The RMS value is a mathematical calculation to determine the power of a voltage/current dissipated in a resistive load, regardless of the wave form.
The basis of the root-mean-square is the area under your wave form.
To obtain the RMS value, you need to integrate the wave form over a half cycle to calculate the area. This is what your Fluke will do for you.
For instance, the RMS value of a 50% duty cycle square wave will be equal to the peak voltage. You're somewhere in between a sine and a square wave.
This is very wrong. RMS is not related to area (integral). Integrating then dividing by time can give average, but not RMS.
RMS is literally the square-Root of the Mean(average) of the Square of the input.
You missed the point about power into a resistive load. RMS is equal to the DC voltage/current that will give the same dissipation in the load.
You were absolutely correct on that point. I should have noted that, instead of implying that your entire statement was inaccurate.
The integral is the mean, which you'll need to compute first for an arbitrary wave form, the root/square part is obvious.
The key is that you do not compute the
mean of the voltage waveform (nor integrate voltage waveform). Instead, you compute the
mean of the Voltage squared. The 'squared' part is critical since power is not proportional to voltage, but to voltage squared. So,
first square the voltage,
then compute mean/average, then take square root.
Consider two signals: one is a DC voltage of 1 VDC. The other is a signal that toggles on/off at some rate, with 50 % duty cycle, and a peak voltage of 2 V (off voltage 0 V). If you integrate both for 1 second, you get 1 V*second. The area under the voltage curve is the same for either, which also implies that the average voltage for either is 1 V. This in intuitive.
The RMS voltage of the 1 VDC signal is 1 Vrms.
The RMS voltage of the 50% positive pulse is 1.4142 Vrms. It will drive twice the power into a resistive load (P=V
2/R). This is not intuitive, but consider that a 2 VDC signal will drive 4x as much power than 1 VDC, by P=V
2/R. Now consider that at 50% duty cycle, you halve the power to only 2x as much as 1 VDC. Manipulating P=V
2/R for a constant R and two voltages V
1 and V
2, we can get P
2/P
1=V
22/V
12. With P
2/P
1 = 2, then V
2 = V
1*1.4142.
The "area under the voltage curve" idea gives average voltage but not RMS.
For more technical information on the non-trivial nature of measuring and calculating RMS voltage, I'd encourage you to read:
https://www.analog.com/media/en/technical-documentation/application-notes/AN-268.pdfand
https://www.analog.com/en/education/education-library/rms-to-dc-application-guide.htmlA side note:
All digital voltmeters (or ADCs) which use a dual slope or multi-slope technique do in fact actually integrate the input voltage over some period of time (usually called the "integration time" for obvious reasons) then de-integrate with a known fixed reference, back to zero, in order to measure the integrated charge. The de-integration time is proportional to the input voltage. This is results in an average voltage over the integration period, and provides effective noise reduction on a DC input, or gives the average level of a (rectified) AC input voltage.