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Exploring Various Control Systems and Its Application
Electrical and Automation Engineering, 2022
The control system is a system that delivers the desired response by controlling the output. The control system manages the behavior of other devices or systems using control rings, executes or regulates commands. It is used to control the first processes of a home heat controller that uses a thermostat to control the boiler in the home. Maybe up to large industrial control systems. Control system a set of mechanical or electronic devices that control other devices or systems through control loops. Typically, control systems are computerized. Control systems are an integral part of industry and automation. Continuous modulated control is a feedback controller used to automatically control a process or process. The control system uses the plant process variable a to compare the process value variable or position (PV) with the desired value or set point (SP). Control signal. Output to the same value. A variable size or set variable sizes are made according to the recommended rule. This will keep the values of the controlled quantities constant or vary as recommended. A control system can be operated by electric, mechanical mixing mechanisms, liquid pressure (liquid or gas) or mechanisms. Although interruptions are most common when a computer is engaged in control circuits, it is generally more convenient to operate all control systems on electricity.
A Comprehensive Study of the Functions and Operations of Control Systems
World Academy of Research in Science and Engineering, 2019
A control system is used to achieve the required feedback by controlling the output. Control systems are widely used in everyday devices such as traffic lights and cars or even processes. These systems help in further understanding and improving the process that is being utilized. It also represents the process in a much more straightforward manner that makes it easier to comprehend. This paper aims to broaden the insight on control systems and to discuss different control systems that are being used.
Introduction to control engineering
Throughout history mankind has tried to control the world in which he lives. From the earliest days he realized that his puny strength was no match for the creatures around him. He could only survive by using his wits and cunning. His major asset over all other life forms on earth was his superior intelligence. Stone Age man devised tools and weapons from flint, stone and bone and discovered that it was possible to train other animals to do his bidding -and so the earliest form of control system was conceived. Before long the horse and ox were deployed to undertake a variety of tasks, including transport. It took a long time before man learned to replace animals with machines.
Control system
A control system manages, commands, directs or regulates the behavior of other devices or systems. It can range from a home heating controller using a thermostat controlling a domestic boiler to large Industrial control systems which are used for controlling processes or machines.
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Copyright © 2006, New Age International (P) Ltd., Publishers
Published by New Age International (P) Ltd., Publishers
All rights reserved.
No part of this ebook may be reproduced in any form, by photostat, microfilm,
xerography, or any other means, or incorporated into any information retrieval
system, electronic or mechanical, without the written permission of the publisher.
All inquiries should be emailed to [email protected]
ISBN (13) : 978-81-224-2484-3
PUBLISHING FOR ONE WORLD
NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS
4835/24, Ansari Road, Daryaganj, New Delhi - 110002
Visit us at www.newagepublishers.com
I
Dedicated this book
to
enkateswara’
‘T
‘Too Lord Sr
Srii V
Venkateswara’
(vi)
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PREFACE
Control Systems Engineering is an exciting and challenging field and is a
multidisciplinary subject. This book is designed and organized around the concepts of control
systems engineering using MATLAB, as they have been developed in the frequency and time
domain for an introductory undergraduate or graduate course in control systems for engineering students of all disciplines.
Chapter 1 presents a brief introduction to control systems. The fundamental strategy of
controlling physical variables in systems is presented. Some of the terms commonly used to
describe the operation, analysis, and design of control systems are described.
An introduction to MATLAB basics is presented in Chapter 2. Chapter 2 also presents
MATLAB commands. MATLAB is considered as the software of choice. MATLAB can be used
interactively and has an inventory of routines, called as functions, which minimize the task of
programming even more. Further information on MATLAB can be obtained from: The
MathWorks, Inc., 3 Apple Hill Drive, Natick, MA 01760. In the computational aspects, MATLAB
has emerged as a very powerful tool for numerical computations involved in control systems
engineering. The idea of computer-aided design and analysis using MATLAB with the Symbolic
Math Tool box, and the Control System Tool box has been incorporated.
Chapter 3 consists of many solved problems that demonstrate the application of MATLAB
to the analysis and design of control systems. Presentations are limited to linear, time-invariant continuous time systems.
Chapters 2 and 3 include a great number of worked examples and unsolved exercise
problems to guide the student to understand the basic principles and concepts in control systems engineering.
I sincerely hope that the final outcome of this book helps the students in developing an
appreciation for the topic of analysis and design of control systems.
An extensive bibliography to guide the student to further sources of information on control systems engineering is provided at the end of the book. All the end-of chapter problems are
fully solved in the Solution Manual available only to Instructors.
Rao V. Dukkipati
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ACKNOWLEDGEMENTS
I am grateful to all those who have had a direct impact on this work. Many people working in the general areas of analysis and design of feedback control systems have influenced the
format of this book. I would also like to thank and recognize all the undergraduate students in
mechanical and electrical engineering program at Fairfield University, over the years with
whom I had the good fortune to teach and work, and who contributed in some ways and feedback to the development of the material of this book. In addition, I greatly owe my indebtedness
to all the authors of the articles listed in the bibliography of this book. Finally, I would very
much like to acknowledge the encouragement, patience, and support provided by my family
members: my wife, Sudha, my family members, Ravi, Madhavi, Anand, Ashwin, Raghav, and
Vishwa who have also shared in all the pain, frustration, and fun of producing a manuscript.
I would appreciate being informed of errors, or receiving other comments about the
book. Please write to the authors’ Fairfield University address or send e-mail to
[email protected].
Rao V. Dukkipati
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CONTENTS
Preface
(vii)
Acknowledgement
1. Introduction to Control Systems
...
(ix)
1
1.1 Introduction
1.2 Control Systems
...
...
1
1
1.2.1 Examples of Control Systems
1.3 Control System Configurations
...
...
2
3
1.4 Control System Terminology
1.5 Control System Classes
...
...
5
6
1.6 Feedback Systems
1.7 Analysis of Feedback
...
...
8
8
1.8 Control System Analysis and Design Objectives
1.9 Summary
...
...
9
10
...
...
10
12
...
26
...
...
26
27
2.1.2 Display Windows
2.1.3 Entering Commands
...
...
27
27
2.1.4 MATLAB Expo
2.1.5 Abort
...
...
27
27
2.1.6 The Semicolon
2.1.7 Typing %
...
...
27
27
2.1.8 The clc Command
2.1.9 Help
...
...
27
27
2.1.10 Statements and Variables
2.2 Arithmetic Operations
...
...
28
28
2.3 Display Formats
2.4 Elementary Math Built-in Functions
...
...
28
29
2.5 Variable Names
2.6 Predefined Variables
...
...
31
31
2.7 Commands for Managing Variables
2.8 General Commands
...
...
32
32
2.9 Arrays
...
34
References
Glossary of Terms
2. MATLAB Basics
2.1 Introduction
2.1.1 Starting and Quitting MATLAB
(xii)
2.9.1 Row Vector
...
34
2.9.2 Column Vector
2.9.3 Matrix
...
...
34
34
2.9.4 Addressing Arrays
2.9.5 Adding Elements to a Vector or a Matrix
...
...
35
35
2.9.6 Deleting Elements
2.9.7 Built-in Functions
...
...
35
35
...
...
37
37
2.10.2 Dot Product
2.10.3 Array Multiplication
...
...
37
37
2.10.4 Array Division
2.10.5 Identity Matrix
...
...
37
37
2.10.6 Inverse of a Matrix
2.10.7 Transpose
...
...
38
38
2.10.8 Determinant
2.10.9 Array Division
...
...
38
38
2.10.10 Left Division
2.10.11 Right Division
...
...
38
38
2.10.12 Eigenvalues and Eigenvectors
2.11 Element-by-Element Operations
...
...
38
39
2.11.1 Built-in Functions for Arrays
2.12 Random Number Generation
...
...
40
41
2.12.1 The Random Command
2.13 Polynomials
...
...
42
42
2.14 System of Linear Equations
2.14.1 Matrix Division
...
...
44
44
2.14.2 Matrix Inverse
2.15 Script Files
...
...
44
49
2.15.1 Creating and Saving a Script File
2.15.2 Running a Script File
...
...
49
50
2.15.3 Input to a Script File
2.15.4 Output Commands
...
...
50
50
2.16 Programming in MATLAB
2.16.1 Relational and Logical Operators
...
...
51
51
2.16.2 Order of Precedence
2.16.3 Built-in Logical Functions
...
...
52
52
2.16.4 Conditional Statements
2.16.5 Nested if Statements
...
...
53
54
2.10 Operations with Arrays
2.10.1 Addition and Subtraction of Matrices
(xiii)
2.16.6 else AND else if Clauses
...
54
2.16.7 MATLAB while Structures
2.17 Graphics
...
...
54
57
2.17.1 Basic 2-D Plots
2.17.2 Specialized 2-D plots
...
...
57
57
2.17.3 3-D Plots
2.17.4 Saving and Printing Graphs
...
...
58
65
2.18 Input/Output in MATLAB
2.18.1 The fopen Statement
...
...
65
65
2.19 Symbolic Mathematics
2.19.1 Symbolic Expressions
...
...
66
66
...
...
68
69
2.20 The Laplace Transforms
2.20.1 Finding Zeros and Poles of B(s)/A(s)
...
...
71
72
2.21 Control Systems
2.21.1 Transfer Functions
...
...
72
72
2.21.2 Model Conversion
2.22 The Laplace Transforms
...
...
72
75
2.23 Summary
Problems
...
...
111
113
...
125
3.1 Introduction
3.2 Transient Response Analysis
...
...
125
125
3.3 Response to Initial Condition
3.4 Second Order Systems
...
...
125
127
3.5 Root Locus Plots
3.6 Bode Diagrams
...
...
127
129
3.7 Nyquist Plots
3.7.1 Polar Plots
...
...
136
136
3.7.2 Nyquist Plot
3.8 Nichols Chart
...
...
137
138
3.8.1 db Magnitude-Phase Angle Plots
3.9 Gain Margin, Phase Margin, Phase Crossover Frequency,
...
138
...
...
139
139
...
140
2.19.2 Solution to Differential Equations
2.19.3 Calculus
3. MATLAB Tutorial
and Gain Crossover Frequency
3.10 Transformation of System Models
3.10.1 Transformation of System Model from Transfer Function
to State Space
(xiv)
3.10.2 Transformation of System Model from State Space to
Transfer Function
3.11 Bode Diagrams of Systems Models Defined in State-Space
...
...
140
140
3.12 Nyquist Plots of a System Defined in State Space
3.13 Transient Response Analysis in State-Space
...
...
141
141
3.13.1 Unit Step Response
3.13.2 Unit Ramp Response
...
...
141
142
3.13.3 Unit Ramp Response
3.13.4 Response to Arbitrary Input
...
...
142
143
...
...
143
143
Summary
Problems
...
...
241
241
Bibliography
...
251
3.14 Response to Initial Condition in State Space
Example Problems and Solutions
Chapter
1
INTRODUCTION
TO
CONTROL SYSTEMS
1.1 INTRODUCTION
Control systems in an interdisciplinary field covering many areas of engineering and
sciences. Control systems exist in many systems of engineering, sciences, and in human body.
Some type of control systems affects most aspects of our day-to-day activities. This chapter
presents a brief introduction and overview of control systems. Some of the terms commonly
used to describe the operation, analysis, and design of control systems are presented.
1.2 CONTROL SYSTEMS
Control means to regulate, direct, command, or govern. A system is a collection, set, or
arrangement of elements (subsystems). A control system is an interconnection of components
forming a system configuration that will provide a desired system response. Hence, a control
system is an arrangement of physical components connected or related in such a manner as to
command, regulate, direct, or govern itself or another system.
In order to identify, delineate, or define a control system, we introduce two terms: input
and output here. The input is the stimulus, excitation, or command applied to a control system,
and the output is the actual response resulting from a control system. The output may or may
not be equal to the specified response implied by the input. Inputs could be physical variables or
abstract ones such as reference, set point or desired values for the output of the control system.
Control systems can have more than one input or output. The input and the output represent
the desired response and the actual response respectively. A control system provides an output
or response for a given input or stimulus, as shown in Fig. 1.1.
Input: stimulus
Desired response
Output: response
Control system
Actual response
Fig. 1.1 Description of a control system
The output may not be equal to the specified response implied by the input. If the output
and input are given, it is possible to identify or define the nature of the system’s components.
Broadly speaking, there are three basic types of control systems:
(a) Man-made control systems
(b) Natural, including biological-control systems
(c) Control systems whose components are both man-made and natural.
1
2
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
An electric switch is a man-made control system controlling the electricity-flow. The
simple act of pointing at an object with a finger requires a biological control system consisting
chiefly of eyes, the arm, hand and finger and the brain of a person, where the input is precisedirection of the object with respect to some reference and the output is the actual pointed direction with respect to the same reference. The control system consisting of a person driving an
automobile has components, which are clearly both man-made and biological. The driver wants
to keep the automobile in the appropriate lane of the roadway. The driver accomplishes this by
constantly watching the direction of the automobile with respect to the direction of road. Fig.
1.2 is an alternate way of showing the basic entities in a general control system.
Objectives
Results
Control system
Fig. 1.2 Components of a control system
In the steering control of an automobile for example, the direction of two front wheels
can be regarded as the result or controlled output variable and the direction of the steering
wheel as the actuating signal or objective. The control-system in this case is composed of the
steering mechanism and the dynamics of the entire automobile. As another example, consider
the idle-speed control of an automobile engine, where it is necessary to maintain the engine idle
speed at a relatively low-value (for fuel economy) regardless of the applied engine loads (like
air-conditioning, power steering, etc.). Without the idle-speed control, any sudden engine-load
application would cause a drop in engine speed that might cause the engine to stall. In this
case, throttle angle and load-torque are the inputs (objectives) and the engine-speed is the
output. The engine is the controlled process of the system. A few more applications of controlsystems can be found in the print wheel control of an electronic typewriter, the thermostatically controlled heater or furnace which automatically regulates the temperature of a room or
enclosure, and the sun tracking control of solar collector dish.
Control system applications are found in robotics, space-vehicle systems, aircraft autopilots
and controls, ship and marine control systems, intercontinental missile guidance systems, automatic control systems for hydrofoils, surface-effect ships, and high-speed rail systems including the magnetic levitation systems.
1.2.1 Examples of Control Systems
Control systems find numerous and widespread applications from everyday to extraordinary in science, industry, and home. Here are a few examples:
(a) Residential heating and air-conditioning systems controlled by a thermostat
(b) The cruise (speed) control of an automobile
(c) Manual control:
(i) Opening or closing of a window for regulating air temperature or air quality
(ii) Activation of a light switch to regulate the illumination in a room
(iii) Human controlling the speed of an automobile by regulating the gas supply to
the engine
(d) Automatic traffic control (signal) system at roadway intersections
(e) Control system which automatically turns on a room lamp at dusk, and turns it off in
daylight
(f) Automatic hot water heater
3
INTRODUCTION TO CONTROL SYSTEM
(g) Environmental test-chamber temperature control system
(h) An automatic positioning system for a missile launcher
(i) An automatic speed control for a field-controlled dc motor
(j) The attitude control system of a typical space vehicle
(k) Automatic position-control system of a high speed automated train system
(l) Human heart using a pacemaker
(m) An elevator-position control system used in high-rise multilevel buildings.
1.3 CONTROL SYSTEM CONFIGURATIONS
There are two control system configurations: open-loop control system and closed-loop
control system.
(a) Block. A block is a set of elements that can be grouped together, with overall characteristics described by an input/output relationship as shown in Fig. 1.3. A block diagram is a
simplified pictorial representation of the cause-and-effect relationship between the input(s)
and output(s) of a physical system.
Physical components
Inputs
Outputs
within the block
Block
Fig. 1.3 Block diagram
The simplest form of the block diagram is the single block as shown in Fig. 1.3. The input
and output characteristics of entire groups of elements within the block can be described by an
appropriate mathematical expressions as shown in Fig. 1.4.
Mathematical
Inputs
Outputs
expression
Fig. 1.4 Block representation
(b) Transfer Function. The transfer function is a property of the system elements only,
and is not dependent on the excitation and initial conditions. The transfer function of a system
(or a block) is defined as the ratio of output to input as shown in Fig.1.5.
Input
Output
Transfer function
Fig. 1.5 Transfer function
4
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Transfer function =
Output
Input
Transfer functions are generally used to represent a mathematical model of each block in
the block diagram representation. All the signals are transfer functions on the block diagrams.
For instance, the time function reference input is r(t), and its transfer function is R(s) where t is
time and s is the Laplace transform variable or complex frequency. Transfer functions can be
used to represent closed-loop as well as open-loop systems.
(c) Open-loop Control System. Open-loop control systems represent the simplest form
of controlling devices. A general block diagram of open-loop system is shown in Fig. 1.6.
Fig. 1.6 General block diagram of open-loop control system
(d) Closed-loop (Feedback Control) System. Closed-loop control systems derive their
valuable accurate reproduction of the input from feedback comparison. The general architecture of a closed-loop control system is shown in Fig. 1.7. A system with one or more feedback
paths is called a closed-loop system.
Disturbance
input 2
D2(s)
Disturbance
input 1
D1(s)
+
Reference
Input
Input
transducer
–
R(s)
Summing
junction
+
Ea(s)
Gc(s)
+
+
Output
Controlled
Summing variable
Plant or
junction
C(s)
process
Gp(s)
Controller
Forward
path
+
H(s)
Feedback
path
Output
transducer or
sensor
Fig. 1.7 General block diagram of closed-loop control system
5
INTRODUCTION TO CONTROL SYSTEM
1.4 CONTROL SYSTEM TERMINOLOGY
The variables in Figs. 1.6 and 1.7 are defined as follows:
C(s) controlled output, transfer function of c(t)
D(s) disturbance input, transfer function of d(t)
Ea(s) actuating error, transfer function of ea(t)
Ga(s) transfer function of the actuator
Gc(s) transfer function of the controller
Gp(s) transfer function of the plant or process
H(s) transfer function of the sensor or output transducer = Gs(s)
R(s) reference input, transfer function of r(t).
A
R
R+B
+
R
+
R–B
+
R
R–B+A
+
+
–
–
B
B
B
(a) Two inputs
(b) Two inputs
(c) Three inputs
Fig. 1.8 Summing point
A
A
A
A
A
A
Takeoff point
A
A
(a)
Takeoff point
(b)
Fig. 1.9 Takeoff point
Actuating or Error Signal. The actuating or error signal is the reference input signal
plus or minus the primary feedback signal.
Controlled Output C(s). The controlled output C(s) is the output variable of the plant
under the control of the control system.
Controller. The elements of an open-loop control system can usually be divided into
two parts: controller and the controlled process. The controller drives a process or plant.
Disturbance or Noise Input. A disturbance or noise input is an undesired stimulus or
input signal affecting the value of the controlled output.
Feed Forward (Control) Elements. The feed forward (control) elements are the components of the forward path that generate the control signal applied to the plant or process. The
feed forward (control) elements include controller(s), compensator(s), or equalization elements,
and amplifiers.
6
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Feedback Elements. The feedback elements establish the fundamental relationship
between the controlled output C(s) and the primary feedback signal B(s). They include sensors
of the controlled output, compensators, and controller elements.
Feedback Path. The feedback path is the transmission path from the controlled output
back to the summing point.
Forward Path. The forward path is the transmission path from the summing point to
the controlled output.
Input Transducer. Input transducer converts the form of input to that used by the
controller.
Loop. A loop is a path that originates and terminates on the same node , and along
which no other node is encountered more than once.
Loop Gain. The loop gain is the path gain of a loop.
Negative Feedback. Negative feedback implies that the summing point is a subtractor.
Path. A path is any collection of a continuous succession of branches traversed in the
same direction.
Path Gain. The product of the branch gains encountered in traversing a path is called
the path gain.
Plant, Process or Controlled System Gp(s). The plant, process, or controlled system
is the system, subsystem, process, or object controlled by the feedback control system. For example, the plant can be a furnace system where the output variable is temperature.
Positive Feedback. Position feedback implies that the summing point is an adder.
Primary Feedback Signal. The primary feedback signal is a function of the controlled
output summed algebraically with the reference input to establish the actuating or error signal.
An open-loop system has no primary feedback signal.
Reference Input R(s). The reference input is an external signal applied to the control
system generally at the first summing input, so as to command a specified action of the process
or plant. It typically represents ideal or desired process or plant output response.
Summing Point. As shown in Fig. 1.8 the block is a small circle called a summing point
with the appropriate plus or minus sign associated with the arrows entering the circle. The
output is the algebraic sum of the inputs. There is no limit on the number of inputs entering a
summing point.
Takeoff Point. A takeoff point allows the same signal or variable as input to more than
one block or summing point, thus permitting the signal to proceed unaltered along several
different paths to several destinations as shown in Fig. 1.9.
Time Response. The time response of a system, subsystem, or element is the output as
a function of time, generally following the application of a prescribed input under specified
operating conditions.
Transducer. A transducer is a device that converts one energy form into another.
1.5 CONTROL SYSTEM CLASSES
Control systems are sometimes divided into two classes : (a) Servomechanisms and
(b) Regulators.
INTRODUCTION TO CONTROL SYSTEM
7
(a) Servomechanisms. Feedback control systems used to control position, velocity, and
acceleration are very common in industry and military applications. They are known as
servomechanisms. A servomechanism is a power-amplifying feedback control system in which
the controlled variable is a mechanical position or a time derivative of position such as velocity
or acceleration. An automatic aircraft landing system is an example of servomechanism. The
aircraft follows a ramp to the desired touchdown point. Another example is the control system
of an industrial robot in which the robot arm is forced to follow some desired path in space.
(b) Regulators. A regulator or regulating system is a feedback control system in which
the reference input or command is constant for long periods of time, generally for the entire
time interval during which the system is operational. Such an input is known as set point. The
objective of the idle-speed control system is known as a regulator system. Another example of a
regulator control system is the human biological system that maintains the body temperature
at approximately 98.6ºF in an environment that usually has a different temperature.
1.5.1 Supplementary Terminology
(a) Linear System. A linear system is a system where input/ output relationships may be
represented by a linear differential equation. The plant is linear if it can be accurately described using a set of linear differential equations. This attribute indicates
that system parameters do not vary as a function of signal level. For linear systems,
the equations that constitute the model are linear.
Similarly, the plant is a lumped-parameter (rather than distributed parameter) system if it can be described using ordinary (rather than partial) differential equations.
This condition is generally accomplished if the physical size of the system is very
small in comparison to the wavelength of the highest frequency of interest.
(b) Time-Variant System. A time-variant is a system if the parameters vary as a function
of time. Thus, a time-variant system is a system described by a differential equation
with variable coefficients. A linear time variant system is described by linear differential equations with variable coefficients. Its derivatives appear as linear combinations, but a coefficient or coefficients of terms may involve the independent variable.
A rocket-burning fuel system is an example of time variant system since the rocket
mass varies during the flight as the fuel is burned.
(c) Time-Invariant System. A time-invariant system is a system described by a differential equation with constant coefficients. Thus, the plant is time invariant if the parameters do not change as a function of time. A linear time invariant system is described by linear differential equations with constant coefficients. A single degree of
freedom spring mass viscous damper system is an example of a time-invariant system provided the characteristics of all the three components do not vary with time.
(d) Multivariable Feedback System. The block diagram representing a multivariable feedback system where the interrelationships of many controlled variables are considered is shown in Fig. 1.12.
8
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Fig. 1.12 Multivariable control system
1.6 FEEDBACK SYSTEMS
Feedback is the property of a closed-loop system, which allows the output to be compared
with the input to the system such that the appropriate control action may be formed as some
function of the input and output.
For more accurate and more adaptive control, a link or feedback must be provided from
output to the input of an open-loop control system. So the controlled signal should be fed back
and compared with the reference input, and an actuating signal proportional to the difference
of input and output must be sent through the system to correct the error. In general, feedback
is said to exist in a system when a closed sequence of cause-and-effect relations exists between
system variables. A closed-loop idle-speed control system is shown in Fig. 1.13. The reference
input Nr sets the desired idle-speed. The engine idle speed N should agree with the reference
value Nr and any difference such as the load-torque T is sensed by the speed-transducer and the
error detector. The controller will operate on the difference and provide a signal to adjust the
throttle angle to correct the error.
Error
Nr
+
–
T
N
Control
Engine
+
N
Speed
Fig. 1.13 Closed-loop idle-speed control system
1.7 ANALYSIS OF FEEDBACK
The most important features, the presence of feedback impacts to a system are the following:
(a) Increased accuracy: its ability to reproduce the input accurately.
(b) Reduced sensitivity of the ratio of output to input for variations in system characteristics and other parameters.
(c) Reduced effects of nonlinearties and distortion.
(d) Increased bandwidth (bandwidth of a system that ranges frequencies (input) over
which the system will respond satisfactorily).
9
INTRODUCTION TO CONTROL SYSTEM
(e) Tendency towards oscillation or instability.
(f) Reduced effects of external disturbances or noise.
A system is said to be unstable, if its output is out of control. Feedback control systems
may be classified in a number of ways, depending upon the purpose of classification. For instance, according to the method of analysis and design, control-systems are classified as linear
or non-linear, time-varying or time-variant systems. According to the types of signals used in
the system, they may be: continuous data and discrete-data system or modulated and
unmodulated systems.
Consider the simple feedback configuration shown in Fig. 1.14, where R is the input
signal, C is the output signal, E is error, and B is feedback signal.
The parameters G and H are constant-gains. By simple algebraic manipulations, it can
be shown that the input-output relation of the system is given by
M=
G
C
=
1 + GH
R
...(1.1)
The general effect of feedback is that it may increase or decrease the gain G. In practical
control-systems, G and H are functions of frequency, so the magnitude of (1 + GH) is greater
than 1 in one frequency range, but less than 1 in another. Thus feedback affects the gain G of a
nonfeedback system by a factor (1 + GH).
+
–
R
+
B
+
E
–
G
C
+
–
H
Fig. 1.14 Feedback system
If GH = – 1, the output of the system is infinite for any finite input, such a state is called
unstable system-state. Alternatively feedback stabilizes an unstable system and the sensitivity
of a gain of the overall system M to the variation in G is defined as:
M
SG
=
Percentage change in M
∂M/M
=
Percentage change in G
∂G/G
...(1.2)
where ∂ M denotes incremental change in M due to incremental change in G(∂G). One can write
sensitivity-function as:
M
SG
=
1
∂M/M
=
1 + GH
∂G/G
...(1.3)
By increasing GH, the magnitude of the sensitivity-function is made arbitrarily small.
1.8 CONTROL SYSTEM ANALYSIS AND DESIGN OBJECTIVES
Control systems engineering consists of analysis and design of control systems configurations. Control systems are dynamic, in that they respond to an input by first undergoing a
10
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
transient response before attaining a steady-state response which corresponds to the input.
There are three main objectives of control systems analysis and design. They are:
1. Producing the response to a transient disturbance which is acceptable
2. Minimizing the steady-state errors: Here, the concern is about the accuracy of the
steady-state response
3. Achieving stability: Control systems must be designed to be stable. Their natural response should decay to a zero values as time approaches infinity, or oscillate.
System analysis means the investigation, under specified condition, of the performance
of a system whose mathematical model is known. Analysis is investigation of the properties and
performance of an existing control system.
By synthesis we mean using an explicit procedure to find a system that will perform in a
specified way. System design refers to the process of finding a system that accomplishes a given
task. Design is the selection and arrangement of the control system components to perform a
prescribed task. The design of control systems is accomplished in two ways : design by analysis
in which the characteristics of an existing or standard system configuration are modified, and
design by synthesis, in which the form of the control system is obtained directly from its specifications.
1.9 SUMMARY
A basic control system has an input, a process, and an output. The basic objective of a
control system is of regulating the value of some physical variable or causing that variable to
change in a prescribed manner in time. Control systems are typically classified as open loop or
closed-loop. Open-loop control systems do not monitor or correct the output for disturbances
whereas closed-loop control systems do monitor the output and compare it with the input. In a
closed-loop control system if an error is detected, the system corrects the output and thereby
corrects the effects of disturbances. In closed-loop control systems, the system uses feedback,
which is the process of measuring a control variable and returning the output to influence the
value of the variable.
Block diagrams display the operational units of a control system. Each block in a component block diagram represent some major component of the control system, such as measurement, compensation, error detection, and the plant itself. It also depicts the major directions of
information and energy flow from one component to another in a control system.
A block can represent the component or process to be controlled. Each block of a control
system has a transfer function (represented by differential equations) and defines the block
output as a function of the input.
Control system design and analysis objectives include: producing the response to a transient disturbance follows a specified pattern (over-damped or under damped), minimizing the
steady-state errors, and achieving the stability.
REFERENCES
Anand, D.K., Introduction to Control Systems, 2nd ed., Pergamon Press, New York, NY, 1984.
Bateson, R.N., Introduction to Control System Technology, Prentice Hall, Upper Saddle River,
NJ, 2002.
INTRODUCTION TO CONTROL SYSTEM
11
Beards, C.F., Vibrations and Control System, Ellis Horwood, 1988.
Bode, H.W., Network Analysis and Feedback Design, Van Nostrand Reinhold, New York,
NY, 1945.
Bolton, W., Control Engineering, 2nd ed., Addison Wesley Longman Ltd., Reading,
MA, 1998.
Chesmond, C.J., Basic Control System Technology, Edward Arnold, 1990.
D’Azzo, J.J., and Houpis, C.H., Linear Control System Analysis and Design: Conventional and Modern, 4th ed., McGraw Hill, New York, NY, 1995.
Dorf, R.C., and Bishop, R.H., Modern Control Systems, 9th ed., Prentice Hall, Upper
Saddle River, NJ, 2001.
Dorsey, John., Continuous and Discrete Control Systems, McGraw Hill, New York,
NY, 2002.
Doyle, J.C., Francis, B.A., and Tannenbaum, A., Feedback Control Theory,
Macmillan, New York, NY, 1992.
Dukkipati, R.V., Control Systems, Narosa Publishing House, New Delhi, India, 2005.
Dukkipati, R.V., Engineering System Dynamics, Narosa Publishing House, New Delhi,
India, 2004.
Franklin, G.F., David Powell, J., and Abbas Emami-Naeini., Feedback Control
of Dynamic Systems, 3rd ed., Addison Wesley, Reading, MA, 1994.
Godwin, Graham E., Graebe, Stefan F., and Salgado, Maria E., Control System
Design, Prentice Hall, Upper Saddle River, NJ, 2001.
Grimble, Michael J., Industrial Control Systems Design, Wiley, New York, NY, 2001.
Gupta, S., Elements of Control Systems, Prentice Hall, Upper Saddle River, NJ, 2002.
Johnson, C., and Malki, H., Control Systems Technology, Prentice Hall, Upper Saddle
River, NJ, 2002.
Kailath, T., Linear Systems, Prentice Hall, Upper Saddle River, NJ, 1980.
Kuo, B.C., Automatic Control Systems, 6th ed., Prentice Hall, Englewood Cliffs, NJ,
1991.
Leff, P.E.E., Introduction to Feedback Control Systems, McGraw Hill, New York,
NY, 1979.
Levin, W.S., Control System Fundamentals, CRC Press, Boca Raton, FL, 2000.
Lewis, P., and Yang, C., Basic Control Systems Engineering, Prentice Hall, Upper Saddle River, NJ, 1997.
Nise, Norman, S., Control Systems Engineering, 3rd ed., Wiley, New York, NY, 2000.
Ogata, K., Modern Control Engineering, 3rd ed., Prentice Hall, Englewood Cliffs,
NJ, 1997.
Ogata, K., System Dynamics, 3rd ed., Prentice Hall, Upper Saddle River, NJ, 1998.
Palm III, W.J., Control Systems Engineering, Wiley, New York, NY, 1986.
Paraskevopoulos, P.N., Modern Control Engineering, Marcel Dekker , Inc., New York,
NY, 2003.
Phillips, C.L., and Harbour, R.D., Feedback Control Systems, 4th ed., Prentice Hall,
Upper Saddle River, NJ, 2000.
12
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Raven, F.H., Automatic Control Engineering, 4th ed., McGraw Hill, New York, NY,
1987.
Richards, R.J., Solving Problems in Control, Longman Scientific & Technical, Wiley,
New York, NY, 1993.
Rohrs, C.E., Melsa, J.L., and Schultz, D.G., Linear Control Systems, McGraw Hill,
New York, NJ, 1993.
Rowell, G., and Wormley, D., System Dynamics, Prentice Hall, Upper Saddle River,
NJ, 1999.
Shearer, J.L., Kulakowski, B.T., and Gardner, J.F., Dynamic Modeling and Control
of Engineering Systems, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 1997.
Shinners, S. M., Modern Control System Theory and Design, 2nd ed., Wiley Inter Science,
New York, NY, 1998.
Sinha, N.K., Control Systems, Holt Rinehart and Winston, New York, NY, 1986.
Thompson, S., Control Systems: Engineering and Design, Longman, 1989.
Vu, H.V., Control Systems, McGraw Hill Primis Custom Publishing, New York, NY, 2002.
Vukic, Z., Kuljaca, L., Donlagic, D., and Tesnjak, S., Nonlinear Control Systems,
Marcel Dekker , Inc., New York, NY, 2003.
GLOSSARY OF TERMS
Terminology used frequently in the field of control systems is compiled here from various
sources.
Action of the Controller: Another term used to describe the controller operations is
the action of a controller.
Actuating or Error Signal: The actuating or error signal is the reference input signal
plus or minus the primary feedback signal.
Actuator: The device that causes the process to provide the output. The device that
provides the motive power to the process.
Angle of Departure: The angle at which a locus leaves a complex pole in the s-plane.
Asymptote: The path the root locus follows as the parameter becomes very large and
approaches infinity. The number of asymptotes is equal to the number of poles minus the number
of zeros.
Automatic Control System: A control system that is self-regulating, without any human intervention.
Automatic: Self-action without any human intervention.
Bandwidth: The frequency at which the frequency response has declined 3 dB from its
low-frequency value.
Block Diagram: A block diagram is a simplified pictorial representation of the causeand-effect relationship between the input(s) and output(s) of a physical system.
Block: A block is a set of elements that can be grouped together with overall characteristics described by an input/output relationship.
Block-Diagram Representation: In a block-diagram representation, each component
(or subsystem) is represented as a rectangular block containing one input and one output in a
block diagram.
INTRODUCTION TO CONTROL SYSTEM
13
Bode Diagram (Plot): A sinusoidal frequency response plot, where the magnitude response is plotted separately from the phase response. The magnitude plot is dB versus log ω,
and the phase plot is phase versus log w. IN control systems, the Bode plot is usually made for
the open-loop transfer function. Bode plots can also be drawn as straight-line approximations.
Bode Plot: The logarithm of the magnitude of the transfer function is plotted versus the
logarithm of ω, the frequency. The phase, φ, of the transfer function is separately plotted versus
the logarithm of the frequency.
Branches: Individual loci are referred to as branches of the root locus. Also, lines that
represent subsystems in a signal-flow graph.
Break Frequency: A frequency where the Bode magnitude plot changes slope.
Breakaway Point: A point on the real axis of the s-plane where the root locus leaves
the real axis and enters the complex plane.
Break-in Point: A point on the real axis of the s-plane where the root locus enters the
real axis from the complex plane.
Cascade Control: Two feedback controllers arranged in such a fashion that the output
of one feedback controller becomes an input to the second controller.
Characteristic Equation: The resulting expression obtained when the denominator of
the transfer function of the system is set equal to zero is known as the characteristic equation.
Closed-Loop Control System: A control system in which the control (regulating action) is influenced by the output.
Closed-Loop Feedback Control System: A system that uses a measurement of the
output and compares it with the desired output.
Closed-Loop Frequency Response: The frequency response of the closed-loop transfer function T (jω).
Closed-Loop System: A system with a measurement of the output signal and a comparison with the desired output to generate an error signal that is applied to the actuator.
Closed-Loop Transfer Function: For a generic feedback system with G(s) in the forward path and H(s) in the feedback path, the closed-loop transfer function, T(s), is G(s)/[1 ±
G(s)H(s)], where the + is for negative feedback, and the – is for positive feedback.
Compensation: The term compensation is usually used to indicate the process of increasing accuracy and speeding up the response.
Compensator: An additional component or circuit that is inserted into the system to
compensate for a performance deficiency.
Configuration Space: Generally speaking, generalized coordinates, qi (i = 1, 2, …, n)
define an n-dimensional Cartesian space that is referred to as the configuration space.
Constant M Circles: The locus of constant, closed-loop magnitude frequency response
for unity feedback systems. It allows the closed-loop magnitude frequency response to be determined from the open-loop magnitude frequency response.
Constant N Circles: The locus of constant, closed-loop phase frequency response for
unity feedback systems. It allows the closed-loop phase frequency response to be determined
from the open-loop phase frequency response.
Continuous-Time Control Systems: Continuous-time control systems or continuousdata control systems or analog control systems contain or process only continuous-time (or analog) signals and components.
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Contour Map: A contour or trajectory in one plane is mapped into another plane by a
relation F(s).
Control System: A control system is an interconnection of components forming a system configuration that will provide a desired system response.
Control: Control means to regulate, direct, command, or govern.
Controllability: A property of a system by which an input can be found that takes every
state variable from a desired initial state to a desired final state in finite time.
Controllable System: A system is controllable on the interval [t0, t f ] if there exists a
continuous input u(t) such that any initial state x(t0) can be driven to any arbitrary trial state
x(tf) in a finite time interval tf – t0 > 0.
Controlled Output C(s): The controlled output C(s) is the output variable of the plant
under the control of the system.
Controlled Variable: The output of a plant or process that the system is controlling for
the purpose of desired transient response, stability and steady-state error characteristics.
Controller Action: The method by which the automatic controller produces the control
signal is known as the control action.
Controller: The subsystem that generates the input to the plant or process.
Corner Frequency: See break frequency.
Critical Damping: The case where damping is on the boundary between underdamped
and overdamped.
Critically Damped Response: The step response of a second-order system with a given
natural frequency that is characterized by no overshoot and a rise time that is faster than any
possible overdamped response with the same natural frequency.
Damped Frequency of Oscillation: The sinusoidal frequency of oscillation of an
underdamped response.
Damped Natural Frequency: The frequency at which the system oscillates before settling down.
Damped Oscillation: An oscillation in which the amplitude decreases with time.
Damping Ratio: The ratio of the exponential decay frequency to the natural frequency.
dc Motor: An electric actuator that uses an input voltage as a control variable.
Decade: Frequencies that are separated by a factor of 10.
Decibel (dB): The decibel is defined as 10 log PG, where PG is the power gain of a signal.
Equivalently, the decibel is also 20 log VG, where VG is the voltage gain of a signal.
Decoupled System: A state-space representation in which each state equation is a function of only one state variable. Hence, each differential equation can be solved independently of
the other equations.
Delay Time: The delay time td is the time needed for the response to reach half the final
value the very first time. The delay time is interpreted as a time domain specification, is often,
defined as the time required for the response to a unit step input to reach 50% of its final value.
Delayed Step Function: A function of time (F(t – a)) that has a zero magnitude before
t = a and a constant amplitude after that.
Design of a Control System: The arrangement or the plan of the system structure and
the selection of suitable components and parameters.
INTRODUCTION TO CONTROL SYSTEM
15
Design Specifications: A set of prescribed performance criteria.
Design: The term design is used to encompass the entire process of basic system modification so as to meet the requirements of stability, accuracy, and transient response.
Digital Control System: A control system using digital signals and a digital computer
to control a process.
Digital Signal: A signal which is defined at only discrete (distinct) instants of the independent variable t is called a discrete-time or a discrete-data or a sampled-data or a digital
signal.
Digital-to-Analog Converter: A device that converts digital signals to analog signals.
Direct System: See Open-loop system.
Discrete-Time Approximation: An approximation used to obtain the time response of
a system based on the division of the time into small increments, ∆t.
Discrete-Time Control Systems: Discrete-time control system, or discrete-data control
system or sampled-data control system has discrete-time signals or components at one or more
points in the system.
Disturbance or Noise Input: A disturbance or noise input is an undesired stimulus or
input signal affecting the value of the controlled output.
Disturbance Signal: An unwanted input signal that affects the system’s output signal.
Disturbance: An unwanted signal that corrupts the input or output of a plant or process.
Dominant Poles: The poles that predominantly generate the transient response.
Dominant Roots: The roots of the characteristic equation that cause the dominant transient response of the system.
Eigenvalues: Any value, λi, that satisfies Axi = λixi for xi ≠ 0. Hence, any value, λi, that
makes xi an eigenvector under the transformation A.
Eigenvector: Any vector that is collinear with a new basis vector after a similarity
transformation to a diagonal system.
Electric Circuit Analog: An electrical network whose variables and parameters are
analogous to another physical system. The electric circuit analog can be used to solve for variables of the other physical system.
Electrical Impedance: The ratio of the Laplace transform of the voltage to the Laplace
transform of the current.
Element (Component): Smallest part of a system that can be treated as a whole (entity).
Engineering Design: The process of designing a technical system.
Equilibrium: The steady-state solution characterized by a constant position or oscillation.
Error Signal: The difference between the desired output, R(s), and the actual output,
Y(s). Therefore E(s) = R(s) – Y(s).
Error: The difference between the input and output of a system.
Feed Forward (Control) Element: The feed forward (control) elements are the components of the forward path that generate the control signal applied to the plant or process. The
feed forward (control) elements include controller(s), compensator(s), or equalization elements,
and amplifiers.
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Feedback Compensator: A subsystem placed in a feedback path for the purpose of
improving the performance of a closed-loop system.
Feedback Elements: The feedback elements establish the fundamental relationship
between the controlled output C(s) and the primary feedback signal B(s). They include sensors
of the controlled output, compensators, and controller elements.
Feedback Path: The feedback path is the transmission path from the controlled output
back to the summing point.
Feedback Signal: A measure of the output of the system used for feedback to control
the system.
Feedback: Feedback is the property of a closed-loop control system which allows the
output to be compared with the input to the system such that the appropriate control action
may be formed as some function of the input and output.
Flyball Governor: A mechanical device for controlling the speed of a steam engine.
Forced Response: For linear systems, that part of the total response function due to
the input. It is typically of the same form as the input and its derivatives.
Forward Path: A forward path is a path that connects a source node to a sink node, in
which no node is encountered more than once.
Forward-Path Gain: The product of gains found by traversing a path that follows the
direction of signal flow from the input node to the output node of a signal-flow graph.
Fourier Transform: The transformation of a function of time, f(t), into the frequency
domain.
Frequency Domain Techniques: A method of analyzing and designing linear control
systems by using transfer functions and the Laplace transform as well as frequency response
techniques.
Frequency Response Techniques: A method of analyzing and designing control systems by using the sinusoidal frequency response characteristics of a system.
Frequency Response: The steady-state response of a system to a sinusoidal input signal.
Gain Crossover Frequency: The frequency at which the open loop gain drops to 0 dB
(gain of 1).
Gain Margin: The gain margin is the factor by which the gain factor K can be multiplied
before the closed-loop system becomes unstable. It is defined as the magnitude of the reciprocal
of the open-loop transfer function evaluated at the frequency ω2 at which the phase angle is –180º.
Gain: The gain of a branch is the transmission function of that branch when the
transmission function is a multiplicative operator.
Heat Capacitance: The capacity of an object to store heat.
Ideal Derivative Compensator: See proportional-plus-derivative controller.
Ideal Integral Compensator: See proportional-plus-integral controller.
Input Transducer: Input transducer converts the form of input to that used by the
controller.
Input: The input is the stimulus, excitation, or command applied to a control system,
generally from an external source, so as to produce a specified response from the control system.
Instability: The characteristic of a system defined by a natural response that grows
without bounds as time approaches infinity.
INTRODUCTION TO CONTROL SYSTEM
17
Integration Network: A network that acts, in part, like an integrator.
Kirchhoff’s Law: The sum of voltages around a closed loop equals zero. Also, the sum of
currents at a node equals zero.
Lag Compensator: A transfer function, characterized by a pole on the negative real
axis close to the origin and a zero close and to the left of the pole, that is used for the purpose of
improving the steady-state error of a closed-loop system.
Lag Network: See Phase-lag network.
Lag-Lead Compensator: A transfer function, characterized by a pole-zero configuration that is the combination of a lag and a lead compensator, that is used for the purpose of
improving both the transient response and the steady-state error of a closed-loop system.
Laplace Transform: A transformation of a function f(t) from the time domain into the
complex frequency domain yielding F(s).
Laplace Transformation: A transformation that transforms linear differential equations into algebraic expressions. The transformation is especially useful for modeling, analyzing,
and designing control systems as well as solving linear differential equations.
Lead Compensator: A transfer function, characterized by a zero on the negative real
axis and a pole to the left of the zero, that is used for the purpose of improving the transient
response of a closed-loop system.
Lead Network: See Phase-lead network.
Lead-Lag Network: A network with the characteristics of both a lead network and a
lag network.
Linear Approximation: An approximate model that results in a linear relationship
between the output and the input of the device.
Linear Combination: A linear combination of n variables, xi, for i = 1 to n, given by the
following sum, S.
Linear System: A linear system is a system where input/output relationships may be
represented by a linear differential equation.
Linearization: The process of approximating a nonlinear differential equation with a
linear differential equation valid for small excursions about equilibrium.
Locus: Locus is defined as a set of all points satisfying a set of conditions.
Logarithmic Magnitude: The logarithmic of the magnitude of the transfer function,
20 log10 |G|.
Logarithmic Plot: See Bode plot.
Loop Gain: For a signal-flow graph, the product of branch gains found by traversing a
path that starts at a node and ends at the same node without passing through any other node
more than once, and following the direction of the signal flow.
Loop: A loop is a closed path (with all arrowheads in the same direction) in which no
node is encountered more than once. Hence, a source node cannot be a part of a loop, since each
node in the loop must have at least one branch into the node and at least one branch out.
Major-Loop Compensation: A method of feedback compensation that adds a compensating zero to the open-loop transfer function for the purpose of improving the transient response of the closed-loop system.
Manual Control System: A control system regulated through human intervention.
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Marginal Stability: The characteristic of a system defined by a natural response that
neither decays nor grows, but remains constant or oscillates as time approaches infinity as long
as the input is not of the same form as the system’s natural response.
Marginally Stable System: A closed-loop control system in which roots of the characteristic equation lie on the imaginary axis; for all practical purposes, an unstable system.
Mason’s Loop Rule: A rule that enables the user to obtain a transfer function by tracing paths and loops within a system.
Mason’s Gain Formula: Mason’s gain formula is an alternative method of reducing
complex block diagrams into a single block diagram with its associated transfer function for
linear systems by inspection.
Mason’s Rule: A formula from which the transfer function of a system consisting of the
interconnection of multiple subsystems can be found.
Mathematical Model: An equation or set of equations that define the relationship between the input and output (variables).
Maximum Overshoot Mp: The maximum overshoot is the vertical distance between
the maximum peak of the response curve and the horizontal line from unity (final value).
Maximum Value of the Frequency Response: A pair of complex poles will result in a
maximum value for the frequency response occurring at the resonant frequency.
Minimum Phase: All the zeros of a transfer function lie in the left-hand side of the
s-plane.
Minor-Loop Compensation: A method of feedback compensation that changes the poles
of a forward-path transfer function for the purpose of improving the transient response of the
closed-loop system.
Multiple-Input, Multiple-Output (MIMO) System: A multiple-input, multiple-output (MIMO) system is a system where several parameters may be entered as input and output
is represented by multiple variables.
Multivariable Control System: A system with more than one input variable or more
than one output variable.
Multivariable Feedback System: The multivariable feedback system where the interrelationships of many controlled variables are considered.
Natural Frequency: The frequency of oscillation of a system if all the damping is
removed.
Natural Response: That part of the total response function due to the system and the
way the system acquires or dissipates energy.
Negative Feedback: The case where a feedback signal is subtracted from a previous
signal in the forward path.
Neutral Zone: The region of error over which the controller does not change its output;
also known as dead band or error band.
Nichols Chart: Nichols chart is basically a transformation of the M- and N-circles on
the polar plot into noncircular M and N contours on a db magnitude versus phase angle plot in
rectangular coordinates.
Nodes: In a signal-flow graph, the internal signals in the diagram, such as the common
input to several blocks or the output of summing junction, are called nodes.
