To find exponent in power law equation of the form y = ax^m + b
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Matt J
2023년 1월 19일
편집: Matt J
2023년 1월 19일
fminspleas downloadable from
is especially appropriate for power law fits.
a = 0.55;
m = 1.3;
b = -0.78;
% dummy data
x = (1:25)';
y = a*x.^m + b + randn(size(x));
m=fminspleas( {@(m,x)x.^m , 1}, 2,x,y, 1.2,2.5 )
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Matt J
2023년 1월 19일
Probably similar, but with 3 unknowns fminsearch is not guaranteed to converge, so no rigorous predictions are possible.
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Mathieu NOE
2023년 1월 19일
hello
try this
may need some refinement for the initial guess for the parameters depending of your data
a = 0.55;
m = 1.3;
b = -0.78;
% dummy data
x = (1:25);
y = a*x.^m + b + randn(size(x));
% equation model y = a*x^m + b
f = @(a,m,b,x) (a*x.^m + b);
obj_fun = @(params) norm(f(params(1), params(2), params(3),x)-y);
% IC guessed
sol = fminsearch(obj_fun, rand(3,1));
a_sol = sol(1)
m_sol = sol(2)
b_sol = sol(3)
y_fit = f(a_sol, m_sol, b_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('Intensity (arb. unit)', 'FontSize', 14)
xlabel('x(nm)', '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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Matt J
2023년 1월 19일
If you have the Curve Fitting Toolbox,
a = 0.55;
m = 1.3;
b = -0.78;
% dummy data
x = (1:25)';
y = a*x.^m + b + randn(size(x));
fobj=fit(x,y,'power2','Lower',[-inf,1.2,-inf],'Upper',[+inf,2.5,+inf])
plot(fobj,x,y)
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