Geoscience Reference
In-Depth Information
Atmospheric Potentials and Admittances
Let us denote the solution of (7.19) by
ζ
1
(
z
)and
∂ζ
1
(
z
)
/∂z
by
ik
0
ε
a
(
z
)
ζ
2
(
z
).
They obey the ground conditions:
z
=
−h
1
ik
0
ε
a
(
∂ζ
1
(
z
)
∂z
Y
(
e
)
g
ζ
1
(
−
h
)=
−
,
ζ
2
(
−
h
)=
=1
.
(7.31)
−
h
)
Denote also the solution of (7.20) by
ζ
3
(
z
)and
∂ζ
3
(
z
)
/∂z
by
ik
0
ζ
4
(
z
) defined
by the boundary conditions
z
=
−h
1
ik
0
∂ζ
3
(
z
)
∂z
Y
(
m
)
g
ζ
3
(
−
h
)=1
,
ζ
4
(
z
)=
=
−
.
(7.32)
Here
Y
(
e
)
g
=
Y
(
e
)
(
(
m
)
g
=
Y
(
m
)
(
h
)
.
Express solutions of (7.19), (7.20) and their derivatives in terms of
ζ
i
(
z
)
−
h
)
,
−
Φ
0
ζ
1
(
z
)
Y
(
e
)
∂Φ
(
z
)
∂z
ik
0
ε
a
(
z
)
Y
(
e
)
Φ
(
z
)=
−
,
=
−
Φ
0
ζ
2
(
z
)
,
(7.33)
g
g
∂Ψ
(
z
)
∂z
Ψ
(
z
)=
Ψ
0
ζ
3
(
z
)
,
=
ik
0
Ψ
0
ζ
4
(
z
)
.
(7.34)
These formulae are general and are also applicable to an inhomogeneous at-
mosphere. Substituting (7.27) and (7.34) into (7.25) and (7.26), we obtain
ζ
1
(
z
)
ζ
2
(
z
)
,
ζ
4
(
z
)
ζ
3
(
z
)
.
Y
(
e
)
(
z
)=
Y
(
m
)
(
z
)=
−
−
(7.35)
Then components of the admittance matrix under the ionosphere are given
by
ζ
1
ζ
2
−
,
Y
11
=
k
x
k
y
k
τ
ζ
4
ζ
3
Y
22
=
−
(7.36)
k
x
,
1
k
τ
ζ
4
ζ
3
ζ
1
ζ
2
+
k
y
Y
12
=
−
(7.37)
k
y
.
1
k
τ
ζ
4
ζ
1
ζ
2
ζ
3
+
k
x
Y
21
=
(7.38)
Let
0)
.
(7.39)
Substituting (7.33) and (7.34) into (7.15) and (7.16), we obtain the horizontal
components
b
(
e
)
,
(
m
)
ζ
i
=
ζ
i
(
−
=(
b
(
e
)
,
(
m
)
,b
(
e
)
,
(
m
)
) in terms of
ζ
i
:
τ
x
y
k
y
−
,
k
0
Φ
0
ζ
1
(
z
)
Y
(
e
)
b
(
e
)
τ
(
z
)=
−
k
x
g
k
0
Ψ
0
ζ
4
(
z
)
k
x
.
b
(
m
)
τ
(
z
)=
−
(7.40)
k
y
Search WWH ::
Custom Search