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Non-linear control of electromagnetic valves for camlessenginesS. Di Gennaro a; B. Castillo-Toledo b; M. D. Di Benedetto aa Department of Electrical and Information Engineering and Center of ExcellenceDEWS, University of L'Aquila, Poggio di Roio, 67040 L'Aquila, Italyb Centro de Investigación y de Estudios Avanzados-CINVESTAV del IPN UnidadGuadalajara Av., Zapopan, Jalisco, México
Online Publication Date: 01 November 2007To cite this Article: Gennaro, S. Di, Castillo-Toledo, B. and Benedetto, M. D. Di(2007) 'Non-linear control of electromagnetic valves for camless engines',International Journal of Control, 80:11, 1796 - 1813To link to this article: DOI: 10.1080/00207170701519748
URL: http://dx.doi.org/10.1080/00207170701519748
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International Journal of ControlVol. 80, No. 11, November 2007, 1796–1813
Non-linear control of electromagnetic valves
for camless engines
S. DI GENNARO*{, B. CASTILLO-TOLEDOzand M. D. DI BENEDETTO{
{Department of Electrical and Information Engineering andCenter of Excellence DEWS, University of L’Aquila,
Poggio di Roio, 67040 L’Aquila, ItalyzCentro de Investigacion y de Estudios Avanzados – CINVESTAV
del IPN Unidad Guadalajara Av. Cientıfica 1145,45010, Col. El Bajıo, Zapopan, Jalisco, Mexico
(Received in final form 18 June 2007)
Conventional internal combustion engines use mechanical camshafts to command the opening
and closing phases of the intake and exhaust valves. The lift valve profile is designed in orderto reach a good compromise among various requirements of the engine operating conditions.In principle, optimality in every engine condition can be attained by camless valvetrains.
In this context, electromagnetic valves appear to be promising, although there are somerelevant open problems. In fact, in order to eliminate acoustic noises and avoid damage tothe mechanical components, the control specifications require sufficiently low impact velocities
between the valve and the constraints (typically the valve seat), so that ‘‘soft-landing’’ isobtained. In this paper, the soft–landing problem is translated into a regulation problemfor the lift valve profile, by imposing that the valve position tracks a desired reference,while the modelled disturbances are rejected. Both reference and disturbance are generated
by an autonomous system. The submanifold characterized by the zeroing of the trackingerror and the rejection of the disturbance, is determined. Finally, the stabilization problemof the system trajectory on such a manifold is solved.
1. Introduction
Conventional internal combustion engines use
mechanical camshafts to command the opening and
closing phases of the intake and exhaust valves.
The lift valve profile, connected with the crankshaft
angle and obtained with a proper cam profile, is
designed in order to find a compromise among various
requirements, such as the engine efficiency, pollution
emissions, fuel economy, valvetrain noise and vibration,
maximization of the output torque and power. In fact,
the different engine operating conditions would need
different lift valve profiles and valve timings, which
cannot be dynamically changed in a mechanically
driven camshaft. To overcome these limitations and
optimize the aforementioned requirements, a solution
based on variable valve timing can be pursued.In this context, camless valvetrains are devices
recently considered to decouple the camshaft and the
valve lift dynamics (see Gray 1988, Ashhab et al. 1996,
Schechter and Levine 1996, Pischinger et al. 2000).
Their main advantages are fuel saving, increase of max-
imum and low speed torque, flatting of the toque char-
acteristic and improvement of driveability, pollution
and energy consumption reductions, increase of the
burn rate, possible variability of the compression rate,
improving of the combustion stability at low speed
(Montanari et al. 2004). On the other hand, the electro-
hydraulic or electromechanical actuators of camless*Corresponding author. Email: [email protected]
International Journal of ControlISSN 0020–7179 print/ISSN 1366–5820 online � 2007 Taylor & Francis
http://www.tandf.co.uk/journalsDOI: 10.1080/00207170701519748
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engines present considerable problems, which still
remain open. In fact, these actuator dynamics are
highly non-linear and unstable near the valve’s terminal
positions, while the control specifications require that
impact velocities between the valve and the constraints
(typically the valve seat) be sufficiently low in order to
eliminate acoustic noises and avoid damage of the
mechanical components. These problems are compli-
cated by the short time (typically 3–5ms) available at
high engine speed to make a transition between the
two valve’s terminal positions, and the constraint in
terms of actuator cost and space limitations. These last
aspects imply that one typical request is the absence of
the valve position sensor.In Hoffmann and Stefanopoulou (2003) a precise
control of the voltage applied to the coils has been
proposed. A controller with a constant preset voltage,augmented by a voltage command based on a linear
feedback, is determined and then modified by an
iterative learning controller (ILC). In Tai et al. (2001)a control-oriented linear model for an EMCV has been
considered, based on a gray-box approach which com-
bines mathematical modelling and system identification.
