERING
After execution of the above program, the following unit impulse curves for different
values of zeta are obtained in a single diagram (Fig. 5.42).
Impulse-response curves for G(s) = 1/[s^2+2(zeta)s+1]
Fig. 5.42 Unit impulse curves for different values of zeta.
REVIEW QUESTIONS
5.1 The block diagram of a system is shown in Figure 5.43. Find the value of k such that
the damping ratio is 0.5. For the same-value of k, obtain rise time (tr), peak time (tp),
maximum overshoot (Mp) and settling time (ts) in the unit step response.
Fig. 5.43
Chapter 05.indd 70
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
5-71
5.2 The mass of a mechanical system oscillates according to Figures 5.44 (a) and (b) when it
is subjected to a force of 10 N. Determine M, B and K of the system.
Fig. 5.44
5.3 The transfer function of a second-order control system is given by
θ(s)
=
1
T (s)
Js + Bs + K
A step input of 10 N m is applied to the system and the test results are found to
be Mp = 6%, tp = 1 sec. and steady-state output of 0.5 rad. Determine the value of J, B
and K.
2
5.4 For the closed-loop system given by
C (s)
R( s )
=
ω2n
s 2 + 2ζωn s + ω2n
(1) Find the values of ζ and ωn if the system responds to a step input with 5% overshoot
and with a settling time of 2 sec. (use 2% criterion).
(2) Find the value of ζ if the system exhibits a damped oscillation as recorded in the
Figure 5.45.
Fig. 5.45
5.5 Compare the unit step, unit impulse and unit ramp response of the three systems shown in
Figures 5.46 (a), (b) and (c). Identify the best system with respect to the speed of response
and maximum overshoot in the step response.
Chapter 05.indd 71
03/11/11 7:06 PM
5-72
CONTROL SYSTEMS ENGINEERING
Fig. 5.46 (a) Positional servosystem; (b) Positional servosystem with PD control;
and (c) Positional servosystem with velocity feedback.
5.6 The open-loop transfer function of a unity feedback control system is given by
G(s) =
K ( s + 2)
2
s + β s 2 + 4s + 1
Determine the value of K and b if
(1) The closed-loop unit step response has ωn = 3 rad/sec. and ζ = 0.2.
(2) The system exhibits sustained oscillations with ωn = 4 rad/sec.
5.7 The angular position of a flywheel is controlled by an error-actuated closed-loop automatic
control system to follow the motion of input lever. The lever maintains sinusoidal
oscillations through ±60° with an angular frequency of 2 rad/sec. The inclusive moment
of inertia of flywheel is 150 kg m2 and the stiffness of the control is 2400 N m per radian
of misalignment. Find the viscous frictional torque required to produce critical damping
and hence calculate the amplitude of the swing of the flywheel and the time lag between
the flywheel and the control lever.
Chapter 05.indd 72
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
5-73
5.8 The maximum overshoot for a unity feedback control system with forward path transfer
function of K/s (sT + 1) is reduced from 60% to 20%. Determine the factor by which the
gain should be reduced to achieve this. Assume that K and T are positive constants.
5.9 The system parameters of a servomechanism according to Figure 5.47 are as follows:
Sensitivity of synchro, Ks = 1 V/rad
Amplifier gain, Ka = 30 V/V
Torque constant of motor, KT = 1 × 10–5 N m/V
Fig. 5.47
Load inertia J = 1.5 × 10−5 Kgm2
Viscous friction f = 1 × 10−5 N m/rad/sec
Tachometer constants K = 0.2 V/rad/sec
Neglecting motor inertia and friction,
(1) Obtain the value of damping ratio, when tachometer is disconnected. Also find the
steady-state error for an input velocity of 1 rad/sec.
(2) Obtain the value of damping ratio when tachometer is a part of the system.
(3) Compare the steady-state behaviour of the system with that of part (1) if the
tachometer is removed and the amplifier output is given by
(
v = K a e ∫ e dt
)
5.10 The system shown in Figure 5.48 employs proportional plus error-rate control. Find the
value of error-rate constant Ke so that the damping ratio is 0.6. Also determine the values
Chapter 05.indd 73
03/11/11 7:06 PM
5-74
CONTROL SYSTEMS ENGINEERING
of settling time, maximum overshoot and steady-state error (for unit ramp input) with and
without error-rate control.
Fig. 5.48
5.11 Evaluate proportional gain, integral time and derivative time for the following PID
controller shown in Figure 5.49.
Fig. 5.49
5.12 Determine the sensitivity of the overall gain for the system shown in Figure 5.50 and find
the value of K if the sensitivity is 0.2 under steady-state condition.
Fig. 5.50
5.13 A feedback system employing output rate damping is shown in Figure 5.51. Find the
value of K1 and K2 so that the closed-loop system resembles a second-order system with
damping ratio equal to 0.5 and frequency of damped oscillation 9.5 rad/sec.
Fig. 5.51
Chapter 05.indd 74
03/11/11 7:06 PM
TIME RESPONSE ANALYSIS
5.14 Integral control of the plant
Gp(s) =
5-75
K
τs + 1
results in a system that is too oscillatory. Will derivative action improve this situation?
5.15 Using MATLAB, determine the unit step, unit ramp and unit impulse response of the
system, the closed-loop transfer function of which is given by
8
C (s)
= 2
R( s )
s + 2s + 10
5.16 Determine the position, velocity and acceleration error constants for the unity feedback
control systems whose open-loop transfer functions are given as follows.
10
(1)
(1 + 0.4s)(1 + 0.5s)
(2)
(3)
(4)
1
K
(s + 0.1s)(1 + 0.5s)
3
11(s + 30)
s (1 + s)(1 + 0.2s)(s 2 + 5s + 15)
K (1 + 2s)(1 + 4s)
s 2 (s 2 + 2 s + 10)
5.17 Determine ess(t) for a system whose open-loop transfer function is given by
s(s + 4)
G(s) H(s) =
s(s + 1)(s + 5)
The input to the system is defined by
r(t) = 0
for t < 0;
r(t) = 2(1+t)
for t ≥ 0
5.18 A unity feedback control system has an open-loop transfer function given by
G(s) =
400
s(1 + 0.1s)
Determine the steady-state error of the system for the following inputs.
(1) r(t) = t2u(t)/2;
(2) r(t) = (1+ t)2u(t)
5.19 Figure 5.52 shows the block diagram of a system.
Determine the steady-state error for unit ramp input taking the value of K equal to 300.
Also determine the value of K for which the state steady-state error for unit ramp input
will be 0.025.
Chapter 05.indd 75
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5-76
CONTROL SYSTEMS ENGINEERING
Fig. 5.52
5.20 For the system shown in Figure 5.53, calculate the steady-state error for (1) unit step
input; (2) unit ramp input and (3) unit acceleration input.
Fig. 5.53
5.21 Calculate the error constants and steady-state errors for three basic types of inputs for the
following system.
K (s + 3.15)
and H (s) = 1
G(s) =
s(s + 1.5)(s + 0.5)
The open-loop transfer function servosystem with unity feedback is represented as
10
G(s) =
s(0.1s + 1)
Calculate the error constants Kp, Kv, Ka, for the system.
Chapter 05.indd 76
03/11/11 7:06 PM
6
STABILITY ANALYSIS
IN S-DOMAIN
6.1
CONCEPT Of STABILITY
System stability is one of the most important performance specifications of a control system.
A system is considered unstable if it does not return to its initial position but continues to oscillate
after it is subjected to any change in input or is subjected to undesirable disturbance.
For any time-invariant control system to be stable, the following two conditions need to be
satisfied:
(1) The system will produce a bounded output for every bounded input.
(2) For no input, the output should tend to be zero, irrespective of any initial conditions.
Stability of a system may be referred to as absolute stability or in terms of relative stability.
The term relative stability is used in relation to comparative analysis of stability of systems and
their operating conditions.
6.1.1
Pole-Zero Location and Conditions for Stability
A single-input–single-output system’s transfer function can be written as
G(s) =
C (s)
R( s)
=
b0 s m + b1s m −1 +  + bm
a0 s n + a1s n −1 +  + an
;
m<n
The denominator of the transfer function is called characteristic polynomial and its roots
are the poles of the transfer function. The characteristic equation is formed by equating the
characteristic polynomial to zero. Thus, for the above case, the characteristic equation is
a0 s n + a1s n −1 + a2 s n − 2 +  + an = 0;
Chapter 6.indd 1
a0 > 0
07/11/11 3:42 PM
6-2
CONTROL SYSTEMS ENGINEERING
The stability of the closed-loop system can be determined by examining the poles of the
closed-loop system, that is, by the roots of the characteristic equation. As we know, the nature of
time response of a system is related to the location of the roots of characteristic equation in s-plane.
For the system to be stable, the roots should have negative real parts. A system will be stable,
unstable or oscillatory depending upon the positions of the roots of the characteristic equation as
shown in the following. In Figure 6.1 the cross indicates the position of a pole in the s-plane. The
positions of the poles and the corresponding time response have been shown in Figure 6.1.
Fig. 6.1 Positions of the roots of the characteristic equations and the corresponding time
response of the system.
The presence of negativeness of any of the coefficients of the characteristic equation implies
that the system is either unstable or at the most has limited stability.
Therefore, the necessary conditions for stability are as follows:
(1)
The positiveness of the coefficients of characteristic equation is necessary. This is the
condition for stability of the systems of the first and second order. The characteristic equation
for a first-order system is a0s + a1 = 0 which has a single root s1 = –a1/a0. The system is
stable provided a0 and a1 are both positive. The characteristic equation for a second-order
system is a0s2 + a1s + a2 = 0 which has two roots s1, s2 = −a1 ± [ a12 − 4a0 a2 ]/2a0 . Again,
if a0, a1 and a2 are positive, the system will be stable.
(2)
For third and higher order systems, the positiveness of all the coefficients of characteristic
equation does not ensure the negativeness of the real parts of complex roots. Therefore, it
is a necessary but not sufficient condition for the systems of third and higher order.
If all the coefficients are positive, the possibility of stability of the system exists. To examine
the sufficient condition for stability, there is a criterion known as the Routh–Hurwitz’s stability
criterion, which is explained in this chapter.
6.1.2
Routh Stability Criterion and its Application
The Routh–Hurwitz criterion for stability does not require calculation of the actual values of the
roots of the characteristic equation. This criterion tells us about the number of roots on the righthand side of the imaginary axis. Moreover, this criterion gives just a qualitative result. It is the
quickest method if we just want to know whether the system is stable or unstable.
The general characteristic equation is given by,
a0 s n + a1s n −1 +  + an − 2 s + an = 0
Chapter 6.indd 2
07/11/11 3:42 PM
STABILITY ANALYSIS IN S-DOMAIN
6-3
Let a0 be positive. If it is negative, multiply both sides of the above equation by –1.
The following are the steps for applying Routh stability criteria
Step 1: If any of the coefficients, a1, a2, …, an–1, an is negative or zero, there is at least one
root of the characteristic equation which has positive real part and the corresponding system is
unstable. No further analysis is needed.
Step 2: If all the coefficients are positive, then from the first step, we cannot conclude
anything about the location of the roots. Then we have to form the following array
sn
a0
a2
a4
a6
………
n–1
s
a1
a3
a5
a7
………
sn–2
b1
b2
b3
b4
………
n–3
s
c1
c2
c3
c4
………










s2
s1
s0
f1
g1
an
where
b1 =
c1 =
a1a2 − a0 a3
a1
b1a3 − a1b2
b1
,
b2 =
,
c2 =
a1 a4 − a0 a5
a1
b1a5 − a1b3
b1
,
,
... ... ...
... ... ...
Now, examine the elements of the first column of the array.
a0, a1, b1, c1, …, f1, g1, an
The following conclusions are derived from the Routh–Hurwitz criterion.
(1) If any of these elements (a0, a1, b1, c1, …) is negative, we have at least one root on the right
of the imaginary axis and the system is unstable.
(2) The number of sign changes in the elements of the first column is equal to the number of
roots located at the right of the imaginary axis.
Therefore, a system is stable if all the elements in the first column of the array are positive,
that is, there is no change in the sign.
Depending on the coefficients of the equation, the following difficulties may arise.
(1) The first element in any of the rows of the array is zero, but the others are not.
(2) The elements in one row of the array are all zero.
In the first case, replace the zero element in the first column by an arbitrary small positive
number ε, and then proceed with array formation and ultimately let e tend to zero.
The second case of problem indicates that there are symmetrically located roots in the
s-plane. The polynomial whose coefficients are just above the row of zeros in the array is called
Chapter 6.indd 3
07/11/11 3:42 PM
6-4
CONTROL SYSTEMS ENGINEERING
an auxiliary polynomial. The auxiliary polynomial is always an even polynomial; that is, only
even powers of s appear. The roots of the auxiliary equation also satisfy the original equation. To
continue with the array, the following steps should be adopted.
(1) Form the auxiliary equation, A(s) = 0.
(2) Take the derivative of the auxiliary equation with respect to s and equate to zero.
dA(s)
That is,
=0
ds
(3) Replace the row of zeros with the coefficients of dA(s)/ds = 0.
(4) Continue the array in the usual manner with replaced coefficients.
A few examples will clarify the procedure of application of Routh–Hurwitz stability criteria.
Example 6.1 Check the system represented by the following characteristic equation for
absolute stability s3 + 5s2 + 25s + 10 = 0.
Solution
The array is
1
s3
2
5
s
1
23
s
0
10
s
25
10
0
0
b1 =
c1 =
5 × 25 − 1 × 10
5
23 × 10 − 5 × 0
23
= 23, b2 = 0
= 10, c2 = 0
All the elements in the first column are positive. There is no root in the right half of the
s-plane. The system is, therefore, absolutely stable.
Example 6.2 Check the following system for absolute stability
s5 + s4 + 3s3 + 9s2 + 16s + 10 = 0
Solution The array is
1
3
s5
16
s
1
9
10
s
–6
6
0
s
10
10
0
s
12
0
0
4
3
2
1
10
s0
In the first column, the numbers are first positive, then negative and again positive. Hence
there are two sign changes in the first column. Thus, there are two roots in the right half of the
s-plane. The system is unstable.
Chapter 6.indd 4
07/11/11 3:42 PM
STABILITY ANALYSIS IN S-DOMAIN
6-5
Example 6.3 The characteristic equation of a feedback control system is found as
s4 + 9s3 + 11s2 + 6s + K = 0
Determine the value of K for which the system is absolutely stable and marginally stable.
Also determine the frequency of sustained oscillation.
Solution The characteristic equation is s4 + 9s3 + 11s2 + 6s + K = 0.
The array is
s4
1
11
K
s3
9
6
0
s2
10.3
K
0
61.8 − 9K
0
0
s1
10.3
K
0
0
s0
9 × 11 − 6 × 1
9 × k − 0 ×1
b1 =
= 10.3, b2 =
=k
9
9
For stability K > 0
and 61.8 – 9K > 0 or K < 6.8
Hence for absolute stability range of gain K is
0 < K < 6.8
The system will be marginally stable when K = 6.8 with k = 6.8, the s ′ row becomes all
zeros.
The polynomial just above the row of zeros in the array is the auxiliary equation.
The auxiliary equation being an even polynominal with even powers of s, is expressed with
K = 6.8 as
10.3s2 + 6.8 = 0
or
Therefore,
s2 = −
6.8
10.3
= 0.66
s = ± j 0. 82 ± jω
Frequency of sustained oscillation = 0.82 rad/sec.
Example 6.4 Using the Routh–Hurwitz stability criterion, ascertain stability for each of the
following cases.
(1) s6 + 2s5 + 8s4 + 12s3 + 20s2 + 16s + 16 = 0.
Solution The array is
s6
1
8
5
s
2
12
s5
1
6
s4
2
12
Chapter 6.indd 5
20
16
16
8
16
Remarks
Auxiliary equation is formed by
considering the row just above the row
of zeros with even power of s.
07/11/11 3:42 PM
6-6
CONTROL SYSTEMS ENGINEERING
s4
1
6
8
s4 + 6s2 + 8 = 0
3
s
0
0
Its derivative is 4s3 + 12s = 0
s3
1
3
or
2
3
s
3
8
s + 3s = 0
s1
1/3
0
The array of s3 is formed accordingly.
s0
8
0
As the coefficients of s3 row are all zeros, the auxiliary A(s) and its first derivative
are formed as shown under “Remarks”. The s3 row was then replaced by the coefficients of
dA(s)/ds = 0.
The first column of the array shows no sign change. Therefore, no roots are on the righthand side of the s-plane. However, the occurrence of a zero row indicates the presence of
symmetrically located roots in the s-plane. The roots of the auxiliary equation are given by
or
s4 + 6s2 + 8 = 0
(s2 + 4) (s2 + 2) = 0
\
s = ± j 2 and s = ± j 2
As there are non-repeated imaginary roots on the imaginary axis, the system is limitedly
stable. Here the roots of the auxiliary equation are also the roots of the characteristic equation as
stated in the problem.
(2) s6 + s5 – 2s4 – 3s3 – 7s2 – 4s – 4 = 0.
Solution The array is
s6
1
–2
–7
–4
Remarks
5
s
1
–3
–4
Auxiliary equation is
4
s
1
–3
–4
s4 – 3s2 – 4 = 0
s3
0
0
0
Its derivative is
3
s
2
–3
4s3 – 6s = 0
s2
–1.5
–4
or
s1
–25/3
0
2s3 – 3s = 0. The s3 row is formed with the
coefficients of the derivative of the
s0
–4
auxiliary equation.
The first column values are all positive and then all negative. Thus, there is one sign change
in the first column of the array. Therefore, one root is lying on the right-hand side of the s-plane.
The system is not stable as the necessary condition of positiveness of all the coefficients does not
hold good.
(3) s6 + 3s5 + 4s4 + 6s3 + 5s2 + 3s + 2 = 0.
Solution The array is
s6
1
4
* 5
s
1
2
5
2
1
Remarks
A1(s) = s4 + 2s2 + 1 = 0
s4
1
dA1(s)/ds = 4s3 + 4s = 0
Chapter 6.indd 6
1
2
07/11/11 3:42 PM
STABILITY ANALYSIS IN S-DOMAIN
s3
s3
s2
0
1
1
0
1
1
6-7
or
s3 + s = 0
and, A2(s) = s2 + 1 = 0
0
= dA2(s)/ds = 2s = 0
s1
2
1
s0
*Instead of 3
6
3 we have written 1
2
1.
There are two rows i.e. s3 row and s ′ row which become zero and there is no sign change in
the first column of the array. The roots of A1(s) = 0 are also the roots of the original characteristic
equation; so the roots of A1(s) = 0 are given by
s4 + 2s2 + 1 = 0
or
(s2 + 1)2 = 0
\
s = ±j and for the second characteristic equation, s2 + 1 = 0, s = ±j
The multiplicity of the pair of roots on jw-axis makes the system unstable.
Example 6.5 The characteristic equation of a servo system is given by
b0s4 + b1s3 + b2s2 + b3s + b4 = 0.
Determine the conditions for stability using Routh–Hurwitz stability criterion.
Solution The array is
s4
b0
3
s
b1
s2
s1
b2
b3
b1b2 − b0b3
b1
b4
b1b4
b1
b3 (b1b2 − b0 b3 ) − b12 b 4
b1
b1b2 − b0 b3
b1
s0
b4
The conditions for stability are stated as follows:
b1 > 0, b0 > 0, (b1b2 – b0b3) > 0,
(b1b2 b3 − b0 b32 − b12 b4 ) > 0 and b4 > 0
Example 6.6 Determine whether the largest time constant of the following characteristic
equation is less than, greater than or equal to 1.0 sec.
s3 + 5s2 + 8s + 6 = 0.
Solution
Chapter 6.indd 7
The characteristic equation is
s3 + 5s2 + 8s + 6 = 0
07/11/11 3:42 PM
6-8
CONTROL SYSTEMS ENGINEERING
By factorization, we can write
(s + 3) (s2 + 2s + 2) = 0
The roots are
s = – 3 and s = –1 ± j
for
s = – 3, we write ζωn = −3 . As shown in Figure 6.2, time constant, t =
Therefore, by taking the magnitude of ζωn = 3, the time constant is
1
ζωn
t = 1/3 = 0.33 sec.
for
s = – 1 ± j1
τ =?
Refer to diagram below where the complex roots have been plotted.
OB
ζ = cos θ =
OA
1
=
2
Fig. 6.2 Plotting of the complex root s = −1 ± j1.
Also,
∴
∴
ωn = OA =
ζωn = |OB| =
t =
1
ζωn
2
1
2
× 2 =1
= 1 sec
So the largest time constant is equal to 1 sec.
Chapter 6.indd 8
07/11/11 3:42 PM
STABILITY ANALYSIS IN S-DOMAIN
6-9
Example 6.7 A unity feedback control system has an open-loop transfer function
G(s) = K(s + 13)/s(s + 3) (s + 7). Using the Routh–Hurwitz stability criterion, find the range of K
for the system to be stable. If K = 1, check all the poles of the closed-loop transfer function having
damping factor greater than 0.85. Assume unity feedback system.
Solution
The characteristic equation of the system is 1 + G(s) H(s) = 0. Here H(s) = 1
1 + G(s) = 1 +
or
K (s + 13)
s(s + 3)(s + 7)
=0
s 3 + 10 s 2 + (21 + K )s + 13K = 0
The array is
s3
s2
1
10
210 − 3K
s1
10
s0
(1)
(2)
(21 + K)
13K
–
13K
Condition
13K > 0
210 − 3K
10
or K > 0
>0
or K < 70
For stability, the range is 0 < K < 70. When K = 1, the characteristic equation is
or
The roots are
s3 + 10s2 + 22s + 13 = 0
(s + 1) (s2 + 9s + 13) = 0
s s = −1, 1 (−9 ± 29 )
2
All roots are negative and real and hence lie on the negative real axis.
Now, damping ratio ζ = cos θ.
Since all roots are negative and real the z line will lie on the negative real axis, q = 0 and
z = cos 0º = 1, therefore, damping ratio z = 1. Hence, for all roots, damping ratio is greater than
0.5.
Example 6.8 The characteristic equation for a feedback control system is given by
s + a1s3 + a2s2 + a3s + K = 0. If the numerical values of a1 = 22, a2 = 10 and a3 = 2, find the value
K which will keep the system stable.
4
Solution
Chapter 6.indd 9
The characteristic equation is given as
s4 + 22s3 + 10s2 + 2s + K = 0
07/11/11 3:42 PM
6-10
CONTROL SYSTEMS ENGINEERING
The array is
s4
s3
1
22
s2
=
s1
2−
218
K
10
2
K
22
484K
218
K
s0
Applying the condition of stability, we have to check for all the coefficients in the first
column to be positive. Thus,
2−
or
484K
218
484K
218
or
>0
<2
K <
2 × 218
484
or
K < 0.9 and K > 0
Therefore, the range of K for stability is determined as
0 < K < 0.9
Example 6.9 The open-loop transfer function of a unity feedback system is given by
K
( s + 2)( s + 4)( s 2 + 6s + 25)
By applying the Routh–Hurwitz criterion, discuss the stability of the closed-loop system as
a function of K. Determine the value of K which will cause sustained oscillations in the closedloop system. What are the corresponding oscillation frequencies?
Solution
We have
C (s)
R( s )
=
G(s)
1 + G(s) H (s)
=
G(s)
1 + G(s)
( H ( s ) = 1)
Characteristic equation is
1 + G(s) = 0
or
K
1+
=0
(s + 2)(s + 4)(s 2 + 6s + 25)
or
s 4 + 12s 3 + 69s 2 + 198s + (200 + K ) = 0
Chapter 6.indd 10
07/11/11 3:42 PM
STABILITY ANALYSIS IN S-DOMAIN
The array is
s4
s3
s2
s1
198 −
1
12
52.5
12(200 + K )
6-11
69
(200 + K)
198
(200 + K)
52.5
(200 + K)
s
The system will be stable, if
200 + K > 0 or K > –200
and
12(200 + K )
> 0 or K < 666.25
198 −
52.5
0
Oscillations will occur when K = 666.25 and in this case the auxiliary equation is
52.5s2 + (200 + K) = 0
or
52.5s2 + (200 + 666.25) = 0
or
s = ±(4.062)j
The frequency of sustained oscillation wn is therefore 4.062 rad/sec.
Example 6.10 The open-loop transfer function of a closed-loop system with unity feedback
is K(s + 2)(s + 1)/(s + 0.1)(s – 1). Using Routh–Hurwitz criterion, show whether the system is
stable or unstable.
Solution
The characteristic equation is
1 + G(s) =
K ( s + 2)( s + 1)
( s + 0.1)( s − 1)
+1 = 0
or
K(s + 2)(s + 1) + (s + 0.1)(s – 1) = 0
or
(K + 1)s2 + (3K – 0.9)s + (2K – 0.1) = 0
The array is
s2
(K + 1)
(2K – 0.1)
s1
(3K – 0.9)
–
0
s
(2K – 0.1)
–
Comments:
(1) For stability, the coefficients of the first column should be positive. That is,
(a) K > –1
(b) K > 0.9/3, that is, K > 0.3
(c) K > 0.1/2, that is, K > 0.05
So the system will be stable if K > 0.3.
Chapter 6.indd 11
07/11/11 3:42 PM
6-12
CONTROL SYSTEMS ENGINEERING
Example 6.11 Check the following systems for their absolute stability.
(1)
s3 – 3s2 + s + 6 = 0
Solution
s3 – 3s2 + s + 6 = 0
The array is
s3
+1
1
s
–3
6
s
+3
0
s
+6
2
1
0
In the first column, sign changes from positive value
to negative value and then again to positive value.
Let us inspect the first column. There are two changes in sign (positive values to negative
and then to positive value). It implies that there are two roots in the right half of s-plane. So, the
system is unstable.
(2) 2s4 + s3 + 3s2 + 5s + 10 = 0
Solution
s
The array is
4
2
3
10
s3
1
5
0
10
0
0
0
0
0
3 − 10
s2
1
= −7
−35 − 10
s1
−7
s0
10
=
45
7
In the first column, there are two sign changes which imply that there are two roots in the
right half of s-plane. So, the system is not stable.
(3) (s – 1)2 (s + 2) = 0
Solution
s3 – 3s + 2 = 0
or
s3 + 0s2 – 3s + 2 = 0
The array is
s3
1
–3
ε
2
We have replaced the zero element in the first column by e
s2
arbitrarily and have proceeded further.
−3ε − 2
0
s1
ε
s0
Chapter 6.indd 12
2
0
07/11/11 3:42 PM
STABILITY ANALYSIS IN S-DOMAIN
6-13
Since there is one entry in the first column as zero, we have replaced that zero by a very
small positive number e. There are two sign changes when e tends to zero. Thus, the system is
unstable since there are two roots in the right half of s-plane.
(4)
s4 + s3 – 3s2 – s + 2 = 0
Solution The array is
s4
1
–3
2
3
s
1
–1
0
s2
–2
2
0
1
s
0
0
0
1
s
–4
0
0
s0
2
It is observed that in the above problem, all the entries in s1 row are zero. So, for finding
the coefficient of the next row, we write the auxiliary equation and differentiate it. The equation
obtained after differentiation gives the elements of the row which were zero.
In the above problem, the auxiliary equation is
–2s2 + 2 = 0
Differentiating, we have –4s = 0 which will give the coefficients of the s1 row. Then, we are
to complete the array by the usual method.
As there are two sign changes in the first column of the array, the system is not stable
because of two roots being in the right half of the s-plane.
(5)
s6 + s5 – 2s4 – 3s3 – 7s2 – 4s – 4 = 0.
Solution The array is presented as follows
s6
1
–2
–7
–4
1
–3
–4
0
s5
4
1
3
–4
0
s
0
0
0
0
s3
3
4
–6
0
0
s
−
3
s2
–4
0
0
2
s1
−50
3
0
0
0
–4
0
0
0
s0
We see that all the coefficients of the s3 row are zero. These have been replaced by the
coefficient of dA(s)/ds = 0 as follows:
Auxiliary equation A(s) is
A(s) = s4 – 3s2 – 4 = 0
Chapter 6.indd 13
dA(s)
ds
= 4s3 – 6s = 0
07/11/11 3:42 PM
6-14
CONTROL SYSTEMS ENGINEERING
Since there is one sign change in the first column of the array, we conclude that the system
is unstable because of existence of one root in the right half of s-plane.
(6)
s4 + 2s3 + 3s2 + 4s + 5 = 0.
Solution The array is
1
3
5
s4
3
2
4
0
s
2
1
5
0
s
1
–6
0
0
s
0
5
0
0
s
There are two sign changes i.e. first positive, then negative and again positive, in the first
column of the array. Thus, there are two poles in the right half of the s-plane. The system is,
therefore, unstable.
(7)
s3 + 2s2 + s + 2 = 0
Solution The array is
1
1
s3
2
2
2
s
1
0
0
s
1
4
s
s0
2
0
Remarks
Auxiliary equation
2s2 + 2 = 0
Its derivative
d
(2s 2 + 2) = 4s
ds
There is no change in the sign of the entries in the first column. So, there is no root in the
right half of the s-plane. Hence, the system is stable.
Example 6.12 Find the range of values of k for which the system, represented by the
following expression, is stable.
s3 + 30s2 + 600s + 600k = 0
Solution
1
600
s3
2
30
600k
s
1
20(30 – k)
0
s
0
600k
s
For the system to be stable, all the values in the first column should be positive.
20(30 – k) > 0
or
30 – k > 0
or
k < 30 and 600k > 0
Chapter 6.indd 14
07/11/11 3:42 PM
STABILITY ANALYSIS IN S-DOMAIN
6-15
or
k>0
Thus, 0 < k < 30.
The range of values of k for which the system will be stable is 0 < k < 30.
Example 6.13 Determine the range of k for stability of the following system (Fig. 6.3).
Fig. 6.3
Solution
The closed-loop transfer function is
C (s)
R( s )
=
=
K
2
s(s + s + 1)(s + 1)
K1
1+
2
s(s + s + 1)(s + 1)
4
3
K1
s + 2s + 2s 2 + s + K1
Characteristic equation is
s4 + 2s3 + 2s2 + s + K1 = 0
s4
1
3
K1
s
2
1
0
s
3/2
0
s1
3 − 4K1
K1
0
0
s0
K1
0
0
3
2
3
For the system to be stable
3 − 4K1
3
>0
or
K1 < 3/4
and
K1 > 0
For the system to be stable, the range of K1 is 0 < K1 < 3/4.
Chapter 6.indd 15
07/11/11 3:42 PM
6-16
CONTROL SYSTEMS ENGINEERING
Example 6.14 Predict whether the following two given systems are stable or not.
(1)
G(s) =
(2)
G(s) =
12( s + 1)
s ( s − 1)( s + 5)
H (s) = 1
,
12
s ( s − 1)(2s + 3)
,
H (s) = 1
Solution
(1) Closed-loop transfer function is
12(s + 1)
s(s − 1)(s + 5)
=
12(s + 1)
R( s )
1+
s(s − 1)(s + 5)
C (s)
=
=
12(s + 1)
s(s − 1)(s + 5) + 12(s + 1)
3
12(s + 1)
s + 4s 2 + 5s + 12
Characteristic equation is s3 + 4s2 + 5s + 12 = 0
The array is
s3
1
5
2
s
4
12
s1
2
0
s
12
0
0
The system is stable because all the coefficients in the first column of the array are positive.
(2)
Closed-loop transfer function is
12
C (s)
s ( s − 1)(2 s + 3)
=
12
R( s )
1+
s ( s − 1)(2s + 3)
=
12
s ( s − 1)(2s + 3) + 10
12
2s3 + s 2 − 3s + 12
Characteristic equation is 2s3 + s2 – 3s + 12 = 0.
Chapter 6.indd 16
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STABILITY ANALYSIS IN S-DOMAIN
6-17
The array is
s3
2
–3
2
s
1
10
s1
–27
0
s0
12
The system is unstable because there are two changes in sign (positive values to negative
value and then to positive value), there are two roots in the right half of the s-plane.
Example 6.15 Consider the characteristic equation of a system:
s4 + s3 + 2s2 + 2s + 3 = 0
What is the status of stability of the system?
Solution The array is
s4
1
2
3
s3
1
2
0
2
s
0
3
Since the first element of row s2 is zero, the element in the row s1 will become infinite. To
overcome this, replace the zero by a small positive number ε and proceed further.
ε
s2
3
(3 − 2ε)
0
s1 −
ε
3
0
s0
The term –(3–2e)/e will tend to be negative when å will tend to zero.
Since there are two sign changes in the first column, the system is not stable and has two
roots in the right half of s-plane.
Example 6.16 Consider the characteristic equation of a system
s5 + 4s4 + 8s3 + 8s2 + 7s + 4 = 0
Find the status of its stability.
Solution The array is
1
8
s5
7
s4
4
8
4
s
6
6
0
s
4
4
s
0
0
3
2
1
The auxiliary equation is
A(s) = 4s2 + 4 = 0
The derivative of A(s) with respect to s is dA(s)/ds = 8s therefore, we consider 8s = 0.
Chapter 6.indd 17
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6-18
CONTROL SYSTEMS ENGINEERING
Replace zeros of the s1 row with the coefficients of the above equation.
s1
8
0
s0
4
Since there is no sign change, no root is lying in the positive half of s-plane. Therefore, the
system is stable.
6.2limitations of routh’s stability
Routh’s stability analysis provides information about the absolute stability of a system. It
only tells us whether a system is stable or not. This is because the Routh’s test only finds out
whether the roots of the characteristic equation of the control system lie on the left-hand side
of the imaginary axis in the s-plane. However, a control system engineer will not remain satisfied
with this information. He or she would like to know how stable the system is. Degree of stability
would depend on the nearness or farness of the roots i.e. the poles of the characteristic equation from
the imaginary axis. The more the poles are away to the left of the imaginary axis, the more stable
the system is.
To study relative stability by using Routh’s criteria, we could shift the imaginary axis
to the left by some units, write the new characteristic equation taking into consideration the
shift of the imaginary axis and then apply Routh’s stability criteria. If the system still shows
stability, we could then know about the degree of its stability. For example, let the characteristic
equation be
5s2 + 6s + 10 = 0
Let us shift the imaginary axis by 2 units, that is, to s = –2.
Now, we will convert the characteristic equation accordingly by substituting s = z – 2 where
z is the new axis. Thus, in terms of z-axis, the characteristic equation becomes
or
5( z − 2)2 + 6( z − 2) + 10 = 0
5z2 – 14z + 18 = 0
Example 6.17 The characteristic equation of a third-order system is given by
s3 + 7s2 + 25s + 39 = 0
Check whether the roots of the characteristic equation are more negative than –1.
Solution
Let us shift the origin to s = –1 by substituting s = z – 1 (as shown in Fig. 6.4) in the
given characteristic equation.
The characteristic equation in terms of z-axis now becomes
(z – 1)3 + 7(z – 1)2 + 25(z – 1) + 39 = 0
or
z3 + 4z2 +14z + 20 = 0
Chapter 6.indd 18
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STABILITY ANALYSIS IN S-DOMAIN
6-19
Forming the array, we get
z3
1
14
z
2
4
20
z1
9
z0
20
Fig. 6.4 Relation between s-axis and z-axis.
Thus, all the roots of the characteristic equation in terms of z lie in the left half of
the z-plane, because the signs of all the elements in the first column of the array are positive.
This implies that all the roots of the given characteristic equation lie to the left of s = –1 in
the s-plane.
REVIEW QUESTIONS
6.1 Use the Routh–Hurwitz criterion to determine the stability of the systems having the
following characteristic equations. Also find the number of roots of each equation that
may lie in the right half of the s-plane.
(1) s3 + 20s2 + 9s + 200 = 0
(2) 3s4 + 10s3 + 5s2 + s + 2 = 0
(3) s5 + 2s4 + 2s3 + 4s2 + s + 1 = 0
(4) s6 + 3s5 + 5s4 + 9s3 + 8s2 + 6s + 4 = 0
6.2 Determine the values of K so that the systems with the following characteristic equations
will be stable.
(1) s3 + as2 + Ks + b = 0, where a, b are constants
(2) s4 + 20Ks3 + 5s2 + (10 + K)s + 15 = 0
(3) s(2s + 1)(3s + 1) + K(s + 1) = 0
(4) s4 + 6s3 + 11s2 + 6s + K = 0
Chapter 6.indd 19
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6-20
CONTROL SYSTEMS ENGINEERING
6.3 Derive the stability conditions which must be satisfied by the coefficients of the following
characteristic equation.
a3s3 + a2s2 + a1s + a0 = 0, where a3 > 0
6.4 A unity feedback control system has the following forward transfer function
G(s) =
K (2 s + 1)
s (4s + 1)( s + 1)2
Determine the stability of the system for the minimum value of K obtained from ess < 0.1
and r(t) = 1 + t.
6.5 Determine the stability for each of the systems with the following loop transfer functions.
(1) G(s) H(s) =
(2) G(s) H(s) =
(3) G(s) H(s) =
(4) G(s) H(s) =
1
(s + 2)(s + 4)
Ks
(s + 1)2
K ( s − 1)
s( s 2 + 4 s + 4 )
K (s − 2)
s( s + 3)( s 2 − 2 s + 3)
6.6 The open-loop transfer function of a unity feedback control system is given by
G(s) =
3
K ( s + 5 )( s + 40 )
s ( s + 200 )( s 2 + 1000 )
Discuss the stability of the closed-loop system in terms of K. What value of K will cause
sustained oscillation? Find the frequencies of the oscillations.
6.7 A certain system has the characteristic equation s(Ts + 1) + K = 0. It is desired that all the
roots lie to the left of the line s = –a to guarantee a time constant not larger than 1/a. Use
the Routh–Hurwitz criterion to find the values of K and T so that both roots meet this
requirement.
6.8 Given a system with the characteristic equation s3 + 9s2 + 26s + K = 0, determine the value
of K that will give a dominant time constant not larger than 0.5. The design of the system
should be slightly under-damped. Is it possible? If so, find the value of K needed and the
resulting damping ratio.
6.9 Find the value of K for which the system represented is stable
G(s) =
H(s) = 1
Chapter 6.indd 20
K
(s + 1)(s + 2)
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STABILITY ANALYSIS IN S-DOMAIN
6-21
6.10 A control system as given below has the transfer function
G(s) =
K
s(1 + 0.1s)(1 + 0.2s)
Determine the maximum value of K that makes the overall system absolutely stable.
6.11 Determine the range of k such that the characteristic equation s3 + 3(k + 1) s2 + (7k + 5)s
+ (4k + 7) = 0 will have roots greater than s = –1.
(Hint: Put s = z – 1 in the given characteristic equation so as to shift the imaginary axis
towards left by –1).
6.12 Using Routh’s stability criterion, determine the relation between K and T so that the unity
feedback control system whose open-loop transfer function given below is stable.
G(s) =
K
s[ s(5 + 10) + T ]
6.13 Using Routh’s stability criterion, determine the values of K and a so that the following
represented system oscillates at a frequency of 2 rad/sec.
Fig. 6.5
6.14 State the necessary and sufficient conditions of stability for first- and second-order control
systems. Explain why these conditions are necessary but not sufficient for stability of
higher order systems.
6.15 A feedback control system whose G(s) and H(s) are
G(s) =
K (s + 40)
s(s + 10)
and H (s) =
1
s + 20
Find the value of gain K which gives marginal stability. Determine the oscillation
frequency. Also determine the range of K for which the system is absolutely stable.
Chapter 6.indd 21
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Chapter 6.indd 22
07/11/11 3:42 PM
7
THE ROOT LOCUS
TECHNIQUE
7.1
INTROdUCTION
We have observed in the previous chapter that there are two related points with respect to study
of stability of a control system. First, all the poles of the closed-loop system must lie on the
left-hand side of the imaginary axis in the s-plane. Second, the examination has to be made to
see how close the poles are with the jw-axis of the s-plane. The second aspect provides us with
information regarding the relative stability of the control system.
We have also seen that the poles are the roots of the characteristic equation. The characteristic
equation is obtained by substituting the denominator of the closed-loop transfer function to zero.
The characteristic equation is useful in the study of performance of a control system.
From the design viewpoint, by writing the differential equation of the system and solving
the differential equation with respect to a controlled variable, the time response can be found
out. Thus, we will know the accurate solution of the equation and hence the performance of the
system. But this approach of solution may be difficult for even a slightly complex system. Efforts
involved in determining the roots of the characteristic equation have been avoided by applying
Routh–Hurwitz criterion as described in the previous chapter. But this criterion only tells the
designer as to whether a system is stable or unstable. The designer of a control system cannot
remain satisfied with this information because he is unable to indicate the degree of stability of
the system. The degree of stability will tell about the amount of overshoot, settling time, etc. for
an input, say a step or a ramp input.
Root locus technique is a powerful graphical method used for the analysis and design of a
control system. This method of analysis not only indicates whether a system is stable or unstable
but also shows the degree of stability of a stable system.
Root locus is a plot of the roots of the characteristic equation of the closed loop as a
function of gain. The effect of adjusting the closed-loop gain of the system on its stability can be
Chapter 7.indd 1
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7-2
CONTROL SYSTEMS ENGINEERING
studied by root locus method of stability analysis. In root locus method, gain is assumed to be a
parameter which is varied from 0 to ∝ and the movement of the poles in the s-plane is sketched.
7.1.1
Root Locus Concept
Consider a second-order unity feedback control system shown in Figure 7.1.
Fig. 7.1 Block diagram of a second-order system.
The open-loop transfer function of the system is given as
G(s) =
K
s( s + a)
The characteristic equation is s(s + a) = 0.
It has two poles at s = 0 and s = –a.
The closed-loop transfer function of the system is
G ( s)
K
C (s)
=
= 2
1 + G ( s) H ( s) s + as + K
R( s )
The characteristic equation of the system is
s2 + as + K = 0
(7.1)
The second-order system as above will be stable for positive values of a and K. Its dynamic
behaviour will be controlled by the roots of the characteristic equation.
Roots of the characteristic equation are
or
s1, s2 =
s1, s2 =
−a ±
a2 − 4K
2
−a
2
2
±
a
  −K
2
If we vary K from zero to infinity, the two roots (s1, s2) will describe a loci in the s-plane.
The root locations for different values of K will change.
By examining the values of roots, that is s1 and s2, we observe that
(1)
When 0 ≤ K < a 2 /4, the roots are real and distinct.
When K = 0 the roots are s1 = 0 and s2 = –a, which are the open-loop poles.
Chapter 7.indd 2
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tHE rOOT lOCUS tECHNIQUE
(2)
7-3
When K = a2/4, the roots are real and equal, that is, s1 = s2 = –a/2.
When a 2 /4 < K < ∞, the roots are complex conjugate with constant real part equal to
–a/2.
The root loci plotted for changing values of K shown in Figure 7.2 indicates the following
system behaviour.
(3)
Fig. 7.2 Root loci of a second-order closed-loop system with characteristic equation,
s 2 + as + K = 0 and changing values of gain K.
The root locus plot has two branches starting at the two open-loop poles (s1 = 0 and s2 = –a)
when the value of gain K is 0.
(2) As K is increased from 0 to a2/4, the roots move along the real axis towards the point, R
(–a/2, 0) from opposite directions. The system will behave as an over-damped system. At
K = a2/4, the roots are –a/2, –a/2. At this value of K, the system will behave as critically
damped system.
(3) At K > a2/4, the roots become complex with real part equal to –a/2, that is, the roots
break away from the real axis, become complex conjugate and move towards ∞ along the
vertical line at σ = −a/2. Since the loci move away from the real axis, the system becomes
under-damped. For this case, settling time is nearly constant as the real part is constant.
We have drawn the root locus by direct solution of the characteristic equation. For higher
order systems, this procedure will become complicated and time consuming.
Evans in 1948 developed a simplified graphical technique for root locus plot which is
described in the following. The characteristic equation of the closed-loop system shown in
Figure 7.1 is 1 + G(s)H(s) = 0
or
G(s)H(s) = –1 = –1 + j0
(7.2)
(1)
(a)
Chapter 7.indd 3
Magnitude criterion
From Equation (7.2), we see that the magnitude of the open-loop transfer function is equal
to unity for all the roots of the characteristic equation, G(s)H(s) = 1. Magnitude criterion
can determine the value of K for a point to be on the root locus.
03/11/11 7:18 PM
7-4
(b)
CONTROL SYSTEMS ENGINEERING
Angle criterion
The angle of the open-loop transfer function is an odd integral multiple of p.
∠G ( s ) H ( s ) = ±180° (2q + 1);
where, q = 0, 1, 2 …
The gain factor K does not affect the angle criterion.
For any point to be on the root locus in the s-plane, it has to satisfy both angle criterion
and magnitude criterion. The magnitude criterion is checked after confirming the existence of
the point on the root locus by applying the angle criterion. To understand this, let us consider an
example where
k
G(s)H(s) =
s ( s + 1)( s + 2)
Let us examine whether s = –0.5 lies on the root locus or not.
First we apply angle criterion as
∠G(s) H (s) at s = –0.5 = ±180° (2q + 1) where q = 0, 1, 2, …
Here,
K
K
∠ G( s ) H ( s ) =
=
(−0.5)(−0.5 + 1)(−0.5 + 2) (−0.5 + j 0)(0.5 + j 0)(1.5 + j 0)
=
K ∠0°
180° 0° 0°
= −180°
Since angle criterion is satisfied, the point s = –0.5 lies on the root locus. Now, we also will
check by applying the magnitude criterion to find the value of K for which the point s = –0.5 lies
on the root locus.
Using magnitude condition,
G(s) H (s) = 1 at s = –0.5
Here,
K
=1
−0.5 0.5 1.5
or
K = 0.375
For this value of K, point s = –0.5 lies on the root locus.
7.1.2
Root Locus Construction Procedure
As we have seen, root locus is the graphical plot of the poles of a closed-loop system with respect
to change in the gain parameter K of the system as s changes from 0 to ∞. The knowledge of
open-loop poles and zeros are important here as the root locus always starts from open-loop poles
Chapter 7.indd 4
03/11/11 7:18 PM
tHE rOOT lOCUS tECHNIQUE
7-5
and terminate on an open-loop zero or infinity. We will take up an example of plotting the root
locus and along with that write the general rules or guidelines.
Example 7.1 Sketch the root locus of a control system whose transfer function is G(s) =
K/s (s + 1) with unity feedback.
Solution We have
G(s) =
G ( s ) H ( s ) =
K
s ( s + 1)
and H ( s ) = 1
K
s ( s + 1)
By examining the denominator of G(s), we find that the number of open-loop poles n = 2.
They are s = 0 and s = –1. As evident from the numerator, the number of open-loop zeros m = 0.
The closed-loop transfer function of the system is
C (s)
G(s)
K /s ( s + 1)
K
=
=
= 2
R( s )
1 + G ( s ) H ( s ) 1 + K /s( s + 1) s + s + K
The characteristic equation is
s2 + s + K = 0
s1, s2 =
−1 ± 12 − 4 K
2
= −0.5 ± 0.25 − K
For K > 0.25, the roots are complex conjugates.
The root loci start at 0 and –1, that is, at the open-loop poles.
The number of branches or root loci = Number of open-loop poles
Number of asymptotes = Number of open-loop poles – Number of open-loop zeros
In this case, asymptotes = 2 – 0 = 2.
Determination of breakaway points:
Characteristic equation is
Substituting,
1 + G(s)H (s) = 0
K
s( s +1)
= –1
or,
K = –s (s + 1)
or,
K = –s2 –s
Chapter 7.indd 5
We put
dK
=0
ds
∴
dK
= –2s – 1 = 0
ds
03/11/11 7:19 PM
7-6
or,
CONTROL SYSTEMS ENGINEERING
s =−
1
= − 0.5
2
The two root loci, starting at 0 and –1 respectively approach each other and breakaway
asymptotically at –0.5.
The value of K at the breakaway point on the real axis is calculated as
K = –s (s + 1)
 1 1

= −  −  − + 1 

 2 2
 1 1
= − − × 
 2 2
= 0.25
For a value of K = 0.25, the two root loci meet at the real axis at s = –0.5 and breakaway at
φ A = 90° and 270° asymptotically as K increases beyond 0.25 towards infinity, where
angles of asymptotes is
(2q + 1)180°
φA =
n−m
where n = number of open-loop poles and m = number of open-loop zeros.
(2 × 0 + 1) × 180 °
=
2−0
= 90°
q = 0, 1, 2, …, (n – m) –1
Fig. 7.3 Root loci of s 2 + s + K = 0.
and
Chapter 7.indd 6
φA =
(2 × 1 + 1)180°
2 −1
= 270°
03/11/11 7:19 PM
tHE rOOT lOCUS tECHNIQUE
7-7
The root locus has been drawn as in Figure 7.3. It may be observed that the root locus is
symmetrical about the real axis, that is σ axis. There is no existence of the root locus to the left
pole at s = –1 on the real axis.
7.1.3.
Root Locus Construction Rules
A mentioned earlier, root locus is the path of the roots of the characteristic equation, 1 + G(s)
H(s) = 0 traced out in s-plane as the system parameter (gain K) is changed.
The root locus diagram or plot can be completed using the following procedure. The
procedure is presented in the form of certain rules.
(1) Starting and termination of root locus—From the open-loop transfer function, locate
the poles and zeros. Each branch of the root locus originates from an open-loop pole with
K = 0 and terminates either on an open-loop zero or at infinity as the value of K increases
from 0 to ∞. In most cases, we will have more poles than zeros. If we have n poles and m
zeros, and n > m, then n – m branches of the root locus will reach infinity. Because the root
loci originate at the poles, the number of root loci is equal to number of poles.
(2) Root locus on the real axis—The root locus on the real axis always lies in a section of the
real axis to the left of an odd number of poles and zeros.
Let the open-loop transfer function of a control system be G(s) = K(s + 1)/s + 2. The
pole is at s = –2 and the zero is at s = –1 as shown in Figure 7.4 (a). The root locus will start
s = –2 and terminate at zero at s = –1. There is existence of root locus to the left of Z and
no existence to the left of P on the real axis (root locus on real axis exists to the left of odd
number of poles and zeros).
Fig. 7.4 (a) Location of poles and zeros and the root locus on the real axis.
Symmetry of the root locus—The root loci must be symmetrical about the real axis
because the complex roots appear in pairs.
(4) The member of asymptotes and their angles with the real axis—The (n – m) branches
of root loci move towards infinity. They do so along straight line asymptotes. The angle of
asymptotes with respect to the real axis is given by
(2q + 1)
180°, q = 0,1, 2, ...
φA =
n−m
where n is the number of poles and m is the number of zeros.
(3)
Chapter 7.indd 7
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7-8
CONTROL SYSTEMS ENGINEERING
Centroid of the asymptotes—The linear asymptotes are centred at a point on the real axis.
This is called the centroid which is given by the relation
Σ real parts of poles − Σ real parts of zeros
σA =
n−m
(6) Breakaway points—The root locus breakaway from the real axis where a number of roots
are available, normally, where two roots exist. The method of determining the breakaway
point is to rearrange the characteristic equation in terms of K. We then evaluate dK/ds = 0 in
order to find the breakaway point. Since the characteristic equation can have real as well as
complex multiple roots, its root locus can have real as well as complex breakaway points.
However, because of conjugate symmetry of root loci, the breakaway point must either be
on the real axis or must occur in complex conjugate pairs.
(7) Intersection of the root locus with the imaginary axis—The point at which the locus
crosses the imaginary axis, in case it does, is determined by applying Routh–Hurwitz
criterion. The value of K for which the locus crosses the imaginary axis is calculated by
equating the terms in the first column of the Routh array of s1 and s0 to zero.
(8) Angle of departure of the root locus—The angle of departure of the locus from a complex
pole is calculated as
φd = 180º – sum of angles made by vectors drawn from the other poles to this pole
+ sum of angles made by vectors drawn from the zeros to this pole.
Let us consider an example. Let
K
G(s)H(s) =
2
s ( s + 2)( s + 6s + 2s )
(5)
The poles are at s1 = 0, s2 = –2, s3 = –3 + j4, s4 = –3 – j4. There are no zeros. The positions
of poles are shown in Figure 7.4 (b). The angle of departure of the root locus from the complex
pole at P3 is calculated as
φd = 180° − (127° + 104° + 90°) + 0 = 180° − 321° = − 141°
Fig. 7.4 (b) Calculation of angle of departure of the root locus from a complex pole.
Chapter 7.indd 8
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tHE rOOT lOCUS tECHNIQUE
7.1.4
7-9
Root Locus Construction Rules-Illustrated Through Examples
The root locus construction rules have been explained in the previous section. However, for the
sake of better understanding the rules are once again illustrated through examples.
The following rules are applicable in sketching the root locus plot.
Rule 1: Symmetry of root locus—Any root locus must be symmetrical about the real axis,
that is, the upper half of the root locus diagram is exactly the same as the lower half about the real
axis. This can be seen from any root locus diagram.
Rules 2 and 3: Starting and termination of root loci—Root locus will start from an openloop pole with gain K = 0 and terminate either on an open-loop zero or to infinity with K = ∞.
Let us illustrate these rules with an example. Let open-loop transfer functions of control
systems are
K ( s + 1)
(1) G(s)H(s) =
(s + 2)
K
s( s + 1)
For (1), the root locus will start at the pole at s = –2 and terminate at zero at s = –1 as shown
in Figure 7.5 (a).
For (2) the poles are at s = 0 and s = –1. As there is no zero, the locus will originate at s = 0
and s = –1 and approach towards each other and then break away to infinity as the value of gain
K is increased continuously as shown in Figure 7.5 (b).
(2)
G(s)H(s) =
Fig. 7.5 (a) Starting and termination of root locus where number of poles are equal to number
of zeroes i.e. n = m; (b) Starting and termination of root locus when n > m.
For
G(s)H(s) =
K ( s + 1)
s+2
Rule 4: Number of root loci—If P is the number or poles and Z is the number of zeros in
the transfer function G(s)H(s), the number of root loci N will be as follows:
N = P if P > Z
N = P = Z if P = Z
Chapter 7.indd 9
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7-10
CONTROL SYSTEMS ENGINEERING
For example, in the root locus shown under Rules 2 and 3, in Figure 7.5 (a) P = 1,
Z = 1. Therefore, the number of root loci, N = P = Z = 1. And in Figure 7.5 (b), P = 2, Z = 0. Therefore,
N = 2. The two root loci originating from origin and –1 respectively are approaching to ∞ in two
directions.
Rule 5: Root loci on the real axis—The root locus on the real axis will lie in a section of
the real axis to the left of an odd number of poles and zeros.
This rule is illustrated through the following examples:
K ( s + 2)( s + 3)
(1) G(s)H(s) =
s ( s + 1)
Its root locus is shown below.
There is no root locus on the real axis between P2 and Z1 because the number of poles and
zeros to its right is even, that is 2. See Figure 7.6.
Fig. 7.6 Location of root locus on the real axis.
Similarly, beyond, z2 to the left on the real axis there cannot be any root locus as the number
of poles and zeros is even, that is 4 in this case.
Rule 6: The number of asymptotes and their angles with the real axis—As the value of
K is increased to ∞, some branches of root locus from the real axis approach infinity along some
asymptotic lines. These asymptotic lines are straight lines originating from the real axis make
certain angles with the real axis. The total number of asymptotic lines and the angles they would
make are calculated as follows:
Number of asymptotic lines asymptotes
= P – Z
where P is the number of poles and Z is the number of zeros of the open-loop transfer function,
G(s)H(s).
The angle of asymptotes with the real axis is
φA =
Let us consider,
(2 q + 1) 180 °
where q = 0, 1, 2, …
P−Z
K
s( s + 2 )
Here, the number of poles P = 2; they are at s = 0 and s = –2 and number of zeros Z = 0.
Chapter 7.indd 10
G(s)H(s) =
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tHE rOOT lOCUS tECHNIQUE
7-11
Number of asymptotes= P – Z = 2 – 0 = 2
Let the angle of two asymptotes be φ A and φ′A respectively. Then,
φA =
φ′A =
(2 × 0 + 1)180°
2−0
(2 × 1 + 1)180°
2−0
= 90°
= 270°
for q = 0
for q = 1
The root locus with the asymptotes are shown in Figure 7.7. One asymptote, LM is making 90º
with the real axis and another asymptote LN is making an angle of 270º with the real axis. There are
two root loci originating at the poles at s = 0 and s = –2. They approach each other and break away
at L and approach towards infinity along the asymptotic lines as the value of K increases from 0
to ∞.
Fig. 7.7 The root locus with the asymptotes.
Rule 7: Centroid of the asymptotes—The point of intersection of the asymptotes with the
real axis is called the centroid σ A which is calculated as
−σ A =
Σ Real parts of poles − Σ Real parts of zeros
P−z
Let us consider the example of Rule 6 where G(s) = K/s(s + 2)
[0 − 2] − [0]
2
= −1
2−0
2
Therefore, the points L where the two asymptotes start is at a distance of –1 from the origin.
Let us consider another example,
−σ A =
G(s)H(s) =
=−
k
s ( s + 4)( s + 5)
The poles are at s = 0, s = –4 and s = –5, which are shown below. There will be no part of
the root locus to the left of B up to point A as there are even number of poles to the right of B.
Chapter 7.indd 11
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CONTROL SYSTEMS ENGINEERING
There will be part of root locus to the left of point A as there are odd number of poles to the right
of A. There will be three root loci originating at the poles. They will terminate at zeros. If there is
not zero present in G(s)H(s), they will approach towards infinity as the value of K is increased.
Fig. 7.8 Centroid i.e. the point of intersection of the asymptotes.
The root loci from B and C will approach each other along the real axis and break away
towards infinity at a point on the real axis in between points B and C which is the centroid. The
centroid of the asymptotes is calculated as
Sum of real parts of poles − Sum of real parts off zeros
P−Z
−5 − 4 + 0 − 0 −9
=
= −3
−σ A =
3−0
3
Rule 8: Breakaway points—The root locus between two adjacent poles approaching each
other break away on the real axis at a point on the real axis and move towards infinity as the
value of K increases. For determining the breakaway point, we write the characteristic equation
in terms of K and evaluate dK/ds = 0. The breakaway points will be either on the real axis or must
occur as complex conjugate pairs. This is illustrated through examples as follows:
K
G(s)H(s) =
s ( s + 1)( s + 2)
−σ A =
The characteristic equations is 1 + G(s)H(s) = 0,
or
or
s(s + 1)(s + 2) + K = 0
that is,
The roots are
K = –s3 – 3s2 – 2s
dK
= –3s2 – 6s – 2 = 0
ds
3s2 + 6s + 2 = 0
s1, s2 =
−6 ±
62 − 4 × 3 × 2
2×3
= –0.43, –1.57
Chapter 7.indd 12
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tHE rOOT lOCUS tECHNIQUE
7-13
The root locus sketch is shown in the following:
Fig. 7.9 Determination of breakaway point of the root locus on the real axis.
P1, P2 and P3 are the positions of the poles at s = 0, s = –1 and s = –2, respectively. There is
no zero in the numerator of the transfer function. The presence of root locus on the real axis will
be to the left of odd number of poles and zeros. So there will be no part of the root locus between
point –1 and –2 as has been shown in Figure 7.9. So, the breakaway point of root locus will be at
–0.43 and not at –1.57.
The intersection of the root locus on the imaginary axis also has to be found out since the
characteristic equation can have real as well as complex multiple roots.
The intersection of the root locus on the imaginary axis is calculated using the characteristic
equation and forming the Routh’s array as
s(s + 1)(s + 2) + K = 0
or
s3 + 3s2 + 2s + K = 0
The Routh array is represented as
s3
1
2
2
s
3
K
6
−
K
0
s1
3
s0
K
0
Chapter 7.indd 13
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CONTROL SYSTEMS ENGINEERING
The value of K for which the root locus crosses the imaginary axis is calculated by equating
the terms of the first column of Routh array of s1 and s0 to zero.
So we write,
6−K
= 0 and K = 0
3
Thus,
K=6
The auxiliary equation is
3s2 + K = 0
The roots of the auxiliary equation are dominant roots which are close to the imaginary axis
or on the imaginary axis.
Thus,
3s2 = –K = –6
or
s2 = –2 or s = j 2
Number of asymptotes= Number of open-loop poles
= 3.
The angles are
(2q + 1)180°
q = 0,1, 2
φA =
P−z
=
=
=
180
3
= 60°
(2 + 1)180°
3
for q = 0
= 180°
(2 × 2 + 1)180°
3
for q = 1
= 300°
for q = 2 2
Centroid
−σ A =
=
Σ Real parts of poles − Σ Real parts of zeros
P−z
(−1 − 2) − 0
3−0
= −1
Thus, the complete root locus is as shown in Figure 7.10.
Rule 9: Angle of departure of the root locus from a complex pole and the angle of
arrival at a zero—Angle of departure of the root locus from a complex pole is given as
φd = 180º – sum of angles of vectors drawn from other poles to this pole
+ sum of angles of vectors drawn to this pole from other zeros.
Let us illustrate this with an example.
Chapter 7.indd 14
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tHE rOOT lOCUS tECHNIQUE
7-15
Fig. 7.10 Determination of angle of departure of the root locus from a complex pole.
Let
G(s)H(s) =
K
2
s ( s + 2)( s + 6s + 2s )
The poles are at s1 = 0, s2 = –2, s3 = –3 + j4 and s4 = –3 – j4.
The positions of poles are shown in Figure 7.11. The angle of departure φ A is calculated as,
φ A = 180° − (127° + 104° + 90°)
= 180° − 321° = −141°
Additional Techniques
Determination of K on root loci
For determining the value of K at any point on the root locus, we can use the following:
Product of all vector lengths drawn from
the poles of G(s)H
H (s) to that point
K =
m
Product of all vector lengths drawn from
the zeroes of G(s) H (s) to that point
(2) Ascertainment of any point to be on root locus
For any point to be on the root locus in the s-plane, it has to satisfy the angle criterion and
magnitude criterion. First, we apply the angle criterion to check that any point in s-plane
(1)
Chapter 7.indd 15
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7-16
CONTROL SYSTEMS ENGINEERING
which satisfied the angle condition has to be on the root locus. Magnitude condition is used
only after confirming the existence of point on the root locus by angle condition.
Fig. 7.11 Method of determination of any point to be on the root locus.
Let us consider, for example, where
G(s)H(s) =
K
s ( s + 1)( s + 2)
Let us examine whether s = –0.5 lies on the root locus or not.
First we apply angle criterion as
∠G ( s ) H ( s ) at s = –0.5 = ±180°(2q + 1) where q = 0, 1, 2, …
∠G ( s ) H ( s ) =
K
(−0.5 + j 0)(−0.5 + 1)(−0.5 + 2)
=
K ∠0 °
180°0°0°
=
(i)
K
(−0.5 + j 0)(0.5 + j 0)(1.5 + j 0)
= −180°
Thus, the angle condition as in (i) above is satisfied and the point s = –0.5 lies on the root
locus. Now we will use magnitude condition as
G ( s ) H ( s ) = 1 at s = –0.5
K
−0.5 0.5 1.5
=1
or
K = 0.375
for this value of K, point s = –0.5 lies on the root locus.
Chapter 7.indd 16
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tHE rOOT lOCUS tECHNIQUE
7-17
Example 7.2 A block diagram representation of a unity feedback control system is as follows.
Fig. 7.12 Block diagram of a feedback control system.
For this system, sketch the root locus. Also determine the value of K so that the damping
ratio, ξ , of a pair of complex conjugate closed-loop poles is 0.5.
Solution
G(s) =
K
s ( s + 1)( s + 2)
For determining the open-loop poles, we equate the denominator of G(s) to 0.
∴
s(s + 1)(s + 2) = 0
(1)
There are three open-loop poles at s = 0, s = –1 and s = –2.
(2)
We know that the number of root locus asymptotes will be equal to number of open-loop
poles minus the number of open-loop zeros. Here there is no open-loop zero.
There will be three branches of the root locus originating respectively at s = 0, s = –1 and
s = –2.
(3)
The three branches of the root locus will move towards infinity, as K increases, along the
asymptotic lines whose angles with the real axis are
φA =
(2q + 1)180°
n−m
;
q = 0,1, 2
= (2q + 1)180° = 60°, 180°, 300°
(4)
The root locus exist on the real axis between s = 0 and s = –1; and s = –2 moving toward ∞.
(5)
The centroid −σ A is calculated as
−σ A =
=
Σ Real parts of poles − Σ Real parts of zeros
Number of poles − Number of zeros
(−1 − 2) − 0
3−0
= –1
(6) The breakaway points on the real axis is found by substituting dK/ds = 0.
The characteristic equation is
Chapter 7.indd 17
s(s + 1)(s + 2) + K = 0
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7-18
CONTROL SYSTEMS ENGINEERING
or
K = –s3 – 3s2 – 2s
dK
= –3s2 – 6s – 2 = 0
ds
that is,
3s2 + 6s + 2 = 0
s1, s2 =
−6 ±
62 − 4 × 3 × 2
2×3
= –0.43, –1.57
(7)
Intersection of the root locus on the imaginary axis is determined as follows.
The characteristic equation of the system is
s(s + 1)(s + 2) + K = 0
or
s3 + 3s2 + 2s + K = 0
The Routh array is
s3
1
2
s2
3
K
s1
s0
6−K
3
K
0
0
We know that the occurrence of a zero row in the Routh array indicates the presence of
symmetrically located roots in the s-plane.
For this,
6−K
=0
3
or
K=6
The auxiliary equation is
3s2 + K = 0
or
3s2 = –K = –6
or
s =± j 2
The position of poles, the asymptotes and the root locus plot have been shown in
Figure 7.13.
Chapter 7.indd 18
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tHE rOOT lOCUS tECHNIQUE
7-19
Fig. 7.13 Root locus plot for s 3 + 3s 2 + 2s + K = 0.
Note that for breakaway point at s = –1.57, the angle criterion is not satisfied and hence
cannot be considered.
Example 7.3 The block diagram of a unity feedback control system is shown in Figure 7.14.
Fig. 7.14 Block diagram of a feedback control system.
Draw the root locus diagram for the above represented control system.
Solution The open-loop transfer function is
G(s)H(s) =
K ( s + 2)( s + 3)
s ( s + 1)
The number of open-loop poles is 2 at s = 0 and s = –1.
The number of open-loop zeros is 2 at s = –2 and s = –3.
Therefore, the number of root locus asymptotes = 2 – 2 = 0.
The number of root locus branches will be 2 originating at s = 0 and s = –1.
Chapter 7.indd 19
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7-20
CONTROL SYSTEMS ENGINEERING
The root loci will terminate at the zeros at s = –2 and s = –3.
Now let us calculate the breakaway points. The characteristic equation is
1 + G (s)H(s) = 0
Substituting the actual values, we get
K ( s + 2)( s + 3)
s ( s + 1)
= –1
K =−
We have to make
dK
ds
s ( s + 1)
( s + 2)( s + 3)
=
−( s 2 + s )
s 2 + 5s + 6
=0
Thus,
or
or
dK
 −( s 2 + 5s + 6)(2 s + 1) + ( s 2 + 5)(2 s + 5) 
=
=0
ds
( s 2 + 5s + 6) 2


2s2 + 6s + 3 = 0
s1, s2 = –0.63, –2.36
The root loci branches originate at s = 0 and s = –1 and get terminated at zeros at s = –2
and s = –3. Thus, root locus exists between 0 and –1 and between –2 and –3. Thus, both the
breakaway points s = –0.63 and s = –2.36 are valid (breakaway and break-in points).
The values of K at s = –0.634 and s = –2.366 are calculated as
K1 =
(0.634)(0.366)
(1.366)(2.366)
= 0.07
and
K2 =
(−2.366)(−1.366)
(−0.366)(0.634)
= 13.93
The root locus plot has been shown is Figure 7.15. The two root loci originate at s = 0 and
at s = –1 on the real axis. As the value of K is increased, they approach each other and break
away at s = –0.634 with value of K = 0.07. As the value of K increases, the two root loci makes
semicircles and break in at s = –2.366 with value of K = 13.93. They get terminated at the two
zeros at s = –3 and s = –2 as shown.


