信号与系统奥本海姆英文版课后答案chapter5

101Chapter5Answers5.1(a)letx[n]=1(1/2)nu[n-1].UsingtheFouriertransformanalysisequation(5.9).theFouriertransform()jwXeofthissignalis()[]jwjwnnXexne=1n1(1/2)njwne=(1)0(1/2)njwnne=jwe1(1(1/2))jwe(b)Letx[n]=|1|(1/2)n.UsingtheFouriertransformanalysisequation(5.9).TheFouriertransform()jwxeofsignalis()[]jwjwnnxexne=0(1)11(1/2)(1/2)njwnnjwnnneeThesecondsummationintheright—handsideoftheaboveequationisexactlythesameasresultofpart(a).Now,0(1)(1/2)njwnne=10(1/2)njwnne=(1/2)1(1(1/2))jweTherefore()jwxe=(1/2)1(1(1/2))jwe+jwe1(1(1/2))jwe=0.75(1.25cos)jwew5.2(a)let[][1][1]xnnn.UsingtheFouriertransformanalysisequation(5.9).theFouriertransform()jwxeofthissignalis()[]jwjwnnxexne=jwe+jwe=2cosw(b)Let[][2][2]xnnn.usingtheFouriertransformanalysisequation(5.9).theFouriertransform()jwxeofthissignalis()[]jwjwnnxexne=2jwe-2jwe=2sin(2)jw5.3Wenotefromsection5.2thataperiodicsignalwithFourierseriesrepresentationx[n]=(2/)jkNnkkNaehasaFouriertransform()jwXe=22()kkkawN(a)Considerthesignal1[]sin()34xnn.Wenotethatthefundamentalperiodofthesignal1[]xnisN=6.Thesignalmaybewrittenas1021[]xn=()3412jnej()3412jnej=26412jnjeej26412jnjeejFormthis,weobtainthenon-zeroFourierseriescoefficientskaof1[]xntherange23as41(1/2)jaje41(1/2)jajeTherefore,intherangew,weobtain1122()2()2()66jwXeawaw/4/4(/){(2/6)(2/6)}jjjewew(b)considerthesignal2[]2cos()68xnn.wenotethatthefundamentalperiodofthesignal1[]xnisN=12.thesignalmaybewrittenas()()6868122881212[]2(1/2)(1/2)2(1/2)(1/2)jnjnjjjnjnxneeeeeeFormthis,weobtainthenon-zeroFourierseriescoefficientskaof2[]xnintherange56kas02a81(1/2)jae81(1/2)jaeTherefore,intherange,weobtain011/8/822()2()2()2()12124(){()()}66jwjjXeawawawwewew5.4(a)UsingtheFouriertransformsynthesisequation(5.8)11[](1/2)()jwjwnxnXeedw(1/2)[2()(/2)(/2)]jwn0(/2)(/2)(1/2)(1/2)jjnjneee1cos(/2)n(b)Usingthetransformsynthesisequation(5.8)22[](1/2)()jwjwnxnXeedw00(1/2)2(1/2)2jwnjwnjedwjedw11(/)[]jnjneejjnjn2(4/())sin(/2)nn5.5Fromthegiveninformation[](1/2)()jwjwnxnxeedw={()}(1/2)|()|)jwjwjXejwnXeeedw3/42/4(1/2)wjwneedwsin((3/2))4(3/2)nnThesignalx[n]iszerowhen(3/2)4nisanonzerointegermultipleoforwhen|n|.thevalueof(3/2)4ncanneverbesuchthatitisanonzerointegermultipleof.Therefore.x[n]=0onlyforn=5.6Throughoutthisproblem,weassumethatX[n]FT1()jwxe(a)Usingthetimereversalproperty(Sec.5.3.6),wehave103x[-n]FT()jwXeUsingthetimeshiftproperty(Sec.5.3.3)onthis.wehaveFTx[-n+1]()jwnjwexeandFTx[-n-1]()jwnjwexeTherefore1[]xnx[-n+1]+x[-n-1]FT()+()jwnjwjwnjweXeeXeFT2X()cosjwew(b)Usingthetimereversalproperty(Sec.5.3.6),wehaveFTx[-n]X()jweUsingthesameconjugationpropertyonthis,wehaveFT*x[-n]*X()jweThereforeFT*2x[n]=(1/2)(x[-n]+x[n])(1/2)*()()jwjwXeXeFTRe{()}jwXe(c)Usingthedifferentiationfrequencyproperty(Sec.5.3.8),wehave()[]jwFTdXenxnjdwUsingthesamepropertysecondtime,222()[]jwFTdXenxndwTherefore222322()()[][]2[]12()jwjwFTjwdXedXexnnxnnxnjXedwdw5.7(a)Considerthesignal1[]ynwithFouriertransform1011()sin()jwkYekwWeseethat1()jwYeisrealandodd.FromTable5.1,weknowthattheFouriertransformofarealandoddsignalispurelyimaginaryandodd.Therefore,wemaysaythattheFouriertransformofapurelyimaginaryandoddsignalisrealandodd.Usingthisobservation,weconcludethat1[]ynispurelyimaginaryandoddNotenowthat11()()jwjwjwXeeYeTherefore,11[][1]xnyn.therefore,isalsopurelyimaginary.but1[]xnisneitherevennorodd(b)Wenotethat2()jwXeispurelyimaginaryandodd.Therefore,2[]xnhastoberealandodd.(d)©Considerasignal3[]ynwhosemagnitudeoftheFouriertransformis3|()|()jwYeAwandwhosephaseoftheFouriertransformis3{()}(3/2)jwYew.since33|()||()|jwjwYeYeand,wemayconcludethatthesignal3[]ynisreal(seeTable5.1,property5.3.4).Now,considerthesignal3[]xnwithFouriertransform333()()()jwjwjXeYeeYjw.UsingtheresultfrompreviousparagraphandthelinearitypropertyoftheFouriertransform.wemayconcludethathastoreal.sincetheFouriertransform,wemayconcludethathastoreal.sincetheFouriertransform3()jwXeisneitherpurelyimaginarynorpurelyreal.thesignal3[]xnisneitherevennorodd5.8Considerthesignal1,10,[]{xn||1||1nnFromthetable5.2,weknowthat11sin(3/2)[]()sin(/2)FTjwwxnXewUsingtheaccumulationproperty(Table5.1，Property5.3.5),wehave01111[]()()(2)1nFTjwjjwkkxkXeXewkeTherefore,intherangew,104111[]()3()1nFTjwjwkxkXeweAlso,intherangew,12()FTwTherefore,intherangew,111[]1[]()5()1nFTjwjwkxnxkXeweThesignalx[n]hasthedesiredFouriertransform.Wemayexpressx[n]mathematicallyas1[]1[]nkxnxk=134n2112nnn5.9Fromproperty5.3.4inTable5.1,weknowthatforarealsignalx[n],1{[]}Im{()}FTjwOdxnjXeFromthegiveninformation1Im()jwjXesinsin2jwjw22(1/2)()jwjwjwjweeeeTherefore,1{[]}{Im()}(1/2)([1][1][2][2])jwOdxnIFTjXennnnWealsokno