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       Let the signal be:
 
       Let the signal be:
  
       <math>x(t) =e^ {-at} \mathit{u} (t)</math>
+
       <math>x(t) =e^ {-at} \mathit{u} (t).</math>
 
        
 
        
       On doin a Laplace Transform
+
       Here is how to compute the Laplace Transform of <math>x(t)</math>:
  
       <math>X(s)= \int_{-\infty}^{\infty}x(t){e^{-st}}\, dt</math>
+
       <math>
 
+
\begin{align}
      <math>X(s)= \int_{-\infty}^{\infty}{e^{-at}}{e^{-st}}dt  ,\mathit{u} (t)=1,t>0</math>
+
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, \\
      <math>X(s)= \frac{1}{s+a}</math>
+
    &=\frac{1}{s+a}. ~^*
 +
\end{align}
 +
</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)
 +
     
  
  
 
* [[Homework _ECE301Fall2008mboutin#10 Daniel Morris: Properties of the Region of Convergence(ROC)]]
 
* [[Homework _ECE301Fall2008mboutin#10 Daniel Morris: Properties of the Region of Convergence(ROC)]]

Revision as of 07:58, 21 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)


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