Nonminimum Phase: Transfer functions with zeros in the right-hand s-plane.
INTRODUCTION TO CONTROL SYSTEM
19
Nonminimum-Phase System: A system whose transfer function has zeros in the right
half-plane. The step response is characterized by an initial reversal in direction.
Nontouching Loops: Loops that do not have any nodes in common.
Nontouching: Two loops are nontouching if these loops have no nodes in common. A
loop and a path are nontouching if they have no nodes in common.
Nontouching-Loop Gain: The product of loop gains from nontouching loops taken two,
three, and four, and so on at a time.
Number of Separate Loci: Equal to the number of poles of the transfer function, assuming that the number of poles is greater than or equal to the number of zeros of the transfer
function.
Noise Input: A disturbance or noise input is an undesired stimulus or input signal
affecting the value of the controlled output.
Nyquist Criterion: If a contour, A, that encircles the entire right half-plane is mapped
through G(s)H(s), then the number of closed-loop poles, Z, in the right half-plane equals the
number of open-loop poles, P, that are in the right half-plane minus the number of
counterclockwise revolutions, N, around – 1, of the mapping; that is, Z = P – N. The mapping is
called the Nyquist diagram of G(s)H(s).
Nyquist Diagram (Plot): A polar frequency response plot made for the open-loop transfer
function.
Nyquist Path: The locus of the points in the s-plane mapped into G(s)-plane in Nyquist
plots is called Nyquist path.
Nyquist Stability Criterion: The Nyquist stability criterion establishes the number of
poles and zeros of 1 + GH(s) that lie in the right-half plane directly from the Nyquist stability
plot of GH(s).
Observability: A property of a system by which an initial state vector, x(t0), can be
found from u(f) and y(t) measured over a finite interval of time from t0. Simply stated, observability is the property by which the state variables can be estimated from a knowledge of the
input, u(i), and output, y(t).
Observable System: A system is observable on the interval [t0, t f ] if any initial state
x(t0) is uniquely determined by observing the output y(t) on the interval [t0, t f ].
Observer: A system configuration from which inaccessible states can be estimated.
Octave: Frequencies that are separated by a factor of two.
Open-Loop Control System: A system that utilizes a device to control the process
without using feedback. Thus the output has no effect upon the signal to the process.
Open-Loop System: A system without feedback that directly generates the output in
response to an input signal.
Open-Loop Transfer Function: For a generic feedback system with G(s) in the forward path and H(s) in the feedback path, the open-loop transfer function is the product of the
forward-path transfer function and the feedback transfer function, or, G(s)H(s).
Output Equation: For linear systems, the equation that expresses the output variables
of a system as linear combinations of the state variables.
Output: The output is the actual response resulting from a control system.
Overdamped Response: A step response of a second-order system that is characterized by no overshoot.
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Overshoot: The amount by which the system output response proceeds beyond the desired response.
Parameter Design: A method of selecting one or two parameters using the root locus
method.
Partial-Fraction Expansion: A mathematical equation where a fraction with n factors
in its denominator is represented as the sum of simpler fractions.
Path Gain: The path gain is the product of the transfer functions of all branches that
form the path.
Path: A path is a sequence of connected blocks, the route passing from one variable to
another in the direction of signal flow of the blocks without including any variable more than
once.
Peak Time: The peak time tp is the time required for the response to reach the first peak
of the overshoot.
Peak Value: The maximum value of the output, reached after application of the unit
step input after time tp.
Percent Overshoot, %OS: The amount that the underdamped step response overshoots
the steady state, or final, value at the peak time, expressed as a percentage of the steady-state
value.
Performance Index: A quantitative measure of the performance of a system.
Phase Crossover Frequency: The frequency at which the open loop phase angle drops
to – 180°.
Phase Margin: The amount of additional open-loop phase shift required at unity gain to
make the closed-loop system unstable.
Phase Variables: State variables such that each subsequent state variable is the derivative of the previous state variable.
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.
Phase-Margin Frequency: The frequency at which the magnitude frequency response
plot equals zero dB. It is the frequency at which the phase margin is measured.
Phase-Margin: Phase margin of a stable system is the amount of additional phase log
required to bring the system to point of instability.
PI Controller: Controller with a proportional term and an integral term (ProportionalIntegral).
Pickoff Point: A block diagram symbol that shows the distribution of one signal to
multiple subsystems.
PID 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.
Plant, Process or Controlled System Gp(s): The plant, process, or controlled system
is the system, subsystem, process, or object controlled by the feedback control system. For example, the plant can be a furnace system where the output variable is temperature.
INTRODUCTION TO CONTROL SYSTEM
21
Plant: See Process.
Polar Plot: A plot of the real part of G(jω) versus the imaginary part of G(jω).
Pole of a Transfer Function: The root (solution) of the (characteristic) equation obtained by setting the denominator polynomial of the transfer function equal to zero; the value of
s that makes (the value of) the transfer function approach infinity (hence the term pole (rising
to infinity)); complex poles always appear as complex conjugate pairs.
Poles: (1) The values of the Laplace transform variable, s, that cause the transfer function to become infinite, and (2) any roots of factors of the characteristic equation in the denominator that are common to the numerator of the transfer function.
Pole-Zero Map: The s-plane including the locations of the finite poles and zeros of F(s)
is called the pole-zero map of F(s).
Positive Feedback: Positive feedback implies that the summing point is an adder.
Primary Feedback Signal: The primary feedback signal is a function of the controlled
output summed algebraically with the reference input to establish the actuating or error signal.
An open-loop system has no primary feedback signal.
Process Controller: See PID controller.
Process: The device, plant, or system under control.
Productivity: The ratio of physical output to physical input of an industrial process.
Proportional Band: The maximum percent error that will cause a change in controller
output from minimum (0%) to maximum (100%).
Proportional-Plus-Derivative (PD) Controller: A controller that feeds forward to
the plant a proportion of the actuating signal plus its derivative for the purpose of improving
the transient response of a closed-loop system.
Proportional-Plus-Integral (PI) Controller: A controller that feeds forward to the
plant a proportion of the actuating signal plus its integral for the purpose of improving the
steady-state error of a closed-loop system.
Proportional-Plus-Integral-Plus-Derivative (PID) Controller: A controller that
feeds forward to the plant a proportion of the actuating signal plus its integral plus its derivative for the purpose of improving the transient response and steady-state error of a closed-loop
system.
Pulse Function: The difference between a step function and a delayed step function.
Ramp Function: A function whose amplitude increases linearly with time.
Reference Input R(s): The reference input is an external signal applied to the control
system generally at the first summing point, so as to command a specific action of the processor
plant. It typically represents ideal or desired process or plant output response.
Relative Stability: The property that is measured by the relative real part of each root
or pair of roots of the characteristic equation.
Residue: The constants in the numerators of the terms in a partial-fraction expansion.
Resonant Frequency: The resonant frequency of a system is defined as the radian
frequency at which the magnitude value of C(jω)/R(jω) occurs.
Rise Time: The rise time tr is customarily defined as the time required for the response
to a unit step input to rise from 10 to 90% of its final value. For underdamped second-order
system, the 0% to 100% rise time is normally used. For overdamped systems, the 10% to 90%
rise time is common.
22
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Risk: Uncertainties embodied in the unintended consequences of a design.
Robot: Programmable computers integrated with a manipulator. A reprogrammable,
multifunctional manipulator used for a variety of tasks.
Robust Control System: A system that exhibits the desired performance in the presence of significant plant uncertainty.
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 infinity.
Root Locus Segments on the Real Axis: The root locus lying in a section of the real
axis to the left of an odd number of poles and zeros.
Root Sensitivity: The sensitivity of the roots as a parameter changes from its normal
value. The root sensitivity is the incremental change in the root divided by the proportional
change of the parameter.
Root: The term root refers to the roots of the characteristic equation, which are the poles
of the closed-loop transfer function.
Root-Locus Analysis: The root-locus method is an analytical method for displaying the
location of the poles of the closed-loop transfer function G/(1 + GH) as a function of the gain
factor K of the open-loop transfer function GH. The method is called the root-locus analysis.
Root-Locus: Root-locus defines a graph of the poles of the closed-loop transfer function
as the system parameter, such as the gain is varied.
Routh-Hurwitz Stability Criterion: The Routh-Hurwitz stability criterion states that
the dynamic system is stable if both of the following conditions are satisfied: (1) all the coefficients of the characteristic equation are positive, and (2) all the elements of the first column of
the Routh-Hurwitz table are positive.
Self-Loop: A self-loop is a feedback loop consisting of a single branch.
Sensitivity: The sensitivity of a system is defined as the ratio of the percentage change
in the system-transfer function to the percentage-change of the process transfer function. In
practice, the system sensitivity is expressed as the ratio of the percentage-variation in some
specific quantity like gain to the percentage change in one of the system parameters.
Settling Time: The time required for the system output to settle within a certain percentage of the input amplitude.
Signal Flow Graph: A signal flow graph is a pictorial representation of the simultaneous equations describing a system. The signal flow graph displays the transmission of signals
through the system just as in the block diagram.
Similarity Transformation: A transformation from one state-space representation to
another state-space representation. Although the state variables are different, each representation is a valid description of the same system and the relationship between the input and
output.
Single-Input, Single-Output (SISO) System: A single-input, single-output (SISO)
system is a system where only one parameter enters as input and only one-parameter results
as the output.
Sink Node: A sink node is a node for which signals flow only toward the node. Also
known as output node.
Sinusoidal Function: A function of time, which is periodically changing.
INTRODUCTION TO CONTROL SYSTEM
23
Source Node: A source node is a node for which signals flow only away from the node.
Hence, for the branches connected to a source node, the arrowheads are all directed away from
the node. Also known as input node.
Specifications: Statements that explicitly state what the device or product is to be and
to do. A set of prescribed performance criteria.
Stability: That characteristic of a system defined by a natural response that decays to
zero as time approaches infinity.
Stabilization: The term stabilization is used to indicate the process of achieving the
requirements of stability alone.
Stable Closed-Loop System: A system in which the open-loop gain is less than 0 db at
a frequency at which the phase angle has reached –180°.
Stable System: A dynamic system with a bounded system response to a bounded input.
State Differential Equation: The differential equation for the state vector: x = Ax + Bu.
State Equations: A set of n simultaneous, first-order differential equations with n variables, where the n variables to be solved are the state variables.
State of a System: A set of numbers such that the knowledge of these numbers and the
input function will, with the equations describing the dynamics, provide the future state of the
system.
State Space: The n-dimensional space whose axes are the state variables.
State Variable Equations: When a system’s equations of motion are rewritten as a
system of first-order differential equations, each of these differential equations consists of the
time derivative of the one of the state variables on the left-hand side and an algebraic function
of the state variables as well as system outputs, on the right-hand side. These differential
equations are referred to as state-variable equations.
State Variable Feedback: Occurs when the control signal, u, for the process is a direct
function of all the state variables.
State Variables: State variables are the variables, which define the smallest set of variables, which determine the state of a system.
State Vector: State vector is a vector, which completely describes a system’s dynamics
in terms of its n-state variables.
State: The property (condition) of a system.
State-Space Representation: A mathematical model for a system that consists of
simultaneous, first-order differential equations and an output equation.
State-Transition Matrix: The matrix that performs a transformation on x(0), taking x
from the initial state, x(0), to the state x(f) at any time, t ≥ 0.
Static Error Constants: The collection of position constant, velocity constant, and acceleration constant.
Steady-State Error: The difference between the input and output of a system after the
natural response has decayed to zero.
Steady-State Response: The system response after the transients have died and output has settled (time response after transient response).
Step Function: A function of time, which has a zero value before t = 0 and has a constant value for all time t ≥ 0.
Subsystem: A system that is a portion of a larger system.
24
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Summing Junction: A block diagram symbol that shows the algebraic summation of
two or more signals.
Summing Point: The summing point also known as a summing joint is the block used to
represent the addition/subtraction of signals. It is represented as a small circle connected to
arrows representing signal lines.
Synthesis: The process by which new physical configurations are created. The combining of separate elements or devices to form a coherent whole.
System Type: The number of pure integrations in the forward path of a unity feedback
system.
System Variables: Any variable that responds to an input or initial conditions in a
system.
System: A system is a collection, set, or arrangement of elements (subsystems).
Takeoff Point: A takeoff point allows the same signal or variable as input to more than
one block or summing point, thus permitting the signal to proceed unaltered along several
different paths to several destinations. It is represented as a dot (solid circle) with arrows pointing away from it.
The Addition Rule: The value of the variable designated by a node is equal to the sum
of all the signals entering the node.
The Design Specifications: The design specifications for control systems generally
include several time-response indices for a specified input as well as a desired steady-state
accuracy.
The Multiplication Rule: A single cascaded (series) connection of (n – 1) branches with
transmission functions G21, G32, G43, …, Gn(n – 1) can be replaced by a single branch with a new
transmission function equal to the product of the original ones.
The Steady-State Response: The steady-state response is that which exists a long
time following any input signal initiation.
The Transient-Response: The transient-response is the response that disappears with
time.
The Transmission Rule: The value of the variable designated by a node is transmitted
on every branch leaving that node.
Time Delay: A pure time delay, T, so that events occurring at time t at one point in the
system occur at another point in the system at a later time, t + T.
Time Domain: The mathematical domain that incorporates the time response and the
description of a system in terms of time t.
Time Response: The time response of a system, subsystem, or element is the output as
a function of time, generally, following application of a prescribed input under specified operating conditions.
Time-Domain Representation: See state-space representation.
Time-Invariant System: A system described by a differential equation with constant
coefficients.
Time-Variant System: A system described by a differential equation with variable coefficients.
Time-Varying System: A system for which one or more parameters may vary with
time.
Total Response: The response of a system from the time of application of an input to
the point when time approaches infinity.
INTRODUCTION TO CONTROL SYSTEM
25
Trade-off: The result of making a judgment about how much compromise must be made
between conflicting criteria.
Transducer: A device that converts a signal from one form to another, for example,
from a mechanical displacement to an electrical voltage.
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(jω).
Transfer Function: The transfer function of a system (or a block) is defined as the ratio
of output to input.
Transient Response: That parts of the response curve due to the system and the way
the system acquires or dissipates energy. In stable systems, it is the part of the response plot
prior to the steady-state response.
Undamped Response: The step response of a second-order system that is characterized by a pure oscillation.
Underdamped Response: The step response of a second-order system that is characterized by overshoot.
Unit Step Function: A function of time that has zero magnitude before time t = 0 and
unit magnitude after that.
Unstable System: A closed-loop control system in which one or more roots of the characteristic equation lie in the RHP (Right-Hand side of the s-Plane).
Zero of a Transfer Function: The root (solution) of the equation obtained by setting
the numerator polynomial of the transfer function equal to 0; the value of s that makes (the
value of) the transfer function equal to zero (hence the term zero).
Zeros: (1) Those values of the Laplace transform variable, s, that cause the transfer
function to become zero, and (2) any roots of factors of the numerator that are common to the
characteristic equation in the denominator of the transfer function.
Zero-State Response: That part of the response that depends only upon the input and
not the initial state vector.
Chapter
2
MATLAB BASICS
2.1 INTRODUCTION
This Chapter is a brief introduction to MATLAB (an abbreviation of MATrix LABoratory)
basics, registered trademark of computer software, version 4.0 or later developed by the Math
Works Inc. The software is widely used in many of science and engineering fields. MATLAB is
an interactive program for numerical computation and data visualization. MATLAB is supported on Unix, Macintosh, and Windows environments. For more information on MATLAB,
contact The MathWorks.Com. A Windows version of MATLAB is assumed here. The syntax is
very similar for the DOS version.
MATLAB integrates mathematical computing, visualization, and a powerful language to
provide a flexible environment for technical computing. The open architecture makes it easy to
use MATLAB and its companion products to explore data, create algorithms, and create custom
tools that provide early insights and competitive advantages.
Known for its highly optimized matrix and vector calculations, MATLAB offers an intuitive language for expressing problems and their solutions both mathematically and visually.
Typical uses include:
• Numeric computation and algorithm development
• Symbolic computation (with the built-in Symbolic Math functions)
• Modeling, simulation, and prototyping
• Data analysis and signal processing
• Engineering graphics and scientific visualization
In this chapter, we will introduce the MATLAB environment. We will learn how to create, edit, save, run, and debug m-files (ASCII files with series of MATLAB statements). We will
see how to create arrays (matrices and vectors), and explore the built-in MATLAB linear algebra functions for matrix and vector multiplication, dot and cross products, transpose, determinants, and inverses, and for the solution of linear equations. MATLAB is based on the language
C, but is generally much easier to use. We will also see how to program logic constructs and
loops in MATLAB, how to use subprograms and functions, how to use comments (%) for explaining the programs and tabs for easy readability, and how to print and plot graphics both
two and three dimensional. MATLAB’s functions for symbolic mathematics are presented. Use
of these functions to perform symbolic operations, to develop closed form expressions for solutions to algebraic equations, ordinary differential equations, and system of equations was presented. Symbolic mathematics can also be used to determine analytical expressions for the
derivative and integral of an expression.
26
MATLAB BASICS
27
2.1.1 Starting and Quitting MATLAB
To start MATLAB click on the MATLAB icon or type in MATLAB, followed by pressing
the enter or return key at the system prompt. The screen will produce the MATLAB prompt >>
(or EDU >>), which indicates that MATLAB is waiting for a command to be entered.
In order to quit MATLAB, type quit or exit after the prompt, followed by pressing the
enter or return key.
2.1.2 Display Windows
MATLAB has three display windows. They are
1. A Command Window which is used to enter commands and data to display plots and
graphs.
2. A Graphics Window which is used to display plots and graphs
3. An Edit Window which is used to create and modify M-files. M-files are files that
contain a program or script of MATLAB commands.
2.1.3 Entering Commands
Every command has to be followed by a carriage return <cr> (enter key) in order that the
command can be executed. MATLAB commands are case sensitive and lower case letters are
used throughout.
To execute an M-file (such as Project_1.m), simply enter the name of the file without its
extension (as in Project_1).
2.1.4 MATLAB Expo
In order to see some of the MATLAB capabilities, enter the demo command. This will
initiate the MATLAB EXPO. MATLAB Expo is a graphical demonstration environment that
shows some of the different types of operations which can be conducted with MATLAB.
2.1.5 Abort
In order to abort a command in MATLAB, hold down the control key and press c to
generate a local abort with MATLAB.
2.1.6 The Semicolon (;)
If a semicolon (;) is typed at the end of a command the output of the command is not
displayed.
2.1.7 Typing %
When percent symbol (%) is typed in the beginning of a line, the line is designated as a
comment. When the enter key is pressed the line is not executed.
2.1.8 The clc Command
Typing clc command and pressing enter cleans the command window. Once the clc command is executed a clear window is displayed.
2.1.9 Help
MATLAB has a host of built-in functions. For a complete list, refer to MATLAB user’s
guide or refer to the on line Help. To obtain help on a particular topic in the list, e.g., inverse,
type help inv.
28
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
2.1.10 Statements and Variables
Statements have the form
>> variable = expression
The equals (“=”) sign implies the assignment of the expression to the variable. For instance, to enter a 2 × 2 matrix with a variable name A, we write
>> A == [1 2 ; 3 4] <ret>
The statement is executed after the carriage return (or enter) key is pressed to display
A=
1
3
2
4
2.2 ARITHMETIC OPERATIONS
The symbols for arithmetic operations with scalars are summarized below in Table 2.1.
Table 2.1
Arithmetic operation
Symbol
Example
Addition
Subtraction
+
–
6+3=9
6–3=3
Multiplication
Right division
*
/
6 * 3 = 18
6/3=2
Left division
Exponentiation
\
^
6\3=3/6=1/2
6 ^ 3 (63 = 216)
2.3 DISPLAY FORMATS
MATLAB has several different screen output formats for displaying numbers. These
formats can be found by typing the help command: help format in the Command Window. A few
of these formats are shown in Table 2.2 for 2π.
Table 2.2 Display formats
Command
Description
Example
Format short
Fixed-point with
4 decimal digits
>> 351/7
ans = 50.1429
Format long
Fixed-point with
14 decimal digits
>> 351/7
ans = 50.14285714285715
Format short e
Scientific notation with
4 decimal digits
>> 351/7
ans = 5.0143e+001
Format long e
Scientific notation with 15
decimal digits
>> 351/7
ans = 5.014285714285715e001
29
MATLAB BASICS
Command
Description
Example
Format short g
Best of 5 digit fixed
or floating point
>> 351/7
ans = 50.143
Format long g
Best of 15 digit fixed
or floating point
>> 351/7
ans = 50.1428571428571
Format bank
Two decimal digits
>> 351/7
ans = 50.14
Format compact
Eliminates empty lines to allow more lines with information
displayed on the screen
Format loose
Adds empty lines (opposite of compact)
2.4 ELEMENTARY MATH BUILT-IN FUNCTIONS
MATLAB contains a number of functions for performing computations which require the
use of logarithms, elementary math functions, and trigonometric math functions. List of these
commonly used elementary MATLAB mathematical built-in functions are given in Tables 2.3
to 2.8.
Table 2.3 Common Math Functions
Function
abs(x)
sqrt(x)
round(x)
fix(x)
floor(x)
ceil(x)
sign(x)
rem(x,y)
exp(x)
log(x)
log10(x)
Description
Computes the absolute value of x.
Computes the square root of x.
Rounds x to the nearest integer.
Rounds (or truncates) x to the nearest integer toward 0.
Rounds x to the nearest integer toward – ∞.
Rounds x to the nearest integer toward ∞.
Returns a value of – 1 if x is less than 0, a value of 0 if x equals 0, and a
value of 1 otherwise.
Returns the remainder of x/y. for example, rem(25, 4) is 1, and rem(100,
21) is 16. This function is also called a modulus function.
Computes ex, where e is the base for natural logarithms, or approximately 2.718282.
Computes ln x, the natural logarithm of x to the base e.
Computes log10 x, the common logarithm of x to the base 10.
Table 2.4 Exponential functions
Function
Description
exp(x)
Exponential (ex)
log(x)
log10(x)
Natural logarithm
Base 10 logarithm
sqrt(x)
Square root
30
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Table 2.5 Trigonometric and hyperbolic functions
Function
Description
sin(x)
Computes the sine of x, where x is in radians.
cos(x)
tan(x)
Computes the cosine of x, where x is in radians.
Computes the tangent of x, where x is in radians.
asin(x)
Computes the arcsine or inverse sine of x, where x must be between – 1 and
1. The function returns an angle in radians between – π/2 and π/2.
acos(x)
Computes the arccosine or inverse cosine of x, where x must be between – 1
and 1. The function returns an angle in radians between 0 and π.
Computes the arctangent or inverse tangent of x. The function returns an
angle in radians between – π/2 and π/2.
atan(x)
atan2(y, x)
Computes the arctangent or inverse tangent of the value y/x. The function
returns an angle in radians that will be between – π and π, depending on the
signs of x and y.
sinh(x)
Computes the hyperbolic sine of x, which is equal to
cosh(x)
Computes the hyperbolic cosine of x, which is equal to
tanh(x)
e x − e− x
.
2
e x + e− x
.
2
sinh x
.
Computes the hyperbolic tangent of x, which is equal to
cosh x
asinh(x)
Computes the inverse hyperbolic sine of x, which is equal to ln(x +
x2 + 1 ).
acosh(x)
Computes the inverse hyperbolic cosine of x, which is equal to ln(x +
x2 − 1 ).
atanh(x)
Computes the inverse hyperbolic tangent of x, which is equal to ln
for |x| ≤ 1.
Table 2.6 Round-off functions
Function
Description
Example
round(x)
Round to the nearest integer
>> round(20/6) ans = 3
fix(x)
Round towards zero
>> fix(13/6) ans = 2
ceil(x)
Round towards infinity
>> ceil(13/5) ans = 3
floor(x)
Round towards minus infinity
>> floor(– 10/4) ans = – 3
rem(x, y)
Returns the remainder after
x is divided by y
>> rem(14,3) ans = 2
sign(x, y)
Signum function. Returns 1
if x > 0, – 1 if x < 0, and 0 if
x – 0.
>> sign(7) ans = 1
1+ x
1−x
31
MATLAB BASICS
Table 2.7 Complex number functions
Function
Description
conj(x)
Computes the complex conjugate of the complex number x. Thus, if x
is equal to a + i b, then conj(x) will be equal to a – i b.
real(x)
imag(x)
Computes the real portion of the complex number x.
Computes the imaginary portion of the complex number x.
abs(x)
angle(x)
Computes the absolute value of magnitude of the complex number x.
Computes the angle using the value of atan2(imag(x), real(x)); thus,
the angle value is between – π and π.
Table 2.8 Arithmetic operations with complex numbers
Operation
Result
c1 + c2
(a1 + a2) + i(b1 + b2)
c1 + c2
(a1 – a2) + i(b1 – b2)
c1 • c2
(a1a2 – b1b2) + i(a1b2 – a2b1)
c1
c2
a1 a2 + b1 b2
a2 b1 − b2 a1
+ i
2
2
2
2
a2 + b2
a2 + b2
|c1|
c1*
a12 + b12 (magnitude or absolute value of c1)
a1 – ib1 (conjugate of c1)
(Assume that c1 = a1 + ib1 and c2 = a2 + ib2)
2.5 VARIABLE NAMES
A variable is a name made of a letter or a combination of several letters and digits.
Variable names can be up to 63 (in MATLAB 7) characters long (31 characters on MATLAB
6.0). MATLAB is case sensitive. For instance, XX, Xx, xX, and xx are the names of four different
variables. It should be noted here that not to use the names of a built-in functions for a variable.
For instance, avoid using: sin, cos, exp, sqrt, ..., etc. Once a function name is used to define a
variable, the function cannot be used.
2.6 PREDEFINED VARIABLES
MATLAB includes a number of predefined variables. Some of the predefined variables
that are available to use in MATLAB programs are summarized in Table 2.9.
32
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Table 2.9 Predefined variables
Predefined variable
in MATLAB
ans
pi
eps
Description
Represents a value computed by an expression but not stored in
variable name.
Represents the number π.
inf
Represents the floating-point precision for the computer being
used. This is the smallest difference between two numbers.
Represents infinity which for instance occurs as a result of a division by zero. A warning message will be displayed or the value
will be printed as ∞.
i
Defined as
j
Same as i.
NaN
Stands for Not a Number. Typically occurs as a result of an expression being undefined, as in the case of division of zero by
zero.
Represents the current time in a six-element row vector containing year, month, day, hour, minute, and seconds.
clock
date
− 1 , which is: 0 + 1.0000i.
Represents the current date in a character string format.
2.7 COMMANDS FOR MANAGING VARIABLES
Table 2.10 lists commands that can be used to eliminate variables or to obtain information about variables that have been created. The procedure is to enter the command in the
Command Window and the Enter key is to be pressed.
Table 2.10 Commands for managing variables
Command
Description
clear
clear x, y, z
Removes all variables from the memory.
Clears/removes only variables x, y, and z from the memory.
who
whos
Lists the variables currently in the workspace.
Displays a list of the variables currently in the memory and their
size together with information about their bytes and class.
2.8 GENERAL COMMANDS
In Tables 2.11 to 2.15 the useful general commands on on-line help, workspace information, directory information, and general information are given.
33
MATLAB BASICS
Table 2.11 On-line help
Function
Description
help
helpwin
helpdesk
help topic
lookfor string
demo
Lists topics on which help is available.
Opens the interactive help window.
Opens the web browser based help facility.
Provides help on topic.
Lists help topics containing string.
Runs the demo program.
Table 2.12 Workspace information
Function
Description
who
whos
what
clear
clear x y z
clear all
mlock fun
munlock fun
clc
home
clf
Lists variables currently in the workspace.
Lists variables currently in the workspace with their size.
Lists m-, mat-, and mex-files on the disk.
Clears the workspace, all variables are removed.
Clears only variables x, y, and z.
Clears all variables and functions from workspace.
Locks function fun so that clear cannot remove it.
Unlocks function fun so that clear can remove it.
Clears command window, command history is lost.
Same as clc.
Clears figure window.
Table 2.13 Directory information
Function
pwd
cd
dir
ls
path
editpath
copyfile
mkdir
Function
computer
clock
date
more
ver
bench
Description
Shows the current working directory.
Changes the current working directory.
Lists contents of the current directory.
Lists contents of the current directory, same as dir.
Gets or sets MATLAB search path.
Modifies MATLAB search path.
Copies a file.
Creates a directory.
Table 2.14 General information
Description
Tells you the computer type you are using.
Gives you wall clock time and date as a vector.
Tells you the date as a string.
Controls the paged output according to the screen size.
Gives the license and the version information about MATLAB installed
on your computer.
Benchmarks your computer on running MATLAB compared to other computers.
34
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Table 2.15 Termination
Function
Description
c (Control-c)
Local abort, kills the current command execution.
quit
exit
Quits MATLAB.
Same as quit.
2.9 ARRAYS
An array is a list of numbers arranged in rows and/or columns. A one-dimensional array
is a row or a column of numbers and a two-dimensional array has a set of numbers arranged in
rows and columns. An array operation is performed element-by-element.
2.9.1 Row Vector
A vector is a row or column of elements.
In a row vector the elements are entered with a space or a comma between the elements
inside the square brackets. For example,
x = [7 – 1 2 – 5 8]
2.9.2 Column Vector
In a column vector the elements are entered with a semicolon between the elements
inside the square brackets. For example,
x = [7 ; – 1 ; 2 ; – 5 ; 8]
2.9.3 Matrix
A matrix is a two-dimensional array which has numbers in rows and columns. A matrix
is entered row-wise with consecutive elements of a row separated by a space or a comma, and
the rows separated by semicolons or carriage returns. The entire matrix is enclosed within
square brackets. The elements of the matrix may be real numbers or complex numbers. For
example to enter the matrix,
1
A=
0
3
−2
− 4
8
The MATLAB input command is
A = [1 3 – 4 ; 0 – 2 8]
Similarly for complex number elements of a matrix B
− 5x
B=
3i
ln 2 x + 7 sin 3 y
5 − 13i
The MATLAB input command is
B = [– 5 * x log(2 * x) + 7 * sin (3 * y) ; 3i 5 – 13i]
35
MATLAB BASICS
2.9.4 Addressing Arrays
A colon can be used in MATLAB to address a range of elements in a vector or a matrix.
2.9.4.1 Colon for a vector
Va(:) – refers to all the elements of the vector Va (either a row or a column vector).
Va(m : n) – refers to elements m through n of the vector Va.
For instance
>>
V = [2 5 – 1 11 8 4 7 – 3 11]
>>
u = V(2 : 8)
u = 5 – 1 11 8 4 7 – 3 11
2.9.4.2 Colon for a matrix
Table 2.16 gives the use of a colon in addressing arrays in a matrix.
Table 2.16 Colon use for a matrix
Command
Description
A(:, n)
A(n, :)
Refers to the elements in all the rows of a column n of the matrix A.
Refers to the elements in all the columns of row n of the matrix A.
A(:, m : n)
Refers to the elements in all the rows between columns m and n of the
matrix A.
Refers to the elements in all the columns between rows m and n of the
matrix A.
A(m : n, :)
A(m : n, p : q)
Refers to the elements in rows m through n and columns p through q of
the matrix A.
2.9.5 Adding Elements to a Vector or a Matrix
A variable that exists as a vector, or a matrix, can be changed by adding elements to it.
Addition of elements is done by assigning values of the additional elements, or by appending
existing variables. Rows and/or columns can be added to an existing matrix by assigning values
to the new rows or columns.
2.9.6 Deleting Elements
An element, or a range of elements, of an existing variable can be deleted by reassigning
blanks to these elements. This is done simply by the use of square brackets with nothing typed
in between them.
2.9.7 Built-in Functions
Some of the built-in functions available in MATLAB for managing and handling arrays
as listed in Table 2.17.
36
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Table 2.17 Built-in functions for handling arrays
Function
Description
Example
length (A)
Returns the number of elements
in the vector A
>> A = [5 9 2 4] ;
>> length(A)
ans = 4
size (A)
Returns a row vector [m, n], where
m and n are the size m × n of the
array A.
>>A = [2 3 0 8 11 ; 6 17 5 7 1]
A=
2 3 0 8 11
6 17 5 7 1
>> size(A)
ans = 2 5
reshape (A, m, n)
Rearrange a matrix A that has r
rows and s columns to have m
rows and n columns. r times s must
be equal to m times n.
>> A = [3 1 4 ; 9 0 7]
A=
314
907
>> B = reshape(A, 3, 2)
B=
3 0
9 4
1 7
diag (v)
When v is a vector, creates a square
matrix with the elements of v in the
diagonal
>> v = [3 2 1];
>> A = diag(v)
A=
3 0 0
0 2 0
0 0 1
diag (A)
When A is a matrix, creates a vector
from the diagonal elements of A.
>> A = [1 8 3 ; 4 2 6 ; 7 8 3]
A=
1 8 3
4 2 6
7 8 3
>> vec = diag(A)
vec =
1
2
3
37
MATLAB BASICS
2.10
OPERATIONS WITH ARRAYS
We consider here matrices that have more than one row and more than one column.
2.10.1 Addition and Subtraction of Matrices
The addition (the sum) or the subtraction (the difference) of the two arrays is obtained by
adding or subtracting their corresponding elements. These operations are performed with arrays
of identical size (same number of rows and columns).
For example if A and B are two arrays (2 × 3 matrices).
a11
A=
a21
a12
a22
a13
b11
and B =
a23
b21
b12
b22
b13
b23
Then, the matrix addition (A + B) is obtained by adding A and B is
a11 + b11
a21 + b21
a12 + b12
a22 + b22
a13 + b13
a23 + b23
2.10.2 Dot Product
The dot product is a scalar computed from two vectors of the same size. The scalar is the
sum of the products of the values in corresponding positions in the vectors.
For n elements in the vectors A and B:
n
dot product = A • B =
∑ ai bi
i =1
dot(A, B) Computes the dot product of A and B. If A and B are matrices, the dot product
is a row vector containing the dot products for the corresponding columns of A and B.
2.10.3 Array Multiplication
The value in position ci, j of the product C of two matrices, A and B, is the dot product of
row i of the first matrix and column of the second matrix:
n
ci, j =
∑ ai, k bk, j .
k=1
2.10.4 Array Division
The division operation can be explained by means of the identity matrix and the inverse
matrix operation.
2.10.5 Identity Matrix
An identity matrix is a square matrix in which all the diagonal elements are 1’s, and the
remaining elements are 0’s. If a matrix A is square, then it can be multiplied by the identity
matrix, I, from the left or from the right:
AI = IA = A
38
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
2.10.6 Inverse of a Matrix
The matrix B is the inverse of the matrix A if when the two matrices are multiplied the
product is the identity matrix. Both matrices A and B must be square and the order of multiplication can be AB or BA.
AB = BA = I
2.10.7 Transpose
The transpose of a matrix is a new matrix in which the rows of the original matrix are the
columns of the new matrix. The transpose of a given matrix A is denoted by AT. In MATLAB,
the transpose of the matrix A is denoted by A′.
2.10.8 Determinant
A determinant is a scalar computed from the entries in a square matrix. For a 2 × 2
matrix A, the determinant is
|A | = a11 a22 – a21 a12
MATLAB will compute the determinant of a matrix using the det function:
det(A) computes the determinant of a square matrix A.
2.10.9 Array Division
MATLAB has two types of array division, which are the left division and the right division.
2.10.10 Left Division
The left division is used to solve the matrix equation Ax = B where x and B are column
vectors. Multiplying both sides of this equation by the inverse of A, A–1, we have
A–1Ax = A–1 B
or
Hence
Ix = x = A–1 B
x = A–1 B
In MATLAB, the above equation is written by using the left division character:
x=A\B
2.10.11 Right Division
The right division is used to solve the matrix equation xA = B where x and B are row
vectors. Multiplying both sides of this equation by the inverse of A, A–1, we have
or
x • AA–1 = B • A–1
x = B • A–1
In MATLAB, this equation is written by using the right division character:
x=B\A
2.10.12 Eigenvalues and Eigenvectors
Consider the following equation,
AX = λX
...(2.1)
Where A is an n × n square matrix, X is a column vector with n rows and λ is a scalar.
The values of λ for which X are nonzero are called the eigenvalues of the matrix A, and
the corresponding values of X are called the eigenvectors of the matrix A.
MATLAB BASICS
39
Eq. (2.1) can also be used to find the following equation
(A – λI)X = 0
...(2.2)
where I is an n × n identity matrix. Eq. (2.2) corresponding to a set of homogeneous equations
and has nontrivial solutions only if the determinant is equal to zero, or
|A – λI| = 0
...(2.3)
Eq. (2.3) is known as the characteristic equation of the matrix A. The solution to Eq. (2.3)
gives the eigenvalues of the matrix A.
MATLAB determines both the eigenvalues and eigenvectors for a matrix A.
eig(A) Computes a column vector containing the eigenvalues of A.
[Q, d] = eig(A) Computes a square matrix Q containing the eigenvectors of A as columns and a square matrix d containing the eigenvalues (λ) of A on the diagonal. The values of
Q and d are such that Q * Q is the identity matrix and A * X equals λ times X.
Triangular factorization or lower-upper factorization: Triangular or lower-upper
factorization expresses a square matrix as the product of two triangular matrices – a lower
triangular matrix and an upper triangular matrix. The lu function in MATLAB computes the
LU factorization:
[L, U] = lu(A) Computes a permuted lower triangular factor in L and an upper triangular factor in U such that the product of L and U is equal to A.
QR factorization: The QR factorization method factors a matrix A into the product of
an orthonormal matrix and an upper-triangular matrix. The qr function is used to perform the
QR factorization in MATLAB:
[Q, R] = qr(A) Computes the values of Q and R such that A = QR.Q will be an orthonormal
matrix, and R will be an upper triangular matrix.
For a matrix A of size m × n, the size of Q is m × m, and the size of R is m × n.
Singular Value Decomposition (SVD): Singular value decomposition decomposes a
matrix A (size m × n) into a product of three matrix factors.
A = USV
where U and V are orthogonal matrices and S is a diagonal matrix. The size of U is m × m, the
size of V is n × n, and the size of S is m × n. The values on the diagonal matrix S are called
singular values. The number of nonzero singular values is equal to the rank of the matrix.
The SVD factorization can be obtained using the svd function:
[U, S, V] = svd(A) Computes the factorization of A into the product of three matrices,
USV, where U and V are orthogonal matrices and S is a diagonal matrix.
svd(A) Returns the diagonal elements of S, which are the singular values of A.
2.11
ELEMENT-BY-ELEMENT OPERATIONS
Element-by-element operations can only be done with arrays of the same size. Elementby-element multiplication, division, and exponentiation of two vectors or matrices is entered in
MATLAB by typing a period in front of the arithmetic operator. Table 2.18 lists these operations.
40
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Table 2.18 Element-by-element operations
Arithmetic operators
Matrix operators
Array operators
+
Addition
+
Addition
–
*
Subtraction
Multiplication
–
•*
Subtraction
Array multiplication
^
/
Exponentiation
Left division
•^
•/
Array exponentiation
Array left division
\
Right division
•\
Array right division
2.11.1 Built-in Functions for Arrays
Table 2.19 lists some of the many built-in functions available in MATLAB for analyzing
arrays.
Table 2.19 MATLAB built-in array functions
Function
Description
Example
mean (A)
If A is a vector, returns the mean
value of the elements
>> A = [3 7 2 16];
>> mean (A)
ans = 14
C = max (A)
If A is a vector, C is the largest
element in A. If A is a matrix, C is a
row vector containing the largest
element of each column of A.
>> A = [3 7 2 16 9 5 18 13 0 4];
>> C = max (A)
C = 18
[d, n] = max (A)
If A is a vector, d is the largest
element in A, n is the position of
the element (the first if several have
the max value).
>> [d, n] = max (A)
d = 18
min (A)
The same as max(A), but for the
smallest element.
>> A = [3 7 2 16];
>> min (A)
ans = 2
[d, n] = min (A)
The same as [d, n] = max(A), but
n=7
for the smallest element.
sum (A)
If A is a vector, returns the sum of
the elements of the vector.
>> A = [3 7 2 16];
>> sum (A)
ans = 28
41
MATLAB BASICS
Function
sort (A)
Description
If A is a vector, arranges the elements
of the vector in ascending order.
Example
>> A = [3 7 2 16];
>> sort (A)
ans = 2 3 7 16
median (A)
If A is a vector, returns the median
value of the elements of the vector.
>> A = [3 7 2 16];
>> median (A)
ans = 5
std (A)
det (A)
If A is a vector, returns the standard
deviation of the elements of the
>> A = [3 7 2 16];
>> std (A)
vector.
ans = 6.3770
Returns the determinant of a square
matrix A.
>> A = [1 2 3 4];
>> det (A)
ans = – 2
dot (a, b)
cross (a, b)
Calculates the scalar (dot) product of
two vectors a and b. The vector can
>> a = [5 6 7];
>> b = [4 3 2];
each be row or column vectors.
>> dot (a, b)
ans = 52
Calculates the cross product of two
>> a = [5 6 7];
vectors a and b, (a × b). The two
vectors must have 3 elements
>> b = [4 3 2];
>> cross (a, b)
ans = – 9 18 – 9
inv (A)
Returns the inverse of a square
matrix A.
>> a = [1 2 3; 4 6 8; – 1 2 3];
>> inv (A)
ans =
– 0.5000
0.0000 – 0.5000
– 5.0000 1.5000 1.0000
3.5000 – 1.0000 – 0.5000
2.12
RANDOM NUMBERS GENERATION
There are many physical processes and engineering applications that require the use of
random numbers in the development of a solution.
MATLAB has two commands rand and rand n that can be used to assign random numbers to variables.
The rand command: The rand command generates uniformly distributed over the interval [0, 1]. A seed value is used to initiate a random sequence of values. The seed value is
initially set to zero. However, it can be changed with the seed function.
The command can be used to assign these numbers to a scalar, a vector, or a matrix, as
shown in Table 2.20.
42
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Table 2.20 The rand command
Command
Description
Example
rand
Generates a single random number
between 0 and 1.
>> rand
ans =
0.9501
rand (1, n)
Generates an n elements row vector of
random numbers between 0 and 1.
>> a = rand(1, 3)
a=
0.4565 0.0185 0.8214
rand (n)
Generates an n × n matrix with
random numbers between 0 and 1.
>> b = rand(3)
b=
0.7382 0.9355 0.8936
0.1763 0.9165 0.0579
0.4057 0.4103 0.3529
rand (m, n)
Generates an m × n matrix with
random numbers between 0 and 1.
>> c = rand(2, 3)
c=
0.2028 0.6038 0.1988
0.1987 0.2722 0.0153
randperm (n)
Generates a row vector with n
elements that are random permutation
of integers 1 through n.
>> randperm(7)
ans =
5 2 4 7 1 6 3
2.12.1 The Random Command
MATLAB will generate Gaussian values with a mean of zero and a variance of 1.0 if a
normal distribution is specified. The MATLAB functions for generating Gaussian values are as
follows:
randn(n) Generates an n × n matrix containing Gaussian (or normal) random numbers
with a mean of 0 and a variance of 1.
Randn(m, n) Generates an m × n matrix containing Gaussian (or normal) random numbers with a mean of 0 and a variance of 1.
2.13 POLYNOMIALS
A polynomial is a function of a single variable that can be expressed in the following
form:
f(x) = a0xn + a1xn–1 + a2xn–2 + … + an–1x1 + an
where the variable is x and the coefficients of the polynomial are represented by the values a0,
a1, … and so on. The degree of a polynomial is equal to the largest value used as an exponent.
A vector represents a polynomial in MATLAB. When entering the data in MATLAB,
simply enter each coefficient of the polynomial into the vector in descending order. For example, consider the polynomial
43
MATLAB BASICS
5s5 + 7s4 + 2s2 – 6s + 10
To enter this into MATLAB , we enter this as a vector as
>> x = [5 7 0 2 – 6 10]
x=
5 7 0 2 – 6 10
It is necessary to enter the coefficients of all the terms.
MATLAB contains functions that perform polynomial multiplication and division, which
are listed below:
conv(a, b)
Computes a coefficient vector that contains the coefficients of the
product of polynomials represented by the coefficients in a and b.
The vectors a and b do not have to be the same size.
[q, r] = deconv(n, d) Returns two vectors. The first vector contains the coefficients of
the quotient and the second vector contains the coefficients of the
remainder polynomial.
The MATLAB function for determining the roots of a polynomial is the roots function:
root(a) Determines the roots of the polynomial represented by the coefficient vector a.
The roots function returns a column vector containing the roots of the polynomial; the
number of roots is equal to the degree of the polynomial. When the roots of a polynomial are
known, the coefficients of the polynomial are determined when all the linear terms are multiplied, we can use the poly function:
poly(r) Determines the coefficients of the polynomial whose roots are contained in the
vector r.
The output of the function is a row vector containing the polynomial coefficients.
The value of a polynomial can be computed using the polyval function, polyval (a, x). It
evaluates a polynomial with coefficients a for the values in x. The result is a matrix the same
size ad x. For instance, to find the value of the above polynomial at s = 2,
>> x = polyval([5 7 0 2 – 6 10], 2)
x=
278
To find the roots of the above polynomial, we enter the command roots (a) which determines the roots of the polynomial represented by the coefficient vector a.
>>roots([5 7 0 2 – 6 10])
ans =
– 1.8652
– 0.4641 + 1.0832i
– 0.4641 – 1.0832i
0.6967 + 0.5355i
0.6967 – 0.5355i
44
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
% or
>> x = [5 7 0 2 – 6 10]
x=
5 7 0 2 – 6 10
>> r = roots(x)
r=
– 1.8652
– 0.4641 + 1.0832i
– 0.4641 – 1.0832i
0.6967 + 0.5355i
0.6967 – 0.5355i
To multiply two polynomials together, we enter the command conv.
The polynomials are: x = 2x + 5 and y = x2 + 3x + 7
>>x = [2 5];
>>y = [1 3 7];
>>z = conv(x, y)
z=
2 11 29 35
To divide two polynomials, we use the command deconv.
z = [2 11 29 35]; x = [2 5]
>> [g, t] = deconv (z, x)
g=1 3 7
t= 0 0
2.14
0
0
SYSTEM OF LINEAR EQUATIONS
A system of equations is nonsingular if the matrix A containing the coefficients of the
equations is nonsingular. A system of nonsingular simultaneous linear equations (AX = B) can
be solved using two methods:
(a) Matrix Division Method.
(b) Matrix Inversion Method.
2.14.1 Matrix Division
The solution to the matrix equation AX = B is obtained using matrix division, or X = A/B.
The vector X then contains the values of x.
2.14.2 Matrix Inverse
For the solution of the matrix equation AX = B, we premultiply both sides of the equation
by A–1.
A–1AX = A–1B
or
IX = A–1B
where I is the identity matrix.
45
MATLAB BASICS
Hence
X = A–1B
In MATLAB, we use the command x = inv (A) * B. Similarly, for XA = B, we use the
command x = B*inv (A).
The basic computational unit in MATLAB is the matrix. A matrix expression is enclosed
in square brackets, [ ]. Blanks or commas separate the column elements, and semicolons or
carriage returns separate the rows.
>>A = [1 2 3 4 ; 5 6 7 8 ; 9 10 11 12]
A=
1
2
3
4
5
6
7
8
9
10
11
12
The transpose of a simple matrix or a complex matrix is obtained by using the apostrophe
key
>>B = A′
B=
1
5
9
2
3
6
7
10
11
4
8
12
Matrix multiplication is accomplished as follows:
>>C = A * B
C=
30
70
110
70
174
278
110
278
446
107
122
122
140
137
158
152
176
137
152
158
176
179
200
200
224
>>C = B * A
C=
The inverse of a matrix D is obtained as
>>D = [1 2 ; 3 4]
D=
1
2
3
4
>>E = inv (D)
E=
– 2.0000
1.0000
1.5000
– 0.5000
46
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Similarly, its eigenvalue is
>>eig (D)
ans =
– 0.3723
5.3723
Matrix operations require that the matrix dimensions be compatible. If A is an n × m and
B is a p × r then A ± B is allowed only if n = p and m = r. Similarly, matrix product A * B is
allowed only if m = p.
