电路理论-5_4电路系统对任意激励的零状态响应

1、5.4 电路系统对任意激励电路系统对任意激励的零状态响应卷积积分的零状态响应卷积积分1.卷积积分定理：任一卷积积分定理：任一LTIS对任意激励信号对任意激励信号f(t)的零状态响应的零状态响应应该等于该激励信号与电路系统冲击响应的卷积积分。即应该等于该激励信号与电路系统冲击响应的卷积积分。即:)()()()()()()(00thtfdhtfdthftyttttzs5.4.1 卷积积分定理：卷积积分定理：0( )lim( )ntP t )()(thtPnn)(lim)(ththnn0Pn(t)tLTISthn(t)1证明：因为( )()()() ()nnnknf tf kP tkf kP tk

2、( )()()() ()nnnkny tf kh tkf kh tk , ,dkd 积分求和任意信号：dtftf)()()(任意信号产生的零状态响应：dthftyzs)()()(当当 时，时，)0(t0)(thttzsdthfty0)()()(0tt)()(ktkf)()(fkf)()(thf0ttt解解: 1.: 1.（1 1）求得电路的冲击响应：）求得电路的冲击响应：因为电路因为电路KCLKCL：)(3)()()()()(3tUetUeLRthttidttdiRLttLRLL2( )te U t)2(2)2(tUet)()(3) 1(3216632)()()(32020203)(0tUee

3、eeeedeedeedhtftitttttttttttL2.)2()(3)()2(3)2(tUeetittL11f(t)t()100( )( ) ()1(1) ( )ttttzsytfh tdedeU t （2）求U(t-1)激励时的响应： 根据延时不变性可得：) 1()1 ()()1(2tUetytzs)()(tUetht) 1()1 ()()1 ()()()()1(21tUetUetytytyttzszszs3.LTIS的完全响应： 利用卷积求得系统零状态响应，再与系统零输入响应叠加，即求得系统的完全响应为： （设系统特征根互异）nittzpszzpdthfeAtytytyi10)()()

4、()()(设：（1）褶叠： （将横轴t ， 对褶过去）（2）平移（3）相乘积分)(h)()(21tff和tttdtthtf21216144)(211)()(112133( )( )1()2416tf th ttd1224324)(211)()(tttdtthtf0)()(thtf( )( )0f th t(a)(b)(c)(d)(e)( )00( )tAf th tBe其余例：求如图（a）（b）所说函数f(t)和h(t)的卷积积分。解： （1）写出表达式：t00taAaf(t)t00h(t)tBBe-t(a)（b）Aaf(t)0Aaf(t)0t(C)（d）Aaf(t)t00h(t)tBBe-t

5、(a)（b）Aaf(t)0Aaf(t)0t(C)（d）)(*)()(thtfty.t0， 无重叠。.0ta，tl1=0, tl2=-,选tr1=a, tr2=t)()(thf和0()0( )( ) ()(1)tttty tfh tdABABede )(0)()1 ()(ataateeABdABety)()()()(1221tftftftf1212 ( )( )( )()f tf tff tdddt则令t2121 ( )( )() ( )( )f tff tdf tf t证明：证明：)()()()()()(321321tftftftftftf)()()()()()()(3121321tftftf

6、tftftftf3.3.结合律：结合律：证明：证明：)()()()()()()()()()()()()()()(321321321321321tftftfddtfffddtfffdtfdfftftftf 二二.卷积的微分和积分：卷积的微分和积分：1.微分：微分：)()()()()()(212121tfdttdfdttdftftftfdtddttdftfddttdffdtffdtdtftfdtd)()()()()()()()(21212121证明：同理可证第二等式tttdftfdftfdff )()()()()()(122121 ttttttdftfddffddffdff )()()( )()(

7、)()()(11212121证明：)()()(21tftfty则有：)()()()(2)(1)(tftftyjiji推论：tdfdttdftftfty 2121)()()()()(例：例：tttUttUtdttUdttdtUtU)()(*)()(*)()(*)()()(*)(tfttf证明：)()()()()()()(tfdtfdtfttf或写为)()()()(tfdtf抽样性)()(*)(00ttftttf2、与阶跃信号的卷积：tdftftUtUtf)()()(*)()()(tttdfdftdftUdtdtUtf)()(*)()(*)()(*)(证明：)(*)(*)(*)(2)2(tUett

8、tUtUett)()(313)()()()()()()()()()()()()()(22)(2222)2(tUeetUetUedtUeUetUetUetUetttUetUetttUtUettttttttttt解：原式解：原式) 3()()(),1()(tUtUthtthGD) 1()()(tUtUtf)(tyhD(t)hD(t)hD(t)hG(t)f(t)h1h2h3y(t)h4123400 ( ) *: ( ) ( )(1)(1)* (1)* ( )(3) ( )(1)(2)* ( )(3) ( )(3)* ( ) ( )(3)* (1) ( )(3)* (2) ( )* ()(hthhhhhtttttU tU ttttU tU tU tU ttU tU ttU tU ttf tt tf t t 即)( )( )(3)(1)(4)(2)(5)( )( )(1)(2)(3)(4)(5)(2)( ) :( )( ) *( )h tU tU tU tU tU tU th tU tU tU tU tU tU ty ty th tf t求)6()6()3()3(2)()1()1()(*)5()4()3