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} $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn