## FOURIER TRANSFORM

$x(t) = e^{-3|t-2|}$

Noticing that there is an absolute value, we can proceed to divide in tow cases.

When

$t-1 < 0 \rightarrow x_1(t) = e^{3t-3}$

and when,

$t-1 >0 \rightarrow x_2(t) = e^{-3t+3}$

So, we can then compute the Fourier series by adding the integrals of each diferent case.

$\ \mathcal{X}(\omega) = \int_{-\infty}^{\infty} x_1(t)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty} x_2(t)e^{-j\omega t} \,dt$

$\mathcal{X}(\omega) = \int_{-\infty}^{1} e^{3t-3}e^{-j\omega t}\,dt + \int_{1}^{\infty} e^{-3t+3}e^{-j\omega t} \,dt$

$\mathcal{X}(\omega) = \frac{1}{e^{3}} \int_{-\infty}^{1} e^{3t-j\omega t}\,dt + e^{3} \int_{1}^{\infty} e^{-3t-j\omega t} \,dt$

$\mathcal{X}(\omega) = \frac{1}{e^{3}} \int_{-\infty}^{1} e^{t(3-j\omega)}\,dt + e^{3} \int_{1}^{\infty} e^{-t(3+j\omega)} \,dt$

$\mathcal{X}(\omega) = {\left. \frac{e^{t(3-j\omega)}}{3-j\omega} \right]^{1}_{-\infty} } \frac{1}{e^{3}} + {\left. -\frac{e^{-t(3+j\omega)}}{3+j\omega} \right]^{\infty}_1 } e^{3}\,$

$\mathcal{X}(\omega) = \frac{1}{e^{3}} \frac{e^{3-j\omega}}{3-j\omega} + e^{3} \frac{e^{-3-j\omega}}{3+j\omega}$

$\mathcal{X}(\omega) = \frac{e^{-j\omega}}{3-j\omega} + \frac{e^{-j\omega}}{3+j\omega}$

## Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva