fitting to a power function with an additional constant

Question

friet 2018-2-6

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/380961-fitting-to-a-power-function-with-an-additional-constant

评论： Walter Roberson 2018-2-6

Hello Matlab! I have this dataponits t and y. I want to fit those data points to a function y=a(t+c)^b, and get the parameters a,c and b.I have tried below, but not converging. Can anyone suggest me Thanks GH

close all
clear all
clc
t=[0.097
477
934
835
590
645
371
709
249
846
978
64
83
60
96
40
94
93
43
36
71
15
71
90
82
64
92
];
y=[20.96
33
77
96
14
33
24
71
90
98
51
05
37
40
49
05
54
17
82
57
26
87
52
25
08
80
61
];
loglog(t,y,'o')
F = @(x,xdata)x(1)*(x(2)+xdata).^x(3);
x0 =  [80 -0.1 130];
[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,t,y)
hold on
plot(t,F(x,t))
hold off

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Answer 1

Walter Roberson 2018-2-6

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/380961-fitting-to-a-power-function-with-an-additional-constant#answer_303619

You used an unconstrained curve fitting, which did a trial run with some values that were invalid for your function.

Your x0 is not especially close to the best parameters, which are approximately 851.271574244121 -0.604937228501513 436.783547500979

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friet 2018-2-6

在 MATLAB Online 中打开

Thanks for your kind help.

even with your suggested intial values, I cant get it work. Can you check the code.

 close all
  clear all
  clc
t=[0.097
477
934
835
590
645
371
709
249
846
978
64
83
60
96
40
94
93
43
36
71
15
71
90
82
64
92
];
y=[20.96
33
77
96
14
33
24
71
90
98
51
05
37
40
49
05
54
17
82
57
26
87
52
25
08
80
61
];
loglog(t,y,'o')
F = @(x,xdata)x(1)*(x(2)+xdata).^x(3);
x0 =  [851.271574244121 -0.604937228501513 436.783547500979];
[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,t,y)
hold on
plot(t,F(x,t))
hold off

Walter Roberson 2018-2-6

You have x(2)+xdata, where xdata is t. min(t) is 0.097 so if your trial x(2) is less than -0.097 then you would get a negative value for x(2)-min(t), and you would then try to raise that negative to a floating point power. You thus need a constraint that x(2) >-min(t) to avoid that negative raised to a floating point.

I will need to investigate further as to why I did not encounter this problem in my own tests.

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