信号与系统奥本海姆英文版课后答案chapter3

53Chapter3Answers3.1UsingtheFourierseriessynthesiseq.(3.38)(2)(2)3(2)3(2)1133()jTtjTtjTtjTtxtaeaeaeae(28)(28)3(28)3(28)2244jtjtjtjteeee64cos()8sin()48tt34cos()8cos()442tt3.2UsingtheFourierseriessynthesiseq.(3.95)2(2)2(2)4(2)4(2)02244[]jNnjNnjNnjNnxnaaeaeaeae(4)2(25)(4)2(25)1jjnjjneeee4812cos()4cos()5453nn438512sin()4sin()5456nn3.3Thegivensignalis(23)(23)(53)(53)11()22222jtjtjtjtxteejeje2(26)2(26)5(26)5(26)1122222jtjtjtjteejejeFormthiswemayconcludethatthefundamentalfrequencyofx(t)is263.Thenon-zeroFourierseriescoefficientsofx(t)are02a,2212aa,*552aaj3.4Since0,022T,Therefore,201()2jktkaxtedtNow,12001111.51.5022adtdtandfor0k1201111.51.522jktjktkaedtedt3[1]2jkekj(2)3sin()2jkkek3.5Both1(1)xtand1(1)xtareperiodicwithfundamentalperiod112T,Sincey(t)isalinearcombinationof1(1)xtand1(1)xt,itisalsoperiodicwithfundamentalperiod212T,Therefore21.Since1()FSkxta.usingtheresultsinTable3.1wehave1(2)1(1)jkTFSkxtae5411(2)(2)11(1)(1)jkTjkTFSFSkkxtaextaeTherefore111(2)(2)11(1)(1)()jkTjkTjkFSkkkkxtxtaeaeeaa3.6(a)Comparing1()xtwiththeFourierseriessynthesiseq.(3.38),weobtaintheFourierseriescoefficients1()xttobe1()010020,kkkotherwisea1()xtFormTable3.1weknowthatif1()xtisreal,thenkahastobeconjugate-symmetric,i.e,*kkaaSincethisisnottruefor,thesignalisnotrealvalued.Similarly,theFourierseriescoefficientsof2()xtarecos(),1001000,kkkotherwiseaFormTable3.1weknowthatif2()xtisreal,thenkahastobeconjugate-symmetric,i.e,*kkaaSincethisisnottruefor2()xt,thesignalisrealvalued.Similarly,theFourierseriescoefficientsof3()xtaresin(2),1001000,jkkkotherwiseaFormTable3.1weknowthatif3()xtisreal,thenkahastobeconjugate-symmetric,i.e,*kkaaSincethisisnottruefor3()xt,thesignalisrealvalued.(b)Forasignaltobeeven,itsFourierseriescoefficientsmustbeeven.Thisistrueonlyfor2()xt.3.7Giventhat()FSkxtawehave()2()FSkkdxtgtbjkadtTTherefore,(2)kkbajTk0kWhen0k5512()kTaxtdtTTusinggiveninformationTherefore,2,0,0(2)kkTkbkjTka3.8Sincex(t)isrealandodd(clue1),itsFourierseriescoefficientskaarepurelyimaginaryandodd(SeeTable3.1)Therefore,kkaaand00a,Alsosinceitisgiventhat0kafor1k,theonlyunknownFourierseriescoefficientsare1aand1a.UsingParseval’srelation221()kTkxtdtaTforthegivensignalwehave2122011()2kkxtdtaUsingtheinformationgiveninclue(4)alongwiththeaboveequation,22111aa2121aTherefore1112aajor1112aajThetwopossiblesignalswhichsatisfythegiveninformationare(22)(22)111()2sin()22jtjtxteetjjand(22)(22)211()2sin()22jtjtxteetjj3.9Theperiodofthegivensignalis4.Therefore,23401[]4jknknaxne21484jkeThisgives03,a112,aj21a,312aj3.10.SincetheFourierseriescoefficientsrepeateveryN,wehave115aa,216aaand317aaFurthermore,sincethesignalisrealandodd,theFourierseriescoefficientskawillbepurelyimaginaryandodd.Therefore,00aand11aa22aa33aaFinally1aj22aj33aj3.11SincetheFourierseriescoefficientsrepeateveryN=10,wehave1115aaFurthermore,sincex[n]isrealandeven,kaisalsorealandeven.Therefore115aaWearealsogiventhat2901[]5010nxnUsingParseval’srelation,250kkNa28150kka822221102||||||50kkaaaa82202||0kkaa56Therefore0kafork=2,….8,Nowusingthesynthesiseq.(3.94),wehave228101[]jknjknNkkkNkxnaeae22101055jnjnee10cos()5n3.12.Usingthemultiplicationproperty(seeTable3.2),wehave3120[][]FSlkllkllNkxnxnababFS0112233kkkkababababFS123222kkkkbbbbSincekbis1forallvaluesofk,itisclearthatkb+21kb+33kbwillbeforallvaluesofk,Therefore,,621FSnxnxforallk,3.13LetusfirstevaluatetheFourierseriescoefficientsoftx.Clearly,sincetxisealandodd,kaispurelyimaginaryandoddTherefore,0a=0.Now,ka=80)8/2()(81dtetxktj=40)8/2(81dtektj-40)8/2(81dtektj=kjekj11Clearly,theaboveexpressionevaluatestozeroorallevenvaluesofkTherefore.ka=5,3,1,24,2,0,0kkjkWhentxispassedthroughanLTIsystemwithfrequencyresponsejH,theoutputtyisgivenby(seeSection3.8)ktjkkejkHaty00Where420T,Sincekaisnonzeroonlyoroddvaluesofk,weneedtoevaluatetheabovesummationonlyoroddk,Furthermore,notethat4sin40kkjkHjkHisalwayszerooroddvaluesofk,Therefore,,0)(ty3.14ThesignalnxisperiodicwithperiodN=4,ItsFourierseriescoefficientsare304241nknjkenxa,41forallkFromtheresultspresentedinSection3.8,weknowthattheoutputnyisgivenby30)42()42(knjkkjkeeHany=)2/()2/(04141jjjeeHeH+)()()2/3()2/3(4141jjjjeeHeeHFromthegiveninformation,weknowthatnyisny=)425cos(n=)42cos(n57=)42()42(2121njnjee=)423()42(2121njnjeeComparingthiswitheq.(S3.14-1),wehave0)()(0jjeHeHAnd,2)(42jjeeHand,2)(423jjeeH3.15FromtheresultsoSection3.8,ktjkkejkHaty00Where1220T,SincejHiszerofor100,thelargestvalueof|k|orwhichkaisnonzeroshouldbesuchthat100kThisimpliesthat|k|8,Therefore,for|k|8,