信号与系统第二章习题

1、100y10y0ty2D3Db12 )(,)(;)()()()(21023212 ，解解：特特征征方方程程：etetttyhcc 21)()(齐齐次次解解ececyttth221)1(2)( 由初始条件由初始条件: 0122121cccc 1221cc解得：解得：)()()(22tutyeett 2.1 解微分方程解微分方程22)0(, 1)0(, 1)0(,0)()1()()2()1(23 yyytyDDDd110110132,1223 ，）（）（）解解：（ttttthtttthttthectecececectyectececectyectececty32221)2(3221)1(321)(

2、)()( 4/12/34/3221132132132131ccccccccccc)()4/12/34/3()(tueteetyttth*30) 0() 0()()() 34()(122 yytuetyDDbt13844)(,2)()(3)()(3103422222)2(2)1(2321)1(321212 AeAeAeAeAetyAetyAetyecectyecectytttttptptptthtth代代入入方方程程得得：设设特特解解，解解： *2.34)()2/12/1()(2/12/1023)0(01)0(23)()()()(322121)1(212321)1(2321tueeetycccc

3、yccyeecectyeecectytytytttttttttph 完完全全解解：5)()1(1)()()(20)(tueaatdetuettucatttaat t0)()1(12tueaat at 2.5t0)( tae t图图形形设设)()(, 1tuettuaat t0 ae t图图解解法法)()(tuettuat *6 40,)cos1(1,040,)4()(sin,0)4()()(sin)(0ttttdtututtututtuet 其其它它其其它它t0 1427 1)1 (3110) 1()()1(31)(33tedettutuetttt t01)(ty31t011) 1( u)()

4、(3tuet0 t1 tt：解解法法1) 1()()(*3 tutuegt*注意注意8 1)1(3110)1()()1(31033tedettutuettt t011)(3 ue )()(3tuet t01)(ty311 0 t1 tt) 1( tu1 t：解解法法2)(* g9 利用卷积积分求冲击响应为利用卷积积分求冲击响应为h(t)的的LTI系统对系统对x(t)的的响应响应y(t)，并概略地画出，并概略地画出y(t)波形。波形。)2() 1(2)()2() 1(2)(*)()(*)()()2() 1(2)()()( txtxtxttttxthtxtytttthb解：)(tx012t12)2

5、( tx012t34123t)1(2 tx0122 4 *2.610)(ty012t3422 4 ) 4(2) 3(5) 2(3) 1()()( tutututututy)4(2)3(5)2(3)1()()2()1(2)(*)2(2)1()()(*)()(),2(2)1()()()( tutututututtttutututhtxtytutututxtx 则则：分分解解，另另解解：先先把把11)2(1)2(1)2(1)2(1*)()(*)()()( txtxtttxthtxtyd )(tx011 t)(th02 t2 1 1)(ty02 t2 时时12 )(ty02 t2 时时120 12)(

6、*)()()(*thtxtyf 求求10)(txt2110)(th112t10t2 t01 t11 2t10t2 t21 t2t1 110t2 t10 t12t1 10t2 t43 t2t1 34110t2 t32 t2t1 3110t2 t1 tt211 2 t2)( t13 4043)2(32)2()2(21)2(10)2(0110)(22121121121111111ttdttddtdttdddttddttdttytttttttttttt 10t2 t32 t2t1 3110t2 t10 t12t1 t 2 10t2 t21 t2t1 114)(*)()(*)()(*)()(*)()()

7、(*)()()(*)()()()()2() 1()2()1()()()()()2() 1()1() 1()()(:22221121121212121thtxthtxthtxthtxththtxtxthtxtyththtututtututhtxtxtututututtx解0)(th112t10)(txt211150)0(1)0(0)0(0)0(1)0()0(0)(0)0()0()(1)(3)0()0( 4)0()0()()(340)0(0)0()()(34)()()1()()34()1()1()1()1(0000)1()1(00002)1(22 hhhhhhdtthhhthdtthhhhhdtt

8、dtthDDhhtthDDthtxDtyDD，可可得得：，由由初初始始条条件件，为为连连续续函函数数）（，）解解：设设（。求求，已已知知 2.716)(21212123)()()()()1()()(2/12/1)(2/1, 2/10)0(1)0()(13034)()(34333321)1(2312122tueeeeethdtthdththDthtueethcchhececthrrrrtthDDttttttttt 可可得得：，代代入入初初始始条条件件特特征征根根，的的特特征征方方程程：）（ 17).(21,)()(tyTTtxthLTI两两种种情情况况下下的的响响应应和和计计算算当当如如图图所所

