自動控制原理英文版課后全部 答案

ModuIe3

Problem3.1

(a)WhentheinputvariableistheforceF.HieinputvariableFandtheoutputvariableyare

relatedbytheequationobtainedbyequatingthemomentonthestick:

/尸=/工+2紇2

33dt

TakingLaplacetransforms,assuminginitialconditionstobezero,

F=-Y+-csY

33

leadingtothetransferfunction

Y3/k

*l+(4c/Z)s

wherethetimeconstantTisgivenby

4c

T=——

k

(b)WhenF=0

Theinputvariableisx,thedisplacementofthetoppointoftheupperspring.Theinput

variablexandtheoutputvariableyarerelatedbytheequationobtainedbythemomentonthe

stick:

k(q)/3+也2

333dt

TakingLaplacetransforms,assuminginitialconditionstobezero,

3Ax=(2女+4cs)Y

leadingtothetransferfunction

Y_3/2

―—l+(2c/Z)s

wherethetimeconstant丁isgivenby

2c

T=—

Im-

X^ole

*K

_k__k_Rc.

一五.~4c

Problem3.2P54

Determinetheoutputoftheopen-loopsystem

G(s)=」一

1+sT

tOtheinput

r(t)=t

Sketchbothinputandoutputasfunctionsoftime,anddeterminethesteady-stateerror

betweentheinputandoutput.ComparetheresultwiththatgivenbyFig3.7.

Solution：

Whiletheinputr(t)=t,useLaplacetransforms,

Inputr(s)=-^

s,

a

Outputc(s)=r(s)G(s)=2=a+丁

/.(l+Ts)/s?+1

IT)

thetime-domainresponsebecomes

c(t)=at-aT[\-^7j

Problem3.3

3.3ThemasslessbarshowninFig.P3.3hasbeendisplacedadistanceXoandissubjectedto

aunitimpulse§inthedirectionshown.Findtheresponseofthesystemfort>0andsketch

theresultasafunctionoftime.Confirmthesteady-stateresponseusingthefinal-value

theorem.

Solution：

Theequationobtainedbyequatingtheforce:

kx0+cx0=5Q)

TakingLaplacetransforms,assuminginitialconditiontobezero,

KXo+CsXo=l

leadingtothetransferfunction

X_1_1[

~F(^=K+Cs=C~K

SH

c

Thetime-domainresponsebecomes

]—t

x(t)=《e[

Thesteady-stateresponseusingthefinal-valuetheorem:

「「111

-ioK+CSSK

K(XQ+X)+CX=6。)=Kx()+X(K+Cs)=1;

,Yi-/rx0\-KXQ]

"Cs+KKC5+i

K

1—Kx-%

x(t)=，()

C

Accordingtothefinal-valuetheorem:

limx(r)=lims-X=lim——=0

.10ST。KC,

-5+1

Problem3.4

Solution:

Uftheinputisaunitstep,then

/?u)=-

s

leadingto,

C⑻=

(l+rs)s

takingtheinverseLaplacetransformgives,

-i

c(t)=l-er

asthesteady-stateoutputissaidtohavebeenachievedonceitiswithin1%ofthefinalvalue,

wecansolute"t"likethis,

c(t)=l-eT=99%xl(thefinalvalueis1)

hence,

-t

er=0.01(thetimeconstantr=10s)

t=4.605xr=46.05$

2.thenumericalvalueofthenumeratorofthetransferfunctiondoesn'taffecttheanswer.

Seethisequation,

If

C(s)A

G(s)=

R(s)\+TS

then

A

C(s)=

(1+TS)S

givingthetime-domainresponse

-t

c(t)=A(]-er)

asthefinalvalueisA,thesteady-slateoutputisachievedwhen,

-i

c(r)=A(l-e7)=Ax99%

solutetheequation,

t=4.605r=46.05s

theresultmakenodifferentfromthatabove,sowesaidthatthenumericalvalueofthe

numeratorofthetransferfunctiondoesn'taffecttheanswer.

Ifa<l,asthetimeincrease,thetwolineswon'tcross.Inthesteadystatetheoutputlagsthe

inputbyatimebymorethanthetimeconstantT.

Thesteadyerrorwillbenegativeinfinite.

