信號與系統(tǒng)奧本海姆英文版課后答案chapter10

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Chapter10Answers

10.1(a)Thegivensummationmaybewrittenas

,byreplacingzwith.Ifr<,then

Andthefunctionwithinthesummationgrowstowardsinfinitywithincreasingn..Also,thesummationdosenotconverge.

Butifr>,thenthesummationconverges.

(b)Thegivensummationmaybewrittenas

,byreplacingzwith.Ifr>,then

Andthefunctionwithinthesummationgrowstowardsinfinitywithincreasingn,.Also,thesummationdosenotconverge.

Butifr<,thenthesummationconverges.

(c)Thesummationmaybewrittenas.byreplacingzwith.Ifr>,thenthe

functionwithinthesummationgrowstowardsinfinitywithincreasingn,.Also,thesummationdosenotconverge.

Butifr<,thenthesummationconverges.

Thesummationmaybewrittenas

byreplacingzwith.Thefirstsummationconvergesforr>.Thesecondsummationconvergeforr>2.Therefore,thesumofthese

twosummationsconvergesfor1/2<r<2.

10.2Usingeq.910.3).

,

10.18.(a)usingtheanalysisofexample10.18,wemayshowthat

Sinceh(z)=y(z)/x(z),wemaywrite

Takingtheinversez-transformweobtain

(b)H(Z)hasonlytwopoles,thesearebothatz=1/3.Sincethesystemiscausal,theROCofH(Z)willbetheform>1/3.SincetheROCincludestheunitcircle,thesystemisStable.

10.19.(a)Theunilateralz-transformis

(b)Theunilateralz-transformis

Allz

Theunilateralz-transformis

10.20.Applyingtheunilateralz-transformtogivendifferenceequation,wehave

Forthezero-inputresponse,assumethatx[n]=0.Sincewearegiventhaty[-1]=2,

Takingtheinverseunilateralz-transform,

Y[n]=

Forthezero-stateresponsesety[-1]=0.Also,wehave

Therefore,

Weusepartialfractionexpansionfollowedbytheinverseunilateralz-transformtoobtain

(c)Thetotalresponseisthesumofthezero-inputresponse.Thisis

10.21.thepole–zeroplotsareallshowninfigureS10.21.

(a)For,allz.

TheFouriertransformexistsbecausetheROCincludestheunitcircle.

Forx[n]=

X(z)=,allzexpect0.

TheFouriertransformexistsbecausetheROCincludestheunitcircle.

Forx[n]=

TheFouriertransformdoesnotexistbecausetheROCdoesnotincludetheunitcircle

Forx[n]=

Forx[n]=

TheFouriertransformdoesnotexistbecausetheROCdoesnotincludetheunitcircle.

For

TheFouriertransformdoesnotexistbecausetheROCincludestheunitcircle.

(g)Consider

Consider

Thez-transformoftheoverallsequencex[n]=is

TheFouriertransformexistbecausetheROCincludetheunitcircle.

Considerx[n]=

TheFouriertransformexistsbecausetheROCincludestheunitcircle.

1

FigureS10.21

10.22.(a)Usingthez-transformanalysisequation,

Thismaybeexpressas

Thishasfourzeroatz=0and8morezerodistributedonacircleofradius1/2.TheROCistheentirezplane.(AlthoughformaninspectionofexpressionforX(z)itseemsliketheseisapoleat1/2whichcancelswiththispole.)SincetheROCincludestheunitcircle,theFouriertransformexists.

(b)Considerthesequence

Now,

And

Therefore,

Notethatx[n]=n.Therefore,

TheROCis(1/2)<<2.Therefore,theFouriertransformexists.

(c)Writex[n]as

Where

And

Usingthedifferentiationproperty,weget

TheROCis(1/2)<.Therefore,theFouriertransformexists.

(d)Thesequencemaybewrittenas

Now

And

Therefore,

X(z)=+

TheROCis.Therefore,theFouriertransformexists.

10.23（i）Thepartialfractionofthegivenis

SincetheROCis.

.

Performinglong-divisioninordertogetaright-sidedsequence.weobtain

Thismayberewrittenas

Therefore.

.

(ii)Thepartialfractionofthegivenis

SincetheROCis.

. .

Performinglong-divisioninordertogetaleft-sidedsequence.weobtain

Thismayberewrittenas

Therefore.

(iii).Thepartialfractionofthegivenis

SincetheROCis.

Performinglong-divisioninordertogetaright-sidedsequence.weobtain

Thismayberewrittenas

.

Therefore.

(iv)Thepartialfractionofthegivenis

SincetheROCis

Performinglong-divisioninordertogetaleft-sidedsequence.weobtain

Thismayberewrittenas

.

Therefore.

.

(v)Thepartialfractionofthegivenis

.

(vi)Wemaysimilarlyshowthatinthiscase,

.

10.24(a)Wemaywriteas

.

Therefore

.

Ifisabsolutelysummable,thentheROCofhastoincludetheunitcircle

Therefore,theROCis.Itfollowthat

.

(b)Carryingoutlongdivisionon,weget

.

Usingtheanalysisequation(10.3),weget

(c)Wemaywriteas

.

Thepartialfractionofis

.

Sinceisabsolutesummable,theROCmustbeinordertoincludetheunitcircle.Itfollowthat

10.25.(a)Thepartialfractionequationofis

.

Sinceisabsolutesummable,theROCmustbeinordertoincludetheunitcircle.Itfollowthat

.

(b)mayberewrittenas

.

Usingpartialfractionexpansion,wemayrewritethisas

.

Ifisright-side,thentheROCforthissignalis.Usingthisfact,wemay

Findtheinversez-transformofthetermwithinsquarebracketabovetobe.Notethat.Therefore,Thisgives

Notingthat.wemayrewritethisas

Thisistheanswerthatweobtainlyinpart(a).

10.26（a）Formpart(b)ofthepreviousproblem,

.

(b)Formpart(b)ofthepreviousproblem,

.

(c)Ifisleft-side,thentheROCforthissignalis.Usingthisfact,wemay.Findtheinversez-transformofthetermwithinsquarebracketabovetobe.Notethat.Therefore,Thisgives

.

10.27Weperformlong-divisiononsoastoobtainaright-sedersequence.Thisgivesus

Therefore,comparingthiswitheq.(10.3)weget

Andforn<-3

10.28(a)Usingeq(10.3),weget

(b)ThereforeX(z)haszeroslyingonacircleofradius0.95(asshownisFigureS10.28)and6polesatz=0

1

6

Re

FigureS10.28

(c)ThemagnitudeFouriertransformisasshowninFigureS10.28

10.29TheplotsareasshowninFigureS10.29

0

0

-

0

0

0

FigureS10.29

10.30Fromthegiveninformation,wehave

And

Usingthetimeshiftingproperty,weget

Usingthetimereversalandproperties,weget

Now,usingtheconvolutionproperty,weget

Therefore

10.31FromClue1,weknowthatisreal.Thereforethepolesandzerosofhavetooccurinconjugatepairs,SinceClue4tellsusthathasapoleat.wecanconcludethatmusthaveantherpoleat.Nowsincehas

Nomorepoles,wehavetoassumethathas2orlesszeros.Ifhasthan2zeros,thenmusthavepolesatinfinity,SinceClue3tellusthathas2zerosattheorigin,weknowthatmustbeofthefrom

Sinceisright-sided,theROCmustbe.

10.32.(a)Wearegiventhatandtherefore

Now.y[n]maybeevaluatedtobe

Simplifying

(b)UsingTable10.2weget

H(z)=|z|>|a|

And

X(z)=Allz,

Therefore

Y(z)=X(z)H(z)=

TheROCis|z|>|a|,Consider

P(z)=

WithROC|z|>|a|,ThepartialfractionexpansionofP(z)is

P(z)=

Therefore,

P[n]=

Now,notethat

Y(z)=P(z),

Therefore,

Thismaybewrittenas

Thisisthesameastheresultofpart(a).

