Question

if f(x) is an even function then fourier transform of f(x) is same as​

Answer 1

Answer:

Hope it helps

Step-by-step explanation:

When f is even then f(x)cos(px) is even and f(x)sin(px) is odd. It follows that the imaginary part of Fourier transform vanishes. Consequently, F(p) is real. When f is odd then f(x)cos(px) is odd and f(x)sin(px) is even.

Answer 2

Answer: For even function

$f(x)=a_0+\sum \limits^{\infty}_{n=1}a_n cos\ \frac{n \pi }{L} x$

Given: An even function f(x).

To Find: Fourier transform of f(x).

Step-by-step explanation:

Step 1: Fourier series are infinite series that represent periodic functions in terms of cosines and sines. So we can say that all signals are the combination of periodic signal or we can represent every signal into combination of different periodic signals.

Fourier series of a function f(x) is given by

$f(x)=a_0+\sum \limits^{\infty}_{n=1}(a_n cos \ nx+b_n sin\ nx)$

Where,

$a_0=\frac{1}{2\pi} \int\limits^{\pi}_{-\pi} {f(x)} \, dx \\a_n=\frac{1}{\pi} \int\limits^{\pi}_{-\pi} {f(x)} \cos \ nx, dx \\b_n=\frac{1}{\pi} \int\limits^{\pi}_{-\pi} {f(x)} \sin \ nx, dx$

Step 2: As we know that a function f(x) is said to be even if and only if the function have same value for -x.

i.e. f(-x) = f(x)

If $f(x)$ is an even function, that is, its Fourier series reduces to a Fourier cosine series and it will be

$f(x)=a_0+\sum \limits^{\infty}_{n=1}a_n cos\ \frac{n \pi }{L} x$

where,

$a_0=\frac{1}{L} \int\limits^L_0 {f(x)} \, dx \\a_n=\frac{2}{L} \int\limits^L_0 {f(x) cos \ \frac{n \pi }{L} x \, dx$

and period is 2L.

For more questions please follow the link given below.

https://brainly.in/question/21705666

https://brainly.in/question/23746204

#SPJ3