where z = Mx can be viewed as a controlled output. Optimal Control Problem: Find u(t) = Fx(t) to minimize J subject to the constraint Ëx = Ax + Bu. Since J is ...
Linear Control Systems Lecture # 14 Linear Quadratic Regulator (LQR)
– p. 1/2
Consider the system x˙ = Ax + Bu
and suppose we want to design state feedback control u = F x to stabilize the system. The design of F is a tradeoff between the transient response and the control effort. The optimal control approach to this design tradeoff is to define the performance index (cost functional) Z ∞ J = [xT (t)Qx(t) + uT (t)Ru(t)] dt 0
and search for the control u = F x that minimizes this index. Q is an n × n symmetric positive semidefinite matrix and R is an m × m symmetric positive definite matrix
– p. 2/2
The matrix Q can be written as Q = M T M , where M is a p × n matrix, with p ≤ n. With this representation xT Qx = xT M T M x = z T z
where z = M x can be viewed as a controlled output Optimal Control Problem: Find u(t) = F x(t) to minimize J subject to the constraint x˙ = Ax + Bu Since J is defined by an integral over [0, ∞), the first question we need to address is: Under what conditions will J exist and be finite?
– p. 3/2
Write J as
J = lim J¯(tf ) tf →∞
J¯(tf ) =
Z
tf
[xT (t)Qx(t) + uT (t)Ru(t)] dt
0
J¯(tf ) is a monotonically increasing function of tf . Hence, as tf → ∞, J¯(tf ) either converges to a finite limit or diverges to infinity
Under what conditions will limtf →∞ J¯(tf ) be finite?
– p. 4/2
Recall that (A, B) is stabilizable if the uncontrollable eigenvalues of A, if any, have negative real parts Notice that (A, B) is stabilizable if (A, B) is controllable or Re [λ(A)] < 0 Definition: (A, C) is detectable if the unobservable eigenvalues of A, if any, have negative real parts Lemma 1 : Suppose (A, B) is stabilizable, (A, M ) is detectable, where Q = M T M , and u(t) = F x(t). Then, J is finite for every x(0) ∈ Rn if and only if Re [λ(A + BF )] < 0
– p. 5/2
Remarks: 1. The need for (A, B) to be stabilizable is clear, for otherwise there would be no F such that Re [λ(A + BF )] < 0 2. To see why detectability of (A, M ) is needed, consider Z ∞ x˙ = x + u, J = u2 (t) dt 0
A = 1, B = 1, M = 0, R = 1 (A, B) is controllable, but (A, M ) is not detectable F = 0 ⇒ u(t) = 0 ⇒ J = 0
This control is clearly optimal and results in a finite J but it does not stabilize the system because A + BF = A = 1
– p. 6/2
Lemma 2: For any stabilizing control u(t) = F x(t), the cost is given by J = x(0)T W x(0) where W is a symmetric positive semidefinite matrix that satisfies the Lyapunov equation W (A + BF ) + (A + BF )T W + Q + F T RF = 0
Remark: The control u(t) = F x(t) is stabilizing if Re [λ(A + BF )] < 0
– p. 7/2
Theorem: Consider the system x˙ = Ax + Bu and the performance index Z ∞ J = [xT (t)Qx(t) + uT (t)Ru(t)] dt 0
where Q = M T M , R is symmetric and positive definite, (A, B) is stabilizable, and (A, M ) is detectable. The optimal stabilizing control is u(t) = −R−1 B T P x
where P is the symmetric positive semidefinite solution of the Algebraic Riccati Equation (ARE) 0 = P A + AT P + Q − P BR−1 BP
– p. 8/2
Remarks: 1. Since the control is stabilizing, Re [λ(A − BR−1 B T P )] < 0
2. The control is optimal among all square integrable signals u(t), not just among u(t) = F x(t) 3. The Riccati equation can have more than one solution, but there is only one solution that is positive semidefinite
– p. 9/2
Example: x˙ 1 = x2 , x˙ 2 = u Z ∞ [x21 + ρu2 ] dt, ρ > 0 J = 0
"
#
"
#
"
#
0 1 0 1 0 A= , B= , Q= , R=ρ 0 0 1 0 0 " # i h i 1 h Q= 1 0 ⇒ M = 1 0 0 " # 0 1 rank[B, AB] = rank = 2 ⇒ (A, B) is controllable 1 0
– p. 10/2
rank
"
M MA
#
= rank
"
1 0 0 1
#
=2
⇒ (A, M ) is observable 0 = P A + AT P + Q − P BR−1 BP P =
"
p11 p12 p12 p22 p12
p12
#
⇒
0 = 1 − ρ1 p212
0 = p11 − ρ1 p12 p22
0 = 2p12 − ρ1 p222 p √ = ± ρ, p22 = 2ρ p12
√ 3/4 √ 1/4 1 √ = ρ, p22 = 2ρ , p11 = p12 p22 = 2ρ ρ
– p. 11/2
P =
" √
2ρ1/4 ρ1/2 √ 3/4 1/2 ρ 2ρ
#
p11 > 0, det(P ) = ρ > 0 ⇒ P is positive definite i h √ 1 F = −R−1 B T P = − ρ1/2 2ρ3/4 ρ # " 0 1 √ −1/4 A + BF = −1/2 − 2ρ −ρ λ1,2 = ρ
−1/4
1 √ (−1 ± j) 2
– p. 12/2
Notice how F and the eigenvalues depend on ρ i h √ F = − ρ−1/2 2ρ−1/4 λ1,2 = ρ
−1/4
1 √ (−1 ± j) 2
What happens as you change ρ?
– p. 13/2
Matlab Calculations: X = lyap(A,N) solves the Lyapunov equation AX + XAT + N = 0
To solve the equation W (A + BF ) + (A + BF )T W + Q + F T RF = 0
use W = lyap((A+B*F)0 ,Q+F0 *R*F) ( Notice the transpose)
– p. 14/2
X = are(A,S,Q)
solves the Riccati equation
XA + AT X + Q − XSX = 0 S = BR−1 B
[K,P,E] = lqr(A,B,Q,R)
solves the Riccati equation
P A + AT P + Q − P BR−1 B T P = 0 K = R−1 B T P and E is a vector whose elements are the eigenvalues of (A − BR−1 B T P ) Notice that F = −K
– p. 15/2
LQR Design Given the system x˙ = Ax + Bu
where (A, B) is stabilizable, we want to design state feedback control u = F x to stabilize the system while meeting Transient response specifications Magnitude constraints on x and u
– p. 16/2
Procedure: 1. Choose Q and R such that Q = M T M , with (A, M ) detectable, and R = RT > 0 2. Solve the Riccati equation P A + AT P + Q − P BR−1 B T P = 0
and compute F = −R−1 B T P 3. Simulate the initial response of x˙ = (A + BF )x for different initial conditions 4. If the transient response specifications and/or the magnitude constraints are not met, go back to step 1 and re-choose Q and/or R
– p. 17/2
Typical Choice: q1 q2 Q= ... qi =
tsi (ximax
)2
,
ri =
r1 r2
, R = ρ
qn
1
rm
1
(uimax
...
)2
,
ρ>0
tsi is the desired settling time of xi ximax is a constraint on |xi | uimax is a constraint on |ui | ρ is chosen to tradeoff regulation versus control effort
– p. 18/2
Example:
0 0 1 0 A= 0 0 1 , B = 0 1 −0.4 −4.2 −2.1
Open-loop eigenvalues are: −0.1, −1 ± 1.7321j Desired settling time is 4 sec. Q = I3 ; Try R = 0.01, 0.1, and 1 R λ 0.01 −9.7364, −0.9039 ± 0.4593j 0.1 −0.6568, −1.9548 ± 1.0158j 1 −0.2599, −1.1433 ± 1.6840j
– p. 19/2
Q = diag(1,1,1)
2
1.5
R=0.01 R=0.1 R=1
1.5
1
2
1
x
x
1
0.5 0 0.5 0
−0.5
0
1
2
3
4
1
5
0.5
0
0
−5
−0.5
−10
0
1
2
3
4
5
0
0.2
0.4
0.6
0.8
1
u
3
x
−1
5
−1
−15
−1.5
−20
−2
−25
−2.5
0
1
2
3
4
5
−30
– p. 20/2
R=0.1 1.5
Q =1 11 Q = 10
0.5 x2
1
1
x
1
11
0.5
−0.5
0 −0.5
0
−1 0
1
2
3
4
5
1
5
0
0
0
1
2
3
4
5
0
1
2
3
4
5
−5
x
u
3
−1
−10
−2
−15 −3 0
1
2
3
4
5
−20
– p. 21/2
Q =1;R=0.1 11 Q =10;R=0.1 11 Q =10;R=0.4
1.5
1 0.5
0.5
0
2
1
x
x
1
11
−0.5
0
−1 −0.5
0
1
2
3
4
5
1
5
0
0
0
1
2
3
4
5
0
0.2
0.4
0.6
0.8
1
−5
x
u
3
−1
−10
−2
−15 −3 0
1
2
3
4
5
−20
– p. 22/2