Example 2.1. Consider the two matrices ;
1
A= 2
– 1
0
3
6
1
4
7
Using MATLAB, determine the following:
(a) A + B
(b) AB
(c) A2
(d) AT
(e) B–1
(f) BTAT
(g) A2 + B2 – AB
(h) determinant of A, determinant of B and determinant of AB.
Solution.
>> A = [1 0 1; 2 3 4; – 1 6 7]
A=
1
2
0
3
1
4
–1 6 7
>> B = [7 4 2 ; 3
B=
(a)
7
4
2
3
–1
5
2
6
1
>> C = A + B
C=
8
4
3
5
–2
8
8
10
8
5 6 ; – 1 2 1]
47
MATLAB BASICS
(b)
>> D = A * B
D=
6
(c)
6
3
19 31
4 40
26
41
>> E = A ^ 2
E=
0
4
(d)
6
33
8
42
4 60 72
>> % Let F = transpose of A
>> F = A′
F=
1
0
(e)
2
3
–1
6
1 4
7
>> H = inv (B)
H=
0.1111
(f)
0.0000
– 0.2222
0.1429 – 0.1429
– 0.1746
0.2857
0.5714
– 0.3651
>> J = B′ * A′
J=
6
19
4
6
3
31
26
40
41
(g) >> K = A ^ 2 + B ^ 2 – A * B
K=
53
15
(h)
52
51
45
58
– 2 28
det (A) = 12
42
det (B) = – 63
det (A * B) = – 756
Example 2.2. Determine the eigenvalues and eigenvectors of A and B using MATLAB
4
A = – 1
2
2
1
5
– 3
3
7
1
B = 8
5
2
7
3
3
6 .
1
48
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Solution. % Determine the eigenvalues and eigenvectors
A = [4
A=
2 – 3 ; – 1 1 3 ; 2 5 7]
4
–1
2 –3
1
3
2
5
7
eig(A)
ans =
0.5949
3.0000
8.4051
lamda = eig(A)
lamda =
0.5949
3.0000
8.4051
[V, D]=eig(A)
V=
– 0.6713
0.9163 – 0.3905
0.6713 – 0.3984
– 0.3144
0.0398
0.3905
0.8337
D=
0.5949
0
0
0
0
3.0000
0
0
8.4051
Example 2.3. Determine the values of x, y, and z for the following set of linear algebraic
equations :
x2 – 3x3 = – 5
2x1 + 3x2 – x3 = 7
4x1 + 5x2 – 2x3 = 10
Solution. Here
0
A = 2
4
AX = B
–1
A AX = A–1B
IX = A–1B
or
X = A–1B
1
3
5
− 3
− 1
− 2
5
B = 7
10
x1
and X = x2
x3
MATLAB BASICS
49
>> A = [0 1 – 3 ; 2 3 – 1 ; 4 5 – 2];
>> B = [– 5 ; 7 ; 10]
>> x = inv (A) * B
x=
– 1.0000
4.0000
3.0000
>> check = A * x
check =
–5
7
10
% Alternative method
>> x = A\B
x=
–1
4
3
2.15
SCRIPT FILES
A script is a sequence of ordinary statements and functions used at the command prompt
level. A script is invoked the command prompt level by typing the file-name or by using the pull
down menu. Scripts can also invoke other scripts.
The commands in the Command Window cannot be saved and executed again. Also, the
Command Window is not interactive. To overcome these difficulties, the procedure is first to
create a file with a list of commands, save it, and then run the file. In this way the commands
contained are executed in the order they are listed when the file is run. In addition, as the need
arises, one can change or modify the commands in the file, the file can be saved and run again.
The files that are used in this fashion are known as script files. Thus, a script file is a text file
that contains a sequence of MATLAB commands. Script file can be edited (corrected and/or
changed) and executed many times.
2.15.1 Creating and Saving a Script File
Any text editor can be used to create script files. In MATLAB script files are created and
edited in the Editor/Debugger Window. This window can be opened from the Command Window. From the Command Window, select File, New, and then M-file. Once the window is open,
the commands of the script file are typed line by line. The commands can also be typed in any
text editor or word processor program and then copied and pasted in the Editor/Debugger Window. The second type of M-files is the function file. Function file enables the user to extend the
basic library functions by adding ones own computational procedures. Function M-files are
expected to return one or more results. Script files and function files may include reference to
other MATLAB toolbox routines.
50
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
MATLAB function file begins with a header statement of the form:
function (name of result or results) = name (argument list)
Before a script file can be executed it must be saved. All script files must be saved with
the extension “.m”. MATLAB refers to them as m-files. When using MATLAB M-files editor, the
files will automatically be saved with a “.m” extension. If any other text editor is used, the file
must be saved with the “.m” extension, or MATLAB will not be able to find and run the script
file. This is done by choosing Save As… from the File menu, selecting a location, and entering a
name for the file. The names of user defined variables, predefined variables, MATLAB commands or functions should not be used to name script files.
2.15.2 Running a Script File
A script file can be executed either by typing its name in the Command Window and then
pressing the Enter key, directly from the Editor Window by clicking on the Run icon. The file is
assumed to be in the current directory, or in the search path.
2.15.3 Input to a Script File
There are three ways of assigning a value to a variable in a script file.
1. The variable is defined and assigned value in the script file.
2. The variable is defined and assigned value in the Command Window.
3. The variable is defined in the script file, but a specified value is entered in the Command Window when the script file is executed.
2.15.4 Output Commands
There are two commands that are commonly used to generate output. They are the disp
and fprintf commands.
1. The disp command
The disp command displays the elements of a variable without displaying the name of
the variable, and displays text.
disp(name of a variable) or disp(‘text as string’)
>>
A = [1 2 3 ; 4 5 6 ];
>> disp(A)
1 2 3
4 5 6
>> disp(‘Solution to the problem.’)
Solution to the problem.
2. The fprintf command
The fprintf command displays output (text and data) on the screen or saves it to a file.
The output can be formatted using this command.
Example 2.4. Write a function file Veccrossprod to compute the cross product of two
vectors a, and b, where a = (a1, a2, a3), b = (b1, b2, b3), and a × b = (a2b3 – a3b2, a3b1 – a1b3, a1b2
– a2b1). Verify the function by taking the cross products of pairs of unit vectors: (i, j), (j, k), etc.
51
MATLAB BASICS
Solution. Function c = Veccrossprod (a, b) ;
% Veccrossprod : function to compute c = a × b where a and b are 3D vectors
% call syntax:
% c = Veccrossprod(a, b) ;
c = [a(2) * b(3) – a(3) * b(2); a(3) * b(1) – a(1) * b(3); a(1) * b(2) – a(2) * b(1)];
2.16
PROGRAMMING IN MATLAB
One most significant feature of MATLAB is its extendibility through user-written programs such as the M-files. M-files are ordinary ASCII text files written in MATLAB language.
A function file is a subprogram.
2.16.1 Relational and Logical Operators
A relational operator compares two numbers by finding whether a comparison statement is true or false. A logical operator examines true/false statements and produces a result
which is true or false according to the specific operator. Relational and logical operators are
used in mathematical expressions and also in combination with other commands, to make decision that control the flow a computer program.
MATLAB has six relational operators as shown in Table 2.21.
Table 2.21 Relational operators
Relational operator
Interpretation
<
<=
Less than
Less than or equal
>
>=
Greater than
Greater than or equal
==
~=
Equal
Not equal
The logical operators in MATLAB are shown in Table 2.22.
Table 2.22 Logical operators
Logical operator
&
Name
AND
Example: A & B
|
Example: A | B
Description
Operates on two operands (A and B). If both are true,
the result is true (1), otherwise the result is false (0).
OR
Operates on two operands (A and B). If either one, or
both are true, the result is true (1), otherwise (both are
false) the result is false (0).
~
Example: ~ A
NOT
Operates on one operand (A). Gives the opposite of the
operand. True (1) if the operand is false, and false (0) if
the operand is true.
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
2.16.2 Order of Precedence
The following Table 2.23 shows the order of precedence used by MATLAB.
Table 2.23
Precedence
Operation
1 (highest)
Parentheses (If nested parentheses exist, inner have precedence).
2
3
Exponentiation.
Logical NOT (~).
4
5
Multiplication, Division.
Addition, Subtraction.
6
7
Relational operators (>, <, >=, <=, = =, ~=).
Logical AND (&).
8 (lowest)
Logical OR (|).
2.16.3 Built-in Logical Functions
The MATLAB built-in functions which are equivalent to the logical operators are:
and(A, B)
or(A, B)
Equivalent to A & B
Equivalent to A | B
not(A)
Equivalent to ~ A
List the MATLAB logical built-in functions are described in Table 2.24.
Table 2.24 Additional logical built-in functions
Function
xor(a, b)
Description
Example
Exclusive or. Returns true (1) if one
>>xor(8, – 1)
operand is true and the other is false
ans =
0
>>xor(8, 0)
ans =
1
all(A)
Returns 1 (true) if all elements in a vector
A are true (nonzero). Returns 0 (false) if
one or more elements are false (zero). If
A is a matrix, treats columns of A as
vectors, returns a vector with 1’s and 0’s.
>>A = [5 3 11 7 8 15]
>>all(A)
ans =
1
>>B = [3 6 11 4 0 13]
>>all(B)
ans =
0
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MATLAB BASICS
Function
Description
Example
any(A)
Returns 1 (true) if any element in a vector A
is true (nonzero). Returns 0 (false) if all
elements are false (zero).
If A is a matrix, treats columns of A as vectors,
returns a vector with 1’s and 0’s.
>>A = [5 0 14 0 0 13]
>>any(A)
ans =
1
>>B = [0 0 0 0 0 0 ]
>>any(B)
ans =
0
find(A)
If A is a vector, returns the indices of the nonzero elements.
>>A = [0 7 4 2 8 0 0 3 9]
>>find(A)
find(A > d)
If A is a vector, returns the address of the
elements that are larger than d (any relational
operator can be used).
ans =
2 3 4 11 8 9
>>find(A > 4)
ans =
456
The truth table for the operation of the four logical operators, and, or, Xor, and not are
summarized in Table 2.25.
Table 2.25 Truth table
INPUT
OUTPUT
AND
OR
XOR
NOT
NOT
A&B
A|B
(A, B)
~A
~B
false
false
false
false
true
true
false
true
false
true
true
true
false
true
false
false
true
true
false
true
true
true
true
true
false
false
false
A
B
false
2.16.4 Conditional Statements
A conditional statement is a command that allows MATLAB to make a decision of whether
to execute a group of commands that follow the conditional statement or to skip these commands.
If conditional expression consists of relational and/or logical operators
if
a < 30
count = count + 1
disp a
end
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
The general form of a simple if statement is as follows:
if
logical expression
statements
end
If the logical expression is true, the statements between the if statement and the end
statement are executed. If the logical expression is false, then it goes to the statements following the end statement.
2.16.5 Nested if Statements
Following is an example of nested if statements:
if
a < 30
count = count + 1;
disp(a);
if
b>a
b=0;
end
end
2.16.6 else AND elseif Clauses
The else clause allows to execute one set of statements if a logical expression is true and
a different set if the logical expression is false.
%
variable name inc
if
inc < 1
x_inc = inc/10;
else
x_inc = 0.05;
end
When several levels of if-else statements are nested, it may be difficult to find which
logical expressions must be true (or false) to execute each set of statements. In such cases, the
elaseif clause is used to clarify the program logic.
2.16.7 MATLAB while Structures
There is a structure in MATLAB that combines the for loop with the features of the if
block. This is called the while loop and has the form:
while logical expression
This set of statements is executed repeatedly as long as the logical expressions remain
true (equals + 1) or if the expression is a matrix rather than a simple scalar variable, as long as
all the elements of the matrix remain nonzero.
end
In addition to the normal termination of a loop by means of the end statement, there are
additional MATLAB commands available to interrupt the calculations. These commands are
listed in Table 2.26 below:
55
MATLAB BASICS
Table 2.26
Command
break
return
error (‘text’)
Description
Terminates the execution of MATLAB for and while loops. In nested
loops, break will terminate only the innermost loop in which it is placed.
Primarily used in MATLAB functions, return will cause a normal
return from a function from the point at which the return statement
is executed.
Terminates execution and displays the message contained in text on
the screen. Note, the text must be enclosed in single quotes.
The MATLAB functions used are summarized in Table 2.27 below:
Table 2.27
Function
Relational
operators
Description
A MATLAB logical relation is a comparison between two variables x
and y of the same size effected by one of the six operators, <, <=, >, >=,
= =, ~=. The comparison involves corresponding elements of x and y,
and yields a matrix or scalar of the same size with values of “true” or
“false” for each of its elements. In MATLAB, the value of “false” is zero,
and “true” has a value of one. Any nonzero quantity is interpreted as
“true”.
Combinatorial
operators
The operators & (AND) and | (OR) may be used to combine two logical
expressions.
all, any
If x is a vector, all(x) returns a value of one if all of the elements of x
are nonzero, and a value of zero otherwise. When X is a matrix, all(X)
returns a row vector of ones or zeros obtained by applying all to each of
the columns of X. The function any operates similarly if any of the
elements of x are nonzero.
find
If x is a vector, i = find(x) returns the indices of those elements of x
that are nonzero (i.e., true). Thus, replacing all the negative elements
of x by zero could be accomplished by
i = find(x < 0);
x(i) = zeros(size(i));
If X is a matrix, [i, j] = find(X) operates similarly and returns the rowcolumn indices of nonzero elements.
56
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Function
Description
if, else, elseif
The several forms of MATLAB if blocks are as follows:
if variable
block of statements
if variable 1
block of statements
executed if variable
variable 1 executed if variable 1
nonzero
end
variable 2
if variable 1
block of statements
executed if
is “true”, i.e.,
is “true”, i.e., nonzero
else
is “true”,
elseif variable 2
block of statements
executed if variable 1
block of statements
executed if
is “false”, i.e., zero
end
is “true”,
else
block of statements
executed if neither
variable is “true”
end
break
Terminates the execution of a for or while loop. Only the innermost
loop in which break is encountered will be terminated.
return
Causes the function to return at that point to the calling routine.
MATLAB M-file functions will return normally without this statement.
error (‘text’)
Within a loop or function, if the statement error(‘text’) is encountered,
the loop or function is terminated, and the text is displayed.
while
The form of the MATLAB while loop is
while variable
block of statements executed as long as the value of
variable is “true” ; i.e., nonzero
end
Useful when a function F itself calls a second “dummy” function “f ”.
For example, the function F might find the root of an arbitrary function identified as a generic f(x). Then, the name of the actual M-file
function, say fname, is passed as a character string to the function
F either through its argument list or as a global variable, and the
function is evaluated within F by means of feval. The use of
feval(name, x1 , x2, ..., xn), where fname is a variable containing
the name of the function as a character string; i.e., enclosed in single quotes, and x1, x2, ..., xn are the variables needed in the argument list of function fname.
57
MATLAB BASICS
2.17
GRAPHICS
MATLAB has many commands that can be used to create basic 2-D plots, overlay plots,
specialized 2-D plots, 3-D plots, mesh, and surface plots.
2.17.1 Basic 2-D Plots
The basic command for producing a simple 2-D plot is
plot(x values, y values, ‘style option’)
where
x values and y values are vectors containing the x- and y-coordinates of points on the
graph.
style option is an optional argument that specifies the color, line-style, and the pointmarker style.
The style option in the plot command is a character string that consists of 1, 2, or 3
characters that specify the color and/or the line style. The different color, line-style and markerstyle options are summarized in Table 2.28.
Table 2.28 Color, line-style, and marker-style options
Color style-option
Line style-option
Marker style-option
y
m
yellow
magenta
–
––
solid
dashed
+
o
plus sign
circle
c
r
g
cyan
red
green
:
–.
dotted
dash-dot
*
x
.
asterisk
x-mark
point
b
w
blue
white
^
s
up triangle
square
k
black
d
diamond, etc.
2.17.2 Specialized 2-D Plots
There are several specialized graphics functions available in MATLAB for 2-D plots. The
list of functions commonly used in MATLAB for plotting x-y data are given in Table 2.29.
Table 2.29 List of functions for plotting x-y data
Function
Description
area
bar
Creates a filled area plot.
Creates a bar graph.
barh
comet
Creates a horizontal bar graph.
Makes an animated 2-D plot.
compass
contour
Creates arrow graph for complex numbers.
Makes contour plots.
contourf
errorbar
Makes filled contour plots.
Plots a graph and puts error bars.
58
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Function
feather
fill
fplot
hist
loglog
pareto
pcolor
pie
plotyy
plotmatrix
polar
quiver
rose
scatter
semilogx
semilogy
stairs
stem
Description
Makes a feather plot.
Draws filled polygons of specified color.
Plots a function of a single variable.
Makes histograms.
Creates plot with log scale on both x and y axes.
Makes pareto plots.
Makes pseudo color plot of matrix.
Creates a pie chart.
Makes a double y-axis plot.
Makes a scatter plot of a matrix.
Plots curves in polar coordinates.
Plots vector fields.
Makes angled histograms.
Creates a scatter plot.
Makes semilog plot with log scale on the x-axis.
Makes semilog plot with log scale on the y-axis.
Plots a stair graph.
Plots a stem graph.
2.17.2.1 Overlay plots
There are three ways of generating overlay plots in MATLAB, they are :
(a) Plot command
(b) Hold command
(c) Line command
(a) Plot command. Example 2.7(a) shows the use of plot command used with matrix
argument, each column of the second argument matrix plotted against the corresponding column of the first argument matrix.
(b) Hold command. Invoking hold on at any point during a session freezes the current
plot in the graphics window. All the next plots generated by the plot command are added to the
exiting plot. See Example 2.7(a).
(c) Line command. The line command takes a pair of vectors (or a triplet in 3–D)
followed by a parameter name / parameter value pairs as argument. For instance, the command: line (x data, y data, parameter name, parameter value) adds lines to the existing axes.
See Example 2.7(a).
2.17.3 3-D Plots
MATLAB provides various options for displaying three-dimensional data. They include
line and wire, surface, mesh plots, among many others. More information can be found in the
Help Window under Plotting and Data visualization. Table 2.30 lists commonly used functions.
59
MATLAB BASICS
Table 2.30 Functions used for 3-D graphics
Command
Description
plot3
Plots three-dimensional graph of the trajectory of a set of three parametric equations x(t), y(t), and z(t) can be obtained using plot 3(x, y, z).
meshgrid
If x and y are two vectors containing a range of points for the evaluation of a function, [X, Y] = meshgrid(x, y) returns two rectangular
matrices containing the x and y values at each point of a two-dimensional grid.
If X and Y are rectangular arrays containing the values of the x and y
coordinates at each point of a rectangular grid , and if z is the value of
a function evaluated at each of these points, mesh(X, Y, z) will produce a three-dimensional perspective graph of the points. The same
results can be obtained with mesh(x, y, z) can also be used.
mesh(X, Y, z)
meshc, meshz
surf
surfc
colormap
shading
view
axis
If the xy grid is rectangular, these two functions are merely variations
of the basic plotting program mesh, and they operate in an identical
fashion. meshc will produce a corresponding contour plot drawn on
the xy plane below the three-dimensional figure, and meshz will add
a vertical wall to the outside features of the figures drawn by mesh.
Produces a three-dimensional perspective drawing. Its use is usually
to draw surfaces, as opposed to plotting functions, although the actual
tasks are quite similar. The output of surf will be a shaded figure. If
row vectors of length n are defined by x = r cos θ and y = r sin θ, with 0
H
≤ θ ≤ 2π, they correspond to a circle of radius r. If r is a column vector
equal to r = [0 1 2]’; then z = r*ones(size(x)) will be a rectangular, 3
× n, arrays of 0’s and 2’s, and surf(x, y, z) will produce a shaded surface bounded by three circles; i.e., a cone.
This function is related to surf in the same way that meshc is related
to mesh.
Used to change the default coloring of a figure. See the MATLAB reference manual or the help file.
Controls the type of color shading used in drawing figures. See the
MATLAB reference manual or the help file.
view(az, el) controls the perspective view of a three-dimensional plot.
The view of the figure is from angle “el” above the xy plane with the
coordinate axes (and the figure) rotated by an angle “az” in a clockwise direction about the z axis. Both angles are in degrees. The default
values are az = 37½º and el = 30º.
Determines or changes the scaling of a plot. If the coordinate axis limits of a two-dimensional or three-dimensional graph are contained in
the row vector r = [xmin, xmax, ymin, ymax, zmin, zmax], axis will return the
values in this vector, and axis(r) can be used to alter them. The coordinate axes can be turned on and off with axis(‘on’) and axis(‘off ’). A few
other string constant inputs to axis and their effects are given below:
axis(‘equal’) x and y scaling are forced to be the same.
60
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Command
Description
axis(‘square’) The box formed by the axes is square.
axis(‘auto’) Restores the scaling to default settings.
axis(‘normal’) Restoring the scaling to full size, removing any effects
of square or equal settings.
axis(‘image’) Alters the aspect ratio and the scaling so the screen
pixels are square shaped rather than rectangular.
contour
The use is contour(x, y, z). A default value of N = 10 contour lines will
be drawn. An optional fourth argument can be used to control the
number of contour lines that are drawn. contour(x, y, z, N), if N is a
positive integer, will draw N contour lines, and contour(x, y, z, V), if
V is a vector containing values in the range of z values, will draw
contour lines at each value of z = V.
Plots lines or curves in three dimensions. If x, y, and z are vectors of
equal length, plot3(x, y, z) will draw, on a three-dimensional coordinate axis system, the lines connecting the points. A fourth argument,
representing the color and symbols to be used at each point, can be
added in exactly the same manner as with plot.
plot3
grid
grid on adds grid lines to a two-dimensional or three-dimensional
graph; grid off removes them.
Draws “slices” of a volume at a particular location within the volume.
slice
Example 2.5. (a) Generate an overlay plot for plotting three lines
y1 = sin t
y2 = t
y3 = t –
t3
t5
t7
+
+
3!
5!
7!
0 ≤ t ≤ 2π
Use (i) the plot command
(ii) the hold command
(iii) the line command
(b) Use the functions for plotting x-y data for plotting the following functions.
(i) f(t) = t cost
0 ≤ t ≤ 10π
(ii) x = et
y = 100 + e3t
0 ≤ t ≤ 2π
Solution:
(a) overlay plot
(i) % using the plot command
t = linspace(0, 2 * pi, 100);
61
MATLAB BASICS
y1 = sin(t); y2 = t;
y3 = t – (t.^ 3)/6 + (t.^ 5)/120 – (t.^ 7)/5040;
plot(t, y1, t, y2, ‘–’, t, y3, ‘o’)
axis([0 5
xlabel(‘t’)
–1 5])
ylabel(‘sin(t) approximation’)
title(‘sin(t) function’)
text(3.5, 0, ‘sin(t)’)
gtext(‘Linear approximation’)
gtext(‘4-term approximation’)
sin (t) function
5
sin (t) approximation
4
Linear approximation
3
2
1
sin (t)
0
1-term approximation
–1
0 0.5 1 1.5 2 2.5 3
3.5 4 4.5
Fig. E 2.5 (a) (i)
(ii)
% using the hold command
x = linspace(0, 2*pi, 100); y1 = sin(x);
plot(x, y1)
hold on
y2 = x; plot(x, y2, ‘–’ )
y3 = x – (x . ^ 3)/6 + (x . ^ 5)/120 – (t . ^ 7)/5040;
plot(x, y3, ‘o’)
axis([0 5 – 1 5])
hold off
5
62
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
5
4
3
2
1
0
–1
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Fig. E 2.5 (a) (ii)
(iii) % using the line command
t = linspace(0, 2*pi, 100);
y1 = sin(t);
y2 = t;
y3 = t – (t.^ 3)/6 + (t.^5)/120 – (t.^7)/5040;
plot(t, y1)
line(t, y2, ‘linestyle’, ‘–’)
line(t, y3, ‘marker’, ‘o’)
axis([0 5 – 1 5])
xlabel(‘t’)
ylabel(‘sin(t) approximation’)
title(‘sin(t) function’)
legend(‘sin(t)’, ‘linear approx’, ‘7th order approx’)
sin (t) function
5
sin (t) approximation
4
sin (t)
linear approx
7th order approx
3
2
1
0
–1
0 0.5 1 1.5 2 2.5 3
t
3.5 4 4.5
Fig. E 2.5(a) (iii)
5
63
MATLAB BASICS
(b) Using Table 2.29 functions
(i) fplot(‘x.*cos(x)’, [0 10*pi])
This will give the following figure (Fig. E 2.5 (b) (i))
40
30
20
10
0
– 10
– 20
– 30
0
5
10
15
20
25
30
Fig. E 2.5 (b) (i)
(ii)
t = linspace(0, 2*pi, 200);
x = exp(t);
y = 100 + exp(3*t);
loglog(x, y), grid
10
10
10
10
10
10
10
9
8
7
6
5
4
3
2
10
0
10
10
1
10
2
Fig. E 2.5 (b) (ii)
Example 2.6. (a) Plot the parametric space curve of
x(t) = t
y(t) = t2
(b)
z(t) = t3
z = – 7 / (1 + x2 + y2)
0 ≤ t ≤ 2.0
|x| ≤5 , |y|≤5
10
3
64
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Solution. (a) >> t = linspace(0, 2,100) ;
>> x = t ; y = t. ^2 ; z = t. ^3;
>> plot3(x, y, z), grid
The plot is shown in Figure E 2.6 (a).
8
6
4
2
0
4
2
3
1.5
2
1
1
0 0
0.5
Fig. E 2.6 (a)
(b) >> t = linspace(0, 2, 100);
>> x = t ; y = t. ^2 ; z = t. ^3;
>> plot3(x, y, z), grid
>> t = linspace (– 5, 5, 50) ; y = x ;
>> z = – 7./(1 + x . ^ 2 + y . ^ 2);
>> mesh(z)
The plot is shown in Figure E 2.6(b).
0
–2
–4
–6
–8
100
50
0 0
20
Fig. E 2.6(b)
40
60
80
100
65
MATLAB BASICS
2.17.4 Saving and Printing Graphs
To obtain a hardcopy of a graph, type print in the Command Window after the graph
appears in the Figure Window. The figure can also be saved into a specified file in the PostScripter
or Encapsulated PostScript (EPS) format. The command to save graphics to a file is
print – d devicetype – options filename
where device type for PostScript printers are listed in the following Table 2.31.
Table 2.31 Devicetype for Post Script printers
Devicetype
Description
Devicetype
Description
ps
Black and white PostScript
eps
Black and white EPSF
psc
ps2
Color PostScript
Level 2 BW PostScript
epsc
eps2
Color EPSF
Level 2 black and white EPSF
psc2
Level 2 color PostScript
epsc2
Level 2 color EPSF
MATLAB can also generate a graphics file in the following popular formats among others.
–dill
saves file in Adobe Illustrator format.
–djpeg
saves file as a JPEG image.
–dtiff
saves file as a compressed TIFF image.
–dmfile
saves file as an M-file with graphics handles.
2.18
INPUT/OUTPUT IN MATLAB
In this section, we present some of the many available commands in MATLAB for reading data from an external file into a MATLAB matrix, or writing the numbers computed in
MATLAB into such an external file.
2.18.1 The fopen Statement
To have the MATLAB read or write a separate data file of numerical values, we need to
connect the file to the executing MATLAB program. The MATLAB functions used are summarized in Table 2.32.
Table 2.32 MATLAB functions used for input/output
Function
fopen
Description
Connects an existing file to MATLAB or to create a new file from MATLAB.
fid = fopen(‘Filename’, permission code);
where, if fopen is successful, fid will be returned as a positive integer greater
than 2. When unsuccessful, a value of – 1 is returned. Both the file name and the
permission code are string constants enclosed in single quotes. The permission
code can be a variety of flags that specify whether or not the file can be written
to, read from, appended to, or a combination of these. Some common codes are:
Code
Meaning
‘r’
read only
‘w’
write only
‘r+’
read and write
‘a+’
read and append
The fopen statement positions the file at the beginning.
(Contd.)
66
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Function
Description
fclose
Disconnects a file from the operating MATLAB program. The use is fclose(fid),
where fid is the file identification number of the file returned by
fopen.fclose(‘all’) will close all files.
fscanf
Reads opened files. The use is
A = fscanf(fid, FORMAT, SIZE)
where FORMAT specifies the types of numbers (integers, reals with or without
exponent, character strings) and their arrangement in the data file, and optional SIZE determines how many quantities are to be read and how they are to
be arranged into the matrix A. If SIZE is omitted, the entire file is read. The
FORMAT field is a string (enclosed in single quotes) specifying the form of the
numbers in the file. The type of each number is characterized by a percent sign
(%), followed by a letter (i or d for integers, e or f for floating-point numbers with
or without exponents). Between the percent sign and the type code, one can
insert an integer specifying the maximum width of the field.
fprintf
Writes files previously opened.
fprintf(fid, FORMAT, A)
where fid and FORMAT have the same meaning as for fscanf, with the exception that for output formats the string must end with \n, designating the end of
a line of output.
2.19
SYMBOLIC MATHEMATICS
In Secs. 2.1 to 2.18, the capability of MATLAB for numerical computations have been
described. In this section some of MATLAB’s capabilities for symbolic manipulations will be
presented. Specifically, the symbolic expressions, symbolic algebra, simplification of mathematical expressions, operations on symbolic expressions, solution of a single equation or a set of
linear algebraic equations, solutions to differential equations, differentiation and integration of
functions using MATLAB are presented.
2.19.1 Symbolic Expressions
A symbolic expression is stored in MATLAB as a character string. A single quote marks
are used to define the symbolic expression. For instance:
‘sin(y/x)’ ; ‘x ^ 4 + 5*x^3 + 7*x^2 – 7’
The independent variable in many functions is specified as an additional function argument. If an independent variable is not specified, then MATLAB will pick one. When several
variables exist, MATLAB will pick the one that is a single lower case letter (except i and j),
which is closest to x alphabetically.
The independent variable is returned by the function symvar,
symvar(s)
Returns the independent variable for the symbolic expression s.
For example:
Expression s
‘5 * c * d + 34’
‘sin(y/x)’
symvar(s)
d
x
67
MATLAB BASICS
In MATLAB, a number of functions are available to simplify mathematical expressions
by expanding the terms, factoring expressions, collecting coefficients, or simplifying the expression. For instance:
expand(s)
Performs an expansion of s.
A summary of these expressions is given in Table 2.33. A summary of basic operations is
given in Table 2.34. The standard arithmetic operation (Table 2.35) is applied to symbolic expressions using symbolic functions. These symbolic expressions are summarized in Table 2.36.
Table 2.33
Simplification
collect
expand
factor
horner
numden
simple
simplify
subexpr
Collect common terms
Expand polynomials and elementary functions
Factorization
Nested polynomial representation
Numerator and denominator
Search for shortest form
Simplification
Rewrite in terms of subexpressions
Table 2.34
Basic Operations
ccode
conj
findsym
fortran
imag
latex
pretty
real
sym
syms
C code representation of a symbolic expression
Complex conjugate
Determine symbolic variables
Fortran representation of a symbolic expression
Imaginary part of a complex number
LaTeX representation of a symbolic expression
Pretty prints a symbolic expression
Real part of an imaginary number
Create symbolic object
Shortcut for creating multiple symbolic objects
Table 2.35
Arithmetic Operations
+
–
*
.*
/
./
\
.\
^
.^
‘
.‘
Addition
Subtraction
Multiplication
Array multiplication
Right division
Array right division
Left division
Array left division
Matrix or scalar raised to a power
Array raised to a power
Complex conjugate transpose
Real transpose
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Table 2.36
Symbolic expressions
horner(S)
Transposes S into its Horner, or nested, representation.
numden(S)
Returns two symbolic expressions that represent, respectively, the
numerator expression and the denominator expression for the
rational representation of S.
numeric(S)
Converts s to a numeric form (S must not contain any symbolic
variables).
poly2sym(c)
pretty(S)
Converts a polynomial coefficient vector c to a symbolic polynomial.
Prints S in an output form that resembles typeset mathematics.
sym2poly(S)
symadd(A, B)
Converts S to a polynomial coefficient vector. *
Performs a symbolic addition, A + B.
symdiv(A, B)
symmul(A, B)
Performs a symbolic division, A / B.
Performs a symbolic multiplication, A * B.
sympow(S, p)
symsub(A, B)
Performs a symbolic power, S ^ p.
Performs a symbolic subtraction, A – B.
2.19.2 Solution to Differential Equations
Symbolic math functions can be used to solve a single equation, a system of equations,
and differential equations. For example:
solve(f) Solves a symbolic equation f for its symbolic variable. If f is a symbolic expression, this function solves the equation f = 0 for its symbolic variable.
solve(f1, … fn) Solves the system of equations represented by f1, …, fn.
The symbolic function for solving ordinary differential equation is dsolve as shown below:
dsolve(‘equation’, ‘condition’) Symbolically solves the ordinary differential equation
specified by ‘equation’. The optional argument ‘condition’ specifies a boundary or initial condition.
The symbolic equation uses the letter D to denote differentiation with respect to the
independent variable. A D followed by a digit denotes repeated differentiation. Thus, Dy represents dy/dx, and D2y represents d2y/dx2. For example, given the ordinary second order differential equation;
dx
d2 x
+ 3x = 7
2 +5
dt
dt
C
with the initial conditions x(0) = 0 and x (0) = 1.
The MATLAB statement that determine the symbolic solution for the above differential
equation is the following:
x = dsolve(‘D2x = – 5*Dx – 3 * x + 7’, ‘x(0) = 0’, ‘Dx(0) = 1’)
The symbolic functions are summarized in Table 2.37.
69
MATLAB BASICS
Table 2.37
Solution of Equations
compose
Functional composition
dsolve
Solution of differential equations
finverse
Functional inverse
solve
Solution of algebraic equations
2.19.3 Calculus
There are four forms by which the symbolic derivative of a symbolic expression is obtained in MATLAB. They are:
diff(f) Returns the derivative of the expression f with respect to the default independent
variable.
diff(f, ‘t’) Returns the derivative of the expression f with respect to the variable t.
diff(f, n) Returns the nth derivative of the expression f with respect to the default independent variable.
diff(f, ‘t’, n) Returns the nth derivative of the expression f with respect to the variable t.
The various forms that are used in MATLAB to find the integral of a symbolic expression
f are given below and summarized in Table 2.38.
int(f) Returns the integral of the expression f with respect to the default independent
variable.
int(f, ‘t’) Returns the integral of the expression f with respect to the variable t.
int(f, a, b) Returns the integral of the expression f with respect to the default independent variable evaluated over the interval [a, b], where a and b are numeric expressions.
int(f, ‘t’, a, b) Returns the integral of the expression f with respect to the variable t
evaluated over the interval [a, b], where a and b are numeric expressions.
int(f, ‘m’, ‘n’) Returns the integral of the expression f with respect to the default independent variable evaluated over the interval [m, n], where m and n are numeric expressions.
The other symbolic functions for pedagogical and graphical applications, conversions,
integral transforms, and linear algebra are summarized in Tables 2.38 to 2.42.
Table 2.38
Calculus
diff
Differentiate
int
Integrate
jacobian
Jacobian matrix
limit
Limit of an expression
symsum
Summation of series
taylor
Taylor series expansion
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Table 2.39
Pedagogical and Graphical Applications
ezcontour
Contour plotter
ezcontourf
Filled contour plotter
ezmesh
Mesh plotter
ezmeshc
Combined mesh and contour plotter
ezplot
Function plotter
ezplot
Easy-to-use function plotter
ezplot3
Three-dimensional curve plotter
ezpolar
Polar coordinate plotter
ezsurf
Surface plotter
ezsurfc
Combined surface and contour plotter
funtool
Function calculator
rsums
Riemann sums
taylortool
Taylor series calculator
Table 2.40
Conversions
char
Convert sym object to string
double
Convert symbolic matrix to double
poly2sym
Function calculator
sym2poly
Symbolic polynomial to coefficient vector
Table 2.41
Integral Transforms
fourier
Fourier transform
ifourier
Inverse Fourier transform
ilaplace
Inverse Laplace transform
iztrans
Inverse Z-transform
laplace
Laplace transform
ztrans
Z-transform
71
MATLAB BASICS
Table 2.42
Linear Algebra
2.20
colspace
Basis for column space
det
Determinant
diag
Create or extract diagonals
eig
Eigenvalues and eigenvectors
expm
Matrix exponential
inv
Matrix inverse
jordan
Jordan canonical form
null
Basis for null space
poly
Characteristic polynomial
rank
Matrix rank
rref
Reduced row echelon form
svd
Singular value decomposition
tril
Lower triangle
triu
Upper triangle
THE LAPLACE TRANSFORMS
The Laplace transformation method is an operational method that can be used to find
the transforms of time functions, the inverse Laplace transformation using the partial-fraction
expansion of B(s)/A(s), where A(s) and B(s) are polynomials in s. In this chapter, we present the
computational methods with MATLAB to obtain the partial-fraction expansion of B(s)/A(s) and
the zeros and poles of B(s)/A(s).
MATLAB can be used to obtain the partial-fraction expansion of the ratio of two polynomials, B(s)/A(s) as follows:
B( s)
num
b(1)s n + b(2)s n−1 + ... + b(n)
=
=
A( s)
den
a(1) sn + a(2)sn −1 + ... + a(n)
where a(1) ≠ 0 and num and den are row vectors. The coefficients of the numerator and denominator of B(s)/A(s) are specified by the num and den vectors.
Hence
num = [b(1) b(2)
den = [a(1) a(2)
…
…
b(n)]
a(n)]
The MATLAB command
r, p, k = residue(num, den)
is used to determine the residues, poles, and direct terms of a partial-fraction expansion of the
ratio of two polynomials B(s) and A(s) is then given by
r(1)
r(2)
r (n)
B(s)
= k(s) +
+
+…+
s − p(1)
s − p(2)
s − p( n)
A(s)
72
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
The MATLAB command [num, den] = residue(r, p, k), where r, p, k are the output
from MATLAB converts the partial fraction expansion back to the polynomial ratio B(s)/A(s).
The command printsys (num, den, ‘s’) prints the num/den in terms of the ratio of
polynomials in s.
The command ilaplace will find the inverse Laplace transform of a Laplace function.
2.20.1 Finding Zeros and Poles of B(s)/A(s)
The MATLAB command [z, p, k] = tf2zp(num,den) is used to find the zeros, poles, and
gain K of B(s)/A(s).
If the zeros, poles, and gain K are given, the following MATLAB command can be used to
find the original num/den:
[num,den] = zp2tf (z, p, k)
2.21
CONTROL SYSTEMS
MATLAB has an extensive set of functions for the analysis and design of control systems.
They involve matrix operations, root determination, model conversions, and plotting of complex functions. These functions are found in MATLAB’s control systems toolbox. The analytical
techniques used by MATLAB for the analysis and design of control systems assume the processes that are linear and time invariant. MATLAB uses models in the form of transfer-functions
or state-space equations.
2.21.1 Transfer Functions
The transfer function of a linear time invariant system is expressed as a ratio of two
polynomials. The transfer function for a single input and a single output (SISO) system is
written as
H(s) =
b0 s n + b1 s n −1 + ... + bn −1 s + bn
a0 sm + a1 sm−1 + ... + am−1 s + am
when the numerator and denominator of a transfer function are factored into the zero-pole-gain
form, it is given by
H(s) =
(s − z1 )(s − z2 )...(s − zn )
(s − p1 )(s − p2 )...(s − pm )
The state-space model representation of a linear control system s written as
x = Ax + Bu
y = Cx + Du
2.21.2 Model Conversion
There are a number of functions in MATLAB that can be used to convert from one model
to another. These conversion functions and their applications are summarized in Table 2.43.
73
MATLAB BASICS
Table 2.43
Model conversion functions
Function
Purpose
c3d
residue
Continuous state-space to discrete state-space
Partial-fraction expansion
ss3tf
ss2zp
State-space to transfer function
State-space to zero-pole-gain
tf2ss
tf2zp
Transfer function to state-space
Transfer function to zero-pole-gain
zp2ss
zp2tf
Zero-pole-gain to state-space
Zero-pole-gain to transfer function
Residue Function: The residue function converts the polynomial transfer function
H(s) =
b0 s n + b1 s n −1 + ... + bn −1 s + bn
a0 sm + a1 sm−1 + ... + am−1 s + am
to the partial fraction transfer function
H(s) =
r1
r2
rn
+
+…+
+ k(s)
s − p1
s − p2
s − pn
[r, p, k] = residue(B, A) Determine the vectors r, p, and k, which contain the residue
values, the poles, and the direct terms from the partial-fraction expansion. The inputs are the
polynomial coefficients B and A from the numerator and denominator of the transfer function,
respectively.
ss2tf Function: The ss2tf function converts the continuous-time, state-space equations
x′ = Ax + Bu
y = Cx + Du
to the polynomial transfer function
H(s) =
b0 sn + b1 sn −1 + ... + bn −1 s + bn
a0 sm + a1 sm −1 + ... + am−1 s + am
The function has two output matrices:
[num, den] = ss2tf(A, B, C, D, iu) Computes vectors num and den containing the
coefficients, in descending powers of s, of the numerator and denominator of the polynomial
transfer function for the iuth input. The input arguments A, B, C, and D are the matrices of
the state-space equations corresponding to the iuth input, where iu is the number of the input
for a multi-input system. In the case of a single-input system, iu is 1.
ss2zp Function: The ss2zp function converts the continuous-time, state-space equations
x′ = Ax + Bu
y = Cx + Du
74
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
to the zero-pole-gain transfer function
H(s) = k
(s − z1 )(s − z2 )...(s − zn )
(s − p1 )(s − p2 )...(s − pm )
The function has three output matrices:
[z, p, k] = ss2zp(A, B, C, D, iu) Determines the zeros (z) and poles (p) of the zero-polegain transfer function for the iuth input, along with the associated gain (k). The input matrices
A, B, C, and D of the state-space equations correspond to the iuth input, where iu is the number
of the input for a multi-input system. In the case of a single-input system iu is 1.
tf2ss Function: The tf 2ss function converts the polynomial transfer function
H(s) =
b0 s n + b1 s n −1 + ... + bn −1 s + bn
a0 sm + a1 sm−1 + ... + am−1 s + am
to the controller-canonical form state-space equations
x′ = Ax + Bu
y = Cx + Du
The function has four output matrices:
[A,B,C,D] = tf2ss(num,den) Determines the matrices A, B, C, and D of the controllercanonical form state-space equations. The input arguments num and den contain the coefficients, in descending powers of s, of the numerator and denominator polynomials of the transfer function that is to be converted.
tf2zp Function: The tf2zp function converts the polynomial transfer function
H(s) =
b0 s n + b1 s n −1 + ... + bn −1 s + bn
a0 sm + a1 sm−1 + ... + am−1 s + am
to the zero-pole-gain transfer function
H(s) = k
(s − z1 )(s − z2 )...(s − zn )
(s − p1 )(s − p2 )...(s − pm )
The function has three output matrices:
[z, p, k] = tf2zp(num,den) Determines the zeros (z), poles (p) and associated gain (k) of
the zero-pole-gain transfer function using the coefficients, in descending powers of s, of the
numerator and denominator of the polynomial transfer function that is to be converted.
zp2tf Function: The zp2tf function converts the zero-pole-gain transfer function
H(s) = k
(s − z1 )(s − z2 )...(s − zn )
(s − p1 )(s − p2 )...(s − pm )
to the polynomial transfer function
H(s) =
b0 s n + b1 s n −1 + ... + bn −1 s + bn
a0 sm + a1 sm−1 + ... + am−1 s + am
The function has two output matrices:
75
MATLAB BASICS
[num, den] = zp2tf(z, p, k) Determines the vectors num and den containing the coefficients, in descending powers of s, of the numerator and denominator of the polynomial transfer
function. p is a column vector of the pole locations of the zero-pole-gain transfer function, z is a
matrix of the corresponding zero locations, having one column for each output of a multi-output
system, k is the gain of the zero-pole-gain transfer function. In the case of a single-output
system, z is a column vector of the zero locations corresponding to the pole locations of vector p.
zp2ss Function: The zp2ss function converts the zero-pole-gain transfer function
H(s) =
(s − z1 )(s − z2 )...(s − zn )
(s − p1 )(s − p2 )...(s − pm )
to the controller-canonical form state-space equations
x′ = Ax + Bu
y = Cx + Du
The function has four output matrices:
[A, B, C, D] = zp2ss(z, p, k) Determines the matrices A, B, C, and D of the controlcanonical form state-space equations. p is a column vector of the pole locations of the zero-polegain transfer function, z is a matrix of the corresponding zero locations, having one column for
each output of a multi-output system, k is the gain of the zero-pole-gain transfer function. In
the case of a single-output system, z is a column vector of the zero locations corresponding to
the pole locations of vector p.
2.22
THE LAPLACE TRANSFORMS
MATLAB can be used to obtain the partial-fraction expansion of the ratio of two polynomials, B(s)/A(s) as follows:
B( s)
num
b(1) s n + b(2) s n−1 + ... + b(n)
=
=
A( s)
den
a(1) sn + a(2) sn −1 + ... + a( n)
where a(1) ≠ 0 and num and den are row vectors. The coefficients of the numerator and denominator of B(s)/A(s) are specified by the num and den vectors.
Hence
num = [b(1) b(2) …
b(n)]
den = [a(1) a(2) …
The MATLAB command
a(n)]
r, p, k = residue(num, den)
is used to determine the residues, poles, and direct terms of a partial-fraction expansion of the
ratio of two polynomials B(s) and A(s) is then given by
r(1)
r (2)
r( n)
B( s)
= k(s) +
+
+…+
s − p(1)
s − p(2)
s − p(n)
A( s)
The MATLAB command [num, den] = residue(r, p, k) where r, p, k are the output from
MATLAB converts the partial fraction expansion back to the polynomial ratio B(s)/A(s).
The command printsys (num,den, ‘s’) prints the num/den in terms of the ratio of polynomials in s.
The command ilaplace will find the inverse Laplace transform of a Laplace function.
76
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Finding Zeros and Poles of B(s)/A(s)
The MATLAB command [z, p, k] = tf2zp(num,den) is used to find the zeros, poles, and
gain K of B(s)/A(s).
If the zeros, poles, and gain K are given, the following MATLAB command can be used to
find the original num/den:
[num, den] = zp2tf (z, p, k)
Example 2.7. Consider the function
H(s) =
n( s)
d( s)
where n(s) = s4 + 6s3 + 5s2 + 4s + 3
d(s) = s5 + 7s4 + 6s3 + 5s2 + 4s + 7
(a) Find n(– 10), n(– 5), n(– 3) and n(– 1)
(b) Find d(– 10), d(– 5), d(– 3) and d(– 1)
(c) Find H(– 10), H(– 5), H(– 3) and H(– 1)
Solution:
(a) >> n = [1 6 5 4 3];
>> d = [1 7 6 5 4 7];
>> n2 = polyval(n, [– 10])
% n = s ^ 4 + 6s ^ 3 + 5s ^ 2 + 4s + 3
% d = s ^ 5 + 7s ^ 4 + 6s ^ 3 + 5s ^ 2 + 4s + 7
n2 = 4463
>> nn10 = polyval(n, [– 10])
nn10 = 4463
>> nn5 = polyval(n, [– 5])
nn5 = – 17
>> nn3 = polyval(n, [– 3])
nn3 = – 45
>> nn1 = polyval(n, [– 1])
nn1 = – 1
(b) >> dn10 = polyval(d, [– 10])
dn10 = – 35533
>> dn5 = polyval(d, [– 5])
dn5 = 612
>> dn3 = polyval(d, [– 3])
dn3 = 202
>> dn1=polyval(d, [– 1])
dn1 = 8
(c) >> Hn10 = nn10/dn10
Hn10 = – 0.1256
>> Hn5 = nn5/dn5
Hn5 = – 0.0278
77
MATLAB BASICS
>> Hn3 = nn3/dn3
Hn3 = – 0.2228
>> Hn1 = nn1/dn1
Hn1 = – 0.1250
Example 2.8. Generate a plot of
y(x) = e–0.7x sin ωx
where w = 15 rad/s, and 0 ≤ x ≤ 15. Use the colon notation to generate the x vector in increments
of 0.1.
Solution.
