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Fit Powerlaw to Data

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Tobias Bramminge
Tobias Bramminge 2019 年 12 月 6 日
コメント済み: Image Analyst 2019 年 12 月 6 日
Hi all!
I need to fit following Power Law to some experimental data.
y = C(B+x)^n
The data I have is as the following:
STRESS = [0.574, 367.364, 449.112, 531.087, 596.241, 649.097, 695.038, 737.173, 815.008];
STRAIN = [2.8746e-04, 0.00063, 0.0459, 0.0901, 0.1320, 0.1725, 0.2132, 0.2557, 0.3579];
The variables are x and y are STRAIN and STRESS respectively, and I would like estimates for C, B and n.
I'm not sure how to use the fitting tool for this kind of specific model.
Any ideas or suggestions would be greatly appreciated!
Thank you very much!

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John D'Errico
John D'Errico 2019 年 12 月 6 日
One requirement is that you have the same number of values in each vector. You don't meet that basic requirement.
numel(STRESS)
ans =
8
numel(STRAIN)
ans =
9
Once you have actually told us the correct data, then you need to explain which variable is intended to be x, and which is y. If you want help, then make it possible, even easy, for someone to help you.
Tobias Bramminge
Tobias Bramminge 2019 年 12 月 6 日
Thank you very much for your reply!
I am very sorry for the bad explanation, I am quite new at requesting help like this. I've updated the data and the variables are x and y.
Is this sufficient?

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Star Strider
Star Strider 2019 年 12 月 6 日
編集済み: Star Strider 2019 年 12 月 6 日
Try this:
STRESS = [0.574, 367.364, 449.112, 531.087, 596.241, 649.097, 695.038, 737.173, 815.008];
STRAIN = [2.8746e-04, 0.00063, 0.0459, 0.0901, 0.1320, 0.1725, 0.2132, 0.2557, 0.3579];
yfcn = @(b,x) b(1).*(b(2)+x).^b(3);
B0 = [1E-6; 100; 2];
B = fminsearch(@(b) norm(STRAIN - yfcn(b,STRESS)), B0);
xv = linspace(min(STRESS), max(STRESS), 50);
yv = yfcn(B,xv);
figure
plot(STRESS, STRAIN, 'pg')
hold on
plot(xv, abs(yv), '-r')
hold off
grid
text(150, 0.27, sprintf('y = %.3E \\cdot (%.3f + x)^{%.3f}', B))
xlabel('STRESS')
ylabel('STRAIN')
Experiment to get different results.
EDIT (6 Dec 2019 at 17:50)
Added plot image:
1Fit Powerlaw to Data - 2019 12 06.png

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Tobias Bramminge
Tobias Bramminge 2019 年 12 月 6 日
Thank you very much!
This is the method I was searching for.
Have a great weekend!
Star Strider
Star Strider 2019 年 12 月 6 日
As always, my pleasure!
You have a great weekend, too!
Image Analyst
Image Analyst 2019 年 12 月 6 日
Why are you trying to include the left most point in the fit? It basically ruins the fit for all the other points. It would be so much better if you didn't include it, like this:
0000 Screenshot.png
See how much closer the fit is for points 2-9 if you omit the first point? Will you ever really be trying to estimate values down in that region where the training data is nearly vertical? And, if so, it is worth getting a looser, worse fit over the majority of your range when you could avoid it with a piecewise fit?

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その他の回答 (2 件)

Stephan
Stephan 2019 年 12 月 6 日

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Image Analyst
Image Analyst 2019 年 12 月 6 日
I'd use fitnlm() in the Statistics and Machine Learning Toolbox.
I suggest that you don't try to fit the first point. When I try to do that, it's impossible to get a fit. I get error messages that says the Jacobian is not well defined. "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. " and it won't do the fit.
But if I try to fit from the second point onwards, I still get the warning but the fit seems reasonable. See plot below. If you want you could do a piece-wise fit where you fit everything to the left of the second point to a line.
0000 Screenshot.png

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