Fourier Transform of Powers of Sinc

Fourier Transforms of One-Dimensional Apertures

Fourier Transforms of Two-Dimensional Apertures

The Fourier transform of the cardinal sine is known to be a rectangular function and that of its square a triangular function. This prompts the question of the general form for an arbitrary power of the function, which is not difficult to determine.

Begin with the symmetric Fourier transform

$\mathcal{F}\left[{sinc}^{n}x\right]\left(k\right)=\frac{1}{\sqrt{2\pi}}\underset{-\infty}{\overset{\infty}{\int}}dx\phantom{\rule{.3em}{0ex}}{e}^{-ikx}\frac{{sin}^{n}x}{{x}^{n}}$

The power of the sine function can be expanded in terms of exponentials with the binomial theorem:

${sin}^{n}x=(\frac{{e}^{ix}-{e}^{-ix}}{2i}{)}^{n}=\frac{1}{(2i{)}^{n}}\sum _{p=0}^{n}(-1{)}^{p}\left(\genfrac{}{}{0ex}{}{n}{p}\right){e}^{i(n-2p)x}$

When applying the residue theorem to evaluate the integral, the contour must be completed in the direction that produces a negative value in the exponential when a complex component is added to the variable of integration. That means the contour is completed upwards for $n-2p-k>0$ and downwards for $n-2p-k<0$ .

In both cases the residue theorem picks out the term in the expansion of the exponential that is one power of the variable of integration less than the denominator. When the the contour is completed downward there is an additional factor of minus one for the reversed contour, but this can be included with a signum function.

The final complication is that the pole at the origin initially lies on the contour, which must be distorted to avoid it. The contour of integration will then only contain a half path around the pole. Since this could go either above or below the pole, the end result is that the contribution from the pole is only half of that expected.

The result of applying the residue theorem is

$\begin{array}{l}\mathcal{F}\left[{sinc}^{n}x\right]\left(k\right)\\ \phantom{\rule{3em}{0ex}}=\frac{i}{(2i{)}^{n}}\sqrt{\frac{\pi}{2}}\phantom{\rule{.3em}{0ex}}\sum _{p=0}^{n}(-1{)}^{p}\left(\genfrac{}{}{0ex}{}{n}{p}\right)\frac{\left[i\right(n-2p-k){]}^{n-1}}{(n-1)!}sgn(n-2p-k)\\ \phantom{\rule{3em}{0ex}}=\frac{1}{{2}^{n}(n-1)!}\sqrt{\frac{\pi}{2}}\phantom{\rule{.3em}{0ex}}\sum _{p=0}^{n}(-1{)}^{p}\left(\genfrac{}{}{0ex}{}{n}{p}\right)(n-2p-k{)}^{n-1}sgn(n-2p-k)\end{array}$

This is a somewhat curious result: the total function is a linear combination of $n+1$ polynomials of degree $n-1$ . Here is how that looks for the first few powers:

The total function is clearly zero for $absk>n$ . This can be easily understood for the first power, which is

$\mathcal{F}[sincx]=\frac{1}{2}\sqrt{\frac{\pi}{2}}\phantom{\rule{.3em}{0ex}}[sgn(1-k)+sgn(1+k\left)\right]$

For $absk>1$ the two terms have opposite signs and cancel, but in the middle region have the same sign and reinforce. The combinations become progressively more complicated for larger powers. This graphic displays the individual parts of the total function:

A bit bizarre that these sum to a simple-looking curve...

Far away from an array of slits, the intensity generated by an incident plane wave can be described by Fraunhofer diffraction, which is simply the Fourier transform of the physical aperture. Since this model has application in both optics and quantum mechanics, it will be treated as a purely mathematical problem.

For an arbitrary number of slits located at points ${a}_{p}$ with widths ${w}_{p}$ , the Fourier transform is rather simple to write in very general form. An incident wave of unit amplitude is only nonzero over the span of each slit. Summing their individual contributions gives

$\begin{array}{l}\mathcal{F}\left[\mathrm{slits}\right]=\frac{1}{\sqrt{2\pi}}\sum _{p}\underset{{a}_{p}-{w}_{p}/2}{\overset{{a}_{p}+{w}_{p}/2}{\int}}d{x}^{\prime}{e}^{-i{x}^{\prime}x}\\ \phantom{\mathcal{F}\left[\mathrm{slits}\right]}=\frac{1}{\sqrt{2\pi}}\sum _{p}\frac{{e}^{-ix({a}_{p}+{w}_{p}/2)}-{e}^{-ix({a}_{p}-{w}_{p}/2)}}{-ix}\\ \mathcal{F}\left[\mathrm{slits}\right]=\frac{1}{\sqrt{2\pi}}\sum _{p}{e}^{-i{a}_{p}x}{w}_{p}sinc\left(\frac{{w}_{p}}{2}x\right)\end{array}$

which is the total value of field or probability along the line of measurement. The intensity of the diffracted wave is given by the absolute square of this total value.

For a single slit, the intensity is a squared cardinal sine

${I}_{\mathrm{single}}=\frac{{w}^{2}}{2\pi}{sinc}^{2}\left(\frac{w}{2}x\right)$

which looks like this as a function of the width of the slit:

This is the expected diffraction peak with a spread dependent on slit size.

