MATLAB code error, LQR

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Mostafa Sallam
Mostafa Sallam el 1 de Jun. de 2023
Respondida: Nathan Hardenberg el 30 de Jun. de 2023
Hello: for the follwoing code:
% Part 1: Define the system parameters and initial conditions
% Define the system parameters
eta = ones(4,1); N = 2*ones(4,1); A = ones(4,1); Z_m = 10*ones(4,1); m = ones(4,1); H = 0.5*ones(4,1);
% Define the torque disturbance
T_d = @(t) [0.05*sin(t); 0.05*cos(t); 0.05*sin(1.3*t); 0.1*sin(t)];
% Define the communication topology
L = [1 0 0 -1;
-1 1 0 0;
0 -1 1 0;
0 0 -1 1];
% Define the initial conditions
x0 = [0; 1; 0.5;
1; 1; 0;
2; 0.1; 1;
3; 0; 1];
% Part 2: Define the LQR controller and simulate the system
% Define the state-space matrices for the system
A_sys = zeros(12,12);
B_sys = zeros(12,4);
C_sys = eye(4,12);
D_sys = zeros(4,4);
for i = 1:4
% Define the state matrices for the ith link
A_i = [0 1 0; -(N(i)/eta(i))*sin(x0(i,1)) -(A(i)/eta(i)) 0; 0 0 0];
B_i = [0; 1/eta(i); 1/(m(i)*eta(i))];
% Update the system matrices with the ith link matrices
A_sys((3*i-2):(3*i),(3*i-2):(3*i)) = A_i;
B_sys((3*i-2):(3*i),i) = B_i;
end
% Define the LQR controller
Q = eye(12);
R = 0.1*eye(4);
[K,~,~] = lqr(A_sys,B_sys,Q,R);
% Define the reference signal
yd = 2;
% Define the simulation time span and time step
tspan = [0 15];
dt = 0.01;
% Define the simulation function
f = @(t,x) multi_agent_system(t,x,K,L,N,eta,A,Z_m,m,H,T_d,yd);
% Simulate the system using theode45 solver and plot the results
% Simulate the system using ode45
[t, x] = ode45(f, tspan, x0);
% Plot the results
figure;
for i = 1:4
subplot(4,1,i)
plot(t, x(:,(3i-2)), 'r-', 'LineWidth', 2); hold on;
plot(t, ydones(size(t)), 'b--', 'LineWidth', 2); hold off;
xlabel('Time (s)');
ylabel(['Position of Link ' num2str(i)]);
legend(['Link ' num2str(i) ' Position'], 'Reference Signal');
end
% Part 3: Define the system dynamics function**
function dxdt = multi_agent_system(t,x,K,L,N,eta,A,Z_m,m,H,T_d,yd)
% Define the state variables
w = x(:,1:3:end);
dotw = x(:,2:3:end);
T = x(:,3:3:end);
% Define the control input
u = -K*x';
% Define the time derivatives of the state variables
d_w = dotw;
d_dotw = -((N./eta).*sin(w) + (A./eta).*dotw - T_d(t)./eta) + u;
d_T = (1./(m.*eta)).*u - (Z_m./(m.*eta)).*dotw -(H./m).*T;
% Define the consensus term
consensus_term = L*x;
% Define the time derivatives of the state variables for each agent
dxdt = zeros(size(x));
for i = 1:4
dxdt(:,(3i-2):(3i)) = [d_w(:,i), d_dotw(:,i), d_T(:,i)] + consensus_term(:,(3i-2):(3i));
end
% Define the reference signal tracking error
error = x(:,1:3:end) - yd;
% Add the reference signal tracking error to the first agent's torque
dxdt(:,3) = dxdt(:,3) - error./eta;
end
it give me the following error:
Error using lqr (line 42)
Cannot compute the stabilizing Riccati solution S for the LQR design.
This could be because:
* R is singular,
* [Q N;N' R] needs to be positive definite,
* The E matrix in the state equation is singular.
Error in New_Model7 (line 44)
[K,~,~] = lqr(A_sys,B_sys,Q,R);
if anyone can help me?

Respuestas (1)

Nathan Hardenberg
Nathan Hardenberg el 30 de Jun. de 2023
I do not know the exact working behind the "Riccati Solution", but I have had this error before. For me it helped to adjust Q and/or R values slightly.
In your example you could for example try this:
R = 0.1*eye(4);
R(1,1) = 0.09; R(2,2) = 0.99; R(3,3) = 0.101; R(4,4) = 0.102
R = 4×4
0.0900 0 0 0 0 0.9900 0 0 0 0 0.1010 0 0 0 0 0.1020
This works for you example. If you have more Info abour you system you can also adjust Q and R a bit more informed. But for a first shot this should be good enough
Also: There is another error in your code afterwards, but this does not seem to have anything to do with the lqr method

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