10.33.(a)Takingthez-transformofbothsidesofthegivedifferenceequationandsimplifying,

Weget

ThepolesofH(z)areat(1/4)Sinceh[n]iscausal,theROChastobe

(b)Wehave

Therefore,

TheROCofY(z)willbetheintersectionoftheROCsofX(z)andH(z).ThisimpliesthattheROCofY(z)is|z|>1/2.ThepartialfractionexpansionofY(z)is

UsingTable10.2weget

10.34.(a)Takingthez-transformlfbothsidesofthegivedifferenceequationandsimplifying,weget

FigureS10.34

ThepolesofH(z)areatz=H(z)hasazeroatz=0.Thepole-zeroplotforH(z)isasshowninFigureS10.34.sinceh[n]iscausal,ROCforH(z)hastobe|z|>

(b)ThepartialfractionexpansionofH(z)is

Therefore,

(c).NowassumingthattheROCisweget

10.35.Takingthez-transformofbothsidesofthegivendifferenceequationandsimplifying.Weget

ThepartialfractionexpansionofH(z)is

IftheROCis|z|>2,then

IftheROCis1/2<|z|<2,then

IftheROCis|z|<1/2,then

Foreach,wenowneedtoshowthatify[n]=inthedifferenceequation,thenx[n]=.Considersubstitutingintothedifferenceequation.Thisyields

Then,

=0,forn<-1,

=2/3-2/3=0,

=0,forn>0.

Itfollowsthat=.itcansimilarlybeandsatisfythedifferenceequation.

10.36.Takingthez-transformofbothsideofthegivendifferenceequationandsimplifying,weget

ThepartialfractionexpansionofH(z)is

SinceH(z)correspondstoastablesystem,theROChasrobe(1/3)<|z|<3.Therefore,

10.37.(a)Theblock-diagrammayberedrawnasshowinpart(a)ofthefigurebelow.ThismaybetreatedasacascadeofthetwosystemsshownwithinthedottedlinesinFigureS10.37.Thesetowsystemsmaybeinterchangedasshowninpart(b)ofthefigureFigureS10.37withoutchangingthesystemfunctionoftheoverallsystem.Fromthefigurebelow,itisclearthat

SystemI

SystemII

y[n]

-1/3

2/9

9/8

x[n]

x[n]

y[n-1]

y[n-1]

9/8

-1/3

2/9

x[n-1]

FigureS10.37

y[n]

(b)Takingthez-transformoftheabovedifferenceequationandsimplifying,weget

H(z)haspolesatz=1/3andz=-2/3.Sincethesystemiscausal.TheROChastobe|z|>2/3.TheROCincludestheunitcircleandhencethesystemisstable.

10.38.(a)

(b)

Usingtheresultsofpart(a)and(d),wemayredrawtheblock-diagramasshowninFigureS10.38.

y[n]

x[n]

-1/2

1/4

-2

1/4

x[n]

y[n]

-2/3

-z/4

1/8

-1/2

x[n]

y[n]

5/3

-14/3

4

-1/2

1/4

FigureS10.38

(d)Usingtheapproachshownintheexamplesinthetextbookwemaydrawtheblock-diagramofandasshowninthedottedboxesinthefigurebelow.H(z)isthecascadeofthesetwosystems.

(e)Usingtheapproachshowintheexamplesshowninthetextbook,wemaydrawtheblock-diagramof,andasshowninthedottedboxesinthefigurebelow.H(z)istheparallelcombinationof,,.

10.39.(a)Thedirectformblockdiagrammaybedrawnasshowninpart(a-i)ofFigureS10.39bynotingthat

Thecascadeblock-diagramisasshowninpart(a-ii)ofFigureS10.39.

A

B

C

D

x[n]

y[n]

r

x[n]

y[n]

FigureS10.39

Part(a):A=5/3,B=11/36.C=15/54,D=1/36

Part(b):A=3/2,B=-1.C=5/4,D=-1/2

Part(b):A=2,B=-z/4.C=3/4,D=-1/8

Part(a):=1,=-1/4,r=2/3,=-1/9

Part(a):=1,=-1/2,r=1/2=-1

Part(a):=1,=-1/2,r=1,=-1/4

Notethat

Therefore,maybedrawnasacascadeoffoursystemsforwhichthecoefficientmultipliersareallreal.

(b)Thedirectformblockdiagrammaybedrawnasshowninpart(b-i)ofFigureS10.39bynotingthat

Thecascadeblock-diagramisasshowninpart(b-ii)ofFigureS10.39.

Nowthat

Therefore,cannotbedrawnascascadeoffoursystemsforwhichthecoefficientmultipliersareallreal.

(c)Thedirectformblockdiagrammaybedrawnasshowninpart(c-i)oftheFigureS10.39bynotingthat

Thecascadeblock-diagramisasshowninpart(c-ii)oftheFigureS10.39.

Notethat

Therefore,cannotbedrawnasacascadeoffoursystemsforwhichthecoefficient

Multipliersareallreal.

10.40.Thedefinitionoftheunilateralz-transformis

sincex[n]=[n-5]iszerointherange0n,x(z)=0.

TheunilateralofLaplacetransformofx[n]=[n-5]

Is

TheunilateralofLaplacetransformofx[n]=is

,>1

(d)TheunilateralofLaplacetransformofis

,>1/2.

(e)Sincex[n]=(-1/3)u[-n-2]iszerointherange,x(z)=0.

(f)TheunilateralofLaplacetransformofis

,Allz.

(g)TheunilateralofLaplacetransformofis

,Allz.

(h)TheunilateralofLaplacetransformofis

>1/2.

10.41.fromthegiveninformation,

>1/2.

And

>1/4.

UsingTable10.2andthetimeshiftpropertyweget

,>1/2.

And

>1/4.

Wehave

.

TheROCis|z|>(1/2).ThepartialfractionexpansionofG(z)is

.

UsingTable10.2andthetimeshiftproperty,weget

.

Wehave

TheROCofQ(z)is|z|>(1/2).Thepartialfractionexpansionofy(z)is

Therefore,

Clearly,q[n]g[n]forn>0.

(a)Takingtheunilateralz-transformofbothsidesofthegivendifferenceequation.weget

Settingx(z)=0,weget

Theinverseunilateralz-transformgivesthezero-inputresponse

Now,sinceitisgiventhatwehave

|z|>1/2.

Settingy[-1]tobezero,weget

Therefore,

Thepartialfractionexpansionofis

Theinverseunilateralz-transformofbothsidesofthegivendifferenceequation,weget

Setting,weget

IM

Re

Y(z)=0.

Theinverseunilateralz-transformgivesthezero-inputresponse

Now,sinceitisgiventhatwehave

|z|>1.

Settingy[-1]tobezero,weget

FigureS10.46

Therefore,

Theinverseunilateralz-transformgivesthezero-stateresponse

Takingtheunilateralz-transformofbothsidesofthegivendifferenceequation,weget

Setting,weget

10.46.Takingthez-transformofbothsidesofthedifferenceequationrelatingx[n]ands[n]andsimplifying,weget

Thesystemhasan8thorderpoleatz=0and8zerosdistributedaroundacircleofradius.ThisisshowinFigureS10.46.TheROCiseverywhereonthez-planeexceptatz=0.

(b)Wehave

Therefore,

TherearetwopossibleROCsfor:or.IftheROCis,thentheROCdosenotincludetheunitcircle.Thisinturnimpliesthatthesystemwouldbeunstableandanti-causal.IftheROCis,thentheROCincludestheunitcircle.Thisinturnimpliesthatthesystemwouldbestableandcausal.

Wehave

.

WeneedtochoosetheROCtobeinordertogetsstablesystem.Nowconsider

WithROC.Takingtheinversez-transform,weget

.

Now,notethat

.

FromTable10.1weknowthat

0,otherwise

10.47.(a)FromClue1,wehaveH(-2)=0.FromClue2,weknowthatwhen

Wehave

.

Therefore,

Substitutingz=-2intheaboveequationandnothingthatN(-2)=0,weget

.

(b)Theresponsetothesignalx[n]=1=.Therefore,

.

10.48.fromthepole-zerodiagram,wemanywrite

And

WhereAandBareconstants.nownotethat

Usingtheproperty10.53ofthez-transform(seeTable10.1),weget

Wemayrewritethisas

Where.Notethatsincebothandarecausal.Wemayassumethatg[n]=0forn<0.Therefore.

u[n].

