Question 4. Exam
$ x(t)= \left\{ \begin{array}{ll}0&, -2<t<-1\\ 1&, -1\leq t\leq 1\\ 0&, 1<t\leq 2\end{array}\right. $
$ a_{k} = \frac{1}{T} \int_{0}^{T}x(t)e^{-jk\omega _{o}t}dt $
T=4
$ = \frac{1}{4} \int_{0}^{4}x(t)e^{-jk\frac{2\pi}{4}t}dt $
$ = \frac{1}{4} \int_{0}^{1}e^{-jk\frac{\pi}{2}t}dt + \frac{1}{4} \int_{3}^{4}e^{-jk\frac{\pi}{2}t}dt $
$ = \frac{1}{4}\frac{j}{\frac{\pi}{2}k}e^{-jk\frac{\pi}{2}t}|_{0}^{1} + \frac{1}{4}\frac{j}{\frac{\pi}{2}k}e^{-jk\frac{\pi}{2}t}|_{3}^{4} $
$ =\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}} - 1 + e^{-jk2\pi} - e^{-jk\frac{3\pi}{2}}) $
$ =\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}} - 1 + 1 - e^{-jk\frac{4\pi}{2}}e^{jk\frac{\pi}{2}}) $
$ =\frac{j}{2\pi k}(e^{-jk\frac{\pi}{2}}-e^{jk\frac{\pi}{2}}) $
$ =\frac{1}{\pi k}(\frac{e^{jk\frac{\pi}{2}}-e^{-jk\frac{\pi}{2}}}{2j}) $
$ =\frac{1}{\pi k}sin(\frac{\pi}{2}k) $
$ a_{0} $ is not defined.
$ a_{0}=\frac{2}{4} = \frac{1}{2} $
$ a_{k}=\frac{1}{\pi k}sin(\frac{\pi}{2}k) $ for all $ k \neq 0, \; a_{0}=\frac{1}{2} $
