khalil - nonlinear systems slides

Nonlinear Systems and ControlLecture # 1

Introduction

– p. 1/18

Nonlinear State Model

x1 = f1(t, x1, . . . , xn, u1, . . . , up)

x2 = f2(t, x1, . . . , xn, u1, . . . , up)

......

xn = fn(t, x1, . . . , xn, u1, . . . , up)

xi denotes the derivative of xi with respect to the timevariable t

u1, u2, . . ., up are input variables

x1, x2, . . ., xn the state variables

– p. 2/18

x =

x1

x2

...

...

xn

, u =

u1

u2

...

up

, f(t, x, u) =

f1(t, x, u)

f2(t, x, u)

...

...

fn(t, x, u)

x = f(t, x, u)

– p. 3/18

x = f(t, x, u)

y = h(t, x, u)

x is the state, u is the inputy is the output (q-dimensional vector)

Special Cases:Linear systems:

x = A(t)x + B(t)u

y = C(t)x + D(t)u

Unforced state equation:

x = f(t, x)

Results from x = f(t, x, u) with u = γ(t, x)

– p. 4/18

Autonomous System:

x = f(x)

Time-Invariant System:

x = f(x, u)

y = h(x, u)

A time-invariant state model has a time-invariance propertywith respect to shifting the initial time from t0 to t0 + a,provided the input waveform is applied from t0 + a ratherthan t0

– p. 5/18

Existence and Uniqueness of Solutions

x = f(t, x)

f(t, x) is piecewise continuous in t and locally Lipschitz inx over the domain of interest

f(t, x) is piecewise continuous in t on an interval J ⊂ R iffor every bounded subinterval J0 ⊂ J , f is continuous in tfor all t ∈ J0, except, possibly, at a finite number of pointswhere f may have finite-jump discontinuities

f(t, x) is locally Lipschitz in x at a point x0 if there is aneighborhood N(x0, r) = x ∈ Rn | ‖x − x0‖ < rwhere f(t, x) satisfies the Lipschitz condition

‖f(t, x) − f(t, y)‖ ≤ L‖x − y‖, L > 0

– p. 6/18

A function f(t, x) is locally Lipschitz in x on a domain(open and connected set) D ⊂ Rn if it is locally Lipschitz atevery point x0 ∈ D

When n = 1 and f depends only on x

|f(y) − f(x)|

|y − x|≤ L

On a plot of f(x) versus x, a straight line joining any twopoints of f(x) cannot have a slope whose absolute value isgreater than L

Any function f(x) that has infinite slope at some point isnot locally Lipschitz at that point

– p. 7/18

A discontinuous function is not locally Lipschitz at the pointsof discontinuity

The function f(x) = x1/3 is not locally Lipschitz at x = 0since

f ′(x) = (1/3)x−2/3 → ∞ a x → 0

On the other hand, if f ′(x) is continuous at a point x0 thenf(x) is locally Lipschitz at the same point becausecontinuity of f ′(x) ensures that |f ′(x)| is bounded by aconstant k in a neighborhood of x0 ; which implies thatf(x) satisfies the Lipschitz condition L = k

More generally, if for t ∈ J ⊂ R and x in a domainD ⊂ Rn, f(t, x) and its partial derivatives ∂fi/∂xj arecontinuous, then f(t, x) is locally Lipschitz in x on D

– p. 8/18

Lemma: Let f(t, x) be piecewise continuous in t andlocally Lipschitz in x at x0, for all t ∈ [t0, t1]. Then, there isδ > 0 such that the state equation x = f(t, x), withx(t0) = x0, has a unique solution over [t0, t0 + δ]

Without the local Lipschitz condition, we cannot ensureuniqueness of the solution. For example, x = x1/3 hasx(t) = (2t/3)3/2 and x(t) ≡ 0 as two different solutionswhen the initial state is x(0) = 0

The lemma is a local result because it guarantees existenceand uniqueness of the solution over an interval [t0, t0 + δ],but this interval might not include a given interval [t0, t1].Indeed the solution may cease to exist after some time

– p. 9/18

Example:x = −x2

f(x) = −x2 is locally Lipschitz for all x

x(0) = −1 ⇒ x(t) =1

(t − 1)

x(t) → −∞ as t → 1

the solution has a finite escape time at t = 1

In general, if f(t, x) is locally Lipschitz over a domain Dand the solution of x = f(t, x) has a finite escape time te,then the solution x(t) must leave every compact (closedand bounded) subset of D as t → te

– p. 10/18

Global Existence and Uniqueness

A function f(t, x) is globally Lipschitz in x if

‖f(t, x) − f(t, y)‖ ≤ L‖x − y‖

for all x, y ∈ Rn with the same Lipschitz constant L

If f(t, x) and its partial derivatives ∂fi/∂xj are continuousfor all x ∈ Rn, then f(t, x) is globally Lipschitz in x if andonly if the partial derivatives ∂fi/∂xj are globally bounded,uniformly in t

f(x) = −x2 is locally Lipschitz for all x but not globallyLipschitz because f ′(x) = −2x is not globally bounded

– p. 11/18

Lemma: Let f(t, x) be piecewise continuous in t andglobally Lipschitz in x for all t ∈ [t0, t1]. Then, the stateequation x = f(t, x), with x(t0) = x0, has a uniquesolution over [t0, t1]

The global Lipschitz condition is satisfied for linear systemsof the form

x = A(t)x + g(t)

but it is a restrictive condition for general nonlinear systems

– p. 12/18

Lemma: Let f(t, x) be piecewise continuous in t andlocally Lipschitz in x for all t ≥ t0 and all x in a domainD ⊂ Rn. Let W be a compact subset of D, and supposethat every solution of

x = f(t, x), x(t0) = x0

with x0 ∈ W lies entirely in W . Then, there is a uniquesolution that is defined for all t ≥ t0

– p. 13/18

Example:x = −x3 = f(x)

f(x) is locally Lipschitz on R, but not globally Lipschitzbecause f ′(x) = −3x2 is not globally bounded

If, at any instant of time, x(t) is positive, the derivative x(t)will be negative. Similarly, if x(t) is negative, the derivativex(t) will be positive

Therefore, starting from any initial condition x(0) = a, thesolution cannot leave the compact set x ∈ R | |x| ≤ |a|

Thus, the equation has a unique solution for all t ≥ 0

– p. 14/18

Equilibrium Points

A point x = x∗ in the state space is said to be anequilibrium point of x = f(t, x) if

x(t0) = x∗ ⇒ x(t) ≡ x∗, ∀ t ≥ t0

For the autonomous system x = f(x), the equilibriumpoints are the real solutions of the equation

f(x) = 0

An equilibrium point could be isolated; that is, there are noother equilibrium points in its vicinity, or there could be acontinuum of equilibrium points

– p. 15/18

A linear system x = Ax can have an isolated equilibriumpoint at x = 0 (if A is nonsingular) or a continuum ofequilibrium points in the null space of A (if A is singular)

It cannot have multiple isolated equilibrium points , for if xa

and xb are two equilibrium points, then by linearity any pointon the line αxa + (1 − α)xb connecting xa and xb will bean equilibrium point

A nonlinear state equation can have multiple isolatedequilibrium points .For example, the state equation

x1 = x2, x2 = −a sin x1 − bx2

has equilibrium points at (x1 = nπ, x2 = 0) forn = 0, ±1, ±2, · · ·

– p. 16/18

Linearization

A common engineering practice in analyzing a nonlinearsystem is to linearize it about some nominal operating pointand analyze the resulting linear model

What are the limitations of linearization?

Since linearization is an approximation in theneighborhood of an operating point, it can only predictthe “local” behavior of the nonlinear system in thevicinity of that point. It cannot predict the “nonlocal” or“global” behavior

There are “essentially nonlinear phenomena” that cantake place only in the presence of nonlinearity

– p. 17/18

Nonlinear Phenomena

Finite escape time

Multiple isolated equilibrium points

Limit cycles

Subharmonic, harmonic, or almost-periodic oscillations

Chaos

Multiple modes of behavior

– p. 18/18


Examples of Nonlinear Systems

– p. 1/17

Pendulum Equation

θ

mg

l

•

mlθ = −mg sin θ − klθ

x1 = θ, x2 = θ

– p. 2/17

x1 = x2

x2 = − g

lsin x1 − k

mx2

Equilibrium Points:

0 = x2

0 = − g

lsin x1 − k

mx2

(nπ, 0) for n = 0, ±1, ±2, . . .

Nontrivial equilibrium points at (0, 0) and (π, 0)

– p. 3/17

Pendulum without friction:

x1 = x2

x2 = −g

lsin x1

Pendulum with torque input:

x1 = x2

x2 = − g

lsin x1 − k

mx2 +

1

ml2T

– p. 4/17

Tunnel-Diode Circuit

1

PPPPPPPPPPRL

CvC+ JJJ vR+E

s iC iRCC CCiL vL+ XX(a) 0 0.5 1

−0.5

0

0.5

1

i=h(v)

v,V

i,mA

(b)

iC = CdvC

dt, vL = L

diL

dt

x1 = vC , x2 = iL, u = E

– p. 5/17

iC + iR − iL = 0 ⇒ iC = −h(x1) + x2

vC − E + RiL + vL = 0 ⇒ vL = −x1 − Rx2 + u

x1 =1

C[−h(x1) + x2]

x2 =1

L[−x1 − Rx2 + u]

Equilibrium Points:

0 = −h(x1) + x2

0 = −x1 − Rx2 + u

– p. 6/17

h(x1) =E

R− 1

Rx1

0 0.5 10

0.2

0.4

0.6

0.8

1

1.2

Q

Q

Q1

2

3

vR

i R

– p. 7/17

Mass–Spring System

BBB

BBB

BBB

mp

-

- y

F

FfFsp

my + Ff + Fsp = F

Sources of nonlinearity:

Nonlinear spring restoring force Fsp = g(y)

Static or Coulomb friction

– p. 8/17

Fsp = g(y)

g(y) = k(1 − a2y2)y, |ay| < 1 (softening spring)

g(y) = k(1 + a2y2)y (hardening spring)

Ff may have components due to static, Coulomb, andviscous friction

When the mass is at rest, there is a static friction force Fs

that acts parallel to the surface and is limited to ±µsmg(0 < µs < 1). Fs takes whatever value, between its limits,to keep the mass at rest

Once motion has started, the resistive force Ff is modeledas a function of the sliding velocity v = y

– p. 9/17

v v

Ff

(b)

v

(c)

v

Ff

(d)

(a)

Ff

Ff

(a) Coulomb friction; (b) Coulomb plus linear viscous friction; (c) static, Coulomb, and linear

viscous friction; (d) static, Coulomb, and linear viscous friction—Stribeck effect– p. 10/17

Negative-Resistance Oscillator

1

CiC LiL ResistiveElement

i +v

(a)CC CC

XX

v

(b)

i = h(v)

h(0) = 0, h′(0) < 0

h(v) → ∞ as v → ∞, and h(v) → −∞ as v → −∞

– p. 11/17

iC + iL + i = 0

Cdv

dt+

1

L

∫ t

−∞v(s) ds + h(v) = 0

Differentiating with respect to t and multiplying by L:

CLd2v

dt2+ v + Lh′(v)

dv

dt= 0

τ = t/√

CL

dv

dτ=

√CL

dv

dt,

d2v

dτ 2= CL

d2v

dt2

– p. 12/17

Denote the derivative of v with respect to τ by v

v + εh′(v)v + v = 0, ε =√

L/C

Special case: Van der Pol equation

h(v) = −v + 1

3v3

v − ε(1 − v2)v + v = 0

State model: x1 = v, x2 = v

x1 = x2

x2 = −x1 − εh′(x1)x2

– p. 13/17

Another State Model: z1 = iL, z2 = vC

z1 =1

εz2

z2 = −ε[z1 + h(z2)]

Change of variables: z = T (x)

x1 = v = z2

x2 =dv

dτ=

√CL

dv

dt=

√

L

C[−iL − h(vC)]

= ε[−z1 − h(z2)]

T (x) =

[

−h(x1) − 1

εx2

x1

]

, T −1(z) =

[

z2

−εz1 − εh(z2)

]

– p. 14/17

Adaptive Control

Plant : yp = apyp + kpu

ReferenceModel : ym = amym + kmr

u(t) = θ∗1r(t) + θ∗

2yp(t)

θ∗1

=km

kpand θ∗

2=

am − ap

kp

When ap and kp are unknown, we may use

u(t) = θ1(t)r(t) + θ2(t)yp(t)

where θ1(t) and θ2(t) are adjusted on-line

– p. 15/17

Adaptive Law (gradient algorithm):

θ1 = −γ(yp − ym)r

θ2 = −γ(yp − ym)yp, γ > 0

State Variables: eo = yp −ym, φ1 = θ1 −θ∗1, φ2 = θ2 −θ∗

2

ym = apym + kp(θ∗1r + θ∗

2ym)

yp = apyp + kp(θ1r + θ2yp)

eo = apeo + kp(θ1 − θ∗1)r + kp(θ2yp − θ∗

2ym)

= · · · · · · + kp[θ∗2yp − θ∗

2yp]

= (ap + kpθ∗2)eo + kp(θ1 − θ∗

1)r + kp(θ2 − θ∗

2)yp

– p. 16/17

Closed-Loop System:

eo = ameo + kpφ1r(t) + kpφ2[eo + ym(t)]

φ1 = −γeor(t)

φ2 = −γeo[eo + ym(t)]

– p. 17/17


Second-Order Systems

– p. 1/??

x1 = f1(x1, x2) = f1(x)

x2 = f2(x1, x2) = f2(x)

Let x(t) = (x1(t), x2(t)) be a solution that starts at initialstate x0 = (x10, x20). The locus in the x1–x2 plane of thesolution x(t) for all t ≥ 0 is a curve that passes through thepoint x0. This curve is called a trajectory or orbitThe x1–x2 plane is called the state plane or phase planeThe family of all trajectories is called the phase portraitThe vector field f(x) = (f1(x), f2(x)) is tangent to thetrajectory at point x because

dx2

dx1=

f2(x)

f1(x)

– p. 2/??

Vector Field diagram

Represent f(x) as a vector based at x; that is, assign to xthe directed line segment from x to x + f(x)

q

q

*

x1

x2

f(x)

x = (1, 1)

x + f(x) = (3, 2)

Repeat at every point in a grid covering the plane

– p. 3/??

−5 0 5−6

−4

−2

0

2

4

6

x1

x 2

x1 = x2, x2 = −10 sin x1

– p. 4/??

Numerical Construction of the Phase Portrait:

Select a bounding box in the state plane

Select an initial point x0 and calculate the trajectorythrough it by solving

x = f(x), x(0) = x0

in forward time (with positive t) and in reverse time (withnegative t)

x = −f(x), x(0) = x0

Repeat the process interactively

Use Simulink or pplane

– p. 5/??

Qualitative Behavior of Linear Systems

x = Ax, A is a 2 × 2 real matrix

x(t) = M exp(Jrt)M−1x0

Jr =

[

λ1 0

0 λ2

]

or

[

λ 0

0 λ

]

or

[

λ 1

0 λ

]

or

[

α −β

β α

]

x(t) = Mz(t)

z = Jrz(t)

– p. 6/??

Case 1. Both eigenvalues are real: λ1 6= λ2 6= 0

M = [v1, v2]

v1 & v2 are the real eigenvectors associated with λ1 & λ2

z1 = λ1z1, z2 = λ2z2

z1(t) = z10eλ1t, z2(t) = z20e

λ2t

z2 = czλ2/λ1

1 , c = z20/(z10)λ2/λ1

The shape of the phase portrait depends on the signs of λ1

and λ2

– p. 7/??

λ2 < λ1 < 0

eλ1t and eλ2t tend to zero as t → ∞

eλ2t tends to zero faster than eλ1t

Call λ2 the fast eigenvalue (v2 the fast eigenvector) and λ1

the slow eigenvalue (v1 the slow eigenvector)

The trajectory tends to the origin along the curve

z2 = czλ2/λ1

1 with λ2/λ1 > 1

dz2

dz1= c

λ2

λ1z

[(λ2/λ1)−1]1

– p. 8/??

z1

z2

Stable Node

λ2 > λ1 > 0

Reverse arrowheads

Reverse arrowheads =⇒ Unstable Node – p. 9/??

x2

x 1

v1

v2

(b)

x1

x 2

v1

v2

(a)

Stable Node Unstable Node

– p. 10/??

λ2 < 0 < λ1

eλ1t → ∞, while eλ2t → 0 as t → ∞

Call λ2 the stable eigenvalue (v2 the stable eigenvector)and λ1 the unstable eigenvalue (v1 the unstableeigenvector)

z2 = czλ2/λ1

1 , λ2/λ1 < 0

Saddle

– p. 11/??

z1

z2

(a)

x 1

x 2v1v2

(b)

Phase Portrait of a Saddle Point

– p. 12/??

Case 2. Complex eigenvalues: λ1,2 = α ± jβ

z1 = αz1 − βz2, z2 = βz1 + αz2

r =√

z21 + z2

2, θ = tan−1

(

z2

z1

)

r(t) = r0eαt and θ(t) = θ0 + βt

α < 0 ⇒ r(t) → 0 as t → ∞

α > 0 ⇒ r(t) → ∞ as t → ∞

α = 0 ⇒ r(t) ≡ r0 ∀ t

– p. 13/??

z1

z2 (c)

z1

z2 (b)

z1

z2(a)

α < 0 α > 0 α = 0

Stable Focus Unstable Focus Center

x 1

x2(c)

x1

x 2(b)

x 1

x2(a)

– p. 14/??

Effect of Perturbations

A → A + δA (δA arbitrarily small)

The eigenvalues of a matrix depend continuously on itsparameters

A node (with distinct eigenvalues), a saddle or a focus isstructurally stable because the qualitative behavior remainsthe same under arbitrarily small perturbations in A

A stable node with multiple eigenvalues could become astable node or a stable focus under arbitrarily smallperturbations in A

– p. 15/??

A center is not structurally stable[

µ 1

−1 µ

]

Eigenvalues = µ ± j

µ < 0 ⇒ Stable Focus

µ > 0 ⇒ Unstable Focus

– p. 16/??


Qualitative Behavior NearEquilibrium Points

&Multiple Equilibria

– p. 1/??

The qualitative behavior of a nonlinear system near anequilibrium point can take one of the patterns we have seenwith linear systems. Correspondingly the equilibrium pointsare classified as stable node, unstable node, saddle, stablefocus, unstable focus, or center

Can we determine the type of the equilibrium point of anonlinear system by linearization?

– p. 2/??

Let p = (p1, p2) be an equilibrium point of the system

x1 = f1(x1, x2), x2 = f2(x1, x2)

where f1 and f2 are continuously differentiableExpand f1 and f2 in Taylor series about (p1, p2)

x1 = f1(p1, p2) + a11(x1 − p1) + a12(x2 − p2) + H.O.T.

x2 = f2(p1, p2) + a21(x1 − p1) + a22(x2 − p2) + H.O.T.

a11 =∂f1(x1, x2)

∂x1

∣

∣

∣

∣

x=p

, a12 =∂f1(x1, x2)

∂x2

∣

∣

∣

∣

x=p

a21 =∂f2(x1, x2)

∂x1

∣

∣

∣

∣

x=p

, a22 =∂f2(x1, x2)

∂x2

∣

∣

∣

∣

x=p

– p. 3/??

f1(p1, p2) = f2(p1, p2) = 0

y1 = x1 − p1 y2 = x2 − p2

y1 = x1 = a11y1 + a12y2 + H.O.T.

y2 = x2 = a21y1 + a22y2 + H.O.T.

y ≈ Ay

A =

a11 a12

a21 a22

=

∂f1

∂x1

∂f1

∂x2

∂f2

∂x1

∂f2

∂x2

∣

∣

∣

∣

∣

∣

∣

x=p

=∂f

∂x

∣

∣

∣

∣

x=p

– p. 4/??

Eigenvalues of A Type of equilibrium pointof the nonlinear system

λ2 < λ1 < 0 Stable Nodeλ2 > λ1 > 0 Unstable Nodeλ2 < 0 < λ1 Saddle

α ± jβ, α < 0 Stable Focusα ± jβ, α > 0 Unstable Focus

±jβ Linearization Fails

– p. 5/??

Example

x1 = −x2 − µx1(x2

1+ x2

2)

x2 = x1 − µx2(x2

1+ x2

2)

x = 0 is an equilibrium point

∂f

∂x=

[

−µ(3x2

1+ x2

2) −(1 + 2µx1x2)

(1 − 2µx1x2) −µ(x2

1+ 3x2

2)

]

A =∂f

∂x

∣

∣

∣

∣

x=0

=

[

0 −1

1 0

]

x1 = r cos θ and x2 = r sin θ ⇒ r = −µr3 and θ = 1

Stable focus when µ > 0 and Unstable focus when µ < 0

– p. 6/??

For a saddle point, we can use linearization to generate thestable and unstable trajectories

Let the eigenvalues of the linearization be λ1 > 0 > λ2 andthe corresponding eigenvectors be v1 and v2

The stable and unstable trajectories will be tangent to thestable and unstable eigenvectors, respectively, as theyapproach the equilibrium point p

For the unstable trajectories use x0 = p ± αv1

For the stable trajectories use x0 = p ± αv2

α is a small positive number

– p. 7/??

Multiple Equilibria

Example: Tunnel-diode circuit

x1 = 0.5[−h(x1) + x2]

x2 = 0.2(−x1 − 1.5x2 + 1.2)

h(x1) = 17.76x1−103.79x2

1+229.62x3

1−226.31x4

1+83.72x5

1

0 0.5 10

0.2

0.4

0.6

0.8

1

Q2

Q3

Q1

vR

iR

Q1 = (0.063, 0.758)

Q2 = (0.285, 0.61)

Q3 = (0.884, 0.21)

– p. 8/??

∂f

∂x=

[

−0.5h′(x1) 0.5

−0.2 −0.3

]

A1 =

[

−3.598 0.5

−0.2 −0.3

]

, Eigenvalues : − 3.57, −0.33

A2 =

[

1.82 0.5

−0.2 −0.3

]

, Eigenvalues : 1.77, −0.25

A3 =

[

−1.427 0.5

−0.2 −0.3

]

, Eigenvalues : − 1.33, −0.4

Q1 is a stable node; Q2 is a saddle; Q3 is a stable node

– p. 9/??

x ’ = 0.5 ( − 17.76 x + 103.79 x2 − 229.62 x3 + 226.31 x4 − 83.72 x5 + y)y ’ = 0.2 ( − x − 1.5 y + 1.2)

−0.5 0 0.5 1 1.5−0.5

0

0.5

1

1.5

x

y

– p. 10/??

Hysteresis characteristics of the tunnel-diode circuit

u = E, y = vR

0 0.5 10

0.2

0.4

0.6

0.8

1

Q2

Q3

Q1

vR

iR

0 1 2 30

0.2

0.4

0.6

0.8

1

1.2

E AB

FC

D

u

y

– p. 11/??


Limit Cycles

– p. 1/??

Oscillation: A system oscillates when it has a nontrivialperiodic solution

x(t + T ) = x(t), ∀ t ≥ 0

Linear (Harmonic) Oscillator:

z =

[

0 −β

β 0

]

z

z1(t) = r0 cos(βt + θ0), z2(t) = r0 sin(βt + θ0)

r0 =√

z2

1(0) + z2

2(0), θ0 = tan−1

[

z2(0)

z1(0)

]

– p. 2/??

The linear oscillation is not practical because

It is not structurally stable. Infinitesimally smallperturbations may change the type of the equilibriumpoint to a stable focus (decaying oscillation) or unstablefocus (growing oscillation)

The amplitude of oscillation depends on the initialconditions

The same problems exist with oscillation of nonlinearsystems due to a center equilibrium point (e.g., pendulumwithout friction)

– p. 3/??

Limit Cycles:

Example: Negative Resistance Oscillator

1

CiC LiL ResistiveElement

i +v

(a)CC CC

XX

v

(b)

i = h(v)

– p. 4/??

x1 = x2

x2 = −x1 − εh′(x1)x2

There is a unique equilibrium point at the origin

A =∂f

∂x

∣

∣

∣

∣

x=0

=

0 1

−1 −εh′(0)

λ2 + εh′(0)λ + 1 = 0

h′(0) < 0 ⇒ Unstable Focus or Unstable Node

– p. 5/??

Energy Analysis:

E = 1

2Cv2

C + 1

2Li2L

vC = x1 and iL = −h(x1) −1

εx2

E = 1

2Cx2

1+ [εh(x1) + x2]

2

E = Cx1x1 + [εh(x1) + x2][εh′(x1)x1 + x2]

= Cx1x2 + [εh(x1) + x2][εh′(x1)x2 − x1 − εh′(x1)x2]

= C[x1x2 − εx1h(x1) − x1x2]

= −εCx1h(x1)

– p. 6/??

x1−a

b

E = −εCx1h(x1)

– p. 7/??

Example: Van der Pol Oscillator

x1 = x2

x2 = −x1 + ε(1 − x2

1)x2

−2 0 2 4−3

−2

−1

0

1

2

3

(b)

x1

x2

−2 0 2 4

−2

−1

0

1

2

3

4

(a)

x1

x2

ε = 0.2 ε = 1

– p. 8/??

z1 =1

εz2

z2 = −ε(z1 − z2 + 1

3z3

2)

−2 0 2−3

−2

−1

0

1

2

3

(b)

z1

z2

−5 0 5 10

−5

0

5

10

(a)

x1

x2

ε = 5

– p. 9/??

x1

x2

(a)

x1

x2

(b)

Stable Limit Cycle Unstable Limit Cycle

– p. 10/??

Example: Wien-Bridge Oscillator

1

C2v2+ PPPPPPPPPP R2BBB BBB BBB R1 C1v1+ s

"!# g(v2)+

Equivalent Circuit

– p. 11/??

State variables x1 = v1 and x2 = v2

x1 =1

C1R1

[−x1 + x2 − g(x2)]

x2 = −1

C2R1

[−x1 + x2 − g(x2)] −1

C2R2

x2

There is a unique equilibrium point at x = 0

Numerical data: C1 = C2 = R1 = R2 = 1

g(v) = 3.234v − 2.195v3 + 0.666v5

– p. 12/??

x ’ = − x + y − (3.234 y − 2.195 y3 + 0.666 y5) y ’ = − ( − x + y − (3.234 y − 2.195 y3 + 0.666 y5)) − y

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

y

– p. 13/??

x ’ = − x + y − (3.234 y − 2.195 y3 + 0.666 y5) y ’ = − ( − x + y − (3.234 y − 2.195 y3 + 0.666 y5)) − y

−6 −4 −2 0 2 4 6

−6

−4

−2

0

2

4

6

x

y

– p. 14/??

Nonlinear Systems and ControlLecture # 6Bifurcation

– p. 1/??

Bifurcation is a change in the equilibrium points or periodicorbits, or in their stability properties, as a parameter isvaried

Example

x1 = µ − x2

1

x2 = −x2

Find the equilibrium points and their types for differentvalues of µ

For µ > 0 there are two equilibrium points at (√

µ, 0) and(−√

µ, 0)

– p. 2/??

Linearization at (√

µ, 0):[

−2√

µ 0

0 −1

]

(√

µ, 0) is a stable node

Linearization at (−√µ, 0):

[

2√

µ 0

0 −1

]

(−√µ, 0) is a saddle

– p. 3/??

x1 = µ − x2

1, x2 = −x2

No equilibrium points when µ < 0

As µ decreases, the saddle and node approach each other,collide at µ = 0, and disappear for µ < 0

x1

x2

x2

x1

x1

x2

µ > 0 µ = 0 µ < 0

– p. 4/??

µ is called the bifurcation parameter and µ = 0 is thebifurcation point

Bifurcation Diagram

µ(a) Saddle−node bifurcation

– p. 5/??

Examplex1 = µx1 − x2

1, x2 = −x2

Two equilibrium points at (0, 0) and (µ, 0)

The Jacobian at (0, 0) is

[

µ 0

0 −1

]

(0, 0) is a stable node for µ < 0 and a saddle for µ > 0

The Jacobian at (µ, 0) is

[

−µ 0

0 −1

]

(µ, 0) is a saddle for µ < 0 and a stable node for µ > 0An eigenvalue crosses the origin as µ crosses zero

– p. 6/??

While the equilibrium points persist through the bifurcationpoint µ = 0, (0, 0) changes from a stable node to a saddleand (µ, 0) changes from a saddle to a stable node

µ(a) Saddle−node bifurcation

µ(b) Transcritical bifurcation

dangerous or hard safe or soft

– p. 7/??

Examplex1 = µx1 − x3

1, x2 = −x2

For µ < 0, there is a stable node at the origin

For µ > 0, there are three equilibrium points: a saddle at(0, 0) and stable nodes at (

√µ, 0), and (−√

µ, 0)

µ(c) Supercritical pitchfork bifurcation

– p. 8/??

Examplex1 = µx1 + x3

1, x2 = −x2

For µ < 0, there are three equilibrium points: a stable nodeat (0, 0) and two saddles at (±√−µ, 0)

For µ > 0, there is a saddle at (0, 0)

µ(d) Subcritical pitchfork bifurcation

– p. 9/??

Notice the difference between supercritical and subcriticalpitchfork bifurcations

µ(c) Supercritical pitchfork bifurcation

µ(d) Subcritical pitchfork bifurcation

safe or soft dangerous or hard

– p. 10/??

Example: Tunnel diode Circuit

x1 =1

C[−h(x1) + x2]

x2 =1

L[−x1 − Rx2 + µ]

A B

x2 = h(x

1)

x1

x2

(a)A B µ

(b)

– p. 11/??

Example

x1 = x1(µ − x2

1− x2

2) − x2

x2 = x2(µ − x2

1− x2

2) + x1


Linearization:

[

µ −1

1 µ

]

Stable focus for µ < 0, and unstable focus for µ > 0

A pair of complex eigenvalues cross the imaginary axis asµ crosses zero

– p. 12/??

r = µr − r3 and θ = 1

For µ > 0, there is a stable limit cycle at r =√

µ

x2

x1

x2

x1

µ < 0 µ > 0

Supercritical Hopf bifurcation

– p. 13/??

Example

x1 = x1

[

µ + (x2

1+ x2

2) − (x2

1+ x2

2)2

]

− x2

x2 = x2

[

µ + (x2

1+ x2

2) − (x2

1+ x2

2)2

]

+ x1


Linearization:

[

µ −1

1 µ

]

Stable focus for µ < 0, and unstable focus for µ > 0

A pair of complex eigenvalues cross the imaginary axis asµ crosses zero

– p. 14/??

r = µr + r3 − r5 and θ = 1

Sketch of µr + r3 − r5:

r r

µ < 0 µ > 0

For small |µ|, the stable limit cycles are approximated byr = 1, while the unstable limit cycle for µ < 0 isapproximated by r =

√

|µ|

– p. 15/??

As µ increases from negative to positive values, the stablefocus at the origin merges with the unstable limit cycle andbifurcates into unstable focus

Subcritical Hopf bifurcation

µ(e) Supercritical Hopf bifurcation

µ(f) Subcrtitical Hopf bifurcation

safe or soft dangerous or hard

– p. 16/??

All six types of bifurcation occur in the vicinity of anequilibrium point. They are called local bifurcations

Example of Global Bifurcation

x1 = x2

x2 = µx2 + x1 − x2

1+ x1x2

There are two equilibrium points at (0, 0) and (1, 0). Bylinearization, we can see that (0, 0) is always a saddle,while (1, 0) is an unstable focus for −1 < µ < 1

Limit analysis to the range −1 < µ < 1

– p. 17/??

x2

x1

µ=−0.95

x1

x2

µ=−0.88

x1

x2

µ=−0.8645

x1

x2

µ=−0.8

Saddle–connection (or homoclinic) bifurcation

– p. 18/??


Stability of Equilibrium PointsBasic Concepts & Linearization

– p. 1/??

x = f(x)

f is locally Lipschitz over a domain D ⊂ Rn

Suppose x ∈ D is an equilibrium point; that is, f(x) = 0

Characterize and study the stability of x

For convenience, we state all definitions and theorems forthe case when the equilibrium point is at the origin of Rn;that is, x = 0. No loss of generality

y = x − x

y = x = f(x) = f(y + x)def= g(y), where g(0) = 0

– p. 2/??

Definition: The equilibrium point x = 0 of x = f(x) is

stable if for each ε > 0 there is δ > 0 (dependent on ε)such that

‖x(0)‖ < δ ⇒ ‖x(t)‖ < ε, ∀ t ≥ 0

unstable if it is not stable

asymptotically stable if it is stable and δ can be chosensuch that

‖x(0)‖ < δ ⇒ limt→∞

x(t) = 0

– p. 3/??

First-Order Systems (n = 1)

The behavior of x(t) in the neighborhood of the origin canbe determined by examining the sign of f(x)

The ε–δ requirement for stability is violated if xf(x) > 0 oneither side of the origin

f(x)

x

f(x)

x

f(x)

x

Unstable Unstable Unstable

– p. 4/??

The origin is stable if and only if xf(x) ≤ 0 in someneighborhood of the origin

f(x)

x

f(x)

x

f(x)

x

Stable Stable Stable

– p. 5/??

The origin is asymptotically stable if and only if xf(x) < 0in some neighborhood of the origin

f(x)

x−a b

f(x)

x

(a) (b)

Asymptotically Stable Globally Asymptotically Stable

– p. 6/??

Definition: Let the origin be an asymptotically stableequilibrium point of the system x = f(x), where f is alocally Lipschitz function defined over a domain D ⊂ Rn

( 0 ∈ D)

The region of attraction (also called region ofasymptotic stability, domain of attraction, or basin) is theset of all points x0 in D such that the solution of

x = f(x), x(0) = x0

is defined for all t ≥ 0 and converges to the origin as t

tends to infinity

The origin is said to be globally asymptotically stable ifthe region of attraction is the whole space Rn

– p. 7/??

Second-Order Systems (n = 2)

Type of equilibrium point Stability PropertyCenter

Stable NodeStable Focus

Unstable NodeUnstable Focus

Saddle

– p. 8/??

Example: Tunnel Diode Circuit

x ’ = 0.5 ( − 17.76 x + 103.79 x2 − 229.62 x3 + 226.31 x4 − 83.72 x5 + y)y ’ = 0.2 ( − x − 1.5 y + 1.2)

−0.5 0 0.5 1 1.5−0.5

0

0.5

1

1.5

x

y

– p. 9/??

Example: Pendulum Without Friction

x ’ = y y ’ = − 10 sin(x)

−4 −3 −2 −1 0 1 2 3 4

−8

−6

−4

−2

0

2

4

6

8

x

y

– p. 10/??

Example: Pendulum With Friction

x ’ = y y ’ = − 10 sin(x) − y

−4 −3 −2 −1 0 1 2 3 4

−8

−6

−4

−2

0

2

4

6

8

x

y

– p. 11/??

Linear Time-Invariant Systems

x = Ax

x(t) = exp(At)x(0)

P −1AP = J = block diag[J1, J2, . . . , Jr]

Ji =

λi 1 0 . . . . . . 0

0 λi 1 0 . . . 0... . . . ...... . . . 0... . . . 1

0 . . . . . . . . . 0 λi

m×m

– p. 12/??

exp(At) = P exp(Jt)P −1 =r

∑

i=1

mi∑

k=1

tk−1 exp(λit)Rik

mi is the order of the Jordan block Ji

Re[λi] < 0 ∀ i ⇔ Asymptotically Stable

Re[λi] > 0 for some i ⇒ Unstable

Re[λi] ≤ 0 ∀ i & mi > 1 for Re[λi] = 0 ⇒ Unstable

Re[λi] ≤ 0 ∀ i & mi = 1 for Re[λi] = 0 ⇒ Stable

If an n × n matrix A has a repeated eigenvalue λi ofalgebraic multiplicity qi, then the Jordan blocks of λi haveorder one if and only if rank(A − λiI) = n − qi

– p. 13/??

Theorem: The equilibrium point x = 0 of x = Ax is stable ifand only if all eigenvalues of A satisfy Re[λi] ≤ 0 and forevery eigenvalue with Re[λi] = 0 and algebraic multiplicityqi ≥ 2, rank(A − λiI) = n − qi, where n is the dimensionof x. The equilibrium point x = 0 is globally asymptoticallystable if and only if all eigenvalues of A satisfy Re[λi] < 0

When all eigenvalues of A satisfy Re[λi] < 0, A is called aHurwitz matrix

When the origin of a linear system is asymptotically stable,its solution satisfies the inequality

‖x(t)‖ ≤ k‖x(0)‖e−λt, ∀ t ≥ 0

k ≥ 1, λ > 0

– p. 14/??

Exponential Stability

Definition: The equilibrium point x = 0 of x = f(x) is saidto be exponentially stable if

‖x(t)‖ ≤ k‖x(0)‖e−λt, ∀ t ≥ 0

k ≥ 1, λ > 0, for all ‖x(0)‖ < c

It is said to be globally exponentially stable if the inequalityis satisfied for any initial state x(0)

Exponential Stability ⇒ Asymptotic Stability

– p. 15/??

Examplex = −x3

The origin is asymptotically stable

x(t) =x(0)

√

1 + 2tx2(0)

x(t) does not satisfy |x(t)| ≤ ke−λt|x(0)| because

|x(t)| ≤ ke−λt|x(0)| ⇒e2λt

1 + 2tx2(0)≤ k2

Impossible because limt→∞

e2λt

1 + 2tx2(0)= ∞

– p. 16/??

Linearizationx = f(x), f(0) = 0

f is continuously differentiable over D = ‖x‖ < r

J(x) =∂f

∂x(x)

h(σ) = f(σx) for 0 ≤ σ ≤ 1

h′(σ) = J(σx)x

h(1) − h(0) =

∫ 1

0

h′(σ) dσ, h(0) = f(0) = 0

f(x) =

∫ 1

0

J(σx) dσ x

– p. 17/??

f(x) =

∫ 1

0

J(σx) dσ x

Set A = J(0) and add and subtract Ax

f(x) = [A+G(x)]x, where G(x) =

∫ 1

0

[J(σx)−J(0)] dσ

G(x) → 0 as x → 0

This suggests that in a small neighborhood of the origin wecan approximate the nonlinear system x = f(x) by itslinearization about the origin x = Ax

– p. 18/??

Theorem:

The origin is exponentially stable if and only ifRe[λi] < 0 for all eigenvalues of A

The origin is unstable if Re[λi] > 0 for some i

Linearization fails when Re[λi] ≤ 0 for all i, withRe[λi] = 0 for some i

Examplex = ax3

A =∂f

∂x

∣

∣

∣

∣

x=0

= 3ax2∣

∣

x=0= 0

Stable if a = 0; Asymp stable if a < 0; Unstable if a > 0When a < 0, the origin is not exponentially stable

– p. 19/??


Lyapunov Stability

– p. 1/10

Let V (x) be a continuously differentiable function defined ina domain D ⊂ Rn; 0 ∈ D. The derivative of V along thetrajectories of x = f(x) is

V (x) =

n∑

i=1

∂V

∂xixi =

n∑

i=1

∂V

∂xifi(x)

=[

∂V∂x1

, ∂V∂x2

, . . . , ∂V∂xn

]

f1(x)

f2(x)...

fn(x)

=∂V

∂xf(x)

– p. 2/10

If φ(t; x) is the solution of x = f(x) that starts at initialstate x at time t = 0, then

V (x) =d

dtV (φ(t; x))

∣

∣

∣

∣

t=0

If V (x) is negative, V will decrease along the solution ofx = f(x)

If V (x) is positive, V will increase along the solution ofx = f(x)

– p. 3/10

Lyapunov’s Theorem:

If there is V (x) such that

V (0) = 0 and V (x) > 0, ∀ x ∈ D/0

V (x) ≤ 0, ∀ x ∈ D

then the origin is a stable

Moreover, if

V (x) < 0, ∀ x ∈ D/0

then the origin is asymptotically stable

– p. 4/10

Furthermore, if V (x) > 0, ∀ x 6= 0,

‖x‖ → ∞ ⇒ V (x) → ∞

and V (x) < 0, ∀ x 6= 0, then the origin is globallyasymptotically stable

Proof: DB

r

Ωβ

Bδ

0 < r ≤ ε, Br = ‖x‖ ≤ r

α = min‖x‖=r

V (x) > 0

0 < β < α

Ωβ = x ∈ Br | V (x) ≤ β

‖x‖ ≤ δ ⇒ V (x) < β

– p. 5/10

Solutions starting in Ωβ stay in Ωβ because V (x) ≤ 0 in Ωβ

x(0) ∈ Bδ ⇒ x(0) ∈ Ωβ ⇒ x(t) ∈ Ωβ ⇒ x(t) ∈ Br

‖x(0)‖ < δ ⇒ ‖x(t)‖ < r ≤ ε, ∀ t ≥ 0

⇒ The origin is stable

Now suppose V (x) < 0 ∀ x ∈ D/0. V (x(t) ismonotonically decreasing and V (x(t)) ≥ 0

limt→∞

V (x(t)) = c ≥ 0

limt→∞

V (x(t)) = c ≥ 0 Show that c = 0

Suppose c > 0. By continuity of V (x), there is d > 0 suchthat Bd ⊂ Ωc. Then, x(t) lies outside Bd for all t ≥ 0

– p. 6/10

γ = − maxd≤‖x‖≤r

V (x)

V (x(t)) = V (x(0)) +

∫ t

0

V (x(τ )) dτ ≤ V (x(0)) − γt

This inequality contradicts the assumption c > 0

⇒ The origin is asymptotically stable

The condition ‖x‖ → ∞ ⇒ V (x) → ∞ implies that theset Ωc = x ∈ Rn | V (x) ≤ c is compact for every c > 0.This is so because for any c > 0, there is r > 0 such thatV (x) > c whenever ‖x‖ > r. Thus, Ωc ⊂ Br. All solutionsstarting Ωc will converge to the origin. For any pointp ∈ Rn, choosing c = V (p) ensures that p ∈ Ωc

⇒ The origin is globally asymptotically stable

– p. 7/10

TerminologyV (0) = 0, V (x) ≥ 0 for x 6= 0 Positive semidefiniteV (0) = 0, V (x) > 0 for x 6= 0 Positive definiteV (0) = 0, V (x) ≤ 0 for x 6= 0 Negative semidefiniteV (0) = 0, V (x) < 0 for x 6= 0 Negative definite

‖x‖ → ∞ ⇒ V (x) → ∞ Radially unbounded

Lyapunov’ Theorem: The origin is stable if there is acontinuously differentiable positive definite function V (x) sothat V (x) is negative semidefinite, and it is asymptoticallystable if V (x) is negative definite. It is globallyasymptotically stable if the conditions for asymptoticstability hold globally and V (x) is radially unbounded

– p. 8/10

A continuously differentiable function V (x) satisfying theconditions for stability is called a Lyapunov function. Thesurface V (x) = c, for some c > 0, is called a Lyapunovsurface or a level surface

V (x) = c 1

c 2

c 3

c 1<c 2<c 3

– p. 9/10

Why do we need the radial unboundedness condition toshow global asymptotic stability?It ensures that Ωc = x ∈ Rn | V (x) ≤ c is bounded forevery c > 0Without it Ωc might not bounded for large cExample

V (x) =x2

1

1 + x2

1

+ x2

2

cccccccccccccc

hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhx 1

x 2

– p. 10/10


Lyapunov Stability

– p. 1/15

Quadratic Forms

V (x) = xT Px =n

∑

i=1

n∑

j=1

pijxixj , P = P T

λmin(P )‖x‖2 ≤ xT Px ≤ λmax(P )‖x‖2

P ≥ 0 (Positive semidefinite) if and only if λi(P ) ≥ 0 ∀i

P > 0 (Positive definite) if and only if λi(P ) > 0 ∀i

V (x) is positive definite if and only if P is positive definiteV (x) is positive semidefinite if and only if P is positivesemidefiniteP > 0 if and only if all the leading principal minors of P arepositive

– p. 2/15

Linear Systemsx = Ax

V (x) = xT Px, P = P T > 0

V (x) = xT Px + xT Px = xT (PA + AT P )xdef= −xT Qx

If Q > 0, then A is Hurwitz

Or choose Q > 0 and solve the Lyapunov equation

PA + AT P = −Q

If P > 0, then A is Hurwitz

Matlab: P = lyap(A′, Q)

– p. 3/15

Theorem A matrix A is Hurwitz if and only if for anyQ = QT > 0 there is P = P T > 0 that satisfies theLyapunov equation

PA + AT P = −Q

Moreover, if A is Hurwitz, then P is the unique solution

Idea of the proof: Sufficiency follows from Lyapunov’stheorem. Necessity is shown by verifying that

P =

∫ ∞

0

exp(AT t)Q exp(At) dt

is positive definite and satisfies the Lyapunov equation

– p. 4/15

Linearization

x = f(x) = [A + G(x)]x

G(x) → 0 as x → 0

Suppose A is Hurwitz. Choose Q = QT > 0 and solve theLyapunov equation PA + AT P = −Q for P . UseV (x) = xT Px as a Lyapunov function candidate forx = f(x)

V (x) = xT Pf(x) + fT (x)Px

= xT P [A + G(x)]x + xT [AT + GT (x)]Px

= xT (PA + AT P )x + 2xT PG(x)x

= −xT Qx + 2xT PG(x)x

– p. 5/15

V (x) ≤ −xT Qx + 2‖P‖ ‖G(x)‖ ‖x‖2

For any γ > 0, there exists r > 0 such that

‖G(x)‖ < γ, ∀ ‖x‖ < r

xT Qx ≥ λmin(Q)‖x‖2 ⇔ −xT Qx ≤ −λmin(Q)‖x‖2

V (x) < −[λmin(Q) − 2γ‖P‖]‖x‖2, ∀ ‖x‖ < r

Choose

γ <λmin(Q)

2‖P‖V (x) = xT Px is a Lyapunov function for x = f(x)

– p. 6/15

We can use V (x) = xT Px to estimate the region ofattraction

Suppose V (x) < 0, ∀ 0 < ‖x‖ < r

Take c = min‖x‖=r

xT Px = λmin(P )r2

xT Px < c ⊂ ‖x‖ < rAll trajectories starting in the set xT Px < c approach theorigin as t tends to ∞. Hence, the set xT Px < c is asubset of the region of attraction (an estimate of the regionof attraction)

– p. 7/15

Example

x1 = −x2

x2 = x1 + (x2

1− 1)x2

A =∂f

∂x

∣

∣

∣

∣

x=0

=

[

0 −1

1 −1

]

has eigenvalues (−1 ± j√

3)/2. Hence the origin isasymptotically stable

Take Q = I, PA+AT P = −I ⇒ P =

[

1.5 −0.5

−0.5 1

]

λmin(P ) = 0.691

– p. 8/15

V (x) = xT Px = 1.5x2

1− x1x2 + x2

2

V (x) = (3x1 − x2)(−x2) + (−x1 + 2x2)[x1 + (x2

1− 1)x2]

= −(x2

1+ x2

2) − (x3

1x2 − 2x2

1x2

2)

V (x) ≤ −‖x‖2+|x1| |x1x2| |x1−2x2| ≤ −‖x‖2+

√5

2‖x‖4

where |x1| ≤ ‖x‖, |x1x2| ≤ 1

2‖x‖2, |x1 − 2x2| ≤

√5‖x‖

V (x) < 0 for 0 < ‖x‖2 <2

√5

def= r2

Take c = λmin(P )r2 = 0.691 × 2√

5= 0.618

V (x) < c is an estimate of the region of attraction– p. 9/15

Example:x = −g(x)

g(0) = 0; xg(x) > 0, ∀ x 6= 0 and x ∈ (−a, a)

V (x) =

∫ x

0

g(y) dy

V (x) =∂V

∂x[−g(x)] = −g2(x) < 0, ∀ x ∈ (−a, a), x 6= 0

The origin is asymptotically stable

If xg(x) > 0 for all x 6= 0, use

V (x) = 1

2x2 +

∫ x

0

g(y) dy

– p. 10/15

V (x) = 1

2x2 +

∫ x

0

g(y) dy

is positive definite for all x and radially unbounded sinceV (x) ≥ 1

2x2

V (x) = −xg(x) − g2(x) < 0, ∀ x 6= 0

The origin is globally asymptotically stable

– p. 11/15

Example: Pendulum equation without friction

x1 = x2

x2 = − a sin x1

V (x) = a(1 − cos x1) + 1

2x2

2

V (0) = 0 and V (x) is positive definite over the domain−2π < x1 < 2π

V (x) = ax1 sin x1 + x2x2 = ax2 sin x1 − ax2 sin x1 = 0

The origin is stable

Since V (x) ≡ 0, the origin is not asymptotically stable

– p. 12/15

Example: Pendulum equation with friction

x1 = x2

x2 = − a sin x1 − bx2

V (x) = a(1 − cos x1) +1

2x2

2

V (x) = ax1 sin x1 + x2x2 = − bx2

2

The origin is stable

V (x) is not negative definite because V (x) = 0 for x2 = 0irrespective of the value of x1

– p. 13/15

The conditions of Lyapunov’s theorem are only sufficient.Failure of a Lyapunov function candidate to satisfy theconditions for stability or asymptotic stability does not meanthat the equilibrium point is not stable or asymptoticallystable. It only means that such stability property cannot beestablished by using this Lyapunov function candidate

Try

V (x) = 1

2xT Px + a(1 − cos x1)

= 1

2[x1 x2]

[

p11 p12

p12 p22

] [

x1

x2

]

+ a(1 − cos x1)

p11 > 0, p11p22 − p2

12> 0

– p. 14/15

V (x) = (p11x1 + p12x2 + a sin x1) x2

+ (p12x1 + p22x2) (−a sin x1 − bx2)

= a(1 − p22)x2 sin x1 − ap12x1 sin x1

+ (p11 − p12b) x1x2 + (p12 − p22b) x2

2

p22 = 1, p11 = bp12 ⇒ 0 < p12 < b, Take p12 = b/2

V (x) = − 1

2abx1 sin x1 − 1

2bx2

2

D = x ∈ R2 | |x1| < π

V (x) is positive definite and V (x) is negative definite over DThe origin is asymptotically stable

Read about the variable gradient method in the textbook

– p. 15/15


The Invariance Principle

– p. 1/16

Example: Pendulum equation with friction

x1 = x2

x2 = − a sin x1 − bx2

V (x) = a(1 − cos x1) +1

2x2

2

V (x) = ax1 sin x1 + x2x2 = − bx22

The origin is stable. V (x) is not negative definite becauseV (x) = 0 for x2 = 0 irrespective of the value of x1

However, near the origin, the solution cannot stayidentically in the set x2 = 0

– p. 2/16

Definitions: Let x(t) be a solution of x = f(x)

A point p is said to be a positive limit point of x(t) if there isa sequence tn, with limn→∞ tn = ∞, such thatx(tn) → p as n → ∞

The set of all positive limit points of x(t) is called thepositive limit set of x(t); denoted by L+

If x(t) approaches an asymptotically stable equilibriumpoint x, then x is the positive limit point of x(t) and L+ = x

A stable limit cycle is the positive limit set of every solutionstarting sufficiently near the limit cycle

– p. 3/16

A set M is an invariant set with respect to x = f(x) if

x(0) ∈ M ⇒ x(t) ∈ M, ∀ t ∈ R

Examples:

Equilibrium points

Limit Cycles

A set M is a positively invariant set with respect tox = f(x) if

x(0) ∈ M ⇒ x(t) ∈ M, ∀ t ≥ 0

Example: The set Ωc = V (x) ≤ c with V (x) ≤ 0 in Ωc

– p. 4/16

The distance from a point p to a set M is defined by

dist(p, M) = infx∈M

‖p − x‖

x(t) approaches a set M as t approaches infinity, if foreach ε > 0 there is T > 0 such that

dist(x(t), M) < ε, ∀ t > T

Example: every solution x(t) starting sufficiently near astable limit cycle approaches the limit cycle as t → ∞

Notice, however, that x(t) does converge to any specificpoint on the limit cycle

– p. 5/16

Lemma: If a solution x(t) of x = f(x) is bounded andbelongs to D for t ≥ 0, then its positive limit set L+ is anonempty, compact, invariant set. Moreover, x(t)

approaches L+ as t → ∞

LaSalle’s theorem: Let f(x) be a locally Lipschitz functiondefined over a domain D ⊂ Rn and Ω ⊂ D be a compactset that is positively invariant with respect to x = f(x). LetV (x) be a continuously differentiable function defined overD such that V (x) ≤ 0 in Ω. Let E be the set of all points inΩ where V (x) = 0, and M be the largest invariant set in E.Then every solution starting in Ω approaches M as t → ∞

– p. 6/16

Proof:

V (x) ≤ in Ω ⇒ V (x(t)) is a decreasing

V (x) is continuous in Ω ⇒ V (x) ≥ b = minx∈Ω

V (x)

⇒ limt→∞

V (x(t)) = a

x(t) ∈ Ω ⇒ x(t) is bounded ⇒ L+ exists

Moreover, L+ ⊂ Ω and x(t) approaches L+ as t → ∞

For any p ∈ L+, there is tn with limn→∞ tn = ∞ suchthat x(tn) → p as n → ∞

V (x) is continuous ⇒ V (p) = limn→∞

V (x(tn)) = a

– p. 7/16

V (x) = a on L+ and L+ invariant ⇒ V (x) = 0, ∀ x ∈ L+

L+ ⊂ M ⊂ E ⊂ Ω

x(t) approaches L+ ⇒ x(t) approaches M (as t → ∞)

– p. 8/16

Theorem: Let f(x) be a locally Lipschitz function definedover a domain D ⊂ Rn; 0 ∈ D. Let V (x) be a continuouslydifferentiable positive definite function defined over D suchthat V (x) ≤ 0 in D. Let S = x ∈ D | V (x) = 0

If no solution can stay identically in S, other than thetrivial solution x(t) ≡ 0, then the origin is asymptoticallystable

Moreover, if Γ ⊂ D is compact and positively invariant,then it is a subset of the region of attraction

Furthermore, if D = Rn and V (x) is radiallyunbounded, then the origin is globally asymptoticallystable

– p. 9/16

Example:

x1 = x2

x2 = −h1(x1) − h2(x2)

hi(0) = 0, yhi(y) > 0, for 0 < |y| < a

V (x) =

∫ x1

0

h1(y) dy + 1

2x2

2

D = −a < x1 < a, −a < x2 < a

V (x) = h1(x1)x2+x2[−h1(x1)−h2(x2)] = −x2h2(x2) ≤ 0

V (x) = 0 ⇒ x2h2(x2) = 0 ⇒ x2 = 0

S = x ∈ D | x2 = 0

– p. 10/16

x1 = x2, x2 = −h1(x1) − h2(x2)

x2(t) ≡ 0 ⇒ x2(t) ≡ 0 ⇒ h1(x1(t)) ≡ 0 ⇒ x1(t) ≡ 0

The only solution that can stay identically in S is x(t) ≡ 0

Thus, the origin is asymptotically stable

Suppose a = ∞ and∫ y0

h1(z) dz → ∞ as |y| → ∞

Then, D = R2 and V (x) =∫ x1

0h1(y) dy + 1

2x2

2 is radiallyunbounded. S = x ∈ R2 | x2 = 0 and the only solutionthat can stay identically in S is x(t) ≡ 0


– p. 11/16

Example: m-link Robot Manipulator

r

r

HH

q1

q2

Load

Two-link Robot Manipulator

– p. 12/16

M(q)q + C(q, q)q + Dq + g(q) = u

q is an m-dimensional vector of joint positions

u is an m-dimensional control (torque) inputs

M = MT > 0 is the inertia matrix

C(q, q)q accounts for centrifugal and Coriolis forces

(M − 2C)T = −(M − 2C)

Dq accounts for viscous damping; D = DT ≥ 0

g(q) accounts for gravity forces; g(q) = [∂P (q)/∂q]T

P (q) is the total potential energy of the links due to gravity

– p. 13/16

Investigate the use of the (PD plus gravity compensation)control law

u = g(q) − Kp(q − q∗) − Kd q

to stabilize the robot at a desired position q∗, where Kp andKd are symmetric positive definite matrices

e = q − q∗, e = q

Me = Mq

= −C q − D q − g(q) + u

= −C q − D q − Kp(q − q∗) − Kd q

= −C e − D e − Kp e − Kd e

– p. 14/16

Me = −C e − D e − Kp e − Kd e

V = 1

2eT M(q)e + 1

2eT Kpe

V = eT Me + 1

2eT Me + eT Kpe

= −eT Ce − eT De − eT Kpe − eT Kde

+ 1

2eT Me + eT Kpe

= 1

2eT (M − 2C)e − eT (Kd + D)e

= −eT (Kd + D)e ≤ 0

– p. 15/16

(Kd + D) is positive definite

V = −eT (Kd + D)e = 0 ⇒ e = 0

Me = −C e − D e − Kp e − Kd e

e(t) ≡ 0 ⇒ e(t) ≡ 0 ⇒ Kpe(t) ≡ 0 ⇒ e(t) ≡ 0

By LaSalle’s theorem the origin (e = 0, e = 0) is globallyasymptotically stable

– p. 16/16


Exponential Stability&

Region of Attraction

– p. 1/18

Exponential Stability:The origin of x = f(x) is exponentially stable if and only ifthe linearization of f(x) at the origin is Hurwitz

Theorem: Let f(x) be a locally Lipschitz function definedover a domain D ⊂ Rn; 0 ∈ D. Let V (x) be acontinuously differentiable function such that

k1‖x‖a ≤ V (x) ≤ k2‖x‖a

V (x) ≤ −k3‖x‖a

for all x ∈ D, where k1, k2, k3, and a are positiveconstants. Then, the origin is an exponentially stableequilibrium point of x = f(x). If the assumptions holdglobally, the origin will be globally exponentially stable

– p. 2/18

Proof: Choose c > 0 small enough that

k1‖x‖a ≤ c ⊂ D

V (x) ≤ c ⇒ k1‖x‖a ≤ c

Ωc = V (x) ≤ c ⊂ k1‖x‖a ≤ c ⊂ D

Ωc is compact and positively invariant; ∀ x(0) ∈ Ωc

V ≤ −k3‖x‖a ≤ −k3

k2V

dV

V≤ − k3

k2dt

V (x(t)) ≤ V (x(0))e−(k3/k2)t

– p. 3/18

‖x(t)‖ ≤[

V (x(t))

k1

]1/a

≤[

V (x(0))e−(k3/k2)t

k1

]1/a

≤[

k2‖x(0)‖ae−(k3/k2)t

k1

]1/a

=

(

k2

k1

)1/a

e−γt ‖x(0)‖, γ = k3/(k2a)

– p. 4/18

Example

x1 = x2

x2 = −h(x1) − x2

c1y2 ≤ yh(y) ≤ c2y

2, ∀ y, c1 > 0, c2 > 0

V (x) = 12 xT

[

1 1

1 2

]

x + 2

∫ x1

0h(y) dy

c1x21 ≤ 2

∫ x1

0h(y) dy ≤ c2x

21

V = [x1 + x2 + 2h(x1)]x2 + [x1 + 2x2][−h(x1) − x2]

= −x1h(x1) − x22 ≤ −c1x

21 − x2

2

The origin is globally exponentially stable– p. 5/18

Region of Attraction

Lemma: If x = 0 is an asymptotically stable equilibriumpoint for x = f(x), then its region of attraction RA is anopen, connected, invariant set. Moreover, the boundary ofRA is formed by trajectories

– p. 6/18

Example

x1 = −x2

x2 = x1 + (x21 − 1)x2

−4 −2 0 2 4−4

−2

0

2

4

x1

x2

– p. 7/18

Example

x1 = x2

x2 = −x1 + 13x3

1 − x2

−4 −2 0 2 4−4

−2

0

2

4

x1

x2

– p. 8/18

Estimates of the Region of Attraction: Find a subset of theregion of attraction

Warning: Let D be a domain with 0 ∈ D such that for allx ∈ D, V (x) is positive definite and V (x) is negativedefinite

Is D a subset of the region of attraction?

NO

Why?

– p. 9/18

Example: Reconsider

x1 = x2

x2 = −x1 + 13x3

1 − x2

V (x) = 12xT

[

1 1

1 2

]

x + 2∫ x1

0 (y − 13y3) dy

= 32x2

1 − 16x4

1 + x1x2 + x22

V (x) = −x21(1 − 1

3x21) − x2

2

D = −√

3 < x1 <√

3Is D a subset of the region of attraction?

– p. 10/18

The simplest estimate is the bounded component ofV (x) < c, where c = minx∈∂D V (x)

For V (x) = xT Px, where P = P T > 0, the minimum ofV (x) over ∂D is given by

For D = ‖x‖ < r, min‖x‖=r

xT Px = λmin(P )r2

For D = |bT x| < r, min|bT x|=r

xT Px =r2

bT P −1b

For D = |bTi x| < ri, i = 1, . . . , p,

Take c = min1≤i≤p

r2i

bTi P −1bi

≤ minx∈∂D

xT Px

– p. 11/18

Example (Revisited)

x1 = −x2

x2 = x1 + (x21 − 1)x2

V (x) = 1.5x21 − x1x2 + x2

2

V (x) = −(x21 + x2

2) − (x31x2 − 2x2

1x22)

V (x) < 0 for 0 < ‖x‖2 <2

√5

def= r2

Take c = λmin(P )r2 = 0.691 × 2√

5= 0.618

V (x) < c is an estimate of the region of attraction

– p. 12/18

x1 = ρ cos θ, x2 = ρ sin θ

V = −ρ2 + ρ4 cos2 θ sin θ(2 sin θ − cos θ)

≤ −ρ2 + ρ4| cos2 θ sin θ| · |2 sin θ − cos θ|≤ −ρ2 + ρ4 × 0.3849 × 2.2361

≤ −ρ2 + 0.861ρ4 < 0, for ρ2 < 10.861

Take c = λmin(P )r2 =0.691

0.861= 0.803

Alternatively, choose c as the largest constant such thatxT Px < c is a subset of V (x) < 0

– p. 13/18

−2 −1 0 1 2−2

−1

0

1

2

x1

x2

(a)−2 0 2

−3

−2

−1

0

1

2

3

x1

x2

(b)

(a) Contours of V (x) = 0 (dashed)V (x) = 0.8 (dash-dot), V (x) = 2.25 (solid)(b) comparison of the region of attraction with its estimate

– p. 14/18

If D is a domain where V (x) is positive definite and V (x)

is negative definite (or V (x) is negative semidefinite and nosolution can stay identically in the set V (x) = 0 other thanx = 0), then according to LaSalle’s theorem any compactpositively invariant subset of D is a subset of the region ofattraction

Example

x1 = x2

x2 = −4(x1 + x2) − h(x1 + x2)

h(0) = 0; uh(u) ≥ 0, ∀ |u| ≤ 1

– p. 15/18

V (x) = xT Px = xT

[

2 1

1 1

]

x = 2x21 + 2x1x2 + x2

2

V (x) = (4x1 + 2x2)x1 + 2(x1 + x2)x2

= −2x21 − 6(x1 + x2)

2 − 2(x1 + x2)h(x1 + x2)

≤ −2x21 − 6(x1 + x2)

2, ∀ |x1 + x2| ≤ 1

= −xT

[

8 6

6 6

]

x

V (x) is negative definite in |x1 + x2| ≤ 1

bT = [1 1], c = min|x1+x2|=1

xT Px =1

bT P −1b= 1

– p. 16/18

−5 0 5−5

0

5(−3,4)

(3,−4)

x2

x1

V(x) = 10

V(x) = 1

– p. 18/18


Converse Lyapunov Functions&

Time Varying Systems

– p. 1/18

Converse Lyapunov Theorem–Exponential Stability

Let x = 0 be an exponentially stable equilibrium point forthe system x = f(x), where f is continuously differentiableon D = ‖x‖ < r. Let k, λ, and r0 be positive constantswith r0 < r/k such that

‖x(t)‖ ≤ k‖x(0)‖e−λt, ∀ x(0) ∈ D0, ∀ t ≥ 0

where D0 = ‖x‖ < r0. Then, there is a continuouslydifferentiable function V (x) that satisfies the inequalities

– p. 2/18

c1‖x‖2 ≤ V (x) ≤ c2‖x‖2

∂V

∂xf(x) ≤ −c3‖x‖2

∥

∥

∥

∥

∂V

∂x

∥

∥

∥

∥

≤ c4‖x‖

for all x ∈ D0, with positive constants c1, c2, c3, and c4Moreover, if f is continuously differentiable for all x, globallyLipschitz, and the origin is globally exponentially stable,then V (x) is defined and satisfies the aforementionedinequalities for all x ∈ Rn

– p. 3/18

Idea of the proof: Let ψ(t;x) be the solution of

y = f(y), y(0) = x

Take

V (x) =

∫ δ

0ψT (t;x) ψ(t;x) dt, δ > 0

– p. 4/18

Example: Consider the system x = f(x) where f iscontinuously differentiable in the neighborhood of the originand f(0) = 0. Show that the origin is exponentially stableonly if A = [∂f/∂x](0) is Hurwitz

f(x) = Ax+G(x)x, G(x) → 0 as x → 0

Given any L > 0, there is r1 > 0 such that

‖G(x)‖ ≤ L, ∀ ‖x‖ < r1

Because the origin of x = f(x) is exponentially stable, letV (x) be the function provided by the converse Lyapunovtheorem over the domain ‖x‖ < r0. Use V (x) as aLyapunov function candidate for x = Ax

– p. 5/18

∂V

∂xAx =

∂V

∂xf(x) −

∂V

∂xG(x)x

≤ −c3‖x‖2 + c4L‖x‖2

= −(c3 − c4L)‖x‖2

Take L < c3/c4, γdef= (c3 − c4L) > 0 ⇒

∂V

∂xAx ≤ −γ‖x‖2, ∀ ‖x‖ < minr0, r1

The origin of x = Ax is exponentially stable

– p. 6/18

Converse Lyapunov Theorem–Asymptotic Stability

Let x = 0 be an asymptotically stable equilibrium point forx = f(x), where f is locally Lipschitz on a domainD ⊂ Rn that contains the origin. LetRA ⊂ D be the regionof attraction of x = 0. Then, there is a smooth, positivedefinite function V (x) and a continuous, positive definitefunction W (x), both defined for all x ∈ RA, such that

V (x) → ∞ as x → ∂RA

∂V

∂xf(x) ≤ −W (x), ∀ x ∈ RA

and for any c > 0, V (x) ≤ c is a compact subset of RAWhen RA = Rn, V (x) is radially unbounded

– p. 7/18

Time-varying Systems

x = f(t, x)

f(t, x) is piecewise continuous in t and locally Lipschitz inx for all t ≥ 0 and all x ∈ D. The origin is an equilibriumpoint at t = 0 if

f(t, 0) = 0, ∀ t ≥ 0

While the solution of the autonomous system

x = f(x), x(t0) = x0

depends only on (t− t0), the solution of

x = f(t, x), x(t0) = x0

may depend on both t and t0– p. 8/18

Comparison Functions

A scalar continuous function α(r), defined for r ∈ [0, a)is said to belong to class K if it is strictly increasing andα(0) = 0. It is said to belong to class K∞ if it definedfor all r ≥ 0 and α(r) → ∞ as r → ∞

A scalar continuous function β(r, s), defined forr ∈ [0, a) and s ∈ [0,∞) is said to belong to class KLif, for each fixed s, the mapping β(r, s) belongs to classK with respect to r and, for each fixed r, the mappingβ(r, s) is decreasing with respect to s and β(r, s) → 0as s → ∞

– p. 9/18

Example

α(r) = tan−1(r) is strictly increasing sinceα′(r) = 1/(1 + r2) > 0. It belongs to class K, but notto class K∞ since limr→∞ α(r) = π/2 < ∞

α(r) = rc, for any positive real number c, is strictlyincreasing since α′(r) = crc−1 > 0. Moreover,limr→∞ α(r) = ∞; thus, it belongs to class K∞

α(r) = minr, r2 is continuous, strictly increasing,and limr→∞ α(r) = ∞. Hence, it belongs to class K∞

– p. 10/18

β(r, s) = r/(ksr + 1), for any positive real number k,is strictly increasing in r since

∂β

∂r=

1

(ksr + 1)2> 0

and strictly decreasing in s since

∂β

∂s=

−kr2

(ksr + 1)2< 0

Moreover, β(r, s) → 0 as s → ∞. Therefore, it belongsto class KL

β(r, s) = rce−s, for any positive real number c, belongsto class KL

– p. 11/18

Definition: The equilibrium point x = 0 of x = f(t, x) is

uniformly stable if there exist a class K function α and apositive constant c, independent of t0, such that

‖x(t)‖ ≤ α(‖x(t0)‖), ∀ t ≥ t0 ≥ 0, ∀ ‖x(t0)‖ < c

uniformly asymptotically stable if there exist a class KLfunction β and a positive constant c, independent of t0,such that

‖x(t)‖ ≤ β(‖x(t0)‖, t−t0), ∀ t ≥ t0 ≥ 0, ∀ ‖x(t0)‖ < c

globally uniformly asymptotically stable if the foregoinginequality is satisfied for any initial state x(t0)

– p. 12/18

exponentially stable if there exist positive constants c,k, and λ such that

‖x(t)‖ ≤ k‖x(t0)‖e−λ(t−t0), ∀ ‖x(t0)‖ < c

globally exponentially stable if the foregoing inequalityis satisfied for any initial state x(t0)

– p. 13/18

Theorem: Let the origin x = 0 be an equilibrium point forx = f(t, x) and D ⊂ Rn be a domain containing x = 0.Suppose f(t, x) is piecewise continuous in t and locallyLipschitz in x for all t ≥ 0 and x ∈ D. Let V (t, x) be acontinuously differentiable function such that

W1(x) ≤ V (t, x) ≤ W2(x)(1)

∂V

∂t+∂V

∂xf(t, x) ≤ 0(2)

for all t ≥ 0 and x ∈ D, where W1(x) and W2(x) arecontinuous positive definite functions on D. Then, the originis uniformly stable

– p. 14/18

Theorem: Suppose the assumptions of the previoustheorem are satisfied with

∂V

∂t+∂V

∂xf(t, x) ≤ −W3(x)

for all t ≥ 0 and x ∈ D, where W3(x) is a continuouspositive definite function on D. Then, the origin is uniformlyasymptotically stable. Moreover, if r and c are chosen suchthat Br = ‖x‖ ≤ r ⊂ D and c < min‖x‖=rW1(x), thenevery trajectory starting in x ∈ Br | W2(x) ≤ c satisfies

‖x(t)‖ ≤ β(‖x(t0)‖, t− t0), ∀ t ≥ t0 ≥ 0

for some class KL function β. Finally, if D = Rn andW1(x) is radially unbounded, then the origin is globallyuniformly asymptotically stable

– p. 15/18

Theorem: Suppose the assumptions of the previoustheorem are satisfied with

k1‖x‖a ≤ V (t, x) ≤ k2‖x‖a

∂V

∂t+∂V

∂xf(t, x) ≤ −k3‖x‖a

for all t ≥ 0 and x ∈ D, where k1, k2, k3, and a arepositive constants. Then, the origin is exponentially stable.If the assumptions hold globally, the origin will be globallyexponentially stable.

– p. 16/18

Example:

x = −[1 + g(t)]x3, g(t) ≥ 0, ∀ t ≥ 0

V (x) = 12x

2

V (t, x) = −[1 + g(t)]x4 ≤ −x4, ∀ x ∈ R, ∀ t ≥ 0

The origin is globally uniformly asymptotically stable

Example:

x1 = −x1 − g(t)x2

x2 = x1 − x2

0 ≤ g(t) ≤ k and g(t) ≤ g(t), ∀ t ≥ 0

– p. 17/18

V (t, x) = x21 + [1 + g(t)]x2

2

x21 + x2

2 ≤ V (t, x) ≤ x21 + (1 + k)x2

2, ∀ x ∈ R2

V (t, x) = −2x21 + 2x1x2 − [2 + 2g(t) − g(t)]x2

2

2 + 2g(t) − g(t) ≥ 2 + 2g(t) − g(t) ≥ 2

V (t, x) ≤ −2x21 + 2x1x2 − 2x2

2 = − xT

[

2 −1

−1 2

]

x

The origin is globally exponentially stable

– p. 18/18


Perturbed Systems

– p. 1/??

Nominal System:

x = f(x), f(0) = 0

Perturbed System:

x = f(x) + g(t, x), g(t, 0) = 0

Case 1: The origin of the nominal system is exponentiallystable

c1‖x‖2 ≤ V (x) ≤ c2‖x‖2

∂V

∂xf(x) ≤ −c3‖x‖2

∥

∥

∥

∥

∂V

∂x

∥

∥

∥

∥

≤ c4‖x‖

– p. 2/??

Use V (x) as a Lyapunov function candidate for theperturbed system

V (t, x) =∂V

∂xf(x) +

∂V

∂xg(t, x)

Assume that

‖g(t, x)‖ ≤ γ‖x‖, γ ≥ 0

V (t, x) ≤ −c3‖x‖2 +

∥

∥

∥

∥

∂V

∂x

∥

∥

∥

∥

‖g(t, x)‖

≤ −c3‖x‖2 + c4γ‖x‖2

– p. 3/??

γ <c3

c4

V (t, x) ≤ −(c3 − γc4)‖x‖2

The origin is an exponentially stable equilibrium point of theperturbed system

– p. 4/??

Example

x1 = x2

x2 = −4x1 − 2x2 + βx3

2, β ≥ 0

x = Ax + g(x)

A =

[

0 1

−4 −2

]

, g(x) =

[

0

βx3

2

]

The eigenvalues of A are −1 ± j√

3

PA + AT P = −I ⇒ P =

3

2

1

8

1

8

5

16

– p. 5/??

V (x) = xT Px,∂V

∂xAx = −xT x

c3 = 1, c4 = 2 ‖P‖ = 2λmax(P ) = 2 × 1.513 = 3.026

‖g(x)‖ = β|x2|3

g(x) satisfies the bound ‖g(x)‖ ≤ γ‖x‖ over compact setsof x. Consider the compact set

Ωc = V (x) ≤ c = xT Px ≤ c, c > 0

k2 = maxxT P x≤c

|x2| = maxxT P x≤c

|[0 1]x|

– p. 6/??

Fact:max

xT P x≤c‖Lx‖ =

√c ‖LP −1/2‖

Proof

xT Px ≤ c ⇔ 1

cxT Px ≤ 1 ⇔ 1

cxT P 1/2 P 1/2x ≤ 1

y =1

√c

P 1/2x

maxxT P x≤c

‖Lx‖ = maxyT y≤1

‖L√

c P −1/2y‖ =√

c ‖LP −1/2‖

– p. 7/??

k2 = maxxT P x≤c

|[0 1]x| =√

c ‖[0 1]P −1/2‖ = 1.8194√

c

‖g(x)‖ ≤ β c (1.8194)2‖x‖, ∀ x ∈ Ωc

‖g(x)‖ ≤ γ‖x‖, ∀ x ∈ Ωc, γ = β c (1.8194)2

γ <c3

c4

⇔ β <1

3.026 × (1.8194)2c≈ 0.1

c

β < 0.1/c ⇒ V (x) ≤ −(1 − 10βc)‖x‖2

Hence, the origin is exponentially stable and Ωc is anestimate of the region of attraction

– p. 8/??

Alternative Bound on β

V (x) = −‖x‖2 + 2xT Pg(x)

≤ −‖x‖2 + 1

8βx3

2([2 5]x)

≤ −‖x‖2 +√

29

8βx2

2‖x‖2

Over Ωc, x2

2≤ (1.8194)2c

V (x) ≤ −(

1 −√

29

8β(1.8194)2c

)

‖x‖2

= −(

1 − βc

0.448

)

‖x‖2

If β < 0.448/c, the origin will be exponentially stable andΩc will be an estimate of the region of attraction

– p. 9/??

Remark: The inequality β < 0.448/c shows a tradeoffbetween the estimate of the region of attraction and theestimate of the upper bound on β

– p. 10/??

Case 2: The origin of the nominal system is asymptoticallystable

V (t, x) =∂V

∂xf(x)+

∂V

∂xg(t, x) ≤ −W3(x)+

∥

∥

∥

∥

∂V

∂xg(t, x)

∥

∥

∥

∥

Under what condition will the following inequality hold?∥

∥

∥

∥

∂V

∂xg(t, x)

∥

∥

∥

∥

< W3(x)

Special Case: Quadratic-Type Lyapunov function

∂V

∂xf(x) ≤ −c3φ

2(x),

∥

∥

∥

∥

∂V

∂x

∥

∥

∥

∥

≤ c4φ(x)

– p. 11/??

V (t, x) ≤ −c3φ2(x) + c4φ(x)‖g(t, x)‖

If ‖g(t, x)‖ ≤ γφ(x), with γ <c3

c4

V (t, x) ≤ −(c3 − c4γ)φ2(x)

– p. 12/??

Examplex = −x3 + g(t, x)

V (x) = x4 is a quadratic-type Lyapunov function for thenominal system x = −x3

∂V

∂x(−x3) = −4x6,

∣

∣

∣

∣

∂V

∂x

∣

∣

∣

∣

= 4|x|3

φ(x) = |x|3, c3 = 4, c4 = 4

Suppose |g(t, x)| ≤ γ|x|3, ∀ x, with γ < 1

V (t, x) ≤ −4(1 − γ)φ2(x)

Hence, the origin is a globally uniformly asymptoticallystable

– p. 13/??

Remark: A nominal system with asymptotically, but notexponentially, stable origin is not robust to smoothperturbations with arbitrarily small linear growth bounds

Examplex = −x3 + γx

The origin is unstable for any γ > 0

– p. 14/??


PassivityMemoryless Functions

&State Models

– p. 1/17

Memoryless Functions

1

PPPPPPPPPP+u -y

(a)

y

u

(b)

power inflow = uy

Resistor is passive if uy ≥ 0

– p. 2/17

u

y

(a)

u

y

(b)

u

y

(c)

Passive Passive Not passive

y = h(t, u), h ∈ [0,∞]

Vector case:

y = h(t, u), hT =[

h1, h2, · · · , hp

]

power inflow = Σpi=1uiyi = uTy

– p. 3/17

Definition: y = h(t, u) is

passive if uTy ≥ 0

lossless if uTy = 0

input strictly passive if uTy ≥ uTϕ(u) for somefunction ϕ where uTϕ(u) > 0, ∀ u 6= 0

output strictly passive if uTy ≥ yTρ(y) for somefunction ρ where yTρ(y) > 0, ∀ y 6= 0

– p. 4/17

Sector Nonlinearity: h belongs to the sector [α, β](h ∈ [α, β]) if

αu2 ≤ uh(t, u) ≤ βu2

y=αu

y= uβ

u

(a)

y

α > 0

y=αu

y=β

u

y

(b)

u

α < 0

Also, h ∈ (α, β], h ∈ [α, β), h ∈ (α, β)

– p. 5/17

αu2 ≤ uh(t, u) ≤ βu2 ⇔ [h(t, u)−αu][h(t, u)−βu] ≤ 0

Definition: A memoryless function h(t, u) is said to belongto the sector

[0,∞] if uTh(t, u) ≥ 0

[K1,∞] if uT [h(t, u) −K1u] ≥ 0

[0,K2] with K2 = KT2> 0 if

hT (t, u)[h(t, u) −K2u] ≤ 0

[K1,K2] with K = K2 −K1 = KT > 0 if

[h(t, u) −K1u]T [h(t, u) −K2u] ≤ 0

– p. 6/17

Example

h(u) =

[

h1(u1)

h2(u2)

]

, hi ∈ [αi, βi], βi > αi i = 1, 2

K1 =

[

α1 0

0 α2

]

, K2 =

[

β1 0

0 β2

]

h ∈ [K1,K2]

K = K2 −K1 =

[

β1 − α1 0

0 β2 − α2

]

– p. 7/17

Example‖h(u) − Lu‖ ≤ γ‖u‖

K1 = L− γI, K2 = L+ γI

[h(u) −K1u]T [h(u) −K2u] =

‖h(u) − Lu‖2 − γ2‖u‖2 ≤ 0

K = K2 −K1 = 2γI

– p. 8/17

A function in the sector [K1,K2] can be transformed into afunction in the sector [0,∞] by input feedforward followedby output feedback

-

- K−1 - y = h(t, u) -

-

- K1

66

+

+

+

−

[K1,K2]Feedforward

−→[0,K]

K−1

−→[0, I]

Feedback−→

[0,∞]

– p. 9/17

State Models

1

u+ +-iL vC

-y v1+ ?i1 v3+ ?i3

v2+ -i2PPPPPPPPPP i1 = h1(v1)

BBB BBB BBB v2 = h2(i2)LC PPPPPPPPPP i3 = h3(v3)

Lx1 = u− h2(x1) − x2

Cx2 = x1 − h3(x2)

y = x1 + h1(u)

– p. 10/17

V (x) = 1

2Lx2

1+ 1

2Cx2

2

∫ t

0

u(s)y(s) ds ≥ V (x(t)) − V (x(0))

u(t)y(t) ≥ V (x(t), u(t))

V = Lx1x1 + Cx2x2

= x1[u− h2(x1) − x2] + x2[x1 − h3(x2)]

= x1[u− h2(x1)] − x2h3(x2)

= [x1 + h1(u)]u− uh1(u) − x1h2(x1) − x2h3(x2)

= uy − uh1(u) − x1h2(x1) − x2h3(x2)

– p. 11/17

uy = V + uh1(u) + x1h2(x1) + x2h3(x2)

If h1, h2, and h3 are passive, uy ≥ V and the system ispassive

Case 1: If h1 = h2 = h3 = 0, then uy = V ; no energydissipation; the system is lossless

Case 2: If h1 ∈ (0,∞] (uh1(u) > 0 for u 6= 0), then

uy ≥ V + uh1(u)

The energy absorbed over [0, t] will be greater than theincrease in the stored energy, unless the input u(t) isidentically zero. This is a case of input strict passivity

– p. 12/17

Case 3: If h1 = 0 and h2 ∈ (0,∞], then

y = x1 and uy ≥ V + yh2(y)

The energy absorbed over [0, t] will be greater than theincrease in the stored energy, unless the output y isidentically zero. This is a case of output strict passivity

Case 4: If h2 ∈ (0,∞) and h3 ∈ (0,∞), then

uy ≥ V + x1h2(x1) + x2h3(x2)

x1h2(x1) + x2h3(x2) is a positive definite function of x.This is a case of state strict passivity because the energyabsorbed over [0, t] will be greater than the increase in thestored energy, unless the state x is identically zero

– p. 13/17

Definition: The system

x = f(x, u), y = h(x, u)

is passive if there is a continuously differentiable positivesemidefinite function V (x) (the storage function) such that

uTy ≥ V =∂V

∂xf(x, u), ∀ (x, u)

Moreover, it is said to be

lossless if uTy = V

input strictly passive if uTy ≥ V + uTϕ(u) for somefunction ϕ such that uTϕ(u) > 0, ∀ u 6= 0

– p. 14/17

output strictly passive if uTy ≥ V + yTρ(y) for somefunction ρ such that yTρ(y) > 0, ∀ y 6= 0

strictly passive if uT y ≥ V + ψ(x) for some positivedefinite function ψ

Examplex = u, y = x

V (x) = 1

2x2 ⇒ uy = V ⇒ Lossless

– p. 15/17

Example

x = u, y = x+ h(u), h ∈ [0,∞]

V (x) = 1

2x2 ⇒ uy = V + uh(u) ⇒ Passive

h ∈ (0,∞] ⇒ uh(u) > 0 ∀ u 6= 0

⇒ Input strictly passive

Example

x = −h(x) + u, y = x, h ∈ [0,∞]

V (x) = 1

2x2 ⇒ uy = V + yh(y) ⇒ Passive

h ∈ (0,∞] ⇒ Output strictly passive

– p. 16/17

Example

x = u, y = h(x), h ∈ [0,∞]

V (x) =

∫ x

0

h(σ) dσ ⇒ V = h(x)x = yu ⇒ Lossless

Example

ax = −x+ u, y = h(x), h ∈ [0,∞]

V (x) = a

∫ x

0

h(σ) dσ ⇒ V = h(x)(−x+u) = yu−xh(x)

yu = V + xh(x) ⇒ Passive

h ∈ (0,∞] ⇒ Strictly passive

– p. 17/17


Positive Real Transfer Functions&

Connection with Lyapunov Stability

– p. 1/??

Definition: A p× p proper rational transfer function matrixG(s) is positive real if

poles of all elements of G(s) are in Re[s] ≤ 0

for all real ω for which jω is not a pole of any element ofG(s), the matrix G(jω) +GT (−jω) is positivesemidefinite

any pure imaginary pole jω of any element of G(s) is asimple pole and the residue matrixlims→jω(s− jω)G(s) is positive semidefinite Hermitian

G(s) is called strictly positive real if G(s− ε) is positive realfor some ε > 0

– p. 2/??

Scalar Case (p = 1):

G(jω) +GT (−jω) = 2Re[G(jω)]

Re[G(jω)] is an even function of ω. The second conditionof the definition reduces to

Re[G(jω)] ≥ 0, ∀ ω ∈ [0,∞)

which holds when the Nyquist plot of of G(jω) lies in theclosed right-half complex plane

This is true only if the relative degree of the transfer functionis zero or one

– p. 3/??

Lemma: A p× p proper rational transfer function matrixG(s) is strictly positive real if and only if

G(s) is Hurwitz

G(jω) +GT (−jω) > 0, ∀ ω ∈ R

G(∞) +GT (∞) > 0 or

limω→∞

ω2(p−q) det[G(jω) +GT (−jω)] > 0

where q = rank[G(∞) +GT (∞)]

– p. 4/??

Scalar Case (p = 1): G(s) is strictly positive real if and onlyif

G(s) is Hurwitz

Re[G(jω)] > 0, ∀ ω ∈ [0,∞)

G(∞) > 0 or

limω→∞

ω2Re[G(jω)] > 0

– p. 5/??

Example:

G(s) =1

s

has a simple pole at s = 0 whose residue is 1

Re[G(jω)] = Re

[

1

jω

]

= 0, ∀ ω 6= 0

Hence, G is positive real. It is not strictly positive real since

1

(s− ε)

has a pole in Re[s] > 0 for any ε > 0

– p. 6/??

Example:

G(s) =1

s+ a, a > 0, is Hurwitz

Re[G(jω)] =a

ω2 + a2> 0, ∀ ω ∈ [0,∞)

limω→∞

ω2Re[G(jω)] = limω→∞

ω2a

ω2 + a2= a > 0 ⇒ G is SPR

Example:

G(s) =1

s2 + s+ 1, Re[G(jω)] =

1 − ω2

(1 − ω2)2 + ω2

G is not PR

– p. 7/??

Example:

G(s) =

s+2s+1

1s+2

−1s+2

2s+1

is Hurwitz

G(jω) +GT (−jω) =

2(2+ω2)1+ω2

−2jω4+ω2

2jω4+ω2

41+ω2

> 0, ∀ ω ∈ R

G(∞) +GT (∞) =

[

2 0

0 0

]

, q = 1

limω→∞

ω2 det[G(jω) +GT (−jω)] = 4 ⇒ G is SPR

– p. 8/??

Positive Real Lemma: Let

G(s) = C(sI −A)−1B +D

where (A,B) is controllable and (A,C) is observable.G(s) is positive real if and only if there exist matricesP = P T > 0, L, and W such that

PA+ATP = −LTL

PB = CT − LTW

W TW = D +DT

– p. 9/??

Kalman–Yakubovich–Popov Lemma: Let

G(s) = C(sI −A)−1B +D

where (A,B) is controllable and (A,C) is observable.G(s) is strictly positive real if and only if there exist matricesP = P T > 0, L, and W , and a positive constant ε suchthat

PA+ATP = −LTL− εP

PB = CT − LTW

W TW = D +DT

– p. 10/??

Lemma: The linear time-invariant minimal realization

x = Ax+Bu

y = Cx+Du

withG(s) = C(sI −A)−1B +D

is

passive if G(s) is positive real

strictly passive if G(s) is strictly positive real

Proof: Apply the PR and KYP Lemmas, respectively, anduse V (x) = 1

2xTPx as the storage function

– p. 11/??

uTy −∂V

∂x(Ax+Bu)

= uT (Cx+Du) − xTP (Ax+Bu)

= uTCx+ 12uT (D +DT )u

− 12x

T (PA+ATP )x− xTPBu

= uT (BTP +W TL)x+ 12uTW TWu

+ 12x

TLTLx+ 12εx

TPx− xTPBu

= 12(Lx+Wu)T (Lx+Wu) + 1

2εxTPx ≥ 1

2εxTPx

In the case of the PR Lemma, ε = 0, and we conclude thatthe system is passive; in the case of the KYP Lemma,ε > 0, and we conclude that the system is strictly passive

– p. 12/??

Connection with Lyapunov Stability

Lemma: If the system

x = f(x, u), y = h(x, u)

is passive with a positive definite storage function V (x),then the origin of x = f(x, 0) is stable

Proof:

uTy ≥∂V

∂xf(x, u) ⇒

∂V

∂xf(x, 0) ≤ 0

– p. 13/??


x = f(x, u), y = h(x, u)

is strictly passive, then the origin of x = f(x, 0) isasymptotically stable. Furthermore, if the storage functionis radially unbounded, the origin will be globallyasymptotically stable

Proof: The storage function V (x) is positive definite

uTy ≥∂V

∂xf(x, u) + ψ(x) ⇒

∂V

∂xf(x, 0) ≤ −ψ(x)

Why is V (x) positive definite? Let φ(t;x) be the solutionof z = f(z, 0), z(0) = x

– p. 14/??

V ≤ −ψ(x)

V (φ(τ, x)) − V (x) ≤ −

∫ τ

0ψ(φ(t;x)) dt, ∀ τ ∈ [0, δ]

V (φ(τ, x)) ≥ 0 ⇒ V (x) ≥

∫ τ

0ψ(φ(t;x)) dt

V (x) = 0 ⇒

∫ τ

0ψ(φ(t; x)) dt = 0, ∀ τ ∈ [0, δ]

⇒ ψ(φ(t; x)) ≡ 0 ⇒ φ(t; x) ≡ 0 ⇒ x = 0

– p. 15/??


x = f(x, u), y = h(x, u)

is zero-state observable if no solution of x = f(x, 0) canstay identically in S = h(x, 0) = 0, other than the zerosolution x(t) ≡ 0

Linear Systems

x = Ax, y = Cx

Observability of (A,C) is equivalent to

y(t) = CeAtx(0) ≡ 0 ⇔ x(0) = 0 ⇔ x(t) ≡ 0

– p. 16/??


x = f(x, u), y = h(x, u)

is output strictly passive and zero-state observable, thenthe origin of x = f(x, 0) is asymptotically stable.Furthermore, if the storage function is radially unbounded,the origin will be globally asymptotically stable

Proof: The storage function V (x) is positive definite

uTy ≥∂V

∂xf(x, u) + yTρ(y) ⇒

∂V

∂xf(x, 0) ≤ −yTρ(y)

V (x(t)) ≡ 0 ⇒ y(t) ≡ 0 ⇒ x(t) ≡ 0

Apply the invariance principle

– p. 17/??

Example

x1 = x2, x2 = −ax31 − kx2 + u, y = x2, a, k > 0

V (x) = 14ax

41 + 1

2x22

V = ax31x2 + x2(−ax

31 − kx2 + u) = −ky2 + yu

The system is output strictly passive

y(t) ≡ 0 ⇔ x2(t) ≡ 0 ⇒ ax31(t) ≡ 0 ⇒ x1(t) ≡ 0

The system is zero-state observable. V is radiallyunbounded. Hence, the origin of the unforced system isglobally asymptotically stable

– p. 18/??


