For a complex waveform as follows:

enter image description here

One can find the squared rms of the complete waveform by finding rms of individual components and then square multiplying it with the corresponding duty cycle. $$s_n = \frac{1}{T} \int_0^T u(t)^2 dt$$ $$u_{rms} = \sqrt{d_1 s_1 +d_2 s_2 + \cdots + d_n s_n}$$

EDIT1: For example finding the rms of following waveform,

enter image description here

$$ I_{rms} = \sqrt{d_1 I_1^2 + d_2 I_2^2}$$ Is the above method also applicable for sinusoidal wave or any other non-linear waveforms?

EDIT2: If I want to calculate the rms of a sine function defined from 54 degrees to 180 degrees with peak value of 27.44 I get 11.48 but the actual answer is 12.66.

$$ RMS = \sqrt{27.44^2 * 0.35 * 0.5} = 11.48 $$ here, duty cycle = 0.35

  • \$\begingroup\$ I don't see how the \$u_i\$'s are defined. What is \$u_2\$ for example? \$\endgroup\$
    – Curd
    Commented Jan 21, 2017 at 16:40
  • \$\begingroup\$ The diagram is just an idea of a complex waveform. $u_i$ could be any line. \$\endgroup\$
    – Vedanshu
    Commented Jan 21, 2017 at 16:42
  • \$\begingroup\$ but \$u_i\$ is one voltage value while the voltage at section \$i\$ is not (necessarily) constant. So what dos \$u_i\$ mean? The starting voltage, the final voltage, the average voltage... of that section??? \$\endgroup\$
    – Curd
    Commented Jan 21, 2017 at 16:44
  • \$\begingroup\$ @Ansh Kumar: I know that that is "just an idea". Still the symbols used need to have a well defined meaning (I'm not asking for numerical values) \$\endgroup\$
    – Curd
    Commented Jan 21, 2017 at 16:48
  • \$\begingroup\$ \$u_i\$ is the segment extended over the period \$T\$. I have edited and put an example in the question. \$\endgroup\$
    – Vedanshu
    Commented Jan 21, 2017 at 16:52

1 Answer 1


Yes, the formula is correct. If you reverse engineer it a bit, it will make a bit more sense.

For instance, \$d_1s_1\$ is the power contribution from \$s_1\$ if the waveform that produced \$s_1\$ (\$u_1\$, via the integral) was present for the whole of the period from 0 to T.

Ditto all the other contributors.

For this waveform: -

enter image description here

If you put numbers down like \$d_1\$ = \$d_2\$ i.e. 50% duty cycle with T=1, \$I_2\$=5 and \$I_1\$=1, the RMS calculated by the equation in the question would produce this: -

\$\sqrt{\frac{25}{2}+\frac{1}{2}}\$ = \$\sqrt{13}\$

Then if you compared this with the more conventional approach of assuming it was a symetrical square wave with peak values of +2 and -2 (superimposed on a DC level of 3) you would get: -

\$\sqrt{2^2+3^2}\$ = \$\sqrt{13}\$

  • \$\begingroup\$ So, is it valid for sinusoidal and any other non-linear functions also ? Because on calculation I was getting wrong answer(small error). \$\endgroup\$
    – Vedanshu
    Commented Jan 21, 2017 at 17:00
  • 1
    \$\begingroup\$ It is valid as far as I'm concerned - maybe you made some rounding error? \$\endgroup\$
    – Andy aka
    Commented Jan 21, 2017 at 17:01
  • \$\begingroup\$ Okay, I'm putting my calculation in the question above. \$\endgroup\$
    – Vedanshu
    Commented Jan 21, 2017 at 17:07
  • \$\begingroup\$ I don't understand the numbers in your calculation, like, where does the 0.35 come from? Ditto the 0.5? Also what is T given that the partial duration of the sinewave is 0.2 x 360 degrees? \$\endgroup\$
    – Andy aka
    Commented Jan 21, 2017 at 17:32
  • \$\begingroup\$ rms of the sine is \$(\text{peak value})/\sqrt{2}\$ so in the final calculation of rms where we have to square its value, I replaced it with \$ (\text{peak value})^2/2\$ and 0.35 is the duty cycle 126/360 = 0.35 \$\endgroup\$
    – Vedanshu
    Commented Jan 21, 2017 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.