```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 PiLet G(t) = (y(t))^2This is quite straight forward to calculate and plot.Then mean = M = (1/Pi) * Int from 0 to Pi of G(t).dtand RMS = sqrt(M)For example I took b=2 and r=3Plot of y(t) is a squarish wave of max value just < 0.5Plot of G(t) is less square of max value approx 0.24Then RMS = sqrt(0.58067/Pi) = 0.42992Any numerical integration program should be satisfactory.There should be no need for Fourier Series.Regards, Peter Scales.
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