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In-Depth Information
∞
H
x
H
x
H
x
o
(
e
ik
z
e
ik
z
)
−
=
−
=
b
m
sin
mz dm
0
and
(
k
)
2
−
(
k
)
2
2
mH
x
o
π
b
m
=
.
)
2
(
Substituting (6.14) and (6.6) in (6.15) and equating the terms with the same sub-
script, we obtain
a
m
e
−
v
=
a
m
e
−
v
+
a
m
sin h
a
m
cos h
v
−
v
=
b
m
,
0
,
(6
.
16)
whence
(
k
)
2
(
k
)
2
e
w
sin h
v
2
mH
x
o
π
−
a
m
=
v
+
(
)
3
cos h
sin h
v
(6
.
17)
(
k
)
2
(
k
)
2
2
mH
x
o
π
−
1
a
m
=−
.
v
+
(
)
2
cos h
sin h
v
Then, with a glance to (6.13) and (6.14),
⎧
⎨
e
−
(
|
y
|−
v
)
sin
mz dm
cos h
v
+
+
2
H
x
o
(
k
)
2
(
k
)
2
v
−
0
∞
m
sin h
H
x
|
y
| ≥
v
(
)
3
π
sin h
v
H
x
=
(6
.
18)
⎩
−
2
H
x
o
(
k
)
2
(
k
)
2
0
y
sin
mz dm
cos h
v
+
−
∞
m
(
)
2
cos h
H
x
|
y
| ≤
v
π
sin h
v
and
⎧
⎨
e
−
(
|
y
|−
v
)
2
H
x
o
(
k
)
2
−
(
k
)
2
π
∞
0
m
2
sin h
v
(
)
3
dm
E
y
+
|
y
| ≥
v
v
+
cos h
sin h
v
E
y
z
=
0
=
⎩
2
H
x
o
(
k
)
2
−
(
k
)
2
π
∞
0
m
2
(
)
2
cos h
ydm
cos h
v
+
E
y
−
|
y
| ≤
v.
sin h
v
.
(6
19)
