Curve fitting using lsqcurvefit

2022 3 7
1 个回答
回答已采纳
更新时间：2022 3 8
9 次查看（30 天）
Ran in:

0 个投票

Ran in:
Hi currently i am trying to fit my measured data to a model. I have followed this example https://au.mathworks.com/matlabcentral/answers/478835-lsqcurvefit-initial-values-stays-the-same#answer_391692 . I am also attaching the code from the example. i want to know what parameters do i need to change to fit my measured data to the model and extract the values from it. Any help is appreicated. I am also attaching my data.
T = readtable('cole.xlsx');
x = T{:,1}; %freq
y = T{:,2}; %real
%Transpose
freq = x.';
e_real = y.';
options = optimset('MaxFunEvals',10000);
options=optimset(options,'MaxIter',10000);
guess = 4.5;
UB = guess + 1;
LB = guess - 1;
lb = [];
ub = [];
x0 = [6,guess,2.5,0.6,0.09];
x = lsqcurvefit(@flsq,x0,freq,e_real,lb,ub,options)
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
x =
4.8828 - 0.0885i 5.0695 - 0.1843i 1.2278 -92.9851i -1.0163 - 9.3958i -0.0126 - 0.0739i
e_f = x(1);
e_del = x(2)*1e2;
tau1 = x(3)*1e-12;
alf1 = x(4);
sig = x(5);
yfit = real(flsq(x,freq));
%plot e_real against freq
plot(freq,e_real,'k.',freq,yfit,'b-')
legend('Data','Fitted')
title('Data and Fitted Curve')
function y = flsq(x,freq)
x(3)=x(3)*1e-12;
y = x(1) + (x(2)-x(1))./(1 + ((1j*2*pi.*freq*x(3)).^(1-x(4))))+x(5)./(1j*2*pi*freq*8.854e-12);
end

7 个评论
显示 5更早的评论 隐藏 5更早的评论

Walter Roberson
Walter Roberson 2022-3-7
I am not sure what you want done different from what you already have?
Kabit Kishore
Kabit Kishore 2022-3-7
编辑：Kabit Kishore 2022-3-7
Hi Walter thank you. I want the fitted curve to go through maximum number of measured data points. Is it possible.
Image Analyst
Image Analyst 2022-3-7
That's a pretty complicated model. Would you rather just fit it to like 3-6 Gaussians instead? Or does that model have some theoretical significance based on the physics of the situation?
Image Analyst
Image Analyst 2022-3-8
@Kabit Kishore, any response to my questions in my last comment above? From Walter's last comment it looks like the model is no good so why use that particular model?
Kabit Kishore
Kabit Kishore 2022-3-8
Hi @Image Analyst. Yes the model that i am trying to fit has theoretical significance. Curently i am guessing the values which make it pretty dificult to fit the equation to my data. Is there a iterative automatic way which updates the values and provides the best fit. Thank you for your response.
Walter Roberson
Walter Roberson 2022-3-8
lsqcurvefit() and fmincon() both do iterative automatic ways to update values to provide best fits.
Walter Roberson
Walter Roberson 2022-3-8
If this has physical significance, then are there any constraints on the parameters of the model? Any that have to be real-valued? Any that have to be positive?

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Walter Roberson
Walter Roberson 2022-3-7
Ran in:

0 个投票

Ran in:
You can get a better fit by either telling lsqcurvefit() a better starting point, or by using a different optimizer.
But... remember that "better fit" mathematically does not necessarily mean "follows the curve more closely"
format long g
filename = 'https://www.mathworks.com/matlabcentral/answers/uploaded_files/917874/cole.xlsx';
T = readtable(filename);
x = T{:,1}; %freq
y = T{:,2}; %real
%Transpose
freq = x.';
e_real = y.';
options = optimset('MaxFunEvals',10000);
options=optimset(options,'MaxIter',10000);
guess = 4.5;
UB = guess + 1;
LB = guess - 1;
lb = [];
ub = [];
x0 = [6,guess,2.5,0.6,0.09];
[x, fval] = lsqcurvefit(@flsq,x0,freq,e_real,lb,ub,options)
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
x =
4.88278441116842 - 0.0884918454761463i 5.06952748610154 - 0.184297340006755i 1.22778921925455 - 92.9851457164006i -1.01629693596011 - 9.39581846397004i -0.0126281060006031 - 0.073855580530327i
fval =
2.50307155416643
e_f = x(1);
e_del = x(2)*1e2;
tau1 = x(3)*1e-12;
alf1 = x(4);
sig = x(5);
yfit = real(flsq(x,freq));
%plot e_real against freq
plot(freq,e_real,'k.',freq,yfit,'b-')
legend('Data','Fitted')
title('Data and Fitted Curve -- lsqcurvefit')
Residue = @(x) norm(flsq(x,freq) - e_real);
options = optimset('MaxFunEvals', 1E5);
options = optimset(options, 'MaxIter', 1E5);
options = optimset(options, 'Display', 'none');
for K = 2 : 50
guess = x0 + randn(1,5);
[x(K,:), fval(K)] = fmincon(Residue, guess, [], [], [], [], [], [], [], options);
end
[~, idx] = min(abs(fval));
disp('original -- lsqcurvefit')
original -- lsqcurvefit
disp([fval(1), x(1,:)])
Columns 1 through 4 2.50307155416643 + 0i 4.88278441116842 - 0.0884918454761463i 5.06952748610154 - 0.184297340006755i 1.22778921925455 - 92.9851457164006i Columns 5 through 6 -1.01629693596011 - 9.39581846397004i -0.0126281060006031 - 0.073855580530327i
disp('best found in several tries of fmincon')
best found in several tries of fmincon
disp([fval(idx), x(idx,:)])
1.82932684842456 4.25784166349341 -638714.651612716 5141.92675473731 -3.02543789002661 0.000633983073703346
e_f = x(idx,1);
e_del = x(idx,2)*1e2;
tau1 = x(idx,3)*1e-12;
alf1 = x(idx,4);
sig = x(idx,5);
yfit = real(flsq(x(idx,:),freq));
%plot e_real against freq
plot(freq, e_real, 'k.', freq, yfit, 'b-')
legend('Data','Fitted')
title('Data and Fitted Curve -- fmincon')
function y = flsq(x,freq)
x(3)=x(3)*1e-12;
y = x(1) + (x(2)-x(1))./(1 + ((1j*2*pi.*freq*x(3)).^(1-x(4))))+x(5)./(1j*2*pi*freq*8.854e-12);
end

