| Line 25: | Line 25: | ||
=\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{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> | ||
| − | + | :<span style="color:red">Instructor's comment: You need to justify this step (i.e. the previous equality). -pm</span> | |
<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) ] | =\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) ] | ||
| Line 31: | Line 31: | ||
DTFT f(w) is periodic funtion, just need to include one period to be sufficient | DTFT f(w) is periodic funtion, just need to include one period to be sufficient | ||
| + | :<span style="color:green">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</span> | ||
===Answer 2=== | ===Answer 2=== | ||
| Line 42: | Line 43: | ||
</math> | </math> | ||
| − | In order for the input x[n] to have such a value, | + | In order for the input x[n] to have such a value, <span style="color:red">(Please justify! -pm)</span> |
<math> \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) | <math> \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) | ||
| Line 54: | Line 55: | ||
</math> | </math> | ||
| − | + | :<span style="color:red">Instructor's comment: Obviously, the previous line contains a mistake (missing a cosine). -pm)</span> | |
<math> \mathcal{X} (\omega) = \pi (\delta(\omega - \frac{2\pi}{500}) + \delta(\omega + \frac{2\pi}{500})) | <math> \mathcal{X} (\omega) = \pi (\delta(\omega - \frac{2\pi}{500}) + \delta(\omega + \frac{2\pi}{500})) | ||
</math> | </math> | ||
| + | :<span style="color:red">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 </span> | ||
| Line 76: | Line 78: | ||
<math>\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}</math> | <math>\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}</math> | ||
| + | |||
| + | :<span style="color:red">Instructor's comment: What about other values of omega? Also, be careful not to confuse the convolution symbol (*) with a multiplication. -pm </span> | ||
===Answer 5=== | ===Answer 5=== | ||
| Line 85: | Line 89: | ||
<math> F[x[n]] = \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) | <math> F[x[n]] = \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) | ||
</math> repp'ed every <math>2\pi</math> | </math> repp'ed every <math>2\pi</math> | ||
| + | :<span style="color:red">Instructor's comment: And the justification for this last step is ??? -pm </span> | ||
===Answer 6=== | ===Answer 6=== | ||
Revision as of 08:30, 16 November 2011
Contents
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.)
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) ) $
- 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] $
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] $
