Differential equations: impulse function

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Homework Help Overview

The discussion revolves around taking the Laplace transform of the product of the Dirac delta function and the sine function, specifically the expression δ(t-1)sin(t). Participants are exploring the implications of the impulse function in the context of differential equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the meaning and implications of the Laplace transform involving the delta function. There are discussions about whether convolution is necessary and how the properties of the delta function simplify the integration process.

Discussion Status

Some participants have offered guidance regarding the definitions of the Laplace transform and the delta function, while others are exploring their understanding of the integral involving the delta function. There is an ongoing examination of the relationship between the functions involved and the appropriate methods to apply.

Contextual Notes

There is a mention of the integral property of the delta function, which is central to the discussion, and participants are navigating the definitions and properties relevant to the problem without reaching a consensus on the approach.

seang
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How do you go about taking the Laplace transform of something like...

[tex]\delta (t-1)sin(t)[/tex]

What does this even mean?
 
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This is straight-forward, it comes directly from applying the defintions of the laplace transform and the impulse function (also called the Dirac delta "function"). So the first thing you should do is look up these definitions.
 
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So do I have to convolve them? I know the laplace transform of both of the functions, but in order to find the laplace transform of {h(t)x(t)} it seems I'll need convolution.

And I'll have to use the fact that

[tex]\int \delta(t-t_0)f(t)dt = f(t_0)[/tex]

Am I on the right track?
 
No, this problem, at least as you stated it, has nothing to do with convolution: you are given a function, [itex]\delta(t-1)sin(t)[/itex] and are asked to find the Laplace transform. The Laplace transform of that is
[tex]\int_0^\infty \delta(t-1)sin(t)e^{-st}dt[/tex].
That's particularly easy to integrate with that delta function in there! Do you know enough about the delta function to know why that is easy?
 
I don't think so. It seems like the integral I posted above would help me in some way, but I'm not sure how.

EDIT: unless, is the laplace transform

[tex]e^{-s(1)}sin(1)[/tex]

?
 
Last edited:

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