Re: [time-nuts] Calculating frequency differences using Lissajou figures

2009-11-09 Thread paul swed
With Loran C shutting down in Jan, curious if anyone still uses it. I have a z3801 but always liked the Austrons 2100s as another way to check everything. I pulled the old Gertch RLF-1 out of the basement and it still works. (retired it 9 years ago) But a far cry from the Loran chains accuracy and

[time-nuts] Calculating frequency differences using Lissajou figures

2009-11-07 Thread Mark Amos
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

Re: [time-nuts] Calculating frequency differences using Lissajou figures

2009-11-07 Thread Jeffrey Pawlan
or stated in another way: your two sources differ by 5.5 parts in 10^-10 Well within the specification for the Rb. But to be precise, but if you wanted to know the absolute accuracy of the Rb then you would need to know the accuracy of the GPSDO at that moment in time. Jeffrey Pawlan

Re: [time-nuts] Calculating frequency differences using Lissajou figures

2009-11-07 Thread Tom Van Baak
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

Re: [time-nuts] Calculating frequency differences using Lissajou figures

2009-11-07 Thread 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

Re: [time-nuts] Calculating frequency differences using Lissajou figures

2009-11-07 Thread 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.a...@toast.net wrote: Time-Nuts, I recently fired up an Efratum LPRO and have been watching the

Re: [time-nuts] Calculating frequency differences using Lissajou figures

2009-11-07 Thread Lux, Jim (337C)
On 11/7/09 12:29 PM, Bruce Griffiths bruce.griffi...@xtra.co.nz wrote: Tom Va 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

Re: [time-nuts] Calculating frequency differences using Lissajou figures

2009-11-07 Thread J. L. Trantham
-nuts-boun...@febo.com [mailto:time-nuts-boun...@febo.com] On Behalf Of Hal Murray Sent: Saturday, November 07, 2009 3:50 PM To: Discussion of precise time and frequency measurement Subject: Re: [time-nuts] Calculating frequency differences using Lissajou figures So your frequency drift

Re: [time-nuts] Calculating frequency differences using Lissajou figures

2009-11-07 Thread WB6BNQ
Mark Amos wrote: Time-Nuts, snip 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

Re: [time-nuts] Calculating frequency differences using Lissajou figures

2009-11-07 Thread Mark Amos
Bill, I did a two sets of measurements using different methods. In the first, I watched the two waveforms in dual trace mode. The Trimble was on channel 1 (stationary) and the LPRO was on channel 2 (slowly drifting to the left). I started timing when they were phase (Channel 1 and Channel 2