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Homework Help: TM waves in a rectangular waveguide

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Having trouble understanding why it is that inside a waveguide sides x=a,y=b propagating in z, subject to b.c. E parallel= 0 and Bperp=0... that for TE Bz=Bcos(pinx/a)cos(pimy/b) but for TM Ez=Esin(pinx/a)sin(pimy/a)?

    2. Relevant equations
    E parallel= 0 and Bperp=0

    For TM Bz=0 for TE Ez=0

    for TM Bz is proportional to d/dx(Bz)

    ^ hence the cosine solutions
    3. The attempt at a solution

    Why is it that inside a waveguide sides x=a,y=b propagating in z where E parallel= 0 and Bperp=0 is it that Bz=Bcos(pinx/a)cos(pimy/b) but Ez=Esin(pinx/a)sin(pimy/a)?

    I know it is because bz is proportional to d/dx(bz) so we get the cosine derivative to sine and we fit our b.c. there but ez is also proportional to d/dx(ez) however we see that Ez has a sine dependence not cosine dependence..
  2. jcsd
  3. Oct 22, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You're essentially asking for the entire derivation of the TE and TM equations, which is a tall order for a single post.

    An abbreviated answer is that for both TE and TM waves the equations start with the form

    Ez = E0zexp(jwt - kz) = XYexp(jwt - kz)

    X = Acos(Mx) + Bsin(Mx)
    Y = Ccos(Ny) + Dsin(Ny

    X = X(x only)
    Y = Y(y only)
    A, B, C, D are constants subject to boundary conditions.
    M = mπ/a, N = nπ/b etc.
    a,b waveguide dimensions.

    Imposing the boundary conditions on E and the requirement of div E = 0 gives your final expressions for Ex, Ey and Ez.

    Then, using
    H/∂t =-(1/μ) del x E
    gives Hx, Hy and Hz.
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