Designing LQR controller error

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Deepayan Bhadra 2021-4-15

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https://ww2.mathworks.cn/matlabcentral/answers/803556-designing-lqr-controller-error

评论： Deepayan Bhadra 2021-4-16

I am trying to design LQR for a 7x7 system.

A = 0	-0.628971428568773	0	0	0	0	0.999999999999951
-0.0239070502560970	0	0	0	1.26736762437470	0.404734566388548	0
-9.73903523911518e-06	0	1.95035679041857	-0.951171532106457	0.000516288618732924	0.000164877061884321	0
0	0	1	0	0	0	0
0.383799830372633	0	0	0	0.854813237018843	-0.138991455550305	0
-0.00476707865861458	0	0	0	0.0327470424766583	0.967898772564207	0
0	-0.157242857142201	0	0	0	0	1

B = 0.628971428571398
0
0
0
0
0
0.157242857142857

C = 0 0 1.00834336091157 0.991656639088434 0 0 0

D = 0

The system has poles on right half plane but is controllable,

rank(ctrb(A,B))

= 7.

When I try

K = lqr(A,B,Q,R)

, I get below. I fail to see what's happening here in view of the information above. Could someone clarify?

Error using lqr (line 42)
Cannot compute a stabilizing LQR gain (the Riccati solution S and gain matrix K are infinite).
This could be because:
  * A has unstable modes that are not controllable through B,
  * Q,R,N values are too large,
  * [Q N;N' R] is indefinite,
  * The E matrix in the state equation is singular.

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Answer 1

Paul 2021-4-16

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编辑：Paul 2021-4-16

在 MATLAB Online 中打开

The system might be controllable in principle, but in reality seems to be problematic for using state feedback. I had trouble even getting pole placement to work numerically. But it can work symbolically. However, converting the symbolic solution to numerical shows great sensitivity for the selected set of closed loop poles:

K = sym('K',[1 7]);
syms s
[c,t]=(coeffs(det(s*eye(7)-(A-B*K)),s,'All')); % characteristic polynomial
[K1,K2,K3,K4,K5,K6,K7]=solve(c==sym(poly(diag(-7:-1))),K); % solve state feedback gains for closed loop poles at -7:-1
eig(A-B*subs(K)) % verify sympolic solution
eig(A-B*double(subs(K)))  % compare to numeric solution
 
ans =
 
 -7
 -6
 -5
 -4
 -3
 -2
 -1
 
ans =
  -1.5887e+01 + 5.1114e+00i
  -1.5887e+01 - 5.1114e+00i
  -1.4663e+00 + 1.3262e+01i
  -1.4663e+00 - 1.3262e+01i
   5.6211e+00 + 0.0000e+00i
  -1.6455e-02 + 0.0000e+00i
   1.1026e+00 + 0.0000e+00i

So the symbolic solution is perfect, but converting to double precision yields a terrible result. The feedback gain matrix is

>> vpa(subs(K)).'
 
ans =
 
 -52271855.708258191568879324847032
   23249.37317998977835990526684776
 -2733390131189.6388535741683213926
  2711961625327.6659529554924801665
  386661508.02580444722226705634626
  44938181476.829118288777618682236
  209087631.25628197842813619653929

which shows very high gains needed to move all those unstable poles into the LHP. I'm not sure what one could specifically conclude abuout this problem from these results, but it certainly suggests that any state feedback approach could be tricky.