Difference between revisions of "HW5.2 Brian Thomas ECE301Fall2008mboutin" - Rhea

Latest revision as of 11:33, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform

Problem: (From Oppenheim/Wisllsky, 4.21 b.) Find $ F(e^{-3|t|}sin(2t)). $ (F meaning the Fourier Transform of said function.)

Solution: First, observe that $ sin(2t) = \frac{1}{2j} (e^{j2t}-e^{-j2t}) $. Then, applying the forumla for determining the Fourier transform,

$ F(e^{-3|t|}sin(2t)) = \int_{-\infty}^\infty e^{-3|t|}sin(2t) e^{-jwt} dt $

$ = \int_{-\infty}^\infty e^{-3|t|} \frac{1}{2j} (e^{j2t}-e^{-j2t}) e^{-jwt} dt $

The simplest way to integrate an absolute value is to split the absolute value up into its positive and negative sections. (Recall that |t| = {t, t ≥ 0; -t, t < 0}.) Hence,

$ = \int_{-\infty}^0 e^{3t} \frac{1}{2j} (e^{j2t}-e^{-j2t}) e^{-jwt} dt + \int_{0}^\infty e^{-3t} \frac{1}{2j} (e^{j2t}-e^{-j2t}) e^{-jwt} dt $

$ =\frac{1}{2j} ( \int_{-\infty}^0 e^{(3 + j(2-w))t} - e^{(3 + j(-2-w))t} dt + \int_{0}^\infty e^{(-3 + j(2-w))t} - e^{(-3 + j(-2-w))t} dt) $

$ =\frac{1}{2j} ( (\frac{e^{(3 + j(2-w))t}}{3 + j(2-w)} - \frac{e^{(3 + j(-2-w))t}}{3 + j(-2-w)})|_{t=-\infty}^{t=0} + (\frac{e^{(-3 + j(2-w))t}}{-3 + j(2-w)} - \frac{ e^{(-3 + j(-2-w))t}}{-3 + j(-2-w)})|_{t=0}^{t=\infty}) $

$ =\frac{1}{2j} ( \frac{1}{3 + j(2-w)} - \frac{1}{3 + j(-2-w)} + \frac{1}{3 - j(2-w)} - \frac{1}{3 - j(-2-w)}) $

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