Question

Determine Fourier transform function A(k) of the oscillatory function.

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\begin{array}{cc}{\mathbf{e}}^{\mathbf{i}{\mathbf{k}}_{\mathbf{o}}\mathbf{x}}& \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{L}\mathbf{\le }\mathbf{x}\mathbf{\le }\frac{\mathbf{1}}{\mathbf{2}}\mathbf{L}\\ \mathbf{0}& \left|x\right|\mathbf{>}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{L}\end{array}\mathbf{\right\}}$

Short answer

$A\left(k\right)\frac{1}{\pi }\frac{\sin \left[\left(k_0-k\right)\left({\displaystyle \frac{L}{2}}\right)\right]}{k_0-k}$

Step-by-step solution

The Fourier transform

The generalization of the Fourier series is known as Fourier transform and it can also refer to both the frequency domain representation and the mathematical function used. The Fourier transform facilitates the application of the Fourier series to non-periodic functions, allowing every function to be viewed as a sum of simple sinusoids.

The equation of the Fourier transform as,

$\mathit{A}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{+}\mathbf{\infty }}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{k}\mathbf{x}}\mathit{d}\mathit{x}$

Trigonometry relation,$\frac{{\mathbf{e}}^{\mathbf{i}\mathbf{\phi }}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{i}\mathbf{\phi }}}{\mathbf{2}}\mathbf{=}\mathit{i}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\varphi }$

As given function role="math" localid="1658391836564" $f\left(x\right)=\left\{\begin{array}{cc}{\mathrm{e}}^{{\mathrm{ik}}_{\mathrm{o}}\mathrm{x}}& -\frac{1}{2}\mathrm{L}\le \mathrm{x}\le \frac{1}{2}\mathrm{L}\\ 0& \left|\mathrm{x}\right|>\frac{1}{2}\mathrm{L}\end{array}\right\}$

Substitute the given function using equation of Fourier transform

Substitute the given function in the equation for the Fourier transform with proper limits from$-\frac{L}{2}to+\frac{L}{2}$

role="math" localid="1658392312745" $$\begin{gathered} A\left(k\right)=\frac{1}{2\mathrm{\pi }}{\int }_{-\frac{1}{2}\mathrm{L}}^{+\frac{1}{2}\mathrm{L}}{\mathrm{e}}^{i{k}_{\mathrm{o}}\mathrm{x}}{e}^{-\mathrm{ikx}}dx \\ A\left(k\right)=\frac{1}{2\mathrm{\pi }}{\left[\frac{1}{i\left(k_0-k\right)}{\mathrm{e}}^{\mathrm{i}({\mathrm{k}}_{\mathrm{o}}-\mathrm{k})\mathrm{x}}\right]}_{\frac{1}{2}\mathrm{L}}^{\frac{1}{2}\mathrm{L}} \\ A\left(k\right)=\frac{1}{2\mathrm{\pi }}\frac{1}{\mathrm{i}({\mathrm{k}}_0-\mathrm{k})}\left[{\mathrm{e}}^{\mathrm{i}({\mathrm{k}}_{\mathrm{o}}-\mathrm{k})\left(\frac{1}{2}\mathrm{L}\right)}{\mathrm{e}}^{\mathrm{i}({\mathrm{k}}_{\mathrm{o}}-\mathrm{k})\left(\frac{1}{2}\mathrm{L}\right)}\right] \end{gathered}$$

Let, $\varphi =(k_0-k)\left(\frac{1}{2}L\right)$

Then, above equation becomes as

role="math" localid="1658392679524" $$\begin{gathered} A\left(k\right)=\frac{1}{2\mathrm{\pi }}\frac{1}{\mathrm{i}({\mathrm{k}}_0-\mathrm{k})}\left[e^{i\phi }-e^{i\phi }\right] \\ \left[\because \frac{e^{i\phi }-e^{i\phi }}{2}=i\sin \varphi \right] \\ A\left(k\right)=\frac{1}{2\mathrm{\pi }}\frac{1}{\mathrm{i}({\mathrm{k}}_0-\mathrm{k})}\left(2i\sin \varphi \right) \end{gathered}$$

Further simplified the above equation by$\varphi =(k_0-k)\left(\frac{1}{2}L\right)$

$A\left(k\right)\frac{1}{\pi }\frac{\sin \left[\left(k_0-k\right)\left({\displaystyle \frac{L}{2}}\right)\right]}{k_0-k}$

Hence, the Fourier transform for A(k) the given function f(x) is

$A\left(k\right)\frac{1}{\pi }\frac{\sin \left[\left(k_0-k\right)\left({\displaystyle \frac{L}{2}}\right)\right]}{k_0-k}$