Member postings for S K

LOL: Last night I had an extended dream in which I was preparing a 400 page Ph.D. dissertation on ... pendulums!

This is getting crazy!

I'm sure you know what you are doing, but just to be clear: Using two channels, you trigger on the first edge (the input), and observe the variation in time between that edge and the second (the output). You can put your scope on infinite persistence or in envelope mode to see the variation building up over many sweeps. If your scope won't calculate RMS of that variation (not too many do), you can get a crude estimate of the RMS by dividing the peak-to-peak jitter by 6 or so.

That OPB960 is a higher-power device (hence the lower stray light sensitivity, I assume) with a much narrower aperture than the Sharp. Looks interesting!

Joseph: Have you measured the RMS jitter in that 5 us delay? Or what are you getting for your pendulum as a whole?

What are people using or planning to use the TDC for?

It's interesting that the detector is black or near black, presumably forming a white light filter.

Since you are pulsing the transmitter, did you trigger the scope on that pulse signal and observe jitter in the delay between that and your output? That jitter is the real noise of interest, as it would directly contribute to your period measurement S.D.

You want to use whichever output edge has the lowest delay / lowest jitter relative to the input signal (e.g. using your pulser). On the Sharp, that was the falling edge. Either ignore the other edge or only use it for non-critical functions.

Diffraction of a single edge is probably a smaller problem than the shadow moving across the non-zero sensor width. But either may contribute to jitter due to it causing a shallower slope of the photodiode signal, i.e. worse noise performance than if you use the pulser. You have access to the photodiode directly if you are in a position to measure it.

Again, I believe that your wandering in the cumulative timing error plot is substantially a random walk due to the summation of random period deviations. (Other sources will be due to temperature, etc., but that would add to the random walk.) You can simulate this very easily: Just produce a thousand (or whatever) normally-distributed numbers generated using your S.D. value and see how that wanders when accumulating them. No need to program, even: a spreadsheet can do this pretty trivially. Of course, you should re-run that a number of times to see how each run randomly differs.

Edited By S K on 21/08/2023 01:14:33

If I estimate your "half power bandwidth" to be 2 times your standard deviation (that's the number I have, anyway), then after rounding the numbers for simplicity, I get Q=1/(2*0.001)=500. How do you get over 20,000? What numbers do you have?

Also, you expressed concern about how your cumulative time error had wandered. This is due in large part to the accumulation of many small random errors in a classic random walk. If those small errors are 250 times lower (i.e., S.D. = 4us vs. 1ms), then the cumulative deviations should, in an typical run, also be 250 times lower. So yes, you do want as low a standard deviation as you can get.

Edited By S K on 18/08/2023 01:46:06

Why would the vertical vs. horizontal orientation of a same-shaped bob matter? I've always wondered why saucer shaped bobs hang vertically, when that requires the alignment to be perfect, while horizontal has no alignment problems.

Are you really using the simplistic formula from that "bandwidth of a tuned circuit" diagram? That delivers a Q of about 500, doesn't it? What formula are you actually using?

The thing is, with an S.D. of 1ms, I don't see how you can find all that much more than that with any similar bandwidth based formula. You need to focus on improving the S.D. first. If I had to guess, your measurement apparatus is probably to blame, not your pendulum.

Edited By S K on 17/08/2023 18:18:45

Aaaand I misspelled John Haine. Sorry.

Well, I used Dave's method on my pendulum data, as far as I understand it (and I may not), and got a "Q" of 190,000. That's not possible.

Yes, the rule here is "always trust John Haines." 😉

Dave:

Something continues to bother me about your method and your obtained value of Q. I am not doubting the fundamentals of the method (much), but the numbers do not seem to make sense in relative terms.

You appear to be getting about 1ms standard deviation in your period measurements, whereas in my last test I'm getting about 4us; a factor of 250 difference. Not all is equal, but assuming that the bandwidth is strongly tied to those numbers (i.e., 250 times lower S.D. means 250 times narrower bandwidth), then using your method I should get a Q that is very approximately 250 times yours. That would not remotely make sense.

What am I getting wrong or not seeing?

Edited By S K on 16/08/2023 14:37:23

Whatever.

I haven't explicitly checked your math, but the results look reasonable (and of course it shows that diffraction is occurring).

I don't know where I got that formula from, but apparently it is wrong. And that would mean that my measured Q just got down-graded too. 😒

Everyone seems to be using a different formula or different method. This needs to be straightened out.

That Olympus web site makes a false, or at least over-simplified, claim or two. It's not the case that if a slit is wider than the wavelength of light then the light passes without any diffraction. Every photographer, with their macroscopic irises, sees this in practice. In fact, even one edge (no "slit" at all) will cause diffraction.

WOW!

Thank you for the info about the Sharp device. So it's a silicon photodiode operating in the near-infra-red. That's perfectly fine.

No, it's a slit as in a simple aperture whose task is to constrain the beam to a small vertical strip. Yes, each edge will diffract, but as I discussed, that doesn't matter if the the slit is close to the photodiode, since virtually all light will be collected by the much larger lens and photodiode anyway.

But now I see that the response time is quoted as 3us, which means that the 4us RMS that I'm seeing is already at or very near the limit of the device's performance. So narrowing the aperture won't help much or at all. That's too bad, but at least you saved me from continuing this effort with the Sharp devices.

Given the documentation and what I've learned, I'd say that the Sharp device is about as good as it will get as-is. That's aside from the ambient light problem, but for that people are already trying out various filters. In the alternative, stopping down the aperture via a slit or pinhole, and adding a more powerful light source should accomplish the goal too. I did just shine the laser on one, full blast (no slit), with the transmitter covered, and it worked perfectly fine, so that remains an option.

If I wanted to do better, I guess I'll have to get an op-amp, etc., and build something around my "1ns" photodetector. Then, as narrow a slit or pinhole as I could make or get would likely be of benefit to timing, as well as simultaneously mitigating the stray light problem.

* An aside: In the famous double-slit experiment, the slit widths and separation affects the pattern seen, but I'm not aware of any necessity that any of them be precisely a wavelength. My understanding is that as long as they are some order of "small," it will demonstrate the phenomena just fine. I'd rather expect that, if a slit was exactly a wavelength, some additional peculiar quantum effects would happen, but I'm not a physicist.

Edited By S K on 16/08/2023 00:22:19

I ran the 50um slit to complete the picture:

• Without slit: 4.08 us RMS
• With 160 um slit: 12.6 us RMS
• With 50 um slit: 169 us RMS

A pretty bad trend, getting exponentially worse! I'm sure glad I didn't buy that 3um slit I was eying (well over $100)!

Dave: I don't know about the glitches, but the clock vs. NTP time plot looks a lot like a typical random walk due to the accumulation of minor random errors. My suggestion is to direct your attention to improving the period measurements. Get a few of the Sharp optos, maybe. They are pretty good and may improve things a lot.

Edited By S K on 15/08/2023 21:33:34

I redid the RMS period measurements on the Sharp opto, with my 160um slit and without.

• Without slit: 4.08 us RMS
• With 160 um slit: 12.6 us RMS

I should retry the 50 um slit again, but it was even higher, and I'm confident in this result already: using a slit makes it worse, presumably because less light entering the photodetector results in a worse signal to noise ratio.

My pendulum is completely free, so I have to launch the pendulum by hand, and it's difficult to do so without introducing at least some wobble orthogonal to the pendulum's normal swing. Therefore, I normally wait 10 minutes or so for it to settle before taking data, and I check for any remaining wobble before data processing. However, I noted that the detector with slit appeared to show wobble of the pendulum more strongly and clearly as a decaying oscillation. I don't yet understand why this could be more evident with the slit, so if anyone has an idea, please let me know.

As for the laser color, etc.: The "lens" is just a dome of plastic with a rather rough and cloudy surface finish. It's transparent aside from the poor finish, of course, so it will transmit red light with ease. Although the sensor is almost certainly not silicon (silicon photodiodes are completely insensitive to IR light), an IR sensor will be just as sensitive if not more to red light, as the shorter wavelength (higher energy) will cause the liberation of electrons even more easily. I don't think any of this has very much to do with diffraction caused by the slit.

My next step will be to retry the laser with the Sharp receiver. I will probably use a pinhole instead of a slit this time, to reduce the net light exposure a little further, while still presumably being stronger than the IR transmitter.

Edited By S K on 15/08/2023 19:52:01

Because diffraction due to the slit doesn't matter if the slit is directly in front of the opto's lens. The lens is about 1.5mm wide, while the slit is a tiny fraction of that, and is maybe 1mm in front of it. The photodetector is hard to see, but it's also about 1.5mm wide. So any light diffracted by the slit will be collected by the lens and seen by the photodiode anyway.

Diffraction due to the flag does matter (it softens the flag's light cut-off) if it's much further away, but that's a different problem.

By the way, I see you want to focus your lens very, very close. You will probably need to modify it to include an extension to be able to do that.

Edited By S K on 15/08/2023 15:03:58

A day of setbacks.

I made a new, smaller slit, this time about 50 microns wide. This is not very hard to do, at least if you have access to a stereo microscope and some fine tweezers. At that scale, it's not easy to measure, though.

I attached it to half of a Sharp opto (the receiver, of course) and hit it with the laser. Unfortunately, it seems that I destroyed it, either electrically somehow or via the laser. There was a peculiar effect which looked a bit like "latchup" (a phenomena that can happen when a flash of strong light hits an IC), but I can't prove it was ever working, so I can't be sure it was the laser either.

I then rigged up a Sharp opto (the whole opto) with the slit, and ran it normally (no laser). The opto's receiver has a domed lens, and it was not difficult to align the slit such that it worked. My goal was to measure the pendulum's RMS period variation, hoping that the slit may improve timing resolution. However, this did not work either, as the RMS value actually shot up. I have to try again, with and without the slit, with all else being equal, before I can be sure. The reason I'm uncertain is that my Agilent counter is infuriatingly cranky. I don't think it likes the very long periods.

If the Sharp really delivers lower time resolution with the slit, I'd theorize that a drop in signal to noise ratio due to less light getting through may be to blame. In that case, I'll try the laser again, maybe with a pinhole instead of a slit.

Also, it occurred to me that a slit would reduce ambient light impingement on its own, in this case by more than a factor of 10. Of course, it reduces the signal too, possibly to a net disadvantage.

Edited By S K on 15/08/2023 06:18:59