Page 317 - Electromagnetics

ˇ
Assuming E z0 =|E z0 |e jξ E , the time-domain representation is found from (4.126):
|E z0 | −αρ E
E z (ρ, t) = √ e cos[ ˇωt − βρ − π/4 + ξ ]. (4.339)
8πkρ
We can identifya surface of constant phase as a locus of points obeying
E
ˇ ωt − βρ − π/4 + ξ = C P (4.340)
where C P is some constant. These surfaces are cylinders coaxial with the z-axis, and are
called cylindrical wavefronts. Note that surfaces of constant amplitude, as determined
by
e −αρ
√ = C A
ρ
where C A is some constant, are also cylinders.
The cosine term in (4.339) represents a traveling wave. As t is increased the argument
of the cosine function remains fixed as long as ρ is increased correspondingly. Hence the
cylindrical wavefronts propagate outward as time progresses. As the wavefront travels
outward, the field is attenuated because of the factor e −αρ . The velocityof propagation
of the phase fronts maybe computed bya now familiar technique. Differentiating (4.340)
with respect to t we find that
dρ
ˇ ω − β = 0,
dt
and thus have the phase velocity v p of the outward expanding phase fronts:
dρ ˇ ω
v p = = .
dt β
Calculation of wavelength also proceeds as before. Examining the two adjacent wave-
fronts that produce the same value of the cosine function in (4.339), we find βρ 1 =
βρ 2 − 2π or
λ = ρ 2 − ρ 1 = 2π/β.

Computation of the power carried bya cylindrical wave is straightforward. Since a
cylindrical wavefront is infinite in extent, we usually speak of the power per unit length
carried bythe wave. This is found byintegrating the time-average Poynting flux given
in (4.157). For electric polarization we find the time-average power flux densityusing
(4.330) and (4.331):

1 1 j 2 (2) (2)∗
ˇ
ˇ ∗ ˆ
ˇ
S av = Re{E z ˆ z × H φ}= Re ˆρ |E z0 | H 0 (kρ)H 1 (kρ) . (4.341)
φ
2 2 16Z ∗ TM
For magnetic polarization we use (4.333) and (4.334):
1 1 jZ TE 2 (2)
ˇ ˆ
ˇ
ˇ ∗
S av = Re{E φ φ × H ˆ z}= Re −ˆρ |H z0 | H 0 (2)∗ (kρ)H 1 (kρ) .
z
2 2 16
For a lossless medium these expressions can be greatlysimplified. By(E.5) we can write
(2) (2)∗
jH (kρ)H (kρ) = j[J 0 (kρ) − jN 0 (kρ)][J 1 (kρ) + jN 1 (kρ)],
0 1
hence
(2) (2)∗
jH (kρ)H (kρ) = [N 0 (kρ)J 1 (kρ) − J 0 (kρ)N 1 (kρ)] + j[J 0 (kρ)J 1 (kρ) + N 0 (kρ)N 1 (kρ)].
0 1
© 2001 by CRC Press LLC