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1、Signals&Systems(SecondEdition)—LearningInstructions(ExercisesAnswers)DepartmentofComputerEngineering2005.12ContentsChapter1·······················································2Chapter2·······················································17Chapter3············································
2、···········35Chapter4·······················································62Chapter5·······················································83Chapter6·······················································109Chapter7·······················································119Chapter8·············
3、··········································132Chapter9·······················································140Chapter10·······················································1601Chapter1Answers1.1ConvertingfrompolartoCartesiancoordinates:1j111j11ecosecos()222222jje2cos()jjs
4、in()e2cos()jjsin()22225jjjee22j42(cos()jjsin())12e449jj9jj441j22ee441j22eej2e41j1.2convertingfromCartesiantopolarcoordinates:j0jj55e,22e,33je213j2jjje2,12je4,1j2e2221j22jjj(1)e4,412ee1j13j4t11.3.(a)E=0edt,P=0,becauseE4j
5、t(2)2(b)xe()t4,()t1.Therefore,E=x()tdt=dt=,2x2211TT2P=limx2()tdtlimdtlim11TT22TTTTT2(c)()t=cos(t).Therefore,E==2=,x2x3()tdtcos()tdtP=1TT211COSt(2)1limcos()tdtlimdtTT2TTTT222nnn(d)1,21.Therefore,E=214x1[]nun[]x1[]nun[]x1[]n3
6、24n04P=0,becauseE<.n(e)j(),2=1.therefore,E=2=,x[]n=282ex2[]nx[]n211NN2P=limx2[]nlim11.NN2NN1nN21nNn22(f)x[]n.Therefore,E=2==,3=cosx3[]ncos(n)cos(n)444nN1cos(n)P=1Nlim11(2)limcosN2N1nN22N21NnN41.4.(a)Thesignalx
7、[n]isshiftedby3totheright.Theshiftedsignalwillbezeroforn<1,Andn>7.(b)Thesignalx[n]isshiftedby4totheleft.Theshiftedsignalwillbezeroforn<-6.Andn>0.(c)Thesignalx[n]isflippedsignalwillbezeroforn<-1andn>2.(d)Thesignalx[n]isflippe