Root Mean Square Value Of A Function

 

Suppose y = f(x) is the equation of a curve. Then the root mean square (rms) value of the function between x = x_1 ad x = x_2 will be

  \[\text{rms} = \sqrt{\left(\frac{1}{x_2 - x_1} \int_{x_1}^{x_2} y^2~dx\right)}.\]

Now I’ll give some examples of that.

EXAMPLE 1

According to Stroud and Booth (2013)*, “Find the rms value of i = \cos x + \sin xover the range x = 0 to \cfrac{3 \pi}{4}.”

SOLUTION

Now here the given function is i = \cos x + \sin x. And I have to find its rms value over the range x = 0 to \cfrac{3 \pi}{4}. Thus it will be

  \[\text{rms}^2 = \frac{1}{3\pi / 4 - 0} \int_{0}^{3\pi / 4} i^2~dx.\]

And that means

  \[\text{rms}^2 = \frac{4}{3\pi } \int_{0}^{3\pi / 4} (\cos x + \sin x)^2~dx.\]

So this gives 

  \begin{eqnarray*}\text{rms}^2 &=& \frac{4}{3\pi } \int_{0}^{3\pi / 4} (\cos^2 x + \sin^2 x + 2\sin x \cos x)~dx\\ &=& \frac{4}{3\pi } \int_{0}^{3\pi / 4} (1 + \sin 2x )~dx.\end{eqnarray*}

Now I’ll integrate it using the standard formulas in integration. And that means

  \[\text{rms}^2 = \frac{4}{3\pi } \left[x - \frac{\cos 2x}{2}\right]_{0}^{3\pi / 4}.\]

Next, I’ll substitute the limits to get

  \[\text{rms}^2 = \frac{4}{3\pi } \left[\left(\frac{3 \pi}{4} - \frac{\cos 3 \pi /2}{2}\right)- \left(0 - \frac{\cos 0}{2}\right) \right].\]

Then I’ll simplify it. And that gives

  \begin{eqnarray*} \text{rms}^2 &=& \frac{4}{3\pi }\left[\left(\frac{3 \pi}{4} - \frac{0}{2}\right) - \left( - \frac{1}{2}\right) \right]\\ &=& \frac{4}{3\pi } \left[ \frac{3 \pi}{4} - 0 +\frac{1}{2} \right] \\ &=& 1 + \frac{2}{3\pi} = 3.9439.\end{eqnarray*}

Therefore the rms value of the function i is

  \[\text{rms} = \sqrt{3.9439} = 1.759.\]

Hence I can conclude that this is the answer to the given example.

Now I’ll give another example.

EXAMPLE 2

According to Stroud and Booth (2013)*, “Calculate the rms value of i = 20 + 100 \sin 100 \pi t between t = 0 and t = 1/50.”

SOLUTION

Now here the given function is i = 20 + 100 \sin 100 \pi t. And I have to get its rms values between t = 0 and t = 1/50. Thus it will be

  \[\text{rms}^2 = \frac{1}{1/50 - 0} \int_{0}^{1/50} i^2~dt.\]

And that means

  \[\text{rms}^2 = \frac{1}{1/50 - 0} \int_{0}^{1/50} (20 + 100 \sin 100 \pi t)^2~dt.\]

So this gives

  \begin{eqnarray*} \text{rms}^2 &=& 50(20)^2 \int_{0}^{1/50} (1 + 5 \sin 100 \pi t)^2~dt \\ &=& (100)^2\times 2 \int_{0}^{1/50} (1 + 25 \sin^2 100 \pi t + 10 \sin 100 \pi t)~dt \\ &=& 2 (100)^2 \int_{0}^{1/50} (1 + 25/2 (1 - \cos 200 \pi t) + 10 \sin 100 \pi t)~dt \\ &=& 2 (100)^2 \int_{0}^{1/50} (1 + 25/2 - 25/2\cos 200 \pi t + 10 \sin 100 \pi t)~dt .\end{eqnarray*}

Then I’ll integrate it to get

  \[\text{rms}^2 = 2 (100)^2 \left[ t + (25/2) t - 25/2 ~\frac{\sin 200 \pi t}{200 \pi} - 10~ \frac{\cos 100 \pi t}{100 \pi}\right]_{0}^{1/50}.\]

Next, I’ll substitute the limits to get

  \begin{eqnarray*} \text{rms}^2 &=& 2 (100)^2 [ \left(\frac{1}{50} + (25/2) \times\frac{1}{50} - 25/2 ~\frac{\sin 200 \pi / 50}{200 \pi} - 10~ \frac{\cos 100 \pi /50}{100 \pi}\right) \\&& - \left( 0 + (25/2) .0 - 25/2 ~\frac{\sin 200 \pi (0)}{200 \pi} - 10~ \frac{\cos 100 \pi (0)}{100 \pi} \right) ] .\end{eqnarray*}

Then I’ll simplify it. So that gives

  \begin{eqnarray*}\text{rms}^2 &=& 2(100)^2 [\left(\frac{1}{50}+\frac{1}{4}-\frac{1}{16 \pi}\sin 4\pi - \frac{1}{10\pi}\cos 2 \pi\right)\\ &&-\left(0+0-0-\frac{1}{10\pi}\cos 0\right)]\\ &=& 2(100)^2 [\left(\frac{27}{100}-\frac{1}{16 \pi}.0- \frac{1}{10\pi}.1\right) + \frac{1}{10\pi}.1]\end{eqnarray*}

And this is because \cos 0 = 1, \cos 2 \pi = 1, \sin 4 \pi = 0.

Now I’ll simplify it a bit more to get the value of \text{rms}^2 as

  \[\text{rms}^2 = 2 (100)^2\left [ \frac{27}{100} - \frac{1}{10\pi} + \frac{1}{10\pi}\right] = 2 (100)^2 \times \frac{27}{100}.\]

So I can rewrite it as

  \[\text{rms}^2 = \frac{2\times(10)^4\times(3)^2\times(3)}{10^2}.\]

And that means

  \[\text{rms}^2 = 6\times(10)^2\times(3)^2 = (30)^2 \times 6.\]

Therefore the rms value of the function will be

  \[\text{rms} = \sqrt{(30)^2 \times 6} = 30\sqrt{6} = 73.485.\]

Hence I can conclude that this is the answer to the given example.