[time-nuts] GPS W/10KHz

Dennis Ferguson dennis.c.ferguson at gmail.com
Mon Feb 10 16:03:36 EST 2014

On 10 Feb, 2014, at 00:48 , Bruce Griffiths <bruce.griffiths at xtra.co.nz> wrote:
> Dennis Ferguson wrote:
>> On 8 Feb, 2014, at 14:50 , EWKehren at aol.com wrote:
>>> The problem with the PLL analog version is the same as with any digital
>>> GPSDO. The saw tooth is present at 10 KHz just like 1 Hz. To the best of my
>>> knowledge there is no GPS receivers out there for less than $ 1000 with out
>>> saw  tooth. Timing receivers output the correction value and you can either
>>> with  software or a variable delay do correction.
>> This is very true, though the sawtooth at a 10 kpps sample rate is going
>> to a little different than the sawtooth at a 1 pps sample rate.  The frequency
>> of the sawtooth noise will lie somewhere in the Nyquist bandwidth.  At a 1 pps
>> sample rate the frequency of the sawtooth noise will hence be somewhere between
>> 0 Hz and 0.5 Hz, while at 10 kpps the sawtooth frequency will range from 0 Hz
>> to 5 kHz.
>> Noise at less than 0.5 Hz is not easy to filter, so you are going to require
>> the correction from the receiver and/or an integrator with a time constant
>> that can only be realized digitally.  Sawtooth noise over most of a 0 Hz to
>> 5 kHz range, on the other hand, should be eliminated by the analog low pass
>> filter after the phase detector in the PLL, giving you something nice and clean
>> coming out.  It is only if you get unlucky and the beat frequency between GPS
>> time and the receiver's oscillator ends up very close to an integer multiple of
>> 10 kHz that you'll see noise at a low enough frequency to leak through into the
>> control response.
>> This is interesting because it suggests that very simple GPSDOs using 10 kHz
>> from the receiver might at times work worse than you are likely to observe in
>> a single bench measurement as aging (or something) moves the receiver's oscillator
>> frequency through one of the "bad" frequency errors.  Or is there a way to avoid
>> that altogether (maybe if the receiver does dithering)?
>> Dennis Ferguson
> Instead of speculating try reading the specifications.
> 1Hz phase modulation of the 10kHz output is present.
> The receiver sawtooth error sample rate is 1Hz not 10kHz.
> The 10kHz output signal phase is adjusted at a 1Hz rate by the receiver.


I'm not sure which equipment you want me to read the specifications for, though
I'd be very interested in knowing.  What I'm describing is the behaviour of the
timepulse output of the LEA-6T, which can be configured to output edges at any
rate from 1 Hz to 10 MHz.  There the only relevant specification I see is the
timepulse output quantization error, which is a constant 21 ns on every output edge
independent of the rate at which the receiver is configured to generate edges.  This
should cause exactly the behaviour described above, and as best I can measure by
comparing 1 pps and 50 pps outputs to the divided-down 10 MHz output of a GPSDO is
consistent with how the receiver actually behaves (though my best measurement is none
too good; I need to get a TIC with a resolution better than the 10 ns my Beaglebone
has).  If you run the output at 10 kpps you get 10,000 samples of the quantization
error every second and can average it out a lot faster than if you only get one sample
of the quantization error every second.  I don't know what a "sawtooth error sample
rate" is if not this.

You seem to be describing a piece of equipment where the sawtooth error is not a
simple consequence of pulse output quantization caused by generating edges with
the receiver's internal free-running clock.  I'd be really curious to know what
equipment this is.  This page


says he looked for but failed to find any sub-Hz sidebands in the Navman Jupiter
10 kHz output so it doesn't seem like that receiver is the one you are thinking of

Dennis Ferguson

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