basic control system 1

DEPARTMENT OFELECTRICAL AND ELECTRONIC ENGINEERING

EE208: Part II Control Systems

C. P. Jobling

15 February 1997Version 0.99

A printable version of these lecture notes are available fordownloading in PDF and Zipped PostScript formats.

Slide 1

DEPARTMENT OF

ELECTRICAL AND

ELECTRONIC ENGINEERING

EE208

Part II Control Systems

1

CONTENTS 2

Contents

1 Introduction to Control Systems 91.1 Control Systems Applications . . . . . . . . . . . . . . . . . . . . 9

1.1.1 Brief History . . . . . . . . . . . . . . . . . . . . . . . . . 91.1.2 The Challenges of Control Systems . .. . . . . . . . . . . 91.1.3 An Example — The Space Shuttle . . .. . . . . . . . . . . 91.1.4 The Benefits of Studying Control . . .. . . . . . . . . . . 18

1.2 Describing Control Systems . . . . . . . . . . . . . . . . . . . . . 181.2.1 The Control System . . .. . . . . . . . . . . . . . . . . . 181.2.2 Description of Input and Output . . . .. . . . . . . . . . . 181.2.3 Advantages of Control Systems . . . .. . . . . . . . . . . 211.2.4 Open-Loop Systems . . .. . . . . . . . . . . . . . . . . . 221.2.5 Closed-Loop Systems . .. . . . . . . . . . . . . . . . . . 241.2.6 Computer-Controlled Systems . . . . .. . . . . . . . . . . 27

2 Introduction to the Analysis and Design Process 292.1 Control Systems Analysis and Design Objectives . . . . . . . . . . 29

2.1.1 Transient Response . . . .. . . . . . . . . . . . . . . . . . 292.1.2 Steady-state accuracy . . .. . . . . . . . . . . . . . . . . . 302.1.3 Stability . . .. . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2 An Example of a Closed-Loop System . . . . . . . . . . . . . . . . 302.2.1 Antenna Azimuth Position Control . . .. . . . . . . . . . . 302.2.2 Block Schematic Diagram. . . . . . . . . . . . . . . . . . 362.2.3 Transient Performance . .. . . . . . . . . . . . . . . . . . 362.2.4 Steady-state error . . . . .. . . . . . . . . . . . . . . . . . 36

2.3 The Design and Analysis Sequence . . . . . . . . . . . . . . . . . . 392.3.1 Determine a Physical System from the Requirements . . . . 392.3.2 Transform the Physical System into a Schematic . . . . . . 392.3.3 Mathematical Models for the Schematic. . . . . . . . . . . 422.3.4 Block Diagram Reduction. . . . . . . . . . . . . . . . . . 422.3.5 Analysis and Design . . .. . . . . . . . . . . . . . . . . . 42

3 Modelling the Azimuth Position Control System 513.1 The Schematic for the Plant . . . . . . . . . . . . . . . . . . . . . . 513.2 Mechanical Side . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.3 Effect of the Gearbox . . . . . . . . . . . . . . . . . . . . . . . . . 52

3.3.1 An Ideally Matched Gearbox . . . . . .. . . . . . . . . . . 563.4 Electrical Side . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.4.1 Armature-Voltage Controlled DC Motor. . . . . . . . . . . 593.5 Coupling Constants . . . . . . . . . . . . . . . . . . . . . . . . . . 623.6 The Rest of the Control System . . . . . . . . . . . . . . . . . . . . 67

3.6.1 Position sensors . . . . . .. . . . . . . . . . . . . . . . . . 683.6.2 Velocity sensors . . . . .. . . . . . . . . . . . . . . . . . 703.6.3 Pre-amplifier . . . . . . . . . . . . . . . . . . . . . . . . . 703.6.4 Power Amplifier . . . . .. . . . . . . . . . . . . . . . . . 703.6.5 Block Diagram of Plant .. . . . . . . . . . . . . . . . . . 71

3.7 Velocity control system . . . . . . . . . . . . . . . . . . . . . . . . 73

CONTENTS 3

4 Evaluation of System Response 764.1 Poles and Zeros and System Response . . . . .. . . . . . . . . . . 774.2 First-Order System Responses and Specifications . . .. . . . . . . 84

4.2.1 The significance ofa . . . . . . . . . . . . . . . . . . . . . 844.2.2 Time Constant . . . . . .. . . . . . . . . . . . . . . . . . 854.2.3 Rise TimeTr . . . . . . . . . . . . . . . . . . . . . . . . . 854.2.4 Settling TimeTs . . . . . . . . . . . . . . . . . . . . . . . 86

4.3 Types of Second-Order System Responses . . .. . . . . . . . . . . 894.3.1 Overdamped Response . .. . . . . . . . . . . . . . . . . . 894.3.2 Underdamped Response .. . . . . . . . . . . . . . . . . . 904.3.3 Undamped Response . . .. . . . . . . . . . . . . . . . . . 934.3.4 Critically Damped System. . . . . . . . . . . . . . . . . . 93

4.4 The General Second-Order Response . . . . . .. . . . . . . . . . . 984.4.1 Definitions .. . . . . . . . . . . . . . . . . . . . . . . . . 984.4.2 Derivation of Formulae . .. . . . . . . . . . . . . . . . . . 994.4.3 Pole-zero locations . . . .. . . . . . . . . . . . . . . . . . 1004.4.4 Further analysis for underdamped second-order systems . . 102

4.5 The Specification of Second-Order Response .. . . . . . . . . . . 1054.5.1 Evaluation ofTp . . . . . . . . . . . . . . . . . . . . . . . 1064.5.2 Evaluation of%OS . . . . . . . . . . . . . . . . . . . . . . 1074.5.3 Evaluation ofTs . . . . . . . . . . . . . . . . . . . . . . . 108

4.6 Relating Response Specifications to Pole Locations in thes-Plane . 1114.6.1 Effect of moving poles along design curves . .. . . . . . . 112

5 Analysis and Design of Feedback Systems 1225.1 Interpretation of the generalised closed-loop transfer function . . . . 1235.2 Unity-gain feedback . . . . . . . . . . . . . . . . . . . . . . . . . . 1235.3 Closed-loop transient performance . . . . . . . . . . . . . . . . . . 1245.4 The Root-Locus: A Preview . . . . . . . . . . . . . . . . . . . . . 128

6 Stability 1336.1 What is stability? . .. . . . . . . . . . . . . . . . . . . . . . . . . 1336.2 How do we determine if a system is stable? . . . . . . . . . . . . . 134

6.2.1 Stability . . .. . . . . . . . . . . . . . . . . . . . . . . . . 1346.2.2 Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . 1366.2.3 Marginal stability . . . . .. . . . . . . . . . . . . . . . . . 138

6.3 Testing for Stability .. . . . . . . . . . . . . . . . . . . . . . . . . 1386.4 The Hurwitz Criterion . . . . . . . . . . . . . . . . . . . . . . . . . 1386.5 The Routh-Hurwitz Stability Criterion . . . . .. . . . . . . . . . . 141

6.5.1 The Routh array . . . . .. . . . . . . . . . . . . . . . . . 1416.5.2 The Routh-Hurwitz Test .. . . . . . . . . . . . . . . . . . 144

6.6 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1456.6.1 A Zero in the first column. . . . . . . . . . . . . . . . . . 1466.6.2 An Entire row is zero . . .. . . . . . . . . . . . . . . . . . 1476.6.3 Interpretation of a row of zeros . . . . .. . . . . . . . . . . 148

6.7 Use of the Routh-Hurwitz Criterion for Control Systems Design . . 1506.8 Relative Stability . .. . . . . . . . . . . . . . . . . . . . . . . . . 151

CONTENTS 4

7 Steady-state errors 1547.1 Forms of Steady-State Errors . . . . . . . . . . . . . . . . . . . . . 1567.2 Block Diagram Perspective . . . . . . . . . . . . . . . . . . . . . . 1597.3 Steady-State Errors for Unity-Gain Feedback Systems . . . . . . . . 162

7.3.1 Step input . .. . . . . . . . . . . . . . . . . . . . . . . . . 1637.3.2 Ramp input .. . . . . . . . . . . . . . . . . . . . . . . . . 1637.3.3 Parabolic input . . . . . .. . . . . . . . . . . . . . . . . . 164

7.4 Definition of Static Error Constants and System Type .. . . . . . . 1657.4.1 Static error constants . . .. . . . . . . . . . . . . . . . . . 1657.4.2 System type number . . .. . . . . . . . . . . . . . . . . . 165

7.5 Steady-State Error Specifications . . . . . . . . . . . . . . . . . . . 169

8 Dynamic Compensation 1728.1 The trouble with gain compensation . . . . . .. . . . . . . . . . . 1728.2 Velocity Feedback Compensation . . . . . . . . . . . . . . . . . . . 1758.3 Other forms of dynamic compensation . . . . .. . . . . . . . . . . 1778.4 PID Cascade Compensators . . . . . . . . . . . . . . . . . . . . . . 1788.5 Design of cascade compensators . . . . . . . . . . . . . . . . . . . 180

9 The Root Locus 1869.1 Introduction . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . 186

9.1.1 The Control System Problem . . . . . .. . . . . . . . . . . 1879.1.2 Complex numbers and their vector representations . . . . . 1899.1.3 Defining the Root Locus .. . . . . . . . . . . . . . . . . . 197

9.2 Properties of the Root Locus . . . . . . . . . . . . . . . . . . . . . 2019.3 Rules for sketching the root-locus . . . . . . . . . . . . . . . . . . 206

9.3.1 Number of branches . . .. . . . . . . . . . . . . . . . . . 2069.3.2 Symmetry . .. . . . . . . . . . . . . . . . . . . . . . . . . 2069.3.3 Real-axis segments . . . .. . . . . . . . . . . . . . . . . . 2079.3.4 Start and end-points . . .. . . . . . . . . . . . . . . . . . 2099.3.5 Asymptotic behaviour . .. . . . . . . . . . . . . . . . . . 210

9.4 Refining the root locus diagram . . . . . . . . . . . . . . . . . . . . 2159.4.1 Real-axis break-away and break-in points . . .. . . . . . . 2159.4.2 Calculation ofj!-Axis crossing . . . . . . . . . . . . . . . 2189.4.3 Angles of Departure and Arrival . . . .. . . . . . . . . . . 220

9.5 Plotting and calibrating the root-locus . . . . .. . . . . . . . . . . 2209.5.1 Transient response from the root-locus .. . . . . . . . . . . 222

10 Frequency Response Techniques 22610.1 Introduction . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . 22610.2 The Open-Loop Frequency Response . . . . .. . . . . . . . . . . 228

10.2.1 An example .. . . . . . . . . . . . . . . . . . . . . . . . . 22910.2.2 Effect of Gain. . . . . . . . . . . . . . . . . . . . . . . . . 232

10.3 The Simplified Nyquist Criterion .. . . . . . . . . . . . . . . . . . 23210.4 Gain and Phase Margins . . . . .. . . . . . . . . . . . . . . . . . 236

10.4.1 Gain margin . . . . . . . . . . . . . . . . . . . . . . . . . 23610.4.2 Phase margin. . . . . . . . . . . . . . . . . . . . . . . . . 23710.4.3 Design considerations of the use of gain and phase margin . 239

CONTENTS 5

A Solutions to Problems 246

B Second-Order Responses 272B.1 Overdamped system response . . .. . . . . . . . . . . . . . . . . . 272B.2 Underdamped response . . . . . .. . . . . . . . . . . . . . . . . . 274B.3 Undamped response .. . . . . . . . . . . . . . . . . . . . . . . . . 276B.4 Critically damped system response. . . . . . . . . . . . . . . . . . 277B.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

CONTENTS 6

Syllabus

EE208:Control Systemsis concerned with the development of techniques for mod-eling linear dynamic systems with feedback and analysing their performance andstability. The topics to be included are as follows:

Introduction: Control system applications; a brief history; an example. De-scribing control systems: inputs and outputs; advantages; open- and closed-loop control systems; computer controlled systems.

Analysis of design objectives for control systems: transient response; steady-state response; stability. An antenna dish azimuth position control system(course case-study1). The design and analysis sequence.

Mathematical modeling.

Time response analysis and design.

– Evaluation of system response: poles and zeros and system response;the characteristic equation; first-order system responses amd specifica-tions;

– Second order system responses: general second-order system; natu-ral frequency; damping ratio; Transient performance specifications forsecond-order systems.

– Analysis and design of feedback systems: closed-loop transient perfor-mance; introduction to the root-locus. drill problem.

– Stability: the Hurwitz criterion; the Routh-Hurwitz criterion;

– Steady-state errors: steady-state errors for unity-gain feedback systems;static error constant and error type; specifications of steady-state errors;

– A taste of feedback compensation: velocity feedback. Second drill prob-lem.

– Root locus: the control system problem; complex numbers and theirvector representation; defining the root locus; properties of the root lo-cus; rules for sketching the root locus. Third drill problem.

Frequency response analysis and design.

– Nyquist: stability criterion; simplified Nyquist stability criterion.

– Relative stability: gain and phase margin; assessing closed-loop perfor-mance from bode diagrams.

– Closed-loop frequency response: M - andN - circles; Hall and Nicholscharts;Mmax. Final drill problem.

1http://www-ee.swan.ac.uk/Courses/level2/ee208/drillp/

CONTENTS 7

Supporting studies

The course is supported by two experiments in the Part II Laboratory Course. Thefirst is an examination of a position control system, a scaled-down version of thecourse “case-study2”. The second makes use of MATLAB and SIMULINK to simu-late a model of the same system. I strongly encourage you to make use of MATLAB

and SIMULINK (which are installed on the PC network) to help you to model, sim-ulate and understand the dynamics of control systems. There are manuals in thetechnician’s office at the back of room 502 but there are now several books such asthe ones by Saadat [21], Shahian and Hassul [22], Leornard and Levine [16] andBishop [2] which provide good tutorial introductions to the use of MATLAB forcontrol systems design. In addition, many standard textbooks have been recentlyupdated to include tutorial and reference material concerned with the use of MAT-LAB in control systems analysis and design. For those of you who own your ownPC, the Student Editions of MATLAB [23] and SIMULINK [24] are available andshould be sufficient to support the analysis methods covered in the course.

As an alternative to MATLAB , Golten [11] is a basic text on control systemsanalysis and design which makes extensive use of the computer program calledCODAS included in the price. There is also an “electronic handbook” for Math-cad which is part of a new venture by McGraw & Hill and Mathsoft to supplementselected titles in the theSchaum’s Outline Series. The title of this package, whichincludes the book [13], the software and a run-time version of Mathcad is “Interac-tive Feedback and Control Systems” [14].

Reading List

The reference section of these notes is an annotated booklist for the control course.The course notes are based on Nise [18] which is therefore the recommended text.Dorf [8], which was the main text in the past, is still worth considering as an alter-native to Nise particularly because it is supported by a solutions manual [7] and aMATLAB tutorial [2]. For a cheaper alternative you might wish to consider DiSte-fanoet al in the Schaum Series [13] (or the electronic handbook version for Math-cad [14]) which is both a good reference and supply of worked problems. Thoselooking for more comprehensive coverage of the course material might considerD’Azzo and Houpis [5].

The other books included in the list fall into the category of alternative titlesor further reading. There are many books on control that also cover much of thematerial that is taught in Part II and Part III, for example you might prefer one of[10, 19, 20]. Serious students of the topic might consider [15], and those with lessinterest might just consider [4, 17, 3] which are all cheap and cheerful. For thosewith an interest in the practicalities of control system design might care to look atBateson [1] which is very much an applied control textbook/reference manual forcontrol systems practitioners. Doeblin [6] is a text with a similar aim which alsoemphasises computer simulation.

Additional background reading can be found in [9] which emphasizes computermodeling and simulation and [12], which is a good tutorial introduction to MAT-

2http://www-ee.swan.ac.uk/Courses/level2/ee208/drillp/

CONTENTS 8

LAB.

Other Sources of Information

Aside from the on-line version of the course material which is to be found at

http://www-ee.swan.ac.uk/Courses/level2/ee208/ee208.html

this list of additional resources may prove useful.For MATLAB , the main tool used to support this course, you will find a wealth

of material on the home-page of the MathWorks3. Included there is an extensivelist of books4 that use MATLAB , a repository5 of contributed software, Frequentlyasked question6 and lots of other information.

For on-line information about other software tools which could be used in thesupport of this course you should try the web home pages ofMathSoft7 (developersof Mathcad), Wolfram Research’sMathSource8 for information onMathematica,and MapleSoft9 for information onMaple (the symbolic processor that is insideMATLAB ’s Symbolic Toolbox andMathcad).

3http://www.matworks.com/4http://www.mathworks.com/BookList.html5http://www.mathworks.com/ftpindex.html6http://www.mathworks.com/faq.html7http://www.mathsoft.com/8http://www.wri.com/WWWDocs/mathsource/9http://www.maplesoft.com/

1 INTRODUCTION TO CONTROL SYSTEMS 9

Lecture 1: Intro

1 Introduction to Control Systems

Slide 2

Introduction to Control Systems

Aims The purpose of this lecture is to:

Introduce the Topic

– To give an idea of the many applications of the subject

– To give an insight into its history

– To highlight its advantages

– To demonstrate the depth and breadth of the subject

– To illustrate its usefulness as a subject worth studying

Define some Simple Terms

1.1 Control Systems Applications

Control systems are an integral part of modern society. There are numerous appli-cations in industry (see Slide 3), they also exist in nature (see Slide 4) and thereare even some non-physical control systems (see Slide 5).

1.1.1 Brief History

Because they are found in nature, control systems can be considered to have been inuse from the earliest times. However, in terms of Human endeavour, slides 10 and11 illustrate the main milestones in the development of man-made control systems.

1.1.2 The Challenges of Control Systems

1.1.3 An Example — The Space Shuttle

Just taking the case of flight control: the navigation functions are illustrated inSlide 16.


Slide 3

Some Control System Applications

Space shuttle

Automatic machine tools

Automatic parts delivery in a factory

Slide 4

Control Systems in Nature

Pancreas— regulates blood sugar

Adrenelin — automatically generated to increase heart-rate and

oxygen intake in times of flight

Eyes— able to follow a moving object

Hand — able to pick up an object and place it at a predetermined

location


Slide 5

Some ‘Artificial’ Applications of Control

Modern Economies

A Model of Student Performance

– Input is available study time.

– Output is performance/exam mark.

– Such a model could be used to predict time required to improve the

grade.

With such a scheme you could decide whether it is worth spending

more effort to pass the Control Systems Exam!

Slide 6

Control Systems Provide Power Applification

Control systems allow us to move large pieces of equipment withprecision

Radio telescopes can be accurately pointed at far reaches of the

universe.

Lifts stop at the right floor.

We could not perform these tasks ourselves. Motors provide thepowerand

control systemsregulatethe position and speed.


