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Outline Feedback Linearzation Preliminary Mathematics Input-State Linearization Input-Output Linearization

Nonlinear ControlLecture 9: Feedback Linearization

Farzaneh Abdollahi

Department of Electrical Engineering

Amirkabir University of Technology

Fall 2011

Farzaneh Abdollahi Nonlinear Control Lecture 9 1/75

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Feedback LinearzationInput-State Linearization

Input-Output LinearizationInternal Dynamics of Linear Systems

Zero-Dynamics

Preliminary MathematicsDiffeomorphism

Frobenius TheoremInput-State Linearization

Control DesignInput-Output Linearization

Well Defined Relative Degree

Undefined Relative DegreeNormal FormZero-DynamicsLocal Asymptotic StabilizationGlobal Asymptotic Stabilization

Tracking ControlFarzaneh Abdollahi Nonlinear Control Lecture 9 2/75

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Feedback Linearzation

The main idea is: algebraicallytransform a nonlinear system dynamicsinto a (fully or partly) linear one, so thatlinear control techniques can be applied.

In its simplest form, feedback linearizationcancels the nonlinearities in a nonlinearsystem so that the closed-loop dynamicsis in a linear form.

Example: Controlling the fluid level in a

tank Objective: controlling of the level h of

fluid in a tank to a specified level hd ,using control input u

the initial level is h0.


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Example Cont’d The dynamics:

A(h)h(t ) = u − a

2gh

where A(h) is the cross section of the tank and a is the cross section of the outlet pipe.

Choose u = a√ 2gh + A(h)v h = v

Choose the equivalent input v : v = −αh where h = h(t )− hd is errorlevel, α a pos. const.

∴ resulting closed-loop dynamics: h + αh = 0⇒

h→

0 as t → ∞ The actual input flow: u = a√ 2gh + A(h)αh

First term provides output flow a√

2gh Second term raises the fluid level according to the desired linear dynamics

If hd is time-varying: v = hd (t )

−αh

∴ h → 0 as t → ∞Farzaneh Abdollahi Nonlinear Control Lecture 9 4/75

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Canceling the nonlinearities and imposing a desired linear dynamics, canbe simply applied to a class of nonlinear systems, so-called companion

form, or controllability canonical form: A system in companion form:

x (n)(t ) = f (x) + b (x)u (1)

u is the scalar control input x is the scalar output;x = [x , x ,..., x (n−1)] is the state vector. f (x ) and b (x ) are nonlinear functions of the states. no derivative of input u presents.

(1) can be presented as controllability canonical form

d dt

x 1...

x n−1

x n

=

x 2...

x nf (x ) + b (x )u

for nonzero b , define control input: u = 1

b

[v −

f ]


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Example: Feedback Linearization of a Two-link Robot

A two-link robot: each joint equippedwith a motor for providing input torque

an encoder for measuring joint position a tachometer for measuring joint velocity

objective: the joint positions q l and q 2follow desired position histories q dl (t ) andq d 2(t )

For example when a robot manipulator isrequired to move along a specified path,e.g., to draw circles.


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Using the Lagrangian equations the robotic dynamics are:

H 11 H 12

H 21 H 22

q 1

q 2 +

−hq 2 −hq 2 − hq 1

hq 1 0

q 1

q 2 +

g 1

g 2 =

τ 1

τ 2

where q = [q 1 q 2]T : the two joint angles, τ = [τ 1 τ 2]T : the joint inputs, and

H 11 = m1l 2c 1 + l 1 + m2[l 21 + l 2c 2 + 2l 1l c 2 cos q 2] + I 2

H 22 = m2l 2c 2 + I 2H 12 = H 21 = m2l 1l c 2 cos q 2 + m2l 2c 2 + I 2

g 1 = m1l c 1 cos q 1 + m2g [l c 2 cos(q 1 + q 2) + l 1 cos q 1]g 2 = m2l c 2g cos(q 1 + q 2), h = m2l 1l c 2 sin q 2

Control law for tracking, (computed torque):

τ 1τ 2

=

H 11 H 12

H 21 H 22

v 1v 2

+

−hq 2 −

hq 2−

hq 1hq 1 0

q 1q 2

+

g 1g 2

where v = q d − 2λ ˙q − λ2q , q = q − q d : position tracking error, λ: pos. const.

∴ ¨q d + 2λ ˙q + λ2q = 0 where q converge to zero exponentially.

This method is applicable for arbitrary # of linksFarzaneh Abdollahi Nonlinear Control Lecture 9 8/75

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Input-State Linearization

When the nonlinear dynamics is not in a controllability canonical form,use algebraic transformations

Consider the SISO system

x = f (x , u )

In input-state linearization technique:

1. finds a state transformation z = z (x ) and an input transformation

u = u (x , v ) s.t. the nonlinear system dynamics is transformed intoz = Az + bv 2. Use standard linear techniques (such as pole placement) to design v .


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Example: Consider

x 1 = −2x 1 + ax 2 + sin x 1x 2 = −x 2 cos x 1 + u cos(2x 1)

Equ. pt. (0, 0) The nonlinearity cannot be directly canceled by the control input u Define a new set of variables:

z 1 = x 1z 2 = ax 2 + sin x 1

∴ z 1 = −2z 1 + z 2

z 2 = −2z 1 cos z 1 + cos z 1 sin z 1 + au cos(2z 1)

The Equ. pt. is still (0, 0). The control law: u = 1

a cos(2z 1) (v − cos z 1 sin z 1 + 2z 1 cos z 1) The new dynamics is linear and controllable: z 1 = −2z 1 + z 2, z 2 = v By proper choice of feedback gains k 1 and k 2 in v = −k 1z 1 − k 2z 2, place

the poles properly.

Both z 1 and z 2 converge to zero, the original state x converges to zeroFarzaneh Abdollahi Nonlinear Control Lecture 9 10/75

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The result is not global. The result is not valid when x l = (π/4± k π/2), k = 0, 1, 2,...

The input-state linearization is achieved by a combination of a statetransformation and an input transformation with state feedback used inboth.

To implement the control law, the new states (z 1, z 2) must be available. If they are not physically meaningful or measurable, they should be

computed by measurable original state x .

If there is uncertainty in the model, e.g., on a error in the computationof new state z as well as control input u .

For tracking control, the desired motion needs to be expressed in terms of

the new state vector. Two questions arise for more generalizations which will be answered in

next lectures: What classes of nonlinear systems can be transformed into linear systems? How to find the proper transformations for those which can?


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Input-Output Linearization

Considerx = f (x , u )

y = h(x )

Objective: tracking a desired trajectory y d (t ), while keeping the whole

state bounded y d (t ) and its time derivatives up to a sufficiently high order are known

and bounded. The difficulty: output y is only indirectly related to the input u

∴ it is not easy to see how the input u can be designed to control thetracking behavior of the output y .

Input-output linearization approach:

1. Generating a linear input-output relation2. Formulating a controller based on linear control


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Example: Consider

x 1 = sin x 2 + (x 2 + 1)x 3x 2 = x 51 + x 3

x 3 = x 21 + u

y = x 1

To generate a direct relationship between the output y and the input u ,differentiate the output y = x 1 = sin x 2 + (x 2 + 1)x 3

No direct relationship differentiate again: y = (x 2 + 1)u + f (x ), wheref (x ) = (x 51 + x 3)(x 3 + cos x 2) + (x 2 + 1)x 21

Control input law: u = 1x 2+1 (v − f ). Choose v = y d − k 1e − k 2e , where e = y − y d is tracking error, k 1 and k 2

are pos. const.

The closed-loop system: e + k 2e + k 1e = 0

∴ e.s. of tracking errorFarzaneh Abdollahi Nonlinear Control Lecture 9 13/75

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Example Cont’d

The control law is defined everywhere except at singularity points s.t.x 2 = −1

To implement the control law, full state measurement is necessary,because the computations of both the derivative y and the input

transformation need the value of x . If the output of a system should be differentiated r times to generate an

explicit relation between y and u , the system is said to have relativedegree r . For linear systems this terminology expressed as # poles

−# zeros.

For any controllable system of order n, by taking at most ndifferentiations, the control input will appear to any output, i.e., r ≤ n. If the control input never appears after more than n differentiations, the

system would not be controllable.


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Feedback Linearzation Internal dynamics: a part of dynamics which is unobservable in the

input-output linearization. In the example it can be x 3 = x 21 + 1

x 2+1 (y d (t )− k 1e − k 2e + f )

The desired performance of the control based on the reduced-order modeldepends on the stability of the internal dynamics.

stability in BIBO sense Example: Consider

x 1x 2

=

x 32 + u

u

(2)

y = x 1

Control objective: y tracks y d . First differentiations of y linear I/O relation The control law u = −x 32 − e (t ) + y d (t ) exp. convergence of e :

e + e = 0 Internal dynamics: x 2 + x 32 = y d − e

Since e and y d are bounded (y d (t )− e ≤ D ) x 2 is ultimately bounded.Farzaneh Abdollahi Nonlinear Control Lecture 9 15/75

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I/O linearization can also be applied to stabilization (y d (t ) ≡ 0): For previous example the objective will be y and y will be driven to zero

and stable internal dynamics guarantee stability of the whole system. No restriction to choose physically meaningful h(x ) in y = h(x )

Different choices of output function leads to different internal dynamicswhich some of them may be unstable.

When the relative degree of a system is the same as its order: There is no internal dynamics The problem will be input-state linearization


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Summary Feedback linearization cancels the nonlinearities in a nonlinear system s.t.

the closed-loop dynamics is in a linear form.

Canceling the nonlinearities and imposing a desired linear dynamics, canbe applied to a class of nonlinear systems, named companion form, orcontrollability canonical form.

When the nonlinear dynamics is not in a controllability canonical form,input-state linearization technique is employed:

1. Transform input and state into companion canonical form2. Use standard linear techniques to design controller

For tracking a desired traj, when y is not directly related to u , I/O

linearizaton is applied:1. Generating a linear input-output relation (take derivative of y r ≤ n times)2. Formulating a controller based on linear control

Relative degree: # of differentiating y to find explicate relation to u .

If r = n, there are n − r internal dynamics that their stability be checked.Farzaneh Abdollahi Nonlinear Control Lecture 9 17/75

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Internal Dynamics of Linear Systems

In general, directly determining the stability of the internal dynamics isnot easy since it is nonlinear. nonautonomous, and coupled to the“external” closed-loop dynamics.

We are seeking to translate the concept of internal dynamics to the morefamiliar context of linear systems.

Example: Consider the controllable, observable system x 1x 2

=

x 2 + u

u

(3)

y = x 1

Control objective: y tracks y d . First differentiations of y y = x 2 + u The control law u = −x 2− e (t ) + y d (t ) exp. convergence of e : e + e = 0 Internal dynamics: x 2 + x 2 = y d − e e and y d are bounded x 2 and therefore u are bounded.


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Now consider a little different dynamics

x 1x 2

=

x 2 + u

−u

(4)y = x 1

using the same control law yields the following internal dynamics

x 2 − x 2 = e (t )− y d

Although y d and y are bounded, x 2 and u diverge to ∞ as t → ∞ why the same tracking design method yields different results?

Transfer function of (3) is: W 1(s ) = s +1s 2

. Transfer function of (4) is: W 2(s ) = s

−1

s 2 . ∴ Both have the same poles but different zeros The system W 1 which is minimum-phase tracks the desired trajectory

perfectly. The system W 2 which is nonminimum-phase requires infinite effort for

tracking.Farzaneh Abdollahi Nonlinear Control Lecture 9 19/75

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Internal Dynamics

Consider a third-order linear system with one zero

x = Ax + bu , y = c T x (5)

Its transfer function is: y = b 0+b 1s a

0+a

1s +a

2s 2+s 3

u

First transform it into the companion form:

z 1z 2z

3

=

0 1 00 0 1

−a

0 −a

1 −a

2

z 1z 2z

3

+

001

u (6)

y = [b 0 b 1 0]

z 1

z 2z 3


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In second derivation of y , u appears:y = b 0z 3 + b 1(

−a0z 1

−a1z 2

−a2z 3 + u )

∴ Required number of differentiations (the relative degree) is indeed thesame as # of poles- # of zeros Note that: the I/O relation is independent of the choice of state variables two differentiations is required for u to appear if we use (5).

