Hi Joe and all,
> I think Roger solved the "problem" with his email to the Linrad
> reflector, copied at the very bottom of this message.
>
> He has found no significant difference in JT65 *decoding ability* with
> Linrad and with a conventional receiver. What he does see is different
> *S/N levels* reported by WSJT. This I can easily believe.
So far good:-)
> There is still this to consider: with xpol antennas the Linrad receiver
> should do better (on average) at JT65 decoding than any receiver
> connected to either the H or V elements alone. If the received signal
> is at 45 degrees, the polarization-matched Linrad advantage should be 3
> dB. If this can't be observed, isn't it true that something else must
> be wrong?
Absolutely. This difference should be large enough to be easily
observable unless Roger has an antenna big enough to never fail
a decode when selecting the best of either H or V.
I suggest a test. Disconnect DC to the mast-mounted amplifiers.
The noise floor should drop by something like 15 dB both in
the conventional receiver and in your WSE system. If there
is no difference that could explain a lower system NF for the
WSE chain , there can be no difference between the conventional
radio with the H antenna and Linrad set to H polarisation.
Both systems are free of spurious responses that could degrade
the system NF and both systems are by far linear enough to
not suffer from any non-linear effect that could affect the
system noise floor.
- o -
Linrad was not written for wsjt, the AFC needs a time delay of
something like 15 seconds to make a non-decodable jt65 signal
stable enough to decode when the signal is close to the jt65
detect threshold (preliminary result) While this would be ok
when chasing weak CQ callers it is certainly not acceptable in
normal operation. It will not be wery difficult to fix because
in a jt65 mode the program will know when the transmission starts
and ends so all the AFC computations could be done very quickly
after the end of the period. CW calls for the output to be
delivered at the same speed as the input so a constant delay
is needed.
There are two different ways of using Linrad for wsjt that I
think would be optimal - but in different situations.
1) Extremely weak signals that have good frequency stability:
-------------------------------------------------------------
In this case it will be a good idea to disable the AGC and
the automatic polarisation adapting. Set the fft2 bandwidth
to 8 Hz for a total processing delay of 0.4 seconds. This
way the high resolution graph will show the spectrum around
the selected signal with 6 Hz per pixel (sine to power 1
window) so one will see several kHz. Set the output rate to
11025 (or higher) and set the first mixer bandwidth reduction
parameter to 3 to use 96/8=12kHz for the baseband. By expanding
the X-axis and selecting a small window on screen you can
select a 3 kHz wide filter using a 64 point fft (10 ms
contribution to the total delay)
The way to operate is to set a large storage time for fft2 and a
large number to average over in the high resolution graph.
When you are finished at one frequency and click on the frequency
where you think an interesting signal has been present for a while,
Linrad will immediately compute the average high resolution spectrum.
Since the main spectrum is nearly 100 kHz you may find that the
sync tone is not where you want it in the baseband, but by clicking
on the sync tone in the high resolution graph you will immediately
bring it to the center of the passband within the accuracy of
the mouse click (a few pixels * 6Hz/pixel) The high resolution graph
will immediately show the new situation. With an appropriate average
that shows the average spectrum over the latest 15 seconds or so
you will see the green and the magenta peak of the sync tone
corresponding to the selected polarisation and the orthogonal one.
If the green is much stronger than the magenta all is fine.
If the magenta is much stronger, click for a 90 degree pol
shift and you will see the hires graph immediately showing
the new situation while the audio output to the next PC will
continue to present an uninterrupted signal exactly as if you
had rotated your X-yagi array. If the computer is not fast
enough there will be a small glitch in the signal.
In case green and magenta are of similar level, click for 45 degrees
either way and the spectrum will immediately show one of them much
stronger than the other.
With WSE units and 80 Hz and sine to power 3 for fft1, the
fft1 size becomes 4096 for a bandwidth of 56 Hz.
With the bandwidth factor parameter 3 and sin to power 1
the fft2 size becomes 16384 for a bandwidth of 8.8Hz
Each transform spans 0.17 seconds but the transforms overlap
so one has to collect 100 transforms to span 10 seconds.
With these parameters Linrad is very sensitive in the main waterfall
provided that the waterfall is made very slow. No more than
10 lines per second. The reason is that S/N may be lost in the process
of converting the 16384 spectrum to a waterfall of about 100 pixels.
For each waterfall line Linrad computes an average spectrum for
the prescribed number of transforms (100 or so) It then locates the
biggest number for each region of 16 (or whatever) fft bins that
are represented by one pixel. When there is no averaging the noise
floor will cause the different bins to take very different numbers
and the largest of the noise floor bins is likely to be about 7 dB
above the average noise floor (worse if the compression is larger than 16
times) S/N has to be a bit above 7 dB (in 8Hz) for a signal to
be visible in the un-averaged waterfall (10lines per second)
This corresponds to -22dB in the JT65 scale. (Not 25 because we have
noise from both polarisations)
When averaging by 100 times we find that the strongest bin
will be likely to be 0.5 dB above the average noise floor.
It is enough for (S+N)/N to be 1 dB for the signal to be well
visible. This is equivalent to S/N=-6dB, an improvement of
13 dB giving a limit at -35dB on the JT65 scale. This number might
be slightly incorrect because I did not do the math, I just
looked an noise on the Linrad screen. The math is reasonably
straightforward though so maybe someone would be interested in
computing the correct limits for different bandwidths and
X-axis compression.
In real life it is not quite as easy. With full sensitivity
on a 100 kHz waterfall one would see thousands of spurs and
they may mask the JT signal. With a 10 second averaging
on a 8 Hz bandwidth the signal must not drift by more than
about 0.5Hz/second to not loose S/N.
2) Unstable signals that can currently not be decoded with wsjt.
---------------------------------------------------------------
The AFC can be set to compress the bandwidth of wsjt signals down
to very low S/N. The price to pay is a very long time delay.
Something like 10 to 20 seconds. One would have to respond to
the previous transmission, not the most recent one. I do not
want to go into details since I have not verified this on signals
that actually fail with the wsjt program.
-----------------------------------------------------------------
73
Leif / SM5BSZ
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