Antipodal and Unipolar Analytical Curve

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James Manns
James Manns 2024-2-25
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]);
  1 个评论
John D'Errico
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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采纳的回答

David Goodmanson
David Goodmanson 2024-2-25
编辑：David Goodmanson 2024-2-25
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).

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