LQR regulator doesn't regulate the system

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DoctorCrow
DoctorCrow 2023-3-21
评论： Sam Chak 2023-3-22
Hello,
I used the function lqr to get a regulator that helps me to regulate my system.
My function is:
r = lqr(A,B,Q,R)
With this function i get a regulator that gives me two conjugate complex poles. Thus the system is not regulated, but continues to oscillate. I don't get any errors or warnings.
How can I fix this problem? .
  2 个评论
Sam Chak
Sam Chak 2023-3-22
Hi, in order to investigate, could you show the matrices for A, B, Q, R?
Also provide the desired settling time if you do not have any specific constraint for the control input u.
If (A, B) is controllable, then the desired settling time is sufficient for the eigenvalue design.
DoctorCrow
DoctorCrow 2023-3-22
Hi,
i got the followingen matrices:
A = [0 1 0 0 ; 0 0 14.72 0 ; 0 0 0 1 ; 0 0 -3.066 0]
B = [0 ; 0.001 ; 0 ; -0.000125]
Q = [1 0 0 0 ; 0 1 0 0 ; 0 0 1 0 ; 0 0 0 1]
R = [1]
I checked, that (A,B) is controllable.

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回答（1 个）

Sam Chak
Sam Chak 2023-3-22
Ran in:
Hi @DoctorCrow
Your selection of and results in the closed-loop eigenvalues having negative real parts very close to zero. Consequently, although the system states converge to steady-state, they also oscillate for a much longer period of time.
A = [0 1 0 0; 0 0 14.72 0; 0 0 0 1; 0 0 -3.066 0];
B = [0; 0.001; 0; -0.000125];
Q = [1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1];
R = 1;
[K1, P, e] = lqr(A, B, Q, R)
K1 = 1×4
1.0000 70.7226 4.6434 333.8955
P = 4×4
1.0e+06 * 0.0001 0.0025 0.0003 0.0120 0.0025 0.1769 0.0116 0.8491 0.0003 0.0116 0.1401 0.0557 0.0120 0.8491 0.0557 4.1219
e =
-0.0004 + 1.7510i -0.0004 - 1.7510i -0.0141 + 0.0141i -0.0141 - 0.0141i
sys1 = ss(A-B*K1, B, eye(4), 0);
step(sys1)
Increasing Q should improve the system response, as shown in the following plot.
Q = 1e7*eye(4);
[K2, P, e] = lqr(A, B, Q, R)
K2 = 1×4
1.0e+03 * 3.1623 4.9643 1.8176 7.6882
P = 4×4
1.0e+08 * 0.1570 0.0732 0.2431 0.3328 0.0732 0.1397 0.0489 0.7207 0.2431 0.0489 5.3231 0.2455 0.3328 0.7207 0.2455 5.1509
e =
-0.2486 + 1.1213i -0.2486 - 1.1213i -1.3865 + 0.0000i -2.1197 + 0.0000i
sys2 = ss(A-B*K2, B, eye(4), 0);
step(sys2)
  2 个评论
Sam Chak
Sam Chak 2023-3-22
You are welcome, @DoctorCrow. If you find the solution helpful, please consider accepting ✔ and voting 👍 on the answer. Thanks a bunch! 🙏

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