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Computing a DCT using an FFT, Signal analysis
Mustardman
post Dec 9 2005, 06:40
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I am trying to make sense of the output of the FFT when used to calculate a type-2 DCT (DCT-II). I managed to scrape together enough documentation that infomed me that, if I wanted to perform a DCT-II using an FFT algorithm, then the data should be arranged to be real even symmetric, with every even indexed data element set to zero.

No problem there. In addition, the outputs are correct, but half the expected value (I can see the output needs scaling by 2 to compensate for the fact that every second sample of the input is zero).

However, I am tyring to figure out how the output comes to be what it is?
In particular, where the zero comes from (term 4 & term 12) [counting from 0].


Input

Real : 0 x0 0 x1 0 x2 0 x3 0 x3 0 x2 0 x1 0 x0
Imag : 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Output

Real : ya yb yc yd 0 -yd -yc -yb -ya -yb -yc -yd 0 yd yc yb
Imag : 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

[grumble: looks good in fixed width font, but spaces are not turning out here]

In my real application, I am using a 16 point DCT, which equates to a 64 point FFT, but the above is representative.

Any hints?



PS: I know there are better ways of doing DCTs, but in my implementation, I only have access to an FFT routine.


Cheers,Owen.
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QuantumKnot
post Dec 12 2005, 01:13
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I agree that there is often not enough documentation on how to do a simple DCT using an FFT. Though it is not the best, computational-wise, it is however interesting to see the relationship between the two, esp. why the DCT is used more often than the FFT in compression (the even-symmetry creates smoothness that results in less high frequency energy).

Anyway, I was interested in this problem a while ago as well and found this link to be the best I've found.

Not sure if this is any help or not, but here is some MATLAB code I wrote during that time, which calculates the DCT using FFT.

CODE
clear all;
close all;

x=[1 2 3 4];
N=length(x);
d = dct(x);

x2=[4 3 2 1 1 2 3 4];
f=fft(x2);

n = 0:2*N-1;

shift=exp(-j*2*pi*(N-0.5)*n/(2*N));

f2 = real(f./shift);

d2 = f2(1:N)/sqrt(2*N);

d2(1)=d2(1)/sqrt(2);

disp(d);
disp(d2);


Notice how I don't do any zero-setting (no idea why you need to anyway). I just take my original signal, mirror it and then take the fft. Now the thing to note is that the Fourier transform of an even symmetry signal is real but we can't make true even symmetry signals on a computer, so we have to compensate for the phase shift caused by the time-shift of 3.5 samples, hence the division by the FFT of the variable 'shift'. Then I scale the coefficients appropriately (to make them orthonormal), and voila.

Hope that helps in a little way cool.gif

EDIT: fixed a spelling mistake

This post has been edited by QuantumKnot: Dec 12 2005, 04:55
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