# How to make a good fittings

2 visualizaciones (últimos 30 días)
Youngki Yeo el 16 de Feb. de 2023
Comentada: Youngki Yeo el 16 de Feb. de 2023
x =
1.0e-08 *
0
0.0430
0.0770
0.1061
0.1309
0.1522
0.1705
0.1918
0.2123
0.2302
0.2471
0.2611
0.2743
0.2852
0.2962
0.3064
0.3170
0.3271
0.3386
0.3484
0.3609
0.3715
0.3801
0.3789
y =
1.0e-05 *
0
0.0008
0.0031
0.0063
0.0107
0.0159
0.0214
0.0263
0.0315
0.0372
0.0432
0.0498
0.0564
0.0631
0.0698
0.0769
0.0839
0.0910
0.0979
0.1054
0.1124
0.1199
0.1275
0.1347
g=fittype(@(a,x) 4/3.*a.*(10.^(-9)*10).^0.5.*x.^(3/2),'coefficient',{'a'},'independent',{'x'},'dependent',{'y'});
myfit=fit(x,y,g);
plot(myfit,x,y);
These are my codes for fitting my data with the function of a*x^2/3 finding the coefficient 'a'.
However,
the results are gone wrong. My fitting function (Orange) looks really deviated from the data points (Blue).
'a' coefficient should be about 30*10^10 value to be overlapped with each other like the below image.
Could you give me the advice to improve my fitting codes?
Thanks.
Best regards,
qwertypo.
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Mathieu NOE el 16 de Feb. de 2023
hello
sorry I don't have the curve fitting toolbox but such a simple fit can be easily done with fminsearch
the constant you are looking for is
a_sol = 3.5829e+10
clear;
x = 1.0e-08 *[0
0.0430
0.0770
0.1061
0.1309
0.1522
0.1705
0.1918
0.2123
0.2302
0.2471
0.2611
0.2743
0.2852
0.2962
0.3064
0.3170
0.3271
0.3386
0.3484
0.3609
0.3715
0.3801
0.3789];
y = 1.0e-05 *[0
0.0008
0.0031
0.0063
0.0107
0.0159
0.0214
0.0263
0.0315
0.0372
0.0432
0.0498
0.0564
0.0631
0.0698
0.0769
0.0839
0.0910
0.0979
0.1054
0.1124
0.1199
0.1275
0.1347];
const = 4/3.*(10.^(-9)*10).^0.5;
% equation model y = a*const*x^(3/2)
f = @(a,x) (a*const*x.^(3/2));
obj_fun = @(params) norm(f(params(1), x)-y);
% IC guessed
a_ic = (y(end)/(const*x(end).^(3/2)))
a_ic = 4.3315e+10
sol = fminsearch(obj_fun, a_ic);
a_sol = sol(1)
a_sol = 3.5829e+10
y_fit = f(a_sol, x);
Rsquared = my_Rsquared_coeff(y,y_fit); % correlation coefficient
figure(1)
plot(x,y,'rd',x,y_fit,'b-');
title(['Power Fit / R² = ' num2str(Rsquared) ], 'FontSize', 15)
ylabel('y', 'FontSize', 14)
xlabel('x', 'FontSize', 14)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function Rsquared = my_Rsquared_coeff(data,data_fit)
% R² correlation coefficient computation
% The total sum of squares
sum_of_squares = sum((data-mean(data)).^2);
% The sum of squares of residuals, also called the residual sum of squares:
sum_of_squares_of_residuals = sum((data-data_fit).^2);
% definition of the coefficient of correlation is
Rsquared = 1 - sum_of_squares_of_residuals/sum_of_squares;
end
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Mathieu NOE el 16 de Feb. de 2023
hello
thanks for accepting my answer !
As I said before, unfortunately I don't have (and practice) the Curve Fitting Toolbox
maybe someone else in this forum will be keen to answer this specific topic.
Youngki Yeo el 16 de Feb. de 2023
Hi, Mathieu.
Okay. I see. Thanks!

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