南邮信号与系统课后答案第二章

南京邮电大学通信与信息工程学院信息工程系第二章信号与系统的时域分析作业2-1绘出下列信号的波形，注意它们的区别。ttu2tut13116tut17tututt10t110t1-10t1102-5试写出题图2-5各信号的解析表达式。11022ttf1(a)1-3032ttf5(e)3te2312sin3122cos31232cos3:)(2131:)(2525251ttuetftutetftutetfetututututfattt或或解：2-6试写出题图2-6各序列的解析表达式。2122123212:)(21212:)(nnkkfkukukfbkkkkkfa或解：11023kkf2(b)2-1-2211023kkf1(a)2-1-22-12-8已知信号如图所示（1）用阶跃信号表示（2）画出的波形（3）画出的波形（4）画出的波形（5）画出的波形tftf22tf12tfdttdftf12204ttf题图(a)-2-2422422)1(tututukf解：120t1222tftf-2-2（2）2204ttf题图(a)-2-2204tdttdf（4）-2(2)(2)(4)206t22112tftf（3）-2-2-62804ttf1（5）-2-242-12解：kykyifkyki、，求已知011010kfififkykykykikikfififkykykykiki10012-16化简下列各式。122ttt361tet0121tttt原式＝解：33143tettet原式解：2-17计算下列各积分的值。0解：原式＝3025122dtttt6cos2226633jjtteetjtjee解：原式dtttetj33522-19解：。试求系统的冲激响应，已知系统的微分方程为thtxtxtytyty223tuececthtthththtt22102120002102323，特征根：特征方程：设tueethcccccchhtt2021212100111201000tueetueetueethththtttttt2220032222-23解：。求系统的单位脉冲响应分方程为设描述某离散系统的差khkxkxkykyky1225.015.0215.015.0105.05.05.015.02210212000kucckhkkhkhkhkk，特征根：特征方程：设15.0341323432125.005.01201021212100kukhcccccchhkkkukukukkukukkukukhkhkhkkkkkkkkkkkkkkkkkkkkk5.0125.0341345.0311325.0381345.03413215.0341325.03813415.0341325.034132215.034132122112212211221122002-25计算下列卷积tuet32241641tuetuett3233222220030333edetudetutuetuettt原式解：55551554154154141tuettuetudetudeetttttttt原式解：2-32的波形图。试画出所示，如题图和冲激响应系统的激励thtxthtx322(b)220ttx-21(1)0tth-1(1)120tthtx-1-33解：1111txtxtttxthtx2-39计算下列序列的离散卷积212kukuku221322213221322132321321311121121211kukkukkkukkkukkkkkukkukkkkukkkukukukukukukukkkkknkn原式解：2212111kukkukukkku原式另解：kuakuakk5kuakkukakuakuaakkknkknnkn)1()1(100原式解：2-40的零状态响应。如图所示，试求该系统和单位脉冲响应某离散系统的输入信号khkx1,2,1,,42,3,2,4khkx解：1043kkx2-1-23-2211043kkh2-124-12725221216812816232446482324121423242-43解：kukxkxkykyk31,15.01)1(态响应。求下列离散系统的零状15.05.005.05.011000kuckhkkhkhk特征根：特征方程：设15.02215.01100kukhcchkkukukhkhkk5.05.02110kukukhkxkykk5.031kukukukukukkkkkkknnkknnkn314.0216.0322321513213212132215.0311002-44解：kukhkukxkxkxk,21)4(试求其零状态响应。，，单位脉冲响应为入为已知下列离散系统的输kukukhkxkyk21kukukukukkkknn22211221121212111002-45解：。应试求系统的单位脉冲响，，，单位脉冲响应分别为图示系统中各子系统的khkukhkkhkukhk211321kh1kh3kh2kxky122212122122122222112121111132313321kukkukukkukukukukukukkukukukhkhkhkhkhkhkhkhkkhkkkkkkkkk