Interference suppression in DSSS(Direct sequence spread spectrum) using kalman filter

I have written this code for interference suppression in direct sequence spread spectrum using the kalman filter.

please guide me what is the problem in my code that how can i get the desire result.

clc

clear all

close all

% Generating the bit pattern with each bit 8 samples long

b=round(rand(1,2000));

pattern=[];

for k=1:200

if b(1,k)==0

sig=zeros(1,10);

else

sig=ones(1,10);

end

pattern=[pattern sig];

pattern1=pattern';

len=length(pattern1);

end

plot(pattern1);

axis([-1 2020 -.5 1.5]);

title('\bf\it Original Bit Sequence1');

%%%Generating pseudorandom binary sequence

%%

seq = ltePRBS(1,len)';

seq(1:len);

len1=length(seq);

figure

stairs(seq);

axis([-1 2000 -0.5 1.5]);

title('Pseodu random pattern');

%Multiplaying the pseudo sequence and massage signal

bbs=zeros(1,2000);

for i=1:len

bbs(:,i)= pattern1(i).*seq(i);

end

figure

stairs(bbs,'linewidth', 2);axis([-1 2000 -.5 1.5]);

title('Pseodu random pattern*bit sequence');

% Generating noise

noise1=sqrt(0.1)*randn(1,2000);

figure

plot(noise1);

axis([-1 2020 -1.5 1.5]);

title('Noise');

%Generating the interference

phi1=1.98;

phi2=-0.9801;

samples = 2000;

e = randn(1,samples); % e(k)=white gaussian noise with variance 0.01

v = var(e); % check variance

v_target = 0.01;

amp = sqrt(v_target/v); % compute amplitude factor correction to get target variance

e = e*amp;

v = var(e) % now we have the correct variance

% init I

I(1) = e(1); % ....i(-1),i(-2)......<0 are equal to zero

I(2) = phi1*I(1)+ 0 +e(2);

% main loop

for k=3:samples

I(k) = phi1*I(k-1)+phi2*I(k-2)+e(k);

end

figure

plot(I)

title('correlated interference');

%adding noise interference and pseudo sequence to get the resultant signal

z=noise1+bbs+I;

figure

plot(z)

title('desired signal');

%%

%APPLY THE CONDITION

phi=[1.98 -0.9801

1 0];

sig_u=0.01;

Q=[sig_u 0

0 sig_u];

M(:,:,1)=2000*eye(2);% M[-1,-1]

H(:,:,1)=[1 0];

sig_w=0.1; % variance

%I_post(:,1)=[0;0]; %I[-1|-1]

I_post=zeros(2,2000);% h[n|n]

I_pre=zeros(2,2000); %h[n|n-1]

M=zeros(2,2,2000); %M[n|n]

M_pre=zeros(2,2,2000);

K = zeros(2,1,2000);

for k=1:2000

if (z(k)-I(k)+1).^2>=(z(k)-I(k)-1).^2;

bbs(k)=1;

elseif (z(k)-I(k)+1).^2<(z(k)-I(k)-1).^2;

bbs(k)=-1;

else

bbs(k)=0;

end

I_post(:,k)=z(k)-bbs(k);

%apply the kalman fiter

I_pre(:,k)=phi*I_post(:,k);

M_pre(:,:,k)=phi*M(:,:,1)*phi'+Q;

%apply thekalman gain

K(:,:,k)=((M_pre(:,:,k)*H')/((sig_w)+H*M_pre(:,:,k)*H')).^1;

%Minimam correctin MSE

I_post(:,k)=I_pre(:,k)+K(:,:,k)*(z(:,k)-H*I_pre(:,k));

%Minimam correction

M(:,:,k)=(eye(2)-K(:,:,k)*H)*M_pre(:,:,k);

end

figure

plot(bbs)

axis([-1 2000 -1.5 1.5]);

title('subtrated sequence');

SINR=zeros(1,2000);

SINR1=zeros(1,2000);

SINR2=zeros(1,2000);

for k1=1:1000

SINR(:,k1)=sum(bbs(k1)^2)/sum(mode(z(k1)-bbs(k1)^2));

del(:,k1)=z(k1)-H*I_post(:,k1);

%output SINR

SINR1(:,k1)=sum(bbs(k1)^2)/sum(mode(del(k1)-bbs(k1)^2));

%SINR improvement

SINR2(:,k1)=sum(mode(z(k1)-bbs(k1)^2))/sum(mode(del(k1)-bbs(k1)^2));

end

figure

t=20*log10(SINR);

hold on

t1=20*log10(SINR2);

hold off

%plot(t,t1)

plot(t,t1)

%axis([-20 -5 10 60]);

how can i get the result in fig given below by using above code.please guide me.