(→The Fourier Transform) |
(→The Fourier Transform) |
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<math>X(\omega)= \frac{-1}{2 j (j(6 \pi - \omega)-4)} + \frac{1}{2 j (-j(6 \pi + \omega)-4)}</math> | <math>X(\omega)= \frac{-1}{2 j (j(6 \pi - \omega)-4)} + \frac{1}{2 j (-j(6 \pi + \omega)-4)}</math> | ||
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| + | A faster way to solve this problem would be to use the Multiplication Property | ||
Revision as of 15:44, 7 October 2008
The Signal
$ (t e^{-4t} \sin{6 \pi t}) u(t) $
The Fourier Transform
$ X(\omega)=\int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt $
$ X(\omega)=\int_{-\infty}^{\infty} (te^{-4t}\sin{6\pi t})u(t) e^{-j\omega t}dt $
$ X(\omega)=\int_{0}^{\infty} (te^{-4t}\sin{6\pi t}) e^{-j\omega t}dt $
$ X(\omega)=\int_{0}^{\infty} (te^{-4t})(\frac {e^{j 6 \pi t} - e^{-j 6 \pi t}}{2 j}) e^{-j\omega t}dt $
$ X(\omega)=\int_{0}^{\infty} \frac {t e^{-4t} e^{j 6 \pi t} e^{-j\omega t}}{2 j} - \frac {t e^{-4t} e^{-j 6 \pi t} e^{-j\omega t}}{2 j}dt $
$ X(\omega)=\int_{0}^{\infty} \frac {t e^{t(j(6 \pi - \omega)-4)}}{2 j} - \frac {t e^{t(-j(6 \pi + \omega)-4)}}{2 j}dt $
$ X(\omega)= \frac{(t (j(6 \pi - \omega)-4) - 1) e^{t(j(6 \pi - \omega)-4)}}{2 j (j(6 \pi - \omega)-4)} - \frac{(t (-j(6 \pi + \omega)-4) - 1) e^{t(-j(6 \pi + \omega)-4)}}{2 j (-j(6 \pi + \omega)-4)}\bigg]_0^\infty $
$ X(\omega)= \frac{-1}{2 j (j(6 \pi - \omega)-4)} + \frac{1}{2 j (-j(6 \pi + \omega)-4)} $
A faster way to solve this problem would be to use the Multiplication Property