Slide 7

Control finds Applications in Transportation

Engine regulation, active suspension systems and anti-lock braking

systems in automobiles

Steering of missiles, planes, aircraft and ships at sea

For example, modern ships use a combination of electrical, mechanical and

hydraulic components to develop rudder commands in response to desired

heading commands. The rudder commands, in turn, produce a rudder angle,

which steers the ship.

Slide 8

Control finds Applications in Process Industries

In the process industries control is used to regulate level, pressure and

temperature of chemical refinery vessels.

In a steel rolling mill, the position of the rolls is controlled according to

the measured thickness of the steel coming off the finishing line.

Strip

measurethickness

Move

OK?

No

Rollers


Slide 9

Control Systems in the Home

CD Playersthe position of the laser spot in relation to the microscopic

pits in a Compact Disc is controlled

Video Recordersthe tracking of the record and play-back heads is

controlled by controlling the velocity of the tape

Central heating systemsuse thermostats to measure and control the

temperature in the room

Washing machinesuse sequencing controls to provide a variety of

wash cycles and temperature controls to avoid damage to delicate

fabrics

Slide 10

Historical Development of Control Systems

Ancient Greece(circa 3000 BC): water clocks, automatic oil lamps;

‘special effects’ in temples.

17th Century:Cornelis Drebbel — temperature control for an egg

incubator

18th Century:James Watt— Flyball governer for steam engine

Late 19th Century to mid-20th Century — development of “classical

control theory”

1960’s – present “modern control theory”


Slide 11

Heroes and Milestones in the Development of Control Systems

Late 19th Century: Fathers of Stability Theory — J. C. Maxwell,

E. J. Routh and A. M. Lyapunov

Late 1920’s – mid 1930’s: Bell Telephone Labs USA. Discovery of

negative feedback (Black), frequency response analysis (H. W. Bode),

stability theory (H. Nyquist).

1948invention of the Root Locus method (W. R. Evans)

1960’sdevelopment of state-space methods (Kalman and others)

Slide 12

Control Engineering is Challenging

It is a multi-disciplinary subject

cuts across numerous engineering disciplines

covers numerous functions within a discipline


Slide 13


It Covers all Aspects of a Project from High to Low Level

from conception through to

system requirements;

subsystem functions;

interconnection of functions;

interfaces between functions;

hardware and software design;

right up to test plans and procedures.

Slide 14


It is Broad and Diverse

Control engineers typically need to work closely with

biologists,

chemical, mechanical and electrical engineers,

mathematicians and

physicists.

They get involved with sensors and actuator technology, electronics,

pneumatics and hydraulics and and computers.


Slide 15

The Space Shuttle

The space shuttle would be impossible to fly without control systems. All

the shuttle’s many control systems are controlled by on-board computers on

a time-shared basis.

The main control systems in the shuttle are:

Flight control

Orbit control

Life support

Slide 16

Flight Control in the Shuttle

Navigation functions take in data to estimate the shuttle’s position and

velocity.

The position and velocity data is used to steer the shuttle:

– In space by use of pulsed jets of gas;

– In the Earth’s atmosphere by adjusting the geometry of the shuttle’s

air-surfaces.


There are numerous subsystems in the shuttle and many engineering disciplinesare needed to make it fly (see slide 17)

Slide 17

Subsystems and Disciplines Represented in the Shuttle

Numerous subsystems

– flight elevon controls to counteract wind disturbances

– life support systems; power systems; heating.

Many disciplines

orbital mechanics; propulsion; aerodynamics; electrical engineering;

mechanical engineering; hydraulics; temperature and pressure control,

etc., etc.


1.1.4 The Benefits of Studying Control

Slide 18

What will I get out of this course?

Control is a top-down engineering subject. Such subjects are rare in

engineering:

most engineering courses are taught bottom up

– they start with components

– develop circuits

– assemble circuits into products.

The reason for this is that top-down courses are difficult to teach because of

the high-level of mathematics needed for a systems approach.

1.2 Describing Control Systems

In this section we shall describe what a control system is, its characteristics, and itsadvantages. We also present two important classifications of control systems.

1.2.1 The Control System

1.2.2 Description of Input and Output

The input represents a desired response.

The output is the actual response.

For example, the inputs and outputs of a lift control system are illustrated inSlide 23. The fourth-floor button is pressed on the ground floor. The lift-car risesto the fourth floor with a speed and floor levelling accuracy designed for passengercomfort.

The fourth floor button is the input shown by a step command. The lift doesnot mimic the input — this would be undesirable for passenger comfort as well asimpossible with finite power supplied by motor. Instead, the input represents theposition we would like the lift to be in when the lift has stopped moving. The liftitself follows thelift response curve.

Two factors make the output different from the input.


Slide 19

Top down design in Control Systems

design high-level system requirements

choose functions and hardware to implement system to meet

requirements.

Control works from the ‘big picture’. It unifies many other elements. This is

part of the difficulty of the subject, it is also the challenge. Recognition of

the unification, that is being able to use lessons learned in other courses,

will help you to master this course material.

Slide 20

Taking Stock

So far in this lecture we have introduced the Topic of Control. In the next

part of the lecture we shall define some of the terms used to describe control

systems. In this section we will give:

a definition of a control system

a description of typical inputs and outputs for control systems

highlight some advantages of control systems

an account of the difference between open-loop and closed-loop control

and

an introduction to computer controlled systems


Slide 21

A Control system consists ofsubsystems and processes (or plants)

assembled for the purpose of controllingthe output of the processes.

A central heating boiler is a process that produces heat as a result of a

flow of fuel.

This process is assembled from subsystems called fuel valves.

Fuel valve actuators regulate the temperature of a room by controlling

the flow of fuel into the boiler.

Other subsystems, such a thermostats, act as sensors, to measure the

room temperature.

Slide 22ControlSystem

Input;Stimulus

Output;Response

DesiredResponse

ActualResponse


Slide 23

desired response --- input

lift response --- output steady-stateerror

steady-statetransient

Time

Floor

4

First consider the instantaneous change in the input against the gradual changein the output. Physical entities cannot change their position or velocity instanta-neously. The state changes through a path dictated by the physical devices andthe way it aquires and dissipates energy. The lift undergoes a gradual change as itmoves from the ground to fourth floor — called thetransient response.

After the transient response is complete, the physical system approaches itssteady-state response which is an approximation to the commanded ordesired re-sponse. This occurs when the lift reaches the fourth floor. The accuracy of the lift’sfinal level is the second factor that makes the output different from the input. Thedifference is called thesteady-state error.

Steady-state error may also be a feature of the system being controlled and it isone of the features that the control engineer considers when specifying the desiredbehaviour. For example, when tracking a satellite, some error may be tolerated pro-vided that the satellite stays close to the centre of the tracking radar beam. However,if a robot is inserting a chip into a PCB the steady-state error must be zero.

1.2.3 Advantages of Control Systems

We tolerate the differences between desired response and actual response becauseof the many advantages of control systems.

Power Amplification: Satellite dish can be positioned by a low power knobat the input but requires large power to rotate. Power gain is one good reasonfor building control systems.

Dangerous Applications — remote control of a robot arm for handling nu-clear material.


Slide 24

Advantages of Control Systems

Power Amplification

Dangerous Applications

Compensation for Human Deficiencies

Convenience by Change of Form of Input

Compensation of Disturbances

Compensation for Human Deficiencies — e.g. to help handicapped people orthe exo-skeleton used by Ripley inAliens.

Convenience by change of form of Input — Temperature control is by theposition of a dial on a thermostat, output is heat.

Compensation of Disturbances — Typical control variables are temperature,position and velocity, voltage, current or frequency. The control system mustyield the correct output even in the presence of disturbances.

Consider Slide 25. The satellite tracking antenna’s position and azimuth angleis controlled. Internal noise or wind gusts disturb the position. What corrects forthe disturbance?

disturbances must be measured

measurements must be used to reposition the dish.

1.2.4 Open-Loop Systems

An open-loop control system is illustrated in Slide 26.

A subsystem called thecontrollerdrives the process.

the input is called thereference.

the output is called thecontrolled variable.


Slide 25

WindDisturbance

Slide 26

Open-Loop Control System

++

++ProcessController

InputorReference

Outputor

Controlled Variable

Distu

rba

nc

e 1

Distu

rba

nc

e 2


Slide 27

Description of an Open-Loop Temperature Control System

Process is a boiler, input is fuel, output is heat.

Controller is electronics, valves, etc. that control fuel flow into furnace.

Input is thermostat position.

other signals, such as disturbances are shown added to the controller or pro-cess outputs via summing junctions.

Open-loop systems cannot compensate for disturbances added to the controller’sdriving signal (disturbance 1) such as amplifier noise which is just added to the pro-cess demand.

Output is also corrupted by disturbances at the output (e.g. wind in the trackingsystem). These cannot be corrected either.

Open-loop systems do not correct for disturbances and are simply commanded.E.g. an electric toaster: time is input, output is colour but colour is not measured.

1.2.5 Closed-Loop Systems

In Slide 28 the architecture of a closed-loop system is shown. Note that the inputtransducer is now shown explicitly. It converts the input which is thedesired outputinto the form required by the controller. An output transducer measures theactualoutput or response of the plant and also converts it to the form required by thecontroller.

The first summing point substracts the output from the input — the result iscalled theactuating signal— and if there is any difference the controller drives theprocess. If there is no difference the plant is not driven since its output is already atthe desired value.

Closed-loop control systems are accurate because they tend to reduce the dif-ference between input and output. They are also less sensitive to disturbances.

Transient performance and steady-state errors can be controlled more conve-niently and with greater flexibility than with open-loop sytems — often by simple


Slide 28

Closed-Loop Control System

++

-

++

+ProcessControllerInput

Transducer

OutputTransduceror Sensor

Input Output

Distu

rba

nc

e 1

Distu

rba

nc

e 2

Slide 29

Description of Closed-Loop Temperature Control System

Input temperature dial position converted into a voltage by a

potentiometer.

Output temperature converted to a voltage by a thermistor.

Differencing circuit subtracts output from input — result is actuating

signal — controller drives the plant only if there is a difference.

Closed-loop systems are less sensitive to disturbances


adjustment of gains in the loop, and sometimes by redesign of the controller (calledcompensation).

On the other hand, closed-loop control systems are more complex, and thereforemore expensive than open-loop systems, so the designer must balance the cost whendesciding what to use.


1.2.6 Computer-Controlled Systems

Slide 30

Computer-Controlled Systems

The controller or compensator is a computer

many loops can be controlled by time sharing.

adjustment of controller parameters are in software rather than

hardware.

supervisory functions such as scheduling, data logging, error and fault

monitoring, can also be done.

In the next lecture we look at the design objectives for control systems.


Slide 31

Summary

In this lecture we have Introduced the Topic of Control and Given

a definition of a control system

a description of typical inputs and outputs

an introduction to the termssteady-state errorandtransient

performance

some advantages of control systems

an illustration of the difference between open-loop and closed-loop

control

an introduction to computer controlled systems

2 INTRODUCTION TO THE ANALYSIS AND DESIGN PROCESS 29

Lecture 2: Analysis and Design

Preamble

In the last lecture we introduced the idea of a control system and examined the basicfeatures of such systems.

Control Systems are dynamic systems, they respond to an input by undergoing atransient response prior to reaching a steady-state response that generally resemblesthe input.

In this lecture we discuss the importance of transient and steady-state responseand then establish our analysis and design objectives. We also introduce a newconcept called stability.

We then present an example of a closed-loop control system to further illustratesome of these concepts. We finally present the control systems analysis and designsequence and finish with an overall summary of the introductory lectures.

2 Introduction to the Analysis and Design Process

Slide 32

Introduction to the Analysis and Design Process

Control systems analysis and design objectives

– Transient response

– Steady-state accuracy

– Stability

An example of a closed-loop system

The design and analysis squence

Summary of introductory lectures

2.1 Control Systems Analysis and Design Objectives

2.1.1 Transient Response

Another example is a hard-disk drive for a computer. The transient is relatedto the read-write time. Read/write cannot take place until the head is in place over


Slide 33

Transient Response (1)

Transient response is very important. Taking the lift example

if the lift moves too slowly the passengers would get impatient

if on coming to rest, the lift was to oscillate for more than about a

second the passengers would get worried.

if the lift moved too quickly there may be structural damage caused to

the building.

the correct track of the disk. So speed of the read/write head over the surface ofthe disk from one track to another will be important for the control of the hard-diskdrive.

2.1.2 Steady-state accuracy

We also focus control systems analysis and design on this aspect of performance.(Slides 35,36,37)

2.1.3 Stability

See slides 38–42.

2.2 An Example of a Closed-Loop System

We now introduce an example closed-loop control system which will serve as a casestudy for this course. The aim in introducing this example here is to:

to see a physical example

to see how it works

to see how we can affect its performance

2.2.1 Antenna Azimuth Position Control

The output is the azimuth angleo(t) follows the inputi(t) of potentiometer.A model of this system is used in the lab.


Slide 34

Analysis and Design Objectives: Transient Response

In this course:

we shall use quantitative measures of transient response

we analyse a system’s existing transient response

we seek to adjust the design parameters to yield a desired transient

response.

Slide 35

Steady-State

Steady-state is concerned with the state of a system after it arrives at the

desired output

lift system: when the lift car reaches the fourth floor;

hard disk controller: when the read-write head is over the correct track

on the hard disk.


Slide 36

Steady-State Accuracy

We are concerned with theaccuracyof the steady-state.

The floor of the lift must be sufficiently level with the floor of the

corridor to allow passengers to safely enter or leave the car;

the read-write head would yield disk errors if it was not positioned

correctly over a track on the disk surface;

a satellite tracking system must keep the satellite within its beam width.

Slide 37

Analysis and Design Objectives: Steady-State Accuracy

In this course:

We shall define quantitative measures for steady-state accuracy

We shall design corrective measures to reduce steady-state errors.


Slide 38

Stability

Transient response and steady-state error are moot points if the system does

not have the characteristic ofstability.

Slide 39

What is Stability?

Recall that the total system response is equal to the sum of the natural

response and the forced response

Total Response = Forced Response + Natural Response

natural response is obtained from thehomogeneoussolution of a

differential equation

forced response is obtained from theparticular solution.


Slide 40

Natural and Forced Response

The natural response (or homogeneous solution) describes the way a

system acquires or dissipates energy. The form and nature of the

natural response depends only on the system, not its inputs.

The form or nature of the forced response (or particular solution)

depends on the input.

Slide 41

Instability

In some systems, the natural response grows without bound rather than

diminishing or oscillating. Eventually the natural response is so much

bigger than the forced response that the system becomes “out of control.”

This condition is calledinstability.

It could lead to self destruction.


Slide 42Analysis and Design Objectives: Stability

Control systemsmustbe designed to be stable.

Slide 43

Summary of Analysis and Design Objectives

The objectives for a control system are to achieve:

stability

steady-state error

transient performance

(where the order indicates priorities for the designer). The question to be

answered in the course is how do we achieve these?


Slide 44

AzimuthAngle

θi t( )

θo t( )

potentiometer

cable

antenna

Antenna Azimuth Position Control

2.2.2 Block Schematic Diagram

A block schematic diagram of the system is shown in Slide 45.The system normally operates to drive the error to zero.When the input= the output there is no actuating signal, the motor is not driven.

The motor is only driven when the input6= the output.The bigger the error the faster the motor.

2.2.3 Transient Performance

What happens if the gain of the signal amplifier is increased?The motor is driven harder, but the actuating signal is still zero when input=

output. The difference will be in transients — motor driven harder so will movefaster. The increased speed leads to increased momentum so the system may over-shoot the final value and be forced by the system to reverse its direction. The resultmay be a diminishing oscillation. (Slide 46)

2.2.4 Steady-state error

In Slide 46 there is no steady-state error. In some systems there is, and the increasein gain will tend to reduce its value. This leads to a trade-off between transientperformance and steady-state error. To combat this extra components may need tobe added to the system to allow both the gain and the transients to be adjusted. Thisis calledcompensation.


Slide 45

Block Schematic Diagram

-

+ Motor, Load & Gears

Signal &Power AmplifierPotentiometer

Potentiometer

Input:Desired Azimuth Angle

Output:Azimuth Angle

Plantor ProcessError or

Actuating Signal

Voltage proportinal tooutput

Voltage proportinalto input

Sensor

θ i t( ) θo t( )

Controller

Slide 46

Re

spo

nse

Time

high gain

low gain

input


Slide 47

Control Systems Analysis and Design Objectives

The design objectives revolve around transient, steady-state accuracy

and stability.

gain adjustments can affect performance and lead to trade-offs among

performance criteria

compensation may be used to achieve performance without trade-offs.


2.3 The Design and Analysis Sequence

The design and analysis sequence is illustrated in Slide 48. It consists of five basicstages:

1. Determine a physical system from the requirements.

2. Transform the physical system into a schematic.

3. Construct a mathematical model.

4. Perform block-diagram reduction.

5. Analysis and design.

Although the sequence is shown to be linear, it need not be and in practice therewill be iterations between stages. Also, it is often the case that the control engineerdoes not have influence over the first stage and may have to design controllers forexisting plant. Another point to be aware of is that the early stages are often quitedifficult! In the next sections we review each of the stages in the sequence.

Slide 48

The Control Systems Design and Analysis Sequence

Determine a physical

system from the require-ments

Mathemat-ically

model the schematic as a block diagram

Analyze or design the system to meet the require-ments

Transform the

physical system into

a schematic diagram

Reduce the block

diagram to a single block or closed-

loop system

2.3.1 Determine a Physical System from the Requirements

(Slides 49 and 50)

2.3.2 Transform the Physical System into a Schematic

(Slides 51 to 53)


Slide 49

Azimuth Position Control System Example

able to position antenna azimuth angle from a remote location

weight of antenna

physical dimensions

desired transient performance

desired steady-state accuracy

etc.

Slide 50

Functional Description

It is sometimes useful to develop a functional diagram of the system which

will help to define the required hardware. Here is a block-schematic

diagram of the azimuth position control system.

Block Schematic Diagram

-

+ Motor, Load & Gears

Signal &Power AmplifierPotentiometer

Potentiometer

Input:Desired Azimuth Angle

Output:Azimuth Angle

Plantor ProcessError or

Actuating Signal

Voltage proportinal tooutput

Voltage proportinalto input

Sensor

θ i t( ) θo t( )

Controller


Slide 51

Transform the Physical System into a Schematic

Makes relationships more concrete

Enables decisions to be made about what can be neglected in

formulating the mathematical model.

Assumptions made can be easily reviewed and schematic and/or model

adjusted as necessary.

Should be kept as simple as possible:

– Checked by analysis and simulation

– Phenomena added if results do not agree with observed behaviour.

Slide 52

Schematic Diagram of Azimuth Position Control System

Inertia

ViscousDamping

Gear

Gear

Gear

Potentiometer

Potentiometer

MotorAmplifiersArmatureresistance

Fixed fie

ld

Differentialand

PowerAmplifier

+

-

-

+

+

-

K

θo t( )

θ i t( )


Slide 53

Some simplifying assumptions

Neglect friction and inertia of potentiometers (no dynamics).

