(Rhea Brian Thomas HW5.2) |
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| − | '''Problem: (From Oppenheim/Wisllsky, 4.21 b.)''' Find <math>F(e^{-3|t|}sin(2t)).</math> | + | '''Problem: (From Oppenheim/Wisllsky, 4.21 b.)''' Find <math>F(e^{-3|t|}sin(2t)).</math> (F meaning the Fourier Transform of said function.) |
'''Solution:''' First, observe that <math>sin(2t) = \frac{1}{2j} (e^{j2t}-e^{-j2t})</math>. Then, applying the forumla for determining the Fourier transform, | '''Solution:''' First, observe that <math>sin(2t) = \frac{1}{2j} (e^{j2t}-e^{-j2t})</math>. Then, applying the forumla for determining the Fourier transform, | ||
Revision as of 17:07, 7 October 2008
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)}) $
