Four Representative Applications of the Energy Shaping Method for Controlled Lagrangian Systems

Abstract

We provide a step-by-step, easy-to-follow procedure for the method of controlled Lagrangian systems. We apply this procedure to solve the energy shaping problem for four benchmark examples: the inertial wheel pendulum, an inverted pendulum on a cart, the system of ball and beam and the Furuta pendulum.

1. Introduction

The energy shaping method is a way to stabilize a mechanical system by altering its energy function by feedback so that the equilibrium point of interest becomes a non-degenerate minimum of the altered energy function. It has the advantage that it provides a constructive procedure for generating stabilizing control laws and yields large regions of convergence. This method is sometimes called the method of controlled Lagrangians in the Lagrangian approach and it has been actively developed [1-9]. As a result, the criteria and the matching conditions for energy shaping for nonlinear mechanical systems with one degree of underactuation and linear mechanical systems with an arbitrary degree of underactuation are now well understood [7, 8]. However, fully worked-out examples using these results are lacking. In this paper, we illustrate how to apply the method of controlled Lagrangians with four benchmark examples: the inertial wheel pendulum, an inverted pendulum on a cart, the system of ball and beam and the Furuta pendulum, through a step-by-step, easy-tofollow procedure.

 

2. Preliminaries

We first review the basic scenario for the energy shaping problem. Given a mechanical system, the configuration space is denoted as 𝒬 with q and , the position and velocity vectors respectively. We focus on controlled Lagrangian systems, i.e. mechanical system whose law of motion is governed by the Lagrangian of the form:

where m = mq is the positive-definite symmetric mass matrix while and V (q) are the kinetic and potential energy of the system, respectively. A controlled Lagrangian system can be described by a triple (L, F,W) , where L is the Lagrangian, F is the external force, and W is the control bundle along which the control force acts on the system.

In what follows, we call n = dimQ the degree of freedom, n2 = dimW the degree of actuation and n1 = n − n2 the degree of underactuation. We will use Greek alphabetical indices and Roman alphabetical indices over different ranges: unless stated otherwise. By adopting the Einstein summation convention, the equations of motion in local coordinates are given by Where are the Christoffel symbols of the first kind of m. Here, we have assumed

Two controlled Lagrangian systems (L, F,W) and (, , Ŵ ), where and (q, )= 1/2(,)−(q), are feedback equivalent if for any control u∈W , there exists û ∈Ŵ such that the closed loop dynamics are the same, and conversely.

In this paper, we follow the setting as in [7]: Given a controlled Lagrangian system (L, F = 0,W) with no external force, we try to find a feedback equivalent system (, , Ŵ ) in which is a gyroscopic force dependent on velocity of degree two, i.e. the k-th component k of the force is given by

where Ĉijk satisfy the following conditions:

In other words, this external force does no work on the feedback equivalent system.

It can be proved [7] that the existence of a feedback equivalent system (, , Ŵ ) for a given controlled Lagrangian system (L,0,W) is related to the existence of solutions for a system of PDEs that are known as matching conditions:

Theorem 1 [7]: (L,0,W) is feedback equivalent to (, , Ŵ ) with a gyroscopic force of degree 2 if and only if there exists a non-degenerate mass matrix and a potential function such that the following equations are satisfied:

where mij (resp. mij ) is the (i,j)-entry of m (resp. m−1 ), is an n×n symmetric matrix defined by

and where Γ rjk=mri[jk,i] are the Christoffel symbols of the second kind of m.

We introduce since only αk, but not αb, appear in the matching PDEs; if we used then all entries of would appear in the matching PDEs. Suppose that we have obtained a feasible solution (and hence , since =mm−1(6)) and for the matching conditions. Then, we can obtain the Lagrangian for the feedback equivalent system. Also, the corresponding control bundle Ŵ is given by Ŵ =m−1W.

Hence, what is left is to compute the gyroscopic force . Following [8], we introduce

where are the Christoffel symbols of the first kind of . Once is determined, we can compute Ŝijk. Then, determine Âijk in terms of Ŝijk using the following scheme: (a) Âijα = Ŝijα, (b) Âβγa = −Ŝaβγ −Ŝγaβ, (c) Âγab = Âbγa = -1/2Ŝabγ (d) Finally, choose any Âabc such that Âabc + Âbca + Âcab = 0. For simplicity, we can take Âabc = 0.

Notice that under this scheme Âijk satisfy the properties in (3). Once Âijk are determined, we can obtain the gyroscopic force terms Ĉijk by (8), or equivalently,

• Procedure for solving energy shaping problems

We can now summarize the general procedure for getting a nonlinear control force for a given controlled Lagrangian system with degree of underactuation equal to n1 ≥ 1. (This procedure is from [9]):

S1. Check if the linearization of the given controlled Lagrangian is controllable or its uncontrollable subsystem is oscillatory. 1 If neither holds, then stop;otherwise, proceed to the next step. [7]

S2. Get a solution for and the (α,i) entries αi of which solve the matching PDEs (4) and (5), keeping in mind that the n1 × n1 matrix [Tαβ ] is positive definite around q = 0 and has a non-degenerate minimum at q = 0. In particular, 11 should be positive around q = 0 when the degree of underactuation n1 is one.

S3. Choose the rest of the entries ab of so that is positive definite, at least at q = 0. In particular, when the degree of freedom n is two, one should choose 22> (12)2/11.

S4. Obtain the mass matrix of the feedback equivalent system, through the equation: = m −1m .

S5. Compute the gyroscopic force in (2) by computing Ŝijk, Âijk and then Ĉijk by (7), (9) and steps (a) − (d) between (8) and (9).

S6. Compute the control bundle Ŵ , which is given by

S7. Choose a dissipative, Ŵ -valued linear symmetric control force û. 2 In particular, for systems with degree of underactuation equal to n1 , one may choose

where D is any (n − n1 )×(n − n1 ) symmetric positive definite matrix and K is the (n − n1 )× n matrix defined by

S8. Compute the corresponding control force u:

where a = n1 +1,…, n . Note that uα for α = 1,···,n1 is zero by (1).

Notice that in the above procedure, we require to be gyroscopic and û dissipative. This implies that for every , < , > = 0 and < û, > ≤ 0. Hence the time derivative of the total energy Ê of the feedback equivalent system satisfies

As a result, Lyapunov stability of the equilibrium (q, ) = (0,0) is guaranteed.

The crucial part of solving an energy shaping problem is to obtain a solution for the matching PDEs. For degree of underactuation equal to one, the matching conditions in Theorem 1 reduce to two PDEs, one for and the other for

It turns out that for this class of mechanical systems, the conditions for energy shaping are related to the linearization of the given system, summarized as follows:

Theorem 2 ([7]). Given (L,0,W) with one degree of underactuation, let (Ll ,0,Wl ) be its linearized system at equilibrium (q, ) = (0,0) . Then there exists a feedback equivalent (, , Ŵ) with gyroscopic of degree 2 and having a non-degenerate minimum at (0, 0) if and only if the uncontrollable dynamics, if any, of (Ll ,0,Wl ) is oscillatory. In addition if (Ll ,0,Wl) is controllable, then (, , Ŵ) can be exponentially stabilized by any linear symmetric dissipative feedback onto Ŵ .

Notice that for systems with higher degrees of underactuation, “energy-shapability” of the linearization is only necessary, but not sufficient for that of the original nonlinear system. Hence, the existence of solution for the matching PDEs requires further study.

 

3. Example 1: Inertial Wheel Pendulum

We follow the setting in [10], as shown in Fig. 1. The configuration space is

Fig. 1.The inertial wheel pendulum

The moments of inertia of the rod and the wheel are I1 and I2 respectively, and the distance of the center of mass of the rod from the unactuated joint (not shown in the figure) is ℓc1 . The control force u is the torque applied to the inertial wheel. Let and I1=I2, then A > I and the Lagrangian is given by where m0 = m1ℓc1 + m2ℓ1 and g is the gravitational constant. The equilibrium (q, ) = (0,0) is unstable. The linearization of this system at (q, ) = (0,0) is controllable. Hence by Theorem 2 we can apply energy shaping to stabilize the equilibrium. The corresponding matching PDEs are

We first solve (14). If we choose 11 = 1 and 12 = b0 where b0 ∈ ℜ , then (14) is satisfied. Substituting this pair of entries into (15), one can solve for which reads for any smooth function f. Now, for simplicity, we choose a particular set of parameters, say, b0 = 2 and f (x) = x2 so that the potential energy becomes

Since A > I, the critical points of are q = 0 and (q1, q2) = (π, −π) . Notice that q = 0 is the only minimum point for . The total energy function has a nondegenerate minimum at (q, ) = (0,0) provided that the matrix (and hence ) is positive definite. Note that 22 is still free for which we just choose 8. The resulting positive definite is then given by from which we can calculate :

Since is a constant matrix, Ŝijk =0 and Âijk =0 . As a result, all Ĉijk terms are zeros.

We now choose the following dissipative control force for the feedback equivalent system, according to (10) with D = 1:

The corresponding control law u, obtained by (11), reads as follows:

By Theorem 2, local exponential stability is guaranteed around the equilibrium (q, ) = (0,0) . To find the region of attraction, however, one needs to apply the LaSalle invariance principle. First, notice that with û defined as in (10), the time derivative of the total energy function is given by

We now choose an r > 0 so that the set is compact and does not include (q1,q2 , 1, 2 ) = . Note that Ωr is positively invariant since dÊ / dt is non-positive. We then need to consider all (q,) such that dÊ / dt is zero, i.e.