>> x = [0 : 0.1 : 15];
>> w = 15;
>> y = exp(– 0.7*x).*sin(w*x);
>> plot(x, y)
>> title(‘y(x) = e^-^0^.^7^x sin\omega x’)
>> xlabel(‘x’)
>> ylabel(‘y’)
y(x) = e
– 0.7x
sin ωx
1
0.8
0.6
0.4
0.2
0
– 0.2
– 0.4
– 0.6
– 0.8
0
5
x
10
15
Fig. E 2.8.
Example 2.9. Generate a plot of
y(x) = e–0.6x cos ωx
where ω = 10 rad/s, and 0 ≤ x ≤ 15. Use the colon notation to generate the x vector in increments
of 0.05.
Solution.
>> x = [0 : 0.1 : 15];
>> w = 10;
78
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> y = exp(– 0.6*x).*cos(w*x);
>> plot(x, y)
>> title(‘y(x) = e^-^0^.^6^x cos\omega x’)
>> xlabel(‘x’)
>> ylabel(‘y’)
y(x) = e
– 0.6x
1
sin ωx
0.8
0.6
0.4
0.2
0
– 0.2
– 0.4
– 0.6
– 0.8
0
5
x
10
15
Fig. E 2.9.
Example 2.10. Using the functions for plotting x-y data given in Table 2.29 plot the
following functions.
(a) r2 = 5 cos 3t
(b) r2 = 5 cos 3t
0 ≤ t ≤ 2π
0 ≤ t ≤ 2π
x = r cos t, y = r sin t
(c) y1 = e–2x cos x 0 ≤ t ≤ 20
y2 = e2x
(d) y =
cos( x)
x
– 5 ≤ x ≤ 5π
(e) f = e–3t/5 cos t
(f) z = –
0 ≤ t ≤ 2π
1 2
x + 2xy + y2
3
|x| ≤ 7, |y| ≤ 7
Solution.
(a)
t = linspace(0, 2*pi, 200);
r = sqrt(abs(5*cos(3*t)));
polar(t, r)
79
MATLAB BASICS
90 2.5
2
120
60
1.5
1
150
30
0.5
0
180
330
210
300
240
270
Fig. E 2.10(a)
(b)
t = linspace(0, 2*pi, 200);
r = sqrt(abs(5*cos(3*t)));
x = r.*cos(t);
y = r.*sin(t);
fill(x, y, ‘k’),
axis(‘square’)
2
1.5
1
0.5
0
– 0.5
–1
– 1.5
–2
–3
–2
–1
0
1
Fig. E 2.10(b)
(c)
x = 1 : 0.1 : 20;
y1 = exp(– 2*x).*cos(x);
y2 = exp(2*x);
Ax = plotyy(x, y1, x, y2);
hy1 = get(Ax(1), ‘ylabel’);
hy2 = get(Ax(2), ‘ylabel’);
set(hy1, ‘string’, ‘exp(– 2x).cos(x)’)
set(hy2, ‘string’, ‘exp(– 2x)’);
2
3
80
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
x 10
2.5
0.06
2
0.04
1.5
0.02
1
0.5
0
– 0.02
0
2
4
6
8
10
12
14
16
18
0
20
Fig. E 2.10(c)
(d)
x = linspace(– 5*pi,5*pi,100);
y = cos(x)./x;
area(x, y);
xlabel(‘x (rad)’), ylabel(‘cos(x)/x’)
hold on
8
6
cos (x)/x
4
2
0
–2
–4
–6
–8
– 15
– 10
–5
0
x (rad)
Fig. E 2.10(d)
(e)
t = linspace(0, 2*pi, 200);
f = exp(– 0.6*t).*sin(t);
stem(t, f)
5
10
15
exp (– 2x)
exp (– 2x).cos (x)
17
0.08
81
MATLAB BASICS
0.6
0.5
0.4
0.3
0.2
0.1
0
– 0.1
0
1
2
3
4
5
6
7
Fig. E 2.10(e)
(f)
r = – 7 : 0.2 : 7;
[X, Y] = meshgrid(r, r);
Z = – 0.333*X.^2 + 2*X.*Y + Y.^2;
cs = contour(X, Y, Z);
label(cs)
100
6
50
4
2
0
0
0
–2
–4
50
50
–6
100
–6
–4
–2
0
2
4
6
Fig. E 2.10(f)
Example 2.11. Use the functions listed in Table 2.30 for plotting 3-D data for the following.
(a)
z = cos x cos y e
|x| ≤ = 7, | y | ≤ 7
–
x2 + y2
5
82
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(b) Discrete data plots with stems
x = t, y = t cos(t)
z = et/5 – 2
0 ≤ t ≤ 5π
(c) A cylinder generated by
r = sin(5πz) + 3
0≤z≤1
0 ≤ θ ≤ 2π
Solution.
(a)
u = – 7 : 0.2 : 7;
[X, Y] = meshgrid(u, u);
Z = cos (X).*cos (Y).*exp(– sqrt(X.^2 + Y.^2)/5);
surf(X, Y, Z)
1
0.5
0
– 0.5
–1
10
10
5
5
0
0
–5
– 10 – 10
–5
Fig. E 2.11(a)
(b)
t = linspace(0, 5*pi, 200);
x = t ; y = t.*cos(t);
z = exp(t/5) – 2;
stem3(x, y, z, ‘filled’);
xlabel(‘t’), ylabel (‘t cos(t)’), zlabel (‘e^t/5 – 1’)
83
MATLAB BASICS
25
20
e 5–1
15
10
t
5
0
–5
20
20
10
15
0
10
– 10
t cos (t)
– 20
5
0
t
Fig. E 2.11(b)
(c)
z =[0 : 0.2 : 1]’;
r = sin(5*pi*z)+3;
cylinder(r)
1
0.8
0.6
0.4
0.2
0
4
4
2
2
0
0
–2
–4 –4
–2
Fig. E 2.11(c)
Example 2.12. Obtain the plot of the points for 0 ≤ t ≤ 6π when the coordinates x,y,z are
given as a function of the parameter t as follows:
x = t sin(3t)
y = t cos(3t)
z = 0.8 t
84
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Solution.
% Line plots
>> t = [0:0.1:6*pi];
>> x = sqrt(t).*sin(3*t);
>> y = sqrt(t).*cos(3*t);
>> z = 0.8*t;
>> plot3(x, y, z, ‘k’, ‘linewidth’, 1)
>> grid on
>> xlabel (‘x’); ylabel (‘y’) ; zlabel (‘z’)
20
15
z
10
5
0
5
5
y
0
0
x
–5 –5
Fig. E 2.12.
Example 2.13. Obtain the mesh and surface plots for the function z =
domain – 2 ≤ x ≤ 6 and 2 ≤ y ≤ 8.
Solution.
% Mesh and surface plots
x = – 2 : 0.1 : 6;
>> y = 2 : 0.1 : 8;
>> [x, y] = meshgrid(x, y);
>> z = 2*x.*y.^2./(x.^2 + y.^2);
>> mesh(x, y, z)
>> xlabel(‘x’); ylabel(‘y’); zlabel(‘z’)
>> surf(x, y, z)
>> xlabel(‘x’); ylabel(‘y’); zlabel(‘z’)
2xy 2
over the
x2 + y2
85
MATLAB BASICS
10
5
0
–5
8
6
4
2 –2
0
6
4
2
Fig. E 2.13(a)
10
z
5
0
–5
8
6
6
4
2 –2
0
2
4
Fig. E 2.13(b)
Example 2.14. Plot the function z = 2– 1.5
≤ 4 and – 4 ≤ y ≤ 4 using Table 2.30
x 2 + y2
sin(x) cos (0.5y) over the domain – 4 ≤ x
(a) Mesh plot
(b) Surface plot
(c) Mesh curtain plot
(d) Mesh and contour plot
(e) Surface and contour plot
Solution.
(a)
% Mesh Plot
>> x = – 4 : 0.25 : 4;
>> y = – 4 : 0.25 : 4;
>> [x, y] = meshgrid(x, y);
>> z = 2.^(– 1.5*sqrt(x.^2 + y.^2)).*cos(0.5*y).*sin(x);
>> mesh(x, y, z)
>> xlabel(‘x’); y label(‘y’)
>> zlabel(‘z’)
86
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
0.4
0.2
z
0
– 0.2
– 0.4
4
2
4
y
2
0
0
–2
–4 –4
–2
x
Fig. E 2.14(a)
(b)
% Surface Plot
>> x = – 4 : 0.25 : 4;
>> y = – 4 : 0.25 : 4;
>> [x, y] = meshgrid(x, y);
>> z = 2.0.^(– 1.5*sqrt(x.^2 + y.^2)).*cos(0.5*y).*sin(x);
>> surf(x, y, z)
>> xlabel(‘x’); y label (‘y’)
>> zlabel(‘z’)
0.4
z
0.2
0
– 0.2
– 0.4
4
4
2
2
0
y
0
–2
–4–4
–2
Fig. E 2.14(b)
(c)
% Mesh Curtain Plot
>> x = – 4.0 : 0.25 : 4;
>> y = – 4.0 : 0.25 : 4;
>> [x, y] = meshgrid(x, y);
x
87
MATLAB BASICS
0.4
0.2
0
z
(d)
>> z = 2.0.^(– 1.5*sqrt(x.^2 + y.^2)).*cos(0.5*y).*sin(x);
>> mesh z(x, y, z)
>> xlabel(‘x’); ylabel(‘y’)
>> zlabel(‘z’)
% Mesh and Contour Plot
>> x = – 4.0 : 0.25 : 4;
>> y = – 4.0 : 0.25 : 4;
>> [x, y] = meshgrid (x, y);
>> z = 2.0.^(– 1.5*sqrt(x.^2 + y.^2)).*cos(0.5*y).*sin(x);
>> meshc(x, y, z)
>> xlabel(‘x’); ylabel(‘y’)
>> zlabel(‘z’)
– 0.2
– 0.4
4
4
2
2
0
0
–2
–4–4
–2
Fig. E 2.14(c)
0.5
0
– 0.5
5
5
y
0
0
–5 –5
Fig. E 2.14(d)
88
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(e)
% Surface and Contour Plot
>> x = – 4.0 : 0.25 : 4;
>> y = – 4.0 : 0.25 : 4;
>> [x, y] = meshgrid(x, y);
>> z = 2.0. ^ (– 1.5*sqrt (x. ^2 + y. ^2)).*cos (0.5*y).*sin(x);
>> surfc(x, y, z)
>> xlabel(‘x’); ylabel(‘y’)
>> zlabel(‘z’)
0.5
0
– 0.5
5
5
0
0
–5–5
Fig. E 2.14(e)
Example 2.15. Plot the function z = 2
− 1.5 x2+y 2
sin(x) cos (0.5y) over the domain – 4 ≤ x
≤ 4 and – 4 ≤ y ≤ 4 using Table 2.30.
(a) Surface plot with lighting
(b) Waterfall plot
(c) 3-D contour plot
(d) 2-D contour plot
Solution.
(a) % Surface Plot with lighting
>> x = – 4.0 : 0.25 : 4;
>> y = – 4.0 : 0.25 : 4;
>> [x, y] = meshgrid(x, y);
>> z = 2.0.^(– 1.5*sqrt(x.^2 + y.^2)).*cos(0.5*y).*sin(x);
>> surfl(x, y, z)
>> xlabel(‘x’); ylabel(‘y’)
>> zlabel(‘z’)
89
MATLAB BASICS
z
0.5
0
– 0.5
5
4
0
–5 –4
–2
2
0
Fig. E 2.15(a)
(b)
% Waterfall Plot
>> x = – 4.0 : 0.25 : 4;
>> y = – 4.0 : 0.25 : 4;
>> [x, y] = meshgrid(x, y);
>> z = 2.0.^(– 1.5*sqrt(x.^2 + y.^2)).*cos(0.5*y).*sin(x);
>> waterfall(x, y, z)
>> xlabel(‘x’); ylabel(‘y’)
>> zlabel(‘z’)
z
0.5
0
– 0.5
5
4
0
0
y
–5 –4
–2
Fig. E 2.15(b)
(c)
% 3-D Contour Plot
>> x = – 4.0 : 0.25 : 4;
>> y = – 4.0 : 0.25 : 4;
>> [x, y] = meshgrid(x, y);
>> z = 2.0.^(– 1.5*sqrt(x.^2 + y.^2)).*cos(0.5*y).*sin(x);
>> contour3(x, y, z, 15)
>> xlabel(‘x’) ; ylabel(‘y’)
>> zlabel(‘z’)
2
90
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
z
0.5
0
– 0.5
4
4
2
0
y
–2
–2
–4–4
2
0
x
Fig. E 2.15(c)
(d)
% 2-D Contour Plot
>> x = – 4.0 : 0.25 : 4;
>> y = – 4.0 : 0.25 : 4;
>> [x, y] = meshgrid(x, y);
>> z = 2.0.^(– 1.5*sqrt(x.^2 + y.^2)).*cos(0.5*y).*sin(x);
>> contour(x, y, z, 15)
>> xlabel(‘x’); ylabel(‘y’)
>> zlabel(‘z’)
4
y
2
0
–2
–4
–4
–3
–2
–1
0
1
2
3
4
Fig. E 2.15(d)
Example 2.16. Using the functions given in Table 2.29 for plotting x-y data, plot the
following functions:
(a) f(t) = t cos t
(b) x = e–2t, y = t
0 ≤ t ≤ 10π
0 ≤ t ≤ 2π
(c) x = t, y = e2t 0 ≤ t ≤ 2π
(d) x = et, y = 50 + et 0 ≤ t ≤ 2π
(e) r2 = 3 sin 7t
y = r sin t
0 ≤ t ≤ 2π
2
(f) r = 3 sin 4t
y = r sin t
0 ≤ t ≤ 2π
(g) y = t sin t
0 ≤ t ≤ 5π
91
MATLAB BASICS
Solution.
(a)
% Use of plot command
>> fplot(‘x.*cos(x)’, [0, 10*pi])
40
20
0
– 20
– 40
0
5
10
15
20
25
30
Fig. E 2.16(a)
(b)
% Semilog x command
>> t = linspace(0, 2 * pi, 200);
>> x = exp(– 2 * t); y = t;
>> semilog x (x, y),grid
8
6
4
2
0
10
–6
10
–4
10
–2
10
0
Fig. E 2.16(b)
(c)
% Semilog y command
t = linspace(0, 2 * pi, 200);
>> semilogy(t, exp(– 2 * t)), grid
10
10
10
10
0
–2
–4
–6
0
1
2
3
4
Fig. E 2.16(c)
5
6
7
92
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(d)
% Use of loglog command
>> t = linspace(0, 2 * pi, 200);
>> x = exp(t);
>> y = 50 + exp(t);
>> loglog(x, y), grid
3
10
2
10
1
10
0
1
10
10
10
2
3
10
Fig. E 2.16(d)
(e)
%Use of stairs command
>> t = linspace(0, 2*pi, 200);
>> r = sqrt(abs(3*sin(7*t)));
>> y = r.*sin(t);
>> stairs(t, y)
>> axis([0 pi 0 inf]);
1.5
1
0.5
0
0
0.5
1
1.5
Fig. E 2.16(e)
(f)
% Use of bar command
>> t = linspace(0, 2*pi,200);
>> r = sqrt(abs(3*sin(4*t)));
>> y = r.*sin(t);
>> bar(t, y)
>> axis([0 pi 0 inf]);
2
2.5
3
93
MATLAB BASICS
1.5
1
0.5
0
0
1
0.5
1.5
2
3
2.5
Fig. E 2.16(f)
(g)
%use of comet command
>> q = linspace(0, 5*pi, 200);
>> y = q.*sin(q);
>> comet(q, y)
10
5
0
–5
– 10
0
5
Fig. E 2.16(g)
Example 2.17. Consider the two matrices
2π
3
A=
5j 10 + 2
7j
B=
2π
j
– 15j
18
Using MATLAB , determine the following:
(a) A + B
(b)
(c)
AB
A2
(d)
(e)
AT
B–1
(f)
(g)
BTAT
A2 + B2 – AB
10
15
94
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Solution:
>> A = [3 2*pi ; 5j 10 + sqrt(2)*j];
>> B = [7j –15j ; 2*pi 18];
(a)
A+B
ans =
3.0000 + 7.0000i 6.2832 – 15.0000i
6.2832 + 5.0000i 28.0000 + 1.4142i
(b)
>> A * B
ans =
1.0e + 002 *
0.3948 + 0.2100i 1.1310 – 0.4500i
(c)
0.2783 + 0.0889i 2.5500 + 0.2546i
>> A ^ 2
ans =
9.0000 + 31.4159i 81.6814 + 8.8858i
(d)
– 7.0711 + 65.0000i 98.0000 +59.7002i
>> inv(A)
ans =
0.1597 + 0.1917i – 0.1150 – 0.1042i
(e)
0.0829 – 0.0916i 0.0549 + 0.0498i
>> B ^ – 1
ans =
0 – 0.0817i 0.0681
(f)
0 + 0.0285i 0.0318
>> inv(B) * inv(A)
ans =
0.0213 – 0.0193i – 0.0048 + 0.0128i
(g)
– 0.0028 + 0.0016i 0.0047 – 0.0017i
>> (A ^ 2 + B ^ 2) – (A * B)
ans =
1.0e + 002 *
– 0.7948 – 0.8383i 0.7358 – 2.1611i
0.7819 + 1.0010i 1.6700 – 0.6000i
Example 2.18. Find the inverse of the following matrices using MATLAB:
3 2 0
(a) 2 – 1 7
5 4 9
– 4 2 5
(b) 7 – 1 6
3 7
2
– 1 2 – 5
3
7
(c) 4
7 – 6 1
95
MATLAB BASICS
Solution.
>> clear % Clears the workspace
>> A = [3 2 0; 2 –1 7; 5 4 9]; % Spaces separate matrix columns – semicolons separate
matrix rows
>> B = [– 4 2 5; 7 – 1 6; 2 3 7]; % Spaces separate matrix columns – semicolons separate
matrix rows
>> C = [– 1 2 – 5; 4 3 7; 7 – 6 1]; % Spaces separate matrix columns – semicolons separate
matrix rows
>> inv(A); % Finds the inverse of the selected matrix
>> inv(B); % Finds the inverse of the selected matrix
>> inv(C) % Finds the inverse of the selected matrix
% Inverse of A
ans =
0.4805
– 0.2208
– 0.1688
0.2338 – 0.1818
– 0.3506
0.0260
0.2727
0.0909
% Inverse of B
ans =
– 0.1773
0.0071
– 0.2624 – 0.2695
0.1206
0.4184
0.1631
0.1135 – 0.0709
% Inverse of C
ans =
0.1667
0.1667
– 0.1667
0.1037
0.1074
0.1259 – 0.0481
0.0296 – 0.0407
Example 2.19. Determine the eigenvalues and eigenvectors of matrix A using MATLAB
4 – 1 5
1 3
(a) A = 2
6 – 7 9
Solution.
(a) A = [4 – 1 5 ; 2 1 3 ; 6 – 7 9]
A=
4
2
–1
1
5
3
6 –7 9
%The eigenvalues of A
format short e
3 5 7
(b) A = 2 4 8
5 6 10
96
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
eig(A)
ans =
1.0000e + 001
5.8579e – 001
3.4142e + 000
%The eigenvectors of A
[Q, d] = eig(A)
Q=
– 5.5709e – 001 – 8.2886e – 001 – 7.3925e – 001
– 3.7139e – 001 – 3.9659e – 002 – 6.7174e – 001
– 7.4278e – 001 5.5805e – 001 – 4.7739e – 002
d=
1.0000e + 001
0
0
0 5.8579e – 001
0
0
0 3.4142e + 000
(b)
A=
3
5
7
2
5
4
6
8
10
%The eigenvalues of A
format short e
eig(A)
ans =
1.7686e + 001
– 3.4295e – 001 + 1.0066e + 000i
– 3.4295e – 001 – 1.0066e + 000i
%The eigenvectors of A
[Q, d] = eig(A)
Q=
Column 1
5.0537e – 001
4.8932e – 001
7.1075e – 001
Column 2
– 2.0715e – 001 – 5.2772e – 001i
7.1769e – 001
– 3.3783e – 001 + 2.2223e – 001i
97
MATLAB BASICS
Column 3
– 2.0715e – 001 + 5.2772e – 001i
7.1769e – 001
– 3.3783e – 001 – 2.2223e - 001i
d=
Column 1
1.7686e + 001
0
0
Column 2
0
– 3.4295e – 001 + 1.0066e + 000i
0
Column 3
0
0
– 3.4295e – 001 – 1.0066e + 000i
Example 2.20. Determine the eigenvalues and eigenvectors of A and B using MATLAB.
3
1
A=
7
1
0
2
2
5
–1
2
–2
3
1
4
6
4
3
5
1
2 – 1 – 2
B=
3 2
1
0
4 1
Solution.
% MATLAB Program
% The matrix “a” = A * B
>> A = [ 3 0 2 1; 1 2 5 4; 7 – 1 2 6; 1 – 2 3 4 ];
>> B = [ 1 3 5 7; 2 – 1 – 2 4; 3 2 1 1; 4 1 0 6 ];
>> a = A * B
a=
13
14
17
29
36
35
15
32
6
39
44
83
22 15
>> eig (a)
12
26
Ans. =
98.5461
2.2964
– 1.3095
– 6.5329
7
4
1
6
98
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
The eigenvectors are:
>> [Q, d] = eig (a)
Q=
– 0.3263 – 0.2845
0.3908
0.3413
– 0.3619
0.7387 – 0.7816 – 0.9215
– 0.8168 – 0.6026
0.4769
– 0.3089
0.1016 – 0.0950
d=
98.5461
0
0
0
2.2964
0
0.0962
0.1586
0
0
– 1.3095
0
0
0
0
0
0
– 6.5329
Example 2.21. Solve the following set of equations using MATLAB.
(a)
x1 + 2x2 + 3x3 + 5x4 = 21
– 2x1 + 5x2 + 7x3 – 9x4 = 18
5x1 + 7x2 + 2x3 – 5x4 = 25
– x1 + 3x2 – 7x3 + 7x4 = 30
(b)
x1 + 2x2 + 3x3 + 4x4 = 8
2x1 – 2x2 – x3 – x4 = – 3
x1 – 3x2 + 4x3 – 4x4 = 8
2x1 + 2x2 – 3x3 + 4x4 = – 2
Solution. (a)
>> A = [1 2 3 5 ; – 2 5 7 – 9 ; 5 7 2 – 5 ; – 1 – 3 – 7 7];
>> B = [21 ; 18 ; 25 ; 30] ;
>> S = A\B
S=
– 8.9896
14.1285
– 5.4438
3.6128
% Therefore x1 = – 8.9896, x2 = 14.12.85, x3 = – 5.4438, x4 = 3.6128.
(b)
>> A = [1 2 3 4 ; 2 – 2 – 1 1 ; 1 – 3 4 – 4 ; 2 2 – 3 4];
>> B = [8 ; – 3 ; 8 ; – 2];
>> S = A\B
S=
2.0000
2.0000
2.0000
– 1.0000
% Therefore x1 = 2.0000, x2 = 2.0000, x3 = 2.0000, x4 = – 1.0000.
MATLAB BASICS
99
Example 2.22. Use diff command for symbolic differentiation of the following functions:
(a) S1 = ex8
(b) S2 = 3x3 ex5
(c) S3 = 5x3 – 7x2 + 3x + 6
Solution.
(a)
>> syms x
>> S1 = exp(x ^ 8);
>> diff (S1)
ans =
8*x^7*exp(x^8)
(b)
>> S2 = 3* x ^3*exp(x^5);
>> diff (S2)
ans =
9*x^2*exp(x^5) +15*x^7*exp(x^5)
(c)
>> S3 = 5*x^3 – 7*x^2 + 3*x + 6;
>> diff (S3)
ans =
15*x^2 – 14*x + 3
Example 2.23. Use MATLAB’s symbolic commands to find the values of the following
integrals.
(a)
∫
0.7
(b)
∫
π
0.2
0.2
(c)
| x|dx
(cos y + 7 y2 ) dy
x
(d) 7x5 – 6x4 + 11x3 + 4x2 + 8x + 9
(e) cos a
Solution.
(a)
>>syms x, y, a, b
>> S1 = abs(x)
>> int (S1, 0.2, 0.7)
ans =
9/40
100
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(b)
>> S2 = cos (y) + 7*y^2
>> int (S2, 0, pi)
ans =
7/3*pi^3
(c)
>> S3 = sqrt (x)
>> int (S3)
ans =
2/3*x^ (3/2)
>> int (S3, ‘a’, ‘b’)
ans =
2/3*b^ (3/2) – 2/3*a^ (3/2)
>> int (S3, 0.4, 0.7)
ans =
7/150*70^ (1/2) – 4/75*10^ (1/2)
(d)
>> S4 = 7*x^5 – 6*x^4 + 11*x^3 + 4*x^2 + 8 * x – 9
>> int (S4)
ans =
7/6*x^6 – 6/5*x^5 + 11/4*x^4 + 4/3*x^3 + 4*x^2 – 9*x
(e)
>> S5 = cos (a)
>> int (S5)
ans =
sin (a)
Example 2.24. Obtain the general solution of the following first order differential equations:
(a)
dy
= 5t – 6y
dt
(b)
d2 y
dy
+3
+y=0
dt 2
dt
(c)
ds
= Ax3
dt
(d)
ds
= Ax3
dA
Solution.
(a)
>> solve (‘Dy = 5*t – 6*y’)
ans =
5/6*t – 5/36 + exp (– 6*t)*C1
101
MATLAB BASICS
(b)
>> dsolve (‘D2y + 3*Dy + y = 0’)
ans =
C1*exp (1/2*(5^ (1/2) – 3)*t) + C2*exp (– 1/2*(5^ (1/2) +3)*t)
(c)
>> dsolve (‘Ds = A*x^3’, ‘x’)
ans =
1/4*A*x^4 + C1
(d)
>> dsolve (‘Ds = A*x^3’, ‘A’)
ans =
1/2*A^2*x^3 + C1
Example 2.25. Determine the solution of the following differential equations that satisfies the given initial conditions.
(a)
dy
= – 7x2
dx
y(1) = 0.7
(b)
dy
= 5x cos2 y
dx
y(0) = π/4
(c)
dy
= – y + e3x
dx
y(0) = 2
(d)
dy
+ 5y = 35
dt
y(0) = 4
Solution.
(a)
>> dsolve (‘Dy = – 7*x^2’, ‘y (1) = 0.7’)
ans =
– 7*x^2*t + 7* x ^2 + 7/10
(b)
>> dsolve (‘Dy = 5*x*cos (y) ^2’, ‘y (0) = pi/4’)
ans =
atan (5*t*x + 1)
(c)
>> dsolve (‘Dy = – y + exp (3*x)’, ‘y (0) = 2’)
ans =
exp (3*x) + exp (– t)*(– exp (3*x) +2)
(d)
>> dsolve (‘Dy + 5*y = 35’, ‘y (0) = 4’)
ans =
7 – 3*exp (– 5*t)
102
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Example 2.26. Given the differential equation
dx
d2 x
+7
+ 5x = 8 u (t)
2
dt
dt
t≥0
Using MATLAB program, find
(a) x(t) when all the initial conditions are zero
(b) x(t) when x (0) = 1 and x(0)
= 2.
Solution.
(a) x(t) when all the initial conditions are zero
>> x = dsolve (‘D2x = – 7*Dx – 5*x + 8’, ‘x(0) = 0’)
x=
(b)
8/5 + (– 8/5 – C2)*exp (1/2*(– 7 + 29^ (1/2))*t) + C2*exp (– 1/2*(7 + 29^ (1/2))*t)
x(t) when x(0) = 1 and = 2
>> x = dsolve (‘D2x = – 7*Dx – 5*x + 8’, ‘x (0) = 1’, ‘Dx (0) = 2’)
x=
8/5 + (– 3/10 – 1/290*29^ (1/2))*exp (1/2*(– 7+29^ (1/2))*t) – 1/290*(– 1 + 3*29^
(1/2))*29^ (1/2)*exp (– 1/2*(7 + 29^ (1/2))*t)
Example 2.27. Given the differential equation
dx
d2 x
+ 12
+ 15x = 35t ≥ 0
2
dt
dt
Using MATLAB program, find
(a) x(t) when all the initial conditions are zero
(b) x(t) when x (0) = 0 and x(0)
= 1.
Solution.
(a) x (t) when all the initial conditions are zero
>> x = dsolve (‘D2x = – 12*Dx – 15*x +35’, ‘x (0) = 0’)
x=
7/3+ (– 7/3 – C2)*exp ((– 6 + 21^ (1/2))*t) + C2*exp (– (6 + 21^ (1/2))*t)
(b)
C
x (t) when x (0) = 0 and x (0) = 1.
>> x = dsolve (‘D2x = – 12*Dx – 15*x + 35’, ‘x (0) = 0’, ‘Dx (0) = 1’)
x=
7/3 + (– 7/6 – 13/42*21^ (1/2))*exp ((– 6 + 21^ (1/2))*t) – 1/126*(– 39 + 7*21^ (1/2))*21 ^
(1/2)*exp (– (6 + 21^ (1/2))*t)
Example 2.28. Find the inverse of the following matrix using MATLAB.
s
A = 2
3
0
s − 3
0
1
2
103
MATLAB BASICS
Solution.
>> A = [s 2 0 ; 2 s – 3 ; 3 0 1] ;
>> inv (A)
ans =
[s/(s^2 – 22),
– 2/(s^2 – 22),
[– 11/(s^2-22),
[– 3*s/(s^2-22),
– 6/(s^2 – 22)]
s/(s^2 – 22), 3*s/(s^2 – 22)]
6/(s^2 – 22), (s^2 – 4)/(s^2 – 22)]
Example 2.29. Expand the following function F(s) into partial fractions using MATLAB.
Determine the inverse Laplace transform of F(s).
F(s) =
1
s + 5s 3 + 7s 2
4
The MATLAB program for determining the partial-fraction expansion is given below:
Solution.
>> b = [0 0 0 0 1];
>> a = [1 5 7 0 0];
>> [r, p, k] = residue (b, a)
r=
0.0510 – 0.0648i
0.0510 + 0.0648i
– 0.1020
0.1429
p=
– 2.5000 + 0.8660i
– 2.5000 – 0.8660i
0
0
k=[]
% From the above MATLAB output, we have the following expression:
F(s) =
r3
r1
r2
r4
+
+
+
s − p1
s − p2
s − p4
s − p3
F(s) =
(0.0510 + 0.0648i)
0.0510 − 0.0648i
+
s − (− 2.5000 − 0.8660i)
s − (2.5000 + 0.8660i)
+
0.1429
− 0.1020
+
s−0
s−0
% Note that the row vector k is zero implies that there is no constant term in this example problem.
% The MATLAB program for determining the inverse Laplace transform of F(s) is given
below:
104
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> syms s
>> f = 1/(s^4 + 5*s^3 + 7*s^2);
>> ilaplace (f)
ans =
1/7*t – 5/49 + 5/49*exp (–)*cos (1/2*3^ (1/2)*t) +11/147*exp (– 5/2*t)*3^
(1/2)*sin(1/2*3^(1/2)*t)
Example 2.30. Expand the following function F(s) into partial fractions using MATLAB.
Determine the inverse Laplace transform of F(s).
F(s) =
5s 2 + 3s + 6
s 4 + 3s 3 + 7s 2 + 12
Solution.
The MATLAB program for determining the partial-fraction expansion is given below:
>> b = [0 0 5 3 6];
>> a = [1 3 7 9 12];
>> [r, p, k] = residue(b, a)
r=
– 0.5357 – 1.0394i
– 0.5357 + 1.0394i
0.5357 – 0.1856i
0.5357 + 0.1856i
p=
– 1.5000 + 1.3229i
– 1.5000 – 1.3229i
– 0.0000 + 1.7321i
– 0.0000 – 1.7321i
k=[]
% From the above MATLAB output, we have the following expression:
F(s) =
r1
r2
r4
r3
+
+
+
s − p1
s − p2
s − p4
s − p3
F(s) =
(− 0.5357 + 1.0394i)
− 0.5357 − 1.0394i
+
s − (− 1.500 + 1.3229 i)
s − (− 1.5000 − 1.3229i)
+
0.5357 − 0.1856i
0.5357 − 0.1856i
+
s
− (− 0 − 1.7321i)
s − (− 0 + 1.7321i)
% Note that the row vector k is zero implies that there is no constant term in this example problem.
% The MATLAB program for determining the inverse Laplace transform of F(s) is given
below:
105
MATLAB BASICS
>> syms s
>> f = (5*s^2 + 3*s +6)/(s^4 + 3*s^3 + 7*s^2 + 9*s +12);
>> ilaplace(f)
ans =
11/14*exp(– 3/2*t)*7^(1/2)*sin(1/2*7^(1/2)*t) – 15/14*exp
(– 3/2*t)*cos(1/2*7^(1/2)*t) + 3/14*3^(1/2)*sin(3^(1/2)*t)+15/14*cos(3^(1/2)*t)
Example 2.31. For the following function F(s):
F(s) =
s 4 + 3s 3 + 5s 2 + 7s + 25
s 4 + 5s 3 + 20s 2 + 40s + 45
Using MATLAB, find the partial-fraction expansion of F(s). Also, find the inverse Laplace
transformation of F(s).
Solution.
F(s) =
s4 + 3s3 + 5s2 + 7s + 25
s4 + 5 s3 + 20 s2 + 40 s + 45
The partial-fraction expansion of F(s) using MATLAB program is given as follows:
num = [ 1 3 5 7 25];
den = [1 5 20 40 45];
[r, p, k] = residue(num, den)
r=
– 1.3849 + 1.2313i
– 1.3849 – 1.2313i
0.3849 – 0.4702i
0.3849 + 0.4702i
p=
– 0.8554 + 3.0054i
– 0.8554 – 3.0054i
– 1.6446 + 1.3799i
– 1.6446 – 1.3799i
k=
1
From the MATLAB output, the partial-fraction expansion of F(s) can be written as
follows:
F(s) =
r3
r1
r2
r4
+k
+
+
+
(s − p1 ) (s − p2 ) (s − p3 ) (s − p4 )
F(s) =
(− 1.3849 − j1.2313)
(− 1.3849 + j1.2313)
+
(s + 0.8554 + j3.005)
(s + 0.8554 − j3.005)
+
(0.3849 − j0.4702)
(0.3849 + j0.4702)
+
+1
(s + 1.6446 − j1.3799)
(s + 1.6446 + j1.3779)
106
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Example 2.32. Obtain the partial-fraction expansion of the following function using
MATLAB:
F(s) =
8(s + 1)(s + 3)
(s + 2)(s + 4)(s + 6) 2
Solution.
F(s) =
8(s + 1)(s + 3)
(8 s + 8)(s + 3)
2 =
2
(s + 2)(s + 4)(s + 6)
(s + 6 s + 8)(s2 + 12 s + 36)
The partial fraction expansion of F(s) using MATLAB program is given as follows:
EDU>> num = conv([8 8], [1 3]);
EDU>> den = conv([1 6 8], [1 12 36]);
EDU>> [r, p, k] = residue(num, den)
r=
3.2500
15.0000
– 3.0000
– 0.2500
p=
– 6.0000
– 6.0000
– 4.0000
– 2.0000
k=[]
From the above MATLAB result, we have the following expansion:
F(s) =
r3
r1
r2
r4
+k
+
+
+
(s − p1 ) (s − p2 ) (s − p3 ) (s − p4 )
F(s) =
3.25
15
−3
− 0.25
+
+
+
+0
(s + 6)
(s − 15)
(s + 3)
(s + 0.25)
It should be noted here that the row vector k is zero, because the degree of the numerator
is lower than that of the denominator.
F(s) = 3.25e–6t + 15e15t – 3e–3t – 0.25e–0.25t.
Example 2.33. Find the Laplace transform of the following function using MATLAB.
(a) f(t) = 7t3 cos (5t + 60°)
(b) f(t) = – 7te– 5t
(c) f(t) = – 3 cos 5t
(d) f(t) = t sin 7t
(e) f(t) = 5 e–2t cos 5t
(f) f(t) = 3 sin(5t + 45º)
107
MATLAB BASICS
(g) f(t) = 5 e–3t cos(t – 45º)
Solution. % MATLAB Program
>> syms t % tell MATLAB that “t” is a symbol.
>> f = 7 * t^3*cos(5*t + (pi/3)); % define the function.
>> laplace(f)
ans =
– 84/(s^2 + 25)^3*s^2 + 21/(s^2 + 25)^2 + 336*(1/2*s – 5/2*3^(1/2))/(s^2 + 25)^4*s^3
– 168* (1/2*s – 5/2*3^(1/2))/(s^2 + 25)^3*s
>> pretty(laplace(f)) % the pretty function prints symbolic output
% in a format that resembles typeset mathematics.
2
1/2 3
s
21
(1/2 s – 5/2 3 ) s
-84 ---------- + ---------- + 336 --------------------2
3
(s + 25)
2
2
(s + 25)
2
4
(s + 25)
1/2
(1/2 s – 5/2 3 ) s
– 168 -------------------2
3
(b)
(s + 25)
>>syms t x
>>f = – 7*t*exp(– 5*t);
>> laplace(f, x)
ans =
– 7/(x + 5)^2
(c)
>>syms t x
>>f = – 3*cos(5*t);
>> laplace(f, x)
ans =
(d)
– 3*x/(x^2 + 25)
>>syms t x
>>f = t*sin(7*t);
>> laplace(f, x)
ans =
1/(x^2 + 49)*sin(2*atan(7/x))
(e)
>>syms t x
>>f = 5*exp(– 2*t)*cos(5*t);
>> laplace(f, x)
ans =
5*(x + 2)/((x + 2)^2 + 25)
108
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(f)
>>syms t x
>>f = 3*sin(5*t + (pi/4));
>> laplace(f, x)
ans =
3*(1/2*x*2^(1/2) + 5/2*2^(1/2))/(x^2 + 25)
(g)
>>syms t x
>>f = 5*exp(– 3*t)*cos(t – (pi/4));
>> laplace(f, x)
ans =
5*(1/2*(x + 3)*2^(1/2)+1/2*2^(1/2))/((x + 3)^2 + 1)
Example 2.34. Generate partial-fraction expansion of the following function
F(s) =
10 5 (s + 7)(s + 13)
s(s + 25)(s + 55)(s 2 + 7s + 75)(s 2 + 7s + 45)
Solution:
Generate the partial fraction expansion of the following function:
numg = poly[– 7 – 13];
numg = poly([– 7 – 13]);
deng = poly([0 – 25 – 55 roots([1 7 75])’ roots([1 7 45])’]);
[numg, deng] = zp2tf(numg’, deng’, 1e5);
Gtf = (numg, deng);
Gtf = tf(numg,deng);
G = zpk(Gtf);
[r, p, k] = residue(numg,deng)
r=
1.0e – 017 *
0.0000
– 0.0014
0.0254
– 0.1871
0.1621
– 0.0001
0.0000
0.0011
p=
1.0e + 006 *
4.6406
1.4250
0.3029
109
MATLAB BASICS
0.0336
0.0027
0.0001
0.0000
0
k=[]
Example 2.35. Determine the inverse Laplace transform of the following functions using
MATLAB.
(a) F(s) =
s
s(s + 2)(s + 6)
(b) F(s) =
(c) F(s) =
3s + 1
(s + 2s + 9)
(d) F(s) =
2
1
s 2 (s + 5)
s − 25
s(s + 3s + 20)
2
Solution:
(a)
>> syms s
>> f = s/(s*((s + 2)*(s + 6)));
>> ilaplace(f)
ans =
(b)
1/2*exp(– 4*t)*sinh(2*t)
>> syms s
>> f = 1/((s^2)*(s + 5));
>> ilaplace(f)
ans =
1/3*t – 2/9*exp(– 3/2*t)*sinh(3/2*t)
(c)
>>syms s
>> f = (3*s + 1)/(s^2 + 2*s + 9);
>> ilaplace(f)
ans =
3*exp(– t)*cos(2*2^(1/2)*t) – 1/2*2^(1/2)*exp(– t)*sin(2*2^(1/2)*t)
(d) >>syms s
>> f = (s – 25)/(s*(s^2 + 3*s + 25));
>> ilaplace(f)
ans =
5/4*exp(– 3/2*t)*cos(1/2*71^(1/2)*t)+23/284*71^(1/2)*exp(– 3/2*t)*sin
(1/2*71^(1/2)*t) – 5/4
Example 2.36. Find the inverse Laplace transform of the following function using
MATLAB.
G(s) =
(s 2 + 9s + 7)(s + 7)
(s + 2)(s + 3)(s 2 + 12s + 150)
.
110
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Solution:
% MATLAB Program
>> syms s % tell MATLAB that “s” is a symbol.
>>G = (s^2 + 9*s +7)*(s + 7)/[(s + 2)*(s + 3)*(s^2 + 12*s + 150)]; % define the function.
>>pretty(G) % the pretty function prints symbolic output
% in a format that resembles typeset mathematics.
(s + 9 s + 7) (s + 7)
-------------------------------------------(s + 2) (s + 3) (s + 12 s + 150)
>> g = ilaplace(G); % inverse Laplace transform
>>pretty(g)
44
2915
1/2
– 7/26 exp(– 2 t) + --- exp(– 3 t) + ------ exp(– 6 t) cos(114 t)
123
889
1/2
+ ------- exp(– 6 t) 114
20254
3198
1/2
sin(114
t)
Example 2.37. Generate the transfer function using MATLAB.
G(s) =
3(s + 9)(s + 21)(s + 57)
s(s + 30)(s 2 + 5s + 35)(s 2 + 28s + 42)
Using
(a) the ratio of factors
(b) the ratio of polynomials.
Solution.
% MATLAB Program:
‘a. The ratio of factors’
>>Gzpk = zpk([– 9 – 21 – 57] , [0 – 30 roots([1 5 35]) ‘roots([1 28 42])’],3)
% zpk is used to create zero-pole-gain models or to convert TF or
% SS models to zero-pole-gain form.
‘b. The ratio of polynomials’
>> Gp = tf(Gzpk) % generate the transfer function
% Computer response:
ans =
(a) The ratio of factors
Zero/pole/gain:
3 (s + 9) (s + 21) (s + 57)
---------------------------------------------------------s (s + 30) (s + 26.41) (s + 1.59) (s^2 + 5s + 35)
ans =
111
MATLAB BASICS
(b) The ratio of polynomials
Transfer function:
3 s^3 + 261 s^2 + 5697 s + 32319
-----------------------------------------------------------------------------s^6 + 63 s^5 + 1207 s^4 + 7700 s^3 + 37170 s^2 + 44100 s
Example 2.38. Generate the transfer function using MATLAB.
G(s) =
s 4 + 20s 3 + 27s 2 + 17s + 35
.
s 5 + 8s 4 + 9s 3 + 20s 2 + 29s + 32
Using
(a) the ratio of factors
(b) the ratio of polynomials.
Solution.
% MATLAB Program:
% a. the ratio of factors
>>Gtf = tf([1 20 27 17 35] , [1 8 9 20 29 32]) % generate the
% transfer function
% Computer response:
Transfer function:
s^4 + 20 s^3 + 27 s^2 + 17 s + 35
--------------------------------------------s^4 + 8 s^3 + 9 s^2 + 20 s + 29
% b. the ratio of polynomials
>> Gzpk = zpk(Gtf) % zpk is used to create zero-pole-gain models
% or to convert TF or SS models to zero-pole-gain form.
% Computer response:
Zero/pole/gain:
(s +18.59) (s + 1.623) (s^2 – 0.214s + 1.16)
------------------------------------------------------------(s + 7.042) (s + 1.417) (s^2 – 0.4593s + 2.906)
2.23 SUMMARY
In this chapter, the MATLAB environment which is an interactive environment for
numeric computation, data analysis, and graphics was presented. Arithmetic operations, display
formats, elementary built-in functions, arrays, scalars, vectors or matrices, operations with
arrays including dot product, array multiplication, array division, inverse and transpose of a
matrix, determinants, element by element operations, eigenvalues and eigenvectors, random
number generating functions, polynomials, system of linear equation, script files, programming
in MATLAB, the commands used for printing information and generating 2-D and 3-D plots,
input/output in MATLAB was presented with illustrative examples. MATLAB's functions for
symbolic mathematics were introduced. These functions are useful in performing symbolic
112
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
operations and developing closed-form expressions for solutions to linear algebraic equations,
ordinary differential equations and systems of equations. Symbolic mathematics for determining
analytical expressions for the derivative and integral of an expression was also presented.
REFERENCES
Chapman, S.J., MATLAB Programming for Engineers, 2nd ed., Brooks/Cole, Thomson
Learning, Pacific Grove, CA, 2002.
Etter, D.M., Engineering Problem Solving with MATLAB, Prentice-Hall, Englewood
Cliffs, NJ, 1993.
Gilat, Amos., MATLAB-An Introduction with Applications, 2nd ed., Wiley, New York,
2005.
Hanselman, D., and Littlefield, B.R., Mastering MATLAB 6, Prentice Hall, Upper
Saddle River, New Jersey, NJ, 2001.
Herniter, M.E., Programming in MATLAB, Brooks/Cole, pacific Grove, CA, 2001.
Magrab, E.B., An Engineers Guide to MATLAB, Prentice Hall, Upper Saddle River,
New Jersey, NJ, 2001.
Marchand, P., and Holland, O.T., Graphics and GUIs with MATLAB, 3rd ed, CRC
Press, Boca Raton, FL, 2003.
Moler, C., The Student Edition of MATLAB for MS-DOS Personal Computers with 3-1/
2” Disks, MATLAB Curriculum Series, The MathWorks, Inc., 2002.
Palm, W.J. III., Introduction to MATLAB 7 for Engineers, McGraw Hill, New York, NY,
2005.
Pratap, Rudra, Getting Started with MATLAB- A Quick Introduction for Scientists and
Engineers, Oxford University Press, New York, NY, 2002.
Sigman, K., and Davis, T.A., MATLAB Primer, 6th ed, Chapman& Hall/CRCPress,
Boca Raton, FL, 2002.
The MathWorks, Inc., MATLAB: Application Program Interface Reference Version 6,
The MathWorks, Inc., Natick, 2000.
The MathWorks, Inc., MATLAB: Creating Graphical User Interfaces, Version 1, The
MathWorks, Inc., Natick, 2000.
The MathWorks, Inc., MATLAB: Function Reference, The MathWorks, Inc., Natick,
2000.
The MathWorks, Inc., MATLAB: Release Notes for Release 12, The MathWorks, Inc.,
Natick, 2000.
The MathWorks, Inc., MATLAB: Symbolic Math Toolbox User’s Guide, Version 2, The
MathWorks, Inc., Natick, 1993-1997.
The MathWorks, Inc., MATLAB: Using MATLAB Graphics, Version 6, The MathWorks,
Inc., Natick, 2000.
113
MATLAB BASICS
PROBLEMS
1. Compute the following quantity using MATLAB in the Command Window:
17 5 − 1 57 log (e3 )
+
10
+ ln(e4 ) + 11
2
2
15 − 13
π 121
2. Compute the following quantity using MATLAB in the Command Window:
B=
tan x + sin 2 x
+ log |x5 – x2 | + cosh x – 2 tanh x.
cos x
for x = 5π/6.
3. Compute the following quantity using MATLAB in the Command Window:
x = a+
14 b
log10 c
ab (a + b)
+ ca +
+ ln(2) +
3
c
log10 (a + b + c)
c | ab|
e
+ 2 sinh a – 3 tanh b
for a = 1, b = 2 and c = 1.8.
4. Use MATLAB to create
(a) a row and column vectors that has the elements: 11, – 3, e7.8, ln(59), tan(π/3), 5
log10(26).
(b) a row vector with 20 equally spaced elements in which the first element is 5.
(c) a column vector with 15 equally spaced elements in which the first element is – 2.
5. Enter the following matrix A in MATLAB and create:
1
9
A = 17
25
33
2 3 4 5 6 7 8
10 11 12 13 14 15 16
18 19 20 21 22 23 24
23 27 28 29 30 31 32
34 35 36 37 38 39 40
(a) a 4 × 5 matrix B from the 1st, 3rd, and the 5th rows, and the 1st, 2nd, 4th, and
8th columns of the matrix A.
