Re: Very large FFT
- From: "AdriaanB" <adriaan.bredekamp@xxxxxxxxxxxxxxx>
- Date: Thu, 26 Apr 2007 23:43:59 -0500
HI
Yes it is clear to me that this is the same Twiddle array that I previous
calculated that for increasing the speed of the FFT calc, but why is he
making it then again ?
It looks like
- FFT with twiddle
-Twiddle again (???)
- FFT with twiddle
-final answer
I noticed the same in a PhD thesis on radio astronomy in the appendix C
where the author made an extra Twiddle step. In the referenced
presentation from Dillon Engineering there was also this step.
What is the purpose of it? Is the 1D FFT calculated as 2D matrix really
that much different to the image processing 2D FFT ?
R
AdriaanB wrote:then
(snip)
He seems to follow the general thing that was also mentioned here of
splitting it in a row * column and then doing FFTs row wise, then
transpose and repeat the rows again (in effect on the columns) and
transposing it again. New question is: Why is the extra twiddle factor
multiplication needed halfway through the algorithm?
There is no extra twiddle factor, it is the same one that
comes on any step of the FFT. It just looks extra because of
the way you are separating it.
I wondered some time ago about an FFT of a whole CD, so maybe
44100*60*70 = 185220000 samples (per channel).
-- glen
_____________________________________
Do you know a company who employs DSP engineers?
Is it already listed at http://dsprelated.com/employers.php ?
.
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