Applications of passive control to electromechanical systems

Applications of passive control to electromechanical systems
Applications of passive control to electromechanical
systems
Carles Batlle
Technical University of Catalonia
EPSEVG, MAIV and IOC
[email protected]
EURON/GEOPLEX Summer School on
Modelling and Control of Complex Dynamical Systems:
from Ports to Robotics
Bertinoro, FC, Italy, July 6-12 2003
Contents
1 Preamble
2
2 Electromechanical energy conversion
2.1 Electric capacitor . . . . . . . . . . .
2.2 Magnetic stationary system . . . . .
2.3 Elementary electromagnet . . . . . .
2.4 Coenergy . . . . . . . . . . . . . . .
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2
3
5
7
8
3 Port hamiltonian system modelling
3.1 General electromechanical systems . . . . . . . . . . . . . . . . . . . . . .
3.2 Power converters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4 Passive control examples
12
4.1 Magnetic levitation system . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
4.2 Power converters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
5 Connecting systems
16
A Solving quasilinear PDEs
18
1
1
Preamble
This lecture is devoted to some applications of passive control for electromechanical systems in the framework of port controlled hamiltonian systems, which in the simplest,
explicit version are of the form
ẋ = (J (x) − R(x))(∇H(x))T + g(x)u,
where x ∈ Rn , J is antisymmetric, R is symmetric and positive semi-definite and u ∈ Rm
is the control. The function H(x) is the hamiltonian, or energy, of the system. The
natural outputs in this formulation are
y = g T (x)(∇H(x))T .
This kind of systems enjoy nice properties when interconnected, and yield themselves
to passive control strategies quite naturally. It is assumed that the reader is already
acquainted with the essential underpinnings of port controlled hamiltonian systems and
passivity (see, for instance, [21] or elsewhere in this Summer School).
We start with an elementary account of how to compute the generalized force in an
electromechanical system. Next we model several systems in the port hamiltonian framework, and apply the IDA-PBC methodology to them. Finally, we include an explicit
example of interconnection of port hamiltonian systems using some of the models presented. An appendix about how to solve the kind of PDE that appear in the examples
is also included for completeness. Some proposed exercises are scattered around the text.
Most of them are straightforward but some may require consulting the cited literature for
hints. The reader is encouraged to model the systems and controls proposed with 20-sim,
preferably using the bond graph formalism.
2
Electromechanical energy conversion
From the most general point of view, electromechanical systems are electromagnetic systems, i.e. systems with electric and/or magnetic fields, with moving parts. The archetypical electromechanical system is the electric machine, which can be used either as a generator, converting the mechanical energy of a primary mover into electric energy, or as an
electric motor, taking the electric power to yield mechanical work. In this Section we will
learn how to compute the energy contained in an electromechanical system and how to
compute the (generalized) force it can yield, and we will illustrate the results with several
examples.
Our presentation follows [11, 12], which is standard electrical engineering material.
Notice however that there is some discussion in the literature [4] about the way that
Maxwell equations are used when dealing with the concept of magnetic flux in systems
with moving parts, and in particular in the case of rotating machines [2, 17, 10], where
noninertial reference frames are used. We will not enter into this in these lectures and
will assume the standard electrical engineering model.
When we change something in a system we generally change the energy it contains.
If we apply an external force and change the geometry of the system, this change must
overcome the internal forces holding the system, and this changes its internal energy. If we
increase the charges in an electric system, say a capacitor, this increases the mechanical
2
forces needed to keep the system together and, if we let the distance between plates vary,
we can use the displacement to produce mechanical work.
Mathematically, this change in the energy of the system can be expressed as
)
Z (q,λ,x) (X
G
M
E
X
X
fj dx̃j
ij dλ̃j −
(1)
vj dq̃j +
Wf (q, λ, x; q0 , λ0 , x0 ) =
(q0 ,λ0 ,x0 )
j=1
j=1
j=1
where we have G generalized coordinates xj describing the geometry of the system, E
electric elements (capacitors) containing charges qj at voltage vj , and M magnetic elements (inductors) through which currents ij flow and containing magnetic fluxes λj . We
denote by fj the (generalized) force done by the system along the coordinate xj . In
general, the dependent variables vj , ij and fj depend on all of the independent variables
qj , λj and xj . For linear electromagnetic systems, v is a linear function of q and i is