Now,clue3slaostatesthat.Therefore.

Or

Therefore,

g[n]=(-)u[n].

10.49.(a)Wemaywritetheleftsideofeq.(p10.49-1)as

(r)=r()

Sincer;thesequence(r)decayswithincreasingn.i.e.,asn->

(r)->0.Therefore,(r)(r)forn.Substitutingthisin

eq.,weget

r()r

Therefore,A=(r/r)=(r/r).

(b)TheaboveinequalityshowsthatifX(z)hasthefiniteboundBfor|z|=r;thenX(z)hasthefinitebound(r/r)Bfor|z|=r.Thus,X(z)convergesfor|z|=r.andProperty4ofSection10.2follows.

(c)Consideraleft-sidedsequencex[n]suchthat

x[n]=0,n>N

andforwhich

=

Thenweneedtoshowthatifr,

|x[n]|rP(s10.49-2)

WhenPisapositiveconstant.

Wemaywritetheleftsideofeq.(s10.49-2

|x[n]|r=|x[n]|(r)=(s10.49-3)

Since;thesequence(r)decayswithdecreasingn,i.e.,as

n->(r)->0.Therefore,(r)(r)for.n

.Substitutingthisineq(s10.49-3).Weget

|x[n]|r=|x[n]|(r)

Therefore,P=(r/r)=(r/r)

TheaboveinequalityshowthatifX(z)hasifX(z)hasthefiniteboundBfor|z|=r.thenX[z]hasthefinitebound(r/r)Bfor|z|=rr.Thus,X(z)convergesfor|z|=rrandProperty5ofSection10.2follows.

10.50.(a)Fromthegivenpole-zeroplot,weget

H(z)=A,

WhereAissomeconstant.Therefore

H(e)=A

And

|H(e)|=H(e)H=|A|[][]

Therefore

|H(e)|=|A|=|A|.

Thisimpliesthat|H(e)=|A|=constant.

(b)Weget|v|=1+a-2acos().

(c)Weget

|v|=1+-=[a+1+2acos]=|v|

10.51.（a）Weknowthatforarealsequencex[n],x[n]=x.Letusfirstfindthez-transformofxintermsofX[z],thez-transformofx[n].Wehave

Y(z)=y[n]z=xz

=[x[n](z)]=[X(z)].

Nowsincex[n]=x,wehaveZ{x[n]}=z{x}whichinturnimpelsthatX(z)=X(z).

(b)IfX(z)hasopoleatz=z.then1/X(z)=0.Fromtheresultofthepreviouspart,

Weknowthat=0.

Conjugatingbothsides,weget1/=0.thisimpliesthatX(z)hasapoleatz.

IfX(z)hasazeroatz=z.thenX(z)=0.Fromtheresultofthepreviouspart,

Weknowthat

=0.

Conjugatingbothsides,wegetX(z)=0.ThisimpliesthatX(z)hasazeroatz.

(c)(1)Thez-transformofgivensequenceis

X(z)=,|z|>1/2

Clearly,X(z)hasapoleatz=1/2andazeroatz=0andthepropertyofpart(b)holds.

(2)Thez-transformofthegivensequenceis

X(z)=

X(z)hastwozeroatz=1/2andzzeroatz=0andthepropertyofpart(b)stillholds

(d)Now,frompart(b)ofproblem10.43weknowthatifx[n]andX(z)hasapoleatz,thenX(z)hasapoleat(1/z)=

Ifx[n]isrealandX(z)hasapoleatz,thenfrompart(b)weknowthatX(z)musthaveapoleatz=.,thenX(z)musthaveapoleat.

Asimilarargumentmaybeconstructedforzeros.

10.52.Wehave

=

=

=.

Usinganargumentsimilartotheoneusedonpart(b)ofproblem10.43.wemayarguethatifhasapole(orzero)atz=,thenwhichhasapole(orzero)atz=1/z.

10.53.Letusassumethatx[n]isasequencewithz-transformX(z)whichhastheROC.

(a)(1)Thez-transformofthesequencey[n]=[n-n]is

Y(z)=

=

Substitutingm=n-nintheaboveequation,weget

Y(z)====

=X(z).

