Question

如何计算x(n)={4,3,2,1,0}的DFT?

Answer 1

程序

x=[4,3,2,1,0]

N = length(x)

n = [0:N-1]

k = [0:N-1]

w = exp(-j*2*pi/N)

nk = n' * k

wnk = w.^(nk)

Xk = x * wnk

结果

x =

4 3 2 1 0

N =

5

n =

0 1 2 3 4

k =

0 1 2 3 4

w =

0.3090 - 0.9511i

nk =

0 0 0 0 0

0 1 2 3 4

0 2 4 6 8

0 3 6 9 12

0 4 8 12 16

wnk =

1.0000 1.0000 1.0000 1.0000 1.0000

1.0000 0.3090 - 0.9511i -0.8090 - 0.5878i -0.8090 + 0.5878i 0.3090 + 0.9511i

1.0000 -0.8090 - 0.5878i 0.3090 + 0.9511i 0.3090 - 0.9511i -0.8090 + 0.5878i

1.0000 -0.8090 + 0.5878i 0.3090 - 0.9511i 0.3090 + 0.9511i -0.8090 - 0.5878i

1.0000 0.3090 + 0.9511i -0.8090 + 0.5878i -0.8090 - 0.5878i 0.3090 - 0.9511i

Xk =

10.0000 2.5000 - 3.4410i 2.5000 - 0.8123i 2.5000 + 0.8123i 2.5000 + 3.4410i