Error in demodulation QAM-16 signal

Question

Kazantcev 2024-6-7

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2126216-error-in-demodulation-qam-16-signal

评论： Voss 2024-6-7

% This code, after demodulation, produces a sequence of 4 values, although the number of values ​​should be equal to the modulation factor M. What could be the problem?
clc
close all
clear all
M = 16;
L=10000;
T=10;
W=43; %SNR
bits1 = randi([0 M-1], 1, L);
bits2 = randi([0 3], 1, L/T);
mod1 = qammod(bits1, M,'UnitAveragePower', true);
% scatterplot(mod1)
mod2 = qammod(bits2, 4, 'UnitAveragePower',true);
% scatterplot(mod2)
for i = 1:length(bits2)
    sym(i,:)=[mod1((i-1)*T+1:i*T) mod2(i)];
end
f=0;
for k=1:L/T
    for j=1:T+1
        f=f+1;
        symbols(f)=sym(k,j);
    end
end
% scatterplot(symbols)
% symbols = awgn(symbols,W,"measured");
pnoise = comm.PhaseNoise('Level',[-70 -104 -110],'FrequencyOffset',[1e4 1e5 2e5], 'SampleRate', 28e6);
symbols2 = pnoise([zeros(1e5,1);symbols']);
% symbols2 = pnoise(symbols);
fc = 1e6; % Carrier frequency in Hz
fs = 28e6; % Sample rate in Hz.
phNzLevel = [-70 -104 -110]; % in dBc/Hz
phNzFreqOff = [1e4 1e5 2e5]; % in Hz
Nspf = 6e6; % Number of Samples per frame
freqSpan = 400e3; % in Hz, for spectrum computation
sinewave = dsp.SineWave( ...
    Amplitude=1, ...
    Frequency=fc, ...
    SampleRate=fs, ...
    SamplesPerFrame=Nspf, ...
    ComplexOutput=true);
pnoise = comm.PhaseNoise( ...
    Level=phNzLevel, ...
    FrequencyOffset=phNzFreqOff, ...
    SampleRate=fs);
sascopeRBW100 = spectrumAnalyzer( ...
    SampleRate=fs, ...
    Method="welch", ...
    FrequencySpan="Span and center frequency", ...
    CenterFrequency=fc, ...
    Span=freqSpan, ...
    RBWSource="Property", ...
    RBW=100, ...
    SpectrumType="Power density", ...
    SpectralAverages=10, ...
    SpectrumUnits="dBW", ...
    YLimits=[-150 10], ...
    Title="Resolution Bandwidth 100 Hz", ...
    ChannelNames={'signal','signal with phase noise'}, ...
    Position=[79 147 605 374]);
x = sinewave();
y = pnoise(x);
sascopeRBW100(x,y)
symbols2 = symbols2(1e5+1:end);
prim(1,:)=symbols2(1:T);
for i = 1:length(bits2)-1
    prim(i+1,:)=symbols2(i*T+i+1:i*T+i+T);
end
for b = 1:length(mod2)
    qam4(b)=symbols2((T+1)*b);
end
h=0;
for u=1:L/T
    for l=1:T
        h=h+1;
        priem(h)=prim(u,l);
    end
end
for g = 1:L/T
    phase_error(g) = angle(qam4(g) / mod2(g));
    compensated4(g) = qam4(g) .* exp(-1i * phase_error(g));
end
for v = 1:L/T
    phase_errorM((v-1)*T+1:v*T)=phase_error(v);
end
for f = 1:L
    compensatedM(f) = priem(f) .* exp(-1i*phase_errorM(f));
end
demod=qamdemod(compensatedM, M, 'bin','OutputType','bit');
% scatterplot(qam4)
% scatterplot(symbols2)
% scatterplot(compensatedM)
[number,ratio]=biterr(bits1,demod);
evm = lteEVM(demod,bits1);
figure('Position',[200 200 1080 540])
subplot(1,2,1)
scatter(real(symbols2),imag(symbols2),300,".")
subplot(1,2,2)
scatter(real(compensatedM),imag(compensatedM),300,".")

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Answer 1

Voss 2024-6-7

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2126216-error-in-demodulation-qam-16-signal#answer_1468921

"the number of values should be equal to the modulation factor M"

For a modulation order M, the number of bits per symbol is k=log2(M).

Put another way, using k bits per symbol produces M=2^k distinct symbols of length k bits.

So M=16 corresponds to k=log2(16)=4 bits per symbol.

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Kazantcev 2024-6-7

I can't make BER=0, I think, something is wrong with demodulation, in the demodulated sequence there are only numbers 5, 7, 13, 15. Now BER is equal to about half of all bit values, so what can I do to resolve this and make BER=0?