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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 e^{\alpha t} e^{-st}dt
= A \int_0^\infty e^{(\alpha - s) t} dt \\
&=& \left.\frac{A}{\alpha-s}e^{(\alpha - s) t} \right\vert _{0}^{\infty}
= \frac{A}{\alpha-s}e^{(\alpha - \sigma - j\omega) \infty}
- \frac{A}{\alpha-s}\\
&=& \left\{\begin{array}{ll}
\frac{A}{s-\alpha}, & \sigma>\alpha \\ [5pt]
\hbox{(indeterminate)}, & \sigma=\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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``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