Page 109 - Electromagnetics

Thus we also have E · H = 0, and find that E must be perpendicular to H.So E, H,
and ˆ z comprise a mutually orthogonal triplet of vectors. A wave having this property is
said to be TEM to the z-direction or simply TEM z . Here “TEM” stands for transverse
electromagnetic, indicating the orthogonal relationship between the field vectors and the
z-direction. Note that

ˆ p × ˆ q =±ˆ z.
The constant direction described by ˆ p is called the polarization of the plane wave.
We are now ready to solve the source-free wave equation (2.328). If we dot both sides
of the homogeneous expression by ˆ p we obtain

2 2 2
∂ E x ∂ E y ∂(ˆ p · E) ∂ (ˆ p · E)
ˆ p · ˆ x + ˆ p · ˆ y − µσ − µ
= 0.
∂z 2 ∂z 2 ∂t ∂t 2
Noting that
2 2 2 2
∂ E x ∂ E y ∂ ∂
ˆ p · ˆ x + ˆ p · ˆ y = (ˆ p · ˆ xE x + ˆ p · ˆ yE y ) = (ˆ p · E),
∂z 2 ∂z 2 ∂z 2 ∂z 2
we have the wave equation
2
2
∂ E(z, t) ∂E(z, t) ∂ E(z, t)
− µσ − µ
= 0. (2.331)
∂z 2 ∂t ∂t 2
i
Similarly, dotting both sides of (2.326) with ˆ q and setting J = 0 we obtain
2
2
∂ H(z, t) ∂ H(z, t) ∂ H(z, t)
− µσ − µ
= 0. (2.332)
∂z 2 ∂t ∂t 2
In a source-free homogeneous conducting region E and H satisfy identical wave equations.
Solutions are considered in § A.1. There we solve for the total field for all z, t given
the value of the field and its derivative over the z = 0 plane. This solution can be
directly applied to find the total field of a plane wave reflected by a perfect conductor.
Let us begin by considering the lossless case where σ = 0, and assuming the region z < 0
contains a perfect electric conductor. The conditions on the field in the z = 0 plane are
determined by the required boundary condition on a perfect conductor: the tangential
electric field must vanish. From (2.330) we see that since E ⊥ ˆ z, requiring

∂ H(z, t)
= 0 (2.333)
∂z z=0
gives E(0, t) = 0 and thus satisfies the boundary condition. Writing

∂ H(z, t)
H(0, t) = H 0 f (t), = H 0 g(t) = 0, (2.334)
∂z
z=0
and setting = 0 in (A.41) we obtain the solution to (2.332):
H 0 z H 0 z
H(z, t) = f t − + f t + , (2.335)
2 v 2 v
where v = 1/(µ
) 1/2 . Since we designate the vector direction of H as ˆ q, the vector field
is

H 0 z H 0 z
H(z, t) = ˆ q f t − + ˆ q f t + . (2.336)
2 v 2 v

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