ol Systems Engineering, International Adaptation, 8th Edition by Nise, Exams of Systems Engineering

ol Systems Engineering, International Adaptation, 8th Edition by Nise

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Copyrightk©
k
2011kbykJohnkWileyk&kSons,kInc.
O
k
N
k
E
Introduction
ANSWERSkTOkREVIEWkQUESTIONS
1.
Guidedkmissiles,kautomatickgainkcontrolkinkradiokreceivers,ksatellitektrackingkantenna
2.
Yesk-kpowerkgain,kremotekcontrol,kparameterkconversion;kNok-kExpense,kcomplexity
3.
Motor,klowkpasskfilter,kinertiaksupportedkbetweenktwokbearings
4.
Closed-
loopksystemskcompensatekforkdisturbanceskbykmeasuringkthekresponse,kcomparingkitktokthekinputk
responsek(thekdesiredkoutput),kandkthenkcorrectingkthekoutputkresponse.
5.
Underkthekconditionkthatkthekfeedbackkelementkiskotherkthankunity
6.
Actuatingksignal
7.
Multipleksubsystemskcanktimeksharekthekcontroller.kAnykadjustmentsktokthekcontrollerkcankbe
kimplementedkwithksimplyksoftwarekchanges.
8.
Stability,ktransientkresponse,kandksteady-statekerror
9.
Steady-state,ktransient
10.
Itkfollowskakgrowingktransientkresponsekuntilktheksteady-
statekresponsekisknoklongerkvisible.kTheksystemkwillkeitherkdestroykitself,kreachkankequilibriumkstate
kbecausekofksaturationkinkdrivingkamplifiers,korkhitklimitkstops.
11.
Naturalkresponse
12.
Determinekthektransientkresponsekperformancekofktheksystem.
13.
Determineksystemkparametersktokmeetkthektransientkresponsekspecificationskforktheksystem.
14.
True
15.
Transferkfunction,kstate-space,kdifferentialkequations
16.
Transferkfunctionk-kthekLaplacektransformkofkthekdifferentialkequation
State-spacek-krepresentationkofkanknthkorderkdifferentialkequationkasknksimultaneouskfirst-
orderkdifferentialkequations
Differentialkequationk-kModelingkaksystemkwithkitskdifferentialkequation
SOLUTIONSkTOkPROBLEMS
50kvolts
1.
Fivekturnskyieldsk50kv.kThereforekKk=
5kxk2
k
r
ad
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O k N k E

Introduction

ANSWERS k TO k REVIEW k QUESTIONS

1. Guidedkmissiles,kautomatickgainkcontrolkinkradiokreceivers,ksatellitektrackingkantenna

2. Yesk-kpowerkgain,kremotekcontrol,kparameterkconversion;kNok-kExpense,kcomplexity

3. Motor,klowkpasskfilter,kinertiaksupportedkbetweenktwokbearings

4. Closed-

loopksystemskcompensatekforkdisturbanceskbykmeasuringkthekresponse,kcomparingkitktokthekinputk

responsek(thekdesiredkoutput),kandkthenkcorrectingkthekoutputkresponse.

5. Underkthekconditionkthatkthekfeedbackkelementkiskotherkthankunity

6. Actuatingksignal

7. Multipleksubsystemskcanktimeksharekthekcontroller.kAnykadjustmentsktokthekcontrollerkcankbe

kimplementedkwithksimplyksoftwarekchanges.

8. Stability,ktransientkresponse,kandksteady-statekerror

9. Steady-state,ktransient

10. Itkfollowskakgrowingktransientkresponsekuntilktheksteady-

statekresponsekisknoklongerkvisible.kTheksystemkwillkeitherkdestroykitself,kreachkankequilibriumkstate

kbecausekofksaturationkinkdrivingkamplifiers,korkhitklimitkstops.

11. Naturalkresponse

12. Determinekthektransientkresponsekperformancekofktheksystem.

13. Determineksystemkparametersktokmeetkthektransientkresponsekspecificationskforktheksystem.

14. True

15. Transferkfunction,kstate-space,kdifferentialkequations

16. Transferkfunctionk-kthekLaplacektransformkofkthekdifferentialkequation

State-spacek-krepresentationkofkanknthkorderkdifferentialkequationkasknksimultaneouskfirst-

orderkdifferentialkequations

Differentialkequationk-kModelingkaksystemkwithkitskdifferentialkequation

SOLUTIONS k TO k PROBLEMS

50 kvolts

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

5 kxk 2 k r

ad

  • = k 1.

