Question: [![enter image description here](https://i.stack.imgur.com/qe9JA.png)](https://i.stack.imgur.com/qe9JA.png)

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

[![enter image description here](https://i.stack.imgur.com/0mP38.png)](https://i.stack.imgur.com/0mP38.png)

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?**

Answer: 
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.
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.