Neglect dynamics of signal amplifiers whose responses will be rapid

compared to the motor. Assume pure gainK.

Armature voltage controlled motor — assume inductance is negligible.

Load can be modelled as an inertia plus bearing resistance.

There are no losses in the gearbox.

2.3.3 Mathematical Models for the Schematic

(Slides 54 and 55)

2.3.4 Block Diagram Reduction

(Slide 56)

2.3.5 Analysis and Design

(Slides 57–60 and Table 1)


Slide 54

Mathematical Models for the Schematic

Use of basic physical laws.

Construction of circuit diagrams and mechanical mobility diagrams.

Differential equations.

Block diagrams with transfer functions.

Covered in Dynamic Systems CourseResult is a block diagram in this case.

Slide 55

Alternative Mathematical Models for the Schematic

State-space models could be used.

For thesenth order differential equations are converted inton first-order

differential equations which are themselves represented by matrices.


Slide 56

Block Diagram Reduction

This is used to obtain “canonical” representations of the system which is the

same as those used to develop the theories used to analyse and design the

system.

Slide 57

Analysis and Design

Performance characteristics such as

stability

steady-state accuracy, and

transient performance

are determined.


Table 1: Test Waveforms Used in Control Systems

Input Function Description Sketch Use

Impulse (t)

(t) = 1 for 0 < t < 0+

= 0 elsewhere

Z 0+

0(t)dt = 1

f t( )

t

δ( )t

Transient re-sponse modelling

Sinusoid sin!tt

Transient re-sponse mod-elling; Steady-state error

Step u(t)u(t) = 1 for t 0

= 0 elsewhere

f t( )

t Transient re-sponse; Steady-state error

Ramp tu(t)tu(t) = t for t 0

= 0 elsewhere

f t( )

t Steady-state error

Parabola 12t2u(t)

12t2u(t) = 1

2t2 for t 0

= 0 elsewhere

f t( )

t Steady-state error


Slide 58

Use of simple test inputs.

Impulse, sinusoid, step, ramp and parabolic inputs are used to excite the

system in order to determine the actual response of the system.

Impulseis used to inject energy into the system so that its natural

response may be obtained. This can be used to determine what the

transfer function of an unknown system is.

Sinusoidis used to determine the steady-state and transient behaviour

from frequency response measurements. It can also be used to

determine the transfer function of an unknown system.

Stepis used to analyse steady-state and transient performance

RampandParabolicinputs are used to determine steady-state

accuracy..

Slide 59

Choosing Components

The choice of components to be assembled into a control system depends

on factors such as speed and power.

The system must be analysed to see if the requirements can be met.

If the requirements cannot be met then the designer may need to design

additional components or make adjustments to the system’s parameters in

order to meet the requirements.


Slide 60

Other Considerations

Sensitivity analysismay need to be performed in order to determine

how changes in system parameters will affect the performance of the

system. Systems must be built to withstand small changes in

parameters due to causes such as temperature, pressure, etc.

Once the design is complete, time response analysis may need to be

done again to verify that the control system meets all the requirements.


Recap of Introductory Lectures

Slide 61

Summary of Introduction (1)

Control systems are dynamic systems with feedback:

– contribute to many aspects of modern society

– exist naturally and in fields such as economics

– are used where power gain, remote control or conversion of input is

required.


Slide 62


A control systemhas an input, a process and an output.

Open-loop systemsdo not monitor or correct the output for

disturbances; they are simpler and less expensive then closed-loop

systems.

Closed-loop systemsmonitor the output and compare it to the desired

output (input). If an error is detected, the closed-loop system corrects

the output; they can correct the effects of disturbances.

Slide 63


Control systems design focuses on

– transient response

– steady-state response

– stability.

During analysis and design, the engineer tries to achieve stability,

transient performance and steady-state accuracy requirements.


In the next lecture we shall develop the mathematical model of the azimuthposition controller.

3 MODELLING THE AZIMUTH POSITION CONTROL SYSTEM 51

Lecture 3: Mathematical Modelling (1)

Preamble

In the last lecture we considered the analysis and design sequence for control sys-tems. An important part of that sequence is the creation of a mathematical model,in the form of a block diagram, for the system to be analysed and designed.

In the next two lectures we shall present the mathematical model of theAntennaAzimuth Control Systemintroduced in the last lecture. The presentation is split intotwo parts. In part 1, a block diagram model of the plant and actuator components ispresented. In part 2 the additional sensing, comparison and amplification compo-nents needed to complete the closed-loop control system are considered. Both partsassume that a sufficient level of modelling knowledge has been acquired in the pre-and co-requisite courses EE106: Dynamic Systems and EE206: Dynamic Systems

3 Modelling the Azimuth Position Control System

3.1 The Schematic for the Plant

Consider the azimuth control system we have already seen. The plant for this isillustrated in Slide 64. It is anarmature-voltage controlled DC motorwhichdrives the load through a gearbox.

Slide 64

Armature Controlled DC Motor — Schematic Diagram

motorN

J

1

12

10

1

=

= kgm

N

J

2

22

20

2

=

= kgm

N

J

3

32

10

2

=

= kgm

N4 20=Z sl ( )

We shall build up to a full block diagram in stages starting from the mechanicalside, introducing the gearbox and finally adding in the electrical side.


3.2 Mechanical Side

The rotor of the electric motor has inertiaJm kg m2 and bearing resistanceRm N m/(rad/s).Assume that we can lump the load effects into a single rotational mechanical impedance

Ql(s)

l(s)= Zl(s)

then, ignoring the gearbox for the moment, a schematic diagram of a driven shaft isthat shown in Slide 65.

Slide 65

Generalised Driven Rotational Shaft

loadimpedance

driving shaftInertia

driving shaftbearingresistance

ω l t( )q te( ) ωm

qmql

Rm zl

Jm

The impedence of the driving shaft isZm(s) = sJm + Rm andQe(s) =Qm(s) + Ql(s), m(s) = l(s) hence a block diagram for the driven shaft isthat shown in Slide 66, which hasQe(s) as input andl(s) as output. Reducingthe block diagram gives

l(s) =1

Zm(s) + Zl(s)Qe(s)

or

Ql(s) =Zl(s)

Zm(s) + Zl(s)Qe(s)

3.3 Effect of the Gearbox

A gearbox is an example of atransformer: a passive element which couples twosystems of the same kind by transmitting energy without loss from one system tothe other. The element is a sink in one system and a source in the other. The roles of


Slide 66

Block Diagram

+-

1Z sm( )

Z sl ( )

Q se( ) Q sm ( ) Ω Ωm ls s( ) ( )=

Q sl ( )

Slide 67

A Gearbox

loadimpedence

drivingshaftimpedence

idler gear

ω l t( )

q te( ) ωmqm

ql

zl

zm N1

N2


source and sink are interchangeable and depend on the direction of the net energyflow.

A side view of a typical gearbox is shown in Slide 67.In a simple gearbox. the driving shaft is connected to a gear-wheel that has

N1 teeth. This drives a second gear-wheel that is attached to the driven shaft. Thesecond gear hasN2 teeth. One input rotation of the driving gear producesN1=N2

rotations of the driven shaft in the opposite direction. If the direction of rotation isto be in the same direction as the driving shaft, anidler gear is placed between thedriving and driven gear-wheels (as in the illustration).

The ratioN2=N1 is called the gear-ratior and is always> 1. Thus:

l

m=

!l

!m=N1

N2

=1

r: (1)

We assume that there are no losses in the gearbox1 so that the energy into thegearbox equals the energy out (2):

!mqm = !lql (2)

and!m = r!l (3)

henceql

qm=

!m

!l= r: (4)

The purpose of a gearbox, a purerotational transformer, is:

1. to change the speed of a power source, i.e. a motor, to meet the need for adifferent output speed, e.g. car wheels.

2. to change the torque of a power source to meet the need for a different outputtorque.

In most cases, the power source is a high-speed low-torque device and the loadis a low-speed high-torque device. So in generalr > 1 and oftenr 1.

When developing a mathematical model for a gearbox, it is important to recog-nise that the energy equations (2) and (4) are fundamental. As a result of thisequation there are a pair of constraints on the driven and driving speeds and torquesthat must be satisfied. In block diagram terms this means that the gearbox is eitherrepresented by the two blocks shown on the left of Slide 68 or, alternatively as thetwo blocks shown at the right of Slide 68 (the equations represented by these blockdiagrams are easily derived from (4)).

To derive equations for the gearbox coupled mechanical system illustrated inSlide 67, we note that

Ql(s) = Zl(s)l(s) (5)

m(s) =1

Zm(s)Qm(s) (6)

1there are actually several losses in practical gearboxes, including friction, backlash, gear-wheelinertia, etc.


Slide 68

Alternative Block Diagrams of a Gearbox

Q sl ( ) Q sm( )

1

r

1r

Ω l s( )Ωm s( )

Q sl ( )Q sm( )

Ωl s( ) Ωm s( )r

r

Motor speed - load torque Load speed - motor torque

Slide 69

Block Diagram of a Gearbox Coupled Drive and Load

+-

Q se( )

Q sl ( )

Q sm( ) 1

r

1

r

Ωl s( )Ωm s( )1

Z sm( )

Z sl ( )


l(s) =1

rm(s) (7)

Qe(s) = Qm(s) +1

rQl(s) (8)

A block diagram representing these equations is shown in Slide 69.From (6) and (7) we have

l(s) =1

rZm(s)Qm(s) (9)

and from (5) and 8):

Qm(s) = Qe(s)Zl(s)

rl(s) (10)

Hence:

Qe(s) = rZm(s) +Zl(s)

rl(s) (11)

Qe(s)

m(s)= Zm(s) +

Zl(s)

r2(12)

Equation (12) is of particular interest since it tells us that the impedance of theload shaft as seen at the input shaft of the gearbox is reduced byr2. Thus for amotor driving a load through a gearbox we have:

m(s)

Qe(s)=

1

Zm(s) + Zl(s)=r2(13)

which should be compared with the results derived in Section 3.2. The interpre-tation of this result is that a gearbox allows quite large transfers of energy withmodest torques. For example, a gearbox with a gear ratio of20 : 1 is able to movean inertia of 400 times larger than the directly connected driven-shaft intertia withthesameamount of input effort.

3.3.1 An Ideally Matched Gearbox

Transformers are often used to match parameters in one system to those of an-other, normally to optimize some aspect of performance. A gearbox could be usedto maximise the rotational accelerationd!(t)

dtin one system achievable by a given

torque in the other. Taking the model of driving shaft and driven load that is repre-sented by the schematic in Slide 64, then, assuming thatZm(s) = sJm + Rm andZl(s) = sJl +Rl, from equation (13) we have:

m(s) =1

(sJm +Rm) + (sJl +Rl)=r2Qe(s)

Qe(s) = (r(sJm + Rm) + (sJl + Rl)=r)l(s) (14)

= (rJm + Jl=r)sl(s) + (rRm +Rl=r)(s) (15)


Thus to minimize the torque due to acceleration (sl(s)), we must minimize(rJm+Jl=r) with respect tor. That is:

d

dr(rJm + Jl=r) = 0 (16)

1

r2Jl + Jm = 0 (17)

r =

rJl

Jm(18)

Such anideally matched gearboxensures that the inertia in one side matchesthe inertia in the other side because each contributes

pJlJm.

3.4 Electrical Side

The equations for the electrical side are derived from the the basic laws for a DCmotor which are:

qe(t) = Kmia(t)if (t) (19)

ve(t) = Kmif (t)!m(t) (20)

whereif (t) is the field circuit current;ia(t) is the armature circuit current;!m(t)is the rotational speed of the rotor of the motor;qe(t) is the electrically generatedtorque applied to the rotor shaft by the interactions of the electrical fields producedby the field and armature coils;ve(t) is the back-emf generated across the brushesof the motor when the rotor rotates and which opposes the armature circuit voltage;andKm is an electromagnetic coupling constant.

These equations are nonlinear. To make them linear, either the field current orthe armature current is kept constant and the motor speed is then controlled by thecurrent flowing in the other circuit. We thus have four basic configurations for theDC motor. If the armature circuit current is kept constant then the motor is said tobefield-controlled. The basic equation of motion becomes:

qe(t) = Kmf if (t) (21)

whereKmf = Kmia = constant is the field-circuit controlled electromagneticccoupling constant which has units N m/A. If the field current is used to controlthe motor the motor is said to befield-current controlledand (21) suffices. If thefield voltage is used to control the motor we need an extra equation to take intoaccount the field circuit impedance which is taken to be the field coil’s inductanceand resistance in series. The motor is then said to befield-voltage controlled.

If the field circuit current is kept constant then the motor is said to bearmature-controlled. The basic equations of motion become:

qe(t) = Kmaia(t) (22)

ve(t) = Kma!m(t) (23)

whereKma = Kmif = constant is the armature-circuit controlled electromagneticcoupling constant which has units N m/A. If the armature current is used to control


the motor the motor is said to bearmature-current controlledand (22) suffices. Ifthe armature voltage is used to control the motor we need extra equations to takeinto account the armature circuit impedance and the back e.m.f. (23). The motor isthen said to bearmature-voltage controlled.

Block diagrams for the possible DC motor configurations are easy to deriveand they are all illustrated in Slide 70. Note that we have used the driven-loadequations derived in Section 3.2 to model the mechanical side of the motor. Youshould be comfortable with deriving models for all these configurations of motor,and to that end, Exercises 3–1 to 3–4 are provided to give you some practice.

Slide 70

Possible DC Motor Configurations

Kma

1sL Ra a+

Ω m s( )Q se( )I sa ( )V sa( )

+-

V se( )

1

Z s Z sm l( ) ( )+Kma

Ωm s( )Q se ( )I sa( ) 1

Z s Z sm l( ) ( )+

Kmf

1

sL Rf f+

Ωm s( )Q se( )I sf ( )V sf ( ) 1

Z s Z sm l( ) ( )+Kmf

Ωm s( )Q se ( )I sf ( ) 1

Z s Z sm l( ) ( )+

Constant Field Current

Constant Armature Current

Field-current controlled Field-voltage controlled

Armature-current controlled Armature voltage controlled

Kma

As an aside, the same basic equations are used to derive models for electricalgenerators. In that case, the input is the rotor speed!m(t) and the output is theback e.m.fve(t). The model is linearised by either keeping the rotor speed or thefield current constant.

DC Motors

Exercises

3–1A field voltage controlled motor drives a load with resistanceR N m=(rad=s)

and inertiaJ kg m2. Determine the transfer function relating the load speed to thefield voltage.

3–2


An armature current controlled motor drives a load with negligible resistanceand inertiaJ kg m2 through a long shaft with complianceC rad=N m. Determinethe transfer function relating the load speed to the armature current.

3–3An armature voltage controlled motor drives a load with resistanceRN m=(rad=s)

and inertiaJ kg m2. Obtain an electrical network for which the input impedance isthe same as the input impedance of the armature circuit.

3–4In a field voltage controlled motor with field resistance 1, field inductance 5

H, rotor resistance0:5 N m=(rad=s) and rotor inertia2 kg m2, theelectromechanical coupling constant relating torque to field current is10N m=A.

If the motor drives the load with resistance0:5 N m=(rad=s) and inertia8 kg m2,determine the output speed following a step input of20 V applied to the field circuitwhen the motor is at rest.

We are now ready to put together these results to construct a model for the DCmotor used to control the azimuth position of the antenna.

3.4.1 Armature-Voltage Controlled DC Motor

For the azimuth position control system we shall use the armature-voltage con-trolled DC motor shown in schematic form in Slide 64.

By putting together all we know so far, the block diagram for this kind of motoris shown in Slide 71. [This block diagram is available for download as a Simulinkmodel1]

The “gearbox” loop can be reduced to that shown in Slide 72 from which itis clear that the motor speed is related to the electrically generated torque by thetransfer function

m(s)

Qe(s)=

1sJm+Rm

1 + sJl+Rl

r21

sJm+Rm

=1

s(Jm + Jl=r2) + (Rm +Rl=r2)(24)

A shorthand for (24) is

m(s)

Qe(s)=

1

sJe + Re

(25)

whereRe = Rm+Rl=r2 andJe = Jm+Jl=r

2 are the effective resistance and iner-tia as seen at the motor shaft.(sJl +Rl)=r

2 is called thereflectedload impedance.

1http://www-ee.swan.ac.uk/Courses/level2/ee208/models/avcdcm.m


Slide 71

Block Diagram Model

1

sJ Rm m+Kma

Kma

1sL Ra a+

Ωm

s( )Q se( )I sa( )V sa ( )

+ +-

-

V se( )

1

r

1

rsJ Rl l+

Q sm( )

Q sl ( )

Ω l s( )

loadimpedance

motortorque

armaturecircuit

admittance

electromagneticcoupling

electromagneticcoupling

backe.m.f.

motoradmittance

gearbox

gearbox

Slide 72

Gear-box Loop

1

sJ Rm m+Kma

Kma

1sL Ra a+

Ωm s( )

Q se( )I sa( )V sa ( )

+ +-

-

V se( )

1

r

Q sm( ) Ω l s( )

gearbox

sJ Rrl l+

2


The block diagram has now been reduced to that shown in Slide 73 [This blockdiagram is available for download as a Simulink model1] from which it is easy toshow that

m(s)

Va(s)=

Kma

(Ra + sLa)(sJe + Re) +K2ma

(26)

Slide 73

Reduced Block Diagram

1

sJ Re e+Kma

Kma

1sL Ra a+


+-

V se( )

1

r

Ωl s( )

gearbox

Ωm s( )

Now m(s) = sm(s) wherem(s) is the transformed motor shaft positionm(t). If we neglect the armature inductanceLa then:

Va(s) =

Ra(sJe + Re) +K2

ma

Kma

sm

=

Ra

Kma

(sJe + Re) +Kma

sm (27)

Hencem(s)

Va(s)=

Kma

RaJe

ss+ 1

Je

Re +

K2ma

Ra

: (28)

Equation (28) is a relatively simple transfer function of the form:

T (s) =K

s(s+ )(29)

of which, more later.

