Find the FS for a square wave defined over one period as [latex]v(t)=\left\{ \begin{matrix}1 \ for 0<t< \pi \\ 0 \ for \ \pi <t<2\pi\end{matrix} \right.[/latex]

The waveform is shown in Fig. 8.3. The period of the waveform is T = 2π and the fundamental frequency is [latex]w_{0} =1.[/latex] 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. [latex]X(0) = \frac{1}{2π} \int_{0}^{π}\;dt =\frac{1}{2}[/latex] [latex]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.[/latex] The coefficient X(0) is the average value of v(t). [latex]v(t) =\frac{1}{2} +\frac{2}{π} (sin(t) +\frac{1}{3}sin(3t) +\frac{1}{5} sin(5t) −···).[/latex] 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.

Question 8.2: Find the FS for a square wave defined over one period as v(t...

Introductory Circuit Theory

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 "Step-by-Step Explanation" refers to a detailed and sequential breakdown of the solution or reasoning behind the answer. This comprehensive explanation walks through each step of the answer, offering you clarity and understanding.
Our explanations are based on the best information we have, but they may not always be right or fit every situation.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

This Question has been Answered.