How to implement Gaussian Approximation
4 次查看(过去 30 天)
显示 更早的评论
I'm having some trouble trying to implement the following function:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/164142/image.png)
To express the inverse function of rho I have used a numerical approximation for the second equality of the piecewise defined function, as it does not have an analytical solution.
I have the impression that something must have been done wrongly, as the result should be more 'positive' (greater than 0) for each iteration.
I will leave here my code:
l = [0 0.3078 0.27287 0 0 0 0.41933];
r = [0 0 0 0 0 0.4 0.6];
sigma = 0.8747;
mu0 = 2/sigma;
iterations = 50;
% Density evolution algorithm depiction for finding the treshold of irregular LDPC codes
syms x;
l_idle = zeros(1,length(l));
r_idle = zeros(1,length(r));
Q_1 = exp(-0.4527*x^0.86 + 0.0218);
Q_2 = sqrt(pi/x)*exp(-x/4)*(1-20/(7*x));
mu = zeros(1,iterations+1);
for k=2:1:iterations+1
for i = 1:length(l_idle)
if ((mu0 + (i-1)*mu(k-1)) < 10)
l_idle(i) = subs(Q_1,x,(mu0 + (i-1)*mu(k-1)));
else
l_idle(i) = subs(Q_2,x,(mu0 + (i-1)*mu(k-1)));
end
end
lambda = l*transpose(l_idle);
for j = 1:length(r_idle)
b = 1-(1-lambda)^(j-1);
g = subs(Q_1,x,b);
if or(subs(Q_1,x,10) < g, subs(Q_1,x,0) >= g)
r_idle(j) = (1/0.4527*(0.0218-log(g)))^(1/0.86);
else
xx = linspace(0, subs(Q_2,x,10), 10000);
yy = f(xx);
r_idle(j) = interp1(yy,xx,b);
end
end
mu(k) = r*transpose(r_idle);
end
If anyone has any question, I will be pleased to answer you.
Thank you in advance, and may you have a nice week!
0 个评论
回答(0 个)
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!