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(2) y(k) 3y(k 1) 2y(k 2) f(k)y( 1) 1,y( 2) 0,f(k) (k)解： 2 3 2 0 1 1, 2 2yx(k) Cx1( 1)k Cx2( 2)k

1

y( 1) y( 1) C Cxx1 x Cx1 122

1Cx2 4 y( 2) y( 2) C C xx1x

42 yx(k) ( 1)k 4( 2)k k 0

16

由y(k) f(k) 3y(k 1) 2y(k 2)得：令yfp P.则有P0 3P0 2P0 1,P0

yf(0) f(0) yf( 1) 2yf( 2) 1 yf(0) 1

y(1) f(1) 3y(0) 2y( 1) 2ff f yf(1) 2

11

y(0) C C 1C ff1f2 f162

解之得：

y(1) C 2C 1 2 C 4

f2ff1f2 36

141

yf(k) [ ( 1)k ( 2)k ] (k)

236

141

y(k) yx(k) yf(k) [( 1)k 4( 2)k ( 1)k ( 2)k ]

236

181

[( 1)k ( 2)k ] k 0236

1

gi(t) (1 e st) (t)

9

dgi(t)5 st

2.7 (a)解：hi(t) e (t)

dt9

10

hu(t) Rhi(t) e st (t)

3

(b)解:由图知ic ir il is

ducd2iluLdil

Lc2 iR l 其中：ic c dtdtRRdt

故有：LCiL

L12

iL iSi L iL iL is iLR55

55

(p2 2p 5)(p 1)2 4

2iL 5iL 5iS H(p) 故iL