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

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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.

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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');
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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);

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