a linear function of λ, while the dependence on x is generally nonlinear.
Now, (1) is a line integral in RG+E+M (at least locally; the x can live in other manifolds
if rotating parts are present). In general its value will depend not only on the initial and
final values, but also on the trajectory between them. However, we will assume that
the system is conservative or, for the more mathematically inclined, that the 1−form
inside the integral in (1) is exact. Physically, this means that phenomena as eddy current
losses, magnetic hysteresis cycles or mechanical friction are neglected. To include these
into the formalism, the thermal domain should be added.
Under this assumption, Wf is a state function and vj , ij or fj can be obtained by
taking the derivatives with respect to qj , λj or (minus) xj , respectively. In particular, the
generalized force, which is our main objective, is given by
fi = −
∂Wf (q, λ, x)
.
∂xi
(2)
However, it seems that we cannot get any further since Wf , as defined by (1), depends
on fi , which is what we are trying to compute. At this point the fact that Wf is a state
function comes to our help. It allows us to reach the final point (q, λ, x) using a particular
path and in such a way that fi does not appear explicitly. First of all, we can assume
that the initial state has q0 = 0, λ0 = 0, since the change in energy when going from
(0, 0, x0 ) to (q0 , λ0 , x0 ) does not contribute to (2). Then, in the first leg of our path, we
go from x0 to x, without changing the zero initial values of q or λ; since the system has
no electromagnetic field, no work is done and Wf 1 = 0. In the second leg we keep the
geometry fixed, so no mechanical work is done, and go to the final values of charge and
magnetic flux:
)
Z (q,λ,x) (X
E
M
X
Wf 2 =
vj (q̃, λ̃, x)dq̃j +
ij (q̃, λ̃, x)dλ̃j ,
(q0 ,λ0 ,x)
j=1
j=1
and since Wf = Wf 2 we only need to know v and i in terms of (q, λ, x).
2.1
Electric capacitor
Figure 1 shows a general plane capacitor, where the dielectric has a nonlinear constitutive
relation. As usual, we assume that the transversal dimensions of the plates are much
3
Nonlinear dielectric
zero E
constant E
zero E
f
−q
+q
x
Figure 1: A nonlinear capacitor.
bigger than the plate’s separation x, so that the electric field is zero outside the capacitor
and constant inside of it. In the vacuum or in linear isotropic dielectrics, it can be shown
that the voltages of a set of conductors depend linearly on the charges. Here we assume
a more general situation where the voltage difference depends nonlinearly on q. To make
contact with the linear case, we introduce a capacitance-like function C(x, q) so that
v(x, q) =
q
.
C(x, q)
According to our general discussion, we have to compute
Wf (q, x) =
Z
(q,x)
(0,x0 )
v dq − f dx
Going first with q = 0 from x0 to x and then from q = 0 to q with x fixed, one gets
Z q
ξ
Wf (x, q) =
dξ.
0 C(x, ξ)
Finally, we derive with respect to x to obtain the force on the plates. An integration
by parts helps to make contact with the linear case:
Z q
∂
1
∂Wf
(x, q) = −
ξ
dξ
f (x, q) = −
∂x
∂x C(x, ξ)
0
Z q 2 2 q2 ∂
ξ ∂
1
1
= −
+
dξ
2 ∂x C(x, q)
C(x, ξ)
0 2 ∂ξ∂x
Z
1
E(x, q) 1 q 2 ∂ 2
ξ
dξ.
+
= q
2
2 0
∂ξ∂x C(x, ξ)
4
For a linear system, C(x, q) = C(x),
f (x, q) = q
E(x, q)
2
which can be easily obtained by simple arguments from the electric field seen by the +q
plate, and
Z q
ξ
1 q2
Wf (x, q) =
dξ =
,
2 C(x)
0 C(x)
which is the well-known elementary result.
2.2
Magnetic stationary system
Figure 2 shows two coils coupled by the magnetic field they generate. Although this is
not an electromechanical system (there are no moving parts) it allows us to introduce
several concepts which will be used in the next example.
The current of any of the coils produces a magnetic field which goes through the other
coil. Variations in any of the currents change the flux of the magnetic induction field
and this, according to Lenz’s law, originates an induced voltage in both coils. The iron
core is introduced to “bend” the magnetic field lines so that the coupling is tighter. The
coils have N1 and N2 turns, respectively. Any turn of any of the two coils has three
contributions to the magnetic flux that goes through it:
• the flux due to the lines of the induction magnetic field generated by the current of
the same coil and which close through the iron core,
• the flux due to the lines of the induction magnetic field generated by the current of
the same coil and which do not close through the iron core, and
• the flux due to the lines of the induction magnetic field generated by the current of
the other coil and which close through the iron core.
This can be written as
Φ1 = Φl1 + Φm1 + Φm2
Φ2 = Φl2 + Φm2 + Φm1
Φm1
Φm2
i1
i2
+
v1
+
N1
Φl1 Φl2
N2 v2
−
−
Figure 2: Magnetically coupled coils.
5
Φl1 and Φl2 are known as the leakage fluxes, while Φm1 and Φm2 are the magnetizing
fluxes. The total flux through the coils, which is the quantity which enters the electrical
equations, is then
λ1 = N1 Φ1 ,
λ2 = N2 Φ2 .
For a linear magnetic system, fluxes can be expressed in terms of path reluctances,
