Fitting data in x,y to a known function

6 次查看(过去 30 天)
Hi everyone. I have a function that is f = 1/(a+b*x) where a and b are the values to obtain and I have some data:
x1 = linspace(1,32,32);
y1 = [0.01 0.02 0.02 0.02 0.02 0.02 0.03 0.03 0.03 0.04 0.04 0.05 0.05 0.06 0.07 0.07 0.08 0.10 0.11 0.14 0.14 0.17 0.17 0.16 0.21 0.31 0.31 2.43 2.43 29.53 29.53 29.53];
I need to fit these two variables x1 and y1 into the function above to obtain a and b. How can I do this?
I'm sorry I'm pretty new in Matlab. Thank you.

采纳的回答

Star Strider
Star Strider 2019-12-12
Use the fminsearch function to fit your data:
x1 = linspace(1,32,32);
y1 = [0.01 0.02 0.02 0.02 0.02 0.02 0.03 0.03 0.03 0.04 0.04 0.05 0.05 0.06 0.07 0.07 0.08 0.10 0.11 0.14 0.14 0.17 0.17 0.16 0.21 0.31 0.31 2.43 2.43 29.53 29.53 29.53];
f = @(p,x) 1./(p(1) + p(2).*x);
P = fminsearch(@(p) norm(y1 - f(p,x1)), rand(2,1));
x1v = linspace(min(x1), max(x1));
figure
plot(x1, y1, 'p')
hold on
plot(x1v, f(P,x1v), '-r')
hold off
grid
producing:
P =
0.845464263829100
-0.025490734971736
where ‘P(1)=a’, and ‘P(2)=b’.
The plot appears correct, even though it looks a bit strange.

更多回答(1 个)

Nicolas B.
Nicolas B. 2019-12-12
If you have it, I would recommend to use the Curve Fitting Toolbox. You can give it the expected function and it fits the parameters.
Otherwise, the only solution would be to yourself try to fit the function on the samples based on an optimization method but that requires to investigate mathematic fitting.
  1 个评论
Raúl Alonso Merino
Raúl Alonso Merino 2019-12-12
I tried and I get a fitted curve, thank you.
General model:
myfit(x) = 1/(a+b*x)
Coefficients (with 95% confidence bounds):
a = 0.9058 (0.4335, 1.378)
b = -0.02738 (-0.04219, -0.01256)
My question now is: is it correct to fit it directly to that function or do I have to fit it to the integral of 1/(a+b*x) which is log(a+b*x)/b + c?

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Get Started with Curve Fitting Toolbox 的更多信息

产品


版本

R2018b

Community Treasure Hunt

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

Start Hunting!

Translated by