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===Answer 1=== | ===Answer 1=== | ||
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| + | <math> x[n] = u[n+2]-u[n-1]</math>. | ||
<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> | ||
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<math>= \sum_{n=-2}^{0} x[n] e^{-j\omega n}</math> | <math>= \sum_{n=-2}^{0} x[n] e^{-j\omega n}</math> | ||
| − | <math>= e^{ | + | <math>= 1+ e^{j\omega} + e^{2j\omega} </math> |
Revision as of 14:11, 12 September 2013
Contents
Practice Problem on Discrete-time Fourier transform computation
Compute the discrete-time Fourier transform of the following signal:
$ x[n]= u[n+2]-u[n-1] $
See these Signal Definitions if you do not know what is the step function "u[n]".
(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
$ x[n] = u[n+2]-u[n-1] $.
$ X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $
$ = \sum_{n=-2}^{0} x[n] e^{-j\omega n} $
$ = 1+ e^{j\omega} + e^{2j\omega} $
Answer 2
alec green
$ X_{2\pi}(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $
$ = \sum_{n=-2}^{0} x[n] e^{-j\omega n} $
$ = e^{2j\omega} + e^{j\omega} + 1 $
Answer 3
Write it here.
