Page 403 - Electromagnetics

Example: fields of a rectangular waveguide. Consider a rectangular waveguide
with a cross-section occupying 0 ≤ x ≤ a and 0 ≤ y ≤ b. The material within the guide
is assumed to be a lossless dielectric of permittivity and permeability µ. We seek the
modal fields within the guide.
Both TE and TM fields exist within the guide. In each case we must solve the differ-
ential equation

2
˜
2
˜
∇ ψ + k ψ = 0.
t
c
A product solution in rectangular coordinates may be sought using the separation of
variables technique (§ A.4). We find that
˜

ψ(x, y,ω) = [A x sin k x x + B x cos k x x] A y sin k y y + B y cos k y y
2
2
2
where k + k = k . This solution is easily verified by substitution.
x y c
For TM modes the solution is subject to the boundary condition (5.143):
˜
ψ e (ρ,ω) = 0, ρ ∈ .
Applying this at x = 0 and y = 0 we find B x = B y = 0. Applying the boundary condition
at x = a we then find sin k x a = 0 and thus
nπ
k x = , n = 1, 2,....
a
˜
Note that n = 0 corresponds to the trivial solution ψ e = 0. Similarly, from the condition
at y = b we find that
mπ
k y = , m = 1, 2,....
b
Thus
nπx mπy

˜
ψ e (x, y,ω) = A nm sin sin .
a b
From (5.137)–(5.139) we find that the fields are
nπx mπy ∓ jk z z

˜
2
E z = k A nm sin sin e ,
a b
c nm
nπ nπx mπy mπ nπx mπy
˜
E t =∓ jk z A nm ˆ x cos sin + ˆ y sin cos e ∓ jk z z ,
a a b b a b
mπ nπx mπy nπ nπx mπy
˜
H t = jk z Y e A nm ˆ x sin cos − ˆ y cos sin e ∓ jk z z .
b a b a a b
Here
1
Y e = !
η 1 − ω 2 /ω 2
c nm
with η = (µ ) 1/2 .
Each combination of m, n describes a different field pattern and thus a different mode,
designated TM nm . The cutoff wavenumber of the TM nm mode is
%
nπ mπ
2 2
= + , m, n = 1, 2, 3,...
k c nm
a b
and the cutoff frequency is
%
nπ mπ
2 2
= v , m, n = 1, 2, 3,...
ω c nm +
a b
© 2001 by CRC Press LLC