SolutionsManual-Statistical and Adaptive Signal Processing 外文学习材料

SolutionsManual-Statistical and Adaptive Signal Processing 外文学习材料

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时间:2019-06-27

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1、SolutionsManualforStatisticalandAdaptiveSignalProcessingThispageisintentionallyleftblank.iiPrefaceThissolutionsmanualprovidessolutionstotheproblemscontainedintheÞrsteditionofourbookSTATISTICALANDADAPTIVESIGNALPROCESSING.ThesolutionshavebeenpreparedandwrittenbyDavidMardenandourselves.Wehaveattempt

2、edtoprovideverydetailedsolutionstotheproblemswithnotationconsistentwiththatusedinthebook.Whereapplicable,wehavealsogivenaprintoutofMatlabcodeandÞguresfortheproblemsolution.Despiteourefforts,however,youmayÞndthatsomeofthesolutionsmaybelessdetailedandreÞnedthanothers.Inevitablythroughtheuseofthisso

3、lutionsman-ual,omissionsanderrorswillbediscovered.Ourgoalistocontinuallyimproveuponthismanualusingyourcomments.Periodically,wewillissuechangestothesolutionsmanualandnewproblemstoyouuponrequest.Inthisrespect,wewouldappreciateanyfeedbackincludingimprovedsolutions,suggestionsfornewprob-lems,correcti

4、onswhichcanbesenttoProf.VinayK.Ingleatvingle@ece.neu.eduoratProf.VinayK.IngleDepartmentofElectricalandComputerEngineeringNortheasternUniversity360HuntingtonAvenueBoston,MA02115Thankyou.DimitrisG.ManolakisVinayK.IngleStephenK.KogoniiiThispageisintentionallyleftblank.ivChapter2Discrete-TimeSignalsa

5、ndSystems2.1SamplingfrequencyFs=100sam/sec(a)Continuous-timesignalxc(t)=2cos(40πt+π/3)hasfrequencyof20Hz.Hence40πnx(n)=xc(t)

6、t=n/Fs=2cos+π/3100whichimpliesthatω=40π=2π.01005(b)Steady-stateresponsey(t):Giventhath(n)=0.8nu(n),thefrequencyresponsefunctionisc,ss1jωH(e)=1−0.8e−jωSinceω0=2π/5,thesyst

7、emresponseatω0is1jω−j0.2517πH(e)==0.9343e1−0.8e−j2π/5Henceyss(n)=2(0.9343)cos(2πn/5+π/3−0.2517π),oryc,ss(t)=1.8686cos(40πt+0.585π)(c)Anyxc(t)thathasthesamedigitalfrequencyω0aftersamplingandthesamephaseshiftasabovewillhavethesamesteadystateresponse.SinceFs=100sam/sec,thetwootherfrequenciesare120an

8、d220Hz.2.2Thediscrete-timesignalisx(n)=Acos(ω0n)wR(n)wherewR(n)isanN-pointrectangularwindow.(a)TheDTFTofx(n)isdeterminedasjωX(e)=F[Acos(ω0n)wR(n)]=(A/2)Fejω0w(n)+(A/2)Fe−jω0w(n)(1)RRUsingtheDTFTofwR(n)asN−1sin(ωN/

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