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

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).

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

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