Page 312 - Electromagnetics

The first term on the right-hand side is zero, and thus using (B.76) we have

˜
˜
− jβz
∇× ∇× H = − jβe ˆ z × (ˆ z × H 0 ) (− jβ)
or, using (B.7),
˜
2 − jβz ˜
∇× ∇× H = β e H 0
˜
since ˆ z · H 0 = 0. With this (4.320) becomes
2 ˜ 2
˜
β H 0 = ω ˜ ¯µ · H 0 . (4.323)
We can solve (4.323) for β bywriting the vector equation in component form:
2
2

β H 0x = ω µ 1 H 0x + jµ 2 H 0y ,
2
2

β H 0y = ω − jµ 2 H 0x + µ 1 H 0y .
In matrix form these are

2 2 2
β − ω µ 1 − jω µ 2 H 0x 0
2 2 2 = , (4.324)
jω µ 2 β − ω µ 1 H 0y 0
and nontrivial solutions occur onlyif
2 2 2

β − ω µ 1 − jω µ 2

2 2 2
= 0.
jω µ 2 β − ω µ 1

Expansion yields the two solutions
√
(4.325)
β ± = ω µ ±
where

ω M
µ ± = µ 1 ± µ 2 = µ 0 1 + . (4.326)
ω 0 ∓ ω
So the propagation properties of the plane wave are the same as those in a medium with
an equivalent scalar permeabilitygiven by µ ± .
Associated with each of these solutions is a relationship between H 0x and H 0y that can
be found from (4.324). Substituting β + into the first equation we have
2
2
ω µ 2 H 0x − jω µ 2 H 0y = 0
or H 0x = jH 0y . Similarly, substitution of β − produces H 0x =− jH 0y . Thus, by(4.321)
the magnetic field maybe expressed as
˜
H(r,ω) = H 0y [± j ˆ x + ˆ y]e − jβ ± z .
By(4.322) we also have the electric field

˜
E(r,ω) = Z TE M H 0y [ˆ x + e ∓ j π 2 ˆ y]e − jβ ± z .
This field has the form of (4.248). For β + we have φ y − φ x =−π/2 and thus the wave
exhibits RHCP. For β − we have φ y − φ x = π/2 and the wave exhibits LHCP.

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