number of turns and currents as
N 1 i1
N 1 i1
N 2 i2
N 2 i2
Φl1 =
, Φm1 =
, Φl2 =
, Φm2 =
,
Rl1
Rm
Rl2
Rm
and one finally gets the well-known linear relation in terms of inductances:
λ1 = L11 i1 + L12 i2 ,
λ2 = L21 i1 + L22 i2 .
For a system with a variable geometry, the mutual inductances L12 and L21 depend on
the geometrical parameters. If the system is nonlinear, the inductances also depend on
the various currents. Usually, only the core experiences nonlinear effects (i.e. saturation),
and hence only the mutual inductances are current-dependent.
The dynamical equations of the system are now given by
dλ2
dλ1
, v 2 = r2 i 2 +
dt
dt
where r1 , r2 are the electric resistances of the respective windings.
An equivalent circuit interpretation, which is often used in the electrical engineering
literature, can be given if we introduce
2
N2
N1
N1
N1
0
0
0
0
r2 ,
i2 =
i2 , v 2 =
v 2 , λ2 =
λ 2 , r2 =
N1
N2
N2
N2
2
N1
N1
0
Ll2 =
L12 .
Ll2 , Lm1 =
N2
N2
This is called referring the winding 2 to winding 1. The equations keep the same form as
before the change
0
dλ1
dλ2
0
0 0
v 1 = r1 i 1 +
, v 2 = r2 i 2 +
,
dt
dt
but now
v 1 = r1 i 1 +
0
λ1 = Ll1 i1 + Lm1 (i1 + i2 )
0
0
0
0
λ2 = Ll2 i2 + Lm1 (i1 + i2 )
which can be seen as a T-circuit (see Figure 3).
As an special case we consider the ideal transformer. Set r1 = r2 = Ll1 = Ll2 = 0.
0
Then v1 = v2 , i.e.
N2
v1 .
v2 =
N1
0
If, additionally, Lm1 = N12 /Rm is large enough so that i1 + i2 is negligible, we get
i2 = −
6
N1
i1 .
N2
0
Ll1
r1
Ll2
0
r2
+
v1
+
0
i1
i2
Lm1
−
0
v2
−
Figure 3: An equivalent circuit for the magnetic system.
r
Φm
i
Φl
v
x
Figure 4: An elementary electromagnet: a magnetic system with a moving part.
2.3
Elementary electromagnet
Figure 4 shows an elementary electromagnet, a magnetic system with a moving part the
flux linkage λ through the coil depends on a geometry variable, the “air gap” x.
The electrical equation of motion is
v = ri +
dλ
,
dt
where the flux linkage can be computed from the number of turns, N , and the magnetic
induction flux, Φ, as
λ = N Φ.
In turn, Φ has a leakage, Φl , and a magnetizing, Φm , parts, Φ = Φl + Φm , which can be
computed in terms of the reluctances of the respective paths:
Φl =
Ni
,
Rl
Φm =
Ni
.
Rm
The reluctance of the magnetizing path has a fixed contribution, the part of the iron path,
and a variable one, the part of the air gap:
Rm =
2x
li
+
,
µri µ0 Ai µ0 Ag
7
where µri , the relative magnetic permeability of the iron core, is of the order of 103 .
Assuming that the sections of the iron and air gap paths are the same, Ai = Ag = A, one
gets
1
li
+ 2x .
Rm =
µ0 A µri
The relation between the current and the flux linkage can finally be written as
2
N2
N
λ=
+
i = (Ll + Lm )i
Rl
Rm
with
Lm =
N2
N 2 µ0 A
b
= li
.
≡
Rm
c
+
x
+
2x
µri
EXERCISE Compute the electromechanical energy Wf and the force necessary to
move x.
2.4
Coenergy
Assume that, from v = ∂q Wf and i = ∂λ Wf , we can express q and λ as functions of v and
i (and x). Then the coenergy, Wc , is the Legendre transform of the energy Wf :
!
E
M
X
X
Wc (v, i, x) =
qj v j +
λj ij − Wf (q, λ, x)
.
j=1
j=1
q=q(v,x),λ=λ(i,x)
For linear electromagnetic systems, i.e. q = C(x)v, λ = L(x)i, energy equals coenergy:
Wc (v, i, x) = Wf (q, λ, x)|q=q(v,x),λ=λ(i,x) .
(3)
In particular, for linear magnetic systems,
1
1
Wf (λ, x) = λT L−1 (x)λ, Wc (i, x) = iT L(x)i.
2
2
Notice that the coupling force can be obtained from the coenergy as
∂x Wc = ∂x λT i − ∂x λT ∂λ Wf + ∂x Wf = −∂x Wf = f.
EXERCISE Prove (3).
EXERCISE Consider an electromechanical system with a nonlinear magnetic material such that
λ = (a + bx2 )i2 ,
where a and b are constants and x is a variable geometric parameter. Compute Wf , Wc
and f , and check all the relevant relations.
3
Port hamiltonian system modelling
In this Section we will introduce the port hamiltonian model of a wide class of electromechanical systems, for which a passive controller will be discussed in the next Section.
We also consider power converters which, although not by themselves electromechanical
systems, are an integral part of the control of electromechanical devices [13].
8
3.1
General electromechanical systems
We consider the class of electromechanical systems [18] described by
λ̇ + Ri = Bv
J θ̈ = −rm θ̇ + Te (i, θ) + Tm
(4)
where, assuming a linear magnetic system (with no permanent magnet), λ = L(θ)i. The
generalized electrical force is given by
1
Te (i, θ) = iT ∂θ L i.
2
Both i and θ can contain an arbitrary number of degrees of freedom and, although we
are adopting a “rotational notation”, translational motion is also included. Tm is the
(external) mechanical force, J is the moment of inertia (or the mass) of the moving parts
and rm is a friction coefficient. The matrix B describes which parts of the electrical system
are actuated by the control voltages v. This general model includes most of the classical
electrical machines, as well as linear motors and levitating systems.
Introducing p = J θ̇ and x = (λ θ p)T , the port hamiltonian description is given by




