STATE TRANSFORMATION FOR EULER-LAGRANGE SYSTEMS
M. Mabrouk
Inria-Lorraine, projet CONGE, ISGMP Bt. A, Ile du Saulcy, 57 045 Metz Cedex 01, France.
J. C. Vivalda
Inria-Lorraine, projet CONGE, ISGMP Bt. A, Ile du Saulcy, 57 045 Metz Cedex 01, France.
Keywords:
Euler-Lagrange systems, state transformation, affine forms.
Abstract:
The transformation of an Euler-Lagrange system into a state affine system in order to solve some interesting
problem as the design of observer, the output tracking control, is considered in this paper. A necessary and a
sufficient condition is given as well as a method to compute this transformation.
1 INTRODUCTION
Euler- Lagrange systems with n generalized configu-
ration coordinates q = (q
1
, ..., q
n
)
are described by
equations of the form
˙q = v,
M(q) ˙v + C(q, v)v + V (q) = τ,
(1)
where M(q) denotes the inertia matrix, while
C(q, v)v, with v = ˙q = ( ˙q
1
, ..., ˙q
n
)
the gener-
alized velocities, denotes the centrifugal and Corio-
lis forces, V (q) consists of the gravity terms and τ
is the vector of input torques. This celebrated fam-
ily of systems has been the subject of an important
literature over half a century, because the equations
of many physical devices belong to this family: (M.
Spong and Vidyasagar, 1989). When these systems
are fully-actuated, they are globally feedback lineariz-
able. But feedback linearization can be performed
only when all the variables are measured. Unfortu-
nately in practice, very often the variables of veloc-
ity cannot be measured. Therefore, the global output
feedback stabilization of these systems with y = q as
output is challenging from a practical point of view.
But, from a theoretical point of view, it is one of the
most difficult problems in the field of nonlinear con-
trol: Indeed, the matrix C(q, v)v is a nonaffine func-
tion of the unmeasured part of the state v: this fact
precludes from applying most of the classical tech-
niques. For instance, the methods of (L. Praly and
Z.P. Jiang, 1993), (J.P. Gauthier and I. Kupka, 1994)
and (J-B. Pomet, R. M. Hirschorn, W. A. Cebuhar,
1993). For more explanations on the obstacles which
are due to the presence of terms nonaffine with respect
to the unmeasured variables, see the introduction of
(F. Mazenc and J.C. Vivalda, 2002).
Most recently, in (Besancon, G. 2000) an elegant
alternative for one-degree-of -freedom systems was
reported. The author presented a reduced order ob-
server converging exponentially . This observer is
based upon a global nonlinear change of coordinates
which makes the system affine in the unmeasured part
of the state. This is crucial to define a very simple
controller to solve the problem of tracking trajectory.
So a very natural question arises: which conditions
ensure that an Euler -Lagrange systems (1) can be
transformed, with the help of a change of coordinates,
into some structure affine in the unmeasured part of
the state.
This question has been addressed in (Besancon, G.
2000) and (Loria, A and Pantely, E, 1999). However
the questions of existence and computation of the re-
quired solution were not answered. In the present
paper, we address these question: we show that this
problem can be brought back to the resolution of a set
of partial differential equation for which an explicit
solution is given.
The paper is organized as follows. The main result
is stated and proved in Section 3. Section 4 is devoted
to example. Section 5 contains concluding remarks.
Preliminary
Euler-lagrange systems are such that:
The matrix M(q) is symmetric positive definite for
all q.
43
Mabrouk M. and C. Vivalda J. (2005).
STATE TRANSFORMATION FOR EULER-LAGRANGE SYSTEMS.
In Proceedings of the Second International Conference on Informatics in Control, Automation and Robotics - Robotics and Automation, pages 43-48
DOI: 10.5220/0001169200430048
Copyright
c
SciTePress
The equality
˙
M(q) = C
(q, v) + C(q, v). holds.
Also, throughout the paper,
M
n
(R) denotes the set of n-square real matrices.
GL
m
(R) denotes the set of n-square real invertible
matrices.
For S M
n
(R) symmetric positive definite S
1
2
denotes the square root of S.
