# $V_{RMS}$ with sinusoidal voltage source seems unintuitive

$V_{RMS}$ with sinusoidal voltage source is $V \sqrt2$. To me, it seems like the frequency should be involved somehow because that changes the shape of the wave. Is $\sqrt2$ somehow tied to the $\sin$ function?

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The frequency doesn't change the shape of the wave: it just scales it linearly in the time domain. But since the M of RMS is "mean", it's invariant under that linear scaling. – Peter Taylor Mar 9 '12 at 16:32
@Peter - that's the thing I couldn't visualize. – uncle brad Mar 9 '12 at 16:44

The $V_{RMS}$ of a sinusoidal voltage $V(t) = V_* \sin(\omega t)$ is defined to be

$$V_{RMS} = \sqrt{\frac{1}{T} \int_0^T V_*^2 \sin^2(\omega t) dt} = \frac{V_*}{\sqrt{T}} \sqrt{\int_0^T \sin^2(\omega t) dt}$$

where the interval $[0,T]$ is a whole number of cycles. hence we need to know the integral of the function $\sin^2(\omega t)$ between $0$ and $T$. This can be calculated with a trick, noticing that $\sin^2(\omega t) = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$:

$$\int_0^T \sin^2(\omega t) dt = \frac{1}{2} \int_0^T (1-\cos(2\omega t)) dt = \left[ \frac{t}{2} - \frac{1}{4\omega} \sin(2\omega t) \right]_0^T$$

Since the interval is a whole number of cycles, all of the $\sin$ terms will vanish (and these are the only terms that had dependence on the frequency, which is why the final result doesn't depend on the frequency) which gives us

$$\int_0^T\sin^2(\omega t) dt = \frac{T}{2}$$

and hence

$$V_{RMS} = \frac{V_*}{\sqrt{T}} \sqrt{\frac{T}{2}} = \frac{V_*}{\sqrt{2}}$$

There's a difference between the formula I quote here and what you quoted, which possibly comes down to different conventions - I take the amplitude of a wave to be the distance from baseline to peak, but some people take it from peak to trough, which would double the answer to $V_*\sqrt{2}$.

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Drawing a graph is a quick way to see that $\int_0^T \sin^2(\omega t)\mathrm dt$ and $\int_0^T \cos^2(\omega t)\mathrm dt$ have the same value, and then $$\int_0^T \sin^2(\omega t)\mathrm dt + \int_0^T \cos^2(\omega t)\mathrm dt = \int_0^T \sin^2(\omega t) + \cos^2(\omega t)\mathrm dt = \int_0^T 1\mathrm dt = T$$ shows that both integrals have value $T/2$. – Dilip Sarwate Mar 9 '12 at 16:25
You had me at "Since the interval is a whole number of cycles, all of the sin terms will vanish..." – uncle brad Mar 9 '12 at 16:47

If you check the definition of RMS, you will note that it involves an integral as follows:

$$RMS(f) = \lim_{T \to \infty} \sqrt{\frac{1}{T} \int_0^T f(t)^2 dt}$$

Now, if you plug in a sinusoidal voltage source, say $U(t) = V_0 sin(\omega t + \phi)$, what can you say about the resulting integral?

To answer your question, consider what happens when you apply the substitution $u = \omega t + \phi$.

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