3信号与系统-每章课后答案第三章作业

《信号与系统》作业第三章本章作业：目标测评3——1、2、33-1，3-2，3-3，3-4，3-7，3-9，3-11，3-14，3-16，3-19，3-21，3-222020/5/19By谢睿1P76目标测评31、已知，求。分别用定义、性质和图解三种方法计算。)]1()([3)()2()()(21tututftututf，)()()(21tftftf解：性质（一）)()()(ttututu)]1()([3)]2()([)()()(21tututututftftf)3()3(3)2()2(3)1()1(3)(3tuttuttutttu解：性质（二）)]1()([3)()2()()(21tututftututf)1(3)(3)()2()()(21tttftttf)]1(3)(3[)]2()([)()()('2'1tttttftftf)3(3)2(3)1(3)(3tttt)3()3(3)2()2(3)1()1(3)(3)(tuttuttutttutf定义及图解：略Signal&SystemBy谢睿22020/5/19解：性质2、。求。分别用定义和性质两种方法计算。)2()()()(2)(21tututftuetft，)()()(21tftftfdtuuetutuet)()()()()0(0tdet)0(10teett)()1(tuet)]2()([)()(2)()()(21tttutuetftftft)2(]1[2)(]1[2)2(tuetuett定义：略Signal&SystemBy谢睿32020/5/19证明：3、试证明：)()()()()1(020121ttfttftftf)()()()()2(12212211ttfttfttfttfdtfftftf)()()()()1(2121dxtxtftxftx)()(02010)()(0201ttfttfdttftfttfttf)()()()()2(22112211可得证。令同理21txtSignal&SystemBy谢睿42020/5/193-1已知系统微分方程、起始条件以及激励信号分别为P82习题三基础与提高题)()(2)-0()(4)(3)(2tuetxytxtydttdyt，，3032)0()(12)-0(31CteCtyytx设02)(3tetytx及其各阶导数不含零状态响应)()(01)2(tthmn)()(32tueCtht设)(4)(3)(tthth带入原方程：试求解该系统的全响应。解：(1)零输入响应)(4)(3])([3232ttueCtueCtt42C)(4)(3tuetht)(4)()()()(32tuetuethtxtyttf则)(4-)(432tuetuett0)(4-)(42)()()()3(323ttuetueetytytytttfx全响应)(2-)(432tuetuettSignal&SystemBy谢睿52020/5/193.1另解：(1)通解303)0()(33teCtytc设带入原方程设tpteCtytuetx242)()()((2)特解(3)完全解：043][22424teeCeCtttttpceeCtytyty2334)()()()(4)(3)(txtydttdy44C)0()-0()()(01yytthmn及其各阶导数不含则又22433CC042)()()(23teetytytyttpcSignal&SystemBy谢睿62020/5/1932340)-0(2)-0(21CCyy，带入初值3-2描述某LTI系统的微分方程为)(6)()(2)(3)(22tfdttdftydttdydttyd21023212、)0()(221teCeCtyttx设及其各阶导数不含零状态响应)()(12)2(tthmn)()()(243tueCtueCthtt设)(6)()()(3)(ttththth带入原方程：解：(1)零输入响应4-543CC)(4-)(5)(2tuetuethtt)](4-)(5[)()()()(2tuetuetuthtftyttf则)(2)(5)(32tuetuetutt0)(2)(5)(33234)()()()3(22ttuetuetueetytytyttttfx全响应)(38)(311)(33tuetuetutt，求系统的全响应。)()(0)-0(2)-0(tutfyy，，已知03234)(2teetyttxSignal&SystemBy谢睿72020/5/193-3电路如题图3-3所示，已知，求的零状态响应。)]2()([)(2tutuetvt)()()(tvtRidttdiL带入设)()(tuCethtLRLC1解：(1)系统方程：(2)单位冲激响应)(tiLRRL0)()()(ttRhthL及其各阶导数不含则)()(01tthmn)(1)(tueLthtLR(3)零状态响应)(1)]2()([)()()(2tueLtutuethtvtytLRtf(若假设L=1，R=1))]2()([]22[)()]2()([22tutueetuetutuettttSignal&SystemBy谢睿82020/5/193-4已知一LTI系统对激励为时的完全响应为，对激励为时的完全响应为，试求)()(1tutf)(tyx(1)该系统的零输入响应(2)该系统的阶跃响应)(2)(1tuetyt)(tg)(3)(2tutf)(2)(3)(22tuetuetytt解：设系统的零输入相应为)(tyx输入为系统的零状态相应为)()(1tutf)()(tgtyf)(2)(3)(3)()()(2)()()(221tuetuetytytytuetytytyttfxtfx)()(21)()()(23)(22tuetuetytuetuetyttfttx)(tgSignal&SystemBy谢睿92020/5/193-5求如图所示函的卷积积分。)()(tytx与)(ty)2(2)2(2)(tututx)2()()(tututh)]2()([)()]2()2([)(2)()()(tttutttuthtxty)()()(ttututu)]2()([)]2()2([)(2ttttttu)]4()2()()2([)(2ttttttu)4()4(2)2()2(2)(2)2()2(2tuttutttutut解：（1）性质Signal&SystemBy谢睿102020/5/19（2）图解法：（过程图略）)()()(thtxty40422202022202222ttdtdtdttttt404228204024220tttttttSignal&SystemBy谢睿112020/5/193-6已知，分别利用图解法和性质计算。)3()1()()]1()()[1()(21tututftututtf，)()(22tftf解：（1）性质)(21)()()()()(2tuttuttuttututu)]3()1-([)]}1()([)]1()({[)()(21tututututtututftf)4()3()2()1()(21)(2tttttutttu)4()4(21)4()4()3()3(21)3()3()2()2(21)2()2()1()1(21)1()1(2222tuttuttuttuttuttuttuttutSignal&SystemBy谢睿122020/5/194043)1(32)1(21)1(10)()(13101021ttdtdtdttftftt（2）图解：步骤图略4043)3(21)3(2323123210)1(21)1(20021321022102tttttttttttSignal&SystemBy谢睿132020/5/193-7已知一个线性时不变系统的输入信号及单位冲激响应如图所示，求零状态响应。)(tf)(th)]2()([sin)()(tututtfa)3()1()()(tututhb)()()(thtftyf410412sin210sin0102310ttdtdttt5053cos131cos1102310tttttt5053]1)3([cos131)]1(cos1[110tttttt步骤图：见下页Signal&SystemBy谢睿142020/5/19Signal&SystemBy谢睿152020/5/193-8已知某系统的微分方程为求此系统的单位冲激响应。)(2)(3)(6)(5)(22tfdttdftydttdydttyd)(th32065212、解：系统的特征方程：)()()(3221tueCtueCthtt设)(2)(3)(6)(5)(ttththth带入原方程：22332121CCCC)(7)(4)(32tuetuethtt及其各阶导数不含)()(12tthmn7421CCSignal&SystemBy谢睿162020/5/193-9已知和的波形如题图3-9所示。试用图解法求。)(tf)(th)()(thtf)2()1(2)()()(tutututfa)1()()()(tututhb解：法一)]1()([)()]2()1(2)([)()()(tttuttttuthtf)]3()2(3)1(3)([)(ttttttu)3()3()2()2(3)1()1(3)(tuttuttutttuSignal&SystemBy谢睿172020/5/19解：法二（图解）图略30321211110100)()(211110ttdttdtdttdttthtftttt3