lecture 16: geometric nonlinear control - umu.se...lecture 16: geometric nonlinear control •...

Lecture 16: Geometric Nonlinear Control

• Background and Useful Concepts

c©Anton Shiriaev. 5EL158: Lecture 16 – p. 1/21



• Feedback Linearization





• Feedback Linearization for Single Link Flexible Joint


A Smooth Manifold in Rn

A smooth manifold M of dimension m is a subset of Rn

defined as the zero set of p = (n − m) smooth scalar functions

h1(x1, x2, . . . , xn) = 0...

hp(x1, x2, . . . , xn) = 0

such that the differentials

dh1(·), dh2(·), . . . , dhp(·)

are linear independent for any point of M.


A Smooth Manifold in Rn

A smooth manifold M of dimension m is a subset of Rn

defined as the zero set of p = (n − m) smooth scalar functions

h1(x1, x2, . . . , xn) = 0...

hp(x1, x2, . . . , xn) = 0

such that the differentials

dh1(·), dh2(·), . . . , dhp(·)

are linear independent for any point of M.

At each point x ∈ M we can attach a tangent space TxM ,

which is an m-dimensional space that specifies the set ofpossible directions of velocities at this point


Example

Consider a unit sphere S2 in R3 defined by

h(x, y, z) = x2 + y2 + z2 − 1 = 0

What is the dimension of this manifold? Is it smooth?


Example

The differential dh(·) of the function

h(x, y, z) = x2 + y2 + z2 − 1 = 0

at the point (x0, y0, z0) ∈ S2 is the row vector

dh(x0, y0, z0) =[

∂∂x

h, ∂∂y

h, ∂∂z

h]∣∣∣x=x0, y=y0, z=z0


Example


h(x, y, z) = x2 + y2 + z2 − 1 = 0


dh(x0, y0, z0) =[

∂∂x

h, ∂∂y

h, ∂∂z

h]∣∣∣x=x0, y=y0, z=z0

=[2x0, 2y0, 2z0

]


Example


h(x, y, z) = x2 + y2 + z2 − 1 = 0


dh(x0, y0, z0) =[

∂∂x

h, ∂∂y

h, ∂∂z

h]∣∣∣x=x0, y=y0, z=z0

=[2x0, 2y0, 2z0

]

How to compute a tangent space Tp0S2 to the sphere S2

at the point p0 = (x0, y0, z0) ∈ S2?


Example


h(x, y, z) = x2 + y2 + z2 − 1 = 0


dh(x0, y0, z0) =[

∂∂x

h, ∂∂y

h, ∂∂z

h]∣∣∣x=x0, y=y0, z=z0

=[2x0, 2y0, 2z0

]

By definition

Tp0S2 =

V = [v1, v2, v3]

T : dh(p0)⊥V

=

V = [v1, v2, v3]T : 2x0 · v1 + 2y0 · v2 + 2z0 · v3 = 0


Example


h(x, y, z) = x2 + y2 + z2 − 1 = 0


dh(x0, y0, z0) =[

∂∂x

h, ∂∂y

h, ∂∂z

h]∣∣∣x=x0, y=y0, z=z0

=[2x0, 2y0, 2z0

]

By definition

Tp0S2 =

V = [v1, v2, v3]

T : dh(p0)⊥V

=

V = [v1, v2, v3]T : 2x0 · v1 + 2y0 · v2 + 2z0 · v3 = 0

~V1 =[1, 0, −x0/z0

], ~V2 =

[0, 1, −y0/z0

]


Smooth Vector and Co-vector Fields

A smooth vector field on a manifold M is a smooth function

x ∈ M → f(x) =

f1(x)...

fm(x)

∈ TxM


Smooth Vector and Co-vector Fields

A smooth vector field on a manifold M is a smooth function

x ∈ M → f(x) =

f1(x)...

fm(x)

∈ TxM

A smooth co-vector field on a manifold M is a smooth function

x ∈ M → w(x) =[w1(x), · · · , wm(x)

]∈ T ∗

xM

Here T ∗

xM is co-tangent space to the manifold M at its point x.


