Question 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.)
Solution
We first started by finding $ a_0 \, $. We notice that the $ x(t) = 1\, $ for half of the time, while $ x(t) = 0\, $ for half of the time. Thus, the average of the signal, $ a_0 = \frac{1}{2}\, $
Mathematical Proof:
$ a_0=\frac{1}{4}\int_{-2}^{2}x(t) dt\, $
$ a_0=\frac{1}{4}\int_{-1}^{1}1 dt\, $
$ a_0=\frac{1}{4} t |^1_{-1}\, $
$ a_0=(\frac{1}{4})2 \, $
$ a_0=\frac{1}{2}\, $
Next, we procede to calculate $ a_k\, $ Notice that the fourier transformation of a square signal is made up from sin waves with infinite number of $ a_k\, $ to produce a perfect signal. Thus $ -\infty < k < \infty \, $
$ \omega _o = \frac{2 \pi} {4} = \frac{\pi}{2}\, $
$ a_k=\frac{1}{4}\int_{-2}^{2}x(t)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}\int_{1}^{2}0 dt + \frac{1}{4}\int_{-2}^{-1}0 dt\, $
The offset of the graph is zero, thus:
$ a_k=\frac{1}{4}\int_{-1}^{1}e^{-jk\frac{\pi}{2}t} dt \, $
$ a_k=\frac{1}{4} \left.\frac{1}{-jk\frac{\pi}{2}}e^{-jk\frac{\pi}{2}t}\right|_{-1}^{1}\, $
$ a_k=\frac{-1}{jk2\pi}(e^{-jk\frac{\pi}{2}} - e^{jk\frac{\pi}{2}})\, $
$ a_k=\frac{1}{k\pi}(\frac{e^{jk\frac{\pi}{2}} - e^{-jk\frac{\pi}{2}}}{2j})\, $
$ a_k=\frac{1}{k\pi}\sin(k\frac{\pi}{2})\, $
