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Differentiation

The differentiation theorem for Laplace transforms:

$\displaystyle \zbox{{\dot x}(t) \leftrightarrow s X(s) - x(0)}
$

where $ {\dot x}(t) \mathrel{\stackrel{\mathrm{\Delta}}{=}}\frac{d}{dt}x(t)$ , and $ x(t)$ is any differentiable function that approaches zero as $ t$ goes to infinity.

Operator notation:

$\displaystyle \zbox{{\cal L}_{s}\{{\dot x}\} = s X(s) - x(0).}
$



Proof: Immediate from integration by parts:

\begin{eqnarray*}
{\cal L}_{s}\{{\dot x}\} &\mathrel{\stackrel{\mathrm{\Delta}}{=}}& \int_{0}^\infty {\dot x}(t) e^{-s t} dt\\ [10pt]
&=& \left. x(t)e^{-s t}\right\vert _{0}^{\infty} -
\int_{0}^\infty x(t) (-s)e^{-s t} dt\\ [10pt]
&=& s X(s) - x(0)
\end{eqnarray*}

since $ x(\infty)=0$ by assumption



Corollary: Integration Theorem:

$\displaystyle \zbox{{\cal L}_{s}\left\{\int_0^t x(\tau)d\tau\right\} = \frac{X(s)}{s}}
$


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``The Laplace Transform'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2014-03-24 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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