Imagine you have a periodic signal F which is nothing but a sinewave (S) plus some DC offset (D).

Here is an illustration of such signal:

enter image description here

Imagine I want to measure the RMS value of this signal F.

The rms value of this signal mathematically can be written as:

rms_of_F = sqrt[ (rms_of_S)^2 + D^2 ]

So far so good.

First of all, if I use the DC settings of a voltmeter I would obviously not read the rms value.

Then I may use AC settings.. But recently I came to a question: What value does ammeter or voltmeter measures (RMS, Average or Peak )? where the answer is chosen as follows:

Measuring RMS values is a bit more expensive than measuring average values, so most multimeters avoid the former. Instead they presume your signal is a sine and measure the average value for the rectified sine or the peak value, after which they apply a conversion factor to find the presumed RMS value.

Does that mean the rms value of my signal F can never be mesured accurately by using a voltmeter?

I'm asking because in case the voltmeter in AC settings doesn't measure the rms value: if one calibrates an instrument and relate the rms value of such signal output to the device reading(any parameter like temperature, flow ect.); and if another person would use a multimeter/voltmeter to use the calibration expression, this could lead him a wrong conclusion because he will not read the true rms voltage but something else. Is that right?

  • 3
    \$\begingroup\$ This is true for "cheap" voltmeters. True RMS meters can be bought which give you the correct RMS for any measured wave. \$\endgroup\$
    – Adam Z
    Commented Jun 10, 2016 at 9:39
  • \$\begingroup\$ Added to the accuracy problem is the frequency - a cheap meter will be good for 50/60 Hz sinewaves but if the frequency is in the hundreds of Hz or above then it will progressively worsen its accuracy. \$\endgroup\$
    – Andy aka
    Commented Jun 10, 2016 at 9:39
  • \$\begingroup\$ @Andyaka Thanks for additional info. But I wanted to be sure what I wrote and conclusion is right. Do you agree the concerns and info in the question? \$\endgroup\$
    – user16307
    Commented Jun 10, 2016 at 9:53

1 Answer 1


You can easily measure the RMS value of a sine wave + offset using a cheap multimeter.

Most cheap (non-RMS-reading) multimeters are AC-coupled on the AC voltage ranges, so you would read the correct RMS value of the ripple only on an AC range, assuming the ripple is sinusoidal, due to the correction factor which is applied. If you happen to have one that's not AC coupled, just stick a large value film cap in series (such that Xc << input resistance of your meter).

Reading the voltage on a DC-volts range will give you the correct average value of the total signal. That also happens to be the RMS value of the offset, assuming a sinusoidal ripple.

The RMS value of the total (refer to this derivation if you need proof) is just the quadrature sum of the two components:

\$V_{RMS} = \sqrt{V_{DC}^2 + V_{AC}^2}\$

In your example, you would read 10.00V on the DC range and 0.707V on the AC range, so the RMS value would be 10.03V.

  • \$\begingroup\$ So basically the AC settings calculate Rms of the ripple which is the standard deviation around the mean? \$\endgroup\$
    – user16307
    Commented Jun 10, 2016 at 12:45
  • \$\begingroup\$ The meter displays a value equal to the RMS value of the ripple, provided it's a sine wave- because what is actually displayed is 1.11 * |Vin - Vav|, averaged. 1.11 is pi/(2*sqrt(2)). It does not actually calculate the RMS value. \$\endgroup\$ Commented Jun 10, 2016 at 12:58

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