1http://www-ee.swan.ac.uk/Courses/level2/ee208/models/avcdcm2.m


3.5 Coupling Constants

[This section is not examinable, but is included for completeness] How do we de-termine the coupling constants and hence find a suitable motor for a given load?Recall that for the motor

Ve(s) = Kmam(s) (30)

= Kmas(s) (31)

andQe(s) = KmaIa(s): (32)

Hence, for the armature circuit

Ia =Va Ve

sLa +Ra

(Ra + sLa)Ia = Va Ve

= Va Kmasm

Va = (Ra + sLa)Qe

Kma

+Kmasm: (33)

Assuming that the armature winding’s inductance,La, is negligible then

Va =Ra

Kma

Qe +Kmasm: (34)

In the time domain:

va(t) =Ra

Kma

qe(t) +Kma!m(t): (35)

If a constant DC voltageva is applied to a given motor, the motor will run at aconstant speed!m with a constant torqueqe, hence in the steady-state

va =Ra

Kma

qe +Kma!m: (36)

Solving forqe:

qe = K2ma

Ra

!m +Kma

Ra

va: (37)

This is a straight-line relationshipqe versus!m as shown in Slide 74.We use adynameterto measure this torque-speed characteristic for a givenva

using a set-up as shown in Slide 75When the speed!m is zero, the curve intercepts the torque axis at a value that

is called thestall-torqueqstall.

qstall =Kma

Ra

va: (38)

When the torque is zero we have a speed called!noload

!noload =va

Kma

(39)


Slide 74

Torque-Speed Relationship for a DC Motor

q ( )Nm

ωm ( )rad / s

va1va2

qstall

ωno load−

Torq

ue

Speed

Slide 75

Dynameter

Tension

The disk is attached to the motor under test. The belt is held against the edge of the disk under tension. As the motor rotates, the friction between the disk and the belt due to the torque generated by the motor causes the tension in the belt to increase. This increase is measured by the spring balance. At the same time the speed of the disk can be measured by using a tachometer or a stroboscope.

A Dynameter

springbalance

torque


Hence the electrical constants are:

Kma

Ra

=qstall

va(40)

Kma =va

!noload(41)

Example

For the motor with the torque-speed characteristic shown in Figure 1 find the trans-fer functionl(s)=Va(s) for an armature-voltage controlled DC motor which drivesa load with inertia 700 kg m2 and bearing resistance 800 N m/(rad/s) through a gear-box with gear ratior = 10. The rotor inertia of the motor is 5 kg m2 and bearingresistance is 2 N m/(rad/s).

Torq

ue (

Nm

)

Speed (rad/s)

500

50

va = 100 V

Figure 1:

Solution

Je = Jm + Jl=r2 = 5 +

700

102= 12

Re = Rm +Rl=r2 = 2 +

800

102= 10

qstall = 500 N m,!noload = 50 rad/s,va = 100 V. Hence

Kma

Ra

=qstall

va=

500

100= 5:

Kma =va

!noload=

100

50= 2:


Given thatm(s)

Va(s)=

Kma

Ra

: 1Je

ss+ 1

Je

Re +Kma

Kma

Ra

then

m(s)

Va(s)=

5:12

ss+ 1

12(10 + 2 5

=

0:417

s(s+ 1:1667):

To findl(s)=Va(s) we note that!l = !m=r hencel = m=r so

l(s)

Va(s)=

0:0417

s(s+ 1:1667):

A More Difficult Problem

Exercises

This problem is similar to the example except that the gearbox has non-negligibleinertia.

3–5An armature voltage controlled DC motor whose torque-speed characteristics

are shown in Figure 2 drives a load with inertia 16 kgm2 and bearing resistance32 N m/(rad/s) through the gearbox, illustrated in Figure 3, in which some of thegears have non-neglible inertia. Find the transfer function relating the load speed tothe armature voltage.

Hint: reflect all the inertias and resistances of the gear-wheels and the loadshaft to the drive shaft using the rule “equivalent impedance= impedance=r2”.

Recap

We have now derived a mathematical model, in block diagram form, of the DCmotor and load which forms theactuatorandplant of the azimuth position controlsystem. We are still some way from a mathematical model of the complete closed-loop control system. To create this we need to add sensors for actual and demandedposition (and perhaps velocity), signal and power amplifiers. This will be the topicof the next lecture at the end of which we shall be able to describe, in block diagramform, suitable control systems for both position and speed control.


Torq

ue (

Nm

)

Speed (RPM)

5

va = 5 V

600

π

Figure 2: Torque-speed characteristics

motorN

J

1

12

10

1

=

= kgm

N

J

2

22

20

2

=

= kgm

N

J

3

32

10

2

=

= kgm

N4 20=Z sl ( )

Figure 3: Gear-box


Lecture 4: Mathematical Modelling (2)

Preamble

In the last lecture we developed a block diagram model for the armature voltagecontrolled DC motor that is used in the antenna azimuth position control system.The block diagram that resulted is shown in Fig. 4 and the corresponding transferfunction is:

m

Va=

Kma

(Ra + sLa)(sJe + Re) +K2ma

1

sJ Re e+Kma

Kma

1sL Ra a+


+-

V se( )

1

r

Ωl s( )

gearbox

Ωm s( )

Figure 4: Block diagram of armature-voltage-controlled DC motor

If the armature circuit inductanceLa is assumed negligible, and it is noted thatm = sm (wherem is the motor shaft position), an equivalent transfer functionrelating the motor-shaft position to the armature voltage is:

m

Va=

Kma

s(Ra(sJe + Re) +K2ma)

=Kma=(RaJe)

s(s+ (1=Je)(Re + (K2ma)=Ra))

which is of the form

G(s) =K

s(s+ )

In this lecture, we will add sensing elements and amplifiers to this model tocreate block-diagram models of the closed-loop control systems for position con-trol (called aservomechanism) and speed control (called avelodyne). We will alsoshow that the steady-state response of these two types of system are different. Theposition control system has zero steady-state error whilst the velocity control sys-tem has a non-zero steady-state error which can be made small if there is sufficientgain in the system.

3.6 The Rest of the Control System

The motor fits into the azimuth position control system as shown in Fig. 5. Notethat the potentiometers, amplifiers, gearbox, inertias and bearing resistances havebeen given physical values according to Table 2 [A Matlab script file that can be


used to load this data into the workspace for the purposes of simulation of one ofthe Simulink models is available1.]

+10

-10

Pre-Amp.

Pot. 10 turn

Pot. 10 turn

Power Amp.

+10

-10

-

+

θ i t( )

θo t( )Rm = 0 01. Nm rad s

Rl = 1N m rad s

Jm = 0 02 2. kg m

Jl =1 2kg m

8Ω100

100s+K

N2 250=

N1 25=

N3 250=

v t1( )

v t2( )

Figure 5: Position Control System Schematic

The various new components are described below and then a complete mathe-matical model of the system, in block diagram form, will be developed.

3.6.1 Position sensors

A potentiometer (Fig. 6) producesvout / in. There are no dynamics.

vout

in=vmax vmin

2n:

In this case 10 turns produces 20 V hence

vout

in=

20

20=

1

V=rad

n turn pot.

θin ( )t

v tout( )

vmax

vmin

Figure 6: Potentiometer

1http://faith.swan.ac.uk/Courses/level2/ee208/models/ee208dat.m


Name Symbol Value UnitsLoad inertia Jl 1 kg m2

Load bearing resistance Rl 1 N m/(rad/s)Motor shaft gear wheel N1 25 teethLoad shaft gear wheel N2 250 teethGear ratio r N2=N1 noneMotor rotor inertia Jm 0.02 kg m2

Motor shaft bearing resistance Rm 0.01 N m/(rad/s)Equivalent inertia Je = Jm + Jl=(r

2) kg m2

Equivalent resistance Re = Rm + Rl=(r2) N m/(rad/s)

Electromagnetic coupling constant(forward)

Kma 0.5 N m/A

Electromagnetic coupling constant(feedback)

Kma 0.5 V/(rad/s)

Armature winding inductance (ne-glected in some models)

La 0.8 H

Armature winding resistance Ra 8 Ten turn potentiometer suppliedwith 10 V

Kp V/rad

Tacho generator Kt V/(rad/s)Voltage pre-amplifier gain K nonePower voltage amplifier (unity DCgain)

100=(s+ 100) none

Table 2: Component Parameters for the Position Control System


3.6.2 Velocity sensors

A tacho-generator (Fig. 7) is used to sense the speed of a motor.vout / !in. Pro-vided that the load-circuit impedance is high and the tacho-generator is physicallysmall with respect to the driven shaft then the device can be assumed to have nodynamics.

vout

!in= KT V=(rad=s):

ω in ( )tv tout( )

Figure 7: Tacho-generator

3.6.3 Pre-amplifier

The pre-amplifier is assumed to be a small current differencing op-amp circuit asshown in Fig. 8.

vp(t) =Rf

Ri

fvi(t) vo(t)g= K fvi(t) vo(t)g

+-v to( )

v ti( )

v tp( )

Ri

Ri

Rf

Rf

Figure 8: Differencing pre-amplifier

3.6.4 Power Amplifier

This produces a high-current voltage output from a low-current voltage input. Suchan amplifier will often have dynamics which cannot be neglected. In this case we


assumeVa(s)

Vp(s)=

100

s+ 100

indicating that the amplifier has unity DC gain and a time constant of1=100 sec-onds. This means that the amplifier would reach63% of its final output voltage in0:01 seconds following a step change in the input voltage (see Fig. 9).

Am

plif

ier r

esp

on

se: 1

V d

em

an

d

63% final value

1/100 sec

Time (seconds)

Figure 9: Response of the power amplifier

3.6.5 Block Diagram of Plant

For the system shown in Fig 5 we have derived a block diagram for the DC motor,gearbox and load which on reduction becomes that shown in Fig. 10.

K R J

s sJ

RK

R

ma a e

ee

ma

a

( )

+ +1 2 1

r

V sa( ) Θm s( ) Θ l s( )

Figure 10: Reduced block diagram of armature volatge controlled DC motor usedin position control system. Output of gearbox is load shaft positionl

The given system parameters are listed in Table 2. Some still need to be calcu-lated. The motor-load gearbox ratio is

r =N2

N1

=250

25= 10:

The equivalent motor-load inertia as reflected back to the motor shaft is

Je = Jm +Jl

r2= 0:02 + 1 1

100= 0:03:


The equivalent bearing resistance is:

Re = Rm +Rl

r2= 0:01 + 1 1

100= 0:02:

Plugging these values into transfer function for the DC motor and load gives:

m(s)

Va(s)=

Kma=(RaJe)

ss+ 1

Je

Re +

K2ma

Ra

(42)

m(s)

Va(s)=

0:5=(8 0:03)

ss+ 1

0:03

0:02 + 0:52

8

=

2:083

s(s+ 1:71)

The transfer function relating the load positionl(s) = o(s) is

o(s)

Va(s)=

1

r

Kma=(RaJe)

ss+ 1

Je

Re +

K2ma

Ra

(43)

=0:2083

s(s+ 1:71):

The load-pot gearbox has unity gear-ratio so that the pot moves at the samespeed as the load. So, putting everything together we end up with the block diagramshown in Fig. 11. [This block diagram is available for download as a Simulinkmodel2]

1

r

V sa( ) Θm s( ) Θo s( )2 083

1 71

.

( . )s s+100

100s+K

1π

1

π

-

+Θ i s( ) V si ( )

V so( )

V sp( )

pot

pot

pre-amppoweramp

motor&

load gearbox

Figure 11: Complete position control system.

It is convenient to reduce the block diagram to theunity-gain feedback canoni-cal formshown in Fig. 12 which we can do because the potentiometer in the forwardpath and the feedback path have the same gain.

Theopen-looptransfer function is then:

Go(s) =o(s)

E(s)=

66:3K

s(s+ 100)(s+ 1:71);

2http://faith.swan.ac.uk/Courses/level2/ee208/models/srvomech.m


1

r

V sa( ) Θm s( ) Θo s( )2 083

1 71

.

( . )s s+100

100s+K

1π

-

+Θ i s( ) V sp( )E s( )

Figure 12: Reduced block diagram

and theclosed-looptransfer function is

Gc(s) =o(s)

i(s)=

66:3K

s3 + 101:7s2 + 171s+ 66:3K:

Note that in the steady-state, i.e. when all terms ins are removed

i = o:

So the steady-state position error is zero!Another name for this type of control system is a “servomechanism”.

3.7 Velocity control system

A “velodyne” is a load-speed control system. A velodyne can be made from thecomponents seen so far if a tacho-generator with gainKT v/(rad/s) is used in thefeedback loop and a potentiometer with gainKT is used as a demanded speedsensor. Such a set up is shown in Fig. 13. [This block diagram is available fordownload as a Simulink model3]

1

r

V sa( ) 2 0831 71

..s+

100

100s+K

-

+V si ( )

V so( )

V sp( )

pot

tacho

pre-amppoweramp

motor&

load gearbox

KT

KT

Ω i s( ) Ωo s( )Ωm s( )

Figure 13: A velocity control system (Velodyne).

On manipulating this block diagram we get the unity-gain feedback control sys-tem shown in Fig. 14.

The closed-loop transfer function for the velodyne is

Gc(s) =o(s)

i(s)=

66:3KKT

s2 + 101:71s+ (171 + 66:3KKT )

In the steady-state!o

!i=

66:3KKT

171 + 66:3KKT

3http://faith.swan.ac.uk/Courses/level2/ee208/models/velodyne.m


1

r

2 0831 71

..s+

100

100s+K

-

+ KT

Ω i s( ) Ωo s( )E s( )

Figure 14: Reduced block diagram

that is!o 6= !i but !o !i if 66:3KKT 171. This can be achieved if thepre-amplifier gainK can be made sufficiently large.

3–6In a control system for rotating a radar aerial assembly, an electric motor with

inertia0:05 g m2 and resistance0:02 N m=(rad=s) is required to drive the inertiawith inertia500 kg m2 and resistance50 N m=(rad=s) through a gearbox. Deter-mine the gear ratio that minimises the torque needed to accelerate the load, and thetransfer function relating aerial speed to motor torque if such a gearbox is used.What is the motor power required to rotate the aerial at10 rev=min?

3–7In a servomechanism using a field voltage controlled motor, the ratio of the mo-tor torque to the error between the demanded and actual load position is100 N m=rad.If the inertia and resistances refererred to the motor shaft are20 kg m2 and3 N m=(rad=s),and the motor drives the load through a gearbox with ratio 50:1, determine the over-all transfer function of the system.

3–8A position control system is illustrated in Figure 3–8. Evaluate the transfer func-

tion of each subsystem and determine the closed-loop transfer functiono(s)=i(s).

+10

-10

Pre-Amp.

Pot. 1 turn

Pot. 1 turn

Power Amp.

+10

-10

-

+

θ i t( )

θo t( )Rm = 0 01. Nm rad s

Rl = 3N m rad s

Jm = 0 05 2. kgm

Jl = 5 2kg m

5Ω150

150s+K

N2 250=

N1 50=

N3 250=

v t1( )

v t2( )

Kma = 1 N m A

(Adapted fromChapter 2 Objective Problem, Nise, Exercise 48, page 109).

Recap

In this lecture we have developed block diagram models for a position control sys-tem and the speed control system created around an armature voltage controlled


DC motor. The position control system (servomechanism) was found to have zerosteady-state error which means that it is ideal for the antenna azimuth position con-trol system. The same mechanism, when used in a speed control system (velodyne),was found to have finite non-zero steady-state error which can be decreased by in-creasing the pre-amplifier gain.

In the next sequence of lectures called collectively “Evaluation of System Re-sponse”, we shall turn our attention to transient response. Beginning in the nextlecture with an evaluation of how poles and zeros effect the system response andbuilding up to an understanding of first and second-order system response and mea-sures which can be used to evaluate and ultimately control them.

4 EVALUATION OF SYSTEM RESPONSE 76

Lecture 5: Evaluation of System Response (1)

Preamble

In the last lecture we developed block diagram models for a position control systemand a speed control system created around an armature voltage controlled DC mo-tor. The position control system (servomechanism) was found to have zero steady-state error which means that it is ideal for the antenna azimuth position controlsystem. The same mechanism, when used in a speed control system (velodyne),was found to have finite non-zero steady-state error which can be decreased byincreasing the pre-amplifier gain.

In this sequence of lectures called collectively “Evaluation of System Response”,we shall turn our attention to the transient response of systems. We begin with anevaluation of how poles and zeros effect the system response and build up to anunderstanding of first-order and second-order system response and measures whichcan be used to evaluate and ultimately to control them.

4 Evaluation of System Response

This section of the course is devoted to the evaluation of the transient response fromthe system model. After introducing the useful concept of poles and zeros, we beginanalyzing models to find the step response of first and second order systems (wherethe termorder refers to the polynomial order of the denominator of the transferfunction).

Slide 76

Evaluation of System Response

ObjectivesGiven the antenna azimuth-angle control system developed in the section on

modelling, at the end of this section we shall be able to:

Predict, by inspection, the form of the open-loop angular velocity

response of the load to a step-voltage input to the power amplifier;

Find the damping ratio and natural frequency of the open-loop system;

Derive the complete analytical expression for the open-loop response

using transfer functions.


4.1 Poles and Zeros and System Response

The output response of a linear system is the sum of the forced response and thenatural response

c(t) = cf (t) + cn(t):

Solving differential equations or using inverse Laplace transforms allow us to eval-uate the output response. But this is laborious and time consuming. By havingan analysis technique that is quick and easy to apply we will increase productiv-ity. Knowledge of the effects of poles and zeros gives us a means of qualitativelyevaluating the response of a system by inspection.

Poles of a Transfer Function: The poles of a transfer function arethose values of the transfer function variables that cause the transferfunction to become infinite.

Zeros of a Transfer Function: The zeros of a transfer function arethose values of the transfer function variables that cause the transferfunction to become zero.

LetG(s) be a transfer function

G(s) =b(s)

a(s)=sm + bm1s

m1 + + b1s+ b0

sn + an1sn1 + + a1s+ a0

The poles ofG(s) are solutions of

a(s) = sn + an1sn1 + + a1s+ a0 = 0

that is, they arezerosof thedenominator polynomiala(s). If the denominator ofG(s) is zero thenG(s) =1.

The zeros ofG(s) are solutions of

b(s) = sm + bm1sm1 + + b1s+ b0 = 0

that is, they arezerosof thenumerator polynomialb(s). If the numerator ofG(s)is zero thenG(s) is also0.

Example 4.1 Determine the step response of a transfer function.

Solution: The step response is

C(s) =s+ 2

s(s+ 5)

=A

s+

B

s+ 5

=2=5

s+

3=5

s+ 5

thus

c(t) =2

5+

3

5e5t:


Slide 77

Example G(s) = s+2s+5

R ss

( ) =1

C s( )

σ

jωs− plane

−5 −2

s

s

++

2

5

poles at

zeros at

s

s

= −= −

5

2

The step response of

G(s) =s+ 2

s+ 5

is shown in Slide 78. A Plot ofjG(s)j for s = x + jy, x = 6; : : : ; 0;y = 3; : : : ; 3, indicating why a pole is called a pole is shown in Slide 79.

Discussion

1. A pole in the input function generates the form of the forced responsecf (t)(a pole at the origin generated a step function in the output).

2. A pole of the transfer function generates the form of the natural response (i.e.a pole ats = 5 generatedcn(t) = e5t).

3. A pole on the real-axis in thes-plane generates an exponential response ofthe formet where is the pole location on the real axis. Thus the furtherleft a pole is on the negative axis, the faster the exponential transient decaysto zero.

4. The zero helps generate the amplitudes for both the steady-state and transientperformances. (This can be seen from the calculation ofA andB in thepartial fraction expansion.)


