second order data fitting using least squares

30 vues (au cours des 30 derniers jours)
Sangmin
Sangmin le 17 Jan 2019
Modifié(e) : Torsten le 18 Jan 2019
hi
I am trying to fitting the data
I want to fitting the data 1 by least square fitting it to a quadratic function around the position of maximum data2(*)
f(x) = a(x-x0)^2 + b(x-x0) + c
where C is an additive constant C = f(x0) = 1.
I used several method (ex data fitting tool...) but failed
If you konw how to solve, pleast let me know
untitled.jpg
  2 commentaires
Torsten
Torsten le 17 Jan 2019
x0 is a fitting parameter or set to a fixed value ?
Sangmin
Sangmin le 18 Jan 2019
Modifié(e) : Sangmin le 18 Jan 2019
x0 is one of the data with the largest value.
x = [0.81 0.85 0.91 1.00 1.17 1.33 1.36 1.37 1.39 1.40 1.42]
y = [0.58 0.69 0.81 0.93 1 0.91 0.84 0.80 0.74 0.67 0.59]
x0 is maximum point (1.17 1)
I want to fit the quadratic curve through x0

Connectez-vous pour commenter.

Réponses (2)

Torsten
Torsten le 18 Jan 2019
Modifié(e) : Torsten le 18 Jan 2019
x = [0.81 0.85 0.91 1.00 1.17 1.33 1.36 1.37 1.39 1.40 1.42];
y = [0.58 0.69 0.81 0.93 1 0.91 0.84 0.80 0.74 0.67 0.59];
x0 = x(5);
y0 = y(5);
xtrans = x - x0;
ytrans = y - y0;
xtrans = xtrans.';
ytrans = ytrans.';
%mat = [sum(xtrans.^4) sum(xtrans.^3);sum(xtrans.^3) sum(xtrans.^2)];
%rhs = [sum(xtrans.^2.*ytrans); sum(xtrans.*ytrans)];
mat = [xtrans.^2 xtrans];
rhs = ytrans;
sol = mat\rhs;
a = sol(1);
b = sol(2);
fun = @(x)a*(x-x0).^2+b*(x-x0)+y0;
yfit = fun(x);
plot(x,y,x,yfit)

Akira Agata
Akira Agata le 18 Jan 2019
Another possible solution:
x = [0.81 0.85 0.91 1.00 1.17 1.33 1.36 1.37 1.39 1.40 1.42]';
y = [0.58 0.69 0.81 0.93 1 0.91 0.84 0.80 0.74 0.67 0.59]';
x0 = x(5);
y0 = y(5);
modelfun = @(a,x) a(1)*(x - x0).^2 + a(2)*(x - x0) + y0;
beta0 = [-1 1]; % Initial guess
mdl = fitnlm(x,y,modelfun,beta0);
xq = linspace(min(x),max(x))';
figure
scatter(x,y)
hold on
plot(xq,predict(mdl,xq))
fitting.png

Catégories

En savoir plus sur Least Squares dans Help Center et File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by