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