Answered: Using complex form, find the Fourier…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Using complex form, find the Fourier series of the function

f(x)=signx={−1,−π≤x≤01,0<x≤π.

Expert Solution

Step 1

Fourier theorem states that any function can be decomposed in terms of trigonometric functions like sine and cosine functions. It is used in the analysis of systems whose eigenfunctions are sinusoidal functions.

Given :

$f (x) = sign (x) = \{\begin{cases} - 1 & - π \leq x \leq 0 \\ 1 & 0 \leq x \leq π \end{cases}$

Step 2

Fourier Transform is defined as:

$f (x) = a_{0} + \sum_{n = 1}^{\infty} (a_{n} \cos (\frac{nπx}{L}) + b_{n} \sin (\frac{nπx}{L})) w h e r e a_{0} = \frac{1}{2 L} \int_{T} f (x) d x a_{n} = \frac{1}{L} \int_{T} f (x) \cos (\frac{nπx}{L}) d x b_{n} = \frac{1}{L} \int_{T} f (x) \sin (\frac{nπx}{L}) d x$

let A is a positive integer between zero and pi:

$\begin{array}{rcl} f (A) & = & 1 \\ f (- A) & = & - 1 \\ s o, f (- A) & = & - f (A) \end{array}$

So given function is ODD because it satisfies $f (- x) = - f (x)$ this property in its domain:

so by property of integral:

$\int_{- a}^{a} (o d d f u n c t i o n) d x = 0$

so a₀ and a_n will be zero now b_n

$\begin{array}{rcl} b_{n} & = & \frac{1}{π} \int_{- π}^{π} f (x) \sin (\frac{nπx}{π}) d x \\ = & \frac{2}{π} \int_{0}^{π} 1 \sin (n x) d x \\ = & \frac{2}{π} \frac{{[- \cos (n x)]}_{0}^{π}}{n} \\ = & \frac{2 (1 - \cos (n π))}{n π} \end{array}$

for even values of n, b_n will be 0.

for odd values of n $\begin{array}{rcl} b_{n} & = & \frac{4}{n π} \end{array}$

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