3 Root-mean-square value of a function

If $f\left(t\right)$ is defined on the interval $a\le t\le b$ , the mean-square value is given by the expression:

$\frac{1}{b-a}{\int \; <!--nolimits\; -->}_a^b{\left[f\left(t\right)\right]}^{2}dt$

This is simply the mean value of ${\left[f\left(t\right)\right]}^2$ over the given interval.

The related quantity: the root-mean-square (r.m.s.) value is given by the following formula.

The r.m.s. value depends upon the interval chosen. If the values of $a$ or $b$ are changed, then the r.m.s. value of the function across the interval from $a$ to $b$ will in general change as well. Note that when finding an r.m.s.value the function must be squared before it is integrated.

Example 3

Find the r.m.s. value of $f\left(t\right)=t^2$ across the interval from $t=1$ to $t=3$ .

Solution

$\text{r.m.s}=\sqrt{\frac{1}{b-a}{\int \; <!--nolimits\; -->}_a^b{\left[f\left(t\right)\right]}^{2}dt}=\sqrt{\frac{1}{3-1}{\int \; <!--nolimits\; -->}_1^3{\left[t^2\right]}^{2}dt}=\sqrt{\frac{1}{2}{\int \; <!--nolimits\; -->}_1^3{t}^{4}dt}=\sqrt{\frac{1}{2}{\left[\frac{t^5}{5}\right]}_1^3}\approx 4.92$

Example 4

Calculate the r.m.s value of $f\left(t\right)=sint$ across the interval $0\le t\le 2\pi$ .

Solution

Here $a=0$ and $b=2\pi$ so $\text{r.m.s}=\sqrt{\frac{1}{2\pi }{\int \; <!--nolimits\; -->}_0^{2\pi }{sin}^{2}tdt}$ .

The integral of ${sin}^{2}t$ is performed by using trigonometrical identities to rewrite it in the alternative form $\frac{1}{2}\left(1-cos2t\right)$ . This technique was described in HELM booklet  13.7.

$\text{r.m.s. value}=\sqrt{\frac{1}{2\pi }{\int \; <!--nolimits\; -->}_0^{2\pi }\frac{\left(1-cos2t\right)}{2}dt}=\sqrt{\frac{1}{4\pi }{\left[t-\frac{sin2t}{2}\right]}_0^{2\pi }}=\sqrt{\frac{1}{4\pi }\left(2\pi \right)}=\sqrt{\frac{1}{2}}=0.707$

Thus the r.m.s value is 0.707 to 3 d.p.

In the previous Example the amplitude of the sine wave was 1, and the r.m.s. value was 0.707. In general, if the amplitude of a sine wave is $A$ , its r.m.s value is 0.707 $A$ .