Revision as of 17:26, 6 September 2011

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.)

Share your answers below

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

$ x[n] = \frac{2\pi}{500}n = \frac{e^{j\frac{2\pi}{500}n}}{2}+\frac{e^{-j\frac{2\pi}{500}n}}{2} $

$ \mathcal{X} (\omega) = \pi (\delta(\omega - \frac{2\pi}{500}) + \delta(\omega + \frac{2\pi}{500})) $

Answer 4

$ \begin{align} x[n]= \cos \left( \frac{2 \pi }{500} n \right) \\ = \frac{ e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2} \end{align} $

$ 0 < \frac{2\pi}{500}n < \pi $

$ -\pi < -\frac{2\pi}{500}n < \pi $

consider $ -\pi < \omega < \pi $

Back to ECE438 Fall 2011 Prof. Boutin