I'm new to electronics and I have a question about oscilloscope measurements. I use a Keysight InfiniiVision 2000X oscilloscope in the lab but it doesn't really matter because my question is quite general.

As I pressed the measurements button, I noticed two measurement options that seemed quite ambiguous to me- "AC RMS" and "DC RMS".

I'm already acquainted with the formula to calculate the rms value of a waveform:

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If I go by simple logic, DC RMS calculates the RMS value of a signal as it is and AC RMS eliminates the DC component of a signal and then calculates the RMS value with the above equation.

My question is whether my assumptions make sense. If they do, I would like to know whether there is more to it.

In addition, if my assumptions are right, can AC RMS be useful at times when I would like to measure the amplitude of a sine wave signal for a linear circuit (deriving the amplitude from RMS value of a sine wave is an easy direct calculation) or is the standard Vpp measurment the most accurate method of determining the amplitude of the input/output signal?

  • 1
    \$\begingroup\$ Vpp and AC rms measure different things, so neither is 'most accurate'. It depends what you want to measure. Signal power, then AC rms. Peak to peak amplitude, then Vpp. If you know the input waveform accurately, be it sinewave, square wave, triangle wave, then you can compute one from the other. \$\endgroup\$
    – Neil_UK
    Commented Jul 26, 2022 at 8:41
  • \$\begingroup\$ @Neil_UK, im pretty sure "DC RMS" and "Vpp" are two different measurements on the scope. He is not refering to a Vpp measurement. \$\endgroup\$
    – Linkyyy
    Commented Jul 26, 2022 at 8:45
  • 1
    \$\begingroup\$ @Linkyyy I agree with your first sentence. Check out the last two lines of his question or is the standard Vpp measurment the most accurate method of determining the amplitude of the input/output signal? for my comment on your second sentence. \$\endgroup\$
    – Neil_UK
    Commented Jul 26, 2022 at 8:56
  • \$\begingroup\$ @Neil_UK, youre right, i missed that. \$\endgroup\$
    – Linkyyy
    Commented Jul 26, 2022 at 9:07
  • \$\begingroup\$ Every "signal" is characterized by different "measurements". No one is false, no one is true. All have their own utility. \$\endgroup\$
    – Antonio51
    Commented Jul 26, 2022 at 9:09

2 Answers 2


TLDR: trust you (cheap) oscilloscope interpretation with suspicion.
Here's an example from my oscilloscope.
A sine wave was measured with a DC offset. Cursors were set on top peak and bottom peak of the sine wave. Cursor measurement:

  • top peak: +10.6V
  • bottom peak: +0.48V

Calculating average value (DC voltage) +5.54V, with peak-to-peak voltage 10.12V RMS of the sinusoidal component is \$10.12\over{2\sqrt{2}}\$=3.58V
When the oscilloscope channel was switched from DC-coupled to AC-coupled the DC offset disappeared, and the internal CyclicRMS measurement display was 3.52V, similar to my eyeball measurement of 3.58V.

Calculating true RMS where the individual contributors of the wave are separated out: \$ \sqrt{5.54^2+({10.12\over{2\sqrt2}})^2}\$=6.595V. This would perhaps correspond to OP's term "DC RMS"

My oscilloscope has a "measure" function which does interpretations of the samples stored in its waveform buffer... oscilloscope waveform sinewave with dc offset

Note that the oscilloscope's interpretation differ from my own eyeball mesurements, using the cursors:

  • peak-to-peak voltage 10.2V versus my 10.12V
  • DC offset (mean) +5.68V versus my +5.54V. A BIG error perhaps caused by the displayed wave starting and ending at a different voltage.
  • Maximum peak +10.2V versus my 10.1V
  • Minimum peak +0.4V versus my +0.48V

Using these Measure values, RMS voltage is \$ \sqrt{5.68^2+({10.2\over{2\sqrt2}})^2}\$=6.728V, similar to the display of 6.72 for CyclicRMS So it would seem that any DC offset is included in RMS measurement.

Why be suspicious of internal measurements?
Internal DC offsets are often a contributor to error. Scale offsets too. Typically, a cheap oscilloscope measurements have 3% error, and that's only valid soon after a calibration.

Noise is included in measurements. Notice that the measure function over-estimated peak-to-peak voltage (10.2V versus my 10.12V). If even one noisy sample in the buffer is above the smooth sinusoidal peak, it is taken as the "top peak". Similarly, noise peaks extend the minimum sinusoidal peak. Oscilloscopes very often over-estimate peak-to-peak amplitude. Your eyeball can do a better job of estimating noisy peaks, averaging out noise.
Similarly, overshoot on a square wave causes over-estimation of peak-to-peak measurements.

When timescale is changed so that a fraction of a wave is displayed, RMS measurements of repetitive waves become invalid and should indicate a suspect RMS measurement. My oscilloscope does not, and dutifully displays a bogus (too small) RMS value.

To make an RMS measurement of a sine wave, I would suggest using cursors (if available) to estimate top-peak and bottom peak, yielding a peak-to-peak voltage difference. Any DC offset should affect both, and its error disappears. Any scale error still affects your result. Then calculate RMS from that...
You must also judge if the wave is truly sinusoidal. An experienced eyeball might detect 5% distortion.


If you want to measure RMS ripple/noise on a power supply, or the RMS value of an AC voltage with a DC offset, then the DC offset needs to be removed before doing the RMS calculation. If a power supply outputs 12VDC + 10mV RMS ripple, then measuring the RMS voltage without removing the DC component first would simply result in 12V RMS.

I'm not sure why there is an "AC RMS" mode though, as setting the scope input to AC then using the usual RMS mode should do the same. Perhaps the manual contains useful details about this.


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