Calculating suitable sampling rate and bandwith - Page 1
Forum section: electronics / beginners

author: ggeorgiev
author_rank: Contributor
timestamp: May 06, 2023, 09:00:05 am
content: Hi all,I want to measure analog signal with arising edge of around 1 µsand choose a suitable ADC or oscilloscope for it. For this, I am calculating the minimum Bandwidth and the Sampling rate to resolve the signal satisfactorily. For me, the most important part of the data acquisition, in that case, is the rising edge, so maybe this influences the calculated specs. So I want to make sure that the captured rising edge is not distorted because of my bad choice of ADC. I have calculated them as follows:Bandwidth * Trise = 0.35; therefore, the minimum Bandwidth should be 350 kHz.The sampling rate must be at least twice as fast, so a minimum of 0.7 MS/s.Out of curiosity, I asked ChatGPT the same question:"I want to measure a signal with a rise time of 1 microsecond. What should the bandwidth be? What does the sampling rate should be ?"and the answer was magnitudes higher.Here the answer:"To measure a signal with a rise time of 1 microsecond accurately, the bandwidth of the measurement system should be at least 350 MHz. This is because the bandwidth should be at least five times higher than the reciprocal of the rise time, as per the rule of thumb known as the "5 times rule".To accurately capture the behavior of the signal, the sampling rate should be at least twice the bandwidth of the signal, according to the Nyquist-Shannon sampling theorem. Therefore, the minimum sampling rate required would be 2 * 350 MHz = 700 MS/s (mega-samples per second).However, this is the minimum sampling rate required to accurately capture the behavior of the signal. In practice, the sampling rate should be chosen to be higher than the minimum sampling rate to ensure that the signal is accurately captured and to reduce the effects of noise and distortion. Therefore, a higher sampling rate may be required depending on the desired level of accuracy, noise characteristics of the signal, and the specifications of the measurement equipment."Can this be correct, and my calculation above be wrong? Or the truth lies somewhere in between?As the difference in the needed equipment is quite high, It would be nice to know the real minimum specification.Thanks for the support!Georg

author: wasedadoc
author_rank: Super Contributor
timestamp: May 06, 2023, 09:17:00 am
content: The ChatGPT answer would mean 700 samples between the 10% and 90% levels of the edge. Obviously that is nonsense. Any number greater than about 10 samples in that time will be adequate.

author: alm
author_rank: Super Contributor
timestamp: May 06, 2023, 09:30:38 am
content: A system with a Gaussian response and a bandwidth of 350 kHz will have a rise time of 1 us. That means that if you feed it a step with an extremely fast rise time, let's say 1 ns, the system will show this as a rise time of 1 us. If you feed it a signal of 1 us, it will show a rise time of 1.4 us. So as you might imagine this system is not very useful if you want a detailed look at the rising edge. I'd say aiming for a rise time of one fifth the signal rise time is reasonable, but a faster rise time will give you an even better view of the signal.Nyquist-Shannon's sampling theorem is talking about an absolutely bandwidth-limited system with an infinite reconstruction filter. Since these things don't exist in real life, having a larger ratio between sampling rate and bandwidth makes life easier and reduces the chance of artifacts. So I'd say at these low sample rates where more sample rate isn't very expensive, a sample rate of 5-10x the bandwidth makes sense.