2. 366 − 0. 634
The root locus is a circle with its centre at –1.5  i.e. =
+ 0 .634 


2
Chapter 7.indd 20
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tHE rOOT lOCUS tECHNIQUE
Fig. 7.15 Root locus of G(s) H(s) =
K (s + 2 )(s + 3 )
s(s + 1)
7-21
⋅
Example 7.4 The transfer function of a system with unity feedback has been shown in
Figure 7.16.
Fig. 7.16
A unity feedback control system.
Draw the root locus diagram for the above system.
Solution The open-loop poles are calculated from the equation s2 + 2s + 3 = 0.
s 1, s 2 =
−2 ±
22 − 4 × 1 × 3
2
= −1 ± j 2
So, there are two open-loop poles. The open-loop zero is at s = –2.
The number of branches of the root locus is equal to the number of open-loop poles, that is 2.
Number of asymptotes = Number of poles – Number of zeros
= 2 – 1
= 1
Angle of the asymptote
(2q + 1)180°
q=0
φA =
n−m
=
Chapter 7.indd 21
180°
1
= 180°
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CONTROL SYSTEMS ENGINEERING
Now, let us determine the break-in point. For this, we take the characteristic equation of the
closed-loop system as
1 + G(s)H(s) = 0
or
G(s)H(s) = –1
Substituting the values,
K ( s + 2)
2
s + 2s + 3
= –1
or
K =−
s 2 + 2s + 3
s+2
Now let us calculate dK/ds and equate it to zero.
dK
 ( s + 2)(2 s + 2) − ( s 2 + 2 s + 3) 
= −
=0
ds
( s + 2)2


 2 s 2 + 6 s + 4 − s 2 − 25 − 3 
= −
=0
2
(
s
+
2)


or
s2 + 4s + 1 = 0
s1, s2 = −2 ±
3 = −3.732, − 0.268
The value of K at s = –3.732 is
K =
s 2 + 2s + 3
s+2
=
(3.732)2 + 2 × 3.732 + 3
3.732 + 2
= 5.46
The break-in point at s = –3.732 lies between the position of zero at s = –2 and the infinity.
At s = –2, the value of K = ∞.
Thus, the two root locus branches originate respectively from −1 + j 2 and −1 − j 2
and break in at the real axis at s = –3.732 as K increases to a value of 5.46 as shown in
Figure 7.17.
The radius of the circle is 3 , that is 1.732. The angle of take-off can be calculated as 145°
as shown in Figure 7.17. Thus, the two branches starting from the open-loop zeros follow a circular
path with the increase of gain factor K. As the value of K becomes 5.46, they meet at the real axis at
Chapter 7.indd 22
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tHE rOOT lOCUS tECHNIQUE
7-23
s = –3.732 and then break away towards s = –2 and infinity respectively with increasing values
of K.
Fig. 7.17 The root locus plot of
K (s + 2 )
s2 + 2s + 2
⋅
Example 7.5 A unity feedback control system has an open-loop transfer function.
G(s) =
2
K
s(s + 4s + 13)
Draw the root locus as the value of K changes from 0 to ∞. Also find the value of K and the
frequency at which the root loci crosses the jω axis.
Solution We have
G(s) =
2
K
s(s + 4s + 13)
Number of open-loop poles = 3
Number of open-loop zeros = 0
Poles are at s1 = 0 and
s 2, s 3 =
−4 ±
42 − 4 × 13
2
= −2 ± j 3
The root locus has three branches.
The centroid of the asymptotes is
Chapter 7.indd 23
σA =
−2 − j 3 − 2 + j 3
3
=−
4
3
= −1.33
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CONTROL SYSTEMS ENGINEERING
The angles the asymptotes make with the real axis are
φA =
and
=
φA =
and
φA =
(2q + 1)180°
q = 0,1, 2
n−m
(2 × 0 + 1)180 °
= 60°
3−0
(2 × 1 + 1)180 °
3−0
= 180°
(2 × 2 + 1)180 °
3−0
= 300°
Let us examine if there is any breakaway point or not.
The closed-loop characteristic equation is 1 + G(s)H(s) = 0
or
K
2
s ( s + 4 s + 13)
= –1
or
or
K = −s 3 − 4s 2 − 13s
dK
ds
= −3s 2 − 8s − 13 = 0
3s2 + 8s + 13 = 0
s1, s2 =
−8 ±
64 − 156
6
=−
4
3
± j 92
This is an imaginary axis and hence there does not exist any breakaway point on the real
axis. Angle of departure from a pole is calculated as
φd = ±(2q + 1)180° − φ,
q = 0,1, 2, 
where φ is the angle contribution at a pole of all other poles and zeros.
The root locus plot has been shown in Figure 7.18 (a).
Angle of departure is [Figure 7.18 (b)]
φd = 180° − φ
φd = 180° – 213.7° = – 33.7°
Since
Chapter 7.indd 24
φ = (90° + 123.7°) = 213.7°
03/11/11 7:19 PM
tHE rOOT lOCUS tECHNIQUE
Fig. 7.18 (a) Root locus plot of G(s ) =
7-25
K
; (b) Determination of angle of departure
s(s + 4s + 13 )
of the root locus from a complex pole.
2
To calculate the value of K.
The characteristic equation is s3 + 4s2 + 13s + K = 0.
Chapter 7.indd 25
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CONTROL SYSTEMS ENGINEERING
The Routh array is
s3
1
13
s2
4
K
52 − K
0
s1
4
K
0
s0
Applying condition for stability
52 − K
4
>0
and
K>0
\
0 < K < 52
For a maximum value of K = 52, the root locus will cut the imaginary axis. If the value
of K equals 52, the system will be oscillatory. For values of K more than 52, the system will be
unstable. When K = 52, the frequency of oscillation is calculated as
4s2 + K = 0
or
52
= −13
s2 = −
4
Therefore,
s = ± j 13 = ± jω
ω = 13 = 3.6 rad/sec.
Example 7.6 The open-loop transfer function of a control system is given as
K
G(s)H(s) =
( s + 1)( s + 10)( s + 30)
Draw the root locus. Determine the value of K for which the system is critically damped
and also the value of K for which the system becomes unstable.
Solution From G(s)H(s) we find, P = 3 and Z = 0. The poles are at s = –1, s = –10 and
s = –30.
Since there are three poles, there will be three root locus branches.
(−1 − 10 − 30) − 0
41
σA =
Centroid
=−
= −13.67
3−0
3
The asymptotes originate at s = –13.67 and make angles of
(2q + 1)180 °
σA =
= 60°, 180° and 300°
n−m
The breakaway point is calculated using the characteristic equation as
(s + 1)(s + 10)(s + 30) + K = 0
Chapter 7.indd 26
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tHE rOOT lOCUS tECHNIQUE
7-27
or
s3 + 41s2 + 340s + 300 + K = 0
or
K = s3 – 41s2 – 340s – 300
or
dK
ds
= –3s2 – 82s – 340 = 0
3s2 + 82s + 340 = 0
s1, s2 =
−82 ±
(82)2 − (4 × 3 × 340)
2×3
= −5.1 and − 22.2
The poles are at s = –1, s = –10 and s = –30. The asymptotes are at angles of 60°, 180°
and 300°. They originate from s = –13.67. Two root loci originating from s = –1 with K = 0 and
s = –10 with K = 0 approach each other and break away at s = –5.1. The breakaway point at
s = –22.2 is discarded because this is not possible.
The root locus plot has been drawn as shown in Figure 7.19.
Fig. 7.19 Root locus for G(s)H(s) =
Chapter 7.indd 27
K
.
(s + 1)(s + 10 )(s + 30 )
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CONTROL SYSTEMS ENGINEERING
We write the Routh array for the equation s3 + 41s2 + 340s + 300 + K = 0
s3
1
340
s2
41
300 + K
340 × 41 − (300 + K )
s1
41
s0
300 + K
Condition for stability is that
13, 640 − 300 − K
>0
41
or
K < 13,640
Thus, the value of K for which the system will be unstable is 13,640 and above.
At s = –5.1, the loci leave the real axis at the breakaway point. The value of K corresponding
to this point of the root locus is the one at which the system is critically damped. Putting s = –5.1
in the expression for K,
K = –s3 – 41s2 – 340s + 300
= –(–5.1)3 – [41 (–5.1)2] – 340( –5.1) – 300
= 500.2
Thus, at K = 500.2, the system is critically damped.
Example 7.7 The open-loop transfer function of a system is given as
K
G(s)H(s) =
s( s + 1)( s + 3)
Draw the root locus diagram and establish the condition for stability with respect to the
value of gain K. Calculate the phase margin and gain margin when the value of gain is reduced,
say equal to 6.
Solution Open-loop poles are at s = 0, s = –1 and s = –3. There are no zeros. The number of
root locus branches are (P – Z) = 3 – 0 = 3.
The centroid σ A is calculated as
Σ (−1 − 3) − Σ 0
4
σA =
= − = −1.33
3−0
3
Angle of asymptotes φ A are calculated as
2q + 1
at q = 0,1, 2
φA =
n−m
= 60°, 180°, 300°
The characteristic equation is s(s + 1)(s + 3) + K = 0
or
s3 + 4s2 + 3s + K = 0
K = –[s3 + 4s2 + 3s]
Chapter 7.indd 28
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tHE rOOT lOCUS tECHNIQUE
or
dK
ds
7-29
= – [3s2 + 8s + 3] = 0
3s2 + 8s + 3 = 0
s =
−8 ± 82 − 4 × 3 × 3
2×3
s1, s2 = –1.33 ± 0.86
= –2.19, –0.47
Break away at s = –2.19 is not possible. The break away will be at s = –0.47. The Routh
array using the characteristic equation is written as
s3 + 4s2 + 3s + K = 0
3
s
1
3
2
s
4
K
12 − K
0
s1
4
K
0
s0
The condition for stability is K > 0
12 − K
>0
4
or
K < 12
Therefore, 0 < K < 12
At K = 12, the system is oscillating in nature. The value of s at K = 12 is calculated using
the Routh array as
4s2 + K = 0; or 4s2 + 12 = 0
or
s2 + 3 = 0
or
s = ±j3 = ±j1.732
using the above calculated values, the root locus plot is made as shown in Figure 7.20.
The root loci cross the jω axis at 1.732 when the value of K = 12 and the system becomes
oscillatory. If the value of K is reduced to 6, the system will be conditionally stable and we have
to locate a point on the root loci somewhat below j1.732. Let this point be at P. By trial and
error, the point P on the root loci has to be fixed such that its distance from all the poles when
multiplied equals gain K = 6. The distances of the three poles from the point P are 1.2, 1.6 and
3.15 respectively.
\
K = 6 = 1.2 × 1.6 × 3.15
Chapter 7.indd 29
03/11/11 7:19 PM
7-30
CONTROL SYSTEMS ENGINEERING
Fig. 7.20 Root locus of G(s) H(s) =
Gain margin =
Value of limiting K
Value of derived K
K
s(s + 1) (s + 3 )
=
12
6
.
=2
Gain margin in dB = 20log10(2) = 6 dB
To determine the phase margin for K = 6, we have to locate a point on the imaginary axis of
the root locus plot such that the magnitude of the transfer function for a frequency corresponding
to that point is 1.
That is,
G ( jω1 ) H ( jω1 ) = 1 for s = jω1
By trial and error, that is by repeated calculation, we will find that frequency ω1
Let ω1 = 2,
6
|G( j2)H( j2)| =
j2 2 + 1 j2 + 3
=
Chapter 7.indd 30
6
1.4 × 2.25 × 3.8
[This value is less than 1].
03/11/11 7:19 PM
tHE rOOT lOCUS tECHNIQUE
Let ω = 1
=
Let ω = 1.25,
6
G ( j1) H ( jω) =
=
j1 ( j1 + 1) j1 + 3
6
1 × 2 × 10
6
= 1. 35t
1 × 1.41 × 3. 1
6
G ( j1.25) H ( j1.25) =
=
7-31
j1.25 ( j1.25 + 1) ( j1.25 + 3)
6
1.25 × 2.56 × 10.56
= 1
Therefore, ω1 = 1.25 is calculated as the gain cross-over frequency.
Phase margin= 180G ( j1.25)H( j1.25)
= 180° + (angle contributed by the three poles at a point on
root having ω1 = 1.25)
Let θp1 , θp2 , θp3 be the angle contributed by the poles respectively. If angles contributed
by the poles are taken as negative,
Phase margin = 180° + (−θp1 − θp2 − θp3 )
1.25
1.25 

= 180° +  −90° − tan −1
− tan −1

1
3 

= 180° − (90 + 50° + 22°)
= 18°
Example 7.8 The transfer function of a unity feedback system is G(s) = K/s(s + 5). Draw the
root locus plot and determine the value of K for a damping ratio of 0.707.
Solution The two poles are at s = 0 and s = –5
There is no zero. Number of root locus branches are P – Z = 2 – 0 = 2. Since there is
no open-loop zero, both the root loci will terminate at ∞. We are to calculate the angle of the
asymptotes, φ A .
φA =
(2q + 1) × 180°
0−2
q = 0, 1
= 90º, 270º
Chapter 7.indd 31
03/11/11 7:19 PM
7-32
CONTROL SYSTEMS ENGINEERING
The asymptotes intersect at the negative real axis as the centroid, σ A .
Σ Real parts of poles − Σ Real parts of zeros
P−Z
(0 − 5) − 0
=
= −2.5
2 −1
To calculate the breakaway point, that is the point on the negative real axis where from the
root loci depart the real axis and become asymptotic and approach ∞, we write the characteristic
equation which is
s(s + 5) + K = 0
s2 + 5s + K = 0
or
K = – (s2 + 5s)
σA =
dK
ds
= – [2s + 5] = 0
5
= −2.5
2
This shows that the breakaway point and the point of intersection of asymptotes on real
axis is the same.
We have,
ωn cos θ = ξωn (see Fig. 8.20)
cos θ = ξ = 0.707
s =−
θ = cos−1 0.707 = 45°
We will draw the x-line at 45° with the negative real axis as shown in Figure 7.21. The
x-line intersects the root locus at R.
Fig. 7.21 Determination of gain K for a particular value of damping ratio.
Chapter 7.indd 32
03/11/11 7:19 PM
tHE rOOT lOCUS tECHNIQUE
7-33
At point R, the value of s = –2.5 + j2.5
Since R is on the root locus, the magnitude criterion of G ( s ) H ( s ) = 1 will be satisfied.
Therefore,
G(s) H (s) = 1
or
K
s ( s + 5)
=1
or
K
(−2.5 + j 2.5) [(−2.5 + j 2.5) + 5]
=1
or
K
(−2.5 + j 2.5)(−2.5 + j 2.5)
=1
or
or
that is,
K
[(6.25 + 6.25)(6.25 + 6.25)]1/2
K
12.5
=1
=1
K = 12.5
Example 7.9 The transfer function of a unity feedback control system is given as
K
G(s) =
s(s + 2)(s + 3)
Calculate the value of K for a damping ratio of 0.707. Also calculate the value of K for
which the system will become oscillatory. Calculate the value of K at s = –4 and –5.
Solution The poles are at s = 0, –2 and –3. There is no zero. Number of branches of the root
locus is, P – Z = 3 – 0 = 3. We know that the poles terminate at zeros or infinity. As there is no
zero, all the three root loci originating from the poles will terminate at ∞.
The characteristic equation is
1 + G(s) = 0
or
K
1+
=0
s ( s + 2)( s + 3)
or
s(s + 2)(s +3) + K = 0
Chapter 7.indd 33
03/11/11 7:19 PM
7-34
CONTROL SYSTEMS ENGINEERING
or
K = –[s(s + 2)(s + 3)]
= – [s3 + 5s2 + 6s]
Substituting dK/ds = 0,
dK
= –[3s2 + 10s + 6] = 0
ds
3s2 + 10s + 6 = 0
s =
−10 ± 10 2 − 4 × 3 × 6
10
2×3
5.3
= −0.8, − 2.54
6
6
The angle of asymptotes, φ A , is calculated as
(2q + 1)180°
φA =
q = 0,1, 2
p−2
(2 × 0 + 1)180°
=
= 60° for q = 0
3
for q = 1 and 2, φ A = 180° and 300°
s =−
±
The point of intersection of the asymptotes with the real axis, that is, the centroid σ A , is
calculated as
Σ Poles − Σ Zeros (0 − 2 − 3) − 0
σA =
=
P−Z
3−0
5
= − = −1. 67
3
The value of K for which the system is stable is calculated by applying Routh–Hurwitz
criterion using the characteristic equation.
Consider the characteristic equation,
s(s + 2)(s + 3) + K = 0
or
s3 + 5s2 + 6s + K = 0
The Routh array is
s3
s2
s1
1
5
30 − K
6
K
0
5
s0
K
0
For stability, all the terms of the first column must be positive.
Hence, K > 0
Chapter 7.indd 34
03/11/11 7:19 PM
tHE rOOT lOCUS tECHNIQUE
and
7-35
30 − K
> 0, that is, K < 30.
5
Therefore, the system is stable if 0 < K < 30.
At K = 30, the system is oscillatory. For K < 30 all the roots will lie on the left-hand side
of s-plane. If K exceeds 30, the roots will lie on the right-hand side of the s-plane. At K = 30, the
locus will intersect the imaginary axis. The value of s at which the locus cuts the imaginary axis
is calculated from the Routh array’s second row as
5s2 + K = 0
or
5s2 + 30 = 0
or
s2 = 6
or
s = ± j 6 = ± j 2.45 = ± jωn
The frequency of oscillation is equal to the distance from origin the point of intersection of
the root locus with the jω axis.
The frequency of oscillation, ωn = 2.45 rad/sec.
The root locus plot has been shown in Figure 7.22.
Fig. 7.22 Root locus of unity feedback control system with G(s) H(s) =
Chapter 7.indd 35
K
s(s + 2 ) (s + 3 )
⋅
03/11/11 7:19 PM
7-36
CONTROL SYSTEMS ENGINEERING
Now we have to calculate the value of K for damping ratio of 0.707.
ωn cos θ = ξωn
We have
that is,
cos θ = ξ = 0.707
q = cos−1 0.707 = 45°
We draw the x line with an angle of 45° with the negative real axis. The x-line cuts the root
locus at a point R. The value of s for that point is –0.6 + j0.8.
Since point R for which s = –0.6 + j0.8 lies on the root locus, the magnitude condition must
be satisfied which is
|G(s)H(s)| = 1
Substituting,
K
=1
s ( s + 2)( s + 3)
K
=1
(−0.6 + j 0.8)(−0.6 + j 0.8 + 2)(−0.6 + j 0.8 + 3)
or
(0.6) + (0.8)
2
2
(
K
(1.4) + (0.8)
2
2
)(
( 2.4) + (0.8)
2
2
)
=1
or
K
1 × 1.6 × 2.5
=1
or
K=4
The values of K at s = –4 and –5 can be calculated by substituting the value of s in the
transfer function as
K
=1
s ( s + 2)( s + 3)
s = –4
Substitute
K
−4(−4 + 2)(−4 + 3)
K=8
s = –5,
or
Substitute
or
Chapter 7.indd 36
=1
K
−5(−5 + 2)(−5 + 3)
=1
K = 30
03/11/11 7:20 PM
tHE rOOT lOCUS tECHNIQUE
7-37
We note that at s = –3, K is 0 as the root loci originating from the pole at s = –3 approach
infinity along the asymptote at 180°, the value of K increases, becoming K = 8 at s = –4 and
K = 30 at s = –5.
Example 7.10 The open-loop transfer function of a control system is given as
K
G(s)H(s) =
2
s( s + 4 )( s + 4 s + 15 )
Draw the root locus plot and calculate the value of K for which the system becomes
oscillatory.
Solution There are four poles at s = 0, s = –4 and
s =
−4 ±
4 2 − 4 × 15 × 1
2
= − 2 ± j 3.3
There is no zero. Hence the number of branches of the root locus is 4. There are four
asymptotes whose angles are
φA =
( 2q + 1) 180°
P−Z
,
q = 0, 1, 2, 3
= 45°, 135°, 225°, 315°
The intersection of the asymptotes on the negative real axis is calculated as
Σ Poles − Σ Zeros (0 − 4 − 2 − 2) − 0
σA =
=
= −2
P−Z
4
We have now to calculate the breakaway point and the intersection of the root locus with
the imaginary axis and the corresponding value of K.
The characteristic equation is
s(s + 4)(s4 + 4s +15) + K = 0
or
s4 + 8s3 + 31s2 + 60s + K = 0
First, we will find the value of K when the root locus crosses the imaginary axis, by applying
Routh’s criterion.
s4 + 8s3 + 31s2 + 60s + K = 0
4
1
31
K
s
3
s
8
60
2
s
23.5
K
s1
23.5 × 60 − 8K
23.5
s0
K
For stability, all the terms of the first column must be positive.
Hence, K > 0
Chapter 7.indd 37
03/11/11 7:20 PM
7-38
CONTROL SYSTEMS ENGINEERING
and
or
or
23.5 × 60 − 8K
23.5
=0
60 – 0.33K > 0
K > 180
At K = 180, the system will be oscillatory.
For K = 180, the value of s is calculated from the third row of the Routh array as
23.5s2 + K = 0
or
23.5s2 + 180 = 0
or
s = ± j 7.7 = ± j 2.7
To locate the breakaway point, we have to start with the characteristic equation as
Substituting dK/ds = 0,
K = −[ s 4 + 8s 3 + 31s 2 + 60 s ]
dK
= −[4s 3 + 24s 2 + 62s + 60]
ds
4s3 + 24s2 + 62s + 60 = 0
From this equation, we have to calculate the three values of s. From the root locus plot
shown in Figure 8.23 (b), we notice that one of the breakaway points is at s = –2. To determine
the other two breakaway points, we divide the above equation by (s + 2).
We get
or
4s 3 + 2424s 2 + 62s + 60
s+2
=0
4s2 + 16s + 30 = 0
s =
−16 ± 162 − 4 × 4 × 30
2×4
= −2 ± j 3.5 = −2 ± j1.87
With the above calculated values, the complete root locus plot has been drawn as shown
in the second diagram given on next page. We have, however, also to calculate the angle of
departure of the root locus from the complex poles at (–2 + j3.3) and (–2 – j3.3). For this, we have
to refer to the location of the poles and zeros.
Since there is no zero in this case, the location of the poles is as shown in Figure 7.23 (a).
Chapter 7.indd 38
03/11/11 7:20 PM
tHE rOOT lOCUS tECHNIQUE
Fig. 7.23 (a) Location of poles; (b) Root locus diagram of G(s) H(s) =
φd + (θp1 + θ2 + θp3 ) = 180º
Here,
3.3
θp1 = tan −1
θp2 = 180° − tan −1
2
7-39
K
s(s + 4) (s 2 + 4s + 15 )
⋅
= 59 °
3.3
2
= 180° – 59° = 121°
θp3 = 90º
Therefore,
φd = 180° – (59° + 121° + 90°)
= 180° – 270°
= –90°
The angle of departure of the root loci from the other conjugate pole will be 90°.
Chapter 7.indd 39
03/11/11 7:20 PM
7-40
CONTROL SYSTEMS ENGINEERING
Example 7.11 The open-loop transfer function of a control system is given as
K
G(s)H(s) = 2
s ( s + 4)
Draw the root locus plot for the system.
Determine the value of K for which the system is stable.
Solution Given
G(s)H(s) =
K
2
s ( s + 4)
Number of zeros= 0
Number of poles= 3 at s = 0 and s = –4
There would be three asymptotes for three root locus branches.
(−4)
σA =
Centroid
= −1.33
3
φA =
(2q + 1)180°
n−m
q = 0,1, 2
= 60°, 180°, 300°
Closed-loop characteristic equation is
s2(s + 4) + K = 0
or
s3 + 4s2 + K = 0
or
K = – s3 – 4s2
dK
ds
= – 3s2 – 8s = 0
8
s1, s2 = 0, − , that is, − 2.67
3
The two roots originating at s = 0 break away at s = 0 and move towards ∞ as K → ∞. The
other root at s = –4 approaches ∞ as K → ∞.
The root locus plot has been shown in Figure 7.24. Using the characteristic equation, the
Routh array is written as
s3
1
0
2
s
4
K
K
0
s1
−
4
K
s0
From the above, it is seen that the system is unstable for every value of K.
Chapter 7.indd 40
03/11/11 7:20 PM
tHE rOOT lOCUS tECHNIQUE
Fig. 7.24 Root locus plot for G(s ) H (s ) =
K
s 2 (s + 4 )
7-41
⋅
Example 7.12 Plot the root locus for a system having open-loop transfer function of
G(s) =
K
s(s + 1)2
Determine the value of K for which the system response will become oscillatory. Also
calculate the frequency of oscillation at that value of K.
Solution
Given
G(s) =
K
s(s + 1)2
Number of zeros= Nil
Number of poles= 3 at s1 = 0, s2,3 = –1, –1
There are three branches of the root locus.
The centroid is
σA =
φA =
(−1 − 1)
3
= −0.67
(2q + 1)180°
3
q = 0,1, 2
= 60°, 180°, 300°
Chapter 7.indd 41
03/11/11 7:20 PM
7-42
CONTROL SYSTEMS ENGINEERING
Breakaway point of the root locus is calculated by using the characteristics equation, which
is
s(s +1)2 + K = 0
or
s3 + 2s2 + s + K = 0
or
K = –s3 – 2s2 – s
dK
= −3s 2 − 4s − 1 = 0
ds
or
3s2 + 4s + 1 = 0
s1, s2 =
−4 ±
42 − 4 × 3
=
2×3
= –1, –0.33
Breakaway point of –0.33 is to be considered.
The root locus plot has been shown in Figure 7.25.
−4 ± 2
6
Fig. 7.25 Root locus plot for G(s ) =
K
s(s + 1)2
.
One root locus starts from s = –1 and goes towards ∞ as K increases. The other two
branches, one from s = –1 and s = 0, approach each other on the real axis and break away at
s = –0.33, crosses the jω axis and moves towards ∞ with increase in value of K.
Chapter 7.indd 42
03/11/11 7:20 PM
tHE rOOT lOCUS tECHNIQUE
7-43
To determine the value of K for stable operation, we write the Routh array using the
characteristic equation.
The characteristic equation is
s3 + 2s2 + s + K = 0
The Routh array is
s3
1
1
s2
2
K
2−K
0
s1
2
K
0
s0
For stable operation,
K>0
and
2−K
>0
2
or
K<2
Thus, the condition for stability is 0 < K < 2.
At K = 2, the root locus crosses the jω axis where the system is oscillatory.
We write the auxiliary equation from the Routh array
2s2 + K = 0
or
2s2 + 2 = 0
s2 = –1
or
s = ± j1
The root locus cuts the jω axis at +j1 and at –j1.
Frequency of oscillation s = ± jω. Therefore, w = 1 rad/sec.
Example 7.13 A feedback control system has an open-loop transfer function
K (s + 1)
G(s)H(s) = 2
s (s + 4)
Draw the root locus as K varies from zero to infinity.
Solution Given
G(s)H(s) =
The poles are at s = 0, 0, –4
The zero is at s = –1
Chapter 7.indd 43
K (s + 1)
s 2 (s + 4)
03/11/11 7:20 PM
7-44
CONTROL SYSTEMS ENGINEERING
There are three branches of the root locus.
Centroid
Σ Real parts of poles − Σ Real parts of zeros
σA =
Number of poles − Number of zeros
−4 − (−1)
=
= −1.5
3 −1
The angle φ A of the asymptotes are
(2q + 1)180°
φA =
n−m
at q = 0,1
= 90° and 270°
The breakaway point of the root loci is calculated using the characteristic equation,
which is
s2(s + 4) + K(s + 1) = 0
or
s3 + 4s2 + Ks + K = 0
K =−
s3 + 4s 2
s +1
 (s + 1) (3s 2 + 8s) − (s 3 + 4s 2 ) × 1 
= −
=0
ds
(s + 1)2


dK
or
or
Therefore,
and
2s3 + 7s2 + 8s = 0
s(2s2 + 7s + 8) = 0
s1 = 0
s2 =
Break away at s2 = −
7
4
±
−7 ±
49 − 64
4
=−
7
4
±
−15
−15 is not applicable.
The two root loci break away at s = 0 and become asymptotic as the value of K increases as
shown in Figure 7.26. The value of K for stability is found using Routh array as
s3 + 4s2 + Ks + K = 0
Chapter 7.indd 44
s3
1
K
s2
4
K
03/11/11 7:20 PM
tHE rOOT lOCUS tECHNIQUE
s1
s0
4K − K
4
K
For stability
and
7-45
0
0
K>0
4K − K
4
> 0, that is,
3K
4
> 0 or K > 0
The system is, therefore, stable for all values of K.
Fig. 7.26 Root locus plot for G(s )H (s ) =
K (s + 1) .
s 2 (s + 4 )
Example 7.14 A feedback control system has an open-loop transfer function
G(s)H(s) =
K (s + 4 )
s 2 (s + 4 )
Draw the root locus as K varies from zero to infinity and comment on its stability (Note the
difference in TF of this problem with respect to the TF of the previous problem).
Solution Given
Chapter 7.indd 45
G(s)H(s) =
K (s + 4 )
s 2 (s + 4 )
03/11/11 7:20 PM
7-46
CONTROL SYSTEMS ENGINEERING
The number of zeros z = 1 at s = –4; The number of poles p = 3 at s = 0, 0, –1.
(s 3 + s 2 )
s 2 ( s + 1)
=−
( s + 4 ) at 0, 0, and –1 in the s-plane. One root locus will terminate at
s+4
the zero at s = –4. The other two will terminate at infinity.
=−
The centroid is
σA =
φA =
−1 − (−4)
3 −1
= +1.5
(2q + 1)180°
n−m
at q = 0,1
= 90° and 270°
The characteristic equation is
s2(s +1) + K(s + 4) = 0
or
s3 + s2 + Ks + 4K = 0
or
K =−
(s 3 + s 2 )
s 2 ( s + 1)
=−
(s + 4 )
s+4
Breakaway point
dK
 ( s + 4)(3s 2 + 2 s ) − ( s 3 + s 2 ) 
= −
=0
ds
( s + 4)2


or
2s3 +13s2 + 8s = 0
s=0
\
Consider the characteristic equation and apply Routh–Hurwitz criterion,
s3 + s2 + Ks + 4K = 0
The Routh array is
s3
1
K
s2
1
4K
s1
–3K
0
s0
4K
0
This shows that the system is unstable. The root locus plot has been shown in Figure 7.27.
Chapter 7.indd 46
03/11/11 7:20 PM
tHE rOOT lOCUS tECHNIQUE
Fig. 7.27 Root locus plot for G(s )H (s ) =
7-47
K (s + 4 ) .
s 2 (s + 1)
Example 7.15 A feedback control system has open-loop transfer function
G(s)H(s) =
K ( s + 30)
( s + 1)( s + 10)( s + 20)
Draw the root locus as K varies from 0 to ∞.
Solution On examining G(s)H(s), we find
P = 3 at s = –1, –10, and –20
Z = 1 at s = –30
Number of root locus branches = P = 3.
Centroid σ A is calculated as
σA =
Σ (−1 − 10 − 20) − Σ (−30)
3 −1
= –0.5
φA =
(2q + 1)180°
n−m
at q = 0,1, 2
= 90°, 270°
The characteristic equation is
(s + 1)(s + 10)(s + 20) + K(s + 30) = 0
Chapter 7.indd 47
03/11/11 7:20 PM
7-48
CONTROL SYSTEMS ENGINEERING
or
s3 + 31s2 + (230 + K)s + (200 + 30K) = 0
or
K =−
s 3 + 31s 2 + 230s + 200
s + 30
Equating dK/ds = 0, breakaway point is calculated as s = –5.2.
The Routh array using the characteristic equation,
s3 + 31s2 + (230 + K) s + (200 + 30K) = 0, is
s3
1
230 + K
s2
31
200 + 30K
s1
s0
6930 + K
31
200 + 30K
0
0
This shows that system is inherently stable for all values of K. The root locus is shown in
Figure 7.28.
Fig. 7.28 Root locus of G(s )H (s ) =
K (s + 30 )
(s + 1)(s + 10 )(s + 20 )
.
Example 7.16 The open-loop transfer function of a feedback control system is given as
K
G(s)H(s) =
s ( s + 3)( s 2 + 2 s + 2)
Chapter 7.indd 48
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tHE rOOT lOCUS tECHNIQUE
7-49
Draw the root locus as K varies from 0 to ∞ . Also calculate the value of K for which the
system becomes oscillatory.
Solution Number of open-loop poles = 4. They are at s = 0, –3, –1 ± j1. There is no finite
open-loop zero.
There are four root locus branches originating at the four poles with value of K = 0. These
branches will terminate at ∞ (i.e., open-loop zeros). This is because there are no finite zeros.
The four branches will tend to reach infinity with the value of K increasing towards an
infinite value along asymptotic path.
The angles the asymptotes with the real axis are calculated as
φA =
=
(2q + 1)180°
m− m
at q = 0,1, 2,3
(2q + 1)180°
= 45°,135°, 225°, 315°
4−0
The centroid of the asymptotes is calculated as
σA =
Σ (−3 − 1 − 1) − Σ (0)
= −
5
4
4−0
= −1.25
Therefore, the asymptotes will be originating at s = –1.25 on the real axis.
The breakaway point or points are calculated using the characteristic equation, which is
1+ G(s)H(s) = 0
or
K
= –1
s (2 + 3)( s 2 + 2 s + 2)
or
K = –s(s + 3)(s2 + 2s + 2)
= – (s4 + 5s3 + 8s2 + 6s)
To calculate s, we have to equate
dK
ds
dK
ds
=0
= −(4s3 +15s2 +16s + 6) = 0
s = –2.3, –0.725 ± j0.365
The breakaway point must be at s = –2.3 as it lies on the real axis. The two root locus
branches originating at s = 0 and s = –3 approach each other on the real axis and break away at
s = –2.3. The root locus plot is shown in Figure 7.29.
Chapter 7.indd 49
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CONTROL SYSTEMS ENGINEERING
Fig. 7.29 Root locus plot of G(s )H (s ) =
K
s(s + 3 )(s 2 + 2 s + 2 )
.
For calculating the value of K at which the root locus intersects the imaginary axis and the
system becomes oscillatory, we start with the characteristic equation and the Routh array.
The characteristic equation is
or
s(s + 3)(s2 + 2s +2) + K = 0
s4 + 5s3 + 8s2 + 6s + K = 0
Applying Routh criterion,
s4
1
8
s3
5
6
2
s
34/5
K
K
0
0
s1
 204/5 − 5K 


5


34
5
0
0
s0
K
0
0
For stability, 0 < K < 8.16.
At a value of K = 8.16, the two root loci intersect the jω axis.
The point of intersection is calculated from the auxiliary equation formed from the
coefficients of s2 row when K = 8.16 as
Chapter 7.indd 50
34
5
s2 + K = 0
03/11/11 7:20 PM
tHE rOOT lOCUS tECHNIQUE
or
s2 = −
K
34/5
=−
8.16 × 5
34
=−
40.8
34
7-51
= −1.21
s = ± j1.1
The angle of departure of the two root loci originating at s = –1 ± j1 is calculated as
∴
φd = 180° − φ
where f is the angle contribution by other poles to this pole.
φd = 180° – (90 + 135° + 27°)
= –72°
Fig. 7.30 Calculation of angle of departure of the root locus from complex poles.
Similarly, the angle of departure of the root locus from the other pole at (–1 –j1) can be
calculated to be +72°.
Example 7.17 Find the roots of the following polynomial by root locus method.
3s4 + 10s3 + 21s2 + 24s – 16 = 0
Solution We will rearrange the polynomial into P(s)/Q(s) form in the following way.
or
3s4 + 10s3 + 21s2 = –24s + 16 = –(24s – 16)
4
24s − 16
3s + 10 s 3 + 21s 2
= –1
or
Chapter 7.indd 51

2
8s − 
3

s 2 (3s 2 + 10s + 21) ×
1
3
= –1
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CONTROL SYSTEMS ENGINEERING
Let us assume K = 8
Therefore,

2
K s − 
3

 2 10
 = –1
2
s s +
s + 7
3


This is of the form,G(s)H(s) = –1
Number of open-loop poles = 4 at s = 0, 0, –1.67 ± j2.06
Number of zeros is equal to 1 at s = 0.67
Number of branches of the root locus = 4
Number of asymptotes = P – Z = 4 – 1 = 3
Centroid
σA =
Σ (−1.67 − 1.67) − Σ (−0.67)
3
= –1.33
Angle of asymptotes
φA =
(2q + 1)180°
= 60°, 180°, 300°
3
Angle of departure of the root locus from –1.67 ± j2.06 is calculated as
φd = ±[180° – S angles from other poles + S angles from zeros]
= ±[180° – (90° + 128° + 128°) + 138°] = ± 30°
To calculate the value of K when the root locus intersects the imaginary axis and the points
of intersection is calculated using the characteristic equation as
10 3
2
s4 +
s + 7 s 2 + Ks − K = 0
3
3
The Routh array is shown below
s4
1
7
–2/3K
s3
10/3
K
0
2
3 

− K
0
s2
K
7 −
3
10 

s1
s0
Chapter 7.indd 52
 3 2 83 
K +
K
−
9 
 10
3 

K
7 −
10


−
2
3
K
0
0
0
0
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tHE rOOT lOCUS tECHNIQUE
7-53
The value of K is calculated from s1 term as
3 2 83
−
K +
K =0
10
9
or
K = 30.7
Condition for stability 0 < K < 30.7
The value of s for K = 30.7 is calculated from the auxiliary equation as
3  2 2

Ks − K =0
7 −
10 
3

or
s = ±j3.04.
The root locus has been drawn using the data obtained as shown in Figure 7.31.
Fig. 7.31 Root locus plot of Example 7.17.
The above Solution has been done assuming K = 8. We have to by trial and error locate the
points on the root loci where K = 8 and from there the roots can be located. From the root locus,
Chapter 7.indd 53
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CONTROL SYSTEMS ENGINEERING
the roots are found as –2.25, –0.75 ± j2.15 for value of K = 8 as shown by using dots on the root
locus.
Example 7.18 The simplified form of open-loop transfer function of an aeroplane with an
autopilot is represented as
K ( s + a)
G(s)H(s) =
s ( s − b)( s 2 + 2ξωn s + ω2n )
Sketch the root locus plot when a = b = 1, x = 0.5 and ωn = 4. For what value of K the
system will remain stable?
Solution The open-loop transfer function after substituting the values of a, b, x and ωn is
G(s)H(s) =
K ( s + 1)
s ( s − 1)( s 2 + 4s + 16)
Number of open-loop poles = 4 at s = 0, +1, –2 ± j3.5, Number of zeros = 1 at s = –1
Number of root locus branches = Number of poles = 4
Number of asymptotes = Number of poles – Number of zeros = 4 – 1 = 3.
Angles of asymptotes with the real axis,
φA =
(2q + 1)180°
3 −1
q = 0,1, 2
= 60°, 180°, 300°
The asymptotes intersect on the real axis at the centroid σ A
Σ Real parts of poles − Σ Real parts of zeros
σA =
Numbber of poles − Number of zeros
Σ (0 + 1 − 2 − 2 ) − Σ (−1)
4 −1
2
= − = −0.67
3
The root locus diagram with the position of poles and zeros and the asymptotes are shown
in Figure 7.32.
We will now find breakaway points using the characteristic equation, which is
s(s – 1)(s2 + 4s + 16) + K(s + 1) = 0
or
σA =
K =−
dK
ds
=−
s ( s − 1)( s 2 + 4s + 16)
( s + 1)
3s + 10s 3 + 21s 2 + 24s − 16
4
( s + 1)2
=0
From which,
s1, s2, s3, s4 are calculated as equal to 0.45, –2.26, –0.76 + j2.16 and –0.76 – j2.16,
respectively. Breakaway points are on the real axis and are at 0.45 and –2.26.
Chapter 7.indd 54
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tHE rOOT lOCUS tECHNIQUE
7-55
Fig. 7.32 Root locus plot of Example 7.18.
Now we locate the breakaway points on Figure 7.32.
Now let us determine the value of K for which the root locus intersects the imaginary axis.
The characteristic equation is
s4 + 3s3 + 12s2 + (K – 16)s + K = 0
Routh array is represented as
s4
1
12
K
s3
3
K – 16
0
K
0
(52 − K )
0
0
k
0
0
s2
s1
s0
52 − K
3
(−K 2 + 59K − 832)
The value of K at the intersection on the imaginary axis by the root locus is calculated as
Chapter 7.indd 55
− K 2 + 59 K − 832
52 − K
=0
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7-56
CONTROL SYSTEMS ENGINEERING
or
K2 – 59K + 832 = 0
K =
59 ±
(59) 2 − 4 × 432
2
59 ± 153
= 35.75, 23.25
2
The shows that the system is stable for
35.75 < K < 23.25
For calculating the crossing points on the imaginary axis, we will use the auxiliary equation
from Routh array of s2 as
=
(52 − K ) s 2
3
+K =0
or
s = ±j2.25 for K = 35.75
and
s = ±j1.55 for K = 23.25
The complete root locus diagram can now be plotted as shown in Figure 7.32.
7.2effect of adding poles and zeros to a(s) h(s) on the root
loci
Often the desired performance specifications of a closed-loop system are not met by merely
adjusting the gain parameter. Once the effects on the root locus of addition of poles and/or zeros
are fully understood, we can readily determine the poles and zeros of a compensator that will
reshape the root locus to meet the desired performance specifications.
The addition of a pole to the open-loop transfer function has the effect of shifting the root
locus to the right, thereby decreasing relative stability and increasing settling time (slow). This
effect may be compared with that of integral control which makes the system less stable by
adding a pole at the origin.
Figure 7.33 (b) and (c) show, respectively, the effects of adding one pole and two poles to
the single-pole system of Figure 7.33 (a). The root locus gradually bends towards right.
The addition of a zero to the open-loop transfer function has the effect of shifting the root
locus to the left, thereby increasing stability and decreasing settling time (fast). This effect may
be compared with that of derivative control which makes the system fast and more anticipatory
by adding a zero in the feed forward transfer function. Figure 7.34 (a) shows the root locus of
a system which is unstable for large gain. Figure 7.34 (b) through (d) show root locus plots for
different locations of a zero added to the open-loop transfer function. Here it is evident that the
inclusion of a zero makes the system of Figure 7.34 (a) stable for all values of gain.
Chapter 7.indd 56
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tHE rOOT lOCUS tECHNIQUE
7-57
Fig. 7.33 Root locus plots of single-, double- and triple pole systems.
The effect of adding poles and zeros to G(s)H(s) can be understood by solving few problems.
Fig. 7.34
7.7
Root locus plots for different locations of a zero added to the three-pole system in (a).
Root Locus Plot with MATLAB
For plotting root loci with MATLAB, the characteristic equation must be written in the form
given by
num
=0
1+ K
den
where num and den are the numerator and denominator polynomials, respectively.
A commonly used MATLAB command of plotting root locus is ‘rlocus (num, den)’ with
the gain vector K automatically determined. For plotting root loci with marks ‘0’ or ‘x’, the
commands to be followed are
r = rlocus (num, den);
plot (r, ‘0’) or plot (r, ‘x’);
Chapter 7.indd 57
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CONTROL SYSTEMS ENGINEERING
MATLAB uses the automatic axis scaling feature of plot command.
For manual axis scaling, the curve plotting limits are to be specified by
v = [x-min
x-max
y-min
y-max]
‘axis (v)’ is the command to be entered next. Typing ‘axis’ resumes auto scaling. Entering
‘axis (‘square’)’ command sets the plotting region to square. Entering axis (‘normal’) will get
back into normal mode. The common axis (‘square’) will set a line with slope 1 at a true 45°, not
skewed by irregular shape of the screen.
Example 7.19 For a unity feedback control system with
K ( s + 1)
G(s) =
s ( s − 1)( s 2 + 4s + 16)
plot the root loci with a square aspect ratio so that a line with slope 1 is a true 45° line.
Solution The following program will generate the plot as shown in Figure 7.35.
MATLAB PROGRAM 7.1
% ————— Root-locus plot ————————
num = [0 0
0
1
1];
3
12
–16
0];
den = [1
rlocus (num, den)
v = [–6
6
–6
6]; axis (v); axis (‘square’)
grid
title (‘Root-locus plot of G(s) = K(s + 1)/[s(s – 1)(s2 + 4s + 16)])
Fig. 7.35 Root locus plot.
Note that the hand-copy plot may not be a square one as it depends on the printer.
Chapter 7.indd 58
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tHE rOOT lOCUS tECHNIQUE
7-59
Example 7.20 Using MATLAB, plot the root loci and their asymptotes for the system whose
open-loop transfer function is given by
G(s)H(s) =
K
s( s + 1)( s + 2 )
Solution The open-loop transfer function may be written as
G(s)H(s) =
K
3
s + 3s 2 + 2 s
It may be noted that
lim
s →∞
K
3
2
s + 3s + 2s
lim
s →∞
K
3
2
s + 3s + 3s + 1
= lim
s →∞
K
( s + 1)3
So the equation for the asymptotes may be given by
Ga(s)Ha(s) =
K
( s + 1)3
The required MATLAB program is as follows.
MATLAB PROGRAM 7.2
% ————— Root-locus plot ————————
% ** for the given system**
num = [0 0
0
1];
den = [1
2
0];
0
0
1];
den a = [1 3
3
1];
3
% ** for the asymptotes **
num a = [0
% *** gain constant K is included in the command to ensure that
% number of rows of r and that of a are the same***
K1 = 0 : 0.1 : 0.3;
K2 = 0.3 : 0.005 : 0.5;
K3 = 0.5 : 0.5 : 10;
K4 = 10 : 5 : 100;
K = [K1 K2 K3 K4];
r = rlocus (num, den, K);
a = rlocus (num a , den a, K);
y = [r a];
plot (y, ‘-‘)
v = (–4 4 –4 4]; axis (v)
grid
title (‘Root-locus plot of G(s) = K/[s(s+1)(s+2] and asymptotes’)
x label (‘Real Axis’)
y label (‘Imng Axis’)
Chapter 7.indd 59
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CONTROL SYSTEMS ENGINEERING
The above program will generate the root locus plot of Figure 7.36.
Fig. 7.36 Root locus plot.
REVIEW QUESTIONS
7.1 Use the Routh–Hurwitz criterion to determine the stability of the systems with characteristic
equations as follows. Also find the number of roots of each equation that lie on the right
half of the s-plane.
(1) s3 + 20s2 + 9s + 200 = 0
(2) 3s4 + 10s3 + 5s2 + s + 2 = 0
(3) s5 + 2s4 + 2s3 + 4s2 + s + 1 = 0
(4) s6 + 3s5 + 5s4 + 9s3 + 8s2 + 6s + 4 = 0
7.2 Determine the values of K so that the systems with the following characteristic equations
will be stable.
(1) s3 + as2 + Ks + b = 0, where a, b are constants
(2) s4 + 20Ks3 + 5s2 + (10 + K)s + 15 = 0
(3) s(2s + 1)(3s + 1) + K(s + 1) = 0
(4) s4 + 6s3 + 11s2 + 6s + K = 0
7.3 Derive the stability conditions which must be satisfied by the coefficients of the following
characteristic equation.
a3s3 + a2s2 + a1s + a0 = 0, where a3 > 0.
7.4 A unity feedback control system has the forward transfer function
K (2 s + 1)
G(s) =
s( 4 s + 1)( s + 1)2
Chapter 7.indd 60
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tHE rOOT lOCUS tECHNIQUE
7-61
Determine the stability condition of the system for the minimum value of K obtained from
ess ≤ 0.1 and r(t) = 1+ t.
7.5 Determine the stability condition for each of the systems with the following loop transfer
functions:
1
(1) G(s)H(s) =
( s + 2 )( s + 4 )
(2) G(s)H(s) =
(3) G(s)H(s) =
(4) G(s)H(s) =
Ks
( s + 1)2
K ( s − 1)
s( s 2 + 4 s + 4 )
K (s + 2)
s( s + 3)( s 2 + 2 s + 3)
7.6 The open-loop transfer function of a unity feedback control system is given by
K ( s + 5 )( s + 40 )
G(s) = 3
s ( s + 200 )( s + 1000 )
Discuss the stability of the closed-loop system in terms of K. What value of K will cause
sustained oscillation? Find the frequencies of oscillations.
7.7 A certain system has the characteristic equation
s(Ts + 1) + K = 0
It is desired that all the roots lie to the left of the line s = –a to guarantee a time constant of
not larger than 1/a. Use the Routh–Hurwitz criterion to find the values of K and T so that
both roots meet this requirement.
7.8 Given is a system with a characteristic equation
s3 + 9s2 + 26s + K = 0
Determine the value of K that will give a dominant time constant not larger than 0.5. The
design of the system must be slightly under-damped. Is it possible? If yes, find the value
of K needed and the resulting damping ratio.
7.9 G(s) = K/(s + 1)(s + 2) is given for a two-input–two-output control system as shown in the
signal flow graph in Figure 7.37.
Fig. 7.37
Chapter 7.indd 61
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CONTROL SYSTEMS ENGINEERING
Use the Routh–Hurwitz criterion to find the range of K (positive constant) so that the
system is asymptotically stable.
7.10 A control system as shown in Figure 7.38 has a transfer function
K
G(s) =
s (1 + 0.1s )(1 + 0.2s )
Fig. 7.38
and its amplifier exhibits a non-linear saturation characteristic. Determine the maximum
value of K that keeps the whole system absolutely stable.
7.11 Plot a root locus diagram with MATLAB for a system the open-loop transfer function of
which is given by
K
G(s)H(s) = 2
( s + 2 s + 2)( s 2 + 2 s + 5)
7.12 With the help of a suitable example, explain the effect of adding poles and zeros to the
open-loop transfer function of a control system.
7.13 The forward path transfer function of a unity feedback system is given below. Construct
the root locus diagram for all values of gain K. Find the value of K at all the breakaway
points.
7.14 Obtain G(s)H(s) from the following equation and draw the root locus for all values of K
s(s + 4)(s2 + 4s + 20) + K = 0
7.15 Draw the root locus for a system having
K
G(s)H(s) =
s ( s + 2)( s 2 + 2 s + 2)
Determine the value of K for which the system is critically damped.
Chapter 7.indd 62
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8
FREQUENCY RESPONSE ANALYSIS
8.1
INTROdUCTION
In the previous chapters we have studied the response of a system to step input, ramp input,
etc. Since most of the input signals in practice are sinusoidal inputs, in this chapter; we will
consider the steady-state response of a system to a sinusoidal input test signal. The response will
also be sinusoidal for a linear constant coefficient system once the transient die out. The output
sinusoidal signal will also have the same frequency as the input signal. However, the magnitude
and phase of the output will differ from the input signal once the frequency of the input signal
is changed. The frequency response analysis deals with the study of steady-state response of a
system to sinusoidal input of variable frequency.
Thus, we can define the frequency response of a system as the steady-state response of the
system to a sinusoidal input signal. For example, let us take up frequency response analysis of a
first-order system.
The analysis involves examining the transfer function G(s) when s = jw and graphically
displaying G( jw) as w varies.
As mentioned, for analysis of many systems, frequency response is of importance since most
of the input signals are either sinusoidal or composed of sinusoidal components (harmonics). The
starting point for frequency response analysis is the determination of system transfer function.
Then in the frequency response, the transfer function is expressed as M(s) = M φ.
Let us consider the transfer function of a first-order system as an example.
C (s)
1
=
G(s) =
R( s ) 1 + s τ
For frequency response analysis we replace s by jw.
Therefore,
1
G( jw) =
1 + jωτ
The system is subjected to sinusoidal input and therefore, we take R(s) = Ai sin ω t
Chapter 8.indd 1
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8-2
CONTROL SYSTEMS ENGINEERING
The output is
C ( jω) =
Ai sin ω t
1 + jωτ
The magnitude of output is
Ao = C ( jω) =
Ai
1 + ω2 τ2
The magnitude of input= Ai
The dimensionless ratio of output to input is given as
A
1
M = o =
Ai
1 + ω2 τ 2
and the phase angle is
φ = tan −1 (−ωτ) = − tan −1 ωτ
(8.1)
(8.2)
1
Time lag
= tan −1 ωτ
ω
Now let us plot M versus w and f versus w.
When w = 0, M = Ao/Ai = 1 and φ = 0
When ω = ∞, M = Ao /Ai = 0 and φ = −90 °
As w increases from 0 to ∞, the magnitude gradually decreases from 1 to 0 and the angle
of lag increases from 0 to –90°. Thus, higher the frequency, higher is the attenuation (decay)
of the output and greater is the angle of lag between output and input. The frequency response
characteristic of a first-order system has been shown in Figure 8.1.
Fig. 8.1 Frequency response of a first-order system.
8.1.1
Polar Plot
Let us consider the first-order system described earlier. The transfer function is G(s).
G( jω) = R(ω) + j X (ω)
Chapter 8.indd 2
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FREQUENCY RESPONSE ANALYSIS
8-3
R(ω) = Re[G ( j ω)] and X (ω) = I m [G ( j ω)]
where
we had
G(s) =
C (s)
=
1
R( s ) 1 + s τ
The polar plot is obtained from the relation
G( jω) =
Writing G( jω) as
1
1 + jωτ
G( jω) = R(ω) + j X (ω)
 ω
1− j 
 ω1 
=
2
 ω
1+  
 ω1 
=
1
 ω
1+ 
 ω1 
2
where ω1 =
− j
 ω
 
 ω1 
 ω
1+ 
 ω1 
2
1
τ
= R(ω) + j X (ω)
To draw the polar plot, we will determine R(ω) and X (ω) at ω = 0 and ω = ∞.
Again at ω = ω1 , the real part R(ω) is equal to the imaginary part X (ω) , that is, φ(ω) = 45°
as shown in Figure 8.2.
Fig. 8.2
Chapter 8.indd 3
Polar plot for a first-order system (e.g., a RC filter).
07/11/11 3:44 PM
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CONTROL SYSTEMS ENGINEERING
Let us consider polar plot of another system, the transfer function of which is given as
G(s) =
K
s(sτ + 1)
Substituting s = jω,
G( jω) =
K
jω( jωτ + 1)
=
K
jω − ω2 τ
The magnitude is
G( jω) =
and phase angle is
K
2
ω + ω4 τ2
 1 
φ(ω) = − tan −1 

 ωτ 
The magnitude and phase angle at ω = 0, w = 1/t, 1/2t and ω = +∞ are calculated as:
For ω = 0, G( jω) = ∞ and φ(ω) = −90°
For ω = 1/2 τ , G ( jω) = 4 K τ / 5 and φ(ω) = −117 °
For ω = 1/τ , G ( jω) = K τ / 2 and φ(ω) = −135°
For ω = ∞, G( jω) = 0 and φ(ω) = −180°
The polar plot can now be drawn using the above values as shown in Figure 8.3.
Fig. 8.3 Polar plot G( j ω) =
Chapter 8.indd 4
K
j ω( j ωτ + 1)
.
07/11/11 3:44 PM
FREQUENCY RESPONSE ANALYSIS
8-5
The limitation of polar plot is that in case of adding poles or zeros to an existing system,
we have to recalculate the values.
By comparing Figures 8.2 and 8.3, we observe that due to addition of a pole, the polar plot
gets clockwise rotation by 90° as w tends to infinity. Nyquist in 1932 related study of polar plots
with system stability.
8.2 F requency d omain S pecifications — B O D E D I A G R A M ,
DETERMINATION OF FREQUENCY DOMAIN
The frequency response specifications are the following:
(1) Resonant peak Mr: This is the maximum value of M and is denoted as Mr. The magnitude
of resonant peak Mr provides us information about the relative stability of the system. Large
resonant peak corresponds to large overshoot in the transient response.
(2) Resonant frequency w r: It is the frequency at which resonant peak occurs, that is, the
maximum value of Mr occurs. High value of ωr indicates that the time response of the
output is faster as peak time is inversely proportional to wr.
(3) Bandwidth BW: This is the range of frequencies in which the value of magnitude M
expressed in dB does not drop more than –3 dB. As shown in Figure 8.4, it is the band of
frequencies at –3 dB points.
Fig. 8.4 Bode plot, frequency response.
8.2.1
Correlation Between Time Response and Frequency Response
For comparative study of time response and frequency response of a system, let us consider a
second-order system. We had seen earlier that the transfer function of a second-order system can
be compressed as
C (s)
R( s )
=
ω2n
s 2 + 2ζωn s + ω2n
ζ is the damping ratio and ωn is the undamped natural frequency of oscillations.
where
Chapter 8.indd 5
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8-6
CONTROL SYSTEMS ENGINEERING
For the sinusoidal transfer function, we will put s = jω in the above expression.
C ( jω)
Dividing by ω2n ,
R( jω)
C ( jω)
R( jω)
=
=
ω2n
( j ω)2 + 2 ζωn j ω + ω2n
ω2n
=
(ω2n − ω2 ) + j 2 ζωn ω
1

ω2 
ω
1 − 2  + j2 ζ
ωn
ωn 

Let ω /ω n = u, then
M ( jω) =
C ( jω)
R( jω)
M ( jω) = M =
M ( jω) = tan −1
=
2
1
(1 − u ) + j2 ζu
= M ∠φ
1
2 2
(1 − u ) + (2ζ u)
2
2ζ u
1 − u2
From Equations (8.3) and (8.4), it is seen that
if
u = 0, M = 1 and φ = 0
1
π
u = 1, M =
and φ = −
2ζ
2
(8.3)
(8.4)
u → ∞, M → 0 and φ → −π
As u changes from 0 to ∞, M changes from 1 to 0 and φ changes from 0 to −π.
The frequency at which M has maximum value is called the resonant frequency ωr .
Let, ur = ωr / ωn where ur is called the normalized resonant frequency.
We will differentiate M with respect to u and substitute u = ur and then equate to zero.
d 
2 2
2 −1/2 = 0
(1 − ur ) + (2 ζur ) 
du
or
2
1

