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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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HotshotGG
post Jan 13 2006, 06:49
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http://www.ee.columbia.edu/~marios/courses...%20(Thesis).pdf

Here is another paper for reference describing faster way's to compute a DCT using an FFT.


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Garf
post Jan 13 2006, 10:03
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QUOTE (HotshotGG @ Jan 13 2006, 07:49 AM)
http://www.ee.columbia.edu/~marios/courses...%20(Thesis).pdf

Here is another paper for reference describing faster way's to compute a DCT using an FFT.
*


The paper is about the MDCT, not the DCT. And the MDCT is based on a type 4 DCT, while he needs a type 2 DCT.

This post has been edited by Garf: Jan 13 2006, 10:04
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