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Difference in proof between TE and TM modes

  1. Mar 8, 2015 #1
    I'm a little confused regarding the maths of TE and TM modes.

    Solving the following system for TE (which derives from Ey(x, z, t) = Em(x) = exp[i(ωt-βz)] ):
    Asin(px) + Bcos(px) , -d/2 < x <d/2
    Cexp(-qx), x>d/2
    Dexp(qx), x<-d/2

    we conclude in two types of solutions for TE modes: symmetric: ptan(pd/2) = q and asymmetric: pcot(pd/2) = -q.

    What is different for the above solutions for TM? I have a hard time determining that. I think it has something to do with the dielectric constant, but I'm not quite sure.

    I can elaborate on any maths you ask for. Also sorry for the quality of my post, it is my first one.
     
  2. jcsd
  3. Mar 8, 2015 #2
    Hi, when you write the solution Ey =exp[i(\omega t - \beta z)], this is a wave traveling in the negative z direction.

    If you want to break this wave as the sum of a TE mode and a TM mode, then for the TE wave, only Ez is 0. Ex, Ey, Hx, Hy, and Hz are not zero. for the TM wave, Hz is 0. Ex, Ey, Ez, Hx, Hy are not zero.

    I cannot relate your question to the definitions of the TE and TM mode.

    elgen
     
  4. Mar 8, 2015 #3
    Elgen, thanks for your answer! That's the very first part of the proof, I'm trying to understand beyond that.
    Fortunately, I found some insight from http://ece562web.groups.et.byu.net/notes/slab_waveguide.pdf.
    TM Modes solutions are found at pages 8-9. The problem is that I don't know how to reach the eigen equations, which are the same as TE's (described earlier in the .pdf), only with the dielectric constant n^2 added. If someone can briefly describe what is different from TE, I'd be grateful!
     
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