奥本海姆《信号与系统》第二版信号与系统答案

Signals&Systems(SecondEdition)—LearningInstructions(ExercisesAnswers)DepartmentofComputerEngineering2005.12ContentsChapter1·······················································2Chapter2·······················································17Chapter3·······················································35Chapter4·······················································62Chapter5·······················································83Chapter6·······················································109Chapter7·······················································119Chapter8·······················································132Chapter9·······················································140Chapter10·······················································1601Chapter1Answers1.1ConvertingfrompolartoCartesiancoordinates:1j111j11coscos()2e222e22jj2cos()jjsin()2cos()jjsin()e22e225jjjee22j42(cos()jjsin())12e4499jjjj444422ee1j22ee1jj2e41j1.2convertingfromCartesiantopolarcoordinates:j0j55e,22ej,33je2213jjjj2,12j4,1j2222eee4,1j22jjj(1)e4121je13je4t11.3.(a)=,=0,becauseEdtP0e4jt(2)2(b)()t4,(t)1.Therefore,=dt=dt=,xe2x2x2()t11TT2=lim()tdtlimdtlim11TT22TTTTx2T2(c)()t=cos(t).Therefore,==2=,x2()tdtcos(t)dtx3=1TT211COS(2t)1limcos(t)dtlimdtTT2TTTT222nnn(d)1,21.Therefore,=214[][]nun[]nun[][]nx1x1x124n043=0，because<.nj()2(e)[]n=28,=1.therefore,=2=,x2ex2[]nx2[]n11NN2=limlim11.NNx2[]n2NN1nN21nNn22(f)[]n=.Therefore,=2==,x3cosx3[]ncos(n)cos(n)444n1cos(n)=11N1Nlim(2)limcosNN2N1nN2221NnN41.4.(a)Thesignalx[n]isshiftedby3totheright.Theshiftedsignalwillbezeroforn<1,Andn>7.(b)Thesignalx[n]isshiftedby4totheleft.Theshiftedsignalwillbezeroforn<-6.Andn>0.(c)Thesignalx[n]isflippedsignalwillbezeroforn<-1andn>2.(d)Thesignalx[n]isflippedandtheflippedsignalisshiftedby2totheright.ThenewSignalwillbezeroforn<-2andn>4.(e)Thesignalx[n]isflippedandtheflippedandtheflippedsignalisshiftedby2totheleft.Thisnewsignalwillbezeroforn<-6andn>0.1.5.(a)x(1-t)isobtainedbyflippingx(t)andshiftingtheflippedsignalby1totheright.Therefore,x(1-t)willbezerofort>-2.(b)From(a),weknowthatx(1-t)iszerofort>-2.Similarly,x(2-t)iszerofort>-1,Therefore,x(1-t)+x(2-t)willbezerofort>-2.(c)x(3t)isobtainedbylinearlycompressionx(t)byafactorof3.Therefore,x(3t)willbezerofort<1.2(d)x(t/3)isobtainedbylinearlycompressionx(t)byafactorof3.Therefore,x(3t)willbezerofort<9.1.6(a)x1(t)isnotperiodicbecauseitiszerofort<0.(b)x2[n]=1foralln.Therefore,itisperiodicwithafundamentalperiodof1.(c)x3[n]isasshownintheFigureS1.6.111x3[n……]…-35…-40n…14……………Therefore,itisperiodic-1withafundamental-1period-1of4.……1.7.