Question

A function $\mathit{f}\left(\alpha \right)$is nonzero only in the region of width $\mathbf{2}\mathit{\delta }$centered at$\mathit{\alpha }\mathbf{=}\mathbf{0}$

$\mathbf{f}\left(\alpha \right)\mathbf{=}\{\begin{array}{ll}C& \left|\alpha \right|\le \delta \\ 0& \left|\alpha \right|\ge \delta \end{array}$

where C is a constant.

(a) Find and plot versus $\mathbf{\beta }$the Fourier transform $\mathbf{A}\left(\beta \right)$of this function.

(b) The function $\mathit{\rho }\mathit{\alpha }$) might represent a pulse occupying either finite distance (localid="1659781367200" $\mathit{\alpha }\mathbf{=}$position) or finite time ($\mathbf{\alpha }\mathbf{=}$time). Comment on the wave number if $\mathbf{\alpha }\mathbf{=}$is position and on the frequency spectrum if $\mathit{\alpha }$is time. Specifically address the dependence of the width of the spectrum on $\mathbf{\delta }$.

Short answer

(a). The graph plot between$A\left(\beta \right)$and$\beta$

(b). There is inverse relation between the width of the pulse $\left(\delta \right)$, and the wave-number or frequency range.

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,

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

The Euler formulas will also be used

$$\begin{gathered} {\mathit{e}}^{\mathbf{i}\mathbf{\theta }}\mathbf{=}\mathbf{cos}\mathbf{}\mathit{\theta }\mathbf{+}\mathit{i}\mathbf{}\mathbf{sin}\mathbf{}\mathit{\theta } \\ {\mathit{e}}^{\mathbf{-}\mathbf{i\theta }}\mathbf{=}\mathbf{cos}\mathbf{}\mathit{\theta }\mathbf{+}\mathit{i}\mathbf{}\mathbf{sin}\mathbf{}\mathit{\theta } \end{gathered}$$

Integrate the equation (2)

The equations (2) separated into two integrals

$$\begin{gathered} A\left(\beta \right)=\frac{C}{2\mathrm{\pi }}{\int }_{-\delta }^{+\delta }{e}^{-i\beta \alpha }d\alpha \\ A\left(\beta \right)=\frac{C}{2\mathrm{\pi }}{\int }_{-\delta }^{+\delta }\left[\cos \left(\beta \alpha \right)-i\sin \left(\beta \alpha \right)\right]d\alpha \\ A\left(\beta \right)=\frac{C}{2\mathrm{\pi }}\left[{\int }_{-\delta }^{+\delta }\cos \left(\beta \alpha \right)d\alpha -i{\int }_{-\delta }^{+\delta }\sin \left(\beta \alpha \right)d\alpha \right] \end{gathered}$$

When taking the integral of an odd function with equal but opposite limits of integration, the result will be zero, due to the symmetry of odd functions about the origin and since sine is an odd function, that integral will be zero:

$$\begin{gathered} A\left(\beta \right)=\frac{C}{2\mathrm{\pi }}\left[{\int }_{-\sigma }^{+\delta }\cos \left(\beta \alpha \right)d\alpha -i{\int }_{-\delta }^{+\delta }\sin \left(\beta \alpha \right)d\alpha \right] \\ A\left(\beta \right)=\frac{C}{2}\left[{\int }^{+\delta }\cos \left(\beta \alpha \right)d\alpha -i\left(0\right)\right] \\ A\left(\beta \right)=\frac{C}{2\mathrm{\pi }}{\int }_{-\delta }^{+\delta }\cos \left(\beta \alpha \right)d\alpha \end{gathered}$$

Since the cosine is an even function with equal but opposite limits of integration, it can be written as just twice the integral from zero to the upper limit of integration (due to the symmetry of the even function about y-axis).

$$\begin{gathered} A\left(\beta \right)=\frac{C}{2\mathrm{\pi }}{\int }_{-\delta }^{+\delta \int }\cos \\ A\left(\beta \right)=\frac{C}{2\mathrm{\pi }}\left[2{\int }_0^{\sigma \int }\cos \left[\right]\right] \\ A\left(\beta \right)=\frac{C}{2\mathrm{\pi }}{\int }_0^{\sigma \int }\cos \end{gathered}$$

And then just integrate that from zero to $\delta$

$$\begin{gathered} A\left(\beta \right)=\frac{C}{\mathrm{\pi }}{\int }_0^{\delta }\cos \left(\beta \alpha \right)d\alpha \\ A\left(\beta \right)=\frac{C}{\mathrm{\pi }}{\left[\frac{\sin \left(\beta \alpha \right)}{\beta }\right]}_0^{\delta } \\ A\left(\beta \right)=\frac{C}{\mathrm{\pi }}\left\{\left[\frac{\sin \left(\beta \alpha \right)}{\beta }\right]-0\right\} \\ A\left(\beta \right)=\frac{C}{\mathrm{\pi }}\frac{\sin \left(\beta \alpha \right)}{\beta } \end{gathered}$$

So, the Fourier transform of$f\left(\alpha \right)$is

$$\begin{gathered} A\left(\beta \right)=\frac{C}{\mathrm{\pi }}\frac{\sin \left(\beta \alpha \right)}{\beta } \\ A\left(\beta \right)=\frac{C}{\mathrm{\pi }}\frac{\sin \left(\beta \alpha \right)}{\beta } \end{gathered}$$

Hence,$$\begin{gathered} A\left(\beta \right)=\frac{C}{\mathrm{\pi }}\frac{\sin \left(\beta \delta \right)}{\beta } \\ A\left(\beta \right)=\frac{C}{\mathrm{\pi }}\frac{\sin \left(\beta \delta \right)}{\beta } \end{gathered}$$.

The graph plot between$A\left(\beta \right)$and$\beta$

Comment wave number and frequency spectrum

(b)

Well, as $\delta$or the interval gets smaller, which means less uncertainty in space or time, the more the uncertainty in the wave-numbers or frequencies respectively (more spreading). The width of the pulse in the frequency domain is found to be$\frac{2\mathrm{\pi }}{\delta }$(inversely proportional with $\delta$).

Therefore, there is inverse relation between the width of the pulse $\left(\delta \right)$, and the wave-number or frequency range.