The control law: u = (a0z 1 + a1z 2 + a2z 3 − b 0

b 1 z 3) +

1

b 1 (−k 1e − k 2e + y d ) ∴ an exp. stable tracking is guaranteed The internal dynamics can be described by only one state equation

z 1 can complete the state vector,( z 1, y , and y are related to z 1, z 2 and z 3through a one-to-one transformation).

z 1 = z 2 = 1b 1 (y − b 0z 1) y is bounded stability of the internal dynamics depends on −b 0

b 1 If the system is minimum phase the internal dynamics is stable

(independent of initial conditions and magnitude of desired trajectory)


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Zero-Dynamics

For linear systems the stability of the internal dynamics is determined bythe locations of the zeros.

To extend the results for nonlinear systems the concept of zero should bemodified.

Extending the notion of zeros to nonlinear systems is not trivial In linear systems I/O relation is described by transfer function which zeros

and poles are its fundamental components. But in nonlinear systems wecannot define transfer function

Zeros are intrinsic properties of a linear plant. But for nonlinear systemsthe stability of the internal dynamics may depend on the specific controlinput.

Zero dynamics: is defined to be the internal dynamics of the system whenthe system output is kept at zero by the input.(output and all of itsderivatives)


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For dynamics (2), the zero dynamics is x 2 + x 32 = 0 we find input u to maintain the system output at zero uniquely (keep x 1

zero in this example), By Layap. Fcn V = x 22 it can be shown it is a.s

For linear system (5), the zero dynamics is z 1 + (b 0/b 1)z 1 = 0

∴ The poles of the zero-dynamics are exactly the zeros of the system.

In linear systems, if all zeros are in LHP g.a.s. of the zero-dynamics g.s. of

internal dynamics.

In nonlinear systems, no results on the global stability only local stability is guaranteed for the internal dynamics even if the

zero-dynamics is g.e.s.

Zero-dynamics is an intrinsic feature of a nonlinear system, which does notdepend on the choice of control law or the desired trajectories.

Examining the stability of zero-dynamics is easier than examining the stability of internal dynamics,But the result is local. Zero-dynamics only involves the internal states

Internal dynamics is coupled to the external dynamics and desired trajs.Farzaneh Abdollahi Nonlinear Control Lecture 9 23/75

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Zero-Dynamics Similar to the linear case, a nonlinear system whose zero dynamics is

asymptotically stable is called an asymptotically minimum phase system,

If the zero-dyiamics is unstable, different control strategies should besought

As summary control design based on input-output linearization is in three

steps:1. Differentiate the output y until the input u appears2. Choose u to cancel the nonlinearities and guarantee tracking convergence3. Study the stability of the internal dynamics

If the relative degree associated with the input-output linearization is the

same as the order of the system the nonlinear system is fully linearized satisfactory controller

Otherwise, the nonlinear system is only partly linearized whether ornot the controller is applicable depends on the stability of the internal

dynamics.Farzaneh Abdollahi Nonlinear Control Lecture 9 24/75

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Preliminary Mathematics

Vector function f : R n → R n is called a vector field in R n. Smooth vector field: function f (x ) has continuous partial derivatives of

any required order.

Gradient of a smooth scalar function h(x ) is denoted by a row vector

∇h = ∂ h

∂ x , where (∇h) j = ∂ h

∂ x j

Jacobian of a vector field f (x):an n × n matrix ∇f = ∂ f ∂ x , where

(∇f )ij = ∂ f i ∂ x j

Lie derivative of h with respect to f is a scalar function defined by

Lf h = ∇hf , where h : R n

→ R : a smooth scalar, f : R n

→ R n

: a smoothvector field.

If g is another vector field: LgLf h = ∇(Lf h)g

L0f h = h; Li

f h = Lf (Li −1f h) = ∇(Li −1

f h)f


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Example: For single output system x = f (x), y = h(x) then

y = ∂ h

∂ x

x = Lf h

y = ∂ [Lf h]

∂ x x = L2

f h

If V is a Lyap. fcn candidate, its derivative V can be written as Lf V .

Lie bracket of f and g is a third vector field defined by[f , g] = ∇gf −∇f g, where f and g two vector field on R n.

The Lie bracket [f, g] is also written as ad f g (ad stands for ”adjoint”).

ad 0f g = g ; ad i f g = [f , ad i −1f g ], i = 1,...

Example: Consider x = f (x) + g(x)u wheref =

−2x 1 + ax 2 + sin x 1−x 2 cos x 1

, g =

0

cos(2x 1)

So the Lie bracket is:

[f, g] = −a cos(2x 1)

cos x 1cos (2x 1)− 2 sin(2x 1)(−2x 1 + ax 2 + sin x 1)


O li F db k Li i P li i M h i I S Li i i I O Li i i

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Lemma: Lie brackets have the following properties:

1. bilinearity:

[α1f 1 + α2f 2, g] = α1[f 1, g] + α2[f 2, g]

[f , α1g1 + α2g2] = α1[f , g1] + α2[f , g2]

where f , f 1, f 2, g g1, g2 are smooth vector fields and α1 and α2 are constant scalars.2. skew-commutativity:

[f , g] = −[g, f ]

3. Jacobi identity

Lad f gh = Lf Lgh − LgLf h

where h is a smooth fcn.


O tli F db k Li ti P li i M th ti I t St t Li i ti I t O t t Li i ti

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Diffeomorphism

The concept of diffeomorphism can be applied to transform a nonlinearsystem into another nonlinear system in terms of a new set of states.

Definition: A function φ :

Rn

→Rn defined in a region Ω is called a

diffeomorphism if it is smooth, and if its inverse φ−

1 exists and is smooth. If the region Ω is the whole space Rn

φ(x ) is global diffeomorphism

Global diffeomorphisms are rare,we are looking for local diffeomorphisms .

Lemma: Let φ(x ) be a smooth function defined in a region Ω in

Rn. If

the Jacobian matrix ∇φ is non-singular at a point x = x 0 of Ω, then φ(x )defines a local diffeomorphism in a subregion of Ω


O tline Feedback Linear ation Preliminar Mathematics Inp t State Lineari ation Inp t O tp t Lineari ation

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Diffeomorphism Consider the dynamic system described by

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

Let the new set of states z = φ(x ) z = ∂φ∂ x

x = ∂φ∂ x

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

The new state-space representation

z = f ∗

(z ) + g ∗

(z )u , y = h∗

(z )

where x = φ−1(z ). Example of a non-global diffeomorphism: Consider

z 1

z 2

= φ(x ) = 2x 1 + 5x 1x 22

3sin x 2

Its Jacobian matrix: ∂φ∂ x

=

2 + 5x 22 10x 1x 2

0 3 cos x 2

.

rank is 2 at x = (0, 0) local diffeomorphism around the origin whereΩ =

(x 1, x 2),

|x 2

|< π/2

.

outside the region, the inverse of φ does not uniquely exist.Farzaneh Abdollahi Nonlinear Control Lecture 9 29/75

Outline Feedback Linearzation Preliminary Mathematics Input State Linearization Input Output Linearization

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Frobenius Theorem An important tool in feedback linearization

Provide necess. and suff. conditions for solvability of PDEs.

Consider a PDE with (n=3):

∂ h

∂ x 1f 1 +

∂ h

∂ x 2f 2 +

∂ h

∂ x 3f 3 = 0

∂ h∂ x 1

g 1 + ∂ h∂ x 2

g 2 + ∂ h∂ x 3

g 3 = 0 (7)

where f i (x 1, x 2, x 3), g i (x 1, x 2, x 3) (i = 1, 2, 3) are known scalar fcns andh(x 1, x 2, x 3) is an unknown function.

This set of PDEs is uniquely determined by the two vectorsf = [f 1 f 2 f 3]T , g = [g 1 g 2 g 3]T .

If the solution h(x 1, x 2, x 3) exists, the set of vector fields f , g iscompletely integrable.

When the equations are solvable?Farzaneh Abdollahi Nonlinear Control Lecture 9 30/75


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Frobenius Theorem Frobenius theorem states that Equation (7) has a solution h(x 1, x 2, x 3) iff

there exists scalar functions α1(x 1, x 2, x 3) and α2(x 1, x 2, x 3) such that

[f , g ] = α1 f + α2 g

i.e., if the Lie bracket of f and g can be expressed as a linear combination of f and g

This condition is called involutivity of the vector fields f , g .

Geometrically, it means that the vector field [f , g ] is in the plane formed by thetwo vectors f and g

The set of vector fields f , g is completely integrable iff it is involutive.

Definition (Complete Integrability): A linearly independent set of vector fields f 1, f 2,..., f m on R n is said to be completely integrable, iff, there exist n −m scalar fcns h1(x ), h2(x ),..., hn−m(x ) satisfying the system of PDEs:

∇hi f j = 0

where 1 ≤ i ≤ n −m, 1 ≤ j ≤ m and ∇hi are li nearly i nde p e nd ent.Farzaneh Abdollahi Nonlinear Control Lecture 9 31/75


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Outline Feedback Linearzation Preliminary Mathematics Input State Linearization Input Output Linearization

Number of vectors: m, dimension of the vectors: n, number of unknownscalar fcns hi : (n-m), number of PDEs: m(n-m)

Definition (Involutivity): A linearly independent set of vector fields f 1, f 2, ..., f m on R n is said to be involutive iff, there exist scalar fcns αijk : R N −→ R s.t.

[f i , f j ](x ) =

m

k =i

αijk (x ) f k (x ) ∀ i , j

i.e., the Lie bracket of any two vector fields from the set f 1, f 2,..., f mcan be expressed as the linear combination of the vectors from the set. Constant vector fields are involutive since their Lie brackets are zero A set composed of a single vector is involutive:

[f , f ] = (∇f )f − (∇f )f = 0

Involutivity means:

rank (f 1(x ) .... f m(x )) = rank (f 1(x ) .... f m(x ) [f i , f j ](x ))

for all x and for all i , j .Farzaneh Abdollahi Nonlinear Control Lecture 9 32/75


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Out e eedbac ea at o e a y at e at cs put State ea at o put Output ea at o

Frobenius Theorem Theorem (Frobenius): Let f 1, f 2, ...., f m be a set of linearly

independent vector fields. The set is completely integrable iff it is involutive.

Example: Consider the set of PDEs:

4x 3∂ h

∂ x 1 − ∂ h

∂ x 2= 0

−x 1∂ h

∂ x 1+ (x 23 − 3x 2)

∂ h

∂ x 2+ 2x 3

∂ h

∂ x 3= 0

The associated vector fields are f 1, f 2

f 1 = [4x 3 − 1 0]T f 2 = [−x 1 (x 23 − 3x 2) 2x 3]T

We have [f 1, f 2] = [−12x 3 3 0]T

Since [f 1, f 2] = −3f 1 + 0f 2, the set f 1, f 2 is involutive and the set of

PDEs are solvable.Farzaneh Abdollahi Nonlinear Control Lecture 9 33/75


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y p p p

Input-State Linearization Consider the following SISO nonlinear system

x = f (x ) + g (x )u (8)

where f and g are smooth vector fields

The above system is also called “linear in control” or “affine”

If we deal with the following class of systems:

x = f (x ) + g (x )w (u + φ(x ))

where w is an invertible scalar fcn and φ is an arbitrary fcn

We can use v = w (u + φ(x )) to get the form (8). Control design is based on v and u can be obtained by inverting w :

u = w −1

(v )− φ(x )

Now we are looking for Conditions for system linearizability by an input-state transformation A technique to find such transformations A method to design a controller based on such linearization technique



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Input-State Linearization Definition: Input-State Linearization The nonlinear system ( 8 ) where f (x )

and g (x ) are smooth vector fields in R n is input-state linearizable if there exist region Ω in R n, a diffeomorphism mapping φ : Ω −→ R n, and a control law:

u = α(x ) + β (x )v

s.t. new state variable z = φ(x ) and new input variable v satisfy an LTI relation:

z = Az + Bv

A =

0 1 0 . . . 00 0 1 . . . .. . . . . . .. . . . . . .

0 0 0 . . . 10 0 0 . . . 0

B =

00...

01

(9)

The new state z is called the linearizing state and the control law u is called thelinearizing control law

Let z = z (x )Farzaneh Abdollahi Nonlinear Control Lecture 9 35/75


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(9) is the so-called linear controllability or companion form

This linear companion form can be obtained from any linear controllablesystem by a transformation if u leads to a linear system, (9) can be

obtained by another transformation easily. This form is an special case of Input-Output linearization leading to

relative degree r = n.

Hence, if the system I/O linearizable with r = n, it is also I/S linearizable.

On the other hand, if the system is I/S linearizable, it is also I/Olinearizable with y = z , r = n.



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Lemma: An nth order nonlinear system is I/S linearizable iff there exists a scalar fcn z 1(x ) for which the system is I/O linearizable with r = n.

Still no guidance on how to find the z 1(x ). Conditions for Input-State Linearization:

Theorem: The nonlinear system ( 8 ) with f (x ) and g (x ) being smoothvector field is input-state linearizable iff there exists a region Ω s.t. the following conditions hold:

The vector fields g , ad f g , ... ad f n−1g are linearly independent in Ω

The set g , ad f g , ... ad f n−2g is involutive in Ω

The first condition: can be interpreted as a controllability condition For linear system, the vector field above becomes B , AB , ... An−1B Linear independency ≡ invertibility of controllability matrix



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The second condition is always satisfied for linear systems since the vector fields are constant, but

for nonlinear system is not necessarily satisfied. It is necessary according to Ferobenius theorem for existence of z 1(x ).

Lemma: If z (x ) is a smooth vector field in Ω, then the set of equations

Lg z = Lg Lf z = ... = Lg Lf k z = 0

is equivalent to Lg z = Lad f g z = ... = Lad f

k g z = 0

Proof: Let k = 1, from Jacobi’s identity, we have

Lad f g z = Lf Lg z − Lg Lf z = 0− 0 = 0

When k = 2, we have from Jacobi’s identity:

Lad f 2g z = Lf 2Lg z

−2Lf Lg Lf z + Lg Lf

2z = 0

−0 + 0 = 0



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Proof of the linearization theorem:

Necessity: Suppose state transformation z = z (x ) and input transformation

u = α(x ) + β (x )v s.t. z and v satisfy (9), i.e.

z 1 = ∂ z 1

∂ x (f + gu ) = z 2

similarly:∂ z 1∂ x

f + ∂ z 1

∂ x gu = z 2

∂ z 2∂ x

f + ∂ z 2

∂ x gu = z 3

...∂ z n∂ x

f + ∂ z n

∂ x gu = v



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z 1, ..., z n−1 are independent of u ,

Lg z 1 = Lg z 2 = ... Lg z n−1 = 0, Lg z n = 0

Lf z i = z i +1, i = 1, 2, ..., n − 1

Use, z = [z 1 Lf z 1, ... Lf n−1z 1]T to get

z k = z k +1, k = 1, ... n − 1

z n = Lf nz 1 + Lg Lf

n−1z 1u

The above equations can be expressed in terms of z 1 only∇z 1ad f

k g = 0, k = 0, 1, 2, ..., n − 2 (10)

∇z 1ad f n−1g = (−1)n−1Lg z n (11)

First note that for above eqs to hold, the vector field g , ad f g , ..., ad f n−

1g must be linearly independent.

If for some i (i ≤ n − 1) there exist scalar fcns α1(x ), ... αi −1(x ) s.t.

ad f i g =

i −1

k =0

αk ad f k g



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We, then have:

∴ ad f n−1g =

n−2

k =n−i −1

αk ad f k g

∴ ∇z 1ad f n−1g =

n−2k =n−i −1

αk ∇z 1ad f k g = 0 (12)

∴ Contradicts with (11). The second property is that ∃ a scalar fcn z 1 that satisfy n − 1 PDEs∇z 1ad f

k g = 0

∴ From the necessity part of Frobenius theorem, we conclude that the set

of vector field must be involutive. Sufficient condition

Involutivity condition =⇒ Frobenius theorem, ∃ a scalar fcn z 1(x ):

Lg z 1 = Lad f g z 1 = ... Lad f k g z 1 = 0, implying

Lg z 1 = Lg Lf z 1 = ... Lg Lf k z 1 = 0



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Define the new sets of variable as z = [z 1 Lf z 1 ... Lf n−1z 1]T , to get

z k = z k +1 k = 1, ..., n − 1

z n = Lf nz 1 + Lg Lf n−

1z 1u (13)

The question is whether Lg Lf n−1z 1 can be equal to zero.

Since g , ad f g , ..., ad f n−1g are linearly independent in Ω:

Lg Lf

n−1z 1

= (−

1)n−1Lad f

n−1

g z

1

We must have Lad f n−1g z 1 = 0, otherwise the nonzero vector ∇z 1satisfies

∇z 1 [g , ad f g , ..., ad f n−1g ] = 0

i.e. ∇z 1 is normal to n linearly independent vector =⇒ impossible

Now, we have:

z n = Lf nz 1 + Lg L

n−1f z 1u = a(x ) + b (x )u



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Now, select u = 1b (x ) (−a(x ) + v ) to get:

z n = v

implying input-state linearization is obtained. Summary: how to perform input-state Linearization

1. Construct the vector fields g , ad f g , ... ad f n−1g

2. Check the controllability and involutivity conditions

3. If the conditions hold, obtain the first state z 1 from:∇z 1ad f i g = 0 i = 0, ..., n − 2

∇z 1ad f n−1g = 0

4. Compute the state transformation z (x ) = [z 1 Lf z 1 ... Lf n−1z 1]T and the

input transformation u = α(x ) + β (x )v :α(x ) = − Lf

nz 1Lg Lf

n−1z 1

β (x ) = 1

Lg Lf n−1z 1



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Example:A single-link flexible-joint manipulator:

The link is connected to the motor shaft via a torsional spring Equations of motion:

I q 1 + MgLsinq 1 + K (q 1 − q 2) = 0

J q 2 − K (q 1 − q 2) = u



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Example:A single-link flexible-joint manipulator:

Equations of motion:I q 1 + MgLsinq 1 + K (q 1 − q 2) = 0J q 2 − K (q 1 − q 2) = u

nonlinearities appear in the first equation and torque is in the second equation

Let:

x =

q 1q 1q 2q 2

, f =

x 2−MgL

I sinx 1 − K

I (x 1 − x 3)

x 4K J (x 1 − x 3)

, g =

0001J

Controllability and involutivity conditions:

[g ad f g ad f 2g ad f

3g ] =

0 0 0 −K IJ

0 0 K IJ

00 − 1

J 0 K

J 21J

0 − K J 2

0



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Example: Cont’d

It’s full rank for k > 0 and IJ < ∞ =⇒ vector fields are linearlyindependent

Vector fields are constant =⇒ involutive

The system is input-state linearizable

Computing z = z (x ), u = α(x ) + β (x )v

∂ z 1∂ x 2

= 0, ∂ z 1∂ x 3

= 0, ∂ z 1∂ x 4

= 0, ∂ z 1∂ x 1

= 0

Hence, z 1 is the fcn of x 1 only. Let z 1 = x 1, then

z 2

= ∇

z 1

f = x 2

z 3 = ∇z 2f = −MgL

I sinx 1 − K

I (x 1 − x 3)

z 4 = ∇z 3f = −MgL

I x 2cosx 1 − K

I (x 2 − x 4)



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Example: Cont’d The input transformation is given by:

u = (v −∇z 4f )/(∇z 4g ) = IJ

K (v − a(x ))

a(x ) = MgL

I sinx 1(x 22 +

MgL

I cosx 1 +

K

I )

+ K

I (x 1 − x 3)(

K

I +

K

J +

MgL

I cosx 1)

As a result, we get the following set of linear equationsz 1 = z 2, z 2 = z 3

z 3 = z 4, z 4 = v

The inverse of the state transformation is given by:x 1 = z 1, x 2 = z 2

x 3 = z 1 + I

K

z 3 +

MgL

I sinz 1

x 4 = z 2 + I

K z 4 +

MgL

I

z 2c o sz 1Farzaneh Abdollahi Nonlinear Control Lecture 9 47/75


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Example Cont’d

State and input transformations are defined globally In this example, transformed state have physical meaning, z 1 : link

position, z 2 : link velocity, z 3 : link acceleration, z 4 : link jerk.

It could be obtained by I/O linearization, i.e. by differentiating the output

q 1. (4 times) We can transform the inequality (11) to a normalized equation by setting∇z 1ad f

n−1g = 1 resulting in:

[ad f 0g ad f

1g ... ad f n−2g ad f

n−1g ]

∂ z 1∂ x 1

...∂ z 1∂ x n

=

0

0...01



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Control Design

Once, the linearized dynamics is obtained, either a tracking orstabilization problem can be solved

For instance, in flexible-joint manipulator case, we have

z (4)1 = v

Then, a tracking controller can be obtained asv = z

(4)d 1 − a3z

(3)1 − a2

¨z 1 − a1˙z 1 − a0z 1

where z 1 = z 1 − z d 1.

The error dynamics is then given by:z

(4)1 + a3z

(3)1 + a2

¨z 1 + a1˙z 1 + a0z 1 = 0

The above dynamics is exponentially stable if ai are selected s.t.

s 4 + a3s 3 + a2s 2 + a1s + a0 is Hurwitz



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Input-Output Linearization

Consider the system:

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

y = h(x ) (14)

Input-output linearization yields a linear relationship between the outputy and the input v (similar to v in I/S Lin.) How to generate a linear I/O relation for such systems? What are the internal dynamics and zero-dynamics associated with this I/O

linearization How to design a stable controller based on this technique?

Performing I/O Linearization The basic approach is to differentiate the output y until the input u

appears, then design u to cancel nonlinearities Sometime, cancelation might not be possible due to the undefined relative

degree.Farzaneh Abdollahi Nonlinear Control Lecture 9 50/75


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Well Defined Relative Degree

Differentiate y and express it in the form of Lie derivative:

y = ∇h(f + gu ) = Lf h(x ) + Lg h(x )u

if Lg h(x ) = 0 for some x = x 0 in Ωx , then continuity implies thatLg h(x ) = 0 in some neighborhood Ω of x 0. Then, the input

transformationu =

1

Lg h(x )(−Lf h(x ) + v )

results in a linear relationship between y and v , namely y = v .

If Lg h(x ) = 0 for all x ∈ Ωx , differentiate y to obtainy = Lf

2h(x ) + Lg Lf h(x )u

If Lg Lf h(x ) = 0 for all x ∈ Ωx , keep differentiating until for someinteger r , Lg Lf

r −1h(x )

= 0 for some x = x 0

∈Ωx



H h

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Hence, we have

y (r ) = Lf r h(x ) + Lg Lf

r −1h(x )u (15)

and the control law

u = 1

Lg Lf r −1h(x )

(−Lf r h(x ) + v )

yields a linear mapping:

y (r ) = v

The number r of differentiation required for u to appear is called therelative degree of the system.

r ≤ n, if r = n, the input-state realization is obtained with z 1 = y . Definition: The SISO system is said to have a relative degree r in Ω if:

Lg Lf i h(x ) = 0 0 ≤ i ≤ r − 2

Lg Lf r −1h(x ) = 0



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Undefined Relative Degree

Sometimes, we are interested in the properties of a system about aspecific operating point x 0.

Then, we say the system has relative degree r at x 0 if

Lg Lf r −1h(x 0) = 0

However, it might happen that Lg Lf r −1h(x ) is zero at x 0, but nonzero in

a close neighborhood of x 0.

The relative degree of the nonlinear system is then undefined at x 0.

Example:

x = ρ(x , x ) + u

where ρ is a smooth nonlinear fcn. Define x = [x x ]T and let y = x =⇒the system is in companion form with r = 2.



However if we define y = x 2 then:

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However, if we define y = x , then:

y = 2x x

y = 2x x + 2x 2 = 2x ρ(x , x ) + 2xu + 2x 2 =⇒Lg Lf h = 2x (16)

The system has neither relative degree 1 nor 2 at x 0 = 0.

Sometime, change of output leads us to a solvable problem. We assume that the relative degree is well defined. Normal Forms

When, the relative degree is defined as r ≤ n, using y , y , ..., y (r −1), wecan transform the system into the so-called normal form.

Normal form allows a formal treatment of the notion of internal dynamicsand zero dynamics.

Let µ = [µ1 µ2 ... µr ]T = [y y ... y (r −1)]T

in a neighborhood Ω of a point x 0.



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Normal Form

The normal form of the system can be written as

µ =

µ2

...µr

a(µ, Ψ) + b (µ, Ψ)u

(17)

Ψ = w (µ, Ψ) (18)y = µ1

The µi and Ψ j are called normal coordinate or normal states .

