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How can I fit an exponential curve?

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Below is an example of finding a fit with only one term of exponential term but I dont know how to find the fit of the curve when it has 2 degree of exponential term, i.e.[y = a*e^(bx) + c*e^(dx)]
example for y = a*e^(bx)
phi =[ones(size(xx)),xx];
yfit = exp(phi*aa);
plot(xx, yy, ro, xx, yfit, k-) ;

Accepted Answer

Star Strider
Star Strider on 15 Nov 2018
Edited: Star Strider on 15 Nov 2018
Try this:
filename1 = 'x2.mat';
D1 = load(filename1);
x = D1.x2;
filename2 = 'y2.mat';
D2 = load(filename2);
y = D2.y2;
fcn = @(b,x) b(1).*exp(b(2).*x) + b(3).*exp(b(4).*x);
[B,fval] = fminsearch(@(b) norm(y - fcn(b,x)), ones(4,1));
plot(x, y, 'p')
hold on
plot(x, fcn(B,x), '-')
hold off
The fitted parameters are:
B =
How can I fit an exponential curve - 2018 11 14.png
EDIT — Added plot figure.
Star Strider
Star Strider on 15 Nov 2018
This line:
fcn = @(b,x) b(1).*exp(b(2).*x) + b(3).*exp(b(4).*x);
is the objective function, the expression that describes the function to fit to the data.
This line:
[B,fval] = fminsearch(@(b) norm(y - fcn(b,x)), ones(4,1))
calls the fminsearch function to fit the function to the data. The norm function compares the function output to the data and returns a single scalar value (the square root of the sum of squares of the difference between the function evaluation and the data here), that fminsearch uses. I refer you to the documentation on fminsearch (link) for details on how it works.
The value would be calculated as:
Rsq = 1 - sum((y - fcn(B,x)).^2) / sum((y - mean(y)).^2)
Rsq =
We are not comparing models, so this is the only statistic available. There are several ways to compare models, a subject much more involved than I will go into here. See any good text on nonlinear parameter estimation for details.
@fengsen huang —
If my Answer helped you solve your problem, please Accept it!

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More Answers (2)

Image Analyst
Image Analyst on 15 Nov 2018
Here's another way using fitnlm(). I get
coefficients =
formulaString =
'Y = 0.012 * exp(6.048 * X) + -1.141 * exp(-12.678 * X)'
Quite a bit different than Star's numbers.
0000 Screenshot.png

Arturo Gonzalez
Arturo Gonzalez on 8 Sep 2020
Per this answer, you can do it with the following matlab code
clear all;
% get data
dx = 0.001;
x = (dx:dx:1.5)';
y = -1 + 5*exp(0.5*x) + 4*exp(-3*x) + 2*exp(-2*x);
% calculate n integrals of y and n-1 powers of x
n = 3;
iy = zeros(length(x), n);
xp = zeros(length(x), n+1);
iy(:,1) = cumtrapz(x, y);
xp(:,1) = x;
for ii=2:1:n
iy(:, ii) = cumtrapz(x, iy(:, ii-1));
xp(:, ii) = xp(:, ii-1) .* x;
xp(:, n+1) = ones(size(x));
% get exponentials lambdas
Y = [iy, xp];
A = pinv(Y)*y;
Ahat = [A(1:n)'; [eye(n-1), zeros(n-1, 1)]];
lambdas = eig(Ahat);
% get exponentials multipliers
X = [ones(size(x)), exp(lambdas'.*x)];
P = pinv(X)*y;
% show estimate
y_est = X*P;
plot(x, y); hold on;
plot(x, y_est, 'r--');

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