How do I fit an exponential curve to my data?

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Say, I have the following data: x=[1,2,4,6,8],y=[100,140,160,170,175]. How do I fit an exponential curve of the form y=a-b*exp(-c*x) to my data? Is there any Matlab function to do that? Thanks in advance.
  2 个评论
Arturo Gonzalez
Arturo Gonzalez 2020-9-1
编辑:Arturo Gonzalez 2020-9-1
Per this answer, you can do it as follows:
clear all;
clc;
% get data
dx = 0.02;
x = (dx:dx:1.5)';
y = -1 + 5*exp(0.5*x) + 4*exp(-3*x) + 2*exp(-2*x);
% calculate integrals
iy1 = cumtrapz(x, y);
iy2 = cumtrapz(x, iy1);
iy3 = cumtrapz(x, iy2);
% get exponentials lambdas
Y = [iy1, iy2, iy3, x.^3, x.^2, x, ones(size(x))];
A = pinv(Y)*y;
lambdas = eig([A(1), A(2), A(3); 1, 0, 0; 0, 1, 0]);
lambdas
%lambdas =
% -2.9991
% -1.9997
% 0.5000
% get exponentials multipliers
X = [ones(size(x)), exp(lambdas(1)*x), exp(lambdas(2)*x), exp(lambdas(3)*x)];
P = pinv(X)*y;
P
%P =
% -0.9996
% 4.0043
% 1.9955
% 4.9999
H. Sh. G.
H. Sh. G. 2022-7-28
编辑:H. Sh. G. 2022-7-28
Nice solution!
I implemented a generalized function to handle n exponential functions.
function [lambdas, c, yhat] = ExpFunFit(Data, n, intcpt)
%Fits an exponential summation function to Data = [t, y].
% y = c0 + Sum{c_i * exp(lambda_i * t)}; i = 1, ..., n.
% H.Sh.G. - 2022
% See Prony's method for exponential function fitting.
if nargin<3, intcpt = 0; end
x = Data(:,1);
nx = size(x, 1);
y = Data(:,2);
% Calculate integrals
yi = [y, zeros(nx, n)];
xi = [zeros(nx, n-1), x];
for i = 2:n+1
yi(:,i) = cumtrapz(x, yi(:,i-1));
if i<=n
xi(:,i-1) = x.^(n+2-i);
end
end
% Get exponentials' lambdas
Y = [yi(:,2:n+1), xi];
if intcpt, Y = [Y, ones(size(x))]; end
A = pinv(Y)*y;
lambdas = eig([A(1:n)'; ...
eye(n-1), zeros(n-1,1)]);
% Get exponentials' multipliers
X = ones(nx, n+1);
for i = 1:n
X (:, i+1) = exp(lambdas(i)*x);
end
if ~intcpt, X = X(:, 2:end); end
c = pinv(X)*y;
yhat = X * c;
Thanks Arturo.
Hamed.

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2013-10-23
编辑:Chad Greene 2021-1-14
[EDIT] Please read about fit and try:
x=[1,2,4,6,8]';
y=[100,140,160,170,175].';
g = fittype('a-b*exp(-c*x)');
f0 = fit(x,y,g,'StartPoint',[[ones(size(x)), -exp(-x)]\y; 1]);
xx = linspace(1,8,50);
plot(x,y,'o',xx,f0(xx),'r-');
  13 个评论
Michael Solonenko
Michael Solonenko 2021-5-14
This requires MATLAB Curve Fit package. Is there a way to do a simple one exponential fit without it?
Image Analyst
Image Analyst 2021-5-14
@Michael Solonenko, of course. My code below doesn't use the Curve Fitting Toolbox. Mine uses the Statistics and Machine Learning Toolbox, which is much more common. You probably have that toolbox. type "ver" on the command line to find out what toolboxes you have.

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更多回答(2 个)

Image Analyst
Image Analyst 2018-11-21
You could also use fitnlm (Fit Non Linear Model):
0000 Screenshot.png

Tamara Schapitz
Tamara Schapitz 2018-11-20
I tried it with the form
'a*exp(-((x/b)^c))'
I got this warning messsage:
Warning: Rank deficient, rank = 1, tol = 4.019437e-14.
and it only plots the data, but not the fit... What am I doing wrong?
  9 个评论
Image Analyst
Image Analyst 2020-4-9
That code requires the Curve Fitting Toolbox, which I don't have, so I can't run it. But I did plot(x,y) and noticed that several of your x all have the same value. Perhaps it could be fixed by making the x all unique values by adding a very tiny amount of random noise to them (but not enough to affect the fit), like
x=[7.48673807584469;7.48673807584469;9.52211367803178;9.52211367803178;9.52211367803178;11.2093975148163;11.2093975148163;11.2093975148163;15.8637542852664;15.8637542852664;15.8637542852664;17.5649842553087;17.5649842553087;17.5649842553087;26.3442681704923;26.3442681704923;26.3442681704923];
x = x + 0.001 * rand(size(x))
If it complains that the x must be sorted in ascending order, you can sort x and y like this:
x=[7.48673807584469;7.48673807584469;9.52211367803178;9.52211367803178;9.52211367803178;11.2093975148163;11.2093975148163;11.2093975148163;15.8637542852664;15.8637542852664;15.8637542852664;17.5649842553087;17.5649842553087;17.5649842553087;26.3442681704923;26.3442681704923;26.3442681704923];
y=[61.8;78.6;63.1;53.8;52.5;31.4;20.8;27.2;28.4;17.2;25.2;16.8;9.5;12.7;13.6;9.2;11.5];
x = x + 0.001 * rand(size(x))
[x, sortOrder] = sort(x, 'ascend');
y = y(sortOrder);
Tamara Schapitz
Tamara Schapitz 2020-4-9
Thanks for your answer! But the similar x-values are not the problem, there is no errror message or such. It DOES fit the data (as I can see in the plot), but the coefficients that are found, are not the correct ones...
I tried it with your fitNonLinearModel.m et voilà! With some small modifications, it works! And the coefficients are totally different from the other ones...
The above code returns:
p= 31.3705881793848 97.355156245024 6.39477241747793
and the adapted fitNonLinearModel.m returns:
coefficients =
10.6643318631924
398.728521237987
0.248005408824638
But I still don't see my mistake... Very strange... also because it worked for the old case two years ago... the only difference is the form of the exponential fit
'a-b*exp(-c*x)'
instead of
'a*exp(-((x/b)^c))'
Maybe I have to change some Syntax, too... Or change the starting value, like last time? Anyway... I have a solution for my plot, but if someone finds the error, I still would like to know...
Thanks!

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