Digital Signal Processing Reference
In-Depth Information
given in Chapter 2, where
1
−v
1
v
1
v
2
01
−
(
v
1
+
v
2
)
00
U
3
=
(4.3)
1
and
1
0
0
1
1
0
L
3
=
,
(4.4)
v
1
− v
2
v
2
− v
1
1
(
v
1
− v
2
)(
v
1
− v
3
)
1
(
v
2
− v
1
)(
v
2
− v
3
)
1
(
v
3
− v
1
)(
v
3
− v
2
)
we find that the inverse of
V
3
is
v
2
v
3
v
1
v
3
v
1
v
2
−
(
v
2
− v
1
)(
v
3
− v
1
)
(
v
2
− v
1
)(
v
3
− v
2
)
(
v
3
− v
1
)(
v
3
− v
2
)
v
2
+
v
3
v
1
+
v
3
v
1
+
v
2
−
−
−1
3
V
=
.
(
v
2
− v
1
)(
v
3
− v
1
)
(
v
2
− v
1
)(
v
3
− v
2
)
(
v
3
− v
1
)(
v
3
− v
2
)
1
(
v
2
− v
1
)(
v
3
− v
1
)
1
(
v
2
− v
1
)(
v
3
− v
2
)
1
(
v
3
− v
1
)(
v
3
− v
2
)
−
(4.5)
This inverse will be of great help in the rest of this study to find the filter
h
(
ω
) for any desired second-order DMA.
4.2 Second-Order Dipole
The dipole has a one at 0
◦
, a null at 90
◦
, and a one at 180
◦
. Hence, its
corresponding coe„cients to solve (4.1) are
α
2,1
= 0,
α
2,2
=
−
1,
β
2,1
= 0,
and
β
2,1
= 1. The solution to (4.1) is then
1
−
1
− e
ωτ
0
e
ωτ
0
1
h
(
ω
)=
(1
− e
ωτ
0
)
1
− e
2
ωτ
0
1
−
1
− e
1
−
ωτ
0
+
.
(4.6)
1
− e
−
ωτ
0
−
2
ωτ
0
−
ωτ
0
1
− e
e
Using the approximation
e
x
≈
1+
x
in (4.6), we get
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