R(t)

t

Ifa=l,asthetimeincrease,thetwolineswillbeparallel.ItisassameasFig3.7.

Ifa>l,asthetimeincrease,thetwolineswillcross.Inthesteadystatetheoutputlagsthe

inputbyatimebylessthanthetimeconstantT.

Thesteadyerrorwillbepositiveinfinite.

Kt)

C(t)

Problem3.5

Solution:

R(S)=4+-.Y(S)=--s-+-6--=--29-1-6-1--1-45-

sss2-(55+1)ss25s+1

...)，(，)=6/—29+29e"5

sothesteady-stateerroris29(-30).TDconformtheresult:

limy(r)=lim(6r-29+29')=oo;

5+6

limy(r)=limy(6)=limY(s)=lim=oo.

foes->0S->0S->0S(5S+1)

e”=lime(f)=lim5-E(S)=lim5[F(5)-/?(S)]

TOOST。5—>0

=limS(G(S)-l]/?(S)=limS-(———1)-(4-+-)

sf。5S+1S2S

=-30

Therefore,thesolutionisbasicallycorrect.

Problem3.6

2y+3y=x

sinceinputisofconstantamplitudeandvariablefrequency,itcanberepresentedas:

X=AJ

asweknow,theoutputshouldbeasinusoidalsignalwiththesamefrequencyoftheinput,it

canalsoberepresentedas:

"DI

y=y()e

hence

2汝匕，“+3兒/皿=4膽網(wǎng)

A=_L_

A.3+2/。

2vv

69=-tan——

3

ItsDC(w—>0)valueis

Requirement熱卜捌

3

—>vv=—

2

whilephaselagoftheinput:0=-tan*11=

4

Problem3.7

Onedefinitionofthebandwidthofasystemisthefrequencyrangeoverwhichtheamplitude

oftheoutputsignalisgreaterthan70%oftheinputsignalamplitudewhenasystemis

subjectedtoaharmonicinput.Findarelationshipbetweenthebandwidthandthetime

constantofafirst-ordersystem.Whatisthephaseangleatthebandwidthfrequency?

Solution：

Fromtheequation3.41

A=Q,20.7%(1)

and69>0(2)

sothebandwidthB(o=1221

T

fromtheequation3.43

乃

1-1

thephaseangleZc0=-tantyr=-tan1.02=—

Problem3.8

3.8Solution

AccordingtogeneralizedtransferfunctionofFirst-OrderFeedbackSystems

C_KGK

R1+KGH1+K+TS

<*ethesteadystateoftheoutputofthissystemis2.5V.

ifs—>0,——>=—.Fromthis,wecangetthevalueofK,thatisK=~-

R1043

Sinceweknowthatthestepinputis10V,takingLaplacetransforms,theinputis_.

S

Thentheoutputisfollowed

10x%

C(s)=

S1+%+TS

TakingreverseLaplacetransforms,

T

C=2.5-2.5e%r=2.5(1-e-生)

Fromthefigure,wecanseethatwhenthetimereached3s,thevalueofoutputis86%ofthe

steadystate.Sowecanknow

/-3=—2=—3=(—2)*4—n匯=82

外3r9,

/3r

4,/3r

l-e-=0.864|,=3=>r=2

31

Thetransferfunctionis

12+8s4+65

Let12+8s=0,wecangetthepole,thatiss=-1.5-2/3

Problem3.9Page55

Solution:

Thetransferfunctioncanberepresented,

G(s)=――—=―~—?―_-

斗(s)匕“(s)匕(S)

While,

一(S)1

匕“(s)1+sRC

Leadingtothefinaltransferfunction,

__________]

G(s)=

1+3SRC+(SRC)2

Andthereason:

thesecondsimplelagcompensationnetworkcanberegardedastheloadofthefirstone,and

accordingtoLoadEffect,theloadaffectstheprimaryrelationship;sothetransferfunctionof

thecombinationdoesn'tequaltheproductofthetwoindividuallagtransferfunction

ModuIe4

Problem4.1

4.1Theclosed-looptransferfunctionis

1()

C——1()

R-l+s(^6)-S2+65+10

Comparingwiththegeneralizedsecond-ordersystem,weget

叱,=加

2EWn=6

?3710

E=

10

Problem4.3

4.3Consideringthespringrisexandthemassrisey.UsingNewton'ssecondlawofmotion

d(x-y)

iny=K(x-y)+c

dt

TakingLaplacetransforms,assumingzeroinitialconditions

mYs2=KX-KY+csX-csY

resultinginthetransferfuncitionwhere

Ycs+K

—=

Xms2+cs+K

And

c=1.26*IO,

Problem4.4

Solution:

Theclosed-looptransferfunctionis

—K?--1--

cSS+2K

~R工K.1-S2+2S+K

1?H■

SS+2

Comparingtheclosed-looptransferfunctionwiththegeneralizedform.

