ok, this afternoon , I tested the Quad , here is what i found.
what is the bandwidth of the analog input channel?
From the connector to U5_pin13 and U16 _pin13 ,it is 10M .
lots of reasons for this .for example, the parasitic capacity around the U5 and U16, it is very distinct.
what is the sample rate ?
The max sample rate is 72M/s when T/div is 0.1us. Although we are seeking to make the sample rate to 144M when the other channel is unenabled, but now ,it seems not work .
as in this picture , the signal is 5M. and you can see about 14 points in a period. so the sample rate is 72M ,and if the CHA was hiden, this wave mess up.
as the 10M signal , the gain becomes about -6db. so , i believe we can say the bandwidth is 5M now
HugeMan, I don’t know whether your Quad has the same hardware as the mine, but of sure the mine has a -3dB cutoff around 2-3MHz.
Please, specify whether the probe could be so relevant in the test, because I have considered it always in my tests.
I don’t understand what you stated also: at the very beginning of your post you are saying that the analog bandwidth is 10MHz, then at the end you say is 5MHz. What is the correct value?
I’d love to detect a 10-15MHz of real bandwidth. It means inputting a 10MHz @1Vpp sine on the probe and reading around 700mVpp on the display. Being step-shaped, no matter if I read 700mVpp or 750mVpp. On my test I read a supposed 400mVpp!
If you have a Quad granting 10MHz of overall bandwidth, I’ll swap with the mine as soon. I will pay the shipping charge also!
About the sampling rate, I have also detected 72Ms/s and it could be fully acceptable from my viewpoint.
At the risk of insulting you, I disagree with everything you have said in these two sentences. I feel compelled to attack and destroy wild statements that dilute and disrupt positive and valid discussions. We are trying to keep these discussions factual and not based upon opinions. Please provide evidence that supports what you have said. Show us in the source code where these processes that you describe, take place, and what the display buffer has to do with sample rate. Or at a minimum show us some waveform analysis that supports your assertions.
I agree with your statement about separating analog bandwidth from DSO results (sample rate). But then you use the findings of DSO display to talk about 2Mhz bandwidth limit that was discussed in previous post. Cant do that here, cant use DSO results to talk about analog bandwidth.
Agree here also. This is a valid statement.
These are wild and unsubstantiated claims. BenF increased the capture buffer update rate by increasing the sample rate at the expense of decreasing the capture buffer size for each acquisition. This was done for situations where the T/Div sample rate being used was below the ADC max sample rate capability. BenF also never attempted to provide enhanced virtual sample rate for repeating pattern signals. Please stop posting made up thoughts which simply distract from these thread discussions.
Thank you for providing some waveform analysis for this thread. Was the -3db roll-off used to determine 10Mhz analog bandwidth?
While looking at the schematic, U-5 and U-16, I noticed something that appears to be odd. C-21 and C-22 appear to provide a signal ground for the ADC /NOT inputs, yet there are no 105 decoupling capacitors for high frequencies. It is quite possible that these missing decoupling capacitors are contributing to the parasitic issues in the front-end circuits. This may also adversely affect the interleaved ADC operation. Correct me if I am wrong, but these ADC /NOT inputs must be high frequency decoupled if they are to represent signal ground plane. Looking back on the Nano V2 schematic, the ADC portion of the STM used unbalanced input, and the Vref was decoupled inside the STM. Here in the Quad we have the ADC Vref being applied via external traces to the ADC /NOT inputs without high frequency decoupling.
I do believe that this could be a smoking gun for the ADC interleave issues, and may also play a role in the parasitic issues.
--------------- correction edit 16May2011 -----------------------------------
Further examination of the FPGA (U6) caps C21 and C22, finds that I was mistaken when I thought both caps were 22ufd. Instead, both caps are connected in parallel and C22 is a 105 cap which will provide the necessary decoupling. I think my eyes crossed on that one.
I understand that English is not your primary language, but I have to say that you have communicated very well on this topic. I wish I could write in Chinese as well as you can in English.
Technically, you are being conservative with the sine wave signal. Both channels enabled, single channel bandwidth would probably be closer to 7Mhz sample band width (still 10Mhz analog bandwidth) in my opinion. But in real life, because the Quad does not support the SinX display function (it only connects sample points with straight lines), then your conservative value would provide a better looking sine wave that is displayed with straight line connection of the dots.
If the above 10Mhz analog bandwidth measurement that you made was based upon a -3db roll-off, and if the interlaced ADC operation were fixed, then you could get about 10Mhz bandwidth for one channel disabled and 144Mhz sample rate, of course the straight line display dot connections will distort. If the smoking gun fix above is implemented, then that may increase the front-end analog bandwidth above 10Mhz and that would increase the current 10Mhz limit.