Let S be the set of points at which dÊ / dt =0 in Ωr , viz

Define M to be the largest invariant subset of S. Let (q(t), (t)) be a trajectory in M. In what follows, the argument t is suppressed for the sake of brevity. Then the trajectory should also satisfy

for some fixed C. Substituting (16) into the equations of motion for the feedback equivalent system (, , Ŵ ) , we have the following systems of differential equations:

Eliminating in the above equations, we can immediately see that q1 must be a constant, say C1 , then by (16) q2= Substituting (q1, q2 ) = into the equations of motion, we can show that C = 0 and C1 = 0 or π both of which imply 1 = 2 = 0 . Since , we conclude the largest invariant subset M of S is {(0, 0, 0, 0)} only. Hence, by LaSalle invariance principle, asymptotic stability is achieved in Ωr . Furthermore, since Ωr is compact, we have exponential stability in Ωr by Lemma 1 in the Appendix.

 

4. Example 2: Inverted Pendulum on a Cart

In this system in Fig. 2, we assume the rod has negligible mass in order to simplify our model. The configuration space is Q = {(q1,q2 ) q1 ∈(−π / 2,π / 2), q2 ∈ℜ}. which considers the pendulum only above the horizontal line. The Lagrangian is given by where g is the gravitational constant. The potential energy V (q) = m1 gℓcosq1 does not attain a minimum at q = 0, and hence the equilibrium point (q, ) = (0,0) is unstable. The linearization of this system at (0, 0) is controllable. Hence, by Theorem 2 we can use the energy shaping method to stabilize this system around the equilibrium.

Fig. 2.An inverted pendulum on a running cart

The matching conditions are

We now try to obtain closed-form solutions for and . First, we may start with the following possible choice: where A0 and A1 ≠ 0 are constants to be determined. Putting this ansatz into the first matching condition, we can obtain

Or

Notice that the second solution for 12 will lead to a potential energy function whose Hessian is not positive definite at q = 0. Hence, we should resort to the first solution of 12 as in (17) and solve the matching condition for to obtain where f = f(x) is any smooth function. Since we require to have a nondegenerate minimum at q = 0, we may set f (x) = x2 so that

To have positive definiteness of at least around q = 0, we may impose A1 > 0 > A0 so that 11 > 0 at least around q = 0 and take any 22 such that det > 0 . In particular, if we take A1 = 2 and A0 = −ε , where ε is fixed and ε∈(0,2) , then which makes det = 1/(m12ℓ2 ) > 0 for all q. In short, we have the following matrix and potential energy :

Define a subset ℜε of Q as follows:

Then is positive definite over ℜε . Furthermore, (0, 0, 0, 0) is the only critical point of within the region ℜε . The resulting mass matrix is = [ij] where

Following the procedure, we can compute the gyroscopic force = [1 ,2]T for the shaped system: where Expr(q,) = (2ℓ1 + ε2){2ε(m12cos2 q1 −(m1 + m2)2) + 2ε2m1(m1sin2q1 +m2)−1}. The control bundle Ŵ is equal to

We now choose a control force û as in (10):

One can then compute u by (11). Note that by Theorem 2, local exponential stability is guaranteed around (q,) = (0,0). To compute the region of attraction, one applies the LaSalle invariance principle. As in the case of inertial wheel pendulum, we start by choosing r > 0 so that the set is compact. Then we define the set

We note that the total energy function Ê has a zero time derivative if and only if from which we have

where C is a constant. Let M be the largest invariant subset of S and consider an arbitrary trajectory (q(t), (t)) in M. This trajectory should satisfy the equations of motion of the feedback equivalent system together with (18). Substituting (18) into those equations of motion, we have

Multiplying (20) by cos q1 and subtracting it from (19), one can obtain

Then by integration twice with respect to t, we have where C1 ,C2 are constant. Now, since sin q1 is always bounded, the above equation holds only if C = C1 = 0 , implying that q1 must be a constant. As C = 0 and 1 = 0 , (20) implies sin q1 = 0 , i.e. q1 = 0 or π . When q1 = 0 , so is q2 . In other words, M ={(0,0,0,0)}. Hence, by LaSalle invariance principle, every trajectory in Ωr will approach (0, 0, 0, 0) asymptotically. Note that when ε→0+ , ℜε →(−π / 2,π / 2)×ℜ . As a result, we can enlarge the region of attraction by letting ε→0+. Since Ωr is chosen to be compact, we also have exponential stability over Ωr by Lemma 1 in the Appendix.

