Revision as of 01:22, 5 December 2008

HW10

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                       == Fundamentals of Laplace Transform ==

     Let the signal be:

     $ x(t) =e^ {-at} \mathit{u} (t). $
     
     Here is how to compute the Laplace Transform of $ x(t) $:

     $  \begin{align} X(s) &= \int_{-\infty}^{\infty}x(t){e^{-st}}\, dt, \\      &= \int_{-\infty}^{\infty}{e^{-at}}{e^{-st}}dt   ,\text{ since }\mathit{u} (t)=1,\text{ for }t>0, \text{ else }\mathit{u} (t)=0, \\      &=\frac{1}{s+a}. ~^*  \end{align}  $

Note: the last equality (with a *) is untrue. Please do not write this on the test or you will get points marked off. I really appreciate this mistake being on Rhea, please do not erase it --Mboutin 11:58, 21 November 2008 (UTC)

Correction of above:

$ \begin{align} X(s) &= \int_{-\infty}^{\infty}x(t){e^{-st}}\, dt, \\ &= \int_{0}^{\infty}{e^{-at}}{e^{-st}}dt ,\text{ let } s=b+j\omega, \\ &=\int_{0}^{\infty}{e^{-(a+b+j\omega)t}}dt, \\ \end{align} $

If $ a+b\leq 0 $, then the integral Diverges

Else,

$ \begin{align} X(s) &=\frac{e^{-(a+b)t}e^{-j\omega t}}{-(a+b+j\omega)}|_0^\infty, \\ &=0-\frac{-1}{s+a}, \\ &=\frac{1}{s+a} \end{align} $

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