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Laplace Transform Theorems

Linearity

The Laplace transform is a linear operator. To show this, let $ w(t)$ denote a linear combination of signals $ x(t)$ and $ y(t)$,

$\displaystyle w(t) = \alpha x(t) + \beta y(t), $

where $ \alpha$ and $ \beta$ are real or complex constants. Then we have

\begin{eqnarray*} W(s) &\isdef & {\cal L}_s\{w\} \isdef {\cal L}_s\{\alpha x(t) ... ..._0^\infty y(t) e^{-st} dt\\ &\isdef & \alpha X(s) + \beta Y(s). \end{eqnarray*}

Thus, linearity of the Laplace transform follows immediately from the linearity of integration.

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)... ...y} - \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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Relation to the z Transform