(New page: Category:ECE301Spring2011Boutin Category:Problem_solving ---- = Practice Question on Computing the Fourier Transform of a Discrete-time Signal = Compute the Fourier transform of ...) |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 1: | Line 1: | ||
[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
---- | ---- | ||
| − | = Practice Question on Computing the Fourier Transform of a Discrete-time Signal = | + | = [[:Category:Problem_solving|Practice Question]] on Computing the Fourier Transform of a Discrete-time Signal = |
Compute the Fourier transform of the signal | Compute the Fourier transform of the signal | ||
| Line 12: | Line 12: | ||
---- | ---- | ||
=== Answer 1 === | === Answer 1 === | ||
| − | + | ||
| + | <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> | ||
| + | |||
| + | <math>\mathcal X (\omega)=\sum_{m=-\infty}^\infty 2\pi \delta (\omega-k\omega_0+2\pi m)</math> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | --[[User:Cmcmican|Cmcmican]] 19:51, 28 February 2011 (UTC) | ||
| + | |||
=== Answer 2 === | === Answer 2 === | ||
Write it here. | Write it here. | ||
Latest revision as of 09:27, 11 November 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.
