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<math>a_k = \frac{1}{T} \int_{0}^{T} x(t) e^{-jkw_0t} dt</math> | <math>a_k = \frac{1}{T} \int_{0}^{T} x(t) e^{-jkw_0t} dt</math> | ||
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| + | [[Image:X(t)_ECE301Fall2008mboutin.jpg]] | ||
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| + | From diagram, <math>a_0 = \frac{1}{2}</math> because it is the average of the signal over one period. | ||
| + | |||
| + | Then, using the formula: | ||
| + | |||
| + | <math>a_k = \frac{1}{4} \int_{0}^{4} x(t) e^{-jk\frac{\pi}{2}t} dt</math> | ||
Revision as of 10:50, 15 October 2008
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. \, $
Answer:
$ T = \frac{2\pi}{w_0} = 4 $
$ w_0 = \frac{\pi}{2} $
$ a_k = \frac{1}{T} \int_{0}^{T} x(t) e^{-jkw_0t} dt $
File:X(t) ECE301Fall2008mboutin.jpg
From diagram, $ a_0 = \frac{1}{2} $ because it is the average of the signal over one period.
Then, using the formula:
$ a_k = \frac{1}{4} \int_{0}^{4} x(t) e^{-jk\frac{\pi}{2}t} dt $
_ECE301Fall2008mboutin.jpg){kind=link}