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TM and TE mode

  1. Jun 20, 2009 #1
    Why is the lowest order TM and TE mode TM11 and TE10(or TE01)? What is the physical meaning of the different orders of the modes?
    Thanks.
     
  2. jcsd
  3. Jun 20, 2009 #2

    Born2bwire

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    I'm going to guess this belongs in the homework section and definitely need more information needs to be provided, like what is the waveguide?
     
  4. Jun 20, 2009 #3
    no, this is not homework, my lecture notes said the lowest order needs to be TM11 and TE10(or TE01), but I don't really understand it.
     
  5. Jun 20, 2009 #4
    If you consider the expression for the superposition of two uniform plane waves propagating symmetrically with respect to the z-axis, you will see that it contains a factor in the form of [tex]sin(\beta x \ cos\ \theta)[/tex]. This factor describes the the standing wave character (here, in the x-direction). When this factor is zero (i.e., when [tex]\beta x \ cos\ \theta =\ m\pi[/tex] where [tex]m = 0, 1, 2, 3, ...[/tex]) the electric field is, of course, also zero.

    This is interesting because it means we can place two perfectly conducting sheets in the planes [tex]x = 0[/tex] and [tex]x = m\lambda /(2\ cos\ \theta)[/tex], without violating the boundary conditions (i.e., zero tangential electric field etc).

    The fields will have m number of one-half apparent wavelengths in the x-direction between the plates.

    This line of reasoning can be extended to a three-dimensional case where a quick glance at the field expressions for TE and TM waves will reveal why TM1,1 and TE1,0 or 0,1 are the lowest possible modes. Plug in m=n=0 for in the expression for TE waves or m = 0 or n = 0 for TM waves and see what happens.
     
    Last edited: Jun 20, 2009
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