3.3 x%1(n) = x(n + 8r) 说明x%1(n)的周期是8 ∑
r=?∞
∞
= e ∴X% 1(k) = x%1(n)WNkn = e ∑ ∑ ∑
N?1 7
2π
? j kn
8
7
? jπ kn 4
X1(0)X1(1)X1(2)X1(4)X1(6)n=0
n=0
n=0
? jπ 0×n
∑7
e 4
= 3
n=0
∑7
? jπ n
e 4
= (1+ 2 / 2) (1? j) n=0
7
2n
∑
? jπe 4
= ? j;X1(3) = ∑7
? jπ 3n
e 4
= (1? 2 / 2) (1+ j)n=0
n=0 ? jπ 4n
∑7
e 4
=1;X1(5) = ∑7
? jπ 5n
e 4
=(1? 2 / 2) (1? j) n=0 n=0 7
jπ 6n
7
∑
?e 4
= j;X1(7) = ∑? jπ 7n
e 4
= (1+ 2 / 2) (1+ j)
n=0
n=0
= = =
= = 解：(1)x(n) =δ(n)
N?1
N?1
根据定义有：X(k) =∑ x (n)W kn
N
= ∑ δ(n)WNkn n=0
n=0
(2)x(n) =δ(n ?n0),0 < n0 < N
X (k) = ∑N?1
x ( n)W kn = ∑N?1
N
δ (n? n0)WN kn =WN kn
0 n=0
n=0
(3)x(n) = an
0 < n < N ?1
X (k) = ∑N ? 1
x(n)W1?(aWN k )N 1?aN
N kn = n=0
1?(aWN k ) = 1?aWN k
=1 0< n0 <
3.6 N
(4)x(n) = nRN (n)
1? zN?1
令x = ∑ z?N
1(n) = RN(n),则X1(z)?n = n=0
1? z?1
根据线性加权性质可得:
X(z) = ?z dX1(z) = ? Nz?N(1? z?1) ? z?1(1? z?N)
dz (1? z?1)2 ∴ X (k) = X (z) N
z=WN
?k = ? 1?Wk ≠ 0
N k
X(k) = N ? nW1
N ?1
Nkn = n =1+ 2 +3+L+ (N ?1) = N(N ?k=0
∑ ∑ 1)
n=0 n=0
2 .9 图略
f (n) = x(n)*y(n) = {
1,2,3,4,5,3,1,?1,?3,?5,?4,?3,?2,?1 f10(n) = x(n)? y(n) = { ?3,?1,1,3,5,3,1,?1,?3,?5 }
3}
数字信号处理_吴镇扬_第二版_第三章习题答案