C

R2+2匏/+而

itisseenthat

K=

Andthat

2g=2；§=蘇

Thepercentageovershootistherefore

PO=100^=100/,^

Where

PO<10%

Whensolved,gives1.2WK(2.86)

WhenKtakesthevalue1.2,thepolesofthesystemaregivenby

?+2^+1.2=0

Whichgives

=-l±O.45js=-l±1.36j

Im

0.45

Re

-0.45

O

Problem4.5

4.5Aunity-feedbackcontrolsystemhastheforward-pathtransferfunction

S(s+10)

Findtheclosed-looptransferfunction,anddevelopexpressionsfbrthedampingratio

AnddampednaturalfrequencyintermofKPlottheclosed-looppolesonthecomplex

PlanefbrK=0,10,25,50,1OO.ForeachvalueofKcalculatethecorrespondingdamping

ratioanddampednaturalfrequency.Whatconclusionscanyoudrawfromtheplot?

Solution:

KS)

SubstituteG(s)=intothefeedbackformula:①(s)=.Andinunit

s(s+10)1+”G(S)

feedbacksystemH=1.

K

Resultin:0(s)=—：

d+iOs+K

Sothedampednaturalfrequency①“=yfK,

.105

dJampingratio^=-=-y=.

Thecharacteristicequationiss2+10S+K=0.

WhenK<25,S=-5±V25-/C;

WhileK>25,s=-5±WK-25;

ThevalueofCOnand(correspondingtoKarelistedasfollows.

K0102550100

Pole1S,0-5+V15-5-5+5i-5+5?

-10-5-Ji?-5-5-5i

Pole2S2

-5-5-73)

0而55應(yīng)10

qooV151V050.5

PlotthecomplexplanefbreachvalueofK:

im

axis

V-5+逃

k-X0O

n

k=50

T。-5-后-5-5+而

—X-X——?real

k"0k"10k-lOk-0axis

k-25

Xk-50

X

Wecanconcludefromtheplot.

Whenk<25,polesdistributeontherealaxis.ThesmallervalueofKis,thefartherpolesis

awayfrompoint-5.ThelargervalueofKis,thenearerpolesisawayfrompoint-5.

Whenk>25,polesdistributeawayfromtherealaxis.ThesmallervalueofKis,thefurther

(nearer)polesisawayfrompoint-5.ThelargervalueofKis,thenearer(farther)polesis

awayfrompoint-5.Andallthepolesdistributeonalineparallelsimaginaryaxis,intersect

realaxisonthepole-5.

Probleni4.6

TakingLaplacetransforms,assumingzeroinitialconditions,reducesthisequationto

Vb_RLs

Ls+R+RLCs?

Sincetheinputisaconstantcurrenti(),so

1

then,

RL

C（s）=匕=

Ls+R+RLCs2

Applyingthefinal-valuetheoremyields

limc（/）

r—>QO'Z

indicatingthatthesteady-statevoltageacrossthecapacitorCeventuallyreachesthe

zero.resultinginfullerror.

Problem4.7

4.7Provethatforanunderdampedsecond-ordersystemsubjecttoastepinput,the

percentageovershootabovethesteady-stateoutputisafunctiononlyofthedamping

ratio.

Fig.4.7

Solution

Theoutputcanbegivenby

2

C（s）=--~3-

s（s+2血6+0）

________s+2g（1）

丁（s+S>+q2（Y）

thedampednaturalfrequency%canbedefinedas

=⑵

substitutingaboveresultsin

l_____s+g____________眄____.

