RMS values

  • Thread starter derek181
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Could someone explain to me the intuition behind RMS values? I understand how you calculate them; you take the average of the squares and square root them. I am just wondering why. For example when it comes to voltage, how did someone just think up of this magical way to compute an effective value that is equivalent to a DC value? I hope this question makes sense.
 

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  • #2
analogdesign
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The main thing is you can't just take an arithmentic average of a sinewave to find its mean, because the average value of an ideal sine is 0! Now how would you find the "average" of a sinewave? You need to be able to compare sinewave amplitudes because a 100 V sinewave has more strength than a 1 V sinewave, but the average is both is zero! So what do you do?

So you have to square it to make sure the average is > 1. (thats the square part) Then you average it. (that's a mean). But now the units are wrong, so you need to take the square root (that's the root part). So there you have it. Not too bad, is it?
 
  • #3
CWatters
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Try this..
http://www.mei.org.uk/files/Industry/Resources/MEIRMSValuesStudentTrial.pdf

It's to do with the equivalent power. The problem is that power is proportional to V2 not V (eg P = V2/R). Basically you are in effect calculating the instantaneous power (squaring), then working out the average (integral), then convert that back to an equivalent voltage (square root).
 
  • #4
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In some cases I have to compare motion measurements on stuctures subjected to irregular loads with simulated data. Because the response is very irregular and fuzzy I can only compare statistical values. The RMS value is such a statistical value, and is the first thing I usually will compare.
 
  • #5
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So because P is proportional to the voltage squared we take the integral of the voltage squared with respect to time and then square root it at the end to just get the voltage and the basis of this is off of the fact that we need to find equivalent heating effects between ac and dc? Also what is this talk about the RMS being a statistical calculation?
 
  • #6
sophiecentaur
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So because P is proportional to the voltage squared we take the integral of the voltage squared with respect to time and then square root it at the end to just get the voltage and the basis of this is off of the fact that we need to find equivalent heating effects between ac and dc? Also what is this talk about the RMS being a statistical calculation?
One way of looking at is could be that it's just another example where we use the same Maths to deal with two different aspects of life. You could also think in terms of a situation where the Power is delivered at random by a series of ' rectangular spikes'. The Mean Power arriving would be sum of the squares of the spike voltages arriving per second and the equivalent DC voltage would be the RMS value. From a statistical point of view, this standard deviation of the Voltage values would be the same as the RMS value (same calculation, same numerical answer)
 

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