信号与系统第二版课后习题解答(3-4)奥本海姆

SignalandSystem1Chap33.1Acontinuous-timeperiodicsignalx(t)isrealvalueandhasafundamentalperiodT=8.ThenonzeroFourierseriescoefficientsforx(t)arejaaaa4,2*3311.Expressx(t)intheform)cos()(0kkkktAtxSolution:Fundamentalperiod8T.02/8/400000000033113333()224434cos()8sin()44jktjtjtjtjtkkjtjtjtjtxtaeaeaeaeaeeejejett3.2Adiscrete-timeperiodicsignalx[n]isrealvaluedandhasafundamentalperiodN=5.ThenonzeroFourierseriescoefficientsforx[n]are10a,4/2jea,4/2jea,3/*442jeaaExpressx[n]intheform)sin(][10kkkknAAnxSolution:for,10a,4/2jea,4/2jea,3/42jea,3/42jeanNjkkNkeanx)/2(][njnjnjnjeaeaeaeaa)5/8(4)5/8(4)5/4(2)5/4(20njjnjjnjjnjjeeeeeeee)5/8(3/)5/8(3/)5/4(4/)5/4(4/221)358cos(4)454cos(21nn)6558sin(4)4354sin(21nnSignalandSystem23.3Forthecontinuous-timeperiodicsignal)35sin(4)32cos(2)(tttxDeterminethefundamentalfrequency0andtheFourierseriescoefficientskasuchthattjkkkeatx0)(.Solution:fortheperiodof)32cos(tis3T,theperiodof)35sin(tis6Tsotheperiodof)(txis6,i.e.3/6/20w)35sin(4)32cos(2)(tttx)5sin(4)2cos(21200tt0000225512()2()2jtjtjtjteejeethen,20a,2122aa,ja25,ja253.5Let1()xtbeacontinuous-timeperiodicsignalwithfundamentalfrequency1andFouriercoefficientska.Giventhat211()(1)(1)xtxtxtHowisthefundamentalfrequency2of2()xtrelatedto?Also,findarelationshipbetweentheFourierseriescoefficientskbof2()xtandthecoefficientskaYoumayusethepropertieslistedinTable3.1.Solution:(1).Because)1()1()(112txtxtx,then)(2txhasthesameperiodas)(1tx,thatis21TTT,12wwSignalandSystem3(2).212111()((1)(1))jkwtjkwtkTTbxtedtxtxtedtT111111(1)(1)jkwtjkwtTTxtedtxtedtTT111)(jkwkkjkwkjkwkeaaeaea3.8Supposegiventhefollowinginformationaboutasignalx(t):1.x(t)isrealandodd.2.x(t)isperiodicwithperiodT=2andhasFouriercoefficientska.3.0kafor1||k.41|)(|21202dttx.Specifytwodifferentsignalsthatsatisfytheseconditions.Solution:0()jktkkxtaewhile:)(txisrealandodd,thenkaispurelyimaginaryandodd，00a,kkaa,.2T,then02/2and0kafor1kso0()jktkkxtae00011jtjtaaeae)sin(2)(11taeeatjtjfor12)(2121212120220aaaadttxja2/21)sin(2)(ttx3.13Consideracontinuous-timeLTIsystemwhosefrequencyresponseis)4sin()()(dtethjHtjIftheinputtothissystemisaperiodicsignalSignalandSystem484,140,1)(tttxWithperiodT=8,determinethecorrespondingsystemoutputy(t).Solution:Fundamentalperiod8T.02/8/40()jktkkxtae00()()jktkkytaHjke0004,.......0sin(4)()0,.......0kkHjkkk000()()4jkwtkkytaHjkeaBecause48004111()1(1)088TaxtdtdtdtT另：x(t)为实奇信号，则ak为纯虚奇函数，也可以得到a0为0。So()0yt.3.15Consideracontinuous-timeideallowpassfilterSwhosefrequencyresponseis1,.......100()0,.......100HjWhentheinputtothisfilterisasignalx(t)withfundamentalperiod6/TandFourierseriescoefficientska,itisfoundthat)()()(txtytxS.Forwhatvaluesofkisitguaranteedthat0ka?Solution:for0()jktkkxtae00()()jktkkytaHjke即对于所有的k，1)(0jkHSignalandSystem5for1,.......100()0,.......100Hj也就是说1000k,126/0T即12k100，k=8，故当k8时，ak＝0。3.35.Consideracontinuous-timeLTIsystemSwhosefrequencyresponseis1,||250()0,HjotherwiseWhentheinputtothissystemisasignalx(t)withfundamentalperiod/7TandFourierseriescoefficientska,itisfoundthattheoutputy(t)isidenticaltox(t).Forwhatvaluesofkisitguaranteedthat0ka?Solution:T=/7,02/14T.ktjwkkeatx0)(tjkwkkejkwHaty0)()(00()kkbaHjkwforotherwisewjwH,.......0250,.......1)(,01,.......17()0,.......kHjkwotherwisethatis0250250,........14kk,andkisinteger,so18....17kork.Let()()ytxt,kkba,itneeds0ka,for18....17kork.SignalandSystem6Chap44.1UsetheFouriertransformanalysisequation(4.9)tocalculatetheFouriertransformsof;(a))1()1(2tuet(b)|1|2teSketchandlabelthemagnitudeofeachFouriertransform.Solution:(a).()()jtXjxtedt2(1)(1)tjteutedt2(1)(2)211tjtjteedteedt(2)(2)221222jtjjeeeeejjj(b).()()jtXjxtedt21)2(21)2(2122122124422|21|21jjjtjtjtjttjttjtejejeejeejedteedteedtee4.2UsetheFouriertransformanalysisequation(4.9)tocalculatetheFouriertransformsof:(a))1()1(tt(b))}2()2({tutudtdSketchandlabelthemagnitudeofeachFouriertransform.Solution:(a).()()jtXjxtedt-jt-[(t1)(t-1)]edt-jt-jt-(t1)e(t-1)edtdt2cosjjee(b).()()jtXjxtedt{(2)(2)}jtdututedtdtSignalandSystem7{(2)(2)}jtttedt(2)(2)jtjttedttedt222sin2jjeej4.5UsetheFouriertransformsynthesisequation(4.8)todeterminetheinverseFouriertransformof)()()(jXjejXjX,where|()|2{(3)(3)}Xjuu23)(jXUseyouranswertodeterminethevaluesoftforwhichx(t)=0.Solution:dweejwXdwejwXtxjwtjwXjjwt)()(21)(21)(dweewuwujwtwj)23()}3()3({221dweedweewtjjjwtwj)23(33)23(331)23(1)23(1)23(3)23(333)23(