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Hi guys I'm currently studying for finals and I came across a square pulse problem.

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I ended up getting this:

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I'm wondering if I can simplify this to a sin function. If so how? And please correct me if I did the problem wrong in the first place.

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    $\begingroup$ Hint: Euler's formula $\sin \theta = \frac{e^{j\theta}-e^{-j\theta}}{2j}$ $\endgroup$ Dec 16, 2013 at 12:50
  • $\begingroup$ Another hint: simplify in the time domain first. Draw the signal or just list out a few samples. $\endgroup$
    – Hilmar
    Dec 16, 2013 at 22:51

1 Answer 1

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You could rewrite $x(t)$ as a rectangular function before doing the Fourier transform. You could deduce the width and center of the rect signal directly from the given expression but it would be a easier to just draw the signal and deduce the values from the graph.

Your rect signal will be in the following form:

$x(t) = rect(\frac{t-t_o}{W})$

where:

$t_0$ - center of the rectangular signal

W - width(duration) of the pulse

From this point you would use a Fourier transform table and some Fourier transform theorems (involving scaling and delay) to get the $sinc$ version of the $X(\omega)$.

Or, if you prefer, you could just modify a little bit the expression you already have to get the expression where you could use Euler's formula as Dilip Sarwate commented.

$X(\omega)=1 \cdot \frac{1}{j\omega}(e^{-j\omega} - e^{-j2\omega})= e^{-j1.5\omega}e^{+j1.5\omega} \frac{1}{j\omega}(e^{-j\omega} - e^{-j2\omega}) = e^{-j1.5\omega}\frac{1}{j\omega}(e^{+j0.5\omega} - e^{-j0.5\omega})$

From this point it should be easy to get the result.

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