SOVA的实现（3）

function L_A = sova(Ly,G,Lu,ind_dec) % Copyright: Yufei Wu, 1998, MPRG lab, Virginia Tech for academic use% This implements Soft Output Viterbi Algorithm in trace back mode % Input: Ly     : Scaled received bits Ly=0.5*L_c*y=(2*a*rate*Eb/N0)*y%        G      : Code generator for the RSC code in binary matrix form%        Lu     : Extrinsic information from the previous decoder.%        ind_dec: Index of decoder=1/2%                  (assumed to be terminated in all-zero state/open)% Output: L_A  : Log-Likelihood Ratio (soft-value) of %                   estimated message input bit u(k) at each stage, %                 ln (P(u(k)=1|y)/P(u(k)=-1|y))lu = length(Ly)/2; % Number of y=[ys yp] in Lylu1 = lu+1; Infty = 1e2;[N,L] = size(G); Ns = 2^(L-1); % Number of statesdelta = 30;  % SOVA window size%Make decision after 'delta' delay. Tracing back from (k+delta) to k,% decide bit k when received bits for bit (k+delta) are processed.  % Set up the trellis defined by G.[nout,ns,pout,ps] = trellis(G);% Initialize the path metrics to -InftyMk(1:Ns,1:lu1)=-Infty; Mk(1,1)=0; % Only initial all-0 state possible% Trace forward to compute all the path metricsfor k=1:lu   Lyk = Ly(k*2-[1 0]); k1=k+1;   for s=1:Ns  % Eq.(9.4.44), Eq.(9.4.45)      Mk0 = Lyk*pout(s,1:2).' -Lu(k)/2 +Mk(ps(s,1),k);      Mk1 = Lyk*pout(s,3:4).' +Lu(k)/2 +Mk(ps(s,2),k);      if Mk0>Mk1, Mk(s,k1)=Mk0; DM(s,k1)=Mk0-Mk1; pinput(s,k1)=0;       else       Mk(s,k1)=Mk1; DM(s,k1)=Mk1-Mk0; pinput(s,k1)=1;      end   endend% Trace back from all-zero state or the most likely state for D1/D2% to get input estimates uhat(k), and the most likely path (state) shatif ind_dec==1, shat(lu1)=1;  else [Max,shat(lu1)]=max(Mk(:,lu1));  endfor k=lu:-1:1   uhat(k)=pinput(shat(k+1),k+1); shat(k)=ps(shat(k+1),uhat(k)+1);end% As the soft-output, find the minimum DM over a competing path %  with different information bit estimate.for k=1:lu   LLR = min(Infty,DM(shat(k+1),k+1));   for i=1:delta      if k+i<lu1        u_=1-uhat(k+i); % the information bit        tmp_state = ps(shat(k+i+1),u_+1);        for j=i-1:-1:0           pu=pinput(tmp_state,k+j+1); tmp_state=ps(temp_state,pu+1);        end        if pu~=uhat(k), LLR = min(LLR,DM(shat(k+i+1),k+i+1)); end      end   end   L_A(k) = (2*uhat(k)-1)*LLR; % Eq.(9.4.46)end