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Existence of the Laplace Transform

A function $ x(t)$ has a Laplace transform whenever it is of exponential order. That is, there must be a real number $ B$ such that

$\displaystyle \lim_{t\to\infty} \left\vert x(t)e^{-Bt}\right\vert=0
$

As an example, every exponential function $ Ae^{\alpha t}$ has a Laplace transform for all finite values of $ A$ and $ \alpha$. Let's look at this case more closely.

The Laplace transform of a causal, growing exponential function

$\displaystyle x(t) = \left\{\begin{array}{ll}
A e^{\alpha t}, & t\geq 0 \\ [5pt]
0, & t<0 \\
\end{array}\right.,
$

is given by

\begin{eqnarray*}
X(s) &\isdef & \int_0^\infty x(t) e^{-st}dt
= \int_0^\infty A...
...alpha \\ [5pt]
\infty, & \sigma<\alpha \\
\end{array} \right.
\end{eqnarray*}

Thus, the Laplace transform of an exponential $ Ae^{\alpha t}$ is $ A/(s-\alpha)$, but this is defined only for re$ \left\{s\right\}>\alpha$.


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Previous: Introduction to Laplace Transform Analysis
Next: Analytic Continuation

written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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