1. No category

CONTENTS
1
INTRODUCTION
Lecture 1: Feedback control
Jorge Cort´es
March 30, 2015
Abstract
The treatment corresponds to selected parts from Chapter 6 in [1] and Chapter 12 in [2].
Contents
1 Introduction
1
1.1
Stabilization with respect to an arbitrary point . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
Regional, global and semiglobal stabilization . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.3
Tracking in the presence of disturbance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.4
Tidbit: plotting phase portraits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
2 Stabilization by state feedback control
4
3 Stabilization by output feedback control
8
4 Integral control
4.1
1
10
Integral control via linearization: state feedback integral controller . . . . . . . . . . . . . . .
12
Introduction
Consider the system on Rn ,
x˙ = f (t, x, u)
(1)
The state feedback stabilization problem for the system (1) is the problem of designing a feedback control
law u = γ(t, x) such that the origin is a uniformly asymptotically stable equilibrium point of the closed-loop
system
x˙ = f (t, x, γ(t, x))
The control law u = γ(t, x) is called static feedback because it is a memoryless function of x. One can also
use a dynamic feedback control
u = γ(t, x, z)
1
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1.1
Stabilization with respect to an arbitrary point
1
INTRODUCTION
where z is the solution of a dynamical system driven by x, that is,
z˙ = g(t, x, z)
Common examples of dynamic state feedback include integral control and adaptive control. In the case of
dynamic feedback control, the origin to be stabilized is x = 0, z = 0.
Consider the system
x˙ = f (t, x, u)
y = h(t, x, u)
(2a)
(2b)
The output feedback stabilization problem for the system (1) is the problem of designing a static output
feedback control law u = γ(t, y) or a dynamic output feedback control law u = γ(t, y, z), z˙ = g(t, y, z), such
that the origin is a uniformly asymptotically stable equilibrium point of the closed-loop system. In the case
of dynamic feedback control, the origin to be stabilized is x = 0, z = 0.
Dynamic controllers are more common in output feedback schemes, since the lack of measurement of some
state variables makes it useful to introduce “observer”-like components.
1.1
Stabilization with respect to an arbitrary point
Given a desired x∗ ∈ Rn , assume there exists a value of the input u∗ such that f (t, x∗ , u∗ ) = 0, for all t ≥ 0
(i.e., u∗ makes the state x∗ an equilibrium). The change of variables
x
¯ = x − x∗ ,
u
¯ = u − u∗ ,
results in
x
¯˙ = f (t, x∗ + x
¯ , u∗ + u
¯) = f¯(t, x
¯, u
¯)
with f¯(t, 0, 0) = 0 for all t ≥ 0. Therefore, there is no loss of generality in reasoning for the origin as
equilibrium. For output feedback problems, the output is redefined as
¯ x
y¯ = h(t, x∗ + x
¯ , u∗ + u
¯) − h(t, x
¯, u
¯) = h(t,
¯, u
¯).
1.2
Regional, global and semiglobal stabilization
If the feedback control guarantees that a certain set is included in the region of attraction or if an estimate
of the region of attraction is given, we say that the feedback control achieves regional stabilization.
If the origin of the closed-loop system is globally asymptotically stable, we say that the control achieves global
stabilization.
If the feedback control does not achieve global stabilization, but can be designed such that any given compact
set (no matter how large) can be included in the region of attraction, we say that the feedback control achieves
semiglobal stabilization.
Example 1.1 Consider the scalar system
x˙ = x2 + u
Linearization around the origin results in x˙ = u. This can be stabilized with u = −kx, k > 0. This results in
x˙ = −kx + x2
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1.3
Tracking in the presence of disturbance
1
INTRODUCTION
Therefore, u = −kx achieves local stabilization. In this example, one can see that the region of attraction
is {x ∈ R | x < k}. Hence, u = −kx achieves regional stabilization. Given any compact set, we can choose
r > 0 such that the set is included in the ball B(0, r). This ball can be included in the region of attraction by
choosing k > r. Therefore, u = −kx achieves semiglobal stabilization. However, it does not achieve global
stabilization, no matter which k we use. Global stabilization is achieved with the control law u = −x2 − kx.•
1.3
Tracking in the presence of disturbance
Consider the system
x˙ = f (t, x, u, w)
(3a)
y = h(t, x, u, w)
(3b)
The goal is to design the control input u so that the controlled output y tracks a reference signal r, that is
e(t) = y(t) − r(t) ≃ 0,
t ≥ t0 ,
where t0 is the time when the control starts.
Note that the initial value of y(t0 ) depends on the initial state x(t0 ). Therefore, meeting this requirement for
all t ≥ t0 will require either pre-setting x(t0 ) or pre-setting the initial value of the reference signal by assuming
knowledge of x(t0 ). Therefore, we usually seek an asymptotic output tracking or asymptotic regulation goal,
i.e.,
e(t) → 0,
t → ∞.
If asymptotic output tracking is achieved in the presence of input disturbance w, we say that asymptotic
disturbance rejection has been achieved.
For a general time-varying disturbance input t 7→ w(t) it might not be possible to achieve asymptotic
disturbance rejection. In such cases, we attempt to achieve disturbance attenuation, i.e., something like
ke(t)k ≤ ε,
t ≥ t0 .
Alternatively, we may consider attenuating the closed-loop input-output map from the disturbance input w
to the tracking error e.
Feedback control laws for the tracking problem are characterized in the same way as for the stabilization
problem. The only note to this is that the definitions refer not only to the size of the initial state x, but also
to the size of the exogenous signals r and w.
1.4
Tidbit: plotting phase portraits
Both Mathematica and Matlab have commands or routines that you can invoke to plot phase portraits. These
are often useful to get a rough idea of the behavior of planar systems, or to disprove/confirm/contradict some
intuition about the solutions of the system.
In Mathematica, you can use VectorPlot in 2D and, for those with really good spatial vision, VectorPlot3D
in 3D. In Matlab, you can use pplane.m and dfield.m provided by Prof. Polking from Rice University at
http://math.rice.edu/~dfield.
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2 STABILIZATION BY STATE FEEDBACK CONTROL
2
Stabilization by state feedback control
Consider the nonlinear control system
x˙ = f (x, u)
where f (0, 0) = 0 and f is continuously differentiable in a domain Dx × Du ⊂ Rd × Rp that contains (0, 0).
Our objective is to design a state feedback control law u = γ(x) that stabilizes the origin x = 0.
Linearization of the system around x = 0, u = 0 yields
x˙ = Ax + Bu,
A=
∂f
(x, u)x=0,u=0 ,
∂x
B=
∂f
(x, u)x=0,u=0 .
∂u
Assume that a matrix K exists such that A − BK has all its eigenvalues in the left half complex plane (more
on this in the future, in general this is not true). Consider then the control law u = −Kx. The closed-loop
nonlinear system looks now as
x˙ = f (x, −Kx).
Clearly the origin is an equilibrium point of the system. Moreover, the linearization around the origin is
∂f
∂f
(x, −Kx) +
(x, −Kx)(−K)
x = (A − BK)x
y˙ =
∂x
∂u
x=0
Since A − BK is Hurwitz, it follows from the Lyapunov indirect method (see [2, Theorem 4.7]) that the
origin is an asymptotically stable equilibrium point of the closed-loop system. Actually, according to [2,
Theorem 4.13], the origin is exponentially stable. As a byproduct of this approach, we can always find a
Lyapunov function for the closed-loop system. This, for instance, can serve to determine an estimate of the
region of attraction. Given any positive definite symmetric matrix Q, solve the Lyapunov equation
P (A − BK) + (A − BK)t P = −Q
for P . We know that there exists a unique solution to this equation because A − BK is Hurwitz [2, Theorem 4.6]. Once solved, we define the quadratic Lyapunov function V (x) = xt P x.
