数字信 处理 基于计算机的方法 第四版 答案 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,ω<ωc,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,ω<ωc1,1,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,ω<ωc1,0,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,−π<ω<0,−j,0<ω<π.⎧⎨⎩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),−π<ω<0,−jX(ejω),0<ω<π.⎧⎨⎩(a)Let€y[n]=H{H{H{H{H{H{x[n]}}}}}}.Hence:€Y(ejω)=(j)6X(ejω),−π<ω<0,(–j)6X(ejω),0<ω<π,⎧⎨⎩=−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[]{}x[]=–∞∞∑=12πHHT(ejω)−ππ∫X(ejω)X(e–jω)dωwhere€HHT(ejω)=j,−π<ω<0,−j,0<ω<π.⎧⎨⎩Sincetheintegrand€HHT(ejω)X(ejω)X(e–jω)isanoddfunctionof€ω,€HHT(ejω)−ππ∫X(ejω)X(e–jω)dω=0.Notforsale7Asaresult:€Hx[]{}x[]=–∞∞∑=0.10.7Giventhefrequencyresponsefortheideallowpassfilter:€HLP(z)=hLP[n]z−nn=0N∑.ItsfrequencyresponseisshowninFigure(a)below.(a)Aplotofthefrequencyresponseof:€HHT(ejω)=€HLP(ej(ω−π2))+HLP(ej(ω+π2))isshowninFigure(b)below.(b)Itisevidentfromthisfigurethat€HHT(ejω)isthefrequencyresponseofanidealHilberttransformer.Therefore,wehave:€HHT(ejω)=HLP(ej(ω+π2))+HLP(ej(ω−π2))€=hLP[n]e−jn(ω+π2)n=0N∑+hLP[n]e−jn(ω−π2)n=0N∑€=hLP[n]n=0N∑e−jnωe−jnπ/2+ejnπ/2()€=2hLP[n]e−jnωn=0N∑cosnπ2⎛⎝⎜⎞⎠⎟.For€nodd,€cos(nω/2)=0andwecandropalloddtermsintheaboveexpression.Let€N=2Mwith€Nevenandlet€r=2n.Then,wecanrewritetheaboveequationas:€HHT(ejω)=2hLP[2r]r=0M∑cos(rπ)e−j2rω.ThecorrespondingtransferfunctionoftheHilberttransformeristhereforegivenby:Notforsale8€HHT(z)=2hLP[2n]cos(nπ)z−2nn=0M∑=2(−1)nhLP[2n]z−2nn=0M∑.10.8Weknowthatthefrequencyresponseoftheidealdifferentiatorisgivenby:€HDIF(ejω)=jω.Hence:€hDIF[n]=12πjωejωn−ππ∫dω=j2πωejωn−ππ∫dω=j2πωejωnjn+ejωnn2⎛⎝⎜⎜⎞⎠⎟⎟−ππ.Therefore:€hDIF[n]=cos(πn)n−sin(πn)πn2=cos(πn)n,n≠0.For€n=0:€hDIF[0]=12πjωdω−ππ∫=0.Hence:€hDIF[n]=0,n=0,cos(πn)n,|n|>0.