Digital Signal Processing Reference
In-Depth Information
which is the potential at the ground plane and at a point infinitely far away. This
means that we must consider two different solutions: region 1, which exists from
y
=
0to
h
(in the dielectric), and region 2, which exists from
y
=
h
to infinity
(in the air), where the potentials must be equal at the boundary of these regions.
Looking at
X
(
x
) in (3-55), the boundary condition in (3-56) is met if
C
1
n
=
C
2
n
and
β
is chosen so that the cosine term equals zero when
x
=±
d/
2. When
C
1
n
=
C
2
n
,
X
(
x
) is reduced to
X(x)
=
C
1
n
cos
βx
and when
β
=
nπ/d
for odd
n
and
x
=±
d/
2,
X(x)
=
0
yielding an appropriate form for
X
(
x
):
X(x)
=
C
1
n
cos
nπ
d
x
(3-58)
The second term
Y
(
y
) in (3-55) will satisfy the boundary condition in (3-57) at
y
=
0if
C
1
n
=−
C
2
n
, yielding
Y(y)
=
C
2
n
sinh
βy
=
C
2
n
sinh
nπ
d
y
(3-59)
, the boundary condition (3-57) is satisfied when
C
1
n
=
At
y
=∞
0:
Y(y)
=
C
2
n
(
cosh
βy
−
sinh
βy)
=
C
2
n
e
−
βy
(3-60)
This allows us to write equations for the potential that satisfies the boundary
conditions by multiplying
X
(
x
) and
Y
(
y
) for the appropriate boundary conditions,
where the product of the constants have been renamed
A
n
and
B
n
for clarity.
A
n
cos
nπ
d
x
sinh
nπ
d
y
∞
when 0
≤
y<h
(
region 1
)
n
=
1
odd
(3-61a)
(x,y)
=
B
n
cos
nπ
d
x
e
−
(nπ/d)y
∞
when
h
≤
y<
∞
(
region 2
)
n
=
1
odd
(3-61b)
The potential at the signal conductor (
y
=
h
) must be continuous, so
A
n
cos
nπ
d
x
sinh
nπ
d
h
=
B
n
cos
nπ
d
x
e
−
(nπ/d)h
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