信号与系统奥本海姆英文版课后答案chapter6

127Chapter6Answers6.6(b)theimpulseresponseh1[n]isasshowninfigures6.6,aswasincrease,itisclearthatthesignificantcentrallobeofh1[n]becomesmoreconcentratedaroundtheorigin.consequently.h[n]=h1[n](-1)^nalsobecomesmoreconcentratedabouttheorigin.6.7thefrequencyresponsemagnitude|H(jw)|isasshowninfigures6.7.thefrequencyresponseofthebandpassfilterG(jw)willbegivenby(){2()cos(4000)}GjFThtt((4000))((4000))HjHjThisisasshowninfigures6.7-6000-4000-20000200040006000FigureS6.7(a)fromthefigure,itisobviousthatthepassbandedgesareat2000∏rad/secand6000∏rad/sec.thistranslatesto1000HZand3000Hz,respectively.(b)(b)fromthefigure,itisobviousthatthestopbandedgesareat1600∏rad/sec.thistranslatesto800Hzand3200Hz,respectively.6.8takingtheFouriertransformofbothsidesofthefirstdifferenceequationandsimplifying,weobtainthefrequencyresponseH(e^jw)ofthefirstfilter.01()().()1MjkjkjkNjjkkkbeYeHeXeaeTakingtheFouriertransformofbothsidesoftheseconddifferenceequationandsimplifying,weobtainthefrequencyresponseH1(e^jw)ofthesecondfilter.01(1)()().()1(1)MkjkjkjkNjkjkkkbeYeHeXeaeThismayalsobewrittenas()()0()1()()().()1MjkjkjjkNjjkkkbeYeHeHeXeaeTherefore.thefrequencyresponseofthesecondfilterisobtainedbushiftingthefrequencyresponseofthefirstfilterby∏.althoughthefirstfitterhasitspassbandbetween-wpandwp.Therefore,thesecondfilterwillhaveitspassbandbetween∏-wpand∏+wp.6.9takingtheFouriertransformofthegivendifferentialequationandsimplifying.weobtainthefrequencyoftheLTIsystemtobe()2()()5jjjYeHeXejTakingtheinverseFouriertransform,weobtaintheimpulseresponsetobe5()2().thteutUsingtheresultderivedinsection6.5.1,wehavethestepresponseofthesystem-4000-2000-1000100020004000H(j)G（j）12852()()*()[1]().5tsthtuteutThefinalvalueofthestepresponseis2().5sWealsohave052()[1].5tseSubstitutings(t0)=(2/5)[1-1/e^2],intheaboveequation,weobtaint0=2/5sec(a)wemayrewriteH1(jw)tobe1()()(0.1).40Hjjjwemaythentreateachofthetwofactorsasindividualfirstordersystemsanddrawtheirbodemagnitudeplots.thefinalbodewillthenbeasumofthesetwobodeplots.thisisshowninthefigures6.10mathematically.thestraight-lineapproximationofthebodemagnitudeplotis101020,0.120log|()|20log(),0,14032,40HjFigureS6.10(b)Usingasimilarapproachasinpart(a),weobtaintheBodeplottobeasshowninFigureS6.10.Mathematically,thestraight-lineapproximationoftheBODEmagnitudeplotis101020,0.220log20log(),0.25028,50Hj6.10.(a)Wemayrewritethegivenfrequencyresponse1H(j)as12250250()()50.525(0.5)(50)Hjjjjj.WemaythenuseanapproachsimilartotheoneusedinExample6.5andinProblem6.11toobtaintheBodemagnitudeplot(withstraightlineapproximations)showninFigureS6.11.Mathematically,thestraight-lineapproximationoftheBodemagnitudeplotis101-280200.110log|()|xj100500.11400-2032402010log|()|xj50011000.1-68-20-4020400.550100100.10.5020-601010log|()|xj10log|()|xj12910101020,0.520log20log()14,0.55040log()48,50Hj(b)Wemayrewritethefrequencyresponse2H(j)as2H()j=20.02(50)()0.21jjj.AgainusinganapproachsimilartotheoneusedinExample6.5,wemaydrawtheBodemagnitudeplotbytreatingthefirstandsecondorderfactorsseparately.ThisGivensusaBodemagnitudeplot(usingstraightline)approximationsasshownbelow:Mathematically,thestraight-lineapproximationoftheBodemagnitudeplotis1010100,120log40log,15020log()34,50Hj6.12.UsingtheBodemagnitudeplot,specifiedinFigureP6.12(a).wemayobtainanexpressionFor1H(j).Thefigureshowsthat1H(j)hasthebreakfrequencies1=1,2=8,And3=40.Thefrequencyresponserisesas20dB/decadeafter1.At2,thisriseiscanceledbya-20dB/decadecontribution.Finally,at3,anadditional-20dB/decade.Contributionresultsinthesubsequentdecayattherateof-20dB/decade,therefore,wemayconcludethat1H()j=123()()()Ajjj.WenowneedtofindA.Notethatwhen=0,20101log(0)Hj=2.Therefore.1H(0)j=0.05.Fromeq.(S6.12-1),weknowthat1H(0)j=320A.Therefore,A=640.Thisgivesus1H()j=640(1)(8)(40)jjj.UsingasimilarapproachonFigureP6.12(b),weobtainH(j)=26.4(8)j.Sincetheoverallsystem(withfrequencyresponseH(j))isconstructedbycascadingSystemswithfrequencyresponses1H()jand2H()j,H(j)=1H()j2H()j.UsingthepreviouslyobtainedexpressionsforH(j)and1H()j,2H()j=H(j)/1H()j=0.01(40)(1)(8)jjj.6.13.Usinganapproachsimilartotheoneusedinthepreviousproblem,weobtainH(j)=320(2)(8)jj.(a)Letusassumethatwedesiretoconstructthissystembycascadingtwosystemswithfrequencyresponses1H()jand2H()j,respectively.WerequirethatH(j)=1H()j2H()j.Weseethat1H()j=40(2)jand2H()j=8(80)jAnd1H()j=32(2)jand2H()j=10(80)jarebothvalidcombinations.(b)Letusassumethatwedesiretoconstructthissystembyconnectingtwosystemswithfrequencyresponses1H()jand2H()jinparallel.Werequirethat130H(j)=1H()j+2H()jUsingpartialfractionexpansiononH(j),weobtainH(j)=16039(2)j-16039(80)jFromtheaboveexpressionitisclearthatwecandefine1H()jand2H()jinonlyoneway6.14.UsinganapproachsimilartotheoneusedinProblem6.12,wehaveH(j)=25000(0.2)(50)(10)jjj.The