Question

A periodic signal x(t) has a trigonometric Fourier series expansion

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

If x(t) = -x (- t) = -x (t - π/ω₀), we can conclude that

1. a_n are zero for all n and b_n are zero for n even 
2. a_n are zero for all n and b_n are zero for n odd
3. a_n are zero for n even and b_n are zero for n odd 
4. a_n are zero for n odd and b_n are zero for n even

Answer 1

Correct Answer - Option 1 : a_n are zero for all n and b_n are zero for n even 

Concept:

For an odd signal x(t) = -x(-t).

The average value of an odd periodic signal is 0.

For half-wave symmetric signal:

x (t ± T/2) = - x(t)

A half-wave symmetric signal only has odd harmonics in its Fourier series representation.

Calculation:

The given function is odd, hence

a_o (Average Value) = 0 and a_n = 0

Also, since x(t) = -x (t – π/ ω0)

It is also half-wave symmetric.

Hence the signal will contain only odd harmonics.