Solutions Manual: Control Systems Engineering International Adaptation 8th Ed by Nise 2025, Exams of Control Systems

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SOLUTIONlMANUAL

AlllChapterslCovered

O l N l E Introduction

ANSWERS l TO l REVIEW l QUESTIONS

1. Guidedlmissiles,lautomaticlgainlcontrollinlradiolreceivers,lsatelliteltrackinglantenna

2. Yesl-lpowerlgain,lremotelcontrol,lparameterlconversion;lNol-lExpense,lcomplexity

3. Motor,llowlpasslfilter,linertialsupportedlbetweenltwolbearings

4. Closed-

looplsystemslcompensatelforldisturbanceslbylmeasuringlthelresponse,lcomparinglitltolthelinputlre

sponsel(theldesiredloutput),landlthenlcorrectingltheloutputlresponse.

5. Underlthelconditionlthatlthelfeedbacklelementlislotherlthanlunity

6. Actuatinglsignal

7. Multiplelsubsystemslcanltimelsharelthelcontroller.lAnyladjustmentsltolthelcontrollerlcanlbeli

mplementedlwithlsimplylsoftwarelchanges.

8. Stability,ltransientlresponse,landlsteady-statelerror

9. Steady-state,ltransient

10. Itlfollowslalgrowingltransientlresponseluntillthelsteady-

statelresponselislnollongerlvisible.lThelsystemlwillleitherldestroylitself,lreachlanlequilibriumlstatelb

ecauseloflsaturationlinldrivinglamplifiers,lorlhitllimitlstops.

11. Naturallresponse

12. Determineltheltransientlresponselperformanceloflthelsystem.

13. Determinelsystemlparametersltolmeetltheltransientlresponselspecificationslforlthelsystem.

14. True

15. Transferlfunction,lstate-space,ldifferentiallequations

16. Transferlfunctionl-lthelLaplaceltransformlofltheldifferentiallequation

State-spacel-lrepresentationloflanlnthlorderldifferentiallequationlaslnlsimultaneouslfirst-

orderldifferentiallequations

Differentiallequationl-lModelinglalsystemlwithlitsldifferentiallequation

SOLUTIONS l TO l PROBLEMS

50 lvolts

1. Five l turns l yields l 50 l v. l Therefore l K l =

5 lxl 2 l r ad

1 - 2 l l Chapterl1:l l Introduction

Amplifierland Heater

Thermostat l valves

controls

Dancerl dynamics Dancer positionlsen sor transducer Motorla ndldrive Amplifier system

Desiredl t

emperature

Temperature

l difference

Voltagel

difference

Fuell flo

w

Actuall t

emperature

l+

Desired

l rolll

angle

Input

l voltag

e

Error

l voltag

e

Aileron

l positio

n

Roll

l rat

e

Rolll a

ngle

Desiredl speed Inputl voltage

• Speed lErrorlv oltage Actual lspee d Voltagelpro portional tolactuallspeed

1 - 4 l l Chapterl1:l l Introduction

Differential

R

• V R

Float - V

am

plifier

Potentiometer Float Actuatorlan dlvalve Potentiometer Amplifiers

a.

Fluidlinput

+V

Desiredl l

evel

Tank

Drain

b.

Desired llevel voltagel in

• Flowlra telin + - Actualllev el Flowlrat elout voltage lout Displacement Drain Integrate

1 - 5 l SolutionsltolProblems

Desiredl

Controllerl

Desired l force

Actual

Desired

l positio

n

Force Depth

Commandedlb loodlpressure (^) + Isofluranelco ncentration bloodlpr essure

• Vaporizer Patient

1 - 7 l SolutionsltolProblems

Actual

15.

di

a. L

dt

+lRil=lu(t)

b. Assumelalsteady-statelsolutionlissl=lB.lSubstitutinglthislintoltheldifferentiallequationlyieldslRBl=

fromlwhichlBl=

1

.lThelcharacteristiclequationlislLMl+lRl=l0,lfromlwhichlMl=l-

R R

.lThus,ltheltotal

L 14.

Desired +

Gyroscopic

HT’s Amplifier

1 - 8 l l Chapterl1:l l Introduction

24 solutionlisli(t)l =l Ae-(R/L)tl+l l 1 l

.lSolvinglforlthelarbitrarylconstants,li(0)l=lAl+l

l 1

=l 0.lThus,lAl =

R 1 l 1

• .lThelfinallsolutionlisli(t)l= R

--l

l 1 l e-(R/L)tl=l l 1 l (1l−l e −(l R /l^ L ) t l). R R R R

c.

18. (^) a. Writinglthellooplequation,l Ri l+l L l^ ldi l +l^ l^1 l l idt l+l v (^) (0)l=l v ( t ) dt C l^

C

d l^2 i l

di b. Differentiatinglandlsubstitutinglvalues, 2 +l^25 i l=l^0 dt l^2 dt

Writinglthelcharacteristiclequationlandlfactoring,

M l^2 l+l 2 M l +l 25 l=l( M l + 1 + 24 i )( M l + 1 − 24 i )l.

Thelgenerallformloflthelsolutionlandlitslderivativelis

i l=l Ae − t lcos(l 24 t )l+l Be − t lsin(l 24 t ) di l =l(−l A l+l dt 24 B ) e − t lcos(l l 24 t )l−l(l l 24 l A l+l B ) e − t lsin(l l 24 t ) Usingl i (0)l=l0;l di l (0)l=l vL l(0)l =l 1 l =l 1 dt L L

i l 0 l=l A l =

di l (0)l=l−l A l+l dt 24 B l= 1 Thus,l A l=l 0 landl B l=. 24

Thelsolutionlis

i l= 1 e − t lsin(l 24 t )

1 - 10 l l Chapterl1:l l Introduction

xpl=lAsin3tl+lBcos3tlSu

bstitutelintoltheldifferentiallequationlandlobtain

(18Al−lB)cos(3t)l−l(Al+l18B)sin(3t)l =l 5sin(3t)

Therefore,l18Al–lBl=l 0 landl–(Al+l18B)l=l5.lSolvinglforlAlandlBlwelobtain

xpl=l(-1/65)sin3tl+l(-18/65)cos3t

Thelcharacteristiclpolynomiallis

M^2 l+l 6 lM+l 8 l=l M+l 4 M+l 2

Thus,ltheltotallsolutionlis

xl=lCle-^4 l^ tl+lDle-^2 l^ tl+l - l 18 l cosl 3 ltl - l 1 sinl 3 lt 65 65 Solvinglforlthelarbitrarylconstants,l x (0)l=l C l+l D l−l l 18 l =l 0 l. 65

Also,l thelderivativeloflthelsolutionlis

dxl =l-l (^3) cos 3 lt l +l l 54 sin l 3 lt l-l 4 lCle-^4 l^ tl- 2 lDle-^2 l^ t dt (^65 )

. (^) −l^ l^3 ll −l 4 C l−l 2 D l=l 0 l,lorl Cl=l −l^ l^3 l andlDl=l^15 l.

Solvinglforlthelarbitrarylconstants,lx(0)

65 10 26

Thelfinallsolutionlis

xl=l-l 18 l cosl 3 ltl - l 1 sinl 3 ltl - l 3 e

• 4 ltl +l 15 l e - 2 lt 65 65 10 26

c. Assumelalparticularlsolutionlof

xpl=lA

Substitutelintoltheldifferentiallequationlandlobtainl25Al=l10,lorlAl=l2/5.lT

helcharacteristiclpolynomiallis

M^2 l+8lMl+l 25 l=l Ml+l 4 l+3li Ml+l 4 - 3il

Thus,ltheltotallsolutionlis

xl=l 2 l +le-l^4 l^ t 5 Blsinl 3 ltl +lClcosl 3 lt

Solvinglforlthelarbitrarylconstants,lx(0)l=lCl+l2/5l=l0.lTherefore,lCl=l-

2/5.lAlso,lthelderivativeloflthelsolutionlis

1 - 11 l SolutionsltolProblems

dt

dxl =l 3 lBl- 4 lC l cos 3 lt l - l 4 lBl+l 3 lC l sin 3 lt l e-^4 l^ t

Solvinglforlthelarbitrarylconstants,lx(0)l =l3Bl–l4Cl=l0.lTherefore,lBl=l-8/15.lThelfinallsolutionlis

x ( t )l=l 2 l −l e −^4 t l ll^8 l sin(3 t )l+l 2 l cos(3 t l) 5 l 15 5 

a. Assumelalparticularlsolutionlof

Substitutelintoltheldifferentiallequationlandlobtain

Equatingllikelcoefficients,

Fromlwhich,lCl=l-l

1 l l

andlDl=l-l

l 1 ll

Thelcharacteristiclpolynomiallis

Thus,ltheltotallsolutionlis

Solvinglforlthelarbitrarylconstants,lx(0)l=lAl-l

1 l

=l2.lTherefore,lAl=

11

.lAlso,lthelderivativeloflthe

5

solutionlis

Solvinglforlthelarbitrarylconstants,lx(0)l =l-lAl +lBl-l0.2l=l-3.lTherefore,lBl= −l 3 l

.lThelfinallsolution

5

is

x ( t )l=l−l 1 l cos(2 t )l−l l 1 l sin(2 t )l+l e − t l l^11 cos( t )l−l 3 l sin( t ) 5 10 l^5 5 

b. Assumelalparticularlsolutionlof

xpl=lCe-2tl+lDtl+lE

Substitutelintoltheldifferentiallequationlandlobtain

ldxl dt

1 - 13 l SolutionsltolProblems

iruse s

Input voltagel+

Inputltra nsducer Sensor Spring Pantographl dynamics Controller Actuator

Controller^ Patient

Desired l l forcel

Fup Springldispla cement Fout

Desiredl A

mountlofl

HIVlv RTI

lPI

Amountlofl

HIVlviruses

1 - 14 l l Chapterl1:l l Introduction

Aerodynamic

Vehicle Electricl Motor Climbingl& lRollinglRes istances

Motive

lForce

Speed

Aerodynamic

a. Desired Speed Inverterl ControllC ommand Controlledl Voltage

Actual

ECU (^) Inverter

1 - 16 l l Chapterl1:l l Introduction

c. Desired Speed lError Accelerator Power ICE Climbingl&l RollinglResi stances Actual Inverterl ControllC ommand Inverter l&lElect ric Motor Motor TotallMotiv elForce Aerodynamic Speed Aerodynamic Planetary lGearlCo ntrol ECU Accelerator Vehicle

2

• lil 2

=il

T l W l O Modeling l in l the l Frequenc y l Domain

SOLUTIONS l TO l CASE l STUDIES l CHALLENGES

Antenna l Control: l Transfer l Functions

Findingleachltransferlfunction:

Vi(s) 10

Pot:l 

i(s)l^

=l

l l

Vp(s)

Pre-Amp:l V

i(s)l^

=lK;

lEa(s) l 150 l

PowerlAmp:lV

p(s)l^

=ls+150lM

otor:lJml=l0.05l+l 5 (

l 50 ll

) =l0. Dml=0.01l+l 3 (

l 50 ll

) =l0.

Kt 1

Ral l =l 5

KtKb 1

Ra =l 5

m(s)

l Ktll

RaJm 0.

Therefore: Ea(s)l l =^ l 1 l KtKb =ls(s+1.32)

s(s+Jl (Dm+l Rl l ))

o(s) 1 lm(s)

m a

l l 0.

And:lE

a(s)l^

=l 5 l E

a(s)l l^

=ls(s+1.32)

Transfer l Function l of l a l Nonlinear l Electrical l Network

Writingltheldifferentiallequation,

d(i 0 l +l i)l +l2(il +l i)^2 dt 0 −l 5 l=lv(t)l.lLinearizingli^2 aboutli 0 ,

(il +

0

i)

2 0 =l2i l i=i 0

i = l2i 0 i.lThus,l(i^

0

+i)

2 0

+l2i 0 i.