30 May 2011 ... Example 24.1: To introduce the idea of feedback linearization, let us start ...
Therefore, the ability to use feedback to convert a nonlinear state.
Last Update: 30.05.2011
24 Feedback Linearization Similarly to the approach taken in sliding mode control, in this section we wish to exploit the form of the system equations to modify the dynamics to something more convenient. We consider a class of nonlinear systems of the form
x f x G xu
y h x
where f : D n and G : D n p are defined on a domain D n , which contains the origin, and pose the question whether there exists a state feedback control
u x x v and a change of variables
z T x that transforms the nonlinear system into an equivalent linear system. If the answer to this question is positive, we can induce linear behavior in nonlinear systems and apply the large number of tools and the well established theory of linear control to develop stabilizing controllers.
24.1 Motivation Example 24.1: To introduce the idea of feedback linearization, let us start with the problem of stabilizing the origin of the pendulum equation. x1 x2 x2 a sin x1 sin bx2 cu
If we choose the control
u
a v sin x1 sin c c
We can cancel the nonlinear term a sin x1 sin . This cancellation results in the linear system
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x1 x2 x2 bx2 v
Thus, the stabilization problem for the nonlinear system has been reduced to a stabilization problem for a controllable linear system. We can proceed to design a stabilizing linear state feedback control
v k1 x1 k2 x2 to locate the eigenvalues of the closed loop system x1 x2
x2 k1 x1 k2 b x2
in the open left half plane. The overall state feedback control law is given by
1 a u sin x1 sin k1 x1 k2 x2 c c Δ
Clearly, we should not expect to be able to cancel nonlinearities in every nonlinear system. There must be a certain structural property of the system that allows us to perform such cancellation. Therefore, the ability to use feedback to convert a nonlinear state equation into a controllable linear state equation by canceling nonlinearities requires the nonlinear space equation to have the structure x Ax B x u x
Where A is n n , B is n p , the pair A, B is controllable, the functions : n p and : n p p are defined in the domain D n that contains the origin, and the matrix x is nonsingular for every x D . If the state equation takes the form
x Ax B x u x , then we can linearize it via the state feedback u x x v where x 1 x , to obtain the linear state equation
x Ax Bv For stabilizing, we design v Kx such that A BK is Hurwitz. The overall nonlinear stabilizing state feedback control is -2-
u x x Kx
Suppose the nonlinear state equation does not have the required structure. Does this mean we cannot linearize the system via feedback? The answer is no. Even if the state equation does not have the required structure for one choice of variables, it might do so for another choice.
Example 24.2: Consider, for example the system x1 a sin x2 x2 x12 u
We cannot simply choose u to cancel the nonlinear term a sin x2 . However, if we first change the variables by the transformation z1 x1 z 2 a sin x 2 x1
then z1 and z2 satisfy
z1 z 2
z z 2 a cos x 2 .x 2 a cos sin 1 2 u z12 a
and the nonlinearities can be cancelled by the control
u x12
1 v a cos x2
which is well defined for 2 x2 2 . The state equation in the new coordinates can be found by inverting the transformation to express x1 , x2 in the term of z1 , z2 ; that is,
x1 z1 z x2 sin 1 2 a which is well defined for a z2 a . The transformed state equation is given by
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z1 z2 z z2 a cos sin 1 2 z12 u a
which is in the required form to use state feedback Δ
Definition 24.1: A nonlinear system
x f x G xu
where f : D n and G : D n p are sufficiently smooth on a domain D n , is said to be feedback linearizable (or input state linearizable) if there exists a diffeomorphism T : D n such that Dz T D contains the origin and a change of variables z T x transforms the system x f x G x u into the form z Az B x u x
with A, B controllable and x nonsingular for all x D .
Definition 24.2: f is called smooth if f C . That is, f is continuous and all derivatives of all order are continuous. Definition 24.3: T is a diffeomorphism if T is smooth, and the inverse exists and is also smooth.
