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       On doin a Laplace Transform
 
       On doin a Laplace Transform
     
+
 
       <math>X(s) =\int_ {\infty\}^{\infty\}</math>
+
       <math>X(s)= \int_{-\infty}^{\infty}x(t){e^{-sT}}\, dT</math>

Revision as of 11:12, 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 $

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