Effacer les filtres
Effacer les filtres

generate an x-matrix at every iteration by for loop

2 vues (au cours des 30 derniers jours)
Ali Al raeesi
Ali Al raeesi le 2 Nov 2022
Modifié(e) : Bruno Luong le 3 Nov 2022
assume that my iteration is 2000 i want to generate H-matrix at every iteration, soi have 2000 H-matrices for every solution.by using for loop then i need calculate c at every iteration(random because H is a random matrix). also using use for-loop
iterations=2000; % here i assume the iterations
for i=1: iterations
H=sqrt(0.5)*(randn+1i*randn)*(1/sqrt(2)); % my H
end
for i=1: iterations
cdf(idx) = log2(det((eye(Nr(2)))+((SNR(idx)/Nt(2))*abs(H)*(abs(H'))))); % my C
end
%% modify my code because i think it wronge to generate H-matrix at every iteration then i need calculate c at every iteration
%% my H is not a matrix can you show me how to generate the H as matrix

Réponses (1)

KALYAN ACHARJYA
KALYAN ACHARJYA le 2 Nov 2022
Modifié(e) : KALYAN ACHARJYA le 2 Nov 2022
for idx=1: 2000
H=sqrt(0.5)*(randn+1i*randn)*(1/sqrt(2)); % my H
%cdf=...
end
#If you wish to save all generated H matrix, use cell array {} to save the H matrix. For cdf case, provide us the Nr or other typical data, so that we can try
cdf(idx) = log2(det((eye(Nr(2)))+((SNR(idx)/Nt(2))*abs(H)*(abs(H'))))); % my C
  3 commentaires
Bruno Luong
Bruno Luong le 2 Nov 2022
Modifié(e) : Bruno Luong le 3 Nov 2022
If you want to generate a random matrix of size (m x n) you should do
randn(m,n)
not
randn
without argument, which returns a scalar.
Ali Al raeesi
Ali Al raeesi le 3 Nov 2022
Nt = [4 4];
Nr = [1 4];
SNR=-10;
iterations=2000;
for i=1: iterations
H[i]=sqrt(0.5)*(randn(1,2)+1i*randn(1,2))*(1/sqrt(2)); % this way my H will be generated as matrix
cdf1(idx) = log2(det((eye(Nr(1)))+((SNR(idx)/Nt(1))*abs(H)*(abs(H'))))); % c
cdf2(idx) = log2(det((eye(Nr(2)))+((SNR(idx)/Nt(2))*abs(H)*(abs(H'))))); % c
end
B = sort(cdf2(idx)); % to sort c
X-axis (Sorted c)
Y-axis(CDF: between 1/5000 to 5000/5000) by using for-loop

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