RMS - quadratic mean vs. AC waveform

[juz_ad]

Hoping some can explain this to me…

To get RMS/quadratic mean of a particular data set I would use:

sqrt((a^2+b^2+c^2…)/number of values in the set) = RMS. e.g. for a data set of (3, 6, 9, 12) I would have sqrt((3^2+6^2+9^2+12^2)/4) = 8.216 (3 dp).

For an AC waveform though - let's assume a single cycle of a sine at 2.8V Peak, I have to use sqrt(((peak/2)^2)/2) = RMS e.g. sqrt(((2.8/2)^2)/2) = 0.989 (3 dp) which is approximately equal to 1V RMS.

Why is that? If the 'data set' of the AC waveform is only one measurement (2.8V Peak) - why do I need to divide it by 2 before squaring and then divide it by 2 as though there were 2 values in the data set?

Thanks!

 

[albbg]

Have a look to https://en.wikipedia.org/wiki/Root_mean_square

 

[KlausST]

Hi,

Your RMS formula is wrong. It is for peak_to_peak values.

Single peak = peak_to_peak / 2.

Hope this helps.

Klaus