Asymmetric Clipped Waveform - find RMS

Discussion in 'Math' started by pdelaney, Feb 11, 2013.

1. pdelaney Thread Starter New Member

Feb 11, 2013
1
0
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?

Thanks

2. blah2222 Distinguished Member

May 3, 2010
581
38
Is this the function you are describing? Hard to read from your post.

$y(t) = [\frac{1}{b}][\frac{e^{bsin(t)} - e^{-rbsin(t)}}{e^{bsin(t)} + e^{-bsin(t)}}]$

3. WBahn Moderator

Mar 31, 2012
23,160
6,973
That's a pretty strange looking function. Have you plotted it to get a feel for what it looks like? Does it look "clipped"? That's not at all apparent, to me, from the form of the equation.

Are you required to produce a closed-form solution as a function of b and r? That's not going to be fun, either directly or via a Fourier transform (good luck finding an applicable pair). You might try doing it numerically, if only to get a feeling for the behavior. I'd set r=1 and see how the answer responds to changes in b. If that is pretty tame, then look at how it responds to values of r relative to b.

4. The Electrician AAC Fanatic!

Oct 9, 2007
2,620
468
If the variable r is set to 1, you have the tanh function, a function I've used before to create a smoothly clipped waveform.

It appears that the variable b controls the amount of clipping. Here are 3 waveforms with r=1 and b equal to 1, 2 and 3, with 3 giving the most clipping:

The variable r controls the asymmetry. Here is the same family of waveforms but with r=1.1:

I don't think you're going to be able to derive a closed form expression for the RMS value, but numerical evaluation isn't hard. Here's a curve of the RMS value for r=1 and b varying from .05 to 50:

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