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I was analyzing the following code for the MRC scheme from the book MIMO-OFDM Wireless Communications with MATLAB. Here, I am not able to understand based on what logic the sigma value calculated sigma=sqrt(0.5/(10^(SNRdB/10)))

It would be very helpful if anyone could give me any hints on the calculation of sigma from SNRdB

Thanks

clear, clf
L_frame=130; N_packet=4000; 
b=2; % Set to 1/2/3/4 for BPSK/QPSK/16QAM/64QAM
SNRdBs=[0:2:20]; sq2=sqrt(2);
for iter=1:3
   if iter==1, NT=1; NR=1; gs='-kx'; % SISO
    elseif iter==2, NT=1; NR=2; gs='-^'; % Numbers of Tx/Rx antennas
    else  NT=1; NR=4; gs='-ro'; % 
   end
   sq_NT=sqrt(NT);
   for i_SNR=1:length(SNRdBs)
      SNRdB=SNRdBs(i_SNR);  
      sigma=sqrt(0.5/(10^(SNRdB/10)));
      for i_packet=1:N_packet
         symbol_data=randi([0, 1],L_frame*b,NT);
         [temp,sym_tab,P]=modulator(symbol_data.',b);
         X=temp.';   % frlg=length(X);
         Hr = (randn(L_frame,NR)+j*randn(L_frame,NR))/sq2;
         H = reshape(Hr,L_frame,NR); Habs = sum(abs(H).^2,2); Z=0;
         for i=1:NR
            R(:,i) = sum(H(:,i).*X,2)/sq_NT + sigma*(randn(L_frame,1)+j*randn(L_frame,1));
            Z = Z + R(:,i).*conj(H(:,i));
         end
         for m=1:P
            d1(:,m)=abs(sum(Z,2)-sym_tab(m)).^2+(-1+sum(Habs,2))*abs(sym_tab(m))^2;
         end
         [y1,i1] = min(d1,[],2);   Xd=sym_tab(i1).';
         temp1 = X>0;  temp2 = Xd>0;
         noeb_p(i_packet)=sum(sum(temp1~=temp2));
      end
      BER(iter,i_SNR) = sum(noeb_p)/(N_packet*L_frame*b);
   end% end of FOR loop for SNR
   semilogy(SNRdBs,BER(iter,:),gs), hold on, axis([SNRdBs([1 end]) 1e-6 1e0])
end
title('BER perfoemancde of MRC Scheme'), xlabel('SNR[dB]'), ylabel('BER') 
grid on, set(gca,'fontsize',9)
legend('SISO','MRC (Tx:1,Rx:2)','MRC (Tx:1,Rx:4)')
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1 Answer 1

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  • $\text{SNR}$ is commonly defined as $ \text{SNR} = P_s$/$P_n$ where $P_s$ is the power (variance) or the signal and $P_n$ the power (variance) of the noise. You have: $$\begin{align} &\text{SNR}_{db} = 10\cdot \log_{10}(P_s/P_n)\\ \implies & P_n = \cfrac{P_s}{10^{\text{SNR}_{db}/10}} \end{align}$$

  • Standard deviation, commonly referred to as $\sigma$, is the square root of the variance. For example, the standard deviation of the noise samples, $\sigma_{P_n}$, is: $$\sigma_{P_n} = \sqrt{P_n}$$

In your case, assuming the signal variance is $0.5$:

P_s = 0.5; %signal variance
P_n = P_s / 10^(SNRdb/10); %noise variance
sigma = sqrt(P_n); %noise standard deviation
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  • $\begingroup$ Thanks for the answer. I was wondering -- from where the parameter 0.5 has come. Of course, if we assume signal variance is 0.5 that makes sense, although nowhere they mentioned this assumption. $\endgroup$
    – Aragorn
    Commented Oct 18, 2022 at 10:42
  • $\begingroup$ Agreed, but I can only assume that's what they're assuming since I have no access to the text... $\endgroup$
    – Jdip
    Commented Oct 18, 2022 at 10:55

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