− (1 − u 2 ) + (2 ζu )2 
2
or
−3/2
−
Chapter 8.indd 6
−4 u + 4 u 3 + 8 ζ 2 u  = 0
4u 3 − 4u + 8ζ 2u
1
[(1 − u 2 ) + (2 ζu )2 ]3/2
2
=0
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FREQUENCY RESPONSE ANALYSIS
8-7
Substituting u = ur ,
4ur3 − 4ur + 8ζ 2 ur = 0
or
4ur3 = 4ur − 8ζ 2 ur
4ur3 = 4ur (1 − 2ζ 2 )
ur2 = 1 − 2ζ 2
ur = 1 − 2ζ 2
or
ωr
ωn
= ur = 1 − 2ζ 2
or
ωr = ωn 1 − 2ζ 2
From Equation (8.3), and (8.5), we have
Mr =
Mr =
Mr =
Mr =
Mr =
Mr =
From Equation (8.4),
1
(1 − u ) + ( 2ζ ur )2
2 2
r
1
[1 − (1 − 2ζ 2 )]2 + [4ζ 2 (1 − 2ζ 2 )]
1
4ζ 4 + 4ζ 2 − 8ζ 4
1
4ζ 2 − 4ζ 4
1
4ζ 2 (1 − ζ 2 )
1
2ζ 1 − ζ 2
−1
∠M ( j ω) = φ = tan
or
φ = tan
Chapter 8.indd 7
(8.5)
−1
2ζ ur
1 − ur2
= tan −1
2ζ 1 − 2ζ 2
1 − 1 + 2ζ 2
1 − 2ζ 2
ζ
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8-8
CONTROL SYSTEMS ENGINEERING
The characteristics of magnitude M and phase angle f for normalized frequency u for some
value of ζ have been shown in Figure 8.5.
Fig. 8.5 Frequency response magnitude and phase angle characteristics.
8.2.2
Correlation Between Time Domain and Frequency Domain Parameters
We had calculated time domain specification parameters like maximum overshoot Mp, peak time
tp, rise time tr, settling time ts, etc. Mp was calculate as
Mp = e−πζ
1−ζ 2
and in the frequency domain resonant peak Mr has been calculated as
Mr =
1
2ζ 1 − ζ 2
It is observed that both Mp and Mr are the functions of damping ratio ζ.
As ζ is increased, the value of maximum overshoot Mp goes on decreasing. When ζ
is made equal to 1, the overshoot disappears, that is, no overshoot is produced by the system
response. In the frequency domain, resonant peak Mr will disappear when ζ > 1/ 2 , that is,
ζ > 0.707.
For lower value of ζ both Mp and Mr will be large which is undesirable. In practice, the
value of ζ is kept such that both the performance indices, that is, Mp and Mr, are correlated.
Therefore, ζ is generally designed as 0.4 < ζ < 0.707.
We had resonant frequency ωr and damped frequency of oscillation ωd as
ωr = ωn (1 − 2ζ 2 )
ωd = ωn (1 − ζ 2 )
Chapter 8.indd 8
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FREQUENCY RESPONSE ANALYSIS
8-9
By comparing these two values, it can be said that there exists correlation between resonant
frequency and damped frequency of oscillations. The ratio of
ωr
=
(1 − 2ζ 2 )
ωd
(1 − ζ 2 )
is also a function of ζ. When ζ lies between 0.4 and 0.707, both ωd and ωr are comparable to
each other.
The magnitude characteristics in frequency domain and also in the time domain for two
different values of have been shown in Figure 8.6.
Fig. 8.6 Magnitude characteristic in both fretquency domain and in time domain.
8.2.3
Bandwidth
Bandwidth is the range of frequencies over which the value of M is equal or greater than
1/ 2 , that is, 0.707. We had
1
M =
2 2
(1 − u ) + (2ζ u)2
ω
Let
ub = normalized bandwidth = b
ωn
Then,
M =
or
or
or
Chapter 8.indd 9
1
(1 − ub2 )2 + ( 2ζub )2
=
1
2
(1 − ub2 )2 + (2ζ ub )2 = 2
1 + ub4 − 2ub2 + 4ζ 2 ub2 = 2
ub4 − 2ub2 + 4ζ 2 ub2 + 1 = 2
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8-10
CONTROL SYSTEMS ENGINEERING
Let
or
ub2 = x, then
x 2 − 2 x + 4ζ 2 x + 1 = 2
x 2 − 2 x (1 − 2ζ 2 ) − 1 = 0
2(1 − 2ζ 2 ) ±
4(1 − 2ζ 2 )2 + 4
or
x =
or
x = 1 − 2ζ 2 ± 1 − 4ζ 2 + 4ζ 4 + 1
or
x = 1 − 2ζ 2 ±
2 − 4ζ 2 + 4ζ 4
ub2 = 1 − 2ζ 2 ±
2 − 4ζ 2 + 4ζ 4
or
As
ub = 1 − 2 ζ 2 +
ub =
2
2 − 4ζ 2 + 4ζ 4
ωb
ωn
ωb = ub ωn = ωn 1 − 2ζ 2 +
ωb is the denormalized bandwidth.
8.2.4
2 − 4ζ 2 + 4ζ 4
Relative and Absolute Stability
Absolute stability indicates whether the system is stable or not. Relative stability indicates how
stable the system is. Absolute stability can be checked by applying Routh–Hurwitz criterion. For
relative stability Routh–Hurwitz can also be applied by shifting the imaginary axis to the left in
the s-plane successively and each time apply Routh–Hurwitz criterion. There are other tools and
techniques to examine relative stability of a control system.
8.2.5
Presentation of Frequency Response in Graphical Form
Sinusoidal transfer functions can be represented graphically from which the stability of the
system can be studied. Three types of graphical representations are discussed in the following
sections.
By now, we have studied that sinusoidal transfer function is a complex function of frequency
ω. It is characterized by its magnitude and phase angle with frequency as the variable parameter.
The following graphical representations for the study of system performance will be studied:
Chapter 8.indd 10
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FREQUENCY RESPONSE ANALYSIS
1.
2.
3.
8-11
Bode plot or logarithmic plot
Polar plot or Nyquist plot
Use of M-circles and N-circles.
Example 8.1 The forward path transfer function of a unity feedback control system is given as
64
G(s) =
s(s + 5)
Calculate the resonant peak, resonant frequency and bandwidth of the closed-loop system.
Solution
C (s)
R( s )
=
G(s)
1 + G(s) H (s)
=
G(s)
1 + G(s)
[ H (s) = 1]
64
64
s(s + 5)
=
=
64
s(s + 5) + 64
1+
s
(
s
+
5)
64
= 2
s + 5s + 64
Comparing with the standard form of transfer function of a second-order system,
C (s)
ω2n
= 2
R( s )
s + 2 ζωn s + ω2n
and
or
ω2n = 64
or, ωn = 8 rad/sec
2ζωn = 5
ζ =
Resonant frequency is
5
2ωn
=
5
2×8
= 0.3125
ωr = ωn 1 − 2ζ 2 = 8 1 − 2(0.3125)2
= 7.1 rad/sec
Resonant peak is
1
1
Mr =
=
2ζ 1 − ζ 2
2 × 0.3125 1 − (0.3125)2
= 1.78
Bandwidth
wr = ωn 1 − 2ζ 2 + (2 − 4ζ 2 + 4ζ 4 )1/2
= 8 1 − 2(0.3125)2 + (2 − 4 × 0.31252 + 4 × 0.31254 )1/2
= 11.25 rad/sec
Chapter 8.indd 11
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8-12
CONTROL SYSTEMS ENGINEERING
8.5SPECIFICATIONS AND TRANSFER FUNCTION FOR BODE DIAGRAM
Bode plot is one of the powerful graphical methods of analysing and designing control systems.
Introduction of logarithmic plots simplifies the determination of graphical representation
of the frequency response of the system. Such logarithmic plots are popularly called Bode plots
(in the honour of H. W. Bode). In Bode plot, we plot logarithm of magnitude versus frequency;
and phase angle versus frequency. Both these plots are drawn on a semi-log graph paper. In this
form of representation of sinusoidal transfer function, the magnitude G( jω) in dB which is
20 log G( jω) is plotted against log ω.
The transfer function of a system in the frequency domain is expressed as
G( jω) = G ( jω) ∠φ(ω)
By expressing the magnitude in terms of logarithm to the base 10, we can write
Logarithmic gain = 20 log10 G( jω) where the units are in decibels.
For a Bode diagram, logarithmic gain in dB versus ω and phase angle φ(ω) versus ω are
plotted separately on a semi-log paper.
We will now consider a generalized transfer function and explain the Bode plot in detail.
Let
G ( s) H ( s) =
K [(1 + s τ1 )(1 + s τ 2 )(1 + s τ 3 ).....]ω n2
s N [( 2 + s τ a )(1 + s τ b )......( s 2 + 2ζω n s + ω2n )]
(8.6)
The sinusoidal form of the transfer function is obtained by substituting s for jω
G ( jω) H ( jω) =
K (1 + jωτ1 )(1 + jωτ 2 )(1 + jωτ 3 )......ω n2
( jω) N [(1 + jωτ a )(1 + jωτ b )......{( jω)2 + 2ζω n ( jω) + ω2n }]
K [(1 + jωτ1 )(1 + jωτ 2 )(1 + jωτ 3 )......]ω n2
=
( jω) N [(1 + jωτ a )(1 + jωτ b )......{(ω2n − ω2 ) + j 2ζω nω}]
(8.7)
Now we will explain the procedure for drawing the Bode plot as follows.
Similarly, phase angle of G( jω) H(jw) is plotted against log ω. Thus, the horizontal axis
is logarithm of frequency and the ordinate is on a linear scale. Hence, Bode plot is drawn on a
semi-log paper.
Magnitude of G( jω) H( jw) in decibels is given as:
20 log10 G( jω) H ( jω) = 20 log10 K + 20 log10 (1 + jωτ1 )
+20 log10 (1 + jωτ2 ) + 20 log10 1 + jωτ3 + 
−20 N log10 ( jω) − 20 log10 (1 + jωτa )
−20 log10 (1 + jωτb ) − 
Chapter 8.indd 12
07/11/11 3:44 PM
FREQUENCY RESPONSE ANALYSIS
−20 log10
For the phase angle we write,
  ω 2
ω
1 − 

 + j 2ξ
ωn 
  ωn 
8-13
(8.8)
G( jω) H( jw) = tan −1 (0) + tan −1 (ωτ1 ) + tan −1 (ωτ2 ) + 
 2ζω ω 
ω
− N tan −1   − tan −1 ωτ a − tan −1 (ωτ b )...........tan −1  2 n 2 
0
 (ω n − ω ) 
or
G( jω) H( jw) = tan −1 (ωτ1 ) + tan −1 (ωτ2 ) + 
 2ζω ω 
− N × 90 − tan −1 ωτ a − tan −1 (ωτ b ) − tan −1  2 n 2  (8.9)
 (ω n − ω ) 
Bode plot is a graph obtained from Equations (8.8) and (8.9) consisting of two parts as follows.
(1) Plot of magnitude of G( jω) H( jw) in decibels versus log10 ω and
(2) Plot of phase angle G( jω) H( jw) versus log10 ω.
8.5.1
Method of Drawing Bode Plot
For plotting magnitude G( jω) H ( jω) in decibels versus log10 ω, we have to add the plots of all
the individual factors as included in Equation (8.8) which are listed below.
(1) Plot of gain K which is a constant
N
(2)
 1 
Plot of poles at the origin, 

 jω 
(3)
Plot of poles on real axis,
(4)
Plot of zeros on real axis, (1 + jωτ)
(5)
Plot of complex conjugate poles,
(6)
Plot of complex conjugate zeros, if present.
1
1 + jωτ
1
 ω   ω 
1 + j 2ξ 
−

 ωn   ωn 
2
We will now explain the Bode plots of the above individual factors:
(i)
Plot for the constant gain K
The magnitude in decibels of the term K is given as
K(dB) = 20log10 (K)
(8.10)
Equation (8.10) indicates that the magnitude is independent of log10 ω. Assuming K as
positive and real, Bode plot has been drawn as shown in Figure 8.7 (a).
Chapter 8.indd 13
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8-14
CONTROL SYSTEMS ENGINEERING
The phase angle is always zero for any value of ω as shown in Figure 8.7 (b).
Fig. 8.7 Bode plot for gain K.
(ii)
Plot of the term 1/( jw)N representing a pole at the origin
The magnitude of the term 1/( jω) N in decibels is written as
1
20 log10
= −20 N log10 ( jω)
( jω) N
The magnitude curve will have a slope of –20 dB/decade for a pole (N = 1).
For N = 2, the slope will be –40 dB/decade.
The phase angle is given by
1
 ω
= − N tan −1   = − N tan −1 ∞ = − N × 90°
0
( jω)
Therefore, for a zero at the origin, we have a logarithmic magnitude +20 log ω where
the slope is +20 dB/decade; and the phase angle is φ(ω) = 90°. The graphs have been shown
in Figure 8.8.
(iii) Graph for the term 1/1 + jwt, that is, for poles on the real axis
N
Fig. 8.8 Bode diagram for ( j ω)±N .
Chapter 8.indd 14
07/11/11 3:44 PM
FREQUENCY RESPONSE ANALYSIS
8-15
The magnitude is written as
1
1


20 log10
= 20 log10 

2
2
(1 + jωτ)
 1+ω τ 
= −20 log10 (1 + ω2 τ2 )1/2
Let us calculate the magnitude at very low and very high frequencies as
when ω << 1/ τ, magnitude is
20 log10
1
(1 + jωτ)
= –20log10 1 = 0 dB
ωτ >> 1,
when
Magnitude = 20 log10
1
(1 + jωτ)
(8.11)
20 log10 (ωτ)
= −20 log10 (ω) − 20 log10 (τ)
Equation (8.12) is similar to the equation of a straight line y = mx + c
Here,
(8.12)
m(slope) = –20 dB/decade
1
τ
c = −20 log10 (τ) = 20 log10  
Equations (8.11) and (8.12) are two curves. To determine where the two curves are
intersecting on the 0 dB axis, we have to equate Equation (8.12) to zero.
Thus, we write
0 = −20 log10 (ω) − 20 log10 (τ)
or
1
20 log10 (ω) = −20 log10 (τ) = 20 log10  
τ
or
ω =
1
τ
Hence, the two curves intersect on 0 dB axis at ω = ωc = 1/ τ where ωc is called the break
frequency or corner frequency.
The phase angle φ(ω) = − tan −1 (ωτ)
At very low frequency, φ(ω) = − tan −1 (0) = 0 °
At very high frequency, φ(ω) = − tan −1 (∞) = −90 °
At corner frequency,
1
1

ω = ωc = ; φ(ω) = − tan −1  × τ  = − tan −1 (1) = −45 °
τ
τ

Chapter 8.indd 15
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8-16
CONTROL SYSTEMS ENGINEERING
The Bode diagram for the pole factor 1/1 + jωτ has been shown in Figure 8.9.
Fig. 8.9 Bode diagram for a pole factor of (1 + j ωτ)−1 .
The Bode diagram for a factor involving zero in the transfer function (1 + jωτ) is
determined in a similar way as follows:
Magnitude is
20 log10 (1 + jωτ) = 20 log10 (1 + ω2 τ2 )1/2
when ωτ << 1, ωτ is negligible as compared to 1.
Therefore,
(8.13)
20 log10 (1 + jωτ) = 20 log10 (1) = 0 dB
And when ωτ >> 1, ωτ is much higher than 1.
Therefore,
(8.14)
20 log10 (1 + jωτ) = 20 log10 ωτ = 20 log10 (ω) + 20 log10 (τ)
Equation (8.15) is similar to a general straight line equation
y = mx + c
Here,
m(slope) = 20 dB/decade
c = 20 log10 (τ)
(8.15)
Equations (8.14) and (8.15) represent two curves. The curve of Equation (8.14) lies on 0 dB
axis, whereas the curve of Equation (8.15) has a slope of 20 dB/decade. The intersection of the
two curves is found by equating Equation (8.15) to zero.
Chapter 8.indd 16
07/11/11 3:44 PM
FREQUENCY RESPONSE ANALYSIS
8-17
Thus,
or
0 = 20 log10 (ω) + 20 log10 (τ)
ω =
1
= ωc
τ
where ωc is called the corner frequency. The Bode diagram for a zero factor of (1 + jωτ) has
been shown in Figure 8.10. The phase angle has been calculated as
Fig. 8.10 Bode plot for zero factor (1 + j ωτ) .
at ωτ << 1,
at ωτ >> 1,
at ω = ωc = 1/τ ,
φ = tan −1 (ωτ)
φ = tan −1 (0) = 0
φ = tan −1 (∞) = 90 °
1

φ = tan −1  × τ  = 45 °
τ

The Bode plot for the above has been shown in Figure 8.10. The magnitude versus log10 ω
graphs for both (1 + jωτ)−1 and (1 + jωτ) have been shown together in Figure 8.11.
The exact or actual curves and the asymptotic curves differ from each other to some extent.
They are matched at the corner frequency ωc = 1/ τ as shown in Figure 8.11. The maximum error
Chapter 8.indd 17
07/11/11 3:44 PM
8-18
CONTROL SYSTEMS ENGINEERING
Fig. 8.11 Actual Bode diagram for (1 + j ωτ) and (1 + j ωτ)−1 showing the magnitudes.
between the exact plot and the asymptotic plot occurs at the corner frequency. From exact plot,
the magnitude at ω = ωc = 1/ τ is calculated as
1


20 log10 (1 − jωτ) = 20 log10  1 + j × τ 
τ


= 20 log10 2 = 3 dB
For the curve for (1 + jωτ) −1 , the magnitude of exact plot at corner frequency will be –3 dB.
If the poles or zeros on the real axis are of the form (1 + jωτ) ± N , the error in magnitude can be
calculated to be equal to ± 3N dB and the phase angle error will be ± N 45°.
8.5.2
Initial Slope of Bode Plot
The initial slope of the Bode plot depends on the type of the system. Let us consider, for example,
type 0, type 1 and type 2 systems.
General sinusoidal transfer function for initial part of the Bode plot (for frequencies lower
than the lowest corner frequency) is
G( jω) H( jw) = K ( jω) ± N
If N is positive, then N represents the number of zeros at the origin. If N is negative, then N
represents the number of poles at the origin.
A type 0 system is represented as, G(s) H(s) = K/sN with N = 0.
For a type 0 system,
K
20 log10 G( jω) H ( jω) = 20 log10
( jω)0
(8.16)
= 20log10 K
The Bode plot has been shown in Figure 8.12 (a).
Chapter 8.indd 18
07/11/11 3:44 PM
FREQUENCY RESPONSE ANALYSIS
For a type 1 system, N = 1
20 log10 G( jω) = 20 log10
K
( jω)1
= 20 log10(K) – 20 log10(w)
Equation (8.17) is similar to the equation of a straight line,
8-19
(8.17)
(8.18)
y = mx + c
Here, slope
m = –20 dB/decade
c = 20log10 (K)
and
Intersection of the graph with 0 dB axis is calculated by equating Equation (8.17) to 0.
Therefore,
or
20 log10 (K ) − 20 log10 (ω) = 0
K =ω
Hence, the graph will intercept 0 dB axis at ω = K as shown in Figure 8.12 (b).
For a type 2 system, we take N = 2
We can write
20 log G( jω) H ( jω) = 20 log10
K
( jω)2
= 20 log10 (K ) − 40 log10 (ω)
= −40 log10 (ω) + 20 log10 (K )
Equation (8.19) is similar to the straight line equation
(8.19)
y = mx + c
Here,
m = –40 dB/decade
c = 20log10 K
The intercept on the 0 dB axis is calculated by equating Equation (8.19) to zero.
Thus,
or
or
0 = 40 log10 (ω) + 20 log10 (K )
20 log10 K = 20 log10 ω2
ω = K
The Bode plot for a type 2 system (initial part) has been shown in Figure 8.12 (c).
Chapter 8.indd 19
07/11/11 3:45 PM
8-20
CONTROL SYSTEMS ENGINEERING
Fig. 8.12 Initial part of Bode plots for type 0, type 1 and type 2 systems.
Table 8.1 shows the initial slope and intersection with 0 dB axis for different types of systems.
Table 8.1 Initial Slope of Bode Plot of Systems.
Type of System (N)
Initial Slope
(–20N) dB/decade
Intersection with 0 dB axis at
Type 0
0 dB/decade
Parallel to 0 dB axis
Type 1
–20 dB/decade
ω=K
Type 2
–40 dB/decade
ω=
Type 3
–60 dB/decade
ω = K 1/3
Type N
–20 N dB/decade
ω = K 1/N
Chapter 8.indd 20
K = K1/2
07/11/11 3:45 PM
FREQUENCY RESPONSE ANALYSIS
8.5.3
8-21
Bode Plot for Quadratic Form of Transfer Function
Let us consider the transfer function
ω2n
G(s) =
s 2 + 2ξωn s + ω2n
The sinusoidal transfer function is
G( jω) =
=
=
ω2n
( jω)2 + 2ξωn ( jω) + ω2n
(ω
ω2n
2
n
(8.20)
− ω2 ) + j 2ζω nω
1
  ω 2 
ω
1 − 
  + j 2ξ
ωn
  ωn  
The magnitude of the above in decibels form is
= 20 log10
= 20 log10
1
  ω 2 
ω
1 − 
  + j 2ξ
ωn
  ωn  
1
2
2
  ω 2 
ω

1 − 

+
2
ξ



  ωn  
 ωn 
2
= −20 log10
  ω 2 
  ω
1 − 
  +  2ξ 
  ωn  
  ωn


 
2
(8.21)
When ω << ωn , that is, at very low frequency, the term (ω / ωn ), (ω/ ωn )2 can be neglected.
Equation (8.21) becomes equal to –20log 10(1) = 0 dB.
When ω >> ωn , that is, at very high frequency, the term 1 and [2ξ(ω / ωn )]2 can be
neglected when compared to [(ω / ωn )2 ]2 . Equation (8.21) then becomes equal to
 ω 
−20 log10 

 ωn 
2
 ω 
= −40 log10 

 ωn 
= −40 log10 (ω) + 40 log10 (ω)
Chapter 8.indd 21
(8.22)
07/11/11 3:45 PM
8-22
CONTROL SYSTEMS ENGINEERING
Equation (8.22) is similar to the straight line equation y = mx + c
Here, m = – 40 dB/decade.
The magnitude in decibels versus log10 (w)
Case I: Graph lies on 0 dB axis.
Case II: Graph has slope of –40 dB/decade.
Now, we have to find where the two graphs are intersected on the 0 dB axis. For finding the
point on the 0 dB axis, we have to equate Equation (8.22) to zero.
Therefore,
or
−40 log10 (ω) + 40 log10 ωn = 0
40 log10 (ω) = 40 log10 (ωn )
or
ω = ωn , where ωn is the corner frequency
The two graphs will intersect at ω = ωn on the 0 dB axis. ωn is called the natural frequency
of oscillation.
8.5.4
Maximum Magnitude of the Second-order Transfer Function
The transfer function is written as
M=
The magnitude is
M =
(ω
ω2n
2
n
− ω ) + j 2ζω nω
2
=
1
 ω2 
ω
1 − 2  + j 2ζ
ωn
 ωn 
1
2

ω2  
ω 
 1 − 2  +  2ξ

ωn   ωn 

(8.23)
(8.24)
Let the frequency at which magnitude will be maximum be denoted by ωr . ωr will be called
the resonant frequency. The maximum value of M, also called the resonant peak Mr, is determined
by maximizing the expression for M with respect to ω. Therefore, we will differentiate M with
respect to ω and equate it to zero.
Chapter 8.indd 22
d 
1


 =0
2
2
dω  
ω  
ω2 

  1 − 2  +  2ξ ω  
ωn 
n  

 

07/11/11 3:45 PM
FREQUENCY RESPONSE ANALYSIS
8-23
or
ω4
ω2
ω2 
1
− 1 + 4 − 2 2 + 4ξ2 2 
2 
ωn
ωn
ωn 
−3/2

ω
ξ2 ω 
4ω3
0 + 4 − 4 2 + 8 2  = 0
ωn
ωn
ωn 

or
−
or
1
 ω3
ω
ξ2 ω 
 4 4 − 4 2 + 8 2 
ωn
ωn 
 ωn
2 
2 
ω4
ω2
2 ω
 1 + 4 − 2 2 + 4ξ 2 
ωn
ωn
ωn 

=0

4ω  ω2
2
=0
−
1
+
2
ξ


ω2n  ω2n

or
ω2
ω2n
∴
= 1 − 2ξ2
ω = ωr = ωn 1 − ξ2
(8.25)
Now let the magnitude of M at ω = ωr be denoted by Mr, called the resonant peak amplitude.
We can express Mr as
1
Mr =
2
2

ωr2 
 ωr 
1
2
−
+
ξ




ω2n 
 ωn 

Let us substitute ωr = ωn 1 − 2ξ2 in the above expression,
Mr =
=
Therefore,
Mr =
Chapter 8.indd 23
1
2
 2ξω 1 − 2ξ2

ω2n (1 − 2ξ2 ) 
n

+
1 −

ωn

ω2n





2
1
4ξ2 − 4ξ 4
1
2ξ 1 − ξ2
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8-24
CONTROL SYSTEMS ENGINEERING
Expressed in decibels,
1


Mr = 20 log10 

2
 2ζ 1 − ζ 
From Equation (8.23), the phase angle of Mr
= − tan −1
2ξ

ω2 
 1 − 2 
ωn 

At corner frequency, ω = ωn
Phase angle of Mr
= − tan −1
ω
ωn
2ξ
ωn
ωn

ω2n 
−
1


ω2n 

= − tan ∞ = −90 °
Thus, for a second-order system under consideration, we find that
Mr =
1
2ξ 1 − ξ2
ωr = ωn 1 − ξ2
And as discussed in the previous chapter, damped natural frequency ωd = ωn 1 − ξ2
Maximum overshoot Mp
=
−πξ
e
1−ξ2
Example 8.2 Sketch the polar plot of a system whose transfer function is given as
G(s) =
10
s(s + 1)(s + 2)
Solution By substituting s = jω in the transfer function,
10
G( jω) =
jω( jω + 1)( jω + 2)
10
=
2
3
−3ω + j (2ω − ω )
Chapter 8.indd 24
07/11/11 3:45 PM
FREQUENCY RESPONSE ANALYSIS
=
=
10[ −3ω2 − j (2ω − ω3 )]
9ω4 + (2ω − ω3 )2
−30ω2
9ω4 + (2ω − ω3 )2
−30ω2
=
10(ω3 − 2ω)
+ j
9ω4 + (2ω − ω3 )2
+ j
4
2
2
(ω + ω )(4 + ω )
Equating respectively the real and imaginary parts,
−
and
30ω2
(ω4 + ω2 )(4 + ω2 )
10(ω3 − 2ω)
4
2
2
(ω + ω )(4 + ω )
10(ω3 − 2ω)
(ω4 + ω2 )(4 + ω2 )
= 0; from which, ω = ∞
= 0; from which, ω3 − 2ω = 0 or, ω = ± 2
Magnitude of G( jω) is
G ( jω) =
Phase angle of G( jω) is
8-25
10
ω 1 + ω2 4 + ω2
ω

G ( jω) =  −90 ° − tan −1 ω − tan −1 

2
G( jω) acts the real axis at ω = ± 2
The value of G( jω) at ω = ± 2 is calculated as
−
4
30ω2
2
2
(ω + ω )(4 + ω )
=−
30 × 2
(4 + 2)(4 + 2)
=−
60
36
= −1.67
The values of G( jω) , that is, its magnitude and phase at ω → 0 and ω → ∞ are calculated as
lim G( jω) = lim
ω→0
ω→0
lim G ( jω) = lim
ω→∞
ω→∞
10
ω 1 + ω2 4 + ω2
10
ω 1 + ω2 4 + ω2
=∞
=0
ω

lim ∠G ( jω) = lim  −90° − tan −1 ω − tan −1  = −90 ° − 0 ° − 0 ° = −90 °
ω→0
x →0 
2

lim ∠G ( jω) = lim  −90° − tan −1
ω→ 2
ω→ 2 
2 − tan −1
1 
 = −180°
2
Chapter 8.indd 25
07/11/11 3:45 PM
8-26
CONTROL SYSTEMS ENGINEERING
ω

lim ∠G ( jω) = lim  −90° − tan −1 ω − tan −1  = −90 ° − 90 ° − 90 ° = −270 °
ω→∞ 
ω→∞
2
Thus, the informations for drawing the polar plot are shown below. The polar plot is shown
in Figure 8.13.
w
|G ( jw)|
0
∞
2
∞
∠G ( jω)
–90°
–1.67
–180°
0
–270°
Fig. 8.13 Polar plot of G(s) =
10
s(s + 1)(s + 2 )
⋅
8.5.5Determination of Gain Margin and Phase Margin for Stability
Analysis
From the polar plot or from the Bode plot, two quantities namely the gain margin and the phase
margin can be found out. This information can be used to study the stability of the control system.
The stability condition of the system can be improved by use of compensating networks (phaselag or phase-lead networks). Knowledge of gain margin and phase margin becomes useful in the
design and use of such compensating networks. These are described in the following sections.
8.5.5.1 Gain Margin and Phase Crossover Frequency
Consider the polar plot of G ( jω) H ( jω) as shown in Figure 8.14.
The frequency (ωc ) at which the polar plot crosses the negative real axis is known as the
phase crossover frequency. The gain margin is a measure of the closeness of the phase crossover
Chapter 8.indd 26
07/11/11 3:45 PM
FREQUENCY RESPONSE ANALYSIS
8-27
Fig. 9.14 Polar plot of G( j ω)H ( j ω) .
point to the point (–1, j0) in the G(s)H(s)-plane. For a closed-loop system having G(s)H(s) as its
open loop transfer function, the gain margin is the reciprocal of magnitude of G( jw) H( jw) at
phase cross-over frequency and is expressed as
1
GM = 20 log
dB
G ( jωc ) H ( jωc )
The following points may be observed from the above expression of GM
(1) For stable systems as
G ( jωc ) H ( jωc ) < 1, the GM in dB is positive.
(2)
For marginally stable systems as
G ( jωc ) H ( jωc ) = 1, the GM in dB is zero.
(3)
Chapter 8.indd 27
For unstable systems as
G ( jωc ) H ( jωc ) > 1, the GM in dB is negative and the gain has to be reduced to make the
system stable.
07/11/11 3:45 PM
8-28
CONTROL SYSTEMS ENGINEERING
So, gain margin may be defined as the amount of gain in decibels that can be allowed to
increase in the loop before the closed-loop system reaches instability.
8.5.5.2 Phase Margin and Gain Crossover Frequency
Referring to Figure 8.14, the gain crossover frequency (ωg ) is the frequency at which
G ( jω) H ( jω) , the magnitude of the open-loop transfer function, is unity. In other words, the
frequency at which the polar plot crosses the unit circle (whose centre is at origin and radius is
unity) is known as gain crossover frequency (ωg ).
The phase margin of a system is given by
PM = ∠G ( jωg ) H ( jωg ) + 180°
Phase margin is measured positively in a counterclockwise direction from the negative real
axis. The phase angle at the gain crossover frequency is (–180° + PM). If an additional phase-lag
equal to PM is introduced at the gain crossover frequency, the polar plot will then pass through
the point (–1 + j0) driving the closed-loop system to the verge of instability. So, the phase margin
may be defined as the amount of additional phase-lag at the gain crossover frequency required to
bring the system to the verge of instability.
Fig. 8.15 Bode plot of G( j ω)H ( j ω) .
It may be noted that the phase margin is always positive for a stable closed-loop system.
Usually, a GM of about 6 dB or a PM of 30° to 35° results in a reasonably good degree of relative
stability.
Chapter 8.indd 28
07/11/11 3:45 PM
FREQUENCY RESPONSE ANALYSIS
8-29
8.5.5.3 Graphical Methods for Obtaining Gain Margin and Phase Margin
As illustrated in Figure 8.15, the gain margin measured at phase crossover frequency (ωc ) is
given by
GM = − G ( jωc ) H ( jωc ) dB
The phase margin measured at gain crossover frequency (ωg ) is given by
PM = 180° + ∠G ( jωg ) H ( jωg )
Fig. 8.16 Magnitude versus phase plot of G( j ω)H ( j ω) .
The gain margin and phase margin are even better illustrated on the magnitude versus
phase plot as shown in Figure 8.16. On this plot, the locus intersects the –180° axis at the phase
crossover frequency (ωc ) and the locus intersects the 0 dB axis at the gain crossover frequency
(ωg ). So the gain margin is simply the distance in decibels measured from the phase crossover
to the critical point at 0 dB and –180°, and phase margin is the horizontal distance in degrees
measured from the gain crossover to the critical point.
Example 8.3 Sketch the Bode plot and hence determine the gain crossover frequency and
phase crossover frequency for the transfer function of a system represented by
Chapter 8.indd 29
G(s) =
10
s(s + 0.5s)(1 + 0.1s)
07/11/11 3:45 PM
8-30
CONTROL SYSTEMS ENGINEERING
Solution By substituting s = jω in the transfer function, we have
10
G( jω) =
jω(1 + j 0.5ω)(1 + j 0.1ω)
(1) The corner frequencies are
1
ω1 =
= 2 rad/sec
0.5
1
= 10 rad/sec
ω2 =
0.1
(2) As per procedure, the starting frequency of Bode plot is taken as lower than the lowest
corner frequency. Here, the lowest corner frequency is 2 rad/sec. Therefore, we can choose
starting frequency as 1 rad/sec.
(3) The system is type 1 system (type of the system is indicated by the power of s in the
denominator of the transfer function). So the initial slope of the Bode plot is –20 dB/decade.
This initial slope will continue till the lowest corner frequency of ω = 2 rad/sec.
(4) The corner frequency of ω = 2 rad/sec is due to the term 1/(1 + j 0.5ω) of the transfer
function. Therefore, the slope of the Bode plot will change by another –20 dB/decade after
the corner frequency of w = 2 rad/sec. The total slope, therefore, will become –40 dB/dec.
This slope will continue till the next corner frequency of w = 10 rad/sec. The corner
frequency of w = 10 rad/sec is due to the term 1/(1 + j 0.1ω) of the transfer function. The
slope of the Bode plot after the corner frequency of 10 rad/sec will change by another
–20 dB/dec. Therefore, after corner frequency of 10 rad/sec, the total slope of the Bode plot
becomes –60 dB/dec, and this slope will continue for higher frequencies.
(5) The magnitude of the Bode plot at different frequencies are calculated as shown below.
Frequency in rad/sec
Magnitude in dB
w=1
K
= 20 log K − 20 log ω
jω
Here
w=2
= 20 log 10 = 20 log 1
= 20 dB
K = 10, given
K
= 20 log 10 − 20 log ω = 20 log 12 + .25ω2
j ω(1 + j ω0 .5 )
= 20 − 20 log 2 − 10 log 2
= 11 dB
w = 10
K
= 20 log K − 20 log ω − 20 log 12 + .25ω2
j ω(1 + j 0.5ω)(1 + j 0.1ω)
− 20 log 12 + 0.01ω2
= −17.16 dB
Chapter 8.indd 30
07/11/11 3:45 PM
FREQUENCY RESPONSE ANALYSIS
8-31
Now, we will calculate the angle G( jω) for frequencies of ω = 1 rad/sec to ω = 15 rad/sec
(say). The calculations are done using the relation, G( jω) = −90° − tan −1 0.5ω − tan −1 0.1ω
(from the system transfer function)
(6)
w
0
0.1
1.0
2
5
10
15
f
–90°
–93.43°
–122.3°
–146.31°
–184.76°
–213.7°
–228.7°
(7)
We now sketch the Bode plot for magnitude and phase angle as shown in Figures 8.17 and
8.18. From the figures we find,
Gain crossover frequency
= 3.9 rad/sec
Phase crossover frequency
= 4.5 rad/sec
Fig. 8.17 Bode plot for G(s) =
Chapter 8.indd 31
10
.
s(s + j 0.5s )(1 + 0.1s )
07/11/11 3:45 PM
8-32
CONTROL SYSTEMS ENGINEERING
Fig. 8.18 Bode plot for G(s ) =
10
.
s(s + j 0.5s )(1 + 0.1s )
Example 8.4 Find the open-loop transfer function of a system whose approximate Bode plot
is shown in Figure 8.19 below.
Fig. 8.19 Bode plot of a given control system.
Chapter 8.indd 32
07/11/11 3:45 PM
FREQUENCY RESPONSE ANALYSIS
8-33
Solution The initial slope of the Bode plot is – 20 dB/decade.
For a slope of –20 dB/decade, the system is type 1. The system transfer function is expressed
as
K
G(s) H(s) =
s
or
G( jω) H ( jω) =
K
jω
20 log10 G ( j ω) H ( j ω) = 20 log10 K − 20 log10 ω
Substituting values,
54 = 20 log10 K − 20 log10 (0. 1)
or,
20 log10 K = 54 + 20 × (–1)
or,
20 log10 K = 54 – 20 = 34
log10 K = 1.57
or,
or,
K = 52.2
Thus, for the initial part of the Bode plot, we get the transfer function
K 52.2
=
s
s
At corner frequency, w = 1 and the slope has changed by another –20 dB/decade. The slope
is negative. The corresponding factor of the TF is 1/(1 + s).
At the corner frequency w = 2, the slope is increased by another –20 dB/decade. The slope
is negative. Hence the corresponding factor of the TF is 1/(1 + 0.5s).
Thus, the transfer function of the control system is
TF =
G(s)H(s) =
K
s(1 + s)(1 + 0.5s)
=
52.2
s(1 + s)(1 + 0.5s)
Example 8.5 The Bode plot for open-loop transfer function of a system is shown in
Figure 8.20. Find the transfer function.
Fig. 8.20
Chapter 8.indd 33
07/11/11 3:45 PM
8-34
CONTROL SYSTEMS ENGINEERING
Solution Initial slope of the Bode plot is –20 dB/decade. The system is a type 1 system. The
corner frequencies are w = 2.5 rad/sec and w = 40 rad/sec.
G ( j ω) H ( j ω) = 20 log10 K − 20 log10 ω
ω = 2.5 rad/sec, G ( j ω) H ( j ω) = 40 dB
at
or
40 = −20 log10 ω + 20 log10 K
40 = −20 log10 (2. 5 ) + 20 log10 K
or
20log10 K = 40 + 7.95
= 47.95 dB
or
K = 250
The first factor of TF is K/s = 250/s.
At corner frequency of w = 2.5 rad/sec, slope changes by –20 dB/decade and the slope is
negative. The transfer function component is 1/(s/2.5) + 1 = 1/(1 + 0.4s).
At corner frequency of w = 40 rad/sec, slope again changes by –20 dB/decade. Slope is
negative. Hence the transfer function component is 1/(s/40) + 1 = 1/(1 + 0.025s).
The open-loop transfer function of the closed loop system is
G(s)H(s) =
250
s(1 + 0.4s)(1 + 0.025s)
Example 8.6 Draw the Bode plot for a control system having transfer function,
100
G(s)H(s) =
s(s + 1)(s + 2)
Determine the following from the Bode plot:
(1) Gain margin; (2) Phase margin; (3) Gain crossover frequency and (4) Phase crossover
frequency.
Solution Let us substitute s = jw in the transfer function as
100
G ( jω) H ( jω) =
jω(1 + jω)(2 + jω)
=
50
jω(1 + jω)(1 + j 0.5ω)
(1) Corner frequencies are ω1 = 1 rad/sec
Chapter 8.indd 34
ω2 =
1
0.5
= 2 rad/sec
07/11/11 3:45 PM
FREQUENCY RESPONSE ANALYSIS
(2)
(3)
(4)
(5)
8-35
The starting of the Bode plot is taken as lower than the lowest frequency. Since lowest
corner frequency here is 1 rad/sec, we can take starting frequency as 0.1 rad/sec.
By examining the transfer function we see that it represents a type 1 system (power of
s in the denominator is 1). So the initial slope is –20 dB/decade and continues to corner
frequency, w = 1 rad/sec.
Corner frequencies, w = 1 rad/sec is due to term 1/(1 + jω) of the TF. Therefore, the
Bode plot after this frequency will have a further slope of –20 dB/decade. Thus, the total
slope will become –40 dB/dec. This slope will continue till the next corner frequency,
ω2 = 2 rad/sec. This corner frequency of ω2 is due to the term 1/(1 + j 0.5ω) of the TF;
due to which there will be another increase of –20 dB in the slope of the Bode plot at
w = 2 rad/sec. Thus, the total slope at frequencies higher than w = 2 rad/sec will be
–60 dB/decade.
The phase angle φ = G ( jω) H ( jω) for a range of frequencies is calculated as follows.
50
G ( jω) H ( jω) =
(0 + jω)(1 + jω)(1 + 0.5 jω)
φ = G ( jω) H ( jω) = − tan −1
or
ω
0
− tan −1
ω
1
− tan −1
0.5ω
1
φ = −90° − tan −1 ω − tan −1 0.5ω
Phase angle f at different values of w have been calculated as following
w
0
0.1
0.2
0.5
1.0
1.3
1.5
2
4.5
f
–90°
–98.6°
107°
–130°
–161.6°
–175.5°
183.2°
198.4°
233°
The magnitudes in dB at different frequencies, that is, at initial and corner frequencies are
calculated as follows:
ω = 0.1
Magnitude
50
jω
= 20 log 50 − 20 log ω
= 20 log 50 − 20 log(0.1)
= 54 dB
ω = 1.0
ω = 2.0
Magnitude
Magnitude
50
j ω(1 + j ω)
50
= 20 log 50 − 20 log ω − 20 log 1 + ω2
= 20 log 50 − 20 log 1 − 20 log 2
= 30 dB
jω(1 + jω)(1 + j 0.5ω)
= 20 log 50 − 20 log 2 − 20 log 12 + 22
−20 log 12 + (.5 × 2)
= 5 dB
Chapter 8.indd 35
07/11/11 3:45 PM
8-36
CONTROL SYSTEMS ENGINEERING
Fig. 8.21 Bode plot for magnitude and phase angle drawn on log scale.
Figure 8.20 shows the Bode plot for magnitude and phase angle drawn on log scale. The
gain margin is calculated at the phase crossover frequency and phase margin is calculated at gain
crossover frequency. The results as found are
Gain margin = 15 dB; Phase margin = 20°
Gain crossover frequency = 2.2 rad/sec; Phase crossover frequency = 1.45 rad/sec.
Example 8.7 Draw the Bode plot for the unity feedback control system whose transfer
function is given as
G(s) H(s) =
10(s + 10)
s(s + 2)(s + 5)
From the plot, determine the values of gain margin and phase margin. State whether the
system is stable or not.
Solution
Chapter 8.indd 36
G(s) H(s) =
10( s + 10)
s ( s + 2)( s + 5)
07/11/11 3:45 PM
FREQUENCY RESPONSE ANALYSIS
(1)
=
10(0.1s + 1)
s(0.5s + 1)(0.2s + 1)
8-37
[dividing numerator and denominator by 10]
The corner frequencies are
ω1 =
ω2 =
ω3 =
1
0.5
1
0.2
1
0.1
= 2 rad/sec
= 5 rad/sec
= 10 rad/sec
(2)
Starting frequency of the Bode plot is taken as lower than the lowest corner frequency.
Here the lowest corner frequency is 2 rad/sec. We can take starting frequency of say,
1 rad/sec.
(3)
The system is a type 1 system since the power of s in the denominator of the transfer
function is 1. So, the initial slope will be –20 dB/decade. This slope will continue till the
corner frequency of w = 2 rad/sec is reached.
(4)
The corner frequency of w = 2 rad/sec is due to the term 1/(0.5s + 1) for which
T ( j ω) = 1/[1 + j ω(0 .5 )]. The slope of the Bode plot after this frequency will change by
–20 dB/decade. Thus, the Bode plot after w = 2 rad/sec will have a slope of –40 dB/decade
and continue till the next corner frequency of w = 5 rad/sec is reached. Corner frequency,
w = 5 rad/sec is due to the term 1/(0.2s + 1) of the transfer function for which we can write
T ( jω) = 1/[1 + jω(0. 2)]. The slope of the Bode plot after w = 5 rad/sec will change by
another –20 dB/decade, making the total slope equal to –60 dB/decade. This slope will
continue till the next corner frequency of w = 10 rad/sec is reached.
The corner frequency of w = 10 rad/sec is due to the term (1 + 0.1s) appearing at the
numerator of the transfer function which can be written as T ( j ω) = 1/[1 + j ω(0 .1)]. The
Bode plot after this corner frequency will change by +20 dB/decade. The slope of the
Bode plot after w = 10 rad/sec will therefore be –40 dB/dec and will continue for higher
frequencies.
5.
Now we will calculate the phase angle φ(ω) for the transfer function.
G( jω) H( jw) =
Chapter 8.indd 37
10(1 + j 0.1ω)
jω(1 + j 0.5ω)(1 + j 0.2ω)
φ(ω) = G ( j ω) H ( j ω) = −90° − tan −1
0 . 5ω
0 . 2ω
0. 1ω
− tan−1
+ tan−1
1
1
1
07/11/11 3:45 PM
8-38
CONTROL SYSTEMS ENGINEERING
The values of φ at different values of ω are calculated as follows.
w rad/sec
0
0.1
0.5
1
2
5
8
10
20
f
–90º
–93º
–107º
–122º
–145º
–176º
–185.3º
–187º
–205º
The magnitudes are calculated as follows:
Frequency
ω = 1 rad/sec
Magnitude
K
jω
= 20 log K − 20 log ω
= 20 log10 − 20 log(1)
= 20 dB
ω = 2 rad/sec
K
jω(1 + jω0.5)
= 20 log k − 20 log ω − 20 log 12 + .25ω2
= 20 log 10 − 20 log 2 − 20 log 12 + .25 × 22
= 11 dB
ω = 5 rad/sec
K
jω(1 + jω0.5)(1 + jω0.2)
= 20 log K − 20 log ω − 20 log 12 + .25ω2 − 20 log 12 + 0.4ω2
= 20 log 10 − 20 log 5 − 20 log 12 + .25 × 25 − 20 log 12 + .04 × 25
= −20 dB
ω = 10 rad/sec
10(1 + jω0.1)
jω(1 + jω0.5)(1 + jω0.2)
= 20 log 10 − 20 log 10 − 20 log 1 + 25 − 20 log 1 + 4 + 20 log 1 + 1
= −31 dB
The Bode plot has been drawn using the above data (Figure 8.22). The phase margin at gain
crossover frequency = 18°. The gain margin calculated at phase crossover frequency = 24 dB
GM = Initial value – Final value = (0) dB – (–24 dB) = 24 dB
Since both phase margin and gain margin are positive, the system is stable.
Chapter 8.indd 38
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FREQUENCY RESPONSE ANALYSIS
8-39
Fig. 8.22 Bode plot for Example 8.7.
Example 8.8 Find the open-loop transfer function of a control system whose approximate
Bode plot is shown in Figure 8.23.
Fig. 8.23 Bode plot of a feedback control system.
Solution The corner frequencies are
ω1 = 2.5 rad/sec, ω2 = 10 rad/sec and ω3 = 25 rad/sec.
Initial slope of the Bode plot is –20 dB/decade. Thus, the system is a type 1 system whose
transfer function will be K/s.
Chapter 8.indd 39
07/11/11 3:45 PM
8-40
CONTROL SYSTEMS ENGINEERING
We calculate the value of K in the following way. Equation of the straight line is written as
y = mx + c
= −20 log ω + c (since the initial slope is − 20 dB/decade)
= −20 log ω + 20 log K
From the figure, we find that at w = 2.5 rad/sec, y = –12 dB
Substituting these values,
–12 = –20log(2.5) + 20log K
or
20log K = 4.04 dB
or
(1)
K = 0.63
Since the first line of the Bode plot has a slope of –20 dB/decade and starts from a point
–4.04 dB at frequency of w = 1 rad/sec, the factor contributing to the total transfer function
of the system is
K
jω
=
0.63
s
(2) At corner frequency w = 2.5 rad/sec, slope changes by +20 dB/decade. So the factor
contributing this change is
 s

+ 1  = (1 + 0.4s)

 2.5

and be placed at the numerator of the transfer function.
(3)
At corner frequency w = 10 rad/sec, slope changes by +20 dB/decade. Since the slope is
again changing in the positive direction, the factor contributing this has to be placed in the
numerator of the transfer function and the factor is
 s

+ 1  = (1 + 0.1s)

 10

(4) At corner frequency w = 25 rad/sec, the slope now changes in the negative direction by
–20 dB/decade. The factor contributing this change has to be placed in the denominator of
the TF and is equal to
 s

+ 1  = (1 + 0.04s)

 25

(5) Thus, the open-loop transfer function of the control system is
Chapter 8.indd 40
G(s) H(s) =
0.63(1 + 0.4s)(1 + 0.1s)
s(1 + 0.04s)
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FREQUENCY RESPONSE ANALYSIS
8-41
Example 8.9 Determine the open-loop transfer function of a system whose approximate
Bode plot is shown in the diagram in Figure 8.24.
Fig. 8.24 Bode plot of a given control system.
Solution
(1) The first line of the Bode plot as above has a positive slope of +40 dB/decade. Therefore,
this corresponds to s2 term in the numerator of the transfer function this being representing
a type 2 system. Thus, we have a transfer function component, T1(s) = Ks2.
(2) At corner frequency of w = 0.5 rad/sec, there is –20 dB/decade change of slope. This will
contribute a term,
1
1
=
T2(s) =
(1 + 2s)
 s 
1+

 0.5 
(3) At corner frequency w = 1 rad/dec, there is further reduction in slope by –20 dB/sec. This
negative change of slope at w = 1 rad/sec corresponds to a transfer function component of
T3(s) =
1
=
1
(1 + s)
s
1+ 
1
(4) At corner frequency w = 5 rad/sec, there is –20 dB/dec reduction of slope which corresponds to
1
1
T4(s) =
=
(1 + 0.2s)
s
1+ 
5
Thus, the transfer function of the system is equal to G(s) = T1(s) T2(s) T3(s) T4(s). That is,
Chapter 8.indd 41
G(s) =
Ks 2
(1 + 2s )(1 + s )(1 + 0.2s )
07/11/11 3:45 PM
8-42
CONTROL SYSTEMS ENGINEERING
We have now to calculate the value of K.
We write the equation of the second line whose slope is +20 dB/decade. At w = 1 rad/sec,
the magnitude is 32 dB. We write,
y = 20 log ω + c
32 dB = 20 log(1) + c
or
c = 32 dB
Then we have to find the magnitude at w = 0.5 rad
y = 20log w + c
= 20log (0.5) + 32
= 25.97 dB (see Figure 8.25)
Fig. 8.25 Determination of transfer function from the Bode plot.
Now we write the equation for +40 dB line
y = 40 log ω + c
25.97 = 40 log(0.5) + c
or
c = 38.02 dB
c = 20log K
or
or
Chapter 8.indd 42
log K =
38.02
20
= 1.9
K = 79.62
07/11/11 3:45 PM
FREQUENCY RESPONSE ANALYSIS
Thus,
8.5.6
G(s) =
8-43
79.62s 2
(1 + 2s)(1 + s)(1 + .2s)
Plotting Bode Diagrams with MATLAB
When the system is defined in the form given by
G(s) =
num( s )
den( s )
where num(s) and den(s) are, respectively, the numerator and denominator polynomial in
descending powers of s, the command bode (num, den) will draw the Bode diagram.
The command [mag, phase, w] = bode (num, den, w) will return the frequency response
of the system in matrices mag, phase and w. The phase angle is in degrees. The mag and phase
matrices are evaluated at user-specified frequency points.
The command mag db = 20 * log 10 (mag) will convert magnitude to decibels.
For specifying the frequency range, use the command w = log space (d1, d2) to generate a
vector of 50 points logarithmically equally spaced between decades 10d1 and 10d2. To generate n
number of frequency points, use the command
w = log space (d1, d2, n)
For example, the command
w = log space (0, 3, 100) will generate 100 points between 1 rad/sec and 1000 rad/sec. Use
the command bode (num, den, w) to incorporate user-specified frequency points.
Example 8.10 The open-loop transfer function of a unity feedback system is given by
9(s 2 + 0.2s + 1)
G(s) =
s(s 2 + 1.2s + 9)
Plot the Bode diagram with MATLAB.
Solution If we use the command bode (num, den), then the frequency range is automatically
determined to be from 0.1 to 10 rad/sec. Here we will use the frequency range to be 0.1 to
1000 rad/sec and hence the command w = log space (–2, 3, 100) will be used.
The following MATLAB Program 8.1 will generate the Bode diagram as shown in
Figure 8.26.
MATLAB PROGRAM 8.1
num = [0 9 1.8 9];
den = [1 1.2 9 0];
w = log space (–2, 3, 100);
bode (num, den, w)
subplot (2, 1, 1);
title (‘Bode Diagram of G(s) = 9(s ^ 2 + 0.2s + 1)/[s(s ^ 2 + 1.2s + 9]’)
Chapter 8.indd 43
07/11/11 3:45 PM
8-44
CONTROL SYSTEMS ENGINEERING
Fig. 8.26 Bode diagram of G(s ) =
8.6
9(s 2 + 0.2s + 1) .
s(s 2 + 1.2s + 9)
Polar Plot and Nyquist Stability Criterion
Nyquist used polar plots to study to stability of control system way back in 1932 and as a mark of
respect to him, polar plots are also called Nyquist plots. It is interesting to observe that the shapes
of polar plots change due to the addition of poles to the transfer function. As an illustration let us
consider the polar plot of transfer function 1/(1 + jω T1 ) of Figure 8.27.
Let us add poles in this transfer function one more in succession making it
1
1
and
⋅
(1 + jω T1 )(1 + jω T2 )
(1 + jω T1 )(1 + jω T2 )(1 + jω T3 )
Every time we add a pole, the polar plot gets rotated by –90° as shown in Figure 8.27 (b)
and (c).
We can observe that addition of a non-zero pole causes rotation of the polar plot at the high
frequency portion by 90°.
If we add a pole at the negative real axis, the transfer function 1/(1 + jω T1 ) will change to
1
a transfer function of
⋅
(1 + jω T1 )(1 + jω T2 )
Chapter 8.indd 44
07/11/11 3:46 PM
FREQUENCY RESPONSE ANALYSIS
8-45
The polar plots of these have been shown in Figure 8.27 (a) and (b), respectively. We can
notice that addition of a pole has rotated the polar plot by –90° at both zero and infinite frequencies.
Fig. 8.27 Effect of addition of poles on polar plot illustrated.
Figure 8.27 (c) shows the polar plot for adding another pole with total transfer function as
1
. Figure 8.27 (d) is the polar plot for adding a pole at the origin
(1 + jω T1 )(1 + jω T2 )(1 + jω T3 )
1
with TF =
⋅
jω(1 + jω T1 )
Nyquist criterion is based on the principle of argument which states that if there are P poles and
Z zeros of the TF enclosed in s-plane contour, then the corresponding GH plane contour will encircle
the origin Z-times in clockwise direction and P-times in anticlockwise direction.
To illustrate this, let us consider an contour in the s-plane which encloses one zero of G(s)H(s),
say s = α1 . The remaining poles and zeros are placed in the s-plane outside the contour as shown in
Figure 8.28.
Chapter 8.indd 45
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8-46
CONTROL SYSTEMS ENGINEERING
Fig. 8.28 Nyquist stability criterion illustrated.
For any non-singular point in the s-plane contour, there is a corresponding point G(s)H(s)
in the GH-plane contour. Let s follows a prescribed path as shown in the clockwise direction
making one circle and returning to its original position. The phasor (s − α1 ) will cover an angle
of −2π , while the net angle covered by the other two phasors, namely (s − β1 ) and (s − α 2 ) ,
is zero. The phasor representing the magnitude of G(s)H(s) in the GH-plane will undergo a net
phase change of −2π. That is, this phasor will encircle the origin once in the clockwise direction
as shown in Figure 8.28 (b). If the contour in the s-plane encloses two zeros, at α1 and α 2 , in
the s-plane, the contour in the GH-plane will encircle the origin two times. Generalizing this,
it can be stated that for each zero of G(s)H(s) in the s-plane enclosed, there is corresponding
encirclement of origin by the contour in the GH-plane once in the clockwise direction.
When the contour in the s-plane encloses a pole and a zero, say at a and b, respectively,
both the phasors (s − α) and (s − β) will generate an angle of 2π as s traverses the prescribed
path. However, since (s – b) will be in the denominator of G(s)H(s), the contour in the GH-plane
will experience one clockwise and one counterclockwise encirclement of the origin.
If we have p number of poles and z number of zeros enclosed by the contour in the s-plane,
the contour in the GH-plane will encircle the origin by (p – z) times in counterclockwise direction
[since (s – p) is in the denominator of the open-loop transfer function, G(s)H(s)].
For a system to be stable, there should not be any zeros of the closed-loop transfer function
in the right half of s-plane. Therefore, a closed-loop system is considered stable, if the member
of encirclements of the contour in GH-plane around the origin is equal to the open-loop poles of
the transfer function, G(s)H(s).
We can express
G(s)H(s) = [1 + G(s)H(s)] – 1
From this, it can be said that the contour G(s)H(s) in GH-plane corresponding to Nyquist
contour in the s-plane is the same as contour of [1 + G(s)H(s)] drawn from the point (–1 + j0).
Thus, it can be stated that if the open-loop transfer function G(s)H(s) corresponding to
Nyquist contour in the s-plane encircles the point at (–1 + j0) the number of times equal to the
Chapter 8.indd 46
07/11/11 3:46 PM
FREQUENCY RESPONSE ANALYSIS
8-47
right half s-plane poles of G(s)H(s) in the counterclockwise direction, it can then be said that the
closed-loop control system is stable.
8.6.1
Nyquist Path or Nyquist Contour
The transfer function of a feedback control system is expressed as
C (s)
G( s )
=
R( s )
1 + G( s ) H ( s )
The denominator is equated to 0 to represent the characteristic equation as
1 + G(s)H(s) = 0
The study of stability of the closed-loop system is done by determining whether the
characteristic equation has any root in the right half of the s-plane. Or, in other words,
we have to determine whether C(s)/R(s) has any pole located in the right half of the s-plane.
We use a contour in the s-plane which encloses the whole of the right hand half of the s-plane.
The contour will encircle in the clockwise direction having a radius of infinity. If there is any pole on
the jw-axis, these are bypasses with small semicircles taking around them. If the system does
not have any pole or zero at the origin or in the jw-axis, the contour is drawn as shown in
Figure 8.29. A few examples on application of Nyquist criterion for stability study will be taken
up now.
Fig. 8.29 Nyquist contour.
Example 8.11 The open-loop transfer function of a unity feedback control system is given as
G(s) H(s) =
K
(1 + τ1s)(1 + τ2 s)
Examine the stability of the system by applying Routh–Hurwitz criterion, and Nyquist
stability criterion.
Chapter 8.indd 47
07/11/11 3:46 PM
8-48
CONTROL SYSTEMS ENGINEERING
Solution The closed-loop transfer function is
K
C (s)
(1 + τ1s )(1 + τ2 s )
=
K
R( s )
1+
(1 + τ1s )(1 + τ2 s )
=
K
2
s τ1 τ2 + s(τ1 + τ2 ) + (K + 1)
The characteristic equation is
or
s 2 τ1τ2 + s(τ1 + τ2 ) + K + 1 = 0
s2 +
(τ1 + τ2 )
τ1τ2
Routh array is
s2
1
s1
τ1 + τ2
s0
of K.
s+
K +1
τ1τ2
=0
K +1
τ1τ2
τ1τ2
K +1
τ1τ2
0
0
Since t1 and t2 are time constants which are positive, the system is stable for all values
Further, as the poles are lying on the left side in the s-plane, the system is stable.
For drawing the Nyquist plot, we substitute s = jw in the transfer function.
We have,
G(s)H(s) =
K
( s + sτ1 )(1 + sτ2 )
=
K
τ1τ2
1 
1 