5 个评论
显示 3更早的评论 隐藏 3更早的评论

Walter Roberson
Walter Roberson 2022-3-7
Sometimes the model is just not a good fit.
Walter Roberson
Walter Roberson 2022-3-8
Ran in:
Ran in:
An improved run
format long g
filename = 'https://www.mathworks.com/matlabcentral/answers/uploaded_files/917874/cole.xlsx';
T = readtable(filename);
x = T{:,1}; %freq
y = T{:,2}; %real
%Transpose
freq = x.';
e_real = y.';
options = optimset('MaxFunEvals',10000);
options=optimset(options,'MaxIter',10000);
guess = 4.5;
UB = guess + 1;
LB = guess - 1;
lb = [];
ub = [];
x0 = [6,guess,2.5,0.6,0.09];
[x, fval] = lsqcurvefit(@flsq,x0,freq,e_real,lb,ub,options)
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
x =
4.88278441116842 - 0.0884918454761463i 5.06952748610154 - 0.184297340006755i 1.22778921925455 - 92.9851457164006i -1.01629693596011 - 9.39581846397004i -0.0126281060006031 - 0.073855580530327i
fval =
2.50307155416643
e_f = x(1);
e_del = x(2)*1e2;
tau1 = x(3)*1e-12;
alf1 = x(4);
sig = x(5);
yfit = real(flsq(x,freq));
%plot e_real against freq
plot(freq,e_real,'k.',freq,yfit,'b-')
legend('Data','Fitted')
title('Data and Fitted Curve -- lsqcurvefit')
Residue = @(x) norm(flsq(x,freq) - e_real);
options = optimset('MaxFunEvals', 1E5);
options = optimset(options, 'MaxIter', 1E5);
options = optimset(options, 'Display', 'none');
for K = 2 : 50
guess = randn(1,5)*100
;
[x(K,:), fval(K)] = fmincon(Residue, guess, [], [], [], [], [], [], [], options);
end
[~, idx] = min(abs(fval));
disp('original -- lsqcurvefit')
original -- lsqcurvefit
disp([fval(1), x(1,:)])
Columns 1 through 4 2.50307155416643 + 0i 4.88278441116842 - 0.0884918454761463i 5.06952748610154 - 0.184297340006755i 1.22778921925455 - 92.9851457164006i Columns 5 through 6 -1.01629693596011 - 9.39581846397004i -0.0126281060006031 - 0.073855580530327i
disp('best found in several tries of fmincon')
best found in several tries of fmincon
disp([fval(idx), x(idx,:)])
1.68283982412135 3.49149418809273 4.1928928642795 147.870041270639 44.9919207152756 -1.52002099046847e-05
e_f = x(idx,1);
e_del = x(idx,2)*1e2;
tau1 = x(idx,3)*1e-12;
alf1 = x(idx,4);
sig = x(idx,5);
yfit = real(flsq(x(idx,:),freq));
%plot e_real against freq
plot(freq, e_real, 'k.', freq, yfit, 'b-')
legend('Data','Fitted')
title('Data and Fitted Curve -- fmincon')
function y = flsq(x,freq)
x(3)=x(3)*1e-12;
y = x(1) + (x(2)-x(1))./(1 + ((1j*2*pi.*freq*x(3)).^(1-x(4))))+x(5)./(1j*2*pi*freq*8.854e-12);
end
Walter Roberson
Walter Roberson 2022-3-8
Ran in:
Ran in:
Slight improvement. I'm not sure you would be able to do much better -- this does 500 starting points.
format long g
filename = 'https://www.mathworks.com/matlabcentral/answers/uploaded_files/917874/cole.xlsx';
T = readtable(filename);
x = T{:,1}; %freq
y = T{:,2}; %real
%Transpose
freq = x.';
e_real = y.';
options = optimset('MaxFunEvals',10000);
options=optimset(options,'MaxIter',10000);
guess = 4.5;
UB = guess + 1;
LB = guess - 1;
lb = [];
ub = [];
x0 = [6,guess,2.5,0.6,0.09];
[x, fval] = lsqcurvefit(@flsq,x0,freq,e_real,lb,ub,options)
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
x =
4.88278441116842 - 0.0884918454761463i 5.06952748610154 - 0.184297340006755i 1.22778921925455 - 92.9851457164006i -1.01629693596011 - 9.39581846397004i -0.0126281060006031 - 0.073855580530327i
fval =
2.50307155416643
e_f = x(1);
e_del = x(2)*1e2;
tau1 = x(3)*1e-12;
alf1 = x(4);
sig = x(5);
yfit = real(flsq(x,freq));
%plot e_real against freq
plot(freq,e_real,'k.',freq,yfit,'b-')
legend('Data','Fitted')
title('Data and Fitted Curve -- lsqcurvefit')
Residue = @(x) norm(flsq(x,freq) - e_real);
options = optimset('MaxFunEvals', 1E5);
options = optimset(options, 'MaxIter', 1E5);
options = optimset(options, 'Display', 'none');
for K = 2 : 500
guess = rand(1,5)*100;
