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Prooving pseudo-periodicity of diffracted field for gratings

  1. Jul 18, 2013 #1

    for my thesis I wanted to show the complete derivation for the grating equation - case: perfectly conducting. The later steps are all no problem, but I am struggling with the proof of pseudo-periodicity. I found in my opinion a nice summary here: http://www.math.purdue.edu/~lipeijun/math598_f10/notes/notes.pdf

    He starts with Maxwell, states the boundary conditions on the grating profile for TE and TM waves (page 25):

    [itex]n \times (E_1 - E_2) = 0 [/itex]
    [itex]n \times (H_1 - E_2) = 0 [/itex]

    and then deduces the Dirichlet and Neumann boundary conditions for the scalar field u that he has defined before.

    The two homogeneous conditions read (page 26)

    [itex]u(x,f(x)) = 0 [/itex]
    [itex]\partial_n u (x,f(x)) = 0 [/itex]

    He states that the Dirichlet and Neumann boundary conditions directly follow, when E= (0,0,u) and H = (0,0,u). But I don't see how, as the normal vector has both x and y components, when moving along the grating profile.

    If someone could point out how this simple form is derived I would be really grateful.

  2. jcsd
  3. Jul 18, 2013 #2

    Andy Resnick

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    Are you asking how to get from Eq. 2.2 to Eqns 2.3 and 2.4?
  4. Jul 18, 2013 #3
    Thanks for looking into the pdf. Yes..thats exactly the step. To me it seems that he put n=(0,1,0). If you could shed some light...
  5. Jul 22, 2013 #4
    any hints someone can give to tackle this problem?
  6. Jul 22, 2013 #5

    Andy Resnick

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    Maybe I'm missing something, but if the surface normal n = (n1, n2,0), eqns 1.8 and 1.10 (and 2.2) result in 2.3 and 2.4. Can you show any of your work?
  7. Jul 22, 2013 #6
    Thank you andy.

    For the first part I agree:

    (n1, n2, 0) x (0,0,u) = (n2 u, -n1 u, 0) = 0.

    This directly implies that u = 0.

    But to get the boundary condition for TM, I don't see it. Without applying any other equation you directly get the same boundary condition as before, as now H = (0,0,u) and hence u(x1,(f(x1)) = 0

    I don't see how to bring this into the form

    [itex]\nabla_n u = 0[/itex]

    and even if..shouldn't there be only one boundary condition for the one section?
  8. Jul 22, 2013 #7

    Andy Resnick

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    I found a few typos in the pdf (for example, 1.10; the first paragraph in section 2.3...). Anyhow, if you note that curl(H) = -ikE then you can (again) begin with n x E = 0 but substitute E_x = -1/ik (partial H_3/partial y), E_y = 1/ik(partial H_3/partial x), evaluate the cross product and obtain partial u/partial n.
  9. Jul 23, 2013 #8
    True that solves it. Thank you alot. I was wondering though: in general it holds that

    [itex]n \times (H_1 - H_2) [/itex]
    Why does this not work here and just leads to the Dirichlet boundary conditions once again?

    Is the following logic sound to finally prove pseudo-periodicity:

    due to uniqueness of solutions to the Helmholtz equation if we show that

    [itex] w(x1,x2) = u^d(x1 + D, x2) \cdot exp(-i D \alpha) [/itex] is a solution to the Helmholtz equation: [itex]\Delta u^d + \kappa^2 u^d = 0 [/itex] and satisfies all boundary constraints (radiation condition and Dirichlet + Neumann) then w(x1,x2) has to be equal to u^d(x1,x2) and hence u is quasi-periodic.

    Helmholtz equation:
    plugging w into the equation
    [itex] exp(-i D \alpha) \Delta u^d(x1 + D, x2) + \kappa^2 u^d(x1 + D, x2) exp(-i D \alpha) = 0 [/itex]

    As the the Helmholtz equation holds for [itex] u^d [/itex] and [itex] u^d(x1 + D, x2) [/itex] can be brought into the form [itex] u^d(x1',x2) [/itex]. It holds for w(x1,x2) as well.

    Boundary conditions:

    In TE polarization we then have to show that [itex] w + u^{inc} = 0 [/itex] correct? But then we would have to proove that

    [itex] u^d(x_1 + D, x_2) exp(-i D \alpha) + e^{i(\alpha x_1 - \beta x_2)} = 0 [/itex]. I really don't know how to do that one. The whole proof seems a little fishy to me. Am I on the right track here, or is it completely off?

    Thanks in advance for your help.
  10. Jul 23, 2013 #9

    Andy Resnick

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    Offhand, I would say that it's due to the boundary conditions: n x H = K, where K is the surface current density: H = 0 inside a perfect conductor for oscillating fields.

    I haven't gone through the pdf file is great detail, but it seems that eqn 2.7 is substantively different than what you wrote above. In any case, are you just trying to work through the pdf to see if you get the same results?
  11. Jul 24, 2013 #10
    I wanted to prove the periodicity in x. All the resources I've found so far always used these handwaving arguments. But I wanted to see it coherently on paper why the proof works like that. Starting with that I stumbled on some problems.

    [itex] w + u^{inc} = 0 [/itex] was a mere try to somehow get started showing:
    The boundary condition is also satisfied by observing that [itex]u^{inc}[/itex] is a quasiperiodic function and using the boundary condition of [itex]u^d[/itex].

    Ha..that should show it in the end. No?

    [tex]\begin{eqnarray} u^d(x,y) &=& -u^{inc} \\
    u^d(x+\Lambda,y) &=& -u^{inc} (x+\Lambda, y) \\
    {} &=& - u^{inc} e^{i\lambda \Lambda} \\
    u^d(x+\Lambda, y) e^{-i\lambda \Lambda}&=& -u^{inc} \\
    w^d &=& -u^{inc}
    \end{eqnarray} [/tex]

    Likewise it can be shown that

    [tex] \partial_n w^d = - \partial_n u^{inc} [/tex]

    That should complete the proof if I am not mistaken. Thank you for the lead :).
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