
6.6 Circular Dielectric Waveguides and Optical Fibers 361
Figure 6.18: Rotation of skew wave and the rotation of polarized field vector
in circular dielectric waveguide.
H
φ2
= j
β
2
τ
ωµ
0
C
χ+
k
2
β
2
²
r2
2
K
n+1
(τρ)−
χ−
k
2
β
2
²
r2
2
K
n−1
(τρ)
e
jnφ
e
−jβz
, (6.228)
H
z2
= −j
τ
2
βχ
ωµ
0
CK
n
(τρ)e
jnφ
e
−jβz
. (6.229)
By investigating the relations between the transverse components, E
ρ
and
E
φ
or H
ρ
and H
φ
, we find that the transverse field vector is elliptically polar-
ized, consisting of two circularly polarized components in opposite directions.
In the above expressions, n is an integer or zero. The modes with n = 0
represent axially symmetric modes or meridional waves. The modes with
n 6= 0 represents axially asymmetric modes or skew waves or circulating
waves. The modes with positive and negative n represent counterclockwise
and clockwise, i.e., left-handed and right-handed skew waves, respectively.
Note that the direction of rotation of skew waves and the direction of rota-
tion of polarized field vectors are entirely different phenomena, see Fig. 6.18.
From the eigenvalue equation (6.210) or (6.212) we can see that, the
cutoff conditions and the dispersion relations are the same for +n and −n,
τ
−n
= τ
+n
and β
−n
= β
+n
. So the clo ckwise and counterclockwise skew
waves with functions e
−jnφ
e
−jβ
−n
z
and e
jnφ
e
−jβ
n
z
can be composed into two
mutually orthogonal angular standing waves sin(nφ)e
−jβ
n
z
and cos(nφ)e
−jβ
n
z
with stationary polarization direction. This is the same as that for all waveg-
uides made by isotropic material and isotropic boundaries.
(2) Solutions to the Eigenvalue Equation
The cutoff condition of the uniform circular dielectric waveguide is τ = 0 and
T = T
c
. Applying this condition in (6.212), we have T
c
. Then the cutoff
angular frequency can be obtained from (6.213) as follows:
T
2
c
= k
2
1
− k
2
2
= ω
2
c
(µ
1
²
1
− µ
2
²
2
), ω
c
=
T
c
√
µ
1
²
1
− µ
2
²
2
. (6.230)