I have P=[2 3 4 0 5 1 0 2 ]; and I want to create a loop to remove zeros from P and calculate C, description in the code
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%% , for example P=[2 0 4 0 5 1 2 0 ]; and after the first iteration P=[2 4 0 5 1 2 0]; and second iteration P=[2 0 4 5 1 2 0 ]; and third one
P=[2 0 4 0 5 1 2 ]; which one will give me the best C .. if I removed the first zero, the second or the thrid one.
%%parameters
clc;
clear;
Pt=5;
Pn=2;
W=rand(4,8);
HT=rand(8,4);
D_Matrix = HT*W_bar;
D_Square = (abs(diag(D_Matrix))').^2;
P = waterfill(Pt,Pn./D_Square);
Find_Zero=find(~P); %Find Zero
if (~isempty(Find_Zero))
while (~isempty(Find_Zero))
First_Z=Find_Zero(1); %First Zero
W_bar(:,First_Z) = []; %Delete Col
HT(First_Z,:) = []; %Delete Row
D_Matrix = HT*W_bar;
D = (abs(diag(D_Matrix))').^2;
P = waterfill(Pt,Pn./D);
Find_Zero=find(~P); %Find Zero Power
end
R = P.*D/Pn;
C = sum(log2(1+R)); %Capacity
end
Réponse acceptée
Jan
le 9 Juin 2021
P = [2 0 4 0 5 1 2 0 ];
zeroIndex = find(P == 0);
for k = 1:numel(zeroIndex)
Q = P;
Q(zeroIndex(k)) = [];
... Insert your tests of Q here
end
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