Find the Fourier series coefficients for the continuous time periodic signal

\(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1.5,\;\;0 < t < 1}\\ { - 1.5,\;\;1 < t < 2} \end{array}} \right.\)

With fundamental frequency f₀ = π

Concept:

Let f(x) is a periodic function defined in (C, C + 2L) with period 2L, then the Fourier series of f(x) is

\(f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty \left[ {{a_n}\cos \frac{{n\pi x}}{L} + {b_n}\sin \frac{{n\pi x}}{L}} \right]\)

Where the Fourier series coefficients a₀, a_n, and b_n are given by

\({a_0} = \frac{1}{L}\mathop \smallint \limits_C^{C + 2L} f\left( x \right)dx\)

\({a_n} = \frac{1}{L}\mathop \smallint \limits_C^{C + 2L} f\left( x \right)\cos \frac{{n\pi x}}{L}dx\)

\({b_n} = \frac{1}{L}\mathop \smallint \limits_C^{C + 2L} f\left( x \right)\sin \frac{{n\pi x}}{L}dx\)

• If f(x) is an odd function, then only b_n exists where a₀ and b_n are zero.
• If f(x) is an even function, then both a₀ and a_n exists where b_n is zero.

Calculation:

\(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1.5,\;\;0 < t < 1}\\ { - 1.5,\;\;1 < t < 2} \end{array}} \right.\)        

Fundamental period = 2

The given function is an odd function and hence only b_n exists.

\({b_n} = \frac{2}{L}\mathop \smallint \limits_0^L x\left( t \right)\sin \frac{{n\pi t}}{L}dt\)

Here L = 1

\( = \frac{2}{1}\mathop \smallint \limits_0^1 1.5\sin n\pi tdt\)

\( = 3\left[ { - \frac{{\cos n\pi t}}{{n\pi }}} \right]_0^1\)

\( = \frac{3}{{n\pi }}\left[ {1 - \cos n\pi } \right]= \frac{3}{{n\pi }}\left[ {1 - \cos n\pi } \right]\)