数字信号处理-基于计算机的方法(第四版)答案8-11章(20161119180734)

Notforsale1SOLUTIONSMANUALtoaccompanyDigitalSignalProcessing:AComputer-BasedApproachFourthEditionSanjitK.MitraPreparedbyChowdaryAdsumilli,JohnBerger,MarcoCarli,Hsin-HanHo,RajeevGandhi,MartinGawecki,ChinKayeKoh,LucaLucchese,MyleneQueirozdeFarias,andTravisSmithCopyright?2011bySanjitK.Mitra.Nopartofthispublicationmaybereproducedordistributedinanyformorbyanymeans,orstoredinadatabaseorretrievalsystem,withoutthepriorwrittenconsentofSanjitK.Mitra,including,butnotlimitedto,inanynetworkorotherelectronicStorageortransmission,orbroadcastfordistancelearning.Notforsale2Chapter10Part110.1Forthisproblem,wehaveN=71ands-p=0.04andweassumethatp=s.(a)UsingKaisersformulaofEq.(10.3):s=10N14.6()s-p()/2[]-13-20=107114.6()0.04()/2[]-13-20=0.0206.s=-20log10s()=33.7320dB.(b)UsingBellangersformulaofEq.(10.4):s=0.1?10-3(N+1)s-p()/221/2=0.1?10-72?3?0.04/221/2=0.0263.s=-20log10s()=31.6dB.(c)UsingHermannsformulaofEq.(10.5),wefirstfindthat:Fs,s()=b1+b2log10s-log10s()=b1=11.01217.Therefore:Ds,s()=Ns-p()/2+Fs,s()s-p()2/42=710.04()/(2)+11.012170.04()2/(2)2=1.4244.Solvingfor:Ds()=a1log10s()2+a2log10s()+a3log10s()-a4log10s()2+a5log10s()+a6=a1log10s()3+a2log10s()2+a3log10s()-a4log10s()2-a5log10s()-a6=a1log10s()3+a2-a4()log10s()2+a3-a5()log10s()-a6.Letx=log10s(),andthusNotforsale3Ds()=0.005309x3+0.06848x2-1.0702x-0.4278=1.4244.Solvingthis,weget:x=-21.5147,10.2049,-1.589.Themostreasonablesolutionisthelastone,whichyields:s=10x=10-1.5890=0.0258,s=-20log10s()=31.78dB.10.2Forthisproblem,wehaveN=71ands-p=0.04=.UsingEq.(10.45):s=2.285()N+8=2.2850.04()71+8=28.3871dB.10.3(a)FromEq.(10.17):HHP(ej)=0,c,1,c,hHPn[]=12HHPej()ejnd-=12ejnd--c+12ejndc=12ejnjn--c+12ejnjnc=12e-jcnjn-e-jnjn+12ejnjn-ejcnjn=sin(n)n-sin(cn)n.Usingthepropertiesofthesincfunction,wearriveat:hHP0[]=1-c.Therefore:hHPn[]=1-c,n=0,-sin(cn)n,otherwise.(b)FromEq.(10.18):HBP(ej)=0,c1,1,c1c2,0,c2,hBPn[]=12HBP(ej)ejnd-=12ejnd-c2-c1+12ejndc1c2=12ejnjn-c2-c1+12ejnjnc1c2=12e-jc1njn-e-jc2njn+12ejc2njn-ejc1njnNotforsale4=sin(c2n)n-sin(c1n)n.(c)FromEq.(10.20):HBS(ej)=1,c1,0,c1c2,1,c2.hBSn[]=12HBS(ej)ejnd-=12ejnd--c2+12ejndc2+12ejnd-c1c1=12ejnjn--c2+12ejnjnc2+12ejnjn-c1c1=12e-jcnjn-e-jnjn+12ejnjn-ejcnjn12e-jc1njn-e-jc1njn=sin(n)n-sin(c2n)n+sin(c1n)n.Usingthepropertiesofthesincfunction,wearriveat:hBS0[]=1-c2-c1.Therefore:hHPn[]=1-c2-c1,n=0,sin(c1n)n-sin(c2n)n,otherwise.10.4TheidealL-banddigitalfilterHML(z)hasafrequencyresponsegivenby:HML(ej)=Ak,fork-1k,1kL.Itcanbeconsideredassumofidealbandpassfilterswithcutofffrequenciesat:c1k=k-1andc2k=k,wherec10=0andc2L=.FromEqn.