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Knowing that T=4 we can visualize the periodic signal in the range <math>0 \leq t \leq 4</math>. x(t) = 1 for <math>0 \leq t \leq 1</math> and <math>3 \leq t \leq 4</math>. Otherwise, x(t) = 0. Therefore: | Knowing that T=4 we can visualize the periodic signal in the range <math>0 \leq t \leq 4</math>. x(t) = 1 for <math>0 \leq t \leq 1</math> and <math>3 \leq t \leq 4</math>. Otherwise, x(t) = 0. Therefore: | ||
| − | <math>= \frac{1}{4} \int_{0}^{ | + | <math>= \frac{1}{4} \int_{0}^{1}e^{-jk\frac{\pi}{2}}dt + \frac{1}{4} \int_{3}^{4}e^{-jk\frac{\pi}{2}}dt</math> |
Revision as of 18:28, 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}}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}}dt + \frac{1}{4} \int_{3}^{4}e^{-jk\frac{\pi}{2}}dt $
