Question

A periodic signal x(t) has a trigonometric Fourier series expansion $x\left(t\right)=a_0+{\sum }_{n=1}^{\propto }(a_n\cos n{\omega }_{0}t+b_n\sin n{\omega }_{0}t)$ If x(t) = -x(-t) $=-x(t-\frac{\pi }{{\omega }_0})$. We can conclude that

• A. $a_n$ are zero for all n and $b_n$ are zero for n even
• B. $a_n$ are zero for all n and $b_n$ are zero for n odd
• C. $a_n$ are zero for n even and $b_n$ are zero for n odd
• D. $a_n$ are zero for n odd and $b_n$ are zero for n even

Answer 1

The correct option is A

$a_n$ are zero for all n and $b_n$ are zero for n even

Given, $x\left(t\right)=-x(-t)=-x(t-\frac{\pi }{{\omega }_0})$
The given signal has
1. odd function symmetry $\Rightarrow a_n=0$
2. Half - wave symmetry $\Rightarrow f\left(x\right)$ contains only odd harmonics
$\therefore a_n$ are zero for all n and $b_n$ are zero for n even.