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时间:2019-03-08
《信号与系统奥本海姆英文版课后答案chapter8》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、Chapter8Answers8.1UsingTable4.1,taketheinverseFouriertransformofYj((ω−ω)).Thisgivesyt()2()=xtejwtc.cTherefore,mt()2=ejwtc.8.2(a)TheFouriertransformY(jw)ofy(t)isgivenbyYjwXtj()((=ω−ω)).cClearly,Yjw()isjustashiftedversionofX()jw.Therefore,x(t)mayberecoveredfromy(t)simplybymultiplyingy(t)bye−jwtc.There
2、isnoconstraintthatneedstobeplacedonωtoensurecthatx(t)isrecoverablefromy()t.(b)Weknowthatyt()Re()=={yt}xt()cos().ωt1cTheFouriertransformYjw()ofy()tisasshowninFigureS8.21111Yjw()=−Xj((ωωω))++Xj((ω))1cc22Xj((ω+ω))Yj()ωcXj((ω−ω))c−ωc0ωc−+ωπ1000ω−1000πccFigureS8.2IfwewanttopreventthetwoshiftedreplicasofY
3、jw()frommultipliedbycos(2000πt),theoutputwillbe1x()tg==()cos(2000tπtx)()sin(2000tππt)cos(2000tx)=()sin(4000tπt)12TheFouriertransformofthissignalis11Xjw()=−−+Xj((ω4000))πωXj((4000))π.144jjThisimpliesthatX()jwiszerofor
4、ω
5、2000≤π.Whenyt()ispassedthroughalowpassfiter1withcutofffrequency2000π,theoutputwil
6、lclearlybezero.Thereforey()t=0.8.4Considerthesignal3yt()=+gt()sin(400)2sin(400)πtπt23=+sin(200ππtt)sin(400)2sin(400πt)=−sin(200ππtt)1cos(800⎡⎤⎣⎦())/2+−2sin(400)1cos(800ππtt⎡⎤⎣⎦()/2)=−−(1/2)sin(200ππtt)(1/4)sin(1000{})sin(600πt)+−sin(400ππtt)(1/2)sin(1200{})sin(400−πt)Ifthissignalispassedthroughalowp
7、assfilterwithcutofffrequency400π,thentheoutputwillbeyt=sin(200π).18.5Thesignalx()tisasshowninFigureS8.5xt()0tEnvelopofAtt150FigureS8.5Theenvelopeofthesignalω()tisasshowninFigureS8.5.Clearly,iswewanttouseasynchronousdemodulationtorecoverthesignalx(t),weneedtoensurethatAisgreaterthantheheighthofthehig
8、hestsidelobe(seeFigureS8.5).Letusnowdeterminetheheightofthehighestsidelobe.Thefirstzero-crossingofthesignalx()toccursattimetsuchthato1000πtt=⇒=π,1/1000.00Similarly,thesecondzero–crossinghappensattimetsuchthat11000πtt=⇒=2,π2/1000.11Thehighestsidelobeoccursattime()tt01+/2.thatis,attimet2=3/2000.Atthis
9、time,theamplitudeofx()tissin(3/2)π2000xt()==−2π3/20003πTherefore,Ashouldatleastbe2000.Themodulationindexcorrespondingtothesmallestpermissible3πvalueofAisMaxvalue.ofx()t1=10003πm==Minpossible.valueofA2
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