Curve fitting with custom function
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I am trying to do a curve fitting on some experimental data with a custom function. I am more specifically trying to bring closer my function to the data. The function is the following one:
Every parameters of the function, except the shear rate () , can vary in order to fit best the data.
Here is a graph with the data and the function plotted with initial values:
The initial values are the following ones:
I already tried to use the curve fitting toolbox but i wasn't able to keep the look of the function.
Does anyone know how i can do that ?
Thank you for the help !
Star Strider on 17 Nov 2021
It would help to have the data.
I would use the fitnlm function for this —
shearRate = logspace(-3, 4, 100); % Create Data
muv = 0.5-tanh(shearRate)*0.01 + randn(size(shearRate))*0.01; % Create Data
shearRate = shearRate(:);
muv = muv(:);
% muInf=0.0035; %[Pa.s]
% mu0=0.108; %[Pa.s]
B0 = [0.0035; 0.108; 8.2; 0.3; 0.64];
viscCar = @(muInf,mu0,lambda,a,n,shearRate) muInf+(mu0-muInf)./(1+(lambda.*shearRate).^a).^((n-1)/a);
viscCarfcn = @(b,shearRate) viscCar(b(1),b(2),b(3),b(4),b(5),shearRate);
mumdl = fitnlm(shearRate,muv, viscCarfcn, B0)
Beta = mumdl.Coefficients.Estimate
plot(shearRate, muv, '.b')
This works, and would actually make sense with the actual data.
More Answers (1)
Alex Sha on 18 Nov 2021
It is hard to get stable and unique result for Da125's problem, especially for parameters of "lambda" and "n". refer to the result below:
Root of Mean Square Error (RMSE): 0.00032178294087402
Sum of Squared Residual: 2.89923930905092E-6
Correlation Coef. (R): 0.998376698597668
Parameter Best Estimate