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===Answer 1=== | ===Answer 1=== | ||
| − | + | <math> | |
| + | \mathcal{F}(x[n]) = \mathcal{F}(cos(\frac{2\pi}{500}n)) = \mathcal{F}(\frac{ | ||
| + | e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2}) | ||
| + | =\frac{1}{2}( \mathcal{F}(e^{j\frac{2\pi}{500}n})+\mathcal{F}(e^{-j\frac{2\pi}{500}n})) | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | =\frac{1}{2}( \pi\sum_{l=-\infty}^{+\infty}\delta(w-\frac{2\pi}{500}-2\pi l) + \pi\sum_{l=- \infty}^{+\infty}\delta(w+\frac{2\pi}{500}-2\pi l) ) | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | =\frac{\pi}{2} \sum_{l=-\infty}^{+\infty}[ \delta(w-\frac{2\pi}{500}-2\pi l)+\delta(w+\frac{2\pi}{500}-2\pi l) ] | ||
| + | </math> | ||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 08:45, 5 September 2011
Contents
Discrete-time Fourier transform computation
Compute the discrete-time Fourier transform of the following signal:
$ x[n]= \cos \left( \frac{2 \pi }{500} n \right) $
(Write enough intermediate steps to fully justify your answer.)
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
$ \mathcal{F}(x[n]) = \mathcal{F}(cos(\frac{2\pi}{500}n)) = \mathcal{F}(\frac{ e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2}) =\frac{1}{2}( \mathcal{F}(e^{j\frac{2\pi}{500}n})+\mathcal{F}(e^{-j\frac{2\pi}{500}n})) $
$ =\frac{1}{2}( \pi\sum_{l=-\infty}^{+\infty}\delta(w-\frac{2\pi}{500}-2\pi l) + \pi\sum_{l=- \infty}^{+\infty}\delta(w+\frac{2\pi}{500}-2\pi l) ) $
$ =\frac{\pi}{2} \sum_{l=-\infty}^{+\infty}[ \delta(w-\frac{2\pi}{500}-2\pi l)+\delta(w+\frac{2\pi}{500}-2\pi l) ] $
Answer 2
Write it here.
Answer 3
write it here.
