LSQcurvefit does not yield same result for comparable data sets

1 view (last 30 days)
I have used the following code to fit my data. Key is that the fit captures the peak at the start of the curve.
For a lot of data sets, the method works (example are x_good,y_good).
For two sets this method does not work adequate enough (x5, y5 and x10, y10)
Anyone who can see if I am doing something wrong? Or give a different method which works better?
clear all
clc
close all
% Working data set
x_good = [0 0.0025 0.020666667 0.0915 0.129 0.169916667 0.204 0.2415 0.299833333 0.359833333 0.409833333 0.459833333 0.589833333 0.674833333];
y_good = [0.000001 -1.791689626 -2.085814283 -1.254192484 -1.130978553 -0.852663995 -0.650156083 -0.605732792 -0.552092535 -0.466707459 -0.381595899 -0.341840176 -0.112360934 -0.107089368];
%Not working data set
x5 = [0 0.0025 0.020666667 0.0915 0.129 0.169916667 0.204 0.299833333 0.359833333 0.509833333 0.589833333 0.674833333];
y5 = [0.000001 -1.512917459 -1.397221246 -0.861543826 -0.678138048 -0.538930943 -0.440640054 -0.253865251 -0.192352332 0.001791084 0.078138742 0.156925367];
%Not working data set
x10 = [0 0.0025 0.020666667 0.0915 0.129 0.169916667 0.204 0.2415 0.299833333 0.359833333 0.409833333 0.459833333 0.589833333 0.674833333];
y10 = [0.000001 -0.909146404 -1.389416499 -0.736431181 -0.767464076 -0.430784784 -0.298350016 -0.477736703 -0.174485909 -0.10975744 -0.038531763 0.009471926 0.133414928 0.178482903];
% LSQ
t = x5;
y =y10;
xspace = linspace(t(1), t(end), 1000);
options = optimoptions('lsqcurvefit', 'MaxFunctionEvaluations', 100e3)
% fit function y = c(1)*exp(-lam(1)*t) + c(2)*exp(-lam(2)*t)
F = @(x,t)(x(1)*exp(-x(2)*t) + x(3)*exp(-x(4)*t));
x4 = [1 1 1 0];
[x,resnorm,~,~,output] = lsqcurvefit(F,x4,t,y, [], [], options)
figure
plot(t,y,'ro')
title("Least squared method ")
hold on
plot(t,F(x,t), 'r')
hold on
plot(xspace, F(x, xspace), '--r')
set(gca, 'YDir','reverse')
legend("data points", "LSQ, resnorm: " + resnorm, "LSQ continous")

Accepted Answer

Torsten
Torsten on 29 Sep 2022
Use
x4 = [-2.1782 5.0283 2.1782 720.8491];
instead of
x4 = [1 1 1 0];
as initial guess for the parameters.
  2 Comments

Sign in to comment.

More Answers (0)

Products


Release

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by