How to use LQR for setpoint tracking?

100 次查看（过去 30 天）
Pedro Carvalho
Pedro Carvalho 2024-5-1
编辑： khalid 2024-8-26
Ran in:
Initially I was using LQR to regulate the error dynamics, i.e., I computed the gains for de = (A - BK)e, but this basically results in a PI controller since the control law (with integral action) is u = -K*e + ki*z. I have seen many sources teaching how to do it by augmenting the state vector with the integral of the error, expanding the matrices to A = [A 0; C 0], etc. but I still can't understand how that works. I am working on a first order cruise control problem. From my observations the integral action is doing all the tracking and the -Kx term is only getting in the way, trying to regulate the state x to zero. Here is my code:
% Parameters
X_u = 0;
X_uu = 22.7841;
m = 5037.7;
% Equilibrium point x0
x0 = 20*1.852/3.6; % Longitudinal linear velocity in m/s
u0 = X_u*x0 + X_uu*x0^2 % Thrust in N
u0 = 2.4120e+03
% Linearize system around x0
A = -(X_u/m + 2*X_uu/m*x0);
B = 1/m;
C = 1;
D = 0;
% System order
n = size(A,1);
% Open loop system
sys_ol = ss(A,B,C,D);
openloopPoles = eig(A)
openloopPoles = -0.0931
% Augmented system with the integral of the error
A_hat = [A zeros(n,1);...
-C 0 ];
B_hat = [B; 0];
Br = [zeros(n,1); 1];
C_hat = [1 zeros(1,n)];
D_hat = 0;
% Q R matrices
Q = 1000*(C'*C);
R = 0.5e-3;
% Feedback gain
K_hat = lqr(A_hat, B_hat,Q,R)
K_hat = 1x2
1.0e+03 * 3.5893 -1.4142
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
K = K_hat(1)
K = 3.5893e+03
ki = -K_hat(2)
ki = 1.4142e+03
% Scaling matrix
%Nbar = rscale(sys_ol,K)
% Closed loop system
AA = [A - B*K B*ki;-C 0];
BB = Br;
CC = [C 0];
DD = 0;
sys_cl = ss(AA, BB, CC, DD);
% Time vector
t = 0:0.1:30;
% Control input
u = u0*ones(size(t));
% Reference setpoint
r = 25*1.852/3.6*ones(size(t));
% Initial states
x0_hat = [x0,0]
x0_hat = 1x2
10.2889 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
% Simulate the response of the system
[y,t,x_hat] = lsim(sys_cl,r,t,x0_hat);
figure
plot(t, y*3.6/1.852, 'k', 'LineWidth', 1.5,'Color','k')
xlabel('Time (seconds)')
ylabel('Speed (knots)')
title('Closed loop response with integrator')
grid on
% Control effort (Thrust)
u_effort = -K*x_hat(:,1) + ki*x_hat(:,2);
figure
plot(t, u_effort,'Color','k')
xlabel('Time (s)')
ylabel('Thrust (N)')
title('Control effort')