As discussed previously in this post, the square wave is a different animal. As described earlier in this thread, the square wave corners are created by the odd order harmonics, and the first 3 odd harmonics are not sufficient to get a reasonable square wave. So the square wave sample bandwidth would be significantly lower to prevent rise time distortion of the square wave which results in the narrowing of the square wave into a sine wave appearance.
Having conducted similar waveform analysis on the Nano V2, I have found that 1/20th sample rate provides a reasonable display, but that display still has rise time issues because of the relationship of the rise time and the Nano sample time limit. These same DSO issues will also apply to the Quad in direct proportion. Therefore the Quad sample bandwidth for a square wave with short rise time will most likely be 72Mhz/20 = 3.6Mhz and 144Mhz/20= 7.2Mhz respectively. If the square wave rise time is longer, then the Quad display will look better.
Have you measured your bandwidth at U5 and U15 pin-3? HugeMan is not referring to displayed results. He is referring to the inputs to the unity gain op-amps that directly feed the ADC.
HugeMan you should repeat your test looking at the op-amp output (input to ADC U-6, pins 3 and 11) because this is a low impedence signal path and will be less affected by parasitics. Unity gain op-amps provide low impedence output and that is why unity gain op-amps should always drive the ADC input pins.
first of all, i want to apologize to you: my last post was not intended to be a tutorial, so i assumed most of the facts i presented were already known.
You can find it in Benf sourcecode (v3.0), function.c, line 402, function void Measure_Wave(void). If you don’t know C, the function description is self explanatory .
Its also used to help with trigger position (i.e. if trigger is in the middle of screen some points have to be sampled before trigger), signal panning and zoom.
Supposing each pixel is a sample (to oversimplify things) in 0.1us/div and the screen uses 320 pixels for the signal, at 1us/div you will need 3200 samples (using only 1 out of 10) at the same sampling speed to complete the screen. Hence the “just in the limit” of the buffer.
Of course, you can use maximum sample rate even for 10s/div, you only have to decimate signal properly.
unlike you I don’t feel compelled to destroy and attack anything. I’m more the kind of make the love and not the war . But lygra, a forum is a place to post opinions (and when possible facts) in order to discuss them.
As the source code is closed, only Benf can throw some light over this. But i agree with you that’s irrelevant to the QUAD Bandwidth. My intention was to show that sampling speed is not a key factor if you have enough analog bandwidth. Most cheap scopes use that.
Check this: cp.literature.agilent.com/litweb … 8794EN.pdf
WTF? I’m used to post things i’m pretty sure of. My post was an answer to questions posed by Thanh and not solved by you.
And excuse me again, i wasn’t aware that i was distracting you. You have to focus yourself a little
My point was that you are comparing apples to oranges. DSO display results are never the proper way to measure analog bandwidth for all the reasons already posted in this thread. Looking forward to your findings when you do open it up, as validation of HugeMan’s findings.
I finally found time to research this topic some more. There appears to be a mistake in your calculations here. The first and subsequent odd harmonics for a 1Mhz square wave would be 3,5,7,9,11,13,15,17,19,21,23,25 and 27Mhz to pass the 13th odd order harmonic.
Anything over the sixth odd harmonic of this 1Mhz square wave will be at or greater than -3db attenuation due to the 15Mhz DSO Quad analog bandwidth, so a poor waveform is to be expected because of analog bandwidth limiting.
Another assertion claimed by you and others is that the Nyquist theory only requires 2x of the signal for a sampling rate. You failed to include the rest of that definition which says “a signal must be sampled at least twice as fast as it’s highest frequency component.” With the square wave this would require many odd order harmonics, so in your example above, to include the 13th odd harmonic, then you would require a sample rate of 54MSa/s to reproduce that 1Mhz square wave waveform accurately.
So even if the analog bandwidth could pass the 27Mhz harmonic (and it apparently can’t), then the sample rate would still be destroyed according to Nyquist while using a 36MSa/s sample rate.
If you fall back and say, OK, lets just use a signal generator that can only output the 5th odd harmonic and we know that we will have a slower rise time (less than 10%) source signal waveform because of this harmonic limit. The fifth order odd harmonic of 1Mhz is 11Mhz so the DSO Quad bandwidth can pass this harmonic. According to Nyquist, with fifth harmonic (11Mhz) being the highest frequency component, then 22MSa/s should accurately reproduce the poor quality input signal square wave, and the DSO Quad will reproduce that signal with its 36MSa/s capability.
