LINEAR,CONTROLSYSTEMS

with I\{ATI AB Applications

B.S. NIANKEFormerly Prof. of Electrical Engineering,

Maulana Azad National Institute of TechnologyBhopai-462 007 (M.P.)

PUBLISHERS2-8, NATH MARKET, NAI SARAK

DELHI-110006.Phones : 2391 2g 80 ; 2722 4L 79. Fax : 23980311

(i)

Contents

1.1. An Example of Control Action

-..*-2: Open-loop Crntrol Systern)-*,{.Closed-loop Control System"" 1.4. Use of Laplace T?ansformation in Control Systems

1.5. Laplace T?ansform1.5.1. Derivation of Laplace transform!.5.2. Basic Laplace transform theorems

1.6. Solved ExamplesProblerns

I2

2

2

3

3

6

7

l4

2.1. Poles and Zeros of a T?ansfer Function2.2.TransferFunctiorrandItsRelationslripWithIrnpulse2.3. Procedure for Determining The Tlansfer Function of a2.4. Solved Examples

Problems

ResponseControl

t7r910

19

23

3.1. Block Diagrarn Reduction3.2. Solved Examples

Problems

25

32

40

4.L. Rules for Drarving Signal Flow Graphs4.2. Manson's Gain Formula4.3. Drawing Signal Flow Graph From a Given Block Diagram4.4. Solved Examples

Problems

44

47

48

49

5;j

5. 1. Electrical Net*orks5.2. Mechanical Systems

5.2.L. Tlanslation Mechanical Systems5.2.2. Rotational Mechanical System

5.3. Hydraulic System5.4. Pneumatic System5.5. Thermal System5.6. Servo Motors5.7 . Generators5.8. Emor Detectors5.9. Solved Examples

Problems

56

58

58

61

63

65

65

66

73nrytl

81

9B

-

| 1. INTRODUCTION

t ,. II4IqqER FrrNCrroN

E s. nlocx uAcRAIvIs

FLOW GRAPHS

MODELLING A CONTROL SYSTEM

( ,i)

i"l

::::(:r.,li s l-

::{;J

-t

E]\

ior

6.1. Tlansient and Steady State Response6.2. Input Test Signals6.3. Time Response of a First Order Control System

6.3.1. Time response of a first order control system subjected to unit stepinput fu.rction

6.3.2. Demarcation between the transient part and steady state part of the timeresponse in terms of time constant

6.3.3. Time response of a first c:der control system subjected to unit rampinput function

6.3.4. Time response of a frrst order control svstem subjected to unit impulse inputfunction

6.4. Time Response of Second Order Control System6.4.1. Time response of a second order control system subjected to unit step

input function6.4.2. Critical Damping6.4.3. Characteristic Equation6.4.4. Tlansient response specifications of second order control system6.4.5. Time response of a second order control system subjected to unit ramp

function6.4.6. Time response of a second order control system subjected to impulse input

function6.5. Time Response of a Third Order Control System

6.5.1. Effect of first order term time constant on time response of third ordercontrol system lzl

6.6. Time Response of Higher Order Control System 1226.6.1. An example of third order unstable control system 122

6.7. Steady State Error 1246.7.1. Static error coefficients t256.7.2. T),pe oftransfer functions and steady state error 1266.7.3. Generalized error coefficients 1296.7.4. Performance indices 131

6.8. Sensitivity 1366.8. 1. Effect of transfer function parameter variations in an open loop control system. 1366.8.2. Effect of forward path transfer function parameter variations in a closed

loop control system. 1376.8.3. Sensitivity of overall transfer function

transfer function G (s).M (s) with respect to forward path

6.8.4. Sensitivity of overall transfer function M (s) withtransfer function.

138

respect to feedback path

6.8.5. Effect of feedback on time constant of a control system6.9. Control Actions

6.9. l. Proportional control6.9.2. Derivative control6.9.3. Integral Control6.9.4. Proportional Plus Derivative Plus Integral Control (PID Control)6.?.5. Derivative Feedback Control

6.10. Solved ExamplesProblems

102

fi2103

103

104

105

106

106

107

111

111

lr2input

115

117

118

Li

of

Ermi

BT

Oc

139

140

141

141

t4lt46t48148

152

175

I o. tnnn RESPONSE ANIALYSI$OF CONTROL SYSTEMS

::Ijg,

(iii)

