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这个问题似乎微不足道，但在尝试解决它时我只能摸不着头脑。我正在尝试将具有恒定模数技术的分数间隔均衡器应用于 64 QAM 星座。该程序适用于 QPSK 或 4 QAM，但是当我将其应用于 64QAM 时，它会引发错误：

Error using  / 
Matrix dimensions must agree.

Error in Working_FSE_CMA_64QAM (line 68)
sb1=sb/(fh(temp));  % scale the output

我没有通信工具箱，因此使用我之前的问题生成 16 QAM 信号中给出的答案生成了 64QAM 符号

有人可以帮助使代码工作吗？谢谢你。

% Blind channel estimation/equalization
% adpative CMA method in Fractional space

T=1000;    % total number of data
dB=25;     % SNR in dB value

%%%%%%%%% Simulate the Received noisy Signal  %%%%%%%%%%%
N=5; % smoothing length N+1
Lh=5;  % channel length = Lh+1
Ap=4;  % number of subchannels or receive antennas

h=randn(Ap,Lh+1)+sqrt(-1)*randn(Ap,Lh+1);   % channel (complex)
for i=1:Ap, h(i,:)=h(i,:)/norm(h(i,:));    end        % normalize
s = (randi(8,1,T)*2-5)+j*(randi(8,1,T)*2-5);  %64 QAM
%s=round(rand(1,T))*2-1;  % QPSK or 4 QAM symbol sequence
%s=s+sqrt(-1)*(round(rand(1,T))*2-1);

% generate received noisy signal
x=zeros(Ap,T);    % matrix to store samples from Ap antennas
SNR=zeros(1,Ap);
for i=1:Ap
    x(i,:)=filter(h(i,:),1,s);
    vn=randn(1,T)+sqrt(-1)*randn(1,T);   % AWGN noise (complex)
    vn=vn/norm(vn)*10^(-dB/20)*norm(x(i,:));  % adjust noise power
    SNR(i)=20*log10(norm(x(i,:))/norm(vn));   % Check SNR of the received samples
    x(i,:)=x(i,:)+vn;                        % received signal
end
SNR=SNR    % display and check SNR

%%%%%%%%%%%%% adaptive equalizer estimation via CMA
Lp=T-N;   %% remove several first samples to avoid 0 or negative subscript
X=zeros((N+1)*Ap,Lp);  % sample vectors (each column is a sample vector)
for i=1:Lp
    for j=1:Ap
        X((j-1)*(N+1)+1:j*(N+1),i)=x(j, i+N:-1:i).';
    end
end

e=zeros(1,Lp);  % used to save instant error
f=zeros((N+1)*Ap,1); f(N*Ap/2)=1;    % initial condition
%R2=2;                  % constant modulas of QPSK symbols
R2 = 1.380953; %For 64 QAM http://www.google.com/patents/US7433400
mu=0.001;      % parameter to adjust convergence and steady error
for i=1:Lp
   e(i)=abs(f'*X(:,i))^2-R2;                  % instant error
   f=f-mu*2*e(i)*X(:,i)*X(:,i)'*f;     % update equalizer 
   f(N*Ap/2)=1;
%   i_e=[i/10000 abs(e(i))]             % output information 
end   

%sb=f'*X;   % estimate symbols (perform equalization)
sb = filter(f, 1, X);
% calculate SER
H=zeros((N+1)*Ap,N+Lh+1);  temp=0;
for j=1:Ap
    for i=1:N+1, temp=temp+1; H(temp,i:i+Lh)=h(j,:); end  % channel matrix
end

fh=f'*H;    % composite channel+equalizer response should be delta-like 
temp=find(abs(fh)==max(abs(fh)));   % find the max of the composite response

sb1=sb/(fh(temp));  % scale the output
sb1=sign(real(sb1))+sqrt(-1)*sign(imag(sb1));  % perform symbol detection
start=N+1-temp;  % general expression for the beginning matching point
sb2=sb1(10:length(sb1))-s(start+10:start+length(sb1));  % find error symbols
SER=length(find(sb2~=0))/length(sb2)   % calculate SER

if 1
    subplot(221), 
    plot(s,'o');   % show the pattern of transmitted symbols
    grid,title('Transmitted symbols');  xlabel('Real'),ylabel('Image')
    axis([-2 2 -2 2])

    subplot(222),
    plot(x,'o');  % show the pattern of received samples
    grid, title('Received samples');  xlabel('Real'), ylabel('Image')

    subplot(223),
    plot(sb,'o');   % show the pattern of the equalized symbols
    grid, title('Equalized symbols'), xlabel('Real'), ylabel('Image')

    subplot(224),
    plot(abs(e));   % show the convergence
    grid, title('Convergence'), xlabel('n'), ylabel('Error e(n)')
end
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1 回答 1

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主要问题是迭代算法不收敛（或者更具体地说是发散）。结果是fcontains 中的值NaN。的结果

temp=find(abs(fh)==max(abs(fh)))

然后是一个空向量，因此与行上的预期标量大小不匹配

sb1=sb/(fh(temp));  % scale the output

要解决此问题，您可能会注意到您使用的mu和的值R2是基于具有单位方差的信号星座。您可以使用以下命令生成这样的 64-QAM 星座：

s = ((randi(8,1,T)*2-9)+j*(randi(8,1,T)*2-9))/sqrt(42);  %64 QAM

此外，几行代码重新定义了复数常量j=sqrt(-1)以用作索引变量。您应该避免使用j作为索引。此外，在 sript 开头清除所有变量（使用clear all）可以帮助您以一致的干净状态开始执行。

于 2015-11-14T15:49:23.887 回答