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Message-ID: <B2BF8D87E3AB42ACA1D0FBEE3FDAB202@JimPC>
From: "James Moritz" <james.moritz@btopenworld.com>
To: <rsgb_lf_group@blacksheep.org>
References: <AA9545F1DC5A42BFA19CAA41B084C146@AGB> <fc7eccec0909231350y51645512ta48df1fd48f475f4@mail.gmail.com> <053D0B59431F4F798458F8429FDCBA9B@AGB>
In-Reply-To: <053D0B59431F4F798458F8429FDCBA9B@AGB>
Date: Wed, 23 Sep 2009 23:56:37 +0100
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Subject: Re: LF: s/n shown by wspr .... what is it actually showing ?
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Dear Graham, LF Group,

The SNR you see with Spectrum Lab will depend also on the relation between 
the spectrogram bandwidth (depending mostly on the resolution of the Fast 
Fourier Transform process) and the spectrum of the signal being monitored. 
Usually, the signal has a larger bandwidth than the FFT "bin", so varying 
amounts of signal appear in different bins. So the SNR measured from the 
spectrum graph is only useful as a relative measure between signals of the 
same type, without further analysis to total up the signal power in 
different bins.

To get the SNR for a particular mode from the WSPR SNR data for 2500Hz BW, 
you need to take the different bandwidths into account:

mode SNR(dB) = WSPR SNR - 10log(mode bandwidth / 2500Hz)

So if WSPR was giving SNR of -10dB, the SNR for PSK31 with 62Hz bandwidth 
would be +6.1dB.

What SNR is actually needed for successful copy depends on the type of 
modulation, the effectiveness of error correction, how good the demodulator 
is, what the nature of the noise is, etc. I suppose for most modes it is 
between about 0 and +several dB

Cheers, Jim moritz
73 de M0BMU