[time-nuts] Calculating frequency differences using Lissajou figures

[Lux, Jim (337C)]

You got it exactly.  The relative phase between the two signals (as
displayed by the Lissajous) rotates one cycle in 182 seconds.. E.g. 1/182 Hz
difference.


On 11/7/09 11:34 AM, "Mark Amos" <mark.amos at toast.net> wrote:

> Time-Nuts,
> 
> I recently fired up an Efratum LPRO and have been watching the slowly rotating
> Lissajou figure produced when comparing it's output with that of a GPSDO on a
> scope. It's a
> beautiful thing (the weather is too cool to watch paint dry...)
> 
> I thought the period of "rotation" of the Lissajou figure could be used to
> determine the frequency difference of the two oscillators, but the math
> escapes me.
> 
> Is it as simple as calculating the inverse of the period of the "rotation"
> through 360 degrees?
> 
> In this case the period between in-phase and in-phase is 182 seconds yielding
> a rotation frequency of 5.5 mHz.  So, is 5.5 mHz the frequency difference
> between the two 10 MHz
> oscillators? Or am I missing something obvious?
> 
> Thanks, in advance,
> 
> Mark
> 
> 
> 
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