Date: Feb 15, 2013 10:23 AM
Author: Peter Scales
Subject: Re: Asymmetric Clipped Waveform - Find Average
> I have an asymmetrical clipped repeating waveform and

> I want to be able to find the root mean square.

>

> The function is as follows, with r and b constants:

>

> y(t) =

> ((exp(sin(t)*b)-exp(-sin(t)*b*r))/(exp(sin(t)*b)+exp(-

> sin(t)*b)))*(1/b)

>

> This is pretty computationally heavy. What are some

> approaches to use to get to a simpler root mean

> square? Should I use a Fourier transform?

Patrick,

RMS is, by definition, the square root of the mean of the square of the function.

y(t) =0 at t=0 and Pi

Let G(t) = (y(t))^2

This is quite straight forward to calculate and plot.

Then mean = M = (1/Pi) * Int from 0 to Pi of G(t).dt

and RMS = sqrt(M)

For example I took b=2 and r=3

Plot of y(t) is a squarish wave of max value just < 0.5

Plot of G(t) is less square of max value approx 0.24

Then RMS = sqrt(0.58067/Pi) = 0.42992

Any numerical integration program should be satisfactory.

There should be no need for Fourier Series.

Regards, Peter Scales.