Difference between revisions of "HW5.2 Josh Long ECE301Fall2008mboutin" - Rhea

Revision as of 16:44, 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/easier way to solve this problem would be to use the Multiplication Property