| Line 30: | Line 30: | ||
<math>X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n}</math> | <math>X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n}</math> | ||
| − | <math>X_(\omega) = \sum_{n=-\infty}^{+\infty} e^{-j2 \pi /100 n} e^{-j\omega n}</math> | + | <math>X_(\omega) = \sum_{n=-\infty}^{+\infty} e^{-j2 \pi /100 n} e^{-j\omega n} + \sum_{n=-\infty}^{+infty} e^{-j2 \pi /100 n} e^{-j\omega n}</math> |
| + | |||
| + | |||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 17:32, 12 September 2013
Contents
Practice Problem on Discrete-time Fourier transform computation
Compute the discrete-time Fourier transform of the following signal:
$ x[n]= \sin \left( \frac{2 \pi }{100} 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!
No need to write your name: we can find out who wrote what by checking the history of the page.
Answer 1
$ x[n]=\sin \left( \frac{2pi}{100} \right) $
$ x[n] = \frac{1}{2j} \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right) $
$ X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $
$ X_(\omega) = \sum_{n=-\infty}^{+\infty} e^{-j2 \pi /100 n} e^{-j\omega n} + \sum_{n=-\infty}^{+infty} e^{-j2 \pi /100 n} e^{-j\omega n} $
Answer 2
Write it here.
Answer 3
We can set the $ x[n]=\sin \left( frac{2\pi} \100 \right) $
