重邮 信号与系统考试答案 考研必备
解:
y
x
(k) 2y(k 1)
x
y
x
(k 2) 0
2
2
2 1 0
x1
( 1) 0
1,2
1
y
x
(k) (c cx2k)( 1)
k
cx1 1 yx( 1) cx1 cx2 1
2( 2) 2 3 cx2 cx1cx2 yx
故:
y
x
(k) (1 2k)( 1)k
k
(3) y(k) y(k 2) f(k 2),y( 1) 2,y( 2) 1 解: 2 1 0;
1,2 j
yx(k) Acok Bsik
22
y( 1) B 2y( 2) A 1
yx(k) ( cos
2
k 2sin
2
k)
k 63.4 )k 02
2.6 (1) y(k) 2y(k 1) f(k),y( 1) 1,f(k) 2 (k)
kk
解: 2 0, 2 y(k) C(2) yp(k) C2 2
yp(k) p0,p0 2p0 2,p0 2
令k 0,y(0) 2y( 1) 2,y(0) 0 y(0) C 2,C 2
所以 y(k) 2(2) 2,k 0
k
其中yx(k) Cx(2)k 2(2)k,k 0
Cx
1,Cx 2 2
yf(k) Cf(2)k yfp(k)
y(k) yx 2(2)k 2 [ 2(2)k] [4(2)k 2] (k)