signals and systems question

I just did that. You ignored it.

Or just do the math, and then you'll know instead of having a superior sort of wild-ass guess.

For nuclear instrumentation (proportional counter pulse digitization), the classic technique is the Wilkinson converter. An op amp precision rectifier feeding an integrator makes a kind of Wilkinson pulse digitizer. When the trailing edge arrives, the rectifier op amp output slams to the negative rail, and that's when you digitize the hold capacitor. You've got lots of time, that way, to do the digitization.

Isn't that kinda like using a DSO? What kind of peak-location scheme, does someone have to look at the trace and set cursors?

Not generally; if you attenuate some frequencies, that (by Parseval's theorem) means the RMS measure of the signal is different in frequency space, therefore also different in point-by-point sample space.

Ahh, good point. And, since the integral is a linear operator, the solution (to my first-thoughts method) is trivial: don't need it at all!

So another question, perhaps: why should the filter ever be correct?

This, I think, should be a similar reasoning. Suppose we wish to know the integral of an impulsive signal. In general, this will be the integral (from 0 to infty) of the impulse response of the system in question. The question is: is that integral invariant under conditions of the filter, given that it's a low-pass filter? And, furthermore, that it's invariant under sampling as well.

The latter won't be as simple to prove, but the former should be simple.

For the all-poles case, the impulse response is a series of t^n exp(-t/tau) type terms, where the exponential takes real or complex-pair time constants. The integrals of which all exhibit a Heaviside step response type behavior. More simply: in the frequency domain, a time-domain integral is simply tacking on a pole at zero.

Zeros in the transfer function are no problem, as long as the number of them is not greater than the number of poles (as that would be at least an ideal differentiator, not even a high-pass filter!).

What is a problem is one or more poles at 0, which would integrate an integrator, and the result is divergent.

What about poles "near zero"? Like the dominant pole in an op-amp? Ah, but we're assuming unity DC gain here, so that's still fine; it's just a very slow (and noisy and unstable) filter. (Not to mention, such poles are never used directly, but always shifted around as part of a feedback loop.)

So, informally, I don't see any reason that the integral unit impulse response, aka the unit step response, should be anything more than f(t --> infty) = 1, for a real filter.

A band-pass (1 So -- you're mostly right. Gaussian filters tend to have pretty broad

So, it would seem to me, at least roughly speaking -- any real filter should be fine, which means you can filter it as hard as you want -- go ahead and put in an elliptical (with the zero(s) at the Nyquist frequency and harmonics, if possible), if you are able.

It seems intuitively unlikely that the integration process will hold if the Nyquist reconstruction theorem is violated, so the priority should seem to be cutting off that high frequency tail as best you can.

Yeah, make sure to stack your sidebands safely and all that. :)

Tim

This is a slightly heretical approach but if you know the shape of your pulses you can potentially get a better (ie higher SNR) by correlating with that and picking the peak value than just using a boxcar integration. The wings contribute a higher proportion of noise to the integral compared to the few samples near the top of the peak.

Weighting the problem for minimum variance might be helpful here.

On a sunny day (Mon, 09 Mar 2015 11:26:25 +0000) it happened Martin Brown wrote in :

I do not see how a lowpass would preserve say the energy content of a pulse. And, when I look at it from the time domain, any sampling will have to start at the start of the pulse, if you have a finite sampling rate,. /\ / \ _| \______ t0 tn

|| so clearly if you start s0 earlier or later you introduce an error. over time, for the same repeating pulse, you get a varying output. So, level detector starts sampler.

Maybe analog delay line with WIDE bandwidth. Scopes work that way.

Smearing out the pulse over a longer time does NOT preserve its surface area, that would mean the energy into a filter would be the same as the energy out. As a filter usually _filters_, and in this case high frequency components, then the energy out must be lower. But I am not using lasers to ignite fussy ion now, so maybe you in your near infinite mathemagical wisdom can explain what is going on? Or were you saying the same thing?

A low-pass does not conserve the energy of the input pulse, but it *does* conserve the total integrated charge. It just so happens that the charge coming out of, say, a photodiode or a photo-multiplier is proportional to the energy in the original phenomenon, a pulse of light.

Low pass filters conserve charge simply because they have unit gain at zero frequency and the zero-frequency content of an input pulse is its integral over all time, i.e., its total charge.

So yes, the surface area of the pulse *is* preserved after low- pass filtering.

Jeroen Belleman

On a sunny day (Mon, 09 Mar 2015 15:23:58 +0100) it happened Jeroen Belleman wrote in :

OK, thank you, yes, should have relaized that, I use photomuliplier current to charge a cap, essentially that is also a low-pass. Maybe J.L. will be happy to hear that.

You still need the sampling pulse synchrozined I think.

Very interesting thread.

Is thre some way to 'methodically' go through and prove or disprove the points people have made?

For example, use octave, create a signal, add approrpiate noise, either white or any shape one might encounter, the simulate the detection process to find the best way to 'detect' the signal.

Contact me offline at the above address and let's do the simulation.

I'd do the 'math' but have NO idea how to do that and would get lost in formulas, so instead, we can set up a 'simulation' and empirically determine which way is better.

