[time-nuts] FMT

[John Ackermann N8UR]

Hal Murray said the following on 11/10/2006 04:52 PM:
>> I think one of the reasons this method works so well is that the FFT
>> effectively averages the signal over some time, and I use a tool in
>> the  software to derive an average across all the FFT results.  That
>> smooths  out the instantaneous variations that make real-time
>> measurement such a  challenge. 
> 
> What size FFT are you using?  What sort of averaging are you doing?
> 
> What is the bandwidth of the signal you are looking at?  How does that 
> compare to the bin size of your FFT?
> 
> If you are recording the raw data and then post processing the signal, I'd 
> expect you could FFT the whole thing.  It has to fit in memory, but that 
> doesn't look like a problem.  I think that would get the bin size down to 1/N 
> Hz if you had N seconds of data.  (But I'm not a DSP wizard.)
> 
> If you do that, there is only one sample so there is nothing to average in 
> the time dimension.  If the signal is wide enough to end up in several bins, 
> you could average in the frequency dimension.

The signal is a CW wave.

I use a Linux spectrum analysis program called Baudline.  It allows me
to do a couple of neat things.  First is that it can downconvert to
improve resolution by decimating and frequency shifting, so I can work
with a spectrum maybe 100 Hz wide.  Then I run a fairly deep FFT -- I
think I used 8192 bins last year.  You end up with resolution in the
milliHertz, but of course after decimating a bunch and using the deep
FFt, it takes 10 or more seconds to fill the bins.

I end up with a waterfall display showing the reference and unknown
signals over the length of the test.  Baudline has another neat tool,
which allows you to do an average over length of the waterfall.  So, I
end up with a single trace that represents the average over the length
of the test.  Baudline also has a "delta" feature that will calculate
the delta frequency between two signals, so that give me the final
output number that I work against the frequency of the known signal to
get the result.

John