Revision as of 15:40, 7 October 2008 by Bchanyas (Talk)

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Suppose we have a signal:

$ e^{-2(t-1)}u(t-1)\, $

The formula of Fourier Transform is:

$ X(w) = \int_{-\infty}^{ \infty} x(t)e^{-jwt}dt\, $

Substituting:

$ X(w) = \int_{-\infty}^{ \infty} e^{-2(t-1)}u(t-1)e^{-jwt}dt\, $

From the step function, the range becomes 1 to $ \infty $, so the equation becomes:

$ X(w) = \int_{1}^{ \infty} e^{-2(t-1)}e^{-jwt}dt\, $
$ X(w) = \int_{1}^{ \infty} e^{2-(2+jw)t}dt\, $

Integrating yields:

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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