南邮信号与系统课后答案第三章

南京邮电大学通信与信息工程学院信息工程系第三章连续时间信号与系统的频域分析作业3-3解：频谱图。边、双边幅度谱和相位试分别画出该信号的单，周期信号63cos232sin46cos62000ttttf20nA46002030n单边频谱：0n002030n6663cos262cos46cos62000ttttf原信号化为：双边频谱：0nF002030n002031230n002030n60020363-5解：。其三角形式傅立叶级数试画出单边频谱，写出所示，的双边频谱如题图已知周期信号53tf0nF0n2.53132-1-2-3n00n132-1-2-3单边频谱：0nA0n2.56132n00n13232cos232cos45.2tttf三角形式傅立叶级数：3-7解：tuetft152)2(换。求下列信号的傅里叶变tueet55原式jtuet515jetueet5555解：tSatf5)7(7555510125101,105101101010uuggtSaASatg对称性3-8解：变换。，求下列函数的傅里叶已知Ftf)()(频域微分时域微分dFjdjdttdftFjdttdfdttdft)3(ddFFddFjjFj解：jjeddFjtfteddFjtftddFjtftddFjttftftfttftft222222)(222222时域微分原式tft23)4(jjjeFeddFjeFtfFtf2222原式又解：2422222222222FddFjddFjttfFtftfttf原式原式22)6(tft3-9解：变换。求下列信号的傅里叶反2281)2(jFtutetfjdjdjtutejtuettt82288818181即解：jeFj8)6(618181186188tuetfjetuejtuetjtt即3-11解：105210525221525410cos525210cos00154541SaSaFFtfFSatgtfttgtututtf调制定理所示信号的频谱。利用频移性质，求题图113-101ttf1(a)52523-12解：jSaffjGFffGSatgtftg20222又列信号的频谱。用时域微积分性质求下1-101ttf2(b)-11-101ttf2-13-14解：前提下，求的，在不求出所示信号如题图FFtf14300)1(FF23000dttfdtetfFFtj解：dF)2(202212100fdFdFdeFfttj101ttf-13-16解：，的最高频率为某带限信号Hz150tf的带宽；和求33)1(tftfHzffFtftfHzffFtftfHzfmmmmm5031333450333131502211解：斯特取样率。进行时域取样，求奈奎和、对33)2(tftftfHzfftftfHzfftftfHzfftftfmsmsms1002390023300222211100则令则令则令3-17解：特取样率。确定下列信号的奈奎斯tSatSa200100)3(Hzfsradsradsradsradssmsmmmmm1002/2002/100,min/200,/1002121或解：tSatSa60100)5(2Hzfsradsradsradsradssmsmmmmm1202/2402/120,max/120260,/1002121或3-24解：。，求；所示，题图器的频率特性如所示系统中，带通滤波如题图tyttstSatfba20cos)(243)(24320202121220cos2222ggFggtSaHFYttSatfAAA由调制定理得：则令1801H-1822-22(b)(a)ty带通滤波器tftsAttSatyFgtSatfttfFFFgggHggHFYBBBBBBA20cos222220cos202021202021202021444422调制定理令1801Y-1822-22-20203-25解：tyttfjjH112cos111时系统的输出）（，求输入某系统频域系统函数1272cos2sin542cos5322542253254532545322121221211122111ttttyjjjjjjjjjHFY另解：1272cos2cos545321212127ttAtyeAejjjHjj