Statistical signal processing - computer operation simulation - receiver composition diversity system calculation error probability - matlab simulation - with code

1. Calculation of Misjudgment Probability in Receiver Composition Diversity System

1. topic

The same incident electromagnetic wave is received by M receivers, and the signal received by each receiver is:

{ ni ( t )} is a Gaussian white noise with a mean value of 0 and a power spectral density of N0/2 .

Ai , Bi are Rayleigh distributed random variables

It is known that P ( Hi )=1/2, i =0,1. 1/ f 1≪ T , 1/ f 2< T , | ω 1- ω 0| is very large. Define SNR

SNR=

The above M receivers constitute a diversity system, and the minimum error probability criterion is used to calculate M = 1, 2, 4, 6, 8, 16. The range of SNR is 0-60 decibels, and the probability of misjudgment when the RF pulse of each channel is 1, 5, 10.

2. Simulation principle

The average probability of error for a binary hypothesis test is

The minimum error probability criterion is to find the most suitable decision threshold th ', so that the average error probability of wrong judgment P e can be minimized. Let the derivative of P e with respect to th ' be 0, then th ' that makes P e reach the minimum value can be obtained, that is

Obviously, ᵆ5<ᵆ1ℎ′ judges ᵃb0 to be true, and ᵆ5≥ᵆ1ℎ' judges ᵃb1 to be true. So the decision rule under the minimum error probability criterion is

For a multi-receiver system, the internal noise of each receiver is independent and identically distributed. Judgment is made according to the signal received by the M receivers to determine which known signal it is. x (t)=[x1(t),⋯  ,xM(t)]T , the likelihood ratio is

The derived decision criterion is as follows

3. Simulation of Diversity System Composed of Receivers

It can be seen from the above figure that as the number of M increases, the bit error rate tends to decrease, and as the signal-to-noise ratio increases, the bit error rate gradually decreases to 0. This is consistent with the reality, indicating that our simulation is more reasonable.

M = [1,2,4,6,8,16]; 
SNR = 0:60; 
Pf = zeros(length(M),length(SNR)); 

for i = 1:length(M) 
    for j = 1:length(SNR) 
        % 计算概率密度函数
        pdf_A = (A0./(A0.^2)).*exp(-A0.^2./(2.*(SNR(i).*N0).^2));
        pdf_B = (A0./(A0.^2)).*exp(-A0.^2./(2.*(SNR(i).*N0).^2));
        Ax = sqrt((-2*A0)*log(1-rand(1)));
        phix = 2*pi*rand(1);
        x = Ax*sin(w0*i+phix)+randn(1,length(i));
        f0(i) = ((x*sin(w0*i)')*dt)^2+((x*cos(w0*i)')*dt)^2;
        f1(i) = ((x*sin(w1*i)')*dt)^2+((x*cos(w1*i)')*dt)^2;
        % 计算错误概率
        Pf(i,j) =  1 - (1-qfunc(sqrt(2*10^(SNR(j)/10)))).^M(i); 
    end 
end 

plot(SNR,Pf); 
xlabel('SNR(dB)'); 
ylabel('Pf'); 
legend('M=1','M=2','M=4','M=6','M=8','M=16');

(code is not exactly correct)