[time-nuts] Thunderbolt AMU to dBc conversion
Arnold Tibus
Arnold.Tibus at gmx.de
Sat Feb 13 23:35:48 UTC 2010
Magnus & Mark,
for my Thunderbolt system I found a very good correlation
between AMU and dBc/ Hz ( observed on my system using
the ThunderboltMonitor watching over long time stable satellite signals )
when I use the Formula: C/N0(dBHz) = 20 log(AMU) + 23.5.
73,
Arnold
On Sat, 13 Feb 2010 18:12:52 +0100, Magnus Danielson wrote:
>Mark Sims wrote:
>> The data was collected in 0.1 dB / AMU steps. There was a lot of noise in the raw data (maybe +/- 1 dB or AMU), but the general trend of the curve was quite distinct. I smoothed it out by hand and filled in a few (maybe 15) of the missing steps that had no signal.
>>
>> The 0.1 resolution of the raw data causes the piecewise linear steps in the data. I was going to run a smoothing filter over it, but the curve is pretty good as-is (especially considering how it was generated).
>>
>> Heather now uses a table lookup from 0.0 to 25.5 AMU with a linear approximation over 25.5 AMU to convert AMU values to dBc values when drawing the signal level map if the receiver is in AMU mode. The plots collected in AMU mode and dBc mode are virtually indistinguishable (within the normal variation seen between two different runs in dBc mode).
>My current speculation is this:
>It looks like the AMU unit is normalized and biased such that AMU value
>0.0 is a reasonable border limit of 20 dBc.
>Thus, from the correlated amplitude A I think the AMU is calculated like
>this:
>AMU = A/A_norm - A_bias
>A_norm scales linearly with sample-rate of digital sampling.
>A_bias adjusts such that A relating to 20 dBc gives the AMU of 0.0.
>The dBc calculation is more straightforward:
>dBc = 20*log(A/A_norm) + dB_offset
>A_bias of 0.6 to 0.7 and dB_offset of 22.6 isn't too far off.
>Cheers,
>Magnus
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