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LQR controller tuning in a closed loop system problem

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Hi, I made some code for the LQR controller in a closed loop to compare open loop system in a discrete time system.
But the result is not what i expected that state trajectories in a closed loop is worse than the open loop system.
And one state in the closed loop is not converging fast enough.
Please check my code. I think it's a tuning problem.
% LQR DC motor control
% State space model of a DC motor
Ra= 1; %Armature electric resistance[Ohm]
La= 0.5; %Electric inductance[H]
Ki= 0.023; %Motor torque constant= Electromotive force constant[Nm/A]
Jm= 0.01; %Moment of inertia of the rotor[kgm^2]
Kb= 0.023;
Bm= 0.00003;
A=[-Ra/La -Kb/La; Ki/Jm -Bm/Jm ];
B=[1/La; 0];
C=[0 1];
D=0;
Ts= 0.2; %Sampling time
Np= 10; %Prediction horizon
[n,d]=ss2tf(A,B,C,D); %Denominator and Nominator of Transfer function
Ct = tf(n,d); %Continous tf
Cs = ss(tf(n,d)); %Continous ss
Dt = c2d(tf(n,d), Ts,'zoh'); %Discretize tf with sampling time Ts
Ds= c2d(Cs, Ts); %Discretize ss
%% define the desired state
xd=[0; 0];
% compute ud(desired control input)
ud=-inv((Ds.B)'*(Ds.B))*(Ds.B)'*(Ds.A)*xd;
%% simulate the system response for the computed ud
% set initial condition
x0=1*randn(2,1);
%final simulation time
tFinal=100;
time_total=0:Ts:tFinal;
input_ud=ud*ones(size(time_total));
[output_ud_only,time_ud_only,state_ud_only] = lsim(Ds,input_ud,time_total,x0);
% figure(1)
% plot(time_total,state_ud_only(:,1),'k')
% hold on
% plot(time_total,state_ud_only(:,2),'r')
% xlabel('Time [s]')
% ylabel('states')
% legend('i_{a}','\omega_{m}')
% grid
%% Design LQR controller
Q = diag([0,300]); % State weighting matrix
%Q = 1*Ds.C'*Ds.C; %Bryson's Rule to define your initial weighted matrices Q
R = 0.01; % Input weighting matrix
% Compute LQR gain matrix K
[K,S,e] = dlqr(Ds.A, Ds.B, Q,R,0);
% eigenvalues of the closed loop system:AS since eig in unit circle
[T,D] = eig(Ds.A-Ds.B*K);
% eigenvalues of the open-loop system %stable
eig(Ds.A);
%simulate the closed loop system
%closed loop matrix
Acl= Ds.A-Ds.B*K;
lqrClosedLoop=ss(Acl, Ds.B, Ds.C,Ds.D);
closed_loop_input=0.2*ones(size(time_total));
%
%figure(2)
%step(lqrClosedLoop)
[output_closed_loop,time_closed_loop,state_closed_loop] = lsim(lqrClosedLoop,closed_loop_input,time_total,x0);
figure(3)
plot(time_total,state_ud_only(:,1),'k')
hold on
plot(time_total,state_ud_only(:,2),'r')
xlabel('Time [s]')
ylabel('states')
legend('i_{a}','\omega_{m}')
plot(time_closed_loop,state_closed_loop(:,1),'m')
hold on
plot(time_closed_loop,state_closed_loop(:,2),'b')
legend('i_{a}','\omega_{m}','i_{a}-Closed Loop ','\omega{m}-Closed Loop')
grid

Risposta accettata

Sam Chak
Sam Chak il 19 Mar 2023
The LQR in this example is designed in continuous-time. Can you obtain the discretized control system?
% LQR DC motor control
% State space model of a DC motor
Ra = 1; % Armature electric resistance[Ohm]
La = 0.5; % Electric inductance[H]
Ki = 0.023; % Motor torque constant= Electromotive force constant[Nm/A]
Jm = 0.01; % Moment of inertia of the rotor[kgm^2]
Kb = 0.023;
Bm = 0.00003;
A = [-Ra/La -Kb/La; Ki/Jm -Bm/Jm ];
B = [1/La; 0];
C = [0 1]; % Output is state 2 (so tune q2 in Q)
D = 0;
Ts = 1; % Desired settling time
q2 = 1e2; % Tune this value only
Q = diag([1, q2]);
R = 1;
K = lqr(A, B, Q, R)
K = 1×2
3.9932 9.9705
sys = ss(A-B*K, B, C, D);
N = 1/dcgain(sys);
Gcl = ss(A-B*K, N*B, C, D) % closed-loop system
Gcl = A = x1 x2 x1 -9.986 -19.99 x2 2.3 -0.003 B = u1 x1 20 x2 0 C = x1 x2 y1 0 1 D = u1 y1 0 Continuous-time state-space model.
tau = Ts*3*2;
[u, t] = gensig('square', tau, 2*tau, 0.01);
lsim(Gcl, u, t), ylim([-1 2]), grid on
stepinfo(Gcl)
ans = struct with fields:
RiseTime: 0.3306 TransientTime: 0.8615 SettlingTime: 0.8615 SettlingMin: 0.9003 SettlingMax: 1.0327 Overshoot: 3.2714 Undershoot: 0 Peak: 1.0327 PeakTime: 0.6823
  4 Commenti
Sam Chak
Sam Chak il 20 Mar 2023
Glad to hear that, @Junhwi. If you find the solution is helpful, please consider accepting ✔ and voting 👍 the Answer. Thanks a bunch! 🙏

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