Feedback Systems: PassivityTheorems

– p. 1/21

-

- -

6

?

u1

u2

e1

e2

y1

y2

H1

H2

−+

++

xi = fi(xi, ei), yi = hi(xi, ei)

yi = hi(t, ei)

– p. 2/21

Passivity Theorems

Theorem 6.1: The feedback connection of two passivesystems is passive

Theorem 6.3: Consider the feedback connection of twodynamical systems. When u = 0, the origin of theclosed-loop system is asymptotically stable if eachfeedback component is either

strictly passive, or

output strictly passive and zero-state observable

Furthermore, if the storage function for each component isradially unbounded, the origin is globally asymptoticallystable

– p. 3/21

Theorem 6.4: Consider the feedback connection of astrictly passive dynamical system with a passivememoryless function. When u = 0, the origin of theclosed-loop system is uniformly asymptotically stable. if thestorage function for the dynamical system is radiallyunbounded, the origin will be globally uniformlyasymptotically stable

Prove using V = V1 +V2 as a Lyapunov function candidate

Proof of Theorem 6.3: H1 is SP; H2 is OSP & ZSO

eT1y1 ≥ V1 + ψ1(x1), ψ1(x1) > 0, ∀ x1 6= 0

eT2y2 ≥ V2 + yT

2ρ2(y2), yT

2ρ(y2) > 0, ∀y2 6= 0

– p. 4/21

eT1y1+eT

2y2 = (u1−y2)

T y1+(u2+y1)T y2 = uT

1y1+uT

2y2

V (x) = V1(x1) + V2(x2)

V ≤ uT y − ψ1(x1) − yT2ρ2(y2)

u = 0 ⇒ V ≤ −ψ1(x1) − yT2ρ2(y2)

V = 0 ⇒ x1 = 0 and y2 = 0

y2(t) ≡ 0 ⇒ e1(t) ≡ 0 ( & x1(t) ≡ 0) ⇒ y1(t) ≡ 0

y1(t) ≡ 0 ⇒ e2(t) ≡ 0

By zero-state observability of H2: y2(t) ≡ 0 ⇒ x2(t) ≡ 0


– p. 5/21

Example

x1 = x2

x2 = −ax3

1− kx2 + e1

y1 = x2 + e1︸︷︷︸

H1

∣∣∣∣∣∣∣∣∣∣∣∣

x3 = x4

x4 = −bx3 − x3

4+ e2

y2 = x4︸︷︷︸

H2

a, b, k > 0

V1 = 1

4ax4

1+ 1

2x2

2

V1 = ax3

1x2 − ax3

1x2 − kx2

2+ x2e1 = −ky2

1+ y1e1

With e1 = 0, y1(t) ≡ 0 ⇔ x2(t) ≡ 0 ⇒ x1(t) ≡ 0

H1 is output strictly passive and zero-state observable

– p. 6/21

V2 = 1

2bx2

3+ 1

2x2

4

V2 = bx3x4 − bx3x4 − x4

4+ x4e2 = −y4

2+ y2e2

With e2 = 0, y2(t) ≡ 0 ⇔ x4(t) ≡ 0 ⇒ x3(t) ≡ 0

H2 is output strictly passive and zero-state observable

V1 and V2 are radially unbounded


– p. 7/21

Loop Transformations

Recall that a memoryless function in the sector [K1,K2]can be transformed into a function in the sector [0,∞] byinput feedforward followed by output feedback

-

- K−1 - y = h(t, u) -

-

- K1

66

+

+

+

−

– p. 8/21

- h -H1-

H2

6

+

−

H1 is a dynamical system

H2 is a memoryless function in the sector [K1,K2]

– p. 9/21

- h - h -H1-

K1

6

H2h

6

K1

6

+

−

+

−

+

−

– p. 10/21

- h - h -H1- K -

K1

6

K−1H2h

6

K1

6

+

−

+

−

+

−

– p. 11/21

- h - h -H1- K - h -

K1

6

?

hK−1H2h

6

K1

6 6

H1

H2

+

−

+

−

+

+

+

+

+

−

– p. 12/21

Examplex1 = x2

x2 = −h(x1) + bx2 + e1

y1 = x2︸︷︷︸

H1

∣∣∣∣∣∣

y2 = σ(e2)︸︷︷︸

H2

σ ∈ [α, β], h ∈ [α1,∞], b > 0, α1 > 0, k = β−α > 0

x1 = x2

x2 = −h(x1) − ax2 + e1

y1 = kx2 + e1︸︷︷︸

H1

∣∣∣∣∣∣∣

y2 = σ(e2)︸︷︷︸

H2

σ ∈ [0,∞], a = α− b

– p. 13/21

Assume a = α− b > 0 and show that H1 is strictly passive

V1 = k

∫ x1

0

h(s) ds+ xTPx

V1 = k

∫ x1

0

h(s) ds+ p11x2

1+ 2p12x1x2 + p22x

2

2

V = kh(x1)x2 + 2(p11x1 + p12x2)x2

2(p12x1 + p22x2)[−h(x1) − ax2 + e1]

Take p22 = k/2, p11 = ap12

– p. 14/21

V = −2p12x1h(x1) − (ka− 2p12)x2

2

+ kx2e1 + 2p12x1e1

= −2p12x1h(x1) − (ka− 2p12)x2

2

+ (kx2 + e1)e1 − e21+ 2p12x1e1

y1e1 = V + 2p12x1h(x1) + (ka− 2p12)x2

2

+ (e1 − p12x1)2 − p2

12x2

1

≥ V + p12(2α1 − p12)x2

1+ (ka− 2p12)x

2

2

Take 0 < p12 < min

ak

2, 2α1

⇒ p2

12< 2p12

k

2= p11p22

H1 is strictly passive. By Theorem 6.4 the origin is globallyasymptotically stable (when u = 0)

– p. 15/21

-

- H1-

H2

6

+

−

– p. 16/21

-

- H1- W−1(s) -

W (s)H2

6

H1

H2

+

−

– p. 17/21

Example

H1 : x = Ax+Be1, y1 = Cx

A =

[

0 1

−1 −1

]

, B =

[

0

1

]

, C =[

1 0]

H2 : y2 = h(e2), h ∈ [0,∞]

C(sI −A)−1B =1

(s2 + s+ 1)Not PR

W (s) =1

as+ 1⇒

(as+ 1)

(s2 + s+ 1)

H1 : x = Ax+Be1, y1 = Cx =[

1 a]

x

– p. 18/21

(as+ 1)

(s2 + s+ 1)

Re

[1 + jωa

1 − ω2 + jω

]

=1 + (a− 1)ω2

(1 − ω2)2 + ω2

limω→∞

ω2Re

[1 + jωa

1 − ω2 + jω

]

= a− 1

a > 1 ⇒(as+ 1)

(s2 + s+ 1)is SPR

V1 = 1

2xTPx, PA+ATP = −LTL− εP, PB = CT

– p. 19/21

H2 : ae2 = −e2 + e2, y2 = h(e2), h ∈ [0,∞]

H2 is strictly passive with V2 = a∫ e20h(s) ds. Use

V = V1 + V2 = 1

2xTPx+ a

∫ e2

0

h(s) ds

as a Lyapunov function candidate for the original feedbackconnection

– p. 20/21

V = 1

2xTP x+ 1

2xTPx+ ah(e2)e2

= 1

2xTP [Ax−Bh(e2)] + 1

2[Ax−Bh(e2)]

TPx

+ ah(e2)C[Ax−Bh(e2)]

= − 1

2xTLTLx− (ε/2)xTPx− xT CTh(e2)

+ ah(e2)CAx

= − 1

2xTLTLx− (ε/2)xTPx

− xT [C + aCA]Th(e2) + ah(e2)CAx

= − 1

2xTLTLx− (ε/2)xTPx− eT

2h(e2)

≤ −(ε/2)xTPx


– p. 21/21


Circle & Popov Criteria

– p. 1/24

Absolute Stability

-

- -

6

r u yG(s)

ψ(·)

−

+

The system is absolutely stable if (when r = 0) the origin isglobally asymptotically stable for all memorylesstime-invariant nonlinearities in a given sector

– p. 2/24

Circle Criterion

Suppose G(s) = C(sI −A)−1B +D is SPR andψ ∈ [0,∞]

x = Ax+Bu

y = Cx+Du

u = −ψ(y)

By the KYP Lemma, ∃ P = P T > 0, L,W, ε > 0


PB = CT − LTW

W TW = D +DT

V (x) = 1

2xTPx

– p. 3/24

V = 1

2xTPx+ 1

2xTPx

= 1

2xT (PA+ATP )x+ xTPBu

= − 1

2xTLTLx− 1

2εxTPx+ xT (CT − LTW )u

= − 1

2xTLTLx− 1

2εxTPx+ (Cx+Du)Tu

− uTDu− xTLTWu

uTDu = 1

2uT (D +DT )u = 1

2uTW TWu

V = − 1

2εxTPx− 1

2(Lx+Wu)T (Lx+Wu) − yTψ(y)

yTψ(y) ≥ 0 ⇒ V ≤ − 1

2εxTPx

The origin is globally exponentially stable

– p. 4/24

What if ψ ∈ [K1,∞]?

- f -G(s) -

ψ(·)

6

+

−- f - f -G(s) -

K1

6

ψ(·)f

6

K1

6

ψ(·)

+

−+

−

+

−

ψ ∈ [0,∞]; hence the origin is globally exponentially stableif G(s)[I +K1G(s)]−1 is SPR

– p. 5/24

What if ψ ∈ [K1, K2]?

- f -G(s) -

ψ(·)

6

+

−- f - f -G(s) - K - f -

K1

6

?

fK−1ψ(·)f

6

K1

6 6

ψ(·)

+

−+

−

+

+

+

+

+

−

ψ ∈ [0,∞]; hence the origin is globally exponentially stableif I +KG(s)[I +K1G(s)]−1 is SPR

– p. 6/24

I+KG(s)[I+K1G(s)]−1 = [I+K2G(s)][I+K1G(s)]−1

Theorem (Circle Criterion): The system is absolutely stableif

ψ ∈ [K1,∞] and G(s)[I +K1G(s)]−1 is SPR, or

ψ ∈ [K1, K2] and [I+K2G(s)][I+K1G(s)]−1 is SPR

Scalar Case: ψ ∈ [α, β], β > αThe system is absolutely stable if

1 + βG(s)

1 + αG(s)is Hurwitz and

Re

[

1 + βG(jω)

1 + αG(jω)

]

> 0, ∀ ω ∈ [0,∞]

– p. 7/24

Case 1: α > 0By the Nyquist criterion

1 + βG(s)

1 + αG(s)=

1

1 + αG(s)+

βG(s)

1 + αG(s)

is Hurwitz if the Nyquist plot of G(jω) does not intersect thepoint −(1/α) + j0 and encircles it m times in thecounterclockwise direction, where m is the number of polesof G(s) in the open right-half complex plane

1 + βG(jω)

1 + αG(jω)> 0 ⇔

1

β +G(jω)

1

α +G(jω)> 0

– p. 8/24

Re

[

1

β +G(jω)

1

α +G(jω)

]

> 0, ∀ ω ∈ [0,∞]

−1/α −1/β

qD(α,β)

θ2 θ1

The system is absolutely stable if the Nyquist plot of G(jω)does not enter the disk D(α, β) and encircles it m times inthe counterclockwise direction

– p. 9/24

Case 2: α = 0

1 + βG(s)

Re[1 + βG(jω)] > 0, ∀ ω ∈ [0,∞]

Re[G(jω)] > −1

β, ∀ ω ∈ [0,∞]

The system is absolutely stable if G(s) is Hurwitz and theNyquist plot of G(jω) lies to the right of the vertical linedefined by Re[s] = −1/β

– p. 10/24

Case 3: α < 0 < β

Re

[

1 + βG(jω)

1 + αG(jω)

]

> 0 ⇔ Re

[

1

β +G(jω)

1

α +G(jω)

]

< 0

The Nyquist plot of G(jω) must lie inside the disk D(α, β).The Nyquist plot cannot encircle the point −(1/α) + j0.From the Nyquist criterion, G(s) must be Hurwitz

The system is absolutely stable if G(s) is Hurwitz and theNyquist plot of G(jω) lies in the interior of the disk D(α, β)

– p. 11/24

Example

G(s) =4

(s+ 1)(1

2s+ 1)(1

3s+ 1)

−5 0 5−4

−2

0

2

4

6

Re G

Im G

– p. 12/24

Apply Case 3 with center (0, 0) and radius = 4

Sector is (−0.25, 0.25)

Apply Case 3 with center (1.5, 0) and radius = 2.834

Sector is [−0.227, 0.714]

Apply Case 2

The Nyquist plot is to the right of Re[s] = −0.857

Sector is [0, 1.166]

[0, 1.166] includes the saturation nonlinearity

– p. 13/24

Example

G(s) =4

(s− 1)(1

2s+ 1)(1

3s+ 1)

−4 −2 0−0.4

−0.2

0

0.2

0.4

Re G

Im G

G is not Hurwitz

Apply Case 1

Center = (−3.2, 0), Radius = 0.168 ⇒ [0.2969, 0.3298]

– p. 14/24

Popov Criterion

- f -G(s) -

ψ(·)

6

+

−

x = Ax+Bu

y = Cx

ui = −ψi(yi), 1 ≤ i ≤ p

ψi ∈ [0, ki], 1 ≤ i ≤ p, (0 < ki ≤ ∞)

G(s) = C(sI −A)−1B

Γ = diag(γ1, . . . , γp), M = diag(1/k1, · · · , 1/kp)

– p. 15/24

- i - G(s) - (I + sΓ) - i -

- M

?

i(I + sΓ)−1ψ(·)

6

M

6

H1

H2

+

−

+

+

+

+

Show that H1 and H2 are passive– p. 16/24

M + (I + sΓ)G(s)

= M + (I + sΓ)C(sI −A)−1B

= M + C(sI −A)−1B + ΓCs(sI −A)−1B

= M + C(sI −A)−1B + ΓC(sI −A+A)(sI −A)−1B

= (C + ΓCA)(sI −A)−1B +M + ΓCB

If M + (I + sΓ)G(s) is SPR, then H1 is strictly passivewith the storage function V1 = 1

2xTPx, where P is given by

the KYP equations


PB = (C + ΓCA)T − LTW

W TW = 2M + ΓCB +BTCTΓ

– p. 17/24

H2 consists of p decoupled components:

γizi = −zi +1

kiψi(zi) + e2i, y2i = ψi(zi)

V2i = γi

∫ zi

0

ψi(σ) dσ

V2i = γiψi(zi)zi = ψi(zi)[

−zi + 1

ki

ψi(zi) + e2i

]

= y2ie2i +1

ki

ψi(zi) [ψi(zi) − kizi]

ψi ∈ [0, ki] ⇒ ψi(ψi − kizi) ≤ 0 ⇒ V2i ≤ y2ie2i

H2 is passive with the storage functionV2 =

∑pi=1

γi∫ zi

0ψi(σ) dσ

– p. 18/24

Use V = 1

2xTPx+

p∑

i=1

γi

∫ yi

0

ψi(σ) dσ

as a Lyapunov function candidate for the original feedbackconnection

x = Ax+Bu, y = Cx, u = −ψ(y)

V = 1

2xTPx+ 1

2xTPx+ ψT (y)Γy

= 1

2xT (PA +ATP )x+ xTPBu

+ ψT (y)ΓC(Ax+Bu)

= − 1

2xTLTLx− 1

2εxTPx

+ xT (CT +ATCTΓ − LTW )u

+ ψT (y)ΓCAx+ ψT (y)ΓCBu

– p. 19/24

V = − 1

2εxTPx− 1

2(Lx+Wu)T (Lx+Wu)

− ψ(y)T [y −Mψ(y)]

≤ − 1

2εxTPx− ψ(y)T [y −Mψ(y)]

ψi ∈ [0, ki] ⇒ ψ(y)T [y−Mψ(y)] ≥ 0 ⇒ V ≤ −1

2εxTPx


Popov Criterion: The system is absolutely stable if, for1 ≤ i ≤ p, ψi ∈ [0, ki] and there exists a constant γi ≥ 0,with (1 + λkγi) 6= 0 for every eigenvalue λk of A, such thatM + (I + sΓ)G(s) is strictly positive real

– p. 20/24

Scalar case1

k+ (1 + sγ)G(s)

is SPR if G(s) is Hurwitz and

1

k+ Re[G(jω)] − γωIm[G(jω)] > 0, ∀ ω ∈ [0,∞)

If

limω→∞

1

k+ Re[G(jω)] − γωIm[G(jω)]

= 0

we also need

limω→∞

ω2

1

k+ Re[G(jω)] − γωIm[G(jω)]

> 0

– p. 21/24

1

k+ Re[G(jω)] − γωIm[G(jω)] > 0, ∀ ω ∈ [0,∞)

Re[G(j )]ω

Im[G(j )]ω ω

−1/k

slope = 1/γ

Popov Plot

– p. 22/24

Example

x1 = x2, x2 = −x2 − h(y), y = x1

x2 = −αx1 − x2 − h(y) + αx1, α > 0

G(s) =1

s2 + s+ α, ψ(y) = h(y) − αy

h ∈ [α, β] ⇒ ψ ∈ [0, k] (k = β − α > 0)

γ > 1 ⇒α− ω2 + γω2

(α− ω2)2 + ω2> 0, ∀ ω ∈ [0,∞)

and limω→∞

ω2(α− ω2 + γω2)

(α− ω2)2 + ω2= γ − 1 > 0

– p. 23/24

The system is absolutely stable for ψ ∈ [0,∞] (h ∈ [α,∞])

−0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2 slope=1

Re G

Im Gω

Compare with the circle criterion (γ = 0)

1

k+

α− ω2

(α− ω2)2 + ω2> 0, ∀ ω ∈ [0,∞], for k < 1+2

√α

– p. 24/24


Boundedness&

Ultimate Boundedness

– p. 1/18

Definition: The solutions of x = f(t, x) are

uniformly bounded if ∃ c > 0 and for every0 < a < c, ∃ β = β(a) > 0 such that

‖x(t0)‖ ≤ a ⇒ ‖x(t)‖ ≤ β, ∀ t ≥ t0 ≥ 0

uniformly ultimately bounded with ultimate bound b if∃ b and c and for every 0 < a < c, ∃ T = T (a, b) ≥ 0such that

‖x(t0)‖ ≤ a ⇒ ‖x(t)‖ ≤ b, ∀ t ≥ t0 + T

“Globally” if a can be arbitrarily large

Drop “uniformly” if x = f(x)

– p. 2/18

Lyapunov Analysis: Let V (x) be a cont. diff. positivedefinite function and suppose that the sets

Ωc = V (x) ≤ c, Ωε = V (x) ≤ ε, Λ = ε ≤ V (x) ≤ c

are compact for some c > ε > 0

ΩεcΩ

Λ

– p. 3/18

Suppose

V (t, x) =∂V

∂xf(t, x) ≤ −W3(x), ∀ x ∈ Λ, ∀ t ≥ 0

W3(x) is continuous and positive definite

Ωc and Ωε are positively invariant

k = minx∈Λ

W3(x) > 0

V (t, x) ≤ −k, ∀ x ∈ Λ, ∀ t ≥ t0 ≥ 0

V (x(t)) ≤ V (x(t0)) − k(t − t0) ≤ c − k(t − t0)

x(t) enters the set Ωε within the interval [t0, t0 + (c − ε)/k]

– p. 4/18

Suppose

V (t, x) ≤ −W3(x), ∀ µ ≤ ‖x‖ ≤ r, ∀ t ≥ 0

Choose c and ε such that Λ ⊂ µ ≤ ‖x‖ ≤ r

Br

B

Ω

µ

c

Ωε

– p. 5/18

Let α1 and α2 be class K functions such that

α1(‖x‖) ≤ V (x) ≤ α2(‖x‖)

V (x) ≤ c ⇒ α1(‖x‖) ≤ c ⇔ ‖x‖ ≤ α−1

1(c)

c = α1(r) ⇒ Ωc ⊂ Br

‖x‖ ≤ µ ⇒ V (x) ≤ α2(µ)

ε = α2(µ) ⇒ Bµ ⊂ Ωε

What is the ultimate bound?

V (x) ≤ ε ⇒ α1(‖x‖) ≤ ε ⇔ ‖x‖ ≤ α−1

1(ε) = α−1

1(α2(µ))

– p. 6/18

Theorem (special case of Thm 4.18): Suppose

α1(‖x‖) ≤ V (x) ≤ α2(‖x‖)

∂V

∂xf(t, x) ≤ −W3(x), ∀ ‖x‖ ≥ µ > 0

∀ t ≥ 0 and ‖x‖ ≤ r, where α1, α2 ∈ K, W3(x) iscontinuous & positive definite, and µ < α−1

2(α1(r)). Then,

for every initial state x(t0) ∈ ‖x‖ ≤ α−1

2(α1(r)), there is

T ≥ 0 (dependent on x(t0) and µ) such that

‖x(t)‖ ≤ α−1

1(α2(µ)), ∀ t ≥ t0 + T

If the assumptions hold globally and α1 ∈ K∞, then theconclusion holds for any initial state x(t0)

– p. 7/18

Remarks:

The ultimate bound is independent of the initial state

The ultimate bound is a class K function of µ; hence,the smaller the value of µ, the smaller the ultimatebound. As µ → 0, the ultimate bound approaches zero

– p. 8/18

Example

x1 = x2, x2 = −(1 + x2

1)x1 − x2 + M cos ωt, M ≥ 0

With M = 0, x2 = −(1 + x2

1)x1 − x2 = −h(x1) − x2

V (x) = xT

1

2

1

2

1

21

x + 2

∫ x1

0

(y + y3) dy (Example 4.5)

V (x) = xT

3

2

1

2

1

21

x + 1

2x4

1

def= xT Px + 1

2x4

1

– p. 9/18

λmin(P )‖x‖2 ≤ V (x) ≤ λmax(P )‖x‖2 + 1

2‖x‖4

α1(r) = λmin(P )r2, α2(r) = λmax(P )r2 + 1

2r4

V = −x2

1− x4

1− x2

2+ (x1 + 2x2)M cos ω

≤ −‖x‖2 − x4

1+ M

√5‖x‖

= −(1 − θ)‖x‖2 − x4

1− θ‖x‖2 + M

√5‖x‖

(0 < θ < 1)

≤ −(1 − θ)‖x‖2 − x4

1, ∀ ‖x‖ ≥ M

√5/θ

def= µ

The solutions are GUUB by

b = α−1

1(α2(µ)) =

√

λmax(P )µ2 + µ4/2

λmin(P )

– p. 10/18


Perturbed Systems&

Input-to-State Stability

– p. 1/??

Perturbed Systems: Nonvanishing Perturbation

Nominal System:

x = f(x), f(0) = 0

Perturbed System:

x = f(x) + g(t, x), g(t, 0) 6= 0

Case 1: The origin of x = f(x) is exponentially stable

c1‖x‖2 ≤ V (x) ≤ c2‖x‖2

∂V

∂xf(x) ≤ −c3‖x‖2,

∥

∥

∥

∥

∂V

∂x

∥

∥

∥

∥

≤ c4‖x‖

∀ x ∈ Br = ‖x‖ ≤ r

– p. 2/??

Use V (x) to investigate ultimate boundedness of theperturbed system

V (t, x) =∂V

∂xf(x) +

∂V

∂xg(t, x)

Assume‖g(t, x)‖ ≤ δ, ∀ t ≥ 0, x ∈ Br

V (t, x) ≤ −c3‖x‖2 +∥

∥

∥

∂V∂x

∥

∥

∥‖g(t, x)‖

≤ −c3‖x‖2 + c4δ‖x‖= −(1 − θ)c3‖x‖2 − θc3‖x‖2 + c4δ‖x‖

0 < θ < 1

≤ −(1 − θ)c3‖x‖2, ∀ ‖x‖ ≥ δc4/(θc3)def= µ

– p. 3/??

Apply Theorem 4.18

‖x(t0)‖ ≤ α−1

2(α1(r)) ⇔ ‖x(t0)‖ ≤ r

√

c1

c2

µ < α−1

2(α1(r)) ⇔ δc4

θc3

< r

√

c1

c2

⇔ δ <c3

c4

√

c1

c2

θr

b = α−1

1(α2(µ)) ⇔ b = µ

√

c2

c1

⇔ b =δc4

θc3

√

c2

c1

For all ‖x(t0)‖ ≤ r√

c1/c2, the solutions of the perturbedsystem are ultimately bounded by b

– p. 4/??

Example

x1 = x2, x2 = −4x1 − 2x2 + βx3

2+ d(t)

β ≥ 0, |d(t)| ≤ δ, ∀ t ≥ 0

V (x) = xT Px = xT

3

2

1

8

1

8

5

16

x (Lecture 13)

V (t, x) = −‖x‖2 + 2βx2

2

(

1

8x1x2 + 5

16x2

2

)

+ 2d(t)(

1

8x1 + 5

16x2

)

≤ −‖x‖2 +

√29

8βk2

2‖x‖2 +

√29δ

8‖x‖

– p. 5/??

k2 = maxxT P x≤c

|x2| = 1.8194√

c

Suppose β ≤ 8(1 − ζ)/(√

29k2

2) (0 < ζ < 1)

V (t, x) ≤ −ζ‖x‖2 +√

29δ8

‖x‖≤ −(1 − θ)ζ‖x‖2, ∀ ‖x‖ ≥

√29δ

8ζθ

def= µ

(0 < θ < 1)

If µ2λmax(P ) < c, then all solutions of the perturbedsystem, starting in Ωc, are uniformly ultimately bounded by

b =

√29δ

8ζθ

√

λmax(P )

λmin(P )

– p. 6/??

Case 2: The origin of x = f(x) is asymptotically stable

α1(‖x‖) ≤ V (x) ≤ α2(‖x‖)

∂V

∂xf(x) ≤ −α3(‖x‖),

∥

∥

∥

∥

∂V

∂x

∥

∥

∥

∥

≤ k

∀ x ∈ Br = ‖x‖ ≤ r, αi ∈ K, i = 1, 2, 3

V (t, x) ≤ −α3(‖x‖) +∥

∥

∥

∂V∂x

∥

∥

∥‖g(t, x)‖

≤ −α3(‖x‖) + δk

≤ −(1 − θ)α3(‖x‖) − θα3(‖x‖) + δk

0 < θ < 1

≤ −(1 − θ)α3(‖x‖), ∀ ‖x‖ ≥ α−1

3

(

δkθ

)

def= µ

– p. 7/??

Apply Theorem 4.18

µ < α−1

2(α1(r)) ⇔ α−1

3

(

δk

θ

)

< α−1

2(α1(r))

⇔ δ <θα3(α

−1

2(α1(r)))

k

µ < α−1

2(α1(r)) ⇔ α−1

3

(

δk

θ

)

< α−1

2(α1(r))

⇔ δ <θα3(α

−1

2(α1(r)))

kCompare with δ <

c3

c4

√

c1

c2

θr

Example

x = − x

1 + x2

V (x) = x4 ⇒ ∂V

∂x

[

− x

1 + x2

]

= − 4x4

1 + x2– p. 8/??


θα3(α−1

2(α1(r)))

k=

θα3(r)

k=

rθ

1 + r2

rθ

1 + r2→ 0 as r → ∞

x = −x

1 + x2+ δ, δ > 0

δ > 1

2⇒ lim

t→∞x(t) = ∞

– p. 9/??

Input-to-State Stability (ISS)

Definition: The system x = f(x, u) is input-to-state stable ifthere exist β ∈ KL and γ ∈ K such that for any initial statex(t0) and any bounded input u(t)

‖x(t)‖ ≤ β(‖x(t0)‖, t − t0) + γ

(

supt0≤τ≤t

‖u(τ )‖)

ISS of x = f(x, u) impliesBIBS stability

x(t) is ultimately bounded by a class K function ofsupt≥t0 ‖u(t)‖

limt→∞ u(t) = 0 ⇒ limt→∞ x(t) = 0

The origin of x = f(x, 0) is GAS

– p. 10/??

Theorem (Special case of Thm 4.19): Let V (x) be acontinuously differentiable function such that

α1(‖x‖) ≤ V (x) ≤ α2(‖x‖)

∂V

∂xf(x, u) ≤ −W3(x), ∀ ‖x‖ ≥ ρ(‖u‖) > 0

∀ x ∈ Rn, u ∈ Rm, where α1, α2 ∈ K∞, ρ ∈ K, andW3(x) is a continuous positive definite function. Then, thesystem x = f(x, u) is ISS with γ = α−1

1 α2 ρ

Proof: Let µ = ρ(supτ≥t0‖u(τ )‖); then

∂V

∂xf(x, u) ≤ −W3(x), ∀ ‖x‖ ≥ µ

– p. 11/??

Choose ε and c such that

∂V

∂xf(x, u) ≤ −W3(x), ∀ x ∈ Λ = ε ≤ V (x) ≤ c

Suppose x(t0) ∈ Λ and x(t) reaches Ωε at t = t0 + T . Fort0 ≤ t ≤ t0 + T , V satisfies the conditions for the uniformasymptotic stability. Therefore, the trajectory behaves as ifthe origin was uniformly asymptotically stable and satisfies

‖x(t)‖ ≤ β(‖x(t0)‖, t − t0), for some β ∈ KL

For t ≥ t0 + T ,

‖x(t)‖ ≤ α−1

1(α2(µ))

– p. 12/??

‖x(t)‖ ≤ β(‖x(t0)‖, t − t0) + α−1

1(α2(µ)), ∀ t ≥ t0

‖x(t)‖ ≤ β(‖x(t0)‖, t − t0) + γ

(

supτ≥t0

‖u(τ )‖)

, ∀ t ≥ t0

Since x(t) depends only on u(τ ) for t0 ≤ τ ≤ t, thesupremum on the right-hand side can be taken over [t0, t]

– p. 13/??

Examplex = −x3 + u

The origin of x = −x3 is globally asymptotically stable

V = 1

2x2

V = −x4 + xu

= −(1 − θ)x4 − θx4 + xu

≤ −(1 − θ)x4, ∀ |x| ≥(

|u|θ

)1/3

0 < θ < 1

The system is ISS with

γ(r) = (r/θ)1/3

– p. 14/??

Examplex = −x − 2x3 + (1 + x2)u2

The origin of x = −x − 2x3 is globally exponentially stable

V = 1

2x2

V = −x2 − 2x4 + x(1 + x2)u2

= −x4 − x2(1 + x2) + x(1 + x2)u2

≤ −x4, ∀ |x| ≥ u2

The system is ISS with γ(r) = r2

– p. 15/??

Example

x1 = −x1 + x2

2, x2 = −x2 + u

Investigate GAS of x1 = −x1 + x2

2, x2 = −x2

V (x) = 1

2x2

1+ 1

4x4

2

V = −x2

1+ x1x

2

2− x4

2= −(x1 − 1

2x2

2)2 −

(

1 − 1

4

)

x4

2

Now u 6= 0, V = −1

2(x1 − x2

2)2 − 1

2(x2

1+ x4

2) + x3

2u

≤ −1

2(x2

1+ x4

2) + |x2|3|u|

V ≤ −1

2(1 − θ)(x2

1+ x4

2) − 1

2θ(x2

1+ x4

2) + |x2|3|u|

(0 < θ < 1)

– p. 16/??

−1

2θ(x2

1+ x4

2) + |x2|3|u| ≤ 0

if |x2| ≥ 2|u|θ

or |x2| ≤ 2|u|θ

and |x1| ≥(

2|u|θ

)2

if ‖x‖ ≥ 2|u|θ

√

1 +

(

2|u|θ

)2

ρ(r) =2r

θ

√

1 +

(

2r

θ

)2

V ≤ −1

2(1 − θ)(x2

1+ x4

2), ∀ ‖x‖ ≥ ρ(|u|)

The system is ISS

– p. 17/??

Find γ

V (x) = 1

2x2

1+ 1

4x4

2

For |x2| ≤ |x1|, 1

4(x2

1+ x2

2) ≤ 1

4x2

1+ 1

4x2

1= 1

2x2

1≤ V (x)

For |x2| ≥ |x1|, 1

16(x2

1+x2

2)2 ≤ 1

16(x2

2+x2

2)2 = 1

4x4

2≤ V (x)

min

1

4‖x‖2, 1

16‖x‖4

≤ V (x) ≤ 1

2‖x‖2 + 1

4‖x‖4

α1(r) = 1

4min

r2, 1

4r4

, α2(r) = 1

2r2 + 1

4r4

γ = α−1

1 α2 ρ

α−1

1(s) =

2(s)1

4 , if s ≤ 1

2√

s, if s ≥ 1

– p. 18/??


Input-Output Stability

– p. 1/15

Input-Output Models

y = Hu

u(t) is a piecewise continuous function of t and belongs toa linear space of signals

The space of bounded functions: supt≥0 ‖u(t)‖ < ∞

The space of square-integrable functions:∫∞0 uT (t)u(t) dt < ∞

Norm of a signal ‖u‖:

‖u‖ ≥ 0 and ‖u‖ = 0 ⇔ u = 0

‖au‖ = a‖u‖ for any a > 0

Triangle Inequality: ‖u1 + u2‖ ≤ ‖u1‖ + ‖u2‖– p. 2/15

Lp spaces:

L∞ : ‖u‖L∞= sup

t≥0‖u(t)‖ < ∞

L2; ‖u‖L2 =

√

∫ ∞

0uT (t)u(t) dt < ∞

Lp; ‖u‖Lp=

(∫ ∞

0‖u(t)‖p dt

)1/p

< ∞, 1 ≤ p < ∞

Notation Lmp : p is the type of p-norm used to define the

space and m is the dimension of u

– p. 3/15

Extended Space: Le = u | uτ ∈ L, ∀ τ ∈ [0, ∞)

uτ is a truncation of u: uτ (t) =

u(t), 0 ≤ t ≤ τ

0, t > τ

Le is a linear space and L ⊂ Le

Example: u(t) = t, uτ (t) =

t, 0 ≤ t ≤ τ

0, t > τ

u /∈ L∞ but uτ ∈ L∞e

Causality: A mapping H : Lme → Lq

e is causal if the valueof the output (Hu)(t) at any time t depends only on thevalues of the input up to time t

(Hu)τ = (Huτ )τ

– p. 4/15

Definition: A mapping H : Lme → Lq

e is L stable if ∃ α ∈ Kβ ≥ 0 such that

‖(Hu)τ ‖L ≤ α (‖uτ ‖L) + β, ∀ u ∈ Lme and τ ∈ [0, ∞)

It is finite-gain L stable if ∃ γ ≥ 0 and β ≥ 0 such that

‖(Hu)τ ‖L ≤ γ‖uτ ‖L + β, ∀ u ∈ Lme and τ ∈ [0, ∞)

It is small-signal L stable (respectively, finite-gain L stable)if ∃ r > 0 such that the inequality is satisfied for all u ∈ Lm

e

with sup0≤t≤τ ‖u(t)‖ ≤ r

– p. 5/15

Example: Memoryless function y = h(u)

h(u) = a + b tanh cu = a + becu − e−cu

ecu + e−cu, a, b, c > 0

h′(u) =4bc

(ecu + e−cu)2 ≤ bc ⇒ |h(u)| ≤ a+bc|u|, ∀ u ∈ R

Finite-gain L∞ stable with β = a and γ = bc

h(u) = b tanh cu, |h(u)| ≤ bc|u|, ∀ u ∈ R∫ ∞

0|h(u(t))|p dt ≤ (bc)p

∫ ∞

0|u(t)|p dt, for p ∈ [1, ∞)

Finite-gain Lp stable with β = 0 and γ = bc

– p. 6/15

h(u) = u2

supt≥0

|h(u(t))| ≤(

supt≥0

|u(t)|)2

L∞ stable with β = 0 and α(r) = r2

It is not finite-gain L∞ stable. Why?

h(u) = tan u

|u| ≤ r <π

2⇒ |h(u)| ≤

(

tan r

r

)

|u|

Small-signal finite-gain Lp stable with β = 0 and γ = tan r/r

– p. 7/15

Example: SISO causal convolution operator

y(t) =

∫ t

0h(t − σ)u(σ) dσ, h(t) = 0 for t < 0

Suppose h ∈ L1 ⇔ ‖h‖L1 =

∫ ∞

0|h(σ)| dσ < ∞

|y(t)| ≤∫ t0 |h(t − σ)| |u(σ)| dσ

≤∫ t0 |h(t − σ)| dσ sup0≤σ≤τ |u(σ)|

=∫ t0 |h(s)| ds sup0≤σ≤τ |u(σ)|

‖yτ ‖L∞≤ ‖h‖L1‖uτ ‖L∞

, ∀ τ ∈ [0, ∞)

Finite-gain L∞ stable

Also, finite-gain Lp stable for p ∈ [1, ∞) (see textbook)– p. 8/15

L Stability of State Models

x = f(x, u), y = h(x, u)

0 = f(0, 0), 0 = h(0, 0)

Case 1: The origin of x = f(x, 0) is exponentially stable

c1‖x‖2 ≤ V (x) ≤ c2‖x‖2

∂V

∂xf(x, 0) ≤ −c3‖x‖2,

∥

∥

∥

∥

∂V

∂x

∥

∥

∥

∥

≤ c4‖x‖

‖f(x, u)−f(x, 0)‖ ≤ L‖u‖, ‖h(x, u)‖ ≤ η1‖x‖+η2‖u‖∀ ‖x‖ ≤ r and ‖u‖ ≤ ru

– p. 9/15

V = ∂V∂x f(x, 0) + ∂V

∂x [f(x, u) − f(x, 0)]

≤ −c3‖x‖2 + c4L‖x‖ ‖u‖ ≤ − c3

c2V + c4L√

c1‖u‖

√V

W (t) =√

V (x(t)) ⇒ W ≤ −(

c3

2c2

)

W +c4L

2√

c1‖u(t)‖

W (t) ≤ e− tc3

2c2 W (0) +c4L

2√

c1

∫ t

0e

− (t−τ )c32c2 ‖u(τ )‖ dτ

‖x(t)‖ ≤√

c2

c1‖x(0)‖e

− tc32c2 +

c4L

2c1

∫ t

0e

− (t−τ )c32c2 ‖u(τ )‖ dτ

‖y(t)‖ ≤ k0‖x(0)‖e−at+k2

∫ t

0e−a(t−τ )‖u(τ )‖ dτ+k3 ‖u(t)‖

– p. 10/15

Theorem 5.1: For each x(0) with ‖x(0)‖ ≤ r√

c1/c2, thesystem is small-signal finite-gain Lp stable for eachp ∈ [1, ∞]

If the assumptions hold globally, then, for each x(0) ∈ Rn,the system is finite-gain Lp stable for each p ∈ [1, ∞]

Example

x = −x − x3 + u, y = tanh x + u

V = 12x2 ⇒ x(−x − x3) ≤ −x2

c1 = c2 = 12, c3 = c4 = 1, L = η1 = η2 = 1

Finite-gain Lp stable for each x(0) ∈ R and each p ∈ [1, ∞]

– p. 11/15

Case 2: The origin of x = f(x, 0) is asymptotically stable

Theorem 5.3: Suppose that, for all (x, u), f is locallyLipschitz and h is continuous and satisfies

‖h(x, u)‖ ≤ α1(‖x‖) + α2(‖u‖) + η, α1, α2 ∈ K, η ≥ 0

If x = f(x, u) is ISS, then, for each x(0) ∈ Rn, the system

x = f(x, u), y = h(x, u)

is L∞ stable

– p. 12/15

Proof

‖x(t)‖ ≤ β(‖x(0)‖, t)+γ

(

sup0≤t≤τ

‖u(t)‖)

, β ∈ KL, γ ∈ K

‖y(t)‖ ≤ α1

(

β(‖x(0)‖, t) + γ(

sup0≤t≤τ ‖u(t)‖))

+ α2(‖u(t)‖) + η

α1(a + b) ≤ α1(2a) + α1(2b)

‖y(t)‖ ≤ α1 (2β(‖x(0)‖, t)) + α1

(

2γ(

sup0≤t≤τ ‖u(t)‖))

+ α2(‖u(t)‖) + η

‖yτ ‖L∞≤ γ0 (‖uτ ‖L∞

) + β0

γ0 = α1 2γ + α2 and β0 = α1(2β(‖x(0)‖, 0)) + η

– p. 13/15

Theorem (Rephrasing of Thm 5.2): Suppose f is locallyLipschitz and h is continuous in some neighborhood of(x = 0, u = 0). If the origin of x = f(x, 0) isasymptotically stable, then there is a constant k1 > 0 suchthat for each x(0) with ‖x(0)‖ < k1, the system

x = f(x, u), y = h(x, u)

is small-signal L∞ stable

– p. 14/15

Example

x1 = −x31 + x2, x2 = −x1 − x3

2 + u, y = x1 + x2

V = (x21 + x2

2) ⇒ V = −2x41 − 2x4

2 + 2x2u

x41 + x4

2 ≥ 12‖x‖4

V ≤ −‖x‖4 + 2‖x‖|u|= −(1 − θ)‖x‖4 − θ‖x‖4 + 2‖x‖|u|, 0 < θ < 1

≤ −(1 − θ)‖x‖4, ∀ ‖x‖ ≥(

2|u|θ

)1/3

ISS

L∞ stable

– p. 15/15


L2 Gain

&

The Small-Gain theorem

– p. 1/12

Theorem 5.4: Consider the linear time-invariant system

x = Ax + Bu, y = Cx + Du

where A is Hurwitz. Let G(s) = C(sI − A)−1B + D.Then, the L2 gain of the system is supω∈R ‖G(jω)‖

– p. 2/12

Lemma: Consider the time-invariant system

x = f(x, u), y = h(x, u)

where f is locally Lipschitz and h is continuous for allx ∈ Rn and u ∈ Rm. Let V (x) be a positive semidefinitefunction such that

V =∂V

∂xf(x, u) ≤ a(γ2‖u‖2 − ‖y‖2), a, γ > 0

Then, for each x(0) ∈ Rn, the system is finite-gain L2

stable and its L2 gain is less than or equal to γ. In particular

‖yτ ‖L2≤ γ‖uτ ‖L2

+

√

V (x(0))

a

– p. 3/12

Proof

V (x(τ ))−V (x(0)) ≤ aγ2

∫ τ

0

‖u(t)‖2 dt− a

∫ τ

0

‖y(t)‖2 dt

– p. 4/12

Proof

V (x(τ ))−V (x(0)) ≤ aγ2

∫ τ

0

‖u(t)‖2 dt− a

∫ τ

0

‖y(t)‖2 dt

V (x) ≥ 0∫ τ

0

‖y(t)‖2 dt ≤ γ2

∫ τ

0

‖u(t)‖2 dt +V (x(0))

a

– p. 4/12

Proof

V (x(τ ))−V (x(0)) ≤ aγ2

∫ τ

0

‖u(t)‖2 dt− a

∫ τ

0

‖y(t)‖2 dt

V (x) ≥ 0∫ τ

0

‖y(t)‖2 dt ≤ γ2

∫ τ

0

‖u(t)‖2 dt +V (x(0))

a

‖yτ ‖L2≤ γ‖uτ ‖L2

+

√

V (x(0))

a

– p. 4/12

Lemma 6.5: If the system

x = f(x, u), y = h(x, u)

is output strictly passive with

uT y ≥ V + δyT y, δ > 0

then it is finite-gain L2 stable and its L2 gain is less than orequal to 1/δ

– p. 5/12

Lemma 6.5: If the system

x = f(x, u), y = h(x, u)

is output strictly passive with

uT y ≥ V + δyT y, δ > 0

then it is finite-gain L2 stable and its L2 gain is less than orequal to 1/δProof

V ≤ uT y − δyT y

= − 1

2δ (u − δy)T (u − δy) + 1

2δuT u − δ2yT y

≤ δ2

(

1

δ2 uT u − yT y)

– p. 5/12

Example

x1 = x2, x2 = −ax3

1− kx2 + u, y = x2, a, k > 0

– p. 6/12

Example

x1 = x2, x2 = −ax3

1− kx2 + u, y = x2, a, k > 0

V (x) = a4x4

1+ 1

2x2

2

– p. 6/12

Example

x1 = x2, x2 = −ax3

1− kx2 + u, y = x2, a, k > 0

V (x) = a4x4

1+ 1

2x2

2

V = ax3

1x2 + x2(−ax3

1− kx2 + u)

= −kx2

2+ x2u = −ky2 + yu

– p. 6/12

Example

x1 = x2, x2 = −ax3

1− kx2 + u, y = x2, a, k > 0

V (x) = a4x4

1+ 1

2x2

2

V = ax3

1x2 + x2(−ax3

1− kx2 + u)

= −kx2

2+ x2u = −ky2 + yu

The system is finite-gain L2 stable and its L2 gain is lessthan or equal to 1/k

– p. 6/12

Theorem 5.5: Consider the time-invariant system

x = f(x) + G(x)u, y = h(x)

f(0) = 0, h(0) = 0

where f and G are locally Lipschitz and h is continuousover Rn. Suppose ∃ γ > 0 and a continuouslydifferentiable, positive semidefinite function V (x) thatsatisfies the Hamilton–Jacobi inequality

∂V

∂xf(x)+

1

2γ2

∂V

∂xG(x)GT (x)

(

∂V

∂x

)T

+1

2hT (x)h(x) ≤ 0

∀ x ∈ Rn. Then, for each x(0) ∈ Rn, the system isfinite-gain L2 stable and its L2 gain ≤ γ

– p. 7/12

Proof

∂V

∂xf(x) +

∂V

∂xG(x)u =

−1

2γ2

∥

∥

∥

∥

∥

u −1

γ2GT (x)

(

∂V

∂x

)T∥

∥

∥

∥

∥

2

+∂V

∂xf(x)

+1

2γ2

∂V

∂xG(x)GT (x)

(

∂V

∂x

)T

+1

2γ2‖u‖2

– p. 8/12

Proof

∂V

∂xf(x) +

∂V

∂xG(x)u =

−1

2γ2

∥

∥

∥

∥

∥

u −1

γ2GT (x)

(

∂V

∂x

)T∥

∥

∥

∥

∥

2

+∂V

∂xf(x)

+1

2γ2

∂V

∂xG(x)GT (x)

(

∂V

∂x

)T

+1

2γ2‖u‖2

V ≤1

2γ2‖u‖2 −

1

2‖y‖2

– p. 8/12

Examplex = Ax + Bu, y = Cx

– p. 9/12


Suppose there is P = P T ≥ 0 that satisfies the Riccatiequation

PA + AT P +1

γ2PBBT P + CT C = 0

for some γ > 0.

– p. 9/12



PA + AT P +1


for some γ > 0. Verify that V (x) = 1

2xT Px satisfies the

Hamilton-Jacobi equation

– p. 9/12



PA + AT P +1


for some γ > 0. Verify that V (x) = 1

2xT Px satisfies the

Hamilton-Jacobi equation

The system is finite-gain L2 stable and its L2 gain is lessthan or equal to γ

– p. 9/12

The Small-Gain Theorem

-

- -

6

?

u1

u2

e1

e2

y1

y2

H1

H2

−+

++

– p. 10/12


-

- -

6

?

u1

u2

e1

e2

y1

y2

H1

H2

−+

++

‖y1τ ‖L ≤ γ1‖e1τ ‖L + β1, ∀ e1 ∈ Lme , ∀ τ ∈ [0, ∞)

– p. 10/12


-

- -

6

?

u1

u2

e1

e2

y1

y2

H1

H2

−+

++

‖y1τ ‖L ≤ γ1‖e1τ ‖L + β1, ∀ e1 ∈ Lme , ∀ τ ∈ [0, ∞)

‖y2τ ‖L ≤ γ2‖e2τ ‖L + β2, ∀ e2 ∈ Lqe, ∀ τ ∈ [0, ∞)

– p. 10/12

u =

[

u1

u2

]

, y =

[

y1

y2

]

, e =

[

e1

e2

]

– p. 11/12

u =

[

u1

u2

]

, y =

[

y1

y2

]

, e =

[

e1

e2

]

Theorem: The feedback connection is finite-gain L stable ifγ1γ2 < 1

– p. 11/12

u =

[

u1

u2

]

, y =

[

y1

y2

]

, e =

[

e1

e2

]


Proof

e1τ = u1τ − (H2e2)τ , e2τ = u2τ + (H1e1)τ

– p. 11/12

u =

[

u1

u2

]

, y =

[

y1

y2

]

, e =

[

e1

e2

]


Proof

e1τ = u1τ − (H2e2)τ , e2τ = u2τ + (H1e1)τ

‖e1τ ‖L ≤ ‖u1τ ‖L + ‖(H2e2)τ ‖L

≤ ‖u1τ ‖L + γ2‖e2τ ‖L + β2

– p. 11/12

‖e1τ ‖L ≤ ‖u1τ ‖L + γ2 (‖u2τ ‖L + γ1‖e1τ ‖L + β1) + β2

= γ1γ2‖e1τ ‖L

+ (‖u1τ ‖L + γ2‖u2τ ‖L + β2 + γ2β1)

– p. 12/12


= γ1γ2‖e1τ ‖L

+ (‖u1τ ‖L + γ2‖u2τ ‖L + β2 + γ2β1)

‖e1τ ‖L ≤1

1 − γ1γ2

(‖u1τ ‖L + γ2‖u2τ ‖L + β2 + γ2β1)

– p. 12/12


= γ1γ2‖e1τ ‖L

+ (‖u1τ ‖L + γ2‖u2τ ‖L + β2 + γ2β1)

‖e1τ ‖L ≤1

1 − γ1γ2

(‖u1τ ‖L + γ2‖u2τ ‖L + β2 + γ2β1)

‖e2τ ‖L ≤1

1 − γ1γ2

(‖u2τ ‖L + γ1‖u1τ ‖L + β1 + γ1β2)

– p. 12/12


= γ1γ2‖e1τ ‖L

+ (‖u1τ ‖L + γ2‖u2τ ‖L + β2 + γ2β1)

‖e1τ ‖L ≤1

1 − γ1γ2

(‖u1τ ‖L + γ2‖u2τ ‖L + β2 + γ2β1)

‖e2τ ‖L ≤1

1 − γ1γ2

(‖u2τ ‖L + γ1‖u1τ ‖L + β1 + γ1β2)

‖eτ ‖L ≤ ‖e1τ ‖L + ‖e2τ ‖L

– p. 12/12


Normal Form

– p. 1/17

Relative Degree

x = f(x) + g(x)u, y = h(x)

where f , g, and h are sufficiently smooth in a domain Df : D → Rn and g : D → Rn are called vector fields on D

y =∂h

∂x[f(x) + g(x)u]

def= Lfh(x) + Lgh(x) u

Lfh(x) =∂h

∂xf(x)

is the Lie Derivative of h with respect to f or along f

– p. 2/17

LgLfh(x) =∂(Lfh)

∂xg(x)

L2fh(x) = LfLfh(x) =

∂(Lfh)

∂xf(x)

Lkfh(x) = LfLk−1f h(x) =

∂(Lk−1f h)

∂xf(x)

L0fh(x) = h(x)

y = Lfh(x) + Lgh(x) u

Lgh(x) = 0 ⇒ y = Lfh(x)

y(2) =∂(Lfh)

∂x[f(x) + g(x)u] = L2

fh(x) + LgLfh(x) u

– p. 3/17

LgLfh(x) = 0 ⇒ y(2) = L2fh(x)

y(3) = L3fh(x) + LgL

2fh(x) u

LgLi−1f h(x) = 0, i = 1, 2, . . . , ρ− 1; LgL

ρ−1f h(x) 6= 0

y(ρ) = Lρfh(x) + LgL

ρ−1f h(x) u


x = f(x) + g(x)u, y = h(x)

has relative degree ρ, 1 ≤ ρ ≤ n, in D0 ⊂ D if ∀ x ∈ D0

LgLi−1f h(x) = 0, i = 1, 2, . . . , ρ− 1; LgL

ρ−1f h(x) 6= 0

– p. 4/17

Example

x1 = x2, x2 = −x1+ε(1−x21)x2+u, y = x1, ε > 0

y = x1 = x2

y = x2 = −x1 + ε(1 − x21)x2 + u

Relative degree = 2 over R2

Example

x1 = x2, x2 = −x1+ε(1−x21)x2+u, y = x2, ε > 0

y = x2 = −x1 + ε(1 − x21)x2 + u

Relative degree = 1 over R2

– p. 5/17

Example

x1 = x2, x2 = −x1+ε(1−x21)x2+u, y = x1+x

22, ε > 0

y = x2 + 2x2[−x1 + ε(1 − x21)x2 + u]

Relative degree = 1 over x2 6= 0

Example: Field-controlled DC motor

x1 = −ax1+u, x2 = −bx2+k−cx1x3, x3 = θx1x2, y = x3

a, b, c, k, and θ are positive constants

y = x3 = θx1x2

y = θx1x2 + θx1x2 = (·) + θx2u

Relative degree = 2 over x2 6= 0– p. 6/17

Normal Form

Change of variables:

z = T (x) =

φ1(x)...

φn−ρ(x)

− − −

h(x)...

Lρ−1f h(x)

def=

φ(x)

− − −

ψ(x)

def=

η

− − −

ξ

φ1 to φn−ρ are chosen such that T (x) is a diffeomorphismon a domain D0 ⊂ D

– p. 7/17

η =∂φ

∂x[f(x) + g(x)u] = f0(η, ξ) + g0(η, ξ)u

ξi = ξi+1, 1 ≤ i ≤ ρ− 1

ξρ = Lρfh(x) + LgL

ρ−1f h(x) u

y = ξ1

Choose φ(x) such that T (x) is a diffeomorphism and

∂φi

∂xg(x) = 0, for 1 ≤ i ≤ n− ρ, ∀ x ∈ D0

Always possible (at least locally)

η = f0(η, ξ)

– p. 8/17

Theorem 13.1: Suppose the system

x = f(x) + g(x)u, y = h(x)

has relative degree ρ (≤ n) in D. If ρ = n, then for everyx0 ∈ D, a neighborhood N of x0 exists such that the mapT (x) = ψ(x), restricted to N , is a diffeomorphism on N . Ifρ < n, then, for every x0 ∈ D, a neighborhood N of x0

and smooth functions φ1(x), . . . , φn−ρ(x) exist such that

∂φi

∂xg(x) = 0, for 1 ≤ i ≤ n− ρ

is satisfied for all x ∈ N and the map T (x) =

[

φ(x)

ψ(x)

]

,

restricted to N , is a diffeomorphism on N

– p. 9/17

Normal Form: η = f0(η, ξ)

ξi = ξi+1, 1 ≤ i ≤ ρ− 1


ρ−1f h(x) u

y = ξ1

Ac =

0 1 0 . . . 0

0 0 1 . . . 0... . . . ...... 0 1

0 . . . . . . 0 0

, Bc =

0

0...0

1

Cc =[

1 0 . . . 0 0]

– p. 10/17

η = f0(η, ξ)

ξ = Acξ +Bc

[

Lρfh(x) + LgL

ρ−1f h(x) u

]

y = Ccξ

γ(x) = LgLρ−1f h(x), α(x) = −

Lρfh(x)

LgLρ−1f h(x)

ξ = Acξ +Bcγ(x)[u− α(x)]

If x∗ is an open-loop equilibrium point at which y = 0; i.e.,f(x∗) = 0 and h(x∗) = 0, then ψ(x∗) = 0. Takeφ(x∗) = 0 so that z = 0 is an open-loop equilibrium point.

– p. 11/17

Zero Dynamics

η = f0(η, ξ)

ξ = Acξ +Bcγ(x)[u− α(x)]

y = Ccξ

y(t) ≡ 0 ⇒ ξ(t) ≡ 0 ⇒ u(t) ≡ α(x(t)) ⇒ η = f0(η, 0)

Definition: The equation η = f0(η, 0) is called the zerodynamics of the system. The system is said to be minimumphase if zero dynamics have an asymptotically stableequilibrium point in the domain of interest (at the origin ifT (0) = 0)

The zero dynamics can be characterized in thex-coordinates

– p. 12/17

Z∗ = x ∈ D0 | h(x) = Lfh(x) = · · · = Lρ−1f h(x) = 0

y(t) ≡ 0 ⇒ x(t) ∈ Z∗

⇒ u = u∗(x)def= α(x)|x∈Z∗

The restricted motion of the system is described by

x = f∗(x)def= [f(x) + g(x)α(x)]x∈Z∗

– p. 13/17

Example

x1 = x2, x2 = −x1 + ε(1 − x21)x2 + u, y = x2

y = x2 = −x1 + ε(1 − x21)x2 + u ⇒ ρ = 1

y(t) ≡ 0 ⇒ x2(t) ≡ 0 ⇒ x1 = 0

Non-minimum phase

– p. 14/17

Example

x1 = −x1 +2 + x2

3

1 + x23

u, x2 = x3, x3 = x1x3 + u, y = x2

y = x2 = x3

y = x3 = x1x3 + u ⇒ ρ = 2

γ = LgLfh(x) = 1, α = −L2fh(x)

LgLfh(x)= −x1x3

Z∗ = x2 = x3 = 0

u = u∗(x) = 0 ⇒ x1 = −x1

Minimum phase

– p. 15/17

Find φ(x) such that

φ(0) = 0,∂φ

∂xg(x) =

[

∂φ∂x1

, ∂φ∂x2

, ∂φ∂x3

]

2+x2

3

1+x2

3

0

1

= 0

andT (x) =

[

φ(x) x2 x3

]T

is a diffeomorphism

∂φ

∂x1·2 + x2

3

1 + x23

+∂φ

∂x3= 0

φ(x) = −x1 + x3 + tan−1 x3

– p. 16/17

T (x) =[

−x1 + x3 + tan−1 x3, x2, x3

]T

is a global diffeomorphism

η = −x1 + x3 + tan−1 x3, ξ1 = x2, ξ2 = x3

η =(

−η + ξ2 + tan−1 ξ2)

(

1 +2 + ξ22

1 + ξ22ξ2

)

ξ1 = ξ2

ξ2 =(

−η + ξ2 + tan−1 ξ2)

ξ2 + u

y = ξ1

– p. 17/17


Controller Form

– p. 1/18

Definition: A nonlinear system is in the controller form if

x = Ax + Bγ(x)[u − α(x)]

where (A, B) is controllable and γ(x) is a nonsingular

u = α(x) + γ−1(x)v ⇒ x = Ax + Bv

The n-dimensional single-input (SI) system

x = f(x) + g(x)u

can be transformed into the controller form if ∃ h(x) s.t.

x = f(x) + g(x)u, y = h(x)

has relative degree n. Why?

– p. 2/18

Transform the system into the normal form

z = Acz + Bcγ(z)[u − α(z)], y = Ccz

On the other hand, if there is a change of variablesζ = S(x) that transforms the SI system

x = f(x) + g(x)u

into the controller form

ζ = Aζ + Bγ(ζ)[u − α(ζ)]

then there is a function h(x) such that the system

x = f(x) + g(x)u, y = h(x)

has relative degree n. Why?

– p. 3/18

For any controllable pair (A, B), we can find a nonsingularmatrix M that transforms (A, B) into a controllablecanonical form:

MAM−1 = Ac + BcλT , MB = Bc

z = Mζ = MS(x)def= T (x)

z = Acz + Bcγ(·)[u − α(·)]

h(x) = T1(x)

– p. 4/18

In summary, the n-dimensional SI system

x = f(x) + g(x)u

is transformable into the controller form if and only if ∃ h(x)such that

x = f(x) + g(x)u, y = h(x)

has relative degree nSearch for a smooth function h(x) such that

LgLi−1

f h(x) = 0, i = 1, 2, . . . , n−1, and LgLn−1

f h(x) 6= 0

T (x) =[

h(x), Lfh(x), · · · Ln−1

f h(x)]

– p. 5/18

The Lie Bracket: For two vector fields f and g, the Liebracket [f, g] is a third vector field defined by

[f, g](x) =∂g

∂xf(x) −

∂f

∂xg(x)

Notation:

ad0

fg(x) = g(x), adfg(x) = [f, g](x)

adkfg(x) = [f, adk−1

f g](x), k ≥ 1

Properties:

[f, g] = −[g, f ]

For constant vector fields f and g, [f, g] = 0

– p. 6/18

Example

f =

[

x2

− sin x1 − x2

]

, g =

[

0

x1

]

[f, g] =

[

0 0

1 0

] [

x2

− sin x1 − x2

]

−

[

0 1

− cos x1 −1

] [

0

x1

]

adfg = [f, g] =

[

−x1

x1 + x2

]

– p. 7/18

f =

[

x2

− sin x1 − x2

]

, adfg =

[

−x1

x1 + x2

]

ad2

fg = [f, adfg] =[

−1 0

1 1

] [

x2

− sin x1 − x2

]

−

[

0 1

− cos x1 −1

] [

−x1

x1 + x2

]

=

[

−x1 − 2x2

x1 + x2 − sin x1 − x1 cos x1

]

– p. 8/18

Distribution: For vector fields f1, f2, . . . , fk on D ⊂ Rn, let

∆(x) = spanf1(x), f2(x), . . . , fk(x)

The collection of all vector spaces ∆(x) for x ∈ D is calleda distribution and referred to by

∆ = spanf1, f2, . . . , fk

If dim(∆(x)) = k for all x ∈ D, we say that ∆ is anonsingular distribution on D, generated by f1, . . . , fk

A distribution ∆ is involutive if

g1 ∈ ∆ and g2 ∈ ∆ ⇒ [g1, g2] ∈ ∆

– p. 9/18

Lemma: If ∆ is a nonsingular distribution, generated byf1, . . . , fk, then it is involutive if and only if

[fi, fj ] ∈ ∆, ∀ 1 ≤ i, j ≤ k

Example: D = R3; ∆ = spanf1, f2

f1 =

2x2

1

0

, f2 =

1

0

x2

, dim(∆(x)) = 2, ∀ x ∈ D

[f1, f2] =∂f2

∂xf1 −

∂f1

∂xf2 =

0

0

1

– p. 10/18

rank [f1(x), f2(x), [f1, f2](x)] =

rank

2x2 1 0

1 0 0

0 x2 1

= 3, ∀ x ∈ D

∆ is not involutive

– p. 11/18

Example: D = x ∈ R3 | x2

1+ x2

36= 0; ∆ = spanf1, f2

f1 =

2x3

−1

0

, f2 =

−x1

−2x2

x3

, dim(∆(x)) = 2, ∀ x ∈ D

[f1, f2] =∂f2

∂xf1 −

∂f1

∂xf2 =

−4x3

2

0

rank

2x3 −x1 −4x3

−1 −2x2 2

0 x3 0

= 2, ∀ x ∈ D

∆ is involutive

– p. 12/18

Theorem: The n-dimensional SI system

x = f(x) + g(x)u

is transformable into the controller form if and only if there isa domain D0 such that

rank[g(x), adfg(x), . . . , adn−1

f g(x)] = n, ∀ x ∈ D0

and

span g, adfg, . . . , adn−2

f g is involutive in D0

– p. 13/18

Example

x =

[

a sin x2

−x2

1

]

+

[

0

1

]

u

adfg = [f, g] = −∂f

∂xg =

[

−a cos x2

0

]

[g(x), adfg(x)] =

[

0 −a cos x2

1 0

]

rank[g(x), adfg(x)] = 2, ∀ x such that cos x2 6= 0

spang is involutive

Find h such that Lgh(x) = 0, and LgLfh(x) 6= 0

– p. 14/18

∂h

∂xg =

∂h

∂x2

= 0 ⇒ h is independent of x2

Lfh(x) =∂h

∂x1

a sin x2

LgLfh(x) =∂(Lfh)

∂xg =

∂(Lfh)

∂x2

=∂h

∂x1

a cos x2

LgLfh(x) 6= 0 in D0 = x ∈ R2| cos x2 6= 0 if ∂h∂x1

6= 0

Take h(x) = x1 ⇒ T (x) =

[

h

Lfh

]

=

[

x1

a sin x2

]

– p. 15/18

Example (Field-Controlled DC Motor)

x =

−ax1

−bx2 + k − cx1x3

θx1x2

+

1

0

0

u

adfg =

a

cx3

−θx2

; ad2

fg =

a2

(a + b)cx3

(b − a)θx2 − θk

[g(x), adfg(x), ad2

fg(x)] =

1 a a2

0 cx3 (a + b)cx3

0 −θx2 (b − a)θx2 − θk

– p. 16/18

det[·] = cθ(−k + 2bx2)x3

rank [·] = 3 for x2 6= k/2b and x3 6= 0

spang, adfg is involutive if [g, adfg] ∈ spang, adfg

[g, adfg] =∂(adfg)

∂xg =

0 0 0

0 0 c

0 −θ 0

1

0

0

=

0

0

0

⇒ spang, adfg is involutive

D0 = x ∈ R3 | x2 >k

2band x3 > 0

Find h such that Lgh(x) = LgLfh(x) = 0; LgL2

fh(x) 6= 0

– p. 17/18

x∗ = [0, k/b, ω0]T , h(x∗) = 0

∂h

∂xg =

∂h

∂x1

= 0 ⇒ h is independent of x1

Lfh(x) =∂h

∂x2

[−bx2 + k − cx1x3] +∂h

∂x3

θx1x2

[∂(Lfh)/∂x]g = 0 ⇒ cx3

∂h

∂x2

= θx2

∂h

∂x3

h = c1[θx2

2+cx2

3]+c2, LgL2

fh(x) = −2c1cθ(k−2bx2)x3

h(x∗) = c1[θ(k/b)2 + cω2

0] + c2

c1 = 1, c2 = −θ(k/b)2 − cω2

0

– p. 18/18


Observer, Output Feedback&

Strict Feedback Forms

– p. 1/12

Definition: A nonlinear system is in the observer form if

x = Ax + γ(y, u), y = Cx

where (A, C) is observable

Observer:

˙x = Ax + γ(y, u) + H(y − Cx)

x = x − x

˙x = (A − HC)x

Design H such that (A − HC) is Hurwitz

– p. 2/12

Theorem: An n-dimensional single-output (SO) system

x = f(x) + g(x)u, y = h(x)

is transformable into the observer form if and only if there isa domain D0 such that

rank

[

∂φ

∂x(x)

]

= n, ∀ x ∈ D0

where φ =[

h, Lfh, · · · Ln−1

f h]T

and the unique vector field solution τ of

∂φ

∂xτ = b, where b =

[

0, · · · 0, 1]T

– p. 3/12

satisfies

[adifτ, ad

jfτ ] = 0, 0 ≤ i, j ≤ n − 1

and[g, ad

jfτ ] = 0, 0 ≤ j ≤ n − 2

The change of variables z = T (x) is given by

∂T

∂x

[

τ1, τ2, · · · τn

]

= I

whereτi = (−1)i−1ad

i−1

f τ, 1 ≤ i ≤ n

– p. 4/12

Example

x =

[

β1(x1) + x2

f2(x)

]

+

[

0

1

]

u, y = x1

φ(x) =

[

h(x)

Lfh(x)

]

=

[

x1

β1(x1) + x2

]

∂φ

∂x=

[

1 0∂β1

∂x1

1

]

; rank

[

∂φ

∂x(x)

]

= 2, ∀ x

∂φ

∂xτ =

[

0

1

]

⇒ τ =

[

0

1

]

– p. 5/12

adfτ = [f, τ ] = −∂f

∂xτ = −

[

∗ 1

∗ ∂f2

∂x2

] [

0

1

]

= −

[

1∂f2

∂x2

]

[τ, adfτ ] =∂(adfτ )

∂xτ = −

[

0 0∂2f2

∂x1∂x2

∂2f2

∂x2

2

] [

0

1

]

[τ, adfτ ] = 0 ⇔∂2f2

∂x22

= 0 ⇔ f2(x) = β2(x1)+x2β3(x1)

[g, τ ] = 0 (g and τ are constant vector fields)

All the conditions are satisfied

– p. 6/12

τ1 = (−1)0ad0fτ = τ =

[

0

1

]

τ2 = (−1)1ad1fτ = −adfτ =

[

1

β3(x1)

]

∂T

∂x

[

τ1, τ2

]

= I

[

∂T1

∂x1

∂T1

∂x2

∂T2

∂x1

∂T2

∂x2

] [

0 1

1 β3(x1)

]

=

[

1 0

0 1

]

∂T1

∂x2

= 1, ∂T1

∂x1

+ β3(x1)∂T1

∂x2

= 0

∂T2

∂x2

= 0, ∂T2

∂x1

+ β3(x1)∂T2

∂x2

= 1

– p. 7/12

∂T1

∂x2

= 1 ⇒ ∂T1

∂x1

+ β3(x1) = 0

T1(x) = x2 −∫ x1

0β3(σ) dσ

∂T2

∂x2

= 0 ⇒ ∂T2

∂x1

= 1, T2(x) = x1

z1 = x2 −

∫ x1

0

β3(σ) dσ, z2 = x1

y = z2

z =

[

0 0

1 0

]

z +

[

β2(y) − β1(y)β3(y) + u∫ y

0β3(σ) dσ + β1(y)

]

y =[

0 1]

z

– p. 8/12

Definition: A nonlinear system is in the output feedbackform if

x1 = x2 + γ1(y)

x2 = x3 + γ2(y)

...xρ−1 = xρ + γρ−1(y)

xρ = xρ+1 + γρ(y) + bmu, bm > 0

...xn−1 = xn + γn−1(y) + b1u

xn = γn(y) + b0u

y = x1

– p. 9/12

Show that

The output feedback form is a special case of theobserver form

It has relative degree ρ

It is minimum phase if the polynomial

bmsm + · · · + b1s + b0

is Hurwitz

– p. 10/12

Definition: A nonlinear system is in the strict feedback formif

x = f0(x) + g0(x)z1

z1 = f1(x, z1) + g1(x, z1)z2

z2 = f2(x, z1, z2) + g2(x, z1, z2)z3

...zk−1 = fk−1(x, z1, . . . , zk−1) + gk−1(x, z1, . . . , zk−1)zk

zk = fk(x, z1, . . . , zk) + gk(x, z1, . . . , zk)u

x ∈ Rn, z1 to zk are scalars

gi(x, z1, . . . , zi) 6= 0 for 1 ≤ i ≤ k

– p. 11/12

Find the relative degree if y = z1

Find the zero dynamics if y = z1 and

fi(x, 0) = 0, ∀ 1 ≤ i ≤ k

– p. 12/12


Stabilization

Basic Concepts & Linearization

– p. 1/??

We want to stabilize the system

x = 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

xδ = f(xss + xδ, uss + uδ)def= fδ(xδ, uδ)

fδ(0, 0) = 0

uδ = γ(xδ) ⇒ u = uss + γ(x − xss)

– p. 2/??

State Feedback Stabilization: Given

x = f(x, u) [f(0, 0) = 0]

findu = γ(x) [γ(0) = 0]

s.t. the origin is an asymptotically stable equilibrium point of

x = f(x, γ(x))

f and γ are locally Lipschitz functions

– p. 3/??

Linear Systemsx = Ax + Bu

(A, B) is stabilizable (controllable or every uncontrollableeigenvalue has a negative real part)

Find K such that (A − BK) is Hurwitz

u = −Kx

Typical methods:

Eigenvalue Placement

Eigenvalue-Eigenvector Placement

LQR

– p. 4/??

Linearizationx = f(x, u)

f(0, 0) = 0 and f is continuously differentiable in a domainDx × Du that contains the origin (x = 0, u = 0)(Dx ⊂ Rn, Du ⊂ Rp)

x = Ax + Bu

A =∂f

∂x(x, u)

∣

∣

∣

∣

x=0,u=0

; B =∂f

∂u(x, u)

∣

∣

∣

∣

x=0,u=0

Assume (A, B) is stabilizable. Design a matrix K such that(A − BK) is Hurwitz

u = −Kx

– p. 5/??

Closed-loop system:

x = f(x, −Kx)

x =

[

∂f

∂x(x, −Kx) +

∂f

∂u(x, −Kx) (−K)

]

x=0

x

= (A − BK)x

Since (A − BK) is Hurwitz, the origin is an exponentiallystable equilibrium point of the closed-loop system

– p. 6/??

Example (Pendulum Equation):

θ = −a sin θ − bθ + cT

Stabilize the pendulum at θ = δ

0 = −a sin δ + cTss

x1 = θ − δ, x2 = θ, u = T − Tss

x1 = x2

x2 = −a[sin(x1 + δ) − sin δ] − bx2 + cu

A =

[

0 1

−a cos(x1 + δ) −b

]

x1=0

=

[

0 1

−a cos δ −b

]

– p. 7/??

A =

[

0 1

−a cos δ −b

]

; B =

[

0

c

]

K =[

k1 k2

]

A − BK =

[

0 1

−(a cos δ + ck1) −(b + ck2)

]

k1 > −a cos δ

c, k2 > −

b

c

T =a sin δ

c− Kx =

a sin δ

c− k1(θ − δ) − k2θ

– p. 8/??

Notions of Stabilization

x = f(x, u), u = γ(x)

Local Stabilization: The origin of x = f(x, γ(x)) isasymptotically stable (e.g., linearization)

Regional Stabilization: The origin of x = f(x, γ(x)) isasymptotically stable and a given region G is a subset ofthe region of attraction (for all x(0) ∈ G, limt→∞ x(t) = 0)(e.g., G ⊂ Ωc = V (x) ≤ c where Ωc is an estimate ofthe region of attraction)

Global Stabilization: The origin of x = f(x, γ(x)) isglobally asymptotically stable

– p. 9/??

Semiglobal Stabilization: The origin of x = f(x, γ(x)) isasymptotically stable and γ(x) can be designed such thatany given compact set (no matter how large) can beincluded in the region of attraction (Typically u = γp(x) isdependent on a parameter p such that for any compact setG, p can be chosen to ensure that G is a subset of theregion of attraction )

What is the difference between global stabilization andsemiglobal stabilization?

– p. 10/??

Examplex = x2 + u

Linearization:

x = u, u = −kx, k > 0

Closed-loop system:

x = −kx + x2

Linearization of the closed-loop system yields x = −kx.Thus, u = −kx achieves local stabilization

The region of attraction is x < k. Thus, for any setx ≤ a with a < k, the control u = −kx achievesregional stabilization

– p. 11/??

The control u = −kx does not achieve global stabilization

But it achieves semiglobal stabilization because anycompact set |x| ≤ r can be included in the region ofattraction by choosing k > r

The controlu = −x2 − kx

achieves global stabilization because it yields the linearclosed-loop system x = −kx whose origin is globallyexponentially stable

– p. 12/??

Practical Stabilization

x = f(x, u) + g(x, u, t)

f(0, 0) = 0, g(0, 0, t) 6= 0

‖g(x, u, t)‖ ≤ δ, ∀ x ∈ Dx, u ∈ Du, t ≥ 0

There is no control u = γ(x), with γ(0) = 0, that can makethe origin of

x = f(x, γ(x)) + g(x, γ(x), t)

uniformly asymptotically stable because the origin is not anequilibrium point

– p. 13/??


x = f(x, u) + g(x, u, t)

is practically stabilizable if for any ε > 0 there is a controllaw u = γ(x) such that the solutions of

x = f(x, γ(x)) + g(x, γ(x), t)

are uniformly ultimately bounded by ε; i.e.,

‖x(t)‖ ≤ ε, ∀ t ≥ T

Typically, u = γp(x) is dependent on a parameter p suchthat for any ε > 0, p can be chosen to ensure that ε is anultimate bound

– p. 14/??

With practical stabilization, we may have

local practical stabilization

regional practical stabilization

global practical stabilization, or

semiglobal practical stabilization

depending on the region of initial states

– p. 15/??

Example

x = x2 + u + d(t), |d(t)| ≤ δ, ∀ t ≥ 0

u = −kx, k > 0, ⇒ x = x2 − kx + d(t)

V = 1

2x2 ⇒ V = x3 − kx2 + xd(t)

V ≤ −k3x2 − x2

(

k3

− |x|)

− |x|(

k3|x| − δ

)

V ≤ −k3x2, for 3δ

k≤ |x| ≤ k

3

Take 3δk

< ε ⇔ k ≥ 3δε

By choosing k large enough we can achieve semiglobalpractical stabilization

– p. 16/??

x = x2 + u + d(t)

u = −x2 − kx, k > 0, ⇒ x = −kx + d(t)

V = 1

2x2 ⇒ V = −kx2 + xd(t)

V ≤ −k2x2 − |x|

(

k2|x| − δ

)

By choosing k large enough we can achieve global practicalstabilization

– p. 17/??


Stabilization

Feedback Lineaization

– p. 1/??

Consider the nonlinear system

x = f(x) + G(x)u

f(0) = 0, x ∈ Rn, u ∈ Rm

Suppose there is a change of variables z = T (x), definedfor all x ∈ D ⊂ Rn, that transforms the system into thecontroller form

z = Az + Bγ(x)[u − α(x)]

where (A, B) is controllable and γ(x) is nonsingular for allx ∈ D

u = α(x) + γ−1(x)v ⇒ z = Az + Bv

– p. 2/??

v = −Kz

Design K such that (A − BK) is Hurwitz

The origin z = 0 of the closed-loop system

z = (A − BK)z

is globally exponentially stable

u = α(x) − γ−1(x)KT (x)

Closed-loop system in the x-coordinates:

x = f(x) + G(x)[

α(x) − γ−1(x)KT (x)]

– p. 3/??

What can we say about the stability of x = 0 as anequilibrium point of

x = f(x) + G(x)[

α(x) − γ−1(x)KT (x)]

x = 0 is asymptotically stable because T (x) is adiffeomorphism. Show it!

Is x = 0 globally asymptotically stable? In general No

It is globally asymptotically stable if T (x) is a globaldiffeomorphism (See page 508)

– p. 4/??

What information do we need to implement the control

u = α(x) − γ−1(x)KT (x) ?

What is the effect of uncertainty in α, γ, and T ?

Let α(x), γ(x), and T (x) be nominal models of α(x),γ(x), and T (x)

u = α(x) − γ−1(x)KT (x)

Closed-loop system:

z = (A − BK)z + Bδ(z)

δ = γ[α − α + γ−1KT − γ−1KT ]

– p. 5/??

z = (A − BK)z + Bδ(z) (∗)

V (z) = zT Pz, P (A − BK) + (A − BK)T P = −I

Lemma 13.3

If ‖δ(z)‖ ≤ k‖z‖ for all z, where

0 ≤ k <1

2‖PB‖

then the origin of (*) is globally exponentially stable

If ‖δ(z)‖ ≤ k‖z‖ + ε for all z, then the state z isglobally ultimately bounded by εc for some c > 0

– p. 6/??

Example (Pendulum Equation):


x1 = θ − δ, x2 = θ, u = T − Tss = T −a

csin δ

x1 = x2

x2 = −a[sin(x1 + δ) − sin δ] − bx2 + cu

u =1

ca[sin(x1 + δ) − sin δ] − k1x1 − k2x2

A − BK =

[

0 1

−k1 −(k2 + b)

]

is Hurwitz

– p. 7/??

T = u +a

csin δ =

1

c[a sin(x1 + δ) − k1x1 − k2x2]

Let a and c be nominal models of a and c

T =1

c[a sin(x1 + δ) − k1x1 − k2x2]

x = (A − BK)x + Bδ(x)

δ(x) =

(

ac − ac

c

)

sin(x1 + δ1) −

(

c − c

c

)

(k1x1 + k2x2)

– p. 8/??

δ(x) =

(

ac − ac

c

)

sin(x1 + δ1) −

(

c − c

c

)

(k1x1 + k2x2)

|δ(x)| ≤ k‖x‖ + ε

k =

∣

∣

∣

∣

ac − ac

c

∣

∣

∣

∣

+

∣

∣

∣

∣

c − c

c

∣

∣

∣

∣

√

k2

1+ k2

2, ε =

∣

∣

∣

∣

ac − ac

c

∣

∣

∣

∣

| sin δ1|

P =

[

p11 p12

p12 p22

]

, PB =

[

p12

p22

]

k <1

2√

p2

12+ p2

22

sin δ1 = 0 ⇒ ε = 0

– p. 9/??

Is feedback linearization a good idea?

Examplex = ax − bx3 + u, a, b > 0

u = −(k + a)x + bx3, k > 0, ⇒ x = −kx

−bx3 is a damping term. Why cancel it?

u = −(k + a)x, k > 0, ⇒ x = −kx − bx3

Which design is better?

– p. 10/??

Example

x1 = x2

x2 = −h(x1) + u

h(0) = 0 and x1h(x1) > 0, ∀ x1 6= 0

Feedback Linearization:

u = h(x1) − (k1x1 + k2x2)

With y = x2, the system is passive with

V =

∫ x1

0

h(z) dz + 1

2x2

2

V = h(x1)x1 + x2x2 = yu

– p. 11/??

The control

u = −σ(x2), σ(0) = 0, x2σ(x2) > 0 ∀ x2 6= 0

creates a feedback connection of two passive systems withstorage function V

V = −x2σ(x2)

x2(t) ≡ 0 ⇒ x2(t) ≡ 0 ⇒ h(x1(t)) ≡ 0 ⇒ x1(t) ≡ 0

Asymptotic stability of the origin follows from the invarianceprinciple

Which design is better? (Read Example 13.20)

– p. 12/??


Stabilization

Partial Feedback Linearization

– p. 1/11

Consider the nonlinear system

x = f(x) + G(x)u [f(0) = 0]

Suppose there is a change of variables

z =

[

η

ξ

]

= T (x) =

[

T1(x)

T2(x)

]

defined for all x ∈ D ⊂ Rn, that transforms the system into

η = f0(η, ξ)

ξ = Aξ + Bγ(x)[u − α(x)]

(A, B) is controllable and γ(x) is nonsingular for all x ∈ D

– p. 2/11

u = α(x) + γ−1(x)v

η = f0(η, ξ), ξ = Aξ + Bv

Suppose the origin of η = f0(η, 0) is asymptotically stable

v = −Kξ, where (A − BK) is Hurwitz

Lemma 13.1: The origin of

η = f0(η, ξ), ξ = (A − BK)ξ

is asymptotically stable if the origin of η = f0(η, 0) isasymptotically stable

Proof: V (η, ξ) = V1(η)+k√

ξT Pξ

– p. 3/11

If the origin of η = f0(η, 0) is globally asymptotically stable,will the origin of

η = f0(η, ξ), ξ = (A − BK)ξ

be globally asymptotically stable? In general NoExample

η = −η + η2ξ, ξ = v

The origin of η = −η is globally exponentially stable, butthe origin of

η = −η + η2ξ, ξ = −kξ, k > 0

is not globally asymptotically stable. The region ofattraction is ηξ < 1 + k

– p. 4/11

Example

η = − 12(1 + ξ2)η

3, ξ1 = ξ2, ξ2 = v

The origin of η = − 12η3 is globally asymptotically stable

v = −k2ξ1−2kξ2def= −Kξ ⇒ A−BK =

[

0 1

−k2 −2k

]

The eigenvalues of (A − BK) are −k and −k

e(A−BK)t =

(1 + kt)e−kt te−kt

−k2te−kt (1 − kt)e−kt

– p. 5/11

Peaking Phenomenon:

maxt

k2te−kt =k

e→ ∞ as k → ∞

ξ1(0) = 1, ξ2(0) = 0 ⇒ ξ2(t) = −k2te−kt

η = − 12

(

1 − k2te−kt)

η3, η(0) = η0

η2(t) =η2

0

1 + η20[t + (1 + kt)e−kt − 1]

If η20 > 1, the system will have a finite escape time if k is

chosen large enough

– p. 6/11

Lemma 13.2: The origin of

η = f0(η, ξ), ξ = (A − BK)ξ

is globally asymptotically stable if the system η = f0(η, ξ)is input-to-state stable

Proof: UseLemma 4.7: If x1 = f1(x1, x2) is ISS and the origin ofx2 = f2(x2) is globally asymptotically stable, then theorigin of

x1 = f1(x1, x2), x2 = f2(x2)

is globally asymptotically stable

– p. 7/11

u = α(x) − γ−1(x)KT2(x)

What is the effect of uncertainty in α, γ, and T2?

Let α(x), γ(x), and T2(x) be nominal models of α(x),γ(x), and T2(x)

u = α(x) − γ−1(x)KT2(x)

η = f0(η, ξ), ξ = (A − BK)ξ + Bδ(z)

δ = γ[α − α + γ−1KT2 − γ−1KT2]

– p. 8/11

Lemma 13.4

If ‖δ(z)‖ ≤ ε for all z and η = f0(η, ξ) is input-to-statestable, then the state z is globally ultimately bounded bya class K function of ε

If ‖δ(z)‖ ≤ k‖z‖ in some neighborhood of z = 0, withsufficiently small k, and the origin of η = f0(η, 0) isexponentially stable, then z = 0 is an exponentiallystable equilibrium point of the system

η = f0(η, ξ), ξ = (A − BK)ξ + Bδ(z)

– p. 9/11

Proof–First Part: As in Lemma 13.3

‖ξ(t)‖≤cε, ∀ t ≥ t0

‖η(t)‖ ≤ β0(‖η(t0)‖, t − t0) + γ0(supt≥t0

‖ξ(t)‖)

‖η(t)‖ ≤ β0(‖η(t0)‖, t − t0) + γ0(cε)

Proof–Second Part:c1‖η‖2 ≤ V1(η) ≤ c2‖η‖2

∂V1

∂ηf0(η, 0) ≤ −c3‖η‖2

∥

∥

∥

∥

∂V1

∂η

∥

∥

∥

∥

≤ c4‖η‖

– p. 10/11

V (z) = bV1(η) + ξT Pξ

V ≤ −

[

‖η‖

‖ξ‖

]T

Q

[

‖η‖

‖ξ‖

]

Q =

[

bc3 −(k‖PB‖ + bc4L/2)

−(k‖PB‖ + bc4L/2) 1 − 2k‖PB‖

]

b = k

Q is positive definite for sufficiently small k

– p. 11/11


Stabilization

Backstepping

– p. 1/??

η = f(η) + g(η)ξ

ξ = u, η ∈ Rn, ξ, u ∈ R

Stabilize the origin using state feedback

View ξ as “virtual” control input to

η = f(η) + g(η)ξ

Suppose there is ξ = φ(η) that stabilizes the origin of

η = f(η) + g(η)φ(η)

∂V

∂η[f(η) + g(η)φ(η)] ≤ −W (η), ∀ η ∈ D

– p. 2/??

z = ξ − φ(η)

η = [f(η) + g(η)φ(η)] + g(η)z

z = u −∂φ

∂η[f(η) + g(η)ξ]

u =∂φ

∂η[f(η) + g(η)ξ] + v

η = [f(η) + g(η)φ(η)] + g(η)z

z = v

– p. 3/??

Vc(η, ξ) = V (η) + 1

2z2

Vc =∂V

∂η[f(η) + g(η)φ(η)] +

∂V

∂ηg(η)z + zv

≤ −W (η) +∂V

∂ηg(η)z + zv

v = −∂V

∂ηg(η) − kz, k > 0

Vc ≤ −W (η) − kz2

– p. 4/??

Example

x1 = x2

1− x3

1+ x2, x2 = u

x1 = x2

1− x3

1+ x2

x2 = φ(x1) = −x2

1− x1 ⇒ x1 = −x1 − x3

1

V (x1) = 1

2x2

1⇒ V = −x2

1− x4

1, ∀ x1 ∈ R

z2 = x2 − φ(x1) = x2 + x1 + x2

1

x1 = −x1 − x3

1+ z2

z2 = u + (1 + 2x1)(−x1 − x3

1+ z2)

– p. 5/??

Vc(x) = 1

2x2

1+ 1

2z2

2

Vc = x1(−x1 − x3

1+ z2)

+ z2[u + (1 + 2x1)(−x1 − x3

1+ z2)]

Vc = −x2

1− x4

1

+ z2[x1 + (1 + 2x1)(−x1 − x3

1+ z2) + u]

u = −x1 − (1 + 2x1)(−x1 − x3

1+ z2) − z2

Vc = −x2

1− x4

1− z2

2

Vc = −x2

1− x4

1− (x2 + x1 + x2

1)2

– p. 6/??

Example

x1 = x2

1− x3

1+ x2, x2 = x3, x3 = u

x1 = x2

1− x3

1+ x2, x2 = x3

x3 = −x1 − (1 + 2x1)(−x1 − x3

1+ z2) − z2

def= φ(x1, x2)

V (x) = 1

2x2

1+ 1

2z2

2, V = −x2

1− x4

1− z2

2

z3 = x3 − φ(x1, x2)

x1 = x2

1− x3

1+ x2, x2 = φ(x1, x2) + z3

z3 = u −∂φ

∂x1

(x2

1− x3

1+ x2) −

∂φ

∂x2

(φ + z3)

– p. 7/??

Vc = V + 1

2z2

3

Vc =∂V

∂x1

(x2

1− x3

1+ x2) +

∂V

∂x2

(z3 + φ)

+ z3

[

u −∂φ

∂x1

(x2

1− x3

1+ x2) −

∂φ

∂x2

(z3 + φ)

]

Vc = −x2

1− x4

1− (x2 + x1 + x2

1)2

+z3

[

∂V

∂x2

−∂φ

∂x1

(x2

1− x3

1+ x2) −

∂φ

∂x2

(z3 + φ) + u

]

u = −∂V

∂x2

+∂φ

∂x1

(x2

1− x3

1+ x2) +

∂φ

∂x2

(z3 + φ) − z3

– p. 8/??

η = f(η) + g(η)ξ

ξ = fa(η, ξ) + ga(η, ξ)u, ga(η, ξ) 6= 0

u =1

ga(η, ξ)[v − fa(η, ξ)]

η = f(η) + g(η)ξ

ξ = v

– p. 9/??

Strict-Feedback Form

x = f0(x) + g0(x)z1

z1 = f1(x, z1) + g1(x, z1)z2

z2 = f2(x, z1, z2) + g2(x, z1, z2)z3

...zk−1 = fk−1(x, z1, . . . , zk−1) + gk−1(x, z1, . . . , zk−1)zk

zk = fk(x, z1, . . . , zk) + gk(x, z1, . . . , zk)u

gi(x, z1, . . . , zi) 6= 0 for 1 ≤ i ≤ k

– p. 10/??

Exampleη = −η + η2ξ, ξ = u

η = −η + η2ξ

ξ = 0 ⇒ η = −η

V0 = 1

2η2 ⇒ V0 = −η2, ∀ η ∈ R

V = 1

2(η2 + ξ2)

V = η(−η + η2ξ) + ξu = −η2 + ξ(η3 + u)

u = −η3 − kξ, k > 0

V = −η2 − kξ2 Global stabilization

– p. 11/??


Stabilization

Passivity-Based Control

– p. 1/??

x = f(x, u), y = h(x)

f(0, 0) = 0

uT y ≥ V =∂V

∂xf(x, u)

Theorem 14.4: If the system is

(1) passive with a radially unbounded positive definitestorage function and

(2) zero-state observable,

then the origin can be globally stabilized by

u = −φ(y), φ(0) = 0, yT φ(y) > 0 ∀ y 6= 0

– p. 2/??

Proof:

V =∂V

∂xf(x, −φ(y)) ≤ −yT φ(y) ≤ 0

V (x(t)) ≡ 0 ⇒ y(t) ≡ 0 ⇒ u(t) ≡ 0 ⇒ x(t) ≡ 0


A given system may be made passive by

(1) Choice of output,

(2) Feedback,

or both

– p. 3/??

Choice of Output

x = f(x) + G(x)u,∂V

∂xf(x) ≤ 0, ∀ x

No output is defined. Choose the output as

y = h(x)def=

[

∂V

∂xG(x)

]T

V =∂V

∂xf(x) +

∂V

∂xG(x)u ≤ yT u

Check zero-state observability

– p. 4/??

Examplex1 = x2, x2 = −x3

1+ u

V (x) = 1

4x4

1+ 1

2x2

2

With u = 0 V = x3

1x2 − x2x

3

1= 0

Take y =∂V

∂xG =

∂V

∂x2

= x2

Is it zero-state observable?

with u = 0, y(t) ≡ 0 ⇒ x(t) ≡ 0

u = −kx2 or u = −(2k/π) tan−1(x2) (k > 0)

– p. 5/??

Feedback Passivation


x = f(x) + G(x)u, y = h(x)

is equivalent to a passive system if there is

u = α(x) + β(x)v

such that

x = f(x) + G(x)α(x) + G(x)β(x)v, y = h(x)

is passive

– p. 6/??

Theorem [31]: The system

x = f(x) + G(x)u, y = h(x)

is locally equivalent to a passive system (with a positivedefinite storage function) if it has relative degree one atx = 0 and the zero dynamics have a stable equilibriumpoint at the origin with a positive definite Lyapunov function

Example: m-link Robot Manipulator

M(q)q + C(q, q)q + Dq + g(q) = u

M = MT > 0, (M − 2C)T = −(M − 2C), D = DT ≥ 0

– p. 7/??

Stabilize the system at q = qr

e = q − qr, e = q

M(q)e + C(q, q)e + De + g(q) = u

(e = 0, e = 0) is not an open-loop equilibrium point

u = g(q) − φp(e) + v, [φp(0) = 0, eT φp(e) > 0 ∀e 6= 0]

M(q)e + C(q, q)e + De + φp(e) = v

V = 1

2eT M(q)e +

∫ e

0

φTp (σ) dσ

V = 1

2eT (M−2C)e−eT De−eT φp(e)+eT v+φT

p (e)e ≤ eT v

y = e

– p. 8/??

Is it zero-state observable? Set v = 0

e(t) ≡ 0 ⇒ e(t) ≡ 0 ⇒ φp(e(t)) ≡ 0 ⇒ e(t) ≡ 0

v = −φd(e), [φd(0) = 0, eT φd(e) > 0 ∀e 6= 0]

u = g(q) − φp(e) − φd(e)

Special case:

u = g(q) − Kpe − Kde, Kp = KTp > 0, Kd = KT

d > 0

– p. 9/??

How does passivity-based control compare with feedbacklinearization?

Example 13.20

x1 = x2, x2 = −h(x1) + u

h(0) = 0, x1h(x1) > 0, ∀ x1 6= 0

Feedback linearization:

u = h(x1) − (k1x1 + k2x2)

x =

[

0 1

−k1 −k2

]

x

– p. 10/??

Passivity-based control:

V =

∫ x1

0

h(z) dz + 1

2x2

2

V = x2h(x1) − x2h(x1) + x2u = x2u

Take y = x2

With u = 0, y(t) ≡ 0 ⇒ h(x1(t)) ≡ 0 ⇒ x1(t) ≡ 0

u = −σ(x2), [σ(0) = 0, yσ(y) > 0 ∀ y 6= 0]

x1 = x2, x2 = −h(x1) − σ(x2)

– p. 11/??

Linearization:[

0 1

−h′(0) −k

]

, k = σ′(0)

s2 + ks + h′(0) = 0

Sketch the root locus as k varies from zero to infinity

One of the two roots cannot be moved to the left ofRe[s] = −

√

h′(0)

– p. 12/??

Cascade Connection:

z = fa(z) + F (z, y)y, x = f(x) + G(x)u, y = h(x)

fa(0) = 0, f(0) = 0, h(0) = 0

∂V

∂xf(x) +

∂V

∂xG(x)u ≤ yT u

∂W

∂zfa(z) ≤ 0

U(z, x) = W (z) + V (x)

U ≤∂W

∂zF (z, y)y + yT u = yT

[

u +

(

∂W

∂zF (z, y)

)T]

– p. 13/??

u = −

(

∂W

∂zF (z, y)

)T

+ v ⇒ U ≤ yT v

The system

z = fa(z) + F (z, y)y

x = f(x) − G(x)

(

∂W

∂zF (z, y)

)T

+ G(x)v

y = h(x)

with input v and output y is passive with U as the storagefunction

Read Examples 14.17 and 14.18

– p. 14/??


Stabilization

Control Lyapunov Functions

– p. 1/12

x = f(x) + g(x)u, f(0) = 0, x ∈ Rn, u ∈ R

Suppose there is a continuous stabilizing state feedbackcontrol u = ψ(x) such that the origin of

x = f(x) + g(x)ψ(x)

is asymptotically stable

By the converse Lyapunov theorem, there is V (x) such that

∂V

∂x[f(x) + g(x)ψ(x)] < 0, ∀ x ∈ D, x 6= 0

If u = ψ(x) is globally stabilizing, then D = Rn and V (x)is radially unbounded

– p. 2/12

∂V

∂x[f(x) + g(x)ψ(x)] < 0, ∀ x ∈ D, x 6= 0

∂V

∂xg(x) = 0 for x ∈ D, x 6= 0 ⇒ ∂V

∂xf(x) < 0

Since ψ(x) is continuous and ψ(0) = 0, given any ε > 0,∃ δ > 0 such that if x 6= 0 and ‖x‖ < δ, there is u with‖u‖ < ε such that

∂V

∂x[f(x) + g(x)u] < 0

Small Control Property

– p. 3/12

Definition: A continuously differentiable positive definitefunction V (x) is a Control Lyapunov Function (CLF) for thesystem x = f(x) + g(x)u if

∂V

∂xg(x) = 0 for x ∈ D, x 6= 0 ⇒ ∂V

∂xf(x) < 0 (∗)

it satisfies the small control property

It is a Global Control Lyapunov Function if it is radiallyunbounded and (∗) holds with D = Rn

The system x = f(x) + g(x)u is stabilizable by acontinuous state feedback control only if it has a CLFIs it sufficient?

– p. 4/12

Theorem: Let V (x) be a CLF for x = f(x) + g(x)u, thenorigin is stabilizable by u = ψ(x), where

ψ(x) =

−∂V

∂xf+

q

(∂V

∂xf)

2

+(∂V

∂xg)

4

(∂V

∂xg)

, if ∂V∂xg 6= 0

0, if∂V∂xg = 0

The function ψ(x) is continuous for all x ∈ D0 (aneighborhood of the origin) including x = 0. If f and g aresmooth, then ψ is smooth for x 6= 0. If V is a global CLF,then the control u = ψ(x) is globally stabilizing

Sontag’s Formula

– p. 5/12

Proof: For properties of ψ, see Section 9.4 of [88]

∂V

∂x[f(x) + g(x)ψ(x)]

If∂V

∂xg(x) = 0, V =

∂V

∂xf(x) < 0 for x 6= 0

If∂V

∂xg(x) 6= 0

V = ∂V∂xf −

[

∂V∂xf +

√

(

∂V∂xf

)2

+(

∂V∂xg)4

]

= −√

(

∂V∂xf

)2

+(

∂V∂xg)4

< 0 for x 6= 0

– p. 6/12

How can we find a CLF?