Distributions and Co-distributions

Let X1(x), . . . , Xk(x) be vector fields on M that are linearly

independent for each x ∈ M. Then a distribution (·) is a

family of linear subspaces of TM defined for each point x as

(x) = spanX1(x), . . . , Xk(x)






(x) = spanX1(x), . . . , Xk(x)

Let W1(x), . . . , Wk(x) be co-vector fields on M linearly

independent for each x ∈ M. Then a co-distribution Ω(·) is a

family of linear subspaces of T ∗M defined for each point x as

Ω(x) = spanW1(x), . . . , Wk(x)






(x) = spanX1(x), . . . , Xk(x)

Let W1(x), . . . , Wk(x) be co-vector fields on M linearly

independent for each x ∈ M. Then a co-distribution Ω(·) is a

family of linear subspaces of T ∗M defined for each point x as

Ω(x) = spanW1(x), . . . , Wk(x)

ddt

x = f(x) + g1(x)u1 + · · · + gm(x)um


Lie Bracket of Vector Fields

Given two vector fields f(·) and g(·) on M = Rn.

The Lie bracket of f and g is new vector field in Rn denoted by

[f, g] and defined for each x ∈ M = Rn by

[f, g](x) :=[

∂∂x

g(x)]

f(x) −[

∂∂x

f(x)]

g(x)






[f, g](x) :=[

∂∂x

g(x)]

f(x) −[

∂∂x

f(x)]

g(x)

Suppose f = Ax and g = Bx, then

[f, g](x) =[

∂∂x

g(x)]

f(x) −[

∂∂x

f(x)]

g(x)

= B · Ax − A · Bx = (BA − AB)︸︷︷︸

= C

x






[f, g](x) :=[

∂∂x

g(x)]

f(x) −[

∂∂x

f(x)]

g(x)

Notations:

ad0f(g) = g

ad1f(g) =

[f, g

]

ad2f(g) =

[f,

[f, g

] ]

...

adkf(g) =

[

f, ad(k−1)f (g)

]


Lie Derivative of Scalar Function

Given• a scalar function h(·) on R

n, i.e. h : Rn → R

• a vector field f(·) on Rn, i.e. f : R

n → Rn

The Lie derivative of h(·) with respect to f(·) is new scalarfunction denoted as Lfh and defined by

Lfh =[

∂∂x

h(x)]

f(x) =[

∂∂x1

h(x)]

·f1(x)+· · ·+[

∂∂xn

h(x)]

·fn(x)




n, i.e. h : Rn → R


n → Rn


Lfh =[

∂∂x

h(x)]

f(x) =[

∂∂x1

h(x)]

·f1(x)+· · ·+[

∂∂xn

h(x)]

·fn(x)

Recursive notation:

Lkfh = Lf

[

L(k−1)f h

]

, k = 1, 2 . . . , L0fh = h




n, i.e. h : Rn → R


n → Rn


Lfh =[

∂∂x

h(x)]

f(x) =[

∂∂x1

h(x)]

·f1(x)+· · ·+[

∂∂xn

h(x)]

·fn(x)

Lemma: Let h : Rn → R be a scalar function and f , g be

vector fields on Rn. Then the following identity holds

L[f,g]h ≡ Lf

[Lgh

]− Lg

[Lfh

]


Frobenius Theorem

Consider two partial differential equations

∂z

∂x= f(x, y, z),

∂z

∂y= g(x, y, z)


Frobenius Theorem


∂z

∂x= f(x, y, z),

∂z

∂y= g(x, y, z)

We are looking for a scalar variable z in the form

z = φ(x, y)

such that

∂

∂xφ(x, y) = f(x, y, φ(x, y)),

∂

∂yφ(x, y) = g(x, y, φ(x, y))


Frobenius Theorem


∂z

∂x= f(x, y, z),

∂z

∂y= g(x, y, z)

We are looking for a scalar variable z in the form

z = φ(x, y)

such that

∂

∂xφ(x, y) = f(x, y, φ(x, y)),

∂

∂yφ(x, y) = g(x, y, φ(x, y))

If we find the solution, then it corresponds to the surface in R3

M =x, y, z = φ(x, y)

How to compute a tangent plane to that surface (manifold)?


Example (Sphere in R3)


h(x, y, z) = x2 + y2 + z2 − 1 = 0


dh(x0, y0, z0) =[

∂∂x

h, ∂∂y

h, ∂∂x

h]∣∣∣x=x0, y=y0, z=z0

=[2x0, 2y0, 2z0

]

By definition

Tp0S2 =

V = [v1, v2, v3]

T : dh(p0)⊥V

=

V = [v1, v2, v3]T : 2x0 · v1 + 2y0 · v2 + 2z0 · v3 = 0

~V1 =[1, 0, −x0/z0

], ~V2 =

[0, 1, −y0/z0

]


Computing a Tangent Space


h(x, y, z) = z − φ(x, y) = 0

at the point p0 = (x0, y0, z0) ∈ R2 is the row vector

dh(p0) =[

∂∂x

h, ∂∂y

h, ∂∂z

h]∣∣∣x=x0, y=y0, z=z0

=[

− ∂∂x

φ(x, y), − ∂∂x

φ(x, y), 1]∣∣∣x=x0, y=y0, z=z0




h(x, y, z) = z − φ(x, y) = 0


dh(p0) =[

∂∂x

h, ∂∂y

h, ∂∂x

h]∣∣∣x=x0, y=y0, z=z0

=[

− ∂∂x

φ(x, y), − ∂∂y

φ(x, y), 1]∣∣∣x=x0, y=y0, z=z0

=[−f(x0, y0, φ(x0, y0)), −g(x0, y0, φ(x0, y0)), 1

]




h(x, y, z) = z − φ(x, y) = 0


dh(p0) =[

∂∂x

h, ∂∂y

h, ∂∂x

h]∣∣∣x=x0, y=y0, z=z0

=[

− ∂∂x

φ(x, y), − ∂∂y

φ(x, y), 1]∣∣∣x=x0, y=y0, z=z0

=[−f(x0, y0, φ(x0, y0)), −g(x0, y0, φ(x0, y0)), 1

]

Tp0M =

~V = [v1, v2, v3]T : dh(p0)⊥~V

=

~V : −f(·)v1 − g(·)v2 + 1 · v3 = 0




h(x, y, z) = z − φ(x, y) = 0


dh(p0) =[

∂∂x

h, ∂∂y

h, ∂∂x

h]∣∣∣x=x0, y=y0, z=z0

=[

− ∂∂x

φ(x, y), − ∂∂y

φ(x, y), 1]∣∣∣x=x0, y=y0, z=z0

=[−f(x0, y0, φ(x0, y0)), −g(x0, y0, φ(x0, y0)), 1

]

Tp0M =

~V : −f(·)v1 − g(·)v2 + 1 · v3 = 0

Basis for Tp0M : ~V1 =

[1, 0, f(·)

]T

, ~V2 =[0, 1, g(·)

]T


Problem Reformulation

Searching for solution z = φ(x, y) of two partial differentialequations

∂∂x

z = f(x, y, z), ∂∂y

z = g(x, y, z)

is equivalent to search

Surface M in R3 whose tangent space in each point TpM is

spanned by ~V1 =[1, 0, f(x, y, z)

]T and ~V2 =

[0, 1, g(x, y, z)

]T




∂∂x

z = f(x, y, z), ∂∂y

z = g(x, y, z)




]T and ~V2 =

[0, 1, g(x, y, z)

]T

If such surface M can be found then

M is integral manifold of the system




∂∂x

z = f(x, y, z), ∂∂y

z = g(x, y, z)




]T and ~V2 =

[0, 1, g(x, y, z)

]T


M is integral manifold of the system


the equations are called integrable



Search for solution z = φ(x, y) of equations

∂∂x

z = f(x, y, z), ∂∂y

z = g(x, y, z)

can be rewritten as the search for solution of other two equations

L~V1

h = 0, L~V2

h = 0

where h = h(x, y, z) and

~V1 =[1, 0, f(x, y, z)

]T and ~V2 =

[0, 1, g(x, y, z)

]T




∂∂x

z = f(x, y, z), ∂∂y

z = g(x, y, z)


