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1、Chapter1ProblemsProblem1.1TheNyquistintervalis[−fs/2,fs/2]=[−4,4]Hz.The6Hzfrequencyofthewheelliesoutsideit,therefore,itwillbealiasedwithf−fs=6−8=−2Hz.Thus,thewheelwillappeartobeturningat2Hzintheoppositedirection.Iffs=12,theNyquistintervalis[−6,6].Pointsonthewheelwillappeartobemovingupandd
2、own,witheitherpositiveornegativesenseofrotation.INTRODUCTIONTOFortheothertwosamplingfrequencies,theNyquistintervalis[−8,8]or[−12,12]Hz,andSignalProcessingtherefore,theoriginal6Hzfrequencyliesinitandnoaliasingwillbeperceived.Problem1.2Thethreetermsofthesignalcorrespondtofrequenciesf1=1,f2=
3、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,SolutionsManualx(n
4、T)=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)SophoclesJ.Orfanidis=5sin(2πn/5)=xa(nT).DepartmentofElectrical&ComputerEngineeringIffs=10Hz,thentheNyquistintervalis[−5
5、,5]Hzandonlyf3liesoutsideit.ItwillbealiasedRutgersUniversity,Piscataway,NJ08855withf3−fs=6−10=−4resultinginthealiasedsignal:orfanidi@ece.rutgers.eduxa(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)+4
6、sin(2πt)sin(3πt)=cos(5πt)+2[cos(πt)−cos(5πt)]=2cos(πt)−cos(5πt)Copyright©2010bySophoclesJ.OrfanidisThefrequenciesinthesignalaref1=0.5andf2=2.5kHz.TheNyquistintervalis[−1.5,1.5]Webpage:www.ece.rutgers.edu/~orfanidi/i2spkHz,andf2liesoutsideit.Thus,itwillbealiasedwithf2a=2.5−3=−0.5givingrise
7、tothesignal:1xa(t)=2cos(2πf1t)−cos(2πf2at)=2cos(πt)−cos(−πt)=cos(πt)x(nT)=cos(2πn/10)+cos(2πn/10)+cos(2πn/10)+cos(6πn/10)=3cos(2πn/10)+cos(6πn/10)=xa(nT)Aclassofsignalsaliasedwithx(t)andxa(t)isobtainedbyreplacingf1andf2bytheirshiftedversions:f1+mfs,f2+nfsresultingin