(完整版)信号与系统奥本海姆-习题答案

Chapter1Answers1.6(a).NoBecausewhent0,)(1tx=0.(b).NoBecauseonlyifn=0,][2nxhasvaluable.(c).YesBecausekkmnkmnmnx]}414[]44[{]4[kmknmkn)]}(41[)](4[{kknkn]}41[]4[{N=4.1.9(a).T=/5Because0w=10,T=2/10=/5.(b).Notperiodic.Becausejtteetx)(2,whileteisnotperiodic,)(2txisnotperiodic.(c).N=2Because0w=7,N=(2/0w)*m,andm=7.(d).N=10Becausenjjeenx)5/3(10/343)(,thatis0w=3/5,N=(2/0w)*m,andm=3.(e).Notperiodic.Because0w=3/5,N=(2/0w)*m=10m/3,it’snotarationalnumber.1.14A1=3,t1=0,A2=-3,t2=1or-1dttdx)(isSolution:x(t)isBecausekkttg)2()(,dttdx)(=3g(t)-3g(t-1)ordttdx)(=3g(t)-3g(t+1)1.15.(a).y[n]=2x[n-2]+5x[n-3]+2x[n-4]Solution:]3[21]2[][222nxnxny]3[21]2[11nyny]}4[4]3[2{21]}3[4]2[2{1111nxnxnxnx]4[2]3[5]2[2111nxnxnxThen,]4[2]3[5]2[2][nxnxnxny(b).No.Forit’slinearity.therelationshipbetween][1nyand][2nxisthesamein-outrelationshipwith(a).youcanhaveatry.1.16.(a).No.Forexample,whenn=0,y[0]=x[0]x[-2].Sothesystemismemory.(b).y[n]=0.Whentheinputis][nA,then,]2[][][2nnAny,soy[n]=0.(c).No.Forexample,whenx[n]=0,y[n]=0;whenx[n]=][nA,y[n]=0.Sothesystemisnotinvertible.1.17.(a).No.Forexample,)0()(xy.Soit’snotcausal.(b).Yes.Because:))(sin()(11txty,))(sin()(22txty))(sin())(sin()()(2121tbxtaxtbytay1.21.Solution:Wehaveknown:(a).(b).(c).(d).1.22.Solution:Wehaveknown:(a).(b).(e).(g)1.23.Solution:For)]()([21)}({txtxtxEv)]()([21)}({txtxtxOdthen,(a).(b).(c).1.24.Solution:For:])[][(21]}[{nxnxnxEv])[][(21]}[{nxnxnxOdthen,(a).(b).(c).1.25.(a).Periodic.T=/2.Solution:T=2/4=/2.(b).Periodic.T=2.Solution:T=2/=2.(d).Periodic.T=0.5.Solution:)}()4{cos()(tutEtxv)}())(4cos()()4{cos(21tuttut)}()(){4cos(21tutut)4cos(21tSo,T=2/4=0.51.26.(a).Periodic.N=7Solution:N=m*7/62=7,m=3.(b).Aperriodic.Solution:N=mm16*8/12,it’snotrationalnumber.(e).Periodic.N=16Solutionasfollow:)62cos(2)8sin()4cos(2][nnnnxinthisequation,)4cos(2n,it’speriodisN=2*m/(/4)=8,m=1.)8sin(n,it’speriodisN=2*m/(/8)=16,m=1.)62cos(2n,it’speriodisN=2*m/(/2)=4,m=1.So,thefundamentalperiodof][nxisN=(8,16,4)=16.1.31.SolutionBecause)()1()(),2()()(113112txtxtxtxtxtx.AccordingtoLTIproperty,)()1()(),2()()(113112tytytytytytyExtraproblems:Sketchtdttxty)()(.1.SupposeSolution:2.SupposeSketch:(1).)]1(2)1()3()[(ttttg(2).kkttg)2()(Solution:(1).(2).Chapter22.1Solution:Becausex[n]=(120–1)0,h[n]=(202)1,then(a).So,]4[2]2[2]1[2][4]1[2][1nnnnnny(b).accordingtothepropertyofconvolutioin:]2[][12nyny(c).]2[][13nyny2.3Solution:][*][][nhnxny][][knhkxkkkknuku]2[]2[)21(2][211)21()21(][)21(12)2(0222nununnkk][])21(1[21nunthefigureofthey[n]is:2.5Solution:Wehaveknown:elsewherennx....090....1][，，,elsewhereNnnh....00....1][，，,(9N)Then,]10[][][nununx,]1[][][Nnununhkknukhnhnxny][][][*][][kknuknuNkuku])10[][])(1[][(So,y[4]kkukuNkuku])6[]4[])(1[][(4,...14,...1400NNkNk=5,then4NAndy[14]kkukuNkuku])4[]14[])(1[][(14,...114,...11455NNkNk=0,then5N4N2.7Solution:[][][2]kynxkgnk（a）[][1]xnn,[][][2][1][2][2]kkynxkgnkkgnkgn(b)[][2]xnn,[][][2][2][2][4]kkynxkgnkkgnkgn(c)SisnotLTIsystem..(d)[][]xnun,0[][][2][][2][2]kkkynxkgnkukgnkgnk2.8Solution:)]1(2)2([*)()(*)()(tttxthtxty)1(2)2(txtxThen,Thatis,otherstttttttty,........010,....2201,.....41..,.........412,.....3)(2.10Solution:(a).Weknow:Then,)()()(ttth)]()([*)()(*)()(tttxthtxty)()(txtxthatis,So,otherstttttty,.....011,.....11,....0,.....)((b).Fromthefigureof)(ty,onlyif1,)(tywouldcontainmerelytherediscontinuities.2.11Solution:(a).)(*)]5()3([)(*)()(3tuetututhtxtytdtueudtueutt)()5()()3()(3)(3ttttdetudetu5)(33)(3)5()3(5,.......353,.....313.........,.........0315395)(33)(3393)(3teededetedetttttttttt(b).)(*)]5()3([)(*)/)(()(3tuettthdttdxtgt)5()3()5(3)3(3tuetuett(c).It’sobviousthatdttdytg/)()(.2.12Solutionktktkttuekttuety)]3(*)([)3(*)()(kktktue)3()3(Consideringfor30t,wecanobtain303311])3([)(eeeektueetytkktkkt.(Becausekmustbenegetive,1)3(ktufor30t).2.19Solution:(a).Wehaveknown:][]1[21][nxnwnw(1)][]1[][nwnyny(2)from(1),21)(1EEEHfrom(2),EEEH)(2then,212212)21(1)21)(()()()(EEEEEEHEHEH][]2[2]1[)21(][nxnynynybut,][]1[43]2[81][nxnynyny143)21(:....812or141(b).from(a),weknow)21)(41()()()(221EEEEHEHEH21241EEEE][)41()21(2][nunhnn2.20(a).11)0cos()cos()()cos()(0dtttdtttu(b).0dttt)3()2sin(50hasvalueonlyon3t,but]5,0[3dttt)3()2sin(50=0(c).0641551)2cos()()2cos()1(dtttudu64)2cos()(dttt0|)2(scott0|)2sin(20tt2.23Solution:kthkTtthtxty)(*)()(*)()(kkTth)(2.