[time-nuts] GPS W/10KHz

[Bruce Griffiths]

Bruce Griffiths wrote:
> Dennis Ferguson wrote:
>> Bjorn,
>>
>>>> All I was pointing out is that at a higher output frequency, like
>>>> 10 kpps, the frequency of the quantization saw tooth error will
>>>> almost always be much higher as well.  There's no need for the digital
>>>> correction since averaging over a relatively short period, like in
>>>> the loop filter of an appropriate analog PLL, will almost always be
>>>> sufficient to smooth the sawtooth.
>>> The sawtooth correction is the difference between where the receiver 
>>> would
>>> wish to place the edge and where its known limited resolution 
>>> electronics
>>> lets it put the edge.
>> Yes, exactly.  The range of the error in edge placement will be a 
>> constant
>> related to (probably equal to) the period of the internal clock it is 
>> using
>> to generate the edges.  For an LEA-6T this is 21 ns so, assuming it 
>> rounds
>> off, any edge it places will be in error by +/- 10.5 ns.
>>
>>> The receiver wish is based on the timesolution from the last 
>>> measurement.
>>> In the Jupiter this is done at 1Hz maximum. The sawtooth correction 
>>> will
>>> apply the same for all 10k (pos or neg) edges in the 10kHz signal 
>>> during
>>> that one second.
>> Yes, this is true.  Note, however, that the time solution produces not
>> only a phase error of the receiver's internal clock with respect to GPS
>> but also the frequency error of the receiver's clock with respect to 
>> GPS.
>> More than this, since the time solution takes time to compute it will be
>> telling the receiver what the phase error was at some point in the past
>> rather than what it is now, let alone what it will at the point in the
>> future when you want to assert a 1 pps signal.  It can place that future
>> edge because it knows the actual frequency of its clock with some 
>> precision
>> from the time solution, and that plus knowing a phase offset at some 
>> past
>> time is sufficient to allow it to extrapolate to a future edge 
>> placement.
>> With an LEA-6T the precision of the edge placement will be +/- (10.5 +
>> epsilon) ns, with the "epsilon" occurring because it is extrapolating a
>> past measurement to place a future edge.
>>
>> Note, however, that the rate at which it can compute time solutions 
>> doesn't
>> change any of this very much.  The fact that an LEA-6 can compute 5 
>> solutions
>> per second rather than just one will at best just make "epsilon" a bit
>> smaller, and this matters not at all since "epsilon" should be pretty 
>> small
>> already.  If the receiver instead only computed a time solution once 
>> every
>> 3 seconds it also wouldn't make a difference, it could still place a 
>> 1 pps
>> edge every second by extrapolating from whatever the last solution was
>> that it managed to complete.  More than this, if you told the 
>> receiver to
>> place 10,000 edges per second instead of just 1, the placement error of
>> each one of those edges, individually, would still be +/- (10.5 + 
>> epsilon) ns.
>>
>> The rate at which the receiver computes new solutions has about 
>> "epsilon"
>> to do with the precision of edge placement.  The sawtooth doesn't 
>> come from
>> the epsilon, it comes from the +/- 10.5 ns.
>>
>>> There are effects that are not easily filtered away in the analog 
>>> domain.
>>> See the archives and
>>>
>>>     http://www.leapsecond.com/pages/m12/sawtooth.htm
>> This is good.  Notice the amplitude and the frequency of those 
>> sawtooths.
>> The amplitude is the period of the internal clock placing the edges, 
>> i.e.
>> 21 ns for an LEA-6T and what looks like>30 ns for the receiver above.
>>
>> Then there's the frequency.  It varies widely but the highest frequency
>> seen, in the 4th graph down, looks to be about 0.5 Hz.  It isn't an
>> accident that there is no higher frequency, and  I'll just assert
>> that this maximum frequency has nothing to do with the time solution
>> update rate of the receiver.
> Nonsense, it has everything to do with the time solution update rate 
> as long as the output signal frequency is greater than or equal to the 
> update rate.
> Such folding only occurs when the output signal frequency is less than 
> the update rate.
>>    It would be not change if you looked at
>> the 1 pps output of a 5-update-per-second LEA-6.  Instead the highest
>> sawtooth frequency is 0.5 Hz because he's looking at a 1 pps output,
>> getting one sample per second, and if you sample a signal at one sample
>> per second then the frequencies you see in the samples are always going
>> to be in the range 0-0.5 Hz.  Essentially this is integrating a "beat"
>> frequency between the receiver's oscillator and GPS time, which could be
>> very high in frequency, but by sampling at 1 pps the difference, 
>> whatever
>> it is, gets folded into the 0-0.5 Hz Nyquist bandwidth.  The low 
>> frequency
>> of the sawtooth observed at 1 pps makes it a problem for analog filters.
>>
>> All I'm pointing out, then, is that if you increase the pulse rate 
>> output
>> by the receiver from 1 pps to 10 kpps you will still get a sawtooth, 
>> like
>> 1 pps, and the amplitude of the sawtooth will be unchanged from 1 
>> pps, but
>> the frequency of the sawtooth won't be limited to 0-0.5 Hz and will 
>> instead be
>> folded into the 10 kpps Nyquist bandwidth of 0 - 5 kHz.  Unless you are
>> very unlucky this will give you the same sawtooth error at a much higher
>> frequency, making it much more amenable to analog domain filtering.
>>
> Thats only true if the GPS timing error is recalculated for each edge 
> of the 10kpps signal.
> This certainly doesn't happen in the Jupiter GPS receivers.
>> Dennis Ferguson
>>
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> Bruce
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>
Another way of viewing the output is that it consists of 2 cascaded 
samplers.
The first sampler samples the GPS error at the update rate (1Hz , 5Hz, 
10Hz depending on the receiver).
The second sampler samples the output of the first sampler at a rate 
that is either a harmonic or subharmonic of the first sample rate.
When the second sample rate is a subharmonic of the update rate of the 
first sampler, the effective sample rate is that of the second saampler.
When the second sampler samples at a harmonic of the update rate the 
effective sample rate is that of the first sampler.

Brucee