Slide 78

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time t seconds

c(t)

Step Response for G(s) = (s + 2)/(s + 5)

Matlab command: step([1 2],[1 5])

Slide 79

-6-5

-4-3

-2-1

0

-4

-2

0

2

40

5

10

15

20

25

30

35

Res Ims

| G(s

) | =

|(s

+ 2

)/(s

+ 5

)|

Poles and Zeros


Slide 80

Contributions of Poles and Zeros

σ

jω

−5σ

jω

−2σ

jωinput pole system zero system pole

C ss s

( ) = ++

2 5 3 5

5

c t e t( ) = + −2

5

3

55

forced response

naturalresponse


Example 4.2 Use poles to evaluate the system response of the system shown inFig. 15 by inspection.

R ss

( ) =1

C s( )s

s s s

++ + +

3

2 4 5( )( )( )

Figure 15: Poles and System Response

Solution:

C(s) =K1

s|zforced response

+K2

s+ 2+

K3

s+ 4+

K4

s+ 5| z natural response

c(t) = K1|zcf (t)

+K2e2t +K3e

4t +K4e5t| z

cn(t)

:

Transfer Functions in MatlabThere are two forms of transfer function representation in Matlab. The most obviousis the polynomial form where

G(s) =b(s)

a(s)=

s2 + 2s+ 3

s3 + 4s2 + 5s+ 6

is entered as two row vectors with the polynomial coefficients entered in the orderof descending powers ofs.

>> b = [1, 2, 3];>> a = [1, 4, 5, 6];

Missing coefficients, must be entered as zero: soq(s) = s2 + 2s andr(s) =s4 + s2 + 1 are entered as

>> q = [1, 2, 0];>> r = [1, 0, 2, 0, 1];

An alternative form of representing transfer functions is thefactored polyno-mial, for example

G(s) =(s+ 1)(s+ 3)

s(s+ 2)(s+ 4)

The advantage of this formulation is that the roots orzerosof the numerator anddenominator polynomials are obvious by inspection so it is often used in the pre-liminary analysis of the performance of a dynamic system. Thepolesof this transferfunction are therefores = 0; 2; 4 and thezerosares = 1; 3. In Matlab,this form of transfer function is specified by a column vector of the zeros and acolumn vector of the poles:


>> z = [-1; -3];>> p = [0; -2; -4];

A third parameter, the overall gainK, completes the definition of the so calledpole-zero-gainform of transfer function. In this caseK = 1

>> k = 1;

Transformations

Matlab supports the transformation of transfer function between forms. For exam-ple to convert a transfer function from ‘expanded’ form to pole-zero-gain form thecommandtf2zp is used:

>> [z1,p1,k1] = tf2zp(b,a)

z1 =-1.0000 + 1.4142i-1.0000 - 1.4142i

p1 =-3.0000-0.5000 + 1.3229i-0.5000 - 1.3229i

k1 =1

To convert from zero-pole-gain form to expanded form we use the functionzp2tf :

>> [b1,a1] = zp2tf(z,p,k)

b1 =0 1 4 3

a1 =1 6 8 0

Partial Fraction Expansions

Matlab also provides a command calledresidue that returns the partial fractionexpansion of a transfer function. That is, given

G(s) =sm + bm1s

m1 + + b1s+ b0

sn + an1sn1 + + a1s+ a0

it returnsR1

s+ p1+

R2

s+ p2+ Rn

s+ pn+K(s)


wherepi are the poles of the transfer function,Ri are the coefficients of the partialfraction terms (called theresiduesof the poles) andK(s) is a remainder polynomialwhich is usually empty. To use this, the starting point must (rather perversely) bethe expanded form of the transfer function. Thus given

C(s) =5(s+ 2)

s(s+ 3)(s+ 10)

we obtain the partial fraction expansion using the Matlab command sequence:

>> k = 5; z = [-2]; p = [0; -3; -10]; % zero-pole-gain form>> [num,den] = zp2tf(z,p,k)

num =0 0 5 10

den =1 13 30 0

(Note that the leading terms innumarezero).

>> [r,p,k] = residue(num,den)r =

-0.57140.23810.3333

p =-10

-30

k =[]

which we interpret to mean

C(s) =0:3333

s+

0:2381

s+ 3 0:5714

s+ 5:

If C(s) represents the step response of the system

G(s) =5(s+ 2)

(s+ 3)(s+ 10)

then the step response is, by inspection,

c(t) = 0:3333 + 0:2381e3t 0:5714e10t:

You can check this with the command:

>> step([5, 10],[1, 13, 30])


(where the1=s term has been eliminated becausestep provides the forcing func-tion itself). This should give exactly the same results as:

t = 0:.05:1.5; % time vectorc = 0.3333 + 0.2381 * exp(-3*t) - 0.5714 * exp(-10*t);plot(t,c)

Exercise

Use Matlab to determine the actual coefficients of the partial fraction expansion forthe previous example.

4.2 First-Order System Responses and Specifications

A first-order system is illustrated in Slide 81. It has the transfer function

G(s) =a

s+ a

which is also called a “first-order lag”. The system has one pole ats = a asshown in thepole-zero diagram. For the step response, the input signal transformis

R(s) =1

s

so the step response is

C(s) = G(s)R(s) =a

s(s+ a):

In the time domain the step response is:

c(t) = cf (t) + cn(t) = 1 eat:

The forced responsecf (t) = 1 is generated by the pole of1=s. The natural responsecn(t) = eat is generated by the pole ofa=(s+ a).

4.2.1 The significance ofa

The parametera is very important for specifying the performance of a first-ordersystem. It is significant because

eatt=1=a

= e1 = 0:37:

Or, alternatively, the step response

c(t) = 1 eatt=1=a

= 0:63:

That is, we can relate the shape of the time response to the parametera.


Slide 81

First-Order System

R ss

( ) =1

C s( )a

s a+ σ

jωs− plane

−a

4.2.2 Time Constant

The parameter = 1=a is called the “time constant” of the first-order response. Itis the time taken foreat to decay to 37% of its initial value or the time taken forthe step response to reach 63% of its final value. has units s.a has units s1 orfrequency.a is called theexponential frequency. The derivative ofeat at t = 0is a, soa is the initial rate of change of the exponential att = 0. Thus the timeconstant can be considered a transient response specification for a first-order systemsince it is related to the speed of response.

The time constant can be evaluated directly from the pole-zero plot. The pole iss = a. The further to the left ofs-plane, the largera hence the smaller the timeconstant = 1=a and the faster the response.

A plot of a typical first-order response is shown in Slide 82. The parametersdue toa are marked on the plot. Some additional time-response parameters that arein common use can also be identified. These are rise-time and settling-time.

4.2.3 Rise TimeTr

The rise-time (symbolTr units s) is defined as the time taken for the step responseto go from 10% to 90% of the final value. For a first-order system it is rather easilyderived by solvingc(t) = eat for c(t0:1) = 0:1cnal andc(t0:9) = 0:9cnal fromwhich

Tr =2:31

a 0:11

a=

2:2

a:


Slide 82

First-Order System Response

63%

10%

100%

90%

98%

t (seconds)

c(t)

1

a2

a

3

a

4

a

initial slope = 1/(time constant) = a

Tr

Ts

2%

4.2.4 Settling TimeTs

We shall define the settling-time (symbolTs units s) to be the time taken for thestep response to come to within 2% of the final value of the step response.1Fromthe definition,

c(Ts) = 0:98cnal

which gives

Ts =4

a:

In other words the 2% settling-time for a first-order system is4 the time constant.

Recap

In this lecture, we have seen that poles determine the nature of the time response:the poles of the input determine the nature of the forced response; the poles of thetransfer function determine the nature of the natural response.

For the latter, we could say that the poles of a transfer function determine thecharacterof the natural response. For this reason, the equation

a(s) = sn + an1sn1 + + a1s+ a0 = 0

is called the “characteristic equation”.Some further observations are:

1Other definitions are common. For example, some authorities prefer to use 5% of the final valueor 1% of the final value. To be absolutely clear we shall use the term x% settling-time to indicate whatband we are using.


The zeros of a transfer function contribute to the amplitude of the componentparts of the total response.

Poles on the real axis generate exponential responses.

We have also examined the response of the first-order system (or first order lag)

G(s) =a

s+ a

and discovered that:

System has a real pole ats = a

Step response:c(t) = 1 eat.

The parametera is called theexponential decay frequency(units s1)

Time constant = 1=a (units s) is time for step response to reach 63% of thefinal value.

The rise time is time taken for step response to rise from 10% to 90% of thefinal value. The rise timeTr = 2:2=a = 2:2 s

The 2% settling timeTs = 4=a = 4 s.

The larger the value ofa the smaller the value of and hence the faster theresponse of the system.

Second-order systems exhibit a wider range of responses than first-order sys-tems. In the next lecture we shall describe the types of responses exhibited bysecond-order systems.



Preamble

Second-order systems exhibit a wide range of responses which must be analysedand described. Whereas for a first-order system, varying a single parameter changesthe speed of response, changes in the parameters of a second order system canchange theform of the response. For example, a second-order system can displaycharacteristics much like a first-order system or, depending on component values,display damped or pure oscillations for its transient response.

In this lecture we shall use the general second-order transfer function shown inSlide 83. to explore the range of responses that are possible.

Slide 83

General second-order transfer function

R ss

( ) =1

C s( )b

s as b2 + +

2 poles. No zeros.

There are four types of response possible for a second order system. They are

Overdamped Response

Underdamped Response

Undamped Response

Critically Damped Response

To explore these we consider the second order system

G(s) =b

s2 + as+ b

with b = 9 anda selected to illustrate each type of response.


4.3 Types of Second-Order System Responses

4.3.1 Overdamped Response

IF a = 9 we have the system shown in Slide 84.

Slide 84

Overdamped system

R ss

( ) =1

C s( )99 92s s+ +

2 poles. No zeros.

The step response1will be

C(s) =9

s(s2 + 9s+ 9)=

9

s(s+ 7:854)(s+ 1:146):

The pole ats = 0 comes from the input (forced response) and there are two realpoles ats = 7:854 ands = 1:146 respectively. Thus the step response will beof the form

c(t) = K1 +K2e7:854t +K3e

1:146t:

(Slide 85).The response consists of the some of two first-order responses. The response

due to the largest pole ats = 7:854 has a time constant offast = 0:127 s. Left toitself, this pole would have a settling time ofTsfast = 0:51 s. The response due tothe smallest pole ats = 1:146 has a time constant ofslow = 0:873 s. This polehas a settling time ofTsslow = 3:49 s. Settling timeTsslow is about 4 times slowerthanTsfast . Thus the “slow pole” dominates the later stages of the response whilethe “fast pole” dominates the early part of the response. In addition, the initial slopeis zero.

1A full analysis of the step response for a general second-order system is given in an Appendix.


Slide 85

Overdamped response

c(t)

t

σ

jωs− plane

−1 146.−7 854.

4.3.2 Underdamped Response

IF a = 3 we have the system shown in Slide 86.The step response will be

C(s) =9

s(s2 + 3s+ 9):

The pole ats = 0 comes from the input (forced response) and there are two complexpoles ats = 1:5 j2:598. The step response of this system will be of the form

c(t) = K1 + e1:5t(K2 cos 2:598t+K3 sin 2:598t):

(Slide 87).Comparing the time response with the pole locations, we see that the expo-

nential decay term is related to the real-part of the complex pole pair. Real part:d = 1:5; exponential terme1:5t = edt. The frequency of the oscillatoryterm is related to the imaginary part of the complex pole pair. Imaginary part:!d = 2:598; sinusoidal term:K2 cos 2:598t + K3 sin 2:598t. This information issummarised in Slide 88).

The frequency of oscillation!d = 2T

(rad/s) is called thedamped naturalfrequency. The time constant of the decay is = 1=d.


Slide 86

Underdamped system

R ss

( ) =1

C s( )9

3 92s s+ +

2 poles. No zeros.

Slide 87

Underdamped response

c(t)

t

σ

jωs− plane

+ j2 598.

− j2 598.

−1 5.


Slide 88

Decaying oscillation

c(t)

t

T

exponential decaygenerated by real part ofcomplex pole pair

sinusoidal oscillationgenerated by imaginary part ofcomplex pole pair


Example 4.3 Determine by inspection the form of the step response of the systemshown in Fig. 16.

R ss

( ) =1

C s( )200

10 2002s s+ +

Figure 16: Example underdamped system

Solution: The poles are:s = 5 j13:23 and the input is a step hence the stepresponse is

c(t) = K1 + e5t(K2 cos 13:23t+K3 sin 13:23t)

= K1 +K4e5t(cos 13:23t )

where = tan1K3=K2 andK4 =pK2

2 +K23 : K1, K2 andK3 are obtained

from the partial fraction expansion

K1

s+

K2s+K3

s2 + 10s+ 200:

The proof is left as an exercise.

4.3.3 Undamped Response


C(s) =9

s(s2 + 9):

The pole ats = 0 comes from the input (forced response) and there are two imagi-nary poles ats = j3. The step response of this system will be of the form

c(t) = K1 +K2 cos 3t:

(Slide 90). The frequency of oscillation of the undamped response is called thenatural frequency!n. In this case the natural frequency!n = 3 rad.s1.

4.3.4 Critically Damped System


C(s) =9

s(s2 + 6s+ 9)=

9

s(s+ 3)2:


Slide 89

Undamped system

R ss

( ) =1

C s( )9

92s +2 poles. No zeros.

Slide 90

Undamped response

c(t)

t

σ

jωs− plane

+ j3

− j3


Slide 91

Critically Damped System

R ss

( ) =1

C s( )9

6 92s s+ +

2 poles. No zeros.

The pole ats = 0 comes from the input (forced response) and there are two realand equal poles ats = 3. The step response of this system will be of the form

c(t) = K1 +K2e3t +K3te

3t:

(Slide 92).


Slide 92

Critically Damped Response

c(t)

t

σ

jωs− plane

−3


Recap

The natural orcharacteristic responseof a second-order system is governed by thepoles of the denominator of the system transfer functionG(s). These characteristicsare:

1. Overdamped. Poles: two real at1 and2. Transient response: twoexponentials with time response equal to1=. cn(t) = K1e

1t+K2e2t.

2. Underdamped. Poles: two complex atd j!d. Transient response:damped sinusoid with exponential envelope whose time constant is1=d andwhose frequency is equal to!d rad/s.cn(t) = K1e

dt cos(!dt+ ).

3. Undamped. Poles: two imaginary atj!1. Transient response: undampedoscillation with frequency equal to!1 rad/s.cn(t) = K1 cos(!1t+ ):

4. Critically damped . Poles: two real and equal at1. Transient response:cn(t) = K1e

1t +K2te1t.

These results are summarised in Slide 93. It is clear from this picture that criticaldamping is the fastest possible response without overshoot.

Slide 93

Second-order responses

c(t)

t

underdamped

undamped

overdampedcritically damped

In the next lecture we will further generalise the second-order transfer functionso that we can determine the form of the step response by inspection without theneed to find the poles of the characteristic equations2 + as+ b = 0.



Preamble

In the last lecture we discussed the forms of characteristic response that the second-order system can exhibit.

In summary, these are:

1. Overdampedwhen the system has two real distinct poles;

2. Underdampedwhen the system has two complex conjugate poles;

3. Undampedwhen the system has two imaginary poles; and

4. Critically damped when the system has two real but equal poles.

We now seek to generalize the discussion and establish a set of quantitativespecifications which will enable the response to be described to a designer withoutthe need for a sketch of the response.

We develop two parameters that describe second-order response in a similarway to the way time constants can be used to describe a first-order response.

The two new quantities are called

1. natural frequency,

2. damping ratio.

4.4 The General Second-Order Response

4.4.1 Definitions

Natural frequency !n is defined as the frequency of oscillation of a second-ordersystem without damping. E.g. the frequency of oscillation of an RLC circuit withR shorted. It has units of rad.s1.Damping ratio is defined by

=Exponential decay frequency

Natural frequency

=1

2

Natural period

Exponential time constant

The damping ratio has no units. That is the damping ratio of a system thatdecays to zero after 3 oscillations in 1 ms is the same as that of a system that decaysto zero in three oscillations in 1 hour. It is independent of the speed of response orthe rate of the oscillation.


4.4.2 Derivation of Formulae

Let us derive formulae for these quantities from their definitions, given that

G(s) =b

s2 + as+ b:

Without damping, the termas = 0 and we have

G(s) =b

s2 + b:

The poles are imaginary, and the frequency of oscillation of this system is!n bydefinition. Thus

!n =pb;

b = !2n:

Assuming an underdamped system. The complex poles have a real part givenby

s = a2:

The magnitude of the real part is the exponential decay “frequency”d. Thusd =a=2 and from the definition

=d

!n=a=2

!n;

hencea = 2!n:

In general then, the second order system has thecanonical form:

G(s) =!2n

s2 + 2!ns+ !2n

(44)

Knowledge of the parameters!n and may be used to determine the type of motionof any particular second-order system.

Example 4.4 Find the natural frequency!n and damping ratio for the systemwith transfer function:

G(s) =36

s2 + 4:2s+ 36:

Solution: Comparing to the standard form

G(s) =36

s2 + 4:2s+ 36=

!2n

s2 + 2!ns+ !2n

!2n = 36! !n =

p36 = 6:

2!n = 4:2

=4:2

2!n=

4:2

2 6= 0:35:


4.4.3 Pole-zero locations

Having defined!n and, let us relate these quantities to the pole locations in thes-plane.

Solving thecharacteristic equation

s2 + 2!ns+ !2n = 0

givess1;2 = !n !n

p2 1:

The various types of response for a given value of natural frequency!n are a func-tion of and may be summarised as shown in Table 3.

Example 4.5 Describe the nature of the second-order system response via the valueof the damping ratio for the systems with transfer function:

G(s) =12

s2 + 8s+ 12

G(s) =16

s2 + 8s+ 16

G(s) =20

s2 + 8s+ 20:

Solution: In all cases the transfer function is of the form

G(s) =b

s2 + as+ b

soa = 2!n and!n =pb hence = a=(2

pb). a = 8 in all cases.

For

G(s) =12

s2 + 8s+ 12

b = 12 hence = 8=(2p12) = 2=

p3 > 1: system response isoverdamped.

For

G(s) =16

s2 + 8s+ 16

b = 16 hence = 8=(2p16) = 1: system response iscritically damped.

For

G(s) =20

s2 + 8s+ 20

b = 20 hence = 8=(2p20) = 2=

p5 < 1: system response isunderdamped.


Table 3: The relationship between damping ratio, pole location and transient re-sponse for second-order systems with transfer functionG(s) = !2

n=(s2+2!ns+

!2n):

Pole Locations Type of Response

= 0

σ

jωs− plane

+ j nω

− j nω

Oscillatory

< 1

σ

jωs− plane

+ −j nω ζ1 2

− −j nω ζ1 2

−ζωn

Underdamped

= 1

σ

jωs− plane

−ζωn

Critically Damped

> 1

σ

jωs− plane

− − −ζω ω ζn n2 1 − + −ζω ω ζn n

2 1

Overdamped


4.4.4 Further analysis for underdamped second-order systems

The step response for the general second-order system.