s + s + 
τ1  
τ2 

Substitute s = jw,
G ( jω) H ( jω) =
Chapter 8.indd 48
K
τ1τ2
1 
1 

 jω +   jω + 
τ1  
τ2 

07/11/11 3:46 PM
FREQUENCY RESPONSE ANALYSIS
8-49
at w = 0,
G ( jω) H ( jω) =
K
τ1 τ2
× τ1 τ2 = K
at ω = ∞,
G ( jω) H ( jω) = 0
at ω = 1/ τ1 τ2 ,
G ( jω) H ( jω) =
The plot of
K τ1τ2
τ1 + τ2
K
(1 + jωτ1 )(1 + jωτ2 )
as w increases from 0 to ∞ has been shown in Figure 8.30.
The Nyquist plot is symmetrical about the real axis since G ( jω) H ( jω) = G (− jω) H (− jω).
Since the plot of G ( jω) H ( jω) does not encircle (–1 + j0) point, the system is stable for any
positive value of t1, t2 and K.
Fig. 8.30 Nyquist plot of G( j ω)H ( j ω) =
K
.
(1 + j ωτ1)(1 + j ωτ2 )
If there are any open-loop poles in the jw axis or an open-loop pole at the origin, as for
example, in the case of G(s)H(s) = K / s(1 + τs), the Nyquist contour must be intended to bypass
the origin or the point on the jw axis where such a pole exists. This is because the s-plane contour
should not pass through a singularity of 1 + G(s)H(s).
Chapter 8.indd 49
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8-50
CONTROL SYSTEMS ENGINEERING
Example 8.12 The open-loop transfer function of a unity feedback control system is given as
G(s)H(s) =
2
K
s(s + s + 4)
Determine the value of K for which the system is stable by applying (1) Routh–Hurwitz
criterion; (2) Nyquist criterion and (3) Root locus technique.
Solution
(1) The characteristic equation is written as
s(s2 + s + 4) + K = 0
or
s3 + s2 + 4s + K = 0
The Routh array is,
s3
1
4
2
s
1
K
4−K
0
s1
1
(2)
K
0
s0
The condition for stability is, K > 0 and (4 – K) > 0, that is, the system is stable if 0 < K < 4.
We have,
G(s)H(s) =
2
K
s(s + s + 4)
Substituting s = jw,
G ( jω) H ( jω) =
=
=
=
=
K
2
jω [( jω) + jω + 4]
K
jω [4 − ω2 + jω]
K [(4 − ω2 ) − jω]
jω [(4 − ω2 ) + jω][(4 − ω2 ) − jω]
−( j 2 ) K [(4 − ω2 ) − jω]
jω [(4 − ω2 ) 2 + ω2 ]
− K [ω + j (4 − ω2 )]
ω [(4 − ω2 ) 2 + ω2 ]
The crossing of Nyquist plot at the real axis can be found by equating the imaginary part
of the above to 0.
Chapter 8.indd 50
07/11/11 3:46 PM
FREQUENCY RESPONSE ANALYSIS
Thus,
8-51
4 − ω2 = 0
ω = ±2
or
The magnitude G ( jω) H ( jω) at ω = ±2 , that is, at ω2 = 4 is found as
G ( jω) H ( jω) =
− K [ω]
ω [(4 − 4) + 4]
=−
K
4
Fig. 8.31 Nyquist plot of G( j ω)H ( j ω) =
K
j ω(4 − ω2 + j ω)
.
The Nyquist plot for G ( jω) H ( jω) has been drawn as in Figure 8.31. From this figure, we
notice that the intercept on the real axis at ω = ±2 is –K/4 and this −K/4 is greater than –1.
The value of K is found as
K
> −1
4
or
K>4
We know that if there are P poles and Z zeros in the transfer function enclosed by the
s-plane contour, the corresponding contour in the GH-plane must encircle origin Z times in the
clockwise direction and P times in the anticlockwise direction resulting encircling (P – Z) times
in the anticlockwise direction. Again G(s)H(s) contour around the origin is the same as contour
of 1 + G(s)H(s) drawn from (–1 + j0) point on the real axis.
Here the encirclement of –1 + j0 has been in the clockwise direction two times, hence
N = –2. There is no pole on the right-hand side of the s-plane and hence P = 0
Substituting
P–Z=N
we have,
0 – Z = –2 or Z = 2
Thus, for K > 4, the system is unstable. The system will be stable if K < 4.
The reader is advised to try the root-locus technique of solving this problem to arrive at the
same result as above.
−
Chapter 8.indd 51
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8-52
CONTROL SYSTEMS ENGINEERING
Example 8.13 The open-loop transfer function of a control system is given. Determine the
closed-loop stability using Nyquist criteria
K
G(s)H(s) =
s(1 + τs)
Solution The number of poles of G(s)H(s) in the right half of s-plane, that is, with positive
real part is zero, that is, P = 0.
To decide on the stability criterion as N = –P, that is, how many times the Nyquist plot
should encircle –1 + j0 point for absolute stability, we write
N = –P = 0, that is, the Nyquist plot
should not encircle –1 + j0 point for absolute stability in this case.
As there is one pole at the origin, it should be bypassed by small semicircle as shown in
Figure 8.32.
Fig. 8.32 Nyquist plot (a) s-plane contour; (b) contain in G(s) H(s) plane.
Substituting s = jω,
G ( jω) H ( jω) =
=
=
=
Chapter 8.indd 52
K
jω(1 + jωτ)
K (−ω2 τ − jω)
(−ω2 τ + jω)(−ω2 τ − jω)
K (−ω2 τ − jω2 )
ω4 τ2 + ω2
−K τ
2 2
ω τ +1
− j
K
ω(1 + ω2 τ2 )
07/11/11 3:46 PM
FREQUENCY RESPONSE ANALYSIS
8-53
lim G ( jω) H ( jω) = ∞
ω→0
lim ∠G ( jω) H ( jω) = –90°
ω→0
lim G ( jω) H ( jω) = 0
ω→0
lim ∠G ( jω) H ( jω) = –180°
ω→0
From the above calculations, the section I of the Nyquist plot in the s-plane is represented
by section I of the Nyquist plot in the G(s)H(s) plane as shown. This part relates to 0 < ω < +∞.
For −∞ < ω < 0, part I is drawn which is the mirror reflection of part I of the portion of the
Nyquist plot. The semicircle around the origin in s-plane is mapped into semicircular path of
infinite radius. This is represented a change of phase from +π/2 to −π/2. The point –1 + j0 is
not circled by the Nyquist contour and hence N = 0.
We had P = 0, N = 2 – P and hence Z = 0. The system is stable.
Example 8.14 A unity feedback control system is represented by its open loop transfer
function
G(s) H(s) =
8
s(s + 1)(s + 2)
Sketch the polar plot and examine whether the system is stable or not. Also calculate its
gain margin from the polar plot.
Solution First we write the transfer function in frequency domain by substituting s = jω.
For a unity feedback system H(s) = 1
G ( jω) H ( jω) =
8
( jω)(1 + jω)(2 + jω)
The magnitude and phase angle respectively are
G ( jω) H ( jω) =
∠G ( jω) H ( jω) =
8
ω × 1 + ω2 × 4 + ω2
8
∠90 tan −1 ω tan −1
ω
2
= −90° − tan −1 ω tan −1
ω
2
lim G ( jω) H ( jω) = ∞ and ∠G ( jω) H ( jω) = −90°
ω→0
lim G ( jω) H ( jω) = 0 and ∠G ( jω) H ( jω) = −90° − 90° − 90°
ω→∞
= –270°
From the above we notice that the locus starts at –90° with magnitude at ∞ and terminates
at the origin with 0 magnitude at –270°.
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CONTROL SYSTEMS ENGINEERING
We also need to calculate the intersection of the locus on the negative real axis by equating
the imaginary part G ( jω) H ( jω) to 0.
For this, we have to rationalize G ( jω) H ( jω)
8
G ( jω) H ( jω) =
( jω)(1 + jω)(2 + jω)
=
=
8(− jω)(1 − jω)(2 − jω)
( jω)(1 + jω)(2 + jω)(− jω)(1 − jω)(2 − jω)
−8 jω(2 − 3 jω − ω2 )
ω2 (1 + ω2 )(4 + ω2 )
24ω2
j8ω(2 − ω2 )
=− 2
−
ω × (1 + ω2 )(4 + ω2 ) ω2 (1 + ω2 )(4 + ω2 )
Equating the imaginary part to 0,
8ω(2 − ω2 ) = 0
ω = 0,
∴
2
Substituting ω = 2 in the real part of G ( jω) H ( jω), value of G ( jω) H ( jω) at the
crossing of real axis is equal to
−
2
24ω2
2
2
ω × (1 + ω )(4 + ω )
=−
24 × 2
2 × (1 + 2)(4 + 2)
48
= −1.33
36
The intersection of the locus with the real axis is at –1.33 + j0. The polar plot is shown in
Figure 8.33. As the locus is encircling the critical point –1 + j0, the system is unstable.
=−
Fig. 8.33 Study of stability of the system using Polar plot.
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FREQUENCY RESPONSE ANALYSIS
8-55
8.6.2.1Determination of Gain Margin and Phase Margin from the Polar
Plot
Let at w = wpc the polar plot intersects the negative real axis as shown in Figure 8.33. It is
also the frequency at which phase angle of G ( jω) H ( jω) is –180°. Stability determination is
done considering the critical factor, G ( jω) H ( jω) = 1 and ∠G ( jω) H ( jω) = −180° , that is,
the point –1 + j0 on the negative real axis. Gain margin GM is the gain that can be introduced in
the system till it reaches instability.
1
GM =
G ( jω) H ( jω)
at ω=ωpc
In Example 8.14, the gain margin,
1
GM =
OQ
1
=
1.33
= 0.75
The gain in decibels is expressed as
GM = 20 log10
1
OQ
dB
In this problem,
GM = 20 log10
1
= 20 log10 0.75
1.33
Phase margin is obtained by drawing a circle of unit radius and locating the point of
intersection P as shown in Figure 8.33 on the locus. Let the frequency at this point be called ωgc .
Phase margin is ∠G( jω) H ( jω) at ω = ωgc . By definition.
PM = 180° + ∠ G ( jω) H ( jω) at ω=ω
gc
Example 8.15 The open-loop transfer function of a control system is given as
8
G(s)H(s) =
(s + 2)(s + 4)
Draw the complete Nyquist plot and comment on the stability of the system.
Solution First we draw the Nyquist contour which consists of the entire jω -axis and a
semicircle of infinite radius. On the jω axis ω varies from −∞ to +∞. On the infinite semicircle
S = lim Re jθ where θ from +π/2 to −π/2. The contour, therefore, will enclose the whole of
R →∞
the right half of the s-plane encircled in the clockwise direction as shown in Figure 8.34.
The transfer function in the frequency domain for the G(s)H(s) is obtained by substituting
s = jω as
G ( jω) H ( jω) =
8
( jω + 2)( jω + 4)
lim G ( jω) H ( jω) = 1
ω→0
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CONTROL SYSTEMS ENGINEERING
lim ∠ G ( jω) H ( jω) = 0
ω→0
lim G ( jω) H ( jω) = 0
ω→∞
lim ∠G ( jω) H ( jω) =
ω→∞
∠0
∠0
=
= −180 °
ω
ω
90 °90 °
tan −1 tan
2
4
Fig. 8.34 Nyquist plot for transfer function, G(s) H(s) =
8
(s + 2)(s + 4)
⋅
The Nyquist locus in the GH-plane for s = 0 to s = +∞ has been drawn in Figure 8.34 (b).
For the infinite semicircle with radius Re jθ described by s = + j∞ to s = 0 to s = − j∞ for
θ = +π/2 to θ = −π/2 is calculated as
lim
R →∞
10
(2 + Re )(4 + Re jθ )
jθ
= 0∠180° to 0∠ − 180°
Therefore, the infinite semicircular path of Nyquist contour maps into the origin in the
Nyquist plot.
For the portion −∞ to 0, the locus will be the mirror image of portion I and is represented
as shown in Figure 8.34 (b). The point −1 + j 0 has also been shown. The locus does not encircle
the point −1 + j 0 and therefore N = 0. There is no pole of G(s)H(s) in the right-hand side of the
s-plane. Therefore, P = 0
N=P–Z
O=O–Z
that is,
Z=0
There are no zeros of 1 + G(s)H(s). That is, the roots of the characteristic equation with
positive real part is absent and hence the closed-loop system is stable.
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FREQUENCY RESPONSE ANALYSIS
8-57
Example 8.16 Determine the stability of the system using Nyquist criteria whose open-loop
transfer function is
G(s)H(s) =
1
s(1 + s)(1 + 2s)
Solution Since there is a pole at the origin, the Nyquist contour will be as shown in
Figure 8.35 below. The Nyquist contour will form an infinitely small semicircle of the centre
by passing the pole at the origin. The transfer function in the frequency domain is obtained by
substituting s = jω in G(s)H(s) as
G ( jω) H ( jω) =
1
jω(1 + jω)(1 + 2 jω)
Fig. 8.35 Nyquist contour.
For drawing the Nyquist plot for the portion I, that is, for s = 0+ to s = + j∞, we determine
magnitude and phase angle of G ( jω) H ( jω) as
lim G ( jω) H ( jω) =
ω→0
lim ∠G ( jω) H ( jω) =
ω→0
=
=
1
j 0(1 + j 0)(1 + 2 j 0)
1
(0 + jω)(1 + jω)(1 + 2 jω)
1
tan −1
−1
2ω
ω
ω
tan −1 tan −1
0
1
1
1
tan ∞ tan
= –90°
Chapter 8.indd 57
=∞
−1
0 tan
−1
0
=
1
∠90° ∠0° ∠0°
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lim G ( jω) H ( jω) = 0
s →∞
lim ∠G ( jω) H ( jω) = −90° − 90° − 90° = −270°
s →∞
For the portion I of the Nyquist contour, the Nyquist plot as shown in Figure 8.36, will
start from –90° with infinite magnitude and terminate at the origin at –270° with 0 magnitude. To
determine whether the locus crosses the negative real axis, we have to rationalize G ( jω) H ( jω)
and put its imaginary part to 0 as done below.
G ( jω) H ( jω) =
=
=
1
jω(1 + jω)(1 + 2 jω)
(1 − jω)(1 − 2 jω)
jω(1 + jω)(1 + 2 jω)(1 − jω)(1 − 2 jω)
1 − 3 jω − 2ω2
jω(1 + ω2 )(1 + 4ω2 )
=
−3 jω
jω(1 + ω2 )(1 + 4ω2 )
=−
Equating imaginary part to 0,
2
3
2
(1 + ω )(1 + 4ω )
+
+
(1 − 2ω2 )
jω(1 + ω2 )(1 + 4ω2 )
(1 − 2ω2 )
2
2
ω(1 + ω )(1 + 4ω )
[ j 2 = −1]
1 − 2ω2 = 0
or
ω =
1
2
For ω = 0.707,
G ( jω) H ( jω) = −
= 0.707
3
1

 1 +  (1 + 2)

2
− j0
1
= −0.66
1.5
For the infinite semicircle, that is, portion II of the Nyquist contour, we put
=−
lim
R →∞
1
jθ
jθ
Re (1 + Re )(1 + 2 Re jθ )
= 0∠ − 270°
Thus, the infinite semicircle will map at the origin itself. For the portion III of the Nyquist
contour, that is, for s → − j∞ to s = j0–, the locus will be the mirror image of the locus for
portion I as has been shown in Figure 8.36.
Now we will do the mapping of the small semicircle at the origin.
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FREQUENCY RESPONSE ANALYSIS
8-59
Considering the radius of the infinitely small semicircle as ε, and as θ varies from −π/2
to 0 to +π/2
1


lim  jθ
 = ∞∠90° to ∞∠0° to ∞∠ − 90°
ε→0 ε e (1 + ε e jθ )(1 + 2ε e jθ )


This part of the Nyquist locus will be on infinite semicircle spread over the entire righthand side of the GH-plane.
The complete Nyquist plot for parts I, II, III and IV of the Nyquist contour has to be drawn
as shown in the diagram below. Note that OB = –0.66. Since the critical point –1 + j0 is not
encircled by the Nyquist locus, the system is stable.
Fig. 8.36 Nyquist plot for G(s)H(s) =
1
s(1 + s )(1 + 2 s )
⋅
Example 8.17 The open-loop transfer function of a system is
K
G(s)H(s)
s(s + 2)(s + 3)
Determine the value of K for which the system will remain stable with (1) phase margin of
60° and (2) gain margin of 6 dB.
Solution
(1) The sinusoidal transfer function is
G ( jω) H ( jω) =
K
jω(2 + jω)(3 + jω)
Phase margin = ∠G ( jω) H ( jω) + 180°
ω
ω
∠G( jω) H ( jω) = −90° − tan −1 − tan −1
2
3
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CONTROL SYSTEMS ENGINEERING
Given phase margin= 60°
Therefore,
60° = −90° − tan −1
or
30° = tan −1
Taking tangent of both the sides,
tan 30° =
or
ω
2
ω
2
+ tan −1
− tan −1
ω
3
+ 180°
ω
3
 ω ω
 + 
2 3
ω ω

1 − × 
2 3

5ω
0.57 =
6 − ω2
or
ω = 0.5
Magnitude G( jω) at ω = 0.5 has to be equated to 1 to determine the maximum value of
gain K.
We rationalize G ( jω) H ( jω) as
G ( jω) H ( jω) =
=
K
jω(2 + jω)(3 + jω)
K (6 − 5 jω − ω2 )
jω(4 + ω2 )(9 + ω2 )
=−
G ( jω) H ( jω) ω=0.5 =
=−
K (2 − jω)(3 − jω)
jω(4 + ω2 )(9 + ω2 )
jk 5ω
jω(4 + ω2 )(9 + ω2 )
5K
(4 + ω2 )(9 + ω2 )
5K
=
(4 + .25)(9 + .25)
− j
+
K (6 − ω2 )
K (6 − ω2 )
jω(4 + ω2 )(9 + ω2 )
ω(4 + ω2 )(9 + ω2 )
(8.26)
=1
or
5K = 39.3
or
K = 7.86
(2) Given gain margin = 16 dB
Let the Nyquist plot intersect the negative real axis at a distance a from the origin. Then,
gain margin= 20 log10 1/a
or
6 = 20 log10 1/a
or
a = 0.5
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FREQUENCY RESPONSE ANALYSIS
8-61
at the point of intersection the imaginary part of the G ( jω) H ( jω) is equal to 0.
Therefore, from Equation (8.26), we can write
K (6 − ω2 )
or
or
ω(4 + ω2 )(9 + ω2 )
=0
ω2 = 6
ω =
6 = 2.45
G ( jω) H ( jω) at ω = 2.45 has to be calculated and equated to a = 0.5 to get the critical
value of gain K for the gain margin of 6 dB. Thus,
5K
5K
2
at ω = 6 is
2
2
(4 + 6)(9 + 6)
(4 + ω )(9 + ω )
Equating this value to a = 0.5
5K
= 0.5
150
or
5K = 75
or
K = 15
The system will remain stable with a gain margin of 6 dB for K = 15.
Example 8.18 The open-loop transfer function of a unity feedback control system is
G(s)H(s) =
(s + 2)
( s + 1)( s − 1)
Draw the Nyquist plot and examine the stability of the closed-loop system. Check also the
stability condition by Routh–Hurwitz criteria.
Solution For unity feedback system, H(s) = 1
G(s)H(s) =
(s + 2)
( s + 1)( s − 1)
There is one pole on the right-hand side of the s-plane at s = 1. The open-loop system is,
therefore, unstable. Let us examine stability of the closed-loop system, first by Nyquist criterion
and then by Routh–Hurwitz criterion.
Substituting s = jω,
G ( jω) H ( jω) =
Chapter 8.indd 61
( jω + 2)
( jω + 1)( jω − 1)
=−
=−
(2 + jω)
(1 + jω)(1 − jω)
(2 + jω)(1 − jω)(1 + jω)
(1 + ω2 )(1 + ω2 )
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8-62
CONTROL SYSTEMS ENGINEERING
 2(1 + ω2 ) + jω(1 + ω2 ) 
= −

2
2
 (1 + ω )(1 + ω ) 
=−
2
2
− j
ω
1+ω
1 + ω2
We will draw the Nyquist contour and then the Nyquist plot.
Fig. 8.37 Nyquist plot for G(s)H(s) =
(s + 2 )
(s + 1)(s − 1)
.
From the transfer function and the Nyquist contour,
as ω → 0 + , G ( jω) H ( jω) = −2 − j 0
ω → +∞,
G ( jω) H ( jω) = −0 − j 0
ω→0 ,
G ( jω) H ( jω) = 2 + j0
ω → −∞,
G ( jω) H ( jω) = 0 + j0
−
For the infinite semicircle
Re jθ + 2
as q changes from +π/2 to −π/2
( Re jθ + 1)( Re jθ − 1)
= 0
The infinite semicircle will map into the origin in the Nyquist plot as θ will change from
+π/2 to −π/2. The complete Nyquist plot for the Nyquist contour considering parts I, II and III
has been drawn as in Figure 8.37 (b). It is observed that the point –1 + j0 has been encirclea by
the locus once in the anticlockwise direction. Therefore,
N = –1
N = Z – P, P = 1
lim
R →∞
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FREQUENCY RESPONSE ANALYSIS
8-63
Hence,
–1 = Z – 1
or
Z=0
Z = 0 means that there are no zeros of 1 + G(s)H(s) in the right-hand side of s-plane.
Hence, the system in closed-loop configuration is stable.
For applying Routh–Hurwitz criterion, we will write the characteristic equation of the
closed-loop system as
1 + G(s)H(s) = 0
or
(s + 2)
1+
=0
(s + 1)(s − 1)
or
(s + 1)(s – 1) + (s + 2) = 0
or
s2 + s + 1 = 0
The Routh array is
s2
1
1
s1
1
0
s0
1
0
As there is no change of sign in the first column of Routh array, the system is stable.
8.7Drawing Nyquist Plots with MATLAB
Nyquist plots are polar plots, while Bode diagrams are rectangular plots. The command nyquist
(num, den) will draw the nyquist plot of G(s) = num(s)/den(s) whereas the command nyquist
(num, den, w) uses the user-specified frequency vector w in radians per second.
MATLAB returns the frequency response of the system in matrices re, im and w and no
plot is drawn when the following command is invoked.
[re, im, w] = nyquist (num, den)
[re, im, w] = nyquist (num, den, w)
or Example 8.19 Draw a Nyquist plot with MATLAB for the open-loop transfer function given by
1
G(s) =
s(s + 1)
Solution The MATLAB command
num = [0
0
1];
den = [1
1
0];
nyquist (num, den)
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8-64
CONTROL SYSTEMS ENGINEERING
will produce an erroneous Nyquist plot. A corrected Nyquist plot is obtained by entering the axis
command
nyquist (num, den)
v = [–2
2
–5
5]; axis (v)
MATLAB Program 8.2 will produce a correct Nyquist plot as shown in Figure 8.38 even
though a warning message “Divide by Zero” may appear on the screen.
MATLAB PROGRAM 8.2
% ……….. Nyquist plot ……………………..
num = [0
0
1];
den = [1
1
0];
nyquist (num, den)
v = (–2
2
–5
5); axis (v)
grid
title (‘Nyquist plot of G(s) = 1/([s(s+1)]’)
Fig. 8.38 Nyquist plot of G(s ) =
1
.
s(s + 1)
The above Nyquist plot includes the loci for both w > 0 and w < 0. For drawing a Nyquist
plot in the positive frequency region (w > 0), we have to use the command [re, im, w] = nyquist
(num, den, w).
MATLAB Program 8.3 will draw the Nyquist plot only in the positive frequency region
(w > 0) as has been shown in Figure 8.39.
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FREQUENCY RESPONSE ANALYSIS
8-65
MATLAB PROGRAM 8.3
% —————— Nyquist plot ———————
num = [0
0
1];
den = [1
1
0];
w = 0.1:0.1:100;
[re, im, w] = nyquist (num, den, w);
plot (re, im)
v = [-2
2
-5
5]; axis (v)
grid
title (‘Nyquist plot of G(s)=1/[s(s+1)]’)
x label (‘Real Axis’)
y label (‘Imag Axis’)
Fig. 8.39 Nyquist plot of G(s ) =
8.8
1
for w > 0.
s(s + 1)
Relative Stability
In addition to knowing whether a system is stable or not, it is also desirable to know how stable
the system is. It is possible to get this information from the open-loop frequency response of the
system.
The nearness of the open-loop frequency response (the polar plot) to the point –1 + j0 in
the GH-plane provides information regarding the degree of stability or the relative stability of the
system.
Two quantities, namely the gain margin and the phase margin, are used to measure the
relative stability. Gain margin is defined as the additional gain required to make the system just
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CONTROL SYSTEMS ENGINEERING
unstable. Phase margin is defined as the additional phase lag required to make the system just
unstable. From the diagram in Figure 8.40, it is deserved that the gain margin is 2. Location of
point A on the negative real axis is at –0.5. This is to be multiplied by 2 to make it –1, that is, the
point of intersection being shifted from A to B. Here the gain margin in dB is expressed as 20log
2 = 6 dB.
Fig. 8.40 Determination of relative stability.
Phase margin is 180 − φ as determined from point C. An additional phase lag of 180 − φ
is required to make the system just unstable.
The gain margin is determined from the gain at phase crossover frequency, that is, the
frequency at which phase lag is 180°. Phase margin is determined from the phase lag at the gain
crossover frequency. This is the frequency at which the gain is 0 dB.
8.9 Frequency Response of a Closed-loop System Using M -circle
and N -circle
Transfer function of a closed-loop system is written as
C (s)
R( s )
=
G(s)
1 + G(s) H (s)
If it is a unity feedback system, H(s) = 1.
And for frequency response, we substitute s = jω. Therefore,
C ( jω)
=
G( jω)
R( jω)
1 + G( jω)
If we draw the polar plot, this will have magnitude M (ω) and phase angle φ(ω). Frequency
response will consist of magnitude and phase angle at different values of ω.
Constant magnitude loci and constant phase angle loci can be drawn in complex G( jω)
plane and using them, the frequency response of a closed-loop system can be studied. These are
described as follows.
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FREQUENCY RESPONSE ANALYSIS
8.9.1
8-67
Constant Magnitude Loci
The constant magnitude loci is a circle and hence it is also called M-circle. For different values
of M, we can draw constant M-circles.
We write G( jω) = x + jy
M = M (ω) =
or
M2 =
or
G( jω)
1 + G( jω)
=
x + jy
1 + x + jy
=
x 2 + y2
(1 + x )2 + y 2
(8.27)
x 2 + y2
(1 + x )2 + y 2
M 2 [1 + 2 x + x 2 + y 2 ] = x2 + y2
Rearranging,
x 2 (1 − M 2 ) + y 2 (1 − M 2 ) − 2 xM 2 = M 2
Dividing both sides by (1 – M 2),
x 2 + y2 − 2 x
or
1 − M2
=
M2
1 − M2
2
 M2 
 M2 
M2
2
x − 2x
+
+
y
=
+




1 − M2  1 − M2 
1 − M2  1 − M2 
2
or
M2
M2
2
2

M2 
M2
x −
 + ( y − 0)2 =
2

1− M 
(1 − M 2 )2
This equation is of the form
( x − a)2 + ( y − b)2 = c2
This is the equation of a circle with centre at (a, b) and radius as c.
Here, the circle has the centre at M 2/(1 – M 2), 0 and radius as M/(1 – M 2).
For construction of M-circles for different values of M, the calculations of centre of the
circles and their radius have been made as shown.
Value of M


2
 M
, 0
Centre of the M-circle 

2
1− M

 M 
Radius of the M-circle 

1− M 2 
v
(0, 0)
0
0 < M < 1, say M = 0.5
(0.33, 0)
0.66
M = 0.7
(0.96, 0)
1.37
M = 0.8
(1.77, 0)
2.22
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CONTROL SYSTEMS ENGINEERING
When M = 1, from Equation (8.27), the value of x can be calculated as
x 2 + y2
M =1=
or
or
or
or
(1 + x )2 + y 2
(1 + x)2 + y2 = x2 + y2
1 + 2x + x2 + y2 = x2 + y2
2x = –1
x =−
1
2
For M = 1, we, therefore, find that x = –1/2 which is a straight line parallel to the y-axis at a
distance of –1/2 i.e. –0.5 from the origin. Such a straight line can be considered to have an infinite
radius
For M > 1 say M = 1.5
Centre of Circle
Radius
(–1.8, 0)
–1.2
M=2
(–1.33, 0)
–0.66
M=3
(–1.125, 0)
–0.375
M=4
(–1.066, 0)
–0.266
The constant M-circles for different values of M as calculated are shown in Figure 8.41. It
is observed that the loci are symmetrical about the locus of M = 1. For M > 1, the M-circles lie
to the left of M = 1 and for M < 1, the M-circles lie to the right of line for M = 1. For M = 0, the
M-circle is the origin itself.
Fig. 8.41 Constant M-circles.
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FREQUENCY RESPONSE ANALYSIS
8.9.2
8-69
Constant N -circles
G( jω) = x + jy
φ =∠
G( jω)
∠x + jy
=
1 + G( jω) ∠1 + x + jy
Taking tangent of the angles on both the sides,
= tan −1
y
x
− tan −1
y
1+ x
y(1 + x ) − yx
x (1 + x )
tan φ =
=
x (1 + x ) + y 2
y y 
1+ 

x (1 + x )
x 1 + x 
y
y
−
x 1+ x
=
2
y
x + y2 + x
Let us assume
tan φ = N
Then,
N =
2
y
x + y2 + x
or
x 2 + y 2 + x − yN = 0
Adding 1/4 + 1/(2N)2 on both the sides,
x 2 + x + y 2 − yN +
1
4
+
or
x2 + x +
1
4
+ y 2 − yN +
1
(2 N )
1
(2 N )
2
2
=
=
1
4
1
4
+
+
1
(2 N )2
1
(2 N )2
or
2
1
1 


x +  +y−

2
2N 


2
 1
1 
=
+

2
 4 (2 N ) 
2
This is the equation of a circle of the form
( x − a)2 + ( y − b)2 = c2
Thus, the equation represents a family of circles whose centre is at (–1/2, 1/2N) and radius
of
Chapter 8.indd 69
1/4 + 1/(2 N )2 .
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8-70
CONTROL SYSTEMS ENGINEERING
For various values of φ, the centre points and radius can be calculated and the N-circles
drawn as shown in the Figure 8.42.
Fig. 8.42 Constant N-circles.
8.9.3
Uses of M -circles and N -circles
The following are the uses of M-circles and N-circles:
(1) Gain adjustment using M-circle
Consider polar plot in the diagram in Figure 8.43.
Fig. 8.43 Gain adjustment by using M-circle drawn on the polar plot.
Chapter 8.indd 70
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FREQUENCY RESPONSE ANALYSIS
8-71
The circle has been drawn with centre on the negative real axis at (M 2/1 – M 2, 0). The circle
is tangent to the line OT. The circle is also touching the polar plot. The radius of the circle is
M/(1 – M 2). The line OT is drawn at an angle q with the negative real axis where q = sin–1 1/M.
sin θ =
AC
OA
where AC = radius of the circle = M/(1 – M 2)
and OA = distance of the centre of the circle
OA =
8.28
M2
8.29
1 − M2
M
2
1
=
sin θ = 1 − M
2
M
M
8.30
1− M2
∴
OB = OA – OB
= OA − AC sin θ
Substituting the values from 8.28, 8.29 and 8.30,
OB =
M2
1− M
2
−
1
= −1
1− M M
M
2
The point B thus gets located on the negative real axis as (–1, 0). The factor by which gain
can be adjusted to obtain the desired value of M can be ascertained.
The steps of determining the gain are mentioned below.
Draw the polar plot for K = 1 (or any assumed value)
•
Calculate θ from sin θ =
•
1
M
Draw a line from origin at angle θ with negative real axis
Draw a circle with centre on negative real axis, which is tangent to the line OT and to the
polar plot
From point C, draw a line CB perpendicular to the negative real axis
Calculate the required value of gain, K as,
•
•
•
•
K =
Assumed value of gain for which the polar plot was drawn
Leength OB
The above procedure is illustrated through the following example.
Example 8.20 The open-loop transfer function with unity feedback is given by
Chapter 8.indd 71
G(s)H(s) =
K
s( s + 1)( s + 5 )
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CONTROL SYSTEMS ENGINEERING
Determine the value of K for which M will be equal to 2.
We will first draw the polar plot of G(s)H(s) assuming K = 1
1
G ( j ω) H ( j ω) =
( j ω) (1 + j ω) (5 + j ω)
ω = 0,
G ( j ω) H ( j ω) = ∞, ∠G ( j ω) H ( j ω) = −90°
ω = ∞, G ( j ω) H ( j ω) = 0, ∠G ( j ω) H ( j ω) = −270°
To determine the value phase cross-over frequency, we will equate the phase angles of
G ( j ω) H ( j ω) to –180o. Thus
f = −180° = −90 ° − tan−1 ω − tan−1 0 .2ω
or,
tan−1 ω + tan−1 0. 2ω = 90º
Taking tan of both sides,
tan(tan−1 ω + tan−1 0. 2ω) = tan 90° = ∞
We know,
tan (A + B) =
(a)
tan A + tan B
1 − tan A tan B
Therefore, from Equation (a), we can write
tan tan−1 ω + tan tan−1 0 .2ω
1 − tan tan−1 ω tan tan−1 0 .2ω
=∞
or,
ω + 0 . 2ω
=∞
1 − ω(0.2ω)
or,
1 − 0 . 2ω 2 =
or,
ω=
1
=
0.2
1 . 2ω
=0
∞
5 = 2.24
We draw the polar plot as in Figure 8.43. We then draw line OT at an angle of 30º with the
negative real axis. A circle is drawn with the centre on the negative real axis and tangent to both
the polar plot and line OT. Note that the M-circle for M > 1 are on the left side of M = 1.
The circle makes tangent to line OT at point C. From C draw a perpendicular line CB on
the negative real axis. With w = 2.24, calculate length OK. Then measure length OB. In this case
the length OB is approximately calcualted as 0.033. The value of K is determined as
K =
1
1
=
= 30.3
OB 0. 033
The intersection of the polar plot and the constant M-circle provides the value of M for a
particular value of w. When the M-circle is tangent to the polar plot, the value of M is the value
of resonant peak Mr the value of w is the resonance frequency wp as has been shown.
Chapter 8.indd 72
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FREQUENCY RESPONSE ANALYSIS
8-73
Fig. 8.44 Use of M-circle.
Use of M-circles and N-circles to determine the closed loop frequency response
Assume the open-loop transfer function of a control system as
25
G(s) =
s(1 + 2 s ) (1 + 5 s )
We write
G ( j ω) =
25
j ω(1 + 2 j ω) (1 + 5 j ω)
We can draw the polar plot of the system as shown in Figure 8.45. We now draw the
M-circles for different values of M. The circles with M > 1 will lie on the left side of M = 1 line
and circles with M < 1 will lie on the right side of M = 1 line as has been shown. The point of
intersection of the polar plot with the M-circles are noted. The intersections give us the value of
M and the corresponding value of frequency w. The M-circle which becomes tangent to the polar
plot gives the value of resonance peak Mr.
Fig. 8.45 (a) Polar plot and M-circle; (b) Frequency response from polar plot and M-circles.
Chapter 8.indd 73
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8-74
CONTROL SYSTEMS ENGINEERING
We now draw the N-circles and the polar plot as shown in Figure 8.46 (a). The point of
intersection of the N-circles and the polar plot provides the values of phase angle at different
values of w. The frequency response of phase angle versus w has been shown in Figure 8.46 (b).
Fig. 8.46 (a) Polar plot and the N-circles; (b) Frequency response of the system.
If we draw the polar plot superimposed on both constant M and constant N-circles, the
magnitude of M and phase angle f can be noted for different values of frequency, w. This way
using constant M and constant N-circles, the closed loop frequency response of the system can
be drawn.
The Nichols Chart
For a given system, we have seen how by using constant M and N-circles and the polar plot,
the frequency response of the system can be drawn.
However, since drawing of polar plot is a bit a lengthy process than Bode plot, N.B. Nichols
used constant M and N-circles on logarithmic co-ordinates by transformation of constant M and
N-circles into log-magnitude and log-phase angle co-ordinates. The chart prepared is named
after Nichols contribution and is called Nichol’s chart. The Nichols chart can be made use of for
drawing the closed-loop frequency response of any system.
REVIEW QUESTIONS
8.1 Plot the Bode diagram for each of the following transfer functions and find gain crossover
frequency:
75(1 + 0.2s)
(1) G(s) =
s(s 2 + 16s + 100)
(2) G(s) =
10(s 2 + 0.4s + 1)
8.2 Given G(s) =
s(s 2 + 0.8s + 9)
Ks 2
(1 + 0.2s)(1 + 0.02s)
Sketch the Bode diagram and obtain the system gain K for the gain crossover frequency
to be 5 rad/sec.
Chapter 8.indd 74
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FREQUENCY RESPONSE ANALYSIS
8-75
8.3 The open-loop transfer function of a system is given by
G(s) H (s) =
100
s ( s + 1) s + 2
Determine
(1) Gain margin and the corresponding gain crossover frequency.
(2) Phase margin and the corresponding phase crossover frequency.
(3) The stability of the system.
8.4 The test data of a unity feedback control system is given below.
w
G ( jw ) , dB
v
46.02
–179.1
0.50
18.14
–175.8
1.0
6.34
–171.8
2.0
–4.90
–166.1
4.0
–15.0
–162.9
8.0
–24.60
–173.1
10.0
–27.96
–180.0
14.0
–33.56
–192.8
20.0
–40.34
–207.9
40.0
–55.74
–234.2
80.0
–72.92
–251.2
140.0
–87.29
–259.1
200.0
–96.53
–262.4
240.0
–101.3
–236.6
G ( j w ), deg.
(1) Plot the data on a semi-log paper.
(2) Determine the transfer function represented by the above data.
(3) What type of system does it represent?
8.5 The asymptotic Bode plot of an open-loop minimum-phase transfer function for a unity
feedback control system is shown in Figure 8.47.
(1) Evaluate the open-loop transfer function.
(2) What is the frequency at which G ( jω) dB = 1? What is the phase angle at this
frequency?
(3) Draw the polar diagram of the open-loop control system.
Chapter 8.indd 75
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8-76
CONTROL SYSTEMS ENGINEERING
Fig. 8.47
8.6 The open-loop transfer function of a unity feedback control system is
K
G( s ) =
2
s(s + s + .5)
Determine the value of K such that the resonant peak magnitude in the frequency response
is 2 dB.
8.7 The block diagram of a control system is shown in the following figure. Draw a Bode
plot of the open-loop transfer function and find the value of gain K that produces a phase
margin of 50°. What is the gain margin for the same value of K?
Fig. 8.48
8.8 A unity feedback control system has the transfer function
K
G( s ) =
s(s + a)
(1) Find the value of K and a to satisfy the frequency domain specification of Mr = 1.04
and ωr = 11.55 rad/sec.
(2) Evaluate the settling time and bandwidth of the system for the values of K and a
determined in part (1).
8.9 The open-loop transfer function of a system is G(s) H (s) =
K
s(T1 s + 1)(T2 s + 1)
Examine the stability of the system for the following two cases:
(1) The gain K is small
(2) The gain K is large.
Chapter 8.indd 76
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FREQUENCY RESPONSE ANALYSIS
8-77
8.10 Using the Nyquist stability criterion, determine the range of gain K (positive) for the
stability of the following system.
Fig. 8.49
8.11 Using MATLAB draw a Bode diagram for the system with open-loop transfer function
25
G( s ) = 2
s + 4s + 25
8.12 The open-loop transfer function is
1
G( s ) =
2
s(s + 0.8s + 1)
Draw a Nyquist plot with MATLAB.
8.13 Construct the Bode plot for the system whose open-loop transfer function with unity
feedback is given as
1
G( s ) =
s(1 + 0.2s)(1 + 0.02s)
Find phase margin and gain margin of the system.
8.14 The open-loop transfer function of a system is given as
1
G( jω) H ( jω) =
jω(1 + 0.1 jω)(1 + jω)
Draw Bode plot and determine the value of K so that the phase margin is 60° and gain
margin is 15 dB.
8.15 A unity feedback system has loop transfer function
(s + 2)
G( s ) =
(s + 1)(s − 1)
Use Nyquist criterion to determine the system stability in the closed-loop configuration.
Is the open-loop system stable?
Chapter 8.indd 77
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Chapter 8.indd 78
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9
STABILITY ANALYSIS IN
FREQUENCY DOMAIN
9.1
POLAR PLOT AND NYQUIST STABILITY CRITERION
Nyquist used polar plots to study to stability of control system way back in 1932 and as a mark of
respect to him, polar plots are also called Nyquist plots. It is interesting to observe that the shapes
of polar plots change due to the addition of poles to the transfer function. As an illustration let us
consider the polar plot of transfer function 1/(1 + jω T1 ) of Figure 9.1.
Let us add poles in this transfer function one more in succession making it
1
1
and
⋅
(1 + jω T1 )(1 + jω T2 )
(1 + jω T1 )(1 + jω T2 )(1 + jω T3 )
Every time we add a pole, the polar plot gets rotated by –90° as shown in Figure 9.1 (b)
and (c).
We can observe that addition of a non-zero pole causes rotation of the polar plot at the high
frequency portion by 90°.
If we add a pole at the negative real axis, the transfer function 1/(1 + jω T1 ) will change to
1
a transfer function of
⋅
(1 + jω T1 )(1 + jω T2 )
The polar plots of these have been shown in Figure 9.1 (a) and (b), respectively. We
can notice that addition of a pole has rotated the polar plot by –90° at both zero and infinite
frequencies.
Chapter 9.indd 1
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9-2
CONTROL SYSTEMS ENGINEERING
Fig. 9.1 Effect of addition of poles on polar plot illustrated.
Figure 9.1 (c) shows the polar plot for adding another pole with total transfer function as
1
. Figure 9.1 (d) is the polar plot for adding a pole at the origin with
(1 + jω T1 )(1 + jω T2 )(1 + jω T3 )
1
TF =
⋅
jω(1 + jω T1 )
Nyquist criterion is based on the principle of argument which states that if there are P poles and
Z zeros of the TF enclosed in s-plane contour, then the corresponding GH plane contour will encircle
the origin Z-times in clockwise direction and P-times in anticlockwise direction.
To illustrate this, let us consider an contour in the s-plane which encloses one zero of G(s)H(s),
say s = α1 . The remaining poles and zeros are placed in the s-plane outside the contour as shown in
Figure 9.2.
Chapter 9.indd 2
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-3
Fig. 9.2 Nyquist stability criterion illustrated.
For any non-singular point in the s-plane contour, there is a corresponding point G(s)H(s)
in the GH-plane contour. Let s follows a prescribed path as shown in the clockwise direction
making one circle and returning to its original position. The phasor (s − α1 ) will cover an angle
of −2π , while the net angle covered by the other two phasors, namely (s − β1 ) and (s − α 2 ) ,
is zero. The phasor representing the magnitude of G(s)H(s) in the GH-plane will undergo a net
phase change of −2π. That is, this phasor will encircle the origin once in the clockwise direction
as shown in Figure 8.28 (b). If the contour in the s-plane encloses two zeros, at α1 and α 2 , in
the s-plane, the contour in the GH-plane will encircle the origin two times. Generalizing this,
it can be stated that for each zero of G(s)H(s) in the s-plane enclosed, there is corresponding
encirclement of origin by the contour in the GH-plane once in the clockwise direction.
When the contour in the s-plane encloses a pole and a zero, say at a and b, respectively,
both the phasors (s − α) and (s − β) will generate an angle of 2π as s traverses the prescribed
path. However, since (s – b) will be in the denominator of G(s)H(s), the contour in the GH-plane
will experience one clockwise and one counterclockwise encirclement of the origin.
If we have p number of poles and z number of zeros enclosed by the contour in the s-plane,
the contour in the GH-plane will encircle the origin by (p – z) times in counterclockwise direction
[since (s – p) is in the denominator of the open-loop transfer function, G(s)H(s)].
For a system to be stable, there should not be any zeros of the closed-loop transfer function
in the right half of s-plane. Therefore, a closed-loop system is considered stable, if the member
of encirclements of the contour in GH-plane around the origin is equal to the open-loop poles of
the transfer function, G(s)H(s).
We can express
G(s)H(s) = [1 + G(s)H(s)] – 1
From this, it can be said that the contour G(s)H(s) in GH-plane corresponding to Nyquist
contour in the s-plane is the same as contour of [1 + G(s)H(s)] drawn from the point (–1 + j0).
Thus, it can be stated that if the open-loop transfer function G(s)H(s) corresponding to
Nyquist contour in the s-plane encircles the point at (–1 + j0) the number of times equal to the
Chapter 9.indd 3
03/11/11 7:58 PM
9-4
CONTROL SYSTEMS ENGINEERING
right half s-plane poles of G(s)H(s) in the counterclockwise direction, it can then be said that the
closed-loop control system is stable.
9.1.1
Nyquist Path or Nyquist Contour
The transfer function of a feedback control system is expressed as
C (s)
G( s )
=
R( s )
1 + G( s ) H ( s )
The denominator is equated to 0 to represent the characteristic equation as
1 + G(s)H(s) = 0
The study of stability of the closed-loop system is done by determining whether the
characteristic equation has any root in the right half of the s-plane. Or, in other words,
we have to determine whether C(s)/R(s) has any pole located in the right half of the s-plane.
We use a contour in the s-plane which encloses the whole of the right hand half of the s-plane.
The contour will encircle in the clockwise direction having a radius of infinity. If there is any pole on
the jw-axis, these are bypasses with small semicircles taking around them. If the system does
not have any pole or zero at the origin or in the jw-axis, the contour is drawn as shown in
Figure 9.3. A few examples on application of Nyquist criterion for stability study will be taken
up now.
Fig. 9.3 Nyquist contour.
Example 9.1 The open-loop transfer function of a unity feedback control system is given as
Chapter 9.indd 4
G(s) H(s) =
K
(1 + τ1s)(1 + τ2 s)
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-5
Examine the stability of the system by applying Routh–Hurwitz criterion, and Nyquist
stability criterion.
Solution The closed-loop transfer function is
K
C (s)
(1 + τ1s )(1 + τ2 s )
=
K
R( s )
1+
(1 + τ1s )(1 + τ2 s )
=
K
2
s τ1 τ2 + s(τ1 + τ2 ) + (K + 1)
The characteristic equation is
or
s 2 τ1τ2 + s(τ1 + τ2 ) + K + 1 = 0
s2 +
(τ1 + τ2 )
τ1τ2
Routh array is
s2
1
s1
τ1 + τ2
s0
of K.
τ1τ2
K +1
τ1τ2
s+
K +1
τ1τ2
=0
K +1
τ1τ2
0
0
Since t1 and t2 are time constants which are positive, the system is stable for all values
Further, as the poles are lying on the left side in the s-plane, the system is stable.
For drawing the Nyquist plot, we substitute s = jw in the transfer function.
We have,
G(s)H(s) =
K
( s + sτ1 )(1 + sτ2 )
=
K
τ1τ2
1 
1 

s + s + 
τ1  
τ2 

Substitute s = jw,
Chapter 9.indd 5
03/11/11 7:58 PM
9-6
CONTROL SYSTEMS ENGINEERING
G ( jω) H ( jω) =
K
τ1τ2
1 
1 

 jω +   jω + 
τ
τ

1 
2 
at w = 0,
G ( jω) H ( jω) =
at ω = ∞,
at ω = 1/ τ1 τ2 ,
The plot of
K
τ1 τ2
× τ1 τ2 = K
G ( jω) H ( jω) = 0
G ( jω) H ( jω) =
K τ1τ2
τ1 + τ2
K
(1 + jωτ1 )(1 + jωτ2 )
as w increases from 0 to ∞ has been shown in Figure 9.4.
The Nyquist plot is symmetrical about the real axis since G ( jω) H ( jω) = G (− jω) H (− jω).
Since the plot of G ( jω) H ( jω) does not encircle (–1 + j0) point, the system is stable for any
positive value of t1, t2 and K.
Fig. 9.4 Nyquist plot of G( j ω)H ( j ω) =
Chapter 9.indd 6
K
.
(1 + j ωτ1)(1 + j ωτ2 )
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-7
If there are any open-loop poles in the jw axis or an open-loop pole at the origin, as for
example, in the case of G(s)H(s) = K / s(1 + τs), the Nyquist contour must be intended to bypass
the origin or the point on the jw axis where such a pole exists. This is because the s-plane contour
should not pass through a singularity of 1 + G(s)H(s).
Example 9.2 The open-loop transfer function of a unity feedback control system is given as
G(s)H(s) =
2
K
s(s + s + 4)
Determine the value of K for which the system is stable by applying (1) Routh–Hurwitz
criterion; (2) Nyquist criterion and (3) Root locus technique.
Solution
(1) The characteristic equation is written as
s(s2 + s + 4) + K = 0
or
s3 + s2 + 4s + K = 0
The Routh array is,
s3
1
4
2
s
1
K
4
−
K
0
s1
1
(2)
K
0
s0
The condition for stability is, K > 0 and (4 – K) > 0, that is, the system is stable if 0 < K < 4.
We have,
G(s)H(s) =
2
K
s(s + s + 4)
Substituting s = jw,
G ( jω) H ( jω) =
=
=
=
Chapter 9.indd 7
K
2
jω [( jω) + jω + 4]
K
jω [4 − ω2 + jω]
K [(4 − ω2 ) − jω]
jω [(4 − ω2 ) + jω][(4 − ω2 ) − jω]
−( j 2 ) K [(4 − ω2 ) − jω]
jω [(4 − ω2 ) 2 + ω2 ]
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9-8
CONTROL SYSTEMS ENGINEERING
=
− K [ω + j (4 − ω2 )]
ω [(4 − ω2 ) 2 + ω2 ]
The crossing of Nyquist plot at the real axis can be found by equating the imaginary part
of the above to 0.
Thus,
4 − ω2 = 0
ω = ±2
or
The magnitude G ( jω) H ( jω) at ω = ±2 , that is, at ω2 = 4 is found as
G ( jω) H ( jω) =
− K [ω]
ω [(4 − 4) + 4]
=−
K
4
Fig. 9.5 Nyquist plot of G( j ω)H ( j ω) =
K
j ω(4 − ω2 + j ω)
.
The Nyquist plot for G ( jω) H ( jω) has been drawn as in Figure 9.5. From this figure, we
notice that the intercept on the real axis at ω = ±2 is –K/4 and this −K/4 is greater than –1.
The value of K is found as
K
> −1
4
or
K>4
We know that if there are P poles and Z zeros in the transfer function enclosed by the
s-plane contour, the corresponding contour in the GH-plane must encircle origin Z times in the
clockwise direction and P times in the anticlockwise direction resulting encircling (P – Z) times
in the anticlockwise direction. Again G(s)H(s) contour around the origin is the same as contour
of 1 + G(s)H(s) drawn from (–1 + j0) point on the real axis.
Here the encirclement of –1 + j0 has been in the clockwise direction two times, hence
N = –2. There is no pole on the right-hand side of the s-plane and hence P = 0
−
Chapter 9.indd 8
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-9
Substituting
P–Z=N
0 – Z = –2 or Z = 2
we have,
Thus, for K > 4, the system is unstable. The system will be stable if K < 4.
The reader is advised to try the root-locus technique of solving this problem to arrive at the
same result as above.
Example 9.3 The open-loop transfer function of a control system is given. Determine the
closed-loop stability using Nyquist criteria
K
G(s)H(s) =
s(1 + τs)
Solution The number of poles of G(s)H(s) in the right half of s-plane, that is, with positive
real part is zero, that is, P = 0.
To decide on the stability criterion as N = –P, that is, how many times the Nyquist plot
should encircle –1 + j0 point for absolute stability, we write
N = –P = 0, that is, the Nyquist plot
should not encircle –1 + j0 point for absolute stability in this case.
As there is one pole at the origin, it should be bypassed by small semicircle as shown in
Figure 9.6.
Fig. 9.6 Nyquist plot (a) s-plane contour; (b) contain in G(s) H(s) plane.
Substituting s = jω,
Chapter 9.indd 9
G ( jω) H ( jω) =
K
jω(1 + jωτ)
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9-10
CONTROL SYSTEMS ENGINEERING
=
=
=
K (−ω2 τ − jω)
(−ω2 τ + jω)(−ω2 τ − jω)
K (−ω2 τ − jω2 )
ω4 τ2 + ω2
−K τ
2 2
ω τ +1
− j
K
ω(1 + ω2 τ2 )
lim G ( jω) H ( jω) = ∞
ω→0
lim ∠G ( jω) H ( jω) = –90°
ω→0
lim G ( jω) H ( jω) = 0
ω→0
lim ∠G ( jω) H ( jω) = –180°
ω→0
From the above calculations, the section I of the Nyquist plot in the s-plane is represented
by section I of the Nyquist plot in the G(s)H(s) plane as shown. This part relates to 0 < ω < +∞.
For −∞ < ω < 0, part I is drawn which is the mirror reflection of part I of the portion of the
Nyquist plot. The semicircle around the origin in s-plane is mapped into semicircular path of
infinite radius. This is represented a change of phase from +π/2 to −π/2. The point –1 + j0 is
not circled by the Nyquist contour and hence N = 0.
We had P = 0, N = 2 – P and hence Z = 0. The system is stable.
Example 9.4 A unity feedback control system is represented by its open loop transfer
function
G(s) H(s) =
8
s(s + 1)(s + 2)
Sketch the polar plot and examine whether the system is stable or not. Also calculate its
gain margin from the polar plot.
Solution First we write the transfer function in frequency domain by substituting s = jω.
For a unity feedback system H(s) = 1
G ( jω) H ( jω) =
8
( jω)(1 + jω)(2 + jω)
The magnitude and phase angle respectively are
Chapter 9.indd 10
G ( jω) H ( jω) =
8
2
ω × 1 + ω × 4 + ω2
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
∠G ( jω) H ( jω) =
8
∠90 tan −1 ω tan −1
ω
2
= −90 ° − tan −1 ω tan −1
9-11
ω
2
lim G ( jω) H ( jω) = ∞ and ∠G ( jω) H ( jω) = −90°
ω→0
lim G ( jω) H ( jω) = 0 and ∠G ( jω) H ( jω) = −90° − 90° − 90°
ω→∞
= –270°
From the above we notice that the locus starts at –90° with magnitude at ∞ and terminates
at the origin with 0 magnitude at –270°.
We also need to calculate the intersection of the locus on the negative real axis by equating
the imaginary part G ( jω) H ( jω) to 0.
For this, we have to rationalize G ( jω) H ( jω)
8
G ( jω) H ( jω) =
( jω)(1 + jω)(2 + jω)
8(− jω)(1 − jω)(2 − jω)
=
( jω)(1 + jω)(2 + jω)(− jω)(1 − jω)(2 − jω)
−8 jω(2 − 3 jω − ω2 )
= 2
2
2
ω (1 + ω )(4 + ω )
24ω2
j8ω(2 − ω2 )
=− 2
−
ω × (1 + ω2 )(4 + ω2 ) ω2 (1 + ω2 )(4 + ω2 )
Equating the imaginary part to 0,
8ω(2 − ω2 ) = 0
ω = 0,
∴
2
Substituting ω = 2 in the real part of G ( jω) H ( jω), value of G ( jω) H ( jω) at the
crossing of real axis is equal to
−
2
24ω2
2
2
ω × (1 + ω )(4 + ω )
=−
24 × 2
2 × (1 + 2)(4 + 2)
48
= −1.33
36
The intersection of the locus with the real axis is at –1.33 + j0. The polar plot is shown in
Figure 9.7. As the locus is encircling the critical point –1 + j0, the system is unstable.
=−
Chapter 9.indd 11
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9-12
CONTROL SYSTEMS ENGINEERING
Fig. 9.7 Study of stability of the system using Polar plot.
9.1.2.1 Determination of Gain Margin and Phase Margin from the Polar Plot
Let at w = wpc the polar plot intersects the negative real axis as shown in Figure 9.7. It is also
the frequency at which phase angle of G ( jω) H ( jω) is –180°. Stability determination is done
considering the critical factor, G ( jω) H ( jω) = 1 and ∠G ( jω) H ( jω) = −180° , that is, the
point –1 + j0 on the negative real axis. Gain margin GM is the gain that can be introduced in the
system till it reaches instability.
GM =
1
G ( jω) H ( jω)
at ω=ωpc
In Example 9.4, the gain margin,
GM =
1
OQ
The gain in decibels is expressed as
In this problem,
=
GM = 20 log10
GM = 20 log10
1
1.33
1
OQ
= 0.75
dB
1
= 20 log10 0.75
1.33
Phase margin is obtained by drawing a circle of unit radius and locating the point of
intersection P as shown in Figure 9.7 on the locus. Let the frequency at this point be called ωgc .
Phase margin is ∠G( jω) H ( jω) at ω = ωgc . By definition.
Chapter 9.indd 12
PM = 180° + ∠ G ( jω) H ( jω) at ω=ω
gc
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-13
Example 9.5 The open-loop transfer function of a control system is given as
8
G(s)H(s) =
(s + 2)(s + 4)
Draw the complete Nyquist plot and comment on the stability of the system.
Solution First we draw the Nyquist contour which consists of the entire jω -axis and a
semicircle of infinite radius. On the jω axis ω varies from −∞ to +∞. On the infinite semicircle
S = lim Re jθ where θ from +π/2 to −π/2. The contour, therefore, will enclose the whole of
R →∞
the right half of the s-plane encircled in the clockwise direction as shown in Figure 9.8.
The transfer function in the frequency domain for the G(s)H(s) is obtained by substituting
s = jω as
G ( jω) H ( jω) =
8
( jω + 2)( jω + 4)
lim G ( jω) H ( jω) = 1
ω→0
lim ∠ G ( jω) H ( jω) = 0
ω→0
lim G ( jω) H ( jω) = 0
ω→∞
lim ∠G ( jω) H ( jω) =
ω→∞
∠0
∠0
=
= −180 °
ω 90 °90 °
−1 ω
tan
tan
2
4
Fig. 9.8 Nyquist plot for transfer function, G(s) H(s) =
8
(s + 2)(s + 4)
⋅
The Nyquist locus in the GH-plane for s = 0 to s = +∞ has been drawn in Figure 9.8 (b).
For the infinite semicircle with radius Re jθ described by s = + j∞ to s = 0 to s = − j∞ for
θ = +π/2 to θ = −π/2 is calculated as
Chapter 9.indd 13
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9-14
CONTROL SYSTEMS ENGINEERING
lim
R →∞
10
(2 + Re )(4 + Re jθ )
jθ
= 0∠180° to 0∠ − 180°
Therefore, the infinite semicircular path of Nyquist contour maps into the origin in the
Nyquist plot.
For the portion −∞ to 0, the locus will be the mirror image of portion I and is represented
as shown in Figure 9.8 (b). The point −1 + j 0 has also been shown. The locus does not encircle
the point −1 + j 0 and therefore N = 0. There is no pole of G(s)H(s) in the right-hand side of the
s-plane. Therefore, P = 0
N=P–Z
O=O–Z
that is,
Z=0
There are no zeros of 1 + G(s)H(s). That is, the roots of the characteristic equation with
positive real part is absent and hence the closed-loop system is stable.
Example 9.6 Determine the stability of the system using Nyquist criteria whose open-loop
transfer function is
G(s)H(s) =
1
s(1 + s)(1 + 2s)
Solution Since there is a pole at the origin, the Nyquist contour will be as shown in
Figure 9.9 below. The Nyquist contour will form an infinitely small semicircle of the centre
by passing the pole at the origin. The transfer function in the frequency domain is obtained by
substituting s = jω in G(s)H(s) as
G ( jω) H ( jω) =
1
jω(1 + jω)(1 + 2 jω)
Fig. 9.9 Nyquist contour.
Chapter 9.indd 14
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-15
For drawing the Nyquist plot for the portion I, that is, for s = 0+ to s = + j∞, we determine
magnitude and phase angle of G ( jω) H ( jω) as
lim G ( jω) H ( jω) =
ω→0
lim ∠G ( jω) H ( jω) =
ω→0
=
1
=∞
j 0(1 + j 0)(1 + 2 j 0)
1
(0 + jω)(1 + jω)(1 + 2 jω)
1
tan −1
=
2ω
ω
ω
tan −1 tan −1
0
1
1
1
−1
tan ∞ tan
= –90°
−1
0 tan
−1
0
=
1
∠90° ∠0° ∠0°
lim G ( jω) H ( jω) = 0
s →∞
lim ∠G ( jω) H ( jω) = −90° − 90° − 90° = −270°
s →∞
For the portion I of the Nyquist contour, the Nyquist plot as shown in Figure 9.10, will
start from –90° with infinite magnitude and terminate at the origin at –270° with 0 magnitude. To
determine whether the locus crosses the negative real axis, we have to rationalize G ( jω) H ( jω)
and put its imaginary part to 0 as done below.
G ( jω) H ( jω) =
=
=
1
jω(1 + jω)(1 + 2 jω)
(1 − jω)(1 − 2 jω)
jω(1 + jω)(1 + 2 jω)(1 − jω)(1 − 2 jω)
1 − 3 jω − 2ω2
jω(1 + ω2 )(1 + 4ω2 )
=
−3 jω
jω(1 + ω2 )(1 + 4ω2 )
=−
Equating imaginary part to 0,
or
Chapter 9.indd 15
(1 + ω2 )(1 + 4ω2 )
+
(1 − 2ω2 )
jω(1 + ω2 )(1 + 4ω2 )
(1 − 2ω2 )
ω(1 + ω2 )(1 + 4ω2 )
[ j 2 = −1]
1 − 2ω2 = 0
ω =
3
+
1
2
= 0.707
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9-16
CONTROL SYSTEMS ENGINEERING
For ω = 0.707,
G ( jω) H ( jω) = −
3
1

 1 +  (1 + 2)

2
− j0
1
= −0.66
1.5
For the infinite semicircle, that is, portion II of the Nyquist contour, we put
=−
lim
R →∞
1
jθ
jθ
Re (1 + Re )(1 + 2 Re jθ )
= 0∠ − 270°
Thus, the infinite semicircle will map at the origin itself. For the portion III of the Nyquist
contour, that is, for s → − j∞ to s = j0–, the locus will be the mirror image of the locus for
portion I as has been shown in Figure 9.10.
Now we will do the mapping of the small semicircle at the origin.
Considering the radius of the infinitely small semicircle as ε, and as θ varies from −π/2
to 0 to +π/2
1


lim  jθ
 = ∞∠90° to ∞∠0° to ∞∠ − 90°
j
θ
j
θ
ε→0 ε e (1 + ε e )(1 + 2ε e )


This part of the Nyquist locus will be on infinite semicircle spread over the entire righthand side of the GH-plane.
The complete Nyquist plot for parts I, II, III and IV of the Nyquist contour has to be drawn
as shown in the diagram below. Note that OB = –0.66. Since the critical point –1 + j0 is not
encircled by the Nyquist locus, the system is stable.
Fig. 9.10 Nyquist plot for G(s)H(s) =
Chapter 9.indd 16
1
s(1 + s )(1 + 2 s )
⋅
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-17
Example 9.7 The open-loop transfer function of a system is
K
G(s)H(s)
s(s + 2)(s + 3)
Determine the value of K for which the system will remain stable with (1) phase margin of
60° and (2) gain margin of 6 dB.
Solution
(1) The sinusoidal transfer function is
G ( jω) H ( jω) =
K
jω(2 + jω)(3 + jω)
Phase margin = ∠G ( jω) H ( jω) + 180°
ω
ω
∠G( jω) H ( jω) = −90° − tan −1 − tan −1
2
3
Given phase margin= 60°
Therefore,
60° = −90° − tan −1
or
30° = tan −1
Taking tangent of both the sides,
tan 30° =
or
0.57 =
ω
2
ω
2
+ tan −1
− tan −1
ω
3
+ 180°
ω
3
 ω ω
 + 
2 3
ω ω

1 − × 
2 3

5ω
6 − ω2
or
ω = 0.5
Magnitude G( jω) at ω = 0.5 has to be equated to 1 to determine the maximum value of
gain K.
We rationalize G ( jω) H ( jω) as
G ( jω) H ( jω) =
=
Chapter 9.indd 17
K
jω(2 + jω)(3 + jω)
K (6 − 5 jω − ω2 )
jω(4 + ω2 )(9 + ω2 )
=−
=
K (2 − jω)(3 − jω)
jω(4 + ω2 )(9 + ω2 )
jk 5ω
jω(4 + ω2 )(9 + ω2 )
+
K (6 − ω2 )
jω(4 + ω2 )(9 + ω2 )
03/11/11 7:59 PM
9-18
CONTROL SYSTEMS ENGINEERING
=−
G ( jω) H ( jω) ω=0.5 =
5K
(4 + ω2 )(9 + ω2 )
5K
(4 + .25)(9 + .25)
− j
K (6 − ω2 )
ω(4 + ω2 )(9 + ω2 )
(9.26)
=1
or
5K = 39.3
or
K = 7.86
(2) Given gain margin = 16 dB
Let the Nyquist plot intersect the negative real axis at a distance a from the origin. Then,
gain margin= 20 log10 1/a
or
6 = 20 log10 1/a
or
a = 0.5
at the point of intersection the imaginary part of the G ( jω) H ( jω) is equal to 0.
Therefore, from Equation (8.26), we can write
K (6 − ω2 )
or
or
ω(4 + ω2 )(9 + ω2 )
=0
ω2 = 6
ω =
6 = 2.45
G ( jω) H ( jω) at ω = 2.45 has to be calculated and equated to a = 0.5 to get the critical
value of gain K for the gain margin of 6 dB. Thus,
5K
5K
2
at ω = 6 is
2
2
(4 + 6)(9 + 6)
(4 + ω )(9 + ω )
Equating this value to a = 0.5
5K
= 0.5
150
5K = 75
or
or
K = 15
The system will remain stable with a gain margin of 6 dB for K = 15.
Example 9.8 The open-loop transfer function of a unity feedback control system is
G(s)H(s) =
(s + 2)
( s + 1)( s − 1)
Draw the Nyquist plot and examine the stability of the closed-loop system. Check also the
stability condition by Routh–Hurwitz criteria.
Solution For unity feedback system, H(s) = 1
Chapter 9.indd 18
G(s)H(s) =
(s + 2)
( s + 1)( s − 1)
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-19
There is one pole on the right-hand side of the s-plane at s = 1. The open-loop system is,
therefore, unstable. Let us examine stability of the closed-loop system, first by Nyquist criterion
and then by Routh–Hurwitz criterion.
Substituting s = jω,
G ( jω) H ( jω) =
( jω + 2)
( jω + 1)( jω − 1)
=−
=−
(2 + jω)
(1 + jω)(1 − jω)
(2 + jω)(1 − jω)(1 + jω)
(1 + ω2 )(1 + ω2 )
 2(1 + ω2 ) + jω(1 + ω2 ) 
= −

