The Fourier series expansion of a periodic function with half-wave symmetry contains:

(1) only even harmonics

(2) only odd harmonics

(3) both even and odd harmonics

(4) no harmonics

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Option 2 : 2
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Half-Wave Symmetry:

A function is said to have half-wave symmetry if it satisfies the following condition:

\(f\left( t \right) = - f\left( {t - \frac{T}{2}} \right)\)

The equation represents that a periodic function has a half-wave symmetry if, after it has been shifted by one-half of a period and inverted, it is said to be identical to the original periodic function. 

The below figure represents the half-wave symmetry.

F1 Shubham Bhatt 30.3.21 Pallavi D7

Observation:

1) If a given function does possess a half-wave symmetry, both ‘an‘ and ‘bn’ are defined as zero for an even value of ‘n’.

2) , a0 is also zero because an average value of a periodic function with this symmetry is zero.

Expressions for the Fourier coefficients are as follows:

a0 = 0

an = bn = 0 for ‘n’ even.

\({a_n} = \frac{4}{T}\mathop \smallint \limits_0^{\frac{T}{2}} f\left( t \right)cosn{\omega _0}t\;dt\)

\({b_n} = \frac{4}{T}\mathop \smallint \limits_0^{\frac{T}{2}} f\left( t \right)sin{\omega _0}t\;dt\)

Conclusion: For the half-wave symmetry, only odd harmonics are present and all the remaining Fourier coefficients will be zero.

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