Remark: When certain output variables are of interest, as in tracking control problems, the state model is described by state and output equations. Linearizing the state equation does not necessarily linearize the output equation. Example 24.2 (ctd): Consider the previous system x1 a sin x2 x2 x12 u
If the system has an output y x2 , then the change of variables and state feedback control
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z1 x1 , z2 a sin x2 , and u x12
1 v a cos x2
yield
z1 z2 z2 v z y sin 1 2 a Δ
While the state equation is linear, solving a tracking control problem for y is still complicated by the nonlinearity of the output equation.
24.1.1 Transformation Matrix T x With the help of a Matrix T x we want to transform
x f x G xu into the new system
z Az B ( x)u ( x)
Given the previous system with a coordinate transformation z T x we derive
T ( x) T ( x) f ( x) G( x)u x x x Az B ( x)u ( x) AT ( x) B ( x)u ( x)
z
and obtain a system of partial differential equations
T x f ( x) AT ( x) B ( x) ( x) T G ( x) B ( x) x By solving this set of partial differential equations for T(x), the transformation is found.
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Remark: T x is not uniquely defined by this system of differential equations as shown when we consider any linear transformation z Mz . Then T x MT x will also satisfy this system but with A MAM 1 and B MB .
We use this fact to choose the matrix A and B to be in the canonical controllability form. For simplicity we consider this for a single-input system. In this case, the control canonical form is as follows. 1 0 0 0 AC , B 1 C 0 1 0 1 n 1 Where the i are the coefficients of the characteristic polynomial of A:
det(sI A)
n 1
s i 0
i
i
Note that ( AC , BC ) is feedback equivalent to a chain of integrators. i.e. with the mapping n 1
u u i s i , the system is transformed to a chain of integrators. Thus we consider i 0
the following system matrices:
0 0 1 0 , B0 A0 0 1 0 0 1 Then with T1 x T x Tn x
we obtain
T2 ( x) T f ( x) A0T ( x) B ( x) ( x) Tn ( x) x ( x) ( x)
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0 T G ( x) B0 ( x) 0 x ( x) Solving this system of partial differential equations we can find T x . The following theorem, stated without proof, gives necessary and sufficient conditions for feedback linearizability of a single-input affine-in-control nonlinear system. We need some notation first. Let f, g be smooth vector fields on and g is defined as . We also define
. The Lie-bracket of f
A distribution D is a mapping that assigns to each point in the state space a subspace of the tangent space at that point. In other words, a distribution is a family of smooth vector fields that span a subspace of at each point. A distribution is involutive if the Lie bracket of any two vector fields in the distribution lies in the distribution. To wit, if D is spanned by , then for each i, j, x, we have where are smooth functions. Note that this is a natural generalization of the concept of linear dependence of vectors. We define the following special distributions:
Theorem 24.1: A single input system at a point x if and only if is a linearly independent family of vectors, and is involutive.
24.2 Input-Output Linearization Consider the single-input-single-output system
x f x g xu
y h x
where f , g , and h are sufficiently smooth in a domain D n . The mappings
f : D n and g : D n are called vector fields on D .
