数字信号处理-基于计算机的方法(第四版)答案 8-11章

Notforsale1SOLUTIONSMANUALtoaccompanyDigitalSignalProcessing:AComputer-BasedApproachFourthEditionSanjitK.MitraPreparedbyChowdaryAdsumilli,JohnBerger,MarcoCarli,Hsin-HanHo,RajeevGandhi,MartinGawecki,ChinKayeKoh,LucaLucchese,MyleneQueirozdeFarias,andTravisSmithCopyright©2011bySanjitK.Mitra.Nopartofthispublicationmaybereproducedordistributedinanyformorbyanymeans,orstoredinadatabaseorretrievalsystem,withoutthepriorwrittenconsentofSanjitK.Mitra,including,butnotlimitedto,inanynetworkorotherelectronicStorageortransmission,orbroadcastfordistancelearning.Notforsale2Chapter10–Part110.1Forthisproblem,wehave€N=71and€ωs−ωp=0.04πandweassumethat€δp=δs.(a)UsingKaiser’sformulaofEq.(10.3):€δs=10N14.6()ωs−ωp()/2π[]−13−20⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟=107114.6()0.04π()/2π[]−13−20⎛⎝⎜⎞⎠⎟=0.0206.€αs=−20log10δs()=33.7320dB.(b)UsingBellanger’sformulaofEq.(10.4):€δs=0.1⋅10−3(N+1)ωs−ωp()/2π2⎛⎝⎜⎜⎞⎠⎟⎟⎛⎝⎜⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟⎟1/2=0.1⋅10−72⋅3⋅0.04π/2π2⎛⎝⎜⎞⎠⎟⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟1/2=0.0263.€αs=−20log10δs()=31.6dB.(c)UsingHermann’sformulaofEq.(10.5),wefirstfindthat:€Fδs,δs()=b1+b2log10δs−log10δs()=b1=11.01217.Therefore:€D∞δs,δs()=Nωs−ωp()/2π+Fδs,δs()ωs−ωp()2/4π2=710.04π()/(2π)+11.012170.04π()2/(2π)2=1.4244.Solvingfor:€D∞δs()=a1log10δs()2+a2log10δs()+a3⎡⎣⎢⎤⎦⎥log10δs()−a4log10δs()2+a5log10δs()+a6⎡⎣⎢⎤⎦⎥=a1log10δs()3+a2log10δs()2+a3log10δs()−a4log10δs()2−a5log10δs()−a6=a1log10δs()3+a2−a4()log10δs()2+a3−a5()log10δs()−a6.Let€x=log10δs(),andthusNotforsale3€D∞δs()=0.005309x3+0.06848x2−1.0702x−0.4278=1.4244.Solvingthis,weget:€x=−21.5147,10.2049,−1.589.Themostreasonablesolutionisthelastone,whichyields:€δs=10x=10−1.5890=0.0258,€αs=−20log10δs()=31.78dB.10.2Forthisproblem,wehave€N=71and€ωs−ωp=0.04π=Δω.UsingEq.(10.45):€αs=2.285Δω()N+8=2.2850.04π()71+8=28.3871dB.10.3(a)FromEq.(10.17):€HHP(ejω)=0,ωωc,1,ωc≤ω≤π,⎧⎨⎩€hHPn[]=12πHHPejω()ejωndω−ππ∫€=12πejωndω−π−ωc∫+12πejωndωωcπ∫€=12πejωnjn⎡⎣⎢⎢⎤⎦⎥⎥−π−ωc+12πejωnjn⎡⎣⎢⎢⎤⎦⎥⎥ωcπ€=12πe−jωcnjn−e−jπnjn⎡⎣⎢⎢⎤⎦⎥⎥+12πejπnjn−ejωcnjn⎡⎣⎢⎢⎤⎦⎥⎥€=sin(πn)πn−sin(ωcn)πn.Usingthepropertiesofthesincfunction,wearriveat:€hHP0[]=1−ωcπ.Therefore:€hHPn[]=1−ωcπ,n=0,−sin(ωcn)πn,otherwise.⎧⎨⎪⎩⎪(b)FromEq.(10.18):€HBP(ejω)=0,ωωc1,1,ωc1≤ω≤ωc2,0,ωc2≤ω≤π,⎧⎨⎪⎩⎪€hBPn[]=12πHBP(ejω)ejωndω−ππ∫€=12πejωndω−ωc2−ωc1∫+12πejωndωωc1ωc2∫€=12πejωnjn⎡⎣⎢⎢⎤⎦⎥⎥−ωc2−ωc1+12πejωnjn⎡⎣⎢⎢⎤⎦⎥⎥ωc1ωc2€=12πe−jωc1njn−e−jωc2njn⎡⎣⎢⎢⎤⎦⎥⎥+12πejωc2njn−ejωc1njn⎡⎣⎢⎢⎤⎦⎥⎥Notforsale4€=sin(ωc2n)πn−sin(ωc1n)πn.