−R 0
0
B 0 v




0
0
1
0 0
ẋ =
(5)
∂x H +
Tm
0 −1 −rm
0 1
with the Hamiltonian
1
1 2
H(x) = λT L−1 (θ)λ +
p.
2
2J
EXERCISE Show that indeed (5) reduces to (4).
EXERCISE Write the port hamiltonian model (5) for the electromagnet.
3.2
Power converters
Power converters are of great value in many growing application areas such as the control of electromechanical devices, portable battery-operated equipment or uninterruptible
power sources. In particular, dc-to-dc power converters are used to reduce or elevate
a given dc voltage, storing the energy in intermediate ac elements and delivering it as
needed. The essential trick is the use of switches, which are operated in a periodic manner and which make the system to alternate between several dynamics. Generally, the
individual dynamics are linear and can be solved analytically; the action of the switches
yields a nonlinear dynamics which can display a rich behavior (see [8, 1] and references
therein).
Figure 5 shows a functional model of the boost (or elevator) converter (the detailed
electronics of how the switches are implemented is not shown). The switches s1 and s2
are complementary: when s1 is closed (s1 = 1), s2 is open (s2 = 0), and viceversa. Thus,
the different circuit topologies can be described with a single boolean variable S = s2 .
The port hamiltonian modelling of electric circuits can be done in a systematic way
using tools from graph theory [3], but since we are dealing here with a circuit of very
small size we will adopt a more pedestrian approach and concentrate on the problems
9
v1
vL
s1
iE
L
iL
E
+
−
i2
i1
iC
v2
C
s2
iR
R
vC
vR
Figure 5: A functional description of the boost converter.
presented by the switches, using the ideas of [7]. A more in-deep conceptual analysis of
the switches can be found in [9].
The hamiltonian dynamical variables of the boost converter are the magnetic flux at
the coil, φL , and the charge of the capacitor, qC . Hence we have two one-dimensional
hamiltonian subsystems, with a global hamiltonian H = HC + HL ,
and
dqC
= iC ,
dt
vC =
∂H
,
∂qC
(6)
dφL
= vL ,
dt
iL =
∂H
,
∂φL
(7)
connected by Kirchoff’s laws
iL
i1
v2 + vL
vC + v1
vC
iE + iL
=
=
=
=
=
=
i1 + i2
iC + iR
E
v2
vR
0
(8)
Here we treat the switches as ports, with their correspondent effort and flow variables.
For the time being we do not terminate the resistive port, i.e. we do not use vR = RiR .
EXERCISE Write the EF -representation [5] of the Dirac structure associated to (8).
Using (6) and (7), the first four equations of (8) can be written as
∂H
= i1 + i2
∂φL
dqC
+ iR
i1 =
dt
dφL
v2 +
= E
dt
∂H
+ v1 = v2
∂qC
10
(9)
The second and third equations in (9) yield a hamiltonian system with four inputs
and J = R = 02×2 :


