TM waves in a rectangular waveguide

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Homework Statement



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)?

Homework 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

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..
 

Answers and Replies

  • #2
rude man
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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

where
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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