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 -- time-nuts@febo.com To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts and follow the instructions there.
