exp(-b/x) fit, inf problem when fitting

3 vues (au cours des 30 derniers jours)
Basil Eldeeb
Basil Eldeeb le 25 Fév 2021
Commenté : Basil Eldeeb le 2 Mar 2021
I am trying to fit some data with the model: exp(-b/x); When x goes to zero, y should go to zero as well since anything power negative infinity is zero. However Matlab sees the infinity and terminates everything. Here is my code:
vv=data(:,1);
ii=data(:,2);
g = fittype('exp(-b/x)');
f0 = fit(vv,ii,g);
xx = linspace(-1,1);
plot(vv,ii,'o',xx,f0(xx),'r-');
grid('on')
  5 commentaires
Basil Eldeeb
Basil Eldeeb le 26 Fév 2021
supposedly when the exponent is quite small it will behave linearly to a first order. The model is more complex, actually, I am just facing a problem with the exp(-b/x) term. I want matlab to evaluate exp(-Inf) without giving error. The other answer shows promise. However it gave an error, you can see it in my response
Mathieu NOE
Mathieu NOE le 26 Fév 2021
not sure it's really a good model...
data = readmatrix('data.txt');
x = data(:,1);
y = data(:,2);
% exponential fit method
% model : y = exp(-b/x)
f = @(b,x) exp(b./x);
obj_fun = @(params) norm(f(params(1), x)-y);
sol = fminsearch(obj_fun, -0.1);
b_sol = sol(1)
y_fit = f(b_sol, x);
figure
plot(x,y,'r',x,y_fit,'-.k');
legend('data','exp fit');

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Réponse acceptée

Matt J
Matt J le 25 Fév 2021
Modifié(e) : Matt J le 25 Fév 2021
When x goes to zero, y should go to zero as well since anything power negative infinity is zero.
Only if b>=0.
g = fittype('exp(-b/x)', 'options', fitoptions('Lower',0) );
  5 commentaires
Matt J
Matt J le 1 Mar 2021
Modifié(e) : Matt J le 1 Mar 2021
You should divide the data into two sets and fit each one separately, e.g.,
pos=vv>0; neg=~pos;
fpos = fit(vv(pos),ii(pos),gpos,options);
fneg = fit(vv(neg),ii(neg),gneg,options);
Basil Eldeeb
Basil Eldeeb le 2 Mar 2021
I appreciate it, thank you for the help!.

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