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== Answer ==
 
== Answer ==
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<math>\,a_k=\frac{1}{4}\int_{-2}^{2}x(t)e^{-jk\frac{\pi}{2}t}\,dt\,</math>

Revision as of 15:01, 13 October 2008

4. Compute the coefficients $ a_k $ of the Fourier series of the signal $ x(t) $ periodic with period $ T=4 $ defined by

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

(Simplify your answer as much as possible.)


Answer

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

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