author: EPAIII
author_rank: Super Contributor
timestamp: May 06, 2023, 10:29:33 am
content: I have always regarded Nyquist-Shannon as the absolute MINIMUM criteria if you ever expect a FREQUENCY to be reproduced. It assumes that an infinite number of samples are taken of a signal that lasts for an infinite number of cycles. Therefore it says NOTHING about the amplitude of that frequency in the reproduced signal at any given instant in time. For that you need a lot more samples.If you don't believe the above, then draw one cycle of a signal and then plot 2.000001 samples, equally spaced on it. The theorem only says greater than 2 samples in one cycle and that number IS greater than 2. If you start with the first sample on the zero axis, the second one will be at about 0.0001 percent of the full amplitude and the third one, which barely fits on that one cycle, will be around 0.0002 percent. Your initial signal will have been reduced by a factor of about 100,000. It will be there, but only barely.It is only if you continue to take samples for a great many cycles at that rate that ANY filter can reconstruct the original signal with any accuracy. In my example you are going to need at least 100,000 cycles sampled at that rate.The OP asks about measuring a 1 us rising edge. But does NOT say what is to be measured. A rising edge does not have an infinite number of cycles. At best it has ONE. Looking to something like Nyquist-Shannon to design a system to measure something like that is insanity. A much better method would be to FIRST describe, IN DETAIL just what is to be measured. It is hard to get somewhere if you don't know where you are trying to go.Then, determine the accuracy needed in that measurement. Only then can you look at the waveform and determine how many samples are needed. And an analysis on a piece of graph paper or with a CAD program would be much more useful than Nyquist-Shannon.As for bandwidth, again a practical consideration of what needs to pass to the A-D converter is needed. A bandwidth is often express at the 3 db down point. So at 1 MHz bandwidth, a 1 MHz signal will be 3 db down. And that is what the A-D converter will see. And if you want to see any details in that rising edge beyond 1 MHz, then you will need higher frequencies in the bandwidth. 2 MHz, 3 MHz, 5, 10, how far do YOU need to go to measure what you want to measure. And to what degree of accuracy at those higher frequencies.Far too little information has been given for any accurate answer to be formulated. I worked in the audio-video-TV world. The 3 to 4 samples per cycle worked for video. For the most part at least. That's more than just 2+. And distortions are still present. With audio, many systems used only the 2+, Nyquist-Shannon criteria (41 or 44 K samples per second). I never liked that. Undoubtedly much of the reason why it is acceptable is a lack of HF response in the ears of most listeners. Yet still, many audiophiles complain about the quality of digital audio. I wonder why. ..... Not really. Nyquist-Shannon is a corporate accountant's dream.Science says 2+ samples are enough, so why spend money on more?

author: tooki
author_rank: Super Contributor
timestamp: May 06, 2023, 11:06:19 am
content: EPAIII: If that’s what you believe, then you fundamentally do not understand Nyquist-Shannon.As long as the bandwidth limit is respected, then every signal — including the highest frequencies at small amplitudes — are reproducedperfectly.Please watch this video, which not only explains it, butdemonstrates it in practice:https://wiki.xiph.org/Videos/Digital_Show_and_TellI basically had made all the same assumptions as you before I saw that video.In practice, it’s often easier to use a higher sampling rate and a simpler low pass filter and then downsample than it is to create a really steep low pass filter. But the downsampled output has all the frequency content up to Nyquist.In an oscilloscope, where you want to inspect the waveform, you indeed need a higher sampling rate because the signal isn’t a simple sine wave, so it has harmonics, and we haven’t filtered them out with a low pass filter.

author: radiolistener
author_rank: Super Contributor
timestamp: May 06, 2023, 11:25:04 am
content: Quote from: ggeorgiev on May 06, 2023, 09:00:05 amyes, you're needs at least 350 kHz analog bandwidth or higher.Regarding to the sample rate, it should be high enough to keep analog antialiasing filter slope above required analog bandwidth and below sampling bandwidth (Fs/2).Since it's hard to implement analog low pass filter with a flat response in a pass band region and very sharp slope, there is a common rule to use sample rate which is about 10-20 times higher than required working bandwidth. It allows to place flatter filter slope between working bandwidth border and sampling bandwidth and keep more flat bandpass region of the antialiasing filter.In your case min sampling rate should be about 350 * 10 = 3500 kHz or higher.You can try to use lower sample rate, but it increase requirements for antialiasing filter on the ADC input. In practice you will be unable to implement good analog low pass filter with very sharp slope, flat pass band response and good enough stop band attenuation. So, in practice the min sample rate should be about 10 times higher than required bandwidth.If you use lower sample rate it leads to aliasing issues, because antialiasing filter will be unable to reject images with good enough attenuation and they appears in a working bandwidth.PS: this is actual if you're needs to sample signal with at least 1 uS rise time. But if you want to measure rise time which is about 1 uS, then you're needs much higher bandwidth, because sampling with 1 uS rise time will be not enough to precise measure 1 uS rise time. How much it depends on your requirements for measurement error.

author: rstofer
author_rank: Super Contributor
timestamp: May 06, 2023, 02:04:41 pm
content: Quote from: tooki on May 06, 2023, 11:06:19 amThat is an excellent video and well worth the time to view.

author: bdunham7
author_rank: Super Contributor
timestamp: May 06, 2023, 03:19:57 pm
content: Quote from: tooki on May 06, 2023, 11:06:19 amActually, in practice, it is absolutely and completely impossible to filter out ALL of the out-of-bandwidth components on the input signal, so the reproduction is NEVER 'perfect' under any conditions. So the question is 'how close can you get?'. The answer for audio inputs is that in the early days it was very, very difficult to implement a reasonable input filter of the type needed until digital oversampling techniques were used.I watched Monty's video long ago and I don't want to watch it again. I'm not an LP-hoarder or conspiracy theorist, but I can tell you that 'stairsteps' were real and could be readily demonstrated with certain early CD players (some high-end ones too) because their initial solution to the difficulty with a brick-wall filter was to not worry about it. Again, the eventual fix that was introduced fairly quickly was digital oversampling and filtering before the analog output stage.For audio, this issue is fixed and long-forgotten. Performance ofproperly designedsystems today are beyond reproach and probably significantly into the overkill zone. However, when this discussion comes up and I hear or see "You don't understand. Nyquist works perfectly because, math!", my response is "No, it never, ever works perfectly--not even close--because, reality!" At least when it comes to oscilloscopes and signals, any discussion of Nyquist without discussing the errors that make it a highly unstable solution as you start to approach the limit is incomplete and not all that helpful, IMO.

author: bdunham7
author_rank: Super Contributor
timestamp: May 06, 2023, 03:47:22 pm
content: Quote from: ggeorgiev on May 06, 2023, 09:00:05 amChatGPT is just plain wrong (no surprise there) and your other calculations are more on the right track, but you can't ignore noise, quantization and a lot of other things that are going to make for a practical solution. The best way to determine and demonstrate the required sample rate would be with an experiment.First, you'd have to be clear about what your stated 1uS rise time really means. Is that truly BW-limited to 350kHz or is it a ramp with sharp corners? When I instruct my AWG to generate a 200uS pulse with 1uS rise time, I get a ramp with sharp corners and there clearly is BW beyond 350kHz. Assuming you want to just not worry about that and you really just want a reasonably clear and well-timed general idea of what the pulse looks like, you can see that at 0.7MSa/s, you will not always have a sample anywhere in the ramp or transition. This can in theory be reconstructed by the sinx/x filter, but in reality any small error in the sample (noise, jitter, etc) will cause large variations in the reconstructed signal. So how many samples is enough? It depends on how cleanly and accurately you want to see the result. I can't directly set the sample rates on my scope, but I can limit the memory and use the zoom function to get 0.2MSa/s, 0.4MSa/s, 1MSa/s, 2MSa/s, etc. So here it is, a 200uS/1uS rise pulse from an SDS2042X AWG viewed on an SDS2354X+ oscilloscope. I used a 1s hold-off and 5s persistence so you can see about 5 different traces with different captures and reconstructions. If all 5 overlap, you have a clean, reproducible result. If you see 5 different traces, obviously there's some variance. As you can see, 10MSa/s is almost indistinguishable from the full 2GSa/s capability of the scope and 4MSa/s is probably good enough for most cases.

