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 $

$ \begin{align}\mathcal{F}[x[n]] = 2\pi * \frac{1}{2} [\delta(\omega - \frac{2\pi}{500}n) + \delta(\omega + \frac{2\pi}{500}n)] \end{align} $

Answer 5

$ \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} $

$ F[x[n]] = \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) $ repp'ed every $ 2\pi $

Answer 6

$ x[n]= \cos \left( \frac{2 \pi }{500} n \right)=\frac{1}{2}(e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}) $

$ \mathcal{X} (\omega) = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right] $

Answer 7

$ \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} $

$ F[x[n]] = \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) $

Note that the since we are dealing with a DT signal, it repeats every $ 2\pi $

Answer 8

$ \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} $

$ \mathcal{X} (\omega) = F[x[n]] = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) $

    $  ( \omega \in [-\pi,\pi]) $

$ \mathcal{X} (\omega) = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right] $

Answer 9

$ x[n]= \cos \left( \frac{2 \pi }{500} n \right)=\frac{1}{2}(e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}) $

$ F[x[n]] = \mathcal{X} (\omega) = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right] $

Answer 10

$ \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} $

$ \mathcal{X} (\omega) = F[x[n]] = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right] $

Back to ECE438 Fall 2011 Prof. Boutin