数字信号处理-基于计算机的方法(第四版)第九章答案

Notforsale1SOLUTIONSMANUALtoaccompanyDigitalSignalProcessing:AComputer-BasedApproachFourthEditionSanjitK.MitraPreparedbyChowdaryAdsumilli,JohnBerger,MarcoCarli,Hsin-HanHo,RajeevGandhi,MartinGawecki,ChinKayeKoh,LucaLucchese,MyleneQueirozdeFarias,andTravisSmithCopyright©2011bySanjitK.Mitra.Nopartofthispublicationmaybereproducedordistributedinanyformorbyanymeans,orstoredinadatabaseorretrievalsystem,withoutthepriorwrittenconsentofSanjitK.Mitra,including,butnotlimitedto,inanynetworkorotherelectronicStorageortransmission,orbroadcastfordistancelearning.Notforsale2Chapter99.1WeobtainthesolutionsbyusingEq.(9.3)andEq.(9.4).(a)€δp=1−10−αp/20=1−10−0.24/20=0.0273,δs=10−αs/20=10−49/20=0.0035.(b)€δp=1−10−αp/20=1−10−0.14/20=0.016,δs=10−αs/20=10−68/20=0.000398.9.2WeobtainthesolutionsbyusingEqs.(9.3)and(9.4).(a)€αp=−20log101−δp()=−20log10(1−0.04)=0.3546dB,€αs=−20log10δs()=−20log100.08()=21.9382dB.(b)€αp=−20log101−δp()=−20log10(1−0.015)=0.1313dB,€αs=−20log10δs()=−20log100.04()=27.9588dB.9.3€G(z)=H2(z),orequivalently,€G(ejω)=H2(ejω)=H(ejω)2.Let€δpand€δsdenotethepassbandandstopbandripplesof€H(ejω),respectively.Also,let€δp,2=2δp,and€δs,2denotethepassbandandstopbandripplesof€G(ejω),respectively.Then€δp,2=1−(1−δp)2,and€δs,2=(δs)2.Foracascadeofsections,€δp,M=1−(1−δp)M,and€δs,M=(δs)M.9.4HLP(ejω)ωωpωsπ–ωp–ωs–πδs1+δp1–δp0ωHHP(ejω)π–πδs1+δp1–δpπ−ωpπ−ωs–(π–ωs)–(π–ωp)0Therefore,thepassbandedgeandthestopbandedgeofthehighpassfilteraregivenby€ωp,HP=π−ωp,and€ωs,HP=π−ωs,respectively.9.5Notethat€G(z)isacomplexbandpassfilterwithapassbandintherange€0≤ω≤π.Itspassbandedgesareat€ωp,BP=ωo±ωp,andstopbandedgesat€ωs,BP=ωo±ωs.Arealcoefficientbandpasstransferfunctioncanbegeneratedaccordingto€GBP(z)=HLP(ejωoz)+HLP(e–jωoz)whichwillhaveapassbandintherange€0≤ω≤πNotforsale3andanotherpassbandintherange€–π≤ω≤0.Howeverbecauseoftheoverlapofthetwospectraasimpleformulaforthebandedgescannotbederived.HLP(ejω)ωωpωsπ–ωp–ωs–πδs1+δp1–δp0ωG(ejω)π–πδs1+δp1–δp0ωoωo+ωsωo+ωpωo−ωpωo−ωs9.6(a)€hp(t)=ha(t)⋅p(t)where€p(t)=δ(t−nT).n=−∞∞∑Thus,€hp(t)=ha(nT)n=−∞∞∑δ(t−nT)..Wealsohave,€g[n]=ha(nT).Now,€Ha(s)=ha(t)e−st−∞∞∫dtand€Hp(s)=hp(t)e−st−∞∞∫dt=ha(nT)δ(t−nT)e−st−∞∞∫dtn=−∞∞∑=ha(nT)e−snTn=−∞∞∑.Comparingtheaboveexpressionwith€G(z)=g[n]z−nn=−∞∞∑=h(nT)z−nn=−∞∞∑,weconcludethat€G(z)=Hp(s)s=1Tlnz.WecanalsoshowthataFourierseriesexpansionofp(t)isgivenby€p(t)=1Te−j(2πkt/T)k=−∞∞∑.Therefore,€hp(t)=1Te−j(2πkt/T)k=−∞∞∑⎛⎝⎜⎜⎞⎠⎟⎟ha(t)=1Tha(t)e−j(2πkt/T)k=−∞∞∑.Hence,€Hp(s)=1THas+j2πktT⎛⎝⎜⎞⎠⎟k=−∞∞∑.Asaresult,wehave€G(z)=1THas+j2πktT⎛⎝⎜⎞⎠⎟k=−∞∞∑s=1Tlnz.