Non-linear Curve Fitting Issue with Experimental Data
显示 更早的评论
I am using the MATLAB curve fitting tool to fit the following data given in matrices A(:,1) through A(:,7). Data in A(:,1)-A(:,7) are y-axes results while the x-axes, given in matrix B, remains the same for all. I have attached the data as .mat file to this post.
Objective: To fit the data (given A, B in data.mat) with a single function
What I have tried so far: I have used the curve fit to fit the data using power function. Although matric A(:,1)-A(:,4) fit pretty well, I am unable to fit A(:,5)-A(:,7) since they have an exponential decay that is difficult to be captured using power functions. Also, note that the data along the x-axes does not saturate as the x is increased.
Below is an image of the curve fit that I tried and it does not seem to satisfy the fast decaying data given in darker (black) lines in the plot.
Here is the code that I am currently using for reference:
color_array = linspace(1,0,length(VG_array));
for k=1:7
base_floor(min(abs(A(:,k)))<9e-12) = min(abs(A(:,k)));
base_floor(min(abs(A(:,k)))>9e-12) = 0;
semilogy(B,A(:,k),...
'Color',[color_array(k) 0 0]);
hold on;
f = fit( B', A(:,k), 'power1' );
options = fitoptions('power1');
options.Robust = 'Bisquare';
plot(B,(f.a*B.^f.b)+base_floor,'--')
hold on;
parameters_wb_delta(:,k) = f.a;
parameters_wb_gamma(:,k) = f.b;
end
What I need help with: I need a single function fit that can capture all the data in the matrices A(:,1)-A(:,7) with respect to B.
采纳的回答
Mathieu NOE
2021-2-25
HELLO
I don't have the CF Toolbox but I could manage to get a reasonnable good fit by using 4th ordre polynomial fit on log log scale
so the fit function is like : yfit = 10^(poly 4th order)
hope it helps
%
for ci = 1:size(A,2)
Bp = polyfit(log10(B), log10(A(:,ci)), 4);
Yfit_log = polyval(Bp,log10(B));
Yfit(ci,:) =10.^Yfit_log;
end
figure(1),loglog(B,A,B,Yfit, '-.', 'MarkerSize', 10, 'LineWidth', 2)
grid on
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Linear and Nonlinear Regression 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
