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[[Category:discrete time Fourier transform]]
 
[[Category:discrete time Fourier transform]]
  
= [[:Category:Problem_solving|Practice Problem]] on Discrete-time Fourier transform computation =
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'''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]'''
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Topic: Discrete-time Fourier transform computation
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==Question==
 
Compute the discrete-time Fourier transform of the following signal:
 
Compute the discrete-time Fourier transform of the following signal:
  
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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!
  
'''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===
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<math>= 1+ e^{j\omega} + e^{2j\omega} </math>
 
<math>= 1+ e^{j\omega} + e^{2j\omega} </math>
  
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:<span style="color:green"> Instructor's comment: Short and sweet. I like that. </span>
  
 
===Answer 2===
 
===Answer 2===
 
 
 
[[Image:green26_ece438_hmwrk3_rect.png|480x320px]]
 
[[Image:green26_ece438_hmwrk3_rect.png|480x320px]]
  
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<math>= e^{2j\omega} + e^{j\omega} + 1</math>
 
<math>= e^{2j\omega} + e^{j\omega} + 1</math>
  
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:<span style="color:green"> Instructor's comment: It does help to visualize the signal first. </span>
 
===Answer 3===
 
===Answer 3===
 
<math>X(\omega) = \sum_{n=-\infty}^{+\infty} (u[n+2] -u[n-1]) e^{-j\omega n}</math>
 
<math>X(\omega) = \sum_{n=-\infty}^{+\infty} (u[n+2] -u[n-1]) e^{-j\omega n}</math>
  
 
<math>= \sum_{n=-2}^{+\infty} u[n+2] e^{-j\omega n} - \sum_{n=1}^{+\infty} u[n-1] e^{-j\omega n}</math>   
 
<math>= \sum_{n=-2}^{+\infty} u[n+2] e^{-j\omega n} - \sum_{n=1}^{+\infty} u[n-1] e^{-j\omega n}</math>   
 
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<math>= \sum_{n=-2}^{0} (u[n+2] -u[n-1]) e^{-j\omega n}</math>   
 
<math>= \sum_{n=-2}^{0} (u[n+2] -u[n-1]) e^{-j\omega n}</math>   
  
 
<math>= e^{2j\omega} + e^{j\omega} + 1</math>
 
<math>= e^{2j\omega} + e^{j\omega} + 1</math>
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:<span style="color:red"> TA's comment: When you write the summation from -2 to 0, you are implying that you've already used the information of the unit step function. So there's no point leaving that in the equation.</span>
  
 
===Answer 4===
 
===Answer 4===
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<math>x[n] = ((\delta[-2]) +(\delta[-1]) +(\delta[0])) </math>
 
<math>x[n] = ((\delta[-2]) +(\delta[-1]) +(\delta[0])) </math>
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<span style="color:green">( Instructor's comment: The right-hand side does not depend on n. Unless your signal is constant, this means you made a mistake.)</span>
  
 
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>
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===Answer 6===
 
===Answer 6===
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Xiang Zhang
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from the equation we can get that
 
from the equation we can get that
  
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<math> X(\omega) = e^{-j2\omega n}(\delta (n+2)-\delta (n+1)+\delta (n)) </math>
 
<math> X(\omega) = e^{-j2\omega n}(\delta (n+2)-\delta (n+1)+\delta (n)) </math>
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<span style="color:green"> (Instructor's comment: The right-hand-side depends on n, but the left-hand-side does not. This means you made a mistake.)</span> 
  
<math> X(\omega) =  e^{-j2\omega} + e^{-j\omega} + 1 </math>
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<math> X(\omega) =  e^{-j2\omega} + e^{-j\omega} + 1 </math>
  
 
===Answer 8===
 
===Answer 8===
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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^{j\omega 2}+e^{j\omega}+1 </math>
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<math> = e^{j\omega 2}+e^{j\omega}+1 \ </math>
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 +
:<span style="color:red"> TA's comment: When you write the summation from -2 to 0, you are implying that you've already used the information of the inoput x[n]. So there's no point leaving that in the equation.</span>
 
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[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013]]
 
[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013]]

Latest revision as of 12:36, 26 November 2013


Practice Question on "Digital Signal Processing"

Topic: Discrete-time Fourier transform computation


Question

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!


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} $

Instructor's comment: Short and sweet. I like that.

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 $

Instructor's comment: It does help to visualize the signal first.

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 $

TA's comment: When you write the summation from -2 to 0, you are implying that you've already used the information of the unit step function. So there's no point leaving that in the equation.

Answer 4

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


$ x[n] = ((\delta[-2]) +(\delta[-1]) +(\delta[0])) $ ( Instructor's comment: The right-hand side does not depend on n. Unless your signal is constant, this means you made a mistake.)

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

Answer 5

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

$ X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $

$ X_(\omega) = \sum_{n=-2}^{0} e^{-j\omega n} $

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

Answer 6

Xiang Zhang


from the equation we can get that

$ X(n) = u[n+2] - u[n-1] = \left\{ \begin{array}{l l} 1 & \quad when \quad n = -2,-1,0\\ 0 & \quad \text{else} \end{array} \right. $

Hence, substitute into the DTFT equation,

$ X_(\omega) = \sum_{n = -\infty}^{ \infty} x[n] e^{-j\omega n} $

change the limit to

$ X_(\omega) = \sum_{n = -2}^{ 0} e^{-j\omega n} $

Then, we expand to the normal expression.

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


Answer 7

$ X(\omega) = \sum_{n=-\infty}^\infty x[n]e^{-j2\omega n} $

$ X(\omega) = e^{-j2\omega n}(\delta (n+2)-\delta (n+1)+\delta (n)) $ (Instructor's comment: The right-hand-side depends on n, but the left-hand-side does not. This means you made a mistake.)

$ X(\omega) = e^{-j2\omega} + e^{-j\omega} + 1 \ $

Answer 8

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} $

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

TA's comment: When you write the summation from -2 to 0, you are implying that you've already used the information of the inoput x[n]. So there's no point leaving that in the equation.

Back to ECE438 Fall 2013

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