p.dec

im slowly getting to grips with fft... but i have a couple of questions which are probably really obvious...

if your FFT size is 1024 does that mean you receive 1024 frequency bins per second or per sample?

if you have an overlap of 2 in pfft~ does that mean the frequency width of each bin becomes larger to make it up to half the sampling rate?

only half the bins are usable in fft. so if your FFT size is 1024 does that mean the right outlet of fftin~ only sends out 512 bins?

answers would really clear a few things up!

if your FFT size is 1024 does that mean you receive 1024 frequency bins per second or per sample? 

answers would really clear a few things up!
Paul


quick-stupid-fft-questions

You really need to work through a good FFT tutorial. The relevant appendix in Roads' _The Computer Music Tutorial_ is highly recommended, but may be more detail than you want. The relevant MSP tutorials ought to get you to speed.

> if your FFT size is 1024 does that mean you receive 1024 frequency bins per second or per sample?

Neither. It takes 1024 samples to calculate a 1024-bin FFT.

FFT Size == Number of samples needed to calculate == Number of bins.

> if you have an overlap of 2 in pfft~ does that mean the frequency width of each bin becomes larger to make it up to half the sampling rate?

Overlap of two means one FFT is calculated starting at time 0, another one at time N/2 (where N is the FFT size).

> only half the bins are usable in fft. so if your FFT size is 1024 does that mean the right outlet of fftin~ only sends out 512 bins?

All the bins are "usable" and all are calculated and all are sent out the fftin~ inlets. Just because some frequency bins are above Nyquist doesn't make them unusable.

You really need to work through a good FFT tutorial. The relevant appendix in Roads' _The Computer Music Tutorial_ is highly recommended, but may be more detail than  you want. The relevant MSP tutorials ought to get you to speed.

> if your FFT size is 1024 does that mean you receive 1024 frequency bins per second or per sample? 

Neither. It takes 1024 samples to calculate a 1024-bin FFT. 

Overlap of two means one FFT is calculated starting at time 0, another one at time N/2 (where N is the FFT size). 

All the bins are "usable" and all are calculated and all are sent out the fftin~ inlets. Just because some frequency bins are above Nyquist doesn't make them unusable.


people do seem to be giving me different answers. although ive managed to work things out.

yes capture~ displays bins 0 255 with a fftsize of 512. so yes half...

> Overlap of two means one FFT is calculated starting at time 0, another one at time N/2 (where N is the FFT size).

i was presuming this already. but what i dont get is: if your overlap is 2 and then you change it to 4 but keep the same fft size it seems to me that with an overlap of 4 the bins are twice as close together and therefore would take more of them to get up to your sampling rate, or is the actual size of each bin different:

ie... with an overlap of 2 the same number of bins take up as much frequency. so then surely when working out the frequency range of each bin this needs to taken into account????? this isnt really mentioned anywhere.

 firstly: 
yes capture~ displays bins 0 255 with a fftsize of 512. so yes half...

> Overlap of two means one FFT is calculated starting at time 0, another one at time N/2 (where N is the FFT size). 

overlap        2                     4
frequency 
bin1            ____                  ____
bin2              ____                 ____
bin3                ____                ____
bin4                  ____               ____
bin5                    ____              ____
bin6                      ____             ____

my picture didnt quite turn out, but hopefully you get the point.

quick stupid fft questions.