⎧⎨⎪⎩⎪Since€hDIF[n]=−hDIF[−n],thetruncatedimpulseresponseisaType3linear-phaseFIRfilter.Themagnituderesponsesoftheabovedifferentiatorforseveralvaluesof€Maregivenbelow:10.9GiventhatN=2M+1,wecanwritetheimpulseresponseofthehighpassfilteras:€ˆhHP[n]=1−ωcπ,forn=M,−sinωc(n−m)()π(n−m),ifn≠M,0≤n<N,0,otherwise.⎧⎨⎪⎪⎪⎩⎪⎪⎪Next,wecheckforthedelaycomplementaryproperty:€ˆHHP(z)+ˆHLP(z)=ˆhHP[n]z−nn=−∞∞∑+ˆhLP[n]z−nn=−∞∞∑€=ˆhHP[n]z−nn=0N−1∑+ˆhLP[n]z−nn=0N−1∑€=ˆhHP[n]+ˆhLP[n]()z−nn=0N−1∑.Notforsale9But:€ˆhHP[n]+ˆhLP[n]=€ˆhHP[n]+ˆhLP[n]=0,0≤n≤N–1,n≠M,1,n=M.⎧⎨⎩,Hence,€ˆHHP(z)+ˆHLP(z)=z–M,andthetwofiltersaredelay-complementary.10.10Thefrequencyresponseofthezero-phaseideallinearpassbandlowpassfilteris:€HLLP(ejω)=ω,ω<ωc,0,otherwise.⎧⎨⎩Therefore,theimpulseresponseiscomputedas:€hLLP[n]=12π−ωejωndω−ωc0∫+ωejωndω0ωc∫⎛⎝⎜⎜⎞⎠⎟⎟€=12π−ωejωnjn+ejωnn2⎡⎣⎢⎢⎤⎦⎥⎥–ωc0+ωejωnjn+ejωnn2⎡⎣⎢⎢⎤⎦⎥⎥0ωc⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟€=12πωcejωcn−ωce−jωcnjn+ejωcn+e−jωcn−2n2⎛⎝⎜⎜⎞⎠⎟⎟€=ωcπnsin(ωcn)+cos(ωcn)−1πn2.10.11Thefrequencyresponseofthezero-phaseidealband-limiteddifferentiatoris:€HBLDIF(ejω)=ω,ω<ωc,0,otherwise.⎧⎨⎩Hence,theimpulseresponsecanbefoundasfollows:€hBLDIF[n]=12πωejωndω−ωcωc∫=12πωejωnjn+ejωnn2⎡⎣⎢⎢⎤⎦⎥⎥−ωcωc⎛⎝⎜⎜⎜⎞⎠⎟⎟⎟€=12πωcejωcn+ωce−jωcnjn+ejωcn−e−jωcnn2⎛⎝⎜⎜⎞⎠⎟⎟€=–jωcπncos(ωcn)+j1πn2sin(ωcn).10.12Thefrequencyresponseofacausalidealnotchfiltercanthusbeexpressedas:€Hnotch(ejω)=Hnotch(ω)ejθ(ω),where€Hnotch(ω)istheamplituderesponsewhichcanbeexpressedas:Notforsale10€Hnotch(ω)=1,0≤ω≤ωo,−1,ωo<ω<π.⎧⎨⎩Itfollowsthenthat€Hnotch(ω)isrelatedtotheamplituderesponse€HLP(ω)oftheideallowpassfilterwithacutoffat€ωothrough€Hnotch(ω)=±[2HLP(ω)−1].Hence,theimpulseresponseoftheidealnotchfilterisgivenby€hnotch[n]=±2hLP[n]−δ[n][],where:€hLP[n]=sin(ωon)πn,−∞<n<∞.Therefore:€hnotch[n]=±1,n=0,±2sin(ωon)πn,otherwise.⎧⎨⎪⎩⎪10.13First,consideranotherfilterwithafrequencyresponse€G(ejω)givenby€G(ejω)=0,0≤ω≤ωp,−π2Δωsinπ(ω−ωp)Δω⎛⎝⎜⎞⎠⎟,ωp<ω≤ωs,−π2Δωsinπ(ω+ωp)Δω⎛⎝⎜⎞⎠⎟,−ωs≤ω≤−ωp,0,elsewhere.