What is Laplace Inverse of exp(-s) / (s+1) and also for $\displaystyle exp(-s) /[ (s+1)(s+1)]$ in second term, at denominator we are having square of (s+1)
Last edited by mr fantastic; Nov 24th 2011 at 06:02 PM. Reason: Title.
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Originally Posted by moonnightingale What is Laplace Inverse of exp(-s) / (s+1) and also for $\displaystyle exp(-s) /[ (s+1)(s+1)]$ in second term, at denominator we are having square of (s+1) $\displaystyle \displaystyle \mathbf{L}^{-1}\left\{e^{-sT}F(s)\right\} = \begin{cases} 0 \textrm{ if } 0 \leq t < T \\ f(t-T) \textrm{ if } t \geq T \end{cases}$
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