L~V1

h = 0, L~V2

h = 0


~V1 =[1, 0, f(x, y, z)

]T and ~V2 =

[0, 1, g(x, y, z)

]T

Indeed, if the solution z = φ(x, y) exists, then the equation

L~V1

h =[

∂∂x

h(x, y, z)]

· 1 +[

∂∂y

h(x, y, z)]

· 0 +

+[

∂∂z

h(x, y, z)]

· f(x, y, z) = 0




∂∂x

z = f(x, y, z), ∂∂y

z = g(x, y, z)


L~V1

h = 0, L~V2

h = 0


~V1 =[1, 0, f(x, y, z)

]T and ~V2 =

[0, 1, g(x, y, z)

]T


L~V1

h =[

∂∂x

h(x, y, z)]

· 1 +[

∂∂z

h(x, y, z)]

· f(x, y, z) = 0

has a solution h(x, y, z) = z − φ(x, y)c©Anton Shiriaev. 5EL158: Lecture 16 – p. 12/21



∂∂x

z = f(x, y, z), ∂∂y

z = g(x, y, z)


L~V1

h = 0, L~V2

h = 0


~V1 =[1, 0, f(x, y, z)

]T and ~V2 =

[0, 1, g(x, y, z)

]T


L~V2

h =[

∂∂x

h(x, y, z)]

· 0 +[

∂∂y

h(x, y, z)]

· 1 +

+[

∂∂z

h(x, y, z)]

· g(x, y, z) = 0




∂∂x

z = f(x, y, z), ∂∂y

z = g(x, y, z)


L~V1

h = 0, L~V2

h = 0


~V1 =[1, 0, f(x, y, z)

]T and ~V2 =

[0, 1, g(x, y, z)

]T


L~V2

h =[

∂∂y

h(x, y, z)]

· 1 +[

∂∂z

h(x, y, z)]

· g(x, y, z) = 0

has a solution h(x, y, z) = z − φ(x, y)c©Anton Shiriaev. 5EL158: Lecture 16 – p. 12/21

Concepts of Integrability and Involutivity for Distributions

The distribution

= span X1(x), X2(x), . . . , Xm(x) , x ∈ Rn

is said to be completely integrable , if there are functions h1(x),

. . . , hn−m(x) such that

• they are linearly independent for any x ∈ Rn;

• they satisfy the system of partial differential equations

LXihj = 0, ∀ i ∈ 1, . . . , m, ∀ j ∈ 1, . . . , (n−m)


Concepts of Integrability and Involutivity for Distributions

The distribution

= span X1(x), X2(x), . . . , Xm(x) , x ∈ Rn

is said to be completely integrable , if there are functions h1(x),

. . . , hn−m(x) such that

• they are linearly independent for any x ∈ Rn;

• they satisfy the system of partial differential equations

LXihj = 0, ∀ i ∈ 1, . . . , m, ∀ j ∈ 1, . . . , (n−m)

The distribution is said to be involutive , if there are scalarfunctions αijk : R

n → R such that

[Xi, Xj

]=

n∑

k=1

αijkXk ∀ i, j


Frobenius Theorem

A distribution on Rn is integrable

m

A distribution on Rn is involutive


Concept of Feedback Linearization

The control system

ddt

x = f(x) + g(x)u, x ∈ Rn, u ∈ R

1, f(0) = 0

is said to be feedback linearizable if

• there is a invertible change of coordinates y = T (x)

• there is a feedback transform u = α(x) + β(x)v

such that the system dynamics written in new coordinates is

d

dty =

0 1 0 0

0 0 1 0

· · ·. . . ·

0 0 0 1

0 0 0 . . . 0

︸︷︷︸

= A

y +

0

0...0

1

︸︷︷︸

= B

v


How to find T (·), α(·) and β(·)?

If the systemddt

x = f(x) + g(x)u, x ∈ Rn, u ∈ R

1, f(0) = 0

is feedback linearizable, then there are new coordinates ydefined by

y = T (x) ⇒ ddt

y = ddt

[T (x)] = ∂∂x

[T (x)] ddt

x



If the systemddt

x = f(x) + g(x)u, x ∈ Rn, u ∈ R

1, f(0) = 0

is feedback linearizable, then there are new coordinates ydefined by

y = T (x) ⇒ ddt

y = ddt

[T (x)] = ∂∂x

[T (x)] ddt

x

The last relation can be rewritten as the equation on unknownfunctions Ti(·)

Ay+Bv = ∂∂x

[T (x)](f(x) + g(x)u

)=

∂∂x

T1(x)...

