On 09/16/2015 05:25 AM, Joe Subich, W4TV wrote:
...
      N      "S"
----------------------------------
      1     S9 +63 dB (-10 dBm)
      3     S9 +53 dB
     10     S9 +43 dB
     32     S9 +33 dB
    100     S9 +23 dB
    316     S9 +13 dB
   ~450     S9 +10 dB
   1000     S9  +3 dB
  ~1400     S9  +0 dB (-73 dBm)

Since it is the instantaneous peaks that cause problems, increasing the
number of signals decreases the frequency of the ADC overflows.  There
is certainly analysis that can be done to compute the probability of a
peak given a specific number of signals and frequency distribution but
my best guess is that the number of signals involved will be somewhere
between 10 and 100.

The amplitude distribution of a large number of signals of different frequencies and amplitudes closely approximates Gaussian noise (see note 1 below). As a rule of thumb the peak to RMS voltage ratio of Gaussian noise is about 5 or 6. Of course, theoretically it is infinity, but peaks over about 5-6 are statistically rare (note 2), so that's a good practical rule of thumb.

A voltage ratio of 5-6 is 14-15.6 dB. ADC overload should be rare as long as you keep the RMS power 15 dB or so below the ADC full-scale.

A lot depends on what happens when the ADC overloads. If it simply clips the signal, an occasional brief overload would not be objectionable. That's how the ADC in the P3 works. On the other hand, if it "wraps around" (for example, 0x7FFF = 32767 wraps to 0x8000 = minus 32768) then it's obviously much more of a problem. In that case, you would need an external analog clipper to keep the ADC from overloading.

Alan N1AL


Note 1:

The Central Limit Theorem:
https://en.wikipedia.org/wiki/Central_limit_theorem

Note 2:

The cumulative Gaussian probability density function is given in the Wikipedia article:

https://en.wikipedia.org/wiki/Normal_distribution#Numerical_approximations_for_the_normal_CDF

If I calculated right then for x=5, CDF = (1 - 3*10^-8). If the ADC sample rate is 100 Msps, then on average there would be 3 overloaded samples per second. For x=6, CDF = (1 - 2.8*10^-11) so there would be one overload about every 360 seconds.
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