Calculate the range of S using Laplace Transform

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SUMMARY

The discussion centers on calculating the range of values for \( s \) in the function \( f(t) = e^{-t/2}u(t) \) using the Laplace Transform. The initial approach involved integrating \( e^{-st}e^{-t/2}u(t) \) from 0 to infinity, resulting in \( \frac{2}{2s+1} \), which indicated that \( s > -0.5 \) to avoid discontinuity. However, using the Laplace Transform chart provided a different result, suggesting that the transform of \( e^{-t/2}u(t) \) is \( \frac{1}{s + 1/2} \), leading to confusion regarding the consistency of the two methods. The correct interpretation of the Laplace Transform is essential for determining the valid range of \( s \).

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  • Understanding of Laplace Transforms
  • Familiarity with exponential functions and their properties
  • Knowledge of integration techniques from 0 to infinity
  • Ability to interpret transform tables accurately
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  • Review the properties of Laplace Transforms, focusing on exponential decay functions
  • Study the derivation and application of the Laplace Transform for piecewise functions
  • Learn how to use Laplace Transform tables effectively for quick reference
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Homework Statement


find the range of values for s: f(t)=e^(-t/2)u(t)

Homework Equations


The Attempt at a Solution


what I did initially was to take the integral from 0 to infinity of e^(-st)e^(-t/2)u(t) dt this gave me 2/(2s+1) which when I sub in -.5 to find a divide by 0 or discontinuity then s>-.5. Then I realizeed I could do it by using the laplace transform chart so the transform of e^(-t/2)u(t) which gives s+(1/2) which gives (2s+1)/2 this is backwards from what I found earlier? I don't understand how the integral can be different from the transform but anyways if I plug in a negative .5 I would get 0 which would not be a discontinuity so it appears this way wouldn't work hmm any sugestions?
 
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I think you should check your transform table again. The entry for exponential decay should be of the form ##\frac{1}{s + \alpha}##
 

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