Revision as of 11:26, 19 November 2008 by Apanja (Talk)

                             == 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} $

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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