i want to store a 100 x 100 matrix in hi_i for each iteration of irs1 and irs 2 when my sij1*sij2' is of 100x100 multiplied by a scalar in h_i.

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Maimoona Asad 2022-2-18

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https://ww2.mathworks.cn/matlabcentral/answers/1652685-i-want-to-store-a-100-x-100-matrix-in-hi_i-for-each-iteration-of-irs1-and-irs-2-when-my-sij1-sij2-i

评论： Maimoona Asad 2022-2-18

L=12;  % IRS
element=10;
lambda=0.06;
l=lambda/2;
%% channel between IRS and IRS
rr=11;
for irs1 = 1:(L-1)        % IRS1
    thetaa=2*pi*rand(rr) ; 
    thetae=2*pi*rand(rr) ;
    phi_a=2*pi*rand(rr) ; 
    phi_e=2*pi*rand(rr) ;
    ij=1;
    for irs2 = (irs1+1):L    % IRS2 
        %% AoA arrival 
        sij2=array_resp(10,phi_a(ij),phi_e(ij));   %100x1
        sij1=array_resp(10,thetaa(ij),thetae(ij)); %100x1
        hi_i(irs1,irs2)=(sqrt(alpha)/di_i(irs1,irs2))*(exp(-1j*(2*pi*di_i(irs1,irs2)/lambda)))*sij2*sij1'; %% distance between the IRSs
        hi_i(irs2,irs1)=hi_i(irs1,irs2);
        ij=ij+1;
    end
    rr=rr-1;
end

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Arif Hoq 2022-2-18

you can use 'cell array' to store your iteration value. what is the function:array_resp. if you have the code of this function please paste here

Maimoona Asad 2022-2-18

在 MATLAB Online 中打开

function array=array_resp(N,theta_a,theta_e)
lambda=0.06;
l=lambda/4;
s_v1=[];
for nn=0:N-1
    h_d1=exp(-1j*pi*(nn)*((2*l)/lambda)*sin(theta_e)*cos(theta_a));
    s_v1=[s_v1 h_d1];
end
sv_1=(s_v1)';
s_v2=[];
for nn=0:N-1
    h_d2=exp(-1j*pi*(nn)*((2*l)/lambda)*cos(theta_e));
    s_v2=[s_v2 h_d2];
end
sv_2=(s_v2)';
array=kron(sv_1,sv_2);

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Answer 1

KSSV 2022-2-18

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在 MATLAB Online 中打开

WE cannot correct your code as some variables are not defined. But still, your error is simple. You are trying to store a 100x100 matrix at a single element which you cannot.

% Demo
A = zeros(3) ;   % 3x3 matrix intilaized 
A(1,1)   = rand ;   % a single element saved no error 
A(1,:) = rand(1,3) ;  % three elements stored at three places no error
A(3,3) = rand(3) ;  % error, you cannot store 3x3 matrix at a single element 
Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 3-by-3.

S the work around is, initiate the initial matrix properly and use correct indexing.

If no store the matrices into a cell .

A = cell(2,2) ;
A{1,1} = rand(3) ; 
A{1,2} = rand(4) ; 

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Maimoona Asad 2022-2-18

now i am sharing the code would you please help with the bold part.

Maimoona Asad 2022-2-18

clc

clear all

close all

clc

clear all

close all

L=12; % IRS

element=10;

lambda=0.06;

l=lambda/2;

Sigma=db2pow(-90); %% -60dbm

a_db=-46;

alpha=db2pow(a_db);

K=4; % Number of users

radius=15;

X0=0; % x-coordinate of Centre of the circle

Y0=0; % y-coordinate of Centre of circle

%% Base stations

x_b=[0.7298; 3; -5.23; 11.86]; % x coordinates of base station

y_b=[12.48; -1.497; -6.139; -3.23]; %y coordiantes of base station

%% IRSs placement

X3=[-7.655; -6.189; -3.251; 2.926; 6.534; 10.79; 5.203; -4.465; 1.42; -8.66; -0.8748; 6.534] ;

Y3=[1.393; 10.42; 6.639; 5.821; 8.425; 2.924; 1.139; 0.7508; -7.917; -6.928; -10.01; -7.03];

p=0;

%% loop for the elements of irs

N=100:100:500; %% number of elements in an IRS

NN = length(N) ;

for n = 1:length(N)

nn=N(n);

for i=1:L

for j=1:element

for k=1:(nn)/(10)

x3(j,k,i) = X3(i,1) + p;

y3(j,k,i) = Y3(i,1);

p=k*l; % Each time lambda is added to the x-coordinate

x_e(j,k,i,n) = x3(j,k,i); % X coordinate of an element of IRS for plotting

y_e(j,k,i,n)= y3(j,k,i); % Y coordinate of an element of IRS for plotting

end

p=0;

Y3(i,1) = Y3(i,1) + l;

end

% plot(x3,y3,'bo','HandleVisibility','off')

end

%% Distances between the IRSs

for irs1 = 1:(L-1) % IRS1

for irs2 = (irs1+1):L % IRS2

di_i(irs1,irs2)=sqrt((X3(irs2)-X3(irs1))^2+(Y3(irs2)-Y3(irs1))^2); %% distance between the IRSs

di_i(irs2,irs1)=di_i(irs1,irs2);

end

%% Distances between the BS and IRSs

for bs = 1:4 % IRS1

for irs = 1:L % IRS2

db_i(bs,irs)=sqrt((X3(irs)-x_b(bs))^2+(Y3(irs)-y_b(bs))^2); %% distance between bs and IRSs

db_i(irs,bs)=db_i(bs,irs);

end

%% channel between BS and IRS

for bs = 1:4 % IRS1

phi_a=2*pi*rand(L) ;

phi_e=2*pi*rand(L) ;

for irs = 1:L % IRS2

ai_b=array_resp(10,phi_a(irs),phi_e(irs)); %steering vector1

hb_i(bs,irs)=(sqrt(alpha)/db_i(bs,irs))*(exp(-1j*(2*pi*db_i(bs,irs)./lambda)));

hb_i(irs,bs)=db_i(bs,irs);

end

%% channel between IRS and IRS

hi_i= cell(100,100);

rr=11;

for irs1 = 1:(L-1) % IRS1

thetaa=2*pi*rand(rr) ;

thetae=2*pi*rand(rr) ;

phi_a=2*pi*rand(rr) ;

phi_e=2*pi*rand(rr) ;

ij=1;

for irs2 = (irs1+1):L % IRS2

%% AoA arrival

sij2=array_resp(10,phi_a(ij),phi_e(ij));

sij1=array_resp(10,thetaa(ij),thetae(ij));

hi_i(irs1,irs2)=(sqrt(alpha)/di_i(irs1,irs2))*(exp(-1j*(2*pi*di_i(irs1,irs2)/lambda)))*sij2*sij1'; %% distance between the IRSs i want a hi_i 100x100 matrix for each iteration but dimensions error occurs

ij=ij+1;

end

rr=rr-1;

end

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i want to store a 100 x 100 matrix in hi_i for each iteration of irs1 and irs 2 when my sij1*sij2' is of 100x100 multiplied by a scalar in h_i.

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i want to store a 100 x 100 matrix in hi_i for each iteration of irs1 and irs 2 when my sij1*sij2' is of 100x100 multiplied by a scalar in h_i.

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