## Signal

$x(t) = e^{3jt}*(u(t+5) - u(t-5)) + e^{-2t}*u(t)\,$

## Transformed

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

$= \int_{-\infty}^{\infty}e^{3jt}*(u(t+5) - u(t-5))e^{-j\omega t}dt + \int_{-\infty}^{\infty}e^{-2t}u(t)e^{-j\omega t}dt\,$

$= \int_{-5}^{5}e^{3jt}e^{-j\omega t}dt + \int_{0}^{\infty}e^{-2t}e^{-j\omega t}dt\,$

$= \int_{-5}^{5}e^{3jt -j\omega t}dt + \int_{0}^{\infty}e^{-2t -j\omega t}dt\,$

$= \int_{-5}^{5}e^{t*(3j -j\omega )}dt + \int_{0}^{\infty}e^{t*(-2 -j\omega )}dt\,$

$= \frac{e^{3jt - j\omega t}}{3j-j\omega}]_{-5}^{5} + \frac{e^{-2t - j\omega t}}{-2 -j\omega}]_{0}^{\infty}\,$

$= \frac{e^{15j - 5j\omega} - e^{-15j + 5j\omega}}{3j-j\omega} + \frac{e^{-2*\infty - \infty j \omega} - e^{0}}{-j -j\omega}\,$

$= \frac{e^{j(15 - 5\omega )} - e^{-j(15 - 5\omega )}}{j(3-\omega )} + \frac{1}{j(1-\omega )}\,$

$= \frac{2sin(15 - 5\omega )}{3-\omega } + \frac{1}{j(1-\omega )}\,$

## Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett