1.算法仿真效果
matlab2022a仿真結果如下（完整代碼運行后無水咏穸埂）：
2.算法涉及理論知識概要
???????LDPC碼是一種線性錯誤修正碼，以其接近香農極限的優(yōu)良性能而被廣泛應用于現(xiàn)代通信系統(tǒng)中塑悼。NMS譯碼是一種基于最小平方誤差準則的軟判決譯碼方法馋嗜，其目標是找到一個最可能的碼字辐啄，使得接收到的信號與該碼字之間的歐氏距離最小澈魄。NMS譯碼算法中包含一些關鍵參數(shù)，如歸一化因子仲翎，這些參數(shù)的選擇對譯碼性能有顯著影響痹扇。
在應用GWO優(yōu)化LDPC碼NMS譯碼算法的歸一化參數(shù)時，首先需要定義目標函數(shù)溯香，通常為目標函數(shù)最小化鲫构，例如最小化誤碼率(BER)或最大化信噪比(SNR)。設歸一化參數(shù)為γ玫坛，則目標函數(shù)可以表示為：
???????GWO算法通過不斷迭代更新γ的值结笨，尋找使f(γ)達到最小（或最大湿镀，根據(jù)目標函數(shù)定義）的γ?炕吸。最終，γ?即為最優(yōu)歸一化參數(shù)勉痴。
在本課題中赫模，我們選擇BER(γ)。
3.MATLAB核心程序
????a=2*(1-(t/Iters)); ?
????for i=1:Num
????????for j=1:dim ??????
????????????r1 ?????= rand;
????????????r2 ?????= rand;
????????????A1 ?????= 2*a*r1-a;%
????????????C1 ?????= 2*r2; ???%
????????????D_alpha = abs(C1*Alpx(j)-xpos(i,j));%
????????????X1 ?????= Alpx(j)-A1*D_alpha; ??????%
????????????r1 ?????= rand;
????????????r2 ?????= rand;
????????????A2 ?????= 2*a*r1-a; %
????????????C2 ?????= 2*r2; %
????????????D_beta ?= abs(C2*btx(j)-xpos(i,j)); %
????????????X2 ?????= btx(j)-A2*D_beta; % ???
????????????r1 ?????= rand;
????????????r2 ?????= rand;
????????????A3 ?????= 2*a*r1-a; %
????????????C3 ?????= 2*r2; %
????????????D_delta = abs(C3*dltx(j)-xpos(i,j)); %
????????????X3 ?????= dltx(j)-A3*D_delta; % ??????????
????????????xpos(i,j) = (X1+X2+X3)/3;%
????????????if xpos(i,j)>=Lmax(j)
???????????????xpos(i,j)=Lmax(j);
????????????end
????????????if xpos(i,j)<=Lmin(j)
???????????????xpos(i,j)=Lmin(j);
????????????end
????????end
????end
end
aa = Alpx
N = 2016;
K = 1008;
R = K/N;
%H矩陣
[ H, Hp, Hs ] = func_H();
...........................................................
fitness=mean(Ber);
figure
semilogy(SNR, Ber,'-b^',...
????'LineWidth',1,...
????'MarkerSize',6,...
????'MarkerEdgeColor','k',...
????'MarkerFaceColor',[0.2,0.9,0.5]);
xlabel('Eb/N0(dB)');
ylabel('Ber');
title(['歸一化最小和NMS,GWO優(yōu)化后的alpha = ',num2str(aa)])
grid on;
save NMS4.mat SNR Ber ?aa
0X_061m