(b) a 16 elements-row vector C from the elements of the 5th row, and the 4th and 6th
columns of the matrix A.
(
6. Given the function y = x
2 + 0.02
+ ex
)
1.8
ln x. Determine the value of y for the fol-
lowing values of x : 2, 3, 8, 10, – 1, – 3, – 5, – 6.2. Solve the problem using MATLAB by
first creating a vector x, and creating a vector y, using element-by-element calculations.
7. Define a and b as scalars, a = 0.75, and b = 11.3, and x, y and z as the vectors, x = 2, 5,
1, 9, y = 0.2, 1.1, 1.8, 2 and z = – 3, 2, 5, 4. Use these variables to calculate A using
element-by-element computations for the vectors with MATLAB.
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
A=
x1.1 y−2 z5
(a + b)b / 3
z y
+
x 2
+a
za
8. Enter the following three matrices in MATLAB and show that
1 2 3
12 −5 4
7 13 4
A = −8 5 7 B = 7 11 6 C = −2 8 −5
−8 4 6
1 8 13
9 −6 11
(a) A + B = B + A
(b) A + (B + C) = (A + B)C
(c) 7(A + C) = 7(A) + 7(C)
(d) A * (B + C) = A * B + A * C
9. Consider the function
H(s) =
where
n( s)
d( s)
n(s) = s4 + 6s3 + 5s2 + 4s + 3
d(s) = s5 + 7s4 + 6s3 + 5s2 + 4s + 7
(a) Find n (– 10), n (– 5), n (– 3), and n (– 1)
(b) Find d (– 10), d (– 5), d (– 3), and d (– 1)
(c) Find H(– 10), H(– 5), H(– 3), and H(– 1)
10. Consider the polynomials
p1(s) = s3 + 5s2 + 3s + 10
p2(s) = s4 + 7s3 + 5s2 + 8s + 15
p3(s) = s5 + 15s4 + 10s3 + 6s2 + 3s + 9
Determine
(a) p1(2), p2(2), and p3(3)
(b) p1(s) p2(s) p3(s)
(c) p1(s) p2(s)/p3(s)
11. The following polynomials are given:
p1(x) = x5 + 2x4 – 3x3 + 7x2 – 8x + 7
p2(x) = x4 + 3x3 – 5x2 + 9x + 11
p3(x) = x3 – 2x2 – 3x + 9
p4(x) = x2 – 5x + 13
p5(x) = x + 5
Use MATLAB functions with polynomial coefficient vectors to evaluate the expressions at x = 2.
12. Determine the roots of the following polynomials:
(a)
p1(x) = x7 + 8x6 + 5x5 + 4x4 + 3x3 + 2x2 + x + 1
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MATLAB BASICS
(b)
p2(x) = x6 – 7x6 + 7x5 + 15x4 – 10x3 – 8x2 + 7x + 15
(c)
(d)
p3(x) = x5 – 13x4 + 10x3 + 12x2 + 8x – 15
p4(x) = x4 + 7x3 + 12x2 – 25x + 8
(e)
(f)
p5(x) = x3 + 15x2 – 23x + 105
p6(x) = x2 – 18x + 23
(g)
p7(x) = x + 7
13. Consider the two matrices
1 0 2
A = 2 5 4
−1 8 7
and
7 8 2
B = 3 5 9
−1 3 1
Using MATLAB, determine the following:
(a) A + B
(b) AB
(c) A2
(d) AT
(e) B–1
(f) BTAT
(g) A2 + B2 – AB
(h) determinant of A, determinant of B and determinant of AB.
14. Use MATLAB to define the following matrices:
2 1
A = 0 5
7 4
5 3
B=
−2 −4
2 3
C = −5 −2
0 3
D = [1
2]
Compute matrices and determinants if they exist.
(a) (ACT)–1
(b) |B|
(c) |ACT|
(d) (CTA)–1
15. Consider the two matrices
2π
3
A=
5 j 10 + 2
j
7 j −15 j
B=
2 π 18
Using MATLAB, determine the following:
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(a) A + B
(b) AB
(c) A2
(d) AT
(e) B–1
(f) BTAT
(g) A2 + B2 – AB
16. Consider the two matrices
1 0 1
A = 2 3 4
−1 6 7
7 4 2
and B = 3 5 6
−1 2 1
Using MATLAB, determine the following:
(a) A + B
(b) AB
(c) A2
(d) AT
(e) B–1
(f) BTAT
(g) A2 + B2 – AB
(h) det A, det B, and det of AB.
17. Find the inverse of the following Matrices:
1
3 2
−
(a) A = 1 5 4
5 7 −9
1 6 3
(b) B = −4 −5 7
8 4 2
−1 −2 5
(c) C = −4 7 2
7 −8 −1
18. Find the inverse of the following matrices using MATLAB.
3 2 0
(a) 2 −1 7
5 4 9
−4 2 5
(b) 7 −1 6
2 3 7
−1 2 −5
3 7
(c) 4
7 −6
1
1
3 2
−
(d) 1 2 4
5 7 −8
1 2 3
(e) −4 −5 7
8 4 1
−1 −2 5
5 6
(f) −4
7
8 −1
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MATLAB BASICS
19. Determine the eigenvalues and eigenvectors of the following matrices using MATLAB.
1 −2
A=
,
1 5
1 5
B=
−2 7
4 6 2
20. If A = 5 6 7
10 5 8
Use MATLAB to determine the following:
(a) the three eigenvalues of A
(b) the eigenvectors of A
(c) Show that AQ = Qd, where Q is the matrix containing the eigenvectors as columns and d is the matrix containing the corresponding eigenvalues on the main
diagonal and zeros else where.
21. Determine eigenvalues and eigenvector of A using MATLAB.
8 3
(b) A =
− 3 4
0.5 − 0.8
(a) A =
1.0
0.75
22. Determine the eigenvalues and eigenvectors of the following matrices using MATLAB.
1 − 2
(a) A =
3
1
1 5
(b) A =
− 2 4
4 −1 5
(c) A = 2
1 3
6 −7 9
3 5 7
(d) A = 2 4 8
5 6 10
0
3
1
2
(e) A =
7 − 1
1 − 2
1
4
6
4
2
5
2
3
5
1 3
2 − 1 − 2
(f) A =
3
2
1
4
1
0
7
4
1
6
23. Determine the eigenvalues and eigenvectors of A * B using MATLAB.
3
1
A=
7
1
−1
2
2
7
−1
8
−2
3
1
4
6
4
1
2
B=
3
4
2
5
−1
−2
2
5
1
−3
7
4
1
6
24. Determine the eigenvalues and eigenvectors of the following matrices using MATLAB.
1 −2
(a) A =
1 3
1 5
(b) A =
− 2 4
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
4 − 1 5
(c) A = 2
1 3
6 − 7 9
3 5 7
(d) A = 2 4 8
5 6 10
0
3
1
2
(e) A =
7 − 1
1 − 2
3
5 7
1
2 − 1 − 2 4
(f) A =
3
2
1 1
1
0 6
4
1
5 4
2 6
3 4
2
25. Determine the eigenvalues and eigenvectors of A and B using MATLAB
1
B = 8
5
4 5 − 3
3
(a) A = − 1 2
2 5
7
2
3
9
6
3 − 1
26. Determine the eigenvalues and eigenvectors of A = a*b using MATLAB.
6
0
a=
1
2
−3
4
4
2
3
8
2
1
1
6
5
4
0
4
b=
1
2
1
2
5
6
5
4
−3
6
3
− 1
2
7
27. Determine the values of x, y, and z for the following set of linear algebraic equations:
x2 – 3x3 = – 7
2x1 + 3x2 – x3 = 9
4x1 + 5x2 – 2x3 = 15
28. Determine the values of x, y, and z for the following set of linear algebraic equations:
(a) 2x + y – 3z = 11
4x – 2y + 3z = 8
– 2x + 2y – z = – 6
(b) 2x – y = 10
– x + 2y – z = 0
– y + z = – 50
29. Solve the following set of equations using MATLAB.
(a) 2x1 + x2 + x3 – x4 = 12
x1 + 5x2 – 5x3 + 6x4 = 35
– 7x1 + 3x2 – 7x3 – 5x4 = 7
x1 – 5x2 + 2x3 + 7x4 = 21
(b) x1 – x2 + 3x3 + 5x4 = 7
2x1 + x2 – x3 + x4 = 6
– x1 – x2 – 2x3 + 2x4 = 5
x1 + x2 – x3 + 5x4 = 4
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MATLAB BASICS
30. Solve the following set of equations using MATLAB.
(a) 2x1 + x2 + x3 – x4 = 10
x1 + 5x2 – 5x3 + 6x4 = 25
– 7x1 + 3x2 – 7x3 – 5x4 = 5
x1 – 5x2 + 2x3 + 7x4 = 11
(b) x1 – x2 + 3x3 + 5x4 = 5
2x1 + x2 – x3 + x4 = 4
– x1 – x2 + 2x3 + 2x4 = 3
x1 + x2 – x3 + 5x4 = 1
31. Solve the following set of equations using MATLAB.
(a) x1 + 2x2 + 3x3 + 5x4 = 21
– 2x1 + 5x2 + 7x3 – 9x4 = 17
5x1 + 7x2 + 2x3 – 5x4 = 23
– x1 – 3x2 – 7x3 + 7x4 = 26
(b) x1 + 2x2 + 3x3 + 4x4 =9
2x1 – 2x2 – x3 + x4 = – 5
x1 – 3x2 + 4x3 – 4x4 = 7
2x1 + 2x2 – 3x3 + 4x4 = – 6
32. Generate a plot of
y(x) = e–0.7x sin ωx
where ω = 15 rad/s, and 0 ≤ x ≤ 15. Use the colon notation to generate the x vector in
increments of 0.1.
33. Plot the following functions using MATLAB.
(a) r2 = 5 cos 3 t
0 ≤ t ≤ 2π
(b) r2 = 5 cos 3 t
0 ≤ t ≤ 2π
x = r cos t, y = r sin t
0 ≤ x ≤ 20
(c) y1 = e–2x cos x
2x
y2 = e
– 5π ≤ x ≤ 5π
(d) y = cos (x) /x
–3t/5
(e) f = e
0 ≤ t ≤ 2π
cos t
2
(f) z = – (1/3) x + 2xy + y
2
| x | ≤ 7, | y | ≤ 7
34. Use MATLAB for plotting 3.D data for the following functions:
(a) z = cos x cos y e
−
x2 + y2
5
| x | ≤ 7, | y | ≤ 7
(b) Discrete data plots with stems
x = t, y = t cos (t)
z = et/5 – 2
0 ≤ x ≤ 5π
(c) An ellipsoid of radii rx = 1, ry = 2.5 and rz = 0.7 centered at the origin
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(d) A cylinder generated by
r = sin (5 π z) + 3
0≤z≤ 1
0 ≤ θ ≤ 2π
35. Obtain the plot of the points for 0 ≤ t ≤ 6π when the coordinates x, y, and z are given as
a function of the parameter t as follows:
x = t sin (3t)
y = t cos (3t)
z = 0.8t
36. Obtain the mesh and surface plots for the function z =
2xy2
x2 + y2
over the domain
– 2 ≤ x ≤ 6 and 2 ≤ y ≤ 8.
37. Plot the function z = 2−1.5
– 4 ≤ y ≤ 4.
(a) Mesh plot
x2 + y2
sin (x) cos (0.5y) over the domain – 4 ≤ x ≤ 4 and
(b) Surface plot
(c) Mesh curtain plot
(d) Mesh and contour plot
(e) Surface and contour plot
(f) Surface plot with lighting
(g) Waterfall plot
(h) 3-D contour plot
(i) 2-D contour plot
38. Plot the function y = |x| cos (x) for – 200 ≤ x ≤ 200.
39. Plot the following functions on the same plot for 0 ≤ x ≤ 2π using the plot function:
(a) sin2 (x)
(b) cos2 x
(c) cos (x)
40. (a) Generate an overlay plot for plotting three lines
y1 = sin t
y2 = t
y3 = t –
Use
t3
t5
t7
+
+
3!
5!
7!
0 ≤ t ≤ 2π
(i) the plot command
(ii) the hold command
(iii) the line command
(b) Use the functions for plotting x-y data given in Table 6.5(b) for plotting the
following functions.
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MATLAB BASICS
0 ≤ t ≤ 10π
(i) f(t) = t cos t
t
(ii) x = e
y = 100 + e3t
0 ≤ t ≤ 2π
41. (a) Plot the parametric space curve of
x(t) = t
y(t) = t2
z(t) = t3
(b) z =
0 ≤ t ≤ 3.0
−7
1 + x2 + y2
| x | ≤ 10, | y | ≤ 10
42. (a) Plot the parametric space curve of
x (t) = t
y (t) = t2
z (t) = t3
0 ≤ t ≤ 3.0
2
(b) z = – 7 / (1 + x + y2)
| x | ≤ 10, | y | ≤ 10
43. Perform the following symbolic operations using MATLAB. Consider the given symbolic expressions have been defined.
S1 = ‘2/(x – 5)’;
S2 = ‘x ^ 5 + 9 * x – 15’ ;
S3 = ‘(x ^ 3 + 2 * x + 9) * (x * x – 5)’ ;
(a) S1S2/S3 (b) S1/S2S3 (c) S1/(S2)2 (d) S1S3/S2 (e) (S2)2/(S1S3)
44. Solve the following equations using symbolic mathematics.
(a) x2 + 9 = 0
(b) x2 + 5x – 8 = 0
(c) x3 + 11x2 – 7x + 8 = 0
(d) x4 + 11x3 + 7x2 – 19x + 28 = 0
(e) x7 – 8x5 + 7x4 + 5x3 – 8x + 9 = 0
45. Determine the values of x, y, and z for the following set of linear algebraic equations:
2x + y – 3z = 11
4x – 2y + 3z = 8
– 2x + 2y – z = – 6
46. Determine the values of x, y, and z for the following set of linear algebraic equations:
2x – y = 10
– x + 2y – z = 0
– y + z = – 50
47. Determine the solutions of the following first-order ordinary differential equations
using MATLAB's symbolic mathematics.
(a) y′ = 8x2 + 5 with initial condition y(2) = 0.5.
(b) y′ = 5x sin2 (y) with initial condition y(0) = π/5.
(c) y′ = 7x cos2 (y) with initial condition y(0) = 2.
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(d) y′ = – 5x + y with initial condition y(0) = 3.
(e) y′ = 3y + e–5x with initial condition y(0) = 2.
48. (a) Given the differential equation
d2 x
dt
2
+7
dx
+ 5x = 8 u(t)
dt
t≥0
Using MATLAB program, find
(i) x(t) when all the initial conditions are zero
(ii) x(t) when x(0) = 1 and x (0) = 3.
(b) Given the differential equation
d2 x
dt
2
+ 12
dx
+ 15x = 35
dt
t≥0
Using MATLAB program, find
(i) x(t) when all the initial conditions are zero
C
(ii) x(t) when x(0) = 0 and x (0) = 1.
(iii) For the following differential equation, use MATLAB to find x(t) when x(t)
when x(0) = – 1 and x (0) = 1
d2 x
dt
2
+8
dx
– 4x = 18 u(t)
dt
(c) For the following differential equation, use MATLAB to find x(t) when x(t) when
x(0) = – 1 and x (0) = 1
d2 x
dt
2
+ 15
dx
+ 8x = – 9 u(t)
dt
(d) For the following differential equation, use MATLAB to find x(t) when x(t) when
x(0) = – 1 and x (0) = 1
d2 x
dt
2
+ 19
dx
+ 9x = – 3 u(t)
dt
49. For the following differential equations, use MATLAB to find x(t) when (a) all the
initial conditions are zero, (b) x(t) when x(0) = 1 and x (0) = – 1.
(a)
(b)
(c)
d2 x
dt
2
d2 x
dt
2
d2 x
dt
2
+ 10
dx
+ 5x = 11
dt
–7
dx
– 3x = 5
dt
+3
dx
+ 7x = – 15
dt
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MATLAB BASICS
d2 x
(d)
dt
+7
2
dx
+ 7x = 26
dt
50. Obtain the first and second derivatives of the following functions using MATLAB’s
symbolic mathematics.
(a) F(x) = x5 – 8x4 + 5x3 – 7x2 + 11x – 9
(b) F(x) = (x3 + 3x – 8)(x2 + 21)
(c) F(x) = (3x3 – 8x2 + 5x + 9)/(x + 2)
(d) F(x) = (x5 – 3x4 + 5x3 + 8x2 – 13)2
(e) F(x) = (x2 + 8x – 11)/(x7 – 7x6 + 5x3 + 9x – 17)
51. Determine the values of the following integrals using MATLAB’s symbolic functions.
(a)
∫ 5 x7 − lx5 + 3 x3 − 8 x2 + 7
(b)
∫
(c)
∫
x cos x
x2 / 3 sin2 2 x
1.8
(d)
∫ 0.2
(e)
∫ −1
x2 sin x dx
− 0.2
| x| dx
52. Given the differential equation
d2 x
dt
2
+3
dx
+ x = 98
dt
t≥0
Using MATLAB program, find
(a) x(t) when all the initial conditions are zero
(b) x(t) when x(0) = 0 and x (0) = 2
53. Determine the inverse of the following matrix using MATLAB.
2
0
3s
A = 7 s – s – 5
3
0 – 3s
54. Expand the following function F(s) into partial fractions with MATLAB:
F(s) =
5s3 + 7 s2 + 8s + 30
s +15 s3 + 62 s2 + 85 s + 25
4
55. Determine the Laplace transform of the following time functions using MATLAB.
(a) f (t) = u (t + 9)
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(b) f (t) = e5t
(c) f (t) = (5t + 7)
(d) f (t) = 5u (t) + 8e7t – 12e–8t
(e) f (t) = e–t + 9t3 – 7t–2 + 8
(f) f (t) = 7t4 + 5t2 – e–7t
(g) f (t) = 9 u t + 5e–3t
56. Determine the inverse Laplace transform of the following rotational function using
MATLAB.
F(s) =
7
7
=
(s + 2)(s + 3)
s + 5s + 6
2
57. Determine the inverse transform of the following function having complex poles
F(s) =
15
(s3 + 5s2 + 11s + 10)
58. Determine the inverse Laplace transform of the following functions using MATLAB:
(a) F(s) =
s
s(s + 2)(s + 3)(s + 5)
(b) F(s) =
(c) F(s) =
5s + 9
(s + 8 s + 5)
(d) F(s) =
3
1
s ( s + 7)
2
s − 28
s( s + 9s + 33)
2
.
Chapter
3
MATLAB TUTORIAL
3.1 INTRODUCTION
MATLAB has an excellent collection of commands and functions that are useful for solving control engineering problems. The problems presented in this chapter are basic linear
control systems and are normally presented in introductory control courses. The application of
MATLAB to the analysis and design of control systems is presented in this chapter with a
number of illustrative examples. The MATLAB computational approach to the transient response analysis, steps response, impulse response, ramp response, and response to the simple
inputs are presented. Plotting root loci, Bode diagrams, polar plots, Nyquist plot, Nichols plot,
and state space method are obtained using MATLAB.
3.2 TRANSIENT RESPONSE ANALYSIS
Transient responses include the step response, impulse response, and ramp response.
They are often used to investigate the time-domain characteristics of control systems. Transient response characteristics including the rise time, peak time, maximum overshoot, settling
time, and steady state error can be obtained from the step response.
When the numerator and denominator of a closed-loop transfer function are known, the
commands step (num, den), step (num, den, t) in MATLAB can be used to generate plots of
unit- step responses. Here, t is the user specified time.
3.3 RESPONSE TO INITIAL CONDITION
3.3.1 Case 1: State Space Approach
Consider a system defined in state-space given by
x& = Ax
(3.1)
x(0) = xo
Assuming that there is no external input acting on the system, the response x(t) knowing
the initial condition x(0) and that x is an n-vector, is obtained as follows:
Taking Laplace transform of both sides of Eq. (3.1), we obtain
s x(s) – x(0) = AX (s)
Equation (3.2) can be rearranged as
s x (s) = AX (s) + x(0)
(3.2)
(3.3)
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Taking inverse Laplace transform of Eq. (3.3), we get
x& = A x + x(0) δ (t)
(3.4)
Defining z& = x, Eq. (3.4) can be written as
..
z = A z& + x(0) δ (t)
(3.5)
Integrating Eq. (3.5), we obtain
Where
z& = A z + x (0) 1(t) = A z + B u
B = x (0)
and
u = 1(t)
(3.6)
Noting that z = x and x& (t) = z(t), we have
x = z& = A z + B u
(3.7)
The response to initial condition is obtained by solving Eqns. (3.6) and (3.7).
The corresponding MATLAB command used to obtain the response curves are given as
follows:
[x, z, t] = step (A, B, A, B);
x1 = [1 0 0 …0] * x′;
x2 = [1 0 0 …0] * x′;
M
xn = [0 0 0 …1] * x′;
plot (t, x1, x2,…, t, xn)
3.3.2 Case 2: State Space Approach
Consider the system defined in state space is by
x& = A x
x (0) = x0
y=Cx
(3.8)
(3.9)
Where x is an n vector and y is an m vector.
By defining
z& = x
(3.10)
z& = A z + x (0) 1(t) = Az + B u
B = x (0) and u = 1(t)
(3.11)
(3.12)
We obtain
Where
Since x = z, Eq. (3.9) becomes
y = C z&
From Eqns. (3.11) and (3.13) , we obtain
(3.13)
y = C (Az + Bu ) = CAz + CB u
(3.14)
The response of the system is obtained from the Eqns. (3.11) and (3.14) to a given initial
condition
The following MATLAB commands may be used to obtain the response curves:
[y, z, t] = step (A, B, C*A , C*B);
127
MATLAB TUTORIAL
y1 = [1 0 0 …0] * y′;
y2 = [0 1 0 …0] * y′;
(3.15)
M
ym = [0 0 0 …1] * y′;
plot ( t, y1 , t, y2 ,........, t, ym)
where the output curves are y1 , y2,…, ym verses t.
3.4 SECOND ORDER SYSTEMS
The standard form of a second- order system is defined by
G(s) =
ωn2
S2 + 2ξωn S + ω2n
(3.16)
Where ξ is the damping ratio of the system and ωn is the undamped natural frequency of the
system.
The dynamic behavior of the second order system is then described in terms of two parameters ξ and ωn.
If 0 < ξ < 1, the closed loop poles are complex conjugates and lie in the left-half s plane.
The system is called underdamped, and the transient response is oscillatory, If ξ = 0, the transient response does not die out. If ξ = 1, the system is called critically damped. Overdamped
system correspond to ξ = 1.
Given ωn and ξ, then the MATLAB command
printsys (num, den)
or
printsys(num, den, s)
prints the num/den as a ratio of polynomials in s .
The unit – step response of the transfer – function system using MATLAB is obtained
with the use of step – response commands with left – hand arguments.
c = step (num, den, t)
or
[y, x, t] = step (num, den, t)
3.5 ROOT LOCUS PLOTS
Locus is defined as a set of all points satisfying a set of conditions. The term root refers to
the roots of the characteristic equation, which are the poles of the closed-loop transfer function.
These poles define the time response of the system and hence the performance and stability of
the system. Hence, root-locus defines a graph of the poles of the closed-loop transfer function as
the system parameter, such as the gain is varied.
Evan’s root locus method, or simply root-locus method, gives all closed-loop poles graphically, using the knowledge provided by the open-loop poles and open-loop zeros. A root-locus
plot is composed of as many individual loci as there are poles. Individual loci are referred to as
branches of the root locus.
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
The poles of a transfer function can be shown graphically in the s-plane by means of a
pole-zero map. The root locus method is an analytical method for displaying the location of the
poles of the closed-loop transfer function
G
1 + GH
as a function of the gain factor K of the open-loop transfer function GH. The method is called
the root locus analysis. The root locus analysis has the advantage that this method requires
only the location of the poles and zeros of GH known and the factorization of the characteristic
polynomial is not required. The method gives accurate time-domain response as well as frequency response information.
The root-locus method is based on the fact that the values of s that make the transfer
function around the loop equal – 1 must satisfy the characteristic equation of the system. The
locus of roots of the characteristic equation of the closed-loop system as the gain is varied from
zero to infinity gives the root-loci plot. The root locus plot indicates the contributions of each
open-loop pole or zero to the locations of the closed-loop poles.
The root locus method allows the prediction of the effects on the location of the closedloop poles of varying the gain value or adding open-loop poles and/or open-loop zeros. Fig. 3.1
shows a canonical feedback control system whose closed-loop transfer function is given by
G
C
=
1 + GH
R
R
(3.17)
+
G
C
–
H
Fig. 3.1.
If the open-loop transfer function GH is represented by
GH =
KN
D
(3.18)
Where D and N are finite polynomials in the complex variable s, and K is the open-loop gain
factor, then the closed-loop transfer function can be written as
GD
G
C
=
=
D + DN
1 + KN / D
R
(3.19)
The roots of the characteristic equation gives the closed-loop poles. That is
D + KN = 0
(3.20)
As the open-loop gain factor K is varied, the location of these roots in the s-plane changes.
A locus of these roots plotted in the s-plane as a function of K is known as a root locus. We
observe from Eq. (3.4) that when K = 0, the roots of the polynomial D gives the poles of the openloop transfer function GH. On the other hand, as K becomes very large, then the roots will
become those of the polynomial N (the open-loop zeros). Hence, the loci of the closed-loop poles
originate from the open-loop poles and terminate at the open-loop zeros ad K varies from zero to
infinity.
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MATLAB TUTORIAL
Consider the system equation
1+
K (s + z1 )( s + z2 ) ... (s + zn )
=0
(s + p1 )(s + p2 ) ... (s + pn )
(3.21)
Equation (3.17) can be written as
1+K
num
=0
den
(3.22)
Where num is the numerator of the polynomial and den is the denominator polynomial, and K
is the gain (K > 0). The vector K contains all the gain values for which the closed loop poles are
to be computed.
The root loci is plotted by using the MATLAB command
rlocus (num, den)
The gain vector K is supplied by the user.
The matrix r and gain vector K are obtained by the following MATLAB commands:
[r, k] = rlocus (num, den)
[r, k] = rlocus (num, den, k)
[r, k] = rlocus (A, B, C, D)
[r, k] = rlocus (A, B, C, D, K)
(3.23)
[r, k] = rlocus (sys)
In Eqns. (3.23), r has length K rows and length [den – 1] columns containing the complex
root locations.
For plotting the root loci, the MATLAB command plot (r, ‘ ‘) is used.
The following MATLAB command are used for plotting the root loci
with mark ‘0’ or ‘x’:
r = rlocus (num, den)
plot (r, ‘0’) or plot (r, ‘x’)
MATLAB provides its own set of gain values used to compute a root locus plot. It also
uses the automatic axis scaling features of the plot command.
3.6 BODE DIAGRAMS
Polar plot is a plot of the magnitude |G(jω)H(jω)| and phase angle |G(jω)H(jω)| in polar
coordinates for various values of frequencies ranging from 0 to ∞ then – ∞ to 0.
Bode plot is a plot of magnitude |G(jω)H(jω)| in decibels versus logω and phase angle
|G(jω)H(jω) | versus log ω in rectangular coordinates.
Magnitude versus phase angle plot or gain-phase plot is a plot of magnitude |G(jω)H(jω)|
in decibels versus phase angle |G(jω)H(jω) | in rectangular coordinates with frequency as varying parameter.
In practice the frequency-functions of the system are so complex and long that the characteristic of the system cannot be determined at the desired frequency just only by inspection of
the system frequency function. Hence the frequency functions of systems are plotted in graphical forms, which indicate the system characteristics. Any curve giving information regarding
the gain or phase shift of the frequency function is known as the frequency response of the
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
system. In polar-plots the amplitude of G(jω) is plotted as the distance from origin while the
phase-angle is plotted as angular displacement from right hand horizontal axis on the polar
graph. These plots are simple to construct and easily provide the information regarding the
magnitude and phase-angle of G(jω) at any desired frequency as compared to rectangular-plots
because polar-plot contains the ready information of both the parameters, amplitude and phase
angle.
A transfer function G(s) may be represented in the frequency domain as a sinusoidal
transfer function by substituting jω for s in the expression for G(s). The resulting form G(jω) is
a complex function of the single variable ω. Thus it can be plotted in 2-dimensions with ω as a
parameter and written in the following equivalent form:
Polar form: G(jω) = |G(jω)| |φ(ω)|
Euler form: G(jω) = |G(jω)| (cos φ(ω) + j sin φ(ω))
Here |G(jω)| is the magnitude of complex function and φ(ω) is the phase angle = arg p(ω).
Bode diagrams are rectangular plots. Bode diagram are also known as logarithmic plot
and consist of two graphs: the first one is a plot of the logarithmic of the magnitude of a sinusoidal
transfer function, the second one is a plot of the phase angle. Both these graphs are plotted
against the frequency on a logarithmic scale. Bode analysis is similar to Nyquist analysis in
that here also the graphical representation of the open-loop frequency response function G(ω)H(ω),
is employed. Bode-plot consists of two graphs: the magnitude of G(ω)H(ω), and the phase angle
of G(jω)H(jω), both plotted as a function of frequency ω. Logarithmic scales are used for the
frequency axes and for |G(jω)H(jω)|. Bode plots illustrate the relative stability of a system.
The following frequency-domain specifications are used:
If the Nyquist-plot does not cross the critical point (– 1 + j0), the system is found to be
stable. The frequency (ω1) at which the magnitude |G(jω)H(jω)| equals to one is called the gaincross over frequency. If the plot at ω1 is rotated through an angle φ in clockwise direction, the
point ω = ω1 and critical point (– 1 + j0) are coincident and this indicates that the closed-loop
system is marginally stable. Any further rotation leads to instability. The angle φ is known as
phase-margin. Intersection of the plot with negative real axis corresponds to frequency ω = ω2.
The phase-angle at ω = ω2 is
|G(jω2)H(jω2)| = – 180º
(3.24)
Hence, the closeness of Nyquist plot to critical point (– 1 + j0), decides the relative stability of the systems. The closer the Nyquist plot to the critical point (– 1 + j0) the system tends
towards instability.
Gain margin is a factor by which the gain of a stable system is allowed to increase before
the system reaches instability. Gain margin is defined as the magnitude of reciprocal of the
open-loop transfer function evaluated at the frequency ω2 at which the phase angle is – 180º.
GM =
1
|G ( jω2 )H ( jω2 )|
(3.25)
where ω2 is the phase cross over frequency.
Phase margin of a stable system is the amount of additional phase lag required to bring
the system to point of instability. It is defined as
PM = [180 + |G(jω1)H(jω1)]
where |G(jω1)H(jω1)| = 1 and ω1 is called the gain-crossover frequency.
For a stable system both GM and PM should be positive.
(3.26)
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MATLAB TUTORIAL
Bode plot consists of two graphs in rectangular coordinates. These are (i) the magnitude
of G(jω)H(jω) in db versus log10ω, (ii) the phase angle of G(jω)H(jω) versus log10ω.
The general open-loop transfer function of a feedback control system may be represented
by
G(s)H(s) =
k[(1 + ST1 )(1 + ST2 ) ...] ω2n
s [(1 + STa )(1 + sTb ) ...] ( s2 + 2ξωn s + ω2n )
N
(3.27)
Here the highest power of s in the numerator is lower than that of the denominator. Now
substituting s = jω, we get
G(jω)H(jω) =
k[(1 + jωT1 )(1 + jωT2 ) ...] ω2n
( jω) [(1 + jωTa )(1 + jωTb ) ...] (ω2n − ω2 2ξjωn ω)
N
(3.28)
Hence, the magnitude is
|G(jω)H(jω)| =
k|1 + jωT1 ||1 + jωT2 |... ωn2
|( j ω)N ||1 + j ωTa |...| ω2n − ω2 + 2ξj ωn ω|
(3.29)
and the phase is
2ξω ω
|G(jω)H(jω)| = tan −1 ωT1 + tan −1 ωT2 − 90 N − tan −1 ωTa ... tan −1 2 n 2
(ω n − ω )
(3.30)
The magnitude can be represented in decibel form as
20 log10|G(jω)H(jω)| = 20 log10 k + 20 log10|1 + jωT1|
+ 20 log10|1 + jωT2| – 20 N log10|jω| – 20 log10|1
+ jωTa | …… 20 log
ω2n − ω2 2ξωn ω
+
ω2n
ω2n
(3.31)
Equation (3.31) shows that the frequency function of an open-loop transfer function
G(jω)H(jω) has factors as follows:
(a) Constant gain factor k
(b) Poles at origin due to factor
1
with N = 0, 1, 2, …
( jω)N
(c) Zeros on real axis due to (1 + jωT)
(d) Poles on real axis due to
1
(1 + jωT )
(e) Complex conjugate poles due to
(3.32)
ωn
(ω2n − ω2 ) + j 2ξωn ω)
Bode plots of continuous-time frequency response functions can be constructed by summing the magnitude and phase angle contributions of each pole and zero. The asymptotic ap-
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
proximations of these plots are often sufficient. The asymptotic Bode plots for G(jω)H(jω) are
obtained by adding the graphs of each of the terms (a) to (e). For example, the Bode plot for gain
term k is obtained from the expressions
k|db = 20 log10 k
|k = 0
(3.33)
dB magnitude
Thus, the magnitude and phase angle for the term K is independent of the frequency.
Hence, the Bode plot for this term is a horizontal straight line, as shown in Fig. 3.2.
20 log10 |KB|
log10 ω
(a) Gain plot
Phase angle
180
KB > 0
0º
–180
log10 ω
KB < 0
(b) Phase plot
Fig. 3.2
From the frequency response-function for a pole of order N at origin or
1
, the Bode
( jω) N
plots are inclined straight lines. Here the magnitude in decibels (dB) = 20 log10
N log10ω and the phase-angle
1
( jω) N
1
( jω) N
= – 90 N. The Bode plots are shown in Fig. 3.3.
= – 20
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MATLAB TUTORIAL
d B m a gn itu d e
60
40
20
0
N=1
N=2
P hase an g le
N=3
0
N=1
– 90
N=2
– 18 0
N=3
– 27 0
log 10 ω
Fig. 3.3
Similarly, the Bode plots for terms (1 + jωT) and 1/(1 + jωT) are shown in Figs. 3.4 and 3.5.
P h ase a n g le
M ag n itud e
90º
20
ω = (1 /T )
0
45º
ω = (1 /T )
0º
0 .2
log 1 0 ω
1
log 1 0
Fig. 3.4
log10 (1/T)
log10 ω
0
1
0
0.2
–20
1
–45º
–90º
Fig. 3.5
5
log10 ω
5
ω
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Note:
1. Exact plots are different from asymptotic plots. The two plots match each other at
lower and higher values of frequencies with respect to the corner frequency ω = 1/T.
2. The initial slope of the Bode plot for type N system is – 20 N db/decade and intersection with x-axis occurs at ω = K1/N.
The Bode plot for the quadratic term
ω2n
ω2n − ω2 + 2ξ jωn ω
can be drawn from the expressions for magnitude in dB and phase angle.
ω2n
ωn2 − ω2 + 2ξ j ωn ω
|G(jω)H(jω)|dB = 20 log10
= 20 log10
1
2
2
ω2
ω
1 − 2 + 2ξ
ωn
ωn
2
= – 20 log10
ω2
ω
1
−
+ 2ξ
2
ωn
ωn
Using the asymptotic approximations, when
2
(3.34)
ω2
ω
ω
<< 1, the terms 2 and
can be
ωn
ωn
ωn
neglected in comparison with 1.
Hence
and when
20 log10
ω2n
(ωn2 − ω2 )2 + 2ξ jωn ω
ω
>> 1, 20 log10
ωn
= – 20 log10
= – 20 log10
1 =0
ω2n
(ω2n − ω2 )2 + 2ξ j ωn ω
ω2
ω
2 = – 40 log10
ωn
ωn
(3.35)
Magnitude in dB
Hence, the plot has a slope of – 40 db/decade. The Bode plot is shown in Fig. 3.6.
10
Asymptotes
ξ=0.1
ξ=0.3
0
–10
0.1
ξ=1
ξ=0.7
1
2
(a)
4
ω/ωn
6
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MATLAB TUTORIAL
ξ=1.0
0º
ξ=0.1
–90º
ξ=0.3
–180º
0.1
1
(b)
Fig. 3.6
The MATLAB command ‘‘bode’’ obtains the magnitudes and phase angles of the frequency response of continuous – time, linear, time – invariant systems.
The MATLAB bode commands commonly used are:
Bode(num, den)
bode(num, den, W)
bode(A, B, C, D)
(3.36)
bode(A, B, C, D, W)
bode(sys)
where w is the frequency vector.
MATLAB bode commands with left hand arguments commonly used are:
[mag, phase, w] = bode (num, den)
[mag, phase, w] = bode (num, den, w)
[mag, phase, w] = bode (A, B, C, D)
[mag, phase, w] = bode (A, B, C,D, w)
(3.37)
[mag, phase, w] = bode (A, B, C, D, iu, w)
[mag, phase, w] = bode (sys)
The MATLAB commands given in Eq. (3.37) returns the frequency response of the system in matrices mag, phase, and w. The plot is not drawn on the screen. The matrices mag,
phase provide the magnitudes and phase angles of frequency response of the system, computed
at the specified frequency points.
The magnitude may be converted into decibles using the MATLAB statement
magdB = 20 * log 10 (mag)
In MATLAB , the following command
or
logspace (d1, d2 )
logspace(d1, d2, n). logspace(d1, d2)
(3.38)
(3.39)
(3.40)
are used to specify the frequency range that will generate a vector of 50 points logarithmically
equally speed between decades 10d1 and 10d2
The MATLAB command
w = logspace (– 1, 2)
may be used to generate 50 points between 0.1 and 100 rad/sec.
(3.41)
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Similarly, the MATLAB command
logspace (d1, d2, n)
(3.42)
generates n points logarthimatically equally spaced between 10d1 and 10d2 where by the n points
include both the endpoints.
3.7 NYQUIST PLOTS
Nyquist analysis is a frequency response method. It is basically a graphical procedure for
determining the absolute and relative stability of closed-loop control systems. Stability information obtained directly from a graph of the open-loop frequency response function GH(ω).
It should be noted here that the Routh-Hurwitz stability method can only be used for
determining absolute stability and is applicable to systems whose characteristic equation is a
finite polynomial in s.
Nyquist method is also useful for obtaining information about transfer functions of systems from the experimental frequency response data. The Nyquist analysis can be used for
systems with time delays without the need for approximations and gives exact results about
both absolute and relative stability of the system.
3.7.1 Polar Plots
The polar plot is a plot of the magnitude G(jω) and its phase angle as the frequency is
taken over its full range from zero to infinity. Consider a continuous system transfer function
G(s) represented in the frequency domain as a sinusoidal transfer function. The resulting G(jω)
is a complex function of single variable ω. It may be plotted either from the magnitude and
phase obtained directly or by plotting the real and imaginary parts of G(jω) as ω varies, with the
polar values of magnitude and phase angle taken from the Cartesian plot. A positive phase
angle denotes a phase advance through the system while a negative value a phase lag.
We can write the following equivalent forms:
Polar form G(jω) = |G(jω)|| φ(ω)|
Euler form G(jω) = |G(jω)|(cos φ(ω) + j sin φ(ω))
(3.43)
where |G(jω)| is the magnitude of the complex function G(jω), and φ(jω) is its phase angle, or
arg G(jω) |G(jω)| cos φ(ω) is the real part, and |G(jω)| sin φ(ω) is the imaginary part of G(jω).
Hence
Rectangular or complex form G(jω) = Re G(jω) + j ImG(jω)
(3.44)
A polar plot of G(jω) is a plot of Im G(jω) versus ReG(jω) in the finite portion of the G(jω)plane for – ∞ < ω < ∞. The magnitude and phase angle of G(jω) are graphed with ω varying from
– ∞ to + ∞. At singular points of G(jω), that is, poles on the jω-axis, |G(jω)| → ∞. The polar plot
of the transfer function of a time invariant, linear system exhibits conjugate symmetry. Hence,
the plot of – ∞ < ω < 0 will be a mirror image about the horizontal axis of the plot for 0 ≤ ω < ∞.
Also, the polar plot for [G(jω) + a] is identical to the polar plot for G(jω) with the origin of
coordinates shifted to the point – a = – (Re a + j Im a) where a is any complex constant.
For sketching a polar-plot of an open-loop transfer function G(s), the following criteria is
used:
(a) From the transfer function G(s), the frequency function G(jω) is first obtained by
letting
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MATLAB TUTORIAL
G(jω) =
K (1 + jω Ta )(1 + jω Tb ) ...... (1 + j ωTm )
( jω) N (1 + jω T1 ) ...... (1 + jω Tn )
(3.45)
From Eq. (7.29), the magnitude and phase angle at ω → 0 is obtained.
(b) At higher frequencies, i.e., as ω → ∝, the magnitude and phase angle are then
obtained.
3.7.2 Nyquist Plot
We have seen that the points in the s-plane are mapped into points of the G(s)-plane by
the function G. In polar-plots s-plane degenerates into a line and G(jω) is represented in a
G(jω)-plane with ω as a parameter.
In a more general sense, only a specific locus of points in the s-plane is mapped into the
G(s)-plane.
G(s) is plotted with s = (σ + jω) using two graphs. The first graph is a graph of jω versus
σ called the s-plane, the same set of coordinates as those used for plotting pole-zero maps. The
second graph is the imaginary part of G(s) versus real part of G(s) called the G(s)-plane.
The Nyquist stability plot is an extension of the polar plot. The locus of points in the splane mapped into G(s)-plane in Nyquist plots is called Nyquist path. This is a closed contour in
s-plane, which completely encloses the entire right half of the s-plane. In order that the Nyquist
path should not pass through any poles of G(s), small semicircles along the imaginary axis or at
the origin of G(s) are required in the path if G(s) has poles on the jω-axis or at the origin.
Nyquist plot is a mapping of the entire Nyquist path into the P(s)=plane.
The Nyquist plot of a sinusoidal transfer function G(jω) is a plot of the magnitude of G(jω)
versus the phase angle of G(jω) on polar coordinates as ω varied from 0 to ∞. Therefore, the
polar plot is the locus of vector |G(jω)| |G(jω)| as ω varied from 0 to ∞.
It should be noted here that each point on the polar plot of G(jω) represents the terminal
point of a vector at a particular value of ω. The real and imaginary components are given by the
projections of G(jω) on the real and imaginary axes. Nyquist plot shows the frequency response
characteristics of a system over the full frequency range in a single plot.
On the other hand, the plot will not indicate the contributions of each individual factor of
the open-loop transfer function.
Nyquist plots are also used in the frequency – response representation of linear, time
invariant, continuous – time feedback control systems. Nyquist plots are polar plots.
The MATLAB command
nyquist (num, den)
Draw the Nyquist plot of the transfer function
G(s) =
num(s)
den(s)
(3.46)
(3.47)
where num and den contain the polynomial coefficients in descending powers of s. The other
MATLAB command uses for drawing Nyquist plots are:
nyquist (num, den, w)
nyquist (A, B, C, D)
nyquist (A, B, C, D, w)
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
nyquist (A, B, C, D, iu, w)
(3.48)
nyquist (sys)
where w is the frequency vector.
The MATLAB command involving the user – specified vector w in Eq. (3.32) computes
the frequency response at the specified frequency points.
The following MATLAB commands
[re, im, w] = nyquist (num, den)
[re, im, w] = nyquist (num, den, w)
[re, im, w] = nyquist (A, B, C, D)
[re, im, w] = nyquist (A, B, C, D, w)
(3.49)
[re, im, w] = nyquist (A, B, C, D, iu, w)
[re, im, w] = nyquist (sys)
are used to obtain the frequency response of the system in the matrices Re, im, and w. The plot
is not drawn on the screen. The matrices Re and im contain the real and imaginary parts of the
frequency response of the system, computed at the frequency points specified in the vector w.
3.8 NICHOLS CHART
Nichols chart analysis is a modification of the Nyquist and Bode methods. It is a frequency response method. The Nichols chart is a useful technique for determining the stability
and the closed-loop frequency response of a feedback system. Nichols chart is basically a transformation of the M- and N-circles on the polar plot into non-circular M and N counters on a db
magnitude versus phase angle plot in rectangular coordinates.
If the open-loop frequency response function of a continuous-time system is represented
by GH(ω), then GH(ω) is plotted on a Nichols chart is called a Nichols chart plot of GH(ω).
The Nichols chart has two advantages over the polar plot. They are:
(a) since |GH(ω)| is plotted on a logarithmic scale, a much wider range of magnitude can
be graphed
(b) the graph of GH(ω) is essentially a summation of the individual magnitude and phase
angle contributions of its poles and zeros.
The stability is obtained from a plot of the open-loop gain versus phase characteristics.
3.8.1 db Magnitude-Phase Angle Plots
The polar form of a continuous time open-loop frequency response function is
GH(ω) = |GH(ω)|∠arg GH(ω)
(3.50)
The db magnitude-phase angel plot of GH(ω) is a graph of |GH(ω)| in decibels versus
GH(ω) in degrees on rectangular coordinates with ω as a parameter. The db magnitude-phase
angle plot for a continuous-time system is constructed by finding 20 log10|GH(ω)| and arg
GH(ω) in degrees for a sufficient number of values ω or ωT and plotting them in rectangular
coordinates with log magnitude as the ordinate and the phase angle as the abscissa.
Nichols chart analysis is another frequency response method and is a modification of the
Nyquist and Bode plot methods. The Nichols chart is essentially a transformation of the M and
N circles on the polar plot into noncircular M and N contours on a db magnitude versus phase
angle plot in rectangular coordinates. If G(jω)H(jω) represents the open-loop frequency response
MATLAB TUTORIAL
139
function of either continuous time or discrete-time system, then G(jω)H(jω) is plotted on a Nichols
chart is known as Nichols chart plot of G(jω)H(jω). The relative stability of the closed-loop system is easily obtained from this graph.
The chart consisting of the M and N loci in the log magnitude verses phase diagram is
called the Nichols chart. The G(jω) locus drawn on the Nichols chart gives both the gain characteristics and phase characteristics of the closed loop transfer function at the same time. The
Nichols chart contains curves of constant closed loop magnitude and phase angle. The Nichols
chart is symmetric about the 180° axis. The M loci are centered about the critical point (0 dB, – 180).
The Nichols chart is useful in determining the frequency response of the closed loop from that of
the open loop. The Nichols chart is produced by using the MATLAB command nichols(num,
den). The command ngrid creates the dotted lines that allow reading closed – loop gain and
phase from the Nichols chart. In order to customize the axes of the Nichols chart, the MATLAB
command axis is used.
3.9 GAIN MARGIN, PHASE MARGIN, PHASE CROSSOVER FREQUENCY,
AND GAIN CROSSOVER FREQUENCY
The MATLAB command
[Gm, pm, wcp, wcg] = margin (sys)
(3.51)
can be used to obtain the gain margin, phase margin, phase crossover frequency, and gain
crossover frequency.
In Equation (3.51), Gm is the gain margin, pm is the phase margin, wcp is the phase –
crossover frequency, and wcg is the gain crossover frequency.
The following MATLAB command is commonly used for obtaining the resonant peak and
resonant frequency:
[mag, phase, w] = bode (num, den, w)
or
[mag, phase, w] = bode (sys, w)
[Mp, k] = max (mag)
(3.52)
resonant peak = 20 * log 10 (Mp)
resonant frequency = w(k)
The following lines are used in MATLAB program to obtain bandwidth:
n=1
while 20 * log 10 (mag (n)) > – 3
n=n+1
end
bandwidth = w(n)
3.10 TRANSFORMATION OF SYSTEM MODELS
In this section, we consider two cases of transformation of system models.
1. Transformation of system model from transfer function to state space
2. Transformation of system model from state space to transfer function
(3.53)
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
3.10.1 Transformation of System Model from Transfer Function to State Space
The closed – loops transfer function can be written as
Y(s)
numerator of polynomials
num
=
=
U(s)
denominator of polynomials
den
The state-space representation is obtained by the MATLAB command
[A, B, C, D] = tf2ss (num, den)
(3.54)
(3.55)
3.10.2 Transformation of System Model from State Space to Transfer Function
The transfer function from state – space equations is obtained by using the MATLAB
command:
[num, den] = ss2tf (A, B, C, D, iu)
(3.56)
where iu corresponds to the system with more than one input. iu is either 1, 2, or 3, where 1
implies input u1, 2 implies input u2, and 3 implies input u3.
