Description
Implement an adaptive interference cancellation system with the least mean squares (LMS) algorithm described in Chapter 4, the normalized LMS (NLMS) algorithm described in Chapter 5, or the variable-step LMS (VS-LMS) algorithm described in Chapter 5. The adaptive system is used to cancel interference, i(n) , contained in a primary signal d (n) , as depicted in Fig. 1. The primary signal serves as the desired signal for the adaptive system. The input signal is a filtered version of i(n) , i.e., x(n) = i(n) h(n) , where denotes the convolution operator.
primary signal d (n) = s(n) + i(n)
reference
input +
x(n) = i(n)h(n) –
e(n) ◼ Figure 1 Adaptive interference canceling.
(1) For the LMS algorithm, the update equation is
fn+1 =fn +e(n)xn
where
is the step size,
e(n) = d(n) − y(n) , and y(n) = fT x with nn
f=[f(0) f(1) f(L−1)]T nnnn
x =[x (0) x (1) x (L−1)]T nnnn
and L being the adaptive filter length.
- (2) For the NLMS algorithm, the update equation is
f =f + e(n)xn n+1 n c+xTnxn
where c is a small positive constant.
- (3) For the VS-LMS algorithm, the update equation is
fn+1 =fn +e(n)Mnxn where Mn is an LL diagonal matrix with
Mn = diag{0(n),1(n),…,L−1(n)}. 1
Adaptive Processor
COM 5220 Adaptive Signal Processing Fall 2021
The i (n) ’s are adjusted according to the following rule:
For n=0, i(n)=max,i=0,1,…,L−1.
For n0, i(n)=i(n)c1 (with c1 1) if e(n)x(n−i) has N1 successive
sign changes and i(n)=i(n)c2 (with c2 1) if e(n)x(n−i) has no sign changes for N2 successive updates, where min i (n) max .
System Specifications:
⚫Assume f0 =0.
⚫Theinformationsignals(n)=1withP{s(n)=+1}=0.5and P{s(n)=−1}=0.5.
⚫ i(n) is generated from a uniformly distributed random variable with range ⚫ h(n) is a five-point impulse response, i.e., h(0) = 0.227 , h(1) = 0.46 ,
h(3) = 0.46 , and h(4) = 0.227 .
[−1, 1] .
h(2) = 0.688 ,
⚫ The adaptive filter length is Parameter Settings:
L = 6 .
⚫ For the LMS algorithm,
⚫ For the NLMS algorithm,
⚫For the VS-LMS algorithm, c1 =0.9, c2 =1.1, and N1 =N2 =3. Also, max
are determined by yourself.
is determined by yourself.
is determined by yourself and c = 10−3 .
Simulation Assignments:
Generate 12,000 samples for each test data, i.e., s(n) , i(n) , and x(n) , in each trial. The performance of each algorithm is examined by the bit error rate (BER) and the average squared error for different averaging intervals.
- (1) Plot the learning curve, i.e., | e(n) |2 vs. the number of iterations for each algorithm. Note
that the learning curve is obtained by averaging the results over 100 trials.
- (2) Calculate BERs 1 and 2 of the adaptive system for different adaptive algorithms, where BER 1 is evaluated from iterations 101 to 12000 and BER 2 is from iterations 1001 to 12000 in each trial. The final BER for each case is obtained by averaging the results over
100 trials.
P (i)= 1 12000{sgn(e(n))s(n)}, i=1,2,…,100
e1 12000−101 n=101 i
i
and min
1 P (i)=
{sgn(e(n))s(n)},i=1,2,…,100
P (i); P =
e2 12000 −1001 1 100
12000 n=1001
ii
P = e1
i=1
e1
e2
i=1
e2
100
100
2
1 100
P (i)
COM 5220 Adaptive Signal Processing Fall 2021 (3)Calculate the average squared error of the adaptive system for different adaptive
algorithms. That is, calculate (1/ M) [s (n)−e (n)]2 in the ith trial, where M is the
n
averaging interval. The squared errors are averaged from iterations 101 to 12000 and from iterations 1001 to 12000, respectively, in each trial. The final average squared error for each case is obtained by averaging the results over 100 trials.
ii
e(i)= 1 12000[s(n)−e(n)]2, i=1,2,…,100
1 12000−101 n=101
ii
1 12000 2
1 100 1 100
e2(i)=12000−1001
n=1001[s(n)−e(n)] , i=1,2,…,100 ii
1 i=1 i=1
from your simulation results?
(Note: Send your simulation results/answers along with the corresponding m-file(s) to the eeclass before 10:10 AM, Dec. 29, 2021.)
e = e (i); e =
2
12
100
(4) What conclusions can you make about performance comparisons of the three algorithms
3