So, in the end you will get a poor display of a 5th order odd harmonic limited signal input because of the signal itself, and you will also get a poor display of the perfect square wave input because of DSO Quad limitations as described above. As I had mentioned in my earlier posts, my poor quality function generator square wave signal looked pretty good on the DSO Quad because it didn’t have enough odd harmonics present to add the DSO artifacts when the DSO Quad capability is exceeded. When I said it looked reasonable, it did look nearly as good as the input signal with some limited sampling artifacts which resulted in narrowed alternations (10% narrowing at the alternation tops). In the same write up, I also stated that the DSO Quad signal generator signal looked terrible, and now this can be attributed to my other statement about the generator signal looking like a very good square wave up to 8Mhz (which means it has lots of odd harmonics present) on a more expensive o’scope.
The attachments further clarify the above paragraph. You can see that the rise time for the 2nd odd harmonic waveform is about 6% (less than 10% rise time). So we will limit ourselves to the 2nd odd harmonic. This is 5 x 3.5Mhz = 17.5Mhz so it should get through the 15Mhz DSO Quad bandwidth with some amplitude fall off. I also mentioned in previous posts about some drooping of the tops and bottoms and this would indicate some kind of amplitude fall-off. According to Nyquist, the required sample rate would be 17.5Mhz x 2 = 35MSa/s. These numbers clearly show that I could in fact see the 10% rise time 3.5Mhz signal on the DSO Quad (with some amplitude distortion which I never measured).
Now lets look at a 3.5Mhz perfect square wave by your definition (13th odd harmonic is present). 3.5Mhz x 27 = 94.5Mhz analog bandwidth requirement and will not do well with a 15Mhz DSO Quad bandwidth limit. For the same perfect square wave, the Nyquist sampling rate would be 94.5Mhz x 2 = 189MSa/s so that wont happen either. This is why you can’t look at the DSO Quad generator 2Mhz or 4Mhz square wave outputs on the DSO Quad display. It has nothing to do with the DSO Quad being defective, but is simply the result of the DSO signal sampling process while sampling a short rise time square wave.
In summary, in my first post in this thread I stated what I had observed but failed to talk about rise time as a factor for display results. Then Venarim in his first post stated that he didn’t believe that I saw that. And then the thread just rambled on. It is my intention that this post will set the record straight and also explain the reason for differing observations between Venarim and myself.
It also might be pointed out here in the bandwidth discussion, that if the DSO Quad design had a hardware trigger circuit (and it does not) then for repeating waveforms, the firmware could be designed for equivalent time sampling with a magnitude order increase in the maximum observable waveform frequency. That would have made a tremendous improvement to the DSO Quad, but that approach was never taken during the design leap from the Nano to the Quad.
hi, i promise my hardware is the same as the one in your hand.
i say the "10M"bandwidth i means the input(connector) to the ADC input(U5 and U16 pin13), i say this just want to clarify confusion on the problem(is the input channel or the sample rate the limitation of the system bandwidth ).so ,we can say: the input analog channel bandwidth is 10M.
i say the "5M"bandwidth i means the system bandwidth. i input a signal of 5M (sinewave),and found the wave on the Quad screen is less than -3db. and if a signal of 10M(sine) , the wave is more the -6db.that is : a 10M signal can pass the input channle(gain less than -3db),while it came into the ADC ,because of the sample rate,the whole gain became more the -6db, so,we can not say the bandwidth is 10M.instead,just about 5m.
sorry, but you’re mistaken here. The n-th armonic of a wave is defined as: f_n-th = f * n-th, where f is the fundamental frequency of the wave and n-th is the harmonic. So for a 1kHz square wave, its 5th armonic will be 5 kHz. And zero harmonic is the DC component. It’s true that perfectly square waves doesn’t have even harmonics (and you counted to the n-th position of a odd harmonic), but that’s another tale.
I would say not only, but at least. Difference is significant.
A perfectly square wave has infinite odd harmonics, and hence can’t be perfectly sampled (also produced). With the corrected calculations you will require 30 Msps (15 Mhz x 2). With the arguments you gave in the previous messages… why not sampling 13th harmonic with 20x? So now its enough with 2x?. You are mixing theory and practise, and that’s quite difficult without a solid ground. That was the point to stress.
Although wikipedia says exactly what you posted before, in reality the signal has to be sampled with at least twice its bandwidth to be perfectly reconstructed.
I don’t know how you know this since FPGA is closed source . But i can bet my mother (only speaking, don’t take me too serious ) that trigger is hardware (obviously is not an analog one, but digital). So QUAD can use equivalent time sampling for sure.
I’d like this thread not to diverge into sampling theory, because its large and complex. So for me, the theory discussion is over.
@HugeMan: I will check the analog bandwidth by open my Quad, as soon I have spare time to do it.
@lygra: I appreciate your effort to demonstrate (better: trying to do it) with tons of theories, but I guess that you are mixing many concepts together.
A square wave is yes composed by odd harmonics (I agree with slimfish about the numbering), but the shapes you drawn could never be what the analog section outputs. You missed a VERY important parameter: the phase!
Remembers that here, for simplicity, we are talking about the modulus of a signal (i.e. a square), but we MUST take in account the phase-shift caused by the analog section. We “should” take in account the phase-shift introduced by the sampling also, but…never mind.
Just take a simple square-wave generator and a R-C lowpass filter. Assuming the time-constant similar to the period, what do you expect to see on a scope?
The input is a square-wave, composed of an infinite number of odd-harmonics. The output cuts them at -6db/oct.
That would be having an output shaped as you drawn?..Yes, if only the filter won’t shift the phase, smoothly from 0 to -90 degrees.
Of course, by scoping the output signal you will see an exponential fragment, repeated on each period. Exactly the same as having a switch charging then discharging the capacitor, via the resistor.
All that only for a simple R-C filter, being just-one-pole complex function. Imagine what would be having a 10+ poles network!
I never talked about the fifth harmonic, I only referred to the 5th odd harmonic, not the same. If the first harmonic is even (2Mhz) then the pattern after the fundamental would be 2nd=even, 3rd=odd,4th=even,5th=odd; so the fifth harmonic would be odd and is 5Mhz, but I was referring to the second odd harmonic (which is the actually the 4th harmonic if you count the none existent even harmonics too) when I said 5x, so once again you are mistaken. As you have already said, the even harmonics don’t exist, so they can not be counted.
See previous response.
My reference was not Wikipedia, my reference was a professional application note produced by Tektronix engineers. If you would bother to read this reference then you may become more informed. As you have already said, the odd harmonics are part of the square wave, and hence the sample rate must be twice the highest odd harmonic expected, unless Nyquist is wrong and you are right.
This is a fool’s bet for someone who doesn’t know the FPGA source code. But that FPGA source code may not be necessary if you look at the STM source code and find it’s triggers source.
I believe that it was you that started with the equivalent time sampling theory, not myself. Sampling theory is not that complex and is explained quite well in the Tektronix reference that I provided in my earlier post. If you look at the Nano post viewtopic.php?f=12&t=1793&start=80 on page 9, BenF fully explains in his May 8th post just exactly how the Nano firmware trigger detection works without a hardware trigger. That discussion is also not very complex. You can also read my Tektronix reference, that also explain exactly how equivalent time sampling works. So I don’t see anything here that is overly large and complex.
The only thing I find complex is trying to determine a hardware trigger within the FPGA, without first looking at the STM source code (which is available) to see if a software trigger is present. An even simpler approach would be to point out the Quad menu choice which provides enabling of this fictional equivalent sampling mode. If the Quad were in continuous equivalent sampling mode, then you could not look at non-repeating signals, but guess what; you can look at non-repeating signals with the Quad. Therefore if you assertion is true, then there must be a Quad menu choice to switch back and forth between equivalent sampling mode and real-time sampling mode. Please be so kind as to point out this menu choice, or stop wasting everyone’s time with your non-validated theories.
What a relief that is.
Look, my goal here is not to constantly criticize you, my only goal is to provide factual enlightenment for those looking for same. If you agree to stop spewing theory, then I will agree not to be critical in return. Do some research on your own as I have done, and present those research findings that support your claims.
Have you measured this phase shift when a 1Mhz square wave passes through the Quad front-end circuits? If so, how much phase shift did you measure for the fundamental, 1st odd harmonic and the 2nd odd harmonic?
When I present a theory, I always use a “maybe” or “might be” during that theory. The rest is not theory but common knowledge of the trade. If you haven’t measured the above phase shift, then it is yourself that is now presenting not tons but ounces of theory.
It requires no degree in mathematics or foundation courses in laplacian and fourier calculations and transforms for calculating the nth harmonic of a fundamental , buffer sizes, sampling rates, time divs, ADCs, unity gain op-amps, low impedance signal path, parasitics and finally disembowelling my quad to fiddle with its innards.
My question is simple.
Have we been mis sold a device that, from what I can see and extrapolate from all the theory, is only good for about 5Meg bandwidth when is said quite clearly 27Meg when it was sold to me.
This is not half, not a quarter, not even one fifth of what I paid good money for.
