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 = \frac{1}{4} \int_{0}^{4} [u(t+1) - u(t-1)] e^{-jk\frac{\pi}{2}t} dt $
$ a_k = \frac{1}{4} \int_{-1}^{1} e^{-jk\frac{\pi}{2}t} dt = \frac{1}{4} [\frac{e^{-jk\frac{\pi}{2}t}}{-jk\frac{\pi}{2}}]^{1}_{-1} $
$ a_k = \frac{1}{4} [ \frac{e^{-jk\frac{\pi}{2}}}{-jk\frac{\pi}{2}} + \frac{e^{jk\frac{\pi}{2}}}{jk\frac{\pi}{2}} ] $
$ a_k = \frac{1}{2jk\pi} [e^{jk\frac{\pi}{2}} - e^{-jk\frac{\pi}{2}}] = \frac{1}{k\pi} [\frac{e^{jk\frac{\pi}{2}} - e^{-jk\frac{\pi}{2}}}{2j}] $
Therefore, $ a_k = \frac{sin(k\frac{\pi}{2})}{k\pi} $ , where $ -\infty < k < \infty $
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