The first part of the Normal form, (17) is another form of (15), however in (18)

the input u does not appear. The system can be transformed to this form if the state transformation φ(x ) is a

local diffeomorphism: φ(µ1 ... µr Ψ1 ... Ψn−r )T

To show that φ is a diffeomorphism, we must show that the Jacobian isinvertible, i.e.

∇µi and

∇Ψi are all linearly independent.



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Normal Form

∇µi are linearly independent =

⇒ µ can be part of state variables, (µ is

output and its r − 1 derivatives) There exist n − r other vector fields that complete the transformation

Note that u does not appear in (18), hence:

∇Ψ j g = 0 1

≤ j

≤ n

−r

∴ Ψ can be obtained by solving n − r PDE above.

Generally, internal dynamics can be obtained simpler by intuition.

Zero Dynamics System dynamics have two parts:

1. external dynamics µ

2. internal dynamics Ψ

For tracking problems (y −→ y d ), one can easily design v once the linearrelation is obtained.

The question is whether the internal dynamics remain boundedFarzaneh Abdollahi Nonlinear Control Lecture 9 56/75


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Zero-Dynamics

Stability of the zero dynamics (i.e. internal dynamics when y is kept 0)gives an idea about the stability of internal dynamics

u is selected s.t. y remains zero at all time.

y (r )(t ) = Lf r h(x ) + Lg Lf

r −1h(x )u 0

≡0 =

⇒u 0(x ) =

−Lf r h(x )Lg Lf

r −1h(x )

∴ In normal form:

µ = 0Ψ = w (0, Ψ)

u 0(Ψ) = −a(0, Ψ)

b (0, Ψ) (19)



E l

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Example

Considerx =

−x 1

2x 1x 2 + sinx 22x 2

+

e 2x 2

1/20

u

y = h(x ) = x 3

We have y = 2x 2

y = 2x 2 = 2(2x 1x 2 + sinx 2) + u

The system has relative degree r = 2 and

Lf h(x ) = 2x 2

Lg h(x ) = 0

Lg Lf h(x ) = 1



E l C ’d

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Example Cont’d

To obtain the normal form

µ1 = h(x ) = x 3

µ2 = Lf h(x ) = 2x 2

The third function Ψ(x ) is obtained by

Lg Ψ = ∂ Ψ∂ x 1

e 2x 2 + 12

∂ Ψ∂ x 2

= 0

One solution is Ψ(x ) = 1 + x 1 − e 2x 2

Consider the jacobian of state transformation z = [µ1 µ2 Ψ]T . The

Jacobian matrix is

0 0 10 2 01

−2e 2x 2 0



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Example Cont’d The Jacobian is non-singular for any x . In fact, inverse transformation is given

by:

x 1 = −1 + Ψ + e µ2

x 2 = 1

2µ2

x 3 = µ1

State transformation is valid globally and the normal form is given by:

µ1 = µ2

µ2 = 2(

−1 + Ψ + e µ2 )µ2 + 2sin(µ2/2) + u

Ψ = (1−Ψ− e µ2 )(1 + 2µ2e µ2 )− 2sin(µ2/2)e µ2 (20)

Zero dynamics is obtained by setting µ1 = µ2 = 0 =⇒

Ψ = −Ψ (21)

which is clearl known to be loball ex onentiall stable.Farzaneh Abdollahi Nonlinear Control Lecture 9 60/75


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Zero-Dynamics

In order to obtain the zero dynamics, it is not necessary to put the systeminto normal form

since µ is known, we can intuitively find n − r vector to complete thetransformation.

As mention before, zero dynamics is obtained by substituting u 0 for u ininternal dynamics.

Definition: A nonlinear system with asymptotically stable zero dynamics is called asymptotically minimum phase

If the zero dynamics is stable for all x , the system is globally minimumphase, otherwise the results are local.



Local Asymptotic Stabilization

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Local Asymptotic Stabilization

Consider again the nonlinear system

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

y = h(x ) (22)

Assume that the system is I/O linearized, i.e.

y (r ) = Lf r h(x ) + Lg Lf

r −1h(x )u (23)

and the control lawu =

1

Lg Lf r −1h(x )

(−Lf r h(x ) + v ) (24)

yields a linear mapping: y (r ) = v

Now let v be chosen as

v = −k r −1y (r −1) − ... − k 1y − k 0y (25)

where k i are selected s.t. K (s ) = s r + k r −1s r −1 + ... + k 1s + k 0 is Hurwitz



Then, provided that the zero-dynamics is asymptotically stable, the control law(24) and (25) locally stabilize the whole system:

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(24) and (25) locally stabilize the whole system:

Theorem: Suppose the nonlinear system ( 22 ) has a well defined relative degree

r and its associated zero-dynamics is locally asymptotically stable. Now, if k i are selected s.t. K (s ) = s r + k r −1s r −1 + ... + k 1s + k 0 is Hurwitz, then the control law ( 24 ) and ( 25 ) yields a locally asymptotically stable system.

Proof: First, write the closed-loop system in a normal form:

µ =

0 1 0 . . . 00 0 1 . . . .. . . . . . .. . . . . . .0 0 0 . . . 1−k 0 −k 1 −k 2 . . . −k r −1

= Aµ

Ψ = w (µ, Ψ) = A1

µ + A2

Ψ + h.o .t .

h.o.t. is higher order terms in the Taylor expansion about x 0 = 0.The above Eq. can be written as:

d

dt

µΨ

=

A 0A1 A2

µΨ

+ h.o .t .



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Now, since the zero dynamics is asymptotically stable, its linearization Ψ = A2Ψ

is either asymptotically stable or marginally stable. If A2 is asymptotically stable, then all eigenvalues of the above system

matrix are in LHP and the linearized system is stable and the nonlinearsystem is locally asymptotically stable

If A2 is marginally stable, asymptotic stability of the closed-loop system

was shown in (Byrnes and Isidori, 1988). Comparing the above method to local stabilization and using linear control:

the above stabilization method can treat systems whose linearizationscontain uncontrollable but marginally stable modes,

while linear control methods requires the linearized system to be strictly

stabilizable



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For stabilization where state convergence is required, we can freelychoose y = h(x ) to make zero-dynamics a.s.