1 - 2 k k Chapterk1:k k Introduction

Amplifierkand Heater

Thermostat k valves

controls

Dancerk dynamics Dancer positionksen sor transducer Motorka ndkdrive Amplifier system

Desiredk

temperature

Temperature

k difference

Voltagek

difference

Fuelk flo

w

Actualk t

emperature

k+

Desired

k rollk

angle

Input

k voltag

e

Error

k voltag

e

Aileron

k positio

n

Roll

k rat

e

Rollk a

ngle

Desiredk speed Inputk voltage

  • Speed kErrorkv oltage Actual kspee d Voltagekpr oportional tokactualkspeed

1 - 3 k SolutionsktokProblems

Amplifier

Motork

andk dr

ivek sy

stem

power

Transducer

Sensork&k

transducer

Transducer

Desired

Input

k voltag

e

Power

k Error

k voltag

e

Rodk

position

Actual

k pow

er

Voltagek pr

oportional

tokactualkpower

Desired Population

Desired

k stude

nt

Actual

Graduatingk

and

drop-

outk rate

Netkrate

Actualk stude

nt

studentk popula

tion +

error rate^ student^ -

ratek k +

ofkinflux population

Desired Voltagek proportional tokdesired volume Volumek error Voltagekrepresentin g actualkvolume (^) Actual volume volume +

  • Effective volume + Voltagekpro portionalktok speed Transducer

Speed

Administration Integrate

Volume kcontrolkcircuit Radio

Admissions

Reactor

1 - 5 k SolutionsktokProblems

Desiredk + Controllerk

Desired

k force

Actual

Desired

k positio

n

Force Depth

Commandedkb loodkpressure (^) + Isofluranekco ncentration bloodkpr essure

  • Vaporizer Patient

1 - 6 k k Chapterk1:k k Introduction

Retinak+kOptical

Internalkeye

kmuscles

Desired^ Brain

k Light

a.

Nervousk

systemk e

lectricalk

impulses

Retina’s

k Light

Intensity Intensity

b.

Desiredk Lig

htk Intensity

Nervousk

systemk e

lectricalk

impulses

Retina’sk Lig

htk Intensity

External

k Light

Ifk thek narrowk lightk beamk isk modulatedk sinusoidallyk thek pupil’sk diameterk willk alsok varyksinusoidallyk(withkakdelaykseekpartkc)kinkproblem) c. k Ifkthekpupilkrespondedkwithknoktimekdelaykthekpupilkwouldkcontractkonlyktokthekpointkw herekaksmallkamountkofklightkgoeskin.kThenkthekpupilkwouldkstopkcontractingkandkwoul dkremainkwithkakfixedkdiameter.

Retinak+kOpticalkNe

rves

Internalkeye

kmuscles

Brain

1 - 8 k k Chapterk1:k k Introduction

24 solutionkiski(t)k =k Ae-(R/L)tk+k k 1 k

.kSolvingkforkthekarbitrarykconstants,ki(0)k=kAk+k

k 1

=k 0.kThus,kAk =

R 1 k 1

  • .kThekfinalksolutionkiski(t)k= R

--k

k 1 k e-(R/L)tk=k k 1 k (1kk e (k^ R /k^ L ) t k). R R R R

c.

18. (^) a. Writingkthekloopkequation,k Ri kk L k^ kdi k k^ k^1 k^ k idt kk v (^) (0)kk v ( t ) dt C k^ ^ C d k^2 i k  di b. Differentiatingkandksubstitutingkvalues, 2 k^25 i kk^0 dt k^2 dt

Writingkthekcharacteristickequationkandkfactoring,

M k^2 kk 2 M k k 25 kk( M k  1  24 i )( M k  1  24 i )k.

Thekgeneralkformkofktheksolutionkandkitskderivativekis

i kk Aet kcos(k 24 t )kk Bet ksin(k 24 t ) di k k(k A kk dt 24 B ) et kcos(k k 24 t )kk(k k 24 k A kk B ) et ksin(k k 24 t ) Usingk i (0)kk0;k di k (0)kk vL k(0)k k 1 k k 1 dt L L

i k 0 k =k A k = 0

di k (0)kkk A kk dt 24 B k= 1 Thus,k A kk 0 kandk B k. 24

Theksolutionkis

i k 1 et ksin(k 24 t )

1 - 9 k SolutionsktokProblems

c.

a. Assumekakparticularksolutionkof

Substitutekintokthekdifferentialkequationkandkobtain

Equatingklikekcoefficients,

Fromkwhich,kCk=k

35 k k

andkDk=k

10 k

Thekcharacteristickpolynomialkis

Thus,kthektotalksolutionkis

Solvingkforkthekarbitrarykconstants,kx(0)k=kAk+ 53 k =k0.kTherefore,kAk=k-k 53 k .kThekfinalksolutionkis

b. Assumekakparticularksolutionkof

1 - 11 k SolutionsktokProblems

dt

dxk =k 3 kBk- 4 kC k cos 3 kt k - k 4 kBk+k 3 kC k sin 3 kt k e-^4 k^ t

Solvingkforkthekarbitrarykconstants,kx(0)k =k3Bk–k4Ck=k0.kTherefore,kBk=k-8/15.kThekfinalksolutionkis

