信号与系统课后答案第三章作业答案

3-1已知系统微分方程、起始条件以及激励信号分别为2d()3()4(),(0)2,()u()dtytytxtyxtett试求解该系统的全响应。解：由题可知：d()3()4()dytytxtt的特征方程为3=0=-；3（1）零输入响应：31()()txytCeut(0)2y12C3()2()txyteut（2）零状态响应：设该方程的特解为2()()tpytPeut23PP44P2()4()tpyteut又方程的特征根=-3322()()4()ttfytCeuteut初始为0，2-4C32()-4()4()ttfyteuteut全响应=()+()xfytyt234()-2(tteuteut)3-2描述某LTI系统的微分方程为22d()d()d()32()6dddytytft()ytfttttt已知，求系统的全响应。'(0)2,(0)0,()u()yyft解：特征方程为：2+3+2=01=-1,2=-2零输入响应：212()()()ttxytCeutCeut1由,得：-(0)2y-'(0)0y1212=2--2=0CCCC12=4=2CC2()(4-2)()ttxyteeut零状态响应：设特解为()()pytAut=6A即()6()pytut零状态响应方程为：234()(6)()ttfytCeCeut又初始值为零，(0)0fy，()0fdytdt22()()1()03()0fdytatbudtadyttaubdtyt在0时，''(0)''(0)1,''(0)'(0)3,(0)(0)yyyyyy,+-(0)(0)=0ffyy(0)(0)3fdydydtdt代入得，3-9C43C-2()(-936)()ttfyteeut全响应-2()(-56)()ttcyteeut3-3电路如题图3-3所示，已知()it2()eu()u(2)tvttt，求的零状态响应。+‐vtRLit题图3-3解：()()()ditvtRitLdt,2()()[()(2)]tditRjtLeututdt特征方程0RRLL2零状态：设特解2()[()(2)]tPitPeutut2()()[()(2)]RttLRitCeutPeutut22PRL+-(0)(0)=0vv22CLR222()()[()(2)]22-RttLfiteuteututLRRL3-4已知一LTI系统对激励为1()u()ftt时的完全响应为，对激励为1()2eu()tytt2()δ()ftt时的完全响应为，试求2()δ()ytt（1）该系统的零输入响应；()xyt（2）该系统的阶跃响应。()gt解：(1)设1()()ftut，零状态响应为()yt12()2()()()()()()()tziziyteutytytdytyttytdt①②12()()dftftdt()()()2()tdytytteutdt③设特解为tBe忽略③中的()t1B设齐次解为tAe()+ttytAee()()1()dytatbuadtytau+-(0)(0)1yy代入()+ttytAee得=0A()()tyteut将()yt代入1()2()()()tziyteutytyt()=()tziyteut（2）()yt即为阶跃响应。()()()tgtyteut3-7已知一个线性时不变系统的输入信号()ft及单位冲激响应如题图3-7所示，求零状态响应。()ht()fyt3()ftt12110()ht10t123()a()b题图3-7453-9已知()()ftht和的波形如题图3-9所示。试用图解法求()()ftht。11111)(tf)(thtt2OO题图3-9①01t时，()()fthtt②12t时，()()2(1)32fthtttt③623t时，()()[2(1)](1)3fthttt④tt-1y(t)121003tt或时，()()0ftht0001()()321232ttttfthttttt，或，，，33t3-11试计算下列卷积：（1）u()*u()tt解：u()*u()()()tttutd7=()=()tudtut（2）3eu()*eu()tttt（3）3e*eu()ttt解：33e*eu()=()tttteeud=20teed=201[]2ttee1=2te（4）eu()*u()tttt（6）eu()*[u()u(2)]ttttt3-14如题图3-14所示，试求)()(thtx和()*()ftht。8)(th()ft()ft25a)(thb()ft)(th)(th()ft(2)u(1)tet题图3-14(1)解：(t)1u(t1)f(t)(t)fh(2)[1u(1)](1)teud(2)(2)(1)(1)(1)euduteud1(2)(2)11teded11(2e)u(t)tee1(2e)u(t)te（2）(t)[u(t)u(t2)]2sfa(t)[u(t2)u(t3)]hb1111(t)(t)ab[(t)u(t)(t)u(t)tu(t)(t1)u(t1)]2222fh9（3）(t)(t1)[u(t1)u(t)](1t)[u(t)u(t1)]f(t)u(t)u(t1)h22221(t)(t)[(t1)u(t1)(t2)u(t2)3tu(t)3(t1)u(t1)]2fh（4）(t)sinfth(t)(t1)u0(t)(t)'(t)sin()tfhhwud00(t1)sintwd001(t1)[(1cos)]u(t)wtw001[1cos(1)]u(t1)wtw3-16已知某系统满足微分方程，若激励分别为(时，试用卷积分析法分别求系统的零状态响应)(2)()(3)(4)(''''tftftytyty23)()()eu()ttcftt)()u()()()eu(afttbftt，，解：求冲激响应：特征方程为2+4+3=0解得=1=3或冲激响应为312()()()tthtAeAeut21212'()()()(3)()tthtAAtAeAeut312121''()()'()(3)()+(9)()tthtAAtAAtAeAeut(t)(t),h(t),h'(t),h''(t)f将代入微分方程，有1212()'()(3)()'()2()AAtAAttt112122112322AAAAAA131(t)(ee)u(t)2tth10(a)时，(t)u(t)f311(t)f(t)(t)(t)(t)(t)(t)22ttfyheuueuu=3311211(1)(t)(1)(t)(e)(t)26326ttteueueut（b）时，2(t)u(t)tfe23211(t)f(t)(t)(t)(t)(t)(t)22ttttfyheueueueu22311()(t)()22tttteeueeu(t)311(e)(t)22tteu(c)时，3(t)u(t)tfe33311(t)f(t)(t)(t)(t)(t)(t)22ttttfyheueueueu33311()(t)t42111[e()](t)424ttttteeueuteu(t)3-19一个LTI系统，初始状态不祥。当激励为时其全响应为；当激励为时其全响应为。求)(tfin2)ut3(2esin2)u()ttt)(2tf3(e2s()tt（1）初始状态不变，当激励为)1(tf时的全响应，并求出零输入相应、零状态响应；（2）初始状态是原来的两倍、激励为)(2tf时系统的全响应。113-21已知系统的阶跃响应是2()eu()tgtt，求此系统在激励为2()3eu()tftt作用下系统的零状态响应。3-22已知微分方程的冲激响应。求系统的微分方程为)()(2)('tftyty232eu()eu()tt2()eu()thtt'()2()ytyttt时系统的零状态响应。解：232()(t)f(t)(t)(2(t)()(t)tttfyheueueu22()32()0022tttteedeed1213|t222000212(t0)ttttttedeedeee222(ttteee1)32(2t1)ttee32'(t)[(2t1)]u(t)ttfyee