7.1. Stability in Terms of Characteristic Equation of a Control System7.1.1. Definition of stability

\",:l .1.2. Absolute and relative stability7.1.3. Location ofthe roots ofcharacteristic equation in s-plane as related

to time response and prediction of absolute stabilrty therefrom7.2. To Determine the Number of Roots Having Positive Real Parts for a Polynomial7.3. Hurwitz Determinants of a Polynomial7.4. Routh-Hurwitz Criterion7.5. Solved Examples : Routh-Hurwitz Criterion

tf.6. Nyquist Criterion7.6.1. Procedure for mapping from s-plane to G(s) H(s)-plane7.6.2. Determination from the Nyquist plot the number of zeros of G(s)H(s)

which are located inside a specified region in s-plane 1927.6.3. Application of Nyquist criterion to determine stability of a closed-loop

control system I92. 7.7. Gain Margin and Phase Margin 201

7.8. Relative Stability From Nyquist Plot 2037.9. Gain Phase Plot 204

7.10. Closed-loop Frequency Response ofa Unity Feedback Control System from Nyquist Plot 205.,7.11. Constant Magnitude Loci : M-circles 206' 7.L2. Constant Phase Angles Loci : N-circles 207

7.13. Closed-loop Frequency Response ofControl System From M and N-circles 209210,-il:1+. Gain Adjustment Using M-circle

7.15. Nichols Chart 2127.16. CutoffFrequency and Bandwidth 2167.l7..Solved Examples : Nyquist Criterion 217

. 7.18. Bode Plot 2307.18.1. Bode plot (Logarithmic plot) for transfer functions 2307.18.2. Graphs for the Gain Term K 2317.18.3. Graphs for the 1"r* -1-Uo)^7.18.4. Graphs for the Term (1 +jto ?)

2.18.5. Graphs for the rerm (1#?)7.18.6. Initial Slope of Bode Plot7 .L8.7. Determination of static error coefficients from initial slope of Bode plot

7.18.8. Graphs for quadratic term0)o2

[((Dr2 $r\ + j2( ro, ot)and determination of gain marg:in,7.18.9. Procedure for drawing bode plot

phase margin and stability.7.18.10. Minimum phase, non-minimum phase and all pass transfet functions

7.19. Solved Examples : Bode Plot7.20. Correlation Between Tiansient Response and Frequency Response7.21. Root Locus?.22. Salient Features of Root Locus Plot7.23. The Procedure for Plotting Root Locus

180

180

180

180

181

182

1E,3

I84190

i90

23t

232

233

236

238

238

242

245

246

253

256

259

264

7. STABILITY AIYALYSIS OF CONTROL SYSTEMS

( ir')

7 .24. Solved Examples : Root Locus7 .25. Root Contours

Problerns

261286

292

8. 1. Phase-lead Compensation, 8.2" Phase-lag Compensation

8.3. Phase-lag-lead Compensation8.4. Concluding Remarks8.5. Feedback Compensation8.6. Solved Examples

Problems

297

299

300

301

302

305

3t2

9.1. Stnte Space Representation9.2. The Ooncept of State9.3. State Space Representation of Systems9.4. Block Diagram for State Equation9.5. Tlansfer Function f)ecomposition

9.5. 1. Direct decomposition9.5.2. Cascade decomposition9.5.3. Parallel decomposition9.5.4. Parallel decomposition of a transfer function having repeated roots of

its characteristic equation.9.6. Solution of State Equation

9.6.1. Determination of State Tbansition Matrix9.6.2. Properties of State Tlansition Matrix

9.7. Tiansfer Matrix9.8. Controllability9.9. Observability

9.10. Solved ExamplesProblems

314

:l 15

:l16

319

320321

322

323

324

325

329

336

336

337

339

342

372

10.1. Sarnpler1 0.2. Sarnpling Process10.3. Laplace Transform of Sarnpled Function10.4. z-transform10.5. z-transform of Some Useful Functions10.6. Inverse Z-transform10.7. Hold Circuit10.8. Reconstruction of Signal : Minimum Sampling Frequency10.9. Pulse Tlansfer Function (z-transfer Function)

