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[[Category:ECE301Spring2011Boutin]]
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= Practice Question on Computing the Fourier Series discrete-time signal =
[[Category:problem solving]]
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= Practice Question on Computing the Fourier Series discrete-time signal=
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Obtain the Fourier series the DT signal
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<math>
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Obtain the Fourier series the DT signal
x[n] = \left\{  
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<math>x[n] = \left\{  
 
\begin{array}{ll}
 
\begin{array}{ll}
 
1, & \text{ for } -5\leq n \leq 5,\\
 
1, & \text{ for } -5\leq n \leq 5,\\
 
0, & \text{ for } 5< |n| \leq 10.
 
0, & \text{ for } 5< |n| \leq 10.
 
\end{array}
 
\end{array}
\right.  \ </math>
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\right.  \ </math>  
  
x[n] periodic with period 20.
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x[n] periodic with period 20.  
  
 
----
 
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==Share your answers below==
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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!
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== Share your answers below ==
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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!  
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----
 
----
===Answer 1===
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Write it here.
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=== Answer 1 ===
===Answer 2===
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Write it here.
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Solution 1: ω<sub>o</sub> = <math>2*\pi/T = 2*pi/20=\pi/10</math>
===Answer 3===
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a<sub>o</sub> is the DC value of the AC signal and is therefore 1/2
Write it here.
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----
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a<sub>k</sub> = <math> 1/20 * \int_a^b \! x(t)*e^{-j*k*(\pi/10)*t}\,dx </math> = <math>e^{j*k*\pi/2}-e^{-j*k*\pi/2}/(2*\pi*j*k) = sin(k*\pi/2)/(k*\pi)</math>
[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]
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<br> === Answer 2 === Write it here. === Answer 3 === Write it here. ---- [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]
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[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]]

Revision as of 12:47, 8 February 2011

Practice Question on Computing the Fourier Series discrete-time signal

Obtain the Fourier series the DT signal

$ x[n] = \left\{ \begin{array}{ll} 1, & \text{ for } -5\leq n \leq 5,\\ 0, & \text{ for } 5< |n| \leq 10. \end{array} \right. \ $

x[n] periodic with period 20.


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

Solution 1: ωo = $ 2*\pi/T = 2*pi/20=\pi/10 $ ao is the DC value of the AC signal and is therefore 1/2

ak = $ 1/20 * \int_a^b \! x(t)*e^{-j*k*(\pi/10)*t}\,dx $ = $ e^{j*k*\pi/2}-e^{-j*k*\pi/2}/(2*\pi*j*k) = sin(k*\pi/2)/(k*\pi) $


=== Answer 2 === Write it here. === Answer 3 === Write it here. ---- Back to ECE301 Spring 2011 Prof. Boutin

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