2 PROBLEM STATEMENT
We consider the family of Euler-Lagrange systems
described by the equations (1) where the output is
q = (q
1
... q
n
)
R
n
, and the input is τ =
(τ
1
... τ
n
)
R
n
. The unmeasured part of the state
is v = ( ˙q
1
... ˙q
n
)
As it is pointed in the introduction the difficulty
to stabilize or to construct observers for system (1)
mainly stems from the fact that Coriolis and centrifu-
gal forces vector in (1), has a quadratic growth in the
generalized velocities v, which are not measured. The
global change of coordinates introduced in (Besan-
con, G. 2000) for one-degree-of freedom (i.e. n = 1)
systems overcomes this problem by rewriting the dy-
namics with functions which are linear in the unmea-
sured velocities. As it is discussed in (Besancon, G.
2000), the design procedure might be extended to the
case of systems with more degrees of freedom, as
soon as the same kind of change of coordinates can
be found, that is to say if we can select an invertible
matrix T(q) such that
d T(q)
dt
= T(q)M
1
(q)C(q, v) (2)
Condition (2) is necessary but not sufficient. A nec-
essary and sufficient condition is the existence of a
nonsingular matrix T such that
d T(q)
dt
v = T(q)M
1
(q)C(q, v)v (3)
Remak 2.1. Since matrix M (q) is positive definite,
there exists an uniformly bounded matrix ∆(q) such
that M(q) =
(q)∆(q). It can be easily checked
that in the case one degree of freedom, ∆(q) is a so-
lution of (3) and (2). But in the general case, this
decomposition of M(q) does not provide a solution of
(3) or (2).
Looking for some more general factorization of
M(q), as investigated in (Loria, A and Pantely, E,
1999) for robot manipulators, may lead to solutions.
Remak 2.2. One can remark that if (2) admits a solu-
tion, then (3) admits also (the same) solution. But the
converse is false as shown in the following example.
Consider the following inertia matrix M(q)
M(q) =
e
q
2
0
0 1
.
Using the Christoffel symbols of the first kind (M.
Spong and Vidyasagar, 1989), matrix C is given by
C(q, v) =
1
2
e
q
2
v
2
v
1
v
1
0
and an easy calculation shows that the matrix
T(q) =
e
q
2
0
1
2
q
1
e
q
2
1
!
(4)
satisfies equation (3), but not (2). In fact, equation (2)
does not admit any solution (as we will see later).
3 MAIN RESULT
In this section we present the main contribution of the
paper, namely the solution of problem (2) or (3). We
begin with the study of the equation (2). We provide
necessary and sufficient conditions for the existence
of a change of coordinates as well as method to com-
pute this solution (Corollary 3.3).
Theorem 3.1. Consider the nonlinear system (1).
Equation (2) admits a solution if and only if
C
i
q
i
C
j
q
i
= C
j
M
1
C
i
C
i
M
1
C
j
, (5)
for all 1 i, j n. Where the matrices C
i
are such
as
C(q, v) =
i=n
X
i=1
C
i
(q)v
i
(6)
To establish Theorem 3.1, we need to prove the fol-
lowing preliminary lemma, and its corollary.
Lemma 3.2. (Isidori, 1989) Let x
1
, ..., x
m
denote co-
ordinates of a point x in R
m
and y
1
, ..., y
n
coordi-
nates of a point y in R
n
. Let M
1
, ..., M
m
be smooth
functions
M
i
: R
m
R
n×n
(7)
such that
M
i
x
k
M
k
x
i
+ M
i
M
k
M
k
M
i
= 0. (8)
Consider the set of partial differential equations
y(x)
x
i
= M
i
(x)y(x), 1 i m. (9)
Given a point (x
0
, y
0
) R
m
× R
n
, there exist a
neighborhood U of x
0
and a unique smooth function
y(x) which satisfies (9) and is such that y(x
0
) = y
0
ICINCO 2005 - ROBOTICS AND AUTOMATION
44
Corollary 3.3. Let M
1
(q), . . . , M
n
(q) be smooth
functions
Consider the set of partial differential equations
T
q
i
(q) = T(q)M
i
(q), i = 1, . . . n. (10)
Given any matrix T
0
GL
m
(R), q
0
R
n
, there
exists a unique smooth matrix T(q) which satisfies
(10) and is such that T(q
0
) = T
0
if and only if the
functions M
1
(q), . . . , M
n
(q) satisfy the conditions
i < j n; M
j
M
i
M
i
M
j
=
M
j
q
i
M
i
q
j
. (11)
Proof. Necessity:
The necessity follows from the assumption that a
solution T(q) exists. Then from the property
2
T
q
i
q
j
=
2
T
q
j
q
i
(12)
one has
(T M
j
)
q
i
=
(T M
i
)
q
j
(13)
Expanding the derivatives on both sides we obtain
T (M
i
M
j
+
M
j
q
i
) = T (M
j
M
i
+
M
i
q
j
)
(14)
which, due to the fact that T(q) is invertible, yields
the condition (11).
