On 11/24/2016 5:16 AM, Bob Camp wrote:
The biggest challenge is to take out the “early stuff”. One approach is to fit
the same equation twice with the time constant restricted to a range on each.
For most OCXO’s (90%) the equation when fit early represents an upper limit
to the drift. You
Hi
> On Nov 23, 2016, at 11:21 PM, Scott Stobbe wrote:
>
> Hi Lars,
>
> There are a few other pieces I have yet to fully appreciate. One of which
> is that Aln(Bt+1) isn't a time-invariant model. In the most common case
> (for the mfg) the time scale aligns with
Hi Lars,
There are a few other pieces I have yet to fully appreciate. One of which
is that Aln(Bt+1) isn't a time-invariant model. In the most common case
(for the mfg) the time scale aligns with infancy of the OCXO, when it's hot
off the line. However after pre-aging, perhaps some service life,
It really means that B will be harder to get a qualitative value for.
Also, you need to have "clean" data or else you will be far off.
Plot the estimated variant and also plot the difference.
As Jim Barnes used to teach, always check the whiteness of matching
residues!
Cheers,
Magnus
On
Hi Scott.
Here is a textfile with data for the 10 years (As in the graph 2001-2011).
Also the ln(bt+1) fit, as Magnus said, has the derivate b/(b*t+1) that with b*t
>>1 is 1/t. But my data has the aging between 1 and 10 years more like
1/sqrt(t) If I just have a brief look on the aging graph.
Bob,
I have to ask about the B-term. In the paper that Scott started this with I see
that B was 4.45. But if I understand you correct Bt<1 even at 30days is normal?
That would mean a B of <0.033?
Lars
>Bob wrote:
>In a conventional fit situation, you have < 30 days worth of data and the
Hi
In a conventional fit situation, you have < 30 days worth of data and the “time
constant”
is > 30 days. Put another way bt <= 1 in the normal case. It is only when you
go out to years
that bt gets large.
Bob
> On Nov 18, 2016, at 9:58 PM, Scott Stobbe wrote:
>
>
Hi Lars,
I agree with you, that if there is data out there, it isn't easy to find,
many thanks for sharing!
Fitting to the full model had limited improvements, the b coefficient was
quite large making it essentially equal to the ln(x) function you fitted in
excel. It is attached as
Hi
> On Nov 18, 2016, at 4:36 PM, Magnus Danielson
> wrote:
>
> Hi Lars,
>
> Now, consider f(t) = a*log(b*t+1), then the derivate is a*b/(b*t+1) and
> second derivate - a * b^2 / (b*t + 1)^2.
>
> Forming first f'(t) and second f"(t) derivate estimates from data
Hi Lars,
Now, consider f(t) = a*log(b*t+1), then the derivate is a*b/(b*t+1) and
second derivate - a * b^2 / (b*t + 1)^2.
Forming first f'(t) and second f"(t) derivate estimates from data is
trivial. Given that we can estimate a and b using
a = - f('t)^2 / f"(t)
b = - f'(t) / (f'(t) * t -
Bob wrote:
>As mentioned earlier in this thread. The function that has been used in
>several posts
>isn’t the right log function. The proper fit is to ln(bt+1)
You are absolutely right. It was my mistake to use the ln(t) in the graph. As
that was what I know in Excel and I don´t have Stable32
Hi
As mentioned earlier in this thread. The function that has been used in several
posts
isn’t the right log function. The proper fit is to ln(bt+1)
Bob
> On Nov 16, 2016, at 4:36 PM, Peter Vince wrote:
>
> Hello Lars,
>
> Just out of curiosity I yesterday put just
Hello Lars,
Just out of curiosity I yesterday put just the first thirty days (like in
> the pdf mentioned below) and let Excel calculate the logarithmic function.
> If I extrapolate that to 10 years it seems that the drift would be
> 6E-13/day but as can be seen in the aging graph it was more
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