Question

(a) Represent as an exponential Fourier transform the function

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}\begin{array}{ll}\sin x,& 0<x<\pi \end{array}\\ \begin{array}{ll}0,& \mathrm{otherwise}\end{array}\end{array}$

Hint: write $\mathbf{sin}\mathit{x}$in complex exponential form.

(b) Show that your result can be written as

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}\underset{0}{\overset{\infty }{\int }}\frac{\cos \alpha x+\cos \alpha (x-\pi )}{1-{\alpha }^2}d\alpha$.

Short answer

By using the Fourier transform formula for the given function and write the exponential as the sum.

Step-by-step solution

Definition of Fourier series

A periodic function, f(x), is expanded by the Fourier series formula into an infinite sum of sines and cosines. It is used to combine a collection of basic oscillating functions, like as sines and cosines, to deconstruct any periodic function or periodic signal.

Step 2:Given parameters

The given functions are

$f\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}\sin x,& 0<x<\pi \end{array}\\ \begin{array}{ll}0,& \mathrm{otherwise}\end{array}\end{array}\right.$

There need to represent the given function as Fourier transform and also prove that the result can be written as

$f\left(x\right)=\frac{1}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\cos \alpha x+\cos \alpha (x-\pi )}{1-{\alpha }^2}d\alpha$

Represent function as Fourier transform

The Fourier transform is equal to

$\begin{array}{c}g\left(\alpha \right)=\frac{1}{2\pi }{\int }_0^{\pi }\sin x{e}^{-i\alpha x}dx\\ =\frac{1}{4\pi i}{\int }_0^{\pi }[e^{ix(1-\alpha )}-e^{-ix(1+\alpha )}]dx\\ ={\frac{1}{4\pi i}[\frac{e^{ix(1-\alpha )}}{i(1-\alpha )}+\frac{e^{-ix(1+\alpha )}}{i(1+\alpha )}]|}_0^{\pi }\\ =-\frac{1}{4\pi }[\frac{e^{i\pi (1-\alpha )}-1}{1-\alpha }+\frac{e^{-i\pi (1+\alpha )}-1}{1+\alpha }]\end{array}$

Further, solving the Fourier transform

$\begin{array}{c}g\left(\alpha \right)=-\frac{1}{4\pi }[-\frac{e^{-i\pi \alpha }+1}{1-\alpha }-\frac{e^{-i\alpha \pi }+1}{1+\alpha }]\\ =\frac{1}{2\pi }\frac{1+e^{-i\pi \alpha }}{1-{\alpha }^2}\end{array}$

Thus, the function is equal to

$f\left(x\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\frac{1+e^{-i\pi \alpha }}{1-{\alpha }^2}{e}^{i\alpha x}d\alpha$

Write exponential as sum of cosine and sine term

$f\left(x\right)=\frac{1}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\cos \alpha x+\cos \alpha (x-\pi )}{1-{\alpha }^2}d\alpha$Use the previous result and write the sum of sine and cosine terms in exponential form.

$\begin{array}{c}f\left(x\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\frac{(1+\cos (\alpha \pi )-i\sin (\alpha \pi \left)\right)\left(\cos \right(\alpha x)+i\sin (\alpha x\left)\right)}{1-{\alpha }^2}d\alpha \\ =\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\left[\frac{(1+\cos (\alpha \pi \left)\right)\cos \left(\alpha x\right)+\sin \left(\alpha \pi \right)\sin \left(\alpha x\right)}{1-{\alpha }^2}+h_o(\alpha ,x)\right]d\alpha \end{array}$

Here,$h_o(\alpha ,x)$ is the odd part of the function under the integral.

It will integrate to zero over the interval. Thus:

$\begin{array}{c}f\left(x\right)=\frac{1}{\pi }{\int }_0^{\infty }\frac{(1+\cos (\alpha \pi \left)\right)\cos \left(\alpha x\right)+\sin \left(\alpha \pi \right)\sin \left(\alpha x\right)}{1-{\alpha }^2}\\ =\frac{1}{\pi }{\int }_0^{\infty }\frac{\cos \left(\alpha x\right)+\cos \left(\alpha \right(x-\pi \left)\right)}{1-{\alpha }^2}d\alpha \end{array}$

Thus, the exponential fourier series transform of the given function $f\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}\sin x,& 0<x<\pi \end{array}\\ \begin{array}{ll}0,& \mathrm{otherwise}\end{array}\end{array}\right.$is $f\left(x\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\frac{1+e^{-i\pi \alpha }}{1-{\alpha }^2}{e}^{i\alpha x}d\alpha$and this result can also be proved.