I recorded this asymmetric rectangle function:


I calculated the RMS voltage to be equal to the amplitude (4,10 V in this case)

However when I measured this exact signal with a TrueRMS multimeter, it showed 3,54 V.

Where is my error?

  • \$\begingroup\$ "MAX" is maximum and not RMS. The devil's in the small detail. \$\endgroup\$
    – Andy aka
    Apr 22 '21 at 7:18
  • \$\begingroup\$ @Andyaka The oscilloscope is not directly measuring the rms with the Max-function, thats correct. However I did the calculation and for a bipolar rectangle signal, the rms is equal to the amplitude. \$\endgroup\$ Apr 22 '21 at 11:47
  • \$\begingroup\$ You need to detail how you made that calculation numerically. Your scope shot implies an RMS value of 4 volts (not 4.1 volts). \$\endgroup\$
    – Andy aka
    Apr 22 '21 at 11:53
  • \$\begingroup\$ @Andyaka I will add my calculations to the question when I'm home. Wikipedia also lists an rms value of 1 for square waves, confirming my calculation. en.wikipedia.org/wiki/Crest_factor#Examples \$\endgroup\$ Apr 22 '21 at 14:06
  • \$\begingroup\$ You shouldn't use 4.1 volts as the effect peak signal. You should use 4 volts. \$\endgroup\$
    – Andy aka
    Apr 22 '21 at 14:23

Let's accept your peak amplitudes are +/- 4.1 V.

The true RMS of that waveform is exactly 4.1 V, it has the same heating effect as 4.1 V DC.

The mean DC of your signal is 2.05 V. If you remove the DC, you end up with a waveform with peaks of +6.15 V and -2.05 V. The RMS of this signal is 3.55 V.

I wonder if your meter is only measuring the AC component?

  • \$\begingroup\$ I had a look at the multimeters manual (MetraHit 2+) and it does not explicitly state wether it can only measure the AC component. Is it typical that multimeters can only measure the AC Component in AC-mode? \$\endgroup\$ Apr 22 '21 at 14:00
  • \$\begingroup\$ +1 Actually I think this is the most likely answer- the meter is AC-coupled so it has no knowledge of the DC component. \$\endgroup\$ Apr 22 '21 at 14:14
  • \$\begingroup\$ I don't have the exact multimeter on hand that I used for the measurement, but I just tested this with another multimeter and indeed, the AC mode seems to ignore the DC-component. Question solved, thanks everyone. \$\endgroup\$ Apr 22 '21 at 14:30

Well, assuming your wave is symmetric (that is, the positive as well as the negative part have the same magnitude), your calculation is right.

You don't know what that "TrueRMS" multimeter does, or what it does expect in a signal. Quite possibly it expects a signal composed of harmonic signals below some relatively low frequency (i.e. the meter has a bandwidth), whereas a square wave has infinite support in spectrum - all power above the point at which the meter's bandwidth ends is ignored; you get less power, and the square of the average power is thus also lower.

Nothing to see here, really - unspecified measurement devices doing unspecified things.


Regardless of duty cycle the power that can be dissipated into a resistive load for V+= |V-| the Vrms=Vpk.

But in your case if you applied +/- 4V and with noise on your measurement, the scope will read V+ + Vnoise = 4.1 , so eyeballing it, that appears to be closer to 4V or the average of each peak.

Putting the 20MHz filter on the noise may improve the scope reading on this low f signal.

Test your meter at different frequencies and check the manual specifications.


A rectangular wave has high frequency components that extend up to very high frequencies.

You have a fundamental (250Hz) that is pretty high in relation to mains to start with.

Typically cheap true-RMS meters don't bother to tell you that the frequency limit is for sinusoidal waves and non-sinusoidal will have less accuracy. A relatively good meter such as the Fluke 179 lists ACV accuracy as +/-4% + 3 digits up to 500Hz.

A cheaper meter such as the Extech EX411A lists ACV bandwidth as up to 1kHz and claims 1% accuracy (no mention of waveform).

You're seeing an error of -21%, which is large but plausible for an inexpensive meter.

Check the reading at lower frequency (such as 50Hz) and see if it is closer.

Edit: I think Neil's answer is correct- the meter is AC coupled and ignores the DC component. Easy to test, input a DC voltage and if it reads zero you have your answer.

  • 1
    \$\begingroup\$ I checked the multimeters manual and indeed it only has a bandwith of up to 1kHz. (it's a "MetraHit 2+") \$\endgroup\$ Apr 22 '21 at 13:57

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