Line 59: Line 59:
 
<math>x[n] = ((\delta[-2]) +(\delta[-1]) +(\delta[0])) </math>
 
<math>x[n] = ((\delta[-2]) +(\delta[-1]) +(\delta[0])) </math>
  
so <math> X[Z] = e^(2 j \omega) +e^(j \omega) +1  </math>
+
so <math> X[Z] = e^{2 j \omega} +e^{j \omega} +1  </math>
  
  
 
----
 
----
 
[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013]]
 
[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013]]

Revision as of 17:08, 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

$ X(\omega) = \sum_{n=-\infty}^{+\infty} (u[n+2] -u[n-1]) e^{-j\omega n} $

$ = \sum_{n=-2}^{+\infty} u[n+2] e^{-j\omega n} - \sum_{n=1}^{+\infty} u[n-1] e^{-j\omega n} $

$ = \sum_{n=-2}^{0} (u[n+2] -u[n-1]) e^{-j\omega n} $

$ = e^{2j\omega} + e^{j\omega} + 1 $

Answer 4

$ x[n] = u[n+2]-u[n-1] $


$ x[n] = ((\delta[-2]) +(\delta[-1]) +(\delta[0])) $

so $ X[Z] = e^{2 j \omega} +e^{j \omega} +1 $



Back to ECE438 Fall 2013

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett