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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

Nugroho Setiawan
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1 Answers1

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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.

Hunter Jiang
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