Find exponential series for the signal shown in figure,

Solution:

$ T=1, \Omega_0=\frac{2 \pi}{T}=\frac{2 \pi}{1}=2 \pi\\ $

Consider the equation of a straight line,

$ \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\\ $

On substituting the coordinates of points P and Q in eq,

$ \frac{x(t)-0}{1-0}=\frac{t-0}{1-0} \Rightarrow x(t)=t\\ $

$ c_0=\frac{1}{T} \int_0^T x(t) d t=\frac{1}{1} \int_0^1(t) d t=\left[\frac{t^2}{2}\right]_0^1=\frac{1}{2}\\ $

$ \begin{gathered} c_n=\frac{1}{T} \int_0^T x(t) e^{-j n \Omega_0 t} d t=\frac{1}{1} \int_0^1 t e^{-j n 2 \pi t} d t=\left[t \frac{e^{-j n 2 \pi t}}{-j n 2 \pi}\right]_0^1-\int_0^1 \frac{e^{-j n 2 \pi t}}{-j n 2 \pi} d t \\\\ =\frac{e^{-j n 2 \pi}}{-j n 2 \pi}+0+\left[\frac{e^{-j n 2 \pi t}}{-j^2(n 2 \pi)^2}\right]_0^1=j \frac{e^{-j n 2 \pi}}{n 2 \pi}+\frac{e^{-j n 2 \pi}}{n^2 4 \pi^2}-\frac{1}{n^2 4 \pi^2} \\\\ =\frac{j}{n 2 \pi}+\frac{1}{n^2 4 \pi^2}-\frac{1}{n^2 4 \pi^2}=\frac{j}{n 2 \pi} \\\\ c_n=\frac{j}{n 2 \pi}\\ \end{gathered} $

Exponential Fourier series

$ x(t)=\sum_{n=-\infty}^{\infty} c_n e^{j n \Omega_0 t}\\ $

$ \begin{aligned} \therefore x(t)=+\cdots & -\frac{j}{6 \pi} e^{-j 6 \pi t}-\frac{j}{4 \pi} e^{-j 4 \pi t}-\frac{j}{2 \pi} e^{-j 2 \pi t}+\frac{1}{2}+\frac{j}{2 \pi} e^{j 2 \pi t}+\frac{j}{4 \pi} e^{j 4 \pi t}+\frac{j}{6 \pi} e^{j 6 \pi t}+\cdots \\\\ & =\frac{1}{2}+\frac{j}{2 \pi}\left[e^{j 2 \pi t}-e^{-j 2 \pi t}\right]+\frac{j}{4 \pi}\left[e^{j 4 \pi t}-e^{-j 4 \pi t}\right]+\frac{j}{6 \pi}\left[e^{j 6 \pi t}-e^{-j 6 \pi t}\right]+\cdots \\\\ & =\frac{1}{2}+\frac{1}{\pi}\left[\frac{e^{j 2 \pi t}-e^{-j 2 \pi t}}{(-1) 2 j}\right]+\frac{1}{2 \pi}\left[\frac{e^{j 4 \pi t}-e^{-j 4 \pi t}}{(-1) 2 j}\right]+\frac{1}{3 \pi}\left[\frac{e^{j 6 \pi t}-e^{-j 6 \pi t}}{(-1) 2 j}\right] \\\\ & =\frac{1}{2}+\left(\frac{-1}{\pi}\right) \sin 2 \pi t-\frac{1}{2 \pi} \sin 4 \pi t-\frac{1}{3 \pi} \sin 6 \pi t \end{aligned}\\ $

$ =\frac{1}{2}-\frac{1}{\pi}\left[\sin 2 \pi t+\frac{1}{2} \sin 4 \pi t+\frac{1}{3} \sin 6 \pi t+\cdots\right]\\ $