[time-nuts] theoretical Allan Variance question
Bob Camp
kb8tq at n1k.org
Sun Oct 30 09:55:13 EDT 2016
Hi
There are entire (big and heavy) books written on quantization errors…..
In a counter, there are a number of different sub-systems contributing
to the error. Depending on the design, each may ( or may not) be a bit better than
absolutely needed. Toss in things like 10 MHz reference feedthrough
(which is decidedly weird statistically) and you have a real mess.
The simple answer is that there is no single answer for a full blown counter.
For an ADC sampling system, there may indeed be a somewhat more
tractable answer to the question (unless feed through is an issue with
your setup …).
Bob
> On Oct 30, 2016, at 8:32 AM, jimlux <jimlux at earthlink.net> wrote:
>
> On 10/29/16 10:14 PM, Tom Van Baak wrote:
>>> One might expect that the actual ADEV value in this situation would be
>>> exactly 1 ns at tau = 1 second. Values of 0.5 ns or sqrt(2)/2 ns might not
>>> be surprising. My actual measured value is about 0.65 ns, which does not
>>> seem to have an obvious explanation. This brings to mind various questions:
>>>
>>> What is the theoretical ADEV value of a perfect time-interval measurement
>>> quantized at 1 ns? What's the effect of an imperfect measurement
>>> (instrument errors)? Can one use this technique in reverse to sort
>>> instruments by their error contributions, or to tune up an instrument
>>> calibration?
>>
>> 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.
>>
>
> isn't the distribution of quantization more like a rectangular distribution (e.g. like an ADC). so variance of 1/12th?
>
>
>
>
>
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