数字信号处理 答案
当然也可以直接计算 X(k)=X1 *(k)X2(k)的 IDFT。
x(n)= IDFT[ X (k )]= IDFT[ X 1* (k ) X 2 (k )] 1 N 1 =∑ X 1* ( k ) X 2 ( k )W N kn N k=01= N=
N 1 ∑ ∑ x1 (l )WNkl X 2 (k )WN kn k=0 l=0 * x1 (l )
N 1
*
∑l=0N 1 k=0
N 1
1 N
∑Xk=0
N 1
k ((l+ n )) 2 ( k )W N
由于 1
N所以
∑ X 2 (k )WN k (l+n)= N∑ X 2 (k )WN k ((l+n))k=0
1
N 1
N
= x2 ((l+ n) N* x(n)=∑ x1 (l ) x 2 ((l+ n)) N R N (n) l=0 N 1
0≤ n≤ N -1
11.证明：
1 N
1∑| X (k )|= N k=02
N 1
1∑ X (k ) X (
k )= N k=0*
N 1
N 1 kn X (k ) ∑ x(n)W N ∑ n=0 k=0 kn N
N 1
*
=∑ x * ( n)n=0
N 1
1 N
∑ X (k )Wk=0
N 1
=∑ x * ( n) x( n )=∑| x( n)| 2n=0 n=0
N 1
N 1
12.解:
由 DFT的共轭对称性可知x(n) jy(n) X(k)=Fep(k)? jY(k)=Fop(k)
方法一
N N (1 ) F ( k )= 1 a+ j 1 b k k 1 aW N 1 bW N
1 1 aN X ( k )= Fep ( k )=[ F ( k )+ F * ( N k )]= 2 1 aWNk 1 1 bN Y (k )= jFop (k )=[ F (k ) F * ( N k )]= 2j 1 bWNk
1 x ( n)= N
∑ X (k )Wk=0
N 1
kn N
1 N 1 1 a N =∑ W N kn k N k=0 1 aW N
=
1 N
N 1 m km kn∑ m=0 a WN WN∑ k=0 m
N 1
1= a N m=0
∑
N 1
K=0
∑W
N 1
k (m n) N
0≤n≤N -1