信号与系统奥本海姆英文版课后答案chapter10

178Chapter10Answers10.1(a)Thegivensummationmaybewrittenas1111()22jwnnrne,byreplacingzwithjwre.Ifr12,then1112rAndthefunctionwithinthesummationgrowstowardsinfinitywithincreasingn..Also,thesummationdosenotconverge.Butifr12,thenthesummationconverges.(b)Thegivensummationmaybewrittenas1111()22jwnnrne,byreplacingzwithjwre.Ifr12,then21rAndthefunctionwithinthesummationgrowstowardsinfinitywithincreasingn,.Also,thesummationdosenotconverge.Butifr12,thenthesummationconverges.(c)Thesummationmaybewrittenas0()2nnjwnnrre.byreplacingzwithjwre.Ifr1,thenthefunctionwithinthesummationgrowstowardsinfinitywithincreasingn,.Also,thesummationdosenotconverge.Butifr1,thenthesummationconverges.(a)Thesummationmaybewrittenas01011()cos(/4)()cos(/4)22njwnnjwnnnrnernebyreplacingzwithjwre.Thefirstsummationconvergesforr12.Thesecondsummationconvergeforr2.Therefore,thesumofthesetwosummationsconvergesfor1/2r2.10.2Usingeq.910.3).311()()[3]()55nnnnnnXzunzz331011[]()[]1125512515nnnzzzz,15z10.18.(a)usingtheanalysisofexample10.18,wemayshowthat1222168()21139ZZHZZZSinceh(z)=y(z)/x(z),wemaywrite121221()[1]()[168]39YZZZXZZZTakingtheinversez-transformweobtain21[][1][2][]6[1]8[2]39ynynnxnxnxn(b)H(Z)hasonlytwopoles,thesearebothatz=1/3.Sincethesystemiscausal,theROCofH(Z)willbetheformz1/3.SincetheROCincludestheunitcircle,thesystemisStable.10.19.(a)Theunilateralz-transformis01()()[5]4nnNxZunz01()4nnnz111,141()4zz(b)Theunilateralz-transformis17900()([3][]2[])(0[][]2,nnnnnxznnnznnzAllzTheunilateralz-transformis011()()21()211,1212nnnnnxzzzzz10.20.Applyingtheunilateralz-transformtogivendifferenceequation,wehave1()[1]2()().zyzyyzyz(a)Forthezero-inputresponse,assumethatx[n]=0.Sincewearegiventhaty[-1]=2,111()(1)2()0().11()2zyzyyzyzzTakingtheinverseunilateralz-transform,Y[n]=1()[].2nn(b)Forthezero-stateresponsesety[-1]=0.Also,wehave1111(){()[],.12212nxzznzzTherefore,1112()()().1214yzzzWeusepartialfractionexpansionfollowedbytheinverseunilateralz-transformtoobtain1111[]()[]()[].3264nnynnn(c)Thetotalresponseisthesumofthezero-inputresponse.Thisis2111[]()[]()[].3264nnynnn10.21.thepole–zeroplotsareallshowninfigureS10.21.(a)For[][5],xnn5()xzz,allz.TheFouriertransformexistsbecausetheROCincludestheunitcircle.(b)Forx[n]=[5],nX(z)=5z,allzexpect0.TheFouriertransformexistsbecausetheROCincludestheunitcircle.(c)Forx[n]=(1)[],nn()[]nnXZxnz0(1)nnnz11/(1),1zzTheFouriertransformdoesnotexistbecausetheROCdoesnotincludetheunitcircle(d)Forx[n]=11()[3],2nn()[]nnXzxnz18013230311()21()241,12(1())2nnnnnnzzzzz(e)Forx[n]=1()[2],3nn()[]nnxzxnz2222021(1/3)1()31()39/(13),1/33,1/3(1(1/3))nnnnnnnnnzzzzzzzzTheFouriertransformdoesnotexistbecausetheROCdoesnotincludetheunitcircle.(f)For[](1/4)[3],nxnn()[]nnxzxnz33330341(1/4)(1/4)(1/4)(1/64)/(14),1/4(1/16)/(1(1/4)),1/4nnnnnnnnnzzzzzzzzzTheFouriertransformdoesnotexistbecausetheROCincludestheunitcircle.(g)Consider1()2[].nxzn11()[]nnxzxnz0011(2)(2)1/(1(1/2)),22/(12),2nnnnnnzzzzzzzConsider2[](1/4)[1].nxnn22()[]nnxzxnz111011(1/4)(1/4)/4[1/(1(1/4))],1/4nnnnnnzzzzzThez-transformoftheoverallsequencex[n]=12[][]xnxnis11112/4(),(1/4)2(12)1(1/4)zzxzzzzTheFouriertransformexistbecausetheROCincludetheunitcircle.181(g)Considerx[n]=2(1/3)[2].nn()[]nnxzxnz222021(1/3)(1/3)[1/(1(1/3))],1/3nnnnnnzzzzzTheFouriertransformexistsbecausetheROCincludestheunitcircle.FigureS10.2110.22.(a)Usingthez-transformanalysisequation,443322110011223344()(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)XzzzzzzzzzzThismaybeexpressas994411(1/2)()(1/2)[].1(1/2)zXzzzThishasfourzeroatz=0and8morezerodistributedonacircleofradius1/2.TheROCistheentirezplane.(AlthoughformaninspectionofexpressionforX(z)itseemsliketheseisapoleat1/2whichcancelswiththispole.)SincetheROCincludestheunitcircle,theFouriertransformexists.(b)Considerthesequence1[](1/2)[]2[1].nnxnununNow,(1/2)[]nunz11,(1/2)1(1/2)zzAnd(2)[1]nunz11,2.12zzTherefore,11111(),(1/2)2.1(1/2)12xzzzzNotethatx[n]=n1[]xn.Therefore,1111212(1/2)2()().(1(1/2))(12)dZZXzzXZdzZZTheROCis(1/2)z2.Therefore,theFouriertransformexists.(c)Writex[n]as12(1/2)[]2[1][][]nnnunnunnxnnxn053norduIeRMIMIeRnS()a()beRMI()deR()c4nMIMIeR023ndorduu()eMIeR03rdorduku()f1MIeR()geRMI()h182Where1[](1/2)[]nxnunz111(),1/21(1/2)XzzzAnd2[](2)[1]nxnunz211(),212XZZZUsingthedifferentiationproperty,weget11121212(1/2)2()()().(1(1/2))(12)ddzzXZzXzzXzdzdzzzTheROCis(1/2)2z.Therefore,theFouriertransformexists.(d)Thesequencemaybewrittenas[(2/6)(/4)][]4{}[1].2jnnexnunNow/42/611,4214jjezez[(2/6)(/4)]4[1]njneunz/42/611,4214jjezezAnd[(2/6)(/4)]4[1]njneunz/42/611,4214jjezezTherefo