⎧⎨⎪⎪⎪⎩⎪⎪⎪Clearly:€G(ejω)=dHLP(ejω)dω.Findingtheimpulseresponseofthesecondaryfunction:€g[n]=12πG(ejω)ejωndω−ππ∫€=−π8πΔωjejπ(ω−ωp)Δω⎛⎝⎜⎜⎞⎠⎟⎟ejωndωωpωs∫−e−jπ(ω−ωp)Δω⎛⎝⎜⎜⎞⎠⎟⎟ejωndωωpωs∫⎛⎝⎜⎜⎜⎜€+ejπ(ω+ωp)Δω⎛⎝⎜⎜⎞⎠⎟⎟ejωndω−ωs−ωp∫−e−jπ(ω+ωp)Δω⎛⎝⎜⎜⎞⎠⎟⎟ejωndω−ωs−ωp∫⎞⎠⎟⎟⎟⎟€=−1j8Δωe−jπωpΔωejωsn+πΔω⎛⎝⎜⎞⎠⎟−ejωpn+πΔω⎛⎝⎜⎞⎠⎟)jn+πΔω⎛⎝⎜⎞⎠⎟⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥⎛⎝⎜⎜⎜⎜−e−jπωpΔωejωsn−πΔω⎛⎝⎜⎞⎠⎟−ejωpn−πΔω⎛⎝⎜⎞⎠⎟jn−πΔω⎛⎝⎜⎞⎠⎟⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥Notforsale11€+ejπωpΔωe−jωpn+πΔω⎛⎝⎜⎞⎠⎟−e−jωsn+πΔω⎛⎝⎜⎞⎠⎟jn+πΔω⎛⎝⎜⎞⎠⎟⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥−e−jπωpΔωe−jωpn−πΔω⎛⎝⎜⎞⎠⎟−e−jωsn−πΔω⎛⎝⎜⎞⎠⎟jn−πΔω⎛⎝⎜⎞⎠⎟⎡⎣⎢⎢⎢⎢⎤⎦⎥⎥⎥⎥⎞⎠⎟⎟⎟⎟€=−14Δωj−sin(ωsn)−sin(ωpn)n+πΔω⎛⎝⎜⎞⎠⎟−−sin(ωsn)−sin(ωpn)n−πΔω⎛⎝⎜⎞⎠⎟⎛⎝⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟€=sin(ωsn)+sin(ωpn)()4Δωj−2π/Δωn2−π2Δω2⎛⎝⎜⎜⎜⎜⎞⎠⎟⎟⎟⎟€=sin(ωcn)cos(Δωn/2)Δωj1π1−(Δω/π)2n2()⎛⎝⎜⎜⎞⎠⎟⎟.UsingthepropertiesoftheDTFT,weknowthat:€hLP[n]=jng[n].Therefore:€hLP[n]=cos(Δωn/2)1−(Δω/π)2n2⎛⎝⎜⎞⎠⎟sin(ωcn)πn⎛⎝⎜⎞⎠⎟.10.14FromEq.(10.9):€ΦR=12πHtejω()−Hdejω()2dω−ππ∫,,where€Htejω()=htn[]e−jωnn=−MM∑.UsingParseval’srelation,wecanwrite:€ΦR=htn[]−hdn[]2n=−∞∞∑€=htn[]−hdn[]2+hd2n[]n=−∞−M−1∑+hd2n[]n=M+1∞∑n=−MM∑.Therefore:€ΦHaan=hdn[]⋅wHannn[]−hdn[]2n=−∞∞∑€=hdn[]12+12cos2πn2M+1⎛⎝⎜⎞⎠⎟⎛⎝⎜⎞⎠⎟−hdn[]2n=−MM∑+hd2n[]n=−∞−M−1∑+hd2n[]n=M+1∞∑Notforsale12Hence:€ΦExcess=ΦR−ΦHaan€=hdn[]⋅wRn[]−hdn[]2n=−MM∑€−hdn[]⋅12+12cos2πn2M+1⎛⎝⎜⎞⎠⎟⎛⎝⎜⎞⎠⎟−hdn[]2n=−MM∑€=−hdn[]2cos2πn2M+1⎛⎝⎜⎞⎠⎟−hdn[]22n=−MM∑€=121+2M()cos2πM2M+1⎛⎝⎜⎞⎠⎟−12.10.15ForaHammingwindow:€ΦHamm=hdn[]⋅wHammn[]−hdn[]2n=−∞∞∑.Recall€ΦR=htn[]−hdn[]2n=−∞∞∑.Therefore:€ΦExcess=ΦR−ΦHamm€=−hdn[]0.46cos2πn2M+1⎛⎝⎜⎞⎠⎟−0.46⎛⎝⎜⎞⎠⎟2n=−MM∑€=−0.46hdn[]cos2πn2M+1⎛⎝⎜⎞⎠⎟−1⎛⎝⎜⎞⎠⎟2n=−MM∑€=0.462M+1()cos2πM2M+1⎛⎝⎜⎞⎠⎟−12.10.16Theresponsesofanideallowpassfilterwithacutofffrequency€ωc=π/2are:€hHBn[]=sinωcn()πn=sinπn/2()πn,€HHB(ejω)=1,−π/2<ω<π/2,0,otherwise.⎧⎨⎩ThefrequencyresponsesoftheHilbertTransformer(fromEq.(10.24))andtheidealdiscrete-timeDifferentiator(fromEq.