# Practice Question on Computing the Fourier Transform of a Continuous-time Signal

Compute the Fourier transform of the signal

$x(t) = e^{-t} u(t).\$

$\boldsymbol{\chi}(\omega)=\int_{-\infty}^\infty x(t)e^{-j\omega t}dt=\int_{-\infty}^\infty e^{-t} u(t)e^{-j\omega t}dt=\int_{0}^\infty e^{-t(j\omega+1)}dt=\frac{e^{-t(j\omega+1)}}{-(j\omega+1)}\Bigg|_0^\infty=\frac{e^{-\infty}}{-(j\omega+1)}-\frac{e^{0}}{-(j\omega+1)}=\frac{1}{j\omega+1}$

$\boldsymbol{\chi}(\omega)=\frac{1}{j\omega+1}$

--Cmcmican 02:31, 19 February 2011 (UTC)

Instructor's comments: Good! Notice that, in contrast with the first example we did in class, you did not have to worry about the denominator being zero for certain values of the frequency. -pm