16-QAM调制和解调-图中缺少一行

Question

我试图对16 QAM进行调制和解调，然后对理论和仿真误码率进行比较。

我是，而不是，在图中得到了simulation-line。

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我不明白我的代码有什么问题。有人能帮我吗？

以下是代码：

M=16;
SNR_db = [0 2 4 6 8 10 12];
x = randi([0,M-1],1000,1);
hmod = modem.qammod(16);
hdemod = modem.qamdemod(hmod,'SymbolOrder', 'Gray');
tx = zeros(1,1000);
for n=1:1000
    tx(n) = modulate(hmod, x(n));
end
rx = zeros(1,1000);
rx_demod = zeros(1,1000);
for j = 1:7
    err = zeros(1,7);
    err_t = zeros(1,7);
    for n = 1:1000
        rx(n) = awgn(tx(n), SNR_db(j));
        rx_demod(n) = demodulate(hdemod, rx(n));

        if(rx_demod(n)~=x(n))
            err(j) = err(j)+1;
        end
    end
    % err_t = err_t + err;
end
theoryBer = 3/2*erfc(sqrt(0.1*(10.^(SNR_db/10))));
figure
semilogy(SNR_db,theoryBer,'-',SNR_db, err, '^-');
grid on
legend('theory', 'simulation');
xlabel('Es/No, dB')
ylabel('Symbol Error Rate')
title('Symbol error probability curve for 16-QAM modulation') 

Answer 1

rate.m

这可以手动实现您想要的功能，而不需要任何工具箱功能(即高级调制器和解调器)。

你也可以试试

编辑commdoc_mod

制作该文件的副本，您应该能够让它用一个简单的循环来做您想做的事情。

编辑

下面是对该文件的修改，这些修改给出了模拟的EbNo曲线，而不是符号错误率曲线。对于任何实际的目的都是足够好的。

M = 16;                     % Size of signal constellation
k = log2(M);                % Number of bits per symbol
n = 3e4;                    % Number of bits to process
nSyms = n/k;                % Number of symbols

hMod = modem.qammod(M);         % Create a 16-QAM modulator
hMod.InputType = 'Bit';         % Accept bits as inputs
hMod.SymbolOrder = 'Gray';         % Accept bits as inputs
hDemod = modem.qamdemod(hMod);  % Create a 16-QAM based on the modulator

x = randi([0 1],n,1); % Random binary data stream
tx = modulate(hMod,x);

EbNo = 0:10; % In dB
SNR = EbNo + 10*log10(k);

rx = zeros(nSyms,length(SNR));
bit_error_rate = zeros(length(SNR),1);
for i=1:length(SNR)
    rx(:,i) = awgn(tx,SNR(i),'measured');
end
rx_demod = demodulate(hDemod,rx);
for i=1:length(SNR)
    [~,bit_error_rate(i)] = biterr(x,rx_demod(:,i));
end

theoryBer = 3/(2*k)*erfc(sqrt(0.1*k*(10.^(EbNo/10))));
figure;
semilogy(EbNo,theoryBer,'-',EbNo, bit_error_rate, '^-');
grid on;
legend('theory', 'simulation');
xlabel('Eb/No, dB');
ylabel('Bit Error Rate');
title('Bit error probability curve for 16-QAM modulation');

Answer 2

在您的代码中，您混淆了符号错误概率和位错误概率。此外，err = zeros(1,7);被放错了位置。

更正后：

M=16;
SNR_db = 0:2:12;
N=1000;
x = randi([0,M-1],N,1);
k = log2(M); % bits per symbol

tx = qammod(x, M,'Gray');
err = zeros(1,7);
for j = 1:numel(SNR_db)
    rx = awgn(tx, SNR_db(j),'measured');
    rx_demod = qamdemod( rx, M, 'Gray' );
    [~,err(j)] = biterr(x,rx_demod);
end

theorySER = 3/2*erfc(sqrt(0.1*(10.^(SNR_db/10))));

figure
semilogy(SNR_db,theorySER,'-',SNR_db, err*k, '^-');
grid on
legend('theory', 'simulation');
xlabel('Es/No, dB')
ylabel('Symbol Error Rate')
title('Symbol Error Probability curve for 16-QAM modulation')

得到的图表是：

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