126
5 An Electromagnetic Model for Anisotropic Media: Green’s Dyad for Plane-. . .
v 1 =
⎡
⎢
⎢
⎣
−jωμ 0 k y
jωμ 0 k x
λ 1 k x
λ 1 k y
⎤
⎥
⎥
⎦ v 2 =
⎡
⎢
⎢
⎣
−jωμ 0 k y
jωμ 0 k x
−λ 1 k x
−λ 1 k y
⎤
⎥
⎥
⎦ v 3 =
⎡
⎢
⎢
⎣
λ 3 k x
λ 3 k y
jωω t k y
−jωω t k x
⎤
⎥
⎥
⎦ v 4 =
⎡
⎢
⎢
⎣
λ 3 k x
λ 3 k y
−jωω t k y
jωω t k x
⎤
⎥
⎥
⎦ .
(5.13)
When k x = k y = 0, the following are linearly-independent eigenvectors:
v 1 =
⎡
⎢
⎢
⎢
⎢
⎣
1
0
0
−
t
μ 0
⎤
⎥
⎥
⎥
⎥
⎦
v 2 =
⎡
⎢
⎢
⎢
⎢
⎣
1
0
0
t
μ 0
⎤
⎥
⎥
⎥
⎥
⎦
v 3 =
⎡
⎢
⎢
⎢
⎢
⎣
0
μ 0
t
1
0
⎤
⎥
⎥
⎥
⎥
⎦
v 4 =
⎡
⎢
⎢
⎢
⎢
⎣
0
−
μ 0
t
1
0
⎤
⎥
⎥
⎥
⎥
⎦
.
(5.14)
When we substitute v 1 , v 2 of (5.13) into (5.4), with the source currents set to
zero, we find that
E z = 0; hence, v 1 , v 2 are transverse electric (TE) modes, with
respect to z. Similarly, v 3 , v 4 are transverse magnetic (TM) modes. Note that the TE
modes are orthogonal to the TM modes. This will facilitate the computation of the
Green’s dyadic. v 1 and v 3 are downward-traveling waves in the z-direction; i.e., they
represent waves that travel in the negative z-direction. v 2 , v 4 are upward-traveling
waves (in the positive z-direction). We see from (5.4) and (5.5) that all modes are
TEM (transverse electric and magnetic) with respect to z for k x = k y = 0, which
is the condition for infinite plane-waves traveling in the z−direction. Furthermore,
under this condition λ 1 = λ 3 , which means that the anisotropy does not manifest
itself. The effects of anisotropy are most pronounced on the TE and TM modes
when the transverse wave-number, k t >> Ω t and Ω z .
We can justify the interpretation of upward- and downward-traveling waves for
the various eigenvectors by returning to the fundamental differential equation (5.3),
which has the solution when J = 0:
e(z) = exp(zS) · e 0
= I · e 0 + zS · e 0 +
z 2 S 2
2
· e 0 + · · · ,
(5.15)
where e 0 is the solution at z = 0, and the second line defines the matrix exponential
operator, exp(zS).
Let e 0 = v 1 ; then
e(z) = e
zλ 1 v 1 ,
(5.16)
which follows upon substituting v 1 into the second line of (5.15), and then making
use of the fact that v 1 is an eigenvector of S with eigenvalue λ 1 . This result is in the
form of a wave propagating in the negative z-direction, and justifies our calling
v 1 a downward-traveling TE wave. A similar analysis holds for the other three
eigenvectors.
5 An Electromagnetic Model for Anisotropic Media: Green’s Dyad for Plane-. . .
v 1 =
⎡
⎢
⎢
⎣
−jωμ 0 k y
jωμ 0 k x
λ 1 k x
λ 1 k y
⎤
⎥
⎥
⎦ v 2 =
⎡
⎢
⎢
⎣
−jωμ 0 k y
jωμ 0 k x
−λ 1 k x
−λ 1 k y
⎤
⎥
⎥
⎦ v 3 =
⎡
⎢
⎢
⎣
λ 3 k x
λ 3 k y
jωω t k y
−jωω t k x
⎤
⎥
⎥
⎦ v 4 =
⎡
⎢
⎢
⎣
λ 3 k x
λ 3 k y
−jωω t k y
jωω t k x
⎤
⎥
⎥
⎦ .
(5.13)
When k x = k y = 0, the following are linearly-independent eigenvectors:
v 1 =
⎡
⎢
⎢
⎢
⎢
⎣
1
0
0
−
t
μ 0
⎤
⎥
⎥
⎥
⎥
⎦
v 2 =
⎡
⎢
⎢
⎢
⎢
⎣
1
0
0
t
μ 0
⎤
⎥
⎥
⎥
⎥
⎦
v 3 =
⎡
⎢
⎢
⎢
⎢
⎣
0
μ 0
t
1
0
⎤
⎥
⎥
⎥
⎥
⎦
v 4 =
⎡
⎢
⎢
⎢
⎢
⎣
0
−
μ 0
t
1
0
⎤
⎥
⎥
⎥
⎥
⎦
.
(5.14)
When we substitute v 1 , v 2 of (5.13) into (5.4), with the source currents set to
zero, we find that
E z = 0; hence, v 1 , v 2 are transverse electric (TE) modes, with
respect to z. Similarly, v 3 , v 4 are transverse magnetic (TM) modes. Note that the TE
modes are orthogonal to the TM modes. This will facilitate the computation of the
Green’s dyadic. v 1 and v 3 are downward-traveling waves in the z-direction; i.e., they
represent waves that travel in the negative z-direction. v 2 , v 4 are upward-traveling
waves (in the positive z-direction). We see from (5.4) and (5.5) that all modes are
TEM (transverse electric and magnetic) with respect to z for k x = k y = 0, which
is the condition for infinite plane-waves traveling in the z−direction. Furthermore,
under this condition λ 1 = λ 3 , which means that the anisotropy does not manifest
itself. The effects of anisotropy are most pronounced on the TE and TM modes
when the transverse wave-number, k t >> Ω t and Ω z .
We can justify the interpretation of upward- and downward-traveling waves for
the various eigenvectors by returning to the fundamental differential equation (5.3),
which has the solution when J = 0:
e(z) = exp(zS) · e 0
= I · e 0 + zS · e 0 +
z 2 S 2
2
· e 0 + · · · ,
(5.15)
where e 0 is the solution at z = 0, and the second line defines the matrix exponential
operator, exp(zS).
Let e 0 = v 1 ; then
e(z) = e
zλ 1 v 1 ,
(5.16)
which follows upon substituting v 1 into the second line of (5.15), and then making
use of the fact that v 1 is an eigenvector of S with eigenvalue λ 1 . This result is in the
form of a wave propagating in the negative z-direction, and justifies our calling
v 1 a downward-traveling TE wave. A similar analysis holds for the other three
eigenvectors.