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<math>= \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}</math> | <math>= \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}</math> | ||
| − | <math>\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}} - 1 + e^{-jk2\pi} - e^{-jk\frac{3\pi}{2}}</math> | + | <math>\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}} - 1 + e^{-jk2\pi} - e^{-jk\frac{3\pi}{2}})</math> |
Revision as of 17:39, 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}}) $
