Curve fitting to data using fit
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I have data x sampled at times t. I would like to fit my function to this data. Below is a code.
clear; close all
x=[100; 85.4019292604501; 77.9310344827586; 79.3365583966828; 70.3524533;
13.213644524237; 24.5654917953199; 12.6526340272125;
9.71822886716503; 9.99113213124446; 10.525];
t=[0; 24; 24; 24; 24; 48; 48; 48; 72; 72; 72;];
mdl=fittype('A*exp((-C.*(1-exp(-lambda.*t))/lambda)-(D*(exp(-lambda.*t)-1+lambda.*t)/lambda^2))','indep','t');
fittedmdl = fit(t,x,mdl,'start',[0.1 0.1 0.1 0.1])
plot(fittedmdl,'k.')
hold on
plot(t,x,'.m', MarkerSize=20)
And I obtain the following figure:
I am not impressed with the fitting. Can someone please check where I could be going wrong. Thanks in anticipation.
9 个评论
Ghazwan
2022-10-11
Is there a fit quality you are looking for?
That is not a lot a data points to fit a rather complex equation to. Can you confirm that the equation as coded is correct?
syms A C D lambda t
A*exp((-C.*(1-exp(-lambda.*t))/lambda)-(D*(exp(-lambda.*t)-1+lambda.*t)/lambda^2))
Ghazwan
2022-10-11
ok. It looks like this is higher order fitting model.
Matt J
2022-10-11
Your starting guess [0.1 0.1 0.1 0.1] looks very arbitrary. Surely you do not expect lambda to have a similar value to the other parameters like A.
Editor
2022-10-11
The coefficient 'C' shoulb be negative. It should be
John D'Errico
2022-10-11
You CANNOT fit a higher order model to this data. If you do, expect garbage. Why? You have only 4 data points, even though some of them are replicates. So no more than 4 parameters can be estimated. At least if you want them to make any sense. And even at that, expect poor results.
As far as getting poor results, you need to provide good starting values for exponential models. Provide crappy or random starting values, then expect crappy/random results. I don't have time right now to look seriously at your data to decide if the model can fit that data, or what are good starting values. I'll look back in later today though.
Editor
2022-10-11
Cris LaPierre
2022-10-11
Just responding about the missing negative sign. The negative sign before C has been applied to the contents inside parentheses. So it is there.
-C(1-exp(
t)) is the same as C(exp(
t)-1)
t)) is the same as C(exp(t)-1)采纳的回答
x=[100; 85.4019292604501; 77.9310344827586; 79.3365583966828; 70.3524533;
13.213644524237; 24.5654917953199; 12.6526340272125;
9.71822886716503; 9.99113213124446; 10.525];
t=[0; 24; 24; 24; 24; 48; 48; 48; 72; 72; 72;];
mdl=fittype('A*exp((-C.*(1-exp(-lambda.*t))/lambda)-(D*(exp(-lambda.*t)-1+lambda.*t)/lambda^2))','indep','t');
fittedmdl = fit(t,x,mdl,'start',[100 0.1 0.1 0.1],'Lower',[0 0 0 0])
H=plot(fittedmdl,t,x);
H(1).MarkerSize=20; H(1).Color='m';
H(2).Color='k';H(2).LineWidth=2;
2 个评论
Editor
2022-10-11
Alex Sha
2024-12-10
If only for @Editor's 11 sets of data, the result from Matt J's code is a local solution, the best one should be:
Sum Squared Error (SSE): 222.486707072606
Root of Mean Square Error (RMSE): 4.49733968911932
Correlation Coef. (R): 0.991854679874119
R-Square: 0.983775705988192
Parameter Best Estimate
--------- -------------
a 100.656116307162
c 0.00236233265266791
d 0.00102171687180199
lambda 2.87765446399321E-10
The SSE from Matt J's is about 614.5217, much bigger than above which is 222.4867
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