For system with only one input, the MATLAB command
[num , den] = ss2tf (A, B, C, D)
or
3.11
[num, den] = ss2tf (A, B, C, D, 1)
may be used
(3.57)
(3.58)
Bode Diagrams of Systems Defined in State Space
Let the control system defined in State Space be
x = Ax + Bu
where
y = Cx + Du
A = state matrix (nxn matrix)
(3.59)
B = control matrix (nxr matrix)
C = output matrix (mxn matrix)
D = output matrix (mxn matrix)
u = control vector (r – vector)
x = state vector (n – vector)
y = output vector (m – vector)
The MATLAB command bode[A, B, C, D] may be used to obtain the Bode diagram of
this system. In fact, the command bode[A, B, C, D] gives a series of Bode plots , one for each
input of the system, with the frequency range automatically determined.
If we use the scalar iu as an index into the inputs of the control system that specifies
which input is to be used for the Bode plot, Then the MATLAB command Bode [A, B, C, D iu]
produces the Bode plots from the input iu to all the outputs (y1, y2, ....., ym) of the system with
the frequency range automatically determined.
u1
If the system has three inputs, then u = u2
u3
MATLAB TUTORIAL
141
For a system with only one input u, then the MATLAB command
Bode[A, B, C, D]
Bode [A, B, C, D, 1] can be used.
or
3.12
(3.60)
(3.61)
NYQUIST PLOTS OF A SYSTEM DEFINED IN STATE SPACE
Consider the system defined in state space given by Equation (3.39). Nyquist plots of the
system defined in Eq. (3.43) may be obtained by using the MATLAB command
nyquist (A, B, C, D)
(3.62)
The MATLAB command given by Eq. (3.62) produces a series of Nyquist plots one corresponding to each input and output combination of the system, with the frequency range automatically determined.
If we used the scalar iu as an index to the inputs of the control system that specifies
which input is to be used for the Nyquist plot, then the MATLAB command nyquist (A, B, C, D,
iu, w) produces Nyquist plots from the input to all the outputs (y1, y2, ....., ym) of the system with
the frequency range automatically determined .
The MATLAB command
nyquist ( A, B, C, D, iu, w)
(3.63)
considers the user – supplied frequency vector w. The vector w specifies the frequency at which
the frequency response should be determined
3.13
TRANSIENTRESPONSE ANALYSIS IN STATE SPACE
In this section, we present the transient—response analysis of systems in state – space
using MATLAB. Specifically, we present the step response, impulse, ramp response, and responses to other forms of simple inputs.
3.13.1 Unit Step Response
For a control system defined in a state space form as in Eq. (3.59), the MATLAB command
step (A, B, C, D)
(3.64)
will generate plots of unit step responses, with the time vector automatically determined provided t is not explicitly provided in the step commands.
The MATLAB command step (sys) may also be used to obtain the unit—step response of
a system.
The command
step (sys)
can be used where the system is defined by
sys = tf (num, den)
(3.65)
(3.66)
or
sys = ss (A, B, C, D)
(3.67)
The following MATLAB step commands with left hand arguments are used then no plot
is shown on the screen.
142
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
[y, x, t] = step [num, den, t]
[y, x, t] = step (A, B, C, D, iu)
[y, x, t] = step (A, B, C, D, iu, t)
(3.68)
Hence , in order to obtain the response curves, plot commands should be used. The matrices x, and y contain the state response of the system and the output respectively, computed at
the time points t . In Eq. (3.64), iu is a scalar index of the inputs of the system, which specifies
the input to be used for the response, and t is the user specified time. The step command in Eq.
(3.69) can be used to obtain a series of step response plots, one for each input and output combination of
x = Ax + Bu
y = Cx + Du
when the system involves multiple inputs and multiple outputs.
(3.69)
3.13.2 Impulse Response
The following MATLAB commands may be used to obtain the unit impulse response of a
control system:
impulse (num, den)
impulse(A, B, C, D)
(3.70)
(3.71)
[y, x, t] = impulse (num, den)
[y, x, t] = impulse (num, den, t)
(3.72)
(3.73)
[y, x, t] = impulse (A, B, C, D)
[y, x, t] = impulse (A, B, C, D, iu)
(3.74)
(3.75)
[y, x, t] = impulse (A, B, C, D, iu, t)
(3.76)
The command in Eq. (3.70) impulse (num, den) shows the plots of the unit impulse response on the monitor (screen). The command in Eq. (3.71) , impulse (A, B,C, D) produces a
series of unit impulse – response plots one for each input and output combination of the system
defined in Eq. (3.59) with the time vector automatically obtained. The vector t in Eqns. (3.73)
and (3.76) is the user supplied time vector, which specifies the times at which the impulse
response is to be obtained. The scalar iu in Eqns. (3.71) and (3.72) is an index into the inputs of
the system and specifies which input is to be used for the impulse response. The matrices x and
y in Eqs.(3.72) to (3.76) contain the state responses of the system and the output respectively,
evaluated at the time points t.
3.13.3 Unit Ramp Response
Consider the system described in state space as
x = Ax + Bu
y = Cx + Du
(3.77)
where u is the unit – ramp function.
When all the initial conditions are zeros, the unit ramp response is the integral of the
unit step response. Therefore, the unit ramp response is given by
z=
∫
t
y dt
0
(3.78)
143
MATLAB TUTORIAL
or
z = y = x1
(3.79)
z = x3
(3.80)
Defining
Equation (3.79) can be written as
x 3 = x1
(3.81)
Combining Eqns. (3.57) and (3.61) , we can write
x = AAx + BBu
z = CCx + DDu
The MATLAB command
[z, x, t] = step (AA, BB, CC, DD)
(3.82)
(3.83)
can be used to obtain the unit – ramp response curve z(t).
3.13.4 Response to Arbitrary Input
The response to an arbitrary input can be obtained by using the following MATLAB
commands:
lsim (num, den, t)
(3.84)
lsim (A, B, C, D, u, t)
y = lsim (num, den, r, t)
(3.85)
(3.86)
y = lsim (A, B, C, D, u, t)
(3.87)
The MATLAB commands in Eqns. (3.80) to (3.83) will generate the response to input
time function r or u.
3.14
RESPONSE TO INITIAL CONDITION IN STATE SPACE
Consider the system defined in state space by
x = Ax + Bu, x(0) = x0
y = cx + Du
The MATLAB command
initial (A, B, C, D, [initial condition], t)
(3.88)
(3.89)
(3.90)
may be used to provide the response to the initial condition.
EXAMPLE PROBLEMS AND SOLUTIONS
Example 3.1. Reduce the system shown in Fig. 3.1 to a single transfer function, T(s) =
C(s)/R(s) using MATLAB. The transfer functions are given as
G1(s) =
1
(s + 7)
1
G2(s) =
(s + 6s + 5)
G3(s) =
1
(s + 8)
2
144
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
G4(s) =
1
s
G5(s) =
7
(s + 3)
G6(s) =
1
(s 2 + 7s + 5)
G7(s) =
5
(s + 5)
G8(s) =
1
(s + 9)
G 8 (s)
+
R (s )
G 1 (s)
+
+
G 3 (s)
G 6 (s)
+
–
G 2 (s)
G 4 (s)
+
G 5 (s)
Fig. 3.1
The transfer functions are given as:
G1 (s) = 1/(s + 7)
G2 (s) = 1/(s2 + 3s + 5)
G3 (s) = 1/(s + 8)
G4 (s) = 1/s
G5 (s) = 7/(s+3)
G6 (s) = 1/(s2 + 7s + 5)
G7 (s) = 5/(s + 5)
G8 (s) = 1/(s + 9)
Solution. % MATLAB Program
G1 = tf ( [0 0 1], [0 1 7]);
G2 = tf ( [0 0 1], [1 6 5]);
G3 = tf ( [0 0 1], [0 1 8]);
G4 = tf ( [0 0 1], [0 1 0]);
G5 = tf ( [0 0 7], [0 1 3]);
G6 = tf ( [0 0 1], [1 7 5]);
G7 = tf ( [0 0 5], [0 1 5]);
G8 = tf ( [0 0 1], [0 1 9]);
+
+
–
G 7 (s)
C (s )
145
MATLAB TUTORIAL
G9 = tf ( [0 0 1], [0 0 1]);
T1 = append (G1, G2, G3, G4, G5, G6, G7, G8, G9);
Q=[1–2–59]
2180
31 8 0
4180
534–6
6700
734–6
8 7 0 0];
Inputs = 9;
Outputs = 7;
Ts = connect (T1, Q, Inputs, Outputs);
T = Tf (Ts) computer response
Transfer function:
10 s^7 + 290 s^6 + 3350 s^5 + 1.98e004 s^4 + 6.369e004 s^3 + 1.089e005 s^2 + 8.895e004 s +
2.7e004 s^10 + 45 s^9 + 866 s^8 + 9305 s^7 + 6.116e004 s^6 + 2.533e005 s^5 + 6.57e005 s^4 +
1.027e006 s^3 + 8.909e005 s^2 + 3.626e005 s + 4.2e004.
Example 3.2. For each of the second order systems below, find ξ, ωn, Ts, Tp, Tr, % overshoot, and plot the step response using MATLAB.
130
(a) T(s) =
s 2 + 15s + 130
(b) T(s) =
0.045
s + 0.025s + 0.045
(c) T(s) =
108
s2 + 1.325 × 103 s + 108
2
Solution.
(a)
>> clf
>> numa = 130;
>> dena = [1 15 130];
>> Ta = tf(numa, dena)
Transfer function:
130
---------------s^2 + 15 s + 130
>> omegana = sqrt (dena(3))
146
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
omegana =
11.4018
>> zetaa = dena(2) / (2*omegana)
zetaa =
0.6578
>> Tsa = 4/ (zetaa*omegana)
Tsa =
0.5333
>> Tpa = pi/ (omegana*sqrt(1-zetaa^2))
Tpa =
0.3658
>> Tra = (1.76*zetaa^3 – .417*zetaa^2 + 1.039*zetaa + 1)/omegana
Tra =
0.1758
>> percenta = exp(–zetaa*pi/ sqrt(1–zetaa^2))*100
percenta =
6.4335
>> subplot(221)
>> step(Ta)
>> title(‘(a)’)
>> ‘(b)’
ans =
(b) >> numb = .045;
>> denb = [1 .025 .045];
>> Tb = tf(numb,denb)
Transfer function:
0.045
--------------------s^2 + 0.025 s + 0.045
>> omeganb = sqrt(denb(3))
omeganb =
0.2121
>> zetab = denb(2) / (2*omeganb)
zetab =
0.0589
>> Tsb = 4/ (zetab*omeganb)
Tsb =
320
MATLAB TUTORIAL
>> Tpb = pi/ (omeganb*sqrt(1–zetab^2))
Tpb =
14.8354
>> Trb = (1.76*zetab^3 – .417*zetab^2 + 1.039*zetab + 1)/omeganb
Trb =
4.9975
>> percentb= exp(– zetab*pi/ sqrt(1–zetab^2))*100
percentb =
83.0737
>> subplot(222)
>> step(Tb)
>> title(‘(b)’)
>> ‘(c)’
ans =
(c)
>> numc = 10E8;
>> denc = [1 1.325*10E3 10E8];
>> Tc = tf(numc, denc)
Transfer function:
1e009
--------------------s^2 + 13250 s + 1e009
>> omeganc = sqrt(denc(3))
omeganc =
3.1623e+004
>> zetac = denc (2) / (2*omeganc)
zetac =
0.2095
>> Tsc = 4/ (zetac*omeganc)
Tsc =
6.0377e – 004
>> Tpc = pi/(omeganc*sqrt (1 – zetac^2))
Tpc =
1.0160e – 004
>> Trc = (1.76*zetac^3 – .417*zetac^2 + 1.039*zetac + 1)/omeganc
Trc =
3.8439e – 005
>> percentc = exp (– zetac*pi/sqrt (1 – zetac^2))*100
percentc =
51.0123
147
148
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> subplot (223)
>> step (Tc)
>> title (‘(c)’)
Stop Response
Stop Response
2
1.5
Amplitude
Amplitude
1.5
1
0.5
1
0.5
0
0
0
0.2
0.6
0.4
0.8
Time (sec)
0
100
200
300
400
Time (sec)
Stop Response
2
Amplitude
1.5
1
0.5
0
0
2
4
6
8
Time (sec)
Fig. E 3.2
Example 3.3. Determine the pole locations for the system shown below using MATLAB.
s 3 − 6s 2 + 7s + 15
C ( s)
= 5
4
s + s − 5s 3 − 9s 2 + 11s − 12
R( s)
Solution.
>> %MATLAB Program
>> den = [1 1 – 5 – 9 11 – 12];
>> A = roots (den)
A=
– 2.1586 + 1.2396i
– 2.1586 – 1.2396i
2.3339
0.4917 + 0.7669i
0.4917 – 0.7669i
149
MATLAB TUTORIAL
Example 3.4. Determine the pole locations for the unity feedback system shown below
using MATLAB.
G(s) =
150
(s + 5)(s + 7)(s + 9)(s + 11)
Solution.
>> %MATLAB Program
>> numg = 150
numg =
150
>> deng = poly ([– 5 – 7 – 9 – 11]);
>> ‘G(s)’
ans =
G(s)
>> G = tf (numg, deng)
Transfer function:
150
-------------------------------------s^4 + 32 s^3 + 374 s^2 + 1888 s + 3465
>> ‘Poles of G(s)’
ans =
Poles of G(s)
>> pole (G)
ans =
– 11.0000
– 9.0000
– 7.0000
– 5.0000
>> ‘T(s)’
ans =
T(s)
>> T = feedback (G, 1)
Transfer function:
150
-------------------------------------s^4 + 32 s^3 + 374 s^2 + 1888 s + 3615
>> pole (T)
ans =
– 10.9673 + 1.9506i
– 10.9673 – 1.9506i
– 5.0327 + 1.9506i
– 5.0327 – 1.9506i
150
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Example 3.5. A plant to be controlled is described by a transfer function
G(s) =
s+5
2
s + 7s + 25
Obtain the root locus plot using MATLAB.
Solution.
>> %MATLAB Program
>> clf
>> num = [1 5];
>> den = [1 7 25];
>> rlocus(num, den);
Computer response is shown in Fig. E 3.5
Root Locus
4
3
2
Imag Axis
1
0
–1
–2
–3
–4
– 16
– 14
– 12
– 10
–8
–6
–4
–2
0
Real Axis
Fig. E 3.5
Example 3.6. For the unity feedback system shown in Fig. E 3.6, G(s) is given as
R(s)
C(s)
G(s)
Fig. E 3.6.
G(s) =
30(s 2 − 5s + 3)
(s + 1)(s + 2)(s + 4)(s + 5)
Determine the closed-loop step response using MATLAB.
Solution.
>> %MATLAB Program
>> numg = 30*[1 – 5 3];
151
MATLAB TUTORIAL
>> deng = poly([– 1 – 2 – 4 – 5]);
>> G = tf(numg,deng);
>> T = feedback(G,1)
>> step(T)
Computer response:
Transfer function:
30 s^2 – 150 s + 90
---------------------------------s^4 + 12 s^3 + 79 s^2 - 72 s + 130
Fig. E3.6(a) shows the response
Step Response
10
8
Amplitude
6
4
2
0
–2
0
1
2
3
Time (sec)
4
5
Fig. E 3.6(a)
Simulation shows over 30% overshoot and non minimum-phase behavior. Hence the
second-order approximation is not valid.
Example 3.7. Determine the accuracy of the second-order approximation using MATLAB
to simulate the unity feedback system shown in Fig. E 3.7 where
G(s) =
15(s 2 + 3s + 7)
(s 2 + 3s + 7)(s + 1)(s + 3)
R(s)
C(s)
G(s)
Fig. E 3.7.
Solution.
>> %MATLAB Program
>> numg = 15*[1 3 7];
>> deng = conv([1 3 7],poly([– 1 – 3]));
152
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> G = tf(numg,deng);
>> T = feedback(G, 1);
>> step(T)
Computer response [see Fig E 3.7(a)].
Step Response
1
0.9
0.8
Amplitude
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
0.5
1.5
2
2.5
3
Time (sec)
Fig. E 3.7(a).
Example 3.8. For the unity feedback system shown in Fig. E 3.8 with
G(s) =
K(s + 1)
s(s + 1)(s + 5)(s + 6)
determine the range of K for stability using MATLAB.
R(s)
G(s)
Fig. E 3.8
Solution.
>> %MATLAB Program
>> K = [0:0.2:200];
>> for i = 1: length (K);
>> deng = poly ([0 –1 – 5 – 6]);
>> dent = deng + [0 0 0 K (i) K (i)];
>> R = roots (dent);
>> A = real(R);
>> B = max (A);
>> if B > 0
>> R
153
MATLAB TUTORIAL
>> K = K (i)
>> break
>> end
>> end
Computer response:
R=
– 10.0000
– 0.5000 + 4.4441i
– 0.5000 – 4.4441i
– 1.0000
A=
– 10.0000
– 0.5000
– 0.5000
– 1.0000
B=
– 0.5000
Example 3.9. Write a program in MATLAB to obtain the Nyquist and Nichols plots for
the following transfer function for k = 30.
G(s) =
k(s + 1)(s + 3 + 7i)(s + 3 – 7i)
(s + 1)(s + 3)(s + 3 + 7i)(s + 3 – 7i)
Solution.
>> %MATLAB Program
>> %Simple Nyquist and Nichols plots
>> clf
>> z = [– 1 – 3 + 7*i – 3 – 7*i];
>> p = [– 1 – 3 – 5 – 3 + 7*i – 3 – 7*i];
>> k = 30;
>> [num, den] = zp2tf (z’, p’, k);
>> subplot (211), nyquist (num, den)
>> subplot (212), Nichols (num, den)
>> ngrid
>> axis ([50 360 – 40 30])
154
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Computer response: The Nyquist and Nichols plots are shown in Fig. E 3.9.
Imaginary Axis
Nyquist Diagram
1
0.5
0
– 0.5
–1
–1
– 0.5
0.5
Real Axis
0
1
1.5
2
Open-Loop Gain (dB)
Nichols Chart
40
0.25 dB
0.5 dB
1 dB
3 dB
6 dB
20
0
0 dB
– 20
– 40
– 270
– 225
– 180
– 135
– 90
Open-Loop Phase (deg)
Fig. E 3.9.
Example 3.10. A PID controller is given by
Gc(s) = 29.125
(s + 0.57)2
s
Draw a Bode diagram of the controller using MATLAB.
Solution.
Gc(s) =
=
29.125( s2 + 1.14 s + 0.3249)
s
29.125s2 + 33.2025s + 9.4627
s
The following MATALB program produces the Bode diagram
>> %MATLAB Program
>> %Bode diagram
>> num= [29.125 33.2025 9.4627];
>> den= [0 1 0];
>> bode (num, den)
>> title (‘Bode diagram of G(s)’)
– 45
155
MATLAB TUTORIAL
Bode Diagram
Magnitude (dB)
50
45
40
35
Phase (deg)
30
90
45
0
– 45
– 90
–1
10
0
10
Frequency (red/sec)
10
1
Fig. E. 3.10 Bode diagram of G(s)
Example 3.11. For the closed-loop system defined by
C ( s)
1
= 2
R( s)
s + 2ξs + 1
(a) plot the unit-step response curves c (t) for ξ =0, 0.1, 0.2, 0.4, 0.5, 0.6, 0.8, and1.0. ωn is
normalized to 1.
(b) plot a three dimensional plot of (a).
Solution.
>> %Two-dimensional plot and three-dimensional plot of unit-step
>> %response curves for the standard second-order system with wn = 1
>> %and zeta = 0, 0.1, 0.2, 0.4, 0.5, 0.6, 0.8, and 1.0
>> t = 0 : 0.2 : 10;
>> zeta = [0 0.1 0.2 0.4 0.5 0.6 0.8 1.0];
>> for n = 1:8;
>> num = [0 0 1];
>> den = [1 2*zeta (n) 1];
>> [y (1 : 51, n), x, t] = step (num, den, t);
>> end
>> %Two-dimensional diagram with the command plot (t, y)
>> plot (t, y)
>> grid
>> title (‘Plot of unit-step response curves’)
>> xlabel (‘t Sec’)
>> ylabel (‘Response’)
156
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> text (4.1, 1.86, ‘\zeta = 0’)
>> text (3.0, 1.7, ‘0.1’)
>> text (3.0, 1.5, ‘0.2’)
>> text (3.0, 1.22, ‘0.4’)
>> text (2.9, 1.1, ‘0.5’)
>> text (4.0, 1.08, ‘0.6’)
>> text (3.0, 0.9, ‘0.8’)
>> text (4.0, 0.9, ‘1.0’)
>> %For three dimensional plot, we use the command mesh (t, eta, y’)
>> mesh (t, eta, y’)
>> title (‘Three-dimensional plot of unit-step response curves’)
>> xlabel (‘t Sec’)
>> ylabel (‘\zeta’)
>> zlabel (‘Response’)
Plot of unit-step response curves
2
ξ=0
1.8
0.1
1.6
0.2
Response
1.4
1.2
0.4
0.5
0.6
0.8
1.0
3
4
1
0.8
0.6
0.4
0.2
0
0
1
2
5
6
7
8
9
10
1 Sec
Fig. E3.11 (a) Plot of unit-step response curves
Three-dimensional plot of unit-step response curves
2
Response
1.5
1
0.5
0
1
0.5
ξ
6
0 0
2
8
10
4
1 Sec
Fig. E 3.11 (b) Three-dimensional plot of unit-step response curves
157
MATLAB TUTORIAL
Example 3.12. A closed-loop control system is defined by,
C(s)
2ζs
= 2
R( s)
s + 2ζs + 1
where ζ is the damping ratio. For ζ = 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, and 1.0 using
MATLAB. Plot
(a) a two-dimensional diagram of unit-impulse response curves
(b) a three-dimensional plot of the response curves.
Solution. A MATLAB program that produces a two-dimensional diagram of unit-impulse response curves and a three-dimensional plot of the response curves is given below:
>> %To plot a two-dimensional diagram
>> t = 0:0.2:10;
>> zeta = [0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0];
>> for n = 1:10;
>> num = [0 2*zeta (n) 1];
>> den= [1 2*zeta (n) 1];
>> [y (1:51, n), x, t] = impulse (num, den, t);
>> end
>> plot (t, y)
>> grid
>> title (‘Plot of unit-impulse response curves’)
>> xlabel (‘t Sec’)
>> ylabel (‘Response’)
>> text (2.0, 0.85, ‘0.1’)
>> text (1.5, 0.75, ‘0.2’)
>> text (1.5, 0.6, ‘0.3’)
>> text (1.5, 0.5, ‘0.4’)
>> text (1.5, 0.38, ‘0.5’)
>> text (1.5, 0.25, ‘0.6’)
>> text (1.7, 0.12, ‘0.7’)
>> text (2.0, – 0.1, ‘0.8’)
>> text (1.5, 0.0, ‘0.9’)
>> text (.5, 1.5, ‘1.0’)
>> %Three-dimensional plot
>> mesh (‘t, eta, y’)
>> title (‘Three-dimensional plot’)
>> xlabel (‘t Sec’)
>> ylabel (‘\zeta’)
>> zlabel (‘Response’)
158
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
The two-dimensional diagram and three-dimensional diagram produced by this MATLAB
program are shown in Figs. E 3.12 (a) and (b) respectively.
Plot of unit-impulse response Curves
2
1.5
1.0
Response
1
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.9
0.8
0.5
0
– 0.5
–1
0
1
2
3
4
5
t sec
6
7
8
9
10
Fig. E 3.12 (a) Two-dimensional plot.
Three-dimensional plot
2
Response
1.5
1
0.5
0
– 0.5
–1
1
0.5
ξ
0
0
2
4
6
8
10
t sec
Fig. E 3.12 (b) Three-dimensional plot.
Example 3.13. For the systems given below write a program in MATLAB that will use an
open-loop transfer function G(s):
G(s) =
50(s + 1)
s(s + 3)(s + 5)
G(s) =
25(s + 1)(s + 7)
s(s + 2)(s + 4)(s + 8)
(a) Obtain a Bode plot
(b) Estimate the percent overshoot, settling time, and peak time
(b) Obtain the closed-loop step response.
MATLAB TUTORIAL
Solution. (a)
>> %MATLAB Program
>> G = zpk ([– 1], [0 – 3 – 5], 50)
>> G = tf (G)
>> bode (G)
>> title (‘System 1’)
>> %title (‘System 1’)
>> pause
>> %Find phase margin
>> [Gm, Pm, Wcg, Wcp] = margin (G);
>> w = 1:.01:20;
>> [M, P, w] =bode (G, w);
>> %Find bandwidth
>> for k = 1:1: length (M);
>> if 20*log10 (M (k)) +7<=0;
>> ‘Mag’
>> 20*log10 (M (k))
>> ‘BW’
>> wBW = w(k)
>> break
>> end
>> end
>> %Find damping ratio, percent overshoot, settling time, and peak time
>> for z = 0:.01:10
>> Pt = atan (2*z/ (sqrt (– 2*z^2 + sqrt (1 + 4*z^4))))*(180/pi);
>> if (Pm – Pt) <= 0
>> z;
>> Po = exp (– z*pi/sqrt (1 – z^2));
>> Ts = (4/ (wBW*z))*sqrt ((1 – 2*z^2) + sqrt (4*z^4 – 4*z^2 + 2));
>> Tp = (pi/ (wBW*sqrt (1 – z^2)))*sqrt ((1 – 2*z^2) + sqrt (4*z^4 – 4*z^2 + 2));
>> fprintf (‘Bandwidth = %g’, wBW)
>> fprintf (‘Phase margin = %g’, Pm)
>> fprintf (‘, Damping ratio = %g’, z)
>> fprintf (‘, Percent overshoot = %g’, Po*100)
>> fprintf (‘, Settling time = %g’, Ts)
>> fprintf (‘, Peak time= %g’, Tp)
>> break
>> end
>> end
159
160
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> T = feedback (G, 1);
>> step (T)
>> title (‘Step response system 1’)
>>%title (‘Step response system 1’)
Computer response:
Zero/pole/gain:
50 (s + 1)
------------s (s + 3) (s + 5)
Transfer function:
50 s + 50
-----------------s^3 + 8 s^2 + 15 s
The Bode plot is shown in Fig. E 3.13(a)
Bode Diagram
Magnitude (dB)
40
20
0
– 20
– 40
Phase (deg)
– 60
– 45
– 90
– 135
– 180 – 1
10
10
0
10
1
2
10
Frequency (rad/sec)
Fig. E 3.13(a)
ans =
Mag
ans =
– 3.0032
ans =
BW
wBW =
9.7900
Bandwidth = 9.79Phase margin = 53.892, Damping ratio = 0.59, Percent overshoot =
10.0693, Settling time = 0.804303, Peak time = 0.461606
The step response is shown in Fig. E 3.13(b)
161
MATLAB TUTORIAL
Step Response
1.4
1.2
Amplitude
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
Time (sec)
3
3.5
4
4.5
5
Fig. E 3.13(b)
(b) Likewise, for this problem
>> G = zpk ([– 1 – 7], [0 – 2 – 4 – 8], 25)
>> G = tf (G)
The following Bode plot and step response are obtained [see Figs. E 3.13(c) and (d).
Zero/pole/gain:
25 (s + 1) (s + 7)
----------------------------s (s + 2) (s + 4) (s + 8)
Transfer function:
25 s^2 + 200 s + 175
---------------------------s^4 + 14 s^3 + 56 s^2 + 64 s
Bode Diagram
Magnitude (dB)
40
20
0
– 20
– 40
Phase (deg)
– 60
– 45
– 90
– 135
– 180 – 1
10
0
10
Frequency (rad/sec)
Fig. E 3.13(c)
10
1
2
10
162
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
ans =
Mag
ans =
– 7.0110
ans =
BW
wBW =
6.5500
Bandwidth = 6.55Phase margin = 63.1105, Damping ratio = 0.67, Percent overshoot =
5.86969, Settling time = 0.959175, Peak time = 0.679904
Step Response
1.4
1.2
Amplitude
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
Time (sec)
3
3.5
4
4.5
5
Fig. E 3.13(d)
Example 3.14. For a unit feedback system with the forward-path transfer function
G(s) =
K
s(s + 5)(s + 12)
and a delay of 0.5 second, estimate the percent overshoot for K = 40 using a second-order approximation. Model the delay using MATLAB function pade (T, n). Determine the unit step response and check the second-order approximation assumption made.
Solution.
>> %MATLAB Program
>> %Enter G(s)
>> numg1 = 1;
>> deng1 = poly ([0 – 5 – 12]);
>> ‘G1(s)’
>> G1 = tf (numg1, deng1)
>> [numg2, deng2] = pade (0.5, 5);
>> ‘G2(s)’
>> G2 = tf (numg2, deng2)
>> ‘G(s) = G1(s) G2(s)’
163
MATLAB TUTORIAL
>> G = G1*G2
>> %Enter K
>> K = input (‘Type gain, K’);
>> T = feedback (K*G, 1);
>> step (T)
>> title ([‘Step response for K =’, num2str (K)])
Output of this program is as follows:
ans =
G1(s)
Transfer function:
1
-------------------------s^3 + 17 s^2 + 60 s
ans =
G2(s)
Transfer function:
– s^5 + 60 s^4 – 1680 s^3 + 2.688e004 s^2 – 2.419e005 s + 9.677e005
--------------------------------------------------------------------------------------------s^5 + 60 s^4 + 1680 s^3 + 2.688e004 s^2 + 2.419e005 s + 9.677e005
ans =
G(s) = G1(s) G2(s)
Transfer function:
– s^5 + 60 s^4 – 1680 s^3 + 2.688e004 s^2 – 2.419e005 s + 9.677e005
--------------------------------------------------------------------------------------------------s^8 + 77 s^7 + 2760 s^6 + 5.904e004 s^5 + 7.997e005 s^4 + 6.693e006 s^3
+ 3.097e007 s^2 + 5.806e007 s
Type Gain, K 40
The following Fig. E 3.14 is obtained.
Step Response
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
– 0.2
0
1
2
4
3
Time (sec)
Fig. E 3.14
5
6
7
164
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Example 3.15. Write a program in MATLAB to obtain a Bode plot for the transfer
function
(a) G(s) =
(b) G(s) =
15
s(s + 3)(0.7s + 5)
(7s 3 + 15s 2 + 7s + 80)
(s4 + 8s3 + 12s 2 + 70s + 110)
Solution. (a)
>> %MATLAB Program
>> %Bode plot generation
>> clf
>> num = 15;
>> den = conv([1 0], conv([1 3],[0.7 5]));
>> bode(num, den)
Computer response: The Bode plot is shown in Fig. E 3.15
Phase (deg)
Magnitude (dB)
Bode Diagram
20
0
– 20
– 40
– 60
– 80
– 100
– 90
– 135
– 180
– 225
– 270 – 1
10
0
10
10
Frequency (rad/sec)
1
10
Fig. E 3.15
Solution.
>> %MATLAB Program
>> %Bode plot
>> clf
>> num=[0 7 15 7 80];
>> den=[1 8 12 70 110];
>> bode(num,den)
Computer response: The Bode plot is shown in Fig. E 3.15(b)
2
165
MATLAB TUTORIAL
Magnitude (dB)
Bode Diagram
10
0
– 10
– 20
Phase (deg)
– 30
0
– 45
– 90
– 135 – 1
10
0
1
10
10
Frequency (rad/sec)
10
2
Fig. E 3.15(b)
Example 3.16. Write a program in MATLAB for a unity-feedback system with
G(s) =
K(s + 7)
(s + 3s + 52)(s 2 + 2s + 35)
2
(a) plot the Nyquist diagram
(b) Display the real-axis crossing value and frequency.
Solution.
>> %MATLAB Program
>> numg = [1 7]
>> deng = conv ([1 3 52], [1 2 35]);
>> G = tf (numg, deng)
>> ‘G(s)’
>> Gap = zpk (G)
>> inquest (G)
>> axis ([– 3e – 3, 4e – 3, – 5e – 3, 5e – 3])
>> w = 0:0.1:100;
>> [re, im] = nyquis t (G, w);
>> for i =1:1: length (w)
>> M(i) = abs (re (i) + j*im (i));
>> A (i) = atan2 (im (i), re (i))*(180/pi);
>> if 180 – abs (A (i)) <= 1;
>> re (i);
>> im (i);
>> K = 1/abs (re (i));
>> fprintf (‘\nw = %g’, w(i))
>> fprintf (‘, Re = %g’, re (i))
166
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> fprintf (‘, Im = %g’, im (i))
>> fprintf (‘, M = %g’, M (i))
>> fprintf (‘, K = %g’, K)
>> Gm = 20*log10 (1/M (i));
>> fprintf (‘, Gm = &G’, Gm)
>> break
>> end
>> end
Computer response:
numg =
17
Transfer function:
s+7
-----------------------------------------------s^4 + 5 s^3 + 93 s^2 + 209 s + 1820
ans =
G(s)
Zero/pole/gain:
(s + 7)
------------------------------(s^2 + 2s + 35) (s^2 + 3s + 52)
The Nyquist plot is shown in Fig. E 3.16.
5
x 10
–3
Nyquist Diagram
4
Imaginary Axis
3
2
1
0
–1
–2
–3
–4
–5
–3
–2
–1
0
1
Real Axis
2
3
4
x 10
–3
Fig. E 3.16
Example 3.17. Write a program in MATLAB for the unity feedback system with
G(s) =
K
[s(s + 3)(s + 12)]
MATLAB TUTORIAL
167
so that the value of gain K can be input. Display the Bode plots of a system for the input value of
K. Determine and display the gain and phase margin for the input value of K.
Solution.
>> %Enter G(s)
>> numg = 1;
>> deng = poly ([0 – 3 – 12]);
>> ‘G(s)’
>> G = tf (numg, deng)
>> w = 0.01:0.1:100;
>> %Enter K
>> K = input (‘Type gain, K’);
>> bode (K*G, w)
>> pause
>> [M, P] = bode (K*G, w);
>> %Calculate gain margin
>> for i = 1:1: length (P);
>> if P (i) <= – 180;
>> fprintf (‘\nGain K = %g’, K)
>> fprintf (‘, Frequency (180 deg) = %g’, w(i))
>> fprintf (‘, Magnitude = %g’, M (i))
>> fprintf (‘, Magnitude(dB) = %g’,20*log10(M(i)))
>> fprintf(‘, Phase = %g’,P(i))
>> Gm = 20*log10(1/M(i));
>> fprintf(‘, Gain margin(dB) = %g’,Gm)
>> break
>> end
>> end
>> %Calculate phase margin
>> for i = 1:1:length(M);
>> if M(i)< = 1;
>> fprintf(‘\nGain K = %g’, K)
>> fprintf(‘, Frequency(0 dB) = %g’, w(i))
>> fprintf(‘, Magnitude=%g’, M(i))
>> fprintf(‘, Magnitude(dB) = %g’, 20*log10(M(i)))
>> fprintf(‘, Phase = %g’,P(i))
>> Pm = 180 + P(i) ;
>> fprintf(‘, Phase margin(dB) = %g’, Pm)
>> break
>> end
>> end
168
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> ‘Alternate program using MATLAB margin function:’
>> clear
>> clf
>> %Bode plot and find points
>> %Enter G(s)
>> numg = 1;
>> deng = poly([0 – 3 – 12]);
>> ‘G(s)’
>> G = tf(numg, deng)
>> w = 0.01:0.1:100;
>> %Enter K
>> K = input(‘Type gain, K ’);
>> bode(K*G, w)
>> [Gm, Pm, Wcp, Wcg] = margin(K*G)
>> ‘Gm(dB)’
>> 20*log10(Gm)
Computer response:
ans =
G(s)
Transfer function:
1
-------------------------s^3 + 15 s^2 + 36 s
Type gain, K 40
The Bode plot is shown in Fig. E 3.17(a).
Bode Diagmram
Magnitude (dB)
50
0
– 50
Phase (deg)
– 100
– 90
– 135
– 180
– 225
– 270 – 2
10
–1
0
10
10
Frequency (rad/sec)
10
1
Fig. E 3.17(a)
Gain K = 40, Frequency(180 deg) = 6.01, Magnitude = 0.0738277, Magnitude(dB) = – 22.6356,
Phase = – 180.076, Gain margin(dB) = 22.6356
169
MATLAB TUTORIAL
Gain K = 40, Frequency(0 dB) = 1.11, Magnitude = 0.93481, Magnitude(dB) = – 0.585534,
Phase = – 115.589, Phase margin(dB) = 64.4107
Alternate program using MATLAB margin function:
ans =
G(s)
Transfer function:
1
------------------------s^3 + 15 s^2 + 36 s
Type gain, K 40
Gm =
13.5000
Pm =
65.8119
Wcp =
6
Wcg =
1.0453
ans =
Gm(dB)
ans =
22.6067
The Bode plot is shown in Fig. E 3.17(b)
Bode Diagmram
Magnitude (dB)
50
0
– 50
Phase (deg)
– 100
– 90
– 135
– 180
– 225
– 270 – 2
10
–1
0
10
10
Frequency (rad/sec)
10
1
Fig. E 3.17(b)
Example 3.18. Write a program in MATLAB for the system shown below so that the value
of K can be input (K = 40).
C(s)
K (s + 5)
=
R( s) s(s 2 + 3s + 15 )
170
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(a) Display the closed-loop magnitude and phase frequency response for unity feedback
system with an open-loop transfer function, KG(s).
(b) Determine and display the peak magnitude, frequency of the peak magnitude, and
bandwidth for the closed-loop frequency response for the input value of K.
Solution.
>> %MATLAB Program
>> %Enter G(s)
>> numg = [1 5];
>> deng = [1 3 15 0];
>> ‘G(s)’
>> G = tf(numg, deng)
>> %Enter K
>> K = input(‘Type gain, K’);
>> ‘T(s)’
>> T = feedback(K*G,1)
>> bode(T)
>> title(‘Closed-loop frequency response’)
>> [M, P, w] = bode(T);
>> [Mp i] = max(M);
>> Mp
>> MpdB = 20*log10(Mp)
>> wp = w(i)
>> for i = 1:1:length(M);
>> if M(i)<= 0.707;
>> fprintf(‘Bandwidth = %g’, w(i))
>> break
>> end
>> end
Computer response:
ans =
G(s)
Transfer function:
s+5
-----------------------s^3 + 3 s^2 + 15 s
Type gain, K 40
ans =
T(s)
171
MATLAB TUTORIAL
Transfer function:
40 s + 200
--------------------------------s^3 + 3 s^2 + 55 s + 200
Mp =
11.1162
MpdB =
20.9192
wp =
7.5295
Bandwidth = 10.8036
The Bode plot is shown in Fig. E 3.18.
Bode Diagram
Phase (deg)
Magnitude (dB)
40
20
0
– 20
– 40
– 60
100
135
90
45
0
– 45 – 1
10
0
1
10
10
Frequency (rad/sec)
10
2
Fig. E 3.18
Example 3.19. Determine the unit-ramp response of the following system using MATLAB
and lsim command.
C(s)
1
=
R(s) 3s 2 + 2s + 1
Solution.
>> %MATLAB Program
>> %Unit-ramp response
>> num = [0 0 1];
>> den = [3 2 1];
>> t = 0:0.1:10;
>> r = t;
>> y = lsim(num, den, r, t);
>> plot(t, r, ‘–’, t, y, ‘o’)
>> grid
172
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> title(‘Unit-ramp response’)
>> xlabel(‘t Sec’)
>> ylabel(‘Unit-ramp input and output’)
>> text(1.0, 4.0, ‘Unit-ramp input’)
>> text(5.0, 2.0, ‘Output’)
Unit-ramp response
Unit-ramp input and output
10
8
6
4 Unit-ramp input
Output
2
0
0
2
4
6
8
10
1 Sec
Fig. E 3.19 Unit-ramp response.
Example 3.20. A higher-order system is defined by
C(s)
7s 2 + 16s + 10
= 4
R(s) s + 5s 3 + 11s 2 + 16s + 10
(a) plot the unit-step response curve of the system using MATLAB
(b) obtain the rise time, peak time, maximum overshoot, and settling time using MATLAB.
Solution.
>> %Unit-step response curve
>> num = [0 0 7 16 10];
>> den = [1 5 11 16 10];
>> t = 0:0.02:20;
>> [y, x, t] = step(num, den, t);
>> plot(t, y)
>> grid
>> title(‘Unit-step response’)
>> xlabel(‘t Sec’)
>> ylabel(‘Output y(t)’)
173
MATLAB TUTORIAL
Unit-step response
1.6
1.4
Output y(t)
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
t Sec
15
20
Fig. E 3.20 Unit-step response.
>> %Response to rise from 10% to 90% of its final value
>> r1 = 1; while y(r1) < 0.1, r1 = r1 + 1; end
>> r2 = 1; while y(r2) < 0.9, r2 = r2 + 1; end
>> rise_time = (r2 – r1)*0.02
rise_time =
0.5400
>> [ymax,tp] = max(y);
>> peak_time = (tp – 1)*0.02
peak_time =
1.5200
>> max_overshoot = ymax – 1
max_overshoot =
0.5397
>> s = 1001; while y(s) > 0.98 & y(s) < 1.02; s = s – 1; end
>> settling_time = (s – 1)*0.02
settling_time =
6.0200
Example 3.21. Obtain the unit-ramp response of the following closed-loop control system
whose closed-loop transfer function is given by
C(s)
s + 12
= 3
R( s)
s + 5s 2 + 8s + 12
Determine also the response of the system when the input is given by
r = e–0.7t.
174
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Solution.
>> %Unit-ramp response-lsim command
>> num = [0 0 1 12];
>> den = [1 5 8 12];
>> t = 0:0.1:10;
>> r = t;
>> y = lsim(num, den, r, t);
>> plot(t, r, ‘–’, t, y, ‘o’)
>> grid
>> title(‘Unit-ramp response’)
>> xlabel(‘t Sec’)
>> ylabel(‘Output’)
>> text(3.0, 6.5, ‘Unit-ramp input’)
>> text(6.2, 4.5, ‘Output’)
Unit-ramp response
10
8
Output
Unit-ramp input
6
Output
4
2
0
0
2
4
6
8
1 Sec
Fig. E 3.21(a) Unit-ramp response curve.
>> %Input r1 = exp(– 0.7t)
>> num = [0 0 1 12];
>> den = [1 5 8 12];
>> t = 0:0.1:12;
>> r1 = exp(– 0.7*t);
>> y1 = lsim(num, den, r1, t);
>> plot(t, r1, ‘–’, t, y1, ‘o’)
>> grid
>> title(‘Response to input r1 = exp(– 0.7t)’)
>> xlabel(‘t Sec’)
10
175
MATLAB TUTORIAL
>> ylabel(‘Input and output’)
>> text(0.5, 0.9, ‘Input r1 = exp(– 0.7t)’)
>> text(6.3, 0.1, ‘Output’)
Response to input r1 = exp(– 0.7t)
1.2
1
Input r1 = exp (– 0.7t )
Input and Output
0.8
0.6
0.4
0.2
Output
0
– 0.2
0
2
4
6
t Sec
8
10
12
Fig. E 3.21(b) Response curve for input r = e–0.7t.
Example 3.22. Obtain the response of the closed-loop system using MATLAB. The closedloop system is defined by
7
C(s)
= 2
R( s)
s + s+7
The input r(t) is a step input of magnitude 3 plus unit-ramp input, r(t) = 3 + t.
Solution.
>> %MATLAB Program
>> num = [0 0 7];
>> den = [1 1 7];
>> t = 0:0.05:10;
>> r = 3 + t;
>> c = lsim(num, den, r, t);
>> plot(t, r, ‘–’, t, c, ‘o’)
>> grid
>> title(‘Response to input r(t) = 3 + t’)
>> xlabel(‘t Sec’)
>> ylabel(‘Output c(t) and input r(t) = 3 + t’)
176
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Response to input r(t) = 3 + t
Output c(t) and input r(t) = 3 + t
14
12
10
8
6
4
2
0
0
2
4
6
8
10
t Sec
Fig. E 3.22 Response to input r(t) = 3 + t.
Example 3.23. Plot the root-locus diagram using MATLAB for a system whose open-loop
transfer function G(s) H(s) is given by
K(s + 3)
G(s)H(s) =
(s 2 + 3s + 4)(s 2 + 2s + 7)
Solution.
K (s + 3)
(s + 3s + 4)(s2 + 2 s + 7)
G(s)H(s) =
=
>> %MATLAB Program
>> num = [0 0 0 1 3];
>> den = [1 5 17 29 28];
>> K1 = 0:0.1:2;
>> K2 = 2:0.02:2.5;
>> K3 = 2.5:0.5:10;
>> K4 = 10:1:50;
>> K5 = 50:5:800;
>> K = [K1 K2 K3 K4 K5];
2
K (s + 3)
(s + 5s + 17 s2 + 29 s + 28)
4
3
177
MATLAB TUTORIAL
>> r = rlocus(num, den, K);
>> plot(r, ‘o’)
>> v = [– 10 5 – 8 8]; axis(v)
>> grid
>> title(‘Root – locus plot of G(s)H(s)’)
>> xlabel(‘Real axis’)
>> ylabel(‘Imaginary axis’)
Root-locus plot of G(s) H(s)
8
6
Imaginary axis
4
2
0
–2
–4
–6
–8
– 10
–5
Real axis
0
5
Fig. E 3.23 Root-locus diagram.
Example 3.24. A unity-feedback control system is defined by the following feedforward
transfer function
G(s) =
K
s(s + 5s + 9)
2
(a) determine the location of the closed-loop poles, if the value of gain is equal to 3
(b) plot the root loci for the system using MATLAB.
Solution.
>> %MATLAB Program to find the closed-loop poles
>> p = [1 5 9 3];
>> roots(p)
ans =
– 2.2874 + 1.3500i
– 2.2874 – 1.3500i
– 0.4253
>> %MATLAB Program to plot the root-loci
>> num = [0 0 0 1];
178
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> den = [1 5 9 0];
>> rlocus(num, den);
>> axis(‘square’)
>> grid
>> title(‘Root-locus plot of G(s)’)
Root Locus
Imag Axis
5
0
–5
–6
–4
–2
Real Axis
0
Fig. E 3.24 Root-locus plot of G(s).
Example 3.25. The open-loop transfer function of a unity-feedback control system is given
by
G(s) =
1
s + 0.3s 2 + 5s + 1
3
(a) draw a Nyquist plot of G(s) using MATLAB
(b) determine the stability of the system.
Solution.
>> % Open-loop poles
>> p = [1 0.3 5 1];
>> roots(p)
ans =
– 0.0496 + 2.2311i
– 0.0496 – 2.2311i
– 0.2008
>> % Nyquist plot
>> num = [0 0 0 1];
>> den = [1 0.3 5 1];
>> nyquist(num,den)
>> v = [– 3 3 – 2 2]; axis(v); axis(‘square’)
>> grid
>> title(‘Nyquist plot of G(s)’)
179
MATLAB TUTORIAL
Nyquist Diagram
2
Imaginary Axis
1.5
1
0.5
0
– 0.5
–1
– 1.5
–2
–3
–2 –1
0
1
Real Axis
2
3
Fig. E 3.25 Nyquist plot of G(s).
There are two open-loop poles in the right half s plane and no encirclement of the critical
point, the closed-loop system is unstable.
Example 3.26. The open-loop transfer function of a unity-feedback control system is given
by
G(s) =
K(s + 3)
s(s + 1)(s + 7)
Plot the Nyquist diagram of G(s) for K = 1, 10, and 100 using MATLAB.
Solution.
