Error Reduction in Regression Model

3 views (last 30 days)
Chris Dan
Chris Dan on 1 Sep 2020
Hello,
I have this problem, where I have to estimate force coefficients. I have the time based forces, of which I take the FFT, they are my reference forces. Then I build matrices to numerically estimate those FFT forces. To get the closest estimate, I need to multiply the matrices with some coeficients and I need to find those coefficients.
Here is my code:
%Main Script
% Original coefficients are
coeff_original = [800;200;10;10];
%FFT values when time is from 0.05 till 1.2590
b_reference = [465.9498;243.369;516.6983;452.2795;81.5134;179.418;65.7858;130.3574;58.6862;69.4206];
% Matrices for Esimating FFT of 0,1,2,3,4 order
Fourier_coeff = [-0.343774677078494 -0.420221293933872 -7.01727121110308 -5.72957795130823;0.420221293933872 -0.343774677078494 5.72957795130823 -7.01727121110308;-0.0195319807839135 - 0.207374228023668i -0.214334169775653 - 0.203163837145307i 2.22772613453351 - 2.97069401648263i -1.76676739751800 - 2.94900590421752i;0.214334169775653 + 0.203163837145307i -0.0195319807839135 - 0.207374228023668i 1.76676739751800 + 2.94900590421752i 2.22772613453351 - 2.97069401648263i;0.00648700344488058 - 0.0121247033805691i -0.0758111625117073 - 0.0914951429790348i -1.16516765707854 + 0.0624026969049595i -0.722094909290415 - 0.717554549271970i;0.0758111625117073 + 0.0914951429790348i 0.00648700344488058 - 0.0121247033805691i 0.722094909290415 + 0.717554549271970i -1.16516765707854 + 0.0624026969049595i; -0.00654908785905161 - 0.0247682539100665i -0.0704081380252118 - 0.0388380755068885i 0.675644862253467 + 0.0894650519433900i -0.652483947613814 - 0.486314732529423i; 0.0704081380252118 + 0.0388380755068885i -0.00654908785905161 - 0.0247682539100665i 0.652483947613814 + 0.486314732529423i 0.675644862253467 + 0.0894650519433900i;-0.0148715743865258 - 0.00418105270557728i -0.0566466827112845 - 0.00986732211865589i -0.460065129690790 - 0.560042604092437i -0.617705886360175 - 0.0876662356015614i; 0.0566466827112845 + 0.00986732211865589i -0.0148715743865258 - 0.00418105270557728i 0.617705886360175 + 0.0876662356015614i -0.460065129690790 - 0.560042604092437i]
% Multiplication factor for Fourier Coefficients because FFT values are doubled by Matlab
Y =[1;1;2;2;2;2;2;2;2;2];
% Final matrix A
A= bsxfun(@times,Fourier_coeff,Y); % Ax = b
Pre_Conditioner =[2;0.1;0.05;2;2;2;200;2000.5;0.03;5].^(-1);
V = bsxfun(@times,A,Pre_Conditioner);
b_reference_2 = Pre_Conditioner.*b_reference;
%initial value of coefficients
c = inv(A.'*A)*A.'*b_reference
c = abs(c);
%% to solve the system Ax = b
Starting_Angle = 0;
Total_error = 16;
while Total_error >15
fun = @(x)Solve (x(1),x(2),x(3),x(4),A,Starting_Angle)
% Function to reduce error in Least Squares
[coeff_Numerical] = fminsearch(fun,c)
[Total_error,b_numerical]=Solve_2(coeff_Numerical(1),coeff_Numerical(2),coeff_Numerical(3),coeff_Numerical(4),A,Starting_Angle)
end
% the function Solve is as follows:
function [error] = Solve(x1, x2, x3, x4, A,starting_angle)
% Preconditioner
Pre_Conditioner =[105.738;1073.7;467.6315;31.7839;3.7720;256.0062;94.7778;396.1459;155.5473;48.1352].^(-1);
% FFT values when time is form 0 till 1.2590
%b_reference = [485.9;251.975;358.55;313.6565;85.322;193.4422;53.425;106.66;71.88;84.8653];
%FFT values when time is from 0.5 till 1.2590
%b_reference = [293.5914;151.6237;215.7961;188.6205;50.1612;115.8065;26.0603;64.289;43.225;50.8902];
%FFT values when time is from 0.05 till 1.2590
b_reference = [465.9498;243.369;516.6983;452.2795;81.5134;179.418;65.7858;130.3574;58.6862;69.4206];
% A = bsxfun(@times,A,Pre_Conditioner);
% b_reference = Pre_Conditioner.*b_reference;
x = [x1;x2;x3;x4];
% Solve the equation Ax = b using Least Squares Method
% Numerical Method
b_2 = A*x
b_2 = abs(b_2);
% Error calculation
for i =1:1:size(b_reference,1)
error_percent(i,1) = (abs((b_reference(i)-b_2(i)))/b_reference(i))*100
end
error = error_percent(1,1)+error_percent(2,1)+error_percent(3,1)+error_percent(4,1)+error_percent(5,1)+error_percent(6,1)+error_percent(7,1)+error_percent(8,1)+error_percent(9,1)+error_percent(10,1)
%error = sum(error_percent,1)/size(error_percent,1)
end
% the function Solve_2 is as follows:
function [error,b_2 ] = Solve_2(x1, x2, x3, x4, A,starting_angle)
% Preconditioner
%Pre_Conditioner =[105.738;1073.7;467.6315;31.7839;3.7720;256.0062;94.7778;396.1459;155.5473;48.1352].^(-1);
% FFT values when time is form 0 till 1.2590
%b_reference = [485.9;251.975;358.55;313.6565;85.322;193.4422;53.425;106.66;71.88;84.8653];
%FFT values when time is from 0.5 till 1.2590
%b_reference = [293.5914;151.6237;215.7961;188.6205;50.1612;115.8065;26.0603;64.289;43.225;50.8902];
%FFT values when time is from 0.05 till 1.2590
b_reference = [465.9498;243.369;516.6983;452.2795;81.5134;179.418;65.7858;130.3574;58.6862;69.4206];
%
%
% A = bsxfun(@times,A,Pre_Conditioner);
% b_reference = Pre_Conditioner.*b_reference;
x = [x1;x2;x3;x4];
% Solve the equation Ax = b using Least Squares Method
% Numerical Method
b_2 = A*x % Least Square Estimation
b_2 = abs(b_2);
% Error calculation
for i =1:1:size(b_reference,1)
error_percent(i,1) = (abs((b_reference(i)-b_2(i)))/b_reference(i))*100
end
error = error_percent(1,1)+error_percent(2,1)+error_percent(3,1)+error_percent(4,1)+error_percent(5,1)+error_percent(6,1)+error_percent(7,1)+error_percent(8,1)+error_percent(9,1)+error_percent(10,1)
%error = sum(error_percent,1)/size(error_percent,1)
end
The program is running but the error is not reducing, basically it is an optimisation problem. I find the initial estimate of hte coeff and from those I calculate the simulated forces. Then I find the error between the simulated forces and the reference forces and try to reduce the error by finding different values of coeff.
I am trying to optimise the coeff, which will be near to the original values, I know the original values but the program has to find out by reducing the error.
Does anybody has leads, how to do it? Any thing will be appreaciated
I used lsqnonlin as well, now I am using fminsearch....

Answers (0)

Categories

Find more on Electrical Block Libraries in Help Center and File Exchange

Products


Release

R2019a

Community Treasure Hunt

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

Start Hunting!