How to enter a sigmoidal equation for a curve fit with lsqfitting

he tratado de ingresar esta ecuación sigmoidal para un ajunste de curva lsqcurvefit con esta ecuación

y=1/(1+exp(4a(b-X)+2))
fun=@(x,xdata)1./(1+exp(4.*x(1).*(x(2) - xdata)+2)) 
x=lsqcurvefit(fun,x0,xdata,ydata)

he podido leer los datos aunque no he podido hacer los ajustes, al parecer reconoce al exponecial pero no los ajuste con X(1) y x(2).

alquien me podra ayudar con este tema

[Matt J]

"I have tried to enter this sigmoidal equation for a curve fitting with lsqcurvefit with this equation:

y=1/(1+exp(4a(b-X)+2))
fun=@(x,xdata)1./(1+exp(4.*x(1).*(x(2) - xdata)+2)) 
x=lsqcurvefit(fun,x0,xdata,ydata)

I have been able to read the data, although I have not been able to make the adjustments. Apparently, it recognizes the exponential but does not adjust with X(1) and X(2)."

[Matt J]

You haven't provided the data we would need to run the code. Most likely, though, the problem is your x0 has been chosen randomly (and therefore, poorly).

[Wilfredo]

Gracias Matt J, x0= [0 5900] puede ser que aqui este el error entonces.

Answer 1

[Image Analyst]

See my attached demo where I fit data to the rate equation (a sigmoid type of formula).

Adapt as needed, like using a different sigmoid formula. If you have problems, write back and attach your data and m-file.

[Wilfredo]

Gracias por tu pronta respuesta, estare analizando la sugerencia, adjunto el ejemplo en cuestión.

Thank you for your prompt response, I will be analyzing the suggestion, I attach the example in question.

[Torsten]

@Wilfredo

Could you please use the google translator to post your questions and comments in English ? There are not so many people speaking Spanish in the forum.

¿Podría utilizar el traductor de Google para publicar sus preguntas y comentarios en inglés? No hay tanta gente hablando español en el foro.

[Wilfredo]

Thanks for the remark.

[Image Analyst]

@Wilfredo do you have some ballpark values for a and b that might make the curve look approximately right?

X = ...
    [0 270 450 720 990 1170 1440 1710 1890 2520 3150 3690 4320 4950 5850 6750 7650 8550 10350 12150];
Y = ...
    [0.00115 0.00115 0.00115 0.00115 0.00115 0.00115 0.00115 0.00115 0.00115 0.00115 0.000865 0.00202 0.00317 0.00533 0.0098 0.01326 0.01571 0.01744 0.02032 0.02104];
plot(X, Y, 'b.-');
grid on;
xlabel('X');
ylabel('Y');

% Fit to y = 1 / (1 + exp(4*a*(b-X) + 2))

I don't see how it can be done if your first number is 1 -- it would have to be around 0.021 or something. Then what would a and b be approximately?

[Image Analyst]

OK, I adjusted the formula slightly to make some of your constants variables and now the fit is pretty good.

% Demo by Image Analyst.
clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format short g;
format compact;
fontSize = 15;
markerSize = 6;
X = ...
    [0 270 450 720 990 1170 1440 1710 1890 2520 3150 3690 4320 4950 5850 6750 7650 8550 10350 12150];
Y = ...
    [0.00115 0.00115 0.00115 0.00115 0.00115 0.00115 0.00115 0.00115 0.00115 0.00115 0.000865 0.00202 0.00317 0.00533 0.0098 0.01326 0.01571 0.01744 0.02032 0.02104];
plot(X, Y, 'b.-', 'MarkerSize', 30);
grid on;
xlabel('X', 'FontSize', fontSize);
ylabel('Y', 'FontSize', fontSize);
title('Data to fit exponential to', 'FontSize', fontSize);
% Maximize the figure.
hFig2.WindowState = 'maximized';
drawnow;
% Convert X and Y into a table, which is the form fitnlm() likes the input data to be in.  Use (:) to reshape data into column vectors.
tableXY = table(X(:), Y(:));
%-----------------------------------------------------------------------------------------------------------------------------------
% y = c1 / (c2 + exp(c3*(c4 + c5*X))))
% Make a model.  B are the coefficients
modelFunction = @(b, tbl) b(1) ./ (b(2) + exp(b(3) .* (b(4) - b(5) .* tbl(:, 1))));
beta0 = [.021, 1, .0004, 11000, 2]; % Guess values to start with.  Just make your best guess.
% Once you get close and see what the coefficients are, you can set beta0 to those values and
% Make another run to get better coefficients.
% Now the next line is where the actual model computation is done.
model1 = fitnlm(tableXY, modelFunction, beta0);

Warning: The Jacobian at the solution is ill-conditioned, and some model parameters may not be estimated well (they are not identifiable). Use caution in making predictions.

% Now the model creation is done and the coefficients have been determined.
% YAY!!!!
coefficients1 = model1.Coefficients{:, 'Estimate'}

coefficients1 = 5×1

1.0e+00 * 0.020835 0.98745 0.00041504 11474 1.8483

% Create smoothed/regressed data using the model:
yFitted = coefficients1(1) ./ (coefficients1(2) + exp(coefficients1(3) .* (coefficients1(4) - coefficients1(5) .* X)));
% Now we're done and we can plot the smooth model as a red line going through the noisy blue markers.
hold on;
plot(X, yFitted, 'r-', 'LineWidth', 2);
grid on;
title('Fitting Noisy Data to the Equation', 'FontSize', fontSize);

drawnow;

Is there a theoretical basis for why you chose this particular sigmoid formula? Like it models well the physics of your experiment, or is it just one that you heard of?

[Alex Sha]

The effect of fitting function "y = c1/(c2+exp(c3*(c4+c5*X))))" is exact same as "y=c1/(c2+exp(c3*x))", the former geives multi-solution

[Wilfredo]

Hello friend, thank you very much for your help, it helps me a lot, for the values I expect are for A=0.000186865 and B=3524.

[Wilfredo]

throws up the following error.

Warning: The Jacobian at the solution is ill-conditioned, and some model parameters may not be estimated well (they are not

identifiable). Use caution in making predictions.