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" <[email protected]> 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 > > > > _______________________________________________ > time-nuts mailing list -- [email protected] > To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts > and follow the instructions there. > _______________________________________________ time-nuts mailing list -- [email protected] To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts and follow the instructions there.
