信号与系统部分课后题答案第1-8章-第二章习题

PAGEPAGE92.1解微分方

(D2

3D

y(t)

y(0)

y(1)(0)解：特征方2

2

1

2齐次解

yh(t)c1(t)e

c2eeeyceeyc (t

2

c由初始条件c

c1c2

c1c 2c

2y(t)

(2e

e2t)u(t)(d)(D3

D2解：（32（

(t)

cehy(1)(t)h

11h1y(2)(t)ceh1 c1c3cc

c3c1

2c2

3

(D2

4D

y(t)

e2tu(t)

y(0)

y1(0)

3

1

yh(t)

ce

e3th1y(1)(t)h1

e

e3t设特解

yp(t)

Ae2tpy(1)(t)p

2Ae2t

y(2)(t)

4Ae2tp代入方程得4Ae2tp

8Ae2t

3Ae2t

e2tA完全解：y(t)

yh(t)

yp(t)

ce

e3t

e2t1y(1)(t)1

e

e3t

2e2ty(0)c1

11y(1)(0)1

3c22

1/1/y(t)

2et

e

1

2e

)u(t

(c)

tu(t)eatu(t)

tea(t)d

[t a2

(1

eat)]u(tea0tea0tateaea(t0t

t1(1eat)]u(tttu(teatu(t)图解法

设a

a2tu(teatu(t)图形(e)sintu(t)[u(t)

其它tsin[u(t)

u(t

0t

其它

(1cost

0t11t024*(g)

e3tu(t)3t

t解

u(t)

te3(t 1(1e3(t1)

t e3(t

)1

注ttt t

y(t13t *(g)

te3tu(t)

1)t

3(t

(13

tu(t

e3u(1 t

t t13

y(t

e3(t)u(tt

响应y(t)，并概略地画出y(t)

h(t)

(t)2

1)

解：y(t)

x(t)*h(t)

x(t)*[(t)2

1)

x(t)2x(t

1)

x(t

x(t21t0x(t21t01234x(tx(t21t0121230t24 y(t)

u(t)

1)

2)

3)

yy(t201234t另解：先x(t分解，x(t

u(t)

1)

2),则：y(t)

x(t)*h(t)[u(t)

1)

2)]*[(t)2

u(t)

1)

2)

3)

(d

y(t)

x(t)*h(t)

x(t)*[

(t

)2

1(t

1x(t1

)

x(t1h1h(t2t02x(tx(tty(t012t02y(t2t212012*(f

求y(t

x(t)*

(t)2tx(t1

h(t

t

t

11t1tt 21t1t11t2 tt2t1tt2t

00t1t21 t23t1

t

t3t210210tt21 tt21t11t2 tt2 t

2ttt1t

1ty(t)

t t

2

t)

d

t1ttt

0t

2

t)

d

d1

1t11 2

2

t)

d

t1

2

t)d 2

2t0 0

2

t

3tt112

x(t)h(th(tt x(t)

1)

1)][u(t

1)

2)]

x1(t)

x2(t)h(t)

[u(t)

1)](2

1)

2)]

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

x(t)*

[x1(t)

x2(t)]*[h1(t)h2x1(t)*h1(t)

x2(t)*h1(t)

x1(t)*h2(t)

x2(t)*h2(t)2.7已知(

4D

3)y(t)

(D1)x(t

h(t)解：设

4D

）ˆ(t

(t

ˆ(1)

22（D0

4D

）ˆ(t

00

(tˆ(1)

(0)

ˆ(1)

(0)

)]

00

ˆ(t

hˆ(t)为连续函，

ˆ(1)

h(t)dt由初始条件

ˆ(1)

0可得：ˆ(1)

（D2

4D

）ˆ(t

(t的特征方程：r2

3

特征

r1

r2ˆ(t

ce

e1代入初始条件1

ˆ(1)

可得

1

2,

1/ˆ(t

[1

2e3t

1

2et]u(th(t)