G(s) =
K (s + 3)
K (s + 3)
= 3
s + 8 s2 + 7 s
s(s + 1)(s + 7)
>> % MATLAB Program
>> num = [1 3];
>> den = [1 8 7 0];
>> w = 0.1:0.1:100;
>> [re1, im1, w] = nyquist(num, den, w);
>> [re2, im2, w] = nyquist(10*num, den, w);
>> [re3,im3,w] = nyquist(100*num, den, w);
>> plot(re1, im1, re2, im2, re3, im3)
>> v = [– 3 3 – 3 3]; axis(v)
>> grid
>> title(‘Nyquist diagrams’)
>> xlabel(‘Real axis’)
>> ylabel(‘Imaginary axis’)
>> text(– 0.2, – 2, ‘K = 1’)
>> text(– 1.5, – 2.0, ‘K = 10’)
>> text(– 2, – 1.5, ‘K = 100’)
180
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Nyquist diagram
3
Imaginary axis
2
1
0
–1
–2
–3
–3
–2
–1
0
Real Axis
1
2
3
Fig. E 3.26 Nyquist Diagrams.
Example 3.27. The open-loop transfer function of a negative feedback system is given by
G(s) =
5
s(s + 1)(s + 3)
Plot the Nyquist diagram for
(a) G(s) using MATLAB
(b) same open-loop transfer function use G(s) of a positive-feedback system using MATLAB
Solution.
G(s) =
5
5
= 3
s(s + 1)( s + 3)
s + 4 s2 + 3 s
>> %Nyquist diagrams of G(s) and – G(s)
>> num1 = [0 0 0 5];
>> den1 = [1 4 3 0];
>> num2 = [0 0 0 – 5];
>> den2 = [1 4 3 0];
>> nyquist(num1,den1)
>> hold
Current plot held
>> nyquist(num2, den2)
>> v = [– 5 5 – 5 5]; axis(v)
>> grid
>> text(– 3, – 1.8, ‘G(s)’)
>> text(1.9, – 2, ‘– G(s)’)
181
MATLAB TUTORIAL
Nyquist Diagram
5
4
Imaginary Axis
3
2
1
0
–1
–2
–3
–4
–5
–5
0
Real Axis
5
Fig. E 3.27 Nyquist diagrams.
Example 3.28. For the system shown in Fig. E 3.28, design a compensator such that the
dominant closed-loop poles are located at s = – 2 ± j 3 . Plot the unit-step response curve of the
designed system using MATLAB.
+
Gc(s)
7
s( 0.5s + 1)
–
Fig. E 3.28 Control system.
Solution. From Fig. E 3.28(a), for the closed-loop pole is to be located at s = – 2 + j 3 ,
the sum of the angle contributions of the open-loop poles (at s = 0, and s = – 2) is given by – 120º
– 90º = – 210º. For the closed-loop pole at s = – 2 + j 3 we need to add 30º to the open-loop
transfer function. In other words, the angle deficiency of the given open-loop transfer function
at the desired closed-loop pole s = – 2 + j 3 is given by
180º – 120º – 90º = – 30º
The compensator must contribute 30º (lead compensator). The simplest form of a lead
compensator is
Gc(s) = K
s+a
s+b
182
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
jω
s = –2 + j2 3
jω
j2
s = –2 + j 3
j1.5
j1.5
j1
–4
–3
–2
–1 0
j1
30º
j0.5
120º
90º
j2
90º
1 σ
–j0.5
(a) Open-loop poles and a desired
–4
–3
–2
–1 0
j0.5
1 σ
–j0.5
(b) Compensator pole-zero configuration
closed-loop pole.
to contribute phase lead angle of 30º.
Fig. E 3.28(a) and (b).
If we select the zero of the lead compensator at s = – 2, then the pole of the compensator
must be located at s = – 4 in order to have a phase lead angle of 30º (see Fig. E 3.28(b)).
Gc(s) = K
Hence
s+2
s+4
The gain K is obtained from the condition
K
or
or
7
s+2
s + 4 s(0.5s + 1)
K=
s(s + 4)
14
Gc(s) = 0.5
=1
s=−2+ j 3
= 0.5
s=−2+ j 3
s+2
s+4
The open-loop transfer function of the compensated system is given by
Gc(s) .
s+2
7
7
14
= 0.5
=
s(s + 4)
s + 4 s( s + 2)
s(0.5 s + 1)
The closed-loop transfer function of the original system is
14
C(s)
= 2
R( s)
s + 2s + 14
The compensated system’s closed-loop transfer function is
7
C(s)
= 2
s + 4s + 7
R( s)
183
MATLAB TUTORIAL
A MATLAB program to plot the unit-step response curves of the original and compensated systems is given below. The unit-step response curves are shown in Fig. E 3.28(c).
% MATLAB Program
num = [0 0
14];
den = [1
2
numc = [0
0
14];
7];
denc = [1
4
t = 0: 0.01: 5;
7];
c1 = step(num, den, t);
c2 = step(numc, denc, t);
plot(t, c1, ‘.’, t, c2, ‘–’)
xlabel(‘t Sec’)
ylabel(‘Outputs’)
text(1.5, 1.3, ‘Original system’)
text(1.7, 1.14, ‘Compensated system’)
grid
title(‘Unit-Step Responses of Original System and Compensated System’)
Unit-Step Responses of Original system and Compensated System
1.5
Original system
Output
Compensated system
1
0.5
0
0
0.5
1
1.5
2
2.5
t Sec
3
3.5
4
4.5
5
Fig. E 3.28(c)
Example 3.29. For the control system shown in Fig. E 3.29 design a compensator such
that the dominant closed-loop poles are located at s = – 1 + j1. Determine also the unit-step and
unit ramp responses of the uncompensated and compensated systems.
+
Gc(s)
–
Lead compensator
Fig. E 3.29.
1
( 0.5s 2 )
184
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Solution. For a desired closed-loop pole at s = – 1 + j1, the angle contribution of the two
open-loop poles at the origin is given by – 135º – 135º = – 270º. Therefore, the angel deficiency
is given by
180º – 135º – 135º = – 90º
Hence, the compensator must contribute 90º.
We select a lead compensator of the form
Gc(s) = K
s+a
s+b
and choose the zero of the lead compensator at s = – 0.5. In order to obtain the phase lead angle
of 90º, the pole of the compensator must be located at s = – 3 (see Fig. E 3.29(a).
jω
j2
j1
90º
–2
–3
–1
0
1
σ
–j1
Fig. E 3.29(a) Pole-zero location of lead compensator contributing
90º phase lead.
Gc(s) = K
Therefore
s + 0.5
s+3
where K must be obtained from the magnitude condition as
K
or
K=
s + 0.5 2
s + 3 s2
=1
s = − 1 + j1
( s + 3) s2
2( s + 0.5)
=2
s = − 1 + j1
Therefore, the lead compensator becomes
s + 0.5
s+3
The feed forward transfer function is
Gc(s) = 2
Gc(s) =
1
0.5s
2
=
4s + 2
s3 + 3 s2
185
MATLAB TUTORIAL
The root-locus plot of the system is shown in Fig. E 3.29(b).
The closed-loop transfer function is given by
4s + 2
C(s)
= 3
R( s)
s + 3s 2 + 4 s + 2
The closed-loop poles are located at s = – 1 ± j1 and s = – 1.
Now we determine the unit-step and unit-ramp responses of the uncompensated and
compensated systems.
A MATLAB program is written to obtain unit-step response curve. The resulting curves
are shown in Fig. E 3.29(b).
% MATLAB program
num = [0
den = [1
0
0
2];
2];
nume = [0
dene = [1
0
3
4
4
2];
2];
t = 0:0.02:10;
c1 = step(num, den, t);
c2 = step(numc, denc, t);
plot(t, c1, ‘.’, t, c2, ‘-’)
grid
title(‘Unit-Step Responses of Uncompensated and Compensated Systems’)
xlabel(‘t Sec’)
ylabel(‘Outputs’)
text(2, 0.88, ‘Compensated system’)
text(3.1,1.48, ‘Uncompensated system’)
Unit-Step Response of Uncompensated and Compensated System
2
1.8
1.6
Uncompensated system
1.4
Output
1.2
1
Compensated system
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
t Sec
6
7
8
9
Fig. E 3.29(b) Unit step Response.
10
186
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
A MATLAB program to obtain unit-ramp response curve is given below. The resulting
response curves are shown in Fig. E 3.29(c).
% MATLAB program
num = [0
0
0
1];
den = [1
nume = [0
0
0
1
0
0];
4
2];
dene = [1
3
t = 0:0.02:15;
4
2
0];
c1 = step(num, den, t);
c2 = step(numc, denc, t);
plot(,t, c1, ‘.’, t, c2, ‘–’)
grid
title(‘Unit-Ramp Responses of Uncompensated and Compensated Systems’)
xlabel(‘t Sec’)
ylabel(‘Input and Outputs’)
legend(‘.’, ‘uncompensated system’, ‘–’, ‘compensated system’)
Unit-Ramp Responses of Uncompensated and Compensated System
15
Input and Output
Uncompensated System
Compensated System
10
5
0
0
5
t Sec
10
15
Fig. E 3.29(c) Unit Ramp Response.
Example 3.30. The PID control of a second-order plant G(s) control system is shown in
Fig. E 3.30. Consider the reference input R(s) is held constant. Design a control system such that
the response to any step disturbance will be damped out in 2 to 3 secs in terms of the 2% settling
time. Select the configuration of the closed-loop poles such that there is a pair of dominant
closed-loop poles. Obtain the response to the unit-step disturbance input and to the unit-step
reference input.
187
MATLAB TUTORIAL
D(s)
K (as + 1)(bs + 1)
s
+
–
+
+
1
s + 3.4s + 8
2
Fig. E 3.30.
Solution. The transfer function is
Gc(s) =
K ( as + 1)(bs + 1)
s
The closed-loop transfer function is given by
s
Cd ( s)
=
2
D (s )
s(s + 3.4 s + 8) + K (as + 1)(bs + 1)
=
s
s3 + (3.4 + Kab)s2 + (8 + Ka + Kb) s + K
(1)
It is required that the response to the unit-step disturbance be such that the settling time
be 2 to 3s and the system have reasonable damping. Hence, we chose ξ = 0.5 and ωn = 4 rad/s for the
dominant closed-loop poles and the third pole at s = – 10 so that the effect of this real pole on the
response is small. The desired characteristic equation is then given by
(s + 10)(s2 + 2 × 0.5 × 4s + 42) = (s + 10)(s2 + 4s + 16)
= s3 + 14s2 + 56s + 160
The characteristic equation for the system given by Eq. (1) is
s3 + (3.4 + Kab)s2 + (8 + Ka + Kb)s + K = 0
Therefore
3.4 + Kab = 14
8 + Ka + Kb = 56
K = 160
which gives
ab = 0.06625, a + b = 0.3
The PID controller now is given by
Gc(s) =
=
K[ abs2 + (a + b)s + 1]
160(0.06626s2 + 0.3s + 1)
=
s
s
10.6(s2 + 4.528s + 15.09)
s
188
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
With this PID controller, the response to the disturbance is
s
Cd(s) =
s + 14 s + 56 s + 160
3
2
D(s) =
s
D(s)
(s + 10)(s2 + 4 s + 16)
For a unit-step disturbance input, the steady-state output is zero, since
lim cd (t ) = lim sCd (s) = lim
s→0
t→∞
s→0
s2
1
=0
(s + 10)(s + 4 s + 16) s
2
The response to a unit-step disturbance input is obtained with MATLAB program. The
response curve is shown in Fig. E 3.30(a). From the response curve we note that the settling
time is approximately 2.7 s. The response damps out rather quickly. Hence, the system designed is acceptable.
% Response to unit-step disturbance input
numd = [0
0 1
0];
dend = [1 14
t = 0:0.01:5;
56
160];
[c1, x1, t] = step(numd, dend, t);
plot(t, c1)
grid
title(‘Response to Unit-Step Disturbance Input’)
xlabel(‘t Sec’)
ylabel(‘Output to Disturbance Input’)
x 10
–3
Response to Unit-Step Disturbance Input
14
Output to Disturbance Input
12
10
8
6
4
2
0
–2
–4
0
0.5
1
1.5
2
2.5
t Sec
3
Fig. E 3.30(a)
3.5
4
4.5
5
189
MATLAB TUTORIAL
For the reference input r(t), the closed-loop transfer function is
10.6s + 48s + 160
10.6(s2 + 4.528s + 15.09)
Cr ( s)
=
= 3
3
2
s + 14 s2 + 56 s + 160
R( s)
s + 14 s + 56 s + 160
2
The response to a unit-step reference input is obtained by the MATLAB program. The
resulting response curve is shown in Fig. 3.30(b). The response curve shows that the maximum
overshoot is 7.3% and the settling time is 1.7s. Thus, the system has quite acceptable response
characteristics.
% Response to unit-step reference input
numr = [0
10.6 48 160];
denr = [1
14
t = 0:0.01:5;
56
160];
[c2, x2, t] = step(numr, denr, t);
plot(t, c2)
grid
title(‘Response to Unit-Step Reference Input’)
xlabel(‘t Sec’)
ylabel(‘Output to Reference Input’)
Response to Unit-Step Reference Input
1.4
Output to Reference Input
1.2
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
t Sec
Fig. E 3.30(b)
Example 3.31. For the closed-loop control system shown in Fig. E 3.31, obtain the range
of gain K for stability and plot a root-locus diagram for the system.
190
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
K (s 2 + 2s + 5)
+
R(s)
C(s)
s(s + 3)(s + 5)(s 2 + 1.5s + 1)
–
Fig. E 3.31.
Solution. The range of gain K for stability is obtained by first plotting the root loci and
then finding critical points (for stability) on the root loci. The open-loop transfer function G(s) is
K (s2 + 2s + 5)
G(s) =
s(s + 3)(s + 5)(s2 + 1.5 s + 1)
K (s2 + 2s + 5)
s5 + 9.5 s4 + 28 s3 + 20 s2 + 15 s
=
A MATLAB program to generate a plot of the root loci for the system is given below. The
resulting root-locus plot is shown in Fig. E 3.31(a).
% MATLAB program
num = [0
den = [1
0
9.5
0
28
1
20
rlocus(num,den)
v = [– 8
2
–5
2
15
5];
0];
5]; axis(v); axis(‘square’)
grid
title(‘Root-Locus Plot’)
Root Locus
5
4
3
Imag Axis
2
1
0
–1
–2
–3
–4
–5
–8
–6
–4
–2
Real Axis
0
2
Fig. E 3.31(a)
From Fig. E 3.31(a), we notice that the system is conditionally stable. All critical points
for stability lie on the jω axis.
191
MATLAB TUTORIAL
To obtain the crossing points of the root loci with the jω axis, we substitute s = jω into the
characteristic equation
or
or
s5 + 9.5s4 + 28s3 + 20s2 + 15s + K(s2 + 2s + 5) = 0
(jω)5 + 9.5(jω)4 + 28(jω)3 + (20 + K)(jω)2 + (15 + 2K)(jω) + 5K = 0
[9.5ω4 – (20 + K) ω2 + 5K] + j[ω5 – 28ω3 + (15 + 2K) ω] = 0
Equating the real part and imaginary part equal to zero, respectively, we get
9.5ω4 – (20 + K) ω2 + 5K = 0
ω5 – 28ω3 + (15 + 2K) ω = 0
(E.1)
(E.2)
Eq. (2) can be written as
ω=0
or
ω4 – 28ω2 + 15 + 2K = 0
K=
(E.3)
− ω4 + 28ω2 − 15
2
(E.4)
Substituting Eq. (4) into Eq. (1), we obtain
9.5ω4 – [20 + ½(– ω4 + 28ω2 – 15)] ω2 – 2.5ω4 + 70ω2 – 37.5 = 0
or
0.5ω6 – 2ω4 + 57.5ω2 – 37.5 = 0
The roots of the above equation can be obtained by MATLAB program given below.
% MATLAB program
a = [0.5
0
–2
roots(a)
0
57.5 0
– 37.5];
Output is:
ans =
– 2.4786 + 2.1157i
– 2.4786 – 2.1157i
2.4786 + 2.1157i
2.4786 – 2.1157i
0.8155
– 0.8155
The root-locus branch in the upper half plane that goes to infinity crosses the jω axis at ω
= 0.8155. The gain values at these crossing points are given by
K=
− 0.81554 + 28 × 0.81552 − 15
= 1.5894
2
for ω = 0.8155
For this K value, we obtain the range of gain K for stability as
1.5894 > K > 0
Example 3.32. For the control system shown in Fig. E 3.32:
(a) plot the root loci for the system
(b) find the range of gain K for stability.
192
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
+
R(s)
K
–
s+3
s+5
3
C(s)
s (s + 3)
2
Fig. E 3.32.
Solution. The open-loop transfer function G(s) is given by
3
3 K (s + 3)
s+3
= 4
2
s + 5 s (s + 3)
s + 8 s3 + 15 s2
G(s) = K
A MATLAB program to generate the root-locus plot is given below. The resulting plot is
shown in Fig. E 3.32(a).
% MATLAB program
num = [0 0
0
1
3];
den = [1
8
15
rlocus(num,den)
0
0];
v = [– 6
grid
5]; axis(v); axis(‘square’)
4
–5
title(‘Root-Locus Plot’)
Root Locus
5
4
3
Imag Axis
2
1
0
–1
–2
–3
–4
–5
–6
–4
–2
0
Real Axis
2
4
Fig. E 3.32(a).
From Fig. E 3.32(a), we notice that the critical value of gain K for stability corresponds
to the crossing point of the root locus branch that goes to infinity and the imaginary axis.
Therefore, we first find the crossing frequency and then find the corresponding gain value.
The characteristic equation is
s4 + 8s3 + 15s2 + 3Ks + 9K = 0
193
MATLAB TUTORIAL
Substituting s = jω into the characteristic equation, we get
(jω)4 + 8(jω)3 + 15(jω)2 + 3K(jω) + 9K = 0
(ω4 – 15ω2 + 9K) + jω(– 8ω2 + 3K) = 0
or
Equating the real part and imaginary part of the above equation to zero, respectively, we
obtain
ω4 – 15ω2 + 9K = 0
(E.1)
ω(– 8ω + 3K) = 0
Eq. (2) can be rewritten as
(E.2)
2
ω=0
– 8ω + 3K = 0
2
or
(E.3)
Substituting the value of K in Eq.(1), we get
ω4 – 15ω2 + 9 x
8 2
ω =0
3
ω4 + 9ω2 = 0
or
which gives
ω = 0 and ω = ± j 3
Since ω = j3 is the crossing frequency with the jω axis, by substituting ω = 3 into Eq. (E.3)
we obtain the critical value of gain K for stability as
K=
8 2 8
ω =
× 9 = 24
3
3
Therefore, the stability range for K is
24 > K > 0.
Example 3.33. For the control system shown in Fig. E 3.33:
(a) plot the root loci for the system
(b) find the value of K such that the damping ratio ζ of the dominant closed-loop poles is
0.6
(c) obtain all closed-loop poles
(d) plot the unit-step respond curve using MATLAB.
+
–
K
s(s + 3)(s + 5)
Fig. E 3.33
Solution. (a) The MATLAB program given below generates a root-locus plot for the
given system. The resulting plot is shown in Fig. E 3.33(a).
% MATLAB program
num = [0
den = [1
0
8
0
15
rlocus(num, den)
v = [– 6
4
–5
1];
0];
5]; axis(v); axis(‘square’)
194
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
grid
title(‘Root-Locus Plot’)
Root Locus
5
4
3
Imag Axis
2
1
0
–1
–2
–3
–4
–5
–6
–4
–2
0
Real Axis
2
4
Fig. E 3.33(a)
(b) We note that the constant ζ points (0 < ζ < 1) lie on a straight line having angle θ from
the jω axis as shown in Fig. E 3.33(b).
jω
Constant ζ line
ωn
θ
–ζωn
σ
Fig. E 3.33(b)
From Fig. E 3.33(b), we obtain
sin θ =
ζωn
=ζ
ωn
Also that ζ = 0.6 line can be defined by
s = – 0.75a + ja
where a is a variable (0 < a < ∞). To obtain the value of K such that the damping ratio ζ of the
dominant closed-loop poles is 0.6 we determine the intersection of the line s = – 0.75a + ja and
the root locus. The intersection point can be obtained by solving the following simultaneous
equations for a.
s = – 0.75a + ja
s(s + 3)(s + 5) + K = 0
(E.1)
(E.2)
195
MATLAB TUTORIAL
From Eqs. (1) and (2), we obtain
or
(– 0.75a + ja)(– 0.75a + ja + 3)(– 0.75a + ja + 5) + K = 0
(1.8281a3 – 2.1875a2 – 3a + K) + j(0.6875a3 – 7.5a2 + 15a) = 0
Equating the real part and imaginary part of the above equation to zero, respectively, we
obtain
1.8281a3 – 2.1875a2 – 3a + K = 0
3
(E.3)
2
0.6875a – 7.5a + 4a = 0
Eq. (E.4) can be rewritten as
or
a=0
0.6875a2 – 7.5a + 4 = 0
or
or
a2 – 10.90991a + 5.8182 = 0
(a – 0.5623)(a – 10.3468) = 0
(E.4)
Therefore a = 0.5323 or a = 10.3468
From Eq. (E.3) we obtain
K = – 1.8281a3 + 2.1875a2 + 3a = 2.0535
K = – 1.8281a3 + 2.1875a2 + 3a = – 1759.74
for a = 0.5626
for a = 10.3468
Since the K value is positive for a = 0.5623 and negative for a = – 10.3468, we select
a = 0.5623. The required gain K is 2.0535.
The characteristic equation with K = 2.0535 is then
s(s + 3)(s + 5) + 2.0535 = 0
s3 + 8s2 + 15s + 2.0535 = 0
or
(c) The closed-loop poles can be obtained by the following MATLAB program.
% MATLAB Program
p = [1 8
roots(p)
15
2.0535];
ans =
– 5.1817
– 2.6699
– 0.1484
Hence, the closed-loop poles are located at
s = – 5.1817, s = – 2.6699, s = – 4.1565
(d) The unit-step response of the system for K = 2.0535 can be obtained from the following MATLAB program. The resulting unit-step response curve is shown in Fig. E 3.33(c).
% MATLAB program
num = [0
den = [1
0
8
0
15
2.0535];
2.0535];
step(num,den)
grid
title(‘Unit-Step Response’)
xlabel(‘t Sec’)
196
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
ylabel(‘Output’)
Step Response
1
0.9
0.8
Amplitude
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
5
10
15
20
25
Time (sec) t sec
30
35
40
Fig. E 3.33(c)
Example 3.34. For the control system shown in Fig. E3.34, the open-loop transfer function is given by
G(s) =
1
s(s + 2)(0.6s + 1)
Design a compensator for the system such that the static velocity error constant Kv is
5s – 1, the phase margin is at least 50º, and the gain margin is at least 10 dB.
R(s)
1
+
s(s + 2)(0.6s + 1)
–
Fig. E 3.34
Solution. We can use a lag compensator of the form
1
s+
Ts + 1
T
= Kc
β>1
Gc(s) = Kcβ
1
βTs + 1
s+
βT
Defining
and
K cβ = K
G1(s) = KG(s) =
K
s(s + 2)(0.6s + 1)
we adjust the gain K to meet the required static velocity error constant.
C(s)
197
MATLAB TUTORIAL
Hence
Kv = lim sGc(s)G(s) = lim s
s→0
= lim
s→0
or
s→0
Ts + 1
G1(s) = lim sG1(s)
s→0
βTs + 1
K
sK
=
=5
s(s + 2)(0.6s + 1)
2
K = 10
With K = 10, the compensated system satisfies the steady-state performance require-
ment.
We can now plot the Bode diagram of
G1(jω) =
10
jω( jω + 2)(0.6 jω + 1)
The magnitude curve and phase-angle curve of G1(jω) are shown in Fig. E 3.34(a). From
this plot, the phase margin is found to be – 20º, which shows that the system is unstable.
Bode Diagram
Magnitude (dB)
40
20
0
– 20
Phase (deg)
– 40
– 90
– 135
– 180
– 225
– 270 – 1
10
0
10
Frequency (rad/sec)
10
1
Fig. E 3.34(a) Bode diagrams for G1 = KG (gain-adjusted but
uncompensated system), Gc/K (gain-adjusted compensator),
and GcG (compensated system).
The addition of a lag compensator modifies the phase curve of the Bode diagram and
therefore we must allow 5º to 12º to the specified phase margin to compensate for the modification of the phase curve. Since the frequency corresponding to a phase margin of 50º is 0.7 rad/s,
the new gain crossover frequency (of the compensated system) must be selected near this value.
We choose the corner frequency ω = 1/T. Since this corner frequency is not too far below the new
gain crossover frequency, the modification in the phase curve may not be small. Also, we add
about 12º to the given phase margin as an allowance to account for the lag angle introduced by
the lag compensator. The required phase margin is now 52º. The phase angle of the
uncompensated open-loop transfer function is – 128º at about ω = 0.5 rad/s. Hence, we choose
the new gain crossover frequency to be 0.5 rad/s. In order to bring the magnitude curve down to
0 dB, the lag compensator is given the necessary attenuation, which in this case is – 20 dB.
198
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Therefore 20 log
1
= – 20
β
β = 10
or
The other corner frequency ω = 1(βT). This corresponds to the pole of the lag compensator and is obtained as
1
= 0.01 rad/s
βT
Hence, the transfer function of the lag compensator is given by
1
s+
10 s + 1
10
Gc(s) = Kc(10)
= Kc
1
100s + 1
s+
100
Since the gain K was calculated to be 10 and β was determined to be 10, we have
Kc =
K
10
=
=1
β
10
Therefore, the compensator Gc(s) is obtained as
10 s + 1
100s + 1
Gc(s) = 10
The open-loop transfer function of the compensated system is therefore
Gc(s)G(s) =
10(10s + 1)
s(100 s + 1)(s + 2)(0.6 s + 1)
The magnitude and phase-angle curves of Gc(jω)G(jω) are shown in Fig. E 3.34(b).
Bode Diagram
Magnitude (dB)
40
20
0
– 20
Phase (deg)
– 40
– 90
– 135
– 180
– 225
– 270 – 1
10
0
10
Frequency (rad/sec)
Fig. E 3.34(b)
10
1
199
MATLAB TUTORIAL
The phase margin of the compensated system is about 50º(the required value). The gain
margin is about 11 dB (acceptable). The static velocity error constant is 5s–1. Thus, the compensated system satisfies the requirements on both the steady state and the relative stability.
We determine now the unit-step response and unit-ramp response of the compensated
system and the original uncompensated system. The closed-loop transfer functions of the compensated and uncompensated systems are given by
C(s)
100s + 10
=
R( s)
60 s4 + 220.6 s3 + 202.2s2 + 102s + 10
1
C(s)
=
3
R( s)
0.6 s + 2.2s2 + 2s + 1
and
respectively.
A MATLAB program to obtain the unit-step and unit-ramp responses of the compensated and uncompensated systems is given below. The resulting unit-step response curves and
unit-ramp response curves are shown in Fig. E 3.34(c) and E 3.34(d) respectively.
%MATLAB program
%
Unit-step response
num = [0 0
0
1];
den = [0.6
numc = [0
2.2
0
2
0
denc = [60
220.6
t = 0:0.1:40;
1];
100
10];
202.2 102
10];
[c1, x1, t] = step(num, den);
[c2, x2, t] = step(numc, denc);
plot(t, c1, ‘.’, t, c2, ‘–’)
grid
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’)
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Unit-Step Responses of Compensated and Uncompensated Systems
1.4
Compensated System
1.2
1
Output
0.8
Uncompensated System
0.6
0.4
0.2
0
0
5
10
15
20
25
30
35
t sec
Fig. E 3.34(c)
% Unit-ramp response
num1 = [0 0
0
0 1];
den1 = [0.6 2.2 2
num1c = [0 0 0 0 100
1 0];
10];
den1c = [60 220.6 202.2 102 10 0];
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’)
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MATLAB TUTORIAL
Unit-Ramp Responses of Compensated and Uncompensated Systems
20
18
16
14
Output
12
10
8
6
Uncompensated System
4
Compensated System
2
0
0
2
4
6
8
10
t Sec
12
14
16
18
20
Fig. E 3.34(d)
Example 3.35. The open-loop transfer function of a unit-feedback system is given by
G(s) =
K
s(s + 3)(s + 5)
Design a compensator Gc(s) such that the static velocity error constant is 10s– 1, the phase
margin is 50º, and the gain margin is 10 dB or more.
Solution. We consider a lag-lead compensator of the form
s +
Gc(s) = Kc
s +
1
1
s +
T1
T2
β
β
s +
T1
T2
The open-loop transfer function of the compensated system is Gc(s)G(s). For Kc = 1,
lim Gc(s) = 1. For the static velocity error constant, we have
s→0
Kv = lim sGc(s)G(s) = lim sGc(s).
s→0
s→0
K
K
=
= 10
15
s(s + 3)(s + 5)
Therefore K = 150
For K = 150, A MATLAB program e used to plot the Bode diagram is given below and the
diagram obtained is shown in Fig. E 3.35(a).
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
% MATLAB program
num = [0
den = [1
0
8
0
15
150];
0];
bode(num, den, w)
Bode Diagram
Magnitude (dB)
40
20
0
– 20
– 40
– 60
Phase (deg)
– 80
– 90
– 135
– 180
– 225
– 270 – 1
10
10
0
10
1
10
2
Frequency (rad/sec)
Fig. E 3.35(a) Bode diagram of G(s) = 150/[s(s + 3)(s + 5)].
From Fig.3.35 the phase margin of the uncompensated system is – 6º, which shows that
the system is unstable. To design a lag-lead compensator we choose a new gain crossover frequency. From the phase-angle curve for G(jω), the phase crossover frequency is ω = 2 rad/s. We
can select the new gain crossover frequency to be 2 rad/s such that the phase-lead angle required at ω = 2 rad/s is about 50º. A single lag-lead compensator can provide this amount of
phase-lead angle.
We can find the corner frequencies of the phase-lag portion of the lag-lead compensator.
Choosing the corner frequency ω = 1/T2 , corresponding to the zero of the phase-lag portion of
the compensator as 1 decade below the new gain crossover frequency, or at ω = 0.2 rad/s. For
another corner frequency ω = 1/(βT2), we need the value of β. The value of β can be obtained
from the consideration of the lead portion of the compensator.
The maximum phase-lead angle φm is given by Eq. (9.10). For α = 1/β, we have given
sin φm =
β−1
β+1
β = 10 corresponds to φm = 54.9º. We require a 50º phase margin, so we can select β = 10.
Hence
β = 10
Then the corner frequency ω = 1/(βT2) and ω = 0.02.
The transfer function of the phase-lag portion of the lag-lead compensator is
5s + 1
s + 0.2
= 10
s + 0.02
50 s + 1
203
MATLAB TUTORIAL
The new gain crossover frequency is ω = 2 rad/s, and |G(j2)| is found to be 6 dB. Therefore, if the lag-lead compensator contributes – 6 dB at ω = 2 rad/s, then the new gain crossover
frequency is as desired. From this requirement, it is possible to draw a straight line of slope
20 dB/decade passing through the point (– 6 dB, – 2 rad/s). The intersections of this line and the
0-dB line and – 20 dB line gives the corner frequencies. The corner frequencies for the lead
portion are ω = 0.4 rad/s and ω = 4 rad/s. Hence, the transfer function of the lead portion of the
lag-lead compensator is given by
1 2.5s + 1
s + 0.4
=
10 0.25s + 1
s+4
By combining the transfer functions of the lag and lead portions of the compensator, we
can find the transfer function Gc(s) of the lag-lead compensator. For Kc = 1, we get
Gc(s) =
(2.5s + 1)(5s + 1)
s + 0.4 s + 0.2
=
(0.25s + 1)(50s + 1)
s + 4 s + 0.02
The Bode diagram of the lag-lead compensator Gc(s) is obtained by the following MATLAB
program. The resulting plot is shown in Fig. E 3.35(b).
% MATLAB program
num = [1 0.6 0.08];
den = [1
4.02 0.08];
bode(num, den)
Bode Diagram
Magnitude (dB)
0
–5
– 10
– 15
Phase (deg)
– 20
90
45
0
– 45
– 90
–3
10
10
–2
–1
0
10
10
Frequency (rad/sec)
10
1
10
2
Fig. E 3.35(b) Bode diagram of the designed lag-lead compensator.
The open-loop transfer function of the compensated system is
Gc(s)G(s) =
(s + 0.4)(s + 0.2)
150
(s + 4)(s + 0.02) s(s + 3)(s + 5)
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
=
150s2 + 90s + 12
s + 12.02 s + 47.24 s3 + 60.94 s2 + 1.2s
5
4
The magnitude and phase-angle curves of the designed open-loop transfer function
Gc(s)G(s) are shown in the Bode diagram of Fig. E 3.35(c). This diagram is obtained using
following MATLAB program. Note that the denominator polynomial den was obtained using
the conv command, as follows:
a = [1 4.02 0.08];
b = [1
8
15
0];
conv(a, b)
ans =
1.0000 12.0200 47.2400 60.9400
% MATLAB program
num = [0 0 0 150
den = [1
12.02
bode(num, den)
90
1.2000
0
12];
47.24 60.94 1.2 0];
Bode Diagram
Magnitude (dB)
100
50
0
– 50
Phase (deg)
– 100
– 45
– 90
– 135
– 180
– 225
– 270
–3
10
10
–2
–1
0
10
10
Frequency (rad/deg)
10
1
10
2
Fig. E 3.35(c) Bode Diagram of Gc(s).G(s).
From Fig. E 3.35(c), the requirements on the phase margin, gain margin, and static
velocity error constant are all satisfied.
Unit-step response
Now
Gc(s)G(s) =
(s + 0.4)(s + 0.2)
150
(s + 4)(s + 0.02) s(s + 3)( s + 5)
205
MATLAB TUTORIAL
and
C ( s)
Gc (s) G (s)
=
R( s)
1 + Gc (s)G(s)
=
150s 2 + 90s + 12
s5 + 12.02 s4 + 47.24 s3 + 210.94 s2 + 91.2 s + 15.2
The unit-step response is obtained by the following MATLAB program and the unit-step
response curve is shown in Fig. E 3.35(d).
% MATLAB program
num = [0 0
0 150
90
12];
den = [1
12.02 47.24 210.94 91.2 15.2];
t = 0:0.2:40;
step(num, den, t)
grid
title(‘unit-Step Response of Designed System’)
Step Response
1.4
1.2
Amlitude
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
Time (sec)
30
35
40
Fig. E 3.35(d) Unit-step Response of designed system Gc(s)G(s).
Unit-ramp response
The unit-ramp response of this system is obtained by the following MATLAB. The unitramp response of Gc(G/(1 + GcG) converted into the unit-step response of GcG/[s(1 + GcG)]. The
unit-ramp response curve obtained is shown in Fig. 3.35(e).
% MATLAB program
num = [0 0
0 0 150
90
den = [1
12.02 47.24 210.94
t = 0:0.2:20;
c = step(num, den, t)
12];
91.2
15.2 0];
206
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
plot(t, c, t, t, ‘.’)
grid
title(‘Unit-Ramp Response of the Designed System’)
xlabel(‘Time (sec)’)
ylabel(‘Amplitude’)
Unit-Ramp Response of the Designed System
20
18
16
Amplitude
14
12
10
8
6
4
2
0
0
2
4
6
8
10 12
Time (sec)
14
16
18
20
Fig. E 3.35(e)Unit-ramp response of the designed system.
Example 3.36. The open-loop transfer function of a unity-feedback control system is given
by
G(s) =
K
s(s 2 + s + 5)
(a) determine the value of gain K such that the phase margin is 50º
(b) find the gain margin for the gain K obtained in (a).
Solution.
G(s) =
K
s(s + s + 5)
2
The undamped natural frequency is
denominator.
5 rad/s and the damping ratio of 0.1. 5 from the
Let the frequency corresponding to the angle of – 130º (Phase Margin of 50) be ω1 and
therefore
∠G(jω1) = – 130º
The Bode diagram is shown in Fig. E 3.36 from MATLAB program.
207
MATLAB TUTORIAL
Bode Diagram
Magnitude (dB)
20
0
– 20
– 40
Phase (deg)
– 60
– 90
– 135
– 180
– 225
– 270 – 1
10
10
0
10
1
Frequency (rad/sec)
Fig. E 3.36.
From Fig. E 3.36, the required phase margin of 50º and occurs at the frequency ω = 1.06
rad/s. The magnitude of G(jω) at this frequency is then – 7 dB. The gain K must then satisfy
or
20 log K = 7 dB
K = 2.23
Example 3.37. For the control system shown in Fig. E 3.37:
(a) design a lead-compensator Gc(s) such that the phase margin is 45º, gain margin is not
less than 8 dB, and the static velocity error constant Kv is 4 s – 1
(b) plot unit-step and unit-ramp response curves of the compensated system using MATLAB.
+
K
s(0.1s + 1)(s + 1)
Gc(s)
–
Fig. E 3.37.
Solution. Consider the lead compensator
1
Ts + 1
T
Gc(s) = Kcα
= Kc
1
αTs + 1
s+
αT
s+
Since Kv is given as 4 s–1, we have
Kv = lim sKcα
s→0
Ts + 1
K
= Kc ∝ K = 4
αTs + 1 s(0.1s + 1)( s + 1)
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
∧
Let K = 1 and define Kcα = K . Then
∧
K =4
The Bode diagram of
4
4
=
3
s(0.1s + 1)( s + 1)
0.1s + 1.1s2 + s
is obtained by the following MATLAB program. The Bode diagram is shown in Fig. E 3.37(a).
% MATLAB program
num = [0
den = [0.1
0
1.1
0
1
4];
0];
bode(num, den)
Bode Diagram
Magnitude (dB)
50
0
– 50
Phase (deg)
– 100
– 90
– 135
– 180
– 225
– 270 – 1
10
10
0
10
1
10
2
Frequency (rad/sec)
Fig. E 3.37(a).
From Fig. E 3.37(a), the phase and gain margins are 17º and 8.7 dB, respectively. For a
phase margin of 45º, let us select
φm = 45º – 17º + 12º = 40º
The maximum phase lead is 40º. Since
sin φm =
1−α
1+α
(φm = 40º)
α is obtained as 0.2174. Let us choose
α = 0.21
To determine the corner frequencies ω = 1/T and ω = 1/(αT) of the lead compensator we
note that the maximum phase-lead angle φm occurs at the geometric mean of the two corner
209
MATLAB TUTORIAL
frequencies, or ω = 1/( α T). The amount of the modification in the magnitude curve at
ω = 1/(
α T) due to the inclusion of the term (Ts + 1)/(αTs + 1) is then given by
1 + j ωT
1 + jωαT
Since
1
α
=
1
=
ω=
1
1
αT
α
= 2.1822 = 6.7778 dB
0.21
The magnitude of |G(jω)| is – 6.7778 dB which corresponds to ω = 2.81 rad/s. Therefore,
we select this as the new gain crossover frequency ωc.
1
=
T
α ωc =
0.21 × 2.81 = 1.2877
ω
2.81
1
= c =
= 6.1319
αT
0.21
α
Gc(s) = Kc
or
s + 1.2877
s + 6.1319
∧
4
K
=
0.21
α
Kc =
and
Hence
0.7768 s + 1
4 s + 1.2877
=4
0.16308 s + 1
0.21 s + 6.1319
Gc(s) =
The open-loop transfer function is
Gc(s)G(s) = 4
=
0.7768 s + 1
1
0.16308 s + 1 s(0.1s + 1)( s + 1)
3.1064 s + 4
0.01631s + 0.2794 s3 + 1.2631s2 + s
4
The closed-loop transfer function is
3.1064 s + 4
C ( s)
=
4
R( s)
0.01631s + 0.2794 s3 + 1.2631s2 + 4.1064 s + 4
The following MATLAB program produces the unit-step response curve as shown in
Fig. E 3.37(b).
% MATLAB program
num = [0
0
0
den = [0.01631
step(num, den)
3.1064
0.2794
4];
1.2631
4.1064
4];
210
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
grid
title(‘Unit-Step Response of Compensated System’)
xlabel(‘t Sec’)
ylabel(‘Output c(t)’)
Step Response
1.4
1.2
Amplitude
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
t sec Time (sec)
2.5
3
3.5
Fig. E 3.37(b).
The following MATLAB program produces the unit-ramp response curves as shown in
Fig. E 3.37(c).
% MATLAB program
num = [0 0 0
den = [0.01631
0
3.1064
4];
0.2794
1.2631
4.1064
t = 0:0.01:5;
c = step(num, den, t);
plot(t, c, t, t)
grid
title(‘Unit-Ramp Response of Compensated System’)
xlabel(‘t Sec’)
ylabel(‘Unit-Ramp Input and System Output c(t)’)
4
0];
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MATLAB TUTORIAL
Unit-Ramp Response of Compensated System
Unit-Ramp Input and System Output c(t)
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
t Sec
3
3.5
4
4.5
5
Fig. E 3.37(c)
Example 3.38. Obtain the unit-step response and unit-impulse response for the following
control system using MATLAB. The initial conditions are all zero.
0
x1
x
0
2 =
x 3
0
x 4
− 0.0069
1
0
0
1
0
0
− 0.0789
− 0.5784
1
− 1.3852
0
0
x1
0
x
2 + 0 [u]
x3
0
x4
2
x1
x
2
y = [1 0 0 0]
x3
x4
Solution. Unit-step response: The following MATLAB program yields the unit-step response of the given system. The resulting unit-step response curve is shown in Fig. E 3.38(a).
% MATLAB program
A = [0 1 0 0;0 0 1 0; 0 0 0 1; – 0.0069 – 0.0789 – 0.5784 – 1.3852];
B = [0; 0; 0; 2];
C = [1 0 0 0];
D = [0];
step(A, B, C, D);
grid
212
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
xlabel(‘t Sec’)
ylabel(‘Output y(t)’)
The output is shown in Fig. E 3.38(a).
Step Response
350
300
Amplitude
250
200
150
100
50
0
0
10
20
30
40
50
Time (sec) t sec
60
70
80
Fig. E 3.38(a)
Similarly with impulse(A, B, C, D) statement, we obtain the response as shown in
Fig. E 3.38(b).
Impulse Response
20
Amplitude
15
10
5
0
–5
0
10
20
30
40
Time (sec)
50
60
70
Fig. E 3.38(b)
Example 3.39. Obtain the state-space representation of the following system using
MATLAB.
C(s)
35s + 7
= 3
R( s)
s + 5s 2 + 36s + 7
213
MATLAB TUTORIAL
Solution.
A MATLAB program to obtain a state-space representation of this system is given below.
% MATLAB program
>> num = [0 0 35 7];
>> den = [1 5 36 7];
>> g = tf(num,den)
Transfer function:
35 s + 7
----------------------------s^3 + 5 s^2 + 36 s + 7
>> [A, B, C, D] = tf2ss(num, den)
A=
– 5 – 36 – 7
1
0
0
0
1
0
B=
1
0
0
C=
0 35 7
D=
0
From the MATLAB output we obtain the following state space equations:
x1
− 5 − 36 − 7 x1
1
x
0
0 x2 + 0 u
2 = 1
0
0
x3
1
0 x3
x1
y = [0 35 7] x2 + [0]u
x3
Example 3.40. Represent the system shown in Fig. E 3.40 using MATLAB in
(a) state space in phase-variable form
(b) state space in modal form.
10(s + 3) (s + 5)
(s + 1) (s + 4) (s + 6) (s + 8)
R(s) +
–
Fig. E 3.40.
C(s)
214
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Solution.
%
(a)
MATLAB Program
phase-variable form’
‘G(s)’
G = zpk([– 3 – 5], [– 1 – 4 – 6 – 8], 10)
‘T(s)’
T = feedback(G, 1, – 1)
[numt, dent] = tfdata(T, ‘V’);
‘Controller canonical form determination’
[AC, BC, CC, DC] = tf2ss(numt, dent)
A1 = flipud(AC);
‘Phase-variable form representation’
Apv = fliplr(A1)
Bpv = flipud(BC)
Cpv = fliplr(CC)
(b) Modal form’
‘G(s)’
G = zpk([– 3 – 5], [– 1 – 4 – 6 – 8], 10)
‘T(s)’
T = feedback(G, 1, – 1)
[numt, dent] = tfdata(T, ‘V’);
‘Controller canonical form’
[AC, BC, CC, DC] = tf2ss(numt, dent)
‘Modal form’
[A, B, C, D] = canon(AC, BC, CC, DC, ‘modal’)
Computer response:
ans =
(a) phase-variable form
ans =
G(s)
Zero/pole/gain:
10 (s + 3) (s + 5)
-----------------------------------(s + 1) (s + 4) (s + 6) (s + 8)
ans =
T(s)
Zero/pole/gain:
10 (s + 5) (s + 3)
---------------------------------------------------------(s + 1.69) (s + 4.425) (s^2 + 12.88s + 45.73)
215
MATLAB TUTORIAL
ans =
Controller canonical form determination
AC =
– 19.0000 – 132.0000 – 376.0000 – 342.0000
1.0000
0
0
0
0
0
1.0000
0
0 1.0000
0
0
BC =
1
0
0
0
CC =
0 10.0000 80.0000 150.0000
DC =
0
ans =
Phase-variable form representation
Apv =
0
0
1.0000
0
0 1.0000
0
0
0
0
0 1.0000
– 342.0000 – 376.0000 – 132.0000 – 19.0000
Bpv =
0
0
0
1
Cpv =
150.0000 80.0000 10.0000
ans =
(b) Modal form
ans =
G(s)
Zero/pole/gain:
10 (s + 3) (s + 5)
-----------------------------------(s + 1) (s + 4) (s + 6) (s + 8)
0
216
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
ans =
T(s)
Zero/pole/gain:
10 (s + 5) (s + 3)
----------------------------------------(s + 1.69) (s + 4.425) (s^2 + 12.88s + 45.73)
ans =
Controller canonical form
AC =
– 19.0000 – 132.0000 – 376.0000 – 342.0000
1.0000
0
0
0
0
0
1.0000
0
0 1.0000
0
0
BC =
1
0
0
0
CC =
0 10.0000 80.0000 150.0000
DC =
0
ans =
Modal form
A=
– 6.4425 2.0551
– 2.0551 – 6.4425
0
0
0
0
B = – 1.6902
-2.7844
-9.8159
3.9211
0.0811
C=
– 0.2709
D=
0
0.1739
0
0
0
0
– 4.4249
0
0.0921
7.2936
0
217
MATLAB TUTORIAL
Example 3.41. Determine the state-space representation in phase-variable form for
the system shown in Fig. E 3.41.
R(s)
50
s + 7s + 10s 3 + 8s 2 + s + 25
5
4
Fig. E 3.41
Solution. Computer program is as follows:
%
MATLAB Program
‘State-space representation’
num = 50
den = [1
7
10
8
1
G = tf(num, den)
[AC, BC, CC, DC] = tf2ss(num, den);
Af = flipud(AC)
A = fliplr(Af)
B = flipud(BC)
C = fliplr(CC)
Computer response:
num =
50
Transfer function:
50
-------------------------------------------------s^5 + 7 s^4 + 10 s^3 + 8 s^2 + s + 25
Af =
0
0
0
0
0
1
0
1
0
1
0
0
–7 – 10 – 8
A=
0
1
0
0
0
1
0
0
0
0
0
0
–25 – 1 – 8 –
B=
0
0
0
0
1
1
0
0
0
–1 –
0
0
0
0
25
0
0
0
0
1
0
0
1
10 – 7
25];
C(s)
218
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
C=
50
0
0
0
0
Example 3.42. Using MATLAB, write the state equations and the output equation
for the phase-variable representation for the following systems in Fig. E 3.42.
5s + 7
R(s)
s + 7s + 3s 2 + 9s + 8
4
3
C(s)
(a)
s 4 + 3s 3 + 12s 2 + 7s + 8
R(s)
s 5 + 9s 4 + 5s 3 + 7s 2
(b)
Fig. E 3.42.
