----- 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
