[time-nuts] TimeLab phase difference (slope sec/sec)

[Tom Van Baak]

 > I'm just curious if the phase difference slope value can be plugged 
in to this equation.

 > I'm seeing 4.22E-12 as the slope value in the upper right of the
 > TimeLab phase difference plot. Is that telling me that my DUT is
 > within +4.22ps / sec from my reference 1PPS for the 24 hour
 > measurement duration?
 >
 > I have attached a screen capture that will hopefully make its way
 > through for viewing.

Chris,

The answer is yes. But let's consider why instead of plugging numbers 
into equations.

Your measurements are not that different from comparing two wrist 
watches. Since two watches never actually run at the same rate you know 
that on average one is fast and one is slow, relative to each other. We 
often measure clock or oscillator rate using a unit-less number like 
percent. Your left watch may be 0.01% faster than your right watch. Or 
maybe it's 15 ppm slower, or 50 ppb faster. These are all dimensionless 
numbers; ratios. So when we talk about an oscillator being 4.22E-12 fast 
it's just 4.22 ppt, or 0.000000000004 or 0.000000000422%. You get the idea.

So where does the 4.22 ps/s thing come from? Well, it turns out that the 
way that we measure two clocks is not to directly compare their 
frequency. You can't tell with a glance at two wristwatches which is 
fast and which is slow. Sure, one may be ahead and one may be behind. 
But that is the time (phase) of the clocks; not their rate.

The best way to find out which clock is fast or slow is to compare their 
times *over a long time*. Eventually a *trend* will be evident. You 
might have to wait an hour before one watch gets a second ahead of the 
other. This is where the "second/second" thing comes in. If a watch is 
fast by 1 second an hour then it must be running 1s/3600s = 0.028% = 278 
ppm faster. Note how both the units of the amount of time gain (1 s) and 
the amount of time spent doing the measurement (3600 s) cancel and 
you're left with a dimensionless number.

In general a frequency difference measurement is just a phase (aka time) 
difference measurement made over some elapsed time (aka measurement 
duration).

So you are using a TAPR/TICC to measure the phase difference between 
your two clocks. It appears you ran it for an entire day (1 d = 86400 s) 
and the net change in phase between the two clocks was 382 ns. That 
means the frequency difference between the two clocks is 382 ns / 86400 
s = 382e-9 / 86400 = 4.22e-012.

The reason this wasn't obvious is that you had the TimeLab 'r' 
(residual) command in effect. In a sense, this removes the very slope 
you're trying to see. If you undo the 'r' you should see one clock 
gradually gaining time relative to the other. The advantage of the 'r' 
command is that it shows you "what's left" if the two clocks were 
running at the same rate; it shows the wander, the short- and mid-term 
noise between the two. And TimeLab also reports the now-removed slope in 
the upper corner.

Now back to 4.22 ps/s. It is true that 4.22e-12 is equal to 4.22 ps/s. 
Mathematically, it is also equal to 382 ns/d, or 133 us/year for that 
matter. Yet, it may be misleading to claim 133 us/year because the word 
"year" may be interpreted as how long the measurement was. But it wasn't 
a year-long experiment, not even close. Similarly, claiming the clock is 
4.22 ps/s may imply to the reader that both the DUT and the REF and the 
TICC are all clean signals at the sub-ps level. But they aren't, not 
even close. To complete the picture you need both the frequency 
difference value and the duration over which the measurement was made.

So that's why you often see frequency differences reported using 
scientific notation, as in 4.22e-12, rather than ps/s or ns/d or us/y.

/tvb