10.10. Stability Analysis of Sarnpled-data Control Systems10. 11. Sclved Exarnples

Problems

,5 ll

378

379380

381

384

390

392393

400

406

417

MPENSATION OF CONTROL SYSTEMS

9. II\TRODUCTION TO STAIE SPACE ANALYSIS OF COT{TROL SYSTEMS

rO. SAIVIPLED DATACONTROL SYSTEMS

11.1. Introduction to MATLAB1L.2. Running MATLAB

Solved Examples

419

421

422

APPENDIX-I : Answers to Selected ProblemsAPPENDIX-II : Objective 1lpe of Questions with AnswersAPPENDIX-Uf : Classifred Solved ExamplesAPPENDIX-fV : Key Formulae, Tables, Charts

466469470-508509-589590-598

REFERENCES 599

INDEX 601

( t,)

SOLUTION OF PROBLEUIS USING COMPUTER

IntroductionThis chopter deols wifh bosic ldeos about the open-loop ond closed-loop control systems, Thedifferentiol equotions describe the dynomic operotion of control systems. The Loploce tronsformtronsforms the differentiol equotion in to on olgebraic equotion, the solulion is obtoined in thefronsform domoin. The time domoin solufion is determined by toking the inverse Lopiace frons-form.

A control system is a combination of elementsarranged in a planned manner ,vherein each elementcauses an effect to produce a desired output. This causeand effect relationship is governed by a rnathematicalrelation.

If the aforesard mathematical relation is linear thecontrol system is termed as linear control system. For alinear system the cause (independent variable or input)and the effect (dependent variable or output) areproportionally related and principle of superposition is applicable throughout the operatingrange of a system.

In a control system the cause acts through a control process which in turn results intoan effect.

There may be variety of systems based on the principle mentioned above but all thesystems have many features in common and as such common approach for the study andanalysis of control systems is possible.

Control systems are used in many applications for example, systems for the control ofposition, velocity, acceleration, temperature, pressure, voltage and current etc.

I.I. AN EXAMPLE OF CONIROL ACIIONControl of a room temperature is achieved by switching ON and switching OFF of a power

supply to a heating appliance. Thus power supply to an appliance is switched ON, when theroom temperature is felt low and switched OFF, when the desired temperature is reached.

The above system can be modifred, if the duration of application ofpower is predeterminedto achieve the room temperature within desired limits.

However, a further refinement can be made by measuring the difference between theactual room temperature and the desired room temperature and this difference being theerror is used to control the element which in turn controls the output, r.e. room temperature.

The above description indicates that in the former case the output (room temperature)has no control on the input and the control action is purely based on a sort ofpredeterminedcalibration only, where as in the latter case the control action is affected by a feedback receivedfrom the output to the input.

r An Exomple of Control ActionI Open-Loop Control Systemr Closed-Loop Control Systemr Use of Loploce Tronsformotion

in Control Systemsr Loploce Tronsformr Solved Exomples

ONE l-t

Contents

Linear Control Systems

1.2. OPEN-LOOP CONTROL SYSTEMHaving explained the concept of control actioo, a

control system can be described by a block diagram asshown in Fig. 1.2.L

The input r controls the output c through a controlaction process. In the block diagram shown in Fig. L.2.1, itis observed that the output has no effect on the controlaction. Such a system is termed as open loop control system.

Fig. l.z.L Open loop eontrol systern.

In an open loop control system the output is neither measured nor fed-back for comparisonwith the input. Faithfulness of an open loop control system depends on the accuracy of inputcalibration.

I.3. CLOSED-IOOP CONTROT SYSIEMIn a closed loop control system the output has an effect on control action through a

feedback as shown in Fig. 1.3.1 and hence closed loop control systems are also termed asfeedback control systems. The control actionis actuated by an error signal e which is thedifference between the input signal r and theoutput signal c. This process of comparisonbetlveen the output and input maintains theoutput at a desired level through controlaction process.

The control systems without involvinghuman intervention for normal operation arecalled automatic control systems.

Fig. 1.3.1. Closed loop eontrol systern.

A closed-loop (feedback) control system using a power amplifying device prior to controllerand the output of such a system being mechanical l.e. position, velocity, acceleration is calledservomechanism.

Comparison of Open-Loop and Closed Loop-Control System

lnput

lnputControlActio n

Co mp a rato r

MeasuringElement

Open-Loop

1. The accuracy of an open-loop system depends onthe calibration of the input. Any departure frompre-deterrnined calibration affects the output.

