数字信号处理 基于计算机的方法 第四版 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,forσo<1,=1,forσo=1,>1,forσo>1.⎧⎨⎪⎩⎪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−1e−Tcos2T()+e−2Tz−2.SinceT=0.25,weget€Gcz()=31−0.9324z−1()1−1.3669z−1+0.6065z−2−31−0.4629z−1()1−1.5622z−1+0.7788z−2..9.10(a)€Gaz()=2zz−e−1.3+5zz−e−2.0=A1zz−e−α1T+A2zz−e−α2T.SinceT=0.5,€α1=2.6,α2=4,A1=2,A2=5,itfollows€Has()=2s+2.6+5s+4.(b)€Gbz()=ze−1.4sin1.6()z2−2ze−1.4cos1.6()+e−2.6=ze−βTsinλT()z2−2ze−βTcosλT()+e−2βT.SinceT=0.5,€λ=3.2,β=2.6,itfollows€Hbs()=3.2s+2.6()2+3.22.9.11(a)€Has()=Gaz()z=41+s1−s⎛⎝⎜⎞⎠⎟=45s2+18s+9()75s2+154s+91.(b)€Hbs()=Gbz()z=41+s1−s⎛⎝⎜⎞⎠⎟=105s3+385s2+467s+19513s+11()27s2+46s+23().9.12Fortheimpulseinvariancedesign:€ωp=ΩpT=2πFpT=2π0.88×103()0.25×10−3()=0.44π.Forthebilineartransformationmethod:€ωp=2tan−1ΩpT2⎛⎝⎜⎞⎠⎟=2tan−1FpTπ()=2tan−10.88×103⋅0.25×10−3⋅π()=0.385π.9.13Fortheimpulseinvariancemethod:€2πFp=ωpT=0.45π0.4×10−3⇒€Fp=562.5Hz.Forthebilineartransformationmethod:€Fp=tanωp2⎛⎝⎜⎞⎠⎟⋅1πT=tan0.45π2⎛⎝⎜⎞⎠⎟⋅1π0.4×10−3()=679.7Hz.Notforsale79.14Thepassbandandthestopbandedgesoftheanaloglowpassfilterareassumedto€Ωp=0.25πand€Ωs=0.55π.Therequirementstobesatisfiedbytheanaloglowpassfilterarethus€20log10Ha(j0.25π)≥−0.5dBand€20log10Ha(j0.55π)≤−15dB.From€αp=20log10(1+ε2)=0.5weobtain€ε2=0.1220184543.From€αs=10log10(A2)=15weobtain€A2=31.6227766.FromEq.(A.6),theinversediscriminationratioisgivenby€1k1=A2−1ε=15.841979andfromEq.(A.5)theinversetransitionratioisgivenby€1k=ΩsΩp=2.2.SubstitutingthesevaluesinEq.(A.9)weobtain€N=log10(1/k1)log10(1/k)=log10(15.841979)log10(2.2)=3.503885.WechooseN=4.FromEq.(A.7)wehave€ΩpΩc⎛⎝⎜⎞⎠⎟2N=ε2.Substitutingthevaluesof€Ωp,N,and€ε2weget€Ωc=1.3007568(Ωp)=1.021612.Usingthestatement[z,p,k]=buttap(4)wegetthepolesofthe4-thorderButterworthanalogfilterwitha3-dBcutoffat1rad/sas€p1=−0.3827+j0.9239,€p2=−0.3827−j0.9239,€p3=−0.9239+j0.3827,and€p4=−0.9239−j0.3827.Therefore,€Han(s)=1(s−p1)(s−p2)(s−p3)(s−p4)=1(s2+0.7654s+1)(s2+1.8478s+1)..Nextweexpand€Han(s)inapartial-fractionexpansionusingtheM-fileresidueandarriveat€Han(s)=−0.9238729s−0.7071323s2+0.7654s+1+0.9238729s+1.7071323s2+1.8478s+1.Wenextdenormalize€Han(s)tomovethe3-dBcutofffrequencyto€Ωc=1.021612usingtheM-filelp2lpresultingin€Ha(s)=Hans1.021612⎛⎝⎜⎞⎠⎟€=−0.943847s−0.738039s2+0.781947948s+1.0437074244+0.943847s+1.78174665s2+1.887749436s+1.0437074244€=−0.943847s−0.738039(s+0.390974)2+(0.9438467)2+0.943847s+1.78174665(s+0.94387471)2+(0.39090656)2.Notforsale8MakinguseoftheM-filebilinearwefinallyarriveat€G(z)=−0.943847z2+0.68178386zz2−1.363567724z+0.4575139+0.943847z2−0.25640047zz2−0.77823439z+0.1514122.9.15Themappingisgivenby€s=1T(1−z−1)orequivalently,by€z=11−sT.For€s=σo+jΩo,€z=11−σoT−jΩoT.Therefore,€z2=1(1−σoT)2+(ΩoT)2.Hence,€z<1for€σo<0.Asaresult,astable€Ha(s)resultsinastable€H(z)afterthetransformation.However,for€σo=0,€z2=11+(ΩoT)2whichisequalto1onlyfor€Ωo=0.Hence,onlythepoint€Ωo=0onthe€jΩ-axisinthe€s-planeismappedontothepoint€z=1ontheunitcircle.Consequently,thismappingisnotusefulforthedesignofdigitalfiltersviaanalogfiltertransformation.9.16Fornoaliasing€T≤πΩc.Figurebelowshowsthemagnituderesponsesofthedigitalfilters€H1(z)and€H2(z).ωωH1(ejω)H2(ejω)1200–π–πππ(a)Themagnituderesponsesofthedigitalfilters€G1(z)and€G2(z)areshownbelow:ωω100–π–πππ3/2G2(ejω)G1(ejω)(b)Ascanbeseenfromtheabove€G1(z)isamulti-passbandfilter,whereas,€G2(z)isahighpassfilter.9.17€Ha(s)iscausalandstableand€Ha(s)≤1,∀s,Now,€G(z)=Ha(s)s=2T1−z−11−z−1⎛⎝⎜⎞⎠⎟.Thus,€G(z)iscausalandstable.Now,€G(ejω)=Ha(s)s=2T1−ejω1+ejω⎛⎝⎜⎞⎠⎟=Ha(s)s=j2Ttan(ω/2)=Ha(j2Ttan(ω/2).Notforsale9Therefore,€G(ejω)=Ha(j2Ttan(ω/2)≤1forallvaluesof€ω.Hence,€G(z)isaBRfunction.9.18(a)€s=2T1+z−11−z−1⎛⎝⎜⎜⎞⎠⎟⎟⇒z−1=s−2/Ts+2/T,or€z=s+2/Ts−2/T=σ+2/T()+jΩσ−2/T()+jΩ.(b)€z2=σ+2/T()2+Ω2σ−2/T()2+Ω2,so€z2σ+jΩ,σ=0=1and€z2s=σ+jΩ,σ<0<1.ThefirstprovesthatapointonthejΩaxisismappedtoapointontheunitcircle,andthesecondprovesthatapointintheleft-halfs-planeismappedtoapointinsidetheunitcircle(stabilityispreserved).Thismappingdoesindeedhaveallthedesirableproperties.(c)If€sf1(z)()isthebilineartransformationinEq.(9.14)and€sf2(z)()isthebilineartransformationinEq.(9.72),then€sf2(z)()=s−f1(z)().(d)€jΩ=2T1+e−jω1−e−jω⇒ω=−2cot−1ΩT2⎛⎝⎜⎞⎠⎟.