Unable to data fitting with Lsqnonlin and estimate the coefficient. Need help fitting the data.
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Anand Ra
el 26 de Jun. de 2021
Comentada: Matt J
el 2 de Jul. de 2021
I am unable to get a good fit. I have to keep guessing the initial assumption d0 to manually get the better fit. Can I please get some help on what is wrog with my code. Btw, I am trying to determine the best coefficient by fitting the y_obs data array into an arbitrary function. Should I add constrains? If so, how?Thank you.
t1 = [0:300:21600]';
y_obs = [
0
1.6216e-01
2.9813e-01
4.0805e-01
4.9338e-01
5.5928e-01
6.0894e-01
6.5506e-01
6.8876e-01
7.1773e-01
7.4284e-01
7.6433e-01
7.8448e-01
7.9912e-01
8.1484e-01
8.2950e-01
8.3760e-01
8.4707e-01
8.5767e-01
8.6299e-01
8.6862e-01
8.7433e-01
8.7879e-01
8.8736e-01
8.9152e-01
8.9903e-01
9.0343e-01
9.0984e-01
9.1344e-01
9.1615e-01
9.2046e-01
9.2313e-01
9.2640e-01
9.2992e-01
9.3134e-01
9.3242e-01
9.3616e-01
9.3986e-01
9.4201e-01
9.4145e-01
9.4434e-01
9.4252e-01
9.4131e-01
9.4249e-01
9.4283e-01
9.4395e-01
9.4355e-01
9.4690e-01
9.4919e-01
9.5255e-01
9.5626e-01
9.6282e-01
9.6283e-01
9.6672e-01
9.6439e-01
9.6887e-01
9.7099e-01
9.7529e-01
9.8034e-01
9.8302e-01
9.8494e-01
9.8828e-01
9.8769e-01
9.9130e-01
9.9108e-01
9.9422e-01
9.9561e-01
9.9737e-01
1.0002e+00
1.0034e+00
1.0044e+00
1.00329
1.0];
d0 = 1.8055e-10;
fun = @(d) ypred(d, t1) - y_obs;
best_d = lsqnonlin(fun, d0)
predicted_y = ypred(best_d, t1);
plot(t1, y_obs, '*', t1, predicted_y, '-');
legend({'observed', 'predicted'})
title('lsqnonlin data fitting')
function y_pred = ypred(d, t1)
a=0.0011;
gama = 0.01005;
L2 = zeros(14,1);
L3 = zeros(100,1);
L4 = zeros(100,1);
L5 = zeros(100,1);
S= zeros(73,1);
y_pred = zeros(73,1);
% t = 0;
L1 = ((8*gama)/((pi*(1-exp(-2*gama*a)))));
format longE
k =1;
for t = t1(:).'
for n=0:1:100
L2(n+1) = exp((((2*n + 1)^2)*-d*pi*pi*t)/(4*a*a));
L3(n+1) = (((-1)^n)*2*gama)+(((2*n+1)*pi)*exp(-2*gama*a))/(2*a);
L4(n+1)= ((2*n)+1)*((4*gama*gama)+((((2*n)+1)*pi)/(2*a))^2);
L5(n+1) = ((L2(n+1)*L3(n+1))/L4(n+1));
end
S((t/300) +1) = sum(L5);
y_pred((t/300)+1)= 1 -(L1*S((t/300) +1)); % predicted data
end
end
2 comentarios
Matt J
el 26 de Jun. de 2021
Looks like a pretty good fit to me. Also, it is easy to verify via plots that the minimum was coorectly found
load Data
d0 = 1.8055e-10;
fun = @(d) ypred(d, t1) - y_obs;
best_d = lsqnonlin(fun, d0)
pfun=@(d) norm(fun(d));
fplot(pfun,[0,5e-10])
hold on; plot(best_d*[1,1],ylim,'--rx'); hold off
Respuesta aceptada
Matt J
el 26 de Jun. de 2021
Since it's a fairly inexpensive 1D function , I think you should just be able to do a 1D grid search over a suitable interval to find a reasonable d0.
16 comentarios
Matt J
el 2 de Jul. de 2021
There are tools that can give you better control over the shape of the fit, for example,
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