1)ˆ(th(t)

dˆ(t

ˆ(t

e2

1et2

1e2

1e3t2e3tu(t

设一个LTIh(t)和输入x(t计算当

2y(t 1t

x(t)

k

kT

- 解：y(t)

h(t)*

x(tk

h(t)*

kT)

k

kTT1时，y(t

k

k)1tT

y(y(t)h(t2kk1t

y(ty(tx解：h(t)

h1(t)

1()*h2(t)*h3(tu(t)

u(t)*

1)*[

(tu(t)

u(t)*

1)

u(t)

由图写出系统的微程，并计算当输x(t)2etu(t)时系统的零状态响应

y''(t-

y'(t

y''(t)

x(t)

5y'(t)

6y(t解：a）y''(t)

x(t)

5y'(t)

6y(ty''(t)

5y'(t)

6y(t)

x(tx(t)

2et

y(0)

y(1)

特征根

3,

2,特解：y(t

Ae把特解代入方程中

Ae

5Ae

6Ae

2et2Ae

2et

A零状态响应yx

(t)

ce3t

e2t

ey2y2x

e3t

e2t

etyx(0)c2

1xy(1)(0)x2

2c11c1解得c2 yx(t)

(et

2e2t

e3t

)u(thh(t)ce2tce3t12h'(t)2ce2t3ce3t12c21ch'(0)1,h(0)h''(t)5h'(t)6h(t)(t

yy

y'y1y''(t)xy''(t)x(t)4y(t)11y(t)y(t)12'1y(t)1

y''(t)

1[y''(tx(t)

1

(t)2

x'(t)

111

'(tx(t)

x'(t)2

(t)2

y'(t y''(t)4y(t)1/2x'(t)x(t(b)

y''(t)

4y(t)

x(t)1

2x'(t解法1

先求x(t

(t)时的ˆ(tˆ'(t)

4ˆ(t

(t)

ˆ'(0)

特征方程

4

12

2ˆ(t

c1sin

cosˆ'(t)

cos

sin由初始条件得ˆ'(0

1

cos0

sin

1

1/

0

sin0

cos0

0 ˆ(t

1

2sinh(t

ˆ(t)

1ˆ'(t21 cos2t2

2

2t)u(ty(t

h(t)* (1

x(t2cos

1

2sin

)2e(t cos2e(t)d

2e(tet[cos2e0

t（sin2t5

cos2t5

1/

）u(t解法y''(t)

4y(t

1

2x'(t)

x(tˆ'(t)

4ˆ(t)

x(t齐次解ˆh(t)

c1cos

sin特解

p(t)

Aep由ˆp

(t)

4

p(t)

2etAe

4Ae

2et

A

2/ˆ(t)

c1cos

sin

2

5eˆ'ˆ(0) c1

2/5

0

2/

2/5

0

1/ˆ(t

2

5cos2t

1

2

5eˆ'(t)

2

5cos2t

2

5ey(t)

ˆ(t

2ˆ'(t5

sin2t

1cos2t5

1

5et）u(t3：

y''(t)4y(t)1/2x'(t)x(t)x(t)2etu(tx(t)1

2x'(t)

2et

[et

(t)

et(t)

et初始条件：

y'(0)

y(0)求x(t

et，t

0时的全响应设yp(t)

Aet

则：Aet

4Aet

et

A

1/yp(t

1

5et特征方

4

12

2yh(t)

cos2t

siny(t)

c1cos2t

sin

1

5ey'(t)

sin

cos2t

1

5etc11/5

1/2由初始条2

1/5

3/y(t)

(3

5sin

1

5cos2t

1

5et

+-+-

1);h2(t)

u(t)

3),求总系(t解：h(t

[(t)

h1(t)*h1(t)]*h2(t[(t)

1)

1)*

1)]*[u(t)

[(t)

1)

2)]*[u(t)

u(t)

1)

2)

3)

4)

*2.23证明：(a)x(t)*

'(t

x'(t证：

[

x'(t

x(tx'(t

x'(t

注：注：x(t)()]x(t)'(x'(t)d(t)(x(t)(t)x(0)(t) x(t)()x(t)(或：x(t)*

'(t)

x'(t)*(t)

x'(tg(g((b)

g'(0)g(

'(

[

)]'

g'(

[

)]'d

g'(