Why does lsqcurvefit result in complex parameters?
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I want to fit experimental impedance data (real(impedance),imag(impedance),frequency).
The written function for getting the parameters has complex formulas for description of the impedance, but real-valued parameters. Unfortunately after the lsqcurvefit I get complex-valued parameters.
(The resnorm2 has a value of apprax 1e-6, which is not so bad.)
I want only real-valued parameters. What can I do to solve this problem?
My question is related to this question: https://ch.mathworks.com/matlabcentral/answers/320390-how-to-avoid-complex-eigenvalues-of-the-matrix-in-its-non-linear-regression-lsqcurvefit (but it was not helpful enough.) I've tried different things but without success till now and would be glad to get a solution from here.
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In the code file "Fit3_exp.m" I wrote:
% Start Parameters (initial guess)
startPar = [0.45*1e-3, 1.4*1.0846e-7, 0.016*4.111, 0.7*1e-3, 11, 0.00020];
% Make the best fitting
fitPar = lsqcurvefit(@Fit3_f1, startPar, vertcat(f, f), vertcat(real(Z_exp), imag(Z_exp)),[0 0 0 0 0 0],[1e-3 1e6 1e6 1e6 1e6 1e6])
% Vector of fitted Zfit data
zFit3 = Fit3_f1(fitPar, vertcat(f, f));
% Plot Fitted Line
plot(zFit3(1:length(zFit3)/2),-zFit3(1+length(zFit3)/2:end), 'r','DisplayName','computed fitting data')
In the function file "Fit3_f1.m" I wrote:
function Z_fit3 = Fit3_f1(par,f)
file='G:\users\DA\Ueberblick EIS.xlsx';
exp_data_Freq = xlsread(file,'A4','A3:A73');
f = exp_data_Freq(:,1);
f = f(1:length(f)/1); %
w = 2*pi*f ;
R0_fit0 = par(1);
L_fit0 = par(2);
R_bat_fit0 = par(3);
R_ct_fit0 = par(4);
C_dl_fit0 = par(5);
s_w_fit0 = par(6);
L_bat_fit0 = 1i*w*L_fit0;
R_el_fit0 = (L_bat_fit0*R_bat_fit0)./(L_bat_fit0+R_bat_fit0);
Z_d_fit0 = (s_w_fit0 ./ sqrt(w)) * (1-1i);
Z_c_fit0 = -1i ./ (C_dl_fit0*w);
R_dl_fit0 = ((Z_c_fit0 .* (R_ct_fit0 + Z_d_fit0)))./(R_ct_fit0 + Z_d_fit0 + Z_c_fit0);
Z_fit0 = R0_fit0 + R_el_fit0 + R_dl_fit0;
Z_fit3 = vertcat(real(Z_fit0), imag(Z_fit0));
I get now only real-valued Parameters, however the fitting is really poor (resnorm2 = 3.5017e-04).
Are there mistakes in my formulas (e.g. with refering to the correct fit function (with indices) or is something else wrong)?
5 Comments
Torsten
on 26 Sep 2018
Fit real and imaginary part of your data to real and imaginary part of your fit function. This way, you only work with real numbers, and the estimated parameters will remain real-valued.
Torsten
on 27 Sep 2018
I don't see an obvious programming error, but you could test the fitting procedure by producing artificial data from your model, perturb these data slightly and try to reproduce the parameters you have used to generate these data.
If this works, you can be sure, that technically, your code is correct.
Then you can try "multistart" or modify your model.
Best wishes
Torsten.
hamp
on 3 Oct 2018
Torsten
on 4 Oct 2018
If you supply real-valued experimental data (real(Z_exp), imag(Z_exp)), return real-valued model data (real(Z_fit0), imag(Z_fit0)) and start with real-valued parameters, the parameters will remain real-valued throughout the fitting process.
Answers (2)
Walter Roberson
on 27 Sep 2018
0 votes
It is because you are ignoring the second input parameter.
Note: for efficiency, do not read files in the objective function. Read the files beforehand and pass them into the objective function by creating an anonymous function that refers to them.
2 Comments
hamp
on 27 Sep 2018
Walter Roberson
on 3 Oct 2018
I do not have your data to test with so I created some random data and changed to code to work with the complex values directly. The results I got back were real-valued for fitPar.
fit4_driver.m:
file='G:\users\DA\Ueberblick EIS.xlsx';
%f = xlsread(file,'A4','A3:A73');
f = rand(71,1);
%realpart = xlsread(file,'TS40','B3:B73');
%imagpart = xlsread(file,'TS40','C3:C73');
realpart = rand(71,1);
imagpart = randn(71,1);
Z_exp = complex(realpart, imagpart);
% Start Parameters (initial guess)
startPar = [0.45*1e-3, 1.4*1.0846e-7, 0.016*4.111, 0.7*1e-3, 11, 0.00020];
% Make the best fitting
fitPar = lsqcurvefit(@Fit4_f1, startPar, f, Z_exp, [0 0 0 0 0 0], [1e-3 1e6 1e6 1e6 1e6 1e6]);
% Vector of fitted Zfit data
zFit4 = Fit4_f1(fitPar, f);
% Plot Fitted Line
plot(real(zFit4),-imag(zFit4), 'r','DisplayName','computed fitting data')
Fit4_f1.m:
function Z_fit0 = Fit4_f1(par,f)
f = f(1:length(f)/1); %
w = 2*pi*f ;
R0_fit0 = par(1);
L_fit0 = par(2);
R_bat_fit0 = par(3);
R_ct_fit0 = par(4);
C_dl_fit0 = par(5);
s_w_fit0 = par(6);
L_bat_fit0 = 1i*w*L_fit0;
R_el_fit0 = (L_bat_fit0*R_bat_fit0)./(L_bat_fit0+R_bat_fit0);
Z_d_fit0 = (s_w_fit0 ./ sqrt(w)) * (1-1i);
Z_c_fit0 = -1i ./ (C_dl_fit0*w);
R_dl_fit0 = ((Z_c_fit0 .* (R_ct_fit0 + Z_d_fit0)))./(R_ct_fit0 + Z_d_fit0 + Z_c_fit0);
Z_fit0 = R0_fit0 + R_el_fit0 + R_dl_fit0;
%Z_fit4 = vertcat(real(Z_fit0), imag(Z_fit0));
end
hamp
on 3 Oct 2018
4 Comments
Walter Roberson
on 3 Oct 2018
The message is correct. You have
Z_fit4 = vertcat(real(Z_fit0), imag(Z_fit0));
so you are producing a 284 x 1 output in response to a 142 x 1 input. Your fitting function must produce exactly one output for every input.
hamp
on 4 Oct 2018
Walter Roberson
on 4 Oct 2018
function Z_fit4 = Fit4_f1(something)
means that the function name is Fit4_f1 and that whenever it is invoked, the value to be returned is whatever has been assigned to the variable Z_fit4. lsqcurvefit requires that the output be the same size as the input. Your current code is calculating an temporary variable Z_fit0 that is the right size, but splits it into real and imaginary parts and so is returning twice as much data as expected.
The current code is also not taking into account that half of the input corresponds to imaginary parts.
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