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Derive conditions which allow us to transform the system such that the input output map is linear. The derivative y is given by y
h f x g x u L f h x Lg h x u x
where h f x Lf h x x
is called the Lie Derivative of h with respect to f . If Lg h x 0 , then y L f h x , independent of u . If we continue to calculate the second derivative of y , denoted by y , we obtain 2
y 2
Lf h x
f x g x u L2f h x Lg L f h x u
Once again, if Lg L f h x 0 , then y 2 L2f h x , independent of u . Repeating this process, we see that if h x satisfies Lg Lif1h x 0 , i 1, 2,..., 1 ; Lg Lf 1h x 0
then u does not appear in the equations of y , y ,…, y
y
p
1
and appears in the equation of
with a nonzero coefficient: y p Lf h x Lg Lf 1h x u
The forgoing equation shows clearly that the system is input-output linearizable, since the state feedback control
u
1
Lf h x v Lg L f h x 1
reduces the input-output map to
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y v
which is a chain of integrators. In this case, the integer is called the relative degree of the system. Example 24.3: Consider the controlled van der Pol equation x1 x2
x2 x1 1 x12 x2 u
with output y x1 . Calculating the derivatives of the output, we obtain y x1 x2
y x2 x1 1 x12 x2 u
Hence, the system has relative degree two in 2 . For the output y x1 x2 2 ,
y x2 2 x2 x1 1 x12 x2 u
and the system has relative degree one in D0 x 2 x2 0 Δ
Example 24.4: Correspondence to relative degree for Linear Systems Consider a linear system represented by the transfer function b s m bm 1s m 1 ... b0 H s m n s an 1s n 1 ... a0 where m n and bm 0 . A state model for the system can be taken as x Ax Bu y Cx
where
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1 0 0 0 1 A 0 a0 a1 C b0
bm
0 0 0 , B 0 0 1 an 1 nn 1 n1 0
1 0 am 0
01n
This linear space model is a special case of x f x g x u , y h x , where f x Ax , g B , and h x Cx . To check the relative degree of the system, we
calculate the derivative of the output. The first derivative is y CAx CBu
If m n 1 , then CB bn1 0 and the system has relative degree one. Otherwise, CB 0 and we continue to calculate the second derivative y . Noting that CA is a row 2
vector obtained by shifting the elements of C one position to the right, while CA2 is obtained by shifting the elements of C two positions to the right, and so on, we see that CAi 1B 0 , for i 1, 2,..., n m 1 , and CAnm.1B bm 0
Thus, u appears first in the equation of y
nm
, given by
y n m CAn m x CAn m 1Bu
and the relative degree of the system is n m (the difference between the degrees of the denominator and numerator polynomials of H s ). Δ
Now let
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1 x n x x def def z T x h x x L 1h x f
where 1 x to n x are chosen such that
i g x 0 , for 1 i n x This condition ensures that when we calculate
f x g x u x
the term u cancels out. It is now easy to verify that the change of variables z T x transforms the system into
f 0 , AC BC x u x y CC
where p , n , AC , BC , CC is a canonical form representation of a chain of integrators, where f 0 ,
f x x x T 1 z
x Lg L f h x and x 1
Lf h x
Lg Lf 1h x
We have kept and expressed in the original coordinates. These functions are uniquely determined in terms of f , g , and h . They are independent of the choice of . They can be expressed in the new coordinates by setting
0 , T 1 z and 0 , T 1 z
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which, of course, will depend on the choice of . In this case, the equation can be rewritten as
Ac BC 0 , u 0 ,
The three equations of the new system are said to be in the normal form. This form decomposes the system into an external part and an internal part . The external part is linearized by the state feedback control
u x x v where x 1 x , while the internal part is made unobservable by the same control. The internal dynamics are described by f0 , . Setting 0 in that equation results in
f 0 , 0 which is called the zero dynamics of the system. If the zero dynamics of the system are (globally) asymptotically stable, the system is called minimum phase. The linearized system may then be stabilized by the choice of an appropriate state feedback: v K . Theorem 24.2: The origin z 0 is an asymptotically stable equilibrium point of the stabilized, linearized system if 0 is an asymptotically stable equilibrium point of
f 0 , 0 In other words, a minimum phase input-output linearizable system can be stabilized by a feedback law u ( x) ( x) KT ( x) Proof The idea is to construct a special Lyapunov function. By the converse Lyapunov theorem V1 such that V1 f , 0 3
in some neighborhood of 0 . Let P PT 0 be the solution of the Lyapunov equation - 12 -
P( AC BC K ) ( AC BC K ) T P I Construct a second Lyapunov function V2
V2 T P Then use V , V1 k T P , k 0 We want to check that the derivative of V , is negative
V
V1 k f 0 ( , ) T P( AC BC K ) ( AC BC K ) P T 2 P I V1 V1 k f 0 ( , ) f 0 ( ,0) T T f ( ,0) 2 P
3 k1 kk2
Where k1 0 follows from local continuity and differentiability of f 0 , and k 2 0 follows from P. By choosing k large enough, we guarantee V 0 .
