How to compute RMS of three-phase voltage

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Win co
Win co on 4 Apr 2012
Hi, I'm working on one problem of prognostic of machinery failure. I want to use RMS of three-phase voltage as parameter for prediction. I would like to know how to compute the RMS of voltage of a 3-phase motor ? In fact, I have a 1000 samples for each voltage phase ( ) and I found a formula to compute RMS from this reference (Equation 12 or Equation 14) But according to their formula, RMS is a constant finally, I think I misunderstood sth but I don't know exactly what it is. Could somebody explain this computation please? Ragards, Winn

Answers (2)

Sean on 4 Apr 2012
I looked at the article you referenced. They appear to be feeding four parameters into a neural network: voltage@f1, current@f1, voltage@f2, and current@f2 (where f1 is the 'natural lag frequency' of their 'resolver', and f2 is #poles*freq/2).
I know you probably are already doing this, but it bears checking: Make sure that you are using the correct voltage differential for your calculations that is appropriate for the 3-phase configuration you are using (e.g. delta or star). The last time I touched on this stuff was ages ago in school, so I won't try to get to much into details that I'd probably get at least partly wrong.
I also noticed that your data is a 1000x3 matrix. From the paper you referenced, I would have expected it to be 1000x6 (3 voltages and 3 currrents). It also wasn't clear from your data what the sampling rate was used. The paper you referenced was sampling at 100kHz, but it seems like your data was probably sampled at something much much lower (which would prevent you from distinguishing the frequencies specified in the paper).
TO SUMMARIZE: 1) Make sure you are sampling at a high enough rate for your metric of interest. 2) Make sure that you are measuring everything you need to calculate your metric your metric. 3) Make sure you are interpretting your data correctly (i.e. correct voltage differential/reference, correct filters for your sampling rate, etc.)
Good luck, good cheer, god bless, and may success smile on your efforts.

Win co
Win co on 4 Apr 2012
thanks for your answer. I didn't show my data of 3 currents because the computation of RMS for U and for I is independent but their formula is the same. Besides I'm just inspired the method of prognostic in the reference, it's why my data are not the same as theirs. I will follow your advice and post my result soon. Ragards, Winn

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