Antipodal and Unipolar Analytical Curve

Question

James Manns 2024-2-25

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2086708-antipodal-and-unipolar-analytical-curve

编辑： David Goodmanson 2024-2-25

I can't seem to replicate the attached figure. The output from the code below is close but not matching along the curves. I've tried modifying the EB/No but can't quite see where the error is. Can you look at the code below and let me know what I'm doing wrong?

% Parameters

EbN0_dB = -1:0.1:15; % Eb/No range in dB

EbN0 = 10.^(EbN0_dB/10); % Eb/No in linear scale

Pb_antipodal_analytical = erfc(sqrt(2 * EbN0)); % Analytical Pb for antipodal

Pb_unipolar_analytical = erfc(sqrt(EbN0)); % Analytical Pb for unipolar

% Plot analytical results

figure;

semilogy(EbN0_dB, Pb_antipodal_analytical, 'b', 'LineWidth', 2);

hold on;

semilogy(EbN0_dB, Pb_unipolar_analytical, 'r', 'LineWidth', 2);

grid on;

xlabel('Eb/No (dB)');

ylabel('Bit Error Probability');

title('Analytical Bit Error Probability vs. Eb/No');

legend('Antipodal', 'Unipolar (Orthogonal)');

axis([-1, 15, 1e-7, 1]);

% Simulation parameters

numBits = 1e6; % Number of bits

numEbN0 = length(EbN0_dB);

Pb_antipodal_simulated = zeros(1, numEbN0);

Pb_unipolar_simulated = zeros(1, numEbN0);

% Simulation loop

for i = 1:numEbN0

% Generate random bits

txBits = randi([0, 1], 1, numBits);

% Antipodal modulation

antipodal_symbols = 2 * txBits - 1;

% Add noise

noise = randn(1, numBits);

received_antipodal = antipodal_symbols + sqrt(0.5/EbN0(i)) * noise;

% Decision

rxBits_antipodal = received_antipodal >= 0;

% Count errors

num_errors_antipodal = sum(rxBits_antipodal ~= txBits);

Pb_antipodal_simulated(i) = num_errors_antipodal / numBits;

% Unipolar modulation

unipolar_symbols = txBits;

% Add noise

received_unipolar = unipolar_symbols + sqrt(1/EbN0(i)) * noise;

% Decision

rxBits_unipolar = received_unipolar >= 0.5;

% Count errors

num_errors_unipolar = sum(rxBits_unipolar ~= txBits);

Pb_unipolar_simulated(i) = num_errors_unipolar / numBits;

end

% Plot simulation results

figure;

semilogy(EbN0_dB, Pb_antipodal_analytical, 'b', 'LineWidth', 2);

hold on;

semilogy(EbN0_dB, Pb_antipodal_simulated, 'b--', 'LineWidth', 2);

semilogy(EbN0_dB, Pb_unipolar_analytical, 'r', 'LineWidth', 2);

semilogy(EbN0_dB, Pb_unipolar_simulated, 'r--', 'LineWidth', 2);

grid on;

xlabel('Eb/No (dB)');

ylabel('Bit Error Probability');

title('Analytical and Simulated Bit Error Probability vs. Eb/No');

legend('Antipodal (Analytical)', 'Antipodal (Simulated)', 'Unipolar (Analytical)', 'Unipolar (Simulated)');

axis([-1, 15, 1e-7, 1]);

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John D'Errico 2024-2-25

编辑：John D'Errico 2024-2-25

Um, you probably are not going to replicate those curves exactly using a finite size simulation. Are saying you did not get the right result from the "analytical" prediction either?

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Answer 1

David Goodmanson 2024-2-25

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2086708-antipodal-and-unipolar-analytical-curve#answer_1416108

编辑：David Goodmanson 2024-2-25

在 MATLAB Online 中打开

Hello James,

For the analytical part, you get an exact match for the curves by replacing

Pb_antipodal_analytical = erfc(sqrt(2 * EbN0)); % Analytical Pb for antipodal
Pb_unipolar_analytical = erfc(sqrt(EbN0)); % Analytical Pb for unipolar

with the same thing but including some factors of 2:

Pb_antipodal_analytical = (1/2)*erfc(sqrt(EbN0)); % Analytical Pb for antipodal
Pb_unipolar_analytical = (1/2)*erfc(sqrt(EbN0/2)); % Analytical Pb for unipolar

Your unipolar simulation matches the analytic curve quite well although there are still some issues with the anitipodal simulation. (For the simulated line types I used 'b.' and 'r.' since the double dash does not appear to be helping).