2
2
(1
+
ω
)(1
+
ω
)


=−
2
2
− j
ω
1+ω
1 + ω2
We will draw the Nyquist contour and then the Nyquist plot.
Fig. 9.11 Nyquist plot for G(s)H(s) =
(s + 2 )
(s + 1)(s − 1)
.
From the transfer function and the Nyquist contour,
as ω → 0 + , G ( jω) H ( jω) = −2 − j 0
ω → +∞,
G ( jω) H ( jω) = −0 − j 0
ω → 0− ,
G ( jω) H ( jω) = 2 + j0
ω → −∞,
G ( jω) H ( jω) = 0 + j0
For the infinite semicircle
Chapter 9.indd 19
03/11/11 7:59 PM
9-20
CONTROL SYSTEMS ENGINEERING
Re jθ + 2
as q changes from +π/2 to −π/2
( Re jθ + 1)( Re jθ − 1)
= 0
The infinite semicircle will map into the origin in the Nyquist plot as θ will change from
+π/2 to −π/2. The complete Nyquist plot for the Nyquist contour considering parts I, II and III
has been drawn as in Figure 9.11 (b). It is observed that the point –1 + j0 has been encirclea by
the locus once in the anticlockwise direction. Therefore,
N = –1
N = Z – P, P = 1
Hence,
–1 = Z – 1
or
Z=0
Z = 0 means that there are no zeros of 1 + G(s)H(s) in the right-hand side of s-plane.
Hence, the system in closed-loop configuration is stable.
For applying Routh–Hurwitz criterion, we will write the characteristic equation of the
closed-loop system as
1 + G(s)H(s) = 0
or
(s + 2)
=0
1+
(s + 1)(s − 1)
or
(s + 1)(s – 1) + (s + 2) = 0
or
s2 + s + 1 = 0
The Routh array is
s2
1
1
lim
R →∞
s1
1
0
s
1
0
0
As there is no change of sign in the first column of Routh array, the system is stable.
9.2
Drawing Nyquist Plots with MATLAB
Nyquist plots are polar plots, while Bode diagrams are rectangular plots. The command nyquist
(num, den) will draw the nyquist plot of G(s) = num(s)/den(s) whereas the command nyquist
(num, den, w) uses the user-specified frequency vector w in radians per second.
MATLAB returns the frequency response of the system in matrices re, im and w and no
plot is drawn when the following command is invoked.
[re, im, w] = nyquist (num, den)
or Chapter 9.indd 20
[re, im, w] = nyquist (num, den, w)
03/11/11 7:59 PM
STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-21
Example 9.9 Draw a Nyquist plot with MATLAB for the open-loop transfer function given by
1
G(s) =
s(s + 1)
Solution The MATLAB command
0
1];
num = [0
den = [1
1
0];
nyquist (num, den)
will produce an erroneous Nyquist plot. A corrected Nyquist plot is obtained by entering the axis
command
nyquist (num, den)
v = [–2
2
–5
5]; axis (v)
MATLAB Program 9.2 will produce a correct Nyquist plot as shown in Figure 9.12 even
though a warning message “Divide by Zero” may appear on the screen.
MATLAB PROGRAM 9.2
% ……….. Nyquist plot ……………………..
1];
num = [0 0
den = [1
1
0];
nyquist (num, den)
2
–5
5); axis (v)
v = (–2
grid
title (‘Nyquist plot of G(s) = 1/([s(s+1)]’)
Fig. 9.12 Nyquist plot of G(s ) =
Chapter 9.indd 21
1
.
s(s + 1)
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9-22
CONTROL SYSTEMS ENGINEERING
The above Nyquist plot includes the loci for both w > 0 and w < 0. For drawing a Nyquist
plot in the positive frequency region (w > 0), we have to use the command [re, im, w] = nyquist
(num, den, w).
MATLAB Program 9.3 will draw the Nyquist plot only in the positive frequency region
(w > 0) as has been shown in Figure 9.13.
MATLAB PROGRAM 9.3
% —————— Nyquist plot ———————
num = [0 0
1];
den = [1
1
0];
w = 0.1:0.1:100;
[re, im, w] = nyquist (num, den, w);
plot (re, im)
v = [-2
2
-5
5]; axis (v)
grid
title (‘Nyquist plot of G(s)=1/[s(s+1)]’)
x label (‘Real Axis’)
y label (‘Imag Axis’)
Fig. 9.13 Nyquist plot of G(s ) =
1
for w > 0.
s(s + 1)
SUMMARY OF NYQUIST STABILITY CRITERION
Nyquist stability criterion employs a graphical method of analysing a system for its stability. We
know that the closed-loop transfer function of a system is
Chapter 9.indd 22
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-23
C (s)
G(s)
=
R(s) 1+ G (s) H (s)
Putting s = jw,
C ( jω )
=
G ( jω )
R( jω ) 1 + G ( jω ) H ( jω )
The characteristic equation is 1 + G(jw) H(jw) = 0. For the system to be stable, the roots of
the characteristic equation must lie on the left-hand side of the s plane.
The Nyquist stability criterion relates the number of poles and zeros of 1 + G(s) H(s) that
lie on the right-hand side of the s plane to the open-loop frequency response of G(jw) H(jw). This
is explained below.
Suppose the open-loop transfer function G(s) H(s) of a closed-loop system is given. We
draw the polar plot of G(jw) H(jw) as w varies from + ∞ to 0 and its mirror reflection for 0 to –∞.
The number of encirclements by the polar plot of (–1 + j0) is observed. We take clockwise
encirclement as negative and anti-clockwise encirclement as positive.
Nyquist criterion for stability is stated as:
For a closed-loop control system to be stable, the Nyquist plot (polar plot) of G(jw) H(jw)
must encircle the point (–1 + j0) as many times as the number of poles of G(s) H(s) that are in the
right-hand side of the s plane.
If N is the number of encirclements of point (–1 + j0) and P is the number of poles of G(s)
H(s) that are on the RHS of s plane, then for stability of the closed loop,
N=P–Z
For example, suppose the open-loop transfer function of a closed-loop system
G(s) H(s) =
s+2
( s +1) ( s −1)
Here, open-loop pole p =1
Let
N = +1
N=P–Z
1 = 1 – Z
Z=0
Hence the system is stable, i.e., no root of the characteristic equation is lying on the righthand side of the s plane.
Example 9.10 The open-loop transfer function of a closed-loop system (feedback system)
is given as
5
G(s) H(s) =
s (1 − s )
Using, Nyquist stability criterion, check for the stability of the closed-loop system. Also
verify your result by Routh’s criterion.
Chapter 9.indd 23
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9-24
CONTROL SYSTEMS ENGINEERING
Solution We put s = jw
5
G(jw) H(jw) =
µ=
And jω (1− jω )
5
ω 1 + ω2
m = –90° – tan–1
−ω
1
We calculate the value of φ and µ for varying values of w as
w=0
φ = –90°
m=∞
w=1
φ = –90° – (– 45°) = –45°
m = 3.54
w=2
φ = –90° – (– 63.5°) = –26.5°
m = 1.12
m=0
w=∞
φ = –90° – (–90°) = 0
From the G(s) H(s), we find that one pole is on the right-hand side of the s plane and hence
p = 1.
–1+ j 0
Re
ω=∞
ω=0
Fig. 9.14 Nyquist plot of G(s) H(s) =
5 s (1 − s)
We draw the Nyquist plot by considering the values of µ and φ for different values of
w from w = 0 to w = ∞ and to w = –∞ by taking the mirror reflection. The number of encirclement
by the Nyquist plot of (–1 + j0) point is nil. Therefore, N = 0. The stability criterion,
N=P–Z
0=1–Z
or, Z=1
This shows that the closed-loop system is unstable.
Chapter 9.indd 24
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STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-25
To check for stability of the closed-loop system, we write the characteristic equation as
1 + G(s) H(s) = 0
5
1+
=0
or,
s (1 − s )
or, s – s2 + 5 = 0
or,
– s2 + s + 5 = 0
or, s2 – s – 5 = 0
Routh’s array:
s2 1 − 5
s1 −1
s 0 −5
0
0
Since there is one sign change in the first column of the array, the closed-loop system is
unstable.
Example 9.11 The open-loop transfer function of a feedback control system is given as
k
G (s) H (s) =
s ( s 2 + 2 s + 2)
Using the Nyquist stability criterion, determine the maximum value of k for which the
closed-loop system will be stable. Check your answer by Routh’s stability criterion.
Solution
Putting s = jw,
G(jw) H(jw) =
=
µ=
K
jω [( jω ) + 2 jω + 2]
2
k
jω [2 − ω 2 + 2 jω ]
k
ω (2 − ω 2 ) 2 + 4ω 2
2ω
φ = − 90° − tan −1
2 −ω2
Chapter 9.indd 25
w=0
φ = 90°
µ=∞
w=1
φ = −153°
µ=
w=2
φ = −206°
µ=
w=∞
φ = −270°
µ=0
K
5
k
2 18
03/11/11 7:59 PM
9-26
CONTROL SYSTEMS ENGINEERING
The intersection of the locus with negative real axis is to be calculated. We equate the
imaginary part of G(jw) H(jw) to 0.
Alternately, we equate φ with –180° to determine the phase crossover frequency.
–180° = – tan–1
tan −1
or,
2ω
2 −ω2
2ω
2 −ω2
– 90°
= 90°
2ω
or, = tan 90° = ∞
2 −ω2
2 – w2 = 0
Therefore, or, w=
2
wpc =
2
The magnitude of m at wpc is calculated by putting w =
µ=
k
k
=


2  (2 − 2)2 + 4 × 2 
2 8
The Nyquist plot is drawn as shown in the figure.
=
k
16
2 in the expression for µ.
=
k
4
Im
ω =∞
R
Re
ωpc = 2
ω=0
Fig. 9.15 Nyquist plot of
Chapter 9.indd 26
K s( s 2 + 2s + 2)
03/11/11 7:59 PM
STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-27
For the system to remain stable m should be less than 1. That is
k
<1
4
k<4
or The value of gain k should be less than 4 for the closed-loop system to remain stable to
check using Routh’s criterion.
The characteristic equation of the closed-loop system is
1 + G(s) H(s) = 0
k
or,
1+
=0
2
s ( s + 2 s + 2)
or,
s(s2 + 2s + 2) + k = 0
or,
s3 + 2s2 + 2s + k = 0
s3 + 2s2 + 2s + k = 0
Routh’s array:
1
2
s3
2
s2
1
s 4−k / 2
k
s0
k
0
0
k>0
4−k
>0
2
4–k>0
or, k<4
Thus the result is checked to be true.
Example 9.12
Determine the closed-loop stability of a control system whose open-loop
transfer function is
1
G(s) H (s) =
using Nyquist criteria.
s (1 + 2 s ) (1 + s )
Solution
Given
G(s) H (s) =
Put
s = jw
G ( jω ) H ( jω ) =
=
1
s (1 + 2 s ) (1 + s )
1
jω (1+ j ω ) (1+ jω )
2
− jω (1− j 2ω ) (1− jω )
jω (1+ j 2ω ) (1+ jω ) × − jω (1− j 2ω ) (1− jω )
[Rationalizing]
Chapter 9.indd 27
03/11/11 7:59 PM
9-28
CONTROL SYSTEMS ENGINEERING
=
=
=
Separating into real and imaginary parts,
=
=
So
− jω (1− j 3ω − 2ω 2 )
ω 2 (1+ 4ω 2 ) (1+ ω 2 )
− jω − 3ω 2 + j 2ω 3
ω 2 (1+ 4ω 2 ) (1+ ω 2 )
−3ω 2 + jω (ω 2 − 1)
ω 2 (1+ 4ω 2 ) (1+ ω 2 )
−3ω 2
ω 2 (1+ 4ω 2 ) (1+ ω 2 )
−3
(1+ 4ω 2 ) (1+ ω 2 )
Re G ( jω ) H ( jω ) =
Im G ( jω ) H ( jω ) =
+j
+ j
ω (2ω 2 − 1)
ω 2 (1+ 4ω 2 ) (1+ ω 2 )
2ω 2 − 1
ω (1+ 4ω 2 ) (1+ ω 2 )
−3
(1+ 4ω ) (1+ ω 2 )
2
2ω 2 − 1
ω (1+ 4ω 2 ) (1+ ω 2 )
Nyquist plot will cut the real axis at a point at which Im G ( jω ) H ( jω ) = 0
2ω 2 − 1
i.e.,
ω (1+ 4ω 2 ) (1+ ω 2 )
or
or
or
Chapter 9.indd 28
=0
2w2 – 1 = 0
2w 2 = 1
1
w2 =
2
1
or
ω=
w
0
0.707
1
10
∞
Re G ( jω ) H ( jω )
–3
–0.6668
–0.3
–.000074
0
Im G ( jω ) H ( jω )
–∞
0
0.1
.004934
0
= 0.707
2
03/11/11 8:00 PM
STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-29
Im
ω=0
Dm
ω = 0.707
–3
–1+j0
ω=1
– 0.3
ω=∞
Re
– 0.67
ω=0
Fig. 9.16
Since the point (–1 + j0) is not encircled by the Nyquist plot, so there are no roots of closedloop characteristic equation having positive real parts; hence the closed-loop system is table.
Example 9.13
Using Nyquist criterion, examine the closed-loop stability of a system whose
open-loop transfer function is given by
50
G(s) H (s) =
( s + 1) ( s + 2)
Solution
Given G ( s ) H ( s ) =
Put 50
( s + 1) ( s + 2)
s = jw
G ( jω ) H ( jω ) =
=
50
(1+ jω ) (2 + jω )
50(1− jω ) (2 − jω )
(1+ jω ) (2 + jω ) (1− jω ) (2 − jω )
[Rationalizing]
=
Chapter 9.indd 29
50 (2 − j ω − ω )
3
2
(1+ ω 2 ) (4 + ω 2 )
03/11/11 8:00 PM
9-30
CONTROL SYSTEMS ENGINEERING
=
50 ( 2 − ω 2 ) − j 150 ω
(1+ ω 2 ) (4 + ω 2 )
Separating real and imaginary parts
=
so
50(2 − ω 2 )
(1+ ω ) (4 + ω )
Re G ( jω ) H ( jω ) =
and
−150 ω
(1+ ω 2 ) (4 + ω 2 )
(1+ ω 2 ) ( 4 + ω 2 )
–150 ω
(1+ ω 2 ) (4 + ω 2 )
=0
i.e., At w = 0
or
with imaginary axis can be found by,
w=±∞
Re G ( jω ) H ( jω ) = 0
equating 50 ( 2 – ω 2 )
i.e.,
Chapter 9.indd 30
− 150 ω
Im G ( jω ) H ( jω ) = 0
i.e.,
of w.
+j
The intersection of Nyquist plot
with real axis can be found by
equating (b)
2
50 ( 2 − ω 2 )
(1+ ω 2 ) (4 + ω 2 )
Im G ( jω ) H ( jω ) =
Now,
(a)
2
(1+ ω 2 ) (4 + ω 2 )
=0
or or 2 – w2 = 0
w2 = 2
or w=±
2 = ± 1.414
Let us find some values Re G ( jω ) H ( jω ) and Im G ( jω ) H ( jω ) for different value
w
0
1
2
∞
Re G ( jω ) H ( jω )
25
5
0
0
Im G ( jω ) H ( jω )
0
– j15
– j11.7
0
03/11/11 8:00 PM
STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-31
Let us draw Nyquist plot using these values
Im [4 (jω) + H (jω)]
5
(25, 0)
ω = CD
–1+j0
ω=0
Re [G (jω) H (jω) ]
(o – 11.7)
ω=
2
(5 –15)
ω=1
Fig. 9.17
Since the point (–1 + j0) is not encircled by the Nyquist plot, there are no roots of closedloop characteristic equation having positive real parts; so the system is stable.
Example 9.14
Determine the closed-loop stability of the system whose open-loop transfer
function is given as
( s + 0.25)
G(s) H (s) = 2
s ( s + 1) ( s + 0.5)
by applying the Nyquist criterion.
Solution
Given,
G(s) H (s) =
Put s = jw
G ( jω ) H ( jω ) =
=
( s + 0.25)
s ( s + 1) ( s + 0.5)
2
( jω + 0.25)
( jω ) ( jω + 1) ( jω + 0.5)
2
( jω + 0.25) ( jω − 1) ( jω − 0.5)
−ω ( jω + 1) ( jω + 0.5) ( jω − 1) ( jω − 0.5)
2
[Rationalizing]
Chapter 9.indd 31
03/11/11 8:00 PM
9-32
CONTROL SYSTEMS ENGINEERING
=
=
( jω + .25)(− ω 2 − j1.5ω + 0.5)
−ω 2 (1+ ω 2 )(ω 2 + .52 )
1.25ω 2 + .125 − jω 3 + j.125ω
−ω 2 (ω 2 + 1) (ω 2 + .25)
=
1.25(ω 2 + 0.1) − jω (ω 2 − 0.125)
−ω 2 (ω 2 + 1) (ω 2 + 0.25)
=
=
So
and
1.25(ω 2 + 0.1)
−ω 2 (ω 2 + 1) (ω 2 + .25)
−1.25(ω 2 + 0.1)
ω 2 (ω 2 + 1) (ω 2 + 0.25)
Re G ( jω ) H ( jω ) =
Im G ( jω ) H ( jω ) =
The point of intersection of Nyquist plot
(i) with real axis
Put
or
+j
+j
−ω (ω 2 − 0.125)
−ω 2 (ω 2 + 1) (ω 2 + .25)
(ω 2 − 0.125)
ω (ω 2 + 1) (ω 2 + 0.25)
−1.25 (ω 2 + 0.1)
ω 2 (ω 2 + 1)(ω 2 + 0.25)
ω 2 − 0.125
ω (ω 2 + 1)(ω 2 + 0.25)
Im G ( jω ) H ( jω ) = 0
ω 2 − 0.125
ω (ω 2 + 1) (ω 2 + 0.25)
=0
or w 2 – 0.125 = 0
or w 2 = 0.125
(b)
ω = ± 0.125
or with imaginary axis
Re G ( jω ) H ( jω ) =0
Put
−1.25(ω 2 + 0.1)
ω 2 (ω 2 + 1) (ω 2 + .25)
or or or Chapter 9.indd 32
=0
w 2 + 0.1 = 0
w 2 = – 0.1
ω = ± j 0.1
03/11/11 8:00 PM
STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-33
or
w 2 (w 2 + 1) (w 2 + .25) = ∞
or w=±∞
Let us calculate some more values for different values of w
0
w
0.125
1
∞
Re G ( jω ) H ( jω ) =0 ∞
– 5.4
– 0.55
0
Im G ( jω ) H ( jω ) =0 ∞
– j0
j 0.35
0
Let us draw the Nyquist plot.
Im [4 (j ω) H (j ω)]
5,4
(– 5.4,0)
ω =1
– 1 + j0
=
= 0.125
Re [4 (j ω) H (j ω)]
ω=0
Fig. 9.18
Since (–1 + j0) is encircled by Nyquist plot two times, so two roots of the system have
positive real parts; hence the system is unstable.
Example 9.15
whose
Using Nyquist criterion, determine the stability of the closed loop system,
G ( s) H ( s) =
0.5(1− S)
s +1
Solution
Given,
G ( s) H ( s) =
Put
0.5(1− S)
s +1
s = jw
Then G ( jω ) H ( jω ) =
0.5(1 − jω )
jω + 1
Rationalizing it
Chapter 9.indd 33
03/11/11 8:00 PM
9-34
CONTROL SYSTEMS ENGINEERING
0.5(1− jω )
=
( jω + 1)
0.5(ω 2 − 1 + j 2ω )
=
1+ ω 2
0.5(ω 2 − 1)
=
1+ ω
+ j
2
+ j
ω
1 +ω2
ω
So
=
and
0.5(ω 2 − 1)
Re G ( jω ) H ( jω ) =
1 + ω2
1+ ω
( jω − 1)
1 +ω2
2
( jω − 1)
0.5( jω − 1 + ω 2 + jω )
=
0.5(ω 2 − 1)
×
1 +ω2
The intersection of Nyquist plot
(i) with real axis can be found by
Im G ( jω ) H ( jω ) =
Putting
i.e., (ii)
1 + ω2
Im G ( jω ) H ( jω ) =0
or
w=0
with imaginary axis can be found by
ω
Putting
1 + ω2
i.e.,
Chapter 9.indd 34
ω
=0
Re G ( jω ) H ( jω ) =0
or w2 = 1
or Let us find some more values.
w=±1
w
0
1
Re G ( jω ) H ( jω )
–0.5
0
Im G ( jω ) H ( jω )
0
0.5
03/11/11 8:00 PM
STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-35
Let us draw the Nyquist plot.
Im ([4 (j ω) H (j ω)
(0,0.5)
0.5
(– 1 + j0)
Re ([G (j ω) H (j ω)
(– 0.5,0)
Fig. 9.19
Since (–1 + j0) point is not encircled by Nyquist plot, so no root of the closed-loop system
lies on right hand side of s plane; so the system is stable.
Using Nyquist criterion, determine the stability of the closed-loop system,
Example 9.16
whose open-loop transfer function is given by
2(1 − s )
G(s) H (s) =
s +1
Solution
Given,
G(s) H (s) =
put 2(1 − s )
s +1
s = jw
G ( jω ) H ( jω ) =
2(1− jω )
=
=
Chapter 9.indd 35
jω + 1
2(1− jω ) (1− jω )
( jω + 1) (1− jω )
2 (1− jω )2
1+ ω 2
03/11/11 8:00 PM
9-36
CONTROL SYSTEMS ENGINEERING
=
2[1+ j 2ω 2 − j 2ω ]
1+ ω 2
=
2[1− ω 2 − j 2ω )
1+ ω 2
=
2(1− ω 2 )
So
and
1+ ω
2
+ j
Re G ( jω ) H ( jω ) =
2(1− ω 2 )
1+ ω 2
Im G ( jω ) H ( jω ) =
− 4ω
− 4ω
1+ ω 2
1+ ω 2
Nyquist plot will cut the real axis at a point at which
Im G ( jω ) H ( jω ) =0
i.e., − 4ω
1+ ω 2
=0
or w=0
And point of intersection of Nyquist plot with imaginary axis can be found by
Putting
Re G ( jω ) H ( jω ) = 0
2
i.e., 2(1− ω ) = 0
1+ ω 2
or or So
At w2 = 1
w=±1
w=0
2(1− 0)
=2
Re G ( jω ) H ( jω ) =
1+ 02
–2( 0)
=0
Im G ( jω ) H ( jω ) =
1+ 02
At Chapter 9.indd 36
w=1
03/11/11 8:00 PM
STABILITY ANALYSIS IN FREQUENCY DOMAIN
So Nyquist plot is
Re G ( jω ) H ( jω ) =
2(1− 12 )
=0
1+ 12
Im G ( jω ) H ( jω ) =
− 4(1) −4
= −2
=
2
1+ 12
9-37
Im [G (j ω) H (j ω)]
ω = 1–1
(0, –2)
ω= 0–
(2, 0)
(– 2, 0)
(– 1,0)
ω = o+
Re [G (j ω) H (j ω)]
(0, –2)
+
ω=1
Fig. 9.20
Since point (–1 + j0) is encircled by the Nyquist plot, the given system is stable.
REVIEW QUESTIONS
9.1 Plot the Bode diagram for each of the following transfer functions and find gain crossover
frequency:
75(1 + 0.2s)
(1) G(s) =
s(s 2 + 16s + 100)
(2) G(s) =
10(s 2 + 0.4s + 1)
9.2 Given G(s) =
Chapter 9.indd 37
s(s 2 + 0.8s + 9)
Ks 2
(1 + 0.2s)(1 + 0.02s)
03/11/11 8:00 PM
9-38
CONTROL SYSTEMS ENGINEERING
Sketch the Bode diagram and obtain the system gain K for the gain crossover frequency
to be 5 rad/sec.
9.3 The open-loop transfer function of a system is given by
G(s) H (s) =
100
s ( s + 1) s + 2
Determine
(1) Gain margin and the corresponding gain crossover frequency.
(2) Phase margin and the corresponding phase crossover frequency.
(3) The stability of the system.
9.4 The test data of a unity feedback control system is given below.
w
|G ( j w)|, dB
G ( j w), deg.
v
46.02
–179.1
0.50
18.14
–175.8
1.0
6.34
–171.8
2.0
–4.90
–166.1
4.0
–15.0
–162.9
8.0
–24.60
–173.1
10.0
–27.96
–180.0
14.0
–33.56
–192.8
20.0
–40.34
–207.9
40.0
–55.74
–234.2
80.0
–72.92
–251.2
140.0
–87.29
–259.1
200.0
–96.53
–262.4
240.0
–101.3
–236.6
(1) Plot the data on a semi-log paper.
(2) Determine the transfer function represented by the above data.
(3) What type of system does it represent?
9.5 The asymptotic Bode plot of an open-loop minimum-phase transfer function for a unity
feedback control system is shown in Figure 9.21.
(1) Evaluate the open-loop transfer function.
Chapter 9.indd 38
03/11/11 8:00 PM
STABILITY ANALYSIS IN FREQUENCY DOMAIN
9-39
(2) What is the frequency at which G ( jω) dB = 1? What is the phase angle at this
frequency?
(3) Draw the polar diagram of the open-loop control system.
Fig. 9.21
9.6 The open-loop transfer function of a unity feedback control system is
K
G( s ) =
2
s(s + s + .5)
Determine the value of K such that the resonant peak magnitude in the frequency response
is 2 dB.
9.7 The block diagram of a control system is shown in the following figure. Draw a Bode
plot of the open-loop transfer function and find the value of gain K that produces a phase
margin of 50°. What is the gain margin for the same value of K?
Fig. 9.22
9.8 A unity feedback control system has the transfer function
K
G( s ) =
s(s + a)
(1) Find the value of K and a to satisfy the frequency domain specification of Mr = 1.04
and ωr = 11.55 rad/sec.
(2) Evaluate the settling time and bandwidth of the system for the values of K and a
determined in part (1).
9.9 The open-loop transfer function of a system is G(s) H (s) =
Chapter 9.indd 39
K
s(T1 s + 1)(T2 s + 1)
03/11/11 8:00 PM
9-40
CONTROL SYSTEMS ENGINEERING
Examine the stability of the system for the following two cases:
(1) The gain K is small
(2) The gain K is large.
9.10 Using the Nyquist stability criterion, determine the range of gain K (positive) for the
stability of the following system.
Fig. 9.23
9.11 Using MATLAB draw a Bode diagram for the system with open-loop transfer function
25
G( s ) = 2
s + 4s + 25
9.12 The open-loop transfer function is
1
G( s ) =
2
s(s + 0.8s + 1)
Draw a Nyquist plot with MATLAB.
9.13 Construct the Bode plot for the system whose open-loop transfer function with unity
feedback is given as
1
G( s ) =
s(1 + 0.2s)(1 + 0.02s)
Find phase margin and gain margin of the system.
9.14 The open-loop transfer function of a system is given as
1
G( jω) H ( jω) =
jω(1 + 0.1 jω)(1 + jω)
Draw Bode plot and determine the value of K so that the phase margin is 60° and gain
margin is 15 dB.
9.15 A unity feedback system has loop transfer function
(s + 2)
G( s ) =
(s + 1)(s − 1)
Use Nyquist criterion to determine the system stability in the closed-loop configuration.
Is the open-loop system stable?
Chapter 9.indd 40
03/11/11 8:01 PM
10
COMPENSATION TECHNIQUES
10.1 NECESSITY OF COMPENSATION
It has often been observed that the performance of a control system does not satisfy the given
specifications in terms of accuracy, stability and so on. After design and testing, if the system
does not perform satisfactorily some changes may be required to be introduced to achieve the
desired results. The changes could be in the form of adjustment of forward path gain or insertion
of a compensating device in the control system.
To reduce steady-state error, gain can be increased. However, it results in an oscillatory
transient response or even instability. Under such circumstances, it may be necessary to introduce
some kind of corrective sub-systems to force the chosen plant (system) to meet the given
specifications. These sub-systems are known as compensators and their job is to compensate for
the deficiency in the performance of the plant. The desired behaviour of a system is specified in
terms of transient response and steady-state error.
Compensators are required in the following two cases: (i) system is unstable, compensation
is required to stabilize it and also to achieve the desired performance specifications; (ii) system is
stable, compensation is required to achieve improved performance specifications.
System performance is specified in terms of steady-state response as well as transient
response. In time domain, stability criterion is mentioned in terms of damping factor, peak
or maximum overshoot and speed of system response i.e. settling time. In frequency domain,
mention is made of resonant peak, phase margin, resonant frequency, bandwidth, etc.
The design of a feed-back control system involves a compromise between the magnitude
of allowable steady-state error and the degree of stability that is desired in the system
performance.
Chapter 10.indd 1
07/11/11 3:48 PM
10-2
CONTROL SYSTEMS ENGINEERING
10.1.1 Effect of Adjustment of Gain
The performance of a control system is made more satisfactory either by changing the loop gain
or by adding poles and zeros, or by changing time constants so as to reduce the steady-state error
and make the system stable within the limit of specifications.
Let us take the example of a position control servo system. Assume that the limit of steadystate error is met by the system but the damping is too low. This means that the system will be
oscillatory. The immediate suggestion would be to reduce the gain of the amplifier which is
connected to the servo motor as shown in Figure 10.1. The reduction of gain of the amplifier
would increase the damping ratio which can be seen by drawing the root locus plot.
Fig. 10.1 Block diagram of a position control servo system.
It may be discovered that the reduction of gain of the amplifier would cause improved stability
but the velocity error of the system will increase. Thus, we see that gain variation alone may not be
sufficient to achieve the desired result. When a system requires improvement in the steady-state error or
the required degree of stability is greater than what can be obtained by the adjustment of amplifier gain,
inclusion of a network in the loop may be required. Such a network is called a compensating network.
10.1.2 Compensation by Inserting a Network or Compensator
Compensation for improved performance may be provided to the control system by inserting
a network either in series as a cascade or series compensation or in parallel as feedback
compensation as shown in Figures 10.2 (a) and (b), respectively.
Fig. 10.2 Series and parallel compensator for improved stability.
Chapter 10.indd 2
07/11/11 3:48 PM
COMPENSATION techniques
10-3
Let us consider some basic compensating networks which are also called compensators.
Compensators may be made by using electrical, mechanical or pneumatic components. Most
often, electrical components such as resistors and capacitors are used to construct compensators.
In this section, we will discuss three types of electrical compensators—lead compensator, lag
compensator and lag–lead compensator.
10.1.3 Lead Compensator
The lead compensator is a R–C network as shown in Figure 10.3. The transfer function of the lead
compensator is calculated as
Fig. 10.3 A lead network.
Gc(s) =
=
Eo ( s )
=
Ei ( s )
R2
R2
=
R1 /Cs
R1
R2 +
R2 +
( R1Cs + 1)/Cs
R1Cs + 1
R2 ( R1Cs + 1)
R1 R2Cs + R1 + R2
=
R2
( R1Cs + 1)
R2
( R1 + R2 )
1+
R Cs
R1 + R 2 1
Now, considering R1C = τ, and R2/(R1 + R2) = a
1

ατ  s + 
E ( s ) α(1 + τs )
τ

Gc(s) = o
=
=
1
Ei ( s ) (1 + ατs )


ατ  s +

ατ



1
s + 
τ

s + Zc
=
=

1  s + Pc
s +

ατ 

τ>0
α=
Zc
<1
Pc
where Zc and Pc are the zero and pole of the compensator, respectively.
The s-plane representation of the lead compensation with the positions of pole and zero is
shown in Figure 10.4. Note that a lead compensator is essentially a high pass filter.
Chapter 10.indd 3
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10-4
CONTROL SYSTEMS ENGINEERING
Fig. 10.4
Positions of pole and zero of a lead compensator in s-plane.
The lead compensator produces an effect of adding a zero in the system. Since it is difficult
to realize in practice a compensator with one zero only, a compensating network as shown in
Figure 10.3 is generally used. Its transfer function has a zero and a pole as has been depicted in
Figure 10.4.
Such a compensator is used usually in association with an amplifier as shown in
Figure 10.5.
Fig. 10.5 Modified lead compensator with amplifier.
At zero frequency, the network has a gain of α < 1 or an attenuation of 1/α. It is convenient
to cancel the dc attenuation of the lead network with on amplification A = 1/a. This arrangement
will place the compensating pole far from the origin in the left half of the s-plane so that this
pole will have a relatively less effect on the root loci. The effect of adding of lead compensator
to a control system will be studied with examples a little later. It may be noted that in a lead
compensator, the pole is placed to the left of the zero as shown in Figure 10.4.
10.1.4 Lag Compensator
A phase lag compensator has been shown in Figure 10.6 (a). The compensator has a simple pole
and a simple zero in the left half of the s-plane, with the pole to the right of the zero. Note that a
lag compensator is essentially a low pass filter.
The transfer function of the compensator is calculated as follows:
Chapter 10.indd 4
Gc(s) =
Eo ( s )
Ei ( s )
=
R2 +
1
Cs
1
R1 + R2 +
Cs
=
R2Cs + 1
CsR1 + CsR2 + 1
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COMPENSATION techniques
10-5
Fig. 10.6 Phase lag compensator and the positions of pole and zero.
=
R2Cs + 1
( R + R2 )
R2Cs 1
+1
R2
R1 + R2
= β,
R2
1
s+
τs + 1
τ
Gc(s) =
=
τsβ + 1 β( s + 1/ βτ)
Defining,
R2C = τ and
The s-plane representation of a lag compensator has been shown in Figure 10.6 (b).
The lag compensator along with the amplifier of gain Kc has been shown in Figure 10.7.
Fig. 10.7 Modified lag compensator with an amplifier.
1
1


s+ 
s+ 


E (s)
K
τC + 1
τ
τ
Gc(s) = o
= Kc
= c 
= 
1
1
Ei ( s )
(βτC + 1)
β 
 

s +
 s +

βτ
βτ

 

where τ = R2C ,
β=
R1 + R2
R2
,
K =
Kc
β
If the pole zero locations are known, τ and β can be determined directly. The location of
zero is at –1/t and that of pole is at −1/βτ. Thus, we can calculate β.
We have four parameters, R1, R2, C and K, to be determined from three parameters—Kc, τ,
and β—as determined from the design considerations.
Chapter 10.indd 5
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10-6
CONTROL SYSTEMS ENGINEERING
Note that in a lead compensator, the zero is located to the right of the pole whereas in a lag
compensator the zero is located to the left of the pole.
10.1.5 Lag–Lead Compensator
The lag–lead compensator comprises a lead compensator with a pole zero pair and a lag
compensator with another pole zero pair connected in series.
These compensators are used when both transient and steady-state performance of
the system have to be modified. A typical lag–lead compensator is shown in Figure 10.8.
Fig. 10.8 A phase lag–lead compensating network.
The overall transfer function of the network is
Gc(s) =
Eo ( s )
Ei ( s )
= kc
R2 +
1
C2 s
1
1
C1s
+
R2 +
1
C2 s
R1 +
C1s
R1

1 
1 


s +
s +

R1C1  
R2C2 


= kc

1
1 
1
 1
 s2 + 
+
+
s +
R1 R2C1C2
 R1C1 R2C1 R2C2 

 
1 
 s + s +
τ
1 
=k 

1 
 s +
s +
βτ
 
1 
1  
 
τ2  
,
1 

ατ2  






β > 1, α < 1
↑
↑
Lag section Lead section
where τ1 = R1C1, τ2 = R2C2
Chapter 10.indd 6
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COMPENSATION techniques
10-7
The s-plane representation of the lag–lead compensator is shown in Figure 10.9
Gc =
( s + Z c1 )( s + Z c 2 )
( s + Pc1 ) s + Pc 2
Fig. 10.9 s-Plane representation of a phase lag–lead compensator.
Figure 10.10 shows the different types of networks with their transfer functions, pole zero
locations and Bode plot.
Compensating
network
Transfer
function
Pole zero
location
Bode plot
Fig. 10.10 Characteristics of frequently used compensating networks.
Chapter 10.indd 7
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10-8
CONTROL SYSTEMS ENGINEERING
10.1.6 Design Procedure
A control system has been shown in Figure 10.11. Examine the effect of adding a series phase lag
network of transfer function, Gc. Here, we have
30
G(s) =
s( s + 3)
1
H(s) =
( s + 1)
1 + 2s
Gc =
1 + 20s
Fig. 10.11 A feedback control system.
A portion of the Nyquist plot for the system has been drawn in Figure 10.12. The polar
plot intersects the negative real axis at –2.5. The system is therefore unstable (encircles –1). The
system can perhaps be made stable by reducing the gain from 30 to a lower value. However,
keeping the system parameters unchanged, let us use the series phase lag network. The transfer
function of the compensated system becomes equal to
C (s)
R( s)
=
Gc ( s )G ( s ) H ( s )
30(1 + 2 s )
=
1 + G ( s )Gc ( s ) H ( s ) s( s + 3) (1 + 20 s ) + 30 (1 + 2 s )
The polar plot of the compensated system has been shown in Figure 10.12. It can be seen
that the compensated system has made the system perform better in terms of the following
parameters.
Fig. 10.12 Polar plot of uncompensated and compensated systems.
The compensated system is now stable as the polar plot intersects the negative real axis to
the right of –1. Though the stability of the system could be improved by lowering the value of
Chapter 10.indd 8
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COMPENSATION techniques
10-9
gain, the same has been achieved by using the series compensating network. Keeping the value
of gain high helps reduce the steady-state error to certain inputs which is also advantageous from
the point of view of reduced errors, caused due to friction, backlash, etc.
Let us now consider through an example the effect of adding open-loop poles and zeros
which will be helpful in understanding design of feedback control system.
(1) Effect of addition of poles: The root loci of configuration with two real poles at s = 0 and
s = –1 has been drawn in Figure 10.13 (a). The addition of a pole at s = –2 has been shown
in Figure 10.13 (b).
Fig. 10.13 Effect of addition of poles on the root locus.
(2)
The addition of a pole at s = –2 has moved the root loci to the right (breakaway point being
moved to the right) and also causes the root loci to bend towards the right half of the s-plane
as shown in Figure 10.13 (b). The breakaway point has been moved from –0.5 to –0.42. The
addition of another pole at a point further to the left of the real axis will move the root loci
more to the right. The root loci diagram shown in Figure 10.13 (a) shows that the system
is stable, but the addition of poles makes the system unstable even for smaller values of
k. The addition of more poles will make the system unstable even for further reduction of
values of gain k.
Effect of addition of zeros: Addition zeros to the loop transfer function produces a phase
lead which tends to stabilize the closed-loop control system. The root loci of a two-pole
configuration has been shown in Figure 10.14 (a). If a zero is added at s = –b as shown in
Figure 10.14 (b), the resultant root loci will bend towards the left. The stability margin is
therefore increased.
Fig. 10.14 Root loci diagram showing the effect of adding a zero (a) before and (b) after.
Chapter 10.indd 9
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10-10
CONTROL SYSTEMS ENGINEERING
The performance of a control system is generally defined either in terms of time domain
performance measures or frequency domain performance measures. The performances of
a system can be specified in terms of maximum allowable steady-state error for several test
signal inputs, maximum overshoot and settling-time for a step input, and so on. The performance
specifications can be defined in terms of the desirable locations of the poles and zeros of the
closed-loop transfer function.
The location of the s-plane poles and zeros of the system transfer function is first specified.
The locus of the roots of the closed-loop system is then obtained for the variation of one of the
system parameters. On examination, when it is found that the locus of roots does not satisfy
the required configuration, compensating networks are used to alter the locus of the roots of
the uncompensated system. We can, therefore, use the root locus method and use a portable
compensator so that the resultant root locus satisfies the desired closed-loop root configuration.
We can also use frequency performance measures to describe the performance of a feedback
control system.
In such a case, we can describe a system in terms of the peak of the closed-loop frequency
response, resonant frequency, bandwidth and phase margin of the system. We can use a suitable
compensating network to improve and satisfy system performance. As we have already seen, the
compensating network is a R–C circuit.
We shall now take up some design examples using root locus.
The phase lead network has a transfer function
1

s+ 

s + Zc
τ
Gc(s) =
= 
1 
s + Pc

s +

ατ 

where α and τ have been defined for the RC network. The locations of pole and zero are
selected to obtain satisfactory root locus for the compensated control system. The procedure for
compensation design is as follows:
(1) Note the system specifications and translate them into a desired root location for the
dominant roots.
(2) Draw the uncompensated root locus.
(3) If a compensator is necessary, place the zero of the phase lead network directly below the
desired root location or to the left of the first two real poles.
(4) Determine the pole location so that the total angle at the desired root location is 180° and,
therefore, is on the compensated root locus.
(5) Calculate the total system gain at the desired root location and calculate the error constant.
If the error constant is not satisfactory, all the steps from (1) to (4) need to be repeated.
These steps are illustrated as in Figure 10.15.
Figure 10.15 (a) shows the dominant root locations satisfying the specifications in terms of
z and ωn. The root locus of the uncompensated system has been shown in Figure 10.15 (b). A zero
is added at A on the real axis to provide a phase lead as shown in Figure 10.15 (c). The zero has
been placed to the left of the first two real poles. Figure 10.15 (c) shows that the desired root is
Chapter 10.indd 10
07/11/11 3:48 PM
COMPENSATION techniques 10-11
directly above the second pole, the zero, z, is placed somewhat to the left of the second real pole.
The designer must continually see that the compensated system will respond to the influence of
real roots and zeros of the system and that the dominant roots will not by themselves dictate the
response. When the final compensation is achieved, the desired root will obviously be on the root
locus and it is expected that the algebraic sum of the vector angles will be equal to 180° at that
point. The angle θp is calculated such that the total angle is 180°. By locating a line at an angle θp
on the real axis and intersecting the desired root, it is possible to locate the compensator pole p.
Fig. 10.15 Compensation on s-plane using a phase lead network.
The advantage of this root locus method is that the designer is able to specify the location
of dominant roots and hence the dominant transient response.
The disadvantage is that it is not possible to directly specify an error constant. After the
design is complete, the designer has to calculate the gain of the system at the root location which
depends on p and z and also calculate the error constant for the compensated system. He has to
repeat the process if the result is not satisfactory.
Now let us consider the phase lead compensation design using Bode diagram.
Here the frequency response of the series compensation network is added to the frequency
response of the uncompensated system. The procedure is as follows. First, the Bode diagram for
the G( jω)H( jω) is drawn. After examining the plot for G( jω)H( jω), we determine a suitable
location for p and z of Gc( jω) in order to reshape the frequency response. The uncompensated
G( jω)H( jw) is plotted with the desired gain to allow an acceptable steady-state error, the
phase margin, and then the expected Mpw are examined to see whether the system satisfies the
specifications. If the phase margin is not sufficient, phase lead can be provided by adding Gc( jω)
at a suitable location.
Chapter 10.indd 11
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10-12
(1)
(2)
(3)
CONTROL SYSTEMS ENGINEERING
The compensating network is determined through the following steps:
Determine the phase margin of the uncompensated system when the error constants
requirements are satisfied.
Determine the requirement of additional phase lead, θm (allowing for small amount of
safety).
Evaluate α from the relation,
sin qm =
(4)
α −1
α +1
Evaluate 10 log α and determine the frequency where the uncompensated magnitude curve
is equal to –10 log α dB. Since the compensation network provides a gain of 10 log α at ωm,
this frequency is now the zero dB cross-over frequency and also ωm.
Calculate pole p = ωm α and ZP/α
Draw the compensated frequency response and check the phase margin and repeat this
process, if necessary.
We shall now take up design of control systems using compensating networks of different
types discussed earlier.
(5)
(6)
Example 10.1 The characteristic equation of an uncompensated system is given as
1 + G(s)H(s) = 1 +
K
=0
s2
whose root locus is on the jw axis. Design a compensator so that the system meets the specification
of settling time at 2% criterion of Ts ≤ 4 sec and the damping ratio z ≥ 0.4.
Solution We know that settling time is the time required for the system to settle within a
ertain percentage of the input. Here, settling time at 2% criterion, Ts ≤ 4 sec. This occurs when
or
e
or
−ζω n Ts
< 0.02
ζω nTs  4
Ts =
4
ζω n
Ts =
4
=4
ζω n
Here, Ts is given as 4.
\
or
ζωn = 1
Let us choose a dominant root location as –1 ± j2 as shown in Figure 10.16 so that value of
ξ satisfies the specification.
Chapter 10.indd 12
07/11/11 3:48 PM
COMPENSATION techniques 10-13
Referring to Figure 10.16,
ωn cos θ = ζωn = 1
Damping Ratio is
z = cos θ =
1
12 + 22
=
1
= 0 .45 ( ζ ≥ 0 .4)
5
Fig. 10.16 Phase lead compensation of Example 10.1.
We have to follow the procedure for compensation design step by step. After we have found
the desired root location and seen that they satisfy the specifications, we place the zero of the
phase lead compensator just below the desired root on the real axis which is at –1. Therefore,
s = –z = –1.
Now we will calculate qp. For this, we will consider the angle contribution of all the roots
at the desired root and equate it to –180°.
\
–180° = −θp – 115° – 115° + 90°
(115° each for the double
pole at the origin)
or
θp = 270° – 230° = 40°
Angles contributed by the poles are taken as negative.
We now draw a line (shown as dotted line) from the desired root making an angle of 40°
with the negative real axis and intersecting it at point A. The point of intersection is noted as
–3.65. Therefore, s = –p = –3.65.
The lead compensator is
s+z
s +1
Gc(s) =
=
s + p s + 3.65
The transfer function was
Chapter 10.indd 13
G(s)H(s) =
K
s2
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10-14
CONTROL SYSTEMS ENGINEERING
Therefore, for series compensation, the transfer function of the compensated system is
K ( s + 1) s 2 ( s + 3 .65 ). The gain K can be calculated by measuring the distance of the poles and
zeros from the desired root location. The value of K is calculated using the relation,
K =
=
Vector length of poles
Vector length of zeros
3. 23 ×
2
5×
5
=
Lengths DA × DO × DO
Length DZ
= 8. 05
Example 10.2 A control system is represented by open-loop transfer function
G(s) =
K
s ( s + 3)
It is desired that damping ratio of the dominant roots of the system be z = 0.5 and that the
velocity error constant be equal to 10.
Solution We know that velocity error constant Kv = lim s G ( s ). Therefore,
Substituting
lim
s →0
sK
s ( s + 3)
=
s →0
K
3
K
= 10
3
Thus, for satisfying the velocity error constant requirement, K must be 30. With K = 30, we
can calculate the roots of the uncompensated system as
s2 + 3s + 30 = 0
or
−3 ± 32 − 4 × 30
s =
= −1.5 ± j 5.3
2
For the uncompensated system,
ζω n = 1.5
z = cos θ =
OT
=
OD
1.5
= 0. 28
5. 32 + 1. 5 2
Since the damping ratio of the uncompensated network is 0.28 and the desired value is 0.5,
a compensating network has to be used. Let us assume a reasonable settling time of 2 sec.
We know that setting time is
4
Ts =
ζωn
Here Ts = 2 sec (assumed)
Chapter 10.indd 14
Ts = 2 =
4
ζω n
07/11/11 3:48 PM
COMPENSATION techniques 10-15
Fig. 10.17 Phase lead compensation for Example 10.2.
2
ζω n = 2, ωn =
=4
0.5
The location of the desired roots are
s1, s2 = −ζωn ± j ωn 1 − ζ 2
\
= −2 ± j 4 1 − (. 5 )2
= −2 ± j 3.8
The location of the desired roots with z = 0.5, ωn = 4, ζω = 2 has been shown in
n
Figure 10.17. The angle at the desired root location has been calculated approximately and
equated as
–180° = −θ p – 138° – 72° + 90°
or
θ p = 60°
The angle of intercept θ p = 60°. The intersection point of a dotted line drawn from the
desired root location on the negative real axis is at –4.8.
The compensating network is
Gc(s) =
K =
( s + 2)
( s + 4.8)
The value of K is calculated by calculating the distance of the poles and zeros from the
desired pole location as
Chapter 10.indd 15
Distance of the poles 4.2 × 3. 9 × 4. 1
= 18.6
=
Distance of the zeros
3.8
07/11/11 3:49 PM
10-16
CONTROL SYSTEMS ENGINEERING
The compensated system is
G(s)H(s) =
18.6 ( s + 2 )
s( s + 3) ( s + 4.8 )
Example 10.3 A control system is represented by an open-loop transfer function
G(s)H(s) =
K
s ( s + 2)
The damping ratio has to be made equal to 0.5 and the velocity error constant restricted
to 18. Design the lead compensating network for the above. Assume a reasonable settling time.
Solution Given, error constant Kv = 18
We know that
Kv = lim sG ( s ) H ( s )
18 = lim
s →0
s →0
sK
s ( s + 2)
=
K
2
K = 36
or Therefore, to satisfy the requirement of error constant, gain K should be equal to 36.
With this value of K, let us calculate the roots of the uncompensated system from the
characteristic equation.
s(s + 2) + 36 = 0
s2 + 2s + 36 = 0
2 2 − 4 × 36
= −1 ± j 5 . 9
2
From Figure 10.18, damping ratio of the uncompensated system is calculated as
ωn cos θ = ζω n
s1, s2 =
−2 ±
Fig. 10.18 Calculation of damping ratio for the uncompensated system with
36
⋅
G(s)H(s) =
s(s + 2 )
Chapter 10.indd 16
07/11/11 3:49 PM
COMPENSATION techniques 10-17
x = cos θ =
1
(5. 9 )2 + 12
= 0.165
This shows that the value of z of the uncompensated system is 0.165 which is less than the
desired value, that is, 0.5.
We know that settling time
4
Ts =
ζω n
If we consider settling time as 1 sec, then
4
Ts = 1 =
ζω n
Therefore, zωn = 4, which will be on the real axis.
Considering the new value of z = 0.5
4
4
ωn = =
=8
ζ 0.5
For locating the desired roots, we will calculate ζωn and cos q. ζωn is the horizontal
distance on the real axis. A vertical line is drawn at ζωn on the real axis. Then a line is drawn
from the origin at an angle q with the negative real axis. The intersection of the two lines gives
the desired root location.
The desired root location has been shown in Figure 10.19 given below at R and R′.
Fig. 10.19 Determination of the desired root location of the compensated system.
Chapter 10.indd 17
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10-18
CONTROL SYSTEMS ENGINEERING
Now we place the zero of the compensator just below the desired pole location at T on the
negative real axis at s = –z = –4 so that numerator of the compensator transfer function is (s + 4).
After placing the zero at position T on the negative real axis, we will locate the position of
the pole of the compensator.
We have now to calculate the angle qp at the desired pole location due to the other poles
and zeros and equate to 180° by assuming the pole location of the compensator on the real axis
at –p. The line joining R and –p makes an angle qp with the negative real axis as has been shown.
The angle qp of the compensator pole location is determined as
–180° = −θ p − 90° − 81° − 60°
−θ p = −180° + 231°
or,
= 51º
The point of intersection of the dotted line from R to –p on the negative real axis is at –10.8.
Thus, the denominator of the compensator with
s = – p = –10.8 is (s +10.8)
(s + 4 )
The transfer function Gc(s) of the compensator is equal to
( s + 10.8 )
The value of K is calculated as
K =
Distance of poles from R RV × RL × RO 10.4 × 7 .9 × 8
=
=
= 83
Distance of zeros from R
7.8
RT
The compensated system is
83 ( s + 4)
= 83
Gc(s)G(s)H(s) =
s ( s + 2) ( s + 10.8)
The velocity error constant of the compensated system has to be calculated and checked
with the desired value.
Kv = lim
s →0
s × 83( s + 4)
s ( s + 2)( s + 10.8)
= 15.5
This is somewhat less than the desired value of 18. We will have to repeat the design
process by making another choice for the desired root. We may choose ζωn = 4.5 so that ωn
become 9 and continue with the set procedure.
We can calculate the values of the components of the phase lead network considering the
design already made (the students will take values of the repeat design and calculate in the same
way).
1
s+
s+4
τ
Gc(s) =
=
1
s + 10.8
s+
ατ
where τ = R1C and α = R2 ( R1 + R2 ).
Chapter 10.indd 18
07/11/11 3:49 PM
COMPENSATION techniques 10-19
Comparing,
τ =
R2
R1 + R2
=
1
4
= R1C
4
10.8
= 0.37
We have to choose a suitable value of C and calculate values of R and R from the above
1
2
relations.
Example 10.4 The open-loop transfer function of a feedback control system is given as
K
G(s)H(s) =
s ( s + 2)( s + 3)
Design a compensating network so that the steady-steate error is not more than 0.3 for unit
ramp input and the damping ratio is 0.6.
Solution
R(s) =
1
(for unit ramp input)
s2
The steady-state error is
Substituting,
ess = lim SE ( s ) = lim
s →0
s →0
SR( s )
1 + G(s)H (s)
1
2
s
ess =
K
1+
s ( s + 2)( s + 3)
s
1  s( s + 2) ( s + 3) 
( s + 2) ( s + 3)
= 
=
s  s( s + 2) ( s + 3) + K  s( s + 2) ( s + 3) + K
ess = 6
K
The value of K for the uncompensated system for a steady-state error comes to
\
0.3 =
6
K
\
K = 20
Therefore, for the steady-state error to be less than 0.3, the value of K for the uncompensated
system is calculated as 20.
Let us draw the root locus diagram for the system. From the transfer function of the system,
the location of the poles are identified at 0, –2, and –3 on the real axis as has been shown in
Figure 10.20.
Chapter 10.indd 19
07/11/11 3:49 PM
10-20
CONTROL SYSTEMS ENGINEERING
Fig. 10.20 Design procedure for calculation of compensator for a control system
for the desired performance.
To fix the slope of z-line, we calculate b and then find q as
tan b =
=
ζω n
ωn 1 − ξ2
ζ
1 − ζ2
Substituting the given value of z = 0.6,
tan b =
0.6
1 − 0.62
 0.95
b = 43°
\ Slope of z-line,
q = 47°
Therefore, we draw the z-line at an angle of 47° with the negative real axis. Let us place the
zero on the negative real axis to the left of two poles at s = –2.5. The desired root location on the
z-line is shown. Thus, the requirement of z = 0.6 is satisfied and the settling time is calculated as
Ts =
4
4
=
= 1.6 sec
ζω n 2.5
We have
zwn = 2.5 wn =
2.5
= 4.2
0.6
Settling time of Ts = 1.6 sec is quite good. The numerator of the compensating network is
(s + 2.5). We are to fix the pole position by drawing a dotted line from the desired pole position
Chapter 10.indd 20
07/11/11 3:49 PM
COMPENSATION techniques 10-21
to the negative real axis at an angle qp which is calculated by equating the angular distance of the
poles and zeros from the desired pole to –180º as
or
–180° = –qp – 85° – 133° + 90°
qp = 52°
We draw a line (shown as dotted) with qp = 52° and the point of intersection is at –5.1.
The denominator of the compensating network is (s + 5.1)
Thus,
Gc =
(s + 2.5)
( s + 5. 1)
The transfer function of the compensated system is
G(s)H(s)Gc(s) =
K ( s + 2.5)
s ( s + 2)( s + 3)( s + 5.1)
The value of K is calculated as (refer to Fig. 10.20)
K =
G(s)H(s)Gc(s) =
Distance of poles 3.9 × 3 × 4.2 × 3
=
= 54
Distance of zero
2.9
54( s + 2.5)
s ( s + 2)( s + 3)( s + 5.1)
Let us now check for the steady-state error.
ess = lim
s →0
sR( s )
1 + G(s) H (s)
1
2
s
= lim
s →0
54( s + 2.5)
1+
s ( s + 2)( s + 3)( s + 5.1)
s
1
s ( s + 2)( s + 3)( s + 5.1)
=
s s ( s + 2)( s + 3)( s + 5.1) + 54( s + 2.5)
2 × 3 × 5.1
=
= 0.13
54 × 2.5
Thus, the steady-state error is well below 0.3 and is acceptable.
The components of the compensating network can now be calculated using the relation.
Chapter 10.indd 21
1
s+
s + 2.5
τ
Gc(s) =
=
s + 5.1 s + 1
ατ
07/11/11 3:49 PM
10-22
CONTROL SYSTEMS ENGINEERING
where t = R1C and a = R2 ( R1 + R2 ) of the phase lead network shown in Figure 10.21.
Fig. 10.21 Compensating phase lead network designed for a system.
We have to choose a value of C1 say 1 mF to calculate values of R1 and R2.
Example 10.5 The open-loop transfer function of a unity feedback system is given as
G(s) H(s) =
1
s (1 + s )(1 + 0.1s )
Find the gain margin and phase margin. If a phase lag element with transfer function
(1 + s/1 + 0.4s) is added in the forward path, determine by how much the gain must be changed
to maintain the same gain margin.
Solution
G(s)H(s) =
1
s (1 + s )(1 + 0.1s )
The corner frequencies are 1 and 10 rad/sec.
(1) The lowest corner frequency is 1 rad/sec. So, we assume the starting point of the Bode plot
at w = 0.1 rad/sec. The starting slope of the Bode plot is –20 dB/decade due to 1 jω. This
slope continues till the corner frequency of w = 1 rad/sec.
(2) At w = 1 rad/sec, slope changes by another –20 dB/decade due to 1 1 + jω. Thus, the slope
of Bode plot after w = 1 rad/sec becomes –40 dB/decade. This slope continues till the next
corner frequency of w = 10 rad/sec.
(3) At corner frequency w = 10 rad/sec, slope changes by another –20 dB/decade due to
1 1 + j 0 .1ω, making the total slope of –60 dB/sec. This slope continues for higher
frequencies.
The magnitudes of the transfer function components respresented in the Bode plot at
different frequencies are calculated as
w = 0.1 rad/sec
k
1
1
= =
= 10
jω ω 0.1
In terms of dB, 20 log 10 = 20 dB
w = 1 rad/sec
1
1
=
jω (1 + jω) ω 1 + ω2
Contd...
Chapter 10.indd 22
07/11/11 3:49 PM
COMPENSATION techniques 10-23
In terms of dB,
–20 log w – 10 log (1 + w2)
= –20 log (1) – 10 log (2)
w = 10 rad/sec
= 0 – 3.0 = –3 dB
1
jω (1 + jω) (1 + 0.1 jω)
= –20 log w – 10 log (1 + w2) – 10 log (1 + 0.01 w2)
= –20 log 10 – 10 log (101) – 10 log (2)
= –20 – 20.04 – 3.0
= –43.04 dB
The phase angle f, that is, G ( jω)1 + ( jω) is calculated using
q = –90° – tan–1 w – tan–1 0.1w
w
0.1
1.0
5
10
f
–96.28°
–140.71°
–195.25°
–219.28°
With the above values, we can now draw the Bode plot as shown in Figure 10.23 and
determine the gain. With a compensating network introduced in the forward path as shown in
Figure 10.22, we will have
Fig. 10.22 A compensating network introduced in the forward path of the control system.
T(s) = Gc ( s )G ( s ) =
=
1+ s
1
×
1 + 0.4s s (1 + s )(1 + 0.1s )
1
s (1 + 0.1s )(1 + 0.4s )
We will draw the Bode plot for the above T(s). The corner frequencies are
w1 = 2.5 rad/sec and w2 = 10 rad/sec
Let us take starting frequency as the same as 0.1 rad/sec. The slope of the Bode point is
–20 dB/decade till corner frequency of w = 2.5 rad/sec is reached. The slope of the Bode plot will
change by –20 dB/decade at corner frequencies w1 = 2.5 rad/sec and w2 = 10 rad/sec. The total
slope beyond w = 10 rad/sec will be –60 dB/dec.
Chapter 10.indd 23
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10-24
CONTROL SYSTEMS ENGINEERING
Phase angles f of G( jw)H( jw) is calculated as
f = –90° – tan–1 0.1 w – tan–1 0.4 w
w = 0.1, f = –92.80°; w = 1, f = 117.51°;
w = 2.5, f = –149.03°; w = 10 rad/sec, f = –210°
Magnitudes are
(1)
At w = 0.1 rad/sec,
1
1
=
= 10
jω 0.1
In terms of dB, 20 log 10 = 20 dB
(2)
At w = 2.5 rad/sec,
1
jω (1 + j 0.4ω)
In terms of dB,
–20 log w – 20log 1 + (4ω) 2
Substituting w = 2.5
Magnitude = –10 dB
(2)
At w = 10 rad/sec,
1
jω (1 + j 0.4ω)
In terms of dB,
–20 log w – 20log 1 + (0.4ω) 2
Substituting w = 10 rad/sec
Magnitude = –35.05 dB.
We now draw the Bode plot for the compensated network with the above values.
The following are the findings:
Without compensation, gain margin = 0 – (–19.8) = 19.8 dB and Phase margin = 48°.
With the phase lag network introduced, the gain margin is 30.3 dB. Since the gain margin
is required to be kept only at 19.5 dB, we change the value of K
or or
Chapter 10.indd 24
20 log K = (30.3 – 19.8) dB = 19.5 dB
log K = 0.525
K = 3.349
07/11/11 3:49 PM
COMPENSATION techniques 10-25
Fig. 10.23 Bode plot for uncompensated and compensated systems as in Example 10.5.
Example 10.6 The open-loop transfer function of a control system is
G(s) =
K
s ( s + 1)( s + 2)
The system is to be compensated so that the following specifications are met
Damping ratio z = 0.5
Undamped natural frequency wn = 2
Calculate the steady-state error which should be within some reasonable limit.
Also calculate the components of the compensating network.
Solution We have wn = 2, z = 0.5, zwn = 0.5 × 2 = 1.0
ωn 1 − ζ 2 = 2 1 − (0. 5 )2 = 1. 73
The positions of the poles and the desired dominant closed-loop poles are shown as in
Figure 10.24.
We have drawn a line at q = 60°. The vertical line from point D to the real axis must be
equal to 1.73. The location of the dominant pole is thus made. We have now to locate the zero on
the negative part of the real axis just below the dominant root. However, since one pole is lying
just below the dominant root, we locate the zero somewhat to the left of the first two poles. One
pole is at s = 0 and the second one is at s = –1. The position of zero may be located at a point
Chapter 10.indd 25
07/11/11 3:49 PM
10-26
CONTROL SYSTEMS ENGINEERING
say –1.2. The numerator of the compensating network is, thus, decided as (s + 1.2). Now to
locate the position of the new pole, we have to draw a line from the point D to the negative real
axis at an angle qp with that axis. Angle qp is calculated by taking into consideration the angle
contribution of all the roots as
–180° = – qp – 120° – 90° + 85°
qp = 55°
or Fig. 10.24 Location of poles and zeros of the compensated control system.
Distance of p from the origin is calculated as
OP = 1 +
1.73
= 2.35
tan 55
The value of K is calculated as
2.2 × 2 × 1.73 × 2
= 8.4
1.85
The denominator of the compensating network is (s + 2.35). Thus,
K =
s + 1.2
s + 2.35
The open-loop transfer function of the compensated system is
Gc =
Gc(s) G(s) =
( s + 1. 2 )8. 2
( s + 2. 35 )s( s + 1)( s + 2 )
Velocity error constant
Kv = lim sG ( s )
s →0
s ( s + 1.2)8.2
s → 0 ( s + 2.35) s ( s + 1)( s + 2)
= lim
 2.1
Chapter 10.indd 26
07/11/11 3:49 PM
COMPENSATION techniques 10-27
The steady-state error is
ess = lim
s →0
sR( s )
1 + G(s) H (s)
Assuming unit ramp input,
R(s) =
1
s2
1
s2
ess = lim
s →0
( s + 1.2)8.2
1+
( s + 2.35) s ( s + 1)( s + 2)
s
= 0.47
Steady-state error of 0.47 is somewhat on the higher side. The position of zero could be
changed somewhat. This would change the value of K and ess.
The components of the compensating network can be calculated from
1
s+
s + 1.2
τ where t = R C and a = R2
Gc(s) =
=
1
s + 2.35 s + 1
R1 + R2
ατ
Example 10.7 The open-loop transfer function of a unity feedback system is
G(s) H(s) =
4
s ( s + 2)
Design a compensator to meet the following specifications and compare the responses
through MATLAB.
(1) State velocity error constant Kv = 20
(2) Phase margin of at least 50°
(3) Gain margin of at least 10 dB
Solution Here we shall use a lead compensator given by
1
s+
τs + 1
τ
Gc(s) = K c α
= Kc
1
ατs + 1
s+
ατ
where 0 < α < 1
Let us define
G1(s) = KG(s) =
4K
s ( s + 2)
where K = Kcα
Chapter 10.indd 27
07/11/11 3:49 PM
10-28
CONTROL SYSTEMS ENGINEERING
To obtain Kv = 20, we must have
Kv = lim sGc ( s )G (c)
s →0
 τs + 1  4
= lim sK c α 