Sufficiency:
Using Lemma 3.2.
Let T
0
GL
m
(R) and note T
1
0
=
1
0
, ...., Γ
n
0
),
Γ
i
0
being the columns of T
1
0
.
Conditions (11) ensure the existence of a family of
functions Γ
k
such that for all k we have
Γ
k
q
i
= M
i
Γ
k
, Γ
k
(q
0
) = Γ
k
0
(15)
The matrix Γ =
1
, ..., Γ
n
) satisfies
Γ
x
i
= M
i
Γ, Γ(q
0
) = T
1
0
. (16)
Since Γ(q
0
) = T
1
0
which is non singular, we con-
clude that there exists neighborhood U of q
0
such
that Γ is non singular and as a solution of (10), we
take the matrix T(q ) = Γ
1
(q).
The previous proof gives a condition of existence,
but not a method allowing construction of the so-
lution. However, the control implementation needs
the knowledge of a matrix T(q).
In the sequel we will give another proof of the
corollary 3.3, based on a reasoning by induction
which provides an explicit solution of (2). More-
over this solution is defined on the whole domain
of definition of the matrices M
i
.
By induction.
By induction on n we show that if (11) holds, then we
have the following property P(n).
P(n): n 1, there exists an invertible matrix T(q)
such that equations (10) holds.
1. For n = 1: Equation (10) becomes
T(q
1
)
q
1
= T
(q
1
) = T(q
1
)M(q
1
) (17)
which admits solutions for all matrix M
1
M
m
(R).
2. For n = 2 : Let M
1
(q
1
, q
2
), M
2
(q
1
, q
2
) M
m
.(R)
Denote by Φ
q
2
(q
1
) the solution of the non au-
tonomous differential equation
dΦ
q
2
dq
1
(q
1
) = Φ
q
2
M
1
(18)
Now, let us construct a particular solution T(q ) of
the system (10) in the form T(q) = Ψ(q
2
q
2
(q
1
).
Since T(q) is a solution of (10), one can prove that
T
q
2
=
dΨ(q
2
)
dq
2
Φ
q
2
+ Ψ(q
2
)
Φ
q
2
q
2
(19)
It follows that
dΨ(q
2
)
dq
2
= Ψ(q
2
)(Φ
q
2
M
2
Φ
1
q
2
Φ
q
2
q
2
Φ
1
q
2
)
Let
K(q) = Φ
q
2
M
2
Φ
1
q
2
Φ
q
2
q
2
Φ
1
q
2
(20)
Then Ψ(q
2
) exists if we have
K(q)
q
1
= 0.
Now
K(q)
q
1
= Φ
q
2
M
1
M
2
Φ
1
q
2
+ Φ
q
2
M
2
q
1
Φ
1
q
2
Φ
q
2
M
2
M
1
Φ
1
q
2
2
Φ
q
2
q
1
q
2
Φ
1
q
2
+
Φ
q
2
q
2
M
1
Φ
q
2
= Φ
q
2
(M
1
M
2
+
M
2
q
1
M
2
M
1
M
1
q
2
1
q
2
= 0
(21)
So P(2) is true.
STATE TRANSFORMATION FOR EULER-LAGRANGE SYSTEMS
45
3. Assume that P(n) is true, where n1.
Now using the induction hypothesis, we must prove
that P(n + 1) is true ?
Let M
1
, . . . , M
n+1
M
m
(R) such that
M
j
M
i
M
i
M
j
=
M
j
q
i
M
i
q
j
(22)
The induction hypothesis implies that there exists
an invertible matrix T
q
n+1
(q
1
, q
2
, . . . , q
n
) such that
T
q
n+1
q
i
= T
q
n+1
M
i
, i = 1, . . . , n.
Let us show that there are solutions of the form T =
Ψ
1
(q
n+1
)T
q
n+1
.
Observe that
T
q
i
= Ψ
1
(q
n+1
)
T
q
n+1
q
i
= Ψ
1
(q
n+1
)T
q
n+1
M
1
= TM
1
,
and
T
q
n+1
=
dΨ
1
dq
n+1
T
q
n+1
+ Ψ
T
q
n+1
q
n+1
= TM
n+1
= Ψ
1
T
q
n+1
M
n+1
So we get
dΨ
1
dq
n+1
= Ψ
1
(T
q
n+1
M
n+1
T
q
n+1
q
n+1
)T
1
q
n+1
(23)
Ψ
1
(q
2
) exists if the term (T
q
n+1
M
n+1
T
q
n+1
q
n+1
)T
1
q
n+1
is independent of q
1
, q
2
, . . . , q
n
.
