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Rectangular Waveguide centered at the origin

  1. Mar 28, 2013 #1
    Hello,

    If a rectangular waveguide (or square well, etc) is centered at x=a/2, y=b/2, solution (e.g. for TE mode) is:

    H = Ho Cos(m*pi*x/a) * Cos(n*pi*y/b) (n,m = 0,1,2,...)

    So for TEn=0,m=1, H = Ho Cos(pi*x/a).

    If it is centered at the origin, you get even and odd solutions:

    H = Ho Cos(m*pi*x/2a) * Cos(n*pi*y/2b) (n,m odd)
    H = Ho Sin(m*pi*x/2a) * Sin(n*pi*y/2b) (n,m even).

    Now, what is the corresponding values for n,m that give the same mode?
     
  2. jcsd
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