Example 2.1 (Forced pendulum) Consider the pendulum equations
θ¨ = −a sin θ − bθ˙ + cT
(4)
where a = g/l > 0, b = k/m ≥ 0, c = 1/ml2 > 0, θ is the angle subtended by the rod and the vertical
axis, and T is the torque applied to the pendulum. Suppose we want to stabilize the pendulum at an angle
θ = θdes . First of all, we need to apply a torque Tdes to make θdes an equilibrium,
−a sin θdes + cTdes = 0.
˙ and the control variable u = T − Tdes . The
Let us choose the state variables x1 = θ − θdes and x2 = θ,
equations read now
x˙ 1 = x2
x˙ 2 = −a(sin(x1 + θdes ) − sin(θdes )) − bx2 + cu
Note that f (0, 0) = 0. The linearization at the origin results in
0
1
0
A=
, B=
−a cos θdes −b
c
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2 STABILIZATION BY STATE FEEDBACK CONTROL
Taking u = K[x1 , x2 ]T , with K = [k1 , k2 ], one can see
0
A − BK =
−a cos θdes − ck1
1
−b − ck2
with eigenvalues
λ=
−(ck2 + b) ±
p
(ck2 + b)2 − 4(a cos θdes + ck1 )
.
2
Therefore, A − BK is Hurwitz so long as k1 > − ac cos θdes and k2 > − cb . The torque is then given by
T =
a
a
˙
sin θdes − Kx = sin θdes − k1 (θ − θdes ) − k2 θ.
c
c
See Figure 1 for an illustration of this control.
•
x’=y
y ’ = − a (sin(x) − sin(tdesired)) − k1 c (x − tdesired) − (b + k2 c) y
a=1
k2 = 1
k1 = 1
c=1
b=1
tdesired = pi/2
4
3
2
y
1
0
−1
−2
−3
−4
−2
−1
0
1
x
2
3
4
Figure 1: Phase portrait of the pendulum example exponentially stabilized around the point θdes = π/2. The
values of the parameters are a = b = c = 1 and k1 = k2 = 1.
Example 2.2 (Tracking with the Robobrain robot) Consider the Robobrain robot in Figure 2. Its evolution can be modeled by the following unicycle model
x˙ = u cos θ,
y˙ = u sin θ,
θ˙ = v,
where u corresponds to forward velocity, and v corresponds to turning velocity. Assume our objective is to
make the robot track a vertical trajectory x = x0 (and therefore, θ = π/2).
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2 STABILIZATION BY STATE FEEDBACK CONTROL
Figure 2: Robobrain robot
Let us make the change of coordinates x
¯ = x − x0 , y¯ = y, θ¯ = θ − π/2 to obtain
¯
x
¯˙ = u cos(θ¯ + π/2) = −u sin θ,
¯
¯
y¯˙ = u sin(θ + π/2) = u cos θ,
θ¯˙ = v.
Our objective is now to drive the system toward the line (0, y¯, 0). Therefore, we choose a proportional
¯ This choice yields the system
controller of the form u = uc > 0, v = −k1 x
¯ − k2 θ.
x
¯˙ = −uc sin θ¯
y¯˙ = uc cos θ¯
θ¯˙ = −k1 x
¯ − k2 θ¯
The linearization of the system at (0, y¯, 0) is
x
¯˙ = −uc θ¯
y¯˙ = uc
θ¯˙ = −k1 x
¯ − k2 θ¯
The sub matrix corresponding to x
¯ and θ¯ is
0
−k1
−uc
−k2
with eigenvalues
λ=
−k2 ±
p
k22 + 4uc k1
.
2
Therefore, for k2 > 0 and k1 < 0, we deduce that the feedback controller
u = uc ,
v = −k1 (x − x0 ) − k2 (θ −
π
).
2
achieves the tracking of the desired trajectory. Figure 3 shows the resulting (x, θ)-phase portrait.
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•
2 STABILIZATION BY STATE FEEDBACK CONTROL
uc = .5
x0 = 1
k1 = − 1 k2 = 1
x ’ = uc cos(theta)
theta ’ = − k1 (x − x0) − k2 (theta − pi/2)
3
2.5
2
1.5
theta
1
0.5
0
−0.5
−1
−1.5
−2
−1
−0.5
0
0.5
1
x
1.5
2
2.5
3
Figure 3: (x, θ)-phase portrait of the Robobrain example. The values of the parameters are x0 = 1, uc = .5,
k1 = −1 and k2 = 1.
Exercise 2.3 Carry over a discussion similar to the one above for the stabilization of a nonlinear system by
dynamic state feedback control.
Example 2.4 (Estimating the region of attraction) Given a positive definite symmetric matrix of your
choosing, e.g., Q = I, solve the Lyapunov equation
P (A − BK) + (A − BK)t P = −Q
for P . We know that there exists a unique solution to this equation because A − BK is Hurwitz [2, Theorem 4.6]. Define then quadratic Lyapunov function V (x) = xt P x. By closely following the proof of the
Lyapunov indirect method [2, Theorem 4.7], one can find an estimate of the region of attraction. The idea
is as follows. Write x˙ = Ax + g(x), with g(x) = f (x, −Kx) − Ax. Note that kg(x)k/kxk → 0, and hence,
for any γ > 0, there exists r such that kg(x)k ≤ γkxk for kxk ≤ r. Now, the evolution of the quadratic
function V along the dynamical system x˙ = f (x, −Kx) is given by
Lf V (x) = xt P (A − BK)x + xt (A − BK)t P x + 2xt P g(x)
= −xt Qx + 2xt P g(x)
Now, take γ < λmin (Q)/(2kP k). For kxk ≤ r, x 6= 0, we have
Lf V (x) ≤ −λmin (Q)kxk2 + 2γkP kkxk2 < 0.
Therefore, on B(0, r) \ {0}, the function V is strictly decreasing. Now, the last step is to determine a value
c such that the sublevel set Ωc of V is contained in this ball.
•
Remark 2.5 (Determining the appropriate gains) Computing conditions on K such that A − BK is
Hurwitz might get computationally intensive. The direct method is to find the eigenvalues of the matrix and
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3 STABILIZATION BY OUTPUT FEEDBACK CONTROL
determine suitable conditions on the gains in K. Alternative approaches include the Routh criterion and
Gersgorin disk theorem. This latter result states that the eigenvalues of a n × n matrix F are contained in
[
i∈{1,...,n}
3
n
z ∈ C kz − fii kC ≤
n
X
j=1,j6=i
o
|fij | .
•
Stabilization by output feedback control
Consider the system
x˙ = f (x, u)
y = h(x)
where f (0, 0) = 0, h(0) = 0 and f , h are continuously differentiable in a domain Dx × Du ⊂ Rn × Rp that
contains (0, 0). Our objective is to design an output feedback law to stabilize the system.
Linearization of the system around x = 0, u = 0 yields
∂f
(x, u)x=0,u=0 ,
∂x
∂h (x) x=0
y = Cx, C =
∂x
We try a dynamic output feedback controller
x˙ = Ax + Bu,
A=
B=
∂f
(x, u)x=0,u=0
∂u
z˙ = F z + Gy
u = Lz + M y
such that the closed-loop matrix (in the variables (x, z))
A + BM C
GC
BL
F
is Hurwitz. This is not always possible. When (A, B) is stabilizable and (A, C) is detectable, an example of
such design is the observer-based controller
z=x
ˆ,
F = A − BK − HC,
G = H,
L = −K,
M = 0,
with K and H such that A − BK and A − HC are Hurwitz. In this case, we have the closed-loop matrix
A
−BK
(5)
HC A − BK − HC
whose eigenvalues are exactly those of A − BK and A − HC.