Easy calculations show that for i = 1, 2, . . . , n; we
have
[T
q
n+1
M
2
T
1
q
n+1
T
q
n+1
q
n+1
T
1
q
n+1
]
q
i
= 0
(24)
This concludes the proof.
Proof of Theorem 3.1
Returning to the problem of the existence of the
solution of (2), note that we have
˙
T =
i=n
X
i=1
T
q
i
v
i
.
Moreover the matrix C(q, v) is linear with respect
to v
i
(M. Spong and Vidyasagar, 1989). This yields
for all q
C(q, v) =
i=n
X
i=1
C
i
(q)v
i
(25)
where matrices C
i
are such that, C
i
+ C
i
=
M
q
i
.
Therefore one can deduce easily that T satisfies
T
q
i
= TM
1
C
i
. (26)
According to Corollary 3.3, we deduce that a solu-
tion of (2) exists if and only if
T
q
i
= TM
i
, i = 1
, . . . n where M
i
= M
1
C
i
.
Now,
M
j
q
i
M
i
q
j
= M
1
(C
i
+ C
i
)M
1
C
j
+M
1
C
j
q
i
M
1
C
i
q
j
+M
1
(C
j
+ C
j
)M
1
C
i
= M
j
M
i
M
i
M
j
+M
1
(
C
j
q
i
C
i
q
j
C
i
M
1
C
j
+ C
j
M
1
C
i
).
It follows that a necessary and sufficient condition
for the existence of T(q) such that (2) is given by:
C
i
q
i
C
j
q
i
= C
j
M
1
C
i
C
i
M
1
C
j
.
Which allows as to conclude.
The preceding theorem gives an algebraic char-
acterization of a family of Euler-Lagrange systems
which can be transformed, with the help of a change
of coordinates into some structure, affine in the un-
measured part of the state v = ˙q.
The following gives another characterization of
such a solution of the differential equation (2),
Theorem 3.4. Consider an Euler-Lagrange system
(1). The following conditions are equivalent .
1. There exists a matrix T(q) such that ( 2) holds.
2. There exists a matrix N (q) such that M(q) =
N
(q)N(q) and N
(q)
˙
N(q) = C(q, v).
3. There exists a function Θ(q) : R
n
R
n
and N(q)
nonsingular such that M (q) = N
(q)N(q) and
Jacobian(Θ) = N (q).
Proof. Due to the space limitation we only give the
proof of the two first properties.
Suppose that (2) admits a solution.
Calculating
d
¯
M
dt
, where
¯
M = T
⊤−1
MT
1
.
ICINCO 2005 - ROBOTICS AND AUTOMATION
46
We have
d
¯
M
dt
= T
⊤−1
(C
M
1
T
)T
⊤−1
MT
⊤−1
T
⊤−1
MT
1
(TM
1
C)T
1
+T
⊤−1
dM
dt
T
1
= T
⊤−1
(C
+ C
dM
dt
)T
1
= 0
(27)
since C
+ C =
dM
dt
.
So,
¯
M = T
⊤−1
MT
1
= S, where S is a symmet-
ric positive definite matrix .
Let N = S
1
2
T, then we obtain
M(q) = N
(q)N(q) (28)
and
N(q)
d
dt
N(q) = C(q, v). (29)
For sufficiency assume that there exists N (q) such
that M(q) = N
(q)N(q) and N
(q)
˙
N(q) =
C(q, v).
So we get, N
(q)
N
q
i
(q) = C
i
(q).
One can check easily that conditions (31) are satis-
fied, which concludes the proof.
Now, we present our final result on reduction of
Euler-Lagrange systems.
Let us consider the problem of finding T such that
(3) is satisfied.
Observe first that the matrix M
1
C(q, v)v is
quadratic in v with coefficients depending only on q.
i.e there exists R
i
such that
M
1
(q)C(q, v)v =
n
X
i=1
v
i
R
i
v (30)
then one can easily show the following
Theorem 3.5. Consider an Euler-Lagrange system
(1). A necessary and sufficient condition for (3) to
admit a solution is that there exist R
i
such that
i < j n; R
j
R
i
R
i
R
j
=
R
j
q
i
R
i
q
j
. (31)
Proof. The proof follows from similar reasoning as
in Theorem 3.1 and Corollary 3.3 gives an explicit
solution of the problem.
