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=== Answer 1 === | === Answer 1 === | ||
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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 14:51, 28 February 2011
Contents
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).\ $
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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.
