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Modes orthogonality in a dielectric slab

  1. Dec 13, 2015 #1
    A typical mode in a dielectric slab like this, with propagation along [itex]x[/itex], uniformity along [itex]z[/itex] and refractive index variation along [itex]y[/itex], is represented by the following function:

    [itex]f (y) = \begin{cases} \displaystyle \frac{\cos (k_1 y)}{\cos (k_1 d)} && |y| \leq d \\ e^{-j k_2 (y - d)} && |y| \geq d \end{cases}[/itex]

    The slab is symmetrical with respect to the [itex](x,z)[/itex] plane and its thickness is [itex]2d[/itex]. [itex]k_2 = - j \gamma[/itex] is purely imaginary and it causes attenuation out of the core. Another acceptable mode can have [itex]\sin (k_1 y)[/itex].
    If [itex]f(y)[/itex] represents the electric field [itex]E_z[/itex] and the field is maximum for [itex]y = 0[/itex], the mode is said to be "even" and it has a cosine behaviour inside the core; if the electric field is zero for [itex]y = 0[/itex], the mode is said to be "odd" and it has a sine behaviour inside the core.

    According to the solutions [itex](k_1, k_2)[/itex] of the system, modes can have one or more zeros in the [itex]|x| \leq d[/itex] zone.

    I would like to prove that two different modes [itex]f_1 (y)[/itex] and [itex]f_2 (y)[/itex] ([itex]f_1 \neq f_2 [/itex]) are orthogonal. This is a very important proof because it would mean that modes can't bring power jointly and can't exchange their power.

    Each mode satisfies the wave equation

    [itex]\displaystyle \frac{\partial^2 f_1 (y)}{\partial y^2} + [n_1^2 k_0^2 - \beta_1^2] f_1 (y) = 0[/itex]
    [itex]\displaystyle \frac{\partial^2 f_2 (y)}{\partial y^2} + [n_1^2 k_0^2 - \beta_2^2] f_2 (y) = 0[/itex]

    ([itex]n_1[/itex] is not related to the mode 1 or mode 2: it is the refractive index of the core). [itex]\beta[/itex] is the propagation constant along the propagation direction [itex]x[/itex]. Suppose that [itex]\beta_1 \neq \beta_2[/itex]. Let's multiply the first equation by [itex]f_2[/itex] and the second by [itex]f_1[/itex]; then, let's subtract the second equation from the first. The result is

    [itex]f_2 (y) \displaystyle \frac{\partial^2 f_1 (y)}{\partial y^2} - f_1 (y) \displaystyle \frac{\partial^2 f_2 (y)}{\partial y^2} = [\beta_1^2 - \beta_2^2] f_1 (y) f_2 (y)[/itex]

    Now, let's integrate between [itex]0[/itex] and [itex]+ \infty[/itex] (if the mode is odd or even, we can integrate on just half of the [itex]y[/itex] axis and the result for the remaining half is the same - if even - or the opposite - if odd):

    [itex] \displaystyle \int_{0}^{+\infty} \left( f_2 (y) \displaystyle \frac{\partial^2 f_1 (y)}{\partial y^2} - f_1 (y) \displaystyle \frac{\partial^2 f_2 (y)}{\partial y^2} \right) dy = \displaystyle \int_{0}^{+\infty} \left[ (\beta_1^2 - \beta_2^2) f_1 (y) f_2 (y)\right] dy [/itex]

    The LHS can be integrated by parts:

    [itex] \displaystyle \int_{0}^{+\infty} \left( f_2 (y) \displaystyle \frac{\partial^2 f_1 (y)}{\partial y^2} - f_1 (y) \displaystyle \frac{\partial^2 f_2 (y)}{\partial y^2} \right) dy = [/itex]
    [itex]= \left[ f_2 (y) \displaystyle \frac{\partial f_1 (y)}{\partial y} \right]_0^{+\infty} - \displaystyle \int_{0}^{+\infty} \left( \frac{\partial f_1 (y)}{\partial y} \frac{\partial f_2 (y)}{\partial y}\right) dy - \left[f_1 (y) \displaystyle \frac{\partial f_2 (y)}{\partial y} \right]_0^{+\infty} + \displaystyle \int_{0}^{+\infty} \left( \frac{\partial f_1 (y)}{\partial y} \frac{\partial f_2 (y)}{\partial y}\right) dy =[/itex]
    [itex]= \left[ f_2 (y) \displaystyle \frac{\partial f_1 (y)}{\partial y} \right]_0^{+\infty} - \left[f_1 (y) \displaystyle \frac{\partial f_2 (y)}{\partial y} \right]_0^{+\infty}[/itex]

    Both [itex]f_1[/itex] and [itex]f_2[/itex] and their derivatives are 0 for [itex]y \to +\infty[/itex] because of the decreasing exponential. So, just the following term has to be evaluated:

    [itex]- f_2 (0) \displaystyle \frac{\partial f_1 (0)}{\partial y} + f_1 (0) \displaystyle \frac{\partial f_2 (0)}{\partial y}[/itex]

    For [itex]y = 0[/itex] several cases can be distinguished. If [itex]f_1[/itex] and [itex]f_2[/itex] are both even or both odd, the expression vanished as desired. But if [itex]f_1[/itex] is odd and [itex]f_2[/itex] is even, or viceversa, respectively the first and the second addend are not zero.

    How can be proved in these last two cases that both the addends (and so all the LHS) anyway vanish? Any hint or any references? I did not find something useful on Google.
     
  2. jcsd
  3. Dec 17, 2015 #2

    RUber

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    Homework Helper

    As you mentioned, if ##f_1## and ##f_2## are both even or both odd, you have been able to show what you want to show.
    If one is odd and another is even, then they are orthogonal from the start.
     
  4. Dec 18, 2015 #3
    Are you meaning that the integral in the RHS

    [itex]\displaystyle \int_{0}^{+ \infty} f_1(y) f_2(y)dy[/itex]

    is [itex]0[/itex] by itself? I know that sine and cosine are orthogonal functions, but it is difficult to figure out how this integral can vanish when the [itex]k_1, k_2[/itex] in [itex]f_1[/itex] are different from the [itex]k_1, k_2[/itex] in [itex]f_2[/itex].

    Moreover,

    [itex]\displaystyle \int_{d}^{+ \infty} e^{- k_2^{(1)}(y - d)}e^{- k_2^{(2)}(y - d)}dy = \displaystyle - \frac{1}{\left( k_2^{(1)} + k_2^{(2)} \right)}\left[ e^{- \left( k_2^{(1)} + k_2^{(2)} \right) (y - d)} \right]_d^{+\infty} = \frac{1}{\left( k_2^{(1)} + k_2^{(2)} \right)} \neq 0[/itex]

    Note: there was an error in my original post, where [itex]f(y)[/itex] should be [itex]e^{-k_2(y-d)}[/itex] for [itex]|y| \geq d[/itex], with [itex]k_2 \geq 0, k_2 \in \mathbb{R}[/itex], and not [itex]e^{-jk_2(y-d)}[/itex].
     
    Last edited: Dec 18, 2015
  5. Dec 18, 2015 #4

    RUber

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    Homework Helper

    When you say:
    you complicate the integral.
    If ##f ## is odd, then ## \int_{-d}^d f(y) dy =0## and ## \int_d^\infty f(y) dy + \int_{-\infty}^{-d} f(y) dy = 0##.
    And note that, as with integers, an odd function times an even function is an odd function.
     
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