Example: Consider the nonlinear system:x 1 = x 21 x 2

x 2 = 3x 2 + u

System linearization at x = 0:x 1 = 0

x 2 = 3x 2 + u

thus has an uncontrollable mode



Example (cont’d)

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p ( ) Define y = −2x 1 − x 2 =⇒

y = −2x 1 − x 2 = −2x 21 x 2 − 3x 2 − u

Hence, the relative degree r = 1 and the associated zero-dynamics is

x 1 = −2x 31

The zero-dynamics is asymptotically stable, hence the control lawu = −2x 21 x 2 − 4x 2 − 2x 1 locally stabilizes the system

Global Asymptotic Stabilization

Stability of the zero-dynamics only guarantees local stability unless relative

degree is n in which case there is no internal dynamics Can the idea of I/O linearization be used for global stabilization problem?

Can the idea of I/O linearization be used for systems with unstable zerodynamics?




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Global stabilization approach based on partial feedback linearization is tosimply regard the problem as a standard Lyapunov controller designproblem

But simplified by the fact that in normal form part of the system

dynamics is now linear. The basic idea is to view µ as the input to the internal dynamics and Ψ

as its output. The first step: find the control law µ0 = µ0(Ψ) which stabilizes the internal

dynamics with the corresponding Lyapunov fcn V 0. Then: find a Lyapunov fcn candidate for the whole system (as a modified

version of V 0) and choose the control input v s.t. V be a Lyapunov fcn forthe whole closed-loop dynamics.



Example:

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Example:

Consider a nonlinear system with the normal form:

y = v

z + z 3 + yz = 0 (26)

where v is the control input and Ψ = [z z ]T

Considering y as an input to internal dynamics (26), it would beasymptotically stabilized by the choice of y = y 0 = z 2

Let V 0 be a Lyap. fcn:

V 0 =

1

2 z 2

+

1

4 z 4

Differentiating V 0 along the actual dynamics results in

V 0 = −z 4 − z z (y − z 2)



Example Cont’d

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Example Cont d

Consider the Lyap. fcn candidate, obtained by adding a quadratic “error”term in y − y 0 to V 0

V = V 0 + 1

2(y − z 2)2

∴ V = −z 4 + (y − z 2)(v − 3z z )

The following choice of control action will then make V n.d.

v =

−y + z 2 + 3z z

∴ V = −z 4 − (y − z 2)2

Application of Invariant-set theorem shows all states converges to zero



Example: A non-minimum phase system

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Example: A non minimum phase system Consider the system dynamics

y = v z + z 3 − z 5 + yz = 0

where again Ψ = [z z ]T

The system is non-minimum phase since its zero-dynamics is unstable The zero-dynamics would be stable if we select y = 2z 4:

V 0 = 1

2 z 2 +

1

6z 6 V 0 = −z 4 − z z (y − 2z 4)

Consider the Lyap. fcn candidate

V = V 0 + 1

2

(y

−2z 4)2

V =

−z 4 + (y

−2z 4)(v

−8z 3z

−z z )

suggesting the following choice of control law

v = −y + 2z 4 + 8z 3z + z z V = −z 4 − (y − 2z 4)2

Application of Invariant-set theorem shows all states converges to zero



Tracking Control

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Tracking Control I/O linearization can be used in tracking problem

Let µd = [y d y d ... y (r −1)d ]T and the tracking error µ(t ) = µ(t )− µd (t )

Theorem: Assume the system ( 22 ) has a well defined relative degree r and µd is smooth and bounded and that the solution Ψd :

Ψd = w (µd , Ψd ), Ψd (0) = 0

exists and bounded and is uniformly asymptotically stable. Choose k i s.t K (s ) = s r + k r −1s r −1 + ... + k 1s + k 0 is Hurwitz, then by using

u = 1

Lg Lf r −1µ1

[−Lf r µ1 + y

(r )d − k r −1µr − ... − k 0µ1] (27)

the whole system remains bounded and the tracking error µ converge to zero exponentially.

Proof: Refer to Isidori (1989).

For perfect tracking µ(0)

≡µd (0)



Tracking Control for Non-minimum Phase Systems:

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g y

The tracking control (27) cannot be applied to non-minimum phase

systems since they cannot be inverted Hence we cannot have perfect or asymptotic tracking and should seek

controllers that yields small tracking errors One approach is the so-called Output redefinition

The new output y 1 is defined s.t. the associated zero-dynamics is stable y 1 is defined s.t. it is close to the original output y in the frequency range

of interest Then, tracking y 1 also implies good tracking the original output y

Example: Consider a linear system

y =

1− s b

B 0(s )A(s )

u b > 0

Perfect/asymptotic tracking is impossible due to the presence of zero @s = b



Example Cont’d

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p

Let us redefine the output as

y 1 = B 0(s )A(s )

u

with the desired output for y 1 be simply y d A controller can be found s.t. y 1 asymptotically tracks y d . What about

the actual tracking error?

y (s ) =

1− s

b

y 1 =

1− s

b

y d

Thus, the tracking error is proportional to the desired velocity y d :

y (t )− y d (t ) = − y d (t )b

∴ Tracking error is bounded as long as y d is bounded, it is small whenthe frequency content of y d is well below b



Example Cont’d

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p

An alternative output, motivated by (1− s

b ) ≈ 1/(1 + s

b ) for small |s |/b :

y 2 = B 0(s )

A(s )(1 + s b

)u

y (s ) =

1− s

b

1 +

s

b

y d =

1− s 2

b 2

y d

Thus, the tracking error is proportional to the desired acceleration y d :

y (t )− y d (t ) = − y d (t )b 2

Small tracking error if the frequency content of y d is below b



Tracking Control

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g

Another approximate tracking (Hauser, 1989) can be obtained by When performing I/O linearization, using successive differentiation, simply

neglect the terms containing the input Keep differentiating n timed (system order) Approximately, there is no zero dynamics It is meaningful if the coefficients of u at the intermediate steps are “small”

or the system is “weakly non-minimum phase” system The approach is similar to neglecting fast RHP zeros in linear systems.

Zero-dynamics is the property of the plant, choice of input and outputand desired Trajectory. It cannot be changed by feedback: Modify the plant (distribution of control surface on an aircraft or the mass

and stiffness in a flexible robot) Change the output (or the location of sensor) Change the input (or the location of actuator) Change the desired Traj.


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