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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(
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 =
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