Geoscience Reference
In-Depth Information
Z
(
e
)
g
(
k
τ
)
−
1
=
Y
(
e
)
g
(
k
τ
)=
Y
(
e
)
(
−
h
;
k
τ
)
,
Z
(
m
)
g
(
k
τ
)
−
1
=
Y
(
m
)
g
(
k
τ
)=
Y
(
m
)
(
−
h
;
k
τ
)
.
(7.27)
Let the ground be approximated by a half-space with constant geoelectric
permeability
ε
g
.
Then the solution for
Φ
and
Ψ
in the ground (
z<
−
h
)can
be written as
Φ
Ψ
∝
exp(
−
iκ
g
z
)
and electric and magnetic admittances become
1
Z
(
e
)
=
k
0
ε
g
κ
g
1
Z
(
m
)
=
κ
g
Y
(
e
)
g
(
m
)
g
=
,
=
k
0
,
(7.28)
g
g
where
κ
g
=
k
0
ε
g
−
k
τ
.
Assume also that
σ
g
is so large that the inequality holds
k
0
√
ε
g
k
τ
.
Then
κ
g
≈
k
0
ε
g
and (7.28) yields
1
Z
(
e
g
≈
Z
(
m
g
≈
√
ε
g
.
1
Y
(
e
)
g
Y
(
m
)
g
=
=
Spectral Admittance Matrix
In the general case, the horizontal components of total electric and magnetic
fields are connected by an admittance matrix
b
τ
(
z
)=
Y
(
z
)
E
τ
(
z
)
,
(7.29)
where
E
τ
(
z
)and
b
τ
(
z
) are horizontal magnetic and electric fields. We find an
expression for
Y
(
z
) in the atmosphere from (7.12), (7.15), (7.16) and (7.23),
(7.24):
Y
(
z
)=
Y
11
,
Y
12
(7.30)
Y
21
Y
22
where
Y
(
m
)
(
z
)
Y
(
e
)
(
z
)
,
k
x
k
y
k
τ
Y
11
=
−
Y
22
=
−
−
k
x
Y
(
m
)
(
z
)+
k
y
Y
(
e
)
(
z
)
,
1
k
τ
Y
12
=
k
y
Y
(
m
)
(
z
)+
k
x
Y
(
e
)
(
z
)
.
1
k
τ
Y
21
=
−
Search WWH ::
Custom Search