How to use LQR for setpoint tracking?

Question

Pedro Carvalho 2024-5-1

1
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2113951-how-to-use-lqr-for-setpoint-tracking

编辑： khalid 2024-8-26

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

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

% 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')

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Answer 1

Sam Chak 2024-5-1

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2113951-how-to-use-lqr-for-setpoint-tracking#answer_1450881

编辑：Sam Chak 2024-5-1

在 MATLAB Online 中打开

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

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')

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Pedro Carvalho 2024-5-2

编辑：Pedro Carvalho 2024-5-2

Yes, makes sense. Thanks!

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.

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Answer 2

Joshua Levin Kurniawan 2024-5-1

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2113951-how-to-use-lqr-for-setpoint-tracking#answer_1450851

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.