- literal 0 is a 1 x 1 scalar
- ones(1,i-1) is ones(1,0) which is 1 x 0 empty
- literal 0 is 1 x 1 scalar
- ones(1,n-i) is ones(1,3-1) (for this iteration) which is ones(1,2) which is 1 x 2
To write perfectly this command: A(i,:) = A(i,:) + h(i)*[0, ones(1,i-1), 0, ones(1,n-i)];
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Please I need to express this command perfectly:
A(i,:) = A(i,:) + h(i)*[0, ones(1,i-1), 0, ones(1,n-i)];
In order to Run this:
n = 3
h(1) = 0.6
h(2) = 0.8
h(3) = 1
T = 4
% Objective function: minimize total power transmitted by all users
f = ones(n, 1);
% Inequality constraints: signal-to-interference ratio must exceed threshold
A = zeros(n,n);
b = zeros(n,1);
for i = 1:3
A(i,i) = -h(i);
A(i,:) = A(i,:) + h(i)*[0, ones(1,i-1), 0, ones(1,n-i)];
b(i) = -T*h(i);
end
% Lower bounds: power transmitted by each user must be non-negative
lb = zeros(n,1);
% Solve using linprog
[x, fval] = linprog(f, A, b, [], [], lb);
% Display results
disp('Optimal power transmitted by each user:');
disp(x);
disp(['Minimum total power transmitted: ', num2str(fval)]);
0 个评论
采纳的回答
Walter Roberson
2023-5-2
n = 3
A = zeros(n,n);
So A will be initialized as 3 x 3
for i = 1:3
A(i,i) = -h(i);
A(i,:) = A(i,:) + h(i)*[0, ones(1,i-1), 0, ones(1,n-i)];
when i = 1 then:
Those widths are 1 + 0 + 1 + 2 = 4.
So you are trying to add a 1 x 4 vector to a 1 x 3 vector A(i,:)
5 个评论
Torsten
2023-5-3
Matlab needs a multiplication sign (*) between numbers you want to multiply.
Walter Roberson
2023-5-3
Assuming that g and h are row vector
A = - h * g.';
A(1:n+1:end) = h;
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