Hey, I knew that. The questions is what happens when the integral is performed by sampling and digitizing the filtered pulse, and summing the samples.

Not if the Shannon criterion is strictly followed. But that requires an ideal lowpass filter, which I don't have.

The "old" digital telephone system comes close to the Shannon limit, using a good lowpass filter and unsynchronized sampling. Of course, it could tolerate a little aliasing distortion, but my customer is being a huge PITA\\\\\\\\\ prudently cautious.

One of my guys did it in Python. But he pulled up some standard filters (Butterworth, Gaussian) and didn't feel like modeling my

Here's one case: signal is a 1 ns impulse, passed through a 5-pole 60 MHz Bessel filter, sampled at 250 MHz. The lower graph is integral error vs ADC sample phasing. It's not too bad.

I don't expect a 1 ns pulse to make it through my front-end amp, and my transitional filter should be a little better than the Bessel

Besides, the customer elected, over my protests, to run loose wires inside the vacuum vessel, from a dual-BNC feedthru flange to the photodiode; I wanted to use matched-impedance flex. No way he's going to get a 1 ns pulse through loose wires.

I doubt that any mortal could do an exact mathematical solution to this problem. It needs simulation.

Maybe I can presuade Rob to simulate my actual filter; he has all of the rest of the stuff in place. I was going to do it in Spice, and still could, but I'd rather let him do it.

Thanks for the offer. I'll keep you in mind. We get problems like this fairly often.

That's much messier. I have a system which does that, integrating by summing samples of an imperfectly filtered input signal. In my case, I always take the ratio of two such signals and the error should drop out.

The fact is that I *do* see that the successive integrals are varying with the phase between the input signal and the sampling train. Again, my signal bandwidth is not strictly limited to less than half the sampling rate, so it looks like Jan is right that synchronization is good under those conditions.

Now, to quantify this is another matter.

Jeroen Belleman

Speaking of weights, back in the day when I was a Chem major in college, one of the grad assistants showed us how to measure the area under the curve when the curve was on a strip chart. He would cut it out with scissors and weigh it! Of course that assumes the paper is of uniform density, but it would seem that was a fair assumption.

I believe this was the output of a gas chromatograph.

That works is the signals are similar except in amplitude. I am acquising a time series of energy flashes that will be similar in shape but different in amplitude, and the amplitudes are important for tuning the system. Sadly, I can't ratio anything, and the blips are unsynchronized to the ADC clock.

On a sunny day (Mon, 09 Mar 2015 08:45:14 -0700) it happened John Larkin wrote in :

I cant speak for Shannon, I did read that 1948 paper few days ago...

but in the time domain it is 100 % obvious that if your sampling does not start at the start of the pulse, is not locked to it, then you have a sample frequency related amplitude variation.

The new DVB-S2 (digital TV) system is close to the Shannon limit so I have heard. They like that as they can get more programs in the same bandwidth.

Maybe it boils down to what do you want to measure, if charge as Jeroen Belleman points out, it should be OK. If energy / power then it is not. ?

I doubt that anyone can do the math. I have an estimated photodiode signal, a sort of gausian impulse with a long complicated tail; let's ignore transmission line effects for now; a preamp; then a cable to the digitizer box. Let's ignore the cable. The digitizer has a diff line receiver, amp, 5-pole transitional lowpass filter, se to d-ff ADC driver chip, the little RC thing, then the 250 MHz ADC.

Hardly the stuff of closed-form solutions. Looks like a little theory and a lot of simulation to quantify the integral error.

Den mandag den 9. marts 2015 kl. 21.46.29 UTC+1 skrev Jan Panteltje:

why is that obvious?

I would think that if the Shannon criterion is strictly followed the samples should uniquely describe the signal regardless of sampling phase

upsampling by inserting zeros and running through a brickwall filter works

-Lasse

On a sunny day (Mon, 09 Mar 2015 13:33:46 -0700) it happened John Larkin wrote in :

If the shape of the pulses is the same, and you only want amplitude, then that asks for a peak detector. That then gives you time to read the amplitude. Then discharge the C in the peak detector after you have read it. That is what I do in my gamma spectrometer. It was published here some years ago.

On a sunny day (Mon, 9 Mar 2015 14:08:35 -0700 (PDT)) it happened Lasse Langwadt Christensen wrote in :

Because for a I would think that if the Shannon criterion is strictly followed the samples

Do not get what you say here.

Look at it as a mixer perhaps, there will be a fdiff superimposed on the output.

Yes; there is enough information to perfectly reconstruct the waveform (ignoring quantization) so there is enough info to compute the area. There is no need to sync the samples to the waveform, which I can't do anyhow. Of course, we need to sum samples that begin a bit before the flash, and include enough that the tail has decayed to almost zero. I do have enough info to bracket the time interval in which the blip is expected to happen. There might be some 10s of ns of uncertainty as to when the flash actually happens.

One could reconstruct the waveform by lowpass filtering the samples, and then upsample at some insane rate, and then sum. I think that's what you are saying. But I think that just summing the 250 MHz samples gives the same answer.