[time-nuts] Thermal impact on OCXO
scott.j.stobbe at gmail.com
Tue Nov 15 23:58:31 EST 2016
Do you recall if you fitted with true ordinary least squares, or fit with a
recursive/iterative approach in a least squares sense. If the aging curve
is linearizable, it isn't jumping out at me.
If the model was hypothetically:
F = A ln( B*t )
F = A ln(t) + Aln(B)
which could easily be fit as
F = A' X + B', where X = ln(t)
It would appear stable32 uses an iterative approach for the non-linear
"y(t) = a·ln(bt+1), where slope = y'(t) = ab/(bt+1) Determining the
nonlinear log fit coefficients requires an iterative procedure. This
involves setting b to an in initial value, linearizing the equation,
solving for the other coefficients and the sum of the squared error,
comparing that with an error criterion, and iterating until a satisfactory
result is found. The key aspects to this numerical analysis process are
establishing a satisfactory iteration factor and error criterion to assure
both convergence and small residuals."
Not sure what others do.
On Mon, Nov 14, 2016 at 7:15 AM, Bob Camp <kb8tq at n1k.org> wrote:
> If you already *have* data over a year (or multiple years) the fit is
> fairly easy.
> If you try to do this with data from a few days or less, the whole fit
> process is
> a bit crazy. You also have *multiple* time constants involved on most
> The result is that an earlier fit will have a shorter time constant (and
> will ultimately
> die out). You may not be able to separate the 25 year curve from the 3
> curve with only 3 months of data.
> > On Nov 13, 2016, at 10:59 PM, Scott Stobbe <scott.j.stobbe at gmail.com>
> > On Mon, Nov 7, 2016 at 10:34 AM, Scott Stobbe <scott.j.stobbe at gmail.com>
> > wrote:
> >> Here is a sample data point taken from http://tycho.usno.navy.mil/ptt
> >> i/1987papers/Vol%2019_16.pdf; the first that showed up on a google
> >> Year Aging [PPB] dF/dt [PPT/Day]
> >> 1 180.51 63.884
> >> 2 196.65 31.93
> >> 5 218 12.769
> >> 9 231.69 7.0934
> >> 10 234.15 6.384
> >> 25 255.5 2.5535
> >> If you have a set of coefficients you believe to be representative of
> >> OCXO, we can give those a go.
> > I thought I would come back to this sample data point and see what the
> > impact of using a 1st order estimate for the log function would entail.
> > The coefficients supplied in the paper are the following:
> > A1 = 0.0233;
> > A2 = 4.4583;
> > A3 = 0.0082;
> > F = A1*ln( A2*x +1 ) + A3; where x is time in days
> > Fdot = (A1*A2)/(A2*x +1)
> > Fdotdot = -(A1*A2^2)/(A2*x +1)^2
> > When x is large, the derivatives are approximately:
> > Fdot ~= A1/x
> > Fdotdot ~= -A1/x^2
> > It's worth noting that, just as it is visually apparent from the graph,
> > aging becomes more linear as time progresses, the second, third, ...,
> > derivatives drop off faster than the first.
> > A first order taylor series of the aging would be,
> > T1(x, xo) = A3 + A1*ln(A2*xo + 1) + (A1*A2)(x - xo)/(A2*xo +1) + O(
> > (x-xo)^2 )
> > The remainder (error) term for a 1st order taylor series of F would be:
> > R(x) = Fdotdot(c) * ((x-xo)^2)/(2!); where c is some value between
> > and xo.
> > So, take for example, forward projecting the drift one day after the
> > day using a first order model,
> > xo = 365
> > Fdot(365) = 63.796 PPT/day, alternatively the approximate derivative
> > is: 63.836 PPT/day
> > |R(366)| = 0.087339 PPT (more than likely, this is no where near 1
> > DAC LSB on the EFC line)
> > More than likely you wouldn't try to project 7 days out, but considering
> > only the generalized effects of aging, the error would be:
> > |R(372)| = 4.282 PPT (So on the 7th day, a 1st order model starts to
> > degrade into a few DAC LSB)
> > In the case of forward projecting aging for one day, using a 1st order
> > model versus the full logarithmic model, would likely be a discrepancy of
> > less than one dac LSB.
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