"System Identification" Assignment

Assignment 1: Recursive least square method parameter identification

Suppose the mathematical model of the identified system is described by the following formula:

Where x ( k ) is white noise with a variance of 0.1. Claim:

1. When the input signal u ( k ) is a white noise sequence with variance 1 , use the input and output data of the system to identify the parameters of the above model online;

2. When the input signal u ( k ) is an inverse M sequence with an amplitude of 1 , use the input and output data of the system to identify the parameters of the above model online;

3. When the input signal u ( k ) is a unit step signal , use the input and output data of the system to identify the parameters of the above model online;

Analyze and compare the effects of different input signals on the identification accuracy of system model parameters.
 

solution:

 

The code needs to be improved and has not been modified yet

【Reference Code】

① White noise sequence with variance 1

%递推最小二乘参数估计（RLS）
clear all; close all;
 
a=[1 -1.5 0.7 0.1]'; 
b=[1 2 1.5]'; d=2; %对象参数
na=length(a)-1; nb=length(b)-1; %na=3、nb=2为A、B阶次
 
L=400; %仿真长度
uk=zeros(d+nb,1); %输入初值：uk(i)表示u(k-i)
yk=zeros(na,1); %输出初值
u=randn(L,1); %输入采用白噪声序列
xi=sqrt(1)*randn(L,1); %白噪声序列  方差=1
 
theta=[a(2:na+1);b]; %对象参数真值
 
thetae_1=zeros(na+nb+1,1); %thetae初值
P=10^6*eye(na+nb+1); 
for k=1:L
    phi=[-yk;uk(d:d+nb)]; %此处phi为列向量
    y(k)=phi'*theta+xi(k); %采集输出数据
   
    %递推最小二乘法
    K=P*phi/(1+phi'*P*phi);
    thetae(:,k)=thetae_1+K*(y(k)-phi'*thetae_1);
    P=(eye(na+nb+1)-K*phi')*P;
    
    %更新数据
    thetae_1=thetae(:,k);
    
    for i=d+nb:-1:2
        uk(i)=uk(i-1);
    end
    uk(1)=u(k);
    
    for i=na:-1:2
        yk(i)=yk(i-1);
    end
    yk(1)=y(k);
end
plot([1:L],thetae); %line([1,L],[theta,theta]);
xlabel('k'); ylabel('参数估计a、b');
legend('a_1','a_2','a_3','b_0','b_1','b_2'); axis([0 L -2 1.5]);

 

 

②Inverse M sequence with amplitude 1

%递推最小二乘参数估计（RLS）
clear all; close all;
 
a=[1 -1.5 0.7 0.1]'; 
b=[1 2 1.5]'; d=2; %对象参数
na=length(a)-1; nb=length(b)-1; %na=3、nb=2为A、B阶次
 
L=400; %仿真长度
uk=zeros(d+nb,1); %输入初值：uk(i)表示u(k-i)
yk=zeros(na,1); %输出初值
%u=randn(L,1); %输入采用白噪声序列
xi=sqrt(1)*randn(L,1); %白噪声序列  方差=1

theta=[a(2:na+1);b]; %对象参数真值
 
thetae_1=zeros(na+nb+1,1); %thetae初值
P=10^6*eye(na+nb+1); 

%M序列及逆M序列的产生
%M序列长度 L=400; %仿真长度
x1=1; x2=1; x3=1; x4=0; %移位寄存器初值xi-1、xi-2、xi-3、xi-4
S=1; %方波初值

for k=1:L
    M(k)=xor(x3,x4); %进行异或运算,产生M序列
    IM=xor(M(k),S); %进行异或运算,产生逆M序列
    if IM==0
        u(k)=-1;
    else
        u(k)=1;
    end
    
    S=not(S); %产生方波
    
    x4=x3; x3=x2; x2=x1; x1=M(k); %寄存器移位
	
	
	phi=[-yk;uk(d:d+nb)]; %此处phi为列向量
    y(k)=phi'*theta+xi(k); %采集输出数据
   
    %递推最小二乘法
    K=P*phi/(1+phi'*P*phi);
    thetae(:,k)=thetae_1+K*(y(k)-phi'*thetae_1);
    P=(eye(na+nb+1)-K*phi')*P;
    
    %更新数据
    thetae_1=thetae(:,k);
    
    for i=d+nb:-1:2
        uk(i)=uk(i-1);
    end
    uk(1)=u(k);
    
    for i=na:-1:2
        yk(i)=yk(i-1);
    end
    yk(1)=y(k);
	
end

plot([1:L],thetae); %line([1,L],[theta,theta]);
xlabel('k'); ylabel('参数估计a、b');
legend('a_1','a_2','a_3','b_0','b_1','b_2'); axis([0 L -2 4]);

③Unit step signal

%递推最小二乘参数估计（RLS）
clear all; close all;
 
a=[1 -1.5 0.7 0.1]'; 
b=[1 2 1.5]'; d=2; %对象参数
na=length(a)-1; nb=length(b)-1; %na=3、nb=2为A、B阶次
 
L=400; %仿真长度
uk=zeros(d+nb,1); %输入初值：uk(i)表示u(k-i)
yk=zeros(na,1); %输出初值

u=ones(L,1);%输入采用单位阶跃信号

xi=sqrt(1)*randn(L,1); %白噪声序列  方差=1
 
theta=[a(2:na+1);b]; %对象参数真值
 
thetae_1=zeros(na+nb+1,1); %thetae初值
P=10^6*eye(na+nb+1); 
for k=1:L
    phi=[-yk;uk(d:d+nb)]; %此处phi为列向量
    y(k)=phi'*theta+xi(k); %采集输出数据
   
    %递推最小二乘法
    K=P*phi/(1+phi'*P*phi);
    thetae(:,k)=thetae_1+K*(y(k)-phi'*thetae_1);
    P=(eye(na+nb+1)-K*phi')*P;
    
    %更新数据
    thetae_1=thetae(:,k);
    
    for i=d+nb:-1:2
        uk(i)=uk(i-1);
    end
    uk(1)=u(k);
    
    for i=na:-1:2
        yk(i)=yk(i-1);
    end
    yk(1)=y(k);
end
plot([1:L],thetae); %line([1,L],[theta,theta]);
xlabel('k'); ylabel('参数估计a、b');
legend('a_1','a_2','a_3','b_0','b_1','b_2'); axis([0 L -2 1.5]);

 

 

Homework 2 Minimal variance self-tuning control experiment

Assuming that the mathematical model parameters of the controlled object with second-order pure lag are unknown or slowly time-varying, the following models are used in the simulation experiment:

Where x ( k ) is white noise with a variance of 0.1. Claim:

1. When the input yr( k ) is set as a step signal with an amplitude of 10 , the minimum variance direct self-correction control algorithm is designed to perform closed-loop control of the above objects;

2. When the input yr( k ) is set as a square wave signal with an amplitude of 10 , a direct self-correcting control algorithm with minimum variance is designed to perform closed-loop control of the above objects;

3. If the accused object model is changed to:

                                              

     Repeat the above experiments (1) and (2), what is the control result? Analyze the reasons.

 

【Reference Code】

%最小二乘参数估计
clear all;

 

 

Homework 3 Model reference adaptive control experiment

Suppose the model parameters of the controlled object are unknown or slowly time-varying, but its state variables are fully observable,

In the simulation, the state equation is taken as:

 

Select reference model:

 

The model reference adaptive control system with fully observable state is shown in the figure below:

The controller adaptive law is:

 

【Reference Code】

%最小二乘参数估计
clear all;