[time-nuts] ADEV vs. OADEV
Magnus Danielson
magnus at rubidium.dyndns.org
Thu Jan 22 23:29:21 UTC 2009
Ulrich,
Ulrich Bangert skrev:
> Magnus,
>
> the paper http://tf.nist.gov/timefreq/general/tn1337/Tn121.pdf is
> thought-provoking. Not that I would simply say that you are right, but
> because I dont't understand some things.
Well, I was taking on this thread in hope to spread some light and I was
not sure who would learn what. I think I have learned a few things, but
so far it seems my view on things have not changed significantly.
I think I can explain some things.
>> N-2m
>> ___
>> 2 1 \ 2
>> sigma (tau) = ----------- > (x - 2x + x )
>> y 2 /___ i+2m i+m i
>> 2(N-2m)tau i = 1
>>
>> a non-interleaved variant would have to be written as
>> (assuming that m
>> divides N):
>>
>> N
>> - - 2
>> m
>> ___
>> m \ 2
>> ------------ > (x - 2x + x )
>> 2 /___ (i+2)m (i+1)m im
>> 2(N-2m)tau i = 1
>
> No discussion about that, simply correct.
I thought we would agree on those things.
> However the note to figure 8 as well as the note to figure 9 cover the
> non-overlapping case. Indeed formulas (8) and (10) are overlapping and
> to me it is a bit kind of magic where they come from in regard to thise
> two notes.
>
> Do you agree to the fact that the ADEV for Tau = 2 s should be the same,
> regardless if computed from 1 s spaced phase data or from 2 s spaced
> phase data?
In the ideal world... yes... but in the real world limits hits us
differently. The definition only says that it is an average, not how an
average should be formed over samples where tau != tau0. A
straightforward interpretation will skip samples, but require a much
longer sequence and we must recall it is just one of several estimators.
Let's make this clear, there is no way to get the real Allan variation,
we can only get estimates and choose among estimators. This becomes
clear as one reads the Bregni book or look at the ITU-T G.810 standard
(get it, it's available for free download as PDF from ITU).
Now, with this in mind, the "Overlapping Allan Variance" was chosen as
the estimator of choice for the AVAR(n*tau0), and for good reasons.
One point of confusion is that AVAR(tau) should not be directly
interpreted as Allan Variance in general, it is actually already defined
and reserved to mean a chosen Allan Variance estimator. This is an
important point when comparing oscillators and systems such as a
standard for measuring oscillators or telecom standards. For other uses
other estimators may be used, but for precision use the overlapped
estimator has been chosen as it better uses the measurement series.
So, to sum things up. You can't get the "propper" Allan variance. You
can only make estimates. The "Original Allan Variance" is a one of many
estimators but to attain good quality it needs a long time-series. The
"Overlapping Allan Variance" is another estimator which has improved
quality for a limitied time series as compared to the "Original Allan
Variance". The terms AVAR and ADEV as specified in several standards
denotes an estimator which is the overlapping estimator, not just any
Allan variance estimator. Notice also that telecom standards require
tau0 to be less than lowest tau, so tau/tau0 is always > 1 as this
ensures quality and allows for the specified highpass filter of 10 Hz,
which is also being debated as being meaningfull or harmfull (see Bregni).
There is still a few lessons to learn on this, but thats about it I think.
So no, 1 s data and 2 s data does not necessarily give the same result,
it depends on which estimator you are using, because you will be using
an estimator regardless of what you think is "right". Thus, if you feel
"Original Allan variance" is right for you, you are in your right to use
it, but it is not the "propper" one to use, and do not confuse it with
AVAR which is a dedicated name for the "Overlapping Allan Variance".
Cheers,
Magnus
More information about the Time-nuts_lists.febo.com
mailing list