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Solution 1: | Solution 1: | ||
| − | 1) ω<sub>o</sub> = <math>2 | + | 1) ω<sub>o</sub> = <math>\frac{2\pi}{T} = 2\frac{\pi}{20}=\frac{\pi}{10}</math> |
2) a<sub>o</sub> is the DC value of the AC signal and is therefore 1/2 | 2) a<sub>o</sub> is the DC value of the AC signal and is therefore 1/2 | ||
| − | 3) 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 | + | 3) 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) = \frac{sin(k\pi/2)}{(k\pi)}</math>([[User:Clarkjv|Clarkjv]] 18:25, 8 February 2011 (UTC)) |
Solution 2: | Solution 2: | ||
| − | 1)ω<sub>o</sub>=<math>\pi | + | 1)ω<sub>o</sub>=<math>\frac{\pi}{10}</math> (see solution 1) |
From example 3.5 (sec. 3.3 pg 193 Signals and Systems 2nd edition Oppenheim) | From example 3.5 (sec. 3.3 pg 193 Signals and Systems 2nd edition Oppenheim) | ||
| − | <math>a_k=sin(k * w_o * T_1) | + | <math>a_k=\frac{sin(k * w_o * T_1)}{(k*\pi)}</math>, |
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 | ||
| − | From <math>a_k=sin(k * w_o * T_1) | + | From <math>a_k=\frac{sin(k * w_o * T_1)}{(k*\pi)}</math>, |
| − | 3) a<sub>k</sub> = <math>sin(k | + | 3) a<sub>k</sub> = <math>\frac{sin(k\\frac{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 13:43, 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 = $ \frac{2\pi}{T} = 2\frac{\pi}{20}=\frac{\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) = \frac{sin(k\pi/2)}{(k\pi)} $(Clarkjv 18:25, 8 February 2011 (UTC))
Solution 2:
1)ωo=$ \frac{\pi}{10} $ (see solution 1)
From example 3.5 (sec. 3.3 pg 193 Signals and Systems 2nd edition Oppenheim)
$ a_k=\frac{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=\frac{sin(k * w_o * T_1)}{(k*\pi)} $,
3) ak = $ \frac{sin(k\\frac{pi}{2})}{(k\pi)} $(Clarkjv 18:25, 8 February 2011 (UTC))
