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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

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  • \$\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

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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}\$

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  • \$\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
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    \$\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

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