【非扫描 清晰版】 奥本海姆 信号与系统 第二版 习题解答 答案

Signals&Systems(SecondEdition)—LearningInstructions(ExercisesAnswers)DepartmentofComputerEngineering2005.121ContentsChapter1·······················································2Chapter2·······················································17Chapter3·······················································35Chapter4·······················································62Chapter5·······················································83Chapter6·······················································109Chapter7·······················································119Chapter8·······················································132Chapter9·······················································140Chapter10·······················································1602Chapter1Answers1.1ConvertingfrompolartoCartesiancoordinates:111cos222je111cos()222je2cos()sin()22jjje2cos()sin()22jjje522jjjee42(cos()sin())1442jjje944122jjjee944122jjjee412jje1.2convertingfromCartesiantopolarcoordinates:055je,22je,233jje21322jje,412jje,2221jje4(1)jje,411jje122213jje1.3.(a)E=4014tdte,P=0,becauseE(b)(2)42()jttxe,2()1tx.Therefore,E=22()dttx=dt=,P=211limlim222()TTTTTTdtdtTTtxlim11T(c)2()tx=cos(t).Therefore,E=23()dttx=2cos()dtt=,P=2111(2)1limlim2222cos()TTTTTTCOStdtdtTTt(d)1[][]12nnunx,2[]11[]4nunnx.Therefore,E=204131[]4nnnxP=0，becauseE.(e)2[]nx=()28nje,22[]nx=1.therefore,E=22[]nx=,P=211limlim1122121[]NNNNnNnNNNnx.(f)3[]nx=cos4n.Therefore,E=23[]nx=2cos()4n=2cos()4n,P=1limcos214nNNnNN1cos()112lim()2122NNnNnN1.4.(a)Thesignalx[n]isshiftedby3totheright.Theshiftedsignalwillbezeroforn1,Andn7.(b)Thesignalx[n]isshiftedby4totheleft.Theshiftedsignalwillbezeroforn-6.Andn0.(c)Thesignalx[n]isflippedsignalwillbezeroforn-1andn2.(d)Thesignalx[n]isflippedandtheflippedsignalisshiftedby2totheright.ThenewSignalwillbezeroforn-2andn4.(e)Thesignalx[n]isflippedandtheflippedandtheflippedsignalisshiftedby2totheleft.Thisnewsignalwillbezeroforn-6andn0.1.5.(a)x(1-t)isobtainedbyflippingx(t)andshiftingtheflippedsignalby1totheright.Therefore,x(1-t)willbezerofort-2.(b)From(a),weknowthatx(1-t)iszerofort-2.Similarly,x(2-t)iszerofort-1,Therefore,x(1-t)+x(2-t)willbezerofort-2.(c)x(3t)isobtainedbylinearlycompressionx(t)byafactorof3.Therefore,x(3t)willbezerofort1.3(d)x(t/3)isobtainedbylinearlycompressionx(t)byafactorof3.Therefore,x(3t)willbezerofort9.1.6(a)x1(t)isnotperiodicbecauseitiszerofort0.(b)x2[n]=1foralln.Therefore,itisperiodicwithafundamentalperiodof1.(c)x3[n]isasshownintheFigureS1.6.Therefore,itisperiodicwithafundamentalperiodof4.1.7.(a)1[]vnx=1111[][]([][4][][4])22nnununununxxTherefore,1[]vnxiszerofor1[]nx3.(b)Sincex1(t)isanoddsignal,2[]vnxiszeroforallvaluesoft.(c)11311[][][][3][3]221122vnnnnnununxxxTherefore,3[]vnxiszerowhenn3andwhenn.(d)1554411()(()())(2)(2)22vtttttututxxxeeTherefore,4()vtxiszeroonlywhent.1.8.(a)01{()}22cos(0)tttxe(b)02{()}2cos()cos(32)cos(3)cos(30)4tttttxe(c)3{()}sin(3)sin(3)2tttttxee(d)224{()}sin(100)sin(100)cos(100)2tttttttxeee1.9.(a)1()txisaperiodiccomplexexponential.101021()jtjttjxee(b)2()txisacomplexexponentialmultipliedbyadecayingexponential.Therefore,2()txisnotperiodic.（c）3[]nxisaperiodicsignal.3[]nx=7jne=jne.3[]nxisacomplexexponentialwithafundamentalperiodof22.(d)4[]nxisaperiodicsignal.ThefundamentalperiodisgivenbyN=m(23/5)=10().3mBychoosingm=3.Weobtainthefundamentalperiodtobe10.(e)5[]nxisnotperiodic.5[]nxisacomplexexponentialwith0w=3/5.Wecannotfindanyintegermsuchthatm(02w)isalsoaninteger.Therefore,5[]nxisnotperiodic.1.10.x(t)=2cos(10t＋1)-sin(4t-1)PeriodoffirsttermintheRHS=2105.PeriodoffirsttermintheRHS=242.Therefore,theoverallsignalisperiodicwithaperiodwhichtheleastcommonmultipleoftheperiodsofthefirstandsecondterms.Thisisequalto.-3-14……1……-10……-4……111-1n………5………x3[n]40-1-2-3123X[n]nFigureS1.1210-1210-1t1-2g(t)2-3-3tFigureS1.14x(t)1.11.x[n]=1+74jne−25jnePeriodoffirsttermintheRHS=1.PeriodofsecondtermintheRHS=7/42=7（whenm=2）PeriodofsecondtermintheRHS=5/22=5(whenm=1)Therefore,theoverallsignalx[n]isperiodicwithaperiodwhichistheleastcommonMultipleoftheperiodsofthethreetermsinnx[n].Thisisequalto35.1.12.Thesignalx[n]isasshowninfigureS1.12.x[n]canbeobtainedbyflippingu[n]andthenShiftingtheflippedsignalby3totheright.Therefore,x[n]=u[-n+3].ThisimpliesthatM=-1andno=-3.1.13y(t)=tdtx)(=dtt))2()2((=2,022,12,0,tttTherefore224dtE1.14Thesignalx(t)anditsderivativeg(t)areshowninFigureS1.14.Thereforekkktkttg12(3)2(3)()ThisimpliesthatA1=3,t1=0,A2=-3,andt2=1.1.15(a)Thesignalx2[n],whichistheinputtoS2,isthesameasy1[n].Therefore,y2[n]=x2[n-2]+21x2[n-3]=y1[n-2]+21y1[n-3]=2x1[n-2]+4x