Laplace Transform of Shifted Dirac Delta Function

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Theorem

Let $\map \delta t$ denote the Dirac delta function.


The Laplace transform of $\map \delta {t - a}$ is given by:

$\laptrans {\map \delta {t - a} } = e^{-a s}$


Proof 1

\(\displaystyle \laptrans {\map \delta {t - a} }\) \(=\) \(\displaystyle \int_0^{\to +\infty} e^{-s t} \map \delta {t - a} \rd t\) Definition of Laplace Transform
\(\displaystyle \) \(=\) \(\displaystyle e^{-s \times a}\) Integral to Infinity of Shifted Dirac Delta Function by Continuous Function
\(\displaystyle \) \(=\) \(\displaystyle e^{-a s}\)

$\blacksquare$


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