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Could anyone help me to solve the following issue.
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I want to calculate the throughput using the following expression
for v =1:size(B,2)
for u =1:size(A,1)
throughput(u,v) =(Bmax.*log2(1+(((A(u,v)).*(B(u,v))/(noise+sum(B(1:u-1,v)).*A(u,v))))))
end
end
where
Bmax=3000;
noise=10;
and A and B gets generated by the code and the result for A and B was
A=[0 0 0 0 0.0447 0
0 0 0 0 0.0294 0]
B=[0 0 0 0 0.2428 0
0 0 0 0 0.4424 0]
A=[0 0 1.9728 0 0 0
0 0 0.0000 0 0 0]
B=[0 0 0.1971 0 0 0
0 0 0.2703 0 0 0]
A=[0.2373 0.2169 0 0.1979 0 0.2098
0.0174 0.0186 0 0.0214 0 0.0249]
B=[0.2160 0.7904 0 0.4386 0 0.8620
0.3774 0.9493 0 0.8335 0 0.9899]
When the code executes the throughput was calculated with respect to last array of A and B.
But i want to calculate the throughput with respect to all arrays of A and B one by one using the for loop throughput expression.Could anyone help me to fix the issue.
2 Kommentare
Antworten (1)
Walter Roberson
am 12 Apr. 2018
Use the same technique I showed you in https://www.mathworks.com/matlabcentral/answers/393815-could-anyone-tell-me-how-to-store-the-result-of-each-iteration-in-an-array-for-the-following-code#answer_314290 and https://www.mathworks.com/matlabcentral/answers/393813-could-anyone-tell-me-how-to-get-the-iterations-with-different-values-in-the-code-where-rng-is-used#answer_314286
"Store everything you might need later. You can always go back afterwards and clean up the parts you did not turn out to need.
Rule of programming: get it to work first, optimize afterwards."
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