• Subject: Re: Linrad Spur Removal
• From: VK2KU <clearmail.com.au; vk2ku@xxxxxxxxxxxxxxxx>
• Date: Sat, 16 Feb 2008 15:01:24 -0800 (PST)

```Ah, the penny has dropped! A flash of understanding.
Another piece of the jigsaw has fallen into place.
Sorry I am a bit slow at times :-)

FFT2 size 32768
Sample rate 111.111kHz
Window sin^1 giving a factor of 1.5 (or 0.666 depending on your
viewpoint).

So FFT2 resolution with no window is the reciprocal of the length of 1
transform,
or 111111/32768 = 3.39Hz, but wider by 1.5 from windowing,
giving 111111/32768*1.5 = 5.08Hz.

And overlapping the transforms by 1.5 because of windowing means that
the length of each transform is
32768/111111/1.5 = 0.197s,
which corresponds to 111111*1.5/32768 = 5.08 transforms per second.

The transform rate is exactly the same as the FFT2 resolution because
the calculation is identical :-)
And the wide waterfall, set at 50 averages, updates every 50*0.197s =
9.85s as observed.
Finally I understand!  EUREKA.

73  Guy  VK2KU

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