s →0
 ατs + 1  s ( s + 2)
= 2K = 20
or
K = Kca = 10
so,
G1( jw) H( jw) =
40
jω( jω + 2)
Fig. 10.25 Bode plot of G1( j ω)H ( j ω) =
40
j ω( j ω + 2)
⋅
From Figure 10.25, we observe that the phase and gain margins of the system are 17° and
+∞ dB, respectively. The phase margin of 17° indicates that the system is oscillatory.
As the addition of a lead compensator shifts the gain cross-over frequency towards the
right, we must offset the increased phase lag of G1( jω) by an amount of 5°. So the maximum
phase lead (fm) required to achieve a phase margin of 50° is given by
Chapter 10.indd 28
fm = 50° – 17° + 5° = 38°
07/11/11 3:49 PM
COMPENSATION techniques 10-29
\
1− α
sin φm = sin 38°
1 + α
=
a = 0.24
gives
As the maximum phase lead angle fm occurs at the geometric mean of two corner frequencies
of the lead network, we have
1
 1  1 
wm =    =
τ
ατ
   τ a
since
1 + jωτ
1
1
=
=
= 6. 2 dB
1 + jωατ ω=ωm
α 0. 49
and G1( jω) = –6.2 dB corresponds to the new gain cross-over frequency of ωm = 9 rad/sec., we
get the corner frequencies as follows:
1
= ωm α = 4.41
τ
and
ω
1
= m = 18.4
ατ
α
Also,
K
10
=
= 41.7
α 0. 24
So the transfer function of the lead compensator becomes
Kc =
Gc(s) = 10
Since
Gc (s )
K
G(s ) H (s ) =
0.227 s + 1
s + 4.41
= 41.7
0.054s + 1
s + 18.4
Gc ( s )
10 G ( s ) H ( s )
10
= Gc(s)H(s)
4
 s + 4.41 
= 41.7 

 s + 18.4  s ( s + 2)
is the open-loop transfer function of the compensated system, the Bode plots of Gc/10, G1 and
GcG are shown in Figure 10.26.
The lead compensator causes the gain cross-over frequency to increase from 6.3 to
9 rad/sec. So the bandwidth increases and, thus, the speed of response increases. The phase
and gain margin of the compensated system are now 50° and +∞ dB, respectively. Thus, the
Chapter 10.indd 29
07/11/11 3:49 PM
10-30
CONTROL SYSTEMS ENGINEERING
Fig. 10.26
Bode plot of compensated system.
compensated system shown in Figure 10.27 meets both the steady-state and relative stability
requirements.
Fig. 10.27 Compensated system.
10.1.7Use of MATLAB Programming
Now we shall compare through MATLAB the unit step and unit ramp responses of both the
compensated and uncompensated systems, the closed-loop transfer functions of which are as
follows
C (s)
4
= 2
for uncompensated system
s + 2s + 4
R( s)
=
Chapter 10.indd 30
166.8s + 735.588
for compensated system
s + 20.4 s 2 + 203.6 s + 735.588
3
07/11/11 3:49 PM
COMPENSATION techniques 10-31
and MATLAB Program 10.1 gives the unit step and unit ramp response curves of both the
compensated and uncompensated system as shown in Figures 10.28 and 10.29.
Fig. 10.28 Unit step response curves of the compensated and uncompensated systems.
Fig. 10.29 Unit ramp response curves of the compensated and uncompensated systems.
Chapter 10.indd 31
07/11/11 3:49 PM
10-32
CONTROL SYSTEMS ENGINEERING
MATLAB PROGRAM 10.1
% ***** Unit-step responses*****
num = [0
0
4];
den = [1
2
4];
numc = [0
0
166.8
735.588];
denc = [1
20.4
203.6
735.588];
t = 0:0.02:6;
[c1, x1, t] = step (num,den,t);
[c2, x2, t] = step (numc, denc,t);
plot (t,c1,‘.’,t,c2,‘-‘)
grid
title (‘Unit-Step Responses of Compensated and Uncompensated Systems’)
xlabel (‘t Sec’)
ylabel[(‘Outputs’)
text(0.35,1.3,‘Compensated system’)
text(1.55,0.88,‘Uncompensated system’)
%*****Unit-ramp responses*****
num1 = [0
0
0
4];
den1 = [1
2
4
0];
num1c = [0
0
0
166.8
735.588];
den1c = [1
20.4
203.6
735.588
0];
t = 0:0.02:5;
[y1,z1,t] = step(num1,den1,t);
[y2,z2,t] = step(num1c,den1c,t);
plot(t,y1,‘.’,t,y2,‘-’,t,t,‘—‘)
grid
title(‘Unit-Ramp Responses of Compensated and Uncompensated Systems’)
xlabel(‘t, Sec’)
ylabel(‘Outputs’)
text(0.77,3.7,‘Compensated system’)
text(2.25,1.1,‘Uncompensated system’)
Example 10.8 The open-loop transfer function of a unity feedback system is
G(s) =
1
s ( s + 1) (0.5s + 1)
Design a compensator to meet the following specifications and compare the responses
through MATLAB.
(1) Static velocity error constant Kv = 5/sec.
(2) Phase margin of at least 40°
(3) Gain margin of at least 10 dB.
Chapter 10.indd 32
07/11/11 3:49 PM
COMPENSATION techniques 10-33
Solution Here we shall use a lag compensator given by
1
s+
τs + 1
τ
Gc(s) = K c β
= Kc
1
βτs + 1
s+
βτ
where β > 1
Let us define
G1(s) = KG ( s ) =
K
s ( s + 1)(0.5s + 1)
where K = Kcβ
To obtain Kv = 5/sec., we must have
Fig. 10.30 Bode plots of G 1 (jw), G c (jw) and G c (jw)G(jw).
 τs + 1 
1
= lim sK c β 
= K cβ = K = 5

s →0
s
1
s
(
s
1)(0.5
s + 1)
βτ
+
+


or
so
K = Kcb = 5
G1( jw) =
5
jω( jω + 1)(0.5 jω + 1)
From the Bode plot (Fig. 10.30) of G1( jω), it is observed that phase margin is –20° which
implies that the system is unstable.
Chapter 10.indd 33
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10-34
CONTROL SYSTEMS ENGINEERING
Since the frequency of 0.7 rad/sec corresponds to the phase margin of 40°, the corner
frequency (1/τ) = 0.1 rad/sec is chosen to prevent an overly large time constant of the lag
compensator. As the corner frequency is not far below the frequency of 0.7 rad/sec, we may add
12° to the specified phase margin of 40° to compensate for the phase lag due to lag compensator.
In order to meet the specified phase margin of 40°, the required phase angle of the uncompensated
system G1( jω) should be (–180° + 40° + 12°) or –128° which corresponds to the new gain crossover frequency of 0.5 rad/sec. To bring the magnitude curve of G1( jω) down to 0 dB at this
frequency, the attenuation needed by the lag compensator is –20 dB.
\
20 log 1/β = –20
or β = 10
So, the other corner frequency of the compensator is (1/βτ) = 0.01 rad/sec.
Kc =
\
K 5
=
= 0.5
β 10
So the open-loop transfer function of the compensated system is
Gc(s)G(s) =
5(10s + 1)
s (100s + 1) ( s + 1) (0.5s + 1)
From the Bode plot of Figure 10.26, we observe that the compensated system now satisfies
the requirements of both the steady-state and relative stability.
Now we shall compare through MATLAB the unit step and unit ramp responses of both the
compensated and uncompensated system whose closed-loop transfer functions are as follows.
C (s)
50s + 5
for compensated system
=
4
3
50s + 150s + 101.5s 2 + 51s + 5
R( s)
and 1
for uncompensated system
0.5s 3 + 1.5s 2 + s + 1
MATLAB Program 10.2 gives the unit step and unit ramp response curves of both the
compensated and uncompensated systems as shown in Figures 10.30 and 10.31.
=
MATLAB PROGRAM 10.2
% ***** Unit-step response*****
num = [0
0
0
1];
den = [0.5
1.5
1
1];
numc = [0
0
0
50
5];
denc = [50
150.5
101.5
51
5];
t = 0:0.1:40;
[c1, x1,t] = step (num,den,t);
[c2, x2,t] = step (numc, denc,t);
plot (t,c1,‘.’,t,c2,‘-‘)
grid
Chapter 10.indd 34
07/11/11 3:49 PM
COMPENSATION techniques 10-35
title (‘Unit-Step Responses of Compensated and Uncompensated Systems’)
xlabel (‘t Sec’)
ylabel[(‘Outputs’)
text(12.2, 1.27,‘Compensated system’)
text(12.2,0.7,‘Uncompensated system’)
%*****Unit-ramp response*****
num1 =[0
0
0
0
1];
den1 = [0.5
1.5
1
1
0];
num1c = [0
0
0
0
50
den1c = [50
150.5
101.5
51
5 0];
5];
t = 0:0.1:20;
[y1,z1,t] = step(num1,den1,t);
[y2,z2,t] = step(num1c,den1c,t);
plot(t,y1,‘.’,t,y2,‘-’,t,t,‘—‘)
grid
title(‘Unit-Ramp Responses of Compensated and Uncompensated Systems’)
xlabel(‘t Sec’)
ylabel(‘Outputs’)
text(8.4,3,‘Compensated system’)
text(8.4, 5,‘Uncompensated system’)
Fig. 10.31 Unit step response curves for the compensated and uncompensated systems.
Chapter 10.indd 35
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10-36
CONTROL SYSTEMS ENGINEERING
Fig. 10.32 Unit ramp response curves for the compensated and uncompensated systems.
Example 10.9 The open-loop transfer function of a unity feedback system is
G(s) =
K
s ( s + 1)( s + 2)
Design a compensator to meet the following specification and draw the response curves
through MATLAB.
(1) Static velocity error constant Kv = 10/sec
(2) Phase margin of at least 50°
(3) Gain margin of at least 10 dB
Solution Here we shall use a lag–lead compensator given by

1 
1
 s +  s + 
τ1 
τ2 
Gc(s) = K c 

β 
1 
 s +  s +

τ1 
βτ2 

where β > 1
As the gain K of the system is adjustable we may assume Kc = 1.
Thus, to satisfy Kv = 10/sec., we must have
Kv = lim sGc ( s )G ( s )
Chapter 10.indd 36
s →0
= lim Gc lim sG ( s )
s →0
s →0
07/11/11 3:49 PM
COMPENSATION techniques 10-37
= lim s
s →0
=
K
s ( s + 1) ( s + 2)
K
= 10
2
K = 20
Fig. 10.33 Bode diagrams of G, G c and G c G.
From the Bode diagram (Fig. 10.33) of G( jω) with K = 20, the phase margin is found to be
–32° which indicates an unstable system. Also we see that ∠G ( jω) = –180° at ω = 1.5 rad/sec,
which may be considered as the new gain cross-over frequency to provide a phase lead angle of
50°.
So the corner frequency of the phase lag portion of lag–lead compensator is chosen as
(1/τ2) = 0.15 rad/sec. which is 1 decade below the new gain cross-over frequency of 1.5 rad/sec.
As α = 1/β in the phase lead portion of lag–lead compensator, the maximum phase lead
angle fm is given by
1
β β −1
sin fm =
=
1 β +1
1+
β
Since we need a 50° phase margin and β = 10 gives fm = 54.9° we may take β = 10. So the
other corner frequency of the phase lag portion becomes (1/βτ2) = 0.015 rad/sec.
1−
Chapter 10.indd 37
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10-38
CONTROL SYSTEMS ENGINEERING
From Bode diagram (Fig. 10.33) of G( jω), the gain margin is found to be 13 dB at
ω = 1.5 rad/sec. So the lag–lead compensator must contribute −13 dB at ω = 1.5 rad/sec. to make
the new gain cross-over frequency at ω = 1.5 rad/sec.
The corner frequencies of the phase lead portion are, thus, obtained by intersecting the
straight line of slope 20 dB/decade through the point (–13 dB, 1.5 rad/sec.) with −20 dB line at
(1/τ1) = 0.7 rad/sec. and with 0 dB line at (β/τ1) = 7 rad/sec.
With Kc = 1, the open-loop transfer function of the compensated system now becomes
20
 s + 0.7  s + 0.15 
Gc(s)G(s) = 


 s + 7  s + 0.015  s ( s + 1)( s + 2)
The Bode diagrams of Gc( jω) and Gc( jω)G( jω) are also drawn in Figure 10.33. The phase
margin (50°), gain margin (16 dB) and static velocity error constant (10/sec) of the compensated
system now satisfy the specified value.
The closed-loop transfer function of the compensated system is now given by
C (s)
95.381s 2 + 81s + 10
=
5
4.7691s + 47.7287 s 4 + 110.3026 s 3 + 163.724 s 2 + 82 s + 10
R( s)
MATLAB Program 10.3 will produce the unit step and unit ramp response curves of the
compensated system as shown in Figures 10.34 and 10.35.
MATLAB PROGRAM 10.3
% ***** Unit-step response*****
num = [0
0
0
95.381
81
10];
den = [4.7691
47.7287
110.3026
163.724
82
10];
t = 0:0.05:20;
c = step (num,den,t);
plot (t,c,‘-‘)
grid
title (‘Unit-Step Responses of Compensated System’)
xlabel (‘t Sec’)
ylabel[(‘Outputs’)
%*****Unit-ramp response*****
num1 =[
0
0
0
0
95.381
81
den1 = [4.7691
47.7287
110.3026
163.724
82
10
0];
10];
t = 0:0.05:20;
c = step(num1,den1,t);
plot(t,c,‘-‘,t,t,‘.’)
grid
title (‘Unit-Ramp Response of Compensated System’)
xlabel(‘t Sec’)
ylabel(‘Output’)
Chapter 10.indd 38
07/11/11 3:49 PM
COMPENSATION techniques 10-39
Fig. 10.34 Unit step response of compensated system.
Fig. 10.35 Unit ramp response of compensated system.
Chapter 10.indd 39
07/11/11 3:49 PM
10-40
CONTROL SYSTEMS ENGINEERING
10.2 PID CONTROLLERS
The position of a controller in a control system has been shown in Figure 10.36. The controller
modifies the input to the system on the basis of feedback.
Reference
Signal
R(s)
Comparator
E(s)
+
–
Controller
Error
System
G(s)
C(s)
Output
Feed back
H(s)
Fig. 10.36 Controller in a control system.
Controllers are of different types:
(i) ON/OFF controllers
(ii) Proportional controllers (P controllers)
(iii) Proportional derivative controllers (PD controllers)
(iv) Proportional integrating controllers (PI controllers)
(v) Proportional integrating derivative controllers (PID controllers)
These are conventional controlled used mainly in linear control applications. These are not
preferred in nonlinear control system application.
The other state-of-the-art controllers are:
(i) fuzzy logic controllers
(ii) artificial neural network (ANN)-based controllers
(iii) neuro-fuzzy system, expert systems and genetic algorithms
In this section, we discuss PID controllers in detail.
10.2.1 Proportional Controllers
A proportional controller in a closed-loop control system has been shown in the figure. An
electronic P controller made with the help of an op-amp has been also shown in Figure 10.37.
Chapter 10.indd 40
07/11/11 3:49 PM
COMPENSATION techniques 10-41
E (s)
R(s)
+
A (s)
K
P - controller
–
Output
G - (s)
System
B(s)
H (s)
Feedback
Ro
e (t)
Ri
Input
–
+
a(t) = k e (t)
Output
Fig. 10.37 A p-controller introduced in a closed-loop control system.
e(t).
The output of the P controller, a(t) = k e(t), i.e., the output is proportional to the error signal
The output by input of the controller is expressed as
A( s)
E ( s)
= K=
K
s0
Thus a P controller is a type zero system.
10.2.2 Proportional Derivative Controllers
The actuating signal a(t) in terms of the error signal e(t) for a PD controller is expressed as
a(t) = K e(t) + Td
d e(t )
dt
= proportional force + derivative force
Taking Laplace transform,
A(s) = KE(s) + Td SE(s)
A( s)
= ( K + S Td )
E ( s)
This transfer function of the PD controller has no poles. Such a controller is difficult to be
made in practice.
A block diagram of a closed-loop control system with a PD controller has been shown in
Figure 10.38.
Transfer function,
Chapter 10.indd 41
07/11/11 3:49 PM
10-42
CONTROL SYSTEMS ENGINEERING
R(s)
+
E(s)
(K + STd)
–
G(s)
P D controller
B(s)
C(s)
A(s)
System
H(s)
Fig. 10.38 Block diagram of a PD Controller.
The overall transfer function,
C ( s)
R( s )
Let G ( s) =
=
( K + S Td ) G ( s)
1+ ( K + S Td ) G ( s) H ( s)
ω n2
s 2 + 2 s1ω n s
, H ( s) = 1
(assuming a second-order unity feedback system)


ω n2
(1+ S Td )  2

 s + 2 ζ ω n s 
C ( s)
=


R( s )
ω n2
1+ (1+ S Td )  2
 ×1
 s + 2 ζ ω n s 
=
(1+ S Td ) ω n2
s 2 + ( 2ζω n + Td ω n2 ) s + ω n2
The characteristic equation is
s2 + (2 ζ wn + Td wn2) s + wn2 = 0

T ω 
s2 + 2 ζ wn ζ + d n  s + wn2 = 0

2 

By comparing this characteristic equation with the standard equation
s2 + 2 ζ wn s + wn2 = 0
Effective damping ratio without PD controller is ζ.
ζ effective damping ratio with a PD controller = ζ +
Td ω n
2
This shows that with PD controller, the damping ratio increases. As a consequence, the
system becomes less oscillatory.
Chapter 10.indd 42
07/11/11 3:49 PM
COMPENSATION techniques 10-43
10.2.3 Proportional Integral Controllers
The actuating signal, a(t), for a PI controller is expressed as
a(t) = K e(t) + Ki ∫ e(t) dt
= proportional force + integral force
Taking Laplace transform,
K
A(s) = K E(s) + i E(s)
s
K 
A( s) 
= K + i 
Transfer function,
E ( s) 
s 
=
( K i + SK )
S
A PI controller is easily realizable as it contains a pole and a zero. Since the pole is at the origin,
it increases the type of the system by one.
10.2.4 Basic Elements of a PID Controller
It is possible to improve the performance of a control system by introducing a PID controller.
The controller output provided by a PID controller is described by the following mathematical
relation:
ea(t) = K e(t) + Ki ∫ e(t) dt + Kd
in Laplace transform form,
Ea(s) = E(s) [K +
d
e(t)
dt
Ki
+ S Kd]
j
The PID controller connected to the system is shown in block diagram form in Figure 10.39.
Erro r
signal
E(s)
Reference Signal
R(s)
+
–
kp + s kd +
Ki
ss
A(s)
System
G(s)
Transf er
function
C(s)
PID Controller
Unity feedback
Fig. 10.39 PID Controller is a unity feedback control system.
PID controllers are mostly used in a variety of industrial control systems like in precision
position control system, temperature control in a thermal system, motor driving a gear train, in
pressure control, level control, etc. The PID controller consists of three components, namely:
Chapter 10.indd 43
07/11/11 3:49 PM
10-44
CONTROL SYSTEMS ENGINEERING
(i) Kp E(s) which is proportional to the error signal, (ii) S Kd E(s) which is proportional to the
K
derivative of the error and (iii) i E(s) which is proportional to the integral of the error.
S
A PID controller has all the advantages of proportional, integral and derivative control.
Their individual disadvantages also get neutralized as they are used together.
Proportional integral increases the loop gain of the system and hence reduces the sensitivity
to the variation of system parameters.
The integral component increases the order of the system and reduces the steady-state error
of the system since a pole gets added at the origin in the s plane.
The main advantage of PID controller is that the gain constants Kp, Kd, Ki can be adjusted
to improve the system performance when there is large variation in the system parameters. This
is called PID tuning.
The transfer function of a PID controller is expressed as
K
A( S ) 
=  K p + S Kd + i 
E ( s) 
s 
=
s 2 K + S K + K 

d
p
i
s
The transfer function has two zeros and one pole. Since it is difficult to realize such a
system, a filter element having a transfer function of 1 ( XS + 1) is often added as a dummy
element to the PID controller.
Example 10.10 The open-loop transfer function of a system is given as
G ( s) =
5
s( s + 2 )
It is desired to locate the poles of this transfer function at –8 and –3 ± j4 by using a PID
controller. Calculate the values of Kp, Kd and Ki of the controller so as to achieve this modification.
Solution R(s)
Error
+
–
k p + s kd +
Ki
s
PID Controller
5
C(s)
s (s + 2)
Plan t
Unity feedback
Fig. 10.40
Chapter 10.indd 44
07/11/11 3:49 PM
COMPENSATION techniques 10-45
For the unity feedback system with a PID controller, the system transfer function is

K  5
 K p + S Kd + i 
s  s( s + 2 )
C ( s)

=

R( s )
K  5
1+  K p + S K d + i 
s  s( s + 2 )

or,
C ( s)
R( s )
=
5 ( Kd s2 + K p s + K i )
s3 + 2s 2 + 5 K d s 2 + 5 K p s + 5 K i
In order to achieve poles at s = – 8 and s = – 3 ± j4, the denominator of C ( s) R( s) should
be equated as
s3 + s2 (2 + 5 Kd) + s 5 Kp + 5 Ki = (s + 8) (s2 + 6s + 25)
or s3 + (2 + 5 Kd) s2 + 5 Kp s + 5 Ki = s3 + 14 s2 + 73s + 200
Equating the coefficient on both sides,
2 + 5 Kd = 14
14 − 2
or, Kd =
= 2.4
5
5 Kp = 73
And,
73
or, Kp =
5
= 14.6
And,
5 Ki = 200
or, Ki = 40
Thus the modified transfer function is
Ki
)
s
=
R( s )
( s + 8) ( s 2 + 6s + 25)
C ( s)
or,
5( K d s 2 + K p s +
C ( s)
R( s )
=
5( 2.4 s 2 + 14.6 s + 40)
( s + 8) ( s 2 + 6s + 25)
For a unit step function, R(s) = 1 s and hence the output C(s) is
C ( s) =
Chapter 10.indd 45
5( 2.4 s 2 + 14.6s + 40)
s ( s + 8)( s 2 + 6s + 25)
07/11/11 3:50 PM
10-46
CONTROL SYSTEMS ENGINEERING
10.2.5 An Electronic PID Controller
ei
eo
K
ef
R2
C1
Ro
R1
Co
R3
Proportional
part
Integral
part
Derivative
part
Fig. 10.41 An electronic PID Controller.
An electronic PID controller has been shown in Figure 10.41. The transfer function of this PID
controller can be derived as
 R T  1 
E0 ( s) R2 
S T0 + 1+ 0 + 0  +

