Stefan wrote:
> f = 8270.005000 Hz
> Start time: 27.September.2017 06:00:00 UTC (daily)
> Symbol period: 8 s
> Characters: 100
> CRC bits: 16
> Coding 8K19A
Each day:
2017-09-27 Eb/N0: -7.5 dB phase 136.9 deg
2017-09-28 Eb/N0: -6.3 dB phase 138.4 deg
2017-09-29 Eb/N0: -4.7 dB phase 135.9 deg
2017-09-30 Eb/N0: -3.1 dB phase 134.9 deg
Four days stacked gives Eb/N0: +0.7 dB phase 136.3 deg;
First 3 days stacked gives -1.3 dB and does not decode with
list size 23e6 using the known reference phase.
29th and 30th stacked decode at rank 6 Eb/N0: -0.9 dB.
That's 25.3 info bits per hour of transmission and the channel
capacity is 28.3 bits per hour. Remarkable what a weak signal
at audio frequency can do.
So rank 6 is about -0.9 dB and -1.3 dB lies somewhere below
rank 23e6. The 0.4 dB difference corresponds to a lot of list
length! This is typical of long messages, 5072 symbols in this
case, the decoding takes place in a space of 5072 dimensions
and the received signal is a point in that space (a vector of
5072 symbol amplitudes). The decoder turns out a list of the
closest convolutional codewords to that received point.
We need the transmitted point to be in that list so that it
can be found and identified by the CRC check of the outer code.
Euclidean distance between the transmitted and received points
is determined by the noise amplitude. A small increase in the
noise (proportional increase of distance in the code space)
produces a huge increase in the volume that has to be examined
thanks to having 5072 dimensions, and the list has to be long
enough to contain all the codewords in the volume 'generated'
by the noise.
--
Paul Nicholson
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