How can I find the RMS value of a wave in irregular form?

Question

As a sine wave passes through a circuit, it clips and the waveform in the figure is formed.

How can I find the RMS value of a full wave?

Answer 1

First write an equation for one cycle of your waveform, based on what you can see on the graph (and the assumption this is indeed a sinewave with part clipped out).

$$v = \left\{ \begin{array}{ll} 15\sin \omega t & \text{for }0 <\omega t<a \\ 0 & \text{for }a<\omega t< \pi \\ 15\sin \omega t & \text{for }\pi <\omega t< 2\pi \end{array}\right. $$

Now take the definition of RMS.

$$v_\text{rms}=\sqrt{\frac1{T}\int_{0}^{T} v^2\,\mathrm dt}\,,$$

Where \$T\$ is the time we are calculating the rms over, \$\frac{2\pi}{\omega}\$ in this case.

\$\omega\$ is just an annoying constant that doesn't actually effect the final answer (time-stretching a waveform does not change it's rms), so treat it as 1 to make the algebra simpler.

Then chop-up the time interval and substitute.

$$v_\text{rms}=\sqrt{\frac1{2\pi}\left(\int_{0}^{a} (15\sin t)^2\,\mathrm dt+\int_{a}^{\pi} 0^2\,\mathrm dt+\int_{\pi}^{2\pi} (15\sin t)^2\,\mathrm dt\right)}\,,$$

We can simplify this a bit by removing the zero term and combining the other two terms by taking advantage of the fact that sin is periodic, so the integral over the range \$2\pi\$ to \$2\pi+a\$ is the same as over the interval \$0\$ to \$a\$.

$$v_\text{rms}=\sqrt{\frac1{2\pi}\int_{\pi}^{2\pi+a} 15^2\sin^2t\,\mathrm dt}\,,$$

To solve such an integral you make use of the identity.

$$\sin^2x = \frac{1 - \cos(2x)}{2}$$

When you substitute that in, you should get something you can integrate fairly easily with well-known integration rules.

Answer 2

This might sound a bit strange, but, really:

By calculating the RMS. It's the root mean square! So,

$$\text{RMS}=\sqrt{\frac1{T_{\text{observation}}}\int_{0}^{T_{\text{observation}}} V^2(t)\,\mathrm dt}\,,$$

nothing more, nothing less. It's really that simple!

Answer 3

Here is how to do it digitally. I think it can be done analytically, also, if desired since it is just a sine wave. First, sample the waveform. Ideally, sample it for exactly one period. Then compute the discrete RMS value.

Let's call the series of samples Y(t). Sample at a reasonable rate. Like 100 samples over the full period. The sampling period is denoted with T. So Y(0) = the first sample, Y(T) is the second sample, Y(2T) is the third, etc.

First compute the series Y^2(t) by squaring each sample individually.

Next compute the mean of Y^(t). The mean is just the average of all the squared samples. Note that the mean is not a series. It is just a single number.

Take the square root of the mean. This is the RMS value of the waveform. RMS is taken to be always positive.