01-01-2014 03:35 AM
Hello,
I've been reading 4 analog channels at 62,5 kHz rate by NI 6221 instrument.
One of them I want to use to calculate the speed of the voltage change and to determine the accuracy of the speed.
From the absolute accuracy table and many info in the NI web/forum I performed the voltage accuracy already.
But I came in trouble on how to determine the time accuracy in order to find out accuracu of every single dV/dt ?
From the instrument specification I have:
Timing accuracy ......................... 50 ppm of sample rate
Timing resolution ....................... 50 ns
My application wrote the sampling time increment as wf_increment=0.000016s
The easiest way would be to do the calculation:
Timing accuracy*wf_increment=50 ppm*0.000016s=0.8ns
But it is hard to believe the accuracy is so fine.
How to find out the standard uncertainty from all those puzzles?
Regards,
Przemyslaw
01-08-2014 07:49 AM
Szanowny Panie Przemysławie,
temat rozwiązaliśmy w dyskusji bezpośredniej, ale wkleję informację dla potomnych. Podsumowując, należy wziąć za wzór dokument ze strony (http://digital.ni.com/public.nsf/allkb/251AF75D1578F957862576A5007810FE) i przyjąć niedokładność zegara w PPM razy częstotliwość próbkowania jako niepewność pomiaru czasu. Inżyniersko ujmując będzie to:
50 ppm of sample rate - wzięte ze specyfikacji, pomnożone przez 10 kHz
50*10*1000/1000000 = 0,5 Hz
Idąc dalej:
1/10 000,5=9,99950002499875e-5 minus 1/10 000, ostatecznie jest to błąd na poziomie około 0,005 us.
Reasumując, jest to błąd raczej pomijalny w porównaniu do błędu pomiaru napięcia.
01-08-2014 05:24 PM
@n_dakota: This question could be of interest to me and other users: would you mind to translate your answer in English?
01-09-2014 01:32 AM
Dear @RobertoBozzolo,
You are right, sorry for that:
We discussed with Przemysław how the uncertainty of time in meassurement should be counted. He got different results, which were based on different assumptions. Final answer is:
1. As a reference point we should take this document: http://digital.ni.com/public.nsf/allkb/251AF75D1578F957862576A5007810FE
2. We should find uncertainty of clock in PPM and multiply it with sampling frequency.
3. In numbers it would look like here:
50 ppm of sample rate - specified for module, multiplied by 10 kHz
50*10*1000/1000000 = 0,5 Hz
Going further:
1/10 000,5=9,99950002499875e-5 and at the end we decuct 1/10 000, finally the result is 0,005 us.
This result is low enough to assume that it could be skipped if we would like to compare it with voltage meassurement uncertainty.
Hope it helps and have a great day!
Tomasz Kachnic
Certified LabVIEW Architect (CLA)
01-09-2014 01:45 AM
OK thanks for that.
It means, if I'm not wrong, that actual period between samples can be 100 µs ± 0.005, right?
Now I understand how to compute it and how to explain to my customers another aspect regarding the accuracy of measuring system.
Now, this applies to accuracy of AI sampling rate and I have two more questions:
01-10-2014 07:16 AM - edited 01-10-2014 07:17 AM
Dear @Roberto Bozzolo,
Below please find my answers to your questions:
It means, if I'm not wrong, that actual period between samples can be 100 µs ± 0.005, right?
Yes.
1. does it applies to AO? That is, is the clock for analog generation the same used by measurement?
AE has different clock with different uncertainty. Looking at it you can see that here we also should be aware of channel latency: http://digital.ni.com/public.nsf/allkb/FA4C741619B95082862568F10072768E
2. will a similar method apply to accuracy of the counters in general (i.e. when used to measure frequency or set pulse width and so on)?
If it is about frequency meassurements, you can find a little more about it here: http://www.ni.com/white-paper/3619/en/
Have a great weekend!
Tomasz Kachnic
Certified LabVIEW Architect (CLA)