[time-nuts] Calculating frequency differences using Lissajou figures

[Bruce Griffiths]

Tom Van Baak wrote:
> Mark,
>
> Your 5.5 mHz is correct for the frequency difference But
> note that's out of 10 MHz so the *relative* frequency error
> is 5.5e-3 Hz / 1e7 Hz, or  5.5e-10 (unit-less).
>
> The other way to look at it is this:
>
> The nominal frequency is 10 MHz, so one period is 100 ns.
> Your Lissajous pattern is seen to repeat every 182 seconds.
> Frequency difference is time drift over elapsed time, so the
> relative frequency difference is 100 ns / 182 s = 5.5e-10.
>
> ---
>
> Also, see if you can repeat this a month from now. If instead
> of 182 s you get, say, 188 s then you have a way to compute
> the frequency drift rate.
>
> 100 ns / 182 s = 5.495e-10 on November 7
> 100 ns / 188 s = 5.319e-10 on December 7
>
> So your frequency drift in this example is 1.7e-11 / month.
>
> /tvb
>
Not quite, you need to take the sign of the frequency difference into 
account.
ie in your example above the sign of the error measured on November 7 
has to be identical to the sign of the error measured on December 7 for 
your calculation to be correct.

Bruce