How to fit the data with 2 term power law function?

6 次查看（过去 30 天）
peipei feng
peipei feng 2020-6-1
评论： peipei feng 2020-6-2
Hi there,
I have the followig data and I want to fit them in the form of z = r^a * v^b.
I tried to obtain log values of the function: log(z) =a*log(r)+b*log(v), but I'm confused as there are so many vectors.
r = [4.242640687119285e-06,5.656854249492381e-06,7.071067811865476e-06,8.485281374238570e-06,9.899494936611665e-06,1.131370849898476e-05,1.272792206135786e-05];
v = [0.3656,0.6093,0.9748,1.2185];
z = [2.20075780727420 2.34800174967587 2.40633263085150 2.46928643488236 2.55551444168423 2.66309682650254 2.78591291730793
1.20828256541234 1.57243262679317 1.78594063849408 1.94976208027465 2.10251979263655 2.25584172701771 2.41163339237766
0.472116287092821 0.855613921769617 1.14739856522263 1.38120431719641 1.58720801177653 1.78025160940097 1.96628391586011
0.243938638287455 0.563375613060710 0.851155979561504 1.09774989977420 1.31858268912292 1.52483525677172 1.72202346894929
];
the plot is also attached, where different curves are from different v values.

请先登录，再回答此问题。

采纳的回答

Ameer Hamza
Ameer Hamza 2020-6-1
编辑：Ameer Hamza 2020-6-2
taking log() of the equation make it very easy to find the parameter (because the equation is linear in term of a and b). Try the following code
r = [4.242640687119285e-06,5.656854249492381e-06,7.071067811865476e-06,8.485281374238570e-06,9.899494936611665e-06,1.131370849898476e-05,1.272792206135786e-05];
v = [0.3656,0.6093,0.9748,1.2185];
z = [2.20075780727420 2.34800174967587 2.40633263085150 2.46928643488236 2.55551444168423 2.66309682650254 2.78591291730793
1.20828256541234 1.57243262679317 1.78594063849408 1.94976208027465 2.10251979263655 2.25584172701771 2.41163339237766
0.472116287092821 0.855613921769617 1.14739856522263 1.38120431719641 1.58720801177653 1.78025160940097 1.96628391586011
0.243938638287455 0.563375613060710 0.851155979561504 1.09774989977420 1.31858268912292 1.52483525677172 1.72202346894929
];
a = zeros(1, 4);
b = zeros(1, 4);
for i=1:numel(a)
V = repmat(v(i), size(r));
Z = log(z(i, :).');
X = [log(r.') log(V.')];
sol = X\Z;
a(i) = sol(1);
b(i) = sol(2);
end
figure;
axes();
hold on
for i=1:numel(v)
z_est = r.^a(i).*v(i).^b(i);
plot(r, z(i,:), '+--', 'DisplayName', ['v = ' num2str(v(i)) ' real']);
plot(r, z_est, 'DisplayName', ['v = ' num2str(v(i)) ' est']);
end
legend('Location', 'best');
  5 个评论
显示 3更早的评论
Ameer Hamza
Ameer Hamza 2020-6-2
In that case, the original solution was correct. As you can see, the value of a and b vary very much to fit these lines, so it might not be possible to get a good fit with single 'a' and 'b'.
Here is the original code for reference
r = [4.242640687119285e-06,5.656854249492381e-06,7.071067811865476e-06,8.485281374238570e-06,9.899494936611665e-06,1.131370849898476e-05,1.272792206135786e-05];
v = [0.3656,0.6093,0.9748,1.2185];
z = [2.20075780727420 2.34800174967587 2.40633263085150 2.46928643488236 2.55551444168423 2.66309682650254 2.78591291730793
1.20828256541234 1.57243262679317 1.78594063849408 1.94976208027465 2.10251979263655 2.25584172701771 2.41163339237766
0.472116287092821 0.855613921769617 1.14739856522263 1.38120431719641 1.58720801177653 1.78025160940097 1.96628391586011
0.243938638287455 0.563375613060710 0.851155979561504 1.09774989977420 1.31858268912292 1.52483525677172 1.72202346894929
];
[V, R] = ndgrid(v, r);
Y = log(z(:));
X = [log(R(:)) log(V(:))];
sol = X\Y; % least square solution
a = sol(1);
b = sol(2);

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Get Started with Curve Fitting 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