Solution. (a)
num = [5 7]
num =
5
7
den = [1 7 3 9 8]
den =
1
7
3
9
G = tf(num,den)
Transfer function:
8
5s+7
------------------------------------s^4 + 7 s^3 + 3 s^2 + 9 s + 8
[Ac, Bc, Cc, Dc] = tf2ss(num, den);
Af = flipud(Ac)
Af =
0
0
0
1
1
0
0
0
1
–7
0
–3
0
–9
0
–8
A = fliplr(Ac)
A=
–8
0
–9
0
–3
0
–7
1
0
0
0
1
1
0
0
0
C(s)
219
MATLAB TUTORIAL
B = flipud(Bc)
B=
0
0
0
1
C = fliplr(Cc)
C=
7
Part 2
5
0
0
num = [1 3 10 5 6];
den = [1 7 8 6 0 0];
G = tf(num, den)
Transfer function:
s^4 + 3 s^3 + 10 s^2 + 5 s + 6
-----------------------------------s^5 + 7 s^4 + 8 s^3 + 6 s^2
[Ac, Bc, Cc, Dc] = tf2ss(num,den);
Af = flipud(Ac);
A = fliplr(Af)
A=
0
1
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
–6 –8 –7
B = flipud(Bc)
B=
0
0
0
0
1
C = fliplr(Cc)
C=
6
5
10
3
1
Example 3.43. Find the transfer function for the following system using MATLAB.
x1
0
x
2 = – 5
x 3
0
1
–2
2
0
0
– 6
0
x1
x
2 + 3
5
x3
0
– 1 u
0
220
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
1
y=
0
x1
0
x2
1
x3
0
0
Solution. The transfer function matrix is given by
G(s) = C[sI – A]–1B
0
A = − 5
0
where
Hence
1
−2
2
0
0
0 ; B = 3
5
− 6
−1
s+2
s
1 0 0
G(s) =
5
0 0 1 0
0
1 0 0
− 1 ; C =
0 0 1
0
0
0
s + 6
−2
0
3
5
0
− 1
0
>> %MATLAB Program
>> syms s
>> C = [1 0 0; 0 0 1];
>> M = [s – 1 0; 5 s + 2 0; 0 – 2 s + 6];
>> B = [0 0; 3 – 1; 5 0];
>> C*inv(M)*B
ans =
[
3/(s^2 + 2*s + 5),
– 1/(s^2 + 2*s + 5)]
[ 6*s/(s^3 + 8*s^2 + 17*s + 30) + 5/(s + 6),
– 2*s/(s^3 + 8*s^2 + 17*s + 30)]
Example 3.44. A control system is defined by the following state space equations:
− 4
x1
x = 2
2
y = [1
− 1
− 3
1
x1
x + 3 u
2
x
2] 1
x2
Find the transfer function G(s) of the system using MATLAB.
Solution.
− 4
A=
2
− 1
1
; B = ; C = [1
− 3
3
2]
The transfer function G(s) of the system is
G(s) = C(sI – A)–1 B = [1
s + 4
2]
−2
1 1
s + 3 3
221
MATLAB TUTORIAL
= [1
=
1
s + 3 − 1 2
[(s + 4)( s + 3) + 2] 2
s + 4 5
2]
1
[ s + 7 s + 14]
2
[1
[12s + 49]
2s + 1
2]
= 2
[ s + 7 s + 14]
5s + 24
>> %MATLAB Program
>> A = [– 4 – 1; 2 – 3];
>> B = [1; 3];
>> C = [1 2];
>> D = 0;
>> [num, den] = ss2tf(A, B, C, D)
num =
0
7.0000 28.0000
den =
1.0000
7.0000 14.0000
The result is same as the one derived above.
Example 3.45. Determine the transfer function G(s) = Y(s)/R(s), for the following
system representation in state space form.
x =
−
0
3
7
0
0
1
0
0
0
5
−6
9
0
0
5
0
x+ r
7
1
5
2
y = [1 3 6 5] x
Solution.
A = [0 3 5 0; 0 0 1 0; 0 0 0 1; – 5 – 6 8 5];
B = [0; 5; 7; 2];
C = [1 3 7 5];
D = 0;
statespace = ss(A, B, C, D)
a=
x1
x2
x3
x4
x1
x2
0
0
3
0
5
1
0
0
x3
x4
0
–5
0
–6
0
8
1
5
222
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
b=
x1
x2
x3
x4
u1
0
5
7
2
y1
x1
1
c=
x2
3
x3
7
x4
5
d=
u1
y1
0
Continuous-time model.
[A, B, C, D] = tf2ss(num,den);
G = tf(num, den)
Transfer function:
s^4 + 3 s^3 + 10 s^2 + 5 s + 6
--------------------------------------s^5 + 7 s^4 + 8 s^3 + 6 s^2
Example 3.46. Determine the transfer function and poles of the system represented
in state space as follows using MATLAB.
9
x = − 3
6
−3
2
8
− 1
1
0 + 2 u(t)
3
− 2
0
y = [2 9 – 12]x; x(0) = 0
0
Solution.
% MATLAB Program
>> A = [8 – 3 4; – 7 1 0; 3 4 – 7]
A=
8 –3
4
–7
1
0
3
4 –7
>> B = [1; 3; 8]
B=
1
3
8
>> C = [1 7 – 2]
223
MATLAB TUTORIAL
C=
1
7
>> D = 0
–2
D=
0
>> [numg, deng] = ss2tf(A, B, C, D, 1)
numg =
1.0e + 003 *
0 0.0060
0.0730 – 2.8770
deng =
1.0000 – 2.0000 – 88.0000 33.0000
>> G = tf(numg, deng)
Transfer function:
6 s^2 + 73 s – 2877
----------------------------s^3 – 2 s^2 – 88 s + 33
>> poles = roots(deng)
poles =
10.2620
– 8.6344
0.3724
Example 3.47. Represent the system shown in Fig. E 3.47 using MATLAB in
(a) state space in phase-variable form
(b) state space in model form.
R(s)
10(s + 4)(s + 6)
(s + 2)(s + 5)(s + 7 )(s + 9)
+
–
Fig. E 3.47
Solution.
% MATLAB Program
‘(a) Phase-variable form’
‘G(s)’
G = zpk ([– 4 – 6] , [– 2 – 5 – 7 – 9], 10)
‘T(s)’
T = feedback (G, 1, – 1)
[numt, dent] = tfdata (T, ‘V’);
‘controller canonical form determination’
C(s)
224
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
[AC, BC, CC, DC] = tf2ss (numt, dent)
A1 = flipud (AC);
‘Phase – variable form representation’
APV = fliplr (A1)
BPV = flipud (BC)
CPV = fliplr (CC)
‘(b) Modal form’
‘G(s)’
G = zpk ([– 4 – 6] , [– 2 – 5 – 7 – 9], 10)
‘T(s)’
T = feedback (G, 1, – 1)
[numt, dent] = tfdata (T, ‘V’);
‘controller canonical form’
[AC, BC, CC, DC] = tf2ss (numt, dent)
‘Modal form’
[A, B, C, D] = canon (AC, BC, CC, DC, ‘modal’)
Computer response:
(a) Phase – variable form
Ans. =
G(s)
Zero/pole/gain:
10 (s + 4) (s + 6)
-----------------------------------(s + 2) (s + 5) (s + 7) (s + 9)
Ans. =
T(s)
Zero/pole/gain:
10 (s + 6) (s + 4)
-----------------------------------------------------(s + 2.69) (s + 5.425) (s^2 + 14.88s + 59.61)
Ans =
Controller canonical form determination
AC =
– 23.0000
1.0000
0
0
– 195.0000
0
1.0000
0
– 701.0000
0
0
1.0000
– 870.0000
0
0
0
225
MATLAB TUTORIAL
BC =
1
0
0
0
CC =
0 10.0000 100.0000 240.0000
DC =
0
Ans. =
Phase – variable form representation
APV =
0
1.0000
0
0
0
1.0000
0
0
1.0000
– 195.0000
– 23.0000
0
0
– 870.0000
BPV =
– 701.0000
0
0
0
0
1
CPV =
240.0000
100.0000
10.0000
(b) Modal form
Ans.=
G(s)
Zero/pole/gain:
10 (s + 4) (s + 6)
----------------------------------(s + 2) (s + 5) (s + 7) (s + 9)
Ans.=
T(s)
Zero/pole/gain:
10 (s + 6) (s + 4)
-------------------------------------------------(s + 2.69) (s + 5.425) (s^2 + 14.88s + 59.61)
Ans.=
Controller canonical form
0
226
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
AC =
– 23.0000
1.0000
– 195.0000
0
– 701.0000
0
1.0000
0
0
1.0000
0
0
– 870.0000
0
0
0
BC =
1
0
0
0
CC =
0
10.0000
100.0000
240.0000
DC =
0
Ans.=
Modal form
A=
– 7.4425
– 2.0551
0
0
2.0551
– 7.4425
0
0
0
0
– 5.4249
0
0
0
0
– 2.6902
B=
– 5.8222
– 13.9839
7.1614
0.2860
C=
– 0.1674
D=
0.1378
0.0504
2.0676
0
Example 3.48. Plot the step response using MATLAB for the following system represented in state space, where u(t) is the unit step.
2
− 5
x = 0 − 9
0
0
0
0
1 + 2 u(t)
− 3
1
0
y = [0 1 1]x; x(0) = 0
0
227
MATLAB TUTORIAL
Solution.
>> A = [– 5 2 0; 0 – 9 1; 0 0 – 3];
>> B = [0; 2; 1];
>> C = [0 1 1];
>> D = 0;
>> S = ss(A, B, C, D)
a=
x1
x1
–5
x2
x3
0
0
x2
2
x3
0
–9
0
1
–3
x2
1
x3
1
b=
u1
x1
x2
0
2
x3
1
y1
x1
0
c=
d=
u1
y1
0
Continuous-time model.
>> step(S)
Step Response
0.7
0.6
Amplitude
0.5
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
1.2
Time (sec)
Fig. E 3.48.
1.4
1.6
1.8
2
228
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Example 3.49.
A control system is defined by
1
x1
0
x = − 30 − 7
2
y1
1
y = 0
2
0
1
x1
1 1 u1
x + 0 1 u
2
2
x1
x
2
Plot the four sets of Bode diagrams for the system [two for input1, and two for input 2]
using MATLAB.
Solution. There are 4 sets of Bode diagrams (2 for input1 and 2 for input 2)
>> %Bode Diagrams
>> A = [0 1; – 30 – 7];
>> B = [1 1; 0 1];
>> C = [1 0; 0 1];
>> D = [0 0; 0 0];
>> bode(A, B, C, D)
Bode Diagram
To Y(1)
To Y(1)
To Y(2)
To Y(2)
Phase (deg) ; Magnitude (dB)
From U(1)
From U(2)
0
– 50
– 100
0
– 45
– 90
– 135
100
0
– 100
180
0
– 180
10
0
10
1
2
3
0
1
10
10 10
10
Frequency (rad/sec)
10
2
10
3
Fig. E 3.49 Bode diagrams.
Example 3.50. Draw a Nyquist plot for a system defined by
0 1 x1
0
x1
x = − 30 7 x + 30 u
2
2
y = [1
x
0] 1 + [0]u
x2
using MATLAB.
Solution. Since the system has a single input u and a single output y, a Nyquist plot
can be obtained by using the command nyquist (A, B, C, D) or nyquist (A, B, C, D, 1).
229
MATLAB TUTORIAL
>> %MATLAB Program
>> A = [0 1; – 30 7];
>> B = [0; 30];
>> C = [1 0];
>> D = [0];
>> nyquist(A, B, C, D)
>> grid
>> title(‘Nyquist plot’)
Nyquist Diagram
0.8
0.6
Imaginary Axis
0.4
0.2
0
– 0.2
– 0.4
– 0.6
– 0.8
–1
– 0.5
0
Real Axis
0.5
1
Fig. E 3.50 Nyquist plot.
Example 3.51. A control system is defined by
x1
1
− 1 − 1 x1
+
x = 7
0 x2
2
1
y1
1
y = 0
2
0 x1
0
+
1 x2
0
0
0
1 u1
0 u2
u1
u
2
The system has two inputs and two outputs. The four sinusoidal output-input relationships are given by
y (jω )
y1 (jω ) y2 (jω ) y1 (jω )
,
,
, and 2
u2 (jω )
u1 (jω ) u1 (jω ) u2 (jω )
Draw the Nyquist plots for the system by considering the input u1 with input u2 as zero
and vice versa.
Solution. The four individual plots are obtained by using the MATLAB command nyquist
(A, B, C, D).
>> %MATLAB Program
>> A = [– 1 – 1; 7 0];
230
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> B = [1 1 ; 1 0];
>> C = [1 0 ; 0 1];
>> D = [0 0 ; 0 0];
>> nyquist(A, B, C, D)
Nyquist Diagram
From U(1)
From U(2)
To Y(1)
0
– 0.5
2
To Y(2)
Imaginary Axis
0.5
0
–2
–1
0
1
2
–1
Real Axis
0
1
Fig. E 3.51 Nyquist plots.
Example 3.52. Obtain the unit-step response, unit-ramp response, and unit-impulse
response of the following system using MATLAB
x1
− 1 − 1.5
x = 2
0
2
y = [1
1.5
x1
x + 0 u
2
x
0] 1
x2
where u is the input and y is the output.
Solution.
>> %Unit-step response
>> A = [– 1 – 1.5 ; 2 0];
>> B = [1.5 ; 0];
>> C = [1 0];
>> D = [0];
>> [y, x, t] = step(A, B, C, D);
>> plot(t, y)
>> grid
>> title(‘Unit-step response’)
>> xlabel(‘t Sec’)
>> ylabel(‘Output’)
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MATLAB TUTORIAL
Unit-Step response
0.6
0.5
0.4
Output
0.3
0.2
0.1
0
– 0.1
– 0.2
– 0.3
0
2
4
6
t Sec
8
10
Fig. E 3.52(a) Unit-step response.
>> %Unit-ramp response
>> A = [– 1 – 1.5 ; 2 0];
>> B = [1.5 ; 0];
>> C = [1 0];
>> D = [0];
>> % New enlarged state and output equations
>> AA = [A zeros (2, 1); C 0];
>> BB = [B; 0];
>> CC = [0 1];
>> DD = [0];
>> [z, x, t] = step (AA, BB, CC, DD);
>> x3= [0 0 1]*x’; plot (t, x3, t, t, ‘–’)
>> grid
>> title (‘Unit-ramp response’)
>> xlabel (‘t Sec’)
>> ylabel (‘Output and unit-ramp input’)
>> text (12, 1.2, ‘Output’)
12
232
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Unit-ramp response
20
18
Output and unit-ramp input
16
14
12
10
8
6
4
2
0
Output
0
5
10
t Sec
15
20
Fig. E 3.52(b) Unit-ramp response.
>> %Unit-impulse response
>> A = [– 1 – 1.5 ; 2 0];
>> B = [1.5; 0];
>> C = [1 0];
>> D = [0];
>> impulse (A, B, C, D)
Impulse Response
1.5
1
Amplitude
0.5
0
– 0.5
–1
0
2
4
6
Time (sec)
8
10
12
Fig. E 3.52(c) Unit-impulse response.
Example 3.53. Obtain the unit-step curves for the following system using MATLAB.
x1
− 1 − 1 x1
1 1 u1
x = 7
x +
0
2
2
1 0 u2
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MATLAB TUTORIAL
y1
0 0 u1
1 0 x1
y =
x + 0 0 u
2
2
0 1 2
Solution.
>> %MATLAB Program
>> A = [– 1, – 1; 7 0];
>> B = [1 1 ; 1 0];
>> C = [1 0 ; 0 1];
>> D = [0 0 ; 0 0];
>> step(A, B, C, D)
Bode Diagram
From U(1)
From U(2)
0.4
To Y(1)
0.2
0
– 0.4
2
1.5
To Y(2)
Amplitude
– 0.2
1
0.5
0
0
2
4
6
8
10 12 0 2
Time (sec)
4
6
8
10 12
Fig. E 3.53 Step Response.
Example 3.54. Obtain the unit-step response and unit-ramp response of the following
system using MATLAB.
x1
− 5
x
1
=
2
x 3
0
y = [0
25
Solution.
>> %MATLAB Program
>> A = [– 5 – 25 – 5; 1 0 0; 0 1 0];
>> B = [1; 0; 0];
>> C = [0 25 5];
>> D = [0];
− 25
0
1
− 5
0
0
x1
1
x
0 u
+
2
x3
0
x1
5] x2 + [0]u
x3
234
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> [y, x, t] = step(A, B, C, D);
>> plot(t, y)
>> grid
>> title(‘Unit-response’)
>> xlabel(‘t Sec’)
>> ylabel(‘Output y(t)’)
Unit-response
1.4
1.2
Output y(t)
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
t Sec
6
7
8
9
10
Fig. E 3.54(a) Unit-step response.
Unit-ramp response:
− 5
1
AA =
0
0
− 25
−5
0
0
1
0
25
5
0
0
=
0
0
0
1
0 B
=
BB = 0
0 0
CC = [0 25
>> %MATLAB Program
5 0] = [C 0]
>> A = [– 5 – 25 – 5 ; 1 0 0 ; 0 1 0];
>> B = [1 ; 0 ; 0];
>> C = [0 25 5];
>> D = [0];
>> AA = [A zeros(3, 1); C 0];
>> BB = [B; 0];
A
25
5
0
0
= A zeros(2, 1); C 0]
0
0
235
MATLAB TUTORIAL
>> CC = [C 0];
>> DD = [0];
>> t = 0:0.01:5;
>> [z, x, t] = step(AA, BB, CC, DD, 1, t);
>> P = [0 0 0 1]*x’;
>> plot(t, P, t, t)
>> grid
>> title(‘Unit-ramp response’)
>> xlabel(‘t Sec’)
>> ylabel(‘Input and output’)
Unit-ramp response
5
4.5
4
Input and output
3.5
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
t Sec
3
3.5
4
4.5
5
Fig. E 3.54(b) Unit-ramp response.
Example 3.55. A control system is given by
3
x1
x
0
=
2
x 3
0
0
1
4
0
0
5
0
x1
x
2
+
2
x3
0
2
u1
0
u2
1
x1
0
x2
1
0
x3
Determine the controllability and observability of the system using MATLAB.
1
y=
0
Solution:
>> %MATLAB Program
>> A = [3 0 0; 0 1 0; 0 4 5];
>> B = [0 2; 2 0; 0 1];
>> C = [1 2 0; 0 1 0];
2
236
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
>> D = [0 0; 0 0];
>> rank ([B A*B A^2*B])
ans =
3
>> rank ([C’ A*C’ A^2*C’])
ans =
3
>> rank ([C*B C*A*B C*A^2*B])
ans =
2
From the above, we observe that the system is state controllable but not completely
observable. It is output controllable.
Example 3.56. Consider the system
3
x1
x
2 = 0
x 3
0
0
1
3
0
0
2
x1
x
2
x3
The output is given by
x1
y = [1 1 1] x2
x3
(a) determine the observability of the system using MATLAB
(b) show that the system is completely observable if the output is given by
x1
y1
1 1 1
y =
x2
2
1 3 2 x
3
using MATLAB.
Solution.
>> %MATLAB Program
>> A = [3 0 0; 0 1 0; 0 3 2];
>> C = [1 1 1];
>> rank ([C′ A′*C′ A′^2*C′])
ans =
3
>> A = [3 0 0; 0 1 0; 0 3 2];
>> C = [1 1 1; 1 3 2];
>> rank ([C′ A′*C′ A′^2*C′])
ans =
3
From the above, we observe that the system is observable and controllable.
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MATLAB TUTORIAL
Example 3.57. Consider the following state equation and output equation
x1
− 1
x
2 = 0
1
x 3
−3
−2
0
− 2
1
− 1
3
x1
x
2 + 0 u
1
x3
x1
y = [1 1 0] x2
x3
Determine if the system is completely state controllable and completely observable using
MATLAB.
Solution. The controllability and observability of the system can be obtained by examining the rank condition of
[B AB A2B] and [C′′ A′′C′′ (A′′)2C′′]
>> %MATLAB Program
>> A = [– 1 – 3 – 2; 0 – 2 1; 1 0 – 1];
>> B = [3; 0; 1];
>> C = [1 1 0];
>> D = [0];
>> rank ([B A*B A^2*B])
ans =
3
>> rank ([C′ A′*C′ A′^2*C′])
ans =
3
We observe the rank of [B AB A2B] is 3 and the rank of [C′′ A′′*C′′ (A′′) 2*C′′] is 3, the
system is completely state controllable and observable.
Example 3.58. Diagonalize the following system using MATLAB.
− 7
x = 15
− 8
y = [1 – 3
−5
6
−3
5
− 12 x +
4
5] x
Solution.
% MATLAB Program:
A = [– 7 – 5 5 ; 15 6 – 12 ; – 8 – 3 4];
B = [– 1; 4; 2];
C = [1; – 3; 5];
[P, D] = eig (A);
− 1
4
r
2
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Ad = inv (P) * A * P
Bd = inv (P) * B
Cd = inv (P)* C
Computer response:
>> Ad = inv (P)*A*P
Ad =
2.4555 + 6.0296i – 0.0000 – 0.0000 – 0.0000 + 0.0000i
– 0.0000 + 0.0000i 2.4555 – 6.0296i – 0.0000 – 0.0000i
0.0000 + 0.0000i 0.0000 – 0.0000i – 1.9110 – 0.0000i
>> Bd = inv (P)*B
Bd =
1.8397 + 2.1026i
1.8397 – 2.1026i
3.3254 + 0.0000i
Cd =
– 2.4324 + 5.4360i
– 2.4324 – 5.4360i
2.6093 + 0.0000i
Example 3.59. Determine the eigenvalues of the following system using MATLAB.
0
x = 0
− 2
2
2
2
0
− 9 x +
5
0
0
r
2
y = [0 0 1] x
Solution.
>> A = [0 2 0; 0 2 – 7; – 2 2 5]; %Define the matrix above
>> eig (A) % Calculate the eigenvalues of matrix A.
ans =
2.0000
2.5000 + 3.4278i
2.5000 - 3.4278i
Example 3.60. For the following forward path of a unity feedback system in state space
representation, determine if the closed-loop system is stable using the Routh-Hurwitz criterion
and MATLAB.
0
x = 0
− 2
2
1
−4
y = [0 1 1] x.
0
9 x +
− 6
0
0
r
2
239
MATLAB TUTORIAL
Solution.
>> A = [0 2 0; 0 1 9; – 2 – 4 – 6]; % Define the matrix
>> B = [0; 0; 2]; % Define the matrix.
>> C = [0 1 1]; % Define the matrix
>> D = 0;
>> ‘G’;
>> G = ss (A, B, C, D); %Create a state-space model
>> ‘T ’;
>> T = Feedback (G, 1);
>> ‘Eigenvalues of T are’;
>> ssdata (T); % Create a state-space model
>> eig (T) % Determine Eigenvalues
ans =
ans =
– 0.8872
– 3.0564 + 5.5888i
– 3.0564 – 5.5888i
The closed loop system is stable as the numbers are all negative with regards to the axis
coordinate system used for Routh-Hurwitz. Negative values are stable, positive values are unstable.
Example 3.61. For the following path of a unity feedback system in state space representation, determine if the closed-loop system is stable using the Routh-Hurwitz criterion and
MATLAB.
0
x = 0
− 3
1
1
−4
0
7 x +
− 6
0
0
u
2
y = [0 1 1] x.
Solution.
% MATLAB Program
A = [0 1 0;
01
B = [0 0 2];
C = [0 1 1];
D = 0;
‘G’
G = ss (A, B, C, D)
‘T ’
T = feedback (G, 1)
‘Eigenvalues of T are’
7;
–3
–4
– 6];
240
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
ssdata (T);
eig (T)
Computer response:
ans =
G
a=
x1
x2
x3
x1
x2
0
0
1
1
0
5
x3
–3
–4
–5
x1
u1
0
x2
x3
0
1
b=
c=
x1
x2
x3
y1
0
1
1
y1
u1
0
d=
Continuous-time model.
ans =
T
a=
x1
x1
0
x2
1
x3
0
x2
x3
0
–3
1
–6
7
–8
b=
u1
x1
x2
0
0
x3
c=
2
y1
d=
y1
x1 x2 x3
0 1 1
u1
0
241
MATLAB TUTORIAL
Continuous-time model:
ans =
Eigenvalues of T are
ans =
– 0.7112
– 3.1444 + 4.4317i
– 3.1444 – 4.4317i
SUMMARY
The classical methods of control systems engineering using MATLAB including the root
locus analysis and design, frequency response methods of analysis, Bode, Nyquist, and Nichols
plots, second order systems approximations, phase and gain margin and bandwidth, state space
variable method, and controllability and observability are covered in this chapter. With this
foundation of basic application of MATLAB, the Chapter provides opportunities to explore advanced topics in control systems engineering.
Extensive worked examples are included with a great number of exercise problems to
guide the student to understand and as an aid for learning about the analysis and design of
control systems using MATLAB.
PROBLEMS
1. [Reduction of multiple subsystems]
Reduce the system shown in Fig. P 3.1 to a single transfer function, T(s) = C(s)/R(s)
using MATLAB. The transfer functions are given as
G1(s) = 1/(s + 3)
G2(s) = 1/(s2 + 3s + 5)
G3(s) = 1/(s + 7)
G4(s) = 1/s
G5(s) = 7/(s + 5)
G6(s) = 1/(s2 + 3s + 5)
G7(s) = 5/(s + 6)
G8(s) = 1/(s + 8)
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
G8(s)
+
R(s) +
G1(s)
+
G3(s)
G6(s)
+
–
–
G7(s)
+
G2(s)
+
C(s)
G4(s)
+
G5(s)
Fig. P 3.1
2. Obtain the unit-step response plot for the unity-feedback control system whose open
loop transfer function is
G(s) =
8
s(s + 1)( s + 3)
using MATLAB. Determine also the rise time, peak time, maximum overshoot, and
settling time in the unit-step response plot.
3. Obtain the unit-acceleration response curve of the unity-feedback control system whose
open loop transfer function is given by
G(s) =
8(s + 1)
s2 (s + 3)
using MATLAB. The unit-acceleration input is defined by
r(t) =
1 2
t
2
(t ≥ 0)
4. The feed forward transfer function G(s) of a unity-feedback system is given by
G(s) =
k(s + 3)2
(s2 + 5)(s + 4)2
Plot the root loci for the system using MATLAB.
5. For the unity feedback shown in Fig. P 3.5, where
G(s) =
K
s(s + 3)(s + 4)(s + 5)
Obtain the following:
(a) display a root locus and pause
(b) draw a close-up of the root locus where the axes go from – 2 to 0 on the real axis
and – 2 to 2 on the imaginary axis
(c) overlay the 15% overshoot line on the close-up root locus
243
MATLAB TUTORIAL
(d) allow you to select interactively the point where the root locus crosses the 15%
overshoot line, and respond with the gain at that point as well as all of the closedloop poles at that gain
(e) find the step response at the gain for 15% overshoot.
R(s) +
G(s)
C(s)
–
Fig. P 3.5
6. For the system shown in Fig. P 3.6, determine the following using MATLAB
(a) display a root locus and phase
(b) display a close-up of the root locus where the axes go from – 2 to 2 on the real axis
and – 2 to 2 on the imaginary axis
(c) overlay the 0.707 damping ratio line on the close-up root locus
(d) obtain the step response at the gain for 0.707 damping ratio.
K
s(s + 3) (s + 5) (s + 7)
R(s) +
C(s)
–
(s + 25)
(s + 10s + 100)
2
Fig. P 3.6
7. Write a program in MATLAB to obtain a Bode plot for the transfer function
G(s) =
(5s3 + 51s2 + 20s + 400)
(s4 + 12s3 + 60s2 + 300 s + 250)
8. Write a program in MATLAB for the unity feedback system with G(s) = K/[s(s + 7)
(s + 15)] so that the value of gain K can be input. Display the Bode plots of t a system
for the input value of K. Determine and display the gain and phase margin for the
input value of K.
9. Write a program in MATLAB for the system shown in Fig. P 3.9 so that the value of K
can be input (K = 40).
R(s) +
E(s)
K (s + 3)
s(s + 4s + 20)
C(s)
2
–
Fig. P 3.9
(a) Display the closed-loop magnitude and phase frequency response for unity feedback
system with an open-loop transfer function, KG(s).
244
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
(b) Determine and display the peak magnitude, frequency of the peak magnitude, and
bandwidth for the closed-loop frequency response for the input value of K.
10. Write a program in MATLAB for a unity feedback system with the forward-path transfer function given by
G(s) =
7(s + 3)
s(s + 4 s + 12)
2
(a) Draw a Nichols plot of an open-loop transfer function.
(b) The user can read the Nichols plot display and enter the value of Mp.
(c) Obtain the closed-loop magnitude and phase plots.
(d) Display the expected values of percent overshoot, settling time, and peak time.
(e) Plot the closed-loop step response.
11. For the system shown in Fig. P 3.11, write a program in MATLAB that will use an
open-loop transfer function G(s).
80(s + 2)
s(s + 1) (s + 3)
R(s) +
C(s)
–
System 1
R(s) +
E(s)
40(s + 3) (s + 5)
s(s + 2) (s + 4) (s + 6)
–
System 2
Fig. P 3.11
(a) Obtain a Bode plot
(b) Estimate the percent overshoot, settling time, and peak time
(c) Obtain the closed-loop step response.
12. Write a program in MATLAB for a unity-feedback system with
G(s) =
K (s + 3)
(s + 5s + 80)(s2 + 4 s + 20)
2
(a) plot the Nyquist diagram
(b) display the real-axis crossing value and frequency.
13. Write a program in MATLAB to obtain the Nyquist and Nichols plots for the following
transfer function for k = 30.
G(s) =
k(s + 1)(s + 2 + 5i) (s + 2 − 5i)
(s + 2)(s + 5)(s + 7)(s + 2 + 7i)(s + 2 − 7i)
245
MATLAB TUTORIAL
14. Write a program in MATLAB for a unity feedback system with the forward-path transfer function given by
G(s) =
(a)
(b)
(c)
(d)
(e)
15. For
7(s + 3)
s(s + 4 s + 12)
2
Draw a Nichols plot of an open-loop transfer function.
The user can read the Nichols plot display and enter the value of Mp.
Obtain the closed-loop magnitude and phase plots.
Display the expected values of percent overshoot, settling time, and peak time.
Plot the closed-loop step response.
a unit feedback system with the forward-path transfer function.
G(s) =
K
s(s + 3)(s + 10)
and a delay of 0.5 second, estimate the percent overshoot for K = 40 using a secondorder approximation. Model the delay using MATLAB function pade(T, n). Determine the unit step response and check the second-order approximation assumption
made.
16. For the control system shown in Fig. 3.16:
(a) plot the root loci of the system
(b) find the value of gain K such that the damping ratio ξ of the dominant closed-loop
poles is 0.5
(c) obtain all the closed-loop poles using MATLAB
(d) plot the unit-step response curve using MATLAB.
K
Input
Output
s(s 2 + 5s + 7 )
Fig. P 3.16
17. Fig. P 3.17 shows a position control system with velocity feedback. What is the response c(t) to the unit step input ?
R(s)
+
+
–
–
80
s(s + 3)
C(s)
1/s
0.15
Fig. P 3.17
18. The open-loop transfer function G(s)H(s) of a control system is
G(s)H(s) =
K
K
= 4
3
2
s + s + 8.25s2 + 4 s
s(s + 0.5)(s + 0.5s + 8)
Plot the root loci for the system using MATLAB.
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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
19. Design a compensator for the system shown in Fig. P 3.19 such that the dominant
closed-loop poles are located at s = – 1 ± j 3 .
+
1
Gc(s)
s2
–
Fig. P 3.19
20. For the control system shown in Fig.3.20:
(a) design a PID control Gc(s) such that the dominant closed-loop poles located at s = – 1
± j1.
(b) select a = 0.6 for the PID controller and find the values of K and b
(c) root-locus plot using MATLAB.
+
R (s)
–
P ID c o n tro ller
( s + a )( s + b )
K
s
G c (s)
P la n t G (s)
1
C (s)
(s2 + 0 . 8 )
Fig. P 3.20
21. Draw a Bode diagram of the open-loop transfer function G(s) of the closed-loop system shown in Fig. P 3.21 and obtain the phase margin and gain margin.
R (s)
1 8 ( s +1)
s ( s + 3 )( s 2 + 2 s + 9 )
C (s)
Fig. P 3.21
22. A block diagram of a process control system is shown in Fig. P 3.22. Find the range of
gain for stability.
Ke−s
s +1
Fig. P 3.22
23. For the control system shown in Fig. P 3.23:
(a) draw a Bode diagram of the open-loop transfer function
(b) find the value of the gain K such that the phase margin is 50º
(c) find the gain margin of the system with the gain obtained in (b).
247
MATLAB TUTORIAL
+
K
–
s + 0. 3
s + 0. 7
12
s(s + 2)
Fig. P 3.23
24. Obtain the unit-step response and unit-ramp response of the following system using
MATLAB.
− 5
x1
x
2 = 1
0
x3
− 25
0
1
− 5
0
0
1
x1
x
2 + 0 u
0
x3
x1
y = [0 25 5] x2 + [0]u.
x3
25. For the mechanical system shown in Fig. P 3.25, the input and output are the displacement x and y respectively. The input is a step displacement of 0.4 m. Assuming
the system remains linear throughout the transient period and m = 3 kg, c = 3 N-s/m,
and k = 1 N/m, determine the response of the system using MATLAB.
y
k
c
x
m
Fig. P 3.25
26. Using MATLAB, write the state equations and the output equation for the phasevariable representation for the following systems in Fig. P 3.26.
R(s)
3s + 7
4
3
s + s + 2s 2 + 7s + 5
C(s)
(a)
R(s)
s 4 + 3s 3 + 10s 2 + 5s + 6
s 5 + 7s 4 + 8s 3 + 6s 2
C(s)
(b)
Fig. P 3.26
27. Determine the transfer function and poles of the system represented in state space as
following using MATLAB.
248
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
−5
1
5
9
x = − 4
3
2
0 x +
− 7
2
5 u(t)
7
0
y = [1 7 – 2] x ; x(0) = 0
0
28. Obtain the root locus diagram of a system defined in state space using MATLAB. The
system equations are
x = Ax + Bu and y = Cx + Du and
where r is the input and y is the output.
The matrices A, B, C, and D are:
0
0
A=
− 150
1
0
− 50
0
1 ; B =
− 15
u=r–y
0
1
− 15
C = [1 0 0] ; D = [0]
29. Obtain the Bode diagram of the following system using MATLAB.
x1
0 1
x = − 30 7
2
x1
0
x + 30 u
2
x
y = [1 0] 1
x2
The input of the system is u and the output is y.
30. A control system is defined by
x1
− 1
x = 7.5
2
− 2 x1
1 1 u1
+
0 x2
1 0 u2
y1
1 0 x1
y = 0 1 x
2
2
0 0 u1
0 0
u2
Write a MATLAB program to obtain the following plots:
(a) two Nyquist plots for the input u1 in one diagram
(b) two Nyquist plots for the input u2 in one diagram.
31. Obtain the unit-ramp response of the system defined by
x1
0
x = − 3
2
y = [1
2 x1
0
+ u
− 1 x2
2
x
0] 1
x2
where u is the unit-ramp input. Use lsim command to obtain the response.
249
MATLAB TUTORIAL
32. Obtain the response curves y(t) using MATLAB for the following system.
x1
– 1 1 x1
0
x = – 1 0 x + 2 u
2
2
x
0] 1
x2
y = [1
The input u is given by:
(a) u= unit-step input
(b) u = e–t
The initial state x(0) = 0.
33. Plot the step response using MATLAB for the following system represented in state
space, where u(t) is the unit step.
− 3
x = 0
0
2
−7
0
0
1 x +
− 4
0
1 u(t)
1
0
y = [0 1 1] x ; x(0) = 0
0
34. Diagonalize the following system using MATLAB.
− 10
x = 15
− 8
−5
4
−3
7
− 12 x +
6
1
2 r
3
y = [1 – 2 3]x
35. Determine to unit-ramp response of the system defined by
2 x1
x1
0
0
x = − 3 − 3 x + u
2
2
2
y = [1
x
0] 1
x2
Using MATLAB where u is the unit-ramp input. Use lsim command in MATLAB.
36. Obtain the unit-impulse response of the following system using MATLAB
1 x1
0
0
x1
x = − 1 − 2 x + 1 u
2
2
x
y = [1 1] 1 + [0]u
x2
250
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
37. A control system is defined by
x1
− 1
x
0
=
2
x3
2
−3
−2
− 3
1
− 1
0
x1
3
x
0
2 + u
x3
1
x1
y = [1 2 0] x2
x3
Determine the controllability and observability of the system using MATLAB.
38. Determine the eigenvalues of the following system using MATLAB.
0
x = 0
− 2
1
1
1
0
− 5 x +
3
0
0 u
1
y = [0 0 1]x.
39. For the following path of a unity feedback system in state space representation, determine if the closed-loop system is stable using the Routh-Hurwitz criterion and
MATLAB.
0
x = 0
− 3
1
1
−4
0
5 x +
− 5
0
0 u
1
y = [0 1 1]x.
40. Consider the differential equation system given by
y = 4 y + 3y = 0 ; y(0) = 0.2 ;
y (0) = 0.1
Find the state-space equation for the system. Also, obtain the response y(t) of the
system subject to the given initial conditions using MATLAB.
251
BIBLIOGRAPHY
BIBLIOGRAPHY
There are several outstanding text and reference books on feedback control systems and
MATLAB that merit consultation for those readers who wish to pursue these topics further.
The following list is but a representative sample of the many excellent references on analysis
and design of feedback control systems and MATLAB.
Control Systems
Anand, D.K., Introduction to Control Systems, 2nd ed., Pergamon Press, New York, NY,
1984.
Atkinson, P., Feedback Control Theory for Engineers, 2nd ed., Heinemann, 1977.
Bateson, R.N., Introduction to Control System Technology, Prentice Hall, Upper Saddle
River, NJ, 2002.
Bayliss, L.E., Living Control Systems, English Universities Press Limited, London, UK,
1966.
Beards, C.F., Vibrations and Control System, Ellis Horwood, 1988.
Benaroya, H., Mechanical Vibration-Analysis, Uncertainties, and Control, Prentice Hall,
Upper Saddle River, NJ, 1998.
Bode, H.W., Network Analysis and Feedback Design, Van Nostrand Reinhold, New York,
NY, 1945.
Bolton, W., Control Engineering, 2nd ed., Addison Wesley Longman Ltd., Reading, MA,
1998.
Brogan, W.L., Modern Control Theory, Prentice Hall, Upper Saddle River, NJ, 1985.
Buckley, R.V., Control Engineering, Macmillan, New York, NY, 1976.
Burghes, D., and Graham, A., Introduction to Control Theory Including Optimal Control,
Ellis Horwood, 1980.
Cannon, R.H., Dynamics of Physical Systems, McGraw Hill, New York, NY, 1967.
Chesmond, C.J., Basic Control System Technology, Edward Arnold, 1990.
Clark, R.N., Introduction to Automatic Control Systems, Wiley, New York, NY, 1962.
D’Azzo, J.J., and Houpis, C.H., Linear Control System Analysis and Design: Conventional
and Modern, 4th ed., McGraw Hill, New York, NY, 1995.
Dorf, R.C., and Bishop, R.H., Modern Control Systems, 9th ed., Prentice Hall, Upper Saddle
River, NJ, 2001.
Dorsey, John., Continuous and Discrete Control Systems, McGraw Hill, New York, NY,
2002.
Douglas, J., Process Dynamics and Control, Volumes I and II, Prentice Hall, Englewood
Cliffs, NJ, 1972.
Doyle, J.C., Francis, B.A., and Tannenbaum, A., Feedback Control Theory, Macmillan,
New York, NY, 1992.
Dransfield, P., and Habner, D.F., Introducing Root Locus, Cambridge University Press,
Cambridge, 1973.
Dukkipati, R.V., Control systems, Narosa Publishing House, New Delhi, India, 2005.
252
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Dukkipati, R.V., Engineering System Dynamics, Narosa Publishing House, New Delhi, India,
2004.
Dukkipati, R.V., Vibration Analysis, Narosa Publishing House, New Delhi, India, 2004.
Evans, W.R., Control System Dynamics, McGraw Hill, New York, NY, 1954.
Eveleigh, V.W., Control System Design, McGraw Hill, New York, NY, 1972.
Franklin, G.F., David Powell, J., and Abbas Emami-Naeini., Feedback Control of
Dynamic Systems, 3rd ed., Addison Wesley, Reading, MA, 1994.
Friedland, B., Control System Design, McGraw Hill, New York, NY, 1986.
Godwin, Graham E., Graebe, Stefan F., and Salgado, Maria E., Control System Design,
Prentice Hall, Upper Saddle River, NJ, 2001.
Grimble, Michael J., Industrial Control Systems Design, Wiley, New York, NY, 2001.
Gupta, S., Elements of Control Systems, Prentice Hall, Upper Saddle River, NJ, 2002.
Guy, J.J., Solution of Problems in Automatic Control, Pitman, 1966.
Healey, M., Principles of Automatic Control, Hodder and Stoughton, 1975.
Jacobs, O.L.R., Introduction to Control Theory, Oxford University Press, 1974.
Johnson, C., and Malki, H., Control Systems Technology, Prentice Hall, Upper Saddle
River, NJ, 2002.
Kailath, T., Linear Systems, Prentice Hall, Upper Saddle River, NJ, 1980.
Kuo, B.C., Automatic Control Systems, 6th ed., Prentice Hall, Englewood Cliffs, NJ, 1991.
Leff, P.E.E., Introduction to Feedback Control Systems, McGraw Hill, New York, NY, 1979.
Levin, W.S., Control System Fundamentals, CRC Press, Boca Raton, FL, 2000.
Levin, W.S., The Control Handbook, CRC Press, Boca Raton, FL, 1996.
Lewis, P., and Yang, C., Basic Control Systems Engineering, Prentice Hall, Upper Saddle
River, NJ, 1997.
Marshall, S.A., Introduction to Control Theory, Macmillan, 1978.
Mayr, O., The Origins of Feedback Control, MIT Press, Cambridge, MA, 1970.
Mees, A.J., Dynamics of Feedback Systems, Wiley, New York, NY, 1981.
Nise, Norman, S., Control Systems Engineering, 3rd ed., Wiley, New York, NY, 2000.
Ogata, K., Modern Control Engineering, 3rd ed., Prentice Hall, Englewood Cliffs, NJ, 1997.
Ogata, K., State Space Analysis of Control Systems, Prentice Hall, Upper Saddle River, NJ,
1967.
Ogata, K., System Dynamics, 3rd ed., Prentice Hall, Upper Saddle River, NJ, 1998.
Palm III, W.J., Control Systems Engineering, Wiley, New York, NY, 1986.
Paraskevopoulos, P.N., Modern Control Engineering, Marcel Dekker , Inc., New York,
NY, 2003.
Phillips, C.L., and Harbour, R.D., Feedback Control Systems, 4th ed., Prentice Hall, Upper
Saddle River, NJ, 2000.
Power, H.M., and Simpson, R.J., Introduction to Dynamics and Control, McGraw Hill,
New York, NY, 1978.
Raven, F.H., Automatic Control Engineering, 4th ed., McGraw Hill, New York, NY, 1987.
Richards, R.J., An Introduction to Dynamics and Control, Longman, 1979.
253
BIBLIOGRAPHY
Richards, R.J., Solving Problems in Control, Longman Scientific & Technical, Wiley, New
York, NY, 1993.
Rohrs, C.E., Melsa, J.L., and Schultz, D.G., Linear Control Systems, McGraw Hill, New
York, NJ, 1993.
Rowell, G., and Wormley, D., System Dynamics, Prentice Hall, Upper Saddle River, NJ,
1999.
Schwarzenbach, J., and Jill, K.F., System Modeling and Control, 2nd ed., Arnold, 1984.
Shearer, J.L., Kulakowski, B.T., and Gardner, J.F., Dynamic Modeling and Control of
Engineering Systems, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 1997.
Shinners, S. M., Modern Control System Theory and Design, 2nd ed., Wiley Inter science,
New York, NY, 1998.
Sinha, N.K., Control Systems, Holt Rinehart and Winston, New York, NY, 1986.
Smith, O.J.M., Feedback Control Systems, McGraw Hill, New York, NY, 1958.
Stefano, D. III., Stubberud, A.R., and Williams, I.J., Schaum’s Outline Series Theory
and Problems of Feedback and Control Systems, McGraw Hill, New York, NY, 1967.
Thompson, S., Control Systems: Engineering and Design, Longman, 1989.
Truxal, J.G., Control System Synthesis, McGraw Hill, New York, NY, 1955.
Umez-Eronini, E., System Dynamics and Control, Brooks/Cole Publishing Company,
Pacific Grove, CA, 1999.
Vu, H.V., Control Systems, McGraw Hill Primis Custom Publishing, New York, NY, 2002.
Vukic, Z., Kuljaca, L., Donlagic, D., and Tesnjak, S., Nonlinear Control Systems, Marcel
Dekker , Inc., New York, NY, 2003.
Welbourn, D.B., Essentials of Control Theory, Edward Arnold, 1963.
Weyrick, R.C., Fundamentals of Automatic Control, McGraw Hill, New York, NY, 1975.
MATLAB
Chapman, S.J., MATLAB Programming for Engineers, 2nd ed., Brooks/Cole, Thomson
Learning, Pacific Grove, CA, 2002.
Dabney, J.B., and Harman, T.L., Mastering SIMULINK 4, Prentice Hall, Upper Saddle
River, NJ, 2001.
Djaferis, T.E., Automatic Control-The Power of Feedback using MATLAB, Brooks/Cole,
Thomson Learning, Pacific Grove, CA, 2000.
Etter, D.M., Engineering Problem Solving with MATLAB, Prentice-Hall, Englewood Cliffs,
NJ, 1993.
Gardner, J.F., Simulation of Machines using MATLAB and SIMULINK, Brooks/Cole,
Thomson Learning, Pacific Grove, CA, 2001.
Harper, B. D., Solving Dynamics Problems in MATLAB, 5th ed, Wiley, New York, 2002.
Harper, B. D., Solving Statics Problems in MATLAB, 5th ed, Wiley, New York, 2002.
Herniter, M.E., Programming in MATLAB, Brooks/Cole, pacific Grove, CA, 2001.
Karris, S.T., Signals and Systems with MATLAB Applications, Orchard Publications,
Fremont, CA, 2001.
Leonard, N.E., and Levine, W.S., Using MATLAB to Analyze and Design Control Systems,
Addison-Wesley, Redwood City, CA, 1995.
254
ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB
Lyshevski, S.E., Engineering and Scientific Computations Using MATLAB, Wiley, New
York, 2003.
Moler, C., The Student Edition of MATLAB for MS-DOS Personal Computers with 3-1/2’’
Disks, MATLAB Curriculum Series, The MathWorks, Inc., 2002.
Ogata, K., Designing Linear Control Systems with MATLAB, Prentice Hall, Upper Saddle
River, NJ, 1994.
Ogata, K., Solving Control Engineering Problems with MATLAB, Prentice Hall, Upper Saddle
River, NJ, 1994.
Pratap, Rudra., Getting Started with MATLAB-A Quick Introduction for Scientists and
Engineers, Oxford University Press, New York, NY, 2002.
Saadat, Hadi., Computational Aids in Control Systems using MATLAB, McGraw Hill, New
York, NY, 1993.
Sigman,K., and Davis, T.A., MATLAB Primer, 6th ed, Chapman& Hall/CRCPress, Boca
Raton, FL, 2002.
The MathWorks, Inc., SIMULINK, Version 3, The MathWorks, Inc., Natick, MA, 1999.
The MathWorks, Inc., MATLAB: Application Program Interface Reference Version 6, The
MathWorks, Inc., Natick, 2000.
The MathWorks, Inc., MATLAB: Control System Toolbox User’s Guide, Version 4, The
MathWorks, Inc., Natick, 1992-1998.
The MathWorks, Inc., MATLAB: Creating Graphical User Interfaces, Version 1, The
MathWorks, Inc., Natick, 2000.
The MathWorks, Inc., MATLAB: Function Reference, The MathWorks, Inc., Natick, 2000.
The MathWorks, Inc., MATLAB: Release Notes for Release 12, The MathWorks, Inc., Natick,
2000.
The MathWorks, Inc., MATLAB: Symbolic Math Toolbox User’s Guide, Version 2, The
MathWorks, Inc., Natick, 1993-1997.
The MathWorks, Inc., MATLAB: Using MATLAB Graphics, Version 6, The MathWorks,
Inc., Natick, 2000.
The MathWorks, Inc., MATLAB: Using MATLAB, Version 6, The MathWorks, Inc., Natick,
2000.
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