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Author Topic:   Block FFT delays?
David McClain
Member
posted 08 May 2001 04:55         Edit/Delete Message   Reply w/Quote
Hi, I measured the overall delay from input to an FFT directly to final output from an inverse FFT at 7 block sizes worth.

I am a bit mystified by this particular size. I would have expected perhaps 2 blocks worth, or maybe some other even number. But why 7 (= 3.5 blocks per FFT)?

- DM

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SSC
Administrator
posted 08 May 2001 09:17         Edit/Delete Message   Reply w/Quote
It's double-buffered so that the entire FFT does not have to be computed in a single sample tick between each block of samples (but can be spread out over the block).

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David McClain
Member
posted 08 May 2001 11:22         Edit/Delete Message   Reply w/Quote
Okay... that gets us to 4 blocks with overlap in input FFT with output inverse FFT. (3 stages per FFT Sound - input, compute, output)


. . input_fwd
. . . . . . . compute_fwd
. . . . . . . . . . . . output_fwd
. . . . . . . . . . . . input_inv
. . . . . . . . . . . . . . . . . compute_inv
. . . . . . . . . . . . . . . . . . . . . . . output_inv

Why 7?

- DM

[This message has been edited by David McClain (edited 08 May 2001).]

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