Best fitting curve for variable data

2022 4 23
2 个回答
回答已采纳
更新时间：2022 4 23
8 次查看（30 天）

0 个投票

Hello
I was trying fitting to fit the following data
x = 1.0779 1.2727 1.4700 1.5766 1.6471 1.7396 1.7828 1.8208 1.8370
y = 7.9511 9.8400 12.8838 15.1925 31.0055 36.0292 62.5528 87.9648 176.4142
for the equation:
y = a exp(b*(1/x)^c)
where a ba nd c are fitting paramters.
I triesd the fiting but unfortinately i was not able to magae it. It will be really helpful experties can guide me in this context.
Thanks in advance!

请先登录，再回答此问题。

 采纳的回答

Matt J
Matt J 2022-4-23
编辑：Matt J 2022-4-23
Ran in:

1 个投票

Ran in:
Using fminspleas from the File Exchange
https://www.mathworks.com/matlabcentral/fileexchange/10093-fminspleas
% data
x = [1.0779 1.2727 1.4700 1.5766 1.6471 1.7396 1.7828 1.8208 1.8370]' ;
y = [7.9511 9.8400 12.8838 15.1925 31.0055 36.0292 62.5528 87.9648 176.4142]' ; ;
funlist={1,@(c,xx) (1./xx).^c};
[c,ab]= fminspleas(funlist,-3,x,log(y),-inf,0);
a=exp(ab(1));
b=ab(2);
a,b,c
a = 8.6541
b = 0.0088
c = -9.4295
xx=linspace(min(x),max(x),100);
fun=@(x)a*exp(b*(1./x).^c);
plot(x,y,'o',xx,fun(xx))
ylim([0,max(y)])

更多回答（1 个）

KSSV
KSSV 2022-4-23
编辑：KSSV 2022-4-23
Ran in:

1 个投票

Ran in:
% data
x = [1.0779 1.2727 1.4700 1.5766 1.6471 1.7396 1.7828 1.8208 1.8370]' ;
y = [7.9511 9.8400 12.8838 15.1925 31.0055 36.0292 62.5528 87.9648 176.4142]' ; ;
eqn = @(a,b,c,x) a*exp(b*(1./x).^c) ; % equation to fit
% Define Start points
x0 = [1 1 1];
% fit-function
fitfun = fittype(eqn);
% fit curve
[fitted_curve,gof] = fit(x,y,fitfun,'StartPoint',x0)
fitted_curve =
General model: fitted_curve(x) = a*exp(b*(1./x).^c) Coefficients (with 95% confidence bounds): a = 15.61 (1.495, 29.73) b = 1.14e-06 (-1.433e-05, 1.661e-05) c = -23.93 (-46.14, -1.715)
gof = struct with fields:
sse: 879.6251 rsquare: 0.9635 dfe: 6 adjrsquare: 0.9513 rmse: 12.1080
% Save the coeffiecient values for a,b,c
coeffvals = coeffvalues(fitted_curve);
% Plot results
scatter(x, y, 'bs')
hold on
plot(x,fitted_curve(x),'r')
legend('data','fitted curve')

3 个评论
显示 1更早的评论 隐藏 1更早的评论

Somnath Kale
Somnath Kale 2022-4-23
Thank you @KSSV for your propmt response!!
Actully im trying to fit the follwing x and y and wanted to plot in following fasion:
x = [0.0970 0.1190 0.1323 0.1419 0.1515 0.1566 0.1605 0.1639 0.1653]';
y = [7.9511 9.8400 12.8838 15.1925 31.0055 36.0292 62.5528 87.9648 176.4142]';
semilogy(1./x,y,'o')
can you pleses help me for this
thanks,
somnath
KSSV
KSSV 2022-4-23
You have the code... Go ahead.
Somnath Kale
Somnath Kale 2022-4-23
编辑：Somnath Kale 2022-4-23
I tried using your code but did not get the expected results for fitting parameters
I have wrote following code uning diffenrt function fminsearch:
% fitting
x = [0.0970 0.1190 0.1323 0.1419 0.1515 0.1566 0.1605 0.1639 0.1653]';
y = [7.9511 9.8400 12.8838 15.1925 31.0055 36.0292 62.5528 87.9648 176.4142]';
semilogy(1./x,y,'o')
f = @(a,b,c,x) a*exp((b./x).^c);
obj_fun = @(params) norm(f(params(1), params(2), params(3), x) -y);
sol = fminsearch(obj_fun, [1,10,1]);
a = sol(1)
b = sol(2)
c = sol(3)
hold on
semilogy(1./x,f(sol(1),sol(2),sol(3),x),'-')
hold off
ylabel('y')
xlabel('1/x')
but im getting following issue:
Exiting: Maximum number of function evaluations has been exceeded
- increase MaxFunEvals option.
Current function value: 103.027992
how take care of this one?

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Spline Postprocessing 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