Adaptive signal processing simon haykins

EC 313: Adaptive Signal Processing Mayank Awasthi(10531005)
MATLAB Assignment – 1 M.Tech , Communication Systems
IIT Roorkee

Chapter -5, Ques-20
function []= ques20()
%Given
a=.99;
uu=.05;

%Initializing v,u matrices to zero
v=zeros(100,101);
u=zeros(100,101);
%White noise generation with mean =0, variance= 0.02
for i= 1:100
v(i,:)= sqrt(0.02)*randn(1,101);
end

%Generating AR process with variance variance=1 considering 100 independent
%realizations
for i=1:100
u(i,:)=filter(1 ,[1 a] ,v(i,:));
end

for n=1:100
w(1)=0;
for i=1:100
f(n,i)=u(n,i)-w(i)*u(n,i+1);
w(i+1)=w(i)+0.05*u(n,i+1)*f(n,i);
end
end

for i=1:100
for j=1:100
f(i,j)=f(i,j)^2;
end
end

%Ensemble averaging over 100 independent realization of the squared value
%of its output.
for n=1:100
J(n)=mean(f(:,n));
end
%Plotting Learning Curve
plot(J)
end

Mayank Awasthi(10531005), M.Tech , Communication Systems, IIT Roorkee

Output:

Ensemble averaging over 100 independent realization of the squared value
of its output vs time 1≤n≤100


Chapter -5, Ques-21
%Given
a1= 0.1;
a2= -0.8;
uu=0.05;
varv= 1- (a1*a1)/(1+a2)- (a2*a2) + (a1*a1*a2)/(1+a2)

v=zeros(1,102);
u=zeros(1,102);
%White noise generation with mean =0, variance= vvarv
v= sqrt(varv)*randn(1,102);
%Generating AR process with variance variance=1
u=filter(1 ,[1 a1,a2] ,v);

%Applying the LMS algorithm to calculate a1 and a2

w1(1)=0;
w2(1)=0;
for i=1:100
f(i)=u(i)-w1(i)*u(i+1)-w2(i)*u(i+2);
w1(i+1)=w1(i)+0.05*u(i+1)*f(i);
w2(i+1)=w2(i)+0.05*u(i+2)*f(i);
end
a1_lms= -w1(100)
a2_lms= -w2(100)

%calculating eigen values
eig1= 1- a1/(1+a2);
eig2= 1+ a1/(1+a2);
%maximum step size
uumax= 2/eig2;
%calculating correlation matx and cross correlation matrix
R=[1, -a1/(1+a2); -a1/(1+a2), 1];
r1= -a1/(1+a2);
r2= (a1*a1)/(1+a2)- a2;
p= [r1;r2];
%Calculating optimum weights and minimum mean square error
wo= inv(R)*p;
Jmin= 1-transpose(p)*wo;

%Estimating expected value of weights and variance using small step size theory with
w(0)=0 so we have
v1(1)= -1/sqrt(2)*(a1+a2);
v2(1)= -1/sqrt(2)*(a1-a2);
for n=1:100
val=1;


for j=1:n
val= (1-uu*eig1)*val;
end
meanv1(n)=v1(1)*val;
meansqv1(n)= uu*Jmin/(2-uu*eig1)+ val*val*(v1(1)*v1(1)- uu*Jmin/(2-
uu*eig1));
end

for n=1:100
val=1;
for j=1:n
val= (1-uu*eig2)*val;
end
meanv2(n)=v2(1)*val;
meansqv2(n)= uu*Jmin/(2-uu*eig2)+ val*val*(v2(1)*v2(1)- uu*Jmin/(2-
uu*eig2));
end

for n=1:100
w1(n)= -a1-(meanv1(n)+meanv2(n))/sqrt(2);
w2(n)= -a2-(meanv1(n)-meanv2(n))/sqrt(2);
theoritical_J(n)= Jmin + eig1*meansqv1(n)*meansqv1(n) +
eig2*meansqv2(100)*meansqv2(n);
end

expectedvalue_weights=[w1(100);w2(100)];
a1_ss= -w1(100)
a2_ss= -w2(100)

