First Translation Property of Laplace Transforms

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Theorem

Let $\map f t: \R \to \R$ or $\R \to \C$ be a function of exponential order $a$ for some constant $a \in \R$.

Let $\laptrans {\map f t} = \map F s$ be the Laplace transform of $f$.

Let $e^t$ be the exponential function.


Then:

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

everywhere that $\laptrans f$ exists, for $\map \Re s > a$


Proof

\(\ds \laptrans {e^{a t} \map f t}\) \(=\) \(\ds \int_0^{\to +\infty} e^{-s t} e^{a t} \map f t \rd t\) Definition of Laplace Transform
\(\ds \) \(=\) \(\ds \int_0^{\to +\infty} e^{-s t + a t} \map f t \rd t\) Exponent Combination Laws
\(\ds \) \(=\) \(\ds \int_0^{\to +\infty} e^{-\paren {s - a} t} \map f t \rd t\)
\(\ds \) \(=\) \(\ds \map F {s - a}\) Definition of Laplace Transform

$\blacksquare$


Examples

Example $1$

$\laptrans {e^{-t} \cos 2 t} = \dfrac {s + 1} {s^2 + 2 s + 5}$


Example $2$

$\laptrans {t^2 e^{3 t} } = \dfrac 2 {\paren {s - 3}^3}$


Example $3$

$\laptrans {e^{-2 t} \sin 4 t} = \dfrac 4 {s^2 + 4 s + 20}$


Example $4$

$\laptrans {e^{4 t} \cosh 5 t} = \dfrac {s - 4} {s^2 - 8 s - 9}$


Example $5$

$\laptrans {e^{-3 t} \paren {3 \cos 6 t - 5 \sin 6 t} } = \dfrac {3 s - 24} {s^2 + 4 s + 40}$


Also known as

The first translation property is also known as the first shifting property.


Also see


Sources

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