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.
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.
