Subject : Signals & Systems

Topic : Fourier Series.

Difficulty: Medium

From the figure,

Period of the signal x(t) = T x(t)= $$\begin{cases} A, & \text{0<x<t 2}="" \[2ex]="" -a,="" &amp;="" \text{t="" 2<x<t}="" \end{cases}$$ <="" p="">

The trigonometric Fourier series equation is given by

$x(t) = a(0) + \sum_{k=1}^∞ a(k) cos⁡(kω_O t)\ +\sum_{k=1}^∞ b(k) sin⁡(kω_O t)\ $

$where a(0) = 1T\int_{\ltT\gt} x(t) \, dt$

$a(k) = 2T\int_{\ltT\gt} x(t) cos⁡(kω_O t) \, dt$

$b(k) = 2T\int_{\ltT\gt} x(t) sin⁡(kω_O t) \, dt$

Step 1: To calculate a(0) $a(0) = 1T\int_{\ltT\gt} x(t) \, dt$

$\ a(0) = 1T[\int_0^{T/2} A \, dt + \int_{T/2}^{T} -A \, dt]$

$\ a(0) = AT[\int_0^{T/2} \, dt + \int_{T/2}^{T} \, dt]$

$=\frac{A}{T} [\frac{T}{2} - T + \frac{T}{2}]$

∴a(0) = 0 ----------- 1

Step 2: To calculate a(k)

$a(k) = 2T\int_{\ltT\gt} x(t) cos⁡(kω_O t) \, dt$

$a(k) = 2T\int_0^T x(t) cos⁡(kω_O t) \, dt$

$= 2T\int_0^T x(t) cos⁡(kω_O t) \,dt - \int_{T/2}^T A cos⁡(kω_O t) \, dt$

Here ω_O= 2π/T

$∴a(k) = 2T\int_0^{T/2} x(t) cos⁡(k 2π/T t) \,dt - \int_{T/2}^T A cos⁡(k 2π/T t) \, dt$

$= 2T\int_0^{T/2} A cos⁡(k 2π/T t) \,dt - \int_{T/2}^T A cos⁡(k 2π/T t) \, dt$

$= 2T\int_0^{T/2} cos⁡(k 2π/T t) \,dt - \int_{T/2}^T cos⁡(k 2π/T t) \, dt$