# How to calculate RMS power of an offset sine wave

I know when it is centered at 0, Vpeak/sqrt(2) gives the RMS.

When I do FFT on LTspice, it gives a peak at -17dB which is equal to 0.141254

On the FFT page, program says that it shows RMS values. But, I don't think the RMS of this wave is 0.14 W. You are also welcome to answer for this question involving LTspice.

However, main question is : I actually don't think this question is specific to power calculation, anyway, what is the method to calculate RMS of such sine waves?

• RMS power does not have any physical meaning. You should be interested in average power (real power). And LTspice can show you the average power. You can display it by holding down the Ctrl key and left-clicking the plot name V(V_speaker)*I(I_speaker)
– G36
Nov 30, 2019 at 15:40
• @G36 then what is the thing with Vrms*Irms equations? Nov 30, 2019 at 16:00
• Vrms*Irms = Average power (real power).
– G36
Nov 30, 2019 at 16:01
• @G36 hmm, I was thinking that it was equal to rms power. It seems rms and average power are different things. Nov 30, 2019 at 16:09
• Term "RMS" refers only to the effective value of an AC voltage and current. And we should not use the term "RMS" to describe the real (true) power (average power/active power). The true power is a simple average value of instantaneous power. And for this, we don't need roots and the square. So RMS power is an inaccurate term, that has no logical justification. And I'm aware that especially in audio communities, they like to use "RMS power" but we know from the definition of a power that this is a wrong term
– G36
Nov 30, 2019 at 16:22

## 4 Answers

You want to know the rms of a waveform $$\ f(t) \$$.

There are two approaches here direct application of the formula:

$$RMS = \sqrt{\dfrac{1}{T} \cdot \int_0^T (f(t))^2 \text{d}t}$$

Where T is the period of the signal. However there is another useful function:

$$RMS = \sqrt{RMS_1^2 + RMS_2^2}$$

Where $$\RMS_1\$$ and $$\RMS_2\$$ are the RMS values two signals that when added together give $$\f(t)\$$.

This only works provided the two signals do not share any frequency content in common. In this case a sine wave only contains one frequency as does DC (0 Hz which is different) so you can use this formula.

The RMS of a sin wave is $$\ \dfrac{\sqrt{2}}{2} \cdot V_{pk} = \dfrac{\sqrt{2}}{4} \cdot V_{pkpk}\$$ where $$\V_{pkpk}\$$ is the peak to peak voltage. The RMS of a DC signal is the same as the DC value.

• Shouldn't your last line read $\frac {\sqrt 2}{2}\cdot V_{pk} = \frac {\sqrt 2}{4}\cdot V_{pkpk}$? Nov 30, 2019 at 15:27
• @Transistor yes it should, and now does. Thanks for pointing out the typing error. I should have read my answer before posting. Nov 30, 2019 at 15:38

As the signal is not totally recognizable in the first figure, I'll assume your sine wave equation to be $$\ f(t) = 0.25 + 0.25\,sin(5000\,\pi\,t)\$$ and $$\f(t)\$$ has Watt unit.

Thus, the RMS value is given by $$\\sqrt{\dfrac{1}{T}\int_{t_0}^{t_0+T} f(t)^2dt}\$$ being T the period of the function, which I have assumed to be $$\\approx .4\,ms\$$. Thus, $$\T = .4\,ms\$$ and as the function has its peak at $$\\approx .5\,W\$$ I have thus reached the equation given above. Figuring out the integral value we end up with :

$$\\sqrt{\dfrac{1}{0.0004}\int_{0}^{0.0004}(0.25 + 0.25\,sin(5000\,\pi\,t))^2dt} = 306.18\,m\,W\$$ as the approximate RMS value.

The RMS value you are seeing is a combination of the DC and AC components. Your signal is the sum of those two components and the RMS value is taken of the sum.

If you want only the RMS value of the AC portion, i.e., the sinusoidal portion, then using the SPICE controls to subtract the DC component and then take the RMS value of that result, which will now be only the AC portion of the signal.

Make sense?

• If it was a voltage or current it would make sense seperating it to DC and AC, however with power, I can't say I am totally into it. Nov 30, 2019 at 15:51

I am assuming the waveform has a peak value of 500mW and an average of 250mW. The rms value of dc portion is 250mW, same as the average value. For the 1kHz ac part, rms is: $$\frac{500-250}{\sqrt{2}} = 177mW$$

So the net rms will be: $$\sqrt{DC rms^{2} + AC rms^{2}} = 306mW$$ To get rms value from LTSpice, CTRL + left-click the waveform name at the top of the screen (waveform name- I(speaker)*V(V_speaker).) $$\ \$$ FFT should give 20*log(ac_rms) which is:

$$20*log(0.177)=-15dB \ at \ 1kHz$$ This is pretty close to what you have got. So looks like your simuation is giving the right results.

That being said, I am not sure if rms value of power has any significance. For most purposes we use average power only.

• Some others referred the value taken by ctrl+left click as "average power", however you call it "rms" so which one? Nov 30, 2019 at 17:48
• Actually, it shows both rms and average seperately, if you do ctrl+left click. Nov 30, 2019 at 18:17
• I forgot to write that actually for power plots, it does not serve an rms value by ctrl+left click Dec 4, 2019 at 4:40
• Yes, it gives the average power as well as the energy for the time period shown in the plot. For sinusoidal signals, the measurement period should be an integral number of half-cycles. Jul 28, 2022 at 4:13