%Generating AR process with variance variance=1 considering 100 independent
%realizations
v=zeros(100,102);
u=zeros(100,102);
%White noise generation with mean =0, variance= 0.02
for i= 1:100
v(i,:)= sqrt(0.02)*randn(1,102);
u(i,:)=filter(1 ,[1 a1,a2] ,v(i,:));
end

for n=1:100
w1(1)=0;
w2(1)=0;
for i=1:100
f(n,i)=u(n,i)-w1(i)*u(n,i+1)-w2(i)*u(n,i+2);
w1(i+1)=w1(i)+0.05*u(n,i+1)*f(n,i);
w2(i+1)=w2(i)+0.05*u(n,i+2)*f(n,i);
end
end


for i=1:100
for j=1:100
f(i,j)=f(i,j)^2;
end
end
%of its output.
for n=1:100
J(n)=mean(f(:,n));
end
%Plotting Learning Curve
subplot(2,1,1);
plot(J)

%ploting learning curve theoritically
subplot(2,1,2);
plot( theoritical_J);

end

Output

>> ques21 ()

varv = 0.2700

a1_lms = 0.980

a2_lms = -0.7591

a1_ss = 0.1276

a2_ss = -0.7720


Comparision of Theoritical learning curve and measured result of part d.


Chapter-5 Ques 21- Part c

%Given
a1= 0.1;
a2= -0.8;

varv= 1- (a1*a1)/(1+a2)- (a2*a2) + (a1*a1*a2)/(1+a2);
v=zeros(1,102);
u=zeros(1,102);
%White noise generation with mean =0, variance= vvarv
v= sqrt(varv)*randn(1,102);
%Generating AR process with variance variance=1
u=filter(1 ,[1 a1,a2] ,v);

w1(1)=0;
w2(1)=0;
for i=1:100
f(i)=u(i)-w1(i)*u(i+1)-w2(i)*u(i+2);
w1(i+1)=w1(i)+0.05*u(i+1)*f(i);
w2(i+1)=w2(i)+0.05*u(i+2)*f(i);
e1(i)= -a1-w1(i);
e2(i)= -a2-w2(i);
end

Fs = 32e3;

Pxx = periodogram(f);
Hpsd = dspdata.psd(Pxx,'Fs',Fs); % Create a PSD data object.
plot(Hpsd); % Plot the PSD.

Pxx = periodogram(e1);

Pxx = periodogram(e2);

End


PSD Plot of f (n)


PSD Plot of e1 (n)


PSD Plot of e2 (n)


Chapter -5, Ques-22
function []= ques22(h1,h2,h3)

iter=500; %no pf iterations
M=21; %no of tap inputs
P=100; %no of independent experiments
uu=0.001; %learning parameter
h=[h1 h2 h3]; %channel impulse response
D1=1; %channel impulse response is symmetric about n=1
D2=(M-1)/2;
D=D1+D2; %total delay

for j=1:P
%Generating Bernoulli sequence
x=sign(rand(1,530)-0.5);
%channel output
u1 =conv(x,h);
v=sqrt(0.01)*randn(1,length(u1));
%adding AWGN noise with channel output
u=u1+v;

%Applying LMS Algoritm
w=zeros(1,M); %initializing weight vector
for i=1:iter
f(j,i) = x(D+i)- w*transpose(u(i:M-1+i));
w= w+ uu*u(i:M-1+i)*f(j,i);
end
end
subplot(2,2,3);
stem(w); %Plotting the weight vector
subplot(2,2,4);
stem(h); %Plotting the impuse response of channel

%of its output.
for j=1:P
f(j,:)= f(j,:).^2;
end
for n=1:iter
J(n)=mean(f(:,n));
End

%Plotting the learning curve
subplot(2,2,1:2);
plot(J);
end


Output:

(i) For h1= 0.25, h2= 1, h3= 0.25

a. Learning curve of the equalizer by averaging the square value of error signal
over ensemble of 100 independent trials of experiment
b. Impulse response of the optimum transversal equalizer
c. Impulse response of channel.


(ii) For h1= 0.25, h2= 1, h3=- 0.25



(iii) For h1=- 0.25, h2= 1, h3= 0.25



Adaptive signal processing simon haykins

Recommended

Recommended

More Related Content

What's hot

What's hot (19)

Similar to Adaptive signal processing simon haykins

Similar to Adaptive signal processing simon haykins (20)

More from Department of Telecommunications, Ministry of Communication & IT (INDIA)

More from Department of Telecommunications, Ministry of Communication & IT (INDIA) (9)

Adaptive signal processing simon haykins