Question 8.2: Find the FS for a square wave defined over one period as v(t...
Find the FS for a square wave defined over one period as
v(t)=\left\{ \begin{matrix}1 \ for 0<t< \pi \\ 0 \ for \ \pi <t<2\pi\end{matrix} \right.
The waveform is shown in Fig. 8.3. The period of the waveform is T = 2π and the fundamental frequency is w_{0} =1. In contrast to the rectified waveform in the last example, this waveform is odd- and odd half-wave symmetric. Therefore, it is composed of odd-indexed sine components only, in addition to the DC component.
X(0) = \frac{1}{2π} \int_{0}^{π}\;dt =\frac{1}{2}
X(k) = \frac{2}{2π} \int_{0}^{π}{sin(k t)}\;dt =\left\{ \begin{matrix}\frac{2}{k\pi } \ for k odd \\ 0 r \ for k even and k \neq 0\end{matrix} \right.
The coefficient X(0) is the average value of v(t).
v(t) =\frac{1}{2} +\frac{2}{π} (sin(t) +\frac{1}{3}sin(3t) +\frac{1}{5} sin(5t) −···).
In exponential form, the magnitude of the components, except DC, gets divided by 2. The FS magnitude spectrum and the phase spectrum of the signal in exponential form are shown, respectively, in Fig. 8.4a, b.
