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=== Answer 1  ===
 
=== Answer 1  ===
Write it here.
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<math>\cos \left( \frac{\pi}{6}n \right)=\frac{1}{2}e^{j\frac{\pi}{6}n}+\frac{1}{2}e^{-j\frac{\pi}{6}n}</math>
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<math>\mathcal X (\omega)=\sum_{m=-\infty}^\infty 2\pi \delta (\omega-k\omega_0+2\pi m)</math>
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<math>\mathcal X (\omega)=\sum_{m=-\infty}^\infty 2\pi \delta (\omega-\frac{\pi}{6}+2\pi m)+\sum_{m=-\infty}^\infty 2\pi \delta (\omega+\frac{\pi}{6}+2\pi m)</math>
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--[[User:Cmcmican|Cmcmican]] 19:51, 28 February 2011 (UTC)
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=== Answer 2  ===
 
=== Answer 2  ===
 
Write it here.
 
Write it here.

Revision as of 15:51, 28 February 2011


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

Compute the Fourier transform of the signal

$ x[n] = \cos \left( \frac{\pi}{6}n \right).\ $


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Answer 1

$ \cos \left( \frac{\pi}{6}n \right)=\frac{1}{2}e^{j\frac{\pi}{6}n}+\frac{1}{2}e^{-j\frac{\pi}{6}n} $

$ \mathcal X (\omega)=\sum_{m=-\infty}^\infty 2\pi \delta (\omega-k\omega_0+2\pi m) $

$ \mathcal X (\omega)=\sum_{m=-\infty}^\infty 2\pi \delta (\omega-\frac{\pi}{6}+2\pi m)+\sum_{m=-\infty}^\infty 2\pi \delta (\omega+\frac{\pi}{6}+2\pi m) $

--Cmcmican 19:51, 28 February 2011 (UTC)

Answer 2

Write it here.

Answer 3

Write it here.


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