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Author | Topic: Block FFT delays? |
David McClain Member |
![]() ![]() ![]() 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 IP: Logged |
SSC Administrator |
![]() ![]() ![]() 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). IP: Logged |
David McClain Member |
![]() ![]() ![]() Okay... that gets us to 4 blocks with overlap in input FFT with output inverse FFT. (3 stages per FFT Sound - input, compute, output)
Why 7? - DM [This message has been edited by David McClain (edited 08 May 2001).] IP: Logged |
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