author: wasedadoc
author_rank: Super Contributor
timestamp: May 06, 2023, 04:35:53 pm
content: Quote from: bdunham7 on May 06, 2023, 03:47:22 pm10Msample/s is 10 samples in 1 microsecond which was given in the first reply.

author: ggeorgiev
author_rank: Contributor
timestamp: May 06, 2023, 04:36:27 pm
content: Quote from: bdunham7 on May 06, 2023, 03:47:22 pmIt is actually just a ramp, so the signal needs 1 us to reach 100%.The signal comes basically from a TIA circuit, and I have to resolve the signal good.This is maybe the other problem. I am not quite sure what good in that case means. I have uploaded a picture with what i expect as a signal. Maybe good would mean that i have at leas 10 data points during the rising edge.

author: tooki
author_rank: Super Contributor
timestamp: May 06, 2023, 04:48:04 pm
content: Quote from: bdunham7 on May 06, 2023, 03:19:57 pmI think any reasonable reader of my response would deduce that I am aware of practical limitations, since I mention how real-world implementations often do it differently instead.Anyhow, the point of my response was not to be a comprehensive treatise on it, but rather a concise primer on the theory to dispel EPAIII’s belief that as you approach Nyquist, you lose amplitude information. That is categorically untrue, and the video demonstrates that exact point beautifully.I did also mention that oscilloscopes need more samples.

author: tooki
author_rank: Super Contributor
timestamp: May 06, 2023, 04:53:45 pm
content: Quote from: ggeorgiev on May 06, 2023, 04:36:27 pmJust a little FYI: rise time is defined not as the time it takes to go from 0% to 100%, but rather as the time to go from 10% to 90%. So expect formulas, etc to be based on that assumption.With your linear ramp, that means your actual rise time isn’t 1 μs, but 0.8 μs.

author: bdunham7
author_rank: Super Contributor
timestamp: May 06, 2023, 05:16:45 pm
content: Quote from: tooki on May 06, 2023, 04:48:04 pmEPAIII's statement that the theorem is only mathematically valid (to perfection) with an infinite number of samples is a valid point and different than the concerns I raised--and his point is probably relevant to sampling transient events. The only question is how closely you can approach the Nyquist limit before this becomes an issue and IIRC the video only shows that if everything else is nearly perfect perhaps you can get pretty close. But if you gettooclose, something gives.Again, to reiterate and rephrase the last sentence of my previous reply to you, any analysis of a sampling system is incomplete unless you demonstrate what happens if you perturb samples slightly. So take your perfect reconstruction with a sample rate of 2.001X your sine wave frequency, perturb one sample in either amplitude or time (the two are equivalent in this case) by a small amount and see what it does to your result. There's a reason that 8-bit sinx/x scopes all pretty much lose it below 2.5X sampling rate, if not sooner.

