Page 394 - Electromagnetics

˜
Similar steps can be used to find the TE representation. Substitution of Π e = 0 and
˜
˜
Π h = ˆ u h into (5.65)–(5.66) gives the fields
˜
˜
E = jω ˜µˆ u ×∇ t h , (5.124)
˜
∂ 2
˜ ∂ h 2 ˜
H =∇ t + ˆ u 2 + k h , (5.125)
∂u ∂u
˜
while h must satisfy
2
˜
2
(∇ + k ) h = 0. (5.126)
Hertzian potential representation of TEM fields. An interesting situation occurs
when a field is both TE and TM to a particular direction. Such a field is said to be
˜ ˜
transverse electromagnetic (or TEM ) to that direction. Unfortunately, with E u = H u =
0 we cannot use (5.115) or (5.116) to find the transverse field components. It turns out
that a single scalar potential function is sufficient to represent the field, and we may use
˜
˜
either e or h .
For the TM case, equations (5.122) and (5.123) showthat we can represent the electro-
˜
magnetic fields completely with e . Unfortunately (5.122) has a longitudinal component,
˜
and thus cannot describe a TEM field. But if we require that e obey the additional
equation
∂ 2

2
˜
+ k e = 0, (5.127)
∂u 2
˜
then both E and H are transverse to u and thus describe a TEM field. Since e must
also obey
˜
2 2
∇ + k e = 0,
using (B.7) we can write (5.127) as
˜
2
∇ e = 0.
t
Similarly, for the TE case we found that the EM fields were completely described in
˜
˜
(5.124) and (5.125) by h . In this case H has a longitudinal component. Thus, if we
require
∂ 2

2
˜
+ k h = 0, (5.128)
∂u 2
˜
˜
then both E and H are purely transverse to u and again describe a TEM field. Equation
(5.128) is equivalent to
2
˜
∇ h = 0.
t
˜
˜
We can therefore describe a TEM field using either e or h , since a TEM field is
˜
both TE and TM to the longitudinal direction. If we choose e we can use (5.122) and
(5.123) to obtain the expressions
˜
˜ ∂ e
E =∇ t , (5.129)
∂u
˜
˜
c
H =− jω˜ ˆ u ×∇ t e , (5.130)
˜
where e must obey

2
∂
2 2
˜
˜
∇ e = 0, + k e = 0. (5.131)
t 2
∂u
© 2001 by CRC Press LLC