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<math> = \int_{-\infty}^{0} e^{2|t|}cos(8t) e^{-j\omega t} dt \! + \int_{0}^{\infty} e^{-2|t|}cos(8t) e^{-j\omega t} dt \!</math>
 
<math> = \int_{-\infty}^{0} e^{2|t|}cos(8t) e^{-j\omega t} dt \! + \int_{0}^{\infty} e^{-2|t|}cos(8t) e^{-j\omega t} dt \!</math>
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after quite a bit of math I get the answer to be
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<math>\frac{1}{2}(\frac{1}{2 + j8 - jw} + \frac{1}{2 -j8 -jw} + \frac{1}{2 - j8 - jw} \frac{1}{2 + j8 + jw})</math>
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I'm not sure if I'm right though because when I checked it in matlab the answer I got was
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<pre> 4*(68+w^2)/(68+w^2-16*w)/(68+w^2+16*w) <\pre>

Revision as of 12:07, 8 October 2008

$ \ x(t) = e^{-2|t|}cos(8t) $

$ X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \! $

$ = \int_{-\infty}^{\infty} e^{-2|t|}cos(8t) e^{-j\omega t} dt \! $

$ = \int_{-\infty}^{0} e^{2|t|}cos(8t) e^{-j\omega t} dt \! + \int_{0}^{\infty} e^{-2|t|}cos(8t) e^{-j\omega t} dt \! $


after quite a bit of math I get the answer to be


$ \frac{1}{2}(\frac{1}{2 + j8 - jw} + \frac{1}{2 -j8 -jw} + \frac{1}{2 - j8 - jw} \frac{1}{2 + j8 + jw}) $


I'm not sure if I'm right though because when I checked it in matlab the answer I got was

 4*(68+w^2)/(68+w^2-16*w)/(68+w^2+16*w) <\pre>

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett