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Fourier Transform of non-centered circular aperture

  1. Jul 11, 2012 #1
    Hi there,
    I have a little problem in wave optics: I have a wave function \psi_{ap} that depends on some geometric parameters, but that has no units itself (as one would expect). But unfortunately when I calculate the Fourier transform of this wave function the Fourier transform has a unit.

    Now I'd like to explain my problem a bit more in detail: The wave function \psi_{ap}, that is to be transformed, is a plane wave traveling in z-direction through two circular apertures that lie in the xy-plane and that are displaced from the origin of this plane by d_i along the x-axis. Additionally I assume a phase shift \varphi_i for each aperture.

    For convenience I will have only a look on the wave functions along the x-axis with y=0.

    \begin{align}
    A(x) =& \left\{
    \begin{array}{lcc}
    1 & \mathrm{for} & x\le 1\\
    0 && \mathrm{else}
    \end{array}
    \right.\\
    \psi_i =& A\left(\frac{x-d_i}{R}\right)*e^{-i\varphi_i}\\
    \psi_{ap} =& \psi_1+\psi_2\\
    \mathrm{with} :& d_1=-33.2\mu m,\,d_2=33.2\mu m,\,R=29\mu m,\, \varphi_1=0,\,\varphi_1=\pi/2\nonumber
    \end{align}

    \psi_ap is the wave function to be Fourier transformed and A is the function describing an aperture. Using the linearity of the Fourier transform I can calculate the transforms for each aperture separately. As the phase shift in each aperture is constant I can put it in front of the Fourier transform.

    \begin{align}
    \psi_{sp} =& \mathcal{F}(\psi_{ap}) = \mathcal{F}(\psi_1)+\mathcal{F}(\psi_2)\\
    \mathcal{F}(\psi_i) =& e^{-i\varphi_i}*\mathcal{F}\left(A\left(\frac{x-d_i}{R}\right)\right)
    \end{align}
    This far everything is fine, but now I have (referring to the remarks on the Wikipedia article) a shift in the "time" and "frequency" domain (Equations 102 and 104 in the tables of the article). By using first 104 and then 102 my Fourier Transform looks like this:
    \begin{align}
    \mathcal{F}\left(A\left(\frac{x-d_i}{R}\right)\right) =& R*e^{-2\pi i d_iq}\frac{\mathrm{J}_1(2\pi Rq)}{Rq}
    \end{align}
    where q is the coordinate in Fourier space, that by definition should have a unit of "1/m".

    My problem now with this expression is that
    • this expression has a unit of "m"
    • this expression -- except for the exponential term -- looks quite different than expected (especially the leading R); I would have expected J_1(2\pi Rq)/R/q

    I would appeciate any remarks.
     
  2. jcsd
  3. Jul 11, 2012 #2

    Mute

    User Avatar
    Homework Helper

    The extra dimension of length comes from the Fourier transform integral. You are integrating over a variable with dimensions of length, so the units of your fourier transform will be the units of your original function times a unit of length. If you want your fourier transformed function to be dimensionless, you should choose a dimensionless variable to integrate over (e.g., x/R rather than just x).
     
  4. Jul 12, 2012 #3
    Thanks for your remark, but I'm afraid having a wave function with a unit is not the problem, but only the symptom.

    Many sources on the net [1],[2] calculate the Fourier transform of a circular aperture to be
    \begin{equation}
    \mathcal{F}(A) = \frac{J_1(2\pi Rq)}{Rq},
    \end{equation}
    but when I do the calculation on my own (see above) I get
    \begin{equation}
    \mathcal{F}(A) = \frac{J_1(2\pi Rq)}{q}.
    \end{equation}
    So, I'm wondering where the error might be...
     
  5. Jul 13, 2012 #4

    Mute

    User Avatar
    Homework Helper

    When I do the 2d Fourier transform with [itex]A(r) = \Theta(a-r)[/itex], where a is the radius of the aperture and Theta is the Heaviside step function, I find

    [tex]\mathcal F[A(r)](k) = \frac{2\pi a}{k} J_1(ka),[/tex]

    where I used the fourier transform convention

    [tex]\mathcal F[A(r)](k) = \int_{\mathbb{R}^2} d\mathbf{r}~A(r) \exp\left(i \mathbf{k} \cdot \mathbf{r} \right) .[/tex]

    I did the integral by choosing my coordinates such that [itex]\mathbf{k} \cdot \mathbf{r} = kr\cos \phi[/itex], and I did the angular integral using the identity

    [tex]e^{iz\cos\phi} = \sum_{n=-\infty}^\infty i^n J_n(z) e^{in\phi}.[/tex]

    The radial integral is easy to do using the identity

    [tex] \left( \frac{1}{x} \frac{d}{dx} \right)^m \left[ x^\alpha J_\alpha(x) \right] = x^{\alpha - m} J_{\alpha -m }(x).[/tex]

    Note that I did not non-dimensionalize anything, so the final units of my answer are a factor of length^2 more than the original aperture function's dimensions. (I made a mistake in my previous post by saying that the increase is a factor of length because I forgot that it should be a 2d fourier transform).

    If I non-dimensionalized the variables as u = x/a and v = y/a and Fourier transformed with respect to u and v (instead of the dimensionful variables x and y), then setting q = ka, the fourier transform would be

    [tex]2\pi \frac{J_1(q)}{q},[/tex]

    which is what your other sources find. So, I would guess that your other sources are fourier transforming with respect to dimensionless variables rather than dimensionful variables. (Also note that my q is equal to 2 pi times your q times your R).
     
  6. Jul 17, 2012 #5
    Thanks for the help. I also found this article describing the transformation in detail.

    Thanks again.
     
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