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<math>a_{k}= \frac{1}{4} \int_{0}^{4}x(t)e^{-jk\frac{\pi}{2}t}dt</math>
 
<math>a_{k}= \frac{1}{4} \int_{0}^{4}x(t)e^{-jk\frac{\pi}{2}t}dt</math>
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 +
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By the definition of <math>a_{0} \!</math>, we know that it is the average of the signal over the period in this case,
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<math>a_{0} = \frac{1}{2} \!</math>

Revision as of 16:08, 14 October 2008

Most Difficult Problem on First Test

The problem that I found most difficult was problem number 4.

4. Compute the coefficients $ a_{k} \! $ of the Fourier series signal $ x(t) \! $ periodic with period $ T = 4 \! $ defined by

$ x(t)= \left\{ \begin{array}{ll}0&, -2<t<-1\\ 1&, -1\leq t\leq 1\\ 0&, 1<t\leq 2\end{array}\right. $


Solution

We know that

$ a_{k} = \frac{1}{T} \int_{0}^{T}x(t)e^{-jk\omega _{o}t}dt $, and since T = 4,

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


By the definition of $ a_{0} \! $, we know that it is the average of the signal over the period in this case, $ a_{0} = \frac{1}{2} \! $

Alumni Liaison

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

Buyue Zhang