[time-nuts] theoretical Allan Variance question

[Anders Wallin]

FWIW this example in AllanTools generates white pm and compares to the
IEEE1139 ADEV formula:
https://github.com/aewallin/allantools/blob/master/examples/ieee1139_white_pm.py

two comments:
1. Theoretical white noise has infinite power, so limiting the RMS to some
value already implies you have limited the measurement bandwidth (f_h in
the formulas)
2. Interestingly my understanding of the phase-meters (Timepod/3120A and/or
DIY USRP SDR) is that they don't measure raw phase of the signal much
better than a 20ps counter. The trick is to measure the phase between
REF/DUT at a high sample-rate, and then dramatically reduce the bandwidth
by low-pass filtering numerically. So an ADEV(1s)<1e-13 (100-fold better
than a typical counter) is not done via some "magic" 100fs
phase-measurement, but instead by doing plain old ~20 ps phase measurements
at a high rate and averaging lots of them together.
I've been playing with an USRP phase-meter and hope to release some
software/results soonish.

Anders
(ITSF2016/Prague next week)


Hi Stu,
> If you have white phase noise with standard deviation of 1 then the ADEV
> will be sqrt(3). This is because each term in the ADEV formula is based on
> the addition/subtraction of 3 phase samples. And the variance of normally
> distributed random variables is the sum of the variances. So if your
> standard deviation is 0.5 ns, then the AVAR should be 1.5 ns and the ADEV
> should be 0.87 ns, which is sqrt(3)/2 ns. You can check this with a quick
> simulation [1].
>
> Note this assumes that 1 ns quantization error has a normal distribution
> with standard deviation of +/- 0.5 ns. Someone who's actually measured the
> hp 5334B quantization noise can correct this assumption.
>
> /tvb
>
> [1] Simulation:
>
> C:\tvb> rand 100000 0.5e-9 0 | adev4 /at 1
> rand 100000(count) 5e-010(sdev) 0(mean)
> ** tau from 1 to 1 step 1
>        1 a 8.676237e-010 99998 t 5.009227e-010 99998
>
> In this 100k sample simulation we see ADEV is close to sqrt(3)/2 ns. The
> TDEV is 0.5 ns. This is because TDEV is based on tau * MDEV / sqrt(3). In
> other words, the sqrt(3) is eliminated in definition of TDEV.
>