Remak 3.6. One can remark that when n = 1, a
solution always exists (See (Besancon, G. 2000)).
In the case of higher order system, the problem (2)
can admit no solution. Indeed, taking again the ex-
ample of the Remark 2.2, a solution exists if and only
if e
q
2
= 0 which is impossible.
But if we take
R
1
=
0 0
1
2
e
q
2
0
, R
2
=
1
2
e
q
2
0
0 0
,
(32)
one can check easily that conditions (30) and (31) are
satisfied. So (3) admits a solution.
Observe also that there is no function Θ(q) such
that Jacobian(Θ) = T(q).
4 EXAMPLE
As an example, consider the inverted pendulum. The
dynamics obtained by Euler-Lagrange formulation
are:
(M + m)¨x + ml
¨
θ cos θ ml
˙
θ
2
sin θ = τ
1
,
ml¨x cos θ + ml
2
¨
θ mlg sin θ = 0,
where (M, x) are mass and position of the cart which
is moving horizontally, (m, l; θ) are mass, length
and angular derivation from the upward vertical po-
sition for the pendulum which is pivoting around a
point fixed on the cart. We denote the state vec-
tor (θ x
˙
θ ˙x)
as (q
1
q
2
v
1
v
2
)
. The output is
y = (q
1
, q
2
)
.
The inertia matrix is:
M(q) =
a
1
a
2
cos q
2
a
2
cos q
2
a
3
with a
1
= M + m, a
2
= ml and a
3
= ml
2
.
Using Christoffel symbols, we obtain
C
1
=(0)
C
2
=
0 a
2
sin(q
2
)
0 0
One can check easily that condition (11) are verified
so according to Theorem 3.1, equation (2) admits a
solution.
Let us consider Φ
q
2
(q
1
) =
1 0
0 1
as a solution
of the differential equation
dΦ
q
2
dq
1
(q
1
) = Φ
q
2
M
1
= 0. (33)
Let
Ψ(q
2
) =
Ψ
11
(q
2
) Ψ
12
(q
2
)
Ψ
21
(q
2
) Ψ
22
(q
2
)
(34)
STATE TRANSFORMATION FOR EULER-LAGRANGE SYSTEMS
47
the solution of the differential equation
dΨ(q
2
)
dq
2
= Ψ(q
2
)(Φ
q
2
M
2
Φ
1
q
2
Φ
q
2
q
2
Φ
1
q
2
)
= Ψ(q
2
)M
1
C
2
(35)
Using Mathematica 4.1, one can check easily that
a solution of Equation (35) with initial condition
Ψ(0) = I
2
=
1 0
0 1
is
Ψ(q
2
) =
1
a
2
β(0) cos(q
2
)a
2
β(q
2
)
a
1
β(0)
0
β(q
2
)
β(0)
!
(36)
where β(q
2
) =
p
a
1
a
3
a
2
2
cos(q
2
)
2
.
Thanks to the Corollary 3.3, a solution T(q
2
) of
Equation (2) with T(0) = T
0
= I
2
, is
T(q
2
) = Ψ(q
2
q
2
(q
1
)
=
1
a
2
β(0) cos(q
2
)a
2
β(q
2
)
a
1
β(0)
0
β(q
2
)
β(0)
!
(37)
Therefore, the following change of coordinates
Θ
1
= q
1
+
Z
q
2
0
a
2
β(0) cos(s) a
2
β(s)
a
1
β(0)
ds,
Θ
2
=
Z
q
2
0
β(s)
β(0)
ds, p = T (q) ˙q,
transforms the dynamics of the Cart-Pendulum into a
double integrator
˙
Θ = p, (38)
˙p = T(q )M
1
(q)τ = u. (39)
Clearly This system is linear in the unmeasured
part of the state and a high gain observer can be con-
structed (M. Mabrouk, F. Mazenc and J.C. Vivalda,
2004).
5 CONCLUSION
A necessary and a sufficient condition for determin-
ing a state change of coordinate which transform an
Euler-Lagrange system into an affine system in the
unmeasured part of state was given. Obviously in the
case of one degree of freedom , a solution always ex-
ists. A case of higher order system, is for instance,
that of the cart-pendulum system (F. Mazenc and J.C.
Vivalda, 2002), the Tora system (Z. P. Jiang and I.
Kanellakopoulos, 2000) and the overhead crane (B. d
Andra-Novel and J. Lvine, 1990). We conjecture the
result several others problems in nonlinear control.
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