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3
three-segment layer (
3
,
h
3
) with the side segments of resistivity
and the central
3
segment of resistivity
and width 2
v
. The sediments rest on the highly resistive
lithosphere (
4
>>
1
,
h
4
>>
h
1
+
h
2
+
h
3
) underlaid with highly conductive mantle
(
.
Let us begin with analytical solution of the problem for the TM-mode. Following
(Berdichevsky and Jakovlev,1990), we ignore the influence of the conductive mantle
and set
5
=
0)
. Then, using Dmitriev's thin-sheet approximation (7.16)
and taking into account that on the surface of the perfect insulator
H
x
=
4
→∞
,
h
4
→∞
0, we write
o
h
1
H
x
E
y
(
y
,
0)
=
E
y
(
y
,
h
1
)
+
i
R
2
d
2
H
x
(
y
,
h
1
)
E
y
(
y
,
h
1
)
=
E
y
(
y
,
h
12
)
+
i
o
h
2
H
x
(
y
,
h
1
)
+
dy
2
H
x
H
x
(
y
,
h
1
)
=
+
S
1
E
y
(
y
,
0)
H
x
(
y
,
h
12
)
=
H
x
(
y
,
h
1
)
+
S
2
E
y
(
y
,
h
1
)
H
x
(
y
,
h
12
)
=−
S
3
(
y
)
E
y
(
y
,
h
12
)
,
where
h
12
=
h
1
+
h
2
and
S
1
=
h
1
/
1
,
S
2
=
h
2
/
2
,
S
3
(
y
)
=
h
3
/
3
(
y
)
,
R
2
=
h
2
2
.
Eliminating
E
y
(
y
h
12
) from these equations,
we get the equation for the transverse impedance at the Earth's surface,
z
=0:
,
h
1
)
,
E
y
(
y
,
h
12
) and
H
x
(
y
,
h
1
)
,
H
x
(
y
,
S
1
R
2
d
2
S
3
(
y
)
Z
⊥
(
y
)
dy
2
o
S
1
h
2
S
3
(
y
)]
Z
⊥
(
y
)
−
[
S
(
y
)
−
i
(7
.
83)
=−{
1
−
i
o
[
h
1
S
2
+
h
12
S
3
(
y
)]
}
,
where
Z
⊥
(
y
)
=−
/
H
x
=
S
1
+
S
2
+
o
S
1
h
2
<<
1
E
y
(
y
)
and
S
(
y
)
S
3
(
y
). With
and
o
[
h
1
S
2
+
h
12
S
3
(
y
)]
<<
1 we are in the
S
-interval. Here (7.83) reduces to the
equation
S
1
R
2
d
2
S
3
(
y
)
Z
⊥
(
y
)
dy
2
S
(
y
)
Z
⊥
(
y
)
−
=−
1
,
(7
.
84)
which falls into two equations with constant coefficients:
d
2
Z
⊥
(
y
)
dy
2
S
1
R
2
S
3
S
Z
⊥
(
y
)
−
=−
1
|
y
| ≥
v
(7
.
85)
d
2
Z
⊥
(
y
)
dy
2
S
1
R
2
S
3
S
Z
⊥
(
y
)
−
=−
1
|
y
| ≤
v,
where
S
3
=
h
3
/
3
,
S
3
h
3
/
3
and
S
=
S
3
,
S
=
S
3
.
=
S
1
+
S
2
+
S
1
+
S
2
+
General solutions of equations (7.85) are
Z
N
Ae
−
g
|
y
|
+
|
y
| ≥
v
Z
⊥
(
y
)
=
(7
.
86)
Z
N
B
cos h
g
y
+
|
y
| ≤
v,