i
1

d
1 0 −1 0 
qC
T
 v2  .
= 0(∇H) +
(10)

0 −1 0 1
iR 
dt φL
E
Next we will use the constraints imposed by the switches to absorb the ports s1 and
s2 into the hamiltonian structure:
• S = 0 ⇒ s1 = 1, s2 = 0 ⇒ v1 = 0, i2 = 0,
• S = 1 ⇒ s1 = 0, s2 = 1 ⇒ i1 = 0, v2 = 0.
Hence, when S = 1 we already have the values of the port variables i1 , v2 in (10),
while if S = 0, using the first and fourth equations in (9),
i1 =
∂H
∂H
, v2 =
.
∂φL
∂qC
We can put together both results as
∂H
,
∂φL
∂H
.
= (1 − S)
∂qC
i1 = (1 − S)
v2
(11)
Now


(1
−
S)
∂H/∂φ
L

d
qC
1 0 −1 0 
 (1 − S) ∂H/∂qC 
=


0 −1 0 1
iR
dt φL
E
−1 0
iR
0
1−S
∂H/∂qC
+
, (12)
=
E
0 1
∂H/∂φL
−(1 − S)
0
which is a port hamiltonian system with outputs
y=
−1 0
0 1
T ∂H/∂qC
∂H/∂φL
=
−vC
iL
=
−vR
−iE
.
Finally, we terminate the resistive port using
iR =
vC
1 ∂H
vR
=
=
R
R
R ∂qC
and get our final port hamiltonian representation of the boost converter
d
qC
0
1−S
1/R 0
∂H/∂qC
0
=
−
+
E, (13)
−(1 − S)
0
∂H/∂φL
0 0
1
dt φL
11
vL
v1
L
s1
iE
E
i1
+
i2
−
s2
iR
iC
iL
C
v2
R
vC
vR
Figure 6: The buck converter.
with natural output
y=
0
1
T ∂H/∂qC
∂H/∂φL
= iL = −iE .
Notice that the interconnection structure J is modulated by the boolean variable S.
Designing a control for this system means choosing S as a function of the state variables.
EXERCISE Figure 6 shows an scheme of the buck (or step-down) power converter.
Show that the final port hamiltonian structure is
d
0
0 1
1/R 0
∂H/∂qC
qC
+
E.
=
−
1−S
∂H/∂φL
−1 0
0 0
dt φL
4
Passive control examples
Remember ([16] and elsewhere in this Summer School) that the central objective of the
IDA-PBC methodology for a given port hamiltonian system
ẋ = (J (x, u) − R(x))(∇H)T (x) + g(x, u),
is to find additional structure Ja and resistive Ra matrices, a control function u = β(x)
and a vector K(x), gradient of a function Ha (x), such that
(J (x, β(x))+Ja (x)−R(x)−Ra (x))K(x) = −(Ja (x)−Ra (x))(∇H)T (x)+g(x, β(x)) (14)
is satisfied, and such that Hd (x) = H(x)+Ha (x) has a minimum at the desired equilibrium
point. Equation (14) is a complicated PDE for Ha and several techniques have been
proposed in the literature [16, 18, 15] to facilitate its solution. Here we are going to
illustrate some of the basic techniques with concrete examples. Appendix A contains a
very simple account of the method of characteristics, which is the basic tool for solving
the kind of PDE that arises from (14).
4.1
Magnetic levitation system
Figure 7 shows a very simplified model of a magnetic levitation system.
The flux lines generated by the current at the coil close through the air gap and the
iron ball. Since the air gap has a variable reluctance, the system tries to close it, and this
counteracts the gravity.
12
u
i
φ
m
y
g
Figure 7: A magnetic levitation system.
The equations of motion are
φ̇ = −Ri + u
ẏ = v
mv̇ = Fm + mg
with φ = L(y)i the linkage flux, R the resistance of the coil, and Fm the magnetic force,
given by
∂Wc
Fm =
,
∂y
where the magnetic coenergy is (we assume a linear magnetic system)
Wc =
1 ∂L 2
i.
2 ∂y
In general, L is a complicated function of the air gap, y. A classical approximation for L
for this kind of systems for small y is
L(y) =
k
a+y
with k, a constants, which is essentially the relation used for our model of the electromagnet, without the leakage term.
Taking x1 = φ, x2 = y, x3 = mv, this can be written as a PCH

