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<math>=\frac{1}{\pi k}(\frac{e^{jk\frac{\pi}{2}}-e^{-jk\frac{\pi}{2}}}{2j})</math>
 
<math>=\frac{1}{\pi k}(\frac{e^{jk\frac{\pi}{2}}-e^{-jk\frac{\pi}{2}}}{2j})</math>
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 +
<math>=\frac{1}{\pi k}sin(\frac{\pi}{2}k)</math>

Revision as of 18:46, 8 October 2008

Test Problem 4

$ a_{k} = \frac{1}{T} \int_{0}^{T}x(t)e^{-jk\omega _{o}t}dt $

From the problem statement we know that T=4

$ = \frac{1}{4} \int_{0}^{4}x(t)e^{-jk\frac{2\pi}{4}t}dt $

Knowing that T=4 we can visualize the periodic signal in the range $ 0 \leq t \leq 4 $. x(t) = 1 for $ 0 \leq t \leq 1 $ and $ 3 \leq t \leq 4 $. Otherwise, x(t) = 0. Therefore:

$ = \frac{1}{4} \int_{0}^{1}e^{-jk\frac{\pi}{2}t}dt + \frac{1}{4} \int_{3}^{4}e^{-jk\frac{\pi}{2}t}dt $

$ = \frac{1}{4}\frac{j}{\frac{\pi}{2}k}e^{-jk\frac{\pi}{2}t}|_{0}^{1} + \frac{1}{4}\frac{j}{\frac{\pi}{2}k}e^{-jk\frac{\pi}{2}t}|_{3}^{4} $

$ =\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}} - 1 + e^{-jk2\pi} - e^{-jk\frac{3\pi}{2}}) $

$ =\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}} - 1 + 1 - e^{-jk\frac{4\pi}{2}}e^{jk\frac{\pi}{2}}) $

$ =\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}) $

$ =\frac{1}{\pi k}(\frac{e^{jk\frac{\pi}{2}}-e^{-jk\frac{\pi}{2}}}{2j}) $

$ =\frac{1}{\pi k}sin(\frac{\pi}{2}k) $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang