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From: "Talbot Andrew" <ACTALBOT@dstl.gov.uk>
To: rsgb_lf_group@blacksheep.org
Cc: "Holtby Linda C" <LCHOLTBY@dstl.gov.uk>
Subject: LF: RE: Re: Jason Tests signalling compariso	n
Date: Thu, 24 Jan 2002 14:54:48 -0000
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>BTW, does anyone know how to compare the weak signal performance of Jason 
>and QRSS or DFCW? It seems clear that the increased bandwidth must result 
>in some loss of SNR, but there will also be gains because the data is being

>transmitted faster for a given dot length.
>73 de M0BMU

All three of these modes use the same basic technique, looking for the
presence or absence of a signal in a given bandwidth - in the case of JASON
doing this 17 times in parallel.  The bandwidth that matters is not the
total bandwidth of the whole signal, just that occupied by one of the tones,
and that is defined in the decoding software.   I think Alberto is using a
2^17 long FFT on an 11025 Hz sampling rate, so each FFT bin occupies
0.084Hz.  Windowing of the FFT data is usually done which widens the
effective bandwidth - a Hamming window gives a factor of 1.85 - so the Rx
bandwidth is now 0.16 Hz.  

What you are doing is trading off the ability to send 3 bits at a time
(actually 4 per symbol but one is used for framing) against total signal
bandwidth.  6 bits giving the full alphabet of 64 characters needs two
symbol lengths.  

QRSS sends characters using variable length coding but an average of
something like 10 dot intervals per character is probably not unreasonable
(E = 2, T = 4, J = 14, 0(zero) = 20 etc.)
DFCW shortens this by a factor of around 2, to 5 dot intervals per
character.

So, assuming equal symbol lengths :
QRSS 10 symbols / character
DFCW  5 symbols / charracter
JASON 2 symbols per character

So for a given signalling, or noise bandwidth, JASON is 5 times quicker than
QRSS, and DFCW twice as fast.  

All three assume a symbol can have the same arbitrary length, and can
therefore be received in the same signal (or noise) bandwidth.  The tone
spacing in Jason is set at three FFT bins to allow independdant (orthogonal)
detection, so for the 17 tone version total bandwidth is 16 units so
effective bandwidth of each of the modes is :

QRSS 1 unit of bandwidth * 10 symbols/character  = 10
DFCW 2 units of bandwidth * 5 symbols/character  = 10
JASON 16 units of bandwidth * 2 symbols/character = 32

So in terms of effective bandwidth CW and DFCW are roughly equal,  Jason is
worse by a factor of 3.  So there is the trade off.  Signalling speed vs.
effective bandwidth..........

But that is not even the whole story .....

A machine can decide on the presence or absence of a signal in a defined
bandwidth a lot better than a human eye can on a waterfall - it most
definitely can, believe it, its true, even if this fact makes you feel
uncomfortable :-((
How much better is debatable, but probably there is 3 - 6dB to be gained by
using a machine for decoding rather than the.   We are now into the realms
of the non linear relationship between error rate and signal to noise, so it
is more difficult to quantify this better decoding in terms of noise
bandwidth, but it is easy to believe the factor of 3 (= 5dB) could be
recovered, in the case of weak and marginal signalling even more so -
meaning JASON matches the performance of QRSS and DFCW for a given symbol
length.

Note, however, that as JASON fixes the symbol length at around 11 seconds,
so to compare like with like, we should only be comparing with 10s QRSS /
DFCW.

Hope that explains the process of how to compare these modes.  All rely on
the same fundamental decoding mechanism, incoherent detection of signal
power in a fixed bandwidth.  Now, if instead of power, we operate with known
phase and look for signal voltage we immediately gain another 6dB, but
that's another story.  Suffice to say,  G3PLX at 393km distance can get 100%
copy of my 30s BPSK transmissions when I am radiating less than 1W of RF,
equating to probably only 100-200 microwatts ERP

Andy  G4JNT


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