(a)……11[]n=[]n[]n([]unu[n4]u[]nu[n4])1xxx11v22Therefore,iszerofor[]n>3.x1(b)Sincex1(t)isanoddsignal,[]niszeroforallvaluesoft.vx2nn(c)1111[]n[][]nnu[3]nu[3]n3xx11xv2222Therefore,iszerowhenn<3andwhenn.3[n]vx1155tt(d)()t(()t())tu(2)tu(t2)44x1xxeev22Therefore,iszeroonlywhent.4()tvx1.8.(a){(tt)}220tcos(0)xe10t(b){(t)}2cos()cos(3t2)cos(3t)cos(3t0)xe24tt(c){(t)}sin(3t)sin(3t)x3ee222ttt(d){(t)}sin(100t)sin(100t)cos(100t)x4eee21.9.(a)()tisaperiodiccomplexexponential.x1j10tj10t()tj2x1ee(b)()tisacomplexexponentialmultipliedbyadecayingexponential.Therefore,x2()tx2isnotperiodic.jn7jn（c）[]nisaperiodicsignal.[]n==.x3x3ee2isacomplexexponentialwithafundamentalperiodof.22(d)[]nisaperiodicsignal.ThefundamentalperiodisgivenbyN=m()x43/510=m().Bychoosingm=3.Weobtainthefundamentalperiodtobe10.3(e)[]nisnotperiodic.isacomplexexponentialwith=3/5.Wecannotfindanyintegermx5w0suchthatm(2)isalsoaninteger.Therefore,[]nisnotperiodic.x5w01.10.x(t)=2cos(10t＋1)-sin(4t-1)PeriodoffirsttermintheRHS=2.105PeriodoffirsttermintheRHS=2.42Therefore,theoverallsignalisperiodicwithaperiodwhichtheleastcommonmultipleoftheperiodsofthefirstandsecondterms.Thisisequalto.372jnjn+41.11.x[n]=1e−e5PeriodoffirsttermintheRHS=1.PeriodofsecondtermintheRHS=2=7（whenm=2）4/7PeriodofsecondtermintheRHS=2=5(whenm=1)2/5Therefore,theoverallsignalx[n]isperiodicwithaperiodwhichistheleastcommonMultipleoftheperiodsofthethreetermsinnx[n].Thisisequalto35.1.12.Thesignalx[n]isasshowninfigureS1.12.x[n]canbeobtainedbyflippingu[n]andthenShiftingtheflippedsignalby3totheright.Therefore,x[n]=u[-n+3].ThisimpliesthatM=-1andno=-3.X[n]-3-2-10123n1.13FigureS1.12t0,t2ty(t)=x()dt=((2)(2))dt=,1,2t20,t22ThereforeEdt421.14Thesignalx(t)anditsderivativeg(t)areshowninFigureS1.14.x(t)g(t)1-11-10120t2t-3-3-2FigureS1.14Thereforeg(t)3(t2k)3(t2k1)kkThisimpliesthatA1=3,t1=0,A2=-3,andt2=1.1.15(a)Thesignalx2[n],whichistheinputtoS2,isthesameasy1[n].Therefore,1y[n]=x[n-2]+x[n-3]22=y[n-2]+y[n-3]1=2x[n-2]+4x[n-3]+(2x[n-3]+4x[n-4])121=2x[n-2]+5x[n-3]+2x[n-4]Theinput-outputrelationshipforSisy[n]=2x[n-2]+5x[n-3]+2x[n-4]4(b)Theinput-outputrelationshipdoesnotchangeiftheorderinwhichS1andS2areconnectedseriesreversed..WecaneasilyprovethisassumingthatS1followsS.Inthiscase,thesignalx1[n],whichistheinputtoSisthesameasy2[n].Thereforey1[n]=2x[n]+4x1[n-1]=2y[n]+4y[n-1]1=2(x[n-2]+x[n-3])+4(x[n-3]+x[n-4])22=2x[n-2]+5x[n-3]+2x[n-4]Theinput-outputrelationshipforSisonceagainy[n]=2x[n-2]+5x[n-3]+2x[n-4]1.16(a)Thesystemisnotmemorylessbecausey[n]dependsonpastvaluesofx[n].