Page 392 - Electromagnetics

Splitting the vectors into longitudinal and transverse parts, and using (5.100) and (5.102),
we equate the longitudinal components of the wave equations to obtain

1 ∂ ˜ρ i
˜ i
˜ i
2 2 ˜

∇ + k E u = + jω ˜µJ +∇ t × J , (5.113)
u
mt
c
˜ ∂u
1 ∂ ˜ρ i
˜ i
2 2 ˜ m c ˜ i

∇ + k H u = + jω˜ J mu −∇ t × J . (5.114)
t
˜ µ ∂u
˜
˜ i
˜ i
We note that if J = J = 0, then H u = 0 and the fields are TM to the u-direction; these
m
t
˜
˜
˜ i
˜ i
fields may be determined completely from E u . Similarly, if J = J mt = 0, then E u = 0
and the fields are TE to the u-direction; these fields may be determined completely from
˜
H u . These properties are used in § 4.11.7, where the fields of electric and magnetic line
sources aligned along the z-direction are assumed to be purely TM z or TE z , respectively.
5.4 TE–TM decomposition
5.4.1 TE–TM decomposition in terms of fields
A particularly useful field decomposition results if we specialize to a source-free region.
˜ i
˜ i
With J = J = 0 in (5.111)–(5.112) we obtain
m
∂ 2 ∂ H u c

2 ˜
˜
˜
+ k H t =∇ t − jω˜ ˆ u ×∇ t E u , (5.115)
∂u 2 ∂u

2 ˜
∂ 2 ∂E u
˜
˜
+ k E t =∇ t + jω ˜µˆ u ×∇ t H u . (5.116)
∂u 2 ∂u
Setting the sources to zero in (5.113) and (5.114) we get
2 2 ˜

∇ + k E u = 0,
2 2 ˜

∇ + k H u = 0.
Hence the longitudinal field components are solutions to the homogeneous Helmholtz
equation, and the transverse components are specified solely in terms of the longitudinal
˜
components. The electromagnetic field is completely specified by the two scalar fields E u
˜
and H u (and, of course, appropriate boundary values).
We can use superposition to simplify the task of solving (5.115)–(5.116). Since each
equation has two forcing terms on the right-hand side, we can solve the equations using
˜
˜
one forcing term at a time, and add the results. That is, let E 1 and H 1 be the solutions to
˜
˜
˜
˜
(5.115)–(5.116) with E u = 0, and E 2 and H 2 be the solutions with H u = 0. This results
in a decomposition
˜
˜
˜
E = E 1 + E 2 , (5.117)
˜
˜
˜
H = H 1 + H 2 , (5.118)
with
˜
˜
˜
˜
˜
E 1 = E 1t , H 1 = H 1t + H 1u ˆ u,
˜
˜
˜
˜
˜
H 2 = H 2t , E 2 = E 2t + E 2u ˆ u.
© 2001 by CRC Press LLC