(10.19)theimpulseresponseofanidealbandpassfilterisgivenby:hBP[n]=sin(c2n)n-sin(c1n)n.Therefore:hBPk[n]=sin(kn)n-sin(k-1n)n..Hence:hML[n]=hBPk[n]k=1L=Aksin(kn)n-sin(k-1n)nk=1L=A1sin(1n)n-sin(0n)n+Aksin(kn)nk=2L-1-Aksin(kn)n-sin(k-1n)nk=2L-1+ALsin(Ln)n-sin(L-1n)nNotforsale5=A1sin(1n)n+Aksin(kn)nk=2L-1-Aksin(k-1n)nk=2L-1-ALsin(L-1n)n=Aksin(kn)n-Aksin(k-1n)nk=2Lk=1L-1.SinceL=,sin(Ln)=0.Wecanaddatermof:ALsin(Ln)ntothefirstsumintheaboveexpressionandchangetheindexrangeofthesecondsum,resultingin:hML[n]=Aksin(kn)n-Ak+1sin(kn)nk=1L-1k=1L.Finally,sinceAL+1=0,wecanaddaterm:AL+1sin(Ln)ntothesecondsum.Thisleadsto:hML[n]=Aksin(kn)n-Ak+1sin(kn)nk=1Lk=1L=(Ak-Ak+1)sin(kn)nk=1L.10.5TheimpulseresponsefortheHilbertTransformerisgivenby:HHT(ej)=j,-0,-j,0.Therefore:hHT[n]=12HHT(ej)ejnd-0+12HHT(ej)ejnd0=12jejnd-0-12jejnd0=22n1-cos(n)()=2sin2(n/2)n,n0.Forn=0,hHT[0]=12jd-0-12jd0=0.Hence:hHT[n]=0,ifn=0,2sin2(n/2)n,ifn0.SincehHT[n]=-hHT[-n],andthelengthofthetruncatedimpulseresponseisodd,itisaType3linear-phaseFIRfilter.Notforsale6Fromthefrequencyresponseplotsgivenabove,weobservethepresenceofripplesatthebandedgesduetotheGibbsphenomenoncausedbythetruncationoftheimpulseresponse.10.6First,wenotethat:?H{x[n]}=hHT[n-k]x[k]k=-..Hence,?FHx[n]{}{}=HHT(ej)X(ej)=jX(ej),-0,-jX(ej),0.(a)Let?y[n]=H{H{H{H{H{H{x[n]}}}}}}.Hence:Y(ej)=(j)6X(ej),-0,(j)6X(ej),0,=-X(ej).Therefore,y[n]=-x[n].(b)Define?g[n]=Hx[n]{},andh*[n]=x[n].Then:?Hx[]{}x[]==g[]h*[]=.ButfromParseval'srelationinTable3.4:?g[]h*[]==12G(ej)G(ej)d.Therefore:?Hx[]{}x[]==12HHT(ej)-X(ej)X(ej)dwhereHHT(ej)=j,-0,-j,0.SincetheintegrandHHT(ej)X(ej)X(ej)isanoddfunctionof,HHT(ej)-X(ej)X(ej)d=0.Notforsale7Asaresult:?Hx[]{}x[]==0.10.7Giventhefrequencyresponsefortheideallowpassfilter:HLP(z)=hLP[n]z-nn=0N.ItsfrequencyresponseisshowninFigure(a)below.(a)Aplotofthefrequencyresponseof:HHT(ej)=HLP(ej(-2))+HLP(ej(+2))isshowninFigure(b)below.(b)ItisevidentfromthisfigurethatHHT(ej)isthefrequencyresponseofanidealHilberttransformer.Therefore,wehave:HHT(ej)=HLP(ej(+2))+HLP(ej(-2))=hLP[n]e-jn(+2)n=0N+hLP[n]e-jn(-2)n=0N=hLP[n]n=0Ne-jne-jn/2+ejn/2()=2hLP[n]e-jnn=0Ncosn2.Fornodd,cos(n/2)=0andwecandropalloddtermsintheaboveexpression.LetN=2MwithNevenandletr=2n.Then,wecanrewritetheaboveequationas:HHT(ej)=2hLP[2r]r=0Mcos(r)e-j2r.ThecorrespondingtransferfunctionoftheHilberttransformeristhereforegivenby:Notforsale8HHT(z)=2hLP[2n]cos(n)z-2nn=0M=2(-1)nhLP[2n]z-2nn=0M.10.8Weknowthatthefrequencyresponseoftheidealdifferentiatorisgivenby:HDIF(ej)=j.Henc