∂∂x

Tn(x)

(f(x) + g(x)u

)



The equations on unknown functions Ti(·) are

A

T1(x)...

Tn(x)

+ Bv =

∂∂x

T1(x)...

∂∂x

Tn(x)

(f(x) + g(x)u

)

with

A =

0 1 0 0

0 0 1 0

· · ·. . . ·

0 0 0 1

0 0 0 . . . 0

, B =

0

0...0

1




A

T1(x)...

Tn(x)

+ Bv =

∂∂x

T1(x)...

∂∂x

Tn(x)

(f(x) + g(x)u

)

with

A =

0 1 0 0

0 0 1 0

· · ·. . . ·

0 0 0 1

0 0 0 . . . 0

, B =

0

0...0

1

LfT1 +[LgT1

]u = T2, LfT2 +

[LgT2

]u = T3, . . .

LfTn +[LgTn

]u = v




A

T1(x)...

Tn(x)

+ Bv =

∂∂x

T1(x)...

∂∂x

Tn(x)

(f(x) + g(x)u

)

or the same

LfT1 +[LgT1

]u = T2, LfT2 +

[LgT2

]u = T3, . . .

LfTn +[LgTn

]u = v

If we assume that T1(·), . . . , Tn(·) are independent on u, then

LgT1 = LgT2 = · · · = LgTn−1 = 0 , LgTn 6= 0




LfT1 +[LgT1

]u = T2, LfT2 +

[LgT2

]u = T3, . . .

LfTn +[LgTn

]u = v


LgT1 = LgT2 = · · · = LgTn−1 = 0 , LgTn 6= 0

The equations to solve become

LfTi = Ti+1, i ∈ 1, . . . , n − 1, LfTn +[LgTn

]u = v




LfT1 +[LgT1

]u = T2, LfT2 +

[LgT2

]u = T3, . . .

LfTn +[LgTn

]u = v


LgT1 = LgT2 = · · · = LgTn−1 = 0 , LgTn 6= 0


LfTi = Ti+1, i ∈ 1, . . . , n − 1, LfTn +[LgTn

]u = v

L[f,g]T1 = Lf [LgT1] − Lg [LfT1] = 0 − LgT2 = 0




LfT1 +[LgT1

]u = T2, LfT2 +

[LgT2

]u = T3, . . .

LfTn +[LgTn

]u = v


LgT1 = LgT2 = · · · = LgTn−1 = 0 , LgTn 6= 0


L[f,g]T1 = 0 , L[f,[f,g]]T1 = 0 , . . . , Lad

n−2

f gT1 = 0 , L

adn−1

f gT1 6= 0

We need to find only T1(·), then T2(·), . . . , Tn(·) are computed!



The partial differential equations

LXiT1 = 0 , i = 1, . . . , n − 1, LXn

T1 6= 0

with

X1 = ad0fg, X2 = ad1

fg, . . . , Xn−1 = ad(n−2)f g, Xn = ad

(n−1)f g

must satisfy properties of Frobenius Theorem :

The distribution defined as spanX1, X2, . . . , Xn should

• be of constant rank, i.e. Xi are to be linear independent• be involutive

If both conditions are satisfied, then the solution T1(x) exists!


Feedback Linearization for Single Link Flexible Joint

The equations of motion are

Iq1 + Mgl sin(q1) + k(q1 − q2) = 0

Jq2 + k(q2 − q1) = u


Feedback Linearization for Single Link Flexible Joint

The equations of motion are

Iq1 + Mgl sin(q1) + k(q1 − q2) = 0

Jq2 + k(q2 − q1) = u

The system is feedback linearizable if k 6= 0, I 6= 0, J 6= 0!


lecture 16: geometric nonlinear control - umu.se...lecture 16: geometric nonlinear control •...

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