Why does my for loop only output the last-time result

I want to calculate different received power for different receiver coordinates, but why does my loop only get the same result?

rx = 4;
ry = 8;
rz = 3;
h = 2;
% Lambertian emission order
n = 1;
ne = 1;
% detector physical area of a PD
A = 1e-4;
% FOV of a receiver
FOV = 95;
%% first transmitter calculation
TP1=[1 1 3]; % transmitter position
% Distance between transmitter and receiver
% reciver
x=linspace(1,rx,4);
y=linspace(1,ry,8);
z=linspace(1,rz,3);
for ii = 1:rx %1 to 4
    for jj = 1:ry %1 to 8
        RP=[x(ii) y(jj) 1]; % receiver position vector 
        h_a1(ii,jj)=0; % reflection from North face
        DLOS = sqrt(dot(TP1-RP,TP1-RP));% distance between transmitter and receiver
        cos_alphaLOS = abs(RP(3)-TP1(3))/DLOS; 
        cos_betaLOS = abs(TP1(3)-RP(3))/DLOS;
        if abs(acosd(cos_betaLOS)) <= FOV
        %receiver power
        h_a1(ii,jj) = h_a1(ii,jj) + ((n+1) * A * (cos_alphaLOS ^ n) * cos_betaLOS * Ps) / 2 * pi * (DLOS ^ 2);
        end
    end
end

Answer 1

[DGM]

Observe this expression:

h_a1(ii,jj) = h_a1(ii,jj) + ((n+1) * A * (cos_alphaLOS ^ n) * cos_betaLOS * Ps) / 2 * pi * (DLOS ^ 2); 

Reduce it to the terms which are ostensibly variable within the scope of the loop:

h_a1(ii,jj) = cos_alphaLOS * cos_betaLOS * DLOS^2;

Substitute the expressions for the first two variables:

h_a1(ii,jj) = abs(RP(3)-TP1(3))/DLOS * abs(TP1(3)-RP(3))/DLOS * DLOS^2;

Simplify:

h_a1(ii,jj) = abs(RP(3) - TP1(3))^2;

Which is constant within the scope of the loop. The entire expression is constant.

[QIN YIBO]

Thank you for your reply.

Do you mean to replace the variable name with a formula? So does the formula become the following?

h_a1(ii,jj) = h_a1(ii,jj) + ((n+1) * A * ((abs(RP(3)-TP1(3))/DLOS) ^ n) * (abs(TP1(3)-RP(3))) * Ps)...
            / 2 * pi * ((sqrt(dot(TP1-RP,TP1-RP))) ^ 2);

[DGM]

Unfortunately, no. I was merely identifying the reason why the output does not change. Fixing the issue requires a conceptual understanding of the task at hand, which is something I don't have.

[QIN YIBO]

I changed my code with the above formula, and now it can output different results. Thank you very much.