Question

Consider the following function:

$\mathit{f}\left(x\right)\mathbf{=}\{\begin{array}{ll}C{e}^{+\alpha x}& -\infty <x<0\\ B{e}^{-\alpha x}& 0\le x\le +\infty \end{array}$

(a) Sketch this function. (Without loss of generality, assume that C is greater than B.) Calculate the Fourier transform $\mathit{A}\left(k\right)$.

(b) Show that for large $\mathit{k}\mathbf{,}\mathit{A}\left(k\right)$is proportional to $\frac{\mathbf{1}}{\mathbf{k}}$.

(c) In general,$\mathit{f}\left(x\right)$is not continuous. Under what condition will it be, and how$\mathit{A}\left(k\right)$does behave at large values ofk if this condition holds?

(d) How does a discontinuity in a function affect the Fourier transform for large values of k?

Short answer

(a) The Fourier transform is $\frac{1}{2\mathrm{\pi }}\left(\frac{\left(C+B\right)\alpha +\left(C-B\right)ik}{{\alpha }^2+k^2}\right)$.

(b) It is proved that for large k,$A\left(k\right)$ is proportional to $\frac{1}{k}$.

(c)$f\left(x\right)$ is continuous at $C=B,A\left(k\right)$, falls off rapidly if this condition holds.

(d) A discontinuity affects the content to fall more slowly.

Step-by-step solution

Fourier transform:

The Fourier transform is a mathematical function that decomposes a waveform that is a function of time into the frequencies that make it up. The result produced by the Fourier transform is a complex valued function of frequency.

The Fourier transform is given by $\mathit{A}\left(k\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathit{\psi }\left(x\right){\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{k}\mathbf{x}}\mathit{d}\mathit{x}$.

(a) Find Fourier transform:

The Fourier transform can be obtained as follows:

$$\begin{gathered} A\left(k\right)=\frac{1}{2\mathrm{\pi }}{\int }_{-\infty }^{\infty }f\left(x\right)e^{-ikx}dx \\ =\frac{C}{2\mathrm{\pi }}{\int }_{-\infty }^0{e}^{\alpha x}{e}^{-ikx}dx+\frac{B}{2\mathrm{\pi }}{\int }_0^{\infty }{e}^{\alpha x}{e}^{-ikx}dx \\ =\frac{C}{2\mathrm{\pi }}{\int }_{-\infty }^0{e}^{\left(\alpha -ik\right)x}dx+\frac{B}{2\mathrm{\pi }}{\int }_0^{\infty }{e}^{-\left(\alpha +ik\right)x}dx \end{gathered}$$

Solve the above equation as follows:

$$\begin{gathered} A\left(k\right)=\frac{C}{2\mathrm{\pi }}\left(\frac{1}{\alpha -ik}\right)+\frac{B}{2\mathrm{\pi }}\left(\frac{1}{\alpha -ik}\right) \\ =\frac{1}{2\mathrm{\pi }}\left(\frac{\left(C+B\right)\alpha \left(C-B\right)ik}{{\alpha }^2+k^2}\right) \end{gathered}$$

The sketch of the given function$f\left(x\right)$is shown below:

(b) Required Proof:

As the Fourier transform $A\left(k\right)$obtained is $\frac{1}{2\mathrm{\pi }}\left(\frac{\left(C+B\right)\alpha +\left(C-B\right)ik}{{\alpha }^2+K^2}\right)$. Now, if the value of k becomes very large, then the first term in the expression becomes negligible because the value of $\alpha$ become negligible compared to k. So, if the value of k becomes

very large then the Fourier transform becomes $A\left(k\right)=\frac{1}{2\mathrm{\pi }}\left(\frac{\left(C-B\right)ik}{k^2}\right)$which is equivalent to $A\left(k\right)=\frac{1}{2\mathrm{\pi }}\left(\frac{\left(C-B\right)i}{k}\right)$.

Hence, it is proved that for large k, A(K) is proportional to $\frac{1}{k}$.

(c) Continuity of function:

The graph in the part (a) implies that the graph is continuous at $C=B$. So, at $C=B$, the expression for$A\left(k\right)$ becomes $\frac{1}{2\mathrm{\pi }}\left(\frac{\left(C+B\right)\alpha }{{\alpha }^2+k^2}\right)$. Here, the value of$A\left(k\right)$ falls down rapidly if the value ofk increases and approaches to very large values.

Thus,$f\left(x\right)$ is continuous at $C=B$,$A\left(k\right)$ falls off rapidly if this condition holds.

(d) Discontinuity affects:

Compare part (b) and (c) to get the condition where the function has a lag in the function decay when becomes very large. The function in part (c) falls more rapidly than in part (b).

Thus, the discontinuity affects the content to fall more slowly.