Page 395 - Electromagnetics

˜
If we choose h we can use (5.124) and (5.125) to obtain
˜
˜
E = jω ˜µˆ u ×∇ t h , (5.132)
˜
˜ ∂ h
H =∇ t , (5.133)
∂u
˜
where h must obey

2
∂ 2
˜
2
˜
∇ h = 0, + k h = 0. (5.134)
t 2
∂u
5.4.3 Application: hollow-pipe waveguides
A classic application of the TE–TM decomposition is to the calculation of waveguide
fields. Consider a hollowpipe with PEC walls, aligned along the z-axis. The inside is filled
with a homogeneous, isotropic material of permeability ˜µ(ω) and complex permittivity
c
˜ (ω), and the guide cross-sectional shape is assumed to be independent of z. We assume
that a current source exists somewhere within the waveguide, creating waves that either
propagate or evanesce away from the source. If the source is confined to the region
−d < z < d then each of the regions z > d and z < −d is source-free and we may
decompose the fields there into TE and TM sets. Such a waveguide is a good candidate
for TE–TM analysis because the TE and TM fields independently satisfy the boundary
conditions at the waveguide walls. This is not generally the case for certain other guided-
wave structures such as fiber optic cables and microstrip lines.
˜
˜
We may represent the fields either in terms of the longitudinal fields E z and H z ,or
in terms of the Hertzian potentials. We choose the Hertzian potentials. For TM fields
˜
˜
˜
˜
˜
˜
we choose Π e = ˆ z e , Π h = 0; for TE fields we choose Π h = ˆ z h , Π e = 0. Both of the
potentials must obey the same Helmholtz equation:
˜
2 2
∇ + k z = 0, (5.135)
˜
˜
˜
where z represents either e or h . We seek a solution to this equation using the
separation of variables technique, and assume the product solution
˜
˜
˜
z (r,ω) = Z(z,ω)ψ(ρ,ω),
where ρ is the transverse position vector (r = ˆ zz + ρ). Substituting the trial solution
into (5.135) and writing
∂ 2
2 2
∇ =∇ +
t
∂z 2
we find that
1 2 2 1 ∂ 2
˜
∇ ψ(ρ,ω) + k =− 2 Z(z,ω).
t
˜
ψ(ρ,ω) Z(z,ω) ∂z
Because the left-hand side of this expression has positional dependence only on ρ while
the right-hand side has dependence only on z, we must have both sides equal to a constant,
2
say k . Then
z
2
∂ Z 2
+ k Z = 0,
z
∂z 2
which is an ordinary differential equation with the solutions
Z = e ∓ jk z z .
© 2001 by CRC Press LLC