Converting from T-domain to S-domain?

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Discussion Overview

The discussion revolves around converting a function from the time domain (T-domain) to the Laplace domain (S-domain), specifically focusing on the function F(t) = e^-4t * sin(10t). Participants explore the application of Laplace transforms to achieve this conversion.

Discussion Character

  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant presents the function F(t) and seeks assistance in converting e^-4t to the S-domain.
  • Another participant suggests using a table of Laplace transforms for the conversion process.
  • A participant questions whether the parameter 'a' in the Laplace transform formula is a constant that can be replaced with a specific value.
  • A later reply clarifies the Laplace transform of e^-at, stating it is given by 1/(s+a), but does not resolve the earlier question about the parameter 'a'.
  • One participant expresses gratitude and indicates that they have resolved their query, but does not provide further details on the resolution.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on the interpretation of the parameter 'a' in the Laplace transform, as participants express varying levels of understanding and clarity on this point.

Contextual Notes

There is some ambiguity regarding the use of the parameter 'a' in the Laplace transform formula, as well as the specific application to the function presented. The discussion does not fully clarify these aspects.

mhnassif
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Homework Statement



convert T-domain to S-domain

F(t)= e^-4t *sin10t

Homework Equations





The Attempt at a Solution



sin10t converted to 10/[(S^2)+100]

but how can i convert e^-4t ?
 
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Thank you,

I know exp(-at) on s-domain will be
L2p9a.jpg


But is the Alpha sign a constant, to replace it with a value?
 
I just took a closer look at that table, I'm not sure why they used a and alpha. Basically this should be:

\mathcal{L}\left(e^{-at})\right = \frac{1}{s+a}

Hope that helps,

Ryan
 
ryan88 said:
I just took a closer look at that table, I'm not sure why they used a and alpha. Basically this should be:

\mathcal{L}\left(e^{-at})\right = \frac{1}{s+a}

Hope that helps,

Ryan

Solved ,, Thank You ,

Have a nice day
 

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