=
Ei ( s) R3 
 R1 T1  S T1 
Where T0 = R0 C0, T1 = R1C1
The proportional controller is a simple system which is an amplifier with adjustable gain
control and is placed in the forward path in series with G(s). The output of the controller is
proportional to the error. Increase in gain reduces steady-state error but increases oscillations.
With PD controller, the effective damping ration increases. This reduces the maximum overshoot
and reduces the oscillation with an integral controller. The Controller output and settles only
when the error is zero. The steady-state error will decrease. Using a PID controller, we can
achieve the following advantages: (i) oscillations are reduced to zero, (ii) transient response and
(iii) steady-state response is increased.
REVIEW QUESTIONS
10.1 For the system shown below, design a lead compensator such that wn = 4 rad/sec and z 0.5
for the compensated system shown in Figure 10.41.
Fig. 10.41
Chapter 10.indd 46
07/11/11 3:50 PM
COMPENSATION techniques 10-47
Consider a plant whose open-loop transfer function is
G(s)H(s) =
1
s ( s + 4s + 13)
2
The complex poles near the origin give slightly damped oscillations which are undesirable.
Insert a gain Kc and a compensator Gc(s) in series to speed up the closed-loop response.
Using the root locus, plot for the compensated system, determine which of the following
compensators will give the best performance.
1
aT
, a >1
(1) Gc ( s ) =
1
s+
T
s+
1
aT , a > 1
(2) Gc ( s ) = a
1
s+
T
s+
(3) Gc ( s ) =
1 − T1 s
1 + T2 s
10.3 A unity feedback control system has the open-loop transfer function of
G(s) =
K
s ( s + 2)
with K = 4, damping ratio ζ = 0.5, natural frequency ωn = 2 rad/sec and Kv = 2/sec.
(1) Design a compensator to obtain ωn = 4 rad/sec. while keeping ζ = 0.5. Find the
compensator’s resistances if C = 1 μF.
(2) Design a compensator that increases Kv to 20/sec, while the original system gives a
satisfactory transient response with K = 4.
10.4 A system with velocity feedback is shown in Figure 10.42.
Fig. 10.42
Determine the value of K and Kh so that the following specifications are satisfied.
(1) Damping ratio of closed-loop poles is 0.5
Chapter 10.indd 47
07/11/11 3:50 PM
10-48
CONTROL SYSTEMS ENGINEERING
(2) Settling time < 2 sec
(3) Static velocity error constant ≥ 50/sec
(4) Static velocity error constant 0 < Kh < 1
10.5 A unity feedback system has an unstable plant transfer function of
G(s) =
1
1000 ( s 2 − 1.1772)
Design a series compensator with Gc(s) = Kp(1 + Tds) such that the damping ratio of the
closed-loop system is 0.7 the and undamped natural frequency is 0.5 rad/sec.
10.6 The open-loop transfer function of a plant is given as
G(s) =
5
s (0.5s + 1)
Design a lead compensator in series so that the dominant closed-loop poles are located at
s = –2 ± j2√3.
10.7 The dominant closed-loop poles of an angular position control system shown in
Figure 10.43 are at s = –3.6 ± j4.8. The damping ratio z of dominant closed-loop poles
is 0.6. The static velocity error constant Kv is 4.1/sec, which means that a ramp input of
360°/sec will produce a steady-state error of 87.8°.
Fig. 10.43
A small change in the undamped natural frequency of the dominant closed-loop pole is
allowed. Design a lag compensator to decrease the steady-state error to one-tenth of the
present value or to increase the static velocity error constant to 41/sec while the damping
ratio is kept at 0.6.
10.8 The open-loop transfer function of a unity feedback system is given by
G(s) =
0.625
s (0.5s + 1)(0.125s + 1)
Design a compensator that gives dominant closed-loop poles at s = –2 ± j2√3 and static
velocity error constant Kv = 80/sec.
10.9 The open-loop transfer function of a unity feedback system is
G(s) =
2s + 1
s ( s + 1)( s + 2)
Design a series compensator so that the unit step response will exhibit a maximum
overshoot of 30% or less and a settling time of 3 sec or less.
Chapter 10.indd 48
07/11/11 3:50 PM
COMPENSATION techniques 10-49
10.10 The open-loop transfer function of a unity feedback system is
G(s) =
1
s ( s + 1)( s + 5)
Design a lag–lead compensator so that Kv = 20/sec., phase margin is 60° and gain margin
is not less than 8 dB. Draw the unit step and unit ramp response curve of the compensated
system with MATLAB.
10.11 Design a log compensating network for a system represented by
G(s) =
K
s ( s + 2)( s + 20)
So that the system performance meets the following specifications.
Phase margin ≥ 3.5° and velocity error constant ku = 20/sec.
Chapter 10.indd 49
07/11/11 3:50 PM
This page is intentionally left blank.
Chapter 10.indd 50
07/11/11 3:50 PM
11
STATE SPACE ANALYSIS OF
CONTINOUS SYSTEM
11.1 INTRODUCTION
The concept of modelling, analysis and design of control systems discussed, so far was based on
their transfer functions which suffer from some drawbacks as stated below.
(i) The transfer function is defined only under zero initial conditions.
(ii) It is only applicable to linear time-invariant systems and generally restricted to singleinput singleoutput (SISO) systems.
(iii) It gives output for a certain input and provides no information about the internal state of
the system.
To overcome these drawbacks in the transfer function, a more generalised and powerful
technique of state variable approach in the time domain was developed. The state variable
method of modelling, analysis and design is applicable to linear and non-linear, time-invariant
or time varying multi-input multi-output (MIMO) systems. The placement of closed-loop
poles for improvement of system performance can be done with state feedback. We shall
now discuss the state variable techniques for linear continuous time systems.
11.2 CONCEPTS OF STATE, STATE VARIABLES AND STATE MODEL
We will define first the concepts of state, state variables, state vector and state space before
proceeding for modelling in state space.
State: The state of a dynamic system is the smallest set of variables (called state variables) such
that the knowledge of these variables at t = t0 together with the knowledge of the inputs for t ≥ t0,
completely determines the behaviour of the system for t ≥ t0.
M11_CSE_ISBN_C11.indd 1
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11-2
CONTROL SYSTEMS ENGINEERING
State Variables: The state variables of a dynamic system are the smallest set of variables that
completely determines the state of the dynamic system.
State Vector: This is a vector consisting of n number of state variables that completely determine
the behaviour of a dynamic system.
State Space: The state space is an n-dimensional space whose coordinate axes are the n number
of state variables that completely determine the behaviour of a dynamic system. Any state can be
represented by a point in the state space.
If x1, x2, ….., xn are the n state variables of a dynamic system, then its state vector (x) is ­written
by an n x1 matrix as follows.
;
ª [ º
«[ »
« »
«»
« »
«»
«¬ [Q »¼
where x, x1, x2, …….., xn all are functions of time t.
State Space Equations: The input variables, output variables and state variables are the three
types of variables used in state space modelling of dynamic systems.
Let us now assume that there are r inputs u1(t), u2(t), ……, ur(t), m outputs y1(t), y2(t),
……., ym(t) and n state variables x1(t), x2(t), …….., xn(t) that completely describe a multipleinput multiple-output (MIMO) system. Then the state equations of a time-invariant system
are given by the following n differential equations of first order.
RU
G[L W
GW
I L [ [ [Q X X XU
[ W
L
I L [ [ [Q X X XU
L
IRU
««Q
.
The above state equations may be written in terms of vector equation as X (t) = f(x(t) u(t)), where
x(t), u(t) and f(x(t), u(t)) are state vector, input vector and function vector, ­respectively, and can
be written as
; W
M11_CSE_ISBN_C11.indd 2
ª [ W
«[ W
« « «
« «¬ [Q W
º
»
»
» X W
»
»
»¼
ª X W
«X W
« « «
« «¬ XU W
º
»
»
» DQG
»
»
»¼
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-3
ª I [ [ [Q X X XU
« I [ [ [ X X X
Q U
« «
«
«
«¬ I Q [ [ [Q X X XU
º
»
»
»
I [ W X W
»
»
»¼
The output equations of the above time invariant system are given by the following m ­equations.
\M W
JM [[«[QXX«XU
IRU
M
«P
7KHVHRXWSXWHTXDWLRQVPD\DOVREHZULWWHQLQWHUPVRIYHFWRUHTXDWLRQVDV
\W
J [ W X W
where y(t) and g(x(t), u(t) are output vector and function vector respectively and can be ­written as
< W
ª \ W
«\ W
« « «
« «¬ \P W
º
»
»
» DQG
»
»
»¼
ª J [ [ [Q X X XU
« J [ [ [ X X X
Q U
« «
«
«
«¬ J P [ [ [Q X X XU
º
»
»
»
J [ W X W
»
»
»¼
Similarly, the state equations and output equations of a time-varying system may be written by
vector equations as
I [ W X W W DQG
; W
< W
J [ W X W W The state equations and output equations together form the state model of a system.
11.2.1 State Model of Linear Systems
The state model of a linear time-invariant system can be written as
DL[DL[«DLQ[Q
[L
DQG
IRU
\M
IRU
M11_CSE_ISBN_C11.indd 3
L
M
ELXELX«ELUXU
«Q
FM[FM[«FMQ[Q
GMXGMX«GMUXU
«P
21/10/11 11:35 AM
11-4
CONTROL SYSTEMS ENGINEERING
DL[DL[«DLQ[Q
[L
L
IRU
DQG
ELXELX«ELUXU
«Q
FM[FM[«FMQ[Q
GMXGMX«GMUXU
«P
\M
M
IRU
The set of state equations and output equations of the above state model may be written
in a vector-matrix form as given below.
; W
< W
and
$[ W %X W
&[ W 'X W
where x(t) is n × 1 state vector, u(t) is r × 1 input vector, A is the n × n state matrix of
constant elements, B is the n × r input matrix of constant elements, C is the m × n output
matrix of constant elements and D is the m × r transmission matrix of constant elements. The
matrices A, B, C, D are defined as follows
ª D D DQ º
« D D D »
Q »
« » %
$ «
«
»
«
»
«¬ DQ DQ DQQ »¼
&
ªF F FQ º
«F F F »
Q »
« «
» '
«
»
«
»
«¬FP FP FPQ »¼
ªE E EU º
«E E E »
U »
« «
»
«
»
«
»
«¬EQ EQ EQU »¼
ª G G GU º
« G G G »
U »
« «
»
«
»
«
»
«¬ G P G P G PU »¼
So, the block diagram of the above linear time-invariant MIMO (r input and m output)
system can be drawn as follows.
'
˜
X W
%
[ W
∫
[ W
&
\ W
$
Fig. 11.1 Block diagram of a linear time-invariant continuous time MIMO
control system in state space
M11_CSE_ISBN_C11.indd 4
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STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-5
In the above figure, the broad arrows represent vector quantities and state vector is the
output of the integrator block.
11.2.2State Model of Single-Input Single-Output Linear Systems
The state model for a linear time-invariant SISO system can be written as follows by taking
r = 1 and m = 1.
[ W
$[ W %X
DQG
\W
&[ W GX
where d is a constant and u is a scalar control variable. Also B and C are row n × 1 and 1 × n
matrices respectively. So the block diagram of the linear time-invariant SISO system can be
drawn as follows
'
˜
X
%
[ W
[ W
&
∫
±
\
$
Fig. 11.2 Block diagram of a linear time-invariant continuous time SISO
control system in state space
It may be noted that state vector as well as state model of a control system in state space
are not unique, although the minimal number (n) of elements in the state vector of the given
system is constant and the system is said to be of the order n.
11.3 STATE MODELS OF LINEAR CONTINUOUS TIME SYSTEMS
Linear time-invariant continuous time dynamic systems are described by linear differential
equations with constant coefficients. The general form of an nth-order linear differential equation
relating the input u(t) and the output y(t) of a linear time invariant continuous time system is
given by
\
Q
D \
Q DQ \ DQ \
E X
P
EX
P EP X EP X
where ai (for i = 1, 2, ...., n) and bj (for j = 1, 2, ......... ,m) are constants; m and n are integers such
that m ≤ n; y(n) and u(m) denote respectively the nth and mth order derivatives with respect to time
.
..
t. The initial conditions are y(0), y(0), y (0) , .......... y(n-1)(0).
M11_CSE_ISBN_C11.indd 5
21/10/11 11:35 AM
11-6
CONTROL SYSTEMS ENGINEERING
The state model of a given differential equation is not unique and any two models can
be uniquely related by a linear transformation of x = Pz where ‘x’ and ‘z’ are two different
sets of state variables belonging to two different state model of the same system and ‘P’ is
a non-singular (that is, |P| ≠ 0) constant matrix. We shall now discuss some commonly used
forms of state models in the following sub-sections.
11.3.1 State Space Representation Using Physical Variables
The state model obtained from physical variables as state variables can be realised through an
example on real physical system. Let us consider a simple electrical system of RLC ­network
shown in Fig. 11.3.
/
H W
5
L W
&
YF W
Fig. 11.3 RLC network
From the figure it is evident that if we have a knowledge of the initial conditions of
v(0), i(0) and the input signal e(t) for t ≥ 0, then the behaviour of the network is completely
­specified for t ≥ 0. We may now assign the current i(t) through the inductor and the voltage
v(t) across the capacitor as state variables x1(t) and x2(t) respectively.
LL WW
7KDWLV
[[ WW
7KDWLV
7KDWLV
[ W
....(11.1)
DQG
YLY WWW
DQG
[ W
[
W
Y
W
DQG
....(11.2)
)URP)LJWKHVWDWHHTXDWLRQRIWKH5/&QHWZRUNFDQEHZULWWHQDV
)URP)LJWKHVWDWHHTXDWLRQRIWKH5/&QHWZRUNFDQEHZULWWHQDV
From
Fig. 11.3, the state equation of the RLC network can be written as
)URP)LJWKHVWDWHHTXDWLRQRIWKH5/&QHWZRUNFDQEHZULWWHQDV
GL WW
5 L W Y W H W
/ GL
5 L W Y W H W
GLGWW
/
5 L W Y W H W
/ GW
GW
G
G YY WW
&
DQG
LL WW YW
& G GW
DQG
& GW
DQG
LW
GW
8VLQJHTXDWLRQV
DQG
WKHDERYHVWDWHHTXDWLRQVDUHDVIROORZV
Using
equations (11.1)
(11.2)
above state equations are as follows
8VLQJHTXDWLRQV
and
DQG
the
WKHDERYHVWDWHHTXDWLRQVDUHDVIROORZV
8VLQJHTXDWLRQV DQG WKHDERYHVWDWHHTXDWLRQVDUHDVIROORZV
G[
5
G[ WW
5 [ W [ W HW
....(11.3)
G[GW
W
5
[ W /
HW
[
W
/
/
/ [ W / [ W / H W
GW
GW
/
/
/
G[
W
....(11.4)
G[ W
[ W
G[GW
[ W
W
&
& [ W
GW
&
GW
M11_CSE_ISBN_C11.indd 6
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-7
If we assume the voltage across the capacitor as the output variable y, then the output ­equation
can be written as
\\
[[ WW
....(11.5)
6RWKHVWDWHPRGHOXVLQJSK\VLFDOYDULDEOHVDVVWDWHYDULDEOHVFDQ
6RWKHVWDWHPRGHOXVLQJSK\VLFDOYDULDEOHVDVVWDWHYDULDEOHVFDQ
So the state model using physical variables as state variables can be written in a vector-matrix
DVIROORZV
DVIROORZV
form as follows.
5
ºº
ªª 5
ºº
»» ª [ º ªª «« / ªª [[ ºº
/ ª [ º «« »»
/
/
«
»
/
«
» «« »» « / » X
«« »»
««
»» «« »» «« »» X
«« »»
«
»» ««¬ [[ »»¼ «« »»
« ««¬ [[ »»¼
¬ ¼
»¼¼
««
»» ¬ ¼ «¬¬ ¬¬ &
¼
&
¼
ª [ º
>@ «« »»
«¬ [ »¼
\
ZKHUHX H W LVWKHLQSXW
11.3.2 State Space Representation Using Phase Variables
Let us consider a dynamic, linear, time-invariant and continuous time system the nth order linear
differential equation of which is given by
\
Q
D \
Q DQ \ DQ \
E X
Q
EX
Q EQ X EQ X ....(11.6)
.
.
where b0, bi and ai are constants for i = 1, 2, …., n and y(0), y (0), …., y(n–1) (0) and u(0), u(0), …,
u(n–1) (0) are the initial conditions to completely specify the system.
In order to obtain the state model of the above linear differential equation, the following
state variables are chosen in such a way as to eliminate the derivatives of the forcing function
u in the state equations.
[
\E X
[
\ E X
[
[ E X \ E X E X
[
[ EX \ E X EX
[
[ E X
[
[ E X
\ E X E X E X
\ E X EX E X
....(11.7)
[ E X \ Q E X Q E X Q E X E X
[Q±
[Q±
[QQ E QQX \ Q E X Q EX Q E QQX E QQX
[
[ E X
[QQ
[QQ E QQX
\ Q E X Q E X Q E X E X
\ Q E X Q EX Q E QQX E QQX
M11_CSE_ISBN_C11.indd 7
21/10/11 11:35 AM
11-8
CONTROL SYSTEMS ENGINEERING
. ..
Putting the values of y, y, y ,.....,y(n−1), y(n) from equations (11.7) into equation (11.6) we
.
will get the value of xn which can be made free from derivatives of forcing function u by
writing the values of β0, β1, β2, …., βn-1, βn as follows.
β
β
β
β
βQ±
βQ
E
E±Dβ
E±Dβ±Dβ
E±Dβ±DE±Dβ
EQ±±DβQ±±«±DQ±β ±DQ±β
EQ±DβQ±±«±DQ±β±DQβ
....(11.8)
So from equations (11.7) and (11.8) the state equations and output equation can be
­written as
[
[ EX
[ E X
[
[Q [Q
DQG
\
[Q E Q X
DQ [ DQ [ D [Q E Q X
[ βX
Hence the state space representation of equation (11.6) in
written as
ª [ º
ª « « [ »
«
« »
« « »
«
« »
« «
»
«
« [Q »
«
« [ »
¬« DQ DQ DQ ¬ Q ¼
DQG
RU
M11_CSE_ISBN_C11.indd 8
\
a vector-matrix form can be
º ª [ º ª E º
»» «« [ »» «« E »»
» « » « »
» «
»«
» X
» « » « »
» « [Q » « E Q »
» «
» «
»
D ¼» ¬« [Q ¼» ¬« E Q ¼»
ª [ º
« [ »
« »
« »
>@ «
» E X
« »
« [Q »
«
»
¬« [Q ¼»
; $[%X ½
< &['X ¾¿
21/10/11 11:35 AM
« [ »
« »
« »
\
>@ «
» E X
« »
« [Q »
«
»
STATE SPACE ANALYSIS
«¬ [Q »¼ OF CONTINOUS SYSTEM
DQG
; $[%X ½
< &['X ¾¿
RU
ª [ º
«
»
« [ »
«
»
« »
ZKHUH[ «
» $
« »
« [ »
« Q »
« [ »
¬ Q ¼
ª « «
« «
« « «
¬« DQ
DQ DQ º
»»
»
» %
»
»
»
D ¼»
11-9
....(11.9)
ª E º
«
»
« E »
«
»
« »
«
»
« »
«E »
« Q »
«E »
¬ Q ¼
& >«@DQG' β
The form of matrix ‘A’ is known as the Bush or Companion form.
The initial conditions of state variables can be written in a vector-matrix form as ­follows.
The block diagram of equation (11.9) is shown in Fig. 11.4.
βQ±
X
β
±
∫
[Q
D
β
∫
[
β
∫
DQ±
[
\
DQ
Fig. 11.4 Block diagram of linear differential equation (11.9) in state space
Phase variables are defined as those particular state variables which are obtained from one (often
the system output) of the system variables and its derivatives. The state model in phase variables
form for the linear differential equation
\
Q
D \
Q DQ \ DQ \
EQ ....(11.10)
with zero initial conditions may be obtained from equation (11.9) by putting β0 = β1 = β2 =
.............. = βn–2 = βn–1 = 0 and βn = b. The corresponding phase variables will be obtained from
equation (11.7).
M11_CSE_ISBN_C11.indd 9
21/10/11 11:35 AM
11-10
CONTROL SYSTEMS ENGINEERING
Two other ways of phase variable formulation of differential equation (11.6) with zero
initial conditions are discussed in sections (a) and (b).
(a) Observable phase variable form: The transfer function Y(s)/U(s) can be
written as
E < V
8 V
Q
¦
L Q
¦
L RU
EL V L
DL V L
E8 V <V
Q
¦
L E8
V DL< V
L
VL
....(11.11)
1RZOHWXVGHILQHWKHVWDWHYDULDEOHVDVIROORZV
EL8 V ±DL< V
V;L V ±;L V
L
IRU
DQG
;Q V
....(11.12)
««Q
From (11.11) and (11.12) we get
<V
<V
Q
V; V ; L V
V L ; L V
V
VL
E8 V E8 V Q
V; LL
¦
¦
L E8 V E8 V L
Q
§¨ ; LL V
¦
© V L ¦
L ¨
© V
L Q
L §; V
; V ·
; LL L V ¸·
V L ¸¹
V
¹
;
V
E8 V ; V ; QQ Q V
E8 V ; V V Q
V
∴
<V
E8 V ; V
∴
<V
E8 V ; V
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
\
EX[
\
EX[
7DNLQJLQYHUVH/DSODFHWUDQVIRUPRI
ZHJHW
Taking
inverse Laplace transform of (11.12)
get
7DNLQJLQYHUVH/DSODFHWUDQVIRUPRI weZHJHW
[L [L EL X DL \
[L [L EL X DL \
RU
[L [L EL X DL EX [
RU
[L [L EL X DL EX [
∴
[L
[L DL [ EL DL E X
∴
[L
[L DL [ EL DL E X
IRU
L
««Q
IRU
L
««Q
DQG
[Q
DQG
[Q
M11_CSE_ISBN_C11.indd 10
....(11.13)
....(11.14)
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-11
From (11.13) and (11.14) we may write the observable phase variable form of representation of
transfer function in state space in a vector-matrix form as follows.
ª D
« D
« «
«
« DQ «¬ DQ
ª [ º
« [ »
« »
« »
«
»
« [Q »
«¬ [Q »¼
ª [ º
« [ »
« »
> @ « » E X
«
»
« [Q »
«¬ [Q »¼
\
DQG
ªE DQ E º
º ª [ º « Q
E D E »
»» «« [ »» « Q Q »
«
»
»« » «
» X
»«
» «
»
» « [Q »
«E D E »
»¼ «¬ [Q »¼ «
»
¬«E DE
¼»
(b) Controllable phase variable form: The transfer function Y(S)/U(S) can be
written as
Q
< V
8 V
E ¦
EL DL E V Q L
L VQ Q
¦ DL V Q L
L <V
∴
b0 U ( s ) n
¦( bi ai b0 ) sni
Q( s ) ....(11.15)
i 1
8 V
4V
ZKHUH
Q
V ¦ DL V
Q
Q L
....(11.16)
L /HWXVQRZWDNHWKHVWDWHYDULDEOHVDVIROORZV
V Q L 4 V
; Q L V
L
IRU
RU
Q Q± «
V ;Q±L V
VQ±L4 V
V ;Q±L V
;Q±L V
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHPD\ZULWH
[Q L [Q L IRUL Q Q± «««
(TXDWLRQ PD\EHZULWWHQDV
V Q4 V Q
¦ DL V Q L 4 V
....(11.17)
....(11.18)
8 V
L M11_CSE_ISBN_C11.indd 11
21/10/11 11:35 AM
RU
V ;Q±L V
VQ±L4 V
V ;Q±L V
;Q±L V
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHPD\ZULWH
[Q L [Q L 11-12 CONTROL SYSTEMS ENGINEERING
IRUL Q Q± «««
(TXDWLRQ PD\EHZULWWHQDV
Equation (11.16) may be written as
V Q4 V Q
¦ DL V Q L 4 V
8 V
L Using (11.17), the above equation may be written as
V; Q V V; Q V Q
Q
DL ; Q L V
¦
D;
V
¦
L L Q L 8V
8V
L 7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
[Q
[Q
)URP DQG ZHJHW
)URP(11.15)
and
DQG(11.17)
we
ZHJHW
From
get
<V
<V
Q
¦
¦
Q DL [Q L X
L DL [Q L X
....(11.19)
L E8 V E8 V 7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
Q
Q
EL DL E
¦
EL DL E
¦
L L ; Q L V
; Q L V
Q
Q
E X ¦
EL DL E [Q L ....(11.20)
\
E X ¦
EL DL E [Q L \
L From (11.18), (11.19) and (11.20) we mayL write the controllable phase variable form of
­representation of transfer function in state space in a vector-matrix form as follows.
ª [ º
« [ »
« »
« »
«
»
« [Q »
«¬ [Q »¼
ª
«
«
«
«
«
«¬ DQ DQ DQ º
»»
»
»
»
D »¼
ª [ º ª º
« [ » « »
« » « »
« » « » X
«
» « »
« [Q » « »
«¬ [Q »¼ «¬ »¼
ª [ º
«[ »
« »
E
D
E
E
D
E
E
D
E
DQG
\
> Q Q Q Q @ « » E X
« [Q »
«
»
¬ [Q ¼
Thus phase variables provide a powerful method of state variable formulation. But they are not a
practical set of state variables from a measurement, control and analysis point of view.
11.3.3 State Space Representation Using Canonical Variables
The linear differential equation (11.6) with zero initial conditions has the transfer function given
by
M11_CSE_ISBN_C11.indd 12
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
< V
8 V
11-13
E V Q E V Q EQ V EQ
V Q D V Q D Q V D Q
The state space representations of transfer function discussed so far are not in a canonical or
normal form. Since the canonical form of representation of transfer function is most suitable
from an analysis point of view, we will discuss them in the following cases.
Case I: When the poles of a transfer function are distinct
Let us now assume that the transfer function Y(s)/U(s) has n distinct poles of λ1, λ2, …, λn and
is of the following form
< V
8 V
RU
E V Q E V Q EQ V EQ
V O V O V OQ
<V
E8 V Q
¦
L &L8 V
V OL
....(11.21)
where Ci’s (for i = 1, 2, …., n) are constants called residues of the poles at s = λi.
Now let us define the state variables as follows.
8 V
; V
V OLL
IRU
L
««Q
RU
V;L V
λL;L V 8 V
L
L
L
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
[LL
OLL [LL X
IRU
L
«Q
)URP
DQG
ZHJHW
From (11.21) and (11.22) we get
<V
E8 V 7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
\
....(11.22)
E X ....(11.23)
QQ
¦ &LL ; LL
L V
L Q
Q
&LL [LL
¦
L L ....(11.24)
From (11.23) and (11.24) we may write the diagonal canonical form of representation of transfer
function in state space in a vector-matrix form as follows.
ª [ º
« [ »
« »
« »
«
»
« [Q »
«¬ [Q »¼
DQG
M11_CSE_ISBN_C11.indd 13
\
ªO
«
«
«
«
«
«¬
>&
O
OQ &
º ª [ º ªº
»» «« [ »» ««»»
» « » « » X
» «
» «»
» « [Q » «»
OQ »¼ «¬ [Q »¼ «¬»¼
&Q
ª [ º
«[ »
« »
&Q @ « » EX
«
»
« [Q »
21/10/11 11:35 AM
ª [ º
ªO « [ »
« O « »
«
« »
« «
»
«
« [Q »
« OQ «¬ [QENGINEERING
»¼
«¬ 11-14 CONTROL SYSTEMS
DQG
\
>&
&
º ª [ º ªº
»» «« [ »» ««»»
» « » « » X
» «
» «»
» « [Q » «»
OQ »¼ «¬ [Q »¼ «¬»¼
&Q
ª [ º
«[ »
« »
&Q @ « » EX
«
»
« [Q »
«¬ [Q »¼
Case II: When the poles of a transfer function are not distinct
Let us assume that the transfer function Y(s)/U(s) has a pole at s = λr of order r and the other
poles are assumed to be distinct and located at s = λj for j = (r + 1), (r + 2), ….., n. So the transfer
function can be written as
< V
E V QQ E V QQ EQ V EQ
< V
E V EU V EQ V EQ
8 V
V OU U V OU V O Q
8 V
V OU V OU V O Q
U Q & 8 V
&8 V
M
E8 V U &LL8 VU L Q & M8 V
RU
< V
OM V
V
O
E8 V L RU
< V
U U L M U
L V OU
M U V OM
¦
¦
/HWXVQRZGHILQHWKHVWDWHYDULDEOHVDVIROORZV
/HWXVQRZGHILQHWKHVWDWHYDULDEOHVDVIROORZV
8 V
;L V
8 V
V OU UU LL ;L V
V OU
IRU
L
«U
IRU
L
«U
8 V
DQG
;M V
8 V
V OM
DQG
;M V
V OM
IRU
M
U U «Q
IRU
M
U U «Q
5HSODFLQJLE\
L±
LQ(11.26),
we
ZHJHW
Replacing
i by (i1) in
get
5HSODFLQJLE\ L± LQ ZHJHW
8 V
;L± V
8 V
V OU UU LL ;L± V
V OU
8 V
RU
;L± V
8 V
V O U V O U U L RU
;L± V
V O U V O U U L ;L V
RU
;L± V
; V
V L OU
RU
;L± V
V OU
RU
V;L± V
λU;L± V ;L V
RU
V;L± V
λU;L± V ;L V
IRU
L
«U
IRU
L
«U
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
[L OU [L [L
[L OU [L [L
IRU
L
«U
IRU
L
«U
)URP ZHJHWV[M V λM[M V 8 V
)URP ZHJHWV[M V λM[M V 8 V
IRU
M
U U «Q
IRU
M
U U «Q
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
M11_CSE_ISBN_C11.indd 14
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
¦
¦
....(11.28)
21/10/11 11:35 AM
L±
V OU
λU;L± V ;L V
«U
V;L± V
RU
IRU
L
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
[L STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
OU [L [L
11-15
IRU
L
«U
ZHJHWV[
VU
8
)URP(11.27),
From
we get sxj (Ms)V= λj λxMj [
(sM) +
(s ) V
IRU
M
U U «Q
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
OM[M X
[ M
M
IRU
....(11.29)
U U «Q
From (11.25), (11.26) and (11.27) we get
<V
U E8 V ¦
E8 V ¦ &L [L
L Q
& L [L V Q
¦&M[M
V
M U
V
L 7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
E X Q
¦
&L [L
\
....(11.30)
L ª [ º
ª [ º
"
theJordan
canonical
" form
º of
«
» and (11.30)
« representation
» ª º
ª OU we may
From (11.28), (11.29)
write
of
« [ »
« O " » « [ » « »
"
transfer function«in state
form as follows.
U
» space« in a vector-matrix
«
»
»
« »
ǻ [ ȼ
«
» ª«[ º» « »
«
»
» « »
" º» «
ª« OU " [
" » « [[U » ª º
OU « [ U »
« O "
« » X
" » «
« [ »
» « »
« U "
"
"
O
[
U
« U »
«
» « U » « »
" OU " » «
« [ »
» « »
«
" » « [[UU » « »
« [ UU »
« " OU « »
« " OU » «[ » « » X
"
[
« U »
«¬ " " O » « U » «¬ »¼
" OU " Q »¼ «¬ [Q »¼ « »
«¬ [ Q »¼
«
[
[
« U »
«
» « U » « »
« »
«
«
» » « »
«
»
» «¬ »¼
«¬ " " OQ ¼» «
«¬ [ Q »¼
«¬ [Q »¼
DQG
\
>&
&
DQG
\
>&
&
M11_CSE_ISBN_C11.indd 15
ª [ º
« [ »
« »
« »
ǻ [ ȼ
[
&U &U &U &Q @ « [U » E X
« [U »
« »
« [U »
«[ »
&U &U &U &Q @ « U » E X
[
«¬ [QU »¼
« [U »
« »
«
»
¬« [Q ¼»
21/10/11 11:35 AM
11-16
CONTROL SYSTEMS ENGINEERING
The n × n matrix, whose diagonal elements are poles at s = λr, λr + 1, ….., λn of the ­transfer
function, is known as the Jordan canonical matrix. This matrix also has the ­property that all
the elements below the principal diagonal are zero and certain elements to the ­immediate
right of the principal diagonal are unity when the adjacent elements in the principal diagonal
are equal. The square submatrix, the diagonal elements of which are all equal to λr, is known
as the Jordan Block.
11.4 CO-RELATION BETWEEN STATE MODEL AND TRANSFER FUNCTION
Let us consider a single-input single-output system the transfer function of which is given by
< V
...(11.31)
*V
8
< VV
<
* VV
*
8
7KHVWDWHPRGHORIWKHDERYHV\VWHPPD\EHJLYHQE\WKHIROORZLQJHTXDWLRQV
8 VV
[
$[ %X
7KHVWDWHPRGHORIWKHDERYHV\VWHPPD\EHJLYHQE\WKHIROORZLQJHTXDWLRQV
....(11.32)
7KHVWDWHPRGHORIWKHDERYHV\VWHPPD\EHJLYHQE\WKHIROORZLQJHTXDWLRQV
\
&[
....(11.33)
$[G8
%X
%X
[[
$[
where
x is the state vector, u is the\ input&[and
y
is
the
output
and
all
are
functions
of
time
t
.
ZKHUH[LVWKHVWDWHYHFWRUXLVWKHLQSXWDQG\LVWKHRXWSXWDQGDOODUHIXQFWLRQVRIWLPHW
G8
\
&[ G8
The
Laplace
transform
of
(11.32)
and
(11.33)
are
given
by
7KH/DSODFHWUDQVIRUPRI
DQG
DUHJLYHQE\
ZKHUH[LVWKHVWDWHYHFWRUXLVWKHLQSXWDQG\LVWKHRXWSXWDQGDOODUHIXQFWLRQVRIWLPHW
ZKHUH[LVWKHVWDWHYHFWRUXLVWKHLQSXWDQG\LVWKHRXWSXWDQGDOODUHIXQFWLRQVRIWLPHW
V; V ± [ $; V %8
V
....(11.34)
7KH/DSODFHWUDQVIRUPRI
7KH/DSODFHWUDQVIRUPRI DQG
DQG DUHJLYHQE\
DUHJLYHQE\
DQG <V
&;
V G8 VV
....(11.35)
V;
$;
V; VV ±
± [[ $; VV %8
%8 V
$VWUDQVIHUIXQFWLRQLVGHILQHGZLWK]HURLQLWLDOFRQGLWLRQSXWWLQJ[
LQ
ZHJH
As transfer function is defined
with
zero initial
condition, putting x(0) = 0 in (11.34), we
DQG
<
&;
DQG
< VV
&; VV G8
G8 VV
V; V
$; V %8 V
get,
$VWUDQVIHUIXQFWLRQLVGHILQHGZLWK]HURLQLWLDOFRQGLWLRQSXWWLQJ[
LQ
ZHJH
$VWUDQVIHUIXQFWLRQLVGHILQHGZLWK]HURLQLWLDOFRQGLWLRQSXWWLQJ[ LQ ZHJH
RU
V; V ±$;
V
%8
V
V;
$;
V; VV
$; VV %8
%8 VV
RU
V,±$
;
V
%8
V
RU
V;
%8
VV
RU
V; VV ±$;
±$; VV
%8
±ZHJHW
3UHPXOWLSO\LQJERWKVLGHVE\
V,±$
RU
V,±$
;
V
%8
RU
V,±$ ; V
%8 VV
;
V
V,±$ ±%8 V
±
3UHPXOWLSO\LQJERWKVLGHVE\
3UHPXOWLSO\LQJERWKVLGHVE\ V,±$
V,±$ ±ZHJHW
ZHJHW
3XWWLQJWKHYDOXHRI; V LQ ;
ZHJHW
....(11.36)
V,±$
VV
; VV
V,±$ ±%8
±%8
±%G@8
<
V
>&
V,±$
V
Putting
the
value
of
X(s)
in
(11.35),
we
get,
3XWWLQJWKHYDOXHRI;
V
LQ
ZHJHW
3XWWLQJWKHYDOXHRI; V LQ ZHJHW
&RPSDULQJ DQG ±%G@8 V
< VVZHJHW
>& V,±$
V,±$ ±
<
>&
%G@8 V ....(11.37)
&RPSDULQJ
DQG
ZHJHW
&
>
DGM
V
,
$
@
%
&RPSDULQJ(11.31)
and
DQG(11.37),
Comparing
get& V,±$ ±%G * we
VZHJHW
G
V,VV
$$
&
>
DGM
,
$
@
%
&
>
DGM
,
@
%
±
*
&
GG ....(11.38)
* VV
& V,±$
V,±$ ±%G
%G $
$
VV,, So the transfer function is expressed in terms of the parameters A, B, C and d of the state
model. It is also evident that |(sI-A)| is the characteristic polynomial of G(s), that is, in other
words, the poles of G(s) are the eigen values of matrix ‘A’. It may also be noted that while
the state model of a system is non-unique, its transfer function is unique.
M11_CSE_ISBN_C11.indd 16
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-17
11.5 DIAGONALISATION OF STATE MATRIX
Let us consider the state model of a system given by
[
$[ %X
[
$[ %X
\
&['X
\
&['X
WHPDWUL[
WHPDWUL[
where ‘A’ is a non-diagonal
state matrix.
DUPDWUL[µ3¶VXFKWKDWWKHWU
If there exists a non-singular
matrix ‘P’ such that the transformation x = Pz transforms
DUPDWUL[µ3¶VXFKWKDWWKHWU
RQLFDOIRUPJLYHQE\
the given state model intoRQLFDOIRUPJLYHQE\
a diagonal canonical form given by
]
3 $3 ] 3 % X
]
3 $3 ] 3 % X
\
&3 ]'X
then the matrix ‘P’ is called diagonalising or modal matrix and the technique used in transforming
the state matrix ‘A’ into a diagonal matrix (P–1 AP) is called diagonalisation.
For determining modal matrix ‘P’, let λi (for i = 1, 2, ….., n) be the distinct roots (known
as the characteristic roots or eigen values of matrix ‘A’) of the characteristic equation given
by
O, $ RU O Q DO Q DQ O DQ
Referring to equation (11.38) it can be concluded that the eigenvalues of the state matrix ‘A’ are
in fact the poles of the system transfer function.
Let x = pi (for i = 1, 2, ….., n) which are (n × 1) vectors be the solution of the matrix
equation
λL , ± $ [
λL , ± $ [
HLJHQYHFWRURIµ$¶F
HLJHQYHFWRURIµ$¶F
Then pi are called the eigenvector
of ‘A’ corresponding to the eigenvalue λi and
$3L
λ L SL
$3L
λ L SL ....(11.39)
p ….. pn) Sbe
formed by
the eigenvectors
of ‘A’. Then
Let the (n × n) matrix
$3P = [p$>S
S
1 2
Q@
>$S
$S@ Q@
$3
$>S
S$S
S
Q
>>$S
λ S λ
S
λQ S@Q@
$S
$S
Q
>λ S λ S
λQ SQ@
ªO º
º»»
ª««O O «
»
using
equation 11.39,
SQ @ «« O »»
XVLQJHTXDWLRQ
> S S «
»
SQ @ «« XVLQJHTXDWLRQ
»»
> S S ««
»
OQ »»¼
«¬ «¬ OQ »¼
∴
∴
M11_CSE_ISBN_C11.indd 17
$3
$3
ªO ª««O O
«
3 «« O
«
3 « º
º»
»
»»»
»
»
21/10/11 11:35 AM
ªO « O
«
«
SQ @ XVLQJHTXDWLRQ
> S S «
«
«
11-18 CONTROL SYSTEMS ENGINEERING
¬ ∴
$3
ªO « O
«
3 « «
«
«¬ º
»»
»
»
»
OQ »¼
º
»»
»
»
»
OQ »¼
3UHPXOWLSO\LQJERWKVLGHVE\3±ZHJHW
3±$3
ªO « O
«
«
«
«
«¬ º
»»
»
»
»
OQ »¼
So P = [p1p2 .... pn] is the required diagonalising or modal matrix. Also, it may be noted that
the characteristic equation and eigenvalues are invariant under non-singular t­ransformation.
If the state matrix ‘A’ is of the Bush or Companion form as shown below
$
ª « «
« «
« «¬ DQ
DQ DQ º
»»
»
»
»
D »¼
....(11.40)
WKHQWKHPRGDOPDWUL[µ3¶LVNQRZQDVWKH9DQGHU0RQGHPDWUL[DQGLVJLYHQE\
3
ª « O
O
« «O
O
« « « Q OQ ¬O
º
OQ »»
OQ »
»
»
»
OQQ ¼
In general, when the state matrix ‘A’ has eigenvalues λ1 of order m and all other
eigenvalues of λm+1, λm+2, ..... λn are distinct, then diagonalisation is not possible, but the
transformation x = pz will yield the matrix (P–1AP) in Jordan canonical form.
The column vectors associated with eigenvectors λ1 of order m are determined from the
transformation
M11_CSE_ISBN_C11.indd 18
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
> S
> S
ª O «ª O O
«« O
SP @ «« SP @ «« ««
««¬ «¬ S S RU
RU
RU
º
»º»
»»»
»
O »»
»
O O »»¼
O »¼
SL O SL
SOL , $
O SSL
$ > S
$ > S
S S SP @
SP @
$SL
$SSL
L 11-19
....(11.41)
RU
O , $ S
SL IRUL PDQGS IRUL
PDQGS
7KHFROXPQYHFWRUVDVVRFLDWHGZLWKGLVWLQFWHLJHQYHFWRUV
λP
The
column vectors associated
with distinct eigenvectors
λm+1λ
, λP
,
....., λn are ­determined
m+2
7KHFROXPQYHFWRUVDVVRFLDWHGZLWKGLVWLQFWHLJHQYHFWRUV
λ
λ
from
P P
O,$ S
L
L
....(11.42)
L , $ SL
IRUL P POQ
IRUL
P P Q
6RWKHPDWUL[3FDQEHIRUPHGDVIROORZV
6RWKHPDWUL[3FDQEHIRUPHGDVIROORZV
3
> S S SP SP SQ @
3
> S S SP SP SQ @
In particular, if the state matrix ‘A’ is in Bush or Companion form (11.40), then the
modified Vander Monde matrix can be written as
3
ª « O
« « O
« « «
« O Q «¬ OP º
OQ »»
OQ »
»
»
»
Q »
OQ
»¼
O
OP G
OQ G O
G OQ G O
OPQ 11.6 SOLUTION OF STATE EQUATION
The state equation of a linear time-invariant system is of the form given below.
[
$[%X
7DNLQJ/DSODFHWUDQVIRUPRQERWKVLGHVRI
ZHJHW
Taking Laplace transform on both sides of (11.43), we get
V; V [ $; V %8 V
RU
V,±$ ; V
[ %8 V
±
3UHPXOWLSO\LQJERWKVLGHVE\ V,±$ ZHJHW
;V
V,±$ ±[ V,±$ ±%8 V
RU
;V
) V [ ) V ) V
ZKHUH
) V
V,±$ ±
DQG
) V
%8 V
V,±$ ±LVGHILQHGDVIROORZV
1RZ
) V
) V
V,±$ ±
M11_CSE_ISBN_C11.indd 19
....(11.43)
....(11.44)
21/10/11 11:35 AM
[
$[%X
7DNLQJ/DSODFHWUDQVIRUPRQERWKVLGHVRI ZHJHW
V; V [ $; V %8 V
RU
V,±$ ; V
[ %8 V
±
3UHPXOWLSO\LQJERWKVLGHVE\ V,±$ ZHJHW
11-20 CONTROL SYSTEMS;
ENGINEERING
V
V,±$ ±[ V,±$ ±%8 V
RU
;V
) V [ ) V ) V
ZKHUH
) V
V,±$ ±
DQG
) V
%8 V
V,±$ ±LVGHILQHGDVIROORZV
1RZ
) V
) V
V,±$ ±
, $ $ $
V V V V
7DNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
/±> V,±$ @±
I W
$ W $W H$W φ W DVVXPHG
, $W ∴
φW
H$W /±> V,±$
±@
f
LL
¦ $W L L
....(11.45)
ZKHUHφ W LVNQRZQDVWKHVWDWHWUDQVLWLRQPDWUL[DQG$R ,
$OVRZHNQRZWKDW
W
/ ª ³ I W W I W GW º
¬« ¼»
∴
/ > ) V ) V
@
) V ) V
W
³ R I W W
W
³H
$ W W
I W GW
%X W GW ....(11.46)
Now taking inverse Laplace transform on both sides of equation (11.44), we get
±
/
[[ WW
II WW [
/±>)
>)± VV )
) VV @@
[
W
[
/
>)
[
W
I
V ) V @
8VLQJ(11.45)
and
DQG(11.46)
we
ZHPD\ZULWH
8VLQJ
DQG
ZHPD\ZULWH
Using
may write
8VLQJ DQG ZHPD\ZULWH
WW $ W W
$W[
W GW
[[ WW
HH$W
%X
[ $W
³³ HH $ WWW $%X
W W W GW
H
%X W GW
[W
H [ ³
WW
W W %X W GW
II WW [[ I
I WW W %X W GW
I W [ ³³ ³ I W W %X W GW
....(11.47)
6R
LVWKHVROXWLRQRIVWDWHHTXDWLRQ
7KHILUVWWHUPRQWKH
So (11.47)
is the solution of state equation
(11.43).
The first term on the right-hand side
6R LVWKHVROXWLRQRIVWDWHHTXDWLRQ
7KHILUVWWHUPRQWKH
6R
LVWKHVROXWLRQRIVWDWHHTXDWLRQ
7KHILUVWWHUPRQWKH
WKH]HURLQSXWUHVSRQVHDQGWKHVHFRQGWHUPLVWKH]HURVWDWHUHVSRQVH
of
(11.47)
is
the
zero
input
response
and
the
second
term
is the zero state response of the
WKH]HURLQSXWUHVSRQVHDQGWKHVHFRQGWHUPLVWKH]HURVWDWHUHVSRQVH
WKH]HURLQSXWUHVSRQVHDQGWKHVHFRQGWHUPLVWKH]HURVWDWHUHVSRQVH
*HQHUDOLVLQJWKHLQLWLDOWLPHDVW
≠VROXWLRQ
PD\EHZULWW
system.
*HQHUDOLVLQJWKHLQLWLDOWLPHDVW≠VROXWLRQ PD\EHZULWW
*HQHUDOLVLQJWKHLQLWLDOWLPHDVW
PD\EHZULWW
≠VROXWLRQ
Generalising
the initial time as t0 ≠ 0,
solution (11.47)
may
be written as
WW
$
W
W
WW
$ WWW $
W
W
$
HH
%X
[ W
[[ WW
[ W HH
%X WW G
GWW
H $ W W [ WW³³ ³ H $ W W %X W GW
[W
W
WW
W
W
II WW WW [[ WW I
I WW WW %X
%X WW GGWW
I W W [ WW³³ ³ I W W %X W GW
W
W
,WPD\EHQRWHGIURP
,WPD\EHQRWHGIURP WKDWLI%
WKDWLI% WKHQ
WKHQ
,WPD\EHQRWHGIURP WKDWLI%
WKHQ
$W
$W
[
W
H
[
[W
H [ $W
φ
φ WW [
[ M11_CSE_ISBN_C11.indd 20
[W
H [ φ W [ 21/10/11 11:35 AM
6R
6R LVWKHVROXWLRQRIVWDWHHTXDWLRQ
LVWKHVROXWLRQRIVWDWHHTXDWLRQ 7KHILUVWWHUPRQWKH
7KHILUVWWHUPRQWKH
WKH]HURLQSXWUHVSRQVHDQGWKHVHFRQGWHUPLVWKH]HURVWDWHUHVSRQVH
WKH]HURLQSXWUHVSRQVHDQGWKHVHFRQGWHUPLVWKH]HURVWDWHUHVSRQVH
*HQHUDOLVLQJWKHLQLWLDOWLPHDVW
*HQHUDOLVLQJWKHLQLWLDOWLPHDVW≠VROXWLRQ
≠VROXWLRQ PD\EHZULWW
PD\EHZULWW
$ W W
W
W
[ W ³³ H
$ W W
H $ W W ANALYSIS
[ W H $OF
%X W GW
[W
W W CONTINOUS
STATE
SPACE
SYSTEM
H
[W
W
W
%X W GW
W
W
II WW WW [[ WW ³³ I
I WW WW %X
%X WW GGWW 11-21
....(11.48)
W
W
It
may be noted from (11.47) WKDWLI%
that if B = 0, then
,WPD\EHQRWHGIURP
,WPD\EHQRWHGIURP WKDWLI%$W WKHQ
WKHQ
[[ WW
HH$W[
[ φ
φ WW [
[ ....(11.49)
is the solution of the state equation x = Ax which is homogeneous. So, it is clear that φ(t) = eAt is
called the state transition matrix because it relates the system state at time t to that at time zero.
11.6.1 Computation of State Transition Matrix
The state transition matrix φ(t) can be computed by several methods. Two of them are given in
formula (11.45). The third one, based on the Cayley-Hamilton theorem, will be discussed here.
If A be the state matrix of order (n × n), then its characteristic equation is given by
O , $ O Q DO Q DQ O DQ ....(11.50)
λt
YHUJHVIRUDOOILQLWHYDOXHVRI
λ
DQGWDQGOHWLWEHGHQRWHGE\
Q all finite
for
λ and t and let it be denoted by
The power series
OλW, $ O
DO Q values
DQ of
T λof e converges
O DQ
λ
H
I
f(λ).
YHUJHVIRUDOOILQLWHYDOXHVRI λDQGWDQGOHWLWEHGHQRWHGE\
Q Q
Q Q ∴
HλW O W O W O W O W WR f
Iλ
Tλ
Q
OW O W O QW Q
Q
Q O Q W Q Q WR f ....(11.51)
If Q(λ) be the quotient polynomial and R(λ) be the remainder polynomial when f(λ) is
divided by the characteristic polynomial q(λ), then we have
....(11.52)
II λ
4λ
λ
4
λ T
T λ
λ 5
5 λ
λ
ZKHUH5 λ LVRIWKHIRUPJLYHQE\
ZKHUH5 λ LVRIWKHIRUPJLYHQE\
.....(11.53)
Q±
α
αλ
α λ
5
α λ
Q±
λ
α
5λ
α
λ α
λ αQ±
Q±λ
λiL(for
IRUL
Q
EHWKHGLVWLQFWHLJHQYDOXHVRIPDWUL[$WKHQIUR
,Iλλ
If
= 1,Q
2, ........., n) beEHWKHGLVWLQFWHLJHQYDOXHVRIPDWUL[$WKHQIUR
the distinct eigenvalues of matrix A, then from (11.50) we may write
,I
i
L IRUL
VRIURP
ZHKDYHT
so
from (11.94),
we have q(λiλ
for all λi.λ
λ) LL= 0IRUDOO
IRUDOO
λLL
VRIURP
ZHKDYHT
4λ
5 λ
II λ
λLL
4
λLL T
T λ
λLL 5
λLL
RU
I
λ
5
λ
5 λLL
RU
I λLL
Q HHOOLL WW
D
RU
....(11.54)
D D
DO
OL D
D O
OL D
D Q O
OLQ RU
L
L
Q L
for i = 1, 2, ....., n. Solving the above equations we obtain the coefficients α0, α1, ......, αn–1 of the
remainder polynomial R(λ).
It should be noted that if igenvalues of A are not distinct and if λ1 be an eigenvalue of
order m, then only one independent equation can be obtained by putting i = 1 into equation
M11_CSE_ISBN_C11.indd 21
21/10/11 11:35 AM
11-22
CONTROL SYSTEMS ENGINEERING
(11.54). The remaining (m-1) linear equations can be obtained by differentiating both sides
of equation (11.54) as follows.
G MM I OL
G MM 5 OL
G M I OM L
G M 5 OM L
G GIOLOM L L G G5OLOM L L GO
MG OL M
G GIOLOL L G MG5OLLOM L L L L M
GGM MGIO O
GMGMM5O§OM QQ OM L W
N ·
G
L
L
H
RU
L
L
L
LDN OL ·
§
¨
G M M OLW
G MM ¦
....(11.55)
N ¸
Q GGOL GM OHOM LW L GGOGL O
RU
§¨©M ¦
N D N OLN ·
¸¹ L M
L
L
H
RU
D
O
L G OML M
G OML M ¨© ¦
NQ L N L ¸
¹L GGOL HOLW LL GGOL §© N N ·
¹¸ L RU
D
O
¨
IRUM P±M M
MM ¦
Q N L
G
O
O
G
§
·¹
G
G L ©N L
IRUM
LW L λE\$LQP±
5HSODFLQJ
HOZHJHW
RU
D ON L IRUM P±
M
M ¨¦ N L ¸
λ
by
A
in
(11.52),
we
get
Replacing
ZHJHW
5HSODFLQJλE\$LQG
OL
GO © N I L$
4 $L T
$ 5 ¹$L λE\$LQP±
ZHJHW
5HSODFLQJ
IRUM
I$
4 $ T $ 5 $
....(11.56)
λE\$LQP±
ZHJHW
5HSODFLQJ
7KH&D\OH\+DPLOWRQWKHRUHPVWDWHVWKDWHYHU\VTXDUHPDWUL[VDWLVILHVLWVRZQ
I$
4 $ T $ 5 $
IRUM 7KH&D\OH\+DPLOWRQWKHRUHPVWDWHVWKDWHYHU\VTXDUHPDWUL[VDWLVILHVLWVRZQ
6RIURP
ZHPD\ZULWH
The
Cayley-Hamilton
theorem
that$ every
square
I $states 4
T $ 5
$ matrix satisfies its own characteristic
λE\$LQ
ZHJHW
5HSODFLQJ
7KH&D\OH\+DPLOWRQWKHRUHPVWDWHVWKDWHYHU\VTXDUHPDWUL[VDWLVILHVLWVRZQ
6RIURP So
ZHPD\ZULWH
Q D $Q± D $D , equation.
from
(11.50)
we
may
write
T
$
$
6RIURP ZHPD\ZULWH
7KH&D\OH\+DPLOWRQWKHRUHPVWDWHVWKDWHYHU\VTXDUHPDWUL[VDWLVILHVLWVRZQ
Q±
Q
I$
4
$
T$$Q±
5
$
$ ZHJHW
$Q
D
D
$D , Q D$Q± DQ±$DQ, )URPHTXDWLRQV
DQGTT
$
$
6RIURP
ZHPD\ZULWH
....(11.57)
Q±
Q
7KH&D\OH\+DPLOWRQWKHRUHPVWDWHVWKDWHYHU\VTXDUHPDWUL[VDWLVILHVLWVRZQ
)URPHTXDWLRQV DQG I
ZHJHW
$
5Q$D $Q± D $D , )URPHTXDWLRQV
DQG
ZHJHW
T
$
$
6RIURP
(11.56)
ZHPD\ZULWH
Q±
Q
I $ φ Wwe
5get
$ From
equations
and (11.57),
7KXVWKHVWDWHWUDQVLWLRQPDWUL[
FDQEHREWDLQHGE\UHSODFLQJ
λE\$LQ
$
5Q$D $Q± D $D
)URPHTXDWLRQV
DQGTI
ZHJHW
$
$
7KXVWKHVWDWHWUDQVLWLRQPDWUL[
W FDQEHREWDLQHGE\UHSODFLQJ
λE\$LQ
Q±
Q, φ $W φ
H5$W$
I$ 5 $
7KXVWKHVWDWHWUDQVLWLRQPDWUL[
φ
W
FDQEHREWDLQHGE\UHSODFLQJ
λ
E\$LQ
I
$W
)URPHTXDWLRQV DQG φ
W ZHJHW
H$W I $ 5 $ H5
7KXVWKHVWDWHWUDQVLWLRQPDWUL[
D $, ID$ $ 5
D $$ D QλE\$LQ
$QQ Iφ $W φ W FDQEHREWDLQHGE\UHSODFLQJ
Thus, the state transition matrix
φ
(
t
)
can
be
obtained
by
replacing
λ
by
(11.53) as ­follows.
$W
D
,
D
$
D
$
D
$AQ in
φ W φ W FDQEHREWDLQHGE\UHSODFLQJ
HD , ID$ $ 5
7KXVWKHVWDWHWUDQVLWLRQPDWUL[
D $$ D QQλE\$LQ
$ φW
HD$W, ID$ $ 5
D $$ D Q $Q D , D $ D $ D Q $Q ....(11.58)
11.6.2 Properties of State Transition Matrix
.
The important properties of the state-transition matrix φ(t) for the time-invariant system x = Ax
are stated below and their proofs are left as an exercise to the students.
D
φ
H$ ,
E
φW
H $W
RU
F
G
H
M11_CSE_ISBN_C11.indd 22
φ± W
φ WW
>φ W @ Q
I W W I W W
H $W
>I W @
φ ±W
H $ W W
H $W H $W
I W I W
I W I W
φ QW
I W W
I W W I W W
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-23
11.7 CONCEPT OF CONTROLLABILITY AND OBSERVABILITY
The concepts of controllability and observability were originally introduced by R. Kalman and
play an important role in the design of modern control system in state space. Precise definitions
of controllability and observability are given below.
A system is said to be controllable at time t0 if it is possible by an unconstrained control
vector u(t) to transfer the system from any initial state x(t0) to any other state x(t1) in a finite
interval of time.
A system is said to be observable at time t0 if it is possible to determine the state x(t0)
from the observation of the output y(t) over a finite time interval.
The conditions of controllability, and of observability and the principle of durability will
be discussed in the following subsections.
11.7.1 Controllability
Let us consider the linear continuous time system be given by
[
$[ %X ....(11.59)
The dimension of the matrices and vectors are as stated in Section 11.2.1.
A system is said to be state controllable at t = t0 if it is possible to construct an
­unconstrained control signal that will transfer an initial state x(t0) to any final state x(t1) in a
finite time interval t0 ≤ t ≤ t1. If every state is controllable, then the system will be completely
state controllable.
Now we shall derive the condition of complete state controllability. The solution of the
given state equation may be written from equation (11.48) as
[W
W
³
H $ W W [ W H $ W W %X W GW W
$WWKHILQDOWLPH W W ZHKDYH
[ W
RU
H $W [ W H $W [ W
W
³
H $ W W [ W H $ W W %X W GW
W
W
³H
$W
%X W GW
W
8VLQJ&D\OH\+DPLOWRQWKHRUHPZHPD\ZULWH
H±$τ
Q ¦D N W
$N
N where αk(τ) for k = 0, 1, ......., (n-1) are scalar time functions.
Thus, the previous equation can be written as
M11_CSE_ISBN_C11.indd 23
21/10/11 11:35 AM
11-24
CONTROL SYSTEMS ENGINEERING
Q ¦
H $W [ W H $W [ W
N βN
ZKHUH
³
W
ª E º
« E »
ª % $% $Q % º « »
¬
¼« »
«
»
E Q ¼
¬
W
H $W [ W H $W [ W
RU
W
$ N % D N W X W GW
³DN W X W
GW W
The left-hand side of the above equation is a (n × 1) known constant vector. For the system to be
completely the state controllable, the above equation must be satisfied with the given constant
vector on the left-hand side.
This requires that the (n × m) matrix [B : AB : .... An–1 B] must be of rank n. This matrix is known
as the controllability matrix and we may denote it as Qc.
So, the condition of complete
state controllability of the system
.
x = Ax + Bu
is that the controllability matrix Qc given by
Qc = [B : AB : ..... : An–1B] be of rank n.
In other words, it may be said that the pair (AB) is controllable if the rank of Qc is n.
Alternative condition of complete state controllability: Let λi (for i = 1,
2, ....., n) be the ­distinct eigenvalues of matrix ‘A’. Also assume that the transformation x = Pz
makes P–1AP
into a diagonal matrix given by
3±$3
ª O
«
«
«
«
¬
O
º
»»
»
»
OQ ¼
Substituting x = Pz into equation (11.59), we have
]
3 $3 ] 3 % X
If the elements of any one row of the (n × r) matrix (P–1B) are all zero, then the
­corresponding state variable cannot be controlled by any of the r elements of control vector
u. So for distinct eigenvalues of matrix ‘A’, the condition of complete state controllability of
.
the system x = Ax + Bu is that no row of P–1B has all zero elements.
If the eigenvalues of ‘A’ are of the form λ1, λ2, λ3, λ4, λ5, λ6, λ7, ......, λn with (n - 3)
distinct eigenvalues and if there exists a transformation x = SZ such that AS = J changes to
Jordan canonical form given by
M11_CSE_ISBN_C11.indd 24
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
-
ª O « O
«
« «
«
«
«
«
«
«
«
«
«
¬«
O
O
O
O
11-25
º
»
»
»
»
»
»
»
»
»
»
»
»
»
OQ ¼»
then substituting X = SZ into equation (11.59) we have,
=
6 $6 = 6 % X
-= 6 % X
.
Thus the system x = Ax + Bu is completely state controllable if and only if the following conditions
hold.
1) No two Jordan blocks in J are associated with the same eigenvalues.
2) The elements of any row of S–1B corresponding to the last row of each Jordan block are not
all zero.
3) The elements of each row of S–1B corresponding to distinct eigenvalues are not all zero.
Output controllability: Let us consider the linear continuous time system
[
$[%X
$[%X
\ [ &['X
\
&['X
7KHGLPHQVLRQVRIYHFWRUVDQGPDWULFHVDUHDVVWDWHGLQ6H
The 7KHGLPHQVLRQVRIYHFWRUVDQGPDWULFHVDUHDVVWDWHGLQ6H
dimensions of vectors and matrices are as stated in Section 11.2.1.
7KHV\VWHPLVVDLGWREHFRPSOHWHO\RXWSXWFRQWUROODEOHLI
The 7KHV\VWHPLVVDLGWREHFRPSOHWHO\RXWSXWFRQWUROODEOHLI
system is said to be completely output controllable if it is possible to construct an
FRQWUROYHFWRUX W WKDWZLOOWUDQVIHUDQ\JLYHQLQLWLDORXWSX
unconstrained
controlW WKDWZLOOWUDQVIHUDQ\JLYHQLQLWLDORXWSX
vector u(t) that will transfer any given initial output y(t0) to any final
FRQWUROYHFWRUX
WLPHLQWHUYDOW
a≤W≤W
time
output
y(t
)
in
finite
interval t0 ≤ t ≤ t1.
1
WLPHLQWHUYDOW
,WPD\EHVWDWHGZLWKRXWSURRIWKDWWKHFRQGLWLRQRIFRPS
≤W≤W
It may
be stated without proof that the condition of complete output controllability is that
,WPD\EHVWDWHGZLWKRXWSURRIWKDWWKHFRQGLWLRQRIFRPS
Q UPDWUL[
the m × (n+1) r matrix
Q UPDWUL[
ª &% &$% &$ % &$ Q % ' º
¬
¼
ª &% &$% &$ % &$ Q % ' º
¬
¼
EHRIUDQNP
EHRIUDQNP
2EVHUYDELOLW\
2EVHUYDELOLW\
/HWXVFRQVLGHUWKHOLQHDUFRQWLQXRXVWLPHV\VWHPJLYHQE\
/HWXVFRQVLGHUWKHOLQHDUFRQWLQXRXVWLPHV\VWHPJLYHQE\
[
$[%X
[
$[%X
\
&['X
\
&['X
M11_CSE_ISBN_C11.indd 25
21/10/11 11:35 AM
FRQWUROYHFWRUX W WKDWZLOOWUDQVIHUDQ\JLYHQLQLWLDORXWSX
WLPHLQWHUYDOW≤W≤W
,WPD\EHVWDWHGZLWKRXWSURRIWKDWWKHFRQGLWLRQRIFRPS
Q UPDWUL[
11-26
ª &% &$% &$ % &$ Q % ' º
¼
¬
CONTROL
SYSTEMS ENGINEERING
EHRIUDQNP
11.7.2
Observability
2EVHUYDELOLW\
Let us/HWXVFRQVLGHUWKHOLQHDUFRQWLQXRXVWLPHV\VWHPJLYHQE\
consider the linear continuous time system given by
[
\
$[%X
&['X
where the dimensions of matrices and vectors are as stated in Section 11.2.1. The system is said
to be completely observable if every state x(t0) can be determined from the observation of y(t)
over a finite interval of time t0 ≤ t ≤ t1. Since control function u does not affect the complete
observability, we may assume without loss of generality that u = 0 and t0 = 0. So the system now
reduces to
[ [[ $[ $[
....(11.60)
$[
\ \\ &[ &[
....(11.61)
&[
$ substituting this
At $
6LQFHWKHVROXWLRQRIWKHKRPRJHQHRXVHTXDWLRQ
LV[
W
H
Since
the
solution
of
the
homogeneous
equation
(11.60)
is
x(t)
=
e
x(0),
6LQFHWKHVROXWLRQRIWKHKRPRJHQHRXVHTXDWLRQ
LV[
W
H
$
6LQFHWKHVROXWLRQRIWKHKRPRJHQHRXVHTXDWLRQ LV[ W H
into
equation
(11.61)
we get,
HTXDWLRQ
ZHJHW
HTXDWLRQ
ZHJHW
HTXDWLRQ
ZHJHW
$W [ \ W \\ WW &H $W&H
[ $W
&H
[
....(11.62)
8VLQJWKH&D\OH\±+DPLOWRQWKHRUHPZHPD\ZULWH
8VLQJWKH&D\OH\±+DPLOWRQWKHRUHPZHPD\ZULWH
8VLQJWKH&D\OH\±+DPLOWRQWKHRUHPZHPD\ZULWH
Q H$W
HH$W
$W
Q Q N
N
¦ D¦
N WD
N WW $
D$
$N
¦
N N N N
6XEVWLWXWLQJWKLVLQWRHTXDWLRQ
ZHJHW
6XEVWLWXWLQJWKLVLQWRHTXDWLRQ
ZHJHW
Substituting
this into equation
(11.62)
we get,
6XEVWLWXWLQJWKLVLQWRHTXDWLRQ
ZHJHW
\ W \\ WW
Q Q Q N
[ NN [[ ¦D¦
N WD
N WW &$
D&$
&$
¦
N
N N N ª &ªª º&
& ºº
« &$
» »»
«« &$
« «« &$
» »»
» »
«
«
RU RU
\
W
\
W
>
D
,
D
,
D
,
@
[ » [ >
D
,
D
,
D
,
@
&$
&$
RU
\ W >D , D , Q D QQ« , @ «« &$
» »» [ «
« «« » »»
« ««Q » QQ »»
¬&$¬¬&$
¼ ¼¼
&$
Therefore, knowing the output y(t) over the time interval 0 ≤ t ≤ t1, x(0) is uniquely
determined from the above equation if and only if the mn × n matrix
ª & º
« &$ »
«
»
« &$ »
«
»
« »
« Q »
¬&$ ¼
RULWVFRQMXJDWHWUDQVSRVHLVRIUDQNQ
7KXVWKHFRQGLWLRQRIFRPSOHWHREVHUYDELOLW\RIWKHV\VWHP
M11_CSE_ISBN_C11.indd 26
[
$[%X
21/10/11 11:35 AM
ª & º
« &$ »
«
»
« &$ »
«
»
« »
« QSTATE
»
SPACE ANALYSIS OF CONTINOUS SYSTEM 11-27
¬&$ ¼
RULWVFRQMXJDWHWUDQVSRVHLVRIUDQNQ
7KXVWKHFRQGLWLRQRIFRPSOHWHREVHUYDELOLW\RIWKHV\VWHP
[
$[%X
\
&['X
LVWKDWWKHREVHUYDELOLW\PDWUL[4JLYHQE\
4
ª& $ & $
¬
& $
Q & º¼
is of rank n, where A* and C* are conjugate transpose of matrices ‘A’ and ‘C’
­respectively.
In other words, it may be said that pair (AC) is observable if rank of Q0 is n.
Alternative condition of complete observability: Let λi (for i = 1, 2, .......,
n) be the distinct eigenvalues of matrix ‘A’. Also assume that the transformation x = Pz makes A
ª O º
= P–1AP into a diagonal matrix given by
ª« O º»
ǻ O O ȼ
$
3 $3 «« O »»
$
3 $3 « O »
»»
$
3 $3 «« «¬ OQ »¼
«¬ OQ »¼
¬ OQ ¼
6XEVWLWXWLQJ[ 3]LQWRHTXDWLRQV DQG ZHJHW
6XEVWLWXWLQJ[
and
DQG
we
ZHJHW
Substituting x = 3]LQWRHTXDWLRQV
Pz into equations (11.60)
(11.61),
get
6XEVWLWXWLQJ[ 3]LQWRHTXDWLRQV
DQG ] 3 $3]
$] ZHJHW
....(11.63)
]\
3 $3] $]
&3]
]\
3 $3] $]
....(11.64)
&3]
\
&3]
At
(11.63)
equation (11.64),
get
Putting the solution z = eH$Wz](0) equation
3XWWLQJWKHVROXWLRQ]
HTXDWLRQ
into
LQWRHTXDWLRQ
we
ZHJHW
3XWWLQJWKHVROXWLRQ] H$W
]
HTXDWLRQ
LQWRHTXDWLRQ
ZHJHW
$W
W
&3H
3XWWLQJWKHVROXWLRQ] H$W] \\HTXDWLRQ
$W]
W
&3H
] LQWRHTXDWLRQ ZHJHW
$W
\W
&3H ] ª HOOWW ] º
ªHOOWW
º
ª« HOW ] º»
ȼ
ǻHOW
O W
º
««ªH HOOWW »
»
««ª H ] »»º
&3 «« HOW »» &3 «« HOOWW ] »»
&3 « H » &3 « H ] »
»
&3 «« »» &3 ««
OQ W OQ W »
«
»»
«¬ H
]
H
OQ W Q
OQ W ¼
¬
¼»
«
«¬ »
HO W ¼
¬«HOQW ]Q ¼»
Q »
«¬ H
]
H
QCP ¼
¬
If all the elements of the ith (for i = 1, 2, ....., n) column of ¼m × n matrix
are zero, then
the state variable zi(0) cannot be determined from the observation of y(t). Thus x(0) cannot
be determined by the linear transformation x = Pz.
Therefore, for distinct eigenvalues of matrix ‘A’, the condition of complete observability
of the system
$[%X
[
\
&['X
LVWKDWQRFROXPQRIPîQPDWUL[&3KDVDOO]HURH
is that no column of m × n matrix CP has all zero elements.
,QFDVHRIUHSHDWHGHLJHQYDOXHVRIPDWUL[µ$¶
DQG WRREWDLQ
]
\
M11_CSE_ISBN_C11.indd 27
6 $6=
&V]
-]
21/10/11 11:35 AM
$[%X
[
\
&['X
11-28 CONTROL SYSTEMS ENGINEERING
LVWKDWQRFROXPQRIPîQPDWUL[&3KDVDOO]HURH
,QFDVHRIUHSHDWHGHLJHQYDOXHVRIPDWUL[µ$¶
In case of repeated eigenvalues of matrix ‘A’, we may substitute x = Sz into equations
DQGand
WRREWDLQ
(11.60)
(11.61)
to obtain
6 $6= -] ....(11.65)
]
\
&V]
....(11.66)
where, the matrix J = S–1AS is in the Jordan canonical form as stated in the alternative c­ ondition
of controllability. Substituting the solution z = eJtz(0) of equation (11.65) into ­equation (11.66),
we get
\W
&6H-W] [
$[%X
\
&['X
7KXVWKHV\VWHPJLYHQE\
is completely observable if and only if the following conditions hold.
1) No two Jordan blocks in J are associated with the same eigenvalues.
2) The elements of any column of CS corresponding to the first row of each Jordan block are
not all zero.
3) The elements of each column of CS corresponding to distinct eigenvalues are not all zero.
11.7.3 Principle of Duality
The principle of duality, due to Kalman, can be used to establish analogies between ­controllability
and observability.
Let us consider the system S1 and its dual system S2 defined as follows:
6\VWHP6
'XDOV\VWHP6
[
$[%X
\
&[
]
$ ] & Y
Q
% ]
where z is an n–dimensional state vector, v is an m-dimensional control vector, n is a r-dimensional
output vector, A* is the conjugate transpose of matrix ‘A’. The dimensions of vectors and matrices
of system S1 are as stated in Section 11.2.1.
The principle of duality states that the system S1 is completely state controllable if and
only if the dual system S2 is completely observable and vice-versa. In other words, the
­principle of duality may be stated as follows.
1) The pair (AB) is completely state controllable implies that the pair (A* B*) is completely
observable.
2) The pair (AC) is completely observable implies that the pair (A*C*) is completely state
controllable.
M11_CSE_ISBN_C11.indd 28
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-29
This principle can easily be verified from the necessary and sufficient conditions of complete
state controllability and complete observability for systems S1 and S2.
Example 11.1 The mechanical system shown in Fig. 11.5 has u1 and u2 as its inputs and y1
and y2 as its outputs. Obtain a state space representation of the system.
Solution
Here u1 and u2 are the force inputs to the system and its outputs y1 and y2 are the ­displacements
measured from the equilibrium position of rest. Applying Newton’s second law of Motion, we
may write the equations of motion for the two masses m1 and m2 as follows.
X N \ E \ \ P \ ....(11.67)
.
.
X
P
\
E
X
P
\
Fig. 11.5
X N \ E \ \ P \
DQG
1RZZHPD\GHILQHGLVSODFHPHQWDQGYHORFLW\WREHVWDWHYDULDEOHVDVIROORZV
[ \
[ \
[ \ DQG
[
7KXVWKHVWDWHHTXDWLRQVPD\EHZULWWHQDV
[
[
7KHRXWSXWHTXDWLRQVDUH
M11_CSE_ISBN_C11.indd 29
....(11.68)
\ [
[
[
N
X
E
E
[ [ [ P
P
P
P
[
N
X
E
E
[ [ [ P
P
P
P
\
\
[
[
21/10/11 11:35 AM
[ \ DQG
[
7KXVWKHVWDWHHTXDWLRQVPD\EHZULWWHQDV
[
[
[
11-30
\ [
N
CONTROL SYSTEMS ENGINEERING
[
[
P
[ X
E
E
[ [ P
P
P
N
X
E
E
[ [ [ P
P
P
P
7KHRXWSXWHTXDWLRQVDUH
\
[
\
[
6RWKHVWDWHVSDFHUHSUHVHQWDWLRQRIWKHV\VWHPLQDYHFWRUPDWUL[IRUPFDQEHZULWWHQDV
ª [ º
« [ »
« »
« [ »
«¬ [ »¼
ª « «
« N
«
« P
«
« ¬
ª \ º
«\ »
¬ ¼
ª [ º
« »
ª º « [ »
« » « [ »
¬
¼
« »
¬ [ ¼
N
P
º
ª
»
» ª [ º «« E
E » «« [ »» « P P »» « [ » «« P
« »
E
E » ¬ [ ¼ «
»
«
P
P ¼
¬
º
»»
» ªX º
» « »
» ¬X ¼
»
»
P ¼
Example 11.2 For the liquid-level system shown in Fig. 11.6, q1, q2, q3, q4 are the deviations
of flow rate from their respective steady-state value, h1 and h2 are the deviations of heads
from their respective steady-state heads, C1 and C2 are the capacitances of two tanks and R1,
R2 are the resistances for liquid flow at the two valves shown in the diagram.
Assuming all the deviations to be small, obtain a state space representation for the system
the two inputs of which are q1 and q2 and two outputs are h1 and h2.
T
T
K
K
5
&
T
5
&
T
Fig. 11.6
M11_CSE_ISBN_C11.indd 30
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-31
Solution
Since all the deviations are small, the system can be considered to be linear and hence the
equations for the system can be written as
&
GK
GW
T T
&
GK
GW
T T T
....(11.70)
K K
5
T
....(11.71)
K
5
T ....(11.72)
DQG
....(11.69)
Eliminating
q3 from
(11.69) and (11.71), we get
(OLPLQDWLQJT
IURP DQG ZHJHW
GK
K
K
T
GK
KK KK T
RU
GK
&
T
5& 5& &
RU
GW
GW
GK
K 5 5K&
T
5&
&
RU
5K& 5K&
&
GW GK
T
(OLPLQDWLQJT
DQGTIURP DQG
ZHJHW
RU
....(11.73)
DQGT
IURP
DQG
ZHJHW (OLPLQDWLQJT
5& 5& &
GW
DQG (OLPLQDWLQJT DQGT IURP GK (11.71)
K Kand
K ZHJHW
Eliminating q3and q4 from&(11.70),
T K we get
K K (11.72),
GK
& T 5 ZHJHW
(OLPLQDWLQJTDQGTIURP
5 DQG
GW GK
K
K 5
K
& GW
T 5
5
GW
GK
K
K 5
K
& GK
K § T ·
T
RU
GK
K5 §¨ 5 ·¸ K T
GW
RU
K
5& ©¨ 5& 5 & ¹¸ &
GW
§ 5 &
GK
T
5K&
5 & ·
&
GW
....(11.74)
©¨ ¹¸ K RU
5K&
5
&
5
&
&
GW
§
·
GK
T
© ¹
LQSXWYDULDEOHVX
DQGRXWSXWYDULDE
/HWXVQRZGHILQHWKHVWDWHYDULDEOHV[
RU
[
¨
¸ K X
/HWXVQRZGHILQHWKHVWDWHYDULDEOHV[
[
LQSXWYDULDEOHVX
X&DQGRXWSXWYDULDE
5
&
5
&
5
&
x
,
x
,
input
variables
u
,
u
and
Let
us
now
define
the
state
variables
GW
© 1¹ 2
output variables y1, y2 as
IROORZV
1 2
/HWXVQRZGHILQHWKHVWDWHYDULDEOHV[[LQSXWYDULDEOHVXXDQGRXWSXWYDULDE
IROORZV
follows.
K [ K
[
IROORZV
[
K [LQSXWYDULDEOHVX
[ KX
DQGRXWSXWYDULDE
/HWXVQRZGHILQHWKHVWDWHYDULDEOHV[
X
T
X
T
K
[
K
[
IROORZV
X T
X T
KTK
X\\[
[
X[ K
T
K
[
[[
X\\\
X T
KTKK
[
8VLQJWKHDERYHGHILQLWLRQVHTXDWLRQV
FDQEHZULWWHQDV
\
K
[[ DQG
8VLQJWKHDERYHGHILQLWLRQVHTXDWLRQV
DQG
FDQEHZULWWHQDV
Using
the above definitions, equations
(11.73)
and
(11.74)
can
be written as
\
K
[
8VLQJWKHDERYHGHILQLWLRQVHTXDWLRQV
DQG FDQEHZULWWHQDV
[ [ X [
8VLQJWKHDERYHGHILQLWLRQVHTXDWLRQV
5
[ DQG
5&
[ FDQEHZULWWHQDV
[
& X & 5& [ 5& [ & X [
5&
5&
&
[ § [ ·X [
[
5 &[ ¨§ 5& & ¸· [ X [
5& [ ©¨ 5& 5 & ¹¸ [ & X § 5 &
5&
5 & ·
&
[ ©¨ ¹¸ [ X [
5&
5& ·¹
&
©§ 5&
6R WKH VWDWH PRGHO LQ WKH YHFWRUPDWUL[
IRUP
UHSUHVHQWDWLRQ
WKH
FDQ E
[ ¨RI
X V\VWHP
[
¸ [ RI
6R WKH VWDWH PRGHO LQ WKH YHFWRUPDWUL[
IRUP
RI
UHSUHVHQWDWLRQ
RI
WKH
V\VWHP FDQ E
5
&
5
&
5
&
&
IROORZV
© ¹
IROORZV
6R WKH VWDWH
PRGHO LQ WKH YHFWRUPDWUL[ IRUP RI UHSUHVHQWDWLRQ RI WKH V\VWHP FDQ E 21/10/11
M11_CSE_ISBN_C11.indd
31
11:35 AM
\
K [[
\
K
[ DQG FDQEHZULWWHQDV
8VLQJWKHDERYHGHILQLWLRQVHTXDWLRQV
8VLQJWKHDERYHGHILQLWLRQVHTXDWLRQV DQG FDQEHZULWWHQDV
[ [ X [
5&
5&
&
[ [ X [
11-32 CONTROL SYSTEMS ENGINEERING
5&
5&
&
§ ·
[ ¨
X [
¸ [ 5&
&
§ 5& 5& ¹·
©
[ ¨
X [
¸ [ 5&
&
© 5& 5 & ¹
6R WKH VWDWH PRGHO LQ WKH YHFWRUPDWUL[ IRUP RI UHSUHVHQWDWLRQ RI WKH V\VWHP FDQ E
So,
inLQ
theWKH
vector-matrix
form
of representation
of theRIsystem
can beFDQ
written
IROORZV
6Rthe
WKHstate
VWDWHmodel
PRGHO
YHFWRUPDWUL[
IRUP
RI UHSUHVHQWDWLRQ
WKH V\VWHP
E as
follows.
IROORZV
ª º
ª
º
»
«5&
»
«
5
&
ª« º» ª [ º ª« &
ª [ º
º ªX º
« » «
»»« »
«« 5 &
»
« [ »
»
5&
¬ª ¼º
«« §¨ ·¸»»ª¬ [ º¼ ««« & »»»ª¬X º¼
« [ »
««¬ 5&
& »¼ «¬X »¼
§© 5& 5& ·¹»»¼ «¬ [ »¼ «¬
¬ ¼
¨
«
»
¸
«
& »¼
¬
© 5& 5 & ¹¼»
¬« 5&
ª º ª [ º
ª \ º
« » « [ »
«\ »
ª¬ º¼
¬ª ¼º ¬ª ¼º
DQG
« » « [ »
«\ »
¬
¼ ¬ ¼
¼
¬
Example 11.3 Consider the electrical circuit shown in Fig. 11.7. Obtain a state model of
the system with zero initial condition where ei(t) and e0(t) are input and output voltages
respectively.
DQG
Fig. 11.7
Solution
)URP.LUFKRII ¶VYROWDJHODZZHJHW
From Kirchoff ’s voltage law, we get
GL
/ L5 ³ L GW
HL
GW
&
L5 L GW
&³
....(11.75)
....(11.76)
HR
7DNLQJ/DSODFHWUDQVIRUPRIDERYHHTXDWLRQVZHJHW
·
§
¨ /V 5 ¸ , V
&V ¹
©
(L V
·
§
¨5 ¸, V
&V
©
¹
(R V
....(11.77)
'LYLGLQJ E\ ZHJHW
M11_CSE_ISBN_C11.indd 32
21/10/11 11:35 AM
L5 ³ L GW
L5 & ³ L GW
&
HR
HR
7DNLQJ/DSODFHWUDQVIRUPRIDERYHHTXDWLRQVZHJHW
7DNLQJ/DSODFHWUDQVIRUPRIDERYHHTXDWLRQVZHJHW
·
§
(L V
¨§ /V 5 ¸· , V STATE
ANALYSIS OF CONTINOUS SYSTEM 11-33
(L SPACE
V
¨© /V 5 &V ¸¹ , V
&V ¹
©
·
§
(R V
¨§ 5 ¸· , V
&V
(R V
¨© 5 ¹¸ , V
&V ¹
©
'LYLGLQJ E\ ZHJHW
Dividing
we get
'LYLGLQJ(11.78)
by
E\(11.77),
ZHJHW
....(11.78)
5 &V
5
&V /V 5 /V 5 &V
&V
·
·
§
§
RU
¨§ /V 5 ¸· ( V
¨§ 5 ¸· (L V
RU
©¨ /V 5 &V ¹¸ ( V
©¨ 5 &V ¹¸ (L V
&V ¹
&V ¹
©
©
·
·
§5
§ 5
RU
¨§ 5 V ¸· (L V
¨§ V 5 V ¸· ( V
RU
©¨ / V &/ ¹¸ (L V
©¨ V / V /& ¹¸ ( V
&/ ¹
/
/& ¹
©/
©
1RZWDNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
1RZWDNLQJLQYHUVH/DSODFHWUDQVIRUPZHJHW
5
5
H 5 H H
5 HL HL
/ HL &/ HL
H / H /& H
/
/&
/
&/
RU
\ D \ D \
EX EX E X RU
\ D \ D \
EX EX E X
ZKHUH
H
\
HL X
(
(
(L
(L
D
5
/
V
V
V
V
D /&
5
E
/
Now
from equation (11.8)
may write
we
1RZIURPHTXDWLRQ
ZHPD\ZULWH
E
....(11.79)
/&
E
β
E β
E±Dβ E
β
E±Dβ±Dβ
5
/
§5·
¨ ¸
/& © / ¹
From
Section 11.4 the state variables x1 and
x2 areDUHJLYHQE\
given by
)URP6HFWLRQWKHVWDWHYDULDEOHV[
DQG[
E D E
[
\±EX \
[
[ EX
5
[ X
/
)URP WKHVWDWHPRGHORIWKHJLYHQFLUFXLWPD\EHZULWWHQDV
M11_CSE_ISBN_C11.indd 33
ª [ º
« [ »
¬ ¼
ª « D
¬ º ª [ º ª E º
X
D »¼ «¬ [ »¼ «¬ E »¼
21/10/11 11:35 AM
E D E
¨ ¸
/& © / ¹
)URP6HFWLRQWKHVWDWHYDULDEOHV[DQG[DUHJLYHQE\
)URP6HFWLRQWKHVWDWHYDULDEOHV[
DUHJLYHQE\
DQG[
[
\±E
X \
[
\±EX \
5
[ EX [ X
[
5
[ EX [ / X
[
11-34 CONTROL SYSTEMS ENGINEERING
/
)URP WKHVWDWHPRGHORIWKHJLYHQFLUFXLWPD\EHZULWWHQDV
)URP WKHVWDWHPRGHORIWKHJLYHQFLUFXLWPD\EHZULWWHQDV
From (11.9) the state model of [the given circuit
be written as
may
ª º ª [ º ª E º
ª º
ǻ D D ȼ ǻ [[ ȼ ǻ EE ȼ X
ǻ [[ ȼ
¼ ¬ ¼ ¬ ¼
¬ ¬ ¼
« D D » « [ » « E » X
« [ »
¼ ¬ ¼ ¬ ¼
¬ ¬ ¼
5
ª
º
º
ª «
ȼ
5
/
ª [ º ª
»
5 º» « » ««
ª« »X
«« 5 »» ª¬ [ º¼ «« /§ 5 · »» X
/¼ « »
¬« /&
¸ »
» ¬ [ ¼ «¬« /&
¨©§ 5
/ ¹· ¼»
/¼
¬ /&
¨ ¸ »
«
¬ /& © / ¹ ¼
\
\
ª[ º
ª[ º
> @ ǻ [[ ȼ E X > @ ǻ [[ ȼ
> @ ¬« [ ¼» E X > @ ¬« [ ¼»
¬ ¼
¬ ¼
([DPSOH &RQVLGHUWKHIROORZLQJWUDQVIHUIXQFWLRQRIDV\VWHP
Example
11.4
the following transfer function of a system.
([DPSOH
Consider
&RQVLGHUWKHIROORZLQJWUDQVIHUIXQFWLRQRIDV\VWHP
< V
V <
V
8
V V V 8
V
V
V in
Obtain the state space representation of this system
a) controllable phase variable form and
b) observable phase variable form.
Solution
The given transfer function can be written as
< V
E V E V E
8 V
V D V D
ZKHUH E E D E D a) The state space representation of the given system in controllable phase variable form
can be written as follows from Section 11.3.2(b).
ª [ º
« [ »
¬ ¼
ª « D
¬ º ª [ º ª º
X
D »¼ «¬ [ »¼ «¬ »¼
º ª [ º ª º
ª « » « [ » « » X
¬
¼ ¬ ¼ ¬ ¼
\
ª[ º
>E DE E DE @ « [ » EX
¬
>
M11_CSE_ISBN_C11.indd 34
¼
ª[ º
@ « »
¬ [ ¼
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-35
The state variables here are defined by
8 V
V V V8 V
V V ; V
; V
V; V
b) From Section 11.3.2(a), the state space representation of the system in observable
phase variable form is given as follows.
ª [ º
« [ »
¬ ¼
ª D º ª [ º ª E DE º
«D » « [ » «E D E » X
¬ ¼ ¬ ¼ ¬ ¼
ª º ª [ º ª º
« » « [ » «» X
¬
¼ ¬ ¼ ¬ ¼
\
ª[ º
> @ « [ » EX
¬
¼
ª[ º
@ « »
¬ [ ¼
>
The state variables here are defined by
V; V ±; V
DQG
RU
RU
V; V
; V
>8 V < V
V
; V
V V 8 V < V
V
V
; V
; V
RU
M11_CSE_ISBN_C11.indd 35
E8 V ±D< V
>8 V ±< V @
E8 V ± D< V
>8 V ±< V @
@
>8 V ; V
V
@
V V 8 V ; V >DV\ [@
V
V
; V
V 8 V
V V ; V
V 8 V
V V 21/10/11 11:35 AM
V V 8 V < V
V
V
; V
RU
11-36
>8 V ; V
V
; V
V CONTROL SYSTEMS;ENGINEERING
V
RU
V
8 V @
V ; V >DV\ [@
V
V 8 V
V V ; V
V 8 V
V V <V
; V
Example 11.5 Consider the following system with differential equation given by
\ \ \ \
X
Obtain a state model of the given system in a diagonal canonical form.
Solution
Taking Laplace transform of the given differential equation we get
< V
8 V
V V V V V V &
&
&
V V V ∴
ZKHUH
<V
E8 V E
&
λ
&8 V &8 V &8 V
V O
V O
V O
&
&
λ
±
±
±
λ ±
From Case I of Section 11.3.3, the diagonal canonical form of state model of the system is given
by
RU
DQG
M11_CSE_ISBN_C11.indd 36
ª [ º
« [ »
« »
¬« [ ¼»
ªO
«
«
¬« ª [ º
« [ »
« »
«¬ [ »¼
ª º ª [ º ªº
« » « [ » «» X
«
» « » « »
«¬ »¼ «¬ [ »¼ «¬»¼
\
>&
O
&
º ª [ º ªº
»» «« [ »» ««»» X
O ¼» ¬« [ ¼» «¬»¼
ª [ º
& @ «« [ »» E X
«¬ [ »¼
ª [ º
« »
21/10/11 11:35 AM
« [ »
«¬ [ »¼
RU
O
«
«¬ » « [ » «» X
O »¼ «¬ [ »¼ «¬»¼
ª [ º
ª º ª [ º ªº
« [ »
« » « [ » «» X
« »
«
» « » « »
SYSTEM 11-37
«¬ [ »¼STATE«¬ SPACE
»¼ «¬ [ »¼OF «¬CONTINOUS
ANALYSIS
»¼
DQG
RU
ª [ º
& @ «« [ »» E X
«¬ [ »¼
\
>&
\
ª [ º
> @ «« [ »»
«¬ [ »¼
&
where the state variables are defined by
; V
8 V
V O
8 V
V ; V
8 V
V O
8 V
V
; V
8 V
V O
8 V
V
Example 11.6 Find the state space model of the following transfer function using the Jordan
canonical form.
< V
VD V 8 V
V V Solution
From the given transfer function we get
<V
ZKHUH
E
&
λ
E8 V &8 V
V O
& 8 V &8 V
V O
V O
&
&
λ
±
±
±
From Case II of Section 11.3.3 we may write the state model in Jordan canonical form as follows.
ª [ º
« [ »
« »
«¬ [ »¼
ªO
«
«
«¬ O
º ª [ º ª º
»» «« [ »» «« »» X
O »¼ «¬ [ »¼ «¬ »¼
ª º ª [ º ª º
« » « [ » « » X
«
» « » « »
«¬ »¼ ¬« [ ¼» «¬ »¼
M11_CSE_ISBN_C11.indd 37
ª [ º
21/10/11 11:35 AM
ª [ º
ªO « [ »
« O
«
« »
11-38 CONTROL SYSTEMS
«
«¬ [ENGINEERING
»
¼
¬
º ª [ º ª º
»» «« [ »» «« »» X
O »¼ «¬ [ »¼ «¬ »¼
ª º ª [ º ª º
« » « [ » « » X
«
» « » « »
«¬ »¼ ¬« [ ¼» «¬ »¼
>&
\
ª [ º
& @ «« [ »» EX
«¬ [ »¼
&
ª [ º
> @ «« [ »»
«¬ [ »¼
Here the state variables are defined as
8 V
V O
; V
8 V
V ; V
8 V
V O
8 V
V ; V
8 V
V O
8 V
V
Example 11.7 Consider the system described by
ª [ º
« [ »
¬ ¼
\
ª º ª [ º ªº
« » « [ » «» X
¬
¼ ¬ ¼ ¬ ¼
ª[ º
> @ « [ »
¬
¼
Obtain the transfer function of the system.
Solution
The given system may be written as
M11_CSE_ISBN_C11.indd 38
[
$[%X
\
&[GX
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
ZKHUH
DQG
$
ª º
« »
¬
¼
%
ªº
«» & ¬¼
G
11-39
> @
º
ªV « V »
¬
¼
7KXVWKHDGMRLQWRIPDWUL[ V,±$ LVJLYHQE\
∴
V,±$
$GM
V,±$
$OVR
_V,±$_
ª V º
« V »¼
¬
V V VV
We may write the transfer function as
< V
8 V
& > DGM V, $ @%
G
V, $
1RZ&>DGM V,±$ @%
>
ª V º ªº
@ «
V »¼ «¬»¼
¬ ª V º
» V
¬V ¼
> @ «
∴
< V
8 V
V
V V Example 11.8 The state space model of a system is given below.
ª [ º
« [ »
« »
«¬ [ »¼
\
M11_CSE_ISBN_C11.indd 39
º ª [ º ª º
ª « »» «« [ »» «« »» X
«
«¬ »¼ «¬ [ »¼ «¬ »¼
ª [ º
> @ «« [ »»
«¬ [ »¼
21/10/11 11:35 AM
11-40
CONTROL SYSTEMS ENGINEERING
Derive a suitable transformation by which another state space model of the same system can be
written as
ª ] º
ª º ª ] º ª º
« ] »
« »» «« ] »» «« »» X
«
« »
¼» ¬« ] ¼» ¬« ¼»
¬« ] ¼»
¬« Solution
Let x = Pz be the required transformation, where
ª S
«S
« «¬ S
3
S
S
S
S º
S »»
S »¼
The given state model can be written as
[
$[%X
$
º
ª « »»
«
«¬ »¼
DQG
%
ª º
« »
«
»
«¬ »¼
DQG
ZKHUH
\
&
&[
>@
ZKHUH
Applying the above transformation, another state model can be written as
]
\
3 $3
∴
3 %
∴
M11_CSE_ISBN_C11.indd 40
ª S
«S
« «¬ S
S
S
S
S º ª º
S »» «« »»
S »¼ «¬ »¼
3 $3 ] 3 % X
&3 ]
ª º
« »»
«
«¬ »¼
ª º
« » DQG&3 « »
«¬ »¼
>@
ª º
« »
«
»
«¬ »¼
21/10/11 11:35 AM
3 $3
∴
« ¬« »
¼»
ª º
« » DQG&3 >@
3 %
« »
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM 11-41
«¬ »¼
ª S
«S
« «¬ S
∴
S
S
S
S º ª º
S »» «« »»
S »¼ «¬ »¼
ª º
« »
«
»
«¬ »¼
ª S º
«S »
« »
«¬ S »¼
RU
ª S
$JDLQ > @ «« S
«¬ S
RU
∴
DQG
S
S
S
ª º
« »
«
»
«¬ »¼
S º
S »»
S »¼
>
>SSS@
S
S
>@
S S S ±
ª º
3 «« »»
«¬ »¼
$3
$JDLQ
º
ª «
RU « »»
¬« »¼
ª S
«S
« ¬« S
@
S
S º
S »»
S ¼»
S
S
ª S
«S
« «¬ S
º
ª «
RU « »»
«¬ »¼
ª « «
«¬ S
S
ª «¬
M11_CSE_ISBN_C11.indd 41
S
S
S S
S
S
S º ª º
S »» «« »»
S »¼ «¬ »¼
º
S »»
S »¼
ª « «
«¬ RU « «
S
S
S
º ª º
S »» «« »»
S »¼ «¬ »¼
S
S
º
»
»
S S »¼
21/10/11 11:35 AM
RU « » « S
¬« »¼ ¬« S
S
S »
S ¼»
S
ª S S
«S
« S
11-42 CONTROL SYSTEMS ENGINEERING
«¬ S S
º
ª «
RU « »»
«¬ »¼
ª « «
¬« S
S
S º ª º
S »» «« »»
S »¼ «¬ »¼
º
S »»
S »¼
ª « «
«¬ S
S
S
S
ª elements of
Comparing
both the matrices, we
get º»
RU « S
p5 = p6 = 0 S
«
»
«¬ S pS9 = – 160
S S »¼
p8 = p6 – 56 = – 56
Thus the required transformation is as follows.
ª
« S [ « ª º
»
«¬ S«[
« »
«¬ [ »¼
ª [ º
«[ »
« »
«¬ [ »¼
RU
º ª º
S »» «« »»
S »¼ «¬ »¼
º
»
S S
ª S ª ]» º
º
« SS » «» »
« S S » « ]¼ »
«¬ S S S »¼ «¬ ] »¼
º ª ] º
ª « »» «« ] »»
«
«¬ »¼ ¬« ] ¼»
Example 11.9 The state model of a system is given by
[
\
$
ZKHUH
$[%X
&[
º
ª
«
»» % «
«¬ »¼
ªº
«« »»
«¬ »¼
DQG
&
>@
2EW L GL
O
L OI
I WK W W
G OE
LW EO W
Obtain a diagonal canonical form of the state model by a suitable transformation matrix.
Solution
The characteristic equation of the matrix ‘A’ is given by
RU
O O
_λ,±$_
O RU
RU
M11_CSE_ISBN_C11.indd 42
λλλ
λ λ λ
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
RU
_λ,±$_
O O
11-43
O RU
RU
λλλ
λ λ λ
∴ λ1 = –1, λ2 = –2 and λ3 = –3 are the three distinct eigenvalues of matrix ‘A’. So the Vander
Monde transformation matrix can be written as
ª
«
« O
«O ¬ 3
∴
∴
3
DGM 3
3
3 $3
O
O º
»
O »
O »¼
ª º
«
»
« »
«¬ »¼
ª º
« »
«
»
«¬ »¼
3
ª º ª º
« » « »»
«
» «
«¬ »¼ «¬ »¼
º
ª « »
«
»
¬« ¼»
3±%
&3
M11_CSE_ISBN_C11.indd 43
ªº
ª º ªº
« » «» « »
«
» « »
« »
«¬ »¼ «¬ »¼
«¬ »¼
ª º
> @ «« »» >@
«¬ »¼
21/10/11 11:35 AM
11-44
CONTROL SYSTEMS ENGINEERING
Applying the transformation x = Pz we get the diagonal canonical form of the state model as
]
\
RU
ª ] º
« ] »
« »
¬« ] ¼»
3 $3 ] 3 % X
&3 ]
ª º ª ] º ª º
« » « ] » « » X
«
» « » « »
«¬ »¼ «¬ ] »¼ «¬ »¼
ª ] º
\
> @ «« ] »»
«¬ ] »¼
Example 11.10 A system is given by the following vector-matrix equations.
ª [ º
« [ »
¬ ¼
ª º ª [ º ª º
« » « [ » « » X
¬
¼ ¬ ¼ ¬ ¼
where the initial condition is given by x(0) = [1 1]T
Determine (a) State transition matrix, (b) Zero input response, (c) Zero state response for u = 1,
(d) Total response, and (e) Inverse of state transition matrix.
Solution
The given state equation may be written as
[
$[ %X
$
ZKHUH
ª º
«
» DQG% ¬ ¼
ª º
« »
¬ ¼
D 6WDWHWUDQVLWLRQPDWUL[Φ
ΦW
V,±$
∴
∴
_V,±$_
V,±$
±
ª V º ª º
« V » « »
¬
¼ ¬
¼
V V
ª V º
« V »
¬
¼
V V DGM V, $
V, $
ª V º
«
»
V V ¬ V ¼
ª V
« V V «
«
«¬ V V M11_CSE_ISBN_C11.indd 44
º
V V »
»
V
»
V V »¼
21/10/11 11:35 AM
∴
∴∴
∴
∴∴
_V,±$_
_V,±$_
_V,±$_
V,±$
±
V,±$
V,±$
±
±
V V V V V V V V V DGM V, $
DGMDGM
V, V,$ $
V, $
V, V,$ $
ª V ºOF CONTINOUS SYSTEM
ANALYSIS
STATE SPACE
ª V ª« V º º»
V « « » V »
V V V V¬¬¬ V ¼ V ¼¼
11-45
V
ª V
º
º º»
ª ª« VV V
V
V
V
« V «« V V V V V V V» »»
« ««
V » »
V V » »»
« «« V V V »¼
«¬ V ¬«¬ VV V V V V V V»¼ »¼
ª ª ª« V V « ««V
V V V « «« « «« V
V
« «¬V V V V
¬ ¬
º
V V V V V V»
»
»
V V V V V V»¼
º
º»
»»
»»
»»
»¼
¼
)URP Φ W LVJLYHQE\ ) W $WH $W
/ ª V, $ º¼
From
(11.45)
)WisLVJLYHQE\
given by ) )
)URP
ΦΦΦ
W(tLVJLYHQE\
W W H H $W //ª¬Vª¬¬, )URP
V,$ $º¼ º¼
ª § W W ·
§ W W · º
ª § ª« §¨ W H W HW · W ·¸ § §¨ W H W HW · Wº·¸ º»
H
H
H
H
H H H ¸ »¸¹ »
©
¹
« ¨ «« ¨ ¸ ¸ ¨ ¨©H ¹ ¹ © © ¹ »¹ »»
« © ««§© · § ·»
«§ ««§¨ W H WW HW · WW ·¸§ §¨ W H WW HW · »WW ·¸ »»
H
H
H
H
H
H
H
H
¸ ¸¹¨ ¨© ¸ » ¸¹ »¼
«¨ «¬¨© ¬© ¬© ¹ ¹© © ¹ ¼ ¹ ¼
(b) Zero input response is obtained from (11.47) by putting u(τ) = 0 and letting it be x1.
H $W [ Φ W [ [
∴
W
H H W
ª
« H W H W
«
«
«
« H W H W
¬
H W H W
ª W
W
« H H
«
«
«
« H W H W
¬
º
»
»
»
»
»
¼
º
» ª º
» « »
»« »
» «»
» ¬ ¼
¼
(c) Zero state response for u = 1 is obtained from (11.47) by putting x(0) and letting it
be x2.
∴
[
W
³RH
$ W W
ª
W«
³ «« M11_CSE_ISBN_C11.indd 45
«¬ %X W GW
H W W H WW
H W W H W W
W W
H
H W W
H W W H W W
º
»
» %GW
»
»¼
21/10/11 11:35 AM
11-46
∴
W $ W W
CONTROL SYSTEMS ENGINEERING
[
H
%X
³R
ª
W«
³ «« «¬ W GW
H W W H WW
H W W H W W
ª W W
H W W
« H
«
« H W W H W W
«¬ ³
ª
«
«
«
«¬
W
H W W H W W
º
»
» %GW
»
»¼
º
»
» GW
»
»¼
W W
º ª§ · § W W · º
H W W GW » «¨ ¸ ¨ H H ¸ »
H
©
¹ ©
¹»
» ««
»
W
HW W H W W GW » «
»¼ « H W H W »»
¬
¼
³
³
W W
H
H W W
W
ª W W º
« H H »
«
»
«¬ H W H W »¼
(d) Total response is given by
[W
[[
ª W
º
W
« H H
» ª W W º
«
» « H H »
·» « « §
»
« ¨ H W H W ¸ » «¬ HW H W »¼
¸»
« ¨
¹¼
¬ ©
ª W W º
« H H »
«
»
«
»
« »
« H W H W »
¬ ¼
∴
M11_CSE_ISBN_C11.indd 46
ª [ W º
«
»
«
»
«¬ [ W »¼
ª W W º
« H H »
«
»
«
»
« W »
W
« H H
»
¬ ¼
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
(e)
11-47
Inverse of Φ(t) as shown below.
) W
H $W
ª
W
W
« H H
«
« H W HW
¬« ) W
W
H H W
H W H W
º
»
»
»
¼»
Example 11.11 Consider the state matrix ‘A’ given by
$
ª º
« »
«
»
«¬ »¼
If it is not possible to diagonalise matrix ‘A’, determine matrix P so that P–1AP is in the Jordan
canonical form.
Solution
The characteristic
T equation of matrix ‘A’ is
RU
_λ,±$_
º
ª O « »»
O
«
«¬ O »¼
or, λ3 – 7λ2 + 15λ – 9 = 0
or, (λ–1) (λ–3)2 = 0
So matrix ‘A’ has eigenvalues of λ1 = 3 of order 2 and λ3 = 1.
Using equation (11.41), the column vectors p1 and p2 associated with λ1 = 3 of order 2 are
obtained from
(λ1 I – A) p1 = 0
and
....(11.80)
(λ1I – A) p2 = –p1
....(11.81)
From equation (11.80) we get
ª º ª S º
«
»« »
« » « S »
«¬ »¼ ¬« S ¼»
ª º
« »
« »
«¬ »¼
By elementary row transformation this equation can be reduced to
M11_CSE_ISBN_C11.indd 47
21/10/11 11:35 AM
11-48
CONTROL SYSTEMS ENGINEERING
ª º
« »
« »
«¬ »¼
ª º ª S º
«
»« »
« » « S »
«¬ »¼ ¬« S ¼»
S
∴
S S VD\
From equation (11.81), we get (as any constant can be assumed)
ª º
« »
« »
«¬ »¼
ª º ª S º
«
»« »
« » « S »
«¬ »¼ ¬« S ¼»
By elementary row transformation this equation can be reduced to
ª º
«»
« »
«¬ »¼
ª º ª S º
«
»« »
« » « S »
«¬ »¼ ¬« S ¼»
S
S
∴
DQG
S VD\
S Using equation (11.42), the column vector p3 associated with λ3 = 1 is obtained from
λ,±$ S
ª º
« »
« »
«¬ »¼
ª º ª S º
« » « S »
«
» « »
«¬ »¼ «¬ S »¼
RU
By elementary row transformation, this equation can be reduced to
ª º ª S º
« » « S »
«
» « »
«¬ »¼ «¬ S »¼
ª º
« »
« »
«¬ »¼
Assuming p33 = 1, we get
p31 = 0 and p32 = 2 p33 = 2
So matrix ‘P’ can be written as follows
3
1RZ
M11_CSE_ISBN_C11.indd 48
3±
ª S
«S
« ¬« S
S
S
S
S º
S »»
S ¼»
ª º
« »
«
»
«¬ »¼
ª º
$GM $ «« »»
3
«¬ »¼
ª º ª
º
21/10/11 11:35 AM
ª S S S º ª º
«S
» «
»
3
« S S » « »
STATE
SPACE ANALYSIS OF CONTINOUS SYSTEM 11-49
¬« S S S ¼» «¬ »¼
3±
1RZ
3±$3
∴
ª º
$GM $ «« »»
3
«¬ »¼
ª º ª º
« » « » 3
«
» «
»
«¬ »¼ «¬ »¼
ª 2 5 10 º ª1 2 0 º
«
»«
»
« 3 3 6 » «1 1 2 »
«¬ 0 1 1»¼ «¬1 1 1 »¼
ª3 1 0º
«
»
«0 3 0 »
«¬0 0 1 »¼
which is in the Jordan canonical form.
Example 11.12 Find the state transition matrix using Cayley-Hamilton theorem for the state
matrix ‘A’ given below.
$
ª º
« »
¬
¼
Solution
The characteristic equation of state matrix ‘A’ is
Tλ
O, $
λ
O
O
Since ‘A’ is of order 2, the remainder polynomial R(λ) will be of the form given below.
R(λ) = α0 + α1λ
....(11.82)
Since λ1 = –1 is the eigenvalue of order 2, the only one independent equation obtained by putting
i = 1 into equation (11.54) is
....(11.83)
H OW
D DO or, H±W
.....(11.84)
α± α
The other equation is obtained by differentiating both sides of equation (11.83) with respect to
λ1 as follows.
OW
t H = α1D DO
or,
te–tH=±Wα1 α± α
Solving equations (11.84) and (11.85), we get,
α0 = (1 + t)e−t and α1 = t e−t
M11_CSE_ISBN_C11.indd 49
....(11.85)
21/10/11 11:35 AM
11-50
CONTROL SYSTEMS ENGINEERING
Using formula (11.58) we may write the state transition matrix φ(t) as given below.
φ(t) = eAt = R(A)
φ(t) = α0 I + α1A
ª D0
«
¬ D1
D1
º
»
D 0 2 D1 ¼
ª W H W
«
W
«¬ WH
WHW
º
»
W H »¼
W
Example 11.13 Consider the system defined by
[
\
ZKHUH
$[%X
&[
$
º
ª
«
»
«
»
«¬ »¼
%
ª º
« »
« »
«¬ »¼
DQG
&
>
@
Check the system for (a) complete state controllability and (b) complete observability.
Solution
(a) Test for complete state controllability
$%
$%
º ª º
ª
«
»» «« »»
«
«¬ »¼ «¬ »¼
ª º
« »
«
»
«¬ »¼
º ª º
ª « »» «« »»
«
«¬ »¼ «¬ »¼
ª º
« »
«
»
«¬ »¼
So the controllability matrix Qc is given by
M11_CSE_ISBN_C11.indd 50
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
4F
4F
11-51
ª % $% $ % º
ª¬ % $% $ % º¼
¬
¼
º
ª º»
ª« »»
«« »
«« ¬« »¼»
¬
¼
1RZ_4F_ ±≠
1RZ_4F_ ±≠
So the rank of matrix Qc is equal to its order, that is, 3. This indicates that according to Kalman’s
test, the system is completely state controllable.
(b) Test for complete observability
&
$
ªº
ª º
« » DQG$ « »
« »
«
»
«¬ »¼
«¬ »¼
$ &
ª º ª º
« » « »
«
»« »
«¬ »¼ «¬ »¼
ª º
« »
« »
«¬ »¼
&
ª º ª º
« » « »
«
»« »
«¬ »¼ «¬ »¼
ª º
« »
« »
«¬ »¼
6RWKHREVHUYDELOLW\PDWUL[4LVJLYHQE\
4
ª&
¬
$ &
$
& º¼
ª º
« »
«
»
«¬ »¼
1RZ
_4_ ≠
So, the rank of matrix Q0 is equal to its order, that is, 3. This indicates that due to Kalman
the system is completely observable.
M11_CSE_ISBN_C11.indd 51
21/10/11 11:35 AM
11-52
CONTROL SYSTEMS ENGINEERING
REVIEW QUESTIONS
11.1 Develop a state variable model for the mechanical system shown below. The displacements
y1 and y2 represent deviations from equilibrium positions when u1 = u2 = 0.
11.2 Draw the state diagram for the following model in which the state equations are
[
[ [ X
[ [ X
DQGWKHRXWSXWHTXDWLRQVDUH
\ [[X
\ [
$VVXPH]HURLQLWLDOFRQGLWLRQV
11.3 Derive a state space representation of the following systems given by
D
< V
8 V
V V V E
\ \ \ X
M11_CSE_ISBN_C11.indd 52
21/10/11 11:35 AM
STATE SPACE ANALYSIS OF CONTINOUS SYSTEM
11-53
11.4 Determine the transfer function of the system which is represented by
ª º
ª º
« » [ «» X
¬
¼
¬ ¼
DQG\ [
[
11.5 For the system with transfer function
V
< V
V V 8 V
obtain the state space representation in (a) controllable canonical form and (b) observable
canonical form.
11.6 Consider the following state model of a system.
ª [ º
« [ »
« »
«¬ [ ¼»
\
ª º ª [ º ª º
« » « [ » « » X
«
»« » « »
«¬ »¼ ¬« [ ¼» «¬ »¼
ª [ º
>
@ «« [ »»
«¬ [ »¼
Determine a suitable transformation matrix that will transform the state equation in the
form given by
ª ] º
« ] »
« »
«¬ ] »¼
ª º ª ] º ª º
« » « ] » « » X
«
»« » « »
«¬ »¼ «¬ ] »¼ «¬ »¼
Then obtain y in terms of z1, z2 and z3.
11.7 Consider the following state space model of a linear system.
[
ª º
ªº
« » [ «» X
¬
¼
¬¼
DQG \
ª º
« » [
¬
¼
Find the complete solution for y(t) when x1(0) = 1, x2(0) = 0 and u is a unit step function.
M11_CSE_ISBN_C11.indd 53
21/10/11 11:35 AM
11-54
CONTROL SYSTEMS ENGINEERING
11.8 Using the Cayley-Hamilton theorem, find the state transition matrix for the following state
equations.
[
ª º
ªº
« » [ « » X
«
»
« »
«¬ »¼
«¬ »¼
11.9 Consider the system defined by
[
$[ %X
\
&[
ZKHUH
ª º
$ «« »»
«¬ »¼
ª º
ª º
% «« »» DQG& «
»
¬ ¼
¬« »¼
Is the system completely state controllable, completely observable and completely output
controllable?
11.10 Transfer function of a system is given by
* V
& V
5V
V
V V Represent the system by a state space model.
M11_CSE_ISBN_C11.indd 54
21/10/11 11:35 AM
APPENDIX 1
LAPLACE TRANSFORM
A1.1 INTRODUCTION
For a proper understanding of the behaviour of a system, one has to represent the system in
the form of mathematical relations. The voltage equation in algebraic form and a differential
equation of an electrical circuit are a few such mathematical relations.
The Laplace transformation offers a very simple and elegant method of solving linear or
linearized differential equations which result from the mathematical modeling of a process or a
system. The Laplace transform also allows us to develop input–process–output models which are
very useful for the study of control systems. It is because of these reasons that the Laplace
transform is used extensively in control engineering.
A1.2 DEFINITION OF LAPLACE TRANSFORM
Let there be a function f (t). The Laplace transform of f (t), denoted by  [ f (t)], is defined as
 [ f (t)] =
∞
∫e
− st
f (t )dt
(A1.1)
0
It can also be represented as
 [ f (t)] = f (s)
(A1.2)
Laplace transformation is a transformation of a function from the time domain (where time
is the independent variable) to the s domain (where s is the independent variable). s is a variable
defined in the complex plane as s = a + ib.
Laplace transformation of a function f (t) exists if the integral
that is, takes a finite value. For example, consider a function
∞
∫ f (t )e
0
− st
dt remains bound,
f (t) = eat
Appendix 1.indd 1
07/11/11 5:01 PM
A-2
CONTROL SYSTEMS ENGINEERING
The Laplace transform of this function is given as
 [ f (t)] =
∞
∞
0
0
at − st
∫ e e dt =
∫e
( a − s )t
dt
Now, if (a – s) > 0 or a > s, then the integral remains unbound.
However, if s > a then there will be a finite solution. Thus, the Laplace transform of the
function exists only for s > a.
The Laplace transformation is a linear operation. For example,
 [a1 f1(t) + a2 f2(t) = a1 [ f1(t)] + a2 [ f2(t)]
where a1 and a2 are constant parameters.
Proof
∞
 [a1 f1(t) + a2 f2(t)] = ∫ [ a1 f1 (t ) + a2 f2 (t )]e− st dt
0
=
∞
∫
0
∞
f1 (t )e− st dt + a2 ∫ f2 (t )e− st dt
0
= a1 [ f1(t)] + a2 [ f2(t)]
(A1.3)
A1.3 LAPLACE TRANSFORM OF SOME BASIC FUNCTIONS
Consider the following three types of functions:
(1) Exponential functions
(2) Trigonometric functions
(3) Standard test signals
1. Exponential functions: An exponential function, as shown in Figure A1.1, is defined as
Fig. A1.1 Exponential function representation.
f (t) = e− at
Then,
Appendix 1.indd 2
 [e–at] =
for t ≥ 0
1
s+a
07/11/11 5:01 PM
APPENDIX 1: LAPLACE TRANSFORM
A-3
Proof
 [e–at] =
∞
∫e
− at − st
e
0
=
∞
∫e
−( s + a )t
0
=
Similarly,
 [eat] =
dt
dt
∞
−1
1
 e −( s + a )t  =


