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== Laplace Transform ==
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The laplace transform of a general signal <math> x(t) </math> is defined as
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<math> X(s) = \int_{-\infty}^{\infty}x(t)e^{-st}dt </math>
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The complex variable s can be written as <math> s = \sigma + jw</math>
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S means complex plane. So the laplace transform <math> X(s) </math> is on complex plane.
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However, the Fourier Transform <math> X(jw) </math> is on pure imaginary axis.
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Here is an example.
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<math> x(t) = -e^{-qt}u(-t) </math>
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<math> X(s) = -\int_{-\infty}^{\infty}e^{-qt}u(-t)e^{-st}dt</math>
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<math>      = -\int_{-\infty}^{0}e^{-qt}e^{-st}dt</math>
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<math>      = -\int_{-\infty}^{0}e^{-(q+s}t}dt</math>
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<math>      = -\int_{-\infty}^{0}e^{-(q+a+jw}t}dt</math>
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<math>      = -\int_{-\infty}^{0}e^{-(q+a}t}e^{-jwt}dt</math>
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if <math> q+a \geq 0, </math> integral diverges
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if <math> q+a \l 0, X(s) = \frac{1}{q+a} </math>
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Then we can figure out where the laplace transform converges.
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In the example, the lapace transform only converges when <math> q+a \l 0 </math>
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That region is called Region of converge, ROC.

Revision as of 09:12, 23 November 2008

Laplace Transform

The laplace transform of a general signal $ x(t) $ is defined as

$ X(s) = \int_{-\infty}^{\infty}x(t)e^{-st}dt $

The complex variable s can be written as $ s = \sigma + jw $

S means complex plane. So the laplace transform $ X(s) $ is on complex plane.

However, the Fourier Transform $ X(jw) $ is on pure imaginary axis.

Here is an example.

$ x(t) = -e^{-qt}u(-t) $

$ X(s) = -\int_{-\infty}^{\infty}e^{-qt}u(-t)e^{-st}dt $

$ = -\int_{-\infty}^{0}e^{-qt}e^{-st}dt $

$ = -\int_{-\infty}^{0}e^{-(q+s}t}dt $

$ = -\int_{-\infty}^{0}e^{-(q+a+jw}t}dt $

$ = -\int_{-\infty}^{0}e^{-(q+a}t}e^{-jwt}dt $

if $ q+a \geq 0, $ integral diverges

if $ q+a \l 0, X(s) = \frac{1}{q+a} $

Then we can figure out where the laplace transform converges.

In the example, the lapace transform only converges when $ q+a \l 0 $

That region is called Region of converge, ROC.

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood