Nonlinear Control
Lecture # 23
State Feedback Stabilization
Nonlinear Control Lecture # 23 State Feedback Stabilization
Basic Concepts
We want to stabilize the system
ẋ = f (x, u)
at the equilibrium point x = xss
Steady-State Problem: Find steady-state control uss s.t.
0 = f (xss , uss )
xδ = x − xss , uδ = u − uss
def
ẋδ = f (xss + xδ , uss + uδ ) = fδ (xδ , uδ )
fδ (0, 0) = 0
uδ = φ(xδ ) ⇒ u = uss + φ(x − xss )
Nonlinear Control Lecture # 23 State Feedback Stabilization
State Feedback Stabilization: Given
ẋ = f (x, u) [f (0, 0) = 0]
find
u = φ(x) [φ(0) = 0]
s.t. the origin is an asymptotically stable equilibrium point of
ẋ = f (x, φ(x))
f and φ are locally Lipschitz functions
Nonlinear Control Lecture # 23 State Feedback Stabilization
Notions of Stabilization
ẋ = f (x, u), u = φ(x)
Local Stabilization: The origin of ẋ = f (x, φ(x)) is
asymptotically stable (e.g., linearization)
Regional Stabilization: The origin of ẋ = f (x, φ(x)) is
asymptotically stable and a given region G is a subset of the
region of attraction (for all x(0) ∈ G, limt→∞ x(t) = 0) (e.g.,
G ⊂ Ωc = {V (x) ≤ c} where Ωc is an estimate of the region
of attraction)
Global Stabilization: The origin of ẋ = f (x, φ(x)) is globally
asymptotically stable
Nonlinear Control Lecture # 23 State Feedback Stabilization
Semiglobal Stabilization: The origin of ẋ = f (x, φ(x)) is
asymptotically stable and φ(x) can be designed such that any
given compact set (no matter how large) can be included in
the region of attraction (Typically u = φp (x) is dependent on a
parameter p such that for any compact set G, p can be chosen
to ensure that G is a subset of the region of attraction )
What is the difference between global stabilization and
semiglobal stabilization?
Nonlinear Control Lecture # 23 State Feedback Stabilization
Example 9.1
ẋ = x2 + u
Linearization:
ẋ = u, u = −kx, k > 0
Closed-loop system:
ẋ = −kx + x2
Linearization of the closed-loop system yields ẋ = −kx. Thus,
u = −kx achieves local stabilization
The region of attraction is {x < k}. Thus, for any set
{−a ≤ x ≤ b} with b < k, the control u = −kx achieves
regional stabilization
Nonlinear Control Lecture # 23 State Feedback Stabilization
The control u = −kx does not achieve global stabilization
But it achieves semiglobal stabilization because any compact
set {|x| ≤ r} can be included in the region of attraction by
choosing k > r
The control
u = −x2 − kx
achieves global stabilization because it yields the linear
closed-loop system ẋ = −kx whose origin is globally
exponentially stable
Nonlinear Control Lecture # 23 State Feedback Stabilization
Linear Systems
ẋ = Ax + Bu
(A, B) is stabilizable (controllable or every uncontrollable
eigenvalue has a negative real part)
Find K such that (A − BK) is Hurwitz
u = −Kx
Typical methods:
Eigenvalue Placement
Eigenvalue-Eigenvector Placement
LQR
Nonlinear Control Lecture # 23 State Feedback Stabilization
Linearization
ẋ = f (x, u)
f (0, 0) = 0 and f is continuously differentiable in a domain
Dx × Du that contains the origin (x = 0, u = 0) (Dx ⊂ Rn ,
Du ⊂ R m )
ẋ = Ax + Bu
∂f ∂f
A= (x, u) ; B= (x, u)
∂x x=0,u=0 ∂u x=0,u=0
Assume (A, B) is stabilizable. Design a matrix K such that
(A − BK) is Hurwitz
u = −Kx
Nonlinear Control Lecture # 23 State Feedback Stabilization
Closed-loop system:
ẋ = f (x, −Kx)
Linearization:
∂f ∂f
ẋ = (x, −Kx) + (x, −Kx) (−K) x
∂x ∂u x=0
= (A − BK)x
Since (A − BK) is Hurwitz, the origin is an exponentially
stable equilibrium point of the closed-loop system
Nonlinear Control Lecture # 23 State Feedback Stabilization
Example 9.2 (Pendulum Equation)
θ̈ = − sin θ − bθ̇ + cu
Stabilize the pendulum at θ = δ1
0 = − sin δ1 + cuss
x1 = θ − δ1 , x2 = θ̇, uδ = u − uss
ẋ1 = x2
ẋ2 = −[sin(x1 + δ1 ) − sin δ1 ] − bx2 + cuδ
0 1 0 1
A= =
− cos(x1 + δ1 ) −b x1 =0
− cos δ1 −b
Nonlinear Control Lecture # 23 State Feedback Stabilization
0 1 0
A= ; B=
− cos δ1 −b c
K= k1 k2
0 1
A − BK =
−(cos δ1 + ck1 ) −(b + ck2 )
cos δ1 b
k1 > − , k2 > −
c c
sin δ1 sin δ1
u= − Kx = − k1 (θ − δ1 ) − k2 θ̇
c c
Nonlinear Control Lecture # 23 State Feedback Stabilization
Feedback Linearization
Consider the nonlinear system
ẋ = f (x) + G(x)u
f (0) = 0, x ∈ Rn , u ∈ Rm
Suppose there is a change of variables z = T (x), defined for
all x ∈ D ⊂ Rn , that transforms the system into the controller
form
ż = Az + B[ψ(x) + γ(x)u]
where (A, B) is controllable and γ(x) is nonsingular for all
x∈D
u = γ −1 (x)[−ψ(x) + v] ⇒ ż = Az + Bv
Nonlinear Control Lecture # 23 State Feedback Stabilization
v = −Kz
Design K such that (A − BK) is Hurwitz
The origin z = 0 of the closed-loop system
ż = (A − BK)z
is globally exponentially stable
u = γ −1 (x)[−ψ(x) − KT (x)]
Closed-loop system in the x-coordinates:
def
ẋ = f (x) + G(x)γ −1 (x)[−ψ(x) − KT (x)] = fc (x)
Nonlinear Control Lecture # 23 State Feedback Stabilization
What can we say about the stability of x = 0 as an equilibrium
point of ẋ = fc (x)?
∂T
z = T (x) ⇒ (x)fc (x) = (A − BK)T (x)
∂x
∂fc ∂T
(0) = J −1 (A − BK)J, J = (0) (nonsingular)
∂x ∂x
The origin of ẋ = fc (x) is exponentially stable
Is x = 0 globally asymptotically stable? In general No
It is globally asymptotically stable if T (x) is a global
diffeomorphism
Nonlinear Control Lecture # 23 State Feedback Stabilization
Estimate of the region of attraction: If T (x) is a
diffeomorphism on a domain D ⊂ Rn , the equation
ż = (A − BK)z is valid in the domain T (D)
P (A − BK) + (A − BK)T P = −Q, Q = QT > 0
Estimate in the z-coordinates:
Ωc = {z T P z ≤ c} ⊂ T (D)
Estimate in the x-coordinates:
T −1 (Ωc ) = {T T (x)P T (x) ≤ c}
Nonlinear Control Lecture # 23 State Feedback Stabilization
Example 9.3 (Recall Example 8.12)
ẋ1 = a sin x2 , ẋ2 = −x21 + u
x1 z2
z = T (x) = ⇒ ż = p
a sin x2 a − z2 (−z12 + u)
2 2
D = {|x2 | < π/2}, T (D) = {|z2 | < a}
K = σ 2 2σ , σ > 0, ⇒ λ(A − BK) = −σ, −σ
2a2
3σ 2 σ 2σ 3 0
P = , Q= , c < min z T P z =
σ 1 0 2σ |z2 |=a 3
T −1 (Ωc ) = {3σ 2 x21 + 2σax1 sin x2 + a2 sin2 x2 ≤ c}
Nonlinear Control Lecture # 23 State Feedback Stabilization
