重邮 信号与系统考试答案 考研必备
Cx1
1j*
Cx2 Cx1
1j
1( 1 j3)k( 1 j3)k
[ ] 故 yx(k)
jj3
=
2
k 1
e
j
2
k3
3
e2j
j
2 k3
23
2ksin
2
k k>=0 3
(6) y(k) 7y(k 1) 16y(k 2) 12y(k 3) 0
yx(0) 0, yx(1) 1, yx(2) 3
解: 7 16 12 0 即 ( 3)( 2) 0
3
2
2
1 3 2,3 2
yx(k) Cx0(3)k (Cx1 Cx2k)(2)k
带入初始条件有
yx(0) cx0 cx1 0
yx(1) 3cx0 2cx1 2cx2 1
(2) 9cx0 4cx1 8cx2 3 yx
解之得:故:
c c
c 2c 1 5c 8c
x0
x1
x1
x2
x1
x2
c
x
x0
1,
k
c
x1
1,
k
c
x2
1
y
2
(k) 3 (1 k)(2) k>=0
2.5(1) y(k) 3y(k 1) 2y(k 2) f(k),y( 1) 0,y( 2) 1 解:
3 2 0 1 1, 2 2
k
k
y
x
(k) cx1( 1) cx2( 2)
1 1
( 1) )cx2( 2) 0 即: yxcx1( 1 2 2
( 2) cx1( 1) cx2( 2) 1 yx
4cx1 cx2 4
2 0 cx1cx2
cx1 2kk
解之得: 故: y(k) 2( 1 4( 2) k 0 ) cx2 4
(2)y(k) 2y(k 1) y(k 2) f(k) f(k 1) y( 1) 1,y( 2)