 

5. Example 3: Ball and Beam

Consider the ball and beam system as shown in Fig. 3. Given that the length of the beam is ℓ , the configuration space is [−ℓ, ℓ]×[−π / 2,π / 2] after nondimensionalization of the time and torque [11]. The Lagrangian of this system is given by: where g is the gravitational constant. The linearization at the equilibrium point (0, 0) is controllable, so we can apply the energy shaping method. The two matching conditions are

Fig. 3.The ball and beam system

We may try an ansatz for 12 first and then solve for 11 , assuming 1 11 = 11 (q1 ) . We thus have the following general solutions [11] for the kinetic matching PDE:

For simplicity, we now take implying and 12=ℓ2 + (q1)2. The resulting potential energy, by solving the second matching condition, takes the form

Again, we takef (x) = x2 to ensure that has a minimum at q = 0. The positive definiteness requirement for is met by taking 22 = (ℓ2 + (q1)2)2/3 Notice that the resulting is positive definite everywhere. Furthermore, compared to the method in [11], we have freedom over the choice of 22 . Now, the corresponding mass matrix is

With all these at hand, we can calculate the gyroscopic terms. By definition, we have Ŝ111=0, for all i ≠ j. The Âijk terms can be computed as follows:

We thus obtain the gyroscopic force terms as follows:

Combining these gyroscopic force terms together, we can now obtain the expression for the gyroscopic force : Where

Now, for the control force, we first compute the control bundle Ŵ :

Then, we choose the dissipative control force û by from which one can compute the corresponding control law u. Local exponential stability is guaranteed by Theorem 2 and one can find a compact region of attraction by applying LaSalle invariance principle. By Lemma 1 in the Appendix, it becomes a region of exponential convergence as well.

 

6. Example 4: Furuta Pendulum

We now come to study the energy shaping problem for the Furuta pendulum. The configuration space of the Furuta pendulum is Q = (−π,π]×(−π,π]. Following the notation in [12] (with some minor changes), the Lagrangian for the Furuta pendulum is given by where α = mℓ2 ,β = mℓR, γ = (M + m)R2 ,D = mgℓ and the parameters are defined in Fig. 4. In [12], the Furuta pendulum is shaped by observing that it can be transformed via feedback to a system equivalent to an inverted pendulum on a cart with some gyroscopic force terms. Here,we will solve the energy shaping problem, using the standardized method of solving matching PDEs. Since the linearization of the system at (q, ) = (0,0) is a controllable system, the energy shaping method applies by Theorem 2. For the sake of simplicity in later computations, we can divide the equations of motion by the parameter α so as to obtain the following mass matrix and potential energy: for some A,B,C > 0. The resulting matching PDEs are

Fig. 4.The Furuta pendulum

We first solve (21). Notice that solutions of the form 11 = A1 + A2 cos2q and 1 and 12 = B1cos q1 cannot give a potential energy in (22) with a minimum at q = 0. We guess that a possible candidate can be the following

Substitute this pair into (21), and we have the following relation on the coefficients:

With X1 and Z2 defined in (24) and (25), the αi entries become and then one can solve for the potential energy: where f = f (x) is any smooth function which attains its minimum value at q = 0. We can take f (x) = x2 for the sake of simplicity. Recall that Tˆ should be positive definite around q = 0, and has a minimum at q = 0.

by evaluating the expression of 11 at (0, 0). The requirement on implies the following constraints:

Notice that (28) is always true due to (27) and the fact that A, C > 0. Since , inequality (26) implies that

As a result, the expression that appears in 11 has the following bounds: implying that the denominator of 11 is never zero.

Hence, using (27), 11 > 0 if and only if

Furthermore, since 2 X3 / X2 > −AB /(A2 +1) , cos2 q1 is always bounded below by 1/(A2 +1).This implies that to enlarge the region of attraction as much as possible, one may choose X2 and X3 so that X3 / X2 is close to −AB /(A2 +1). For instance, one may choose X2 = A2 +1 and X3 = −rAB where r < 1 is close to 1. Then by (27) Z1 must be negative as the denominator in (27) is positive. After setting , one can solve the energy shaping problem as in the previous examples, which is left to the readers.

 

7. Conclusions and Future Work

In this paper we introduced a standardized procedure for shaping a controlled Lagrangian system, and illustrated this procedure using four examples. Recently we discovered some criteria for shaping a mechanical system with two degrees of underactuation and with more than three degrees of freedom [9], so a similar energy shaping procedure can be proposed in this case. Nevertheless, energy shaping for higher degrees of underactuation still remains largely unsolved. We plan to investigate this issue in the future.