 

 
0 0 0
R 0 0
1
∂H  






0 0 1
0 0 0
0 u
ẋ =
−
+
∂x
0 −1 0
0 0 0
0
with Hamiltonian
1 2
1
(a + x2 )x21 +
x − mgx2 .
2k
2m 3
Note that the gravity term could also have been included as an external mechanical force.
H(x) =
13
The gradient of the hamiltonian is

2
x1 a+x
k
1 2
x1 − mg  .
(∇H)T =  2k

x3
m
Given a desired y ∗ , the equilibrium point is

 √
2kmg
,
y∗
x∗ = 
0
with an equilibrium control
R ∗
x1 (a + x∗2 ).
k
We will first try to compute the IDA-PBC control without using the specific theory
presented for general electromechanical systems.
Taking first Ja = Ra = 0, the IDA-PBC equation (J − R)K(x) = gβ(x) yields in this
case
u∗ =
−RK1 (x) = β(x)
K3 (x) = 0
−K2 (x) = 0,
and we see that Ha (x) = Ha (x1 ), which can be chosen so that
Hd (x) = H(x) + Ha (x1 )
has a critical point at x = x∗ .
Unfortunately
2

∂ Hd
(x) = 
∂x2
1
(a
k
+ x2 ) + Ha00 (x1 )
x1
k
0
x1
k
0
0

0
0 
1
m
has at least one negative eigenvalue no matter what Ha we choose, so x∗ will not be
asymptotically stable. The source of the problem is the lack of coupling between the
mechanical and magnetic part in the interconnection matrix J. To solve this, we aim at




0 0 −α
0 0 −α
1 ,
Jd =  0 0
i.e.
Ja =  0 0 0  .
α −1 0
α 0 0
Taking Ra = 0, the IDA-PBC equation now becomes
α
x3 + β(x)
m
K3 (x) = 0
α
αK1 (x) − K2 (x) = − (a + x2 )x1 ,
k
−αK3 − RK1 (x) =
14
Now Ha = Ha (x1 , x2 ). Using the second equation, the first equation yields the control
u = β(x) = RK1 − α
x3
,
m
while the third equation is a PDE for Ha (x1 , x2 ):
α
x1 (a + x2 )
∂Ha ∂Ha
−
= −α
.
∂x1
∂x2
k
(15)
EXERCISE Solve (15) by the method of characteristics. It is better to give the
initial condition as (0, s, Φ(s)) instead of (s, 0, Φ(s)).
The way in which (14) has been solved in this case seems to be quite model-dependent.
In [18] a more general method for systems of the form (5) with Tm = ∂θ V (θ) is proposed.
The central idea is to aim for a Hd given by
1 2
1
p,
Hd (x) = (λ − µd (θ, p))T L−1 (θ)(λ − µd (θ, p)) + Vd (θ) +
2
2J
where µd (θ, p) is a kind of desired permanent magnet, and consider a hamiltonian structure
of the form


−R
α(x) β(x)
.
0
1
Jd − Ra =  −αT (x)
T
−β (x) −1 −ra (p)
It can be seen then that the method boils down to an algebraic equation for
id = L−1 (θ)µd (θ, p),
namely
p
1 T
id ∂θ L(θ)id + ∂θ Vd (θ) − ∂θ V (θ) − (rm − ra (p)) = 0,
2
J
and that α, β and u can be computed from it.
(16)
EXERCISE Try to derive (16). See [18] for details.
EXERCISE Solve (16) for the levitating ball system. Use Vd (y) = Kp √ (y−y
∗ )2
1+(y−y ∗ )2
,
2
p
ra (p) = ra1 + ra2 1+p
2 , with Kp > 0, ra1 > 0, ra2 > 0.
4.2
Power converters
We retake the boost converter, or rather, its averaged model, where u = 1 − S, instead of
taking values in the set {0, 1}, varies over [0, 1] (see [6] for a discussion of this averaging
process; essentially, S is changed periodically with a frequency much higher than the
highest natural frequency of the system).
We have
0
1/R 0
0 u
,
, g=
, R=
J (u) =
1
0 0
−u 0
and, assuming a linear electromagnetic system,
H(x1 , x2 ) =
1 2
1 2
x1 +
x.
2C
2L 2
15
We set as control objective the regulation of the load voltage at Vd , so that the equilibrium
LV 2
point is x∗ = (CVd , REd ) and the equilibrium value of the control is
u∗ =
E
.
Vd
Notice that this makes sense (i.e. u∗ ∈ [0, 1]) since this is a boost converter and Vd ≥ E.
To solve (14) we take Ja = Ra = 0 and get
−1/R u ∂Ha
= gE,
−u 0
∂x
i.e.
∂Ha
= E,
∂x1
∂Ha
1 ∂Ha
+u
= 0.
−
R ∂x1
∂x2
−u
The standard way to solve this system [16] is to solve for the derivatives of Ha and
then impose the identity of the second order cross derivatives. One gets
∂Ha
E
= − ,
∂x1
u
E 1
∂Ha
.
= −
∂x2
R u2
Imposing the identity of the cross derivatives we arrive at the PDE
2 ∂u
∂u
−u
= 0.
R ∂x1
∂x2
(17)
EXERCISE Solve (17). As initial condition for the characteristic’s method, take
(0, k1 s, as + b), with k1 , a and b constants. In [19, 20] it is discussed how to choose the
values of k1 , a and b to impose the stability of the equilibrium point, as well as how to
improve the resulting controller so that it is robust with respect to E and R.
EXERCISE Repeat the above analysis for the buck converter. See the beginning of
Section 3.2 of [16] for some hints.
EXERCISE Implement the buck and boost systems and controllers in 20-sim. To
test them, use L = 20 mH, C = 20 µF, E = 15 V and R = 30 Ω.
5
Connecting systems
In this Section we will present an example of a hamiltonian system appearing as the result
of interconnecting two of them by means of a Dirac structure (in its Kirchoff’s law form).
The systems we are going to consider are the magnetic levitation system and the boost
converter without the resistive port. The hamiltonian structures are given by
16
i
iR
boost
u
vR
magnetic
levitation
system
Figure 8: Interconnection of the boost converter with the magnetic levitation system.
• magnetic levitation system:

 

 
0 0 0
R 0 0
1
ẇ =  0 0 1  −  0 0 0  (∇H)T +  0  u, (18)
0 −1 0
0 0 0
0
1 2
1
(a + w2 )w12 +
w − mgw2 ,
(19)
H(w1 , w2 , w3 ) =
2k
2m 3
1 0 0 (∇H)T = i.
y =
(20)
• open boost converter (we set η = 1 − S):
0 η
−1 0
iR
T
ż =
(∇H̃) +
,
−η 0
0 1
E
1 2
1 2
H̃(z1 , z2 ) =
z1 +
z2 ,
2C
2L
−vR
−1 0
T
(∇H̃) =
.
ỹ =
−iE
0 1
(21)
(22)
(23)
Now we interconnect both systems as shown in Figure 8. Kirchoff’s laws are
iR = i,
vR = u,
or, using (20) and (23),
iR =
∂H
,
∂w1
u=
∂ H̃
.
∂z1
(24)
These two relations can be substituted into (18) and (21), so that the input terms get
some hamiltonian gradients. The general theory of interconnection of port hamiltonian
systems tells us that these terms can be absorbed into the structure part if both systems
are taken together, and this is indeed what happens:

 

 
0 0 0 −1 0
R 0 0 0 0
0
 0 0 1 0 0   0 0 0 0 0 
 0 

 

 
T

 