author: MrAl
author_rank: Super Contributor
timestamp: May 06, 2023, 05:17:05 pm
content: Quote from: ggeorgiev on May 06, 2023, 09:00:05 amHello there,Before i say anything else i should say that i would buy the best scope i can get for the money i want to spend, as long as it can do what i want it to do. If i find one for the money i want to spend and it's better than what i need, i would get the better one.I have found ChatGPT to be more like the Comedy Channel when it comes to technical specifications. It tends to mix specs up a lot for some reason. For example, a particular op amp had a bandwidth of 20MHz and slew rate of 18v/us and it reported 18MHz and 20v/us, and after i corrected it the first time it thanked me and then reported 20MHz and 20v/us, so it only corrected one thing i told it about, and then after the second time i corrected it the reply was finally correct. I've found it to mix things up a lot and that means it renders answers that are way, way off, more than most humans would ever do.With that, if we look at the numbers it produced this time, the 350MHz is awfully close if we change the "M" to a "k", and i have seen this happen before too. That would make it 350kHz which is better for that part of the result, although not complete by far.The next thing we see is the "5 times rule" which is also correct, but it was never applied to the 350MHz or anything else for that matter. It was just mentioned an apparently never used for anything.Taking these two together, if we start from 350kHz and multiply that by 5 we get 1.75MHz, which is much better. This would be the -3db cutoff frequency of a first order filter that would reproduce a 1us rise time to a decent degree of accuracy. If we take the 350kHz alone we get a bad display of a 1us rise time signal. It's only after we adjust it by that factor of 5 do we come up with a reasonable result for the required bandwidth.Nothing says this better than a plot.In the diagram attached, the red plot is the signal with the (approximate) 1us rise time, and the green plot is what we would see with a bandwidth of 350kHz, which is clearly not very good. As we increase the bandwidth by 5 times, we get the blue plot, which is a lot closer do what we actually have. Now since a wave with a risetime of 1us does not perfectly ramp up and suddenly flatten out as the red plot does, if we take the blue plot and use that as the input to the scope (a more reasonable wave to test with) we would get something like the green line again with a 350kHz bandwidth although it would be delayed a little more, and that is still not very reasonable. If we again increase the bandwidth by a factor of 5 and again use the blue wave for the input, we would get the violet plot, which is not bad at all when it comes to reproducing the blue plot.So the rule has to be BW=5*0.35/Trise or better.It's also interesting that if you take a 350kHz sine wave and run it through a filter with BW of five times that, you get a reasonable display of the sine wave. Some people though quote the 10 times factor, which would require a BW of 10 times 350kHz, but that could be an overkill. It's also interesting that a 350kHz sine wave has a rise time of roughly 1us when we look at the major part of the rise.Note that i take the view of all this mostly in the time domain because that is what we actually see on the scope. We can look at it in the frequency domain if we use a spectrum analyzer but usually to see the rise time we would be working in the time domain.

author: bdunham7
author_rank: Super Contributor
timestamp: May 06, 2023, 05:27:35 pm
content: Quote from: ggeorgiev on May 06, 2023, 04:36:27 pmIf your signal looks like that you have BW very far above 1MHz. 10MSa/s might be good enough, but since you don't know exactly what the signal does look like, you won't know unless you experiment a bit. Is there some reason you might need to use the least number of samples possible?

author: BillyO
author_rank: Super Contributor
timestamp: May 06, 2023, 05:58:20 pm
content: Not sure if it was mentioned, but chatGPT needs to go back and practice some simple arithmetic. 5/1us = 5MHz, not 350 MHz. That right there should have sent you on your way.

author: tooki
author_rank: Super Contributor
timestamp: May 06, 2023, 06:59:00 pm
content: Quote from: BillyO on May 06, 2023, 05:58:20 pmIt is a text prediction engine. It doesn’t know mathat all.

author: BillyO
author_rank: Super Contributor
timestamp: May 06, 2023, 07:01:05 pm
content: Quote from: tooki on May 06, 2023, 06:59:00 pmSurely you are aware I was joking.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: May 06, 2023, 07:01:09 pm
content: There is not enough information in your orignal post to give a decent answer.For example, when your signal (with 1us rise time) has a 50% duty cycle, a base frequency of 1kHz, and you only want to mearsure that base frequency, then a sample rate of a few kHz can be enough.However, if you want to measure the exact rise time, over and undershoot and ringing on that signal, you may need a sample rate of 100Msps or more.It's been a long time since I learned Shannon, but I think it's only valid for repetitive (periodic) signals, and it gets "abused" for single shot because the prediction is "good enough" for most cases.But if it's important to you, then study the Shannon stuff and then draw your own conclusions, instead of trusting some numbers posted by some random guy on the internet.

