Page 289 - Electromagnetics

and thus we can write (4.265) as

dk
S av · = w em .
dω
ω= ˇω
Since for a uniform plane wave in an isotropic medium k and S av are in the same direction,
we have

ˆ
S av = k dω w em
dβ ω= ˇω
and the velocityof energytransport for a plane wave of frequency ˇω is then

v e = k ˆ dω .
dβ
ω= ˇω
Thus, for a uniform plane wave in a lossless medium the velocityof energytransport is
identical to the group velocity.

Nonuniform plane waves. A nonuniform plane wave has the same form (4.216) as a
uniform plane wave, but the vectors k and k described in (4.217) are not aligned. Thus

ˇ

e
E(r) = E 0 e − jk ·r k ·r .
In the time domain this becomes
ˇ

ˆ
E(r) = E 0 e k ·r cos[ ˇωt − k (k · r)]
ˆ

where k = k k . The surfaces of constant phase are planes perpendicular to k and
propagating in the direction of k . The phase velocityis now
ˆ
v p = ˇω/k
and the wavelength is

λ = 2π/k .
In contrast, surfaces of constant amplitude must obey

k · r = C
and thus are planes perpendicular to k .

In a nonuniform plane wave the TEM nature of the fields is lost. This is easilyseen
ˇ
bycalculating H from (4.219):
ˇ
ˇ
ˇ
k × E(r) k × E(r) k × E(r)

ˇ
H(r) = = + j .
ˇ ωµ ˇ ωµ ˇ ωµ
ˇ
Thus, H is no longer perpendicular to the direction of propagation of the phase front. The
power carried bythe wave also differs from that of the uniform case. The time-average
Poynting vector
∗

ˇ
1 k × E
ˇ
S av = Re E ×
2 ˇ ωµ
can be expanded using the identity(B.7):
1 1
ˇ
ˇ
∗
∗
ˇ ∗
ˇ ∗
S av = Re k × (E × E ) + E × (k × E) . (4.266)
2 ˇ ωµ ∗
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