解法二:公式解N 1 7 j nk ~ X (k ) DFS ~ (n) ~ (n)e N ~ (n)W8nk x x x n 0 0 2
en 0
3
j nk 4
1 e
π j k 4 4 π j k 4
e e
π j k 2 π j k 8
(e (e
π j k 2 π j k 8
e e
π j k 2 π j k 8
) )
1 eπ sin k 2 π sin k 8
e
3 j πk 8
解法二:公式解N 1 7 j nk ~ X (k ) DFS ~ (n) ~ (n)e N ~ (n)W8nk x x x n 0 0 2
en 0
3
j nk 4
1 e
π j k 4 4 π j k 4
e e
π j k 2 π j k 8
(e (e
π j k 2 π j k 8
e e
π j k 2 π j k 8
) )
1 eπ sin k 2 π sin k 8
e
3 j πk 8