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where
Y
xx
,
Y
yy
are the components of the admittance tensor determined
by (1.19). Substituting (1.19) into (1.91), we get
Y
xy
,
Y
yx
,
tan
2
1
+
E
Z
E
(
E
)
=
l
1
tan
2
E
−
l
2
tan
E
+
l
3
(1
.
92)
1
=
E
,
l
1
sin
2
−
l
2
sin
E
cos
+
l
3
cos
2
E
E
where
+
Z
xy
Z
yy
+
Z
yx
2
Z
xx
Z
yy
−
Z
xy
Z
yx
2
Z
xx
Z
yy
−
Z
xy
Z
yx
2
2
2Re(
Z
xx
Z
yx
+
Z
yy
Z
xy
)
Z
xx
Z
yy
−
Z
xy
Z
yx
|
Z
xx
|
l
1
=
2
,
l
2
=
l
3
=
2
.
2
E
,at
which the
E-
polarized impedance has a maximum and minimum. The condition
The scalar indicator
Z
E
can be called a
E
-
polarized impedance
.
Determine
dZ
E
d
E
=
0
gives the equation
l
2
l
1
−
tan 2
=
l
3
,
(1
.
93)
E
max
E
min
E
which has two solutions,
and
, differing by
/
2.
Letusplotavalue
Z
H
(
H
) on the polarization axis of the magnetic field. As the
angle
, the resultant point describes a closed curve that is the
polar diagram of H-polarized impedance
. The polar diagram of
Z
H
(
H
varies from 0 to 2
H
) is a regular
oval determined by (1.89). Its inversion
Y
H
(
H
)
=
1
/
Z
H
(
H
) gives an ellipse defined
by equation
k
1
Y
H
sin
2
k
2
Y
H
sin
k
3
Y
H
cos
2
+
H
cos
+
=
1
.
(1
.
94)
H
H
H
Now plot a value
Z
E
(
E
) on the polarization axis of the electric field. As the angle
E
varies from 0 to 2
, the resultant point describes a closed curve that is the
polar
diagram of the E-polarized impedance
. The polar diagram of
Z
E
(
E
) is an ellipse
determined by the equation
l
1
Z
E
sin
2
l
2
Z
E
sin
l
3
Z
E
cos
2
E
−
E
cos
E
+
E
=
1
.
(1
.
95)
Following (Counil et al., 1986), we introduce an
angular skew parameter
in
accordance with (1.90) and (1.93):
