Digital Signal Processing Reference
In-Depth Information
Fig. 5.4
Obtained beam
pattern for a linear array using
adaptive filtering and the
optimal solution (
5.80
)
0
RLS 100 iterations
−10
−20
−30
RLS 1000 iterations
−40
Optimum
beamformer
−50
RLS 100 iterations
RLS 1000 iterations
Optimum beamformer
−60
−70
−80
−60
−40
−20
0
20
40
60
80
Incidence angle
10 antennas
capable of sensing electromagnetic waves. The separation between the elements
of the array is 10 cm, and the carrier frequency
f
Let us consider the case in which we have a linear array of
L
=
2
=
is 1.5 GHz. The wave
π
propagation speed is the speed of the light
16
(
c
10
8
m/s). The signal received
=
3
×
by the array is:
x
(
n
)
=
y
(
n
)
s
(θ
0
)
+
y
(
n
)
s
(θ
1
)
+
y
(
n
)
s
(θ
2
)
+
v
(
n
),
(5.79)
θ
θ
θ
0
1
2
2
where
y
(
n
)
,
i
=
0
,
1
,
2 are random signals with variances
σ
θ
i
,
i
=
0
,
1
,
2 respec-
θ
i
tively. The noises
v
k
(
n
)
,
k
=
0
,...,
L
−
1 are i.i.d. and independent from
y
(
n
)
θ
i
i
=
0
,
1
,
2. It will be assumed that the signal coming from
θ
0
is the only one in which
we are interested and that
θ
0
is known. The other signals coming from
θ
1
and
θ
2
are
not wanted. We will assume that the directions
θ
2
are not known. This could
model a situation in which a base station is trying to listen to a user at position
θ
1
and
θ
0
,
while other two users located at
θ
2
are transmitting too, but the base station
is not aware of their positions. In this manner, we cannot use them as restrictions to
set
w
H
s
θ
1
and
w
H
s
0. Problem (
5.73
) can be rewritten as
17
:
(θ
1
)
=
(θ
2
)
=
L
w
H
R
x
w
subject to
s
H
w
opt
=
arg min
w
(θ
0
)
w
=
1
.
(5.80)
∈
C
An adaptive solution for
w
opt
can be expressed, using (
5.75
)
16
The frequency and the velocity of any narrowband wave cannot be arbitrary and are linked by
λ
is the
wavelength
which has units of length. It is common in beamforming to
express the separation between elements in numbers of wavelength. For the example in this section
λ
=
20 cm, so the separation between the array elements is
λ/
2.
17
Although the restrictions
w
H
s
(θ
1
)
=
w
H
s
(θ
2
)
=
0 cannot be imposed, it is clear that the
influence of the signals coming from
θ
1
and
θ
2
will appear on
R
x
.
f
=
c
,where
λ
