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2)a<sub>o</sub> is still the DC value of the AC signal and therefore, | 2)a<sub>o</sub> is still the DC value of the AC signal and therefore, | ||
| − | a<sub>o</sub> = 1/2 | + | a<sub>o</sub> = 1/2 |
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| + | From <math>a_k=sin(k * w_o * T_1)/(k*\pi)</math>, | ||
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3) a<sub>k</sub> = <math>sin(k*\pi/2)/(k*\pi)</math>([[User:Clarkjv|Clarkjv]] 18:25, 8 February 2011 (UTC)) | 3) a<sub>k</sub> = <math>sin(k*\pi/2)/(k*\pi)</math>([[User:Clarkjv|Clarkjv]] 18:25, 8 February 2011 (UTC)) | ||
[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Revision as of 14:28, 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.
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:
1) ωo = $ 2*\pi/T = 2*pi/20=\pi/10 $
2) ao is the DC value of the AC signal and is therefore 1/2
3) 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) $(Clarkjv 18:25, 8 February 2011 (UTC))
Solution 2:
1)ωo=$ \pi/10 $ (see solution 1)
From example 3.5 (sec. 3.3 pg 193 Signals and Systems 2nd edition Oppenheim)
$ a_k=sin(k * w_o * T_1)/(k*\pi) $,
2)ao is still the DC value of the AC signal and therefore, ao = 1/2
From $ a_k=sin(k * w_o * T_1)/(k*\pi) $,
3) ak = $ sin(k*\pi/2)/(k*\pi) $(Clarkjv 18:25, 8 February 2011 (UTC))
