If \(f\left( x \right) = \frac{l}{4} - x\), when \(0 < x < \frac{l}{2}\)

\( = x - \frac{{3l}}{4}\) ,when \(\frac{l}{2} < x < l\)

then the value of aThis question was previously asked in

TN TRB Electrical 2012 Paper

Option 1 : 0

Electric Circuits and Fields

968

30 Questions
40 Marks
40 Mins

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, an, and bn 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 bn exists where a0 and bn are zero.
- If f(x) is an even function, then both a0 and an exists where bn is zero.

Calculation:

\({a_0} = \frac{1}{l}\left[ {\mathop \smallint \limits_0^{\frac{l}{2}} \left( {\frac{l}{4} - x} \right)dx + \mathop \smallint \limits_{\frac{l}{2}}^l \left( {x - \frac{{3l}}{4}} \right)dx} \right]\)

= 0