Root mean square

The root mean square of a time-varying signal is the constant (DC) value that would deliver the same average power into a resistor as the signal itself. It is the single most useful “size” measure for an AC signal, because power — not peak voltage — is usually what you care about.

For a signal $v(t)$ that repeats with period $T$, the RMS value is

$V_{rms}=\sqrt{\frac{1}{T}{\int }_0^T{v}^2(t)dt}$

Read the name backwards and it tells you the recipe: take the signal, square it, take the mean (average) over one period, then take the square root. The squaring is the important step — it is what makes $V_{rms}$ track power rather than just amplitude.

Why this is the right definition

[Background from general knowledge, not the source PDF — the source states $V_{rms}=V_p/\sqrt{2}$ for a sinusoid but does not derive it.]

The instantaneous power dissipated by a voltage $v(t)$ across a resistor $R$ is $p(t)=v^2(t)/R$. The average power over one period is

$P_{avg}=\frac{1}{T}{\int }_0^T\frac{v^2(t)}{R}dt=\frac{1}{R}\cdot \frac{1}{T}{\int }_0^T{v}^2(t)dt$

Now ask: what constant DC voltage $V_{dc}$ would dissipate this same average power? A DC voltage gives $P=V_{dc}^2/R$. Setting the two equal,

$\frac{V_{dc}^2}{R}=\frac{1}{R}\cdot \frac{1}{T}{\int }_0^T{v}^2(t)dt\Rightarrow V_{dc}=\sqrt{\frac{1}{T}{\int }_0^T{v}^2(t)dt}=V_{rms}$

So $V_{rms}$ is defined to be the DC-equivalent heating voltage. A $120\text{V}$ RMS wall outlet heats a kettle exactly as fast as a steady $120\text{V}$ DC supply would, even though its peak voltage is about $170\text{V}$.

RMS of a sinusoid

For $v(t)=V_p\sin (\omega t)$ — a sinusoid of peak amplitude $V_p$ — work the integral over one period:

$V_{rms}^2=\frac{1}{T}{\int }_0^T{V}_p^2{\sin }^2(\omega t)dt$

Use the identity ${\sin }^2\theta =\frac{1}{2}(1-\cos 2\theta )$:

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

Over a full period the $\cos (2\omega t)$ term integrates to zero (its average is zero — it goes through two complete cycles), leaving

$V_{rms}^2=\frac{V_p^2}{T}\cdot \frac{T}{2}=\frac{V_p^2}{2}\Rightarrow \boxed{V_{rms}=\frac{V_p}{\sqrt{2}}\approx 0.707V_p}$

This $1/\sqrt{2}$ factor is specific to a pure sinusoid. A square wave of peak $V_p$ has $V_{rms}=V_p$ (it sits at $\pm V_p$ all the time), and a triangle wave has $V_{rms}=V_p/\sqrt{3}$. The general integral always applies; only the constant changes with waveform shape.

Why it matters

Whenever a circuit’s job is to deliver or dissipate power — heating, audio output, RF transmission — the relevant amplitude is the RMS value, because $P_{avg}=V_{rms}^2/R$. Quoting a sinusoid by its peak amplitude overstates its heating ability by a factor of $\sqrt{2}$. AC voltmeters and the "$120\text{V}$"/"$230\text{V}$" ratings of mains supplies are always RMS for exactly this reason.