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

2021 2 25
1 个回答
回答已采纳
更新时间：2021 3 2
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0 个投票

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 个评论
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Mathieu NOE
Mathieu NOE 2021-2-25
hello
could you share the data as well ?
tx
Basil Eldeeb
Basil Eldeeb 2021-2-26
it's unable to upload, the data is visibly linear, However my model is exponential:
0.000000 -4.333102E-5
0.050000 0.030123
0.100000 0.061024
0.150000 0.092435
0.200000 0.123947
0.250000 0.155877
0.300000 0.187916
0.350000 0.219968
0.400000 0.251753
0.450000 0.283991
0.500000 0.316017
0.550000 0.347719
0.600000 0.379196
0.650000 0.410459
0.700000 0.441492
0.750000 0.472394
0.800000 0.503174
0.850000 0.533827
0.900000 0.564569
0.950000 0.595374
1.000000 0.626224
Mathieu NOE
Mathieu NOE 2021-2-26
yes , your data show a very linear trend
how can you expect to fit an exponential model to these data ? it will never work
Basil Eldeeb
Basil Eldeeb 2021-2-26
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 2021-2-26
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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Matt J
Matt J 2021-2-25
编辑：Matt J 2021-2-25

0 个投票

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) );

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Basil Eldeeb
Basil Eldeeb 2021-2-26
编辑：Basil Eldeeb 2021-2-26
Hello, Thanks for the response. I take it matlab applies limit evaluation on its own. However, the code is giving me this error:
Error using curvefit.basefitoptions/set
The name 'Lower' is not an accessible
property for an instance of class
'basefitoptions'.
now I have two questions:
Why am I receiving this error?
How do I specify which parameter is to be limited if I have more than one?
Matt J
Matt J 2021-2-26
Ran in:
Ran in:
I guess they wanted a different syntax...
data=[0.000000 -4.333102E-5
0.050000 0.030123
0.100000 0.061024
0.150000 0.092435
0.200000 0.123947
0.250000 0.155877
0.300000 0.187916
0.350000 0.219968
0.400000 0.251753
0.450000 0.283991
0.500000 0.316017
0.550000 0.347719
0.600000 0.379196
0.650000 0.410459
0.700000 0.441492
0.750000 0.472394
0.800000 0.503174
0.850000 0.533827
0.900000 0.564569
0.950000 0.595374
1.000000 0.626224];
vv=data(:,1);
ii=data(:,2);
g = fittype('exp(-b/x)');
options=fitoptions('exp(-b/x)','Lower',0);
f0 = fit(vv,ii,g,options);
Warning: Start point not provided, choosing random start point.
plot(f0,vv,ii)
grid('on')
Basil Eldeeb
Basil Eldeeb 2021-3-1
Thank you for the response. I wanted to ask how to specify multiple options for multiple parameters, for example:
g = fittype('a*x^2*exp(-b/x)');
One more question; For my model the exp(-b/x) is only valid for positive x. So is there a way where I can apply two fittings simultaneously to the same data? i.e. one including exp(-b/x) as a multiple for +ve x and excludes it for -ve x. Thanks in advance!
Matt J
Matt J 2021-3-1
编辑：Matt J 2021-3-1
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 2021-3-2
I appreciate it, thank you for the help!.

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