Closed-Loop

2. As the error behveen the reference input and theoutput is continuously measured through feedback,the closed-loop system works lnore accurately.

2. The closecl-loop system is complicated to constructand costiy.

3. The closed-loop systems can become unstableunder certain conditions.

4. In terms of the performance the closed-loop systemadjusts to the effects of non-linearities present inits elements

2.

3.

The open-loopcheap.

The open-loop

system is simple to construct and

systems are generally stable.

4. The operation of open-loop system is affected duethe presence of non-linearities in its elements.

I.4. USE OF LAPIACE IRANSFORMAIION IN CONTROT SYSTEMSThe control action for a dynamic control system whether electrical, mechanical, thermal,

hydraulic etc. can be represented by a differential eq;ation and the output response ofsucha dynamic system to a specified input can be obtained by solving the said differential equation'The system differential equation is derived according to physical laws governing a system inquestion.

Introduction

In order to facilitate the solution of a differential equation describing a control system,the equation is transformed in to an algebraic form. The differential equation wherein timebeing the independent variable is transformed into a corresponding aigebraic equation byusing Laplace transformation technique and the differential equation thus transformed isknown as the equation in frequency domail' Hence, Laplace transform technique transformsa time domain differential equation into a frequency domain algebraic equation.

Lf(t)-F(s) t>0

DT F (s) f (t) , e-'t dt . .. ( 1.1)

The term " Laplace transform of f (t1" is used for the letter tf (t).The time function /(r) is obtained back from the Laplace transform by a process called

inverse Laplace transformation and denoted as ,- I thusL't Vf (t)1 = L-r lF (s)l =f (r)

The time function /(t) and its Laplace transform F'(s) are a transform pair.Table 1.5 gives transform pairs of some commonly used functions and Laplace transform

pairs for some functions are derived here under.1.5.1. Derivation of Laplace transform1. Laplace transform ofe"'

I .5. LAPLACE TRANSFORMIn order to transform a given function

iransform first multiply f (t) by - " , s beingproduct w.r.t. time with limits as zero andrransform of f (t), which is denoted by F (s) or

The mathematical expression for Laplace

2. In the functi on f (t) r e*t put aeot = eot

Therefore, usrn,g F,q. 0.2) L[1] =

L eot = f* no,"o

Leot=; 1(s-o)

e-rtd,t-l* r(a-s)rdt=. 1 :- J0 (s_o)

As the inverse Laplace transform is denoted by the letter .c- 1 and,

Laplace transform of -]. is eo' and expressed as below,(s-n)--r[ t I atL l-l=e

1ts-o)l

of time f (t) into its corresponding Laplacea complex number (s = o +/rrl). Integrate thisinfinity. This integration results in LaplaceLf (t).transform is,

. ..( 1.2)

therefore, the inverse

. ..( 1.3)

=I;

-0= l. Hence, f (t) - 1

1

L[r] -

(s

1S

or

and[t]

=[

. ..( 1.4)

...( 1.5)L-L

-0)

3

Linear Control Systems

the functi on f (f) = e" pu t a - i0)eot = 4rx. Hence, f (t) = ei*t

Therefore, using Eq. (L.2) td't = - 1-

(s -"rco)

d@t - (cos at + j sin or)4(cos tof +7 sin olr) =, 1. . - s +/o)(s -Jt t) - 1rz + 6n2)

Separating into real and imaginary parts,

3. In

4. In the

Hence,

Therefore, using

^at ^(- c +7crl) /e -e'f (t) = e? c

+7to) /

and

and

-6cos o)t= "+'v (t'+ c,l2)Isin0)f= ,' ,''v (r, * c,l2)

L- L[ = t -l=.orrrLtr' * ,')J

- r-uD tr'{'

L_rl o _l =sinorLffil -suru,function f (t) = e" put (r = (- u +/to)

...( 1.6)

...( 1.7 )

...( 1.8)

...( 1.e)

.. .( 1. 10)

.."(1.11)

...(1.t2)

...(1.13)

Eq. (1.2)

L e? u +.yor) f11=-=-s-(-cr+/rrlt (s+cx) -jrn

'.' eto+/ro) t = e- o'lcos at +7 sin ol/)

... k-* lcos ro/ +7 sin t,lr) = # = ffiSeparating into real and imaginary parts,

k-* ' cos (N = (t *^o) =(s+s)2+r,l'k-d . sin 0)f = 'u(s+cr)z+ol2

L-L[ -tt*gl-.-l :e d.cosrrl/u Ltt*a)2*r'l

v vv

rL-tl- to l-^-at ,

L(, + a)2.* ,J..| = e *' ' sin 0)f

In the function / (f) = eo' put a - I5.eo'=rL'' =e'. Hence, f (t)=ei

,1flo

= (s - 1)Therefore, using Eq. (1.2).

