Hey Jonathon

I am copying my reply to the list to expose my own potential ignorance.

The CASPER FFT implements a DFT where each resultant bin is the same as
mixing with a complex exponential and then low pass filtering the result
(as per the DFT definition). The complex exponentials have frequencies
centred at 0, 2B/N, 2(2B/N) etc where B is the Bandwidth of your signal and
N is the number of FFT bins. So the 'DC' FFT bin goes from -(1/2)*(2B/N) to
(1/2)*(2B/N) or -B/N to B/N. For example, a signal sampled at 1024GS/s
gives a B of 512MHz. Assuming a 1k point FFT we have a 'DC' bin from
-0.5MHz to 0.5MHz. The N/2 FFT bin, in our example, contains 510.5MHz to
511.5MHz. So, all of the FFT bins are shifted down by 1/2 bin from where
you might intuitively think they are located. Because we sample a real
signal, and due to aliasing, we discard the last N/2 FFT bins as they
contain the same information, but as shown above, they are not actually the
same as the first N/2.

You probably know all of this but it does help answer your question. We
simply discard the last N/2 channels. So you can never get the last half an
FFT bin's worth of frequency info up to Nyquist (in our example, from
511.5MHz to 512MHz. The DC bin is strange as it contains the aliased band
from 1023.5MHz to 1024 MHz (hopefully nicely filtered out in analogue
stages), as well as 0 MHz to 0.5MHz. So far, we have treated it as
something we throw away and have not tried to extract anything from it.

This might be a problem for people wanting to do a very coarse FFT, as half
an FFT bin might be a lot of bandwidth to discard.

Regards
Andrew

On Fri, May 19, 2017 at 6:17 PM, Jonathon Kocz <jxk...@gmail.com> wrote:

> Hi Andrew,
>
> Sorry to bug you, but I thought it was easier to ask first before looking
> at the FFT in detail.
>
> Do you remember, in the CASPER FFT, is the Nyquist frequency (the purely
> imaginary ch N/2 + 1) put in the imaginary part of the DC or discarded?
>
> Cheers,
> Jonathon
>

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