C(s) =!2n

s(s2 + 2!ns+ !2n)

=K1

s+

K2s+K3

s2 + 2!ns+ !2n

where it is assumed that < 1 (the underdamped case) is now obtained1

Expanding by partial fractions yields

C(s) =1

s+

(s+ !n) +p12

!np1 2

(s+ !n)2 + !2

n(1 2)

from which

c(t) = 1 e!nt

cos!n

p1 2t+

p1 2

sin!np1 2t

!

= 1 1p1 2

e!nt cos(!np1 2t+ ) (45)

where

= tan1p

1 2:

If we define the exponential decay frequency to bed (the size of the real-partof the complex pole pair) then its value is

d = !n

. Similarly, if we define a second quantity, to be called the “decayed natural fre-quency”

!d

, to be the frequency of the decayed sinusoid, then its value is

!d = !np1 2 rad.s1

and equation (45) may be simplified further to:

c(t) = 1 !n

!dedt cos(!dt+ ) (46)

where is nowtan1(d=!d).Slide 94 shows the development of the second-order response from overdamped

to oscillatory response as is decreased from2 to 0. The graph is plotted against“normalized” time!ntwhich means that the time-axis is independent of the naturalfrequency. For an actual second-order system, the response for a given dampingratio will be a scaled version of the response that is shown here.

1A full analysis of the step response for a general second-order system is left as an exercise.


Slide 94

The Family of Step Responses for Second-Order Systems

ωnt

c tn( )ω

ζ = 1 2

ζ = 1

ζ = 1 4

ζ = 1 2

ζ = 1 10

ζ = 0ζ = 3 2ζ = 2

Recap

In this lecture we have developed a general formula for the second-order transferfunction

G(s) =!2n

s+2!ns+ !2n

in which the form of the response is determined by the damping ratio:

1. Overdampedwhen > 1, two real poles ats = !n !np2 1;

2. Underdamped when < 1, two complex conjugate poles ats = !n j!n

p1 2;

3. Undampedwhen = 0, two imaginary poles ats = j!n;

4. Critically damped when = 1, two real and equal poles ats = !n.

The natural frequency!n governs the speed of response. For the underdampedresponse, the

exponential decay frequencyis d = !n s1

damped natural frequencyis!d = !np1 2 rad s1.

The closed-loop poles ares = d j!d and the time response formula (45) canbe simplified to

c(t) = 1 !n

!dedt cos(!dt+ )

where = tan1

d

!d:


In the next lecture we will develop some performance measures based on atypical underdamped response curve.



Preamble

In the last lecture we developed a general formula for the second-order transferfunction

G(s) =!2n

s2 + 2!ns+ !2n

in which the form of the response is determined by the damping ratio:

1. Overdampedwhen > 1;

2. Underdampedwhen < 1;

3. Undampedwhen = 0;

4. Critically damped when = 1.

We also stated that the natural frequency!n governs the speed of response anddefined the termsexponential decay frequency(d) anddamped natural frequency(!d). We also considered the location of the poles of the second-order system anddiscovered, for underdamped systems,s = d j!d.

Finally, we developed a formula for the step response:

c(t) = 1 1p1 2

e!nt cos(!np1 2t+ )

= 1 !n

!dedt cos(!dt+ ) (47)

In this lecture, based on a typical underdamped response curve, we will developequations for the performance measures,percentage overshoot, settling time, andrise timein terms of the generalised second-oder parameters.

4.5 The Specification of Second-Order Response

In order to specify a second-order response we need to define some performancemeasures based on a typical underdamped response curve. The specifications are:

1. Peak TimeTp: the time taken to reach the first, or maximum, peak.

2. Percent overshoot%OS: the the amount that the waveform overshoots thesteady state, or final value at the peak time, expressed as a percentage of thesteady-state value.

3. Settling time Ts: the amount of time required for the transient’s dampedoscillations to stay within2% of the final value.

4. Rise timeTr: the amount of time required to go from 10% to 90% of thefinal value.


Slide 95

Second-Order Response Specifications

c(t)

t

10%

100%102%

90%98%

Tr

Ts

Tp

cmax

%OS 2%cfinal

A summary of these specifications is illustrated in Slide 95.Rise time, settling time and peak time yield information about the speed and

“quality” of the transient response. They can help the designer achieve a givenspeed of response without excessive overshoot or oscillations.

It should be noted that the last two specifications (Tr andTs) are the sameas those used for first-order systems and they may also be used for overdampedand critically damped second-order systems too. In fact, these specifications mayalso used for systems with order higher than two, provided that the response isof the same approximate shape. However, analytical formulae relating the time-response specification parameters to pole-and-zero locations can only be developedfor second order systems.

In the next sections, relationships betweenTp, %OS, Ts andTr and!n and are developed. Following that, they are related to the location of poles in thes-plane.

4.5.1 Evaluation ofTp

In order to find the value of the peak overshoot we must differentiate the time re-sponse with respect to time and then find the maximum value. This can be doneusing Laplace transforms since

Lf _c(t)g = sC(s) =!2n

s2 + 2!ns+ !2n

Completing squares in the denominator gives:

Lf _c(t)g =

!np12

!np1 2

(s+ !n)2 + !2n(1 2)


So_c(t) =

!np1 2

e!nt sin!np1 2t:

Setting the derivative to zero gives

!np1 2t = n

ort =

n

!np1 2

:

Each value ofn yields the value for a local maxima or minima. Lettingn = 0givest = 0 which corresponds to the starting point of the step response. Settingn = 1 gives the time at which the response reaches its first peak, that isTp, thus:

Tp =

!np1 2

(48)

4.5.2 Evaluation of%OS

From Slide 95, the percent overshoot,%OS, is given by

%OS =cmax cnal

cnal 100 (49)

cmax is found by substitutingt = Tp into Equation (47). Thus using (48) we have

cmax = 1 e(=p

12)

cos +

p1 2

sin

!

= 1 + e(=p

12) (50)

For the unit step used in (47)

cnal = 1 (51)

Therefore, substituting (50) and (51) into (49) gives:

%OS = e(=p

12) 100 : (52)

Note that the percentage overshoot is a function only of the damping ratio!The inverse of equation (52) allows us to find the value of the damping ratio

that yields a given%OS thus:

= ln(%OS=100)q2 + ln2(%OS=100)

: (53)

A plot of the relationship between%OS and is given in Slide 96.


Slide 96

Relationship between %OS and

Damping ratio, ζ

Pe

rce

nt

ove

rsh

oo

t, %

OS

4.5.3 Evaluation ofTs

In order to find the settling time we need to find the time for whichc(t) reaches andstays within2% of cnal. From Slide 95, an estimate ofTs is the time for whichthe decaying sinusoid in equation (47) reaches an amplitude of 0.02, or

e!nt1p

1 2= 0:02:

This is a conservative estimate because we are assuming thatcos(!np1 2 t

) = 1 at t = Ts. Nevertheless, solving fort gives

Ts = ln(0:02

p1 2)

!n(54)

It is easy to verify that the numerator of equation (54) yields values between 3.91and 4.74 as varies from 0 to 0.9. We therefore use an approximation forTs whichcan be used for all values of. The approximation is taken to be

Ts =4

!n(55)

Evaluation of Tr

There is no precise analytical relationship between rise time and damping ratio ornatural frequency. However, by experimentation a relationship can be found asshown in Slide 97.


Slide 97

Damping ratio versus normalised rise time

Rise

tim

e x

Na

tura

l Fre

qu

en

cy,

ωnT

r

Damping Ratio, ζ

Dampingratio0.10.20.30.40.50.60.70.80.9

Normalizedrise time

1.1041.2031.3211.4631.6381.8542.1262.4672.883

For0:866 < < 0:5 an approximation to rise time can be taken to be

Tr 1:8

!n(56)

This suggests that rise-time depends on!n, but as we shall see, this is rather a crudeapproximation and must also be taken into account in practice. Nonetheless, weshall use the approximation in equation (56) from time to time.

Example 4.6 Given the transfer function

G(s) =100

s2 + 15s+ 100

findTp, %OS, Ts andTr.

Solution: From the model transfer function!n = 10 and = 0:75. Substitutingthese values into the formulae forTp,%OS andTs givesTp = 0:475 seconds,%OS= 2.838, andTs = 0:533 seconds. From the graph in Slide 97 we find that for =0:75, !nTr 2:3 seconds. Dividing by!n givesTr = 0:23 seconds. All withouttaking inverse-laplace transforms of the step response ofG(s) and measuring theresponse!

Recap

In this section we have developed formulae for the second-order system perfor-mance parameters


Percentage overshoot(%OS) (52), (53)

Time-to-peak(Tp) (48)

Settling time(Ts) (55)

Rise-timeTr (Fig. 97 and Eq. (56))

in terms of the generalized second-order parameters and!n.Although the definitions of settling-time and rise time are the same as used

for first-order systems, percentage overshoot only applies to second-order systemswhich are underdamped or undamped.

The same parameters may be used to categorise the performance of higher-ordersystems provided that they exhibit a response which is similar in shape to that of asecond-order system. However, in that case, there is no direct relationship betweenthe response parameters and the system poles.

In the next lecture we shall conclude the development of performance charac-teristics by finding out how the location of the system poles can be used to specifya particular kind of performance.



Preamble

In the last lecture we developed formulae for the second-order system performanceparametersPercentage overshoot(%OS), Time-to-peak(Tp), Settling time(Ts),andRise-timeTr, in terms of the generalized second-order parameters and!n.

In this lecture we shall conclude the development of performance charateris-tics by finding out how the location of the system poles can be used to specify aparticular kind of performance.

4.6 Relating Response Specifications to Pole Locations in thes-Plane

We now have the relationships between the specifications for peak-time, rise-time,settling-time and percent overshoot to the natural frequency and damping ratio ofan underdamped second-order system. We now need to relate these to the locationsof the poles on thes-plane. Once we have done this we will be in a position tospecify the pole locations that yield a particular response. That is we shall be ableto solve the so-called “synthesis” problem.

The pole plot for an underdamped second-order system is shown again in Slide 98.It is easy to show that the radial distance from the origin to the pole is!n andthe angle subtended by the radial line (measured from the negative real axis) is = cos1 .

Slide 98

Pole-plot for an underdamped second-order system

σ

jωs− plane

+ −j nω ζ1 2

− −j nω ζ1 2

−ζωn

ωn

α

Previously, when considering pole locations, we used the symbol!d (dampednatural frequency) for the imaginary part of the pole andd (exponential decay


frequency) for the real part of the pole. From the pole locations it is clear that

d = !n

and!d = !n

p1 2

Thus, in terms of pole locations

Tp =

!np1 2

=

!d(57)

Ts =4

!n=

4

d(58)

Equation (57) shows thatTp is inversely proportional to the size of the imag-inary part of the pole. Since horizontal lines on thes-plane are lines of con-stant imaginary part, then they are lines of constant peak-time.

Similarly, equation (58) shows thatTs is inversely proportional to the size ofthe real part of the pole. Thus, vertical lines on thes-plane, which are linesof constant real part, are lines of constant settling-time.

Further, since = cos, radial lines are lines of constant. Since percentovershoot is only a function of, radial lines are thus lines of constant over-shoot.

Finally, the approximate equation!n = 1:8=Tr implies that curves of con-stant natural frequency (semicircles of radius= !n) correspond to responseswith constant rise- time. Again,Tr is inversely proportional to!n.

Putting all this together we can show curves of constant peak time, settling time,rise time and percentage overshoot (Slide 99).

4.6.1 Effect of moving poles along design curves

To illustrate the effect of the design parameters in greater detail the following slidesshow the effect of moving the poles along these curves.

In Slide 100, the step responses for a system whose poles are moved keep-ing d constant are shown. As the poles move away from the real axis, thefrequency!d increases, but the exponential envelope given byedt remainsthe same. So the settling time remains virtually the same although the per-centage overshoot increases because the damping reduces.

In Slide 101, the effect of moving the poles to the left with constant!d isshown. Now, the damped natural frequency remains the same so thatTp isconstant. However, as the poles move further to the left, damping is increasedso that percentage overshoot is reduced and the oscillations are damped outquicker.


Slide 99

Curves of constant design parameters

σ

jω

Tp1

Tp2

Ts2Ts1

%OS1

%OS2

Tr1

Tr2

Slide 100

Poles move with constant real part

motion ofpoles

envelope the same

c(t)

t

σ

jωs− plane


Slide 101

Poles move with constant imaginary part

c(t)

t

σ

jωs− plane

motion ofpoles

frequency same

Slide 102

Poles move with constant damping ratio

motion ofpoles

overshoot samec(t)

t

σ

jωs− plane


In Slide 102, the percentage overshoot remains the same as the poles moveaway from the origin along the radial line. The system becomes faster as thenatural frequency is increased.

Finally, in Slide 103 we see that the approximationTr = 1:8=!n is not veryprecise. As the poles move together along the curve of constant!n the risetime varies quite by quite a large amount. This is because the damping ratio ischanging also. Nevertheless, the relationship betweenspeed of responseandnatural frequencyfor a given value of can be obtained from design curveslike that shown in the last lecture.

Slide 103

Poles move with constant natural frequency

motion ofpoles

rise-time differs

c(t)

t

σ

jωs− plane


q t( )

θ( )t

C = 0 2. rad / Nm

J

R

Figure 17: Rotational Mechanical System

Example 4.7 The poles for a second-order systems are located ats = 3 j7,determine the system response.

Solution: From trigonometry = cos = cos(tan1(7=3)) = 0:394. The naturalfrequency is given by!n =

p32 + 72 = 7:616. The peak time is

Tp =

!d=

7= 0:449 seconds:

The percent overshoot is

%OS = e(=p

12) 100 = 26:018%

The approximate settling time is

Ts =4

d=

4

3= 1:333 seconds:

For the rise time we see that!nTr 1:44when = 0:394. ThusTr 1:44=7:616 =0:18 seconds.

Example 4.8 For the rotational mechanical system shown in mobility diagram formin Figure 17, determine the values of the bearing resistanceR and inertiaJ re-quired if the response of the angular position to a step change in torque input of1 N m is to have a 20% overshoot and a settling time of 2 seconds.

Solution: Taking(s) to be the output we have

Q(s) = Js2(s) + Rs(s) +1

C(s)

so the system transfer function is

G(s) =(s)

Q(s)=

1

Js2 +Rs+ 1=C

=1=J

s2 + R=Js+ 1=CJ


For a20% overshoot

= ln(0:2)q2 + ln2(0:2)

= 0:46:

ForTs = 2 seconds!n = 4=2 = 2. Hence!n = 4:35 rad/s.The denominator of the system transfer function must then satisfy

s2 + 4s+ 18:93 = s2 +R=Js+ 1=CJ

Now, sinceC = 0:2 5=J = 18:93 ! J = 0:26 kg m2. R=J = 4 ! R = 1:06Nm/(rad/s).

Example 4.9 Section objective example.We now put our knowledge to use to examine the performance of the open-loop

antenna azimuth control system which can be used as a velocity control system. Theblock diagram is shown in Slide 104.

Slide 104

Open-Loop Velocity Control System

110

2 0831 71

..s+

100

100s+

Ωo s( )V sp( ) V sa( ) Ωm s( )

power amp Motor and load gearbox

For this system:

1. Predict, by inspection the form of the open-loop angular velocity response ofthe load following a step change in the voltage input to the power amplifier.

2. Find the damping ratio and natural frequency of the open-loop system.

3. Derive the complete analytical expression for the open-loop angular veloc-ity response of the load to step-voltage input to the power amplifier, usingtransfer functions.

4. Simulate, using Matlab, the step response of the open-loop system.


Solution:

1. The transfer function relating angular velocity of the load to the voltage inputto the power amplifier is

G(s) =20:83

(s+ 100)(s+ 1:71)

The step response will be of the form

c(t) = A+ Be100t + Ce1:71t

2. The damping ratio and natural frequency can be found by expanding the de-nominator of the transfer function

G(s) =20:83

s2 + 101:71s+ 171

and comparing it tos2 + 2!ns + !2n. Thus!n =

p171 rad/s and =

101:71=(2p171) = 3:89. Thus the system is overdamped, confirming the

response predicted above.

3.

o(s) =20:83

s(s+ 100)(s+ 1:71)

Expanding into partial fractions gives:

o(s) =0:122

s+

2:12 103

s+ 100 0:124

s+ 1:71

Transforming to time domain gives

!o(t) = 0:122 + (2:12 103)e100t 0:124e1:71t

4. To get the step response in Matlab (shown in Fig. 18)

>> num = 20.83;>> den = conv([1 100],[1 1.71]); % polynomial

% multiplicationstep(num,den) title(’Chapter Objective Problem’)

4–1Find the step response for each of the sytems with transfer function:

1. G(s) = 5=(s+ 5)

2. G(s) = 20=(s+ 20)

Also find the time constant, rise time and settling time in each case.


0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 0.5 1 1.5 2 2.5 3

Time (secs)

Am

plitu

de

Chapter Objective Problem

Figure 18: Satellite Azimuth Control System: Open-Loop Speed Response

v to( )v ti ( ) C

R

4–2A simple low-pass filter can be constructed from theRC circuit shown below.Show that the transfer function of such a filter is

G(s) =Vo(s)

Vi(s)=

1

sCR+ 1;

and determine the maximum frequency square wave that can be passed with achange in amplitude (vi to vo) of less than 2%.

4–3For each of the transfer functions shown below, find the location of the polesand zeros, plot them on thes-plane, and then write down the expression for thegeneral form of the step response without solving the inverse Laplace transform.State the nature of each response.

1. G(s) = 2s+2

2. G(s) = 5(s+3)(s+6)

3. G(s) = 10(s+7)

(s+10)(s+20)

4. G(s) = 20s2+6s+144

5. G(s) = s+2s+9


6. G(s) = (s+5)

(s+10)2

4–4 Determine the exact response for each of the systems in Problem 4–3 usingLaplace transform techniques.

4–5For each of the systems in Problem 4–3 find the natural frequency and dampingratio, and hence confirm the type of response predicted.

4–6A system has a damping ratio of 0.5, a natural frequency of 100 rad/s, and a dcgain of 1. Find its response to a unit step input.

4–7For each of the second-order systems below, find, !n, Ts, Tp, Tr, and%OS.

1. G(s) = 120=(s2 + 12s+ 120)

2. G(s) = 0:01=(s2 + 0:002s+ 0:01)

3. G(s) = 109=(s2 + 6280s+ 109)

4–8For each of the second-order system specifications below, find the location ofthe second-order pair of poles.

1. %OS = 10%, Ts = 0:5 second.

2. %OS = 15%, Tp = 0:25 second.

