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Solution 1: | Solution 1: | ||
| − | 1) | + | 1) <math>w_o = \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) | + | 3) <math>a_k= 1/20 * \int_{-5}^{5} \! 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) | + | 1)<math>w_o=\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) | ||
| − | + | where T_1 is half the pulse width, | |
<math>a_k=\frac{sin(k * w_o * T_1)}{(k*\pi)}</math>, | <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=\frac{sin(k * w_o * T_1)}{(k*\pi)}</math> | + | From <math>a_k=\frac{sin(k * w_o * T_1)}{(k*\pi)}</math> |
| − | 3) | + | 3)<math>a_k=\frac{sin(k\pi/2)}{(k\pi)}</math> |
([[User:Clarkjv|Clarkjv]] 18:25, 8 February 2011 (UTC)) | ([[User:Clarkjv|Clarkjv]] 18:25, 8 February 2011 (UTC)) | ||
[[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Revision as of 13:57, 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) $ w_o = \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) $ a_k= 1/20 * \int_{-5}^{5} \! 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)} $(Clarkjv 18:25, 8 February 2011 (UTC))
Solution 2:
1)$ w_o=\frac{\pi}{10} $ (see solution 1)
From example 3.5 (sec. 3.3 pg 193 Signals and Systems 2nd edition Oppenheim) where T_1 is half the pulse width, $ 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)$ a_k=\frac{sin(k\pi/2)}{(k\pi)} $ (Clarkjv 18:25, 8 February 2011 (UTC))
