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<math>a_k=\frac{1}{20}\int_{-10}^{10}x(t)e^{-jkw_0t}dt=\frac{1}{20}\int_{-5}^{5}e^{-jk\frac{\pi}{10}t}dt=\frac{1}{20}\Bigg[\frac{e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}}{-jk\frac{\pi}{10}}\Bigg]=\frac{1}{-jk2\pi}\Bigg(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}\Bigg)</math> | <math>a_k=\frac{1}{20}\int_{-10}^{10}x(t)e^{-jkw_0t}dt=\frac{1}{20}\int_{-5}^{5}e^{-jk\frac{\pi}{10}t}dt=\frac{1}{20}\Bigg[\frac{e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}}{-jk\frac{\pi}{10}}\Bigg]=\frac{1}{-jk2\pi}\Bigg(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}\Bigg)</math> | ||
| − | --[[User:Cmcmican|Cmcmican]] 21: | + | <math>x(t)=\frac{1}{2}e^{-jk\frac{\pi}{10}t}+\sum_{k=-\infty}^\infty\frac{1}{-jk2\pi}\Bigg(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}\Bigg)e^{-jk\frac{\pi}{10}t}</math> |
| + | |||
| + | --[[User:Cmcmican|Cmcmican]] 21:35, 7 February 2011 (UTC) | ||
===Answer 2=== | ===Answer 2=== | ||
Revision as of 17:35, 7 February 2011
Contents
Practice Question on Computing the Fourier Series continuous-time signal
Obtain the Fourier series the CT signal
$ x(t) = \left\{ \begin{array}{ll} 1, & \text{ for } -5\leq t \leq 5,\\ 0, & \text{ for } 5< |t| \leq 10, \end{array} \right. \ $
x(t) 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
$ k=0\, $
$ a_0=\frac{1}{20}\int_{-10}^{10}x(t)e^{-0}dt=\frac{1}{20}\int_{-5}^{5}1dt=\frac{1}{2} $
$ k\ne0 $
$ a_k=\frac{1}{20}\int_{-10}^{10}x(t)e^{-jkw_0t}dt=\frac{1}{20}\int_{-5}^{5}e^{-jk\frac{\pi}{10}t}dt=\frac{1}{20}\Bigg[\frac{e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}}{-jk\frac{\pi}{10}}\Bigg]=\frac{1}{-jk2\pi}\Bigg(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}\Bigg) $
$ x(t)=\frac{1}{2}e^{-jk\frac{\pi}{10}t}+\sum_{k=-\infty}^\infty\frac{1}{-jk2\pi}\Bigg(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}\Bigg)e^{-jk\frac{\pi}{10}t} $
--Cmcmican 21:35, 7 February 2011 (UTC)
Answer 2
Write it here.
Answer 3
Write it here.