Exercise 3.1 Prove that the eigenvalues of the matrix (5) are the eigenvalues of A − BK and A − HC.
[Hint: play with the eigenvectors]
When the controller is applied to the nonlinear system, it results in the system
x˙ = f (x, Lz + M h(x))
z˙ = F z + Gh(x)
The origin x = 0, z = 0 is an equilibrium point of the system, and the linearization has an associated Hurwitz
matrix. Therefore, the origin is exponentially stable. A Lyapunov function for the closed-loop system can be
obtained by solving a Lyapunov equation for the Hurwitz matrix.
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3 STABILIZATION BY OUTPUT FEEDBACK CONTROL
Example 3.2 (Forced pendulum) Consider the pendulum equations
θ¨ = −a sin θ − bθ˙ + cT
where a = g/l > 0, b = k/m ≥ 0, c = 1/ml2 > 0, θ is the angle subtended by the rod and the vertical
axis, and T is the torque applied to the pendulum. Suppose we want to stabilize the pendulum at an angle
θ = θdes . Let us choose the state variables x1 = θ −θdes and x2 = θ˙ and the control variable u = T − ac sin θdes .
˙ i.e.,
Additionally, suppose we can measure the angle θ, but not the velocity θ,
y = x1 = θ − θdes .
The equations read now
x˙ 1 = x2
x˙ 2 = −a(sin(x1 + θdes ) − sin θdes ) − bx2 + cu
y = x1
The matrices A, B, C are given by
A=
1
,
−b
0
−a cos θdes
B=
0
,
c
C= 1 0
We implement the dynamic output feedback controller using the observer
x
ˆ˙ = Aˆ
x + Bu + H(y − x
ˆ1 )
Taking H = [h1 , h2 ]T , it can be verified that A − HC is Hurwitz if
h1 + b > 0,
h1 b + h2 + a cos θdes > 0.
1
−b − ck2
Now, let K = [k1 , k2 ] and consider
0
A − BK =
−a cos θdes − ck1
with eigenvalues
λ=
−(ck2 + b) ±
p
(ck2 + b)2 − 4(a cos θdes + k1 )
.
2
From here, we deduce that A − BK is Hurwitz so long as
a
k1 > − cos θdes ,
c
b
k2 > − .
c
The torque is then given by
T =
a
sin θdes − K x
ˆ.
c
Figure 4 gives an example of the implementation of the controller.
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•
4
theta versus time
INTEGRAL CONTROL
theta - thetaDesired, z1 versus time
1.4
0.4
1.2
0.2
1
0
0.8
0.6
-0.2
0.4
-0.4
0.2
-0.6
2
4
6
8
10
2
4
6
8
10
Figure 4: Implementation of output feedback controller in the forced pendulum example. The parameters
are θdes = π/4, k1 = k2 = h1 = h2 = 1, a = c = 10, b = 1.
4
Integral control
Remember our discussion to design a state feedback controller for the pendulum in Example 2.1. We came
up with the torque
T =
a
a
˙
sin θdes − Kx = sin θdes − k1 (θ − θdes ) − k2 θ.
c
c
The feedback part of the controller can be designed to be robust to a wide range of parameter perturbations.
For instance, if we know a/c ≤ ρ, but do not know the exact value of a and c, we can ensure that A − BK
is Hurwitz by choosing
k1 > ρ,
k2 > 0.
However, the calculation of the steady-state component is quite sensitive to parameter perturbations. Suppose
that we compute it using the nominal values a0 , c0 for a, c, respectively. Then we will apply the torque
T =
a0
sin θdes − k1 (θ − θdes ) − k2 θ˙
c0
in the pendulum equation (4). A little bit of math gives the steady-state of the closed-loop system as the
value θss that verifies
a
0
sin θdes − k1 (θss − θdes ) .
a sin θss = c
c0
For θdes = 0 or θdes = π (the original equilibrium of the system), we deduce θss = θdes . So in this case, the
steady-state component is robust to parameter perturbations (can you explain why?). However, for other
values, this is no longer true. For instance, for θdes = 45 degrees, a = a0 , c = c0 /2 and k1 = 3a0 /c0 , we have
θss = 36 degrees.
Example 4.1 (Cruise control) Consider the following simple dynamical model of the evolution of the
velocity v of your preferred car
v˙ =
1
− bv + u + w .
m
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4
INTEGRAL CONTROL
Here m > 0 stands for the mass of the car, and b > 0 is a friction coefficient. The control u stands for the
engine power. Finally, w accounts for the unknown hill inclination. The cruise controller of your preferred
car certainly does not know about the exact incline of the road at any given instant in time! Assume you set
your desired speed to be v = vdes . Let us then choose the controller
u = bv − k(v − vdes ).
The first term takes care of making vdes an equilibrium. We select the other one to make it stable. In the
absence of external disturbances (i.e., w = 0, flat road), we then have
v˙ = −
k
(v − vdes ).
m
So long as k > 0, we have exponentially stabilized the system. However, what happens if the road incline is
not zero? Then, the real system will really look like
v˙ =
1
(−k(v − vdes ) + w).
m
and the equilibrium will be
vss = vdes +
w
k
The bigger the gain k, the smaller the deviation from the desired velocity. However, your preferred car horse
power is limited, and hence you cannot keep cranking up the engine! It is actually better to use integral
control, as we demonstrate next.
•
In this section, we present an integral control approach that takes care of fixing the mismatch problem at
steady state. The idea is to reformulate our objective as an asymptotic regulation problem as defined in
Section 1.3. We solve the asymptotic regulation under all parameter perturbations that do not destroy the
stability of the closed-loop system. Consider the system
x˙ = f (x, u, w)
y = h(x, w)
where w ∈ Rl is a vector of unknown constant parameters and disturbances. The functions f : Rn × Rp ×
Rl → Rn and h : Rn × Rl → Rp are continuously differentiable in (x, u) and continuous in w in a domain
Dx × Du × Dw ⊂ Rn × Rp × Rl . Let r ∈ Dr ⊂ Rp be a constant reference that is available online. Our
objective is to design a feedback controller such that
y(t) → r,
Rephrasing the control problem.
equilibrium point where y = r. Let
t → ∞.
This regulation task will be achieved by stabilizing the system at an
r
v=
∈ Dv = Dr × Dw
w
Assume that for each v ∈ Dv , there exists a unique pair xss , uss that depends continuously on v and satisfies
the equations
0 = f (xss , uss , w)
r = h(xss , w)
(6a)
(6b)
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4.1
Integral control via linearization: state feedback integral controller
4
INTEGRAL CONTROL
To introduce integral action, we set e = y − r and integrate this regulation error,
σ˙ = y − r
Together with the state equation, we then have
x˙ = f (x, u, w)
σ˙ = h(x, w) − r
(7a)
(7b)
The control task is now to design a feedback controller that stabilizes this system at an equilibrium point
(xss , σss ), where σss produces the desired uss . See Figure 5.
−σ
r
u
y
+
−
Figure 5: Integral control.
The fact that this controller is robust to all parameter perturbations that do not destroy the stability of the
closed-loop system can be explained as follows. The inclusion of the integrator causes the regulation error to
be zero at equilibrium. Parameter perturbations will change the equilibrium point, but the condition e = 0
will still be maintained. Therefore, as long as the perturbed equilibrium point remains asymptotically stable,
regulation will be achieved.
4.1
Integral control via linearization: state feedback integral controller
Let us devise a local solution to the regulation problem via linearization. We consider state feedback. Our
objective is to design a feedback controller u = γ(x, σ) to stabilize the system (7) at (xss , σss ), where
uss = γ(xss , σss ). We consider a linear control law of the form
u = −K1 x − K2 σ
(8)
Substituting into the equations of the system, we get
x˙ = f (x, −K1 x − K2 σ, w)
σ˙ = h(x, w) − r
(9a)
(9b)
whose equilibrium points satisfy
0 = f (¯
x, u
¯, w)
0 = h(¯
x, w) − r
u
¯ = −K1 x
¯ − K2 σ
¯
By assumption (6), we deduce that x
¯ = xss and u
¯ = uss . By choosing K2 invertible, we guarantee that there
is a unique solution σss of the equation uss = −K1 xss − K2 σss .
12
MAE281b – Nonlinear Control
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4.1
Integral control via linearization: state feedback integral controller
4
INTEGRAL CONTROL
We now have to stabilize (xss , σss ). Let ξ = [x − xss , σ − σss ]T . Linearization of (9) yields
ξ˙ = (A − BK)ξ
where
A 0
B
A=
, B=
, K = K1 K2 ,
C 0
0
∂f ∂h ∂f , B=
, C=
A=
∂x x=xss ,u=uss
∂u x=xss ,u=uss
∂x x=xss
(10)
Under appropriate conditions on A, B, C, one can design K such that A − BK is Hurwitz for all v ∈ Dv . For
any such design, K2 will necessarily be invertible. With this controller, (xss , σss ) is an exponentially stable
equilibrium point. One can also add a term of the form K3 e to the control law u in (8) in order to improve
performance.
Example 4.2 (Cruise control revisited) Consider the cruise control equations
v˙ =
1
− bv + u + w .
m
Let us design an integral control to stabilize the system to a desired velocity vdes no matter what unknown
road incline our car faces. We choose the state variable x = v − vdes , and the output function y = x (i.e.,
we can measure the difference between the actual and the intended velocities). The unknown disturbance is
just w ∈ R. Our equations then look like
1
− b(x + vdes ) + u + w
m
σ˙ = x − r
x˙ =
Clearly, given (r, w), there exist a unique pair (xss , uss ) such that
1
− b(xss + vdes ) + uss + w
m
0 = xss − r
0=
Actually, solving these equations, we get
xss = r,
uss = −w + b(xss + vdes ).
We aim for an integral control of the form
u = −k1 x − k2 σ.
We should choose σss in such a way that it produces uss , i.e., such that uss = −k1 xss − k2 σss . From this,
we deduce that
σss =
1
1
(−uss + k1 xss ) = (w − b(r + vdes ) + k1 r)
k2
k2
We want therefore to stabilize (xss , σss ). Note that our reference is r = 0 (so that the output y = x
asymptotically converges to 0). With the same notation as above, we have
ξ˙ = (A − BK)ξ
13
MAE281b – Nonlinear Control
c 2008-2015 by Jorge Cort´
Copyright es. Permission is granted to copy, distribute and modify this file, provided that the
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4.1
Integral control via linearization: state feedback integral controller
4
INTEGRAL CONTROL
where
ξ=
x − xss
,
σ − σss
A=
b
−m
1
0
,
0
B=
1
m
0
K = k1
,
k2
Therefore, we have
A − BK =
b
−
−m
1
k1
m
− km2
0
with eigenvalues
λ=
−(b + k1 ) ±
p
(b + k1 )2 − 4k2 m
.
2m
Therefore, choosing k2 > 0 and k1 > −b guarantees that (xss , σss ) is an exponentially stable equilibrium
point. In the original variables, the control reads
u = −k1 (v − vdes ) − k2 σ,
σ˙ = v − vdes .
Note the difference with respect to the one that we computed previously,
u = bv − k(v − vdes ).
Figure 6 gives a comparison of the performance of both controllers.
velocity versus time
velocity versus time
1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
2
4
6
•
8
10
2
4
6
8
10
Figure 6: Comparison of the performance of the feedback controllers without (left) and with integral action
(right). The desired velocity is vdes = 1. The solid lines represent the evolution under perfect information
of the friction coefficient b = 1 and the road incline w = 0. The dashed lines represent the evolution under
slight perturbations of these parameters (b = .9, w = .1). The control gains are k1 = 2, k2 = 1.
