## FANDOM

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Short for 'Root Mean Square' value. A number used as a measure of the magnitude of an AC signal, either voltage or current. Defined as

$V_{RMS} = \sqrt{\frac{1}{T} \int_0^T V^2 dt}$

This turns out to be the DC value that would produce the same amount of power dissipation in a load. To see this, note that power is proportional to the square of both current and voltage, and that the RMS value squared is equal to the average value of the signal squared.

For a sine wave $A \sin (t)$, it can be shown from the definition that the RMS value is equal to $A/\sqrt{2}$.

For a sine wave plus offset, $B + A \sin (t)$, it can be shown from the definition that the RMS value is equal to $\sqrt{B^2 + \frac{A^2}{2}}$.

## Proofs Edit

### Sine Wave Edit

From the definition

$V_{RMS}^2 = \frac{1}{T} \int_0^T V^2 \sin^2 t\ dt = \frac{V^2}{T} \int_0^T \frac{1 - \cos (2t)}{2} dt$

Evaluating the definite integral,

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

As $T \to \infty$, $\frac{V^2 \sin (2T)}{4T} \to 0$, and therefore $V_{RMS}^2 \to \frac{V^2}{2}$. Taking square roots it follows that for a sinusoidal signal $V_{RMS} = \frac{V}{\sqrt{2}}$.

### Sine Wave with Offset Edit

From the definition

$V_{RMS}^2 = \frac{1}{T} \int_0^T (B + A \sin t )^2 dt = \frac{1}{T} \int_0^T \left( B^2 + 2 A B \sin t + A^2 \sin^2 t \right) dt$

Noting that the last term in the integral is the same as in the previous proof, we evaluate the definite integral to get

$\frac{1}{T} \left[ B^2 T + 2 A B \left(1 - \cos T \right) + A^2 \left( \frac{T}{2} - \frac{\sin (2T)}{2} \right) \right] = B^2 + \frac{A^2}{2} + \frac{2AB}{T} - \frac{2AB\cos T}{T} - \frac{A^2 \sin(2T)}{4T}$

As $T \to \infty$, all except the first two terms go to zero, and therefore $V_{RMS}^2 \to B^2 + \frac{A^2}{2}$. Taking square roots it follows that for a sinusoidal signal with an offset $V_{RMS} = \sqrt{B^2 + \frac{A^2}{2}}$.