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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 Venkateswara’ ‘Too Lord Sr Srii V

(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.

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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.

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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.

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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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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.

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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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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.

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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.

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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.

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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.

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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.

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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.

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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

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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]

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

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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

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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) 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 Computes the hyperbolic cosine of x, which is equal to

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

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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.

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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.

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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.

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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]

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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.

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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

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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 a + b 21 21

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

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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.

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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

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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.

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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

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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

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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

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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

2 –3 1 3 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.

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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.

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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.

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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);

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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

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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

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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

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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

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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.)

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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

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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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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.

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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

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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)

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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.

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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

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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:

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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.

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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

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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

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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

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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)

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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

2 0 s − 3 0 1

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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) =

r1 r2 r3 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:

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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)

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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)

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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

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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) =

1 s 2 (s + 5)

(c) F(s) =

3s + 1 (s + 2s + 9)

(d) F(s) =

s − 25 s(s + 3s + 20)

2

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)

.

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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 =

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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

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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.

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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 10 18 23 34

3 11 19 27 35

4 12 20 28 36

5 13 21 29 37

6 14 22 30 38

7 15 23 31 39

8 16 24 32 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. 0.5 − 0.8 (a) A = 1.0 0.75

8 3 (b) A = − 3 4

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

5 1 3 2 − 1 − 2 (f) A = 3 2 1 4 1 0

2 1 5 4 2 6 3 4

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.

121

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 | x | ≤ 10, | y | ≤ 10 (b) z = – 7 / (1 + x + y2) 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

123

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) =

ω2n 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

G GD C = = 1 + KN / D D + DN 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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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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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 |... ω2n |( 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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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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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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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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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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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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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

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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.

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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 )

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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. (a) T(s) =

130 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

0

100

200

300

400

Time (sec)

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 s + 7s + 25 2

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))

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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

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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 )

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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

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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.

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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’)

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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 G(s)H(s) =

K(s + 3) (s 2 + 3s + 4)(s 2 + 2s + 7)

G(s)H(s) =

K (s + 3) (s + 3s + 4)(s2 + 2 s + 7)

Solution.

= >> %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];

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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

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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)). Hence

Gc(s) = K

s+2 s+4

The gain K is obtained from the condition

K

or

or

s+2 7 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 + 4 s( s + 2) s(s + 4) 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)

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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.

Therefore

Gc(s) = K

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.

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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.

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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.

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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’)

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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’)

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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 Gc(s) = Kcβ = Kc β>1 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’)

200

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’)

201

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).

202

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)

204

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

20 25 15 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];

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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.

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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 or

Gc(s) = Kc

s + 1.2877 s + 6.1319

∧

and

4 K = 0.21 α

Kc =

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];

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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

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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:

− 5 − 36 − 7 x1 1 x1 x = 1 0 0 x2 + 0 u 2 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)

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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

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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)

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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 −2

s 1 0 0 G(s) = 5 0 0 1 0

0 1 0 0 − 1 ; C = 0 0 1 0

0 0 0 3 s + 6 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

=

2]

1 s + 3 − 1 2 [(s + 4)( s + 3) + 2] 2 s + 4 5

1 [1 [ s + 7 s + 14] 2

[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 x1 0 1 x1 0 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 x1 1 + x = 7 0 x2 2 1 1 y1 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];

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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

x1 1.5 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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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

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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 + u 0 2 x3 0

x1 5] x2 + [0]u x3

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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

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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

x1 3 x = 0 2 x 3 0

0 1 4

0 0 5

x1 0 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

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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

x1 3 x = 0 2 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 = 0 2 1 x 3

−3 −2 0

− 2 1 − 1

x1 3 x + 0 u 2 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

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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];

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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

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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

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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).

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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)

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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)

+ –

C(s)

80 s(s + 3)

+ –

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 s(s + 0.5)(s + 0.5s + 8) 2

=

Plot the root loci for the system using MATLAB.

K s + s + 8.25s2 + 4 s 4

3

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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).

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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.

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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.

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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

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ANALYSIS AND DESIGN OF CONTROL SYSTEMS USING MATLAB

37. A control system is defined by

x1 − 1 x = 0 2 x3 2

−3 −2 0

− 3 1 − 1

x1 3 x + 0 u 2 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.

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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.

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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.

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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.

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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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