(b)Theoutputofthesystemwillbey[n]=[n][n2]=0(c)Fromtheresultofpart(b),wemayconcludethatthesystemoutputisalwayszeroforinputsoftheform[nk],kґ.Therefore,thesystemisnotinvertible.1.17(a)Thesystemisnotcausalbecausetheoutputy(t)atsometimemaydependonfuturevaluesofx(t).Forinstance,y(-)=x(0).(b)Considertwoarbitraryinputsx(t)andx(t).x(t)y(t)=x(sin(t))x(t)y(t)=x(sin(t))Letx3(t)bealinearcombinationofx(t)andx(t).Thatis,x(t)=ax(t)+bx(t)Whereaandbarearbitraryscalars.Ifx(t)istheinputtothegivensystem,thenthecorrespondingoutputy(t)isy(t)=x(sin(t))=ax(sin(t))+x(sin(t))=ay(t)+by(t)Therefore,thesystemislinear.1.18.(a)Considertwoarbitraryinputsx[n]andx[n].nn0x[n]y[n]=x1[k]knn0nn0x[n]y[n]=x2[k]knn0Letx[n]bealinearcombinationofx[n]andx[n].Thatis:x[n]=ax[n]+bx[n]whereaandbarearbitraryscalars.Ifx[n]istheinputtothegivensystem,thenthecorrespondingoutputnn0y[n]isy[n]=x3[k]knn0nn0nn0=(ax1[k]bx2[k])=ax1[k]+bknn0knn0=ay1[n]+by[n]Thereforethesystemislinear.(b)Consideranarbitraryinputx[n].Let5nn0y1[n]=x1[k]knn0bethecorrespondingoutput.Considerasecondinputx2[n]obtainedbyshiftingx[n]intime:x[n]=x[n-n1]Theoutputcorrespondingtothisinputisnn0nn0nn1n0y2[n]=x2[k]=x1[kn1]=x1[k]knn0knn0knn1n0Alsonotethaty[n-n]=.Therefore,y2[n]=y[n-n]Thisimpliesthatthesystemistime-invariant.(c)Ifx[n]1,buty(t)=1fort>1.1.41.(a)y[n]=2x[n].Therefore,thesystemistimeinvariant.(b)y[n]=(2n-1)x[n].Thisisnottime-invariantbecausey[n-N0]≠(2n-1)2x[n-N0].(c)y[n]=x[n]{1+(-1)n+1+(-1)n-1}=2x[n].Therefore,thesystemistimeinvariant.1.42.(a)ConsidertwosystemS1andS2connectedinseries.Assumethatifx1(t)andx2(t)aretheinputstoS1..theny1(t)andy2(t)aretheoutputs.respectively.Also,assumethatify1(t)andy2(t)aretheinputtoS2,thenz1(t)andz2(t)aretheoutputs,respectively.SinceS1islinear,wemaywrites1ax1tbx2tay1tby2t,whereaandbareconstants.SinceS2isalsolinear,wemaywrites2ay1tby2taz1tbz2t,Wemaythereforeconcludethata(t)b(t)s1s2a(t)b(t)x1x2z1z2Therefore,theseriescombinationofS1andS2islinear.SinceS1istimeinvariant,wemaywrites1x1tT0y1tT0ands2y1tT0z1tT0Therefore,ss12x1tT0z1tT0Therefore,theseriescombinationofS1andS2istimeinvariant.(b)False,Lety(t)=x(t)+1andz(t)=y(t)-1.Thesecorrespondstotwononlinearsystems.Ifthesesystemsareconnectedinseries,thenz(t)=x(t)whichisalinearsystem.14(c)Letusnametheoutputofsystem1asw[n]andtheoutputofsystem2asz[n].Then11y[]nz[2]nw[2]nw[2n1]w[2n2]2411xnxn1xn224Theoverallsystemislinearandtime-invariant.1.43.(a)Wehavextsy(t)SinceSistime-invariant.xtTsy(tT)Nowifx(t)isperiodicwithperiodT.x{t}=x(t-T).Therefore,wemayconcludethaty(t)=y(t-T).Thisimpliesthaty(t)isalsoperiodicwithT.Asimilarargumentmaybemadeindiscretetime.(b)1.44(a)Assumption:Ifx(t)=0fortt0,x2(t)≠x1(t),Sincethesystemislinear,x1tx2ty1ty2t,Sincex12txt0fort