If we know of any stabilizing control with a correspondingLyapunov function V , then V is a CLF

Feedback Linearization

x = f(x) +G(x)u, z = T (x), z = (A−BK)z

P (A−BK) + (A−BK)TP = −Q, Q = QT > 0

V = zTPz = T T (x)PT (x) is a CLF

Backstepping

– p. 7/12

Example:x = ax− bx3 + u, a, b > 0


u = −ax+ bx3 − kx (k > 0)

x = −kxV (x) = 1

2x2 is a CLF

∂V

∂xg = x,

∂V

∂xf = x(ax− bx3)

– p. 8/12

−∂V∂xf +

√

(

∂V∂xf

)2

+(

∂V∂xg)4

(

∂V∂xg)

= − x(ax− bx3) +√

x2(ax− bx3)2 + x4

x

= −ax+ bx3 − x

√

(a− bx2)2 + 1

ψ(x) = −ax+ bx3 − x

√

(a− bx2)2 + 1

Compare with−ax+ bx3 − kx

– p. 9/12

Method ExpressionFL-u −ax+ bx3 − kx

FL-CLS x = −kxCLF-u −ax+ bx3 − x

√

(a− bx2)2 + 1

CLF-CLS −x√

(a− bx2)2 + 1

Method Small |x| Large |x|FL-u (−a+ k)x bx3

FL-CLS x = −kx x = −kxCLF-u −(a+

√a2 + 1)x −ax

CLF-CLS −√a2 + 1x −bx3

– p. 10/12

Lemma: Let V (x) be a CLF for x = f(x) + g(x)u andsuppose ∂V

∂x(0) = 0. Then, Sontag’s formula has a gain

margin [12,∞); that is, u = kψ(x) is stabilizing for all k ≥ 1

2

Proof: Let

q(x) = 1

2

− ∂V

∂xf +

√

(

∂V

∂xf

)2

+

(

∂V

∂xg

)4

q(0) = 0,∂V

∂xg 6= 0 ⇒ q > 0

∂V

∂xg = 0 ⇒ q = −∂V

∂xf > 0 for x 6= 0

q(x) is positive definite

– p. 11/12

u = kψ(x) ⇒ x = f(x) + g(x)kψ(x)

V =∂V

∂xf +

∂V

∂xgkψ

∂V

∂xg = 0 ⇒ V =

∂V

∂xf < 0 for x 6= 0

∂V

∂xg 6= 0, V = −q + q +

∂V

∂xf +

∂V

∂xgkψ

q +∂V

∂xf +

∂V

∂xgkψ

= −(

k − 1

2

)

∂V

∂xf +

√

(

∂V

∂xf

)2

+

(

∂V

∂xg

)4

– p. 12/12


Stabilization

Output Feedback

– p. 1/12

In general, output feedback stabilization requires the use ofobservers. In this lecture we deal with three simple caseswhere an observer is not needed

Minimum Phase Relative Degree One Systems

Passive systems

System with Passive maps from the input to thederivative of the output

– p. 2/12

Minimum Phase Relative Degree One Systems

x = f(x) + g(x)u, y = h(x)

f(0) = 0, h(0) = 0, Lgh(x) 6= 0, ∀ x ∈ D

Normal Form:

φ(0) = 0, Lgφ(x) = 0,

[

η

y

]

=

[

φ

h(x)

]

η = f0(η, y), y = γ(x)[u− α(x)], γ(x) 6= 0

– p. 3/12

Assumptions:The origin of η = f0(η, 0) is exponentially stable

c1‖η‖2 ≤ V1(η) ≤ c2‖η‖2

∂V1

∂ηf0(η, 0) ≤ −c3‖η‖2

∣∣∣∣

∂V1

∂η

∣∣∣∣≤ c4‖η‖

‖f0(η, y) − f0(η, 0)‖ ≤ L1|y|

|α(x)γ(x)| ≤ L2‖η‖ + L3|y|

γ(x) ≥ γ0 > 0

– p. 4/12

High-Gain Feedback:

u = −ky, k > 0

η = f0(η, y), y = γ(x)[−ky − α(x)]

V (η, y) = V1(η) + 1

2y2

V =∂V1

∂ηf0(η, y) − kγ(x)y2 − α(x)γ(x)y

V =∂V1

∂ηf0(η, 0) +

∂V1

∂η[f0(η, y) − f0(η, 0)]

− kγ(x)y2 − α(x)γ(x)y

V ≤ −c3‖η‖2 + c4L1‖η‖ |y| − kγ0y2 + L2‖η‖ |y| + L3y

2

– p. 5/12

V ≤ −c3‖η‖2 + c4L1‖η‖ |y| − kγ0y2 + L2‖η‖ |y| + L3y

2

V ≤ −

[

‖η‖

|y|

]T [

c3 −1

2(L2 + c4L1)

−1

2(L2 + c4L1) (kγ0 − L3)

]

︸︷︷︸

Q

[

‖η‖

|y|

]

det(Q) = c3(kγ0 − L3) − 1

4(L2 + c4L1)

2

det(Q) > 0 for k >1

c3γ0

[c3L3 + 1

4(L2 + c4L1)

2]

The origin of the closed-loop system is exponentially stable

If the assumptions hold globally, it is globally exp. stable

– p. 6/12

Passive Systems

Suppose the system

x = f(x, u), y = h(x, u)

is passive (with a positive definite storage function) andzero-state observable

Then, it can be stabilized by

u = −ψ(y), ψ(0) = 0, yTψ(y) > 0, ∀ y 6= 0

V ≤ uTy = −yTψ(y) ≤ 0

V = 0 ⇒ y = 0 ⇒ u = 0

y(t) ≡ 0 ⇒ x(t) ≡ 0

– p. 7/12

Systems with Passive Maps from u to y

x = f(x, u), y = h(x)

f(0, 0) = 0, h(0) = 0, h is cont. diff.

Suppose the system

x = f(x, u), y =∂h

∂xf(x, u)

def= h(x, u)

is passive (with a positive definite storage function) andzero-state observable

V ≤ uT y

With u = 0, y(t) ≡ 0 ⇒ x(t) ≡ 0

– p. 8/12

- h - Plant -

bs

s+aψ(·)

6

+

−

u y

z

- h - Plant - s -

b

s+aψ(·)

6

+

−

u y y

z

– p. 9/12

For 1 ≤ i ≤ m

zi is the output ofbis

s+ aidriven by yi

ui = −ψi(zi)

ai, bi > 0, ψi(0) = 0, ziψi(zi) > 0 ∀ zi 6= 0

zi = −aizi + biyi

Use

Vc(x, z) = V (x) +

m∑

i=1

1

bi

∫ zi

0

ψi(σ) dσ

to prove asymptotic stability of the origin of the closed-loopsystem

– p. 10/12

Example: Stabilize the pendulum

mℓθ +mg sin θ = u

at θ = δ1 using feedback from θ

x1 = θ − δ1, x2 = θ

x1 = x2, x2 = 1

mℓ[−mg sin θ + u]

u = mg sin θ − kpx1 + v, kp > 0

x1 = x2, x2 = − kp

mℓx1 + 1

mℓv

y = x1, y = x2

– p. 11/12

V = 1

2kpx

21 + 1

2mℓx2

2

V = kpx1x2 +mℓx2

[

− kp

mℓx1 + 1

mℓv]

V = x2v = yv

With v = 0, x2(t) ≡ 0 ⇒ x1(t) ≡ 0

- bss+a

-y z

ξ = −aξ + y, z = b(−aξ + y)

v = −kdz, kd > 0

u = mg sin θ − kp(θ − δ1) − kdz

– p. 12/12


Robust Stabilization

Sliding Mode Control

– p. 1/17

Example

x1 = x2 x2 = h(x) + g(x)u, g(x) ≥ g0 > 0

Sliding Manifold (Surface):

s = a1x1 + x2 = 0

s(t) ≡ 0 ⇒ x1 = −a1x1

a1 > 0 ⇒ limt→∞

x1(t) = 0

How can we bring the trajectory to the manifold s = 0?

How can we maintain it there?

– p. 2/17

s = a1x1 + x2 = a1x2 + h(x) + g(x)u

Suppose∣∣∣∣

a1x2 + h(x)

g(x)

∣∣∣∣≤ (x)

V = 12s2

V = ss = s[a1x2+h(x)]+g(x)su ≤ g(x)|s|(x)+g(x)su

β(x) ≥ (x) + β0, β0 > 0

s > 0, u = −β(x)

V ≤ g(x)|s|(x) − g(x)β(x)|s|

V ≤ g(x)|s|(x) − g(x)((x) + β0)|s| = −g(x)β0|s|

– p. 3/17

s < 0, u = β(x)

V ≤ g(x)|s|(x) + g(x)su = g(x)|s|(x) − g(x)β(x)|s|

V ≤ g(x)|s|(x) − g(x)((x) + β0)|s| = −g(x)β0|s|

sgn(s) =

1, s > 0

−1, s < 0

u = −β(x) sgn(s)

V ≤ −g(x)β0|s| ≤ −g0β0|s|

V ≤ −g0β0

√2V

– p. 4/17

V ≤ −g0β0

√2V

dV√

V≤ −g0β0

√2 dt

2√

V

∣∣∣∣

V (s(t))

V (s(0))

≤ −g0β0

√2 t

√

V (s(t)) ≤√

V (s(0)) − g0β01

√2

t

|s(t)| ≤ |s(0)| − g0β0 t

s(t) reaches zero in finite time

Once on the surface s = 0, the trajectory cannot leave it

– p. 5/17

s=0

What is the region of validity?

– p. 6/17

x1 = x2 x2 = h(x) − g(x)β(x)sgn(s)

x1 = −a1x1 + s s = a1x2 + h(x) − g(x)β(x)sgn(s)

ss ≤ −g0β0|s|, if β(x) ≥ (x) + β0

V1 = 12x2

1

V1 = x1x1 = −a1x21 + x1s ≤ −a1x

21 + |x1|c ≤ 0

∀ |s| ≤ c and |x1| ≥ c

a1

Ω =

|x1| ≤c

a1, |s| ≤ c

Ω is positively invariant if∣∣∣∣

a1x2 + h(x)

g(x)

∣∣∣∣≤ (x) over Ω

– p. 7/17

Ω =

|x1| ≤c

a1, |s| ≤ c

-

6

HHHHHHHHHHHHHHHHHH

HH

HH

HH

HH

HH

HH

HH

HH

HH

x1

x2s = 0

c/a1

c

∣∣∣∣

a1x2 + h(x)

g(x)

∣∣∣∣≤ k1 < k, ∀ x ∈ Ω

u = −k sgn(s)

– p. 8/17

Chattering

BB

BBB

BB

BBB

BB

BBB

HHYHHHHH

HHYHHH

Sliding manifold

a

s < 0 s > 0

How can we reduce or eliminate chattering?

– p. 9/17

Reduce the amplitude of the signum function

s = a1x2 + h(x) + g(x)u

u = − [a1x2 + h(x)]

g(x)+ v

s = δ(x) + g(x)v

δ(x) = a1

[

1 −g(x)

g(x)

]

x2 + h(x) −g(x)

g(x)h(x)

∣∣∣∣

δ(x)

g(x)

∣∣∣∣≤ (x), β(x) ≥ (x) + β0

v = −β(x) sgn(s)

– p. 10/17

Replace the signum function by a high-slope saturationfunction

u = −β(x) sat

(s

ε

)

sat(y) =

y, if |y| ≤ 1

sgn(y), if |y| > 1

-

6

−1

1

y

sgn(y)

-

6

−1

1

yε

sat(y

ε

)

– p. 11/17

How can we analyze the system?

For |s| ≥ ε, u = −β(x) sgn(s)

With c ≥ ε

Ω =

|x1| ≤ ca1

, |s| ≤ c

is positively invariant

The trajectory reaches the boundary layer |s| ≤ εinfinite time

The boundary layer is positively invariant

– p. 12/17

Inside the boundary layer:

x1 = −a1x1 + s s = a1x2 + h(x) − g(x)β(x)s

ε

x1x1 ≤ −a1x21 + |x1|ε

0 < θ < 1

x1x1 ≤ −(1 − θ)a1x21, ∀ |x1| ≥ ε

θa1

The trajectories reach the positively invariant set

Ωε = |x1| ≤ε

θa1, |s| ≤ ε

in finite time– p. 13/17

What happens inside Ωε?

Find the equilibrium points

0 = −a1x1 + s = x2, 0 = a1x2 + h(x) − g(x)β(x)s

ε

φ(x1) =h(x)

a1g(x)β(x)

∣∣∣∣x2=0

x1 = εφ(x1)

Suppose x1 = εφ(x1) has an isolated root x1 = εk1

h(0) = 0 ⇒ x1 = 0

– p. 14/17

z1 = x1 − x1, z2 = s − a1x1

x2 = −a1x1 +s = −a1(x1 − x1)+s−a1x1 = −a1z1 +z2

z1 = −a1x1 + s = −a1z1 + z2

z2 = a1x2 + h(x) − g(x)β(x)s

ε

= a1(z2 − a1z1) + h(x) − g(x)β(x)z2 + a1x1

ε

z2 = ℓ(z) − g(x)β(x)z2

ε

ℓ(z) = a1(z2 − a1z1) + a1g(x)β(x)

[h(x)

a1g(x)β(x)−

x1

ε

]

– p. 15/17

z1 = −a1z1 + z2, z2 = ℓ(z) − g(x)β(x)z2

ε

ℓ(0) = 0, |ℓ(z)| ≤ ℓ1|z1| + ℓ2|z2|g(x)β(x) ≥ g0β0

V = 12z21 + 1

2z22

V = z1(−a1z1 + z2) + z2

[

ℓ(z) − g(x)β(x)z2

ε

]

V ≤ −a1z21 + (1 + ℓ1)|z1| |z2| + ℓ2z

22 − g0β0

εz22

– p. 16/17

V ≤ −a1z21 + (1 + ℓ1)|z1| |z2| + ℓ2z

22 − g0β0

εz22

V ≤ −[

|z1||z2|

]T [

a1 −12(1 + ℓ1)

−12(1 + ℓ1)

(g0β0

ε − ℓ2

)

]

︸︷︷︸

Q

[

|z1||z2|

]

det(Q) = a1

(g0β0

ε− ℓ2

)

− 14(1 + ℓ1)

2

h(0) = 0 ⇒ limt→∞

x(t) = 0

h(0) 6= 0 ⇒ limt→∞

x(t) =

[

x1

0

]

Read Section 14.1.1

– p. 17/17




– p. 1/15

Regular Form:

η = fa(η, ξ)

ξ = fb(η, ξ) + g(η, ξ)u + δ(t, η, ξ, u)

η ∈ Rn−1, ξ ∈ R, u ∈ R

fa(0, 0) = 0, fb(0, 0) = 0, g(η, ξ) ≥ g0 > 0

Sliding Manifold:

s = ξ − φ(η) = 0, φ(0) = 0

s(t) ≡ 0 ⇒ η = fa(η, φ(η))

Design φ s.t. the origin of η = fa(η, φ(η)) is asymp. stable

– p. 2/15

s = fb(η, ξ) −∂φ

∂ηfa(η, ξ) + g(η, ξ)u + δ(t, η, ξ, u)

u = −1

g

(

fb −∂φ

∂ηfa

)

+ v or u = v

u = −L

(

fb −∂φ

∂ηfa

)

+ v, L =1

gor L = 0

s = g(η, ξ)v + ∆(t, η, ξ, v)

∆ = fb −∂φ

∂ηfa + δ − gL

(

fb −∂φ

∂ηfa

)

∣

∣

∣

∣

∆(t, η, ξ, v)

g(η, ξ)

∣

∣

∣

∣

≤ (η, ξ) + κ0|v|

– p. 3/15

∣

∣

∣

∣

∆(t, η, ξ, v)

g(η, ξ)

∣

∣

∣

∣

≤ (η, ξ) + κ0|v|

(η, ξ) ≥ 0, 0 ≤ κ0 < 1 (Known)

ss = sgv + s∆ ≤ sgv + |s| |∆|

ss ≤ g[sv + |s|( + κ0|v|)]

v = −β(η, ξ) sgn(s)

β(η, ξ) ≥(η, ξ)

1 − κ0

+ β0, β0 > 0

ss ≤ g[−β|s| + |s| + κ0β|s|] = g[−β(1 − κ0)|s| + |s| ]

ss ≤ g[−|s| − (1 − κ0)β0|s| + |s| ]

– p. 4/15

ss ≤ −g(η, ξ)(1 − κ0)β0|s| ≤ −g0β0(1 − κ0)|s|

v = −β(x) sat

(

s

ε

)

, ε > 0

ss ≤ −g0β0(1 − κ0)|s|, for |s| ≥ ε

The trajectory reaches the boundary layer |s| ≤ ε in finitetime and remains inside thereafter

Study the behavior of η

η = fa(η, φ(η) + s)

What do we know about this system and what do we need?

– p. 5/15

α1(‖η‖) ≤ V (η) ≤ α2(‖η‖)

∂V

∂ηfa(η, φ(η) + s) ≤ −α3(‖η‖), ∀ ‖η‖ ≥ γ(|s|)

|s| ≤ c ⇒ V ≤ −α3(‖η‖), for ‖η‖ ≥ γ(c)

α(r) = α2(γ(r))

V (η) ≥ α(c) ⇔ V (η) ≥ α2(γ(c)) ⇒ α2(‖η‖) ≥ α2(γ(c))

⇒ ‖η‖ ≥ γ(c) ⇒ V ≤ −α3(‖η‖) ≤ −α3(γ(c))

The set V (η) ≤ c0 with c0 ≥ α(c) is positively invariant

Ω = V (η) ≤ c0 × |s| ≤ c, with c0 ≥ α(c)

– p. 6/15

α(.)

α(ε)

α(c)

c0

ε c

V

|s|

Ω = V (η) ≤ c0 × |s| ≤ c, with c0 ≥ α(c)

is positively invariant and all trajectories starting in Ω reachΩε = V (η) ≤ α(ε) × |s| ≤ ε in finite time

– p. 7/15

Theorem 14.1: Suppose all the assumptions hold over Ω.Then, for all (η(0), ξ(0)) ∈ Ω, the trajectory (η(t), ξ(t)) isbounded for all t ≥ 0 and reaches the positively invariantset Ωε in finite time. If the assumptions hold globally andV (η) is radially unbounded, the foregoing conclusion holdsfor any initial stateTheorem 14.2: Suppose all the assumptions hold over Ω

(0) = 0, κ0 = 0

The origin of η = fa(η, φ(η)) is exponentially stale

Then there exits ε∗ > 0 such that for all 0 < ε < ε∗, theorigin of the closed-loop system is exponentially stable andΩ is a subset of its region of attraction. If the assumptionshold globally, the origin will be globally uniformlyasymptotically stable

– p. 8/15

Example

x1 = x2 + θ1x1 sin x2, x2 = θ2x22+ x1 + u

|θ1| ≤ a, |θ2| ≤ b

x2 = −kx1 ⇒ x1 = −kx1 + θ1x1 sin x2

V1 = 1

2x2

1⇒ x1x1 ≤ −kx2

1+ ax2

1

s = x2 + kx1, k > a

s = θ2x22+ x1 + u + k(x2 + θ1x1 sin x2)

u = −x1 − kx2 + v ⇒ s = v + ∆(x)

∆(x) = θ2x22+ kθ1x1 sin x2

– p. 9/15

∆(x) = θ2x22+ kθ1x1 sin x2

|∆(x)| ≤ ak|x1| + bx22

β(x) = ak|x1| + bx22+ β0, β0 > 0

u = −x1 − kx2 − β(x) sgn(s)

Will

u = −x1 − kx2 − β(x) sat

(

s

ε

)

stabilize the origin?

– p. 10/15

Example: Normal Form

η = f0(η, ξ)

ξi = ξi+1, 1 ≤ i ≤ ρ − 1


ρ−1

f h(x) u

y = ξ1

View ξρ as input to the system

η = f0(η, ξ1, · · · , ξρ−1, ξρ)

ξi = ξi+1, 1 ≤ i ≤ ρ − 2

ξρ−1 = ξρ

Design ξρ = φ(η, ξ1, · · · , ξρ−1) to stabilize the origin

– p. 11/15

s = ξρ − φ(η, ξ1, · · · , ξρ−1)

Minimum Phase Systems: The origin of η = f0(η, 0) isasymptotically stable

s = ξρ + k1ξ1 + · · · + kρ−1ξρ−1

η = f0(η, ξ1, · · · , ξρ−1, −k1ξ1 − · · · − kρ−1ξρ−1)

ξ1

...

ξρ−1

=

1. . .

1

−k1 −kρ−1

ξ1

...

ξρ−1

– p. 12/15

Multi-Input Systems

η = fa(η, ξ)

ξ = fb(η, ξ) + G(η, ξ)E(η, ξ)u + δ(t, η, ξ, u)

η ∈ Rn−p, ξ ∈ Rp, u ∈ Rp

fa(0, 0) = 0, fb(0, 0) = 0, det(G) 6= 0, det(E) 6= 0

G = diag[g1, g2, · · · , gm], gi(η, ξ) ≥ g0 > 0

Design φ s.t. the origin of η = fa(η, φ(η)) is asymp. stable

s = ξ − φ(η)

s = fb(η, ξ)−∂φ

∂ηfa(η, ξ)+G(η, ξ)E(η, ξ)u+δ(t, η, ξ, u)

– p. 13/15

s = fb(η, ξ)−∂φ

∂ηfa(η, ξ)+G(η, ξ)E(η, ξ)u+δ(t, η, ξ, u)

u = E−1

−L

[

fb −∂φ

∂ηfa

]

+ v

, L = G−1 or L = 0

si = gi(η, ξ)vi + ∆i(t, η, ξ, v), 1 ≤ i ≤ p∣

∣

∣

∣

∆i(t, η, ξ, v)

gi(η, ξ)

∣

∣

∣

∣

≤ (η, ξ) + κ0 max1≤i≤p

|vi|, ∀ 1 ≤ i ≤ p

(η, ξ) ≥ 0, 0 ≤ κ0 < 1 (Known)

β(x) ≥(x)

1 − κ0

+ β0, β0 > 0

– p. 14/15

sisi = sigivi + si∆i ≤ gisivi + |si|[ + κ0 max1≤i≤p

|vi|]

vi = −β sgn(si), 1 ≤ i ≤ p

sisi ≤ gi[−β + + κ0β]|si|

= gi[−(1 − κ0)β + ]|si|

≤ gi[− − (1 − κ0)β0 + ]|si|

≤ −g0β0(1 − κ0)|si|

Now use

vi = −β sat

(

si

ε

)

, 1 ≤ i ≤ p

Read Theorem 14.1 and 14.2 in the textbook

– p. 15/15



Lyapunov Redesign &Backstepping

– p. 1/??

Lyapunov Redesign (Min-max control)

x = f(x) +G(x)[u+ δ(t, x, u)], x ∈ Rn, u ∈ Rp

Nominal Model: x = f(x) +G(x)u

Stabilizing Control: u = ψ(x)

∂V

∂x[f(x) +G(x)ψ(x)] ≤ −W (x), ∀ x ∈ D, W is p.d.

u = ψ(x) + v

‖δ(t, x, ψ(x) + v)‖ ≤ ρ(x) + κ0‖v‖, 0 ≤ κ0 < 1

x = f(x) +G(x)ψ(x) +G(x)[v + δ(t, x, ψ(x) + v)]

V =∂V

∂x(f +Gψ) +

∂V

∂xG(v + δ)

– p. 2/??

wT =∂V

∂xG

V ≤ −W (x) + wT v + wT δ

wT v+wT δ ≤ wT v+‖w‖ ‖δ‖ ≤ wT v+‖w‖[ρ(x)+κ0‖v‖]

v = −η(x)w

‖w‖

(w

‖w‖= sgn(w) for p = 1

)

wT v + wT δ ≤ −η‖w‖ + ρ‖w‖ + κ0η‖w‖

= −η(1 − κ0)‖w‖ + ρ‖w‖

η(x) ≥ρ(x)

(1 − κ0)⇒ wT v + wT δ ≤ 0 ⇒ V ≤ −W (x)

– p. 3/??

v =

−η(x) w‖w‖

, if η(x)‖w‖ ≥ ε

−η2(x)wε , if η(x)‖w‖ < ε

η(x)‖w‖ ≥ ε ⇒ V ≤ −W (x)

For η(x)‖w‖ < ε

V ≤ −W (x) + wT[

−η2 ·w

ε+ δ

]

≤ −W (x) −η2

ε‖w‖2 + ρ‖w‖ + κ0‖w‖‖v‖

= −W (x) −η2

ε‖w‖2 + ρ‖w‖ +

κ0η2

ε‖w‖2

– p. 4/??

V ≤ −W (x) + (1 − κ0)

(

−η2

ε‖w‖2 + η‖w‖

)

−y2

ε+ y ≤

ε

4, for y ≥ 0

V ≤ −W (x) + ε(1 − κ0)

4, ∀ x ∈ D

Theorem 14.3: x(t) is uniformly ultimately bounded by aclass K function of ε. If the assumptions hold globally andV is radially unbounded, then x(t) globally uniformlyultimately boundedCorollary 14.1: If ρ(0) = 0 and η(x) ≥ η0 > 0 we canrecover uniform asymptotic stability

– p. 5/??

Example: Pendulum with horizontal acceleration ofsuspension point

m[

ℓθ + A(t) cos θ]

= T/ℓ−mg sin θ

Stabilize the pendulum at θ = π

x1 = θ − π, x2 = θ, a =g

ℓ, c =

1

mℓ2, h(t) =

A(t)

ℓ

x1 = x2, x2 = a sinx1 + cu+ h(t) cosx1

Nominal Model: x1 = x2, x2 = a sinx1 + cu

ψ(x) = −

(a

c

)

sinx1 −

(1

c

)

(k1x1 + k2x2)

– p. 6/??

[

0 1

−k1 −k2

]

︸︷︷︸

Hurwitz

, V (x) = xTPx

δ =1

c

[(ac− ac

c

)

sinx1 + h(t) cosx1

−

(c− c

c

)

(k1x1 + k2x2)

]

+

(c− c

c

)

v

∣∣∣∣

c− c

c

∣∣∣∣≤ κ0,

∣∣∣∣

ac− ac

c

∣∣∣∣+

∣∣∣∣

c− c

c

∣∣∣∣

√

k21

+ k22

≤ k, |h(t)| ≤ H

|δ| ≤(k‖x‖ +H)

c+ κ0 |v|

def= ρ(x) + κ0 |v|, (κ0 < 1)

– p. 7/??

η(x) =ρ(x)

(1 − κ0), η(x) ≥

H

c(1 − κ0)

w =∂V

∂xG = 2xTP

[

0

1

]

= 2(p12x1 + p22x2)

v =

−η(x)sgn(w), if η(x)|w| ≥ ε

−η2(x)wε, if η(x)|w| < ε

u = −

(a

c

)

sinx1 −

(1

c

)

(k1x1 + k2x2) + v

Will this control stabilize the origin x = 0?

– p. 8/??

Backstepping

z1 = f1(z1) + g1(z1)z2

z2 = f2(z1, z2) + g2(z1, z2)z3...

zk−1 = fk−1(x, z1, . . . , zk−1) + gk−1(z1, . . . , zk−1)zk

zk = fk(z1, . . . , zk) + gk(z1, . . . , zk)u

gi 6= 0, 1 ≤ i ≤ k

– p. 9/??

z1 = f1 + g1z2 + δ1(z)

z2 = f2 + g2z3 + δ2(z)

...zk−1 = fk−1 + gk−1zk + δk−1(z)

zk = fk + gku+ δk(z)

|δ1(z)| ≤ ρ1(z1)

|δ2(z)| ≤ ρ2(z1, z2)

...|δk−1| ≤ ρk−1(z1, . . . , zk−1)

|δk| ≤ ρk(z1, . . . , zk)

The virtual control zi = φi(z1, . . . , zi−1) should be smooth

– p. 10/??

Example:

x1 = x2 + θ1x1 sinx2, x2 = θ2x2

2+ x1 + u

|θ1| ≤ a, |θ2| ≤ b

δ1 = θ1x1 sinx2, δ2 = θ2x2

2

x1 = x2 + θ1x1 sinx2, |θ1x1 sinx2| ≤ a|x1|

x2 = −k1x1

V1 = 1

2x2

1, V1 ≤ −(k1 − a)x2

1; Take k1 = 1 + a

z2 = x2 + (1 + a)x1

– p. 11/??

x1 = −(1 + a)x1 + θ1x1 sinx2 + z2

z2 = ψ1(x) + ψ2(x, θ) + u

ψ1 = x1 + (1 + a)x2, ψ2 = (1 + a)θ1x1 sinx2 + θ2x2

2

Vc = 1

2x2

1+ 1

2z2

2

Vc ≤ −x2

1+ z2[x1 + ψ1(x) + ψ2(x, θ) + u]

First Approach (Example 14.13):

u = −x1 − ψ1(x) − kz2, k > 0

Vc ≤ −x2

1− kz2

2+ z2ψ2(x, θ)

Restrict analysis to the compact set Ωc = Vc(x) ≤ c

– p. 12/??

ψ2 = (1 + a)θ1x1 sinx2 + θ2x2

2

|ψ2| ≤ a(1 + a)|x1| + bρ|x2|, ρ = maxx∈Ωc

|x2|

x2 = z2 − (1 + a)x1

|ψ2| ≤ (1 + a)(a+ bρ)|x1| + bρ|z2|

Vc ≤ −x2

1− kz2

2+ (1 + a)(a+ bρ)|x1| |z2| + bρz2

2

We can make V neg. def. by choosing k large enough

Can this control achieve global stabilization?

Can it achieve semiglobal stabilization?

– p. 13/??

Second Approach (Example 14.14):

u = −x1 − ψ1(x) − kz2 + v

Vc ≤ −x2

1− kz2

2+ z2[ψ2 + v]

|ψ2| ≤ a(1 + a)|x1| + bx2

2

v =

−η(x) sgn(z2), if η(x)|z2| ≥ ε

−η2(x)z2/ε, if η(x)|z2| < ε

η(x) = η0 + a(1 + a)|x1| + bx2

2, η0 > 0, ε > 0

Vc ≤ −x2

1− kz2

2+ε

4

Show that this control is globally stabilizing– p. 14/??


Tracking

Feedback Linearization &Sliding Mode Control

– p. 1/11

SISO relative-degree ρ system:

x = f(x) + g(x)u, y = h(x)

f(0) = 0, h(0) = 0

LgLi−1f h(x) = 0, for 1 ≤ i ≤ ρ − 1, LgL

ρ−1f h(x) 6= 0

Normal form:η = f0(η, ξ)

ξi = ξi+1, 1 ≤ i ≤ ρ − 1


ρ−1f h(x)u

y = ξ1

f0(0, 0) = 0

– p. 2/11

Reference signal r(t)

r(t) and its derivatives up to r(ρ)(t) are bounded for allt ≥ 0 and the ρth derivative r(ρ)(t) is a piecewisecontinuous function of t;

the signals r,. . . ,r(ρ) are available on-line.

Goal: limt→∞

[y(t) − r(t)] = 0

R =

r...

r(ρ−1)

, e =

ξ1 − r...

ξρ − r(ρ−1)

= ξ − R

– p. 3/11

η = f0(η, e + R)

e = Ace + Bc

[

Lρfh(x) + LgL

ρ−1f h(x)u − r(ρ)

]

Ac =

0 1 0 . . . 0

0 0 1 . . . 0... . . . ...... 0 1

0 . . . . . . 0 0

, Bc =

0

0...0

1

u =1

LgLρ−1f h(x)

[

−Lρfh(x) + r(ρ) + v

]

e = Ace + Bcv

– p. 4/11

v = −Ke ⇒ e = (Ac − BcK)︸︷︷︸

Hurwitz

e

limt→∞

e(t) = 0 ⇒ limt→∞

[y(t) − r(t)] = 0

e(t) is bounded ⇒ ξ(t) = e(t) + R(t) is bounded

What about η(t)?η = f0(η, ξ)

Local Tracking (small ‖η(0)‖, ‖e(0)‖, ‖R(t)‖):

Minimum Phase ⇒ The origin of η = f0(η, 0) isasymptotically stable

⇒ η is bounded for sufficiently small‖η(0)‖, ‖e(0)‖, and ‖R(t)‖

– p. 5/11

Global Tracking (large ‖η(0)‖, ‖e(0)‖, ‖R(t)‖):

What condition on η = f0(η, ξ) is needed?

Example 13.21

x1 = x2, x2 = −a sin x1 − bx2 + cu, y = x1

e1 = x1 − r, e2 = x2 − r

e1 = e2, e2 = −a sin x1 − bx2 + cu − r

u =1

c[a sin x1 + bx2 + r − k1e1 − k2e2]

e1 = e2, e2 = −k1e1 − k2e2

See simulation in the textbook

– p. 6/11


x = f(x) + g(x)[u + δ(t, x, u)], y = h(x)

Lgh(x) = · · · = LgLρ−2f h(x) = 0, LgL

ρ−1f h(x) ≥ a > 0

η = f0(η, ξ)

ξ1 = ξ2

......

