47信号与系统奥本海姆英文版课后答案chapter4

80Chapter4Answers4.1(a)Let.112tutxtthentheFouriertransformtofxjxis:jdtdttujXjtjttjt2/111212jXisasshowninfigures4.1.(b)Let.12ttxthentheFouriertransformtofxjXis:2112112124/42/2/jjjtjttjttjtjjdtdtdtjXjXisasshowninfigures4.14.2(a)Let.111tttxthentheFouriertransformtofxjX1is:cos2111jjtjdtttjXjX1isassketchedinfigures4.2.(b)thesignal222tututxisasshowninthefigurebelow.Clearly,2222tttutudtdTherefore)2sin(2)]2()2([)(222jeedtettjxjjtj)(1jxisassketchedinfigures4.2.4.3(a)thesignal)4/2sin()(1ttxisperiodicwithafundamentalperiodicofT=1.Thistranslationstoafundamentalfrequencyof20.thenonzeroFourierseriescoefficientsofthissignalsofthissignalmaybefoundbywritingitintheformtjjtjjtjtjeejeejeejtx2)4/2)4/)4/2()4/2(12121)(21)(therefore,thenonzeroFourierseriescoefficientsof)(1txaretjjtjjeejaeeja2)4/22)4/121,21FormSection4.2weknowthatforperiodicsignals,theFouriertransformsconsistsoftrainofimpulseoccurringat0k.Furthermore,theareaundereachimpulseis2timestheFourierseriescoefficientsjX1/2002(a)(b)jX2Xj20figures4.13/2/2/23/221Xjfigures4.281ka.Therefore,for1()xtthecorrespondingFouriertransforms1()xjisgivenby11010()2()2()xjaa/4/4(/)(2)(/)(2)jjjeje(b)Thesignal2()1cos(6/8)xttisperiodicwithafundamentalperiodof1/3T.Thistranslatestoafundamentalfrequencyof06.ThenonzeroFourierseriescoefficientsofthissignalmaybefoundbywritingitintheform(6/8)(6/8)21()1()2jtjtxtee/86/8611122jjtjjteeeeTherefore,ThenonzeroFourierseriescoefficientsof2()xtare/86/86011111,,.22jjtjjtaaeeaeeFormSection4.2weknowthatforperiodicsignals,theFouriertransformsconsistsoftrainofimpulseoccurringat0k.Furthermore,theareaundereachimpulseis2timestheFourierseriescoefficientska.Therefore,for2()xtthecorrespondingFouriertransforms2()xjisgivenby201010()2()2()2()xjaaa/8/82()(6)(6)jjee4.4(a)TheinverseFouriertransformsis1()(1/2)[2()(4)(4)]jwtxted44(1/2)[2]jtjtjteee441(1/2)(1/2)1cos(4)jtjteet(b)TheinverseFouriertransformsis22()(1/2)()jwtxtXjwedw2002(1/2)22jwtjwtedwedw22(1)/()(1)/()jtjtejtejt2(4sin)/()jtt4.5Formthegiveninformation()(1/2)()jwtxtXjwedw()(1/2)()jwjXewtjwXjweedw32/33(1/2)2wjweedw2sin[3(2/3)](2/3)ttThesignal()xtiszerowhen3(2/3)tisanonzerointegermultipleofthisgives2/3,02ktforkIandk4.6.Throughoutthisproblem,weassumethatx(t)FTX1(j)(a)Usingthetimereversalproperty(Sec.4.3.5),wehavex(-t)FTX1(-j)Usingthetimeshiftingproperty(Sec.4.3.2)onthis,wehavex(-t+1)FTjteX1(-j)andx(-t-1)FTjteX1(-j)therefore82x1(t)=x(-t+1)+x(-t-1)FTjteX1(-j)+jteX1(-j)FT2X(-j)cos(b)Usingthetimescalingproperty(Sec.4.3.5),wehavex(3t)FT1/3X(j/3)Usingthetimeshiftingpropertyonthis,wehavex2(3(t-2))FT2je1/3X(j/3)(c)Usingthedifferentiationintimeproperty(Sec.4.3.4),wehave()dxtdtFT()jXjApplyingthispropertyagain,wehave22()dxtdtFT2()XjUsingthetimeshiftingproperty,wehavex3=22(1)dxtdtFT2()Xjjte4.7(a)SinceX1(-j)isnotconjugatesymmetric,thecorrespondingsignalx1(t)isnotrealSinceX1(-j)isneitherevennorodd,thecorrespondingsignalx1(t)isneitherevennorodd(b)theFTofarealandoddsignalispurelyimaginaryandodd.therefore,wemayconcludethattheFTofapurelyimaginaryandoddsignalisrealandodd.sinceX2(j)isrealandoddwemaythereforeconcludethatthecorrespondingsignalx2(t)ispurelyimaginaryandodd.(c)Considerasignaly3(t)whosemagnitudeoftheFTis|3()Yj|=A(),andwhosephaseoftheFTis3{()}2Yj.Since|3()Yj|and3{()}Yj=-3{()}Yj,wemayconcludethatthesignaly3(t)isreal(d)SinceX4(j)isbothrealandeven,correspondingsignalx4(t)isrealandeven4.8(a)Thesignalx(t)isasshowninFigureS4.8.Wemayexpressthissignalas()xt()tytdtWherey(t)istherectangularpulseshowninS4.8UsingtheintegrationpropertyofFTwehavex(t)FTX(j)=1()(0)()YjYjjweknowfrom4.2that()Yj=2sin(/2)wwThereforeX(j)=22sin(/2)()wjw(b)ifg(t)=x(t)-(1/2)()=22sin(/2)wjw4.9(a)thesignalx(t)isplottedinfigureS4.9x(t)=()(1/2)tytdtutusingtheresultobtainedinpart(a)ofthepreviousproblem,theFTX(j)ofx(t)isX(j)=22sin(/2)()wjw-FT{u(t-1/2)}=2sinjejj(b)theevenpartofx(t)isgivenby{()}vxt=(x(t)+x(-t))/2Thisisasshowninthe4.9Thereforesin{{()}}FTvxtNowtherealpartofanswertopart(a)is1sinRe{}Re{(cossin)}jejjjj(c)theFToftheoddpartofx(t)issameasjtimesimaginarypartoftheanswertopart(a),wehave22sinsincosIm{}jejjTherefore,thedesiredresultisy(t)1/2t1-1tx(t)1-111x(t)1/2-1/2t1-1/21/2ty(t)FigureS4.883FT{Oddpartofx(t)}=2sincosjj4.10（a）weknowfromtable4.2thatsinFTttRectangularfunctiony(j)[seefigures4.10]Therefore2sin()FTtt(1/2)[Rectangularfunctiony(j)*Rectangularfunctiony(j)]Thisisatriangularfunction1()yjasshowinthefigures4.10Usingtable4.1we