| Line 28: | Line 28: | ||
=\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) ] | =\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> | </math> | ||
| + | |||
===Answer 2=== | ===Answer 2=== | ||
| − | + | <math> x[n] = \int_{-\pi}^{\pi} \mathcal{X} (w)e^{j\omega n} dw | |
| + | </math> | ||
| + | |||
| + | The input x[n] can can be written in the exponential form. | ||
| + | |||
| + | <math> x[n] = cos(\frac{2\pi}{500}n) | ||
| + | = \frac{e^{j\frac{2\pi}{500}n} + e^{-j\frac{2\pi}{500}n}}{2} | ||
| + | </math> | ||
| + | |||
| + | In order for the input x[n] to have such a value, | ||
| + | |||
| + | <math> \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) | ||
| + | </math> | ||
| + | |||
| + | |||
===Answer 3=== | ===Answer 3=== | ||
write it here. | write it here. | ||
---- | ---- | ||
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
Revision as of 14:50, 6 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
$ x[n] = \int_{-\pi}^{\pi} \mathcal{X} (w)e^{j\omega n} dw $
The input x[n] can can be written in the exponential form.
$ x[n] = cos(\frac{2\pi}{500}n) = \frac{e^{j\frac{2\pi}{500}n} + e^{-j\frac{2\pi}{500}n}}{2} $
In order for the input x[n] to have such a value,
$ \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) $
Answer 3
write it here.
