Question

Find the exponential Fourier transform of the given f(x) and write f(x)as a Fourier integral [that is, find $\mathbf{\text{g}}\mathbf{\left(}\mathbf{\text{α}}\mathbf{\right)}$in equation (12.2) and substitute your result into the first integral in equation (12.2)].

role="math" localid="1664339656101" $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\begin{array}{rr}\mathbf{-}\mathbf{1}\mathbf{,}& \mathbf{-}\mathbf{\pi }\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{0}\end{array}\\ \begin{array}{rr}\mathbf{1}\mathbf{,}& \mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{\pi }\end{array}\\ \begin{array}{rr}\mathbf{0}\mathbf{,}& \mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{>}\mathbf{\pi }\end{array}\end{array}$

Short answer

The exponential Fourier transform of the given function is $g\left(\alpha \right)=\frac{i}{\pi }\frac{\cos \left(\alpha \pi \right)-1}{\alpha }$and f(x) as a Fourier integral is$f\left(x\right)=\frac{i}{\pi }{\int }_{-\infty }^{\infty }\frac{\cos \left(\alpha \pi \right)-1}{\alpha }{e}^{i\alpha x}d\alpha$.

Step-by-step solution

Given Information.

The given equation is$f\left(x\right)=\{\begin{array}{l}\begin{array}{rr}-1,& -\mathrm{\pi }<\mathrm{x}<0\end{array}\\ \begin{array}{rr}1,& 0<\mathrm{x}<\mathrm{\pi }\end{array}\\ \begin{array}{rr}0,& \left|\mathrm{x}\right|>\mathrm{\pi }\end{array}\end{array}$.

Step 2: Meaning of the Fourier series.

A Fourier series is an infinite sum of sines and cosines expansion of a periodic function. The orthogonality relationships of the sine and cosine functions are used in the Fourier series.

Find the exponential Fourier transform

The following are the formulas for the Fourier series transforms,

$\begin{array}{c}f\left(x\right)=\underset{-\infty }{\overset{\infty }{\int }}g\left(\alpha \right)e^{i\alpha x}d\alpha \\ g\left(\alpha \right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)e^{-i\alpha x}dx\end{array}$

Here $g\left(\alpha \right)$is called the Fourier transform of f(x).

Find the value of$g\left(\alpha \right)$.

$\begin{array}{c}g\left(\alpha \right)=-\frac{1}{2\pi }{\int }_{-\pi }^0{e}^{-i\alpha x}dx+\frac{1}{2\pi }{\int }_0^{\pi }{e}^{-i\alpha x}dx\\ =-\frac{1}{2\pi }{\int }_{\pi }^0{e}^{i\alpha x}d(-x)+\frac{1}{2\pi }{\int }_0^{\pi }{e}^{-i\alpha x}dx\\ =\frac{1}{2\pi }{\int }_0^{\pi }(e^{-i\alpha x}-e^{i\alpha x})dx\\ =-\frac{i}{\pi }{\int }_0^{\pi }\sin \left(\alpha x\right)dx\end{array}$

Further solving

_{$\begin{array}{c}g\left(\alpha \right)={\frac{i}{\pi }\frac{\cos \left(\alpha x\right)}{\alpha }|}_0^{\pi }\\ =\frac{i}{\pi }\frac{\cos \left(\alpha \pi \right)-1}{\alpha }\end{array}$}

Therefore, the exponential Fourier transform of the given equation is$f\left(x\right)=\frac{i}{\pi }{\int }_{-\infty }^{\infty }\frac{\cos \left(\alpha \pi \right)-1}{\alpha }{e}^{i\alpha x}d\alpha$.