Complex value computed by model function, fitting cannot continue. Try using or tightening upper and lower bounds on coefficients.

Hello, I really need some help on data fitting. I would like to fit my data custom equation:a+b*exp((-(x/c))^d. I try to use upper and lower bounds on coefficients but it does not work.If anyone can tell me what to do to resolve this I would greatly appreciate it.

[Torsten]

The expression a+b*exp((-(x/c))^d cannot become complex-valued. My guess is that you use a different expression in your code.

[Matt J]

We do not know what you tried (no code is shown), so we cannot guess what went wrong. I will point out though that the model is over-parametrized. It is enough to use 3-parameters y=a+b*exp(-c*x). No custom equation is needed for this - it falls within the 'exp2' fitType of the fit() command.

[Torsten]

@Matt J

Because of the error message and the overfitting, I suspect

a+b*exp(-(x/c)^d)

instead of

a+b*exp((-(x/c))^d

is meant.

[Daniel Jiao]

@Torsten Thanks lot, I'll check it. This equation is given by an article, so it's a theoretical formula, but yeah, I'll check whether it's correct, thanks so much!

Answer 1

[Matt J]

fminspleas from this FEX download,

https://www.mathworks.com/matlabcentral/fileexchange/10093-fminspleas

is helpful for these kinds of models.

[x,y]=readvars('200ms_SD.xlsx');

flist={1,@(c,x) exp(-c*x)};

[c,ab]=fminspleas(flist,+1,x,y);
a=ab(1);
b=ab(2);

f=@(x)a+b*exp(-c*x);

xf=linspace(min(x),max(x));
plot(x,y,'.c',xf,f(xf)); axis padded; legend Data Fit

[Matt J]

Here is the same approach as applied to y=a+b*exp(-(x/c)^d)

[x,y]=readvars('200ms_SD.xlsx');

flist={1,@(p,x) exp(-(p(1)*x).^p(2))};

[p,ab]=fminspleas(flist,[1;1],x,y,[0;0]);
a=ab(1);
b=ab(2);
c=p(1);
d=p(2);

f=@(x)a+b*exp(-(c*x).^d);

xf=linspace(min(x),max(x));
plot(x,y,'.c',xf,f(xf)); axis padded; legend Data Fit

[Daniel Jiao]

Thanks a lot for this answer, it really give me some ideas and confidence, as the comment you mentioned above, I just try to use fit() and fittype(), like:

ft = fittype( 'a+b*exp((-(x/c))^d)', 'independent', 'x', 'dependent', 'y' );

fitobject = fit(x,y,ft,'lower',[-1,0,0.002,0],'upper',[0,3,0.005,1]);

and I also use the curve fitting toolbox. Just some basic way, since it was first time I try to do curve fitting.

Anyway that's really helpful! Thanks so much for that, and I'll try it!

[Matt J]

You're welcome, but when you've decided upon a solution, please Accept-click the appropriate answer.

Answer 2

[Mathieu NOE]

hello

a very basic code using only fminsearch (no toolbox required )

I preferred to smooth a bit your data (otherwise it looks more like a cloud) but it's not absolutely needed - but you end up with other parameters after the fit

hope it helps !

data = readmatrix('200ms_SD.xlsx');

x =  data(:,1);
y =  data(:,2);

[x,ia,ic] = unique(x);
y = y(ia);
ys = smoothdata(y,'gaussian',100);

% curve fit using fminsearch
% model a+b*exp((-(x/c))^d
f = @(a,b,c,d,x) a + b.*exp(-(x/c).^d);
obj_fun = @(params) norm(f(params(1), params(2), params(3), params(4),x)-ys);

sol = fminsearch(obj_fun, [ys(end),(max(ys)-ys(end)),1,1]);

 
Exiting: Maximum number of function evaluations has been exceeded
         - increase MaxFunEvals option.
         Current function value: 1.159572 

a_sol = sol(1)

a_sol = 0.0903

b_sol = sol(2)

b_sol = 0.5209

c_sol = sol(3)

c_sol = 0.0028

d_sol = sol(4)

d_sol = 0.1143

y_fit = f(a_sol, b_sol, c_sol, d_sol, x);
R2 = my_R2_coeff(ys,y_fit); % correlation coefficient

plot(x,y, 'k.',x,ys,'r',x, y_fit,'b-')
title([' Fit / R² = ' num2str(R2) ], 'FontSize', 15)
legend('raw data','smoothed','fit');

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function R2 = my_R2_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 (R squared) is
    R2 = 1 - sum_of_squares_of_residuals/sum_of_squares;
    
end

[Daniel Jiao]

Hello,thanks a lot for the answer, it's really helpful! I'll try to read and try the code you offered! My original code is much more basic, since I'm just a beginner.

Thanks again that I really got some knowledge from you!