(10.26))areasfollows:€HHT(ejω)=j,−π<ω<0,−j,0<ω<π.⎧⎨⎩€HDIF(ejω)=jω.InordertogetderivetheimpulseresponseoftheHilbertTransformer,wenotethat:€HHT(ejω)=jHHB(ejω−π/2())−jHHB(ejω+π/2()).Therefore,usingthepropertiesoftheDTFT,weget:€hHTn[]=jhHB[n]ejn(π/2)−jhHB[n]e−jn(π/2)€=2hHB[n]12jejn(π/2)−12je−jn(π/2)⎛⎝⎜⎞⎠⎟€=2nπsinnπ2⎛⎝⎜⎞⎠⎟sinnπ2⎛⎝⎜⎞⎠⎟.InordertoderivetheimpulseresponseoftheDifferentiator,wenotethat:€HDIF(ejω)=jωHHB(ejω−π/2())+jωHHB(ejω+π/2()).Therefore,usingthepropertiesoftheDTFT,andinparticular:€∂∂nx[n]⇔jωXejω()Weget(byuseofthedoubleangleformula):€hDIFn[]=∂∂nhHB[n]ejn(π/2)()+∂∂nhHB[n]e−jn(π/2)()Notforsale13€=2∂∂nhHB[n]12ejn(π/2)+12e−jn(π/2)⎡⎣⎢⎤⎦⎥⎛⎝⎜⎞⎠⎟€=2∂∂nhHB[n]cosπn2⎛⎝⎜⎞⎠⎟⎛⎝⎜⎞⎠⎟€=∂∂n2sin(πn/2)cos(πn/2)πn⎛⎝⎜⎞⎠⎟€=∂∂nsin(πn)πn⎛⎝⎜⎞⎠⎟€=cos(πn)πn−sin(πn)π2n2.10.17Foreachproblem,thecodesusedtogeneratetheplotsaregivenbelow,assumingthat€ωcistheaverageofthestopbandandpassbandfrequencies:fRange=-M:M;idealLPF=(wc/pi)*sinc((wc/pi)*fRange);fNum=idealLPF.*hann(L)';[h,w]=freqz(fNum,1,512);plot(w/pi,20*log10(abs(h)));grid;xlabel('\omega/\pi');ylabel('Gain,dB');(a)Given:€ωp=0.42π,ωs=0.58π,δp=0.002,δs=0.008.Thus:€Δω=0.16π,€αs=−20log10δs=41.93dB.FromTable10.2,weseethatforfixed-windowfunctions,wecanachievetheminimumstopbandattenuationbyusingHann,Hamming,orBlackmanwindows.Hannwillhavethelowestfilterlength:Since€M=3.11π0.16π=19.43,€NHaan=2M+1⎡⎤=40.Thecorrespondingfrequencyresponse,generatedwiththecode,isshownbelow:(b)Given:€ωp=0.65π,€ωs=0.76π,€δp=0.002,€δs=0.004.Thus:€Δω=0.11π,€αs=−20log10δs=47.95dB.Notforsale14FromTable10.2,weseethatforfixed-windowfunctions,wecanachievetheminimumstopbandattenuationbyusingHann,Hamming,orBlackmanwindows.Hannwillhavethelowestfilterlength:Since€M=3.11π0.16π=28.27,€NHaan=2M+1⎡⎤=59.Thecorrespondingfrequencyresponse,generatedwiththecode,isshownbelow:10.18Thecodefortheplottingofthefrequencyresponseisasfollows:fRange=-M:M;idealBPF=(wc2/pi)*sinc((wc2/pi)*fRange)...-(wc1/