author: MrAl
author_rank: Super Contributor
timestamp: May 06, 2023, 09:31:39 pm
content: I made a typo in reply #12.I wrote:So the rule has to be BW=5*0.35/freq or better.when it was suposed to be:So the rule has to be BW=5*0.35/Trise or better.It's corrected now.That is close to sqrt(3)/Trise.

author: gf
author_rank: Super Contributor
timestamp: May 06, 2023, 09:57:24 pm
content: Quote from: bdunham7 on May 06, 2023, 05:16:45 pmIt's a matter of DSP resources and number of leading/trailing samples which must be discarded. 2.5x sample rate (i.e. cut-off at 0.8*Nyquist), with 1/256 pass band ripple and stop band rejection at Nyquist by factor 256 (~48dB) can be realized with a reasonably small number of FIR taps (say 30). However, nobody can/will spend 10,000+ taps in order to achieve the same performance at 2.001x sample rate.

author: David Hess
author_rank: Super Contributor
timestamp: May 06, 2023, 10:07:52 pm
content: Quote from: ggeorgiev on May 06, 2023, 09:00:05 amI would generally start at 1/10th the rise time I want to measure, so 100 nanosecond rise time, which is 3.5 MHz bandwidth, and 7 MS/s, which is not particularly demanding. The measurement can still be made with lower performance by adding the rise times to derive what the signal must be.

author: radiolistener
author_rank: Super Contributor
timestamp: May 06, 2023, 11:05:06 pm
content: Quote from: bdunham7 on May 06, 2023, 03:19:57 pmit works in math, but there is no ideal DAC/ADC in reality and no ideal reconstruction/antialiasing filter.So, the reality is much worse than math expectations

author: radiolistener
author_rank: Super Contributor
timestamp: May 06, 2023, 11:21:21 pm
content: Quote from: MrAl on May 06, 2023, 05:17:05 pmThe main issue with chatGpt is that it mix math calculations as he want. For example if you ask him to calculate something it can propose you some formula and calculations, but if you check his calculations you will found that it's wrongAnd if you ask him what is the source of the formula he will propose you some reference on literature but if you check it, it doesn't have such formula at all. If you try to ask why chatGpt lie about formula source, it just says "sorry" and propose another reference, but it's also wrong... and it repeats in a loop, so you will be unable to get formula reference and will be unable to check it if you don't know it wellSo, an answer of chat gpt about formulas and calculatios is a kind of cheating. He just provide you some random formula and random calculations which at a glance looks very similar to a true, but if you check it, you will find that it's wrong...I'm not sure why it happens, may be math analysis requires more context memory or more strict rules which is not supported by chat gpt...

author: wasedadoc
author_rank: Super Contributor
timestamp: May 06, 2023, 11:39:23 pm
content: Quote from: radiolistener on May 06, 2023, 11:05:06 pmWorse? Yes. Much worse? No.

author: MrAl
author_rank: Super Contributor
timestamp: May 07, 2023, 12:33:49 am
content: Quote from: David Hess on May 06, 2023, 10:07:52 pmIf you do a simulation you dont need the BW that high. Not that it hurts of course.

author: MrAl
author_rank: Super Contributor
timestamp: May 07, 2023, 12:36:47 am
content: Quote from: radiolistener on May 06, 2023, 11:21:21 pmYeah i found that too.It also does not know how to properly write out some of the formulas at all in the correct algebraic form.For example, for y=a/(b+c) it might print out a/b+c, although maybe not for that simple but that's the idea. Longer formulas it messes up badly, and when told about it it never comes up with a good correction.I dont know how this is supposed to pass for AI. It's AAI, artificial artificial intelligence (A^2 I ) ha ha. More like AS, Artificial Stupidity.I think it can broaden your thought process but not anything too detailed.This reminds me of the first Chess programs/machines. Almost anyone could beat one, but then as time went on they got much much better so they are hard to beat now. Grandmasters have trouble with some of them. Maybe AI will eventually evolve but for now it's not very detailed.

author: David Hess
author_rank: Super Contributor
timestamp: May 07, 2023, 07:10:39 am
content: Quote from: MrAl on May 07, 2023, 12:33:49 amYou do not, but a higher bandwidth allows direct measurement of the signal's rise time without compensating for the rise time of the system.The calculation can be done to find the bandwidth for a given error in the measurement. For 10% error, the bandwidth required is 764 KHz or 458 nanosecond rise time. For 5% error, 1.1 MHz or 320 nanosecond rise time.

author: Siwastaja
author_rank: Super Contributor
timestamp: May 07, 2023, 07:27:49 am
content: A few practical points:* If your input signal is quite free from noise or other signals above your desired bandwidth, this eases the requirements for anti-alias filter;* On the other hand, if you have strong interfering signals, or just a lot of wideband noise, you need a lot of attenuation on the stop band to stop noise aliasing to your pass band.* High sample rate ADC and DSP is nowadays cheaper (and way better) than steep analog filters.* Therefore, you usually don't want just 2*Nyquist ADC sample rate, but much more. (You can get away with 2*Nyquist if the signals are free of out-band interference/noise.)* Given a lot of oversampling (ADC sampling more than the minimum Nyquist), you might get away with a cheap RC filter for analog anti-alias, or, for maximum group delay flatness, something like 2nd order Bessel filter. Thanks to oversampling, no need for higher order, more expensive and harder to design filters* Now you can design a steep filter on digital domain which is nearly perfect, before decimating the data to the final sample rate you need.* Added bonus: oversampling and digital filtering gives you extra bits and noise reduction (as long the ADC linearity is decent), so say a cheap 12-bit 10MSPS ADC with marginal noise specifications is probably better than a 1 MSPS 16-bit ADC with better noise specs, because with latter, you would need to design a much steeper analog filter, which will again add noise and distortion.* I would assume group delay flatness is important in your application, therefore analog filter with bessel characteristics and on digital domain, FIR instead of IIR.* ChatGPT is language model and cannot calculate even simple equations. It's totally useless in designing anything.

author: David Hess
author_rank: Super Contributor
timestamp: May 07, 2023, 07:38:37 am
content: Quote from: Siwastaja on May 07, 2023, 07:27:49 amThe digitizer non-linearity produces mixing products between the sampling frequency and noise free perfectly bandlimited input signal which cannot be removed by the anti-aliasing filter. These can be seen on DSOs operating close to their Nyquist frequency as "wobbulation" in the sinx/x reconstructed waveform.So you cannot get away with 2*Nyquist, or even 2.5*Nyquist, unless your ADC is also perfect.

author: tggzzz
author_rank: Super Contributor
timestamp: May 07, 2023, 08:22:57 am
content: Quote from: ggeorgiev on May 06, 2023, 09:00:05 amKey question: is your analogue signal a one-off event or repetitive?In either case the bandwidth of the analogue components in front of the digitiser is the same.If it is a one-off event then the minimum sampling frequency is in the MHz range.Butif it is arepetitivesignal then you can choose whatever sampling rate is convenient, provided you can correlate the sampling instant with the waveform.Examples,back in the mid 70s I saw a signal such as you describe captured by an X-Y pen plotter (full-scale frequency response < 1Hz!)I have a 70s era scope that captures 50ps risetimes with a 13kHz sampling ratein the 80s I used high end HP scopes that capture 1GHz signals with a 40MS/s sampling ratemany modern scopes still use such capabilities; look for ETS, RIS and similar terms

author: wasedadoc
author_rank: Super Contributor
timestamp: May 07, 2023, 09:23:08 am
content: Whether real time sampling or ETS the critical point is to have sufficient samples spanning the waveform portion being examined.

