little math thing

On a sunny day (Wed, 01 Dec 2021 08:55:43 -0800) it happened snipped-for-privacy@highlandsniptechnology.com wrote in snipped-for-privacy@4ax.com:

Not sure what you want, but if I wanted to know jitter between 2 signals I would put both into an XOR gate or 'phase compararor' if you want, Changes in the XOR output can then be low passed and the changes in amplitude are then a measure for the jitter. That is as old as the world? I do not trust digital scopes not even for the simplest things. ??

I want to do this in a 10 GHz bandwidth, and that scope will be part of our test system anyhow. We automate testing and archive test reports.

I haven't used an analog scope in a decade or two. The 11801s are hybrids, digital scopes with (vertical raster scan) CRTs.

One of my guys has a 7104 in his office, the 1 GHz microchannel-plate analog scope. He still likes it. That's the last analog scope anybody uses.

In a sampling scope, you get only one point per trigger, so each point on the screen has its own independent instance of the jitter.

If the amplitudes are stable, just threshold both and look at the jitter in the difference of the threshold-crossing times. If the jitter is much less than the rise time, you can adjust the crossing times by taking the nearest point and computing

delta_t = (V(n)-Vth)/(dV/dt).

That's probably what I'd do to start with.

You can refine the threshold crossing times by fitting a line to the nearest 10 samples (say), computing where that line crosses the threshold, and doing as above for several data points per trace. The jitter is the spread of the samples vs. the best fit line near the threshold. Using samples to compute the line will cause the jitter to be underestimated by a little bit.

Cheers

Phil Hobbs

You want a polygon area calculation. Assuming both signals start at zero, and rise to 3.3V, the time between them is the area (in volt-seconds) bounded by the two traces divided by the (3.3V - 0) range. To compute the area of a polygon, you just list the points in clockwise-around-the-edge order, and take the first three points, calculate that triangle area. If the triangle sweeps clockwise from point two to point three, it's positive area, otherwise negative. Then calculate triangle area for points one, three, four... then one, four, five... and when you get back to the origin, you're done.

The 'clockwise-around-the-edge assumes that some zero-size triangles occur when both signals are pre-edge at 0V, and when both signals are post-edge at 3.3V,

I think I'd be inclined to go for pairs of points either side of the steepest point on the curve working outward to make a few independent estimates of the crossing point. That should give you a feel for how much jitter is in the sampling arm as well as in the signal source.

Hopefully one of them will dominate but if not at least you know.

Calibrating it by deliberately slugging the rise time with a small load and seeing if you can recover the right answer would be a good test.

The problem is that a dual-channel sampling scope takes one pair of points per trigger. A matched pair has the same time from trigger but different amplitudes. If you work from pairs, the only information is the voltages.

I guess one could compute the local, average slopes and translate the voltage difference into a time difference, and work with those numbers. Sort of what Phil suggested.

It's easy to compute the RMS jitter of each signal relative to the external trigger - the later rev 11801's did that - and compute channel-to-channel jitter from those measurements. In moderation, like down to maybe half the scope's native jitter.

This is from a little Basic program that slurped a waveform from an

11801 and computed jitter from trigger.

Estimated scope jitter was entered manually, and it mathed that out.