重邮 信号与系统考试答案 考研必备
(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