Clearly,Y(z)convergeswhereX(z)convergesexceptfortheadditionordeletionofz=0

becauseoftheterm.Therefore,theROCis.exceptforthepossibleadditionofz=0intheROC.

(2)Thez-transformerthesequencey[n]=is

Y(z)=

=

=

=X()

SinceX(z)convergesfor,Y(z)convergesfor,Therefore,

theROCofY(z)is.

(b)(1)FromProblem10.51(a);weknowthatthez-transformofthesequencey[n]=isY(z)=.TheROCofY(z)isthesameastheROCofX(z).

(2)SupposethattheROCofx[n]is,Fromsubpart(2)ofpart(a).thez-transformofis

Y(z)=X()

WithROC.Therefore.

10.54.(a)Letx[n]=0forn>0.Then,

=

=x[0]+x[-1]z+x[-2]z+…….

Therefore

.

(b)Letx[n]=0forn<0.Then,

=

=x[0]+x[-1]z+x[-2]z+…….

Therefore,

=x[1].

10.55.(a)Fromtheinitialvaluetheorem,wehave

X(z)=x[0]non-zeroandfinite.

Therefore,asz->,X(z)tendstoafinitenon-zerovalue.ThisimpliesthatX(z)hasneitherpoleszerosatinfinity.

(b)Arationalz-transformismadeupoffactorsoftheform1/(z-a)and(z-b).Notethatthefactor1/(z-b)hasapoleatz=aandazeroatz=.Alsonotethatthefactor(z-b)hasazeroatz=.Fromtheresultofpart(a),weknowthatacausalsequencehasnopoleorzeroatinfinity.Therefore,allzerosatinfinitycontributedbyfactorsoftheform1/(z-a).Consequently,thenumberofzerosinthefinitez-planemustequalthenumberofpolesinthefinitez-plane.

10.56.(a)Thez-transformofis

=

=]

=

=

(b)Usingthetimeshiftingproperty(10.5.2),weget

Where(z)isthez-transformof.Substitutingintheresultofpart(a),weget

Nothingthatthez-transformofmaybewrittenas

=;

Wemayrewritetheresultofpart(b)as

.

10.57.(a)isapolynomialoforderin.isapolynomialoforderin.Therefore,isapolynomialoforderin.thisimpliesthatM=.

(b)Bynothingthaty[0]isthecoefficientofterminY(z).y[1]isthecoefficientofterminY(z),andy[2]isthecoefficientofterminY(z).weget

y[0]=,

y[1]=,

y[3]=.

(c)wenotethepatternthatemergeformpart(b).Thek-thpointinthesequencey[n]isthecoefficientofinY(z),ThetermofY(z)isformedbythefollowingsum:(theproductofthetermofwiththetermof)+(theproductofthetermofwiththetermof)+(theproductofthetermofwiththetermof)+…….+(theproductofthetermofwiththetermof).

Therefore,

y[k]=

Since[m]=0form>andm<0,wemayrewritethisas

.

10.58Consideracausalandstablewithsystemfunction.Letitsinversesystemhavethefunction.Thepolesofarethezerosofandthezerosofarethepolesof.

Fortocorrespondtobeacasualandstablesystem,allitspolesmustbewithintheunitcircle.Similarly,fortocorrespondtobeacasualandstablesystem,allitspolesmustbewithintheunitcircle.Sincethepolesofarethezerosof,thepreviousstatementimpliesthatthezerosofmustbewithintheunitcircle.Therefore,allpolesandzerosofaminimum-phasesystemmustbewithintheunitcircle.

10.59(a)FromFigureS10.59,wehave

==>=

Also

willbe

Finally,

Sincecorrespondstoacausalfilter,theROCwillbe|z|>|k|/3.

(b)Forthesystemtobestable,theROCofmustincludetheunitcircle.Thisispossibleonlyif|k|/3<1.Thisisimpliesthat|k|hastobelessthan3.

(c)Ifk=1,then

=

Theresponsetox[n]=willbeoftheform

y[n]=x[n]H(2/3)=.

10.60.Theunilateralz-transformofy[n]=x[n+1]is