[time-nuts] frequency comparator reading question

Bill Hawkins bill at iaxs.net
Thu Feb 27 12:55:19 EST 2014


The question can't be answered without knowing what the range switch
does.
In my experience, the cycle that is divided into 360 degrees is the
period
of the input signal regardless of the range switch.

You don't say, but the usual GPSDO produces a 10 MHz signal, unless it's
for a telco application. If the phase meter goes through one cycle in a
second
there is a one cps difference between the signals. One cycle at 10 MHz
is
one part in 10E7. One cycle in a kilohertz is 1 part in 1000. 1 cycle in

100 seconds for 10 MHz is one part in 10E9

Does the time for a phase rotation vary with the range switch?

Bill Hawkins

-----Original Message-----
From: Bob Camp
Sent: Wednesday, February 26, 2014 9:16 PM

Not having one here, about all I can guess is that there are 360 degrees
in a cycle. If it's going through 360 degrees in 10 seconds it's 0.1 Hz
off at what ever point it's comparing.  If it takes 100 seconds that's
0.01 Hz.

Yes I get this pesky decimal point stuff wrong from time to time ...

Bob

On Feb 26, 2014, at 9:19 PM, Paul A. Cianciolo <paulc at snet.net> wrote:

> I have a Fluke montronics frequency comparator.
> It has 2 inputs, one from my GPS and one from my DUT.
> After a given oscillator is warmed up, I can read the meter in parts 
> 10 -X
> There are 2 meters, One for phase 0 to 360 degrees, and one for part 
> to the -nth
> In the 10-9th position on the selector switch the a given oscillator 
> will show a 0 to 360 degrees travel lets say in 1 second.




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