Consider the following statements for quarter-wave symmetry :

A periodic function possesses a quarter-wave symmetry, if

1. it has either odd or even symmetry

2. it has half-wave symmetry

Which of the above statements is/are correct?

A periodic function x(t) which has either odd symmetry or even symmetry along with the half-wave symmetry is said to have quarter-wave symmetry.

Mathematically, a periodic function x(t) is said to have quarter wave symmetry, if it satisfy the following condition:

x(t) = x(-t) or x(t) = -x(-t)

and, x(t) = -x(t ± \(T \over 2\))

Hence both options 1 and 2 are correct.

Additional Information

The Fourier Series representation of a wave function xT(t) is

\({x_T}(t) = {a_0} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos \left( {n{\omega _0}t} \right) + {b_n}\sin \left( {n{\omega _0}t} \right)} \right)}\)

Where for a time period of (-T/2 to T/2)

\({a_0} = {2 \over T}\int\limits_{{-{T} \over 2}}^{{T \over 2}} {{x_T}\left( t \right)dt}\)

\({a_n} = {2 \over T}\int\limits_{{-{T} \over 2}}^{{T \over 2}} {{x_T}\left( t \right)\cos \left( {n{\omega _0}t} \right)dt}\)

\({b_n} = {2 \over T}\int\limits_{{-{T} \over 2}}^{{T \over 2}} {{x_T}\left( t \right)\sin \left( {n{\omega _0}t} \right)dt}\)

Even symmetrical wave: a0 ≠ 0, an ≠ 0, bn = 0

Odd symmetrical wave: a0 = 0, an = 0, bn ≠ 0

Half wave symmetry: 

a0 = 0, an = 0 for n = even n and bn = 0 for all n.

an ≠ 0 and bn ≠ 0  for n = odd.

Quarter-Wave Symmetry:

If the periodic function were to be made even, then

a0 = 0, bn = 0 for all n and an = 0 for n = even

If a quarter-wave symmetric periodic function is made odd,

a0 = 0, an = 0 for all n and bn = 0 for n = even