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=== Answer 1 === | === Answer 1 === | ||
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| + | For a square wave, | ||
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| + | <math>a_k=\frac{sin(k\omega_0 T_1)}{\pi k}</math> | ||
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| + | In this case, | ||
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| + | <math>\omega_0=\frac{2\pi}{20}=\frac{\pi}{10} \mbox{ and } T_1 = 5</math> | ||
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| + | Therefore | ||
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| + | <math>\chi(\omega)=\sum_{k=-\infty}^{\infty}2\pi\frac{sin(k\frac{\pi}{10} 5)}{\pi k}\delta(\omega-k\frac{\pi}{10})=\sum_{k=-\infty}^{\infty}2\frac{sin(k\frac{\pi}{2})}{ k}\delta(\omega-k\frac{\pi}{10})</math> | ||
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| + | --[[User:Cmcmican|Cmcmican]] 21:13, 21 February 2011 (UTC) | ||
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=== Answer 2 === | === Answer 2 === | ||
Write it here. | Write it here. | ||
Revision as of 17:13, 21 February 2011
Contents
Practice Question on Computing the Fourier Transform of a Continuous-time Signal
Compute the Fourier transform of the 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
For a square wave,
$ a_k=\frac{sin(k\omega_0 T_1)}{\pi k} $
In this case,
$ \omega_0=\frac{2\pi}{20}=\frac{\pi}{10} \mbox{ and } T_1 = 5 $
Therefore
$ \chi(\omega)=\sum_{k=-\infty}^{\infty}2\pi\frac{sin(k\frac{\pi}{10} 5)}{\pi k}\delta(\omega-k\frac{\pi}{10})=\sum_{k=-\infty}^{\infty}2\frac{sin(k\frac{\pi}{2})}{ k}\delta(\omega-k\frac{\pi}{10}) $
--Cmcmican 21:13, 21 February 2011 (UTC)
Answer 2
Write it here.
Answer 3
Write it here.
