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Differentiation

The differentiation theorem for Laplace transforms states that

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

where $ {\dot x}(t) \isdef \frac{d}{dt}x(t)$ , and $ x(t)$ is any differentiable function that approaches zero as $ t$ goes to infinity. In operator notation,

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



Proof: This follows immediately from integration by parts:

\begin{eqnarray*}
{\cal L}_{s}\{{\dot x}\} &\isdef & \int_{0}^\infty {\dot x}(t) e^{-s t} dt\\
&=& \left. x(t)e^{-s t}\right\vert _{0}^{\infty} -
\int_{0}^\infty x(t) (-s)e^{-s t} dt\\
&=& 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}}
$

Thus, successive time derivatives correspond to successively higher powers of $ s$ , and successive integrals with respect to time correspond to successively higher powers of $ 1/s$ .


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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA