What is the Laplace Transform of |sint|?

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Homework Help Overview

The discussion revolves around finding the Laplace transform of the function |sin(t)|, which involves understanding its behavior over different intervals due to the modulus operation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the piecewise nature of |sin(t)|, noting its positive and negative segments across specified intervals. Some suggest expressing the function in terms of convolutions or using geometric series to facilitate the Laplace transform.

Discussion Status

The conversation is ongoing, with participants exploring various interpretations and approaches to the problem. There is recognition of the need to integrate over different intervals, and some guidance has been offered regarding the use of geometric series.

Contextual Notes

Participants mention the requirement to use Laplace transforms in the context of solving a differential equation, which adds a layer of complexity to the problem.

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Homework Statement

Need to find the Laplace transform of |sint| (modulus).



Homework Equations





The Attempt at a Solution

I am really not sure how to proceed here - any help would be much appreciated.
 
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|sin(t)| = sin(t) on [0, pi], and |sin(t)| = -sin(t) on [pi, 2pi] or on [-pi, 0]
 
I don't know if this is the best way to go about it, but perhaps you can express the function as a convolution of a half wave with a train of delta functions (or something like that).
 
Mark44 said:
|sin(t)| = sin(t) on [0, pi], and |sin(t)| = -sin(t) on [pi, 2pi] or on [-pi, 0]

Thanks - I thought of this as well, but this would mean I have to integrate on each interval, and I get sum(n=0, n=inf) ((1+exp(pi*s)/exp(n*pi*s)*(s^2+1)). Is there a way to simplify this? I'm supposed to be using Laplace transforms to solve a differential equation with |sint| as the inhomogeneous part.
 
Think geometric series where r=exp(-pi*s).
 
vela said:
Think geometric series where r=exp(-pi*s).
Ah of course, thanks :)
 

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