As we know the mean value of a pulse wave (depending on its duty cycle) corresponds to a constant DC.

But when it comes to alternating current through a resistor for DC power correspondance we measure the RMS value i.e. RMS of an alternating current becomes such as the current which lights a bulb as the same intensity as if we have a DC value of the same with that RMS.

I measure pulse wave voltage with Voltmeter's DC settings. I think it gives the mean value.

So my question is if we have a pulse wave does it make sense anymore to talk about its RMS value?

  • \$\begingroup\$ A true-RMS multimeter will give RMS readings. Other multimeters use "averaging" to approximate this - these are inaccurate for non-sinusoidal waveforms. True-RMS multimeters are usually a little more expensive. \$\endgroup\$ Commented Nov 26, 2013 at 11:13

4 Answers 4


So my question is if we have a pulse wave does it make sense anymore to talk about its RMS value?

Yes, of course it does; the rms value of the pulse wave is the effective DC voltage across a resistor that gives the same average power.

Recall that the instantaneous power associated with a resistor is

$$p_R(t) = \dfrac{v^2_R(t)}{R} $$

The average power, over a period \$T\$, is then

$$p_{avg} = \dfrac{1}{T} \int_0^Tp_R(t)\,dt =\dfrac{1}{T} \int_0^T\dfrac{v^2_R(t)}{R}\,dt$$

Thus, the equivalent DC voltage that produces the same average power is

$$V_{eq} = \sqrt{p_{avg}\cdot R} = \sqrt{\dfrac{1}{T} \int_0^Tv^2_R(t)\,dt}$$

But, that last term is precisely the root of the mean of the square (rms) value of \$v_R(t)\$.

So, yes, it makes sense to talk about the rms value of a pulse waveform or any other voltage or current waveform for that matter.

  • \$\begingroup\$ but when controlling dc motor with pulse width modulation. they use duty cycle and find out the mean value not rms. why so? \$\endgroup\$
    – user16307
    Commented Nov 26, 2013 at 14:28
  • \$\begingroup\$ aha so from your derivation it is exactly like in ac sinusoidal case. when it comes to e.g a bulb intensity equivalent we should definitely find rms equivalent then? not mean value. i think u explained it very clear! \$\endgroup\$
    – user16307
    Commented Nov 26, 2013 at 14:36
  • \$\begingroup\$ so when we apply a pwm to a dc motor and obtain a particular speed(rpm), and later if we want to obtain the same speed with a constant dc should we first find rms voltage of that pwm instead of mean value of voltage? is that right? \$\endgroup\$
    – user16307
    Commented Nov 26, 2013 at 14:37
  • \$\begingroup\$ oh i think im wrong.in dc motor case integration of voltages must be equal not integration of powers as in light bulb case. so i think in dc motor control what matters is mean value of pwm to find its constant dc equivalent?.. \$\endgroup\$
    – user16307
    Commented Nov 26, 2013 at 14:52
  • 1
    \$\begingroup\$ That's what I thought. I am asking this because in a peer reviewed paper I am reading the authors use Vrms*Irms to compute the efficiency of a switching amplifier, although there are many harmonics and the laod is complex. Perhaps I should ask a question specifically about this issue. Thank you @AlfredCentauri. \$\endgroup\$
    – Leo
    Commented Jun 26, 2015 at 23:43

If we have a pulse wave does it make sense anymore to talk about its rms value?

Yes it does. If the voltage is 50% duty and rises to +1V at the top and -1V at the bottom it will measure: -

  • Zero volts on a multimeter measuring DC
  • Theoretically 1V on a multimeter measuring AC

You can't rationalize one from the other when they are AC coupled.

And what if they aren't AC coupled - say you measured 2.5V for a 5V square wave - sure, you can predict it is 50% duty but you can't say the heating effect it has is the same as 2.5V DC.

The RMS of a 5V square wave with 50% duty is not 2.5 volts it is \$\sqrt{\dfrac{5^2}{2}}\$ = 3.536 volts: -

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  • \$\begingroup\$ I was not talking about square pulse wave. just a pulse without going negative values. O and 5V for isntance. \$\endgroup\$
    – user16307
    Commented Nov 26, 2013 at 14:25
  • \$\begingroup\$ @user16307 I was just generalizing, covering both AC and DC scenarios. \$\endgroup\$
    – Andy aka
    Commented Nov 26, 2013 at 14:46

The average power going into a system is a meaningful number (it represents the amount of energy per second). If the voltage and current going into a system are always proportional (meaning voltage varies in phase with current) then the average power going into a system will be proportional to the square of voltage and also proportional to square of current. The RMS voltage is essentially the square root of the amount of power that a would be generated across a unit load by a given voltage or current waveform.

In the case of a motor driven by PWM, the voltage and current waveforms are likely to be substantially out of phase, in ways which will vary depending upon various factors including the level of mechanical loading. As such, the power being driven into the motor will be not be proportional to the RMS voltage nor to the RMS current. Within the motor and control system, energy will be shuffled around between places it's put to use (motor's mechanical load), places it's wasted mechanically (bearing friction), places it's wasted electrically (resistance in the motor windings and the control system), and places it's sometimes stored and sometimes harvested (mechanical inertia and electrical inductance). Power dissipation due to winding resistance is proportional to RMS-measured current, pretty much independent of whatever else is going on, but most other kinds of energy transfer will vary in other ways.

With regard to other kinds of loads, the usefulness of RMS voltage as a measure will depend upon the nature of the load. In some cases, simple time-averaged voltage will have a more meaningful relationship with behaviors of interest, while in other cases RMS voltage is what's important.


There are many ways to define "average" when measuring an AC voltage or current. RMS usually makes the most sense, because it is the only way which, for any waveform, preserves equations such as:

$$ E = IR $$

$$ P = IE $$

$$ P = I^2 R $$

$$ P = \frac{E^2}{R} $$

RMS voltage or current answers the question:

What is the equivalent DC voltage or current with the same power into a resistive load as this AC voltage or current?

RMS is the only method that does this for any periodic waveform, be it sinusoidal, square, a pulse train, symmetrical or not, or even totally irregular. You might say that the magic of it is in the square part of root mean square, because power into a resistive load is proportional to the square of current or voltage. Thus, two 50% duty cycle square waves, one from 0V to 5V, and the other from -2.5V to 2.5V, do not carry the same power. Only RMS takes this into account:

$$ \sqrt{\frac{(5V)^2 + (0V)^2}{2}} = \sqrt{\frac{25(V^2)}{2}} = \sqrt{12.5(V^2)} \approx 3.5V $$

$$ \sqrt{\frac{(-2.5V)^2 + (2.5V)^2}{2}} = \sqrt{\frac{6.25(V^2) + 6.25(V^2)}{2}} = \sqrt{6.25(V^2)} \approx 2.5V $$

  • \$\begingroup\$ Phil, I'm pretty sure there's an error in your answer. An ideal rectifier gives the absolute value, i.e., the positive root of the square of the input. Taking ideal ripple filter to mean a filter that passes only the DC component of the input, the result is the mean of the root of the square not the root of the mean of the square. For example, if the input to the rectifier is a sine wave, the DC component is \$\frac{2V_p}{\pi}\$ whilst the rms value is, of course, \$\frac{V_p}{\sqrt{2}}\$. \$\endgroup\$ Commented Nov 26, 2013 at 14:46
  • \$\begingroup\$ @AlfredCentauri I'm not sure I follow that, but I guess there are many ways my statement could have been interpreted. Maybe this revision is less ambiguous? \$\endgroup\$
    – Phil Frost
    Commented Nov 26, 2013 at 17:03

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