4–9For a general second-order system, find the location of the poles if the percent-age overshoot is 30% and the settling time is 0.05 seconds.

4–10FindJ andC in the rotational mechanical system shown below if the systemis to yield a30% overshoot and a settling time of 4 seconds.

q t( )

θ( )t J

1Nm / (rad / s)

C

All problems except problem 4–1 adapted from Chapter 4 of Nise.


Summary of the “Evaluation of System Response” Se-quence

This section of the course was devoted to the evaluation of the transient responsefrom the system model.

It introduced the useful concept of poles and zeros.

Step responses for first and second order systems were analysed.

The generalised parameter for the first order system was found to be thetimeconstant and the performance specificationsrise-timeTr andsettling timeTs were related to .

The generalised parameters for second-order systems were found to bedamp-ing ratio andnatural frequency!n;

Four different types of response are possible for stable second-order systems:

– overdampedwhen > 1,

– critically dampedwhen = 1,

– underdampedwhen0 < < 1,

– undampedwhen = 0.

The time response specificationspercentage overshoot; %OS,peak timeTp;settling timeTs and rise-timeTr can be formulated in terms of and!n.

Curves of constant time response specifications can be drawn in thes-planeand used to design a system which must have a given response.

We also completed the objectives:

Given the antenna azimuth-angle control system developed in thesection on modelling

Predict, by inspection, the form of the open-loop angular velocityresponse of the load to a step-voltage input to the power amplifier;

Find the damping ratio and natural frequency of the open-loopsystem;

Derive the complete analytical expression for the open-loop re-sponse using transfer functions.

In the next lecture we will return to consider the control system design problemand see how feedback enables us to satisfy some of the performance specifications.

5 ANALYSIS AND DESIGN OF FEEDBACK SYSTEMS 122

Lecture 10: Analysis and Design of Feedback Systems

Preamble

In the last section of the course we defined ways to evaluate the transient response offirst and second-order systems from knowledge of their transfer function. We nowapply this knowledge to feedback control systems. In particular we shall introduce:

Transfer functions for general feedback systems;

The unity-gain feedback system, a so-calledcanonical form;

The effect of gain on the system response of a system with the same form oftransfer function as the antenna azimuth position control system;

Introduce the root locus.

5 Analysis and Design of Feedback Systems

The feedback form of a control system topology is illustrated below in Fig. 19. Asimplified model is shown in Figure 20.

-

+

Inputtransducer

Input Actuatingsignal

Controller

Output

Plant

Outputtransducer

Feedback

G s1( ) G s2( ) G s3( )

H s2( ) H s1( )

R s E s( ) C s( )

Figure 19: Feedback form of a control system topology

Now, for the simplified system

E(s) = R(s) C(s)H(s)

C(s) = G(s)E(s)

C(s) = G(s)R(s)G(s)H(s)C(s)

[1G(s)H(s)]C(s) = G(s)R(s)

Gc(s) =C(s)

R(s)=

G(s)

1 +G(s)H(s)(59)

The block diagram of this reduced “closed-loop” control system is shown in Fig. 21.


-

+

Input Actuatingsignal

Plant+

controller

Output

Feedback

G s( )

H s( )

R s E s( ) C s( )

Figure 20: Simplified feedback control system topology

R s C s( )G s

G s H s

( )

( ) ( )1+

Figure 21: Reduced feedback control system

5.1 Interpretation of the generalised closed-loop transfer func-tion

The components of equation (59) are interpreted as follows:

The transfer functionG(s)H(s) is called the “loop transfer function”.

1+G(s)H(s) = 0 is called the “closed-loop characteristic equation” (CLCE).As we shall see, the CLCE is a very important equation in feedback controlsystems analysis and design.

Gc(s) is called the “closed-loop transfer function”.

5.2 Unity-gain feedback

The “unity-gain feedback” canonical form1 is shown in Fig. 22.

Go(s) is called the “open-loop transfer fuction”.

In comparison with the previous model,H(s) = 1, hence, from equation (59)we have:

Gc(s) =Go(s)

1 +Go(s)(60)

This form of system will be used almost exclusively in the remainder of thiscourse.

1canonical— authoritative, standard, accepted


-

+

Input Errorsignal

Output

R s C s( )E s( )G so ( )

Figure 22: Unity-gain feedback canonical form

5.3 Closed-loop transient performance

Consider the example servomechanism shown in Figure 23.

-

+R s C s( )K

s s a( )+

Figure 23: An example system: a position control system orservomechanism

For this system:

Go(s) =K

s(s+ a)

Gc(s) =Go(s)

1 +Go(s)=

K

s2 + as+K:

We see that this is a second order transfer function which will have various formsof damping depending on the value ofK. As K is varied, the closed-loop polesmove through three types of behaviour from overdamped, to critically-damped andon to underdamped response.

At K = 0, the poles are the same as for the open-loop, that isp1;2 = 0;a(labelleds1 in Fig. 24).

For0 < K < a2=4

the poles are real and located at

p1;2 = a2p(a2 4K)

2

(labelleds2 in Fig. 24).


AsK increases, the poles move towards each other along the real axis and theresponse is overdamped (although the rise- and settling-times reduce), untilboth poles come together atp1;2 = a=2, whenK = a2=4, and the responseis critically damped (s3 in Fig. 24).

AsK is further increased, the poles become complex with real part

d = a=2

and imaginary part

!d =

p(4K a2)

2

which increases in size asK increases (s4 in Fig. 24). The real-part remainsconstant whilst the damping ratio is reduced. Thus,%OS inscreases whilstthe settling time remains constant.

These results are summarised in Table 4, and the main classes of pole locations,labelled as above, are illustrated in Figure 24.

Table 4: Pole locations for the example system as a function ofK

K p1 p2 label

0 a 0 s1

0 < K < a2=4 a=2p(a2 4K)=2 a=2 +

p(a2 4K)=2 s2

a2=4 a=2 a=2 s3

a2=4 < K a=2 + jp(4K a2)=2 a=2 j

p(4K a2)=2 s4


σ

jωs− plane

s1 s1s2 s2s3

s4

s4

−a − a

2

Figure 24: Pole locations for example system


Example 5.1 Find Tp, %OS andTs for the feedback control system illustrated inFigure 25 below.

-

+R s C s( )25

5s s( )+

Figure 25:

Solution:

Gc(s) =25

s2 + 5s+ 25=

!2n

s2 + 2!ns+ !2n

Hence!n =p25 = 5, 2!n = 5 and = 0:5. Therefore

Tp =

!np1 2

= 0:726 sec:

%OS = e=p

12 100 = 16:303%:

Ts =4

!n= 1:6 sec:

Example 5.2 For the system shown in Figure 26 below, design the gainK neededto give a%OS of 10%.

-

+R s C s( )K

s s( )+ 5

Figure 26:

Solution:

Gc(s) =K

s2 + 5s+K

2!n = 5, !n =pK, thus

=5

2pK:


For a 10% overshoot

= ln(10=100)q2 + ln2(10=100)

= 0:591:

K =

5

2

2

= 17:892:

Note: for this system the settling time isTs = 4=(!n) = 4=(2:5) = 1:6 seconds.We cannot design for a settling time less that this since the real part of the poles isfixed and is not adjustable by means of the gainK. We would need to add extracomponents to achieve a settling time less than 2 seconds.

5.4 The Root-Locus: A Preview

Consider the system of Figure 23 again. This has closed-loop transfer function

Gc(s) =K

s2 + as+K:

We calculated the locations of the poles ofGc(s) for variations ofK and plottedtheir locations in thes-plane (see Table 4 and Figure 24 above).

We could plot two continuous “curves” through these points to indicate themovement of the poles as a continuous function ofK (see Figure 27).

σ

jωs− plane

s1 s1

−a − a

2

Figure 27: Motion of the closed-loop poles of the control system with open-looptransfer functionGo(s) = K=(s(s+ a))

These curves describe the “locus” of the closed-loop poles asK increases. This“ root locus” can be sketched or generated for any system provided theopen-looptransfer function is known.

In Matlab, it is generated as shown in Slide 105 and the result is that shown inSlide 5.4.


Slide 105

Root Locus

For the system with open-loop transfer functionGo(s) = 1=s(s+ 5) and

unity-gain feedback:

>> num_Go = [1];

>> den_Go = [1, 5, 0] % Go(s) = 1/(sˆ2 + 5s)!

>> rlocus(num_Go, den_Go)

Slide 106

Root Locus (2)

-10

-8

-6

-4

-2

0

2

4

6

8

10

-10 -8 -6 -4 -2 0 2 4 6 8 10

xx

Real Axis

Imag

Axi

s


First Drill ProblemTransient Response of Feedback Control Systems

Aim: Reduce a feedback control system to a single transfer function that relates theoutput to the input in order to analyse and design the closed-loop transient response.Problem: Given the block diagram for the antenna azimuth position control systemwith velocity feedback compensation shown in Figure 281:

1. Find the equivalent unity-gain feedback canonical form by block diagramreduction and hence write down the open-loop transfer functionGo(s), theclosed-loop transfer functionGc(s) and the closed-loop characteristic equa-tion (CLCE).

2. Simplify the system by replacing the power amplifier by its DC gain (=1)and disconnecting the tacho. Evaluate the closed-loop peak timeTp, percentovershoot%OS and settling timeTs when the gainK = 800.

3. Derive the expression for the closed-loop step response of the system ofpart 2.

4. For the simplified system of part 2, find the value ofK that yields a 5%overshoot.

5. With the tacho connected, but the power-amplifier still replaced by its DCgain, show that it is possible to design a closed-loop response in which thedamping ratio and the settling time may be controlled by suitable choice ofthe parametersK andKT . Find suitable values for the parametersK andKT that will give a 10% overshoot and a settling time of 1.5 seconds. Whatis the corresponding rise time?

1Please note the error in the figure. The input signal, shown asi(s) in the diagram should readi(s)


V sa ( )

V sω ( )

0 2083

1 71

.

( . )s s+100

100s+K- -

+ +

V si ( )

V so ( )

V sp( )

pot

pot

tacho

pre-amppoweramp

motor&

load

1

π

1

π

sKT

Ω i s( ) Θo s( )

Figure 28: Antenna azimuth position control system

Answers

1.

Go(s) =6:63K

s((s+ 1:71)(s+ 100) + 20:83KT );

Gc(s) =6:63K

s3 + 101:7s2 + (1:71 + 20:83KT )s+ 6:63K;

CLCE! s3 + 101:7s2 + (1:71 + 20:83KT )s+ 6:63K = 0:

2. Tp = 0:435 seconds;%OS = 69:1%; Ts = 4:7 seconds.

3. o(t) = 1 e0:855t(cos 7:23t+ 0:118 sin 7:23t):

4. K = 23:2.

5. K = 172:64; KT = 10:99; Tr 0:63 seconds.

Note: In the following, unless otherwise specified, assume that the closed-loop isobtained by the application of unity-gain feedback.

5–1A control system has a controller and plant transfer functionG(s) and a feed-back transfer functionH(s). For the closed-loop system, determine the character-istic equation and the closed-loop zeros in terms of the poles and zeros ofG(s)andH(s) for the cases whereH(s) = 1 (unity-gain feedback canonical form) andH(s) = NH(s)=DH(s).

5–2An open-loop system has poles ats = 0;1 and6 and a zero ats = 2. Ifthe open-loop gainK = 18 find the closed-loop step response and compare it to theopen-loop step response. Comment on the result. (Hint: use Matlab to determinethe zeros of the closed-loop characteristic equation.)


5–3A servomechanism has an open-loop transfer funtion

Go(s) =K

s(s+ 20):

Determine the value ofK for which the closed-loop system is

1. critically damped, and

2. ideally damped.

5–4A control system is to satisfy the following performance criteria

1. rise time:Tr 0:5 seconds.

2. settling time:Ts 2 seconds.

3. Percent overshoot in the range1% < %OS < 10%.

Sketch the region of thes-plane in which the dominant poles of the closed-loop system must lie to meet these performance constraints and hence determinethe closed-loop transfer function of a second-order system that satisfies the require-ments. (Note: you should assume thatTr = 1:8=!n).

Recap

In this lecture we have applied our knowledge of second-order system response tothe analysis of closed-loop feedback control systems. We have seen that the open-loop gainK in the transfer function effects the performance, and that by choice ofsuitable values for this parameter we can design closed-loop systems to have certainspecified behaviour. However, there are limitations because gain typically can onlybe used to design one parameter, e.g. overshoot, rise-time, settling-time, and rarelycan it be used to design two or more parameters simultaneously.

We also observed that if we regard the open-loop gain as a continuously vary-ing parameter we can create a locus of movement for the closed-loop poles of thefeedback control system which we call the “root locus”. We will return to this indue course.

Early in the course we stated that there are three important aspects to be consid-ered in the design of control systems:

stability,

steady-state performance, and

transient performance.

We have spent quite some time on the third of these. In the next few lectures weturn our intention to the other, even more important, characteristics of stability andsteady-state behaviour. We begin in the next lecture with a look at stability.

6 STABILITY 133

Lecture 11: Stability (1)

Preamble

Three requirements enter into the design of a feedback control system:

transient response

stability

steady-state errors.

So far, we have considered transient response. We now discuss stability.In this lecture we will:

Definestability for linear time-invariant systems and see that a system whosepoles all lie to the left of the imaginary axis in thes-plane are stable;

Define instability and see that a system will be unstable if it has any polesto the right of the imaginary axis in thes-plane or it has poles of a certainstructure on the imaginary axis;

Definemarginal stabilityand see that a system is marginally stable if it haspoles on the imaginary axis;

Develop a test for stability which does not need the location of the systempoles to be found.

6 Stability

Stability is the most important system specification. If a system is unstable, tran-sient performance and steady-state errors are moot points. An unstable system can-not be designed for a specific transient response or a steady-state error requirement.

6.1 What is stability?

There are many definitions for stability depending upon the kind of system or onespoint of view. In this section, we limit ourselves to a consideration of the stabilityof linear time- invariant systems(LTIs).

Recall that the response of an LTI system is given by

c(t) = cforced(t) + cnatural(t):

An LTI is stableif the natural response approaches zero as time approachesinfinity:

cnatural(t)jt!1 = 0

Only the forced response remains ast!1.

c(t)jt!1 = cforced(t)

6 STABILITY 134

An unstablesystem has a natural response that grows without bound, so that:

cnatural(t)jt!1 =1and therefore

c(t)jt!1 =1 A marginally stablesystem has a natural response that neither grows nor

decays ast!1 but either oscillates or remains at a constant value.

Physically, an unstable system whose natural response grows without bound cancause damage to the system, adjacent property or human life. In practice manysystems are designed with limit stops to prevent runaway. From the time-responsepoint of view, instability is indicated by transients that get bigger and consequentlyby a total response that does not reach a steady state.

6.2 How do we determine if a system is stable?

6.2.1 Stability

Recall from our study of system poles and zeros that poles to the left of the imagi-nary axis in thes-plane—a region called theleft-half plane(LHP)—yield responsesthat are either decaying exponentials or damped sinusoids. These natural responsesdecay to zero as time approaches infinity. Thus:

Closed-loop stability: a closed loop control system is stable if all theclosed-loop poles are located in the left half plane.

Example 6.1 Determine the stability of the closed-loop control system shown inFig. 29.

31 2s s s( )( )+ +

R s( ) C s( )+-

E s( )

Figure 29:

Solution: The closed-loop transfer function is

Gc(s) =3

s3 + 3s2 + 2s+ 3:

The poles are the zeros (roots) of the closed-loop characteristic equation (CLCE)

s3 + 3s2 + 2s+ 3 = 0

That is:s = 2:672;0:164 j1:047:

The pole-locations and the resulting response are illustrated inSlide 107.

6 STABILITY 135

Slide 107

Stable response

LHP

σ

jωs− planec(t)

t

6 STABILITY 136

6.2.2 Instability

A system which has all its poles in the LHP is stable: all the poles will be negativereal or complex with negative real parts. On the other hand, poles in the right-half plane are positive or complex with positive real parts. They produce responseswhich are increasing exponentials or increasing sinusoids. These grow withoutbound and hence yield unstable responses. Thus if there are poles in the right-halfplane the system is unstable.

Example 6.2 Determine the stability of the closed-loop control system shown inFig. 30.

7

1 2s s s( )( )+ +R s( ) C s( )+

-E s( )

Figure 30:

Solution: This time the closed-loop transfer function is

Gc(s) =7

s3 + 3s2 + 2s+ 7

but now the poles are:

s = 3:087;+0:0434 j1:505:

Thus, two poles are in the right-half plane (RHP) and the resulting response isunstable (Slide 108).

If there are double or triple poles on the imaginary axis the response will be ofthe formAtn cos(!t + ) n = 1; 2; : : :. Such responses also grow without boundsince clearlytn !1 whent!1. Thus:

Instability of a closed-loop control system:A system is unstable ifits closed-loop transfer function has at least one pole in the right-halfplane and/or poles ofmultiplicity> 1 on the imaginary axis.

6 STABILITY 137

Slide 108

Unstable response

σ

jωs− planec(t)

t

LHP

6 STABILITY 138

6.2.3 Marginal stability

Finally, if a system has a single pair of poles on the imaginary axis, or a single poleat the origin, then we say that the system ismarginally stable. It will have a naturalresponse containing an undamped oscillation or a constant value ast!1.

6.3 Testing for Stability

It would seem to be a simple matter to test for stability. All we need to know is thatall the closed-loop poles are in the LHP. But, in fact, it is not so simple as it at firstappears to be.

Example 6.3 Consider the closed-loop system shown in Fig. 31. Is the systemstable?

10 24 6 8 10

( )( )( )( )( )

ss s s s s

++ + + +

R s( ) C s( )+-

E s( )

Figure 31:

Solution: The closed-loop transfer function is:

Gc(s) =10(s+ 2)

s5 + 28s4 + 284s3 + 1232s2 + 1390s+ 20:

Although we know the location of the open-loop poles, this doesn’t help us find thethe location of the closed-loop poles. Neither is there an analytical way of findingthe roots of the CLCE. We could use a root finding algorithm such as

>> roots([1, 28, 284, 1232, 1930, 20])

in Matlab (maybe your calculator has a similar function). However, not everyonealways has access to such facilities so we need another way to determine the sortsof roots that a system characteristic equation has. Our requirements are simplifiedbecause we only need to test the sign of the real part of the poles rather than findingtheir location. Namely, we need to answer the question “are there any roots ofthe CLCE which have positive real part?” and, more pertinant perhaps “is there asimple method of checking that all the roots of a polynomial have negative real partwithout resorting to root-finding algorithms?” The answer to the last question isyes. It is called the “Hurwitz test” and we present it now.

