Infinity loop - waterfilling algorithm

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Ahmad Halimi
Ahmad Halimi 2020-4-20
评论： Ahmad Halimi 2020-4-21
I'm trying to write a code to carry out the Waterfilling algorithm for MIMO channels. I've written the code below but there is something wrong with my WHILE loop. I get negative power and also endless loop at the end. I'm trying to write the code for the given formula below:
and here is my code:
P=zeros(1,r);
sq=zeros(1,3);
for i=1:numel(SNR)
p=1;
go = true;
while go
for j=1:r-p+1
K(j)=1/L(j);
T =sum(K);
end;
m =(1/(r-p+1))*(1+ (1/0.1)*T);
for j=1:r-p+1
P(j)= m-(1/(0.1*L(j)));
if P(r-p+1)< 0
P(r-p+1)=0;
end;
sq(j) = sum(P);
if sq(j) <= 1.00005
go = false;
else
p=p+1;
end;
end;
end;
for j=1:r
C_equal_t2(j)=log2(1+SNR(i)*L(j)*P(j));
C_eigen(i)=sum(C_equal_t2);
end;
end;
  2 个评论
Geoff Hayes
Geoff Hayes 2020-4-20
Ahmad - in your code you have
for j=1:r-p+1
P(j)= m-(1/(0.1*L(j)));
if P(r-p+1)< 0
P(r-p+1)=0;
end;
% etc.
end
Why, on each iteration of the loop, do you check to see if
P(r-p+1)< 0
? You are doing the same comparison for each j. Do you mean this to be
if P(j) < 0
P(j)=0;
end;
instead?
Ahmad Halimi
Ahmad Halimi 2020-4-21
No. Actually for each iteration it's needed to check just the last P which is indicated by P(r-p+1)
the SNR range is [0.01 to 100] and the problem is that for low values the loop never ends, it's ok for large values like 20 as an example

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