Re: Very large FFT

AdriaanB wrote:
(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 then
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

.

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