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==Share your answers below==
 
==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!
 
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!
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'''No need to write your name: we can find out who wrote what by checking the history of the page.'''
 
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===Answer 1===
 
===Answer 1===
 
Andrei Henrique Patriota Campos
 
  
 
<math> x[n] = u[n+2]-u[n-1]</math>.
 
<math> x[n] = u[n+2]-u[n-1]</math>.
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===Answer 2===
 
===Answer 2===
alec green
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[[Image:green26_ece438_hmwrk3_rect.png|480x320px]]

Revision as of 15:30, 12 September 2013


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.)


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!

No need to write your name: we can find out who wrote what by checking the history of the page.


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

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.


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