请先登录，再回答此问题。

采纳的回答

Sam Chak
Sam Chak 2024-5-1
编辑：Sam Chak 2024-5-1
Ran in:
Hi @Pedro Carvalho
Your code appears to be error-free. However, the control action you implemented differs from the error-based PI control scheme that was mentioned.
To comprehend why integral control can track the setpoint, it is important to visualize that the plant is a 1st-order transfer function, denoted as , where the plant lacks an integrator (referred to as a Type-0 system).
The state-feedback control term will shape and enhance the transient behavior, resulting in . Nevertheless, a steady-state error will persist since .
To eliminate the steady-state error, introducing an integrator in the cascade compensation path is necessary, transforming it into a Type-1 system. This results in . The closed-loop system can then be expressed as . Consequently, the steady-state error is eliminated since .
Update: In the code, the initial value of the Integrator output (2nd state variable, z0) should be set to a non-zero value. This is necessary because the initial velocity (x0) is non-zero. Therefore, the initial value of the Integrator output can be calculated by solving the control law and considering the initial error (r0 - x0).
% Parameters
X_u = 0;
X_uu = 22.7841;
m = 5037.7;
% Equilibrium point x0
x0 = 20*1.852/3.6; % Longitudinal linear velocity in m/s
u0 = X_u*x0 + X_uu*x0^2; % Thrust in N
% Linearize system around x0
A = -(X_u/m + 2*X_uu/m*x0)
A = -0.0931
B = 1/m
B = 1.9850e-04
C = 1;
D = 0;
% System order
n = size(A,1);
% Open loop system
sys_ol = ss(A,B,C,D);
Gp = tf(sys_ol)
Gp = 0.0001985 ----------- s + 0.09307 Continuous-time transfer function.
openloopPoles = eig(A);
% Augmented system with the integral of the error
A_hat = [A zeros(n,1);...
-C 0 ];
B_hat = [B;
0];
Br = [zeros(n,1); 1];
C_hat = [1 zeros(1,n)];
D_hat = 0;
% Q R matrices
Q = 1000*(C'*C);
R = 0.5e-3;
% Feedback gain
K_hat = lqr(A_hat, B_hat,Q,R)
K_hat = 1x2
1.0e+03 * 3.5893 -1.4142
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
K = K_hat(1);
ki = -K_hat(2);
% Scaling matrix
% Nbar = rscale(sys_ol,K)
% Closed loop system
AA = [A-B*K, B*ki;
-C, 0];
BB = Br;
CC = [C 0];
DD = 0;
sys_cl = ss(AA, BB, CC, DD);
A-B*K
ans = -0.8056
B*ki
ans = 0.2807
Gcl = tf(sys_cl)
Gcl = 0.2807 ----------------------- s^2 + 0.8056 s + 0.2807 Continuous-time transfer function.
% steady-state response
ssr = dcgain(Gcl)
ssr = 1.0000
% Time vector
t = 0:0.1:30;
% Control input
u = u0*ones(size(t));
% Reference setpoint
r0 = 25*1.852/3.6;
r = r0*ones(size(t));
% Initial states
z0 = (u0 + K*x0)/ki + (r0 - x0);
x0_hat = [x0, z0];
% Simulate the response of the system
[y, t, x_hat] = lsim(sys_cl, r, t, x0_hat);
figure
plot(t, y*3.6/1.852, 'k', 'LineWidth', 1.5, 'Color', '#265EF5')
xlabel('Time (seconds)')
ylabel('Speed (knots)')
title('Closed loop response with integrator')
grid on
% Control effort (Thrust)
u_effort = -K*x_hat(:,1) + ki*x_hat(:,2);
figure
plot(t, u_effort, 'LineWidth', 1.5, 'Color', '#F15EF5'), grid on
xlabel('Time (s)')
ylabel('Thrust (N)')
title('Control effort')
  4 个评论
显示 2更早的评论
khalid
khalid 2024-8-26
编辑：khalid 2024-8-26
Hi guys! Could anyone please tell me while choosing the augmented A and B why did you choose zeros(n,1) ? Why that specific size? I know the theory says Aaug= [A 0; -C 0] I just don't understand then how is it that size? Also how do I know which one is my Kx and which one is my Ki. My lqr gain is a 2x5 matrix.
I have 5 states in my system but I only want to track 2 and regulate the other 3. How do I do this? How do I choose the integrator? In your case you had only 1 state judging by your A matrix. Please help in this regard.

请先登录，再进行评论。

更多回答（1 个）

Joshua Levin Kurniawan
Hello Pedro. Regarding the LQR controller it has a unique properties. In this case, without an integral action, the controller only tries to compesate the system into a steady state condition (i.e. it does not necessarily reach ), or in another word, we only want to regulate the system to have a stable behaviour. However, this is different for the case where we added the integral action, where the error e is supressed to reach zero.
Lets say that we want to track a specific state, , which we modelled as y for convenience. Then, as the standard state space term,
.
Then, the control function can be defined as . Here, we want to suppress the error e to equal to zero, naturally, we want to add them into the state matrix, which we called the augmented matrix . Remember that
Hence, the matrix can be defined as
Therefore, by conducting the LQR method to the augmented system as described as above, you can get the optimal control full-state feedback gain matrix for the integrating action.

类别

在 Help Center 和 File Exchange 中查找有关 State-Space Control Design 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by