6.4 The Hurwitz Criterion

Consider a closed-loop characteristic polynomial in factored form:

P (s) =nYi=1

(s pi)

6 STABILITY 139

wherepi is a zero ofP (s) (that is a root of the CLCE). If all poles are in the LHP,then all factors will be of the form(s + pi) (because eachpi will have a negativereal part). The coefficients of the expanded polynomial will therefore only havepositivesign. This is true even if some factorspi are complex

+ j!

because complex factors always appear as conjugate pairs so that

(s+ + j!)(s+ j!) = s2 + 2s+ 2 + !2

which also has positive coefficients. No coefficents can be missing since this wouldimply cancellation between positive and negative roots, or roots on the imaginaryaxis, which we do not allow. Thus a necessary condition for a system to be stableis for its CLCE to have all coefficients of every termsn; sn1; : : : ; s0 present andall positive1.

To be more formal:

The Hurwitz Criterion: The Hurwitz criterion states that a system isunstable if there are any negative or missing coefficients of the closed-loop characteristic equation.

The test of stability, called theHurwitz Testis very simple to apply:

s3 + 27s2 26s + 24 is unstable because the coefficient of thes term isnegative;

s3 + 27s2 + 26s is unstablebecause thes0 term is missing; but

s3 + 27s2 + 26s+ 24 may bestable.

The problem with the Hurwitz criterion is that a system that passes the Hurwitz testis not guaranteedto be stable, as in the final example above. Hence we call theHurwitz criterion anecessarybut notsufficientcriterion for stability. The HurwitzTest provides a useful means of rejecting some closed-loop systems, but we need astronger test to be able to say for definite that a system that passes the Hurwitz testis actually stable.

This more stringent test is based on theRouth-Hurwitzcriterion and is calledtheRouth Test. It is based on the so-calledRouth Arraywhich is constructed fromthe coefficients of the characteristic polynomial and it, and the test based on it, aredescribed in the next lecture.

Recap

In this lecture we have defined stability for linear time-invariant (LTI) systems.

An LTI system isstableif the natural responsecnatural(t) decays to zero ast!1. In terms of system transfer functions, a stable system will only havepoles in thes-plane which lie to the left of the imaginary axis. This region ofthes-plane is called the left-half-plane (LHP).

1or all negative which implies multiplication ofP (s) by1

6 STABILITY 140

An LTI system will beunstableif, for any reason, the natural responsecnatural(t)does not decay to zero ast ! 1. The conditions under which this occursare:

– when the system has at least one pole to the right of the imaginary axisin thes-plane (the right-half-plane), or

– the system has poles of multiplicity> 1 on the imaginary axis.

An LTI system ismarginally stableif it has poles of multiplicity= 1 on theimaginary axis. Such a system will have a natural response that is oscillatory(undamped) or results in a constant value.

In order to test for stability we need to examine the sign of the poles of thecharacteristic equation. Any system that only has poles in the left-half-plane mustproduce a characteristic polynomial that has positive coefficients. TheHurwitz testmakes use of this property so that any system whose characteristic polynomial hasany negative or missing coefficients must be unstable. However, the Hurwitz crite-rion is only anecessary condition, since polynmials that pass the Hurwitz test canstill have roots in the right-half-plane. We therefore need anecessaryandsufficienttest. Such a test is based on theRouth-Hurwitzcriterion and is called theRouth Test.It is based on the so-calledRouth Arraywhich is constructed from the coefficientsof the characteristic polynomial and it, and the test based on it, are described in thenext lecture.

6 STABILITY 141

Lecture 12: Stability (2)

Preamble

In the last lecture we defined stability for linear time-invariant (LTI) systems andshowed how knowledge of the location of the system poles on thes-plane couldtell us if a system was stable or not. However, in order to know the location of thepoles, we need to find the roots of the closed-loop characteristic equation, and thismay not be possible unless we have access to a computer or sophisticated calculator.It turned out, however, that in order to judge a system’s stability we don’t need toknow the actual location of the poles, just their sign—that is whether the poles arein the right-half or left-half plane. The Hurwitz criterion can be used to indicatethat a characteristic polynomial with negative or missing coefficients is unstable.However, a polynomial that passes the Hurwitz test may still have unstable roots.In this lecture we demonstrate theRouth-Hurwitz Crterionwhich can be used totest the stability of polynomials that pass the Hurwitz test.

TheRouth-Hurwitz Criterionis called anecessaryandsufficienttest of stabilitybecause a polynomial that satisfies the criterion is guaranteed to stable. The crite-rion can also tell us how many poles are in the right-half plane or on the imaginaryaxis.

Having obtained a test for stability that is valid for all cases we can use it indesign to ensure that, for example, the system gain is never sufficient to cause in-stability. Using a stability test in design in this way is calleddesign for relativestability.

6.5 The Routh-Hurwitz Stability Criterion

To use the Routh test we first need to construct aRouth array. We present theconstruction of a typical Routh array by means of an example, then present theRouth-Hurwitz Criterion.

6.5.1 The Routh array

Consider the system shown in Fig. 32. The closed-loop characteristic equation is:

a4s4 + a3s

3 + a2s2 + a1s+ a0 = 0: (61)

N sa s a s a s a s a

( )

44

33

22

1 0+ + + +R s( ) C s( )

Figure 32: General closed-loop transfer function for a fourth-order system

6 STABILITY 142

The Routh array is simply a rectangular matrix with one row for each power ofs in the closed-loop characteristic polynomial. In this case there are5 rows, one foreach terms4; s3; : : : ; s0 but in general there will ben+ 1 rows for a system witha characteristic polynomial whose highest power ofs is n. There is a relationshipbetween the rows and the powers ofs which we will make use of later, so to makethe relationship explicit, it is useful to label the rows of the Routh array as shownbelow.

s4 :s3 :s2 :s1 :s0 :

0BBBB@

: : :

: : :

: : :

: : :

: : :

1CCCCA

The next operation is to initialise the Routh array by filling the first two rowswith the coefficients of the characteristic polynomial as follows:

1. For rows4 the first element isa4 the coefficient ofs4. The next element isa2 the coefficient ofs2 and the final element isa0 the coefficient ofs0. Thusthe first row is:

s4 : a4 a2 a0:

Note that in fact what we are doing is skipping the odd coefficients of thecharacteristic polynomial when constructing the row fors4. In general therule is: if n is eventhen thesn row consists of the coefficients of theevenpowers ofs. If n is odd, then thesn row consists of the coefficients of theoddpowers ofs.

2. For the second row —s3 in this case andsn1 in general — the elements aresimply those coefficients skipped in the second row written down in order.Thus:

s3 : a3 a1 0

(where the final zero is added to make the two rows equal in length).

At the end of initialisation, the Routh array for a fourth-order system will be asshown in Table 5 (where we now use a table for clarity in what follows).

Table 5: Starting layout for Routh array

s4 a4 a2 a0s3 a3 a1 0s2

s1

s0

The next stage in the process is to fill in the remaining rows of the table. Thisis done by operating on the two initial rows to create row 3, applying the sameoperations on rows 2 and 3 to generate row 4, and so on until all rows are complete.The operations that are applied to do this are now described.

6 STABILITY 143

1. Starting at the left-most end of the array create a2 2 matrix from the firsttwo rows such that the first column consists of the first element of row 1 and2 and the second column contains the second elements of row 1 and 2. Let’scall this matrixR1. That is, for Table 5 we have:

R1 =

a4 a2a3 a1

The first element of the third row of the Routh array, we’ll call itb1, is thengiven by

b1 = det(R1)=R1(1; 2):

In other words:

b1 =

a4 a2a3 a1

a3

=(a4a1 a2a3)

a3

=a2a3 a4a1

a3

2. The second element of row 3,b2, is calculated in much the same way.R2

is obtained by leaving column 1 as it is and replacing column 2 with theelements of column 3. Since column 1 is the same as inR1, the divisorR2(1; 2) is also unchanged so that:

b2 =

a4 a0a3 0

a3

=a0a3 a4 0

a3

3. This process is continued until the determinant is zero, after which zero is putinto all remaining columns of row 3.

4. The whole process is repeated for the remaining rows except that the top leftelement of the matrix moves down one row at the start of each new row. Thatis, row 4 is created from rows 2 and 3 (using the odd elements ofP (s) andthe newly calculated coefficientsb1, b2, etc.), and row 5 is constructed fromrows 3 and 4. At the end of the process the array looks like that shown inTable 6.

Note: you may find it easier to remember that det(R) is given by theproduct ofthe off diagonal terms the product of the diagonal terms.

Once the Routh array is complete we are ready to perform the Routh test.

6 STABILITY 144

Table 6: Completed Routh array

s4 a4 a2 a0s3 a3 a1 0

s2 b1 =a2a3 a4a1

a3b2 =

a0a3 a4 0

a3= a0 b3 =

0 a3 a4 0

a3= 0

s1 c1 =a1b1 a3b2

b1c2 =

0 b1 a3 0

b1= 0 c3 =

0 b1 a3 0

b1= 0

s0 d1 =b2 c1 b1 0

c1= b2 d2 =

0 c1 b1 0

c1= 0 d3 =

0 c1 b1 0

c1= 0

6.5.2 The Routh-Hurwitz Test

The Routh-Hurwitz criterion enables us to test a system whose closed- loop char-acteristic equation passes the Hurwitz criterion.

The Routh-Hurwitz Criterion: The number of roots of the character-istic polynomial that are in the right-half plane is equal to the numberof sign changes in the first column of the Routh Array. If there are nosign changes, the system is stable.

Example 6.4 Test the stability of the closed-loop system shown in Fig. 33.

10002 3 5( )( )( )s s s+ + +

R s( ) C s( )

100010 31 10303 2s s s+ + +

R s( ) C s( )

+

-

E s( )

Figure 33:

Solution: Since all the coefficients of the closed-loop characteristic equations3 +10s2 + 31s + 1030 are present, the system passes the Hurwitz test. So we mustconstruct the Routh array in order to test the stability further. Here is the openingRouth array:

6 STABILITY 145

s3 1 31 0s2 10 1030 0s1

s0

The Routh array is unchanged by the multiplication of any row by a constant. Thiscan be used to simplify the Routh array between stages. For example, in the secondrow above can be simplified by multiplying each term by1=10, thus:

s3 1 31 0s2 1 103 0s1

s0

We now construct the rest of the table using the rules already described:

s3 1 31s2 1 103

s131 1 1 103

1= 72 0 1 1 0

1= 0

s072 103 1 0

72 = 10372 0 1 0

72 = 0

For clarity, we can rewrite the array:0BB@

1 31 01 103 0

72 0 0103 0 0

1CCA

and now it is clear that column 1 of the Routh array is:0BB@

11

72103

1CCA

and it has two sign changes (from1 to 72 and from72 to 103). Hence thesystem is unstable with two poles in the right-half plane.

6.6 Special Cases

Two special cases can occur when constructing a Routh array:

a zero may appear in the first column of the array;

a complete row can become zero.

6 STABILITY 146

6.6.1 A Zero in the first column

If the first element of a row is zero, division by zero would be required to form thenext row. To avoid this, a small number (epsilon) is used to replace the zero inthe first column. The value is allowed to approach zero from either the positiveor negative side, after which the signs of the entries in the first column can bedetermined.

Example 6.5 Consider the control system with closed-loop transfer function

Gc(s) =10

s5 + 2s4 + 3s3 + 6s2 + 5s+ 3:

Construct the Routh array and interpret the stability of the closed-loop system.

Solution: The characteristic polynomial isP (s) = s5 + 2s4 + 3s3 + 6s2 + 5s+ 3so the Routh array will be:

s5 1 3 5s4 2 6 3s3 0! 7=2 0

s26 7

3 0

s142 49 62

12 140 0

s0 3 0 0

Considering just the sign changes in column 1:

Label First column ! 0+ ! 0

s5 1 + +s4 2 + +s3 +

s26 7

+

s142 49 62

12 14+ +

s0 3 + +

If is chosen positive there are two sign changes. If is chosen negative thereare also two sign changes. Hence the system has two poles in the right-half planeand it doesn’t matter whether we chose to approach zero from the positive or thenegative side. This is always the case!

6 STABILITY 147

6.6.2 An Entire row is zero

This occurs when there is an even polynomial that is a factor of the original poly-nomial.

Example 6.6 Construct the Routh array for the system with closed-loop transferfunction:

Gc(s) =10

s5 + 7s4 + 6s3 + 42s2 + 8s+ 56:

Solution: The Routh array for the example is:

s5 1 6 8s4 7! 1 42! 6 56! 8s3 0 0 0s2

s1

s0

We cannot procede any further because the third row is zero. In order to pro-cede we have to take the so-calledauxillary polynomialQ(s) formed from the rowpreceeding the zero row:

Q(s) = 7s4 + 6s2 + 8:

The auxillary polynomial iseven(only even powers ofs are present) and it is also afactor of the original characteristic polynomial (as you may easily verify for your-self). To progress further, we have to differentiateQ(s) with respect tos:

dQ(s)

ds= 4s3 + 12s+ 0

and replace the zero row with a row formed from the coefficients of the derivative:

s5 1 6 8s4 1 6 8s3 0! 4! 1 0! 12! 3 0s2

s1

s0

The remainder of the Routh array is constructed as usual.

s5 1 6 8s4 1 6 8s3 1 3 0s2 3 8 0s1 1=3 0 0s0 8 0 0

There are no sign changes in the completed Routh array, hence the system is stable.

6 STABILITY 148

6.6.3 Interpretation of a row of zeros

An entire row of zeros will appear in the Routh array when a purely even polynomialis a factor of the characteristic polynomial. For example the polynomials4+5s2+7is a purely even polynomial; it only has even powers ofs. Even polynomials haveroots which are symmetrical about the imaginary axis. This symmetry can occurunder several conditions:

1. the roots are real and symmetric about the imaginary axis;

2. the roots are imaginary and symmetric about the real axis, or

3. the roots are quadrantal.

These cases are illustrated in Fig. 34. each case, or any combination of these cases,will generate an even polynomial.

jω

σ

A: Real and symmetrical about the originB: Imaginary and symmetrical about the originC: Quadrantal and symmetrical about the origin

A A

B

B

CC

C C

Figure 34: Root positions to generate even polynomials: Patterns A, B or C (or anycombination)

It is the even polynomial that causes a row of zeros to appear in the Routh array.Thus the row of zeros tells us that there are roots that are symmetric about theorigin. Someof these roots could be on the imaginary axis (symmetry type B). Onthe other hand, if we do not have a zero row, we cannot possibly have roots on thej! axis.

Another characteristic of the Routh array for the case in question is that the rowprevious to the row of zeros contains the even polynomial that is a factor of theoriginal polynomial. Thus in the previous example, the polynomials4 + 6s2 + 8 isa factor of the original polynomial. Finally, the Routh test from the row containing

6 STABILITY 149

the even polynomial down to the last row of the Routh array tests only the poles inthe even polynomial.

A further example should clarify all this.

Example 6.7 For the system with closed-loop transfer function

Gc(s) =20

s8 + s7 + 12s6 + 22s5 + 39s4 + 59s3 + 48s2 + 38s+ 20

find how many poles are in the right-half plane, the left-half plane and on thej!-axis.

Solution: Construct the Routh array shown below.

s8 1 12 39 48 20s7 1 22 59 38 0s6 10! 1 20! 2 10! 1 20! 2 0s5 20! 1 60! 3 40! 2 0 0s4 1 3 2 0 0s3 0 0 0 0 0s2

s1

s0

For convenience thes6 row has been multiplied by1=10 and thes4 row by1=20. There is a complete row of zeros at thes3 row. Moving back to thes4 row,we extract the even polynomial

Q(s) = s4 + 3s2 + 2

and take its derivative:dQ(s)

ds= 4s3 + 6s+ 0:

The zero row is then replaced by4; 6; 0 = 2; 3; 0 and the Routh array is com-pleted:

s8 1 12 39 48 20s7 1 22 59 38 0s6 1 2 1 2 0s5 1 3 2 0 0s4 1 3 2 0 0s3 2 3 0 0 0s2 3=2! 3 2! 4 0 0 0s1 1=3 0 0 0 0s0 4 0 0 0 0

6 STABILITY 150

Interpretation

Since all the entries from the even polynomial from thes4 to thes0 entry are atest of the even polynomialQ(s), we can draw our first conclusions about thispolynomial. There are no sign changes from thes4 to thes0 row, so there are nopoles in the right-half-plane (this rules out poles that are symmetric about the realaxis or are quadrantal). But since there must be some symmetric poles, there mustbe 4 poles on thej!-axis. The remaining roots are evaluated from the remainingrows of the Routh array. There are two sign changes hence there are two RHP poles.The remaining poles must be in the LHP. To summarise:

Even(4th-order) Rest(4th-order) Total (8th- order)0 RHP 2 RHP 2 RHP0 LHP 2 LHP 2 LHP4 j! 0 j! 4 j!

Drill Problems

Each of the following transfer functions is the open-loop transfer function for acontrol system with unity-gain feedback. In each case, construct the Routh arrayfor the closed-loop characteristic-polynomial and comment on the stability, andwhere appropriate, on the location of the closed-loop poles in thes-plane.

1.

Go(s) =2000

s(s3 + 6s2 + 11s+ 6)

2.

Go(s) =1

s(2s4 + 3s3 + 2s2 + 3s+ 2)

3.

Go(s) =128

s(s7 + 3s6 + 10s5 + 24s3 + 96s2 + 128s+ 192)

6.7 Use of the Routh-Hurwitz Criterion for Control Systems De-sign

To conclude this section, we examine the use of the Routh-Hurwitz criterion in thedesign of feedback control systems.

Example 6.8 For the closed-loop system illustrated in Fig. 35, find the range ofvalues of the gain parameterK for which the closed-loop system is stable.

Solution: The closed-loop transfer function is

Gc(s) =K

s3 + 18s2 + 77s+K:

The Routh array is:

6 STABILITY 151

K

s s s( )( )+ +7 11R s( ) C s( )+

-E s( )

Figure 35:

s3 1 77s2 18 K

s11386K

180

s0 K 0

If K is assumed to be positive then we can use the Routh-Hurwitz criterion toset limits on the design values ofK allowed for stable operation. There will be nosign changes ifK > 0 or if 1386K > 0, that is ifK < 1386.

If K > 1386 there will be two sign changes so there will be two poles in theRHP and the system will be unstable.

If K = 1386 then thes1 row will be zero. The previous row isQ(s) = 18s2 +1386 and the derivative is36s so the new Routh array will be:

s3 1 77s2 18 K ! 1386s1 0! 36 0s0 1386 0

For the even polynomialQ(s) there are no sign changes froms1 to s0 so there mustbe two imaginary roots and the system is marginally stable.

Example 6.9 For the antenna azimuth control problem illustrated in Fig. 36 theclosed-loop transfer function is

Gc(s) =6:63K

s3 + 101:71s2 + 171s+ 6:63K:

Find the range of pre-amplifier gainsK for which the closed-loop system is stable.

Solution: 0 < K < 2623:29.

6.8 Relative Stability

The ability to test a control system for stability against some parameter is veryuseful, but for design purposes we are not so much interested inabsolute stabilityas designing a control system so that it is far removed from instability.

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