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where
10
01
0
Z
xx
Z
xy
Z
yx
Z
yy
1
10
−
[
I
]
=
[
R
(
−
/
2]
=
[
Z
(
x
,
y
)]
=
y
)
=
1
2
S
1
(
x
(
x
,
,
y
) [
R
(
−
π/
2][
Z
(
x
,
y
)]
and
H
int
τ
H
int
τ
H
ext
τ
H
ext
τ
,
=
,
,
,
=
,
,
,
(
x
y
)
(
x
y
0)
(
x
y
)
(
x
y
0)
H
τ
(
x
,
y
)
=
H
τ
(
x
,
y
,
0)
E
τ
(
x
,
y
)
=
E
τ
(
x
,
y
,
0)
.
Now we gain the benefit from the magnetic tensor
[
M
(
x
,
y
|
x
B
,
y
B
)] defin-
ing the relation between the horizontal magnetic fields
H
τ
(
x
,
y
) and
H
τ
(
x
B
,
y
B
)
at
two
sites:
at
a
site
O(
x
,
y
)
and
at
a
base
site
B(
x
B
,
y
B
).
Substituting
[
M
(
x
,
y
|
x
B
,
y
B
)]
H
τ
(
x
B
,
y
B
)for
H
τ
(
x
,
y
), we rewrite (11.33) in the form
=
y
)
[
M
(
x
H
int
τ
(
x
,
y
)
(
x
,
,
y
|
x
B
,
y
B
)]
H
τ
(
x
B
,
y
B
)
a
(11
.
34)
=
[
I
]
−
y
)
[
M
(
x
H
ext
τ
(
x
,
y
)
(
x
,
,
y
|
x
B
,
y
B
)]
H
τ
(
x
B
,
y
B
)
.
b
y
B
) be given at the base site (say, from the
TEM-sounding). Then we can turn to (11.33
b
) and determine the external magnetic
field
H
ext
τ
Let the sediments conductance
S
1
(
x
B
,
y
B
) at the base site. Taking into account that
H
ext
τ
(
x
B
,
is uniform in the
model excited by a plane wave, we get by virtue of (11.33
b
):
=
[
I
]
−
y
B
)
H
τ
(
x
B
H
ext
τ
(
x
B
,
,
y
B
)
.
(11
.
35)
Using (11.35), we obtain the internal magnetic field
H
int
τ
(
x
,
y
) at any of observa-
tion sites:
+
y
B
)
}
H
int
τ
H
ext
τ
|
(
x
,
y
)
=
H
τ
(
x
,
y
)
−
={
[
M
(
x
,
y
x
B
,
y
B
)]
−
[
I
]
(
x
B
,
H
τ
(
x
B
,
y
B
)
.
(11
.
36)
,
y
B
) from (11.33
b
) and (11.36), and derive a redundant
matrix equation that gives unknown
S
1
(
x
Finally we exclude
H
τ
(
x
B
,
,
,
y
|
x
B
,
y
) with known [
Z
]
[
M
(
x
y
B
)]
and
S
1
(
x
B
,
y
B
):
y
)
[
M
(
x
+
y
B
)
.
(
x
,
,
y
|
x
B
,
y
B
)]
=
[
M
(
x
,
y
|
x
B
,
y
B
)]
−
[
I
]
(
x
B
,
(11
.
37)
Solving this equation, we obtain
S
1
(
x
S
1
(
x
,
y
)
+
,
y
)
S
1
(
x
,
y
)
=
,
(11
.
38)
2
