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