■
Remark 1: This is a local result Remark 2: If the origin 0 is globally asymptotically stable for f0 , 0 one might think that system can be globally stabilized. This is not the case. Remark 3: Global stability can be guaranteed if the system is input to state stable and linearizable. Remark 4: One may think that the system can be globally stabilized, or at least semiglobally stabilized, by designing the linear feedback control v K to assign the eigenvalues of A BK far to the left in the complex plane so that the solution of
A BK decay to zero arbitrarily fast. Then, the solution of f0 , will quickly approach the solution of f0 ,0 , which is well behaved, because its origin is globally asymptotically stable. But consider the following example.
Example 24.5: Consider the third-order system
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1 1 2 3 2
1 2 2 v The linear feedback control
def
v k 21 2k 2 K
assigns the eigenvalues of
0 A BK 2 k
1 2k
at k and k . The exponential matrix 1 kt e kt e A BK t 2 kt k te
te kt
1 kt e
kt
shows that as k , the solution t will decay to zero arbitrarily fast. Notice, however, that the coefficient of the 2,1 element of the exponential matrix is a quadratic function of k . It can be shown that the absolute value of this element reaches a maximum value k / e at 1 k . While this term can be made to decay to zero arbitrarily fast by choosing k large, its transient behavior exhibits a peak of the order of k . The interaction of peaking with nonlinear growth could destabilize the system. In particular, for the initial states 0 0 , 1 0 1 , and 2 0 0 , we have 2 t k 2te kt and
1 1 k 2te kt 3 2
During the peaking period, the coefficient of 3 is positive, causing t to grow. Eventually, the coefficient of 3 will become negative, but that might not happen soon enough , since the system might have a finite escape time. Indeed the solution
2 t
0 2
1 0 2 t 1 kt e kt 1
shows that if 0 2 1 , the system will have a finite escape time if k is chosen large enough. Δ - 14 -
Remark: It is useful to know that the zero dynamics can be characterized in the original coordinates. Noting that y t 0 t 0 u t x t
we see that if the output is identically zero, the solution of the state equation must be confined to the set
Z * x D0 h x L f h x
Lf 1h x 0
and the input must be def
u u * x x xZ *
Example 24.6: The system x1 x1
2 x32 u 1 x32
x2 x3 x3 x1 x3 u y x2
has an open-loop equilibrium point at the origin. The derivatives of the output are y x2 x3 y x3 x1 x3 u
Therefore, the system has relative degree two in 3 . Using Lg L f h x 1 and L2f h x x1 x3 , we obtain
1 and x x1 x3
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To characterize the zero dynamics, restrict to Z * x x2 x3 0
and take u u* x 0 . This process yields x1 x1
which shows that the system is minimum phase. To transform it into the normal form, we want to choose a function x such that
0 0 ,
g x 0 x
and T x x x2
x3
T
is a diffeomorphism on some domain containing the origin. The partial differential equation 2 x32 0 x1 1 x32 x3
can be solved by separating variables to obtain
x x1 x3 tan 1 x3 which satisfies the condition 0 0 . The mapping T x is a global diffeomorphism, as can be seen by the fact that for any z 3 , the equation T x z has a unique solution. Thus, the normal form
2 tan 1 x3 1
2 22 2 1 22
1 2
2 2 tan 1 2 2 u y 1 is defined globally. Δ - 16 -
Remark: (Robustness) nonlinearities, which is, only approximations ˆ ,
Feedback linearization is based on exact cancellation of in practice, not often practically possible. Most likely we have ˆ and Tˆ x of the true , , and T x . The feedback control
law has the form
u ˆ ˆ KTˆ2
and the closed loop system
f , A B x ˆ ˆ KTˆ2 Adding and subtracting BK to the equation we obtain
f , A BK B z where
z ˆ ˆ KT2 ˆ K Tˆ2 T2
The local closed loop system differs from the nominal one by an additive perturbation. Thus, it is often assumed that in most cases no serious problems are to be expected.
Remark: In some cases it might be useful not to cancel all nonlinearities. For instance x ax bx 3 u
Take u K a x , K 0 and we obtain x Kx bx 3
which is asymptotically stable.
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