重邮 信号与系统考试答案 考研必备

k>0时,有h(k)-4h(k-1)+8h(k-2)=0

2 4 8 0 12=2 j2=22

4

h(c)= (c)+4 (k-1)-8h(k-2) h(0)= (0)+4h(-1)-8h(-2)=1=p h(1)= (1)+4h(0)-8h(-1)=4=22故：p=1,Q=1. H(k)=(22)(sin

12

Q 22

p2

k k +sin) (k) 44k k k

=2(22)(sin+sin) (k)

44

k

(5)

y(k+2)+2y(k+1)+2y(k)=f(k+1)+2f(k) 解： h0(k+2)+2 h0(k+1)+2 h0(k)= (k)

3k 3k

+Qsin) 44

3k 3k 1

ho(2)=2[pcos+Qsin]=-2Q=1 所以Q=-

222

ho(k)=(

k

)(pcos

ho(1)= =p-

2[ pcos

1113k 3k

+Qsin]=2[p]

44222

11=0 p=+ 22

13k 13k

sin+cos] (k-1) 22244

k 13k 3 sin(-) (k-1) 244

k

所以 ho(k)=（ =

）[-

h(k)= ho(k+1)+2 h0(k)

k 13 3 3k 3

=2sin[(k+1)- ] (k)+ 22sin(-) (k-1)

4444

k 1k3 3 3k 3

=2sin[(k+1)- ](k)+ 22sin(-) (k-1)

4444k3k

(k-1) =-2cos4

2.11 (1) y(k 2) y(k 1) 2f(k 1) f(k) (由图得)

k

移序得：y(k 1) y(k) 2f(k) f(k 1)设h0(k 1) h0(k) (k),有 1 h0(k) C( 1)k (k 1).