(inverse laplace transform)
(inverse laplace transform)
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== inverse laplace transform ==
 
== inverse laplace transform ==
The inverse Laplace transform is given by the following complex integral, which is known by various names (the '''Bromwich integral''', the '''Fourier-Mellin integral''', and '''Mellin's inverse formula'''):
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The inverse Laplace transform is given by the following complex integral
 
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: <math>f(t) = \mathcal{L}^{-1} \{F(s)\} = \frac{1}{2 \pi i} \int_{ \gamma - i \cdot \infty}^{ \gamma + i \cdot \infty} e^{st} F(s)\,ds,</math>
 
: <math>f(t) = \mathcal{L}^{-1} \{F(s)\} = \frac{1}{2 \pi i} \int_{ \gamma - i \cdot \infty}^{ \gamma + i \cdot \infty} e^{st} F(s)\,ds,</math>

Revision as of 16:22, 24 November 2008

definition

The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by:

$ F(s) = \mathcal{L} \left\{f(t)\right\}=\int_{0^-}^{\infty} e^{-st} f(t) \,dt. $

The lower limit of 0 is short notation to mean

$ \lim_{\varepsilon\to 0+}\int_{-\varepsilon}^\infty $

and assures the inclusion of the entire Dirac delta function δ(t) at 0 if there is such an impulse in f(t) at 0.

The parameter s is in general complex number:

$ s = \sigma + i \omega \, $

inverse laplace transform

The inverse Laplace transform is given by the following complex integral

$ f(t) = \mathcal{L}^{-1} \{F(s)\} = \frac{1}{2 \pi i} \int_{ \gamma - i \cdot \infty}^{ \gamma + i \cdot \infty} e^{st} F(s)\,ds, $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood