信号处理导论_Sophocles_J._Orfanidis_参考答案

INTRODUCTIONTOSignalProcessingSolutionsManualSophoclesJ.OrfanidisDepartmentofElectrical&ComputerEngineeringRutgersUniversity,Piscataway,[e-mail]©2010bySophoclesJ.OrfanidisWebpage:~orfanidi/i2spChapter1ProblemsProblem1.1TheNyquistintervalis[−fs/2,fs/2]=[−4,4]Hz.The6Hzfrequencyofthewheelliesoutsideit,therefore,itwillbealiasedwithf−fs=6−8=−2Hz.Thus,thewheelwillappeartobeturningat2Hzintheoppositedirection.Iffs=12,theNyquistintervalis[−6,6].Pointsonthewheelwillappeartobemovingupanddown,witheitherpositiveornegativesenseofrotation.Fortheothertwosamplingfrequencies,theNyquistintervalis[−8,8]or[−12,12]Hz,andtherefore,theoriginal6Hzfrequencyliesinitandnoaliasingwillbeperceived.Problem1.2Thethreetermsofthesignalcorrespondtofrequenciesf1=1,f2=4,andf3=6Hz.Ofthese,f2andf3lieoutsidetheNyquistinterval[−2.5,2.5].Therefore,theywillbealiasedwithf2−fs=4−5=−1andf3−fs=6−5=1andthealiasedsignalwillbe:xa(t)=10sin(2πt)+10sin(2π(−1)t)+5sin(2πt)=5sin(2πt)Toshowthattheyhavethesamesamplevalues,wesett=nT,withT=1/fs=1/5sec.Then,x(nT)=10sin(2πn/5)+10sin(8πn/5)+5sin(12πn/5)But,sin(8πn/5)=sin(2πn−2πn/5)=−sin(2πn/5)andsin(12πn/5)=sin(2πn+2πn/5)=sin(2πn/5)Thus,x(nT)=10sin(2πn/5)−10sin(2πn/5)+5sin(2πn/5)=5sin(2πn/5)=xa(nT).Iffs=10Hz,thentheNyquistintervalis[−5,5]Hzandonlyf3liesoutsideit.Itwillbealiasedwithf3−fs=6−10=−4resultinginthealiasedsignal:xa(t)=10sin(2πt)+10sin(8πt)+5sin(2π(−4)t)=10sin(2πt)+5sin(8πt)Problem1.3Usingthetrigidentity2sinαsinβ=cos(α−β)−cos(α+β),wefind:x(t)=cos(5πt)+4sin(2πt)sin(3πt)=cos(5πt)+2[cos(πt)−cos(5πt)]=2cos(πt)−cos(5πt)Thefrequenciesinthesignalaref1=0.5andf2=2.5kHz.TheNyquistintervalis[−1.5,1.5]kHz,andf2liesoutsideit.Thus,itwillbealiasedwithf2a=2.5−3=−0.5givingrisetothesignal:1xa(t)=2cos(2πf1t)−cos(2πf2at)=2cos(πt)−cos(−πt)=cos(πt)Aclassofsignalsaliasedwithx(t)andxa(t)isobtainedbyreplacingf1andf2bytheirshiftedversions:f1+mfs,f2+nfsresultingin:xmn(t)=2cos(πt+6πmt)−cos(πt−6πnt)Problem1.4Usingatrigidentity,wewritex(t)=cos(8πt)+cos(10πt)+cos(2πt).Thefrequenciescon-tainedinx(t)arethus,1Hz,4Hz,and5Hz.Ifthesamplingrateis5Hz,thentheNyquistintervalis[−2.5,2.5]Hz,andtherefore,the4Hzand5Hzcomponentslieoutsideitandwillbealiasedwiththefrequencies4−5=−1Hzand5−5=0Hz,etc.Problem1.5Usingtrigidentitieswefind:x(t)=sin(6πt)1+2cos(4πt)=sin(6πt)+sin(10πt)+sin(2πt)withfrequencycontent:f1=3,f2=5,f3=1kHz.TheNyquistintervalis[−2,2]kHz,andthealiasedfrequenciesare:f1a=f1−fs=3−4=−1,f2a=f2−fs=5−4=1,f3a=f3=1Thus,xa(t)=sin(−2πt)+sin(2πt)+sin(2πt)=sin(2πt)Problem1.6Usethetrigonometricidentity2cosacosb=cos(a+b)+cos(a−b)threetimestogetx(t)=2[cos(10πt)+cos(6πt)]cos(12πt)=cos(22πt)+cos(2πt)+cos(18πt)+cos(6πt)Thefrequenciespresentinthissignalaref1=11,f2=1,f3=9,andf4=3Hz.Withasamplingrateof10Hz,onlyf1andf3lieoutsidetheNyquistinterval[−5,5]Hz,andtheywillbealiasedwithf1−fs=11−10=1Hzandf3−fs=9−10=−1Hz.Thealiasedsignalwillbe:xa(t)=cos(2π(1)t)+cos(2πt)+cos(2π(−1)t)+cos(6πt)=3cos(2πt)+cos(6πt)Toprovetheequalityofthesamples,replacet=nT=n/10,becauseT=1/fs=1/10.Then,x(nT)=cos(22πn/10)+cos(2πn/10)+cos(18πn/10)+cos(6πn/10)But,cos(22πn/10)=cos(2πn+2πn/10)=cos(2πn/10)andsimilarly,cos(18πn/10)=cos(2πn−2πn/10)=cos(2πn/10).Therefore,thesamplevaluesbecome2x(nT)=cos(2πn/10)+cos(2πn/10)+cos(2πn/10)+cos(6πn/10)=3cos(2πn/10)+cos(6πn/10)=xa(nT)Iffs=12Hz,thenf1andf3willlieoutsideof[−6,6]andwillbealiasedwith11−12=−1and9−12=−3.Thealiasedsignalwillbe:xa(t)=cos(2π(−1)t)+cos(2πt)+cos(2π(−3)t)+cos(6πt)=2cos(2πt)+2cos(6πt)Problem1.7Weusethesametechniqueasinthesquare-waveExample1.4.6.Atasamplingrateof8Hz,thesignalfrequenciesof{1,3,5,7,9,11,13,15,...}willbealiasedwith:{1,3,−3,−1,1,3,−3,−1,...}Thereforeonlysin(2πt)andsin(6πt)termswillappearinthealiasedsignal.Thus,wewriteitintheform:xa(t)=Bsin(2πt)+Csin(6πt)TodetermineBandC,wedemandthatxa(t)andx(t)agreeatthesamplinginstantst=nT=n/8,becauseT=1/fs=1/8sec.Therefore,wedemandBsin(2πn/8)+Csin(6πn/8)=x(n/8)Settingn=1,2,wegettwoequationsBsin(2π/8)+Csin(6π/8)=x(1/8)=0.5Bsin(4π/8)+Csin(12π/8)=x(2/8)=1⇒B1√2+C1√2=0.5B−C=1Thevaluesforx(1/8)andx(2/8)wereobtainedbyinspectingthetriangularwaveform.SolvingforBandC,wefind:B=√2+24,C=√2−24Problem1.8Forfs=5kHz,wehave:xa(t)=sin(2πf1t)Forfs=10kHz,wehave:xa(t)=2sin(2πf1t)+sin(2πf2t)3Fig.P1.1Parts(a,b)ofProblem1.8.Problem1.9Theaudibleandinaudiblepartsare:x(t)=sin(10πt)+sin(20πt)audible+sin(60πt)+sin(90πt)inaudibleThefrequenciesofthefourtermsandtheiraliasedversionsare:fA=5fB=10fC=30fD=45⇒fAa=5fBa=10fCa=30−40=−10fDa=45−40=5a.Whenthereisnoprefilterthealiasedsignalattheoutputofthereconstructorwillbe:ya(t)=sin(10πt)+sin(20πt)+sin(−20πt)+sin(10πt)=2sin(10πt)b.Whenthereisaperfectprefilter,thefC,fDcomponentsareremovedpriortosamplingthus,thesampledandreconstructedsignalwillremainunchanged,thatis,theaudiblepartofx(t):ya(t)=sin(10πt)+sin(20πt)c.TheattenuationsintroducedbythepracticalprefiltershowninFig.P1.2areobtainedbydeterminingthenumberofoctavestothefrequenciesfC,fDandmultiplyingbythefilter’sattenuationof48dB/octave:log2fCfs/2 =log23020=0.585⇒AC=48×0.585=28.08dBlog2