0
s+a
s+a
(A1.4)
1
s−a
(A1.5)
2. Trigonometric functions: Consider the sinusoidal function f (t ) = sin ω t
Then,
 [sin wt] =
ω
s + ω2
(A1.6)
2
Proof
 [sin wt] =
∞
∫ sin ωte
− st
dt
0
∞
∫
0
=
1
∫ 2 j  e
− ( s − j ω) t
dt − e− ( s − jω)t  dt
∞
=
1
2j
 e − ( s − j ω) t e − ( s + j ω) t 
+
−

s + jω  0
 s − jω
=
1
2j
 1
1 
1  s + jω − s + jω 

 s − jω − s + jω  = 2 j 
2
2


 s − ( jω)

=
1  2 jω 
ω
= 2

2
2 
2 j  s + ω  s + ω2
∞
0
Similarly,

e jωt − e − jωt 

sin
ω
=
t


2j


1 jωt − jωt − st
( e − e )e dt
2j
 [cos wt] =
s
s + ω2
2
(A1.7)
1 jωt

− jωt  
 Note: cos ω t =  e + e
 
2

Appendix 1.indd 3
07/11/11 5:01 PM
A-4
CONTROL SYSTEMS ENGINEERING
A1.4 STANDARD TEST SIGNALS
In time-domain analysis, the dynamic response of the system to different types of inputs, which
are a function of time, is analysed at different intervals of time after the application of the input
signal. In practice, the input signals applied to a measurement system are not known in advance.
In most cases, the actual input signals vary in random fashion with respect to time and, therefore,
cannot be mathematically defined. However, for the purpose of analysis and design, it is necessary
to assume some basic types of input signals which can be easily defined mathematically, so that
the performance of a system can be analysed with these standard signals.
In the time-domain analysis, the following standard test signals are used:
(1) Step signal
(2) Ramp signal
(3) Parabolic signal
(4) Impulse signal
1. Step signal: A step signal is one, the value of which changes from one level (usually
zero) to another level, A, in zero time.
Mathematically, it is defined as
r(t) = A;
t≥0
= 0; t < 0
The step signal is plotted in Figure A1.2.
Fig. A1.2 Step signal representation.
In the Laplace form,
 [r(t)] = R(s) =
∞
∫ Ae
0
− st
dt
∞
= A ∫ e− st dt
0
∞
 e− st 

e0 
0
= A
=
A
−



−s 
 −st  0

A
R(s) =
s
If step signal is a unit step, then A = 1.
Appendix 1.indd 4
(A1.8)
07/11/11 5:01 PM
APPENDIX 1: LAPLACE TRANSFORM
A-5
So,
1
s
2. Ramp signal: A ramp signal is one, which starts at a value of zero and increases linearly
with time like velocity and, therefore, is also known as velocity signal as shown in Figure A1.3.
R(s) =
Fig. A1.3 Ramp signal.
Mathematically,
r(t) = At; t > 0
= 0; t < 0
In the Laplace form,
 [r(t)] = R(s) =
∞
∫
0
∞
Ate− st dt = A ∫ te− st dt
0
∞
∞
∞
∞ − st
 te− st 
 te− st 
 e− st 
e
= A
dt = A 
 − A∫
 + A

−s
 −s 0
 −s 0
 − s.s  0
0

 
 du
Vdt  dt 
 ∫ UVdt = U ∫ Vdt − ∫ 
∫
 
 dt

 e−∞ e−0 
A
= 0 + A 2 − 2  = 0 + 2
s
−s 
 −s
 e−∞ = 0 


So,
A
(A1.9)
s2
From Equations (A1.8) and (A1.9) we see that the ramp signal is an integral of the step signal.
3. Parabolic signal: A parabolic signal is one which is proportional to the square of time
and therefore, represents a constant acceleration. Hence the signal is also known as acceleration
signal.
Mathematically,
r(t) = t2; t > 0
= 0; t < 0
Appendix 1.indd 5
 [r(t)] = R(s) =
07/11/11 5:01 PM
A-6
CONTROL SYSTEMS ENGINEERING
Figure A1.4 (a) and (b) show the parabolic signals.
Fig. A1.4 Parabolic signal representation.
The Laplace transform of a unit parabolic signal is
R(s) =
∞
∫t
0
2 − st
e
dt =
2
s3
(A1.10)
if the parabolic signal has the magnitude A as shown in Figure A1.4 (b).
Mathematically,
r(t) = At2; t > 0
= 0; t < 0
The Laplace transform is
R(s) =
∞
∫
0
∞
At 2 e− st dt = A ∫ t 2 e− st dt
0
A
(A1.11)
s3
From expressions (A1.9) and (A1.11) it is clear that the parabolic signal is the integral of
the ramp signal.
4. Impulse signal: A unit impulse is defined as a signal which has zero value everywhere
except at t = 0 where the magnitude is finite. This function is generally called as d (delta) function.
The s function has the following properties:
d(t) = 0; t ≠ 0
=
ε
∫ δ(t )dt
=1
−ε
ε → 0.
where
Since a perfect impulse function is not practically realizable, it is approximated by a pulse
of small width having a unit area as shown in Figure A1.5 (a). The Laplace transform of unit
impulse is
 [d(t)] = 1
(A1.12)
Appendix 1.indd 6
07/11/11 5:01 PM
APPENDIX 1: LAPLACE TRANSFORM
A-7
Fig. A1.5 (a) Unit impulse function; (b) Impulse function of strength A.
Impulse signal representation.
A1.5 THEOREMS OF LAPLACE TRANSFORM
The following are the theorems of Laplace transform;
Final value theorem
1.
lim f (t ) = lim [ sF ( s )]
t →∞
s →∞
F(s) =  [ f (t)]
where
Proof: Using Laplace transform of derivative, we have,
∞
∫
0
df (t ) − st
e dt = sF(s) – f (0)
dt
Taking the limit of both sides as s → ∞
∞
lim ∫
s →∞
0
df (t ) − st
e dt = lim [ sF ( s ) − f (0)]
s →∞
dt
Since variable s is independent of time,
∞
or
∫ slim
→∞
0
df (t ) − st
e dt = lim [ sF ( s ) − f (0)]
s →∞
dt
∞
∫
0
df (t )
dt = lim [ sF ( s ) − f (0)]
s →∞
dt
So,
Appendix 1.indd 7
lim f (t ) − f (0) = lim sF ( s ) − f (0)
t →∞
s →∞
lim f (t ) = lim sF ( s )
t →∞
s →∞
(A1.13)
07/11/11 5:01 PM
A-8
2.
CONTROL SYSTEMS ENGINEERING
Initial value theorem
lim f (t ) = lim sF ( s )
where
t →∞
s →∞
F(s) =
∞
∫ f (t )e
− st
dt =  [ f (t)]
0
Proof: We know that
 df (t ) 

 = sF(s) – f (0)
 dt 
or
∞
∫
0
df (t ) − st
e dt = sF(s) – f (0)
dt
Taking lim to both sides,
s →∞
∞
0
df (t ) − st
e dt = lim [ sF ( s ) − f (0)]
s →∞
dt
∫ slim
→∞
df (t ) − st
e dt = lim [ sF ( s ) − f (0)]
s →∞
dt
lim ∫
s →∞
∞
0
0 = lim sF ( s ) − f (0)
s →∞
f (0) = lim sF ( s )
s →∞
or
lim f (t ) = lim sF ( s )
t →∞
(A1.14)
s →∞
A1.6 Laplace Transform of Varions Functions
Laplace transforms of various functions are shown in Table A1.1.
Table A1.1 Laplace Transforms of Various Functions.
Time Function (t ≥ 0)
Unit impulse, d(t)
Unit pulse, dA(t)
Unit step
Appendix 1.indd 8
Laplace Transform
1
1 1 − e −sA
A
s
1/s
07/11/11 5:01 PM
APPENDIX 1: LAPLACE TRANSFORM
Time Function (t ≥ 0)
Laplace Transform
Ramp, f(t) = t
1/s2
t2
2!/s3
tn
n!/sn+1
e–at
1
s+a
t ne–at
sin (wt)
cos (wt)
sinh (wt)
cosh (wt)
e–at sin (wt)
e–at cos (wt)
A-9
n!
(s + a )n +1
ω
s 2 + ω2
s
2
s + ω2
ω
s 2 − ω2
s
s 2 − ω2
ω
(s + a )2 + ω2
(s + a )
(s + a )2 + ω2
)
Appendix 1.indd 9
07/11/11 5:01 PM
APPENDIX 2
MATLAB FUNDAMENTALS
A2.1 INTRODUCTION
MATLAB, an abbreviation for Matrix Laboratory, is an interactive matrix-based program for
scientific and engineering calculations. The base program together with a toolbox (m-files) meant
for a particular topic extends the capability to use MATLAB in various fields such as Control
Systems, Statistics, Fuzzy Logic, Neural Network, etc.
A MATLAB session can be started on most systems by executing the command MATLAB.
To end the MATLAB session, simply execute the command exit or quit.
A MATLAB program line that begins with ‘%’ is comments or remarks which are not
executed. In the event of comments or remarks being more than one line, each line must begin
with %.
A semicolon at the end of a statement is used to suppress printing, that is, its result is not
displayed without hampering its execution. While entering a matrix, the semicolon is also used
to indicate the end of a row, except at the last row.
The colon operator is used in MATLAB to create vectors, to subscript matrices and to
specify for iteration. As an example, m : n means [m m + 1 …… n], B (: , j) means the jth column
of B, and B (i, :) means ith row of B.
In the following sections we will deal with four objects that allow us to interact with the
programs.
A2.2 STATEMENTS AND VARIABLES
In MATLAB, statements are formed by assigning an expression to a variable as follows:
variable = expression
For example, A = [1 2; 3 4] is a statement whose execution results in the matrix
1.0000
A =
 3.0000
Appendix 2.indd 10
2.0000 
4.0000 
07/11/11 5:05 PM
APPENDIX 2: MATLAB FUNDAMENTALS
But the statement A = [1 2;
A-11
3 4] will produce no output after its execution.
The usual mathematical operators like + for addition, – for subtraction, * for multiplication,
/ for division and ^ for power can be used in expressions, and to alter the order of operation,
parentheses are to be used.
A convenient feature of MATLAB is that the variables need not be dimensioned before
use. A variable name begins with a letter and is followed by letters and numbers comprising
the maximum length of 19 characters. If no variable is assigned to an expression, the result is
assigned to a generic variable.
MATLAB is case sensitive. The variables n and N are recognized as different quantities.
The command ‘who’ in lowercase gives a list of all variables in workspace. The command ‘clear’
will erase all non-permanent variables from the workspace. The command ‘clear’ x will erase a
particular variable x.
MATLAB has many pre-defined variables like pi (for p), Inf (for + ∞), NaN (for Nota-Number), i (for −1 ), j (for −1 ). One may use i as an integer and reserve j for complex
arithmetic. Pre-defined variables can be reset to their default value by using ‘clear’ name (for
example, clear pi).
To continue a long statement in the next line, an ellipsis consisting of three or more periods
(…) followed by the carriage return can be used. Many statements can be placed in one line by
using commas or semicolons.
In MATLAB, all computations are performed in double precision. The default output
format contains four decimal places for non-integers. However, there are four possible ways of
displaying output by using ‘format’ command as follows:
1.
pi or format short; pi will display a 5-digit scaled fixed point number.
ans. = 3.1416.
2.
format long; pi will display a 15-digit scaled fixed point number.
ans. = 3.14159265358979.
3.
format short e; pi will display a 5-digit scaled floating point number.
ans. = 3.1416e + 00.
4.
format long e; pi will display a 15-digit scaled floating point number.
ans. = 3.141592653589793e + 00.
The variables used in the workspace can be stored in a file named matlab.mat by using
‘save’ command.
A2.3 MATRICES
The basic computational unit in MATLAB is the matrix and its functions. No dimension or type
statements are needed in using matrices. The elements of a matrix can contain real and complex
numbers, elementary math functions, as well as trigonometric and logarithmic functions. A typical
Appendix 2.indd 11
07/11/11 5:05 PM
A-12
CONTROL SYSTEMS ENGINEERING
matrix expression is written within square brackets, [.]. The column elements are separated by
blanks or commas and the rows are separated by semicolon or carriage returns. Matrices can
be input across many lines by using a carriage return following the semicolon or in place of the
semicolon.
For an example,
A = [ –2*j
sin (pi/2)
3;
2.5
–6
0.5
carriage return
cos(pi/3)
exp (0.8)
0]
will produce,
1. 0000 3. 0000 
 0 − 2.0000i

A =
2. 5000 −6. 0000 0. 5000 

0. 5000
2 .2255 0.0000 
Basic matrix operations are addition, subtraction, multiplication, transpose, powers and
array operations. Some examples follow.
Example 1 Addition of matrices
A = [1
2;
3
4];
B = [0
2;
–1
1];
A+B
will produce
1.0000 4.0000 
ans. = 

 2.0000 5.0000 
Example 2 Multiplication of matrices
C = [1; 2]
A*C
will generate
5.0000
ans. =
11.0000
Example 3 Transpose of matrix
A
will produce
1.0000 3.0000 
ans. = 

 2.0000 4.0000 
If u and v are m × 1 vectors, then u′ * v is called the inner product (or dot product) which
is a scalar and u * v′ is called the outer product which is a m × m matrix.
The modified element-by-element operations of matrices are known as array operation.
Array multiplication (.*), division (./) and power (.^) operators require the preceding dot.
Appendix 2.indd 12
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APPENDIX 2: MATLAB FUNDAMENTALS
A-13
Example of array multiplication
A = [1;
3;
2];
B = [–1; 2;
–5];
A*B
will produce
 − 1.0000 
ans. =  6.0000 
 −10.0000 
The colon notation is used to generate a row vector x = [xi : dx : xf], where xi is the initial
value, dx is the increment and xf is the final value.
Example
x = [0.2 : 0.2 : 1];
y = x .* sin (x);
[x y]
will produce two vectors x and y
 0.2000
 0.4000

 0.6000
ans. = 
 0.8000
1.0000
0.0397 
0.1558 
0.3388 

0.5793 
0.8415 
If A is an n × n matrix, the n scalars l that satisfy Ax = lx are the eigenvalues of A. They
are found by using the command eig (A). For example,
A = [2
1;
0
3]
eig (A)
will produce
 2.0000 
ans. = 

 3.0000 
Eigenvectors are obtained from the double assignment statement
[X, D] = eig (A)
where the diagonal elements of D are eigenvalues and columns of X are the corresponding
eigenvectors, so that AX = XD. For example, if
1
0
0

0
1 
A =0
 −6 −11 −6 
Then the statement
[X, D] = eig (A)
Appendix 2.indd 13
07/11/11 5:05 PM
A-14
CONTROL SYSTEMS ENGINEERING
will produce
0. 2182 −0. 1048 
 −0. 5774
0. 3145 
X =  0. 5774 −0. 4364
 −0. 5774
0. 8729 −0. 9435 
0
0
 −1.0000

D =
0 −2.0000
0 
0
0 −3.0000 

The coefficients of the characteristic polynomial of a matrix ‘A’ are obtained by using the
statement p = poly (A). For example,
A = [1
2
3
;
4
5
6
;
7
8
0]
p = poly (A)
will produce the characteristic polynomial s3 – 6s2 – 72s – 27 by giving the output
p=
1.0000
–6.0000
–72.0000
–27.0000
3
2
The roots of the characteristic equation s – 6s – 72s – 27 = 0 will be obtained as follows:
r = roots (p)
produces
12.1229 
r =  −5.7345 
 −0.3884 
The following command will get back the original polynomial from the given roots.
q = poly (r)
will produce
q = 1.0000
–6.0000
–72.0000
–27.0000
4
2
To evaluate the polynomial p(s) = s + 5s + 7 at s = 1, the following command
p = [1
0
5
0
7];
polyval (p, 1)
produces
ans. = 13.0000
The command [r, p, k] = residue (num, den) will express the ratio b(s)/a(s) of two
polynomials into partial fraction given by
b(s) = r (1) + r (2) +  + r (n) + k ( s )
s − p(1) s − p (2)
s − p ( n)
a( s)
where
r(1), r(2), …, r(n) are the residues, p(1), p(2), …, p(n) are the poles and k(s) is the direct
(or constant) term. For example,
num = [2
5
3
6];
den = [1
6
11
6];
Appendix 2.indd 14
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APPENDIX 2: MATLAB FUNDAMENTALS
A-15
[r, p, k] = residue (num, den)
will produce
 −6.0000 
r =  −4.0000 
 3.0000 
 −3.0000 
p =  −2.0000 
 −1.0000 
k = [2]
This MATLAB result gives the following partial fraction
2s3 + 5s2 + 3s + 6
−6
−4
3
=
+
+
+2
s + 3 s + 2 s +1
s3 + 6s2 + 11s + 6
The command c = conv (a, b) will produce the multiplication of two polynomials a(s) and
b(s). For example,
a = [1
2
3];
b = [4
5
6];
= 4.0000
13.0000
c = conv (a, b)
produces
c
28.0000
27.0000
18.0000
The command deconv (c, a) will divide polynomial c(s) by a(s) to give quotient and
remainder polynomials of q(s) and r(s), respectively.
A2.4 GRAPHICS
Graphics play an important role in the design and analysis of control systems. MATLAB uses
a graph display to present plots. Plot formats and the customized plots are the two basic groups
of graphic functions which are given in Tables A2.1 and A2.2, respectively. A basic x–y plot is
generated with the combination of functions plot, xlabel, ylabel, title and grid.
Table A2.1 Plot Format.
plot (x, y)
Plots the vector x versus the vector y
semilog x (x, y)
Plots the vector x versus vector y where x-axis is log10 and y-axis is linear
semilog y (x, y)
Plots the vector x versus vector y where x-axis is liner and y-axis is log10
loglog (x, y)
Plots the vector x versus vector y where both axes are log10
Appendix 2.indd 15
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A-16
CONTROL SYSTEMS ENGINEERING
Table A2.2 Functions for Customised Plots.
title (‘text’)
Puts ‘text’ at the top of the plot
xlabel (‘text’)
Labels x-axis with ‘text’
ylabel (‘text’)
Labels y-axis with ‘text’
text (p1, p2, ‘text’)
Puts ‘text’ at (p1, p2) in screen co-ordinates
subplot
Sub-divides the graphics window
grid
Draws grid lines on the current plot
To plot multiple curves on a single graph, the following plot command with multiple
arguments is to be used.
plot (x1, y1, x2, y2, …, xn, yn).
The default line (or point) types can also be altered by the line type chosen from Table A2.3
as given by the following command.
plot (x1, y1, ‘:’, x2, y2, ‘+’)
Table A2.3 Line and Point Types.
Line Types
Point Types
Solid
—
Point
.
Dashed
——
Plus
+
Dotted
:
Star
*
Dash-dot
—.
Circle
O
x-mark
x
The hold command freezes the current plot and inhibits erasure and rescaling. This
command also allows us to plot multiple curves on a single graph. The current plot will be
released by entering the hold command again.
Colour statements are used as plot (x, y, r) which indicates a red line. Similarly, the command
plot (x, y, ‘+g’) indicates that the line is formed by green + marks. Other available colours are b
for blue, w for white and i for invisible.
In MATLAB, the plot is automatically scaled. For manual axis scaling, the following
commands will freeze the current axis scaling for subsequent plots.
v = [x–min
x–max
y–min
y–max]
axis (v)
To resume auto scaling, type axis again. axis (‘square’) sets the plot region on the region to
be square; axis (‘normal’) sets back the plot region to be normal.
A2.5 SCRIPTS
A long sequence of commands to be executed for control system design and analysis is stored in a
file of the form filename.m. These files are called m-files. Scripts are a type of m-files containing
Appendix 2.indd 16
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APPENDIX 2: MATLAB FUNDAMENTALS
A-17
the ASCII text of codes necessary to perform some function. A text editor is required to create
and edit m-files. A script is invoked at the command prompt by simply typing its filename. A
script file should have a header and descriptive comments that begin with ‘%’. For example, the
program stored in a script file plotdata.m is given below.
%
This is a script for plotting
%
the function y = sin (alpha*t).
%
The value of alpha must be
%
given before invoking script.
t
= [0 : 0.01 : 1];
y
= sin (alpha *t);
plot (t, y)
xlabel (‘Time [sec]’)
ylabel (‘y(t) = sin (alpha *t)’)
grid
The command
Alpha = 50;
plotdata
will produce the graph after execution of the above script as has been shown in Figure A2.1.
Fig. A2.1 Graph of y(t) = sin (at) for a = 50 using the script plotdata.m.
Some MATLAB functions used to write script are given in Table A2.4.
Table A2.4 Some MATLAB Functions.
Function Name
Function Description
abs
Computes the absolute value
acos
Computes the arccosine
ans
Variable created for expressions
asin
Computes the arcsine
Contd...
Appendix 2.indd 17
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A-18
CONTROL SYSTEMS ENGINEERING
Function Name
Function Description
atan
Computes the arctangent (2 quadrant)
atan2
Computes the arctangent (4 quadrant)
axis
Specifies the manual axis scaling on plots
bode
Generates Bode frequency response plots
c2dm
Converts a continuous-time state variable system representation to a
discrete-time system representation
clear
Clears the workspace
clg
Clears the graph window
cloop
Computes the closed-loop system with unity feedback
conj
Computes the complex conjugate
conv
Multiplies two polynomials (convolution)
cos
Computes the cosine
ctrb
Computes the controllability matrix
diary
Saves the session in a disk file
d2cm
Converts a discrete-time state variable system representation to a
continuous time system representation
dstep
Computes the unit step response of a discrete-time system
eig
Computes the eigenvalues and eigenvectors
end
Terminates control structures
exp
Computes the exponential with base e
expm
Computes the matrix exponential with base e
eye
Generates an identity matrix
feedback
Computes the feedback interconnection of two systems
for
Generates a loop
format
Sets the output display format
grid
Adds a grid to the current graph
help
Prints a list of HELP topics
hold
Holds the current graph on the screen
i
−1
imag
Computes the imaginary part of a complex number
impulse
Computes the unit impulse response of a system
inf
Represents infinity
j
−1
linspace
Generates linearly spaced vectors
load
Loads variables saved in a file
log
Computes the natural logarithm
log10
Computes the logarithm base 10
loglog
Generates log-log plots
Contd...
Appendix 2.indd 18
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APPENDIX 2: MATLAB FUNDAMENTALS
Function Name
A-19
Function Description
logspace
Generates logarithmically spaced vectors
abs
Computes the absolute value
Isim
Computes the time response of a system to an arbitrary input and initial
conditions
margin
Computes the gain margin, phase margin and associated crossover
frequencies from frequency response data
max
Determines the maximum value
mesh
Creates three-dimensional mesh surfaces
meshdom
Generates arrays for use with the mesh function
min
Determines the minimum value
minreal
Transfer function pole-zero cancellation
NaN
Representation for Not-a-Number
ngrid
Draws grid lines on a Nichols chart
nichols
Computes a Nichols frequency response plot
num2str
Converts numbers to strings
nyquist
Calculates the Nyquist frequency response
obsv
Computes the observability matrix
ones
Generates a matrix of integers where all the integers are 1
pade
Computes an nth order Pade approximation to a time delay
parallel
Computes a parallel system connection
plot
Generates a linear plot
poly
Computes a polynomial from roots
polyval
Evaluates a polynomial
printsys
Prints state variable and transfer function representations of linear
systems in a readable form
pzmap
Plots the pole-zero map of a linear system
rank
Calculates the rank of a matrix
real
Computes the real part of a complex number
residue
Computes a partial fraction expansion
rlocfind
Finds the gain associated with a given set of roots on a root locus plot
rlocus
Computes the root locus
roots
Determines the roots of a polynomial
roots 1
Same as the roots function, but gives more accurate answers when
there are repeated roots
semilogx
Generates an x–y plot using semilog scales with the x-axis log10 and
the y-axis linear
semilogy
Generates an x–y plot using semilog scales with the y-axis log10 and
the x-axis linear
series
Computes a series system connection
shg
Shows graph window
Contd...
Appendix 2.indd 19
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A-20
CONTROL SYSTEMS ENGINEERING
Function Name
Function Description
sin
Computes the sine
sqrt
Computes the square root
ss2tf
Converts state variable form to transfer function form
step
Calculates the unit step response of a system
abs
Computes the absolute value
subplot
Splits the graph window into subwindows
tan
Computes the tangent
text
Adds text to the current graph
title
Adds a title to the current graph
tf2ss
Converts a transfer function to state variable form
who
Lists the variables currently in memory
whos
Lists the current variables and sizes
xlabel
Adds a label to the x-axis of the current graph
ylabel
Adds a label to the y-axis of the current graph
zeros
Generates a matrix of zeros
Appendix 2.indd 20
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APPENDIX 3
KEY CONTROL TERMINOLOGIES
Actuating signal: It is also called error signal (see error signal).
Asymptote: The path the root locus follows as the parameter becomes very large and approaches
infinity.
Bandwidth: The frequency at which the frequency response declines to 3 dB from its lowfrequency value.
Bode plot: The logarithm of the magnitude of the transfer function is plotted versus the logarithm
of w, the frequency. The φ of the transfer function is plotted versus the logarithm of the frequency.
Characteristic equation: The relation formed by equating to zero the denominator of a transfer
function.
Closed-loop feedback control system: A system that uses a measurement of the output and
compares it with the desired output.
Compensation: The alteration or adjustment of a control system in order to provide a suitable
performance.
Compensator: An additional component or circuit that is inserted into the system to compensate
for a performance deficiency.
Control system: An interconnection of components that forms a system configuration providing
a desired response.
Controlled variables: Quantity or condition of the control system which is directly controlled
or measured.
Dominant roots: The roots of the characteristic equation that cause the dominant transient
response of the system.
Damped oscillation: An oscillation in which the amplitude decreases with time.
Damping ratio: A measure of damping. A dimensionless number for the second-order
characteristic equation.
Error signal: The difference between the desired output, R(s), and the actual output, C(s).
Therefore, E(s) = R(s) – C(s).
Appendix 3.indd 21
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A-22
CONTROL SYSTEMS ENGINEERING
Frequency response: The steady-state response of a system to a sinusoidal input signal.
Gain margin: The increase in the system gain when phase = –180° will result in a marginally
stable system with the intersection of the –1 + j0 point on the Nyquist diagram.
Laplace transform: A transformation of a function f (t) from the time domain into the complex
frequency domain, yielding F(s).
Lead–lag network: A network with dual characteristics of a lead network and a lag network.
Linear system: A system that satisfies the properties of superposition and homogeneity.
Natural response: The frequency of natural oscillation that would occur for two complex poles
if the damping were equal to zero.
Negative feedback: The output signal is fed back so that it is subtracted from the input signal.
Nichols chart: A chart displaying the curves for the relationship between the open-loop and
closed-loop frequency response.
Nyquist stability criterion: A feedback system is stable if and only if, the contour in the
G(s)-plane does not encircle the (–1, 0) point when the number of poles of G(s) in the right-hand
s-plane is zero.
If G(s) has P poles in the right-hand plane, then the number of counterclockwise encirclements
of the (–1, 0) point must be equal to P for a stable system.
Open-loop control system: A system that utilizes a device to control the process without using
the feedback. The output has no effect on the signal process.
Optimization: The adjustment of the parameters to achieve the most favourable or advantageous
design.
Optimum control system: A system whose parameters are adjusted so that the performance
index reaches an extreme value.
Overshoot: The amount by which the system output response proceeds beyond the desired
response.
PD controller: A controller with three terms in which the output is the sum of a proportional
term, an integrating term and a differentiating term, with an adjustable gain for each term.
Peak time: The time for a system to respond to a step input and finally rise to a peak response.
Performance index: A quantitative measure of the performance of a system.
Phase margin: The amount of phase shift of the GH( jw) at unity magnitude will result in a
marginally stable system with intersections of the –1 + j0 point on the Nyquist diagram.
Phase-lag network: A network that provides a negative phase angle and a significant attenuation
over the frequency range of interest.
Phase-lead network: A network that provides a positive phase angle over the frequency range
of interest. Thus, phase lead can be used to cause a system to have an adequate phase margin.
PI controller: Controller with a proportional term and an integral term (proportional-integral).
Polar plot: A plot of the real part of G( jw) versus the imaginary part of G( jw).
Appendix 3.indd 22
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APPENDIX 3: Key Control Terminologies
A-23
Positive feedback: The output signal is fed back so that it adds to the input signal.
Resonant frequency: The frequency, wr, at which the maximum value of the frequency response
of a complex pair of poles is attained.
Rise time: The time for a system to respond to a step input and attain a response equal to a
percentage of the magnitude of the input. The 0–100% rise time Tr measures the time to 100% of
the magnitude of the input.
Root locus: The locus or path of the roots traced out on the s-plane as a parameter is changed.
Root locus method: The method for determining the locus of roots of the characteristic equation
1 + KP(s) = 0 as K varies from 0 to µ.
Routh–Hurwitz criterion: A criterion for determining the stability of a system by examining
the characteristic equation of the transfer function. The criterion states that the number of roots
of the characteristic equation with positive real parts is equal to the number of changes of sign of
the coefficients in the first column of the routh array.
Sensors: These are low-powered transducers which produce output signal as a measure of the
controlled variable.
Settling time: The time required for the system output to settle within a certain percentage of the
input amplitude.
Signal-flow graph: A diagram that consists of nodes connected by several direct branches and is
a graphical representation of a set of linear relations.
Simulation: Model of a system that is used to investigate the behaviour of a system by utilizing
actual input signals.
Stability: Performance measurement of a system. A system is stable if all the poles of the transfer
function have negative real parts.
State-variable: The set of variables that describes a system.
Steady-state error: Error when the time period is large and the transient response has decayed,
leaving a continuous response.
System: An interconnection of elements and devices for a desired purpose.
Test input signal: An input signal used as a standard test of a system’s ability to respond
adequately.
Time domain: The mathematical domain that incorporates the time response and the description
of a system in terms of time t.
Transfer function: The ratio of the Laplace transform of the output variable to the Laplace
transform of the input variable.
Transfer function in the frequency domain: The ratio of the output to the input signal where
the input is a sinusoid. It is expressed as G( jw).
Transient response: The response of a system as a function of time.
Appendix 3.indd 23
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Appendix 3.indd 24
07/11/11 5:12 PM
Model Question PaPers
Model Question PaPer - i
Control Systems
Time: 3 hours
Max Marks: 80
Answer any FIVE questions
All questions carry equal marks
1. (a) Find by SFG technique, the transfer function of the control system, the block diagram
repretsentation of which is shown in Figure 1.
(8)
Fig. 1
Block diagram of a control system.
(b) Draw the network diagram for the system in Figure 2 and also draw its analogous circuit.
(8)
Fig. 2
Model Question papers .indd 1
A mechanical translational system.
18/11/11 10:26 AM
Q-2
2.
Model Question Papers
A unity feedback system has an open-loop transfer function
G(s) =
25
s(s + 8)
Determine its damping ratio, peak overshoot and time required to reach the peak output.
Now a derivative component having transfer function of s/5 is introduced in the system.
Discuss its effect on the above-obtained values.
(16)
3.
ketch the Bode plot and hence determine the gain crossover frequency and phase
S
crossover frequency for the transfer function of a system represented by
G(s) =
10
s(s + 0.5s)(1 + 0.1s) (16)
4. (a) The open-loop transfer function of a unity feedback system is given by
K
( s + 2)( s + 4)( s 2 + 6s + 25)
By applying the Routh–Hurwitz criterion, discuss the stability of the closed-loop system
as a function of K. Determine the value of K which will cause sustained oscillations in the
closed-loop system. What are the corresponding oscillation frequencies?
(16)
5. The open-loop transfer function of a system is
G(s)H(s)
K
s(s + 2)(s + 3)
Determine the value of K for which the system will remain stable with (1) phase margin of 60°
(16)
and (2) gain margin of 6 dB.
6. D
etermine whether the largest time constant of the following characteristic equation is less
than, greater than or equal to 1.0 sec.
s3 + 5s2 + 8s + 6 = 0.
(16)
7. A feedback control system has an open-loop transfer function
K (s + 1)
G(s)H(s) = 2
s (s + 4)
Draw the root locus as K varies from zero to infinity.
8. (a) Explain lag compensator.
(b) The open-loop transfer function of a unity feedback system is
Model Question papers .indd 2
G(s) =
(16)
(8)
1
s ( s + 1) (0.5s + 1)
18/11/11 10:26 AM
Model Question Papers
Q-3
Design a compensator to meet the following specifications and compare the responses
through MATLAB.
(1) Static velocity error constant Kv = 5/sec.
(2) Phase margin of at least 40°.
(3) Gain margin of at least 10 dB.
(8)
Model Question papers .indd 3
18/11/11 10:26 AM
Model Question Paper - II
Control Systems
Time: 3 hours
Max Marks: 80
Answer any FIVE questions
All questions carry equal marks
1. (a) Derive the transfer function of an armature controlled DC motor
(b) Find the transfer function for the system shown in Figure 1 by Mason’s gain formula.
(8+8)
Fig. 1 Block diagram of a control system.
2. (a)A hydraulic actuator with incompressible oil inside the cylinder is shown in Figure 2. The
mass of piston and leakage are assumed negligible. Obtain the transfer function relating
displacement x of the valve piston and displacement y of the main piston.
Fig. 2 A simplified hydraulic pump.
Model Question papers .indd 4
18/11/11 10:26 AM
Model Question Papers
Q-5
(b)Derive the transfer function of a simple mercury glass thermometer and draw its electrical
analogous electrical circuit.
(8+8)
3.Find the open-loop transfer function of a system whose approximate Bode plot is shown
in Figure 3 below.
(16)
Fig. 3 Bode plot of a given control system.
4. A block diagram representation of a unity feedback control system is as follows.
Fig. 4 Block diagram of a feedback control system.
For this system, sketch the root locus. Also determine the value of K so that the damping ratio,
ξ , of a pair of complex conjugate closed-loop poles is 0.5.
(16)
5. Discuss in detail the design procedure for a lead compensator network.
(16)
6. (a) Classify the Second-order Control System based on the damping ration.
(b)The block diagram of a servo system is shown in Figure 5. Determine the characteristic
equation of the system. Also, calculate the following when the unit step input is given:
(1) Undamped frequency of oscillation
(2) Damped frequency of oscillation
(3) Damping ratio and damping factor
(4) Maximum overshoot
(5) Time interval after which maximum and minimum output will occur
Model Question papers .indd 5
18/11/11 10:26 AM
Q-6
Model Question Papers
Fig. 5 Block diagram of a servo system.
(6)
(7)
Settling time
Number of cycles completed before the output is settled within 2%, 5% of final
(4+12)
value.
7. (a) What is bandwidth? Derive the expression for normalized and denormalized bandwidth.
(b) Sketch the polar plot of a system whose transfer function is given as
(4+12)
G(s) =
10
s(s + 1)(s + 2)
8. The open-loop transfer function of a unity feedback system is given as
G(s) H(s) =
1
s (1 + s )(1 + 0.1s )
Find the gain margin and phase margin. If a phase lag element with transfer function
(1 + s/1 + 0.4s) is added in the forward path, determine by how much the gain must be
changed to maintain the same gain margin.
(16)
Model Question papers .indd 6
18/11/11 10:26 AM
Model Question Paper - III
Control Systems
Time: 3 hours
Max Marks: 80
Answer any FIVE questions
All questions carry equal marks
1. (a) Explain the linearizing effect of feedback.
(b) The dynamic behaviour of the system is described by the equation, dC
dt + 10C = 40e, where ‘e’ is the input and ‘C’ is the output. Determine the transfer
[10+6]
function of the system. 2.Derive the transfer function for D.C. servomotor. Explain about torque–speed characteristics
of D.C. servomotor. [16]
3. (a) What are the types of controllers that are used in closed loop system? Explain them.
(b) The response of a system subjected to a unit step input is c(t) = 1 + 0.2e – 60t – 1.2e – 10t
Obtain the expression for the closed loop transfer function? Also determine the Un
[8+8]
damped natural frequency and damping ratio of the system? 4.Sketch the root locus plot for the systems whose open loop transfer function is given by G(s)
H(s) = K s(s + 1)(s + 3). Determine
(a) the gain margin
(b) the phase margin for K = 6. [16]
5. Sketch the Bode plots for a system
G(s) = 15(s + 5)
s(s2 + 16s + 100)
Hence determine the stability of the system. [16]
6. (a) Explain the effect of addition of a pole at the origin on the polar plot of a given system.
(b)Sketch the polar plot and hence find the frequency at which the plot intersects the +ve
imaginary axis for the system
G(s) = 0.1
s2(1 + s)(1 + 0.1s)
Also find the corresponding magnitude. [6+10]
Model Question papers .indd 7
18/11/11 10:26 AM
Q-8
Model Question Papers
7. (a) What is compensation? What are the different types of compensators?
(b)What is a lag compensator? Obtain the transfer function of lag compensator and draw
pole-zero plot.
(c) Explain the different steps to be followed for the design of compensator using Bode plot.
[3+3+10]
8. (a) Consider the following state space model of a linear system.


 
x =  −6 4  x +  1  µ
 −2 0 
 1 


and y =  1 0  x
 1 1 
(b) Obtain the state model of armature controlled dc motor?
Model Question papers .indd 8
18/11/11 10:26 AM
Model Question Paper - IV
Control Systems
Time: 3 hours
Max Marks: 80
Answer any FIVE questions
All questions carry equal marks
1. Define and explain the following terms:
(a) Characteristic equation
(b) Order of a transfer function
(c) Type of a transfer function
(d) Poles and zeros of a transfer function.
[16]
2.Represent the following set of equations by a signal flow graph and determine the overall gain
relating x5 and x1:
[16]
x2 = ax1 + fx2
x3 = bx2 + ex4
x4 = cx3 + hx5
x5 = dx4 + gx2
3. (a) What are the types of controllers that are used in closed loop system? Explain them?
(b)The response of a system subjected to a unit step input is c(t) = 1 + 0.2e – 60t – 1.2e – 10t
Obtain the expression for the closed loop transfer function? Also determine the
Undamped natural frequency and damping ratio of the system? [8+8]
4.Explain the stability of the system from the root locus plot in the following situations with
suitable examples:
(a) addition of open loop poles
(b) addition of open loop zeros [16]
5. (a) Write a note on frequency domain specifications
(b)For a system the resonant peak is 1 dB an wn = 10 r/s. Find resonant frequency. Also find
the phase angle of closed loop transfer function at resonant frequency.
[6+10]
6. (a)What is “Nyquist Contour”?
(b) A system is given by
G(s) = 4s + 1s2(s + 1)(2s + 1) Sketch the Nyquist plot and hence determine the stability
[2+14]
of the system. Model Question papers .indd 9
18/11/11 10:26 AM
Q-10
Model Question Papers
7. (a) What is compensation? What are the different types of compensators?
(b)What is a lag compensator? Obtain the transfer function of lag compensator and draw
pole-zero plot.
(c) Explain the different steps to be followed for the design of compensator using Bode plot.
[3+3+10]
8. (a) Discuss the significance of state space analysis.
(b) Define state variables.
(c) Obtain the state variable representation of an armature controlled D.C servomotor?
Model Question papers .indd 10
18/11/11 10:26 AM
Model Question Papers
Q-11
Solutions to Model Question Paper - I
1. (a) Refer Page. No. 4-18 (Example: 4.9).
(b) Refer Page. no. 2-23 (Example: 2.6).
2. (a) Refer Page. No. 5-32 (Example: 5.5).
3. (a) Refer Page. No. 8-29 (Example: 8.3).
4. (a) Refer Page. No. 6-10 (Example: 6.9).
5. Refer Page. No. 8-59 (Example: 8.17).
6. Refer Page. no. 6-7 (Example: 6.6).
7. Refer Page. No. 7-43 (Example: 7.13).
8. (a) Refer Page. No. 10-4.
(b) Refer Page. No. 10-32 (Example:10.8).
Model Question papers .indd 11
18/11/11 10:26 AM
Q-12
Model Question Papers
Solutions to Model Question Paper - II
1. (a) Refer Page. No. 3-31.
(b) Refer Page. No. 4-21 (Example: 4.12).
2. (a) Refer Page. No. 2-28 (Example: 2.9).
(b) Refer Page. No. 2-30 (Example: 2.10).
3. (a) Refer Page. No. 8-32 (Example: 8.4).
4. Refer Page. No. 7-17 (Example: 7.2).
5. Refer Page. No. 10-3.
6. (a) Refer Page. No. 5-18.
(b) Refer Page. No. 5-29 (Example: 5.3).
7. (a) Refer Page. No. 8-9.
(b) Refer Page. No. 8-24 (Example: 8.2).
8. Refer Page. No. 10-22 (Example: 10.5).
Model Question papers .indd 12
18/11/11 10:26 AM
INDEX
A
a.c. synchro motor, 3-44–3-46
transfer function of, 3-44–3-46
absolute stability, 8-10
angle of departure, 7-8
B
bandwidth, 8-5, 8-9–8-10
block diagram simplification, 3-4–3-6
rules of, 3-4–3-6
Bode diagram, 8-12–8-44
specifications and transfer function for,
8-12–8-44
Bode plot, 8-13–8-18
initial slope of, 8-18–8-20
method of drawing, 8-12–8-18
quadratic form of transfer function and,
8-21–8-22
branch, 4-2
break frequency, 8-15
breakaway points, 7-8, 7-12–7-14
C
closed-loop control system, 1-6–1-7
block diagram of, 3-2–3-4
transfer function of, 3-2–3-4
compensation techniques, 10-1–10-46
necessity of, 10-1
Index.indd 1
complete observability, 11-27–11-28
alternative condition of, 11-27–11-28
complete state controllability, 11-24–11-25
alternative condition of, 11-24–11-25
constant magnitude loci, 8-67–8-68
constant N-circle, 8-69–8-70
continuous time control systems, 1-8
control system,1-1–1-6
basic components of, 1-5
characteristic equation of, 3-4
classification of, 1-6–1-10, 5-3–5-6
concept of, 1-1
examples of, 1-2–1-3, 1-10–1-12
sensitivity of, 5-38–5-40
time response of, 5-1–5-70
controllability, 2-34–2-35, 11-23–11-25
corner frequency. See break frequency
D
d.c. motor drive, 3-32–3-37
block diagram of, 3-32–3-37
damped natural frequency, 8-24
delay time, 5-23
derivative control, 7-56
derivative feedback control, 5-53–5-54
deterministic control systems, 1-9
differential equations, 4-14–4-16
SFG for solution of, 4-14–4-16
discrete control system, 1-8
07/11/11 11:19 AM
I-2
INDEX
distributed parameter control systems, 1-8–1-9
disturbance, 1-9–1-10
duality, 11-28–11-29
principle of, 11-28–11-29
dynamic systems, 1-9
E
error constants, 5-40–5-48
error detector, 1-5, 3-2
block diagram representation of, 3-2
error signal, 1-5
F
feedback control system, 5-11–5-21
characteristic equation of, 5-11–5-21
feedback control, 1-12–1-13
feedback element, 1-5
first-order system, 5-4
time response of, 5-6
force–current analogy, 2-13–2-15
force–voltage analogy, 2-12–2-13
forward path gain, 4-2
frequency response, 8-1–8-75
closed-loop system using M-circle and
N-circle and, 8-66–8-74
correlation with time response and, 8-5
presentation in graphical form, 8-10–
8-11
G
gain crossover frequency, 8-28
gain margin, 8-26–8-28
determination from polar plot, 8-55, 9-12
graphical methods for obtaining, 8-29
gear trains, 3-37–3-40
block diagram of, 3-37–3-40
Index.indd 2
H
higher order systems, 5-26–5-38
dominant closed-loop poles of, 5-26–
5-38
hydraulic system, 2-15–2-16
I
impulse input, 5-10–5-11
response of first-order system to, 5-10–
5-11
impulse signal, 5-3
input node, 4-2
input signals, 5-2–5-3
types of, 5-2–5-3
integral control, 7-56
L
lag compensator, 10-4–10-6
lag–lead compensator, 10-6–10-7
lead compensator, 10-3–10-4
linear continuous time systems, 11-5–11-16
state models of, 11-5–11-16
linear control systems, 1-7
linear invariant SISO system, 2-1
impulse response of, 2-1
linear systems, 11-3–11-5
state model of, 11-3–11-5
loop gain, 4-2
lumped parameter control systems, 1-8–1-9
M
MATLAB, 5-66–5-70
drawing Nyquist plots with, 8-63–8-65,
9-20–9-22
plotting bode diagrams with, 8-43
transient response analysis using, 5-66–
5-70
07/11/11 11:19 AM
INDEX
use of, 10-30–10-36
maximum overshoot, 5-12, 5-24–5-25, 8-24
maximum percentage overshoot, 5-22
M-circles, 8-70–8-71
uses of, 8-70–8-71
MIMO. See multi-input-multi-output (MIMO)
control systems
multi-input-multi-output (MIMO) control
systems, 1-8
N
N-circles, 8-70
uses of, 8-70–8-71
non-linear control systems, 1-7
non-touching loops, 4-2
normalized resonant frequency, 8-6
Nyquist contour. See Nyquist path
Nyquist path, 8-47, 9-4–9-5
Nyquist stability criterion, 8-44–8-63,
9-1–9-20
polar plot and, 8-44–8-63, 9-1–9-20
O
observability, 2-34–2-35, 11-26–11-28
open-loop control systems, 1-6–1-7
output controllability, 11-25–11-26
output node, 4-2
P
parabolic signal, 5-3
path, 4-2
peak time, 5-22–5-24
phase crossover frequency, 8-26–8-28
phase margin, 8-28
determination from polar plot, 8-55, 9-12
graphical methods for obtaining, 8-29
Index.indd 3
I-3
PID controllers, 10-40–10-46
basic elements of, 10-43–10-44
pneumatic system, 2-16–2-17
polar plot, 8-44–8-63, 9-1–9-20
process control, 1-9
proportional control, 5-48–5-49
proportional controllers, 10-40–10-41
proportional derivative controllers, 10-41–
10-42
proportional integral controllers, 10-43
proportional plus derivative control, 5-50–
5-51
proportional plus integral control, 5-51–5-52
proportional plus integral plus derivative
control, 5-52–5-53
R
ramp input, 5-8–5-10
response of first-order system to, 5-8–
5-10
ramp signal, 5-3
regulator, 1-9
relative stability, 8-10, 8-65–8-66
resonant frequency, 8-5–8-6, 8-22
resonant peak, 8-5
resonant peak amplitude, 8-23
rise time, 5-22, 5-23
root locus technique, 7-1–7-60
concept of, 7-2–7-4
construction procedures of, 7-4–7-5
construction rules of, 7-7–7-8
MATLAB and, 7-57–7-60
rotational mechanical system, 2-11–2-12
Routh stability criterion, 6-2–6-4
application and, 6-2–6-4
limitations of, 6-18–6-19
Routh–Hurwitz stability criteria, 6-4
07/11/11 11:19 AM
I-4
INDEX
S
second-order control system, 5-4–5-6, 5-12
classification of time response for, 5-18
positional servomechanism and analysis,
5-12–5-16
time response subject to unit step input,
5-18–5-21
transient response of, 5-23–5-26
second-order transfer function
maximum magnitude of, 8-22–8-24
sensitivity, 5-39
servomechanism, 1-9, 3-41–3-42
block diagram of, 3-41–3-42
settling time, 5-23, 5-25–5-26
signal flow graph, 4-1–4-27
construction of, 4-2–4-14
definition of, 4-1
single-input single-output linear systems, 11-5
state model of, 11-5
single-input-single-output (SISO) control
systems, 1-8
SISO. See single-input-single-output (SISO)
control systems
stability, 6-1–6-19
concept of, 6-1–6-18
pole-zero location and conditions for,
6-1–6-2
state equation, 11-19–11-22
solution of, 11-19–11-22
state matrix, 11–17–11-19
diagonalisation of, 11-17–11-19
Index.indd 4
state space, 11-2
state space equations, 11-2–11-3
state transition matrix, 11-21–11-22
properties, 11-22
state variables, 11-2, 2-32
state vector, 2-32, 11-2
static systems, 1-9
steady state-response-steady state error, 5-40–
5-48
step input, 5-6–5-8
response of first-order system to, 5-6–5-8
step signal, 5-3
stochastic control systems, 1-9
T
temperature control system, 1-10
thermal system, 2-17–2-18
time-invariant control systems, 1-8
time-varying control systems, 1-8
transfer function, 2-3–2-6
definition of, 2-4
poles and zeros of, 2-5
procedure for determining of, 2-5
transient response analysis, 5-66–5-70
using MATLAB, 5-66–5-70
translational mechanical system, 2-9–2-11
Z
zero-order system, 5-4
07/11/11 11:19 AM