My question is how are Seeedstudio going to address this issue of mis selling and then saying nothing to their customers apart from …
Well this is my comment and I leave it to you to decide on the action
Ohhh and BTW before the flaming starts … this is not a serous post I just wanted to lighten the mood a little
To be honest firmware issue aside i quite like it and once the niggles are ironed out and my confidence in its abilities is a little higher it will have pride of place on my hobby bench.
If you want to count only odd armonics to fade your mistake is up to you. But fifth armonic of a 1 kHz wave is 5 kHz. And venarim was refering to that.
Your reference is a manual (a.k.a. tutorial - simplify things) which by the way i’ve read. Also i’ve read more than “a couple” of books that support what i’m saying. Even i built a radio (which works by the way which samples @ 1.5 times the carrier frequency. I suppose at this time our world as we know it is collapsing . And by the way, Nyquist is STILL correct (because the signal bandwidth is 3 kHz).
A fool’s bet. Yeah. I did look into STM the code, did you? Of course not, because you are wrong again. Look @ BIOS.h, process.c (__set function) and you will understand (i hope so). And again, what would be the purpose of sampling at 72 Msps when you can not trigger with a precision of more than 1us (in case internal ADC would be used)?.
Of course, the sampling theory is very simple. Read a manual and that’s it. Shannon and Nyquist would commited suicide if they could read you.
I’ve never said that equivalent time sampling is implemented in QUAD. I said it is POSSIBLE for sure. Excuse me but you know we have to wait until Benf took the source code disaster of DSO Nano and made a pretty decent firmware… not expected for seeed programmers to evolve so fast.
Huh, that’s funny. I’m a professional researcher/university teacher (+20 years in real electronic design). I’m proud of being always open to criticism. As Bainesbunch have said, this have gone too far. I’m glad you finally have some relief.
The story is always funnier than expected.
I have done some measurements right now.
As HugeMan/lygra suggested, I feed the ch-A input (without any probe) directly with the signal generator. BTW: the instrumentation is still exactly the same.
I checked the signal either on the Quad display and with the LeCroy scope on U5 pin13.
The amplitude of the input was constant at 2Vpp.
Sine wave test.
A manual sweep from 1 to 15MHz (step by 1MHz) clearly shown that the band is NOT constant (ref U5 pin 13).
Assuming 100% at 1MHz, the amp will decrease briefly to 50% (-6dB) at about 3MHz. Still remain at 50% until you reach about 7-8MHz, then raises to about 75% around 10MHz. Hereinafter will raise even more: at 15MHz (the max I can test) is about 200%.
So, at 10MHz the attenuation is about 75% (i.e. -3dB)…that’s the way HugeMan says the bandwidth is 10MHz…but FOR ME the bandwidth has to be taken as the VERY FIRST POINT where the attenuation falls to -3dB…once again the point is around 2MHz.
Square wave test.
Just a simpler test than before: always scoping the pin 13 U5.
At 1MHz the wave has the rising edge clearly rounded. The displayed wave on Quad looks pretty the same as the LeCroy’s one.
Going higher with the freq, 3MHz, the square is almost a triangle/sine, with a well-visible spike just before the falling edge.
Above 4-5MHz the wave on the LeCroy scope looks as sine.
The spike is clearly due to the unexpected amplitude above 10MHz.
the analog section has a bandwidth of 2-3MHz;
the “supposed” bandpass is very far to be flat (as should be);
the analog switching strategy is faulty: if the signal cannot be shown as it is, the Quad should constrain the user’s selection;
as the current analog section hardware, this Quad is good for signals under 100-200KHz.
I will leave my Quad in the lab, if someone of you is asking for any additional test.
PS: suggestion to HugeMan.
I would think about an external box, embedding a good analog section, that can feed a good bandpass and a reliable way to display signals.
I am not so happy about this toy.
Thanks for providing these very necessary measurements.
Neither you nor HugeMan have specified which range scale you are using for these measurements. As long as the range scale is not changed, then I would suspect that calibration of that range scale would not be a factor. On the other hand, the probe compensation adjustments for that channel could have very dramatic affect.
U5 pin “Y”, is that what you are measuring? The pin numbering seems to be inconsistent on the schematic. It appears that U5 “Y” is the input to the op-amp U-7, and U5 “X” is the output of the op-amp U-7. If so, then U5 “X” will be the low impedence signal into the ADC, and that is where the measurement should take place.
I am still awaiting my replacement Quad. Maybe you could upgrade to the latest firmware, conduct the new probe compensation procedures provided by HugeMan, and then repeat your test to see if the results are different. Hopefully the probe compensation procedure will help to flatten this bandwidth. It may be robbing Peter (2-3Mhz) to pay Paul (15Mhz). If this is true then the bandwidth will probably end up being more than 15Mhz.