C（5）=S（s+血）2+02（s+「y“）2+%2-

takingtheinversetransformyields

13/

C?)=l-sinQ"+0)

where(4)

tan^=￡?

themaximumoutputis

sin(叫+。)

7171(5)

t-=-

sothemaximumis

c(o)=l+e-%/F7

thepercentageovershootistherefore

PO=100"*1必7

Problem4.8

Solutionto4.8:

Consideringthemassmdisplacedadistancexfromitsequilibriumposition,

thefree-bodydiagramofthemasswillbeasshownasfollows.

UsingNewton'ssecondlawofmotion,

p-2kx-cx=tnx

mx+ex+2kx=p

TakingLaplacetransforms,assumingzeroinitialconditions,

X{ms2+cs+2k)=P

resultsinthetransferfunction

X/P=(1//H)/(52+(c/m)s+2攵/m)

=(2/k)(2k/m)(s2+(c/m)s+2khn)

Aswesee

X(ms2+cs+2k)=P

AsPisconstant

SoXoc——

ms+cs+2Z

When5=--=-6.25xIO-5

2m

(ms2+cs+2k}=105

\/min

Xg=k0」

Thisisasecond-ordertransferfunctionwhere

co：=2k/m

and

g=c/2wnm=c/212km

Thedampednaturalfrequencyisgivenby

2

cod=conJi—--N2klmy/l-c/Skni

=42k/m-(c/2m)2

Usingthegivendata,

(o?=V2X5X104/2X106=7(X05=0.2236

250

=2.7950xlO-4

2X2X106XV0X)5

%=0.2236x-(2.7950x1O_4)2=0.2236

Withthesedatawecandrawapicture

71

—=14.050

A

—=16000

7；=4.6^.=73600

mcod

*?*人x仁)~^~「明(一弧sin(odtp+(Ddcoso/p)=0

?*-tan叫=j=7.03-=0.02加

Problem4.10

4.10

solution:

ThesystemissimilartotheoneinthebookonPAGE58toPAGE63.Thedifferenceisthe

connectionofthespring.Sothetransferfunctionis

嗎2

%d+23.5+嗎2

必_k.k&N

2

ed~RJs+RCs+(R+kuk,nN)kp

J=N3+L，C=N&+C「,

NR

dampingratio

2KK.KJ

ButthevalueofJisdifferent,becausethereisaspringconnected.

Becauseoffinal-valuetheorem,

%=紇

為一24

ModuIe5

Problem5.4

5.4Theclosed-looptransferfunctionofthesystemmaybewrittenas

10K

C=S2+6S+10=JL-1°(K+1)

2

R|+10K1+KS+6S+10A-+10

S2+6S+10

Theclosed-looppolesarethesolutionsofthecharacteristicequation

SL±J36-4(10+10K)=_3±Q^

2

叱,=J10(l+K)

2EjlO(l+K)=6

110(1+K)

Inordertostudythestabilityofthesystem,thebehavioroftheclosed-looppoleswhenthe

gainKincreasesfromzerotoinfintewillbeobserved.Sowhen

K.=2E=—S=-3土而/

'10

K,=10E=S=-3±VK)TJ

2110

AT,=20E=^^-S=-3±V20Tj

370

雙擊下面可以看到原圖

Re

Problem5.5

Solution

Theclosed-looptransferfunctionis

K

_K_K

R—K?+A：(l+a.v)s2+aKs+K

S

Comparingtheclosed-looptransferfunctionwiththegeneralizedform,

Rs2+2^co,,s+co^

Leadingto

con=y/~K

EU4K

Thepercentageovershootistherefore

PO=100e"=40%

Producingtheresult

4=0.869(0.28)

Andthepeaktime

=4s

4Jl-產(chǎn)

Leadingto

4=1.586(0.82)

Problems.7

5.7ProvethattherisetimeTrofasecond-ordersystemwithaunitstepinputisgivenby

1CM1_J1_72

。=晟tan“襄=￡tan-1—~~

Plottheriseagainstthedampingratio.

Solution:

n，

Accordingto(4.33):c(t)=l-e~^(cosco.t+,^sincodt).4.33

Whent=Tr,c(t)=l.substituec(t)=1into(4.33)

Producingtheresult

1CM1-J_2

T,，tan-'=晟tan」一%一

Plottherisetimeagainstthedampingratio:

Problem5.9

Solutionto5.9:

Asweknowthatthesystemistheopen-looptransferfunctionofaunity-feedbackcontrol

system.