For a pair of slits of equal width located equal distances from the origin, the intensity acquires an extra trigonometric factor

$\begin{array}{l}{I}_{\mathrm{double}}=\frac{{w}^{2}}{2\pi}{sinc}^{2}\left(\frac{w}{2}x\right)|{e}^{-iax}+{e}^{iax}{|}^{2}\\ \phantom{{I}_{\mathrm{double}}}=\frac{{w}^{2}}{2\pi}{sinc}^{2}\left(\frac{w}{2}x\right)\times 4{cos}^{2}ax=4{I}_{\mathrm{single}}{cos}^{2}ax\end{array}$

and look like this as a function of free parameters:

The second slit adds additional structure to the diffraction peak, which will be true for any higher number of slits. Indeed for an array of *N* slits spaced equally 2*a* apart, the trigonometric term can be summed with a geometric series,

$|\sum _{p=1}^{N}{e}^{2ipax}{|}^{2}=|\sum _{p=0}^{N-1}{e}^{2ipax}{|}^{2}=|\frac{1-{e}^{2iNax}}{1-{e}^{2iax}}{|}^{2}=(\frac{sinNax}{sinax}{)}^{2}$

where phase factors under the absolute square are discarded at each step. The intensity of the array of slits is thus

${I}_{N}=\frac{{w}^{2}}{2\pi}{sin}^{2}\left(\frac{w}{2}x\right)\times (\frac{sinNax}{sinax}{)}^{2}={I}_{\mathrm{single}}(\frac{sinNax}{sinax}{)}^{2}$

which is consistent with the result for a double slit. The array of slits produces this modification to the diffraction peak of the single slit:

This graphic includes the option to exclude diffraction in order to see pure interference by itself.

The two-dimensional Fourier transform of a square aperture is a simple extension of the one-dimensional slit. For a square with sides *w*, the two-dimensional transform is

$\begin{array}{l}\mathcal{F}\left[\mathrm{square}\right]=\frac{1}{2\pi}\underset{-w/2}{\overset{w/2}{\int}}d{x}^{\prime}\underset{-w/2}{\overset{w/2}{\int}}d{y}^{\prime}\phantom{\rule{.3em}{0ex}}{e}^{-i{x}^{\prime}x-{y}^{\prime}y}\\ \phantom{\mathcal{F}\left[\mathrm{square}\right]}=\frac{1}{2\pi}\frac{{e}^{-ixw/2}-{e}^{ixw/2}}{-ix}\frac{{e}^{-iyw/2}-{e}^{iyw/2}}{-iy}\\ \mathcal{F}\left[\mathrm{square}\right]=\frac{{w}^{2}}{2\pi}sinc\left(\frac{w}{2}x\right)sinc\left(\frac{w}{2}y\right)\end{array}$

which is simply the product of two one-dimensional transforms. The intensity for the square aperature is

${I}_{\mathrm{square}}=\frac{{w}^{4}}{4{\pi}^{2}}{sinc}^{2}\left(\frac{w}{2}x\right){sinc}^{2}\left(\frac{w}{2}y\right)$

which apart from the leading multiplicative factor looks like this:

The two-dimensional Fourier transform of a circular aperture is more complicated because the kernel is not separable. For a circle with diamenter *w* one has

$\begin{array}{l}\mathcal{F}\left[\mathrm{circle}\right]=\frac{1}{2\pi}\underset{0}{\overset{w/2}{\int}}{r}^{\prime}d{r}^{\prime}\underset{0}{\overset{2\pi}{\int}}d{\phi}^{\prime}\phantom{\rule{.3em}{0ex}}{e}^{-i{r}^{\prime}cos{\phi}^{\prime}rcos\phi -i{r}^{\prime}sin{\phi}^{\prime}rsin\phi}\\ \phantom{\mathcal{F}\left[\mathrm{circle}\right]}=\underset{0}{\overset{w/2}{\int}}{r}^{\prime}d{r}^{\prime}\frac{1}{2\pi}\oint d{\phi}^{\prime}\phantom{\rule{.3em}{0ex}}{e}^{-i{r}^{\prime}rcos({\phi}^{\prime}-\phi )}\\ \mathcal{F}\left[\mathrm{circle}\right]=\underset{0}{\overset{w/2}{\int}}d{r}^{\prime}\phantom{\rule{.3em}{0ex}}{r}^{\prime}{J}_{0}\left({r}^{\prime}r\right)=\frac{1}{{r}^{2}}\times \frac{w}{2}r\phantom{\rule{.3em}{0ex}}{J}_{1}\left(\frac{w}{2}r\right)\end{array}$

recognizing one form of the definition of the Bessel function of the first kind in the second step. The relation $\frac{d}{dx}\left[x{J}_{1}\right(x\left)\right]=x{J}_{0}\left(x\right)$ can be confirmed with recurrence relations for the Bessel functions. This can be slightly rewritten as

$\mathcal{F}\left[\mathrm{circle}\right]=\frac{{w}^{2}}{4}\frac{{J}_{1}\left({\displaystyle \frac{w}{2}}r\right)}{{\displaystyle \frac{w}{2}}r}$

The intensity for a circular aperture is

${I}_{\mathrm{circle}}=\frac{{w}^{4}}{16}\frac{{J}_{1}^{2}\left({\displaystyle \frac{w}{2}}r\right)}{({\displaystyle \frac{w}{2}}r{)}^{2}}$

which again apart from the leading multiplicative factor looks like this:

The fraction of two functions in this expression is an analog of the cardinal sine function.

*Uploaded 2020.03.25 — Updated 2020.03.26*
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