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Fri Aug 25 09:04:52 EDT 2006


sinusoidal signal we can see that a certain amount of amplitude noise
directly translates into phase noise when it comes to measurements of
zero crossings and that the slope is the "translation factor". In the
first graphics I have made the slope to 1 giving a 1:1 translation of
amplitude noise into phase noise.

Now consider "Abbildung 7" on page 13. Here, almost everything is the
same. The signal has the same amplitude and the same signal to noise
ratio. However it has only half the frequency. Due to that it has only
half of the slope at the zero crossing than the first signal and you see
nicely how 1 part of amplitude noise now roughly translates into 2 parts
of phase noise. It becomes clear, that if everything else stays the same
the slope of the wave at the zero crossings and therefore nothing else
than the wave's frequency decides how precise we can measure the time of
its zero crossings.

Perhaps you do already see what that means for DMTD? Even if we consider
the mixing process as being noise free which is a VERY optimistic
assumption then every factor by which we down mix the signal and by
which we magnify the effect to be measured will be counteracted by a
decrease in slope to noise ratio in the same measure.

For those who feel that this objection is pure academic I suggest the
following experiment: If you own a counter that can do statistics, take
it and lock it to the best frequency reference that you have available.
Now take the best synthesizer generator that you have available, for
example a HP3325 and lock it to the same reference. Set the generator to
1 Hz sine and let the counter make say 100 measurements on period length
and then display the standard deviation of the measurements. I am sure
you will be surprised. 

For those of you who have not this equipment at hand: I just made the
experiment with my HP5370A and my HP3325A. The HP3325A was set to 1 Hz
sine and 1 Vpp output. The HP5370A was set to period measurement and
sample size = 100.  

The first 100 measurements gave a standard variation from sample to
sample of 434 microseconds! 434 MICROSECONDS? Where has the famous 20 ps
resolution of the 5370 gone? The second hundred samples deliver a
standard variation of 289 from sample to sample, the third 363
microseconds, so the first measurement can not have been this wrong. I
do not need to remember you that the standard deviation is kind of
typical error. Assumed a Gaussian distribution of errors the real error
of a single measurement may even be as big as +/- 3 times this value.
Just to check that nothing is defect now set the 3325 to 10 MHz and
watch the counter display a standard variation of 30-40 ps as we are
used from it. This all is taking place just by means of different slope
to noise ratio and nothing else. 

So, while down mixing the OUT signal from 10 MHz to 1 Hz may have
increased the effect to be measured to 10 microseconds which we had
expected to be VERY EASY to measure, the true problem now turns out that
we must measure a resolution of about 10 microseconds on a 1 Hz signal
with a very low slope to noise figure! 

I once computed that on 10 MHz a signal to noise ratio of 20 dB is
sufficient to measure the zero crossings of a sinusoidal signal with an
uncertainity of about 1 ns. With some simple considerations one can
easily compute that for a 1 Hz signal a stunning signal to noise ratio
of 160 dB is necessary for the same precision. With a frequency relation
of 7 orders of magnitude the slope of the 1 Hz signal is 7 orders of
magnitude smaller than that of the 10 MHz signal. In order to get the
same slope to noise ratio the root mean square level of the noise has to
decrease by 7 orders of magnitude which give rise to a necessary
decrease of 140 dB in noise power.

Have you been told that by one of the friendly authors who yarn about
DMTD? There aren't no such things in reality as nanosecond resolution
timing measurements on 1 Hz sinusoidal signals! And yes, the sinusoidal
form of the signal IS part of the problem! But in case you are going to
think about making a "digital" signal out of the sine by some clever
trick, you are on the wrong track because it takes the translation of
amplitude noise to phase noise just from one point to another point in
the apparatus but does not solve it. And before you give it a try on
yourself: The trigger circuits in modern counters are already pretty
tough!

As it turns out there IS a clever way to handle this situation at least
up to the principal limits. Again there ARE a very few specialists who
know about this problem very well but they are rare to find animals. If
you ever want to build a DMTD system by yourself then be sure that you
have read 

Dick / Kuhnle / Sydnor: "Zero-crossing detector with sub microsecond
jitter and crosstalk" 

before! Some people will tell you that you need "low noise zero crossing
detectors" but only these guys will explain you in full detail how to
build them! This text is a bit difficult to get from the net. If you do
not manage to download it please ask me for help. Basically the authors
use a cascaded chain of low pass filters and combinations of
non-limiting and limiting amplifiers to increase the slope of the signal
in several steps while at the same time they try to filter out as much
noise as possible. As you can see from the title anything better than 1
microsecond jitter (!) is considered state of the art.

Third Pitfall of DMTD: Phase corruption due to mutual crosstalk

Given 2 signals in the same circuit there will be a mutual crosstalk
between them. Crosstalk means that a damped version of signal 1 rides on
top of signal 2 and that a damped version of signal 2 rides on top of
signal 1. Similar to the case of noise there may be more or less
crosstalk which documents in the crosstalk damping figure.

Let us consider how crosstalk can influence oscillator stability
measurements. Let us first assume that we have 2 signals of the SAME
frequency and what happens if there is crosstalk between them. 

If both signals have the same phase then "signal 2 riding on the top of
signal 1" means that the amplitudes of both signals add up at every
point. Depending on the crosstalk damping figure the result will be a
wave which's amplitude at every point in time is a little bit greater
than that of signal 1 alone.  

If both signals are 180 degree out of phase then "signal 2 riding on the
top of signal 1" means that the amplitudes of both signals subtract at
every point. Depending on the crosstalk damping figure the result will
be a wave which's amplitude at every point in time is a little bit
smaller than that of signal 1 alone.  

The cases of 90 degree and 270 degree phase shift are a bit more
complex. Signal 2 now has its maximal and minimal amplitude at the
points where signal 1 has its zero crossings. "Riding on top" in this
case means that the zero crossings of the combined signal are shifted a
little bit to left or right just depending on the sign of signal 2. 

And of course the phase shift between signal 1 and signal 2 may be
anything in between creating a mixture of amplitude and phase
deformation effects. All of that deformation effects are not serious in
the sense that they are static in time. Every period of the combined
signal bears the same deformation.

But now let us consider the cases that the two signals are not of the
SAME frequency but have different frequencies which are very close to
each other. This is exactly the situation when we are going to compare
two very good oscillators! Now the situation is different in that signal
1 does not ride "static" on signal 2. Instead of having a constant phase
relation as with same frequency signals, the phase of signal 1 now moves
along the phase of signal 2 with an velocity that is given by the beat
frequency of the two signals and the peaks of signal 1 are sometimes at
the peaks of signal 2 but also sometimes at the zero crossings of signal
2. Due to this crosstalk the combined signals are both amplitude AND
phase modulated by their counterpart. That is what is meant by phase
corruption due to crosstalk. 

Now that we understand how there is phase corruption, let us compute how
big this phase corruption really is. In order to say how big the phase
corruption is, we need to say for which crosstalk damping figure we are
going to compute it. In situations like this I say: 100 dB. 100 dB is a
handy number in that we have real world examples available of what 100
db means: 

If you buy a good coaxial cable, this may have a shielding effectiveness
of 80 dB at radio frequencies. If you spend some bucks more you can get
a shielding effectiveness of 90 dB. 100 dB shielding is top and only
possible with double shielding and I do not remember to have seen a
shielding effectiveness been advertised better than 110 dB. So, 100 dB
are a VERY good isolation of two electrical circuits from each other and
are surely very hard to realize on the same printed circuit board or
within the same device. So, if 100 dB may be considered state of the art
in isolation let us see what -100 dB in mutual crosstalk means.

Assume a signal having the amplitude 1 V and the frequency 1 Hz. A
second signal that is damped 100 dB to this signal has an amplitude 1E-5
V. The 1 Hz signal has a slope of 2*Pi at the zero crossing, so the
damped signal riding on top will shift the zero crossing by
approximately +/-1E-5/(2*Pi) which is app. +/-0.6 milli degree and a +/-
1.6E-13 s effect expressed in absolute time on a 10 MHz carrier. This
should be very easy noticed on high resolution instruments as the TSC
5110!

AllanChart.Pdf shows a simulated oscillator comparison of two sources
that are absolutely stable but have a limited (sn+n)/n of about 80 dB
and a mutual crosstalk of -100 dB. Because the sources itself are stable
the only source of un-stability in the measurement is the jitter due to
amplitude to phase noise conversion and we would expect the tau-sigma
diagram to decline at a -1 slope from some starting point with white
amplitude noise. Instead we receive this! The two frequencies have been
chosen to be 1/16 Hz apart, therefore the big peak in Allan deviation at
approximately 8 s. You may be curious to ask how an effect that is 100
dB down the carrier can have THAT big influence with a (sn+n)/n of 80
dB. One would expect everything below -80 dBc to be "buried" in noise.
But here you must remember that for the (sn+n)/n figure the TOTAL noise
energy is used. If you look at the noise energy as a function of
frequency, which you do if you look at the signal with a spectrum
analyzer then the display for an 80 dB (sn+n)/n signal may look pretty
similar to SpecChart.Pdf showing that a -100 dBc carrier will make a
prominent peak in he spectrum. And with every 20 dB less isolation the
effect will be one order of magnitude more prominent in the
tau-sigma-diagram.



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