In Wang et al. (2002) a physics-based model for anEMCV is derived. Moreover, a sensitivity study has
been conducted to characterize the ability of the control
signal to affect the reduction on contact velocities. This
model has been enriched by Peterson et al. (2002a) byintroducing the impact dynamics, and a self-tuning
non-linear controller has been designed. An observer
based output feedback controller has been proposed inPeterson et al. (2002b), while in Peterson et al. (2003)linear, non-linear and cycle-to-cycle self-tuning control-lers have been considered.
Our work uses the model of an electromechanicalactuator presented in Marchi et al. (2002) and Ronchiet al. (2002) . The soft-landing problem is translatedinto a regulation problem for the lift valve profile,by imposing that the valve position tracks a desiredreference, while the modeled disturbances are rejected.Both reference and disturbance are generated by anautonomous system, usually called exosystem. The sub-manifold characterized by the zeroing of the trackingerror and the rejection of the disturbance, is determined.Finally, the stabilization problem of the system trajec-tory on such a manifold is solved. The resulting control-ler uses measurements of the whole state. Clearly, this isits main limitation. Nevertheless, its utility is representedby the fact that it can be used along with a state observeror as a first step in the development of a dynamiccontroller, which uses only output measurements.
2. Mathematical model of the electromagnetic valves and
problem formulation
In this section we present the mathematical model of anelectromagnetic valve system (EMVS), represented infigure 1. This valve is composed of an anchor movingbetween two electromagnets, connected with the stemof the valve. The valve is opened and closed by meansof the attractive forces applied by the electromagnetson the anchor. In what follows we first present thedynamics of the electromagnets, then the mechanicaldynamics.
2.1 Flux and eddy current dynamics
We first consider the equation describing the dynamicsof the electric windings of the electromagnetic valve,which are (Marchi et al. 2002, Ronchi et al. 2002)
_�mj ¼1
N
�umj � RjImj
�, j ¼ 1, 2; ð1Þ
where �mj are the magnetic fluxes, Rj are the electricalresistances, and N is the number of turns of thewindings, which is the same for both windings for eachmagnet. Moreover, umj, Imj are the voltages and currentsof the electric circuits.
The currents Imj can be given as a function of �mj byexpressing the magnetomotive force Mmj as the sum oftwo terms (Ronchi et al. 2002)
Mmjð�mj,�mjÞ ¼ N jð�mjÞ þ Rjð�mjÞ�mj, j ¼ 1, 2 ð2Þ
Electromagnet 1
Electromagnet 2
Valve
Valve spring
Valve Stem
Anchor
xr
0
r
r
Fel.v
Ffr.v
Fm2
Fg
Fm1Fm1,ma
Vs1
Vs2
V1
Fm2,ma
d2
d1
d
V2
Coils
Coils
Torsion bar
Figure 1. Scheme of a electromagnetic valve system(EMVS).