author: ggeorgiev
author_rank: Contributor
timestamp: May 07, 2023, 10:14:13 am
content: Quote from: tggzzz on May 07, 2023, 08:22:57 amAs the signal comes from a TIA circuit it can be one or several pulses.In most cases, I will take several pulses, but it is possible that I take only one pulse too.So my analog signal can be a one-off event or repetitive.

author: ggeorgiev
author_rank: Contributor
timestamp: May 07, 2023, 10:33:58 am
content: Quote from: bdunham7 on May 06, 2023, 05:27:35 pmNo, there is no reason.Basically, I am now at a point where I have a designed TIA circuit, and I need:Either to choose ADC and design PCB by myself, so that I can sample the signal from the TIAChoose an ADC and directly buy an Evaluation board such ashttps://www.digikey.ch/de/products/detail/texas-instruments/ADC31JB68EVM/5764613for exampleOr choose and buy a scopeAs I got the answer from ChatGPT and checked for the scopes with specs about 500 MHz BW and 1Gs/s the price was around 10k EuroThe second board is around 2kAnd buying just an ADC the IC would be around 400/500 Euro, but with a high-tech fast ADC I am not sure that I will be able to program it myself.So I wanted to be sure which minimal specs do I really need, before just deciding to buy one of the three options above that i have.

author: wasedadoc
author_rank: Super Contributor
timestamp: May 07, 2023, 10:41:43 am
content: As you don't already have a scope I would recommend you choose that option. A Rigol or Siglent in the ballpark of 500 Euro will be more than enough for your current specific need and will serve you well for many years. Of course you may decide to go for a higher spec scope at a higher price.

author: gf
author_rank: Super Contributor
timestamp: May 07, 2023, 11:16:06 am
content: @ggeorgiev, which properties of the edge do you want to measure at the end? Which metrics exactly matter for your particular use case? And what amount of uncertainty can you tolerate for the measured quantities?

author: ggeorgiev
author_rank: Contributor
timestamp: May 07, 2023, 01:03:40 pm
content: Quote from: gf on May 07, 2023, 11:16:06 amIt is the pulse shape of a laser that I want to measure. I want to see how does the pulse over time look like. For example, if it is overshooting or not and if there is a change in the rise time after several consecutive pulses. The metric on the y-axis is basically the attenuated power of the laser. But I am not interested in the laser power, but just the pulse shape. The point with the uncertainty is hard because I am not interested in some metrics, as I said above, but more just how does the pulse behave.

author: bdunham7
author_rank: Super Contributor
timestamp: May 07, 2023, 02:02:47 pm
content: Quote from: ggeorgiev on May 07, 2023, 10:33:58 amIf you only need one instance and a standalone oscilloscope is an option, it will certainly be your best choice for this. 100MHz BW is almost certainly plenty and decent 100MHz 1GSa/s (or even 200MHz) scopes start under $500.

author: tggzzz
author_rank: Super Contributor
timestamp: May 07, 2023, 05:08:35 pm
content: Quote from: ggeorgiev on May 07, 2023, 01:03:40 pmIn that case, make sure you adequately characterise the frequency response of your TIA. Phil Hobb's book is an excellent starting pointhttps://electrooptical.net/building-electrooptical-systems/Quote(Please correct your quoting in your reply #33 above!)In that case it isnota repetitive signal.If you want to capture successive pulses then you may not want to capture the "dead time" in between pulses. In that case a cheap simple entry-level scope probablywon'tbe adequate.Please don't use ChatGPT as a search engine. Would you trust the output from a "statistical parrot" or "keyword bullshitter"?! Neither would I.

author: radiolistener
author_rank: Super Contributor
timestamp: May 22, 2023, 08:05:50 am
content: Quote from: wasedadoc on May 06, 2023, 11:39:23 pmWell, it depends on the expected quality. If you're expect 20-30 dB dynamic range (4-6 bits), the reality will be just a little bit worse than math expectations. But if you're expect 50-100 dB dynamic range and even more (8 bits and higher), then reality will be much worse than math expectations