ξρ−1 = ξρ


ρ−1f h(x)[u + δ(t, x, u)]

y = ξ1

e = ξ − R

– p. 7/11

η = f0(η, ξ)

e1 = e2

......

eρ−1 = eρ

eρ = Lρfh(x) + LgL

ρ−1f h(x)[u + δ(t, x, u)] − r(ρ)(t)

Sliding surface:

s = (k1e1 + · · · + kρ−1eρ−1) + eρ

s(t) ≡ 0 ⇒ eρ = −(k1e1 + · · · + kρ−1eρ−1)

– p. 8/11

η = f0(η, ξ)

e1 = e2

......

eρ−1 = −(k1e1 + · · · + kρ−1eρ−1)

Design k1 to kρ−1 such that the matrix

1. . .

1

−k1 −kρ−1

is Hurwitz

Assumption: The system f0(η, ξ) is BIBS stable

– p. 9/11

s = (k1e1 + · · · + kρ−1eρ−1) + eρ =

ρ−1∑

i=1

kiei + eρ

s =

ρ−1∑

i=1

kiei+1+Lρfh(x)+LgL

ρ−1f h(x)[u+δ(t, x, u)]−r(ρ)(t)

u = −1

LgLρ−1f h(x)

[ρ−1∑

i=1

kiei+1 + Lρfh(x) − r(ρ)(t)

]

+ v

s = LgLρ−1f h(x)v + ∆(t, x, v)

∣∣∣∣∣

∆(t, x, v)

LgLρ−1f h(x)

∣∣∣∣∣≤ (x) + κ0|v|, 0 ≤ κ0 < 1

– p. 10/11

v = −β(x) sat

(s

ε

)

, ε > 0

β(x) ≥(x)

(1 − κ0)+ β0, β0 >

What properties can we prove for this control?

– p. 11/11


Tracking

Equilibrium-to-EquilibriumTransition

– p. 1/12

η = f0(η, ξ)

ξi = ξi+1, 1 ≤ i ≤ ρ − 1

ξρ = Lρfh(x)

︸︷︷︸

fb(η,ξ)

+ LgLρ−1f h(x)

︸︷︷︸

gb(η,ξ)

u

y = ξ1

Equilibrium point:

0 = f0(η, ξ)

0 = ξi+1, 1 ≤ i ≤ ρ − 1

0 = fb(η, ξ) + gb(η, ξ)u

y = ξ1

– p. 2/12

ξ1 = y, ξi = 0 for 2 ≤ i ≤ ρ − 1

0 = f0(η, y, 0, · · · , 0), u = −fb(η, y, 0, · · · , 0)

gb(η, y, 0, · · · , 0)

Assume f0(η, y, 0, · · · , 0) has a unique solution η in thedomain of interest

η = φη(y), u = φu(y)

By assumption (and without loss of generality)

φη(0) = 0, φu(0) = 0

– p. 3/12

Goal: Move the system from equilibrium at y = 0 toequilibrium at y = y, either asymptotically or over a finitetime period

First Approach: Apply a step command

y∗(t) =

0, for t < 0

y for t ≥ 0

Is this allowed ?

Take r = y, for t ≥ 0

r(i) = 0 for i ≥ 2

– p. 4/12

η(0) = 0, e1(0) = −y, ei(0) = 0 for i ≥ 2

The shape of the transient response depends on thesolution of

e = (Ac − BcK)e

in feedback linearizationor the solution of

e1

e2...

eρ−1

=

1. . .

1

−k1 −kρ−1

e1

e2...

eρ−1

+

1

s

in sliding mode controlWhat is the impact of the reaching phase?

– p. 5/12

Second Approach: Take r(t) as the zero-state response ofa Hurwitz transfer function driven by y∗Typical Choice:

aρ

sρ + a1sρ−1 + · · · + aρ−1s + aρ

Choose the parameters a1 to aρ to shape the response of r

r(0) = 0 ⇒ e1(0) = 0 ⇒ e(0) = 0


Sliding Mode Control:

– p. 6/12

The derivatives of r are generated by the pre-filter

z =

1. . .

1

−aρ −a1

z +

aρ

y∗

r =[

1]

z

r = z1, r = z2, . . . . . . r(ρ−1) = zρ

r(ρ) = −

ρ∑

i=1

aρ−i+1zi + aρy∗

Does r(t) satisfy the assumptions imposed last lecture?

– p. 7/12

Example 13.22

x1 = x2, x2 = −10 sin x1 − x2 + 10u, y = x1

Move the pendulum from equilibrium at x1 = 0 toequilibrium at x1 = π

2

Constraint : |u(t)| ≤ 2

y∗ =

0, for t < 0π2 for t ≥ 0

Pre-Filter:1

(τs + 1)2

– p. 8/12

z =

[

0 1−1τ 2

−2τ

]

z +

[

01τ 2

]

y∗

r =[

1 0]

z

r = z1, r = z2, r =1

τ 2(y∗ − r) −

2

τz2

r(t) = π2

[

1 − e− t

τ

(1 + t

τ

)]

u = 0.1(10 sin x1 + x2 + r − k1e1 − k2e2)

– p. 9/12

0 0.5 1 1.5 20

0.5

1

1.5

2

Out

put

τ = 0.05

0 0.5 1 1.5 20

0.5

1

1.5

2

Out

put

τ = 0.25

outputreference

0 0.5 1 1.5 2

−2

−1

0

1

2

τ = 0.05

Con

trol

Time

outputreference

0 0.5 1 1.5 2

−2

−1

0

1

2

Time

Con

trol

τ = 0.25

– p. 10/12

Third Approach: Plan a trajectory (r(t), r(t), . . . , r(ρ)(t))to move from (0, 0, . . . , 0) to (y, 0, . . . , 0) in finite time T

Example: ρ = 2

TT/20

a

−a

r(2)

at a(T−t)

r(1)

at2/2 −aT2/4+aTt−at2/2

aT2/4r

– p. 11/12

r(t) =

at2

2for 0 ≤ t ≤ T

2

−aT 2

4 + aT t − at2

2 for T2 ≤ t ≤ T

aT 2

4 for t ≥ T

a =4y

T 2⇒ r(t) = y for t ≥ T

– p. 12/12


Observers

Linearizationand

Extended Kalman Filter (EKF)

– p. 1/12

Linear Observer via Linearization

x = f(x, u), y = h(x)

0 = f(xss, uss), yss = h(xss)

Linearize about the equilibrium point:

xδ = Axδ + Buδ, yδ = Cxδ

xδ = x − xss, uδ = u − uss, yδ = y − yss

What are A, B, C?

˙xδ = Axδ + Buδ + H(yδ − Cxδ), x = xss + xδ

(A − HC) is Hurwitz

It will work locally for sufficiently small ‖xδ(0)‖, ‖xδ(0)‖,and ‖uδ(t)‖

– p. 2/12

Feedback Control:

˙xδ = Axδ + Buδ + H(yδ − Cxδ)

uδ = −Kxδ, u = uss − Kxδ

Verify that the closed-loop system has an equilibrium pointat

x = xss, x = 0

and linearization at the equilibrium point yields[

xδ

˙x

]

=

[

(A − BK) BK

0 (A − HC)

] [

xδ

x

]

Which theorem would justify this controller locally?

– p. 3/12

Nonlinear Observer via Linearization

x = f(x, u), y = h(x)

0 = f(xss, uss), yss = h(xss)

˙x = f(x, u) + H[y − h(x)]

Equilibrium Point: x = xss, u = uss, x = xss

x = x − x

˙x = g(x, x, u), g(xss, 0, uss) = 0

Verify that linearization at the equilibrium point yields

˙x = (A − HC)x

Investigate the design of H and the use in feedback control

– p. 4/12

Extended Kalman Filter (EKF)

x = f(x, u) + w, y = h(x, u) + v

˙x = f(x, u) + H(t)[y − h(x, u)]

x = x − x

˙x = f(x, u) + w − f(x, u) − H[h(x, u) + v − h(x, u)]

Substitute x = x + x and expand the RHS in a Taylorseries about x = 0

˙x = [A(t) − H(t)C(t)]x + η(x, t) + ξ(t)

A(t) =∂f

∂x(x(t), u(t)), C(t) =

∂h

∂x(x(t), u(t))

η(0, t) = 0, ξ(t) = w(t) − H(t)v(t)

– p. 5/12

Assuming that x(t), u(t), w(t), v(t), and H(t) arebounded and f and h are twice continuously differentiable,show that

‖η(x, t)‖ ≤ k1‖x‖2, ‖ξ(t)‖ ≤ k2

Hint:

f(x, u) − f(x, u) −∂f

∂x(x, u)x

=

∫

1

0

∂f

∂x(σx + x, u) dσx −

∂f

∂x(x, u)x (Exercise 3.23)

=

∫

1

0

[

∂f

∂x(σx + x, u) −

∂f

∂x(x, u)

]

dσ x

– p. 6/12

Kalman Filter Design: Let Q(t) and R(t) be symmetricpositive definite matrices that satisfy

0 < q1I ≤ Q(t) ≤ q2I, 0 < r1I ≤ R(t) ≤ r2I

Let P (t) be the solution of the Riccati equation

P = AP + PAT + Q − PCT R−1CP, P (t0) = P0 > 0

If (A(t), C(t) is uniformly observable, then P (t) exists forall t ≥ t0 and satisfies

0 < p1I ≤ P (t) ≤ p2I ⇒ 0 < p3I ≤ P −1(t) ≤ p4I

See a texbook on optimal control or optimal estimation

H(t) = P (t)C(t)T R−1(t)

– p. 7/12

Compute A(t) and C(t)

A(t) =∂f

∂x(x(t), u(t)), C(t) =

∂h

∂x(x(t), u(t))

Solve the Riccati equation

Compute H(t)

H(t) = P (t)C(t)T R−1(t)

Remark: The Riccati equation and the observer equationhave to be solved simultaneously in real time because A(t)and C(t) depend on x(t) and u(t)

– p. 8/12

Lemma: The origin of

˙x = [A(t) − H(t)C(t)]x + η(x, t)

is exponentially stable and the solutions of

˙x = [A(t) − H(t)C(t)]x + η(x, t) + ξ(t)

are uniformly ultimately bounded by an ultimate boundproportional to k2

Proof:V = xT P −1x

V = xT P −1 ˙x + ˙xTP −1x + xT d

dtP −1x

– p. 9/12

d

dtP −1 = −P −1PP −1

V = xT P −1(A − PCT R−1C)x

+ xT (AT − CT R−1CP )P −1x

− xT P −1PP −1x + 2xT P −1(η + ξ)

V = xT P −1(AP + PAT − PCT R−1CP − P )P −1x

− xT CT R−1Cx + 2xT P −1(η + ξ)

= −xT (P −1QP −1 + CT R−1C)x + 2xT P −1(η + ξ)

≤ −c1‖x‖2 + c2‖x‖3 + c3‖x‖ (c3 ∝ k2)

– p. 10/12

Stochastic Interpretation: When w(t) and v(t) are

zero-mean, white noise stochastic processes,

uncorrelated, i.e., Ew(t)vT (τ ) = 0, ∀t, τ , and

Ew(t)wT (τ ) = Q(t)δ(t − τ )

Ev(t)vT (τ ) = R(t)δ(t − τ )

then x(t) is an approximation of the minimum varianceestimate that minimizes

E

[y(t) − h(x(t), u(t))]T [y(t) − h(x(t), u(t))]

and P (t) is an approximation of the covariance matrix

E

[x(t) − x(t)][x(t) − x(t)]T

– p. 11/12

Feedback Control: What can you say about the closed-loopsystem when x is used in feedback control?

– p. 12/12


Observers

Exact Observers

– p. 1/12

Observer with Linear Error Dynamics

Observer Form:

x = Ax+ γ(y, u), y = Cx

where (A,C) is observable, x ∈ Rn, u ∈ Rm, y ∈ Rp

From Lecture # 24: An n-dimensional SO system

x = f(x) + g(x)u, y = h(x)

is transformable into the observer form if and only if

φ =[

h, Lfh, · · · Ln−1

f h]T

, rank

[

∂φ

∂x(x)

]

= n

b =[

0, · · · 0, 1]T

,∂φ

∂xτ = b

– p. 2/12

[adifτ, adjfτ ] = 0, 0 ≤ i, j ≤ n− 1

[g, adjfτ ] = 0, 0 ≤ j ≤ n− 2

Change of variables:

τi = (−1)i−1adi−1

f τ, 1 ≤ i ≤ n

∂T

∂x

[

τ1, τ2, · · · τn

]

= I

z = T (x)

– p. 3/12

x = Ax+ γ(y, u), y = Cx

˙x = Ax+ γ(y, u) +H(y − Cx)

x = x− x

˙x = (A−HC)x

Design H such that (A−HC) is Hurwitz

What about feedback control?

Let u = ψ(x) be a globally stabilizing state feedbackcontrol

u = ψ(x)

˙x = Ax+ γ(y, u) +H(y − Cx)

– p. 4/12

How would you analyze the closed-loop system?

x = Ax+ γ(Cx, ψ(x− x))

˙x = (A−HC)x

We know that

the origin of x = Ax+ γ(Cx, ψ(x)) is globallyasymptotically stable

the origin of ˙x = (A−HC)x is globally exponentiallystable

What additional assumptions do we need to show that theorigin of the closed-loop system is globally asymptoticallystable?

– p. 5/12

Circle Criterion Design

x = Ax+ γ(y, u) − Lβ(Mx), y = Cx

where (A,C) is observable, x ∈ Rn, u ∈ Rm, y ∈ Rp,Mx ∈ Rℓ, β(η) = [ β1(η1), . . . , βℓ(ηℓ) ]T

˙x = Ax+ γ(y, u) −Lβ(Mx−N(y−Cx)) +H(y−Cx)

x = x− x

˙x = (A−HC)x− L[β(Mx) − β(Mx−N(y − Cx))]

˙x = (A−HC)x− L[β(Mx) − β(Mx− (M +NC)x)]

Definez = (M +NC)x

ψ(t, z) = β(Mx(t)) − β(Mx(t) − z)

– p. 6/12

˙x = (A−HC)x− Lψ(t, z)

z = (M +NC)x

G(s)def= (M +NC)[sI − (A−HC)]−1L

- n - -

6

0 z

G(s)

ψ(·)

−

+

ψ(t, z) =[

ψ1(t, z1), . . . , ψℓ(t, zℓ)]T

– p. 7/12

Main Assumption: βi(·) is a nondecreasing function

(a− b)[βi(a) − βi(b)] ≥ 0, ∀ a, b ∈ R

If βi(ηi) is continuously differentiable

dβi

dηi≥ 0, ∀ηi ∈ R

ziψi(t, zi) = zi[βi((Mx)i) − βi((Mx)i − zi)] ≥ 0

zTψ(t, z) ≥ 0

– p. 8/12

By the circle criterion (Theorem 7.1) the origin of


z = (M +NC)x

is globally exponentially stable if

G(s)def= (M +NC)[sI − (A−HC)]−1L

is strictly positive real

Design Problem: Design H and N such that G(s) isstrictly positive realFeasibility can be investigated using LMI (Arcak &Kokotovic, Automatica, 2001)

– p. 9/12

Example:

x1 = x2, x2 = −x3

1− x3

2+ u, y = x1

A =

[

0 1

0 0

]

, C =[

1 0]

, γ =

[

0

−y3 + u

]

,

L =

[

0

1

]

, M =[

0 1]

, β(η) = η3,dβ

dη= 3η2 ≥ 0

h =

[

h1

h2

]

, N

G(s) = (M +NC)[sI− (A−HC)]−1L =s+N + h1

s2 + h1s+ h2

– p. 10/12

From Exercise 6.7, G(s) is SPR if and only if

h1 > 0, h2 > 0, 0 < N + h1 < h1

h1 = 2, h2 = 1, N = − 1

2

G(s) =s+ 3

2

(s+ 1)2

˙x1 = x2 + 2(y − x1)

˙x2 = −y3 + u−(

x2 + 1

2(y − x1)

)3+ (y − x1)

– p. 11/12

What about feedback control?

Let u = φ(x) be a globally stabilizing state feedback control

Closed-loop system under output feedback:

x = Ax+ γ(y, φ(x− x)) − Lβ(Mx)


z = (M +NC)x

How would you analyze the closed-loop system?

ψ(t, z) depends on x(t). How would you show that ψ iswell defined?

What about the effect of uncertainty?

– p. 12/12


Observers

High-Gain ObserversMotivating Example

– p. 1/14

x1 = x2, x2 = φ(x, u), y = x1

Let u = γ(x) stabilize the origin of

x1 = x2, x2 = φ(x, γ(x))

Observer:

˙x1 = x2 + h1(y − x1), ˙x2 = φ0(x, u) + h2(y − x1)

φ0(x, u) is a nominal model φ(x, u)

x1 = x1 − x1, x2 = x2 − x2

˙x1 = −h1x1 + x2, ˙x2 = −h2x1 + δ(x, x)

δ(x, x) = φ(x, γ(x)) − φ0(x, γ(x))

– p. 2/14

Design H =

[

h1

h2

]

such that Ao =

[

−h1 1

−h2 0

]

is Hurwitz

Transfer function from δ to x:

Go(s) =1

s2 + h1s + h2

[

1

s + h1

]

Design H to make supω∈R ‖Go(jω)‖ as small as possible

h1 =α1

ε, h2 =

α2

ε2, ε > 0

Go(s) =ε

(εs)2 + α1εs + α2

[

ε

εs + α1

]

– p. 3/14

Go(s) =ε

(εs)2 + α1εs + α2

[

ε

εs + α1

]

Observer eigenvalues are (λ1/ε) and (λ2/ε) where λ1 andλ2 are the roots of

λ2 + α1λ + α2 = 0

supω∈R

‖Go(jω)‖ = O(ε)

– p. 4/14

η1 =x1

ε, η2 = x2

εη1 = −α1η1 + η2, εη2 = −α2η1 + εδ(x, x)

Ultimate bound of η is O(ε)

η decays faster than an exponential mode e−at/ε, a > 0

Peaking Phenomenon:

x1(0) 6= x1(0) ⇒ η1(0) = O(1/ε)

The solution contains a term of the form1

εe−at/ε

1

εe−at/ε approaches an impulse function as ε → 0

– p. 5/14

Example

x1 = x2, x2 = x3

2+ u, y = x1

State feedback control:

u = −x3

2− x1 − x2

Output feedback control:

u = −x3

2− x1 − x2

˙x1 = x2 + (2/ε)(y − x1)

˙x2 = (1/ε2)(y − x1)

– p. 6/14

0 1 2 3 4 5 6 7 8 9 10−2

−1.5

−1

−0.5

0

0.5

x 1SFBOFB ε = 0.1OFB ε = 0.01OFB ε = 0.005

0 1 2 3 4 5 6 7 8 9 10−3

−2

−1

0

1

x 2

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−400

−300

−200

−100

0

u

t

– p. 7/14

ε = 0.004

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08−0.6

−0.4

−0.2

0

0.2

x 1

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08−600

−400

−200

0

x 2

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08−1000

0

1000

2000

u

t

– p. 8/14

u = sat(−x3

2− x1 − x2)

0 1 2 3 4 5 6 7 8 9 10−0.05

0

0.05

0.1

0.15x 1

SFBOFB ε = 0.1OFB ε = 0.01OFB ε = 0.001

0 1 2 3 4 5 6 7 8 9 10

−0.1

−0.05

0

0.05

x 2

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−1

−0.5

0

u

t

– p. 9/14

Region of attraction under state feedback:

−3 −2 −1 0 1 2 3−2

−1

0

1

2

x1

x 2

– p. 10/14

Region of attraction under outputfeedback:

−1.5 −1 −0.5 0 0.5 1 1.5

−1

−0.5

0

0.5

1

x1

x 2

ε = 0.1 (dashed) and ε = 0.05 (dash-dot)

– p. 11/14

Analysis of the closed-loop system:

x1 = x2 x2 = φ(x, γ(x − x))

εη1 = −α1η1 + η2 εη2 = −α2η1 + εδ(x, x)

-

6

x

ηO(1/ε)

O(ε)

ΩcΩb -

-

qq

DDDDDDDDW

DDDDDDDDW

– p. 12/14

What is the effect of measurement noise?

The high-gain observer is an approximate differentiator

Transfer function from y to x (with φ0 = 0):

α2

(εs)2 + α1εs + α2

[

1 + (εα1/α2)s

s

]

→

[

1

s

]

as ε → 0

Differentiation amplifies the effect of measurement noise

y = x1 + v, kn = supt≥0

|v(t)| < ∞

– p. 13/14

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.160.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

the high−gain parameter epsilon

the

erro

r bo

und

εopt = O

(

√

kn

kd

)

, kd = supt≥0

|x1(t)|, kn = supt≥0

|v(t)|

– p. 14/14


Observers

High-Gain ObserversStabilization

– p. 1/11

x = Ax+Bφ(x, z, u)

z = ψ(x, z, u)

y = Cx

ζ = q(x, z)

u ∈ Rp, y ∈ Rm, ζ ∈ Rs, x ∈ Rρ, z ∈ Rℓ

A,B,C are block diagonal matrices

Ai =

0 1 · · · · · · 0

0 0 1 · · · 0...

...0 · · · · · · 0 1

0 · · · · · · · · · 0

ρi×ρi

, Bi =

0

0...0

1

ρi×1

– p. 2/11

Ci =[

1 0 · · · · · · 0]

1×ρi

, ρ =m∑

i=1

ρi

Normal form

Mechanical and electromechanical systems

Example: Magnetic Suspension

x1 = x2

x2 = g −k

mx2 −

L0ax2

3

2m(a+ x1)2

x3 =1

L(x1)

[

−Rx3 +L0ax2x3

(a+ x1)2+ u

]

– p. 3/11

Stabilizing (partial) state feedback controller:

u = γ(x, ζ)

ϑ = Γ(ϑ, x, ζ), u = γ(ϑ, x, ζ)

Closed-loop system under state feedback:

X = f(X ), X = (x, z, ϑ)

The origin of X = f(X ) is asymptotically stable

Observer:

˙x = Ax+Bφ0(x, ζ, u) +H(y − Cx)

– p. 4/11

H is block diagonal

Hi =

αi1/ε

αi2/ε2

...αiρi−1

/ερi−1

αiρi/ερi

ρi×1

sρi + αi1sρi−1 + · · · + αiρi−1

s+ αiρi

is Hurwitz and ε > 0 (small)φ0(x, ζ, u) is a nominal model of φ(x, z, u), which isglobally bounded in x

– p. 5/11

Theorem 14.6 (Nonlinear Separation Principle:Suppose the origin of X = f(X ) is asymptotically stableand R is its region of attraction. Let S be any compact setin the interior of R and Q be any compact subset of Rρ.Then,

∃ ε∗1> 0 such that, for every 0 < ε ≤ ε∗

1, the solutions

(X (t), x(t)) of the closed-loop system, starting inS × Q, are bounded for all t ≥ 0

given any µ > 0, ∃ ε∗2> 0 and T2 > 0, dependent on

µ, such that, for every 0 < ε ≤ ε∗2, the solutions of the

closed-loop system, starting in S × Q, satisfy

‖X (t)‖ ≤ µ and ‖x(t)‖ ≤ µ, ∀ t ≥ T2

– p. 6/11

given any µ > 0, ∃ ε∗3> 0, dependent on µ, such that,

for every 0 < ε ≤ ε∗3, the solutions of the closed-loop

system, starting in S × Q, satisfy

‖X (t) − Xr(t)‖ ≤ µ, ∀ t ≥ 0

where Xr is the solution of X = f(X ), starting at X (0)

if the origin of X = f(X ) is exponentially stable, then ∃ε∗4> 0 such that, for every 0 < ε ≤ ε∗

4, the origin of the

closed-loop system is exponentially stable and S × Q isa subset of its region of attraction.

– p. 7/11

Key ideas of the proof:

Representation of the closed-loop system as asingularly perturbed one with X as the slow and η(scaled estimation error) as the fast

Use of a converse Lyapunov theorem to constructpositively invariant sets

Use of global boundedness in x to show that η reachesO(ε) while X is inside a positively invariant set

Nonlocal versus local analysis

Novel Feature: Performance recovery

– p. 8/11

Example 14.19:

mℓ2θ +mg0ℓ sin θ + k0ℓ2θ = u

Stabilization at (θ = π, θ = 0). From Section 14.1

u = −k sat

(

a1(θ − π) + θ

µ

)

Suppose we only measure θ

˙θ = ω + (2/ε)(θ − θ)

˙ω = φ0(θ, u) + (1/ε2)(θ − θ)

φ0 = −a sin θ + cu

– p. 9/11

u = −k sat

(

a1(θ − π) + ω

µ

)

or

u = −k sat

(

a1(θ − π) + ω

µ

)

– p. 10/11

0 2 4 6 8 100.5

1

1.5

2

2.5

3

3.5

θ

(a)

0 2 4 6 8 10−2

−1

0

1

2

ω

(b)

0 2 4 6 8 100.5

1

1.5

2

2.5

3

3.5

θ

Time

(c)

0 2 4 6 8 100

0.5

1

1.5

2

2.5

3

3.5

θ

(d)

Time

SFBOFB ε = 0.05OFB ε = 0.01

– p. 11/11


Integral Control

– p. 1/17

x = f(x, u, w)

y = h(x, w)

ym = hm(x, w)

x ∈ Rn state, u ∈ Rp control input

y ∈ Rp controlled output, ym ∈ Rm measured output

w ∈ Rl unknown constant parameters and disturbances

Goal:y(t) → r as t → ∞

r ∈ Rp constant reference, v = (r, w)

e(t) = y(t) − r

– p. 2/17

Assumption: e can be measured

Steady-state condition: There is a unique pair (xss, uss)that satisfies the equations

0 = f(xss, uss, w)

0 = h(xss, w) − r

Stabilize the system at the equilibrium point x = xss

Can we reduce this to a stabilization problem by shifting theequilibrium point to the origin via the change of variables

xδ = x − xss, uδ = u − uss ?

– p. 3/17

Integral Action:σ = e

Augmented System:

x = f(x, u, w)

σ = h(x, w) − r

Task: Stabilize the augmented system at (xss, σss) whereσss produces uss

- l - - - -

666

r −σ u y

−+

∫ StabilizingController

MeasuredSignals

Plant

– p. 4/17

Integral Control via Linearization

State Feedback:

u = −K1x − K2σ − K3e

Closed-loop system:

x = f(x, −K1x − K2σ − K3(h(x, w) − r), w)

σ = h(x, w) − r

Equilibrium points:

0 = f(x, u, w)

0 = h(x, w) − r

u = −K1x − K2σ

Unique equilibrium point at x = xss, σ = σss, u = uss

– p. 5/17

Linearization about (xss, σss):

ξδ =

[

x − xss

σ − σss

]

ξδ = (A − BK)ξδ

A =

[

A 0

C 0

]

, A =∂f

∂x(x, u, w)

∣

∣

∣

∣

eq

, C =∂h

∂x(x, w)

∣

∣

∣

∣

eq

B =

[

B

0

]

, B =∂f

∂u(x, u, w)

∣

∣

∣

∣

eq

K =[

K1 + K3C K2

]

– p. 6/17

(A, B) is controllable if and only if (A, B) is controllableand

rank

[

A B

C 0

]

= n + p

Task: Design K, independent of v, such that (A − BK) isHurwitz for all v

(xss, σss) is an exponentially stable equilibrium point of theclosed-loop system. All solutions starting in its region ofattraction approach it as t tends to infinity

e(t) → 0 as t → ∞

– p. 7/17

Pendulum Example:


Regulate θ to δ

x1 = θ − δ, x2 = θ, u = T

x1 = x2

x2 = −a sin(x1 + δ) − bx2 + cu

xss =

[

0

0

]

, uss =a

csin δ

σ = x1

– p. 8/17

A =

0 1 0

−a cos δ −b 0

1 0 0

, B =

0

c

0

K1 = [k1 k2], K2 = k3, K3 = 0

(A − BK) will be Hurwitz if

b+k2c > 0, (b+k2c)(a cos δ+k1c)−k3c > 0, k3c > 0

Supposea

c≤ ρ1,

1

c≤ ρ2

k2 > 0, k3 > 0, k1 > ρ1 + ρ2

k3

k2

– p. 9/17

Output Feedback: We only measure e and ym

σ = e = y − r

z = Fz + G1σ + G2ym

u = Lz + M1σ + M2ym + M3e

Task: Design F , G1, G2, L, M1, M2, and M3,independent of v, such that Ac is Hurwitz for all v

Ac =

A + BM2Cm + BM3C BM1 BL

C 0 0

G2Cm G1 F

Cm =∂hm

∂x(x, w)

∣

∣

∣

∣

eq

– p. 10/17

Integral Control via Sliding Mode Design

η = f0(η, ξ, w)

ξ1 = ξ2

......

ξρ−1 = ξρ

ξρ = b(η, ξ, u, w) + a(η, ξ, w)u

y = ξ1

a(η, ξ, w) ≥ a0 > 0

Goal:y(t) → r as t → ∞

ξss = [r, 0, . . . , 0]T

– p. 11/17

Steady-state condition: There is a unique pair (ηss, uss)that satisfies the equations

0 = f0(ηss, ξss, w)

0 = b(ηss, ξss, uss, w) + a(ηss, ξss, w)uss

e0 = y − r

z = η − ηss, e =

e1

e2

...eρ

=

ξ1 − r

ξ2

...ξρ

– p. 12/17

z = f0(η, ξ, w)def= f0(z, e, w, r)

e0 = e1

e1 = e2

......

eρ−1 = eρ

eρ = b(η, ξ, u, w) + a(η, ξ, w)u

Partial State Feedback: e1, . . . , eρ are measured

s = k0e0 + k1e1 + · · · + kρ−1eρ−1 + eρ

k0 to kρ−1 are chosen such that the polynomial

λρ + kρ−1λρ−1 + · · · + k1λ + k0 is Hurwitz

– p. 13/17

s = k0e1 + · · · + kρ−1eρ + b(η, ξ, u, w) + a(η, ξ, w)u

s = ∆(η, ξ, u, w, r) + a(η, ξ, w)u∣

∣

∣

∣

∆(η, ξ, u, w, r)

a(η, ξ, w)

∣

∣

∣

∣

≤ (e) + κ0|u|, 0 ≤ κ0 < 1

u = −β(e) sat

(

s

µ

)

β(e) ≥(e)

(1 − κ0)+ β0, β0 > 0

For |s| ≥ µ, ss ≤ −a0(1 − κ0)β0

What about the other state variables?

– p. 14/17

z = f0(z, e, w, r)

ζ = Aζ + Bs (A is Hurwitz)

s = −a(·)β(e) sat

(

s

µ

)

+ ∆(·)

ζ = [e0, . . . , eρ−1]T

α1(‖z‖) ≤ V1(z, w, r) ≤ α2(‖z‖)

∂V1

∂zf0(z, e, w, r) ≤ −α3(‖z‖), ∀ ‖z‖ ≥ γ(‖e‖)

V2(ζ) = ζT Pζ, PA + AT P = −I

– p. 15/17

Ω = |s| ≤ c ∩ V2 ≤ c2ρ1 ∩ V1 ≤ c0

Ωµ = |s| ≤ µ ∩ V2 ≤ µ2ρ1 ∩ V1 ≤ α2(γ(µρ2))

All trajectories starting in Ω enter Ωµ in finite time and stayin thereafterInside Ωµ there is a unique equilibrium point at

(z = 0, e = 0, e0 = e0), s = k0e0, uss = −β(0)s

µ

Under additional conditions (the origin of z = f0(z, 0, w, r)is exponentially stable), local analysis inside Ωµ shows thatfor sufficiently small µ all trajectories converge to theequilibrium point as time tends to infinity

– p. 16/17

Output Feedback: Only e1 is measured

High-gain Observer:

e0 = e1

u = −β sat

(

k0e0 + k1e1 + k2e2 + · · · + eρ

µ

)

˙ei = ei+1 +

(

αi

εi

)

(e1 − e1), 1 ≤ i ≤ ρ − 1

˙eρ =

(

αρ

ερ

)

(e1 − e1)

β = β(e1, e2, . . . , eρ)

– p. 17/17

khalil - nonlinear systems slides

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