Filter löschen
Filter löschen

Info

Diese Frage ist geschlossen. Öffnen Sie sie erneut, um sie zu bearbeiten oder zu beantworten.

Could anyone help me to solve the following issue.

1 Ansicht (letzte 30 Tage)
Prabha Kumaresan
Prabha Kumaresan am 12 Apr. 2018
Geschlossen: MATLAB Answer Bot am 20 Aug. 2021
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
KSSV
KSSV am 12 Apr. 2018
Similar question is answered.......follow the same here too.
Prabha Kumaresan
Prabha Kumaresan am 12 Apr. 2018
I tried but i am getting the error as the code executes
Bmax=3000;
noise=10;
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]
N = length(A) ;
throughput = cell(N,1) ;
for k =1:N
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
end
with respect to the code i cannot mention A{1},B{1},A{2},B{2},C{3},C{3} as A and B array are generated by the code.could you please help me to solve it.

Antworten (1)

Walter Roberson
Walter Roberson am 12 Apr. 2018
"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."

Diese Frage ist geschlossen.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by