(c)FromEq.(10.20):€HBS(ejω)=1,ωωc1,0,ωc1≤ω≤ωc2,1,ωc2≤ω≤π.⎧⎨⎪⎩⎪€hBSn[]=12πHBS(ejω)ejωndω−ππ∫€=12πejωndω−π−ωc2∫+12πejωndωωc2π∫+12πejωndω−ωc1ωc1∫€=12πejωnjn⎡⎣⎢⎢⎤⎦⎥⎥−π−ωc2+12πejωnjn⎡⎣⎢⎢⎤⎦⎥⎥ωc2π+12πejωnjn⎡⎣⎢⎢⎤⎦⎥⎥−ωc1ωc1€=12πe−jωcnjn−e−jπnjn⎡⎣⎢⎢⎤⎦⎥⎥+12πejπnjn−ejωcnjn⎡⎣⎢⎢⎤⎦⎥⎥12πe−jωc1njn−e−jωc1njn⎡⎣⎢⎢⎤⎦⎥⎥€=sin(πn)πn−sin(ωc2n)πn+sin(ωc1n)πn.Usingthepropertiesofthesincfunction,wearriveat:€hBS0[]=1−ωc2−ωc1π.Therefore:€hHPn[]=1−ωc2−ωc1π,n=0,sin(ωc1n)πn−sin(ωc2n)πn,otherwise.⎧⎨⎪⎩⎪10.4TheidealL-banddigitalfilter€HML(z)hasafrequencyresponsegivenby:€HML(ejω)=Ak,for€ωk−1≤ω≤ωk,1≤k≤L.Itcanbeconsideredassumofidealbandpassfilterswithcutofffrequenciesat:€ωc1k=ωk−1and€ωc2k=ωk,where€ωc10=0and€ωc2L=π.FromEqn.(10.19)theimpulseresponseofanidealbandpassfilterisgivenby:€hBP[n]=sin(ωc2n)πn−sin(ωc1n)πn.Therefore:€hBPk[n]=sin(ωkn)πn−sin(ωk−1n)πn..Hence:€hML[n]=hBPk[n]k=1L∑=Aksin(ωkn)πn−sin(ωk−1n)πn⎛⎝⎜⎞⎠⎟k=1L∑€=A1sin(ω1n)πn−sin(0n)πn⎛⎝⎜⎞⎠⎟+Aksin(ωkn)πnk=2L−1∑€−Aksin(ωkn)πn−sin(ωk−1n)πn⎛⎝⎜⎞⎠⎟k=2L−1∑+ALsin(ωLn)πn−sin(ωL−1n)πn⎛⎝⎜⎞⎠⎟Notforsale5€=A1sin(ω1n)πn+Aksin(ωkn)πnk=2L−1∑€−Aksin(ωk−1n)πnk=2L−1∑−ALsin(ωL−1n)πn€=Aksin(ωkn)πn−Aksin(ωk−1n)πnk=2L∑k=1L−1∑.Since€ωL=π,sin(ωLn)=0.Wecanaddatermof:€ALsin(ωLn)πntothefirstsumintheaboveexpressionandchangetheindexrangeofthesecondsum,resultingin:€hML[n]=Aksin(ωkn)πn−Ak+1sin(ωkn)πnk=1L−1∑k=1L∑.Finally,since€AL+1=0,wecanaddaterm:€AL+1sin(ωLn)πntothesecondsum.Thisleadsto:€hML[n]=Aksin(ωkn)πn−Ak+1sin(ωkn)πnk=1L∑k=1L∑€=(Ak−Ak+1)sin(ωkn)πnk=1L∑.10.5TheimpulseresponsefortheHilbertTransformerisgivenby:€HHT(ejω)=j,−πω0,−j,0ωπ.⎧⎨⎩Therefore:€hHT[n]=12πHHT(ejω)ejωndω−π0∫+12πHHT(ejω)ejωndω0π∫€=12πjejωndω−π0∫−12πjejωndω0π∫€=22πn1−cos(πn)()=2sin2(πn/2)πn,n≠0.For€n=0,€hHT[0]=12πjdω−π0∫−12πjdω0π∫=0.Hence:€hHT[n]=0,ifn=0,2sin2(πn/2)πn,ifn≠0.⎧⎨⎪⎩⎪Since€hHT[n]=−hHT[−n],andthelengthofthetruncatedimpulseresponseisodd,itisaType3linear-phaseFIRfilter.Notforsale6Fromthefrequencyresponseplotsgivenabove,weobservethepresenceofripplesatthebandedgesduetotheGibbsphenomenoncausedbythetruncationoftheimpulseresponse.10.6First,wenotethat:€H{x[n]}=hHT[n−k]x[k]k=−∞∞∑..Hence,€FHx[n]{}{}=HHT(ejω)X(ejω)€=jX(ejω),−πω0,−jX(ejω),0ωπ.⎧⎨⎩(a)Let€y[n]=H{H{H{H{H{H{x[n]}}}}}}.Hence:€Y(ejω)=(j)6X(ejω),−πω0,(–j)6X(ejω),0ωπ,⎧⎨⎩=−X(ejω).Therefore,€y[n]=−x[n].(b)Define€g[n]=Hx[n]{},and€h*[n]=x[n].Then:€Hx[]{}x[]=–∞∞∑=g[]h*[]=–∞∞∑.ButfromParseval'srelationinTable3.4:€g[]h*[]=–∞∞∑=12πG(ejω)–ππ∫G(ejω)dω.Therefore:€Hx