equation of zero forcing receiver
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hi,
can help me in this error , i don't know the solution ?
and i want to sure if this equation is right for zero forcing estimator ?
N_r = 128 ;
K = 8 ;
H = zeros(N_r, K);
SNR_dB = -12:4:20;
SNR = 10.^(0.1.*SNR_dB);
for iter = 1:length(SNR)
S = sqrt((SNR(iter))/K).*dftmtx(K) ;
W_ZF = ((H.^H) .* (H .* (H.^H)).^(-1)) .* S ; %" error in this line , Arrays have incompatible sizes for this operation"
H_ZF = (W_ZF .^ H) .* y1 ;
end
5 Commenti
Dyuman Joshi
il 16 Set 2023
What is the size of variables H and S?
Maha Saif
il 16 Set 2023
Dyuman Joshi
il 16 Set 2023
Undefined parameter iter in your code.
Check above.
Maha Saif
il 16 Set 2023
Maha Saif
il 16 Set 2023
Risposta accettata
Più risposte (2)
Walter Roberson
il 16 Set 2023
W_ZF = ((H.^H) .* (H .* (H.^H)).^(-1)) .* S ;
The first part of that expression results in a 128 x 8 array that is all infinity -- with H being all 0, the H .* gives all 0 and all-zero ^-1 is going to be all infinity.
S is 8 x 8.
You cannot use element-by-element multiplication between a 128 x 8 matrix and an 8 x 8 matrix.
You could potentially use * matrix-multiplication between a 128 x 8 matrix and an 8 x 8 matrix, giving a 128 x 8 result. Which will probably be entirely complex-NaN due to the infinities in the left-hand side...
4 Commenti
Maha Saif
il 17 Set 2023
Walter Roberson
il 17 Set 2023
I do not have any idea what the equations are for zero forcing receiver.
Maha Saif
il 17 Set 2023
N_r = 128 ;
K = 8 ;
H = zeros(N_r, K);
SNR_dB = -12:4:20;
SNR = 10.^(0.1.*SNR_dB);
numSNR = length(SNR);
W_ZF = cell(numSNR,1);
H_ZF = cell(numSNR,1);
for iter = 1:numSNR
S = sqrt((SNR(iter))/K).*dftmtx(K) ;
W_ZF{iter} = ((H.^H) .* (H .* (H.^H)).^(-1)) * S ;
H_ZF{iter} = (W_ZF{iter} .^ H) .* y1 ;
end
However, I have no idea whether the modified expression is a correct expression for zero forcing receiver.
And it's still going to be all complex nan and inf anyhows. The matrix of zeros raised to power -1 is going to produce all infinities, and when you start manipulating infinities nan are a very common result.
Maha Saif
il 17 Set 2023
0 voti
1 Commento
I had to guess about the size and values of y1 and about what might be reasonable "complex" entries.
If you have even a single zero entry in H then you get a NaN output -- at least for the corresponding row.
N_r = 128 ;
K = 8 ;
y1 = randi([0 1], N_r, K);
SNR_dB = -12:4:20;
SNR = 10.^(0.1.*SNR_dB);
numSNR = length(SNR);
H_ZF0 = cell(numSNR,1);
H_ZF1 = cell(numSNR,1);
H_ZFc = cell(numSNR,1);
H0 = zeros(N_r, K);
H1 = ones(N_r, K);
Hc = ones(N_r, K) * (1+1i); Hc(1,1) = 0;
for iter = 1:numSNR
S = sqrt((SNR(iter))/K).*dftmtx(K) ;
W_ZF = ((H0.^H0) .* (H0 .* (H0.^H0)).^(-1)) * S ;
H_ZF0{iter} = (W_ZF .^ H0) .* y1 ;
W_ZF = ((H1.^H1) .* (H1 .* (H1.^H1)).^(-1)) * S ;
H_ZF1{iter} = (W_ZF .^ H1) .* y1 ;
W_ZF = ((Hc.^Hc) .* (Hc .* (Hc.^Hc)).^(-1)) * S ;
H_ZFc{iter} = (W_ZF .^ Hc) .* y1 ;
end
H_ZF0{1}(1:5,:)
H_ZF1{1}(1:5,:)
H_ZFc{1}(1:5,:)
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