unable to match the vector length for left and right side for the e(k) value

clc;
clear all;
close all;
M=3000; % number of data samples
T=2000; %number of training symbols
dB=25 %SNR in dB value

dB = 25

L=30; %length for smoothing
ChL=5; %length of the channel(ChL+1)
EqD=round((L+ChL)/2);  %Delay for equalization

Ch=randn(1,ChL+1)+sqrt(-1)*randn(1,ChL+1); %complex channel
Ch=Ch/norm(Ch); %scale the channel with norm

TxS=round(rand(1,M))*2-1; %QPSK transmitted sequence
TxS=TxS+sqrt(-1)*(round(rand(1,M))*2-1);

x=filter(Ch,1,TxS); %channel distortion
n=randn(1,M); %+sqrt(-1)*randn(1,M); %Additive white gaussian noise
n=n/norm(n)*10^(-dB/20)*norm(x);%scale the noise power in accordance with SNR
x=x+n; %received noisy signal

K=M-L;  %Discarding several starting samples for avoiding 0's and negative
X=zeros(L+1,K); %each vector column is a sample
for i=1:K
    X(:,i)=x(i+L:-1:i).';
end

%Adaptive RLS Equalizer
e=zeros(1,T-10); %initial error 
c=zeros(1,L+1); % initial condition

%Initialization of Rinverse
R_inverse=100*eye(L+1);

for k=1:1:T-10
   % y=X(:,k+10)';
   % disp(y)
e(k)=TxS(k+10+L-EqD)-c*X(:,k+10)*X(:,k+10)';

%finding error
 %RLS Parameters
 filtered_infrmn_vect=R_inverse*X(:,k+10);
 norm_error_power=X(:,k+10)'*filtered_infrmn_vect;
 gain_constant=1/(1+norm_error_power);
 norm_filtered_infrmn_vect=gain_constant*filtered_infrmn_vect;

 %updating weight values of equalizer
 % Weight update using RLS algorithm
 c=c+ (e(k)*norm_filtered_infrmn_vect);

 %updating R inverse
 R_inverse=R_inverse-norm_filtered_infrmn_vect'*norm_filtered_infrmn_vect;
end

Unable to perform assignment because the left and right sides have a different number of elements.

sb= c*X;

%SER(decision part)
sb1=sb/norm(c);
sb1=sign(real(sb1))+sqrt(-1)*sign(imag(ab1));%symbol detection
start=12;
sb2=sb1-TxS(start+1:start+length(sb1)); %error detection
SER=length(find(sb2~=0))/length(sb2); % SER calculation
disp(SER);

%plot of transmitted symbols
subplot(2,2,1),
plot(TxS,'*');
grid,title('Input symbols'); xlabel('real part'),ylabel('imaginary part')
axis([-2 2 -2 2])

%plot of received symbols
subplot(2,2,2),
plot(x,'o');
grid,title('Received symbols'), xlabel('real part'), ylabel('imaginary part')

%plot of the equalized symbols
subplot(2,2,3),
plot(sb,'o');
grid,title('Equalized symbols'), xlabel('real part'), ylabel('imaginary part')

%convergence
subplot(2,2,4),
plot(abs(e));
grid, title('Convergence'), xlabel('n'), ylabel('error signal')

[Satyajit Rajbanshi]

Can anyone please help..how the error can be solved

Answer 1

[Chandra]

Hi,

In this code, tr using a cell when assigning an array and extract the values individually when plotting

So here are some changes that needed for the code to get executed

change at line 25

e=cell(1,(T-10)); %initial error 

change at line 32

e(k)={TxS(k+10+L-EqD)-c*X(:,k+10)*X(:,k+10)'};

change at line 41

c=c+ (cell2mat(e(k))*norm_filtered_infrmn_vect);

change at line 49 (here ab1 is used instead of sb1 Check before correcting)

sb1=sign(real(sb1))+sqrt(-1)*sign(imag(sb1));%symbol detection

change the last plot depending on your requirement

for i = 1:T-10
plot(abs(e{i}));
hold on;
end

Please refer to the the attachment for the complete program in a txt document

Refer to the "cell2mat " documentation for extracting the values from cell in the above program.

Answer 2

[Satyajit Rajbanshi]

@Chandra Thank you so much for your kind help sir..it's working but i'm not getting any plot for figure 3.

[Chandra]

Hi,

change the line 41 to the following code line and give the value of "a" less than 1 like 0.1,0.001,...(prefer a =, 0.01)

 c=c+ a*(cell2mat(e(k))*norm_filtered_infrmn_vect);

change the plot 3 i.e., on line 65 to ( remove abs as we are plotting real and imaginary )

 plot((sb),'o');