LQR issues and converting minreal gains to full-state feedback

I have a 7th order system that I am trying to obtain full-state feedback gains for using LQR.

When I use LQR on the system, I get the following error:

------------------------------------begin error-------------------------------------------------------

The "lqr" command failed to stabilize the plant or find an optimal feedback gain. To remedy this problem:

1. Make sure that all unstable poles of A are controllable through B (use MINREAL to check)

2. Modify the weights Q and R to make [Q N;N' R] positive definite (use EIG to check positivity).

------------------------------------end error-------------------------------------------------------

The unstable poles of the system are controllable and condition 2 is met because I am using Q = I and R = I (of correct size).

If I use minreal on my system with C = I, I am able to use LQR to get a set of gains for the reduced system.

Two questions

1. Why isn't the LQR command working with the full system?
2. Is there a way to convert the set of gains from the reduced system into full state feedback gains?

I have attached a file containing the full system and reduced system.

Answer 1

[Sebastian Castro]

For your first question, the full system is not controllable so LQR isn't expected to work. For the system to be controllable, the rank must be 7 (the total number of states), and it isn't.

>> rank(ctrb(sys))
ans = 4

For your second question, it's a little involved. You can look at the controllability staircase form using

[Abar,Bbar,Cbar,T,k] = ctrbf(sys.A,sys.B,sys.C);

Then, you can find an LQR controller on the controllable portion (in this case, states 4:7).

 K = lqr(Abar(4:7,4:7),Bbar(4:7,:),eye(4),eye(3))

You can pad the resulting K matrix with a bunch of zeros and then use the transformation matrix to get something that will work on the full system.

 Kaug = [zeros(3,3) K];
 Ktrans = Kaug*T;

Then, you can string this together as follows:

 X = ss((Ap-Bp*Ktrans),zeros(7,3),eye(7),Dp);

(If you want this to be a LQR servo, the B matrix could instead be Bp*Ktrans)

Two of the poles of this system above were basically zero, but slightly positive. I was hoping this was a numerical issue, so I tried to simulate this with an initial condition of 1 for the first state and was happy to see a stable response.

t = (0:0.01:100)';
lsim(X,zeros(3,length(t)),t,[1 0 0 0 0 0 0])

- Sebastian

[caidan007]

I used this method for a higher order(n=64) MIMO system dx=Ax+Bu,y=Cx. I have attached a file with matrix A,B,C,D data.

sys=ss(A,B,C,D);

>> rank(ctrb(sys))

ans =

8

so,there are 56 uncontrollable states and 8 controllable states. Then

[Abar,Bbar,Cbar,T,k] = ctrbf(sys.A,sys.B,sys.C);
K = lqr(Abar(57:64,57:64),Bbar(57:64,:),eye(8),eye(4));
Kaug = [zeros(4,56) K];

but I found (A-B*Ktrans) has many eigenvalues with positive real part, that means the full system with feedback gain Ktrans is unstable.

So,for a not fully controllable system,how can I get feedback gain Ktrans.

Thanks for your reply!

Answer 2

[Layne]

Sebastian,

Thanks for the answer. I am confused about the first statement in the error from the LQR command. It reads

"Make sure that all unstable poles of A are controllable through B (use MINREAL to check)"

Not that A and B have to be controllable. This statement implies that as long as the system is stabilizable, which my system is, the LQR command will work. Could you expand on this please?