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Re: Linrad Spur Removal
- Subject: Re: Linrad Spur Removal
- From: Leif Asbrink <sm5bsz.com; leif@xxxxxxxxxxxxxxxx>
- Date: Fri, 15 Feb 2008 10:57:53 +0100
Hi Guy,
> I have not seen discussion here of Linrad Spur Removal in recent
> times.
I am really missing it because the spur routines are quite
a bit different now and I am interested in feedback because
there are probably many things that can be improved.
> There seems to be no good reason not to use the Autospur option, which
> has the advantage that manual removal of a particular spur may not
> work for one of several reasons, whereas Autospur will pick it up
> sooner or later!
> Somehow I expected to see all my computer birdies vanish instantly,
> whereas the actual process took several minutes to eliminate (or
> reduce) between 10 and 20 birdies. Even then some of them reappear,
> perhaps as they or the SDR-IQ calibration drifts, and are then picked
> up again and eliminated.
Elimination should be instant. If it is not, the S/N of the spur
is not good enough due to modulation sidebands or otherwise.
> The little box at the top RH corner of the main waterfall changes from
> yellow (on) to blue (off) when you click it, and the number of treated
> spurs is displayed immediately above that.
>
> The overall effect is very impressive in reducing lots of spurs to a
> manageable level (depending on how dirty they are) and eliminating
> others. It is likely that my settings were not optimal, so to enable
> Leif or anyone else to comment, this is how I had things set up:
> Enable AFC/Spur... 2 (Autospur)
> AFC lock range 150 Hz
> AFC Max Drift 100 Hz/min
> Enable Morse Decoding 0
> Max number of spurs to cancel 100
> Spur time constant (0.1s) 1 (notch BW = 10Hz)
>
> I understand this to be an aggressive setting.
> My FFT2 bandwidth is 5Hz (for JT65B)
> FFT2 size = 32768
> Sampling rate 111 kHz
> so each FFT2 transform (I and Q) takes about 0.6s (I think).
111000/32768=3.387Hz. (0.3s) This would be the bandwidth of an
un-windowed transform.
With a sine squared window, the bandwidth would be 6.7 Hz
while transforms (overlapping) would arrive at a rate of
6.7 Hz.
The spur removal works on fft2 transforms, the minimum number
is 4, so the minimum time over which the phase and amplitude
must behave well is 0.6 seconds. The spur time constant is
0.6 seconds, setting a lower value like 0.1 does not change
anything. I do not know exactly what bandwidth this corresponds
to because that depends on circumstances. For linear frequency
drift or linear amplitude variations, the bandwidth is
several Hz, but for abrupt changes of the phase or amplitude,
the bandwidth is narrower than 1 Hz.
> I recall a comment by Leif on his website that using too aggressive
> settings will actually eliminate a CW signal, in spite of its
> interrupted nature.
This is one of the things that should be greatly improved now.
I am interested in recordings where something goes wrong.
> What about a JT65B sync tone? Is there any danger
> that the very signals I am trying to see on the main waterfall will be
> eliminated as supposedly unwanted spurs?
I have no idea, I never tried it and nobody ever reported anything:-(
Dear Guy, your settings are not aggressive at all in a CW context,
but I do not know enough about JT65 to say anything about that.
The new routines look for a rised noise floor near the spur and
if there is, it is interpreted as modulation (CW keying) and then
the spur is not removed. In JT65, the modulation sidebands are
very close to the carrier and fft2 will not resolve them so all
the energy of the spur will seem to be in one fft bin and removal
will be tried. Whether the JT65 signal survives or not depends
on the libration fading which might change phase and amplitude
rapidly enough to not fit the time constant.
73
Leif / SM5BSZ
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