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

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Satyajit Rajbanshi on 12 May 2022
Commented: Chandra on 26 May 2022
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
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')
##### 1 CommentShowHide None
Satyajit Rajbanshi on 12 May 2022

Chandra on 25 May 2022
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.

### More Answers (1)

Satyajit Rajbanshi on 26 May 2022
@Chandra Thank you so much for your kind help sir..it's working but i'm not getting any plot for figure 3.
##### 1 CommentShowHide None
Chandra on 26 May 2022
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');