SoGH(S)=G(S)

Givenas

/G-2)(S+5)

Theclose-looptransferfunctionofthesystemmaybewrittenas

G(s)________4K______

R\'1+G〃(s)(s-2)(s+5)+4K

Thecharacteristicequationis

(s-2)(s+5)+4K=Ons2+3s+4K-io=o

AccordingtotheRouth'smethod,theRouth'sarraymustbeformedasfollow

5214/C-10

s30

504K-10

Forthereisnoclosed-looppolestotherightoftheimaginaryaxis

4/C-10>0=>K>2.5

GiventhatQ=0.5

con=J4K-10

3

，=./=>K=4.75

2J4K-10

WhenK=0,therootare

s=+2,-5

Accordingtothecharacteristicequation,thesolutionsare

whileK<3.0625,wehaveoneortwosolutions,allareintegralnumber.

Orwewillhavesolutionswithimaginarynumber.

Sowecandraw

□Open-looppoles

Closed-looppoles

Problems.10

5.10

solution:

?=0.6

wn=2rad/s

accordingto

/CMi

(c)=1-I2sin(w/+</>)=-

,1一7z

eT-sin(16+0)=O.4tan*

finally,tisdelaytime:

bl.23s(0.67)

Module6

Problem6.3

FirstweassumethedisturbanceDtobezero:

e=R-C

101

C=Ke

5+1s

Hence:'=")

R10K+s(s+l)

ThenwesettheinputRtobezero:

10e_10

C=(Ke+O)?

s(s+l)~D~~10K+s(s+l)

Addingthesetworesultstogether:

3-R12D

10K+s(s+l)10K+s(s+l)

R(s)=";D(s)W

_5+110_5-9

'~10Ks+V(s+l)-10KS+S2(S+I)-100^+52(5+1)

thesteady-stateerror:

c~—0rc—9

%=lims?c=lim?'、~:=lim—5-^=-0.09

DS->OS'+S~+100SST0S+S+100

Problem6.4

Determinethedisturbancerejectionratio(DRR)forthesystemshowninFigP.6.4

+

fig.P.6.4

solution：

fromthediagramwecanknow:

c=0.05

sowecangetthat

DRR=(△◎”)*=]+=]+2^211

(AO〃)CLcR0.05

21

soc=0.025,DRR=9

O.Lv+0.05~0.055+0.025

Problem6.5

6.5Solution

Forthepurposesofdeterminingthesteady-stateenorofthesystem,weshouldgetto

knowtheeffectoftheinputandthedisturbancealongwhentheotherwillbeassumedtobe

zero.

Firsttosimplifytheblockdiagramtothefollowingpatte亡

Allowingthetransferfunctionfromtheinputtotheoutputpositiontobewrittenas

%_20

6?,+2s+20

Q20020*2二40

01-JS2+2S+20-JS2+2S+20S~(Jr+2s+20).s

AccordingtotheequationE=R-C:

%「($)=lim[%(s)—％(s)]=lim[-(l)]=lim?"+2)=。2

°八〃go%'，s2+2s+20蘇TOJd+2s+20

問題；

1.系統(tǒng)型為2,對于階躍輸入，穩(wěn)態(tài)誤差為0.

2.終值定理寫的不對。

andthetransferfunctionfromtheinputtotheoutputpositiontobewrittenas

。02=__1__

Js2+2s+20

g______'_____T=______-_____

02ZS-2+2.V+20".y(Jr+2.v+20)

Accordingtotheequatione=-c

essf(5)=lim[—％($)]=lim[sJ=lim十--=-0.05

s-八，四。1八￡一。1Jr+2S+20^JS2+2S+20

sothestateerrorshouldbe:ess=essr+ess/=0.2-0.05=0.15

Problem6.9

Solution:

Thetransferfunctionoftheinsideloopis

10

s(s+l)=1()

=一1+10-―1+(10k+1)5

s(s+1)

Andthetransferfunctionofthewholesystemis

C__________10________

R~52+(10^+1)5+10

Wegetthevalueofand0fromtheaboveequation:

wn=Vio

10&+1

AndthevalueofPOis

PO=lOOe-s/G

Accordingtotheformerequation:

