91影视

The given equation is$f\left(x\right)=\{\begin{array}{l}\begin{array}{ll}1,& \mathrm{蟺}/2<\left|\mathrm{x}\right|<\mathrm{蟺}\end{array}\\ \begin{array}{ll}0,& \text{otherwise}\end{array}\end{array}$

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.

The following are the formulas for the Fourier series transforms,

$\begin{array}{c}f\left(x\right)=\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}g\left(伪\right)e^{i伪x}d伪\\ g\left(伪\right)=\frac{1}{2蟺}\underset{鈭�\mathrm{鈭�}}{\overset{\mathrm{鈭�}}{鈭�}}f\left(x\right)e^{鈭�i伪x}dx\end{array}$

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

Find the value of$g\left(伪\right)$.

$\begin{array}{c}g\left(伪\right)=\frac{1}{2蟺}{鈭�}_{鈭�蟺}^{鈭�蟺/2}{e}^{鈭�i伪x}dx+\frac{1}{2蟺}{鈭�}_{蟺/2}^{蟺}{e}^{鈭�i伪x}dx\\ ={\frac{1}{2蟺}\frac{鈭�1}{i伪}{e}^{鈭�i伪x}|}_{鈭�蟺}^{鈭�蟺/2}鈭�{\frac{1}{2蟺}\frac{1}{i伪}{e}^{鈭�i伪x}|}_{蟺/2}^{蟺}\\ =\frac{i}{2蟺伪}(e^{i伪蟺/2}鈭�e^{i伪蟺})+\frac{i}{2蟺伪}(e^{鈭�i伪蟺}鈭�e^{鈭�i伪蟺/2})\\ =\frac{i}{蟺伪}\left(i\sin \frac{伪蟺}{2}鈭�i\sin \left(伪蟺\right)\right)\end{array}$

Further solving

_{$g\left(伪\right)=\frac{\sin \left(伪蟺\right)鈭�\sin (伪蟺/2)}{蟺伪}$}

Therefore, the exponential Fourier transform of the given equation is $f\left(x\right)={鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}\frac{\sin \left(伪蟺\right)鈭�\sin (伪蟺/2)}{伪蟺}{e}^{i伪x}d伪$.