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===Answer 2=== | ===Answer 2=== | ||
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| + | [[Image:green26_ece438_hmwrk3_rect.png|x[n]]] | ||
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| + | <math>X_{2\pi}(\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> | ||
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| + | <math>= e^{2j\omega} + e^{j\omega} + 1</math> | ||
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| + | ===Answer 3=== | ||
Write it here. | Write it here. | ||
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[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013]] | [[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013]] | ||
Revision as of 12:14, 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
$ \sum \sum_n^\infty x[n] \delta[n] $.
Answer 2
![x[n]](/rhea/index.php/File:Green26_ece438_hmwrk3_rect.png)
$ 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.
