Question

Question:

If x(t) is a periodic signal with even symmetry, then, its Fourier series expansion will​

Answer 1

Answer:

(1)not have any sine component

(2)not have any cosine component

(3)not have any dc component

(4)have only even harmomic component

Answer 2

Answer:

1.) will not have any sine term

2.) will contain only cosine terms.

Explanation:

Fourier expansion : $f(x) = \frac{a_0}{2} + \Sigma a_n cos(nx) +\Sigma b_n sin(nx)$

Given: x(t) is an even function.

A function is called even if f(−t)=f(t)

which means, x(-t)=x(t)

we know,

• The product of two even or two odd functions is even.

• The product of an even and an odd function is odd.

Coefficient of fourier expansion: $a_n =\frac{2}{\pi} \int\limits^{\pi}_0 {x(t)cos(nt)} \, dt$

and $b_n =\frac{2}{\pi} \int\limits^{\pi}_0 {x(t)sin(nt)} \, dt$

Here, if $x(t)$ is an even function then, then the integration term becomes  odd; $x(t)sin(nt) = even*odd = odd$

This makes the coefficient $a_n =0$. Hence, all the sine terms diappears.

#SPJ3