Laplace Transform of Function of t minus a/Proof 2
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Theorem
Let $f$ be a function such that $\laptrans f$ exists.
Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of $f$.
Let $a \in \C$ or $\R$ be constant.
Let $g$ be the function defined as:
- $\map g t = \begin{cases} \map f {t - a} & : t > a \\ 0 & : t \le a \end{cases}$
Then:
- $\laptrans {\map g t} = e^{-a s} \map F s$
Proof
\(\ds \laptrans {\map g t}\) | \(=\) | \(\ds \int_0^\infty e^{-s t} \map g t \rd t\) | Definition of Laplace Transform | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^a e^{-s t} \map g t \rd t + \int_a^\infty e^{-s t} \map g t \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^a 0 \times e^{-s t} \rd t + \int_a^\infty e^{-s t} \map f {t - a} \rd t\) | Definition of $\map g t$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^\infty e^{-s t} \map f {t - a} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty e^{-s \paren {u + a} } \map f u \rd u\) | Integration by Substitution: $t = u + a$ | |||||||||||
\(\ds \) | \(=\) | \(\ds e^{-a s} \int_0^\infty e^{-s u} \map f u \rd u\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{-a s} \map F s\) | Definition of Laplace Transform |
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Translation and Change of Scale Properties: $9$