I have an electrosurgical hf generator unit that produces a modulated signal with duty cycle in blend mode. I have the output of this unit connected across a load. With scope connected across the load, I measure peak-peak volt of a signal with a duty cycle in a pulse. The waveform looks like as shown. It has hf sinusoidal oscillation in active part The waveform looks like as shown. It has hf sinusoidal oscillation in active part.

I want to calculate active power delivered by the unit. What would be the correct way of calculating the active power output? How to calculate RMS volt from Vp-p?

  • \$\begingroup\$ Can you express your voltage waveform as a function including the modulation? Since you should be able to, you can then just actually do the RMS integral calculation. \$\endgroup\$ Jul 13, 2016 at 15:46
  • \$\begingroup\$ I've updated the original question with the waveform details. \$\endgroup\$
    – user116646
    Jul 13, 2016 at 17:16

1 Answer 1


From the way you've described your waveform you have a voltage as a function of time that looks something like,

$$ v(t) = 235\sin\left(\frac{2\pi t}{T}\right)\left[ u(t) -u(t - 0.85T) \right] \quad t\in[0,T]. $$

For the RMS voltage calculation you then get: $$ V_{RMS}^2 = \frac{1}{T} \int_0^T 235^2\sin^2\left(\frac{2\pi t}{T}\right)\left[ u(t) -u(t - 0.85T) \right]^2dt \\ = \frac{235^2}{T} \int_0^{0.85T}\sin^2\left(\frac{2\pi t}{T}\right)dt \\ = \frac{235^2}{T} \int_0^{0.85T} \left(\frac {1-\cos\left(\frac{4\pi t}{T}\right)}{2}\right) dt \\ = \frac{235^2}{T}\left[ \frac{0.85T}{2} - \frac{T}{8\pi}\sin\left(0.85\cdot4\pi\right)\right] \\ =235^2\left[\frac{0.85}{2}-\frac{\sin\left(0.85\cdot4\pi\right)}{8\pi}\right]. $$ Then you'll have, $$ V_{RMS} = 235\sqrt{\frac{0.85}{2}-\frac{\sin\left(0.85\cdot4\pi\right)}{8\pi}}\approx 160~\text{V}. $$ So for your power calculation then, $$ P = \frac{V_{RMS}^2}{R}=\frac{160^2}{500}=51.2~\text{W} $$ I hope this makes it clear that for more complex waveforms calculating the RMS voltage is not quite as simple as it is for "nice" sinusoids. What's worse about this waveform is that the RMS voltage depends on whether or not you base your waveform on a sine or cosine waveform!

Update: It looks like the waveform you are supplying is both amplitude and duty cycle modulated, you can follow the same procedure to find the RMS voltage of your actual waveform. You may have to do this over a number of periods (until the modulating and modulated sinusoids are harmonics of each other). Your new waveform is something like, $$ v(t) = 235m\sin\left(\frac{2\pi t}{T_1}\right)\sin\left(\frac{2\pi t}{T_2}\right)\left[u(t) - u(t - 0.85T_1\right], $$ where in this case the waveform with period \$T_1\$ is the modulating waveform with modulation index \$m\$.


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