Revision as of 08:37, 16 November 2011

Practice Problem on 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!

Note from the instructor: There is still not a single solution on this page that would deserve full credit. Please keep trying. -pm

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) ) $

Instructor's comment: You need to justify this step (i.e. the previous equality). -pm

$ =\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) ] $

DTFT f(w) is periodic funtion, just need to include one period to be sufficient

Instructor's comment: Correct, but do not write your answer in such a way that it looks like the FT is zero outside of one period. -pm

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, (Please justify! -pm)

$ \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} $

Instructor's comment: Obviously, the previous line contains a mistake (missing a cosine). -pm)

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

Instructor's comment: You would get close to zero points for this answer, because a) it is not periodic, b) it is just "plugged in" without justification. -pm

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} $

Instructor's comment: What about other values of omega? Also, be careful not to confuse the convolution symbol (*) with a multiplication. -pm

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 $

Instructor's comment: And the justification for this last step is ??? -pm

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] $

Instructor's comment: Please justify this last step! -pm

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 $

Instructor's comment: How did you find out the FT of a complex exponential? Please justify. -pm

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}) $ (Why? -pm )

    $  ( \omega \in [-\pi,\pi]) $ (This should be above the previous equation for clarity. -pm) 

$ \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] $

Instructor's comment: You did not fully justify your answer. -pm

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] $

Instructor's comment: Not enough justification for full credit. -pm

Back to ECE438 Fall 2011 Prof. Boutin