Why does this code give error?

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Sadiq Akbar on 16 Apr 2024
Commented: Sadiq Akbar on 22 Apr 2024
clear;clc
%%
fc = 3e8;% Carrier frequency
Nb = 1000;% Number of snapshots
c = 3e8;
wavelength = c/fc;% The wavelength of the received signal
d = 0.5*wavelength;
theta = [5 40];
M = length(theta);% Number of signals
N = 10;% Number of antennas
% Wavenumber
beta = 2*pi/wavelength;
% Signal amplitude
A = 1;
% SNR (dB)
snr = 5;
% Variance of noise
sigma=sqrt((A^2)/(2*10^(snr/10)));
% Source signal
D = randi(M,Nb,1);
S =A*(2*D - 1);
% The electrical phase shift from element to element along the array
phi=beta*d*cos(theta*pi/180);
% Matrix of steering vectors
for i=1:M
for k=1:N;
VecteurDirectionnel(k,i)= exp(j*(k-1)*phi(i));
end
end
% White Gaussien noise
B = (sigma^2)*(randn(N,Nb)+j*randn(N,Nb))/sqrt(2);
% Array output:signal plus noise
X = VecteurDirectionnel*S+B;
Error using *
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the number of rows in the second matrix. To operate on each element of the matrix individually, use TIMES (.*) for elementwise multiplication.
% Estimation of the spatial correlation matrix of the observed signal
Rxx = X*X'/Nb;
% Eigen decomposition
[Vi,Li] = eig(Rxx);
[L,I] = sort(diag(Li),'descend');
V = Vi(:,I);
Vs = V(:,1:M);
Vs1=Vs(1:N-1,:);
Vs2=Vs(2:N,:);
% Direction Of Arrival
% Least square
xsi=linsolve(Vs(1:N-1,:),Vs(2:N,:));
% DOA estimation
doa=acosd((angle(eig(xsi))/(2*pi*d)))

Matt J on 16 Apr 2024
Edited: Matt J on 16 Apr 2024
As you can see, VecteurDirectionnel and S are the wrong sizes for matrix multiplication,
whos VecteurDirectionnel S
Name Size Bytes Class Attributes S 1000x1 8000 double VecteurDirectionnel 10x2 320 double complex
% Array output:signal plus noise
X = VecteurDirectionnel*S+B;
Error using *
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the number of rows in the second matrix. To operate on each element of the matrix individually, use TIMES (.*) for elementwise multiplication.
Matt J on 19 Apr 2024
Well S is Nb x 1 because D is Nb x 1, from this line,
D = randi(M,Nb,1);
Perhaps you meant to have,
D = randi([low, hi] , [M,Nb]);
but we have no way of knowing what you intended the low and hi limits to be.
Sadiq Akbar on 22 Apr 2024
Thanks a lot for your help. Yes, you are right. It works now.