Re: Waterfall Rate and FFT2 Size

[Leif Asbrink <sm5bsz.com; leif@xxxxxxxxxxxxxxxx>]

Hi Guy,

> My SDR-IQ after decimating by 600 runs at a sample rate of 111.111kHz,
> which I take to mean 111111 samples of I AND 111111samples of Q each
> second. 
Yes.

> My FFT2 size is currently 16384, which at this rate would take
> 0.147s. 
Yes.

> With the waterfall averages set to 50, that makes 7.4s per
> waterfall line, which is about what I observe. FFT2 resolution
> bandwidth would be 111111/16384, or 6.78Hz without windowing, and
> 10.17Hz with a sin^1 window, which is what Linrad reports. So far so
> good, I think. Not necessarily optimal, but that's a different matter.
The transforms overlap at the -3 dB point of the window. You should
get 7.4 s per waterfall line without a window, but with a window the
time becomes shorter:

Win   time factor
 0       1.0
 1       0.666
 2       0.5
 3       0.416
 4       0.364

The window means multiplying with a function that is zero
at the end points. After back transformation one would
have to divide by the window to get the original function
back. We can not divide by zero!! To preserve accuracy
Linrad does not divide by numbers smaller than -3 dB,
that is why the transforms overlap so much with high powers
of sine. 

For fft1 it is a good idea to set a high bandwidth like 200
Hz with a sine 3 window. It gives a rather wide peak but with
very good shape factor. It will filter out even very strong 
signals adequately (which a sine window would not do)

For fft2, a sine window is adequate because there are no
strong signals and then the higher resolution with a low
order filter is advantageous. (The same resolution with a
smaller time delay)

> Now, in senswat.htm Leif gives an example for a Delta-44 card sampling
> at 96KHz, which I take to mean that all 4 channels are sampled at this
> rate, I and Q for 2 polarizations. 
Yes.

> FFT2 bandwidth is given as "about
> 8Hz" in a sin^1 window as above, which would equate to 5.3Hz with no
> window. 
The fft2 size is 16384. Without a window each bin would be 5.85 Hz.

So at a sampling rate of 96kHz the FFT2 size should be
> 96000/5.3, or 18113 which must actually be 2^14 or 16384. 
Yes.

> At 96kHz
> this number of samples would take 0.17s, and 100 averages (as stated
> in the webpage) would take 17s, nearly twice as long as the 10s
> stated.
It would actually take 17*0.666=11.322s (See table above)

73

Leif

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