x ( t )kk 2 k k e ^4 t k kk^8 k sin(3 t )kk 2 k cos(3 t k) 5 k 15 5 

a. Assumekakparticularksolutionkof

Substitutekintokthekdifferentialkequationkandkobtain

Equatingklikekcoefficients,

Fromkwhich,kCk=k-k

1 k k

andkDk=k-k

k 1 kk

Thekcharacteristickpolynomialkis

Thus,kthektotalksolutionkis

Solvingkforkthekarbitrarykconstants,kx(0)k=kAk-k

1 k

=k2.kTherefore,kAk=

11

.kAlso,kthekderivativekofkthe

5

solutionkis

Solvingkforkthekarbitrarykconstants,kx(0)k =k-kAk +kBk-k0.2k=k-3.kTherefore,kBk= k 3 k

.kThekfinalksolution

5

is

x ( t )kkk 1 k cos(2 t )kk k 1 k sin(2 t )kk et k k^11 cos( t )kk 3 k sin( t ) 5 10 k^5 5 

b. Assumekakparticularksolutionkof

xpk=kCe-2tk+kDtk+kE

Substitutekintokthekdifferentialkequationkandkobtain

kdx kdt

1 - 12 k k Chapterk1:k k Introduction

Equatingklikekcoefficients,kCk=k5,kDk=k1,kandk2Dk+kEk=k0.kFr

omkwhich,kCk=k5,kDk=k1,kandkEk=k-k2.

Thekcharacteristickpolynomialkis

Thus,kthektotalksolutionkis

Solvingkforkthekarbitrarykconstants,kx(0)k=kAk+k 5 k-k 2 k=k 2 kTherefore,kAk=k-1.kAlso,k thekderivativekofkthe

solutionkis

dx k k(k A kk B ) e k^ t k k Btet k k 10 e ^2 t k k 1 dt

Solvingkforkthekarbitrarykconstants,kx(0)k =kBk-k 8 k=k1.kTherefore,kBk=k9.kThekfinalksolutionkis

c. Assumekakparticularksolutionkof

xpk=kCt^2 k+kDtk+kE

Substitutekintokthekdifferentialkequationkandkobtain

Equatingklikekcoefficients,kCk=k

1 k

,kDk=k0,kandk2Ck+k4Ek=k0.k

Fromkwhich,kCk=k

4 k ,kDk=k0,kandkEk=k-k 8 k.

Thekcharacteristickpolynomialkis

Thus,kthektotalksolutionkis

Solvingkforkthekarbitrarykconstants,kx(0)k=kAk-k

8 k =k^1 kTherefore,kAk=k 8 k.kAlso,k^ thekderivativekofkthe

solutionkis

Solvingkforkthekarbitrarykconstants,kx(0)k =k2Bk=k2.kTherefore,kBk=k1.kThekfinalksolutionkis

dx k

1 - 14 k k Chapterk1:k k Introduction

Aerodynamic

Vehicle Electrick Motor Climbingk& kRollingkRe sistances

Motive

kForce

Speed

Aerodynamic

a. Desired Speed Inverterk ControlkC ommand Controlledk Voltage

Actual

ECU (^) Inverter

1 - 15 k SolutionsktokProblems

Climbingk& kRollingkRe sistances

Desired Speed

AcceleratorkD

isplacement Motive Actual

_

Aerodynamic

Speed

Aerodynamic

ECU Accelerator, Vehicle b.

2

- kik

2

=i

k

T k W k O

Modeling k in k the k Frequen

cy k Domain

SOLUTIONS k TO k CASE k STUDIES k CHALLENGES

Antenna k Control: k Transfer k Functions

Findingkeachktransferkfunction:

Vi(s) 10

Pot:k 

i(s)k^

=k

k k

Vp(s)

Pre-Amp:k V

i(s)k^

=kK;

kEa(s) k 150 k

PowerkAmp:kV

p(s)k^

=ks+150kM

otor:kJmk=k0.05k+k 5 (

k 50 kk

) =k0.

Dmk=0.01k+k 3 (

k 50 kk

) =k0.

Kt 1

Rak k^ =k 5

KtKb 1

Ra =k 5

m(s)

k Ktkk

RaJm 0.

Therefore: Ea(s)k k =

k 1

k

KtKb =ks(s+1.32)

s(s+Jk (Dm+k Rk k ))

o(s) 1 km(s)

m a

k k 0.

And:kE

a(s)k^

=k 5 k E

a(s)k^ k^

=ks(s+1.32)

Transfer k Function k of k a k Nonlinear k Electrical k Network

Writingkthekdifferentialkequation,

d(i 0 k k i)k

k2(ik k i)^2

dt 0

k 5 kkv(t)k.kLinearizingki^2 aboutki 0 ,

(ik +

0

i)

2 0

=k2i k

i=i 0

i = k2i 0 i.kThus,k

(i

0

+i)

2 0

1 - 18 k k Chapterk1:k k Introduction

+k2i 0 i.