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<center> <math>x(t)= \left\{ \begin{array}{ll}0&, -2<t<-1\\ 1&, -1\leq t\leq 1\\ 0&, 1<t\leq 2\end{array}\right. </math> | <center> <math>x(t)= \left\{ \begin{array}{ll}0&, -2<t<-1\\ 1&, -1\leq t\leq 1\\ 0&, 1<t\leq 2\end{array}\right. </math> | ||
| + | |||
| + | |||
| + | == Solution == | ||
| + | We know that | ||
| + | |||
| + | <math>a_{k} = \frac{1}{T} \int_{0}^{T}x(t)e^{-jk\omega _{o}t}dt</math>, and since T = 4, | ||
| + | |||
| + | <math>a_{k}= \frac{1}{4} \int_{0}^{4}x(t)e^{-jk\frac{2\pi}{4}t}dt</math> | ||
Revision as of 16:07, 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
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{2\pi}{4}t}dt $
