好文档 - 专业文书写作范文服务资料分享网站

南航金城信号与线性系统课后答案 第二章 连续系统的时域分析习题解答

天下 分享 时间: 加入收藏 我要投稿 点赞

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i0(t)和u0(t)。试列写各响应关于激励微分算子方程。

4k? ? f(t) ? + 6k? u0(t) 2?F - (a)

图题2-1

i0(t) 1? f(t) 1H 1? (b) i0(t) 1F + u0(t) - 解:

f(t)(a) (1?0.002p?1)u0(t)??(3p?625)u0(t)?375f(t) ;464 i0?2?10?6pu0? (3p?625)i0?7.5?10?4pf ;(b) u0(t)?f1?p1?p? u0p?11f , i??f ,0p?1p2?p?1p2?p?1 ? (p2?p?1)u0?(p?1)f ; (p2?p?1)i0?f .2-2 求图题2-1各电路中响应i0(t)和u0(t)对激励f(t)的传输算子H(p)。

u0(t)i0(t)7.5?10?4p375? ; Hi f(p)?? ;解:(a) Hu f(p)?f(t)3p?625f(t)3p?625(b) Hu f(p)?u0(t)i(t)p?1p?p?2 ; Hi f(p)?0?2 .f(t)p?p?1f(t)p?p?12

2-3 给定如下传输算子H(p),试写出它们对应的微分方程。

pp?3 ; (2) H(p)? ;p?3p?3

p?3p(p?3)(3) H(p)? ; (4) H(p)? .2p?3(p?1)(p?2)d yd fd yd f解:(1) ?3y? ; (2) ?3y??3f ;

d td td td t(1) H(p)?d yd fd 2yd yd 2fd f(3) 2?3y??3f ; (4) 2?3?2y?2?3 .

d td td td td td t2-4 已知连续系统的输入输出算子方程及0– 初始条件为:

2p?4f(t), y(0-)?2, y?(0-)?1 ;(p?1)(p?3)?(2p?1)(2) y(t)?f(t), y(0-)?0, y?(0-)?1, y??(0-)?0 ; 2p(p?4p?8)3p?1(3) y(t)?f(t), y(0-)?y?(0-)?0, y??(0-)?4 .2p(p?2)(1) y(t)?试求系统的零输入响应yx(t)(t?0)。 解:(1) p1??1, p2??3, y(t)?A1e?t?A2e?3t,

?2?A1?A2?A?3.5 ???1?y(t)?3.5e?t?1.5e?3t , t?0;?1??A1?3A2?A2??1.5(2) p1?0, p2, 3??2?j2 , y(t)?A1?A2e?2tcos(2t?A3),?0?A1?A2cosA3?A1?0?? ?1??2A2(cosA3?sinA3)??A2?0.5?y(t)?0.5e?2tsin2t , t?0 . ?0?4A2cosA?A??90?23?3?(3) p1?0, p2, 3??2 , y(t)?A1?(A2t?A3)e?2t,?0?A1?A3?A1?1?? ?0?A2?2A3??A2??2?y(t)?1?(2t?1)e?2t , t?0 .?4??4A?4A?A??123?3?2-5 已知图题2-5各电路零输入响应分别为:

(a) ux(t)?6e?3t?4e?4tV, t?0 ;(b) ux(t)?2e求u(0-)、i(0-)。

解:(a) u(0?)?ux(0?)?6?4?2V;

?3tcost?6e?3tsintV, t?0 .

1 6 F - u 1? + (a) i 1? i 1H i(0?)?ix(0?)?1(?18?16)?2?5A613 (b) ux(0?)?ux(0?)?2?0?2V; i(0?)?ix(0?)?0.1(?6?6)?0.2-6 图题2-6所示各电路:

(a) 已知i(0-) = 0,u(0-) = 5V,求ux(t); (b) 已知u(0-) = 4V,i(0-) = 0,求ix(t); (c) 已知i(0-) = 0,u(0-) = 3V,求ux(t) .

2解:(a) Z(p)?0?5?p?6?0?p?5p?6?0

6? 1H + u 0.1F - (b) 图题2-5

p

i 5? (a) 1H 1 F 6 + u - 1? i + 1F u 1H 1? - (b) 图题2-6

i + 1F u - 1 ??5 (c)

1 H 4 ?p1??2 , p2??3?ux(t)?A1e?2t?A2e?3tux(0?)?5V, ux'(0?)??5?A1?A2i(0?)?0??C?0??2A1?3A2

?A1?15, A2??10, ux(t)?15e?2t?10e?3tV, t??0? .(b) Y(p)?0?1?p?1?0?p2?2p?2?011?p ?p1??1?j1 , p2??1?j1?ix(t)?A1e?tcos(t?A2)?0?A1cosA2 ix(0?)?0 , ix'(0?)??1?0?4?4???4??A1cosA2?A1sinA2 ?A1?4, A2??π/2, ix(t)?4e?tsint A, t??0? .(c) 同理:p?5?4?0?p2?5p?4?0,p1??1,p2??4,p ux?A1e?t?A2e?4t, ux(0?)?3V, ux'(0?)??5?3?0??15, A1??1, A2?4, ux?4e?4t?e?tV, t??0?.2-7 已知三个连续系统的传输算子H(p)分别为:

(1) 2p?4?(2p?1)3p?1 ; (2) ; (3) . 22(p?1)(p?3)p(p?4p?8)p(p?2)试求各系统的单位冲激响应h(t)。 解:(1) H(p)?1?1?h(t)?(e?t?e?3t)?(t) ;

p?1p?311)p2?(B?1)p?1(A?1Ap?B82(2) H(p)??8?2??A?,B??1.5pp?4p?8p(p2?4p?8)811(p?2)?0.875?2 ?H(p)??8?8 p(p?2)2?22 ?h(t)?(?1?1e?2tcos2t?0.875e?2tsin2t)?(t) ;8811(3) H(p)?4?2.52?4?h(t)?(1?5te?2t?1e?2t)?(t) .p(p?2)p?24242-8 求图题2-8所示各电路中关于u(t)的冲激响应h(t)。 i1 2? 1?f?i1?pu?8pu?u?4f 解:(a)??2i1?2i1?u?4pu?0f 4? 1F 2i 1 +-+ u - ?tu0.54 ?H(p)????h(t)?0.5e8?(t)V. f8p?1p?0.125(a)

(b) H(p)?10.5p1?3?p0.5p0.5H ?2?2?22p?3p?2p?1p?2+ f - 1 ??3 + u 1F - (b) ?h(t)?(2e?t?2e?2t)?(t)V.1 2p11 1? (c) H(p)?1??2F + + 3 311p?7p?6?0.5p?1?f u p11?30.5F 1? - - p(c) ??0.4?2.4 ? h(t)?(2.4e?6t?0.4e?t)?(t)V.p?1p?6图题2-8

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)。

12(1)pp21?4(a) H(p)???解:2p?p?12p?12p2?1p22 ?h(t)?1?(t)?2sint?(t),242+ 1F f 1F - (a)

+ 1H u

-

t?? t g(t)??h(?)d??1?(t)?2?cos???(t)?1cost?(t) . 0_24?22? 0?2+ 1?11f ? 1 tpp?114112 (b) H(p)?????h(t)?δ(t)?eε(t),- 1?22p?12p?124p2? 1τ t? 1 t1112 g(t)??hτ()dτ?ε(t)?[e]ε(t)?(1?e2)ε(t) . 0_+ 222 0?1? + 1? 1F u - (b)

1? + u - 1F (c)

t1H (c) H(p)?2??2?1?h(t)?(2e?2 t?e? t)ε(t), p?21?1p?2p?1p?1pf 2? - 图题2-9

t ? t?2 t g(t)??hτ()dτ?[?e?2τ?e?τ] tε(t)?(e?e)ε(t) .0 0_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h1(t)??(t?1),h2(t)??(t),试求整个系统的冲激响应h(t)。 f(t) ?

图题2-10

h1(t) h2(t) y(t)

解:求和号后的冲激响应为?(t)??(t?1),于是整个系统的冲激响应为:

h(t)??(t)??(t?1)

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。

(1) f1(t)*f2(t) ; (2) f1(t)*f3(t) ; (3) f1(t)*f4'(t) .

f1(t) 1 t

(1) f2(t) (1) t

f3(t) (1) (1) 3 0 2 4 (-1) (c)

图题2-11

1 -4 -2 0 f4(t) 1 t

-1 0 1 (d)

t

-2 0 2 (a) -2 0 2 (b)

解:

f1(t)?1(t?2)?(t?2)?t?(t)?1(t?2)?(t?2)22(1) f1(t)*f2(t)?f1(t?2)?f1(t?2) ?1(t?4)?(t?4)?(t?2)?(t?2)?t?(t)2 ?(t?2)?(t?2)?1(t?4)?(t?4);2(2) f1(t)*f3(t)?f1(t?2)?f1(t?3)?f1(t?4)f1(t)*f2(t) t 2 4 (1)

f1(t)*f3(t) 0.5 0 1 2 3 4 5 6 t (2)

111 ?t?(t)?(t?1)?(t?1)?(t?2)?(t?2)222 ?(t?3)?(t?3)?1(t?4)?(t?4)?1(t?5)?(t?5)?1(t?6)?(t?6);222f1(t)*f4 '(t) (3) f1(t)*f4'(t)?f1(t?1)?f1(t?1)1 3 ?1(t?3)?(t?3)?(t?1)?(t?1)?1(t?1)?(t?1)-3 -1 0 22-1 ?1(t?1)?(t?1)?(t?1)?(t?1)?1(t?3)?(t?3)(3) 22 ?1(t?3)?(t?3)?3(t?1)?(t?1)?3(t?1)?(t?1)?1(t?3)?(t?3) .22221 t

2-12 求下列各组信号的卷积积分。

(1) f1(t)??(t) , f2(t)??(t?1) ; (2) f1(t)??(t) , f2(t)?e?t?(t) ;(3) f1(t)?e?t?(t) , f2(t)?e?2t?(t) ; (4) f1(t)?e?t?(t) , f2(t)?sint?(t) ; (5) f1(t)?sin?t[?(t)??(t?1)] , f2(t)??(t?1)??(t?2) ;(6) f1(t)???(t?nT) , f2(t)?sin?t ?(t) .

Tn?0解:(1) y(t)?(t?1)?(t?1) ;?(2) y(t)?(?e-?d?)ε(t)?(1?e-t)ε(t) ; 0 t

(3) y(t)?1(e-t?e-2t)ε(t)?(e-t?e-2t)ε(t) ;2?1

南航金城信号与线性系统课后答案 第二章 连续系统的时域分析习题解答

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i0(t)和u0(t)。试列写各响应关于激励微分算子方程。4k??f(t)?+6k?u0(t)2?F-(a)图题2-1i0(t)1?f(t)1H1?(b)i0(t)1F+u0(t)-解:<
推荐度:
点击下载文档文档为doc格式
1f8vm0iy8l23x6i11fyp2nsft0iv0l00r2f
领取福利

微信扫码领取福利

微信扫码分享