simple fft code problem

Question

0 个投票

x=rand(1,8);
for k=1:8
    for  m=1:8        
       l(m)=x(m)*exp(-i*2*pi*k*m/8);
    end
    X(k)=sum(l);
end
X=X

I used this code to implement the fast Fourier transform but it didn't work. Is there any help?

2 个评论
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Matt J 2012-12-3

This is closer to the DFT than the FFT.

Azzi Abdelmalek 2012-12-3

It's right, this is 'nt FFT algorithm, but the FFT is just a faster way to calculate a DFT. the result will be the same

Answer 1

Azzi Abdelmalek 2012-12-3

编辑：Azzi Abdelmalek 2012-12-3

0 个投票

%k and m start at 0

 x=rand(1,8);
for k=0:7 
    for m=0:7
       l(m+1)=x(m+1)*exp(-i*k*m*pi/4);
    end
    X(k+1)=sum(l);
end

%or

R=exp(-i*2*pi/8)
k=0:7;
XX=exp(-i*pi/4).^(k'*k)*x'

Remark: for big array, you must use FFT algorithm

Azzi Abdelmalek 2012-12-3

compare the result with fft(x), it's the same

ayman osama 2012-12-3

thank u for help