Existence of the Laplace Transform
A function![$ x(t)$](http://www.dsprelated.com/josimages_new/filters/img1659.png)
![$ B$](http://www.dsprelated.com/josimages_new/filters/img949.png)
![$\displaystyle \lim_{t\to\infty} \left\vert x(t)e^{-Bt}\right\vert=0
$](http://www.dsprelated.com/josimages_new/filters/img1672.png)
![$ Ae^{\alpha t}$](http://www.dsprelated.com/josimages_new/filters/img1673.png)
![$ A$](http://www.dsprelated.com/josimages_new/filters/img122.png)
![$ \alpha$](http://www.dsprelated.com/josimages_new/filters/img406.png)
The Laplace transform of a causal, growing exponential function
![$\displaystyle x(t) = \left\{\begin{array}{ll}
A e^{\alpha t}, & t\geq 0 \\ [5pt]
0, & t<0 \\
\end{array}\right.,
$](http://www.dsprelated.com/josimages_new/filters/img1674.png)
![\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*}](http://www.dsprelated.com/josimages_new/filters/img1675.png)
Thus, the Laplace transform of an exponential
is
, but this is defined only for
re
.
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