Introduction

and

'2 t3et =L

L+t+ n* zs+...11111

(s-1)=;*P+n*J+"'Table 1.5. Table of Laplace transform pairs

S. No. f (t\ f(s)=Llf@l

1

2

6 (r) unit impulse at / = 0

u (t) unit step at t = 0

u (t - ?) unit step at t = T

t

-ate

ate

te- ot

te't

,tt ,- ut

sin cot

cos tof

e- st sin co/

e- qt

cos rof

sinh a/

cosh at

I

1s

1 _s?-es

I-2s

1

Ts

lnJ,,-i

1

s+o

_1s-o

1

,t-2

t'

e\-, u

';4

5

:."".6

7

.8

9

,10

?t1

L2

13

L4

15

16

L7

(s + o)2

1

(s - a)2

ln6 + @Y';i

c,

ffits

ffio)G;ffi

(s+a)

GE+TC[

TAsW

Linear Control Systems

.'. Comparing the terms

rtrl=*, r,ld=1

"lL1=l"lzz)- tz

"l*S=# orrb't=# (1 14)and "-rl ^-1=r, ..(1.1b)a*u L 1""*t-1 -'

1.5.2. Basic Laplace transform theoremsBasic theorems of Laplace transform are given below :1. Laplace transform of linear combination

rla fr(t) + bf2(/)l = oFr (s) + bF2 (s) '..(1.16)where fi(t),fz(f) are functions of time and o, b are constants.

2. If the Laplace transform of f (t) is F (s), then.. df (t\(i\ rw=[sF(s)-f(0+l]

dt

(.iil L*f= 1s2 F (s) - s /(o + ) -/' (o + )l

tiia ttlP= 1s3F(s)-s2l(0 +)-sf'(O+)-/-(0+)l ...(1.17)dto

wheref(0+), f '(O+),f" (o+)... are the values of f (t),ryW... att=(0+)3. If the Laplace transform of f (r) is F(s), then

ta lrra=[49*f-':o+rlLs s I

Introductinn

or hm^ /(r) = .]11

s F (s) ...(1.19)7. Final value theorem

Iim /(r)= Iim s Lf (t) ...(1.20a)i-+* s+0

or lim /(r) = lim s F (s) ...(1.20)t--s* s-+0

The final value theorem gives the final value (t + *; of a time function using its Laplacetransform and as such very useful in the analysis of control systems. However, if thedenominator of s F (s) has any root having real part as zero or positive, then the final valuetheorem is not valid.

I.6. SOLVED EXAMPLESExample 1.6.1. Find the inuerse Laplace trarusform of the following functions :

(i) F(s) =--]^ Ul)F(s)=- "qt!s(s+l) s(s'+4s+3)Gii) F(st= ^ L (id Frs)= ^.]Ls"+4s+8 s"+4s+6(u),F(s) = ,u (ui)F(s)= -s2+-2s+3

s (s"+ 4s+5) s"+6s'+ 12s+8

Solution. (l) F' (s)' s(s+I)Using partial fraction expansion,

k. k,F1s;=?.#The coefficients can be determined as k1 = 1 and hz=-t

F1s;=1- Is s+1Taking inverse Laplace transform on both sides

L-t F(s) = -r'[l - -f.l- \-/ [s s+1]or L-rF(s)=1,-r1_ --r 1s'L

"+1f (t) = (L - e-t) Ans.

(ii) .t,(s)= -"*6s(s"+4s+3)

The term (s2 + 4s + 3) can be factorised as (s + 1) (s + 3)

Using partial fraction expansion,k., k, k,F(s)=;+;;i+r*3

Ttre coefficients can be determined as ht=2,k2= - 2.5 and h=i.