| Line 10: | Line 10: | ||
<math>= \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</math> | <math>= \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</math> | ||
| + | |||
| + | <math>= \frac{1}{4}\frac{j}{\pi /2 k}e^{-jk\frac{\pi}{2}t}|_{0}^{1} + \frac{1}{4}\frac{j}{\pi /2 k}e^{-jk\frac{\pi}{2}t}|_{3}^{4}</math> | ||
Revision as of 18:33, 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}{\pi /2 k}e^{-jk\frac{\pi}{2}t}|_{0}^{1} + \frac{1}{4}\frac{j}{\pi /2 k}e^{-jk\frac{\pi}{2}t}|_{3}^{4} $
