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I have the data:

X = [0; 0.5; ...]

Y = [0; 0.8; ...]

And the function to which I want to model the data:

F = a + exp ^ (- x * b)

How do I find "a" and "b" for a good curve fit?

I should not use cftool.

Image Analyst
on 22 Apr 2017

If you don't have the Curve Fitting Toolbox but have the Statistics and Machine Learning Toolbox, you can use fitnlm():

% Uses fitnlm() to fit a non-linear model (an exponential decay curve) through noisy data.

% Requires the Statistics and Machine Learning Toolbox, which is where fitnlm() is contained.

% Initialization steps.

clc; % Clear the command window.

close all; % Close all figures (except those of imtool.)

clear; % Erase all existing variables. Or clearvars if you want.

workspace; % Make sure the workspace panel is showing.

format long g;

format compact;

fontSize = 20;

% Create the X coordinates from 0 to 20 every 0.5 units.

X = 0 : 0.5 : 20;

% Define function that the X values obey.

a = 10 % Arbitrary sample values I picked.

b = 0.4

Y = a + exp(-X * b); % Get a vector. No noise in this Y yet.

% Add noise to Y.

Y = Y + 0.05 * randn(1, length(Y));

% Now we have noisy training data that we can send to fitnlm().

% Plot the noisy initial data.

plot(X, Y, 'b*', 'LineWidth', 2, 'MarkerSize', 15);

grid on;

% Convert X and Y into a table, which is the form fitnlm() likes the input data to be in.

tbl = table(X', Y');

% Define the model as Y = a + exp(-b*x)

% Note how this "x" of modelfun is related to big X and big Y.

% x((:, 1) is actually X and x(:, 2) is actually Y - the first and second columns of the table.

modelfun = @(b,x) b(1) + exp(-b(2)*x(:, 1));

beta0 = [10, .4]; % Guess values to start with. Just make your best guess.

% Now the next line is where the actual model computation is done.

mdl = fitnlm(tbl, modelfun, beta0);

% Now the model creation is done and the coefficients have been determined.

% YAY!!!!

% Extract the coefficient values from the the model object.

% The actual coefficients are in the "Estimate" column of the "Coefficients" table that's part of the mode.

coefficients = mdl.Coefficients{:, 'Estimate'}

% Create smoothed/regressed data using the model:

yFitted = coefficients(1) + exp(-coefficients(2)*X);

% Now we're done and we can plot the smooth model as a red line going through the noisy blue markers.

hold on;

plot(X, yFitted, 'r-', 'LineWidth', 2);

grid on;

title('Exponential Regression with fitnlm()', 'FontSize', fontSize);

xlabel('X', 'FontSize', fontSize);

ylabel('Y', 'FontSize', fontSize);

legendHandle = legend('Noisy Y', 'Fitted Y', 'Location', 'north');

legendHandle.FontSize = 25;

% Set up figure properties:

% Enlarge figure to full screen.

set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);

% Get rid of tool bar and pulldown menus that are along top of figure.

set(gcf, 'Toolbar', 'none', 'Menu', 'none');

% Give a name to the title bar.

set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off')

Walter Roberson
on 22 Apr 2017

The best I find for the 7-coefficient model, using custom code that I have not released, is

[1.95345951298651754, -0.271402425481958642, 0.114129599568334961, 0.6048791673085121, 0.000138307518530737773, 1.61189969373566888, -0.0124593500324885736]

You can get much the same result using cftool if you adjust the bounds. For example for each positive number above, set the lower bound to 0 and upper to 2; and for each negative number above set the lower bound to -2 and the upper to 0.

I evaluated over a large range of coefficients; there are some other locations that are not bad, but this was the best I found. For example, at

[1.95618718619627652, -0.206949316792977123, 48925.8303936270968, 1.56148395793133599, 0.011815555675624435, 0.601652543952944985, -0.000134723232812764676]

the residue is only about 15% larger.

Walter Roberson
on 23 Apr 2017

For your reduced system with 5 coefficients,

Y= a-b*exp(-c*X)-d*exp(-e*X)

there are two minima that are equivalent, one near

[1.96086280008987 0.596272887303567 0.000128842293439593 1.50536232706941 0.0109031671832912]

and the other near

[1.96086281505995 1.50536227966151 0.0109031664979331 0.596272910300341 0.000128842291110631]

You can get close to the first of those in a fraction of a second using cftool custom equation and using the Fit Options to constrain each of the parameters between 0 and 2.

Alex Sha
on 24 Feb 2020

The below should be the best solution:

Root of Mean Square Error (RMSE): 0.0255893453091926

Sum of Squared Residual: 0.047146650721423

Correlation Coef. (R): 0.999022475612635

R-Square: 0.998045906779199

Adjusted R-Square: 0.997989266395987

Determination Coef. (DC): 0.998045906779199

Chi-Square: 0.0562099503181627

F-Statistic: 5533.08505193596

Parameter Best Estimate

---------- -------------

a 1.95345951949563

b -0.271402458132563

c 0.114129545758927

d 1.61189973355979

e 0.0124593503150589

f 0.604879174590309

g -0.000138307512053813

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