----- Original Message -----
Sent: Wednesday, August 24, 2011 5:12
PM
Subject: Re: LF: WSPR or QRSS: which is
better?
WSPR works in a 1.46Hz signal bandwidth and because of its very high
level of error correction and soft-decision decoding, means that it will
work at a S/N of about 3dB in this bandwidth, and sometimes a bit lower
still (Normally, FSK with no correction at all needs about 10 -
12dB S/N for near error-free performance)
QRSS has to show about 6 - 10dB in its signal bandwidth to be able
to discern fully what is sent, although a slightly lower S/N may be useable
when you 'know' what you should be receiving. (A form of forward error
correction is now in use here as well perhaps :-) So lets say 5dB
S/N is a working value..
So take 3dB in 1.46Hz as a starting point and derive the
bandwidth for QRSS needed to get 5dB S/N with the same signal. This will
have to be narrower to get a 2dB higher S/N and works out as 1.46 /
10^(2/10) = 0.92Hz
So QRSS used with a 0.9Hz bandwidth - which I think means about a 2 - 3s
dot period ought to be decoded at the same S/N as a WSPR signal.
Which is probably the info you wanted.
But now compare source coding efficiencies. WSPR fits a
callsign, locator and power level into a 110 second transmission - and gives
absolutely guaranteed error free decoding, or nothing at all. About
12 characters in actuality, but that is being a bit unfair as the coding
forces certain callsign and locator formatting. So in all
probablility, more like 7 or 8 effective characters (I'm being a bit empirical
here)
Assuming standard QRSS - not DFCW - , which if like standard Morse, then
5 characters takes about 50 dot symbols to send (12WPM = 60 chars in 1 minute,
= 1 char / second, or about 10 dot periods / second. Dot
speed = WPM / 1.2) If we have 2s dots, that is 5 characters can be
sent in the time for a WSPR transmission.
So as a quick estimate, WSPR wins by roughly 2dB in S/N terms for a
given dot period / noise bandwidth. And at similar S/N values, WSPR is
about 1.5 times faster
Andy
