第二章 连续系统的时域分析习题解答

1、 P2-1 X 第二章 连续系统的时域分析习题解答2-1 图题 2-1所示各电路中,激励为 f (t ,响应为 i 0(t 和 u 0(t 。试列写各响应关 于激励微分算子方程。 解 :.1 p ( ; 1(1 p ( , 11 , 11 ( b (; 105. 7 625(3 102 ;(375 ( 6253( ( 002. 0( a (0202200204006000f i p f p u p f p p p u i f p p p pf t u pf i p pu i t f t u p t f t u p =+=+=+=+=+=+=+=+- 2-2 求图题 2-1各电路中响应 i 0(

2、t 和 u 0(t 对激励 f (t 的传输算子 H (p 。解 :. 1( ( ( ; 11 ( ( ( b (; 6253105. 7 ( ( ( ; 6253 ( ( ( a (220 20 40 0 +=+=+=+=-p p pp t f t i p H p p p t f t u p H p p t f t i p H p t f t u p H f i f u f i f u2-3 给定如下传输算子 H (p ,试写出它们对应的微分方程。.2(1(3( ( 4( ; 323 ( 3(; 33 ( 2( ; 3 ( 1( +=+=+=+=p p p p p H p p p H p p

3、 p H p p p H解 :; 3d d 3d d 2( ; d d 3d d 1( f tfy t y t f y t y +=+=+. d 3d d 2d 3d d 4( ; 3d 3d 2 3( 2222f tf y y t y f f y y +=+=+2-4 已知连续系统的输入输出算子方程及 0 初始条件为:. 4 (0y , 0 (0y y(0 , ( 2(13 ( 3(; 0 (0y , 1 (0y , 0 y(0 , ( 84(12( ( 2(;1 (0y , 2 y(0 , ( 3(1(42 ( 1(-2-=''='=+=''=

4、9;=+-='=+=t f p p p t y t f p p p p t y t f p p p t y f (u 0(t (b u 0(t (a图题 2-1 2 试求系统的零输入响应 y x (t (t /0 。解:, e e ( , 3 , 1 1(32121t t A A t y p p -+=-=-=. 0 , e 12(1 (121444200 ,e ( ( , 2 , 0 3(. 0 , 2sin e 5. 0 (905. 00cos 240 sin (cos21cos 0 ,2cos(e ( , 2j 2 , 0 2(; 0 , e 5. 1e 5. 3 (5. 15.

5、 3312 232132323123213 , 2123213233232132213 , 213212121/t t t y A A A A A A A A A A t A A t y p p t t t y A A A A A A A A A A A A t A A t y p p t t y A A A A A A t t t t t t -+-=-=-=+-=-=+=+=-=-=+-=+=+=±-=-=-=-=+= 2-5 已知图题 2-5各电路零输入响应分别为:.0 , V sin e6cos e2 ( (b; 0 , V e 4e 6 ( a (3343x x /t t

6、t t u t t u ttt t -+=-=求 u (0- 、 i (0- 。解 :; V 246 0( 0( a (x =-=+-u u.0 66(1. 0 0( 0( V; 202 0( 0( b (A31 1618(6 0( 0( x x x x =+-=+=+-=+-+-+-i i u u i i 2-6 图题 2-6所示各电路:(a 已知 i (0- = 0, u (0- = 5V,求 u x (t ; (b 已知 u (0- = 4V, i (0- = 0,求 i x (t ; (c 已知 i (0- = 0, u (0- = 3V,求 u x (t .解 :065050 ( (

7、2=+=+=p p p p a Z (b 图题 2-5 (a (a(b (c图题 2-6 P2-3 . 0 , V e 10e 15 ( , 10 , 15320500( 0(' , V 5 0(e e (3 , 2 32212121322121-=-=-=+=+=-=-=/ t t u A A A A A A C i u u A A t u p p t t x x x t t x.0 , V e e 4 , 4 , 1 , 15035 0(' , V 3 0( , e e ,4, 1, 04505: c (. 0 , A sin e 4 ( , 2 , 4 sin cos 4

8、cos 04401 0(' , 0 0( cos(e (11 , 11 02200 ( b (4214212122121212121212-=-=-=+-=+=-=-=+=+=-=-=+-=+=-=+-=+=+=/ / t u A A u u A A u p p p p p t t t i A A A A A A A A i i A t A t i j p j p p p p p t t x x x t t x t x x x t x 同理 /Y 2-7 已知三个连续系统的传输算子 H (p 分别为:. 2(13 3( ; 84( 12( 2( ; 421(22+-+p p p p p

9、 p p p 试求各系统的单位冲激响应 h (t 。 解 :; ( e e ( ( ( 1(3t t h p H t t -+=+=.( e 4e 24( (2 2( ( 3(;( 2sin e 875. 02cos e ( ( 2 2(2875. 0 2( ( 5. 1, 81 84(1 ( (84 ( 2(2222222222t t t h p p p p H t t t t h p p p H B A p p p p B p A p p B Ap p H t t t t -+=+-+=-+-=+-+-=-=+-+-=+-= 2-8 求图题 2-8所示各电路中关于 u (t 的冲激响应 h

10、 (t 解 :(af u pu pu u i i pui f 480422111=+=-+=-e 5. 0 ( ( 81t h p H -= (a 4 .V ( e 4. 0e 4. 2( ( 67215. 01 ( c (.V ( e 2e 2( ( 233 ( b (6222t t h p p p p p H t t h p p p p H t t t t -=+=+=+=-=-=+=+= 2-9 求图题 2-9所示各电路关于 u (t 的冲激响应 h (t 与阶跃响应 g (t 。解 :2cos (0 2cos ( ( ( ,(sin 2 ( ( 2(212 ( a ( t_0 222t

11、 tt d h t g t t t h p p p p p p p H =+=-=+-=+=+=-, (e 4 (2 (212121 ( b ( 1 t t t h p p p p p H t -+=+=+=+=. ( e e ( (0e e ( ( e e 2( (1212 ( c ( e 21( (0e 2 (2 ( ( 2 2_0 2 _0 t tt t t d h t g t h p p p p H t t t t d h t g tt t t t -=+-=-=+-+=+-+=-=-=2-10 如图题 2-10所示系统,已知两个子系统的冲激响应分别为 h 1(t 5(t 21 , h

12、 2(t 5(t 解 :求和号后的冲激响应为 1( (-+t t ,于是整个系统的冲激响应为:(b u(c 图题 2-8u(a(b(c图题 2-9图题 2-10y (t f (t P2-5 1( ( (-+=t t t h 2-11 各信号波形如题图 2-11所示,试计算下列卷积,并画出其波形。. (' ( 3( ; ( ( 2( ; ( ( 1(41 31 21t f t f t f t f t f t f * 解 :.3( 3(2 1( 1(2 1( 1(2 3( 3(2 3( 3(2 1( 1( 1( 1(2 1( 1(21( 1( 3 ( 3(2 1( 1( (' (

13、3(;6( 6( 5( 5( 4( 4( 3( 3( 2( 2( 1( 1( ( 4( 3( 2( ( ( 2(;4( 4( 2( 2( ( 2( 2( 4( 4( 2( 2( ( ( 1(2( 2( ( 2( 2( (11411113111211-+-+=-+-+-+=-+=-+-+-=-+-=-+-+-+=-+=-+-+=t t t t t t t t t t t t t t t t t t t t t f t f t f t f t t t t t t t t t t t t t t t f t f t f t f t f t t t t t t t t t t t f t f t f

14、t f t t t t t t t f *2-12 求下列各组信号的卷积积分。; 2( 1( ( , 1( (sin ( 5(; (sin ( , (e ( 4( ; (e ( , (e ( 3(;(e ( , ( ( 2( ; 1( ( , ( ( 1(2121 2212121+-=-=-=-t t t f t t t t f t t t f t t f t t f t t f t t f t t f t t f t t f t t t t . ( sin ( , ( ( 6(201t t T t f nT t t f n =-=解 :;( e e ( ( e e ( ( 3(;( e 1(

15、 ( d e ( ( 2(;1( 1( ( 1(22 tt t t y t t t y t t t y t -t t -t -t -=-=-=-=图题 2-11(a(b(c(d(1(2ty(t 1 (4 f 2 (t = 1 (e j t e -j t 2j -2 -1 0 -t jt -t -j t 1 1 y (t = (e e (t (e e (t 2 j( j 1 2 j( j 1 t 1 2 2 je -t j(e j t + e -j t + (e j t e -j t (t = 1 (e -t cos t + sin t (t ; 4j 2 (5 y (t = f1 (t 1 +

16、 f1 (t + 2 = = sin(t 1 (t 1 (t 2 + sint (t + 2 (t + 1 ; (6 y (t = f 2 (t nT = sin (t nT (t nT T n =0 n =0 = sin t(sin t (t . T T y(t 1 0 2T 3T 4T t 2 - 13 求图示各组波形的卷积积分 y(t = f1(t* f2(t 。 f1(t f1(t 1 0 2 4 (a f2(t e2 t 2 4 et 1 0 图题 2 - 13 2 (b t -1 0 2 f2(t 1 t 1 0 e t (t - t 解： e e 1 2 t f1( f2(t-

