Modern Control Engineering

Section 10–8 / Quadratic Optimal Regulator Systems 799

Hence, the characteristic equation becomes

The closed-loop poles are located at

Since the pole at s=2is in the right-half s plane, the system is unstable whatever Kmatrix is
chosen. Hence, quadratic optimal control techniques cannot be applied to this system.
Let us assume that matrices QandRin the quadratic performance index are given by

and that we write MATLAB Program 10–18. The resulting MATLAB solution is

K = [NaN NaN]

(NaN means ‘not a number.’) Whenever the solution to a quadratic optimal control problem does
not exist, MATLAB tells us that matrix Kconsists of NaN.

Q= B

1

0

0

1

R, R=[1]

s=- 1 - k 1 , s= 2

=As+ 1 +k 1 B(s- 2 )= 0

∑s I-A+BK∑=^2

s+ 1 +k 1

0

• 1 +k 2
  s- 2

2

MATLAB Program 10–18

% ---------- Design of quadratic optimal regulator system ----------

A = [-1 1;0 2];

B = [1;0];

Q = [1 0;0 1];

R = [1];

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

Warning: Matrix is singular to working precision.

K =

NaN NaN

% If we enter the command [K,P,E] = lqr(A,B,Q,R), then

[K,P,E] = lqr(A,B,Q,R)

Warning: Matrix is singular to working precision.

K =

NaN NaN

P=

-Inf -Inf

-Inf -Inf

E =

-2.0000

-1.4142