9、示示和和输输入入系系统统的的单单位位冲冲击击响响应应设设一一个个 1-101th(t)X(t)(1)-2T-T0T2Tt*2.12 kkTttx)()( )()(*)()(*)()(kTthkTtthtxthtykk 解：解：181)()(1 kthtyTk时时，t1-101y(t)2()(2kthtyTk 1-101ty(t)192.15求求系系统统总总冲冲激激响响应应。图图中中),()(),1()(),()(321tthtthtuth h2(t)h1(t)h3(t)h1(t)(tx)(ty) 1()() 1(*)()()(*) 1(*)()()(*)(*)()()(3211 tututtu

10、tutttutuththththth 解解：20系系统统的的零零状状态态响响应应。时时程程，并并计计算算当当输输入入由由图图写写出出系系统统的的微微分分方方,)(2)(tuetxt y(t)*2.19)(6)(5)()( tytytxty x(t)+-5-6(a)(ty)( ty21122265)(, 2, 30)0(,2)()()(6)(5)()(6)(5)()(21)1( AeAeeAeAeAeAetyyyetxtxtytytytytytxtyatttttttt把把特特解解代代入入方方程程中中特特解解：特特征征根根：时时）（解解： tttxtttxeececyeececty 2132)1(

11、213223)(零状态响应：零状态响应：22 0123)0(01)0(12)1(12ccyccyxx)()2()(123221tueeetycctttx 解解得得：ttttececthececthhhtththth32213221 32)()(0)0(,1)0()()(6)(5)( 1121cc解解得得：23x(t)+-41/2+y(t)(b)()(2/1)(4)( txtxtyty )(21)()()(4)()(111 1tytytytytxty )(21)( 4)(21)()(2)(21)(4)()(21)()(1111 1 1 tytytxtxtytxtytxtytyty 1y1y 1y

12、24tthccchccchtctcthtctcthjhhtthththttxtxtxtytyb2sin2/1)(000cos0sin0)0(2/110sin20cos21)0(2sin22cos2)(2cos2sin)(2040)0(1)0()()(4)()()()(:1)(2/1)()(4)()(2211212121122 由由初初始始条条件件得得特特征征方方程程为为时时的的先先求求解解法法 252sin2cos2sin2cos2)2sin2/12cos2/1()(*)()()()2sin212cos21()(21)()(000)(0)(0)( dedeedededetxthtytuttth

13、ththtttt )(5/12cos512sin53tuettt）（ 26tttttpptphetctctyAeAeAeetytyAetytctctytxtytytxtxtyty 5/22sin2cos)(5/2242)(4)()(2sin2cos)()()(4)()()(2/1)(4)(221 21 得得由由特特解解：齐齐次次解解：解解法法得得代代入入0)()0( tyy27ttetttyetttycccc 5/22cos5/2sin5/4)(5/22sin5/12cos5/2)( 5/105/225/205/22211)(5/12cos512sin53)(2/1)( )(tuetttyty

14、tyt）（ 28tpttttpttttttetyAeAeAeAetytetxyyeteteetxtxtuetx 5/ 1)(5/ 14)(0)(0)0(1)0()()(2)(2/ 1)()(2)(3则则：设设时时的的全全响响应应，求求初初始始条条件件：解解法法 )()(2/ 1)(4)( txtxtyty 29)5/12cos5/12sin5/3()(5/35/115/1205/15/12cos22sin2)(5/12sin2cos)(21212121tttetttyccccetctctyetctcty ：由由初初始始条条件件tctctyjh2sin2cos)(20421122 特特征征方方程

15、程：30。求求总总系系统统图图中中)(),3()()();1()(21thtututhtth )5()4()3()2() 1()()3()(*)2() 1()()3()(*)1(*) 1() 1()()(*)(*)()()()(2111 tututututututututtttututtttththththtth 解解：x(t)(1th)(1th)(1th)(2th+-y(t)2.2131)()()()(*)()(txdtxttxa )()()()()()()()( )()()()(txdtxdtxtdtxdtxdtx 证证：*2.23 证明：证明：)()()()()() 0()()()()()()()( )()( txtxtxttxtdtxtxtx 注注：)()(*)()(*)(txttxttx