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Re: Asymmetric Clipped Waveform  Find Average
Posted:
Feb 15, 2013 10:23 AM


> 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.



