Revision as of 16:44, 7 October 2008 by Longja (Talk)

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

Alumni Liaison

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

Buyue Zhang