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Author Topic: Bizarre low upstream SNRM - line #3 again  (Read 861 times)

re0

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Re: Bizarre low upstream SNRM - line #3 again
« Reply #30 on: November 28, 2018, 09:59:14 PM »

So where am I going wrong?

I thought you could work out the required SNR for the current bits allocated with bits*3+6 where bits are the bits allocated, 3 is 3 dB of SNR (equating to a single bit) and 6 is the target SNRM in dB (any target SNRM figure can be used).

I somehow think this is wrong since the first set of values (when the SNR was ~1.7 dB) seem to show it being sufficient by that calculation (possibly up to around 7 dB margin), but the later stats appear to show the SNR being 6 dB higher (thus almost a 12 dB margin by the calculation).

Can someone correct me? I know the very simple calculation is flawed.
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Weaver

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Re: Bizarre low upstream SNRM - line #3 again
« Reply #31 on: November 29, 2018, 04:19:50 AM »

I would assume that we need to convert log2 bits into log10 for dB and double it because of the squaring operation required to get power from amplitude. Constellations are two-dimensional, so we need to also have a factor of 1/2 in there.

So bits * 2 / 2 * 10 * log10( 2 ) + some constant which I don't know the value of. That would be about bits * 0.3 * 10 + constant, ie bits * 3 + C. But I don't know what the assumed zero dB power reference is. Since SNR is a ratio, the constants will cancel out, so in that case isn't my constant simply zero?
« Last Edit: November 29, 2018, 05:12:22 AM by Weaver »
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