91Ó°ÊÓ

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.

The given functions are

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

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

$f\left(x\right)=\frac{1}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{\cos \mathrm{Î\pm }x+\cos \mathrm{Î\pm }(x−\mathrm{Ï€})}{1−{\mathrm{Î\pm }}^2}d\mathrm{Î\pm }$

The Fourier transform is equal to

$\begin{array}{c}g\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}\sin x{e}^{−i\mathrm{Î\pm }x}dx\\ =\frac{1}{4\mathrm{Ï€}i}{∫}_0^{\mathrm{Ï€}}[e^{ix(1−\mathrm{Î\pm })}−e^{−ix(1+\mathrm{Î\pm })}]dx\\ ={\frac{1}{4\mathrm{Ï€}i}[\frac{e^{ix(1−\mathrm{Î\pm })}}{i(1−\mathrm{Î\pm })}+\frac{e^{−ix(1+\mathrm{Î\pm })}}{i(1+\mathrm{Î\pm })}]|}_0^{\mathrm{Ï€}}\\ =−\frac{1}{4\mathrm{Ï€}}[\frac{e^{i\mathrm{Ï€}(1−\mathrm{Î\pm })}−1}{1−\mathrm{Î\pm }}+\frac{e^{−i\mathrm{Ï€}(1+\mathrm{Î\pm })}−1}{1+\mathrm{Î\pm }}]\end{array}$

Further, solving the Fourier transform

$\begin{array}{c}g\left(\mathrm{Î\pm }\right)=−\frac{1}{4\mathrm{Ï€}}[−\frac{e^{−i\mathrm{Ï€}\mathrm{Î\pm }}+1}{1−\mathrm{Î\pm }}−\frac{e^{−i\mathrm{Î\pm }\mathrm{Ï€}}+1}{1+\mathrm{Î\pm }}]\\ =\frac{1}{2\mathrm{Ï€}}\frac{1+e^{−i\mathrm{Ï€}\mathrm{Î\pm }}}{1−{\mathrm{Î\pm }}^2}\end{array}$

Thus, the function is equal to

$f\left(x\right)=\frac{1}{2\mathrm{Ï€}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}\frac{1+e^{−i\mathrm{Ï€}\mathrm{Î\pm }}}{1−{\mathrm{Î\pm }}^2}{e}^{i\mathrm{Î\pm }x}d\mathrm{Î\pm }$

$f\left(x\right)=\frac{1}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{\cos \mathrm{Î\pm }x+\cos \mathrm{Î\pm }(x−\mathrm{Ï€})}{1−{\mathrm{Î\pm }}^2}d\mathrm{Î\pm }$Use the previous result and write the sum of sine and cosine terms in exponential form.

$\begin{array}{c}f\left(x\right)=\frac{1}{2\mathrm{Ï€}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}\frac{(1+\cos (\mathrm{Î\pm }\mathrm{Ï€})−i\sin (\mathrm{Î\pm }\mathrm{Ï€}\left)\right)\left(\cos \right(\mathrm{Î\pm }x)+i\sin (\mathrm{Î\pm }x\left)\right)}{1−{\mathrm{Î\pm }}^2}d\mathrm{Î\pm }\\ =\frac{1}{2\mathrm{Ï€}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}\left[\frac{(1+\cos (\mathrm{Î\pm }\mathrm{Ï€}\left)\right)\cos \left(\mathrm{Î\pm }x\right)+\sin \left(\mathrm{Î\pm }\mathrm{Ï€}\right)\sin \left(\mathrm{Î\pm }x\right)}{1−{\mathrm{Î\pm }}^2}+h_o(\mathrm{Î\pm },x)\right]d\mathrm{Î\pm }\end{array}$

Here,$h_o(\mathrm{Î\pm },x)$ is the odd part of the function under the integral.

It will integrate to zero over the interval. Thus:

$\begin{array}{c}f\left(x\right)=\frac{1}{\mathrm{Ï€}}{∫}_0^{\mathrm{∞}}\frac{(1+\cos (\mathrm{Î\pm }\mathrm{Ï€}\left)\right)\cos \left(\mathrm{Î\pm }x\right)+\sin \left(\mathrm{Î\pm }\mathrm{Ï€}\right)\sin \left(\mathrm{Î\pm }x\right)}{1−{\mathrm{Î\pm }}^2}\\ =\frac{1}{\mathrm{Ï€}}{∫}_0^{\mathrm{∞}}\frac{\cos \left(\mathrm{Î\pm }x\right)+\cos \left(\mathrm{Î\pm }\right(x−\mathrm{Ï€}\left)\right)}{1−{\mathrm{Î\pm }}^2}d\mathrm{Î\pm }\end{array}$

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