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

Compute the Fourier transform of the signal

$x(t) = \left\{ \begin{array}{ll} 1, & \text{ for } -5\leq t \leq 5,\\ 0, & \text{ for } 5< |t| \leq 10, \end{array} \right. \$

x(t) periodic with period 20.

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For a square wave,

$a_k=\frac{sin(k\omega_0 T_1)}{\pi k}$

In this case,

$\omega_0=\frac{2\pi}{20}=\frac{\pi}{10} \mbox{ and } T_1 = 5$

Therefore

$\chi(\omega)=\sum_{k=-\infty}^{\infty}2\pi\frac{sin(k\frac{\pi}{10} 5)}{\pi k}\delta(\omega-k\frac{\pi}{10})=\sum_{k=-\infty}^{\infty}2\frac{sin(k\frac{\pi}{2})}{ k}\delta(\omega-k\frac{\pi}{10})$

--Cmcmican 21:13, 21 February 2011 (UTC)

TA's comments: Good Job. You may use \sin to produce a $\sin$.