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Revision as of 11:19, 24 November 2008
== 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} $
- Ananya Panja