(7-1)(b)Thetransformationfromthe€s-planeto€z-planeisgivenby€z=esT.Ifweexpress€s=σo+jΩo,thenwecanwrite€z=rejω=eσoTejΩoT.Therefore,Notforsale4€z=1,forσo1,=1,forσo=1,1,forσo1.⎧⎨⎪⎩⎪Orinotherwords,apointintheleft-half-planeismappedontoapointinsidetheunitcircleinthe€z-plane,apointintheright-half-planeismappedontoapointoutsidetheunitcircleinthe€z-plane,andapointonthejω-axisinthe€s-planeismappedontoapointontheunitcircleinthe€z-plane.Asaresult,themappinghasthedesirablepropertiesenumeratedinSection9.1.3.(c)However,allpointsinthe€s-planedefinedby€s=σo+jΩo±j2πkT,k=0,1,2,…,,aaremappedontoasinglepointinthe€z-planeas€z=eσoTejΩo±2πkT⎛⎝⎜⎞⎠⎟T=eσoTejΩoT.Themappingisillustratedinthefigurebelow1σ−1jΩzImzRez-plane-planesT3ππT–T3π–πTNotethatthestripofwidth€2π/Tinthe€s-planeforvaluesof€sintherange€−πT≤Ω≤πTismappedintotheentire€z-plane,andsoaretheadjacentstripsofwidth€2π/T.Themappingismany-to-onewithinfinitenumberofsuchstripsofwidth€2π/T.ItfollowsfromtheabovefigureandalsofromEq.(7-1)thatifthefrequencyresponse€Ha(jΩ)=0for€Ω≥πT,then€G(ejω)=1THa(jωT)for€ω≤π,andthereisnoaliasing.(d)For€z=ejω=ejΩT,orequivalently,€ω=ΩT.9.7Assume€ha(t)iscausal.Now,€ha(t)=Ha(s)estds.∫Hence,€g[n]=ha(nT)=Ha(s)∫esnTds.Therefore,Notforsale5€G(z)=g[n]z−nn=0∞∑=Ha(s)esnTz−n∫n=0∞∑ds=Ha(s)z−nn=0∞∑esnT∫ds=Ha(s)1−esTz−1∫ds.Hence€G(z)=ResiduesHa(s)1−esTz−1⎡⎣⎢⎤⎦⎥allpolesofHa(s)∑.9.8€Ha(s)=As+α.Thetransferfunctionhasapoleat€s=−α.Now€G(z)=Residueats=–αA(s+α)(1−esTz−1)⎡⎣⎢⎤⎦⎥=A1−esTz−1s=–α=A1−e−αTz−1.9.9(a)€Has()=2(s+2)(s+3)(s2+4s+5)=−1s+3+0.5−0.5j(s+2−j)+0.5+0.5j(s+2+j)€=−1s+3+s+3s+2()2+12=−1s+3+s+2s+2()2+12+1s+2()2+12.UsingEq(9.71),weget€Gaz()=−11−e−3Tz−1+1−z−1e−2TcosT()1−2z−1e−2TcosT()+e−4Tz−2+z−1e−2TsinT()1−2z−1e−2TcosT()+e−4Tz−2.SinceT=0.25,weget€Gaz()=−11−0.4724z−1+1−0.4376z−11−1.1754z−1+0.3679z−2..(b)€Hbs()=2s2+s−1(s+4)(s2+2s+10)=1.5s+4+0.25+0.75j(s+1−3j)+0.25−0.75j(s+1+3j)€=1.5s+4+0.5s−8s+1()2+32=−1s+3+0.5s+1s+1()2+32−0.53()3s+1()2+32.UsingEq(9.71),weget€Gbz()=1.51−e−4Tz−1+0.51−z−1e−Tcos3T()1−2z−1e−Tcos3T()+e−2Tz−2−1.5z−1e−Tsin3T()1−2z−1e−Tcos3T()+e−2Tz−2.SinceT=0.25,weget€Gbz()=1.51−0.3679z−1+0.51−2.1624z−1()1−1.1397z−1+0.6065z−2..(c)€Hcs()=−s2+2s+11(s2+2s+5)(s2+s+4)€=1.5+j(s+1−2j)+1.5−j(s+1+2j)+−1.5−0.315j(s+0.5−0.515j)+−1.5+0.315j(s+0.5+0.515j)Notforsale6€=3s−1/3s+1()2+4−3s−1s+0.5()2+0.515()2€=3s+1s+1()2+22+3−2/3()2s+1()2+22−3s+0.5s+0.5()2+0.515()2−3−3/15()0.515s+0.5()2+0.515()2.UsingEq(9.71),weget€Gcz()=31−z−1e−Tcos2T()1−2z−1e−Tcos2T()+e−2Tz−2−2z−1e−Tsin2T()1−2z−