Why is DFT phase different from FFT phase?

If I compute the DFT using the definition given on Wikipedia, real components relate directly to cosine amplitudes and imaginary components relate directly to sine amplitudes.

I.e.

dft(sin(t))= [0, 0-0.5i, 0, 0, 0, 0, 0, 0+0.5i]

where

t=linspace(2*pi/8,2*pi,8);

but using the fft function:

fft(sin(t))/8= [0, 0.3536-0.3536i, 0, 0, 0, 0, 0, 0.3536+0.3536i]

The magnitudes are equal but I can no longer seperate sine and cosine components with FFT. Is there any way I can pre-treat my data so I can use FFT to give the same outputs as DFT?

Answer 1

[Wayne King]

The DFT is a linear operator from \mathbb{C}^N to \mathbb{C}^N.

Your vector sin(t) for the argument values you have defined is

    0.7071
    1.0000
    0.7071
    0.0000
   -0.7071
   -1.0000
   -0.7071
   -0.0000

The 2nd element of the DFT is equal to inner product of your vector and

W = exp(1i*(2*pi)/8*(0:7)');

So

X = [ 0.7071
  1.0000
  0.7071
  0.0000
 -0.7071
 -1.0000
 -0.7071
 -0.0000];
dot(W,X)

Why do you expect any other result? The output of fft() matches exactly what you would expect from the DFT (given the definition used in MATLAB).

[Jeff]

Aaaahhhh yes, I knew something wasn't right. I had used

W=exp(1i*(2*pi)/8*(1:8)');

and it matched up perfectly because I was using

t=linspace(2*pi/8,2*pi,8);

when I should have been using

t=linspace(0,2*pi-2*pi/8,8);

A basis of slightly shifted sine waves to match up to my slightly shifted sine wave. To be fair, I should have probably waited until morning before posting the question.

Thanks for the help, I can sleep happy now