[GPS_Standard] What is the effect on the accuracy and speed stabilization should use the 10th harmonic

[Dave Platt]

On 03/10/2013 09:32 AM, saper wrote:
> I have a 10MHz OCXO MV89. Getting from 10MHz 100MHz, using the 10th
> harmonic
> MV89 sinus -> PECL nomparator make square 45/65 -> high quality
> band-pass filter 100MHz -> ERA amplifer -> high quality band-pass filter
> 100MHz.
> My question is simple
> Somehow affects the accuracy and speed stabilization, when the feedback
> instead of the basics 10MHz will use 10tj harmonics 100MHz?
> I would counter implemented in FPGA.

Well, I'm going to go on intuition here - I may be right or wrong,
but in either case I don't have direct experience with this sort
of setup nor can I offer a mathematical proof of what I suspect.

My suspicion is that this may be a somewhat tricky thing to do,
with some odd quirks.  Getting a strong, clean 10th harmonic out
by this sort of arrangement might not be easy.  The Fourier
expansion of a square wave consists of the sum of *odd* harmonics,
not even.  You'll probably get *some* 10th harmonic content, but
it's not likely to be a strong one - you'll need a good deal of
amplification, and noise may be a significant concern.  Using
a comparator threshold to give you a nonsymmetrical result will
add some second-harmonic content but I'm not sure that you can
count on a useful increase in higher-even-order harmonics such
as 10x.

As far as timing stability goes... if you're planning on running
the FLL counter from the amplified 10x harmonic, I'm concerned
that you may have some problems with jitter.  In effect, what's
going to be happening (I think) is that you've got two high-Q
bandpass filters which will "ring" at 100 MHz, when excited by
an incoming burst of 100 MHz content (the 10th harmonic of the
OXCO) which is likely to occur mostly at the rising and falling
edges of the squared-up comparator output.  The tricky bit is
that the two bursts of 100 Mhz impulse are *not* guaranteed to
be in the same phase relationship with one another (I think).

That is:  on the rising edge of the comparator, the BPF
will be "kicked" into ringing by the 10th overtone of the
OXCO.  This ringing will decay somewhat during the time
when the comparator output remains high (the amplitude
of the 100 MHz signal will decrease).  Then, the comparator
output will go low, "kicking" the BPF again... but this
second "kick" is likely to be out of phase with the
remains of the original "ringing".  This will cause
some amount of jitter (in effect a modulation) of the
100 MHz output from the second BPF.  When the comparator goes
high again, the same thing will happen... a "glitch" in the
amplitude and phase of the 100 MHz signal will occur.

So, if my hunch is correct, your 100 MHz signal is likely
to have a significant amount of phase noise on it... more than
was present in the original OXCO.  This might have some effect
(probably small?) on the FLL pulse-counting system... it'll
probably become negligible if your averaging time is long enough.

Another approach to getting a 100 MHz signal, would be for
you to use a PLL circuit with a 10x divider to generate the
100 MHz signal.  This would give a strong, clean signal - less
chance of low-level noise because you won't need a
harmonic multipler / BPF system.  PLLs add their own types
of jitter, though.

I guess I don't see why pulse-counting the 10th harmonic signal,
would give you any better stability or accuracy than just
pulse-counting the 10 MHz OCXO fundamental.  It might well be worse.
*Anything* you do to create the 100 MHz signal from the
10 MHz OXCO output is necessarily going to involve processing
and manipulation of the 10 MHz signal, adding noise of one
sort or another... and you might have to increase the averaging
period of the FLL to filter out the effect of this noise, thus
nullifying any benefit you might hope to achieve by pulse-counting
a faster-clocked signal.

That's what my intuition is telling me.  If anybody can prove me
wrong, or can confirm that I'm right... I'm very interested in either
case!