Laplace Transform of Dirac Delta Function/Lemma
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Lemma for Laplace Transform of Dirac Delta Function
Let $F_\epsilon: \R \to \R$ be the real function defined as:
- $\map {F_\epsilon} t = \begin{cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le t \le \epsilon \\ 0 & : t > \epsilon \end{cases}$
Then:
- $\laptrans {\map {F_\epsilon} t} = \dfrac {1 - e^{-s \epsilon} } {\epsilon s}$
Proof
\(\ds \laptrans {\map {F_\epsilon} t}\) | \(=\) | \(\ds \int_0^\infty e^{-s t} \map {F_\epsilon} t \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\epsilon e^{-s t} \map {F_\epsilon} t \rd t + \int_\epsilon^\infty e^{-s t} \map {F_\epsilon} t \rd t\) | Sum of Integrals on Adjacent Intervals for Integrable Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\epsilon e^{-s t} \dfrac 1 \epsilon \rd t + \int_\epsilon^\infty e^{-s t} \times 0 \rd t\) | Definition of $F_\epsilon$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 \epsilon \int_0^\epsilon e^{-s t} \rd t\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 \epsilon \intlimits {\dfrac {e^{-s t} } {-s} } 0 \epsilon\) | Primitive of $e^{a x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 \epsilon \paren {\dfrac {e^{-s \epsilon} - e^{-s \times 0} } {-s} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {1 - e^{-s \epsilon} } {\epsilon s}\) | simplification |
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Impulse Functions. The Dirac Delta Function: $41$