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<math>X(s)= \int_{-\infty}^{\infty}{e^{-at}}{e^{-st}}dt ,\mathit{u} (t)=1,t>0</math> | <math>X(s)= \int_{-\infty}^{\infty}{e^{-at}}{e^{-st}}dt ,\mathit{u} (t)=1,t>0</math> | ||
| + | |||
| + | <math>X(s)= \frac{1}{s+a}</math> | ||
Revision as of 11:26, 19 November 2008
== Fundamentals of Laplace Transform ==
Let the signal be:
$ x(t) =e^ {-at} \mathit{u} (t) $
On doin a Laplace Transform
$ X(s)= \int_{-\infty}^{\infty}x(t){e^{-st}}\, dt $
$ X(s)= \int_{-\infty}^{\infty}{e^{-at}}{e^{-st}}dt ,\mathit{u} (t)=1,t>0 $
$ X(s)= \frac{1}{s+a} $
