[time-nuts] Limitations of Allan Variance applied to frequency divided signal?
Tijd Dingen
tijddingen at yahoo.com
Fri May 13 15:28:03 UTC 2011
In trying to put together a way to calculate Allan variance based on a series of timestamps of every Nth cycle, I ran into the following...
Suppose you have an input signal, but it's a bit on the high side. So you use a prescaler to divide it down to a manageable frequency range. And now you want to use that signal to be able to say something useful about the original high frequency signal.
Now taking a look at the part about "Non-overlapped variable tau estimators" in the wikipedia article here:
http://en.wikipedia.org/wiki/Allan_variance#Non-overlapped_variable_.CF.84_estimators
It seems to me that "divide by 4 and then measure all cycles back-to-back" is essentially the same as "measure all cycles of the undivided high frequency signal back-to-back" and decimate. Or "skipping past n − 1 samples" as the wiki article puts it. And that is disregarding /extra/ jitter due to the divider, purely for the sake of simplicity.
Plus, I strongly suspect that all these commercial counters that can handle 6 Ghz and such are not timestamping every single cycle back-to-back either. Especially the models that have a few versions in the series. One cheaper one that can handle 300 MHz for example, and a more expensive one that can handle 6 GHz. That reads like: "All models share the same basic data processing core and the same time interpolators. For the more expensive model we just slapped on an high bandwidth input + a prescaler."
Anyways, any drawbacks to calculating Allan Variance of a divided signal that I am overlooking here?
regards,
Fred
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