[x(K,:), fval(K)] = fmincon(Residue, guess, [], [], [], [], [], [], [], options);
end
[~, idx] = min(abs(fval));
disp('original -- lsqcurvefit')
original -- lsqcurvefit
disp([fval(1), x(1,:)])
Columns 1 through 4 2.50307155416643 + 0i 4.88278441116842 - 0.0884918454761463i 5.06952748610154 - 0.184297340006755i 1.22778921925455 - 92.9851457164006i Columns 5 through 6 -1.01629693596011 - 9.39581846397004i -0.0126281060006031 - 0.073855580530327i
disp('best found in several tries of fmincon')
best found in several tries of fmincon
disp([fval(idx), x(idx,:)])
1.66154803509977 3.41081977527717 4.19802178524268 149.532439789817 25.0006532849926 2.49463043656685e-06
e_f = x(idx,1);
e_del = x(idx,2)*1e2;
tau1 = x(idx,3)*1e-12;
alf1 = x(idx,4);
sig = x(idx,5);
yfit = real(flsq(x(idx,:),freq));
%plot e_real against freq
plot(freq, e_real, 'k.', freq, yfit, 'b-')
legend('Data','Fitted')
title('Data and Fitted Curve -- fmincon')
function y = flsq(x,freq)
x(3)=x(3)*1e-12;
y = x(1) + (x(2)-x(1))./(1 + ((1j*2*pi.*freq*x(3)).^(1-x(4))))+x(5)./(1j*2*pi*freq*8.854e-12);
end
Walter Roberson
Walter Roberson 2022-3-8
Ran in:
Ran in:
ga() is another iterative means for improving fit, but it does not help much. In one of my earlier runs, I got better than the below display, but it was still worse than fmincon() so I am not going to bother to show it.
format long g
filename = 'https://www.mathworks.com/matlabcentral/answers/uploaded_files/917874/cole.xlsx';
T = readtable(filename);
x = T{:,1}; %freq
y = T{:,2}; %real
%Transpose
freq = x.';
e_real = y.';
options = optimset('MaxFunEvals',10000);
options=optimset(options,'MaxIter',10000);
guess = 4.5;
UB = guess + 1;
LB = guess - 1;
lb = [];
ub = [];
x0 = [6,guess,2.5,0.6,0.09];
[x, fval] = lsqcurvefit(@flsq,x0,freq,e_real,lb,ub,options)
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
x =
4.88278441116842 - 0.0884918454761463i 5.06952748610154 - 0.184297340006755i 1.22778921925455 - 92.9851457164006i -1.01629693596011 - 9.39581846397004i -0.0126281060006031 - 0.073855580530327i
fval =
2.50307155416643
e_f = x(1);
e_del = x(2)*1e2;
tau1 = x(3)*1e-12;
alf1 = x(4);
sig = x(5);
yfit = real(flsq(x,freq));
%plot e_real against freq
plot(freq,e_real,'k.',freq,yfit,'b-')
legend('Data','Fitted')
title('Data and Fitted Curve -- lsqcurvefit')
Residue = @(x) norm(flsq(x,freq) - e_real);
options = optimset('MaxFunEvals', 1E5);
options = optimset(options, 'MaxIter', 1E5);
options = optimset(options, 'Display', 'none');
tic
for K = 2 : 200
[x(K,:), fval(K)] = ga(Residue, 5, [], [], [], [], [], [], [], options);
end
toc
Elapsed time is 38.143890 seconds.
[~, idx] = min(abs(fval));
disp('original -- lsqcurvefit')
original -- lsqcurvefit
disp([fval(1), x(1,:)])
Columns 1 through 4 2.50307155416643 + 0i 4.88278441116842 - 0.0884918454761463i 5.06952748610154 - 0.184297340006755i 1.22778921925455 - 92.9851457164006i Columns 5 through 6 -1.01629693596011 - 9.39581846397004i -0.0126281060006031 - 0.073855580530327i
disp('best found in several tries of ga')
best found in several tries of ga
disp([fval(idx), x(idx,:)])
2.0261188723609 4.23046723004257 3.35326686725843 153.911545598562 -0.327809240216987 0.0453959038185919
e_f = x(idx,1);
e_del = x(idx,2)*1e2;
tau1 = x(idx,3)*1e-12;
alf1 = x(idx,4);
sig = x(idx,5);
yfit = real(flsq(x(idx,:),freq));
%plot e_real against freq
plot(freq, e_real, 'k.', freq, yfit, 'b-')
legend('Data','Fitted')
title('Data and Fitted Curve -- ga')
function y = flsq(x,freq)
x(3)=x(3)*1e-12;
y = x(1) + (x(2)-x(1))./(1 + ((1j*2*pi.*freq*x(3)).^(1-x(4))))+x(5)./(1j*2*pi*freq*8.854e-12);
end

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Get Started with Optimization Toolbox 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by