Least Square Parameter Estimation with Matlab

Question

I tried to find a value of a parameter of the system namely(a1 and a2) using Least Square with my own models.

My Model is MISO system: % tz(k) = a1*(((1/2*t)*(to*to(k-1)))- (1/2*t)*(tz*tz(k-1))) + a2*(((1/2*t)*(tn*tn(k-1)))- (1/2*t)*(tx*tx(k-1)));

My method to calculate the parameter : Xθ = (X'X)^(-1)X'Y

I was tried to fit the REGRESSOR with my own models, but it always showed error, where I made a mistake??

 close all;
clear all;
clc;


%% =====================
% MISO system using Least Square
% Model of the system, described by diference equation
% tz(k) = a1*(((1/2*t)*(to*to(k-1)))- (1/2*t)*(tz*tz(k-1))) + a2*(((1/2*t)*(tn*tn(k-1)))- (1/2*t)*(tx*tx(k-1)));


% Sample Data input and output of the system,
t = 15 ; %sampling time 15 minutes
to=[23; 23; 23; 23; 23; 23; 23;]; %input 1 
tz=[26; 27; 27; 27; 26; 27; 26]; %output 
tn=[26; 27; 27; 27; 26; 27; 26]; %input 2
tx=[26; 27; 27; 27; 26; 27; 26]; %input 3

% Problem: Find the value of a1 and a2 with Least Square Algorithm.


%% =====================
% Least Square
% model: Y=Reg*Par; 
% Reg: regresor, Par: parameter.

% Arranging elemen Y, Reg, and Par:
% Y=[y(2); ...; y(N);
% Reg=[-y(1) u(1); ....; -y(N-1) u(N-1)];
% Reg=[-y(1) u(1); ....; -y(N-1) u(N-1)];
% Par=[a1;b1];

% Find value of Parameter:
% Par=(Reg.'*Reg)\Reg.'*Y;


%% =====================
% Program MATLAB

N=length(tz);    % Count Total element vektor tz.

% Register elemen Y
Y=tz(2:end,1); 

% Register elemen Regressor
Reg=[((1/2*t)*to*(to(1:N-1,1)))-((1/2*t)*tz*(tz(1:N-1,1))) ((1/2*t)*tn*(tx(1:N-1,1)))-((1/2*t)*tn*(tx(1:N-1,1)))];

% Least Square for getting the value of  a1 and b4:
Par=(Reg.'*Reg)\Reg.'*Y;

disp('Value of a1 and a2 :');
a1=Par(1)
a2=Par(2)

Thanks in advance

Answer 1

Try:

Reg=zeros(6,2);
for ii=2:length(to)
Reg(ii-1,:) = [(((1/2*t)*(to(ii)*to(ii-1)))- (1/2*t)*(tz(ii)*tz(ii-1))),...
             (((1/2*t)*(tn(ii)*tn(ii-1)))- (1/2*t)*(tx(ii)*tx(ii-1)))];
end

Or:

Reg1 = [(((1/2*t)*(to(1:end-1).*to(2:end)))- (1/2*t)*(tz(1:end-1).*tz(2:end))),...
             (((1/2*t)*(tn(1:end-1).*tn(2:end)))- (1/2*t)*(tx(1:end-1).*tx(2:end)))];

isequal tells me that they get same answer.

However, your data must have some problem since the second column of Reg is all zeros and Par(2) will become NaN .