(New page: == '''Fundamentals of Laplace Transform''' == Let the signal be: <math>x(t) =e^ {-at} \mathit{u} (t).</math> Here is how to compu...) |
|||
| (One intermediate revision by the same user not shown) | |||
| Line 17: | Line 17: | ||
</math> | </math> | ||
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 --[[User:Mboutin|Mboutin]] 11:58, 21 November 2008 (UTC) | 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 --[[User:Mboutin|Mboutin]] 11:58, 21 November 2008 (UTC) | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
Latest 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)
