
Fitting gaussian exponential to logscale
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sandy
el 23 de Nov. de 2020
Comentada: Mathieu NOE
el 23 de Nov. de 2020
Hello!
I have been trying to fit a log scale plot to an exponential function given below:
fun=@(t,x)2*(t(1))^2.*(1-exp(-((x./t(2)).^(2*t(3)))));
But I am unable to get a fit that is even close to the data plot. this is what I tried to do:
function fun = myfun(t,x,y)
t(1) = t(1)*1e-9;
t(2) = t(2)*1e-9;
t(3) = t(3)*1e-9;
xx=log(x)
yy=log(y)
fun=@(t,x)2*(t(1))^2.*(1-exp(-((x./t(2)).^(2*t(3)))));;
t=lsqcurvefit(fun,t,x,y)
plot(xx,yy,'ko',xx,fun(t,xx),'b-')
end
and then caling the function as:
t=[7.25553e-10 2.2790e-9 2.27908e-9]
myfun(t,x,y)
But this is what I get instead:

The x and y data is attached as .txt
I tried playing with different inital values but none of them gives me close to a good fit. I dont understand what is wrong here.
Could someone please help me out ?
I also do not have access to the curve fitting tool or any other tool in matlab just so you know. Any help would be really appreciated!
Thank you!
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Respuesta aceptada
Alan Stevens
el 23 de Nov. de 2020
The logs of the y values are negative, whereas your function will return positive values only! You are probably better off trying the following
% Scale data
xx = 10^7*x;
yy = 10^18*y;
fun=@(t,x) 2*(t(1))^2.*(1-exp(-((x./t(2)).^(2*t(3)))));
t0 = [1 1 0.5];
t = fminsearch(@(t) fcn(t,xx,yy,fun),t0);
disp(t)
Y = fun(t,xx);
% Rescale curve fit
yfit = Y*10^-18;
plot(x,y,'o',x,yfit),grid
xlabel('x [m]'),ylabel('y [m^2]')
legend('data','curve fit')
function F = fcn(t,xx,yy,fun)
Y = fun(t,xx);
F = norm(Y-yy);
end
which results in the following fit

2 comentarios
Alan Stevens
el 23 de Nov. de 2020
I don't think you can get that automatically from fminsearch. However, it's possible that taking the same approach with lsqcurvefit, or lsqnonlin, the error range might be returned automatically (I can't tell as I don't have the Optimization toolbox).
Más respuestas (1)
Mathieu NOE
el 23 de Nov. de 2020
hello
this a code that seems to work
hope it helps
I prefered to simplify as much the work of the optimizer , so first convert x and y in log scale and shift origin so curve start at x = y = 0
clc
clear all
close all
T = readtable('trial.txt');
C = table2array(T);
x = C(2:end,1);
y = C(2:end,2);
% convert x and y in log scale
lx = log10(x);
ly = log10(y);
% center origine so lx and ly first point = 0
mlx = min(lx);
lxc = lx - mlx;
mly = min(ly);
lyc = ly - mly;
% resample for higher points density in the first half of lx
% seems to help get better results
lxxc = linspace(min(lxc),max(lxc),length(lxc));
lyyc = interp1(lxc,lyc,lxxc);
plot(lxc, lyc,'+b',lxxc, lyyc,'+r');
f = @(a,b,x) a.*(1-exp(b.*x));
obj_fun = @(params) norm(f(params(1),params(2),lxxc)-lyyc);
sol = fminsearch(obj_fun, [1,1]);
a_sol = sol(1);
b_sol = sol(2);
figure;
plot(lxc, lyc, '+', 'MarkerSize', 10, 'LineWidth', 2)
hold on
plot(lxxc, f(a_sol, b_sol, lxxc), '-');grid
% finally
% lyc = a.*(1-exp(b.*lxc)); % converted back to :
% ly = mly + a.*(1-exp(b.*(lx - mlx)));
% and in linear scale :
y_fit = 10.^(mly + a_sol.*(1-exp(b_sol.*(log10(x) - mlx))));
figure;
plot(x, y, '+', 'MarkerSize', 10, 'LineWidth', 2)
hold on
plot(x, y_fit, '-');grid

4 comentarios
Mathieu NOE
el 23 de Nov. de 2020
no, it's basically where you define the model fit equation
the fminsearch will use that funtion in it's evaluation
for more details , look at : help fminsearch
tx
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