PO=10%=f=0.6=嗎±1

2V10

Thefinalresultis

^=i.2Vio-i=O28

10

Module7

Problem7.5

Determinetheclosed-looptransferfunctionandthepercentageovershootforasystem

describedbythepole-zeromapshowninFig.P7.5,assumingthesteadystategainofthe

closed-looptransferfunctionisunity.

a1

SOLUTION：

Fromthefigure,wecangettheBodeformfunction

C_l+ST

元=%：+20/?！?1

r=13,=J(-2/+F=V5-g=-2

_?,/-=0.894^0.9/=—=-

網(wǎng)2

FromFig.7.6onpage117,wecanseethat,undertheeffectofclosed-loopzero,while

3

C=0.9and/=—,thePercentovershootislessthan1%.

2

C_____________1___________

?一(l+sr)(s2/o：+27s/q+l)

-l/r=-3=>r=l/3

=2,<y?=75?2.24n7=2/石*0.894,

C_15

--(s+3)(s?+4s+5)

Problem7.7

7.7Asystemhasatransferfunctionthatmaybewrittenintheform

C=1

R(5+l)(52+as+b)

Itisknownthatfbrthesecond-orderterm,4=0.2.Investigationoftheunitstepresponse

revealsanovershootof5%oftheinput.Calculatetheconstantsaandb,plottheclosed-loop

systempolesonthecomplexplane,andcommentonthereducabilityofthesystem.

Solution:

Thethird-ordersystemwithq=0.2wouldhaveanovershootof5%iftheadditionalpole

werelocatedsuchthattheparameterp=2.25.

Andfromtheknownparameters,1/=1,

then

=2.25（看圖似不超過2.1）

Thetime-domainresponseparametersmaybecalculateas

con=4rad/sand?=0.2

So

a=2〃"=1.6andb=冠=16

thenwewillplotclosed-loopsystempolesonthecomplexplane:

Theotherrealpolecannotbeneglected,theeffectofthereal-axispolecontributiontothe

responsewillbetomakeitmoresluggish.

Problem7.9

Solution:

Examinationoftheplantrevealsthefollowingtime-domainparameters:

(o=2radIsJ=1-=/?;

nT

ExaminingFig7.6,thesystemwith=1wouldhaveanovershootof5%ifthezerois

locatedsuchthattheparametery=0.6.Thisrelatedtotheparameterbby

b=l.2

Iftheinputisaunitstep,thesteady-stateoutputmaybeobtainedbyusingthefinal-value

theorem

$+12

limc(r)=limsC(s)=lim；=0.3

mo5->os->os2+4s+4

Asthereisa5%overshoot,themaximumvalueofthesystemis

(l+5%)xlimc(t)=0.3x105%=0.315

Problem7.11

k(s+1)

A0(S)==TT-=?S=_1；

l+GH]?k(s+l)(k+\)s+k

s

.Gk

B(p(s)==;nozeros;

l+GHl+k(s+l)

(s+1)(5+2)

C0($)=G=s(s+3)=______(s+l)(s+2)(s+4)______

°S~l+GH~?(s+l)(s+2)(s+5)-s(s+3)(s+4)+(s+l)(s+2)(s+5)

s(s+3)(s+4)

G5(52+25+10)

D(s+1);s=0,-l+3i;

OG)=2

l+GH1+(S+1)(S2：2S+10)5+(5+l)(5+25+10)

ModuIe8

Problem8.3

C_GG

8.3(1)

不一l+GH-1+G'

Resultingin

GG

G書上有結(jié)果(8.62)G=

1+(H+1)1+G(H-1)

1°01

TakeG=—,H=into

5+2s+55+1

Wegetthat

105+10

G=—：

s'+3A1?—3s+5

⑵AccordingtoTable8.1

K=limG(5)=2

It'satype0system,thesteady-stateerrorduetoaunitstepinput

]

e&=l+K-5

(3)Foraunitstepinput

RG)=L

Accordingto8.59

1+G(H-1)1+2*(1-1)1

◎(f)=1日——

.v->0\+GH1+2*13

Pmblem8.3

Solution

G⑸=7^7?H(S)==

Hence,theclosed-looptransferfunctionis

10

G(s)二s,+2s+510(5+