(e)G(z)isahigh-passfilterbecausesmallΩ(largeω)frequenciespassedwhilelargeΩ(smallω)frequenciesareattenuated.9.19€G(z)=HLP(s)s=2T1+z−11−z−1=1−α21−z−11+αz−1⎛⎝⎜⎜⎞⎠⎟⎟,whereαisdefinedinEq.(9.25).G(z)isahigh-passfilter.€G(z)=GLP(−z).G(z)=GLP(‒zIfβ=‒α,then.Notforsale109.20,whereαisdefinedinEq.(9.25).G(z)isalow-passfilter.€G(z)=GLP(−z).If€β=−α,then€G(z)=1−β21+z−11−βz−1⎛⎝⎜⎜⎞⎠⎟⎟=GLP(z)..9.21Let€y(t)=ddtx(t)beapproximatedby€y(nT)≈x(nT)−x(nT−T)T.Then€Y(z)=1−z−1TX(z),whichsuggeststhemapping€s=1−z−1T,or€z=11−Ts=1(1−Tσ)−j(TΩ).€z2=11−Tσ()2+(TΩ)2,so€zσ+jΩ,σ=0⇒z−0.5=0.5and€z2s=σ+jΩ,σ<0<1.Thistheimaginaryaxisinthes-domainismappedtothecircleofradius0.5centeredatz=0.5,butthetransformationpreservesstability.Ananaloghigh-passfiltercannotbemappedtoadigitalhigh-passfilterbecausethepolesofthedigitalfilterdonotlieintheleft-halfofthez-plane.9.22€G(z)=1+α2⋅1−2βz−1+z−21−β(1+α)z−1+αz−2.For€β=cosωo,thenumeratorof€G(z)becomes€1−2cos(ωo)z−1+z−2=(1−ejωoz−1)(1−e−jωoz−1)whichhasrootsat€z=e±jωo.Thenumeratorof€G(zN)isthengivenby€(1−ejωoz−N)(1−e−jωoz−N)whoserootsareobtainedbysolvingtheequation€zN=e±jωo,andaregivenby€z=ej(2πn±ωo)/N,0≤n≤N−1.Hence€G(zN)hasNcomplexconjugatezero-pairslocatedontheunitcircleatanglesof€2πn±ωoNradians,€0≤n≤N−1.For€ωo=π/2,thereare2Nequallyspacedzerosontheunitcirclestartingat€ω=π/2N.9.23Let€Ha(s)=a0+a1s+a2s2++aNsNb0+b1s+b2s2++bNsNdenotetheanalogtransferfunctionand€G(z)=q0+q1z−1+q2z−2++qNz−Nd0+d1z−1+d2z−2++dNz−Ndenotethedigitaltransferfunctionobtainedafterapplyingthebilineartransformation€s=cz−1z+1where€c=2/T.Alltransferfunctioncoefficientsareassumedtoberealnumbers.Thenumeratorandthedenominatorcoefficientsof€Ha(s)and€G(z)canberepresentedinvectorformasshownbelow:€A=[a0,a1,a2,…,aN],€B=[b0,b1,b2,…,bN],Notforsale11€Q=[q0,q1,q2,…,qN],€D=[d0,d1,d2,…,dN].TodeterminetheelementsofQandDfromAandB,wecomparethenumeratoranddenominatorcoefficientsofbothtransferfunctions.ItcanbeshownthatforN=2,wehave€G(z)=q0+q1z−1+q2z−2d0+d1z−1+d2z−2€=(a0+a1c+a2c2)+(2a0−2a2c2)z−1+(a0−a1c+a2c2)z−2(b0+b1c+b2c2)+(2b0−2b2c2)z−1+(b0−b1c+b2c2)z−2.Bycomparingthecoefficientsoflikepowersof€z−1weget€q0q1q2⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥=11120−21−11⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥a0a1ca2c2⎡⎣⎢⎢⎢⎤⎦⎥⎥⎥.AsimilarrelationbetweentheelementsofBandDcanbederived.Inthegeneralcase,thesematrixequationscanbecompactlyrepresentedintheform€Q=PNAcand€D=PNBcwhere€PNisthePascalmatrixoforder€N+1and€Ac=a0,a1c,a2c2,…,aNcN[]and€Bc=b0,b1c,b2c2,…,bNcN[].Inthegeneralcase,theelementsof€PNaregivenasfollows:(1)Theelementsofthefirstrowareallones,(2)Theelementsofthelastcolumnaregivenby€pi,N+1=(−1)i−1n!(n−i+1)!(i−1)!,1≤i≤N+1,(3)Theremainingelementsaregivenby€pi,j=pi−1,j+pi−1,j+1+pi,j+1,where€2≤i≤N+1,N≥j≥1.9.24(a)€H(z)=12[1+A4(z)]=N(z)D(z).Wecanwrite€A4(z)=z−2D1(z−1)D1(z)⋅z−2D2(z−1)D2(z),where€D1(z)=1−β1(1+α1)z−1+α1z−2and€D2(z)=1−β2(1+α2)z−1+α2z−2.Therefore,€N(z)=12D1(z)D2(z)+z−4D1(z−1)D2(z−1)[].Now,€z−4N(z−1)=12z−4D1(z−1)D2(z−1)+D1(z)D2(z)[]=N(z).Hence,€N(z)isasymmetricpolynomial.Itfollowsthen€P(z)=12[α1−β1(1+α1)z−1+z−2()α2−β2(1+α2)z−1+z−2()€+1−β1(1+α1)z−1+α1z−2()1−β2(1+α2)z−1+zα2−2()]Notforsale12€=1+α1α22[1−(1+α1)(1+α2)(β1+β2)1+α1α2z−1+2[α1+α2+β1β2(1+α1)(1+α2)]1+α1α2z−2€−(1+α1)(1+α2)(β1+β2)1+α1α2z−3+z−4]€=a(1+b1z−1+b2z−2+b1z−3+z−3),where€b1=−(1+α1)(1+α2)(β1+β2)1+α1α2,(7-a)€b2=2[α1+α2+β1β2(1+α1)(1+α2)]1+α1α2,(7-b)(b)€a=1+α1α22.(7-c)(c)for€z=ejω,wecanwrite€N(ejω)=a(1+b1e−jω+b2e−j2ω+b1e−j3ω+e−j4ω)€=ae−j2ω(b2+2b1cosω+2cos2ω).Now,€N(ejω)=0for€i=1,2.For€i=1,weget€b2+2b1cosω1+2cos2ω1=0,(7-d)for€i=2,weget€b2+2b1cosω2+2cos2ω2=0,(7-e)SolvingEqs.(7-d)and(7-e)weget€b1=−2(cosω1+cosω2),(7-f)and€b2=2(2cosω1cosω2+1).(7-g)FromEqs.(7-a)and(7-f)wehave€(1+α1)(1+α2)(β1+β1)1+α1α2=2(cosω1+cosω2),(7-h)andfromEqs.(7-b)and(7-g)wehave€2[α1+α2+β1β2(1+α1)(1+α2)]1+α1α2=2(2cosω1cosω2+1).(7-i)Substituting€α1=1−tan(B1/2)1+tan(B1/2)and€α2=1−tan(B2/2)1+tan(B2/2),andafterrearrangementweget€β1+β2=(cosω1+cosω2)[1+tan(B1/2)tan(B2/2)]=Δθ1,(7-j)and€β1β2=[1+tan(B1/2)tan(B2/2)]cosω1cosω2=Δθ2.(7-k)Theabovetwononlinearequationscanbesolvedyielding€β1=θ1±θ12−4θ22and€β2=θ2θ1.(d)Forthedoublenotchfilterwiththefollowingspecifications:€ω1=0.2π,€ω2=0.6π,€B1=0.2π,and€B2=0.25πwegetthefollowingvaluesfortheparametersofthenotchfiltertransferfunction:€α1=0.5095,α2=0.4142,θ1=0.5673,θ2=−0.1491,&be