Spatial Fourier Transform: Bessel x Sinusoidal

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SUMMARY

The discussion focuses on evaluating the spatial Fourier transform involving Bessel functions and sinusoidal components. The integral in question is expressed as I(k_x, k_y) = ∫∫ J_{m-1}(αρ) sin((m + 1)φ) e^{jρ(k_x cosφ + k_y sinφ)} ρ dρ dφ. Participants suggest using complex exponentials to simplify the φ integral and refer to the integral representation of Bessel functions for further analysis. Numerical integration is proposed as a fallback if closed-form solutions are unattainable, with references to the Digital Library of Mathematical Functions (DLMF) for similar integrals.

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  • Understanding of Bessel functions, specifically J_{m-1} and J_{m+1}
  • Familiarity with complex exponentials and their applications in integrals
  • Knowledge of numerical integration techniques
  • Basic concepts of Fourier transforms in two dimensions
NEXT STEPS
  • Study the integral representation of Bessel functions, particularly J_\ell(β)
  • Learn about Lommel's integral and its applications in solving integrals involving Bessel functions
  • Explore recurrence relations for Bessel functions to facilitate closed-form solutions
  • Investigate numerical integration methods suitable for well-behaved integrands
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Mathematicians, physicists, and engineers working with Fourier transforms and Bessel functions, particularly those involved in signal processing or wave analysis.

tworitdash
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I am trying to formulate an analytical expression for the spectrum of the electric fields on a circular aperture (cylindrical waveguide). The field expressions are a multiplication of Bessel function and sinusoidal function. I am attaching only one kind of integration that I have.
I(k_x, k_y) = \int_{0}^{R} \int_{0}^{2\pi} J_{m-1}(\alpha \rho) \sin((m + 1) \phi) e^{j\rho(k_x \cos\phi + k_y \sin\phi)} \rho d\rho d\phiIs there any way to do it? J is the Bessel function of the first kind. I thought of partially doing only the phi integral as \int_{0}^{2\pi} \sin((m + 1) \phi) e^{j\rho(k_x \cos\phi + k_y \sin\phi)} d\phi but then again I am not able to find any solution.
 
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You can definitely do the ##\phi## integral - I have been doing a lot of similar integrals recently. What you want to do is define ##k_x = k \cos\psi## and ##k_y = k \sin\psi##, and split up the ##\sin ## into complex exponentials. Then you get integrals of the form,
$$ \begin{eqnarray*}
\mathcal{I}^{\pm} & = & \int_0^{2\pi} \, e^{\pm j (m+1) \phi} \, e^{j k \cos(\phi-\psi)} \, d\phi \\
& = & e^{\pm j (m+1) \psi} \int_0^{2\pi} \, e^{\pm j (m+1) (\phi-\psi)} \, e^{j k \cos(\phi-\psi)} \, d\phi \\
& = & e^{\pm j (m+1) \psi} \int_{-\psi}^{2\pi-\psi} \, e^{\pm j (m+1) \xi} \, e^{j k \cos\xi} \, d\xi
\end{eqnarray*}
$$

To proceed you need the integral,
$$ J_\ell(\beta) = \frac{1}{2\pi j^\ell} \int_0^{2\pi} \, e^{j \ell \phi} \, e^{j \beta \cos\phi} \, d\phi
$$
which commonly occurs in these kinds of problems. Note that this is the integral of a periodic function over one period, so the result does not depend on which period you integrate over.

The integral over ##\rho## is harder, because you have an integrand that is basically ##\rho \, J_{m-1}(a\rho) \, J_{\pm(m+1)}(b\rho)##. Similar integrals can be found at DLMF (https://dlmf.nist.gov/10.22), but the exact integral is not there. Perhaps you can use recurrence relations for the ##J_n## to get this into a form where you can do this closed-form? If not, then you may be stuck doing numerical integration, which should be pretty easy as the integrand is well behaved.

Jason
 
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By the way, the integral representation of the Bessel function above follows from the expression
$$
e^{- j z \cos\phi} = \sum_{n=-\infty}^\infty j^{-n} e^{j n \phi}\, J_n(z).
$$
You could have also just used this expansion in your original integral and of course ended up with the same result.
 
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Thank you so much for the inputs @jasonRF . I will try. Previously I was successful with the integration of the product of Bessel functions. So, probably for the \rho variations, I can formulate something with Lommel's integral. I hope I can see some light at the end! :)
 

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