Information Technology Reference
In-Depth Information
( W and has only a minimum,
we can use the steepest gradient method to repeatedly search least mean-square error
)]
which is the quadratic equation of weight coefficient
)
eE corresponding optimal weights. At this time, we can obtain the iterative
equation of weights:
2
[
(
n
min
. (6)
Wn Wn en Xn
(
+=
)
( )
−⋅
α
( )
( )
where α is a parameter of controlling stability and convergence rate, called step fac-
tor. Equation (6) is called LMS algorithm. The smaller α value makes convergence
rate the slower, and the bigger α value leads to divergence.
Definitions of input signal, output signal and reference signal are determined by
specific application requirements when we use filter based on LMS algorithm as shown
in Fig.1. Adaptive notch filter can carry out notch to input signal by adopting the above
adaptive algorithm and filter single frequency signal from input signal. We will regard
input signal ( )
xn that is a single-frequency signal as local reference signal for sin-
gle-frequency adaptive notch filter in Fig.1, and regard desired signal as input signal.
Task of adaptive notch filter is to filter single frequency signal from input signal. Filter
weights can be reduced to two because reference signal has only single-frequency
component. So one weight is in-phase component, and another i s orthogonal compo-
nent. We give the principle diagram of adaptive filter as shown in Fig.2.
noise
( n
)
+
s
( n
)
+
e
( n
)
d
( n
)
+
w
1 n
)
+
y
( n
)
x
( n
)
+
w
2 n
)
90
D
Fig. 2. The principle diagram of adaptive notch filter
3 The Simulation of Adaptive Notch Filter in Matlab and Simulink
In the simulation experiment, assume that input signal is composed of speech signal
and gaussian noise, we take desired signal as input signal according to Fig.1,
e.g.
d
(
n
)
=
s
(
n
)
+
noise
(
n
)
, where
sn A
()
=⋅
sin(2
f πθ
+
)
, A is signal amplitude,
s
0
is frequency of single frequency
θ
is initial phase, and
f
signal;
0
s
, B is noise signal amplitude. Let ( )
noise n
()
=
B
sin(2
π
f t
)
sn frequency
s f z
=
2
,
n
, and
. Assume that reference
A =
10
,
θ
=
0
,
noise n frequency
( )
n f z
=
50
B =
3
0
 
Search WWH ::




Custom Search