 
ẋ = 
 0 −1 0 0 0  −  0 0 0 0 0  (∇HT ) +  0  E
 1 0 0 0 η   0 0 0 0 0 
 0 
0 0 0 −η 0
0 0 0 0 0
1
17
where x = (w1 w2 w3 z1 z2 )T and HT = H + H̃, and where the natural output of the
interconnected system is
yT = (0 0 0 0 1)(∇HT )T = iL = −iE .
EXERCISE Write the port hamiltonian model of the interconnection of the open
(non resistance-terminated) buck converter and the electromagnet.
EXERCISE - OPEN QUESTION Can you write the magnetic levitating ball
model as the interconnection of the mechanical part (the ball) and the electrical one (the
coil)?
A
Solving quasilinear PDEs
Equations of the form
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u),
(25)
where ux = ∂u/∂x, uy = ∂u/∂y, appear frequently in modern control theory, and in
particular in the IDA-PBC scheme. Equation (25) is called a quasilinear PDE because
the derivatives in u appear linearly, although in general the dependence on u is nonlinear.
Geometrically, the solution to (25) is a surface u = u(x, y) whose normal (ux , uy , −1)
is constrained by (25). This simple fact allows the explicit construction of a solution for
this low dimension case, although the resulting method, called characteristics’s method in
the literature, can be generalized to higher order cases as well as to fully nonlinear PDEs.
Let (x0 , y0 .u0 ) be a point on a solution surface and let (x(τ ), y(τ ), u(τ )) be a curve on
the surface through it at τ = 0. This means that the tangent vector (x0 (0), y 0 (0), u0 (0))
must be tangent to the surface at the point. Let us see how can we impose this condition.
Let (p = ux (x0 , y0 ), q = uy (x0 , y0 ), −1) be a normal at the point. According to the
preceding discussion, it must satisfy
c(x0 , y0 , u0 ) = a(x0 , y0 , u0 )p + b(x0 , y0 , u0 )q.
(26)
The set of all planes through the point (x0 , y0 , u0 ) is given by
u − u0 = p(x − x0 ) + q(y − y0 ),
(27)
and p and q must satisfy (26) for this plane to be tangent to the surface. We can be sure
that the tangent vector to the curve is indeed tangent to the surface if we impose that it
belongs to the whole family of planes. Now it can be shown that the equation of the line
common to all the planes in the set is
x − x0
y − y0
u − u0
=
=
.
a(x0 , y0 , u0 )
b(x0 , y0 , u0 )
c(x0 , y0 , u0 )
(28)
EXERCISE Show that (28) yields the equation of the line common to the family of
planes (27).
18
Using the vector along the line of (28), we impose
x0 (0) = a(x0 , y0 , u0 ),
y 0 (0) = b(x0 , y0 , u0 ),
u0 (0) = c(x0 , y0 , u0 ),
or, taking into account that this must be valid for any point on the curve,
x0 (τ ) = a(x(τ ), y(τ ), u(τ )),
y 0 (τ ) = b(x(τ ), y(τ ), u(τ )),
u0 (τ ) = c(x(τ ), y(τ ), u(τ )).
(29)
The solutions to this system of ODE are called characteristic curves of the PDE,
while their projections on the plane u = 0 are simply called characteristics. To generate a
solution surface, one must start with a curve of initial conditions (x(0, s), y(0, s), u(0, s))
and solve (29) for each point on the curve. This way one gets (x(τ, s), y(τ, s), u(τ, s)). If
the curve of initial conditions does not lie on a characteristic curve, it is possible to solve
for τ and s in terms of x and y, and finally get u(x, y).
As a (manifestly trivial) example, let us consider
3ux + 5uy = u,
with an initial curve (s, 0, f (s)) where f is arbitrary. We have to solve
x0 = 3,
y 0 = 5,
u0 = u.
The solution satisfying the initial conditions is
x = 3τ + s,
y = 5τ,
u = f (s)eτ .
From the first two equations we get τ = y/5 and s = x − 3y/5, and the corresponding
solution surface is
3y y
u(x, y) = f (x − )e 5 .
5
EXERCISE For the above example, show that we cannot get the solution surface if
the initial condition is given on (3s, 5s, f (s)).
Finally, let us remark that when several PDE for the same function are involved, some
compatibility conditions must be met for the system to be solvable. See Chapter 2 of [14]
for an account in terms of Fröbenius theorem and the integrability of a distribution of
vector fields.
19
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20
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21
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