Example 4.3 (Forced pendulum) Consider the pendulum equations
θ¨ = −a sin θ − bθ˙ + cT
where a = g/l > 0, b = k/m ≥ 0, c = 1/ml2 > 0, θ is the angle subtended by the rod and the vertical axis,
and T is the torque applied to the pendulum. Suppose we want to regulate θ to θdes . Let us choose the state
˙ the control variable T and y = x1 . The equations read now
variables x1 = θ − θdes and x2 = θ,
x˙ 1 = x2
x˙ 2 = −a sin(x1 + θdes ) − bx2 + cT
y = x1
14
MAE281b – Nonlinear Control
c 2008-2015 by Jorge Cort´
Copyright es. Permission is granted to copy, distribute and modify this file, provided that the
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REFERENCES
REFERENCES
Note that our reference is r = 0 (so that the output y = x1 asymptotically converges to 0). From (6), we get
the equilibrium point xss = (0, 0), with associated torque Tss = ac sin(θdes ). We design an integral controller
of the form
T = −k1 x1 − k2 x2 − k3 σ.
Note that σss = − k13 Tss .
The matrices A, B, C in equation (10) are given by
0
1
A=
,
−a cos θdes −b
B=
0
,
c
C= 1 0
One can deduce that the matrix

0
A − BK = −a cos θdes − ck1
1
1
−b − ck2
0

0
−ck3 
0
is Hurwitz if
b + k2 c > 0,
(b + k2 c)(a cos θdes + k1 c) > k3 c,
k3 c > 0.
Suppose that we do not know the values of the parameters a, b, c, but we have upper bounds a/c ≤ ρ1 and
1/c ≤ ρ2 . Then, the choice
k2 > 0,
k1 > ρ1 +
k3
ρ2 > 0,
k2
k3 > 0
ensures that A − BK is Hurwitz. The feedback control law is given by
T = −k1 (θ − θdes ) − k2 θ˙ − k3 σ
σ˙ = θ − θdes
which is the classical PID (Proportional Integral Derivative) controller. Note the difference with regards to
the one that we computed previously
T =
a
˙
sin θdes − k1 (θ − θdes ) − k2 θ.
c
Stabilization with the PID controller will be achieved under parameter perturbations that satisfy (b +
•
k2 c)(a cos θdes + k1 c) > 0. Figure 7 gives a comparison of the performance of both controllers.
The case of output feedback is similar and slightly more involved, but the design rationale is the same, see [2,
pp. 484-485].
References
[1] S. S. Sastry, Nonlinear Systems: Analysis, Stability and Control, ser. Interdisciplinary Applied Mathematics.
Springer, 1999, no. 10.
[2] H. K. Khalil, Nonlinear Systems, 3rd ed.
Prentice Hall, 2002.
15
MAE281b – Nonlinear Control
c 2008-2015 by Jorge Cort´
Copyright es. Permission is granted to copy, distribute and modify this file, provided that the
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REFERENCES
REFERENCES
theta versus time
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0.5
1
1.5
2
2.5
theta versus time
1
3
3.5
4
0.5
1
1.5
2
2.5
3
3.5
4
Figure 7: Comparison of the performance of the feedback controllers without (left) and with integral action
(right). The desired angle is θdes = π/4. The solid lines correspond to the evolution of a system with a = 10,
b = 1 and c = 10, and perfect information on these parameters. The dashed lines correspond to the evolution
of a system with a = 10, b = .5 and c = 5, but with the non-integral controller designed for the system with
values a = 10, b = 1 and c = 10.
16
MAE281b – Nonlinear Control
c 2008-2015 by Jorge Cort´
Copyright es. Permission is granted to copy, distribute and modify this file, provided that the
original source is acknowledged.