17、2 2 2 - 14 已知 f (t * t ( t = (t + e t 1 (t ，求 f (t 。 f (t 0 t+1 t 0 t 解：微分： f (t * (t = (1 e (t + ( 0 + e 1 (t = (1 e ( t ； 再微分： f (t * ( t = e t ( t + (1 e 0 = e t (t = f (t . 2 - 15 6 1 0 2 某 LTI 系统的激励 f (t和冲激响应 h(t如图题 2-15 t 所示，试求系统的零状态响应 yf (t，并画出波形。 解： y f ( t = h(t 1 0 2 图题 2 - 15 yf (t 1 0 2

18、4 1 ( ( 2d * (t (t 2 0 2 t t =1 t 2 (t (t 2+ (t 2* (t (t 2 4 = 1 t 2 (t (t 2+ (t 2 1 (t 22 (t 2 (t 4 (t 4 4 4 t 2 - 16 图题 2-16 表示一个 LTI 系统的输入-输出关系。试求出该系统的冲激响应。 f (t 4 2 0 1 2 3 t 0 24 图题 2 - 16 1 2 3 4 t y(t (a 输入 (b 输出 解： y ( t = 2 f (t 2 f (t 2 h( t = 2 ( t 2 (t 2 2 - 17 已知某系统的微分方程为 y ( t + 3 y (

19、t + 2 = f (t + 3 f (t ，0 - 初始条件 y (0 = 1 , y (0 = 2 ，试求： (1 系统的零输入响应 yx(t； (2 激励 f (t5 (t 时，系统的零状态响应 yf (t和全响应 y(t； (3 激励 f (t5 e23t (t 时，系统的零状态响应 yf (t和全响应 y(t。 解：(1 算子方程为： ( p + 1( p + 2 y (t = ( p + 3 f (t 1 = A1 + A2 A = 4 y x (t = A1e t + A2 e 2 t 1 2 = A1 2 A2 A2 = 3 y x (t = 4e t 3e 2 t , t/

20、0 ; ( 2 H ( p = p+3 = 2 1 h (t = ( 2e t e 2t (t p + 3p + 2 p + 1 p + 2 y f (t = h (t * (t = ( 3 2e t + 1 e 2t (t 2 2 y (t = y x (t + y f (t = ( 3 + 2e t 5 e 2t (t 2 2 2 (3 y f (t = h (t * e 3t (t = (e t e 2t (t y (t = y x (t + y f (t = (5e t 4e 2t (t 2 - 18 图题 2-18 所示的系统，求当激励 f (t5 e2t (t 时，系统的零状态响应

21、。 P2-7 2 f (t - x1 x2 - y (t f (t x1 -3 (b x2 -2 x3 - y (t (a 图题 2 - 18 解：(a 令 f (t = (t，则 y(t = h(t， 显然 : x1 = px2 = f (t x2 , h(t = 2 x1 x 2 h (t = ( 2p 1 (t = ( 2 3 (t = 2 (t 3e t (t p +1 p +1 p +1 y f (t = 2 (t 3e t (t * e t (t = (2 3t e t (t (b 令 f (t = (t，则 y(t = h(t， 显然 : x1 = px2 = p 2 x3 =

22、f (t 3 px3 2 x3 , h(t = x 2 + x3 = (1 + p x3 h (t = p +1 f (t = 1 (t = e 2 t (t p+2 p + 3p + 2 2 y f (t = e 2t (t * e t (t = (e t e 2t (t t 2 - 19 图题 2-19 所示电路， < 0 时 S 在位置 a 且电路已达稳态；t = 0 时将 S 从 a 板到 b, 求 t > 0 时的零输入响应 ux(t、 零状态响应 uf (t和全响应 u(t。 解：i 先求零状态响应 u f (t： + a S b + 1V 2V 1 F + uC 2H 2 + u 1 1 iL p+22 H ( p = 1 1 = 0.5 = 1 + 0.5 + 2 2 p + 1 1 + 2 p + 0.5 p+2 p + 0.5 p + 2 p h(t = (t + (0.5e 0.5t + 2e 2 t (t , t 图题 2 - 19 u f (t = ( + (0.5e 0.5 + 2e 2 1 + (t d (t = 2(1 e 0.5t e 2t (t 0 ii 求零输入响应 ux(t： u x (t = A1e 0.5t + A2 e 2 t ux (0 + = 1 × 1 + 1 = 2V, u L x (0 + = 1V,