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with N jð�mjÞ � 0 a non-linear function of the flux �mj,describing the drop of the magnetomotive force in themagnetic flux path in the iron, while Rjð�mjÞ�mj describesthe drop in the air-gap, linear in the flux �mj due to theabsence of saturation in the air, and non-linear in theair-gap �mj through the magnetic reluctanceRjð�mjÞ � 0. In what follows, we assume that the forcesclosing the valve are positive. It is worth noting thatwe suppose that on the anchor no magnetic material(permanent magnets) is present; therefore, each electro-magnet can only attract the anchor. We also supposethat only one electromagnet is supplied at each timeinstant.The functions N jð�mjÞ, j¼ 1, 2, can be described
by some exponential functions obtained by a best fitprocedure from experimental data
N jð�mjÞ ¼ N 0j enj�m j�mj, j ¼ 1, 2
N 0j, nj>0. The magnetic reluctance Rjð�mjÞ is anon-linear function of the air-gap thickness �mj � 0.Under the hypothesis that the anchor is connectedwith the valve stem tip, and that the arc can be approxi-mated by the cord, �mj can be expressed as functions ofthe anchor tip position xa 2 ½��, ��
�mj ¼
��þ ð�1Þjxa
�rl,mj, j ¼ 1, 2; ð3Þ
where 2� is the anchor tip displacement (coinciding withthe valve vertical stroke), 0 < rlm, j ¼ dj=d < 1, j¼ 1, 2,are the lever ratios, and xa is positive in the samedirection of xv, see figure 1. Experimentally, one obtainsthe following expressions
Rjð�mjÞ ¼ pj1
�1� e�pj2�mj
�þ pj3�mj, j ¼ 1, 2;
where pj1, pj2, pj3 are positive parameters obtained byidentification. It is worth noting that Rjð�mjÞ � 0 forall �mj.Denoting by Ipj the eddy currents in parasitic circuits
with resistance Rpj and inductance Lpj coupled with themagnetic circuits, having dynamics (Marchi et al. 2002)
�RpjIpj ¼ _�mj þ Lpj_Ipj, j ¼ 1, 2 ð4Þ
the expression of the magnetomotive force is (Marchiet al. 2002)
NImj þ Ipj ¼ Mmjð�mj,�mjÞ
¼ N jð�mjÞ þ Rjð�mjÞ�mj, j ¼ 1, 2; ð5Þ
where (2) has been used. Therefore, by substituting (5)into (1), one gets the flux dynamics
_�mj ¼Rj
N2
�Ipj �Mmjðxa,�mjÞ
�þ
1
Numj, j ¼ 1, 2; ð6Þ
where
Mmjðxa,�mjÞ ¼ N jð�mjÞ þ RjðxaÞ�mj, j ¼ 1, 2
RjðxaÞ ¼ Rjð�mjÞ
����mj¼ð�þð�1Þ
jxaÞrl,mj
¼ �ajbje�bjxa þ cjxa þ dj
aj ¼ bjpj1e�pj2rl,mj�, bj ¼ ð�1Þ
jrl,mjpj2,
cj ¼ ð�1Þj rl,mjpj3, dj ¼ pj1 þ �rl,mjpj3:
Moreover, using (4), (6), one works out the eddy currentdynamics
_Ipj ¼ �1
LpjRpj þ
Rj
N2
� �Ipj þ
Rj
LpjN2Mmjðxa,�mjÞ
�1
LpjNumj, j ¼ 1, 2: ð7Þ
2.2 Mechanical dynamics
The mechanical dynamics of the actuator can beobtained by writing the balance of the forces acting onthe anchor, of mass ma, and on the valve, of mass mv.In order to consider a specific situation, as anticipated,we suppose that the anchor remains always connectedwith the valve during its stroke. Moreover, the contribu-tion due to the gravity is supposed to be negligible, andall the mechanical components of the valve are consid-ered rigid bodies.Anchor dynamics. One first considers the (attractive)forces developed by the electromagnets on the movinganchor. Choosing the magnetic axis as reference, thesemagnetic forces can be written as
Fmj,mað�mj,�mjÞ ¼ �ð�1Þj 1
2
dRjð�mjÞ
d�mj�2mj, j ¼ 1, 2
with �mj the flux and �mj the air-gap thickness. Herewe considered the force Fm1,ma closing the valve aspositive force, and Fm2,ma opening the valve as anegative force (see figure 1). In order to write all themechanical equations in the same reference, theseforces can be replaced by the forces Fmj, j¼ 1, 2,acting on the anchor’s intersection point with thevalve stem axis (see figure 1), made equivalent to theforces Fmj,ma by equating the generated momenta.
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Considering relations (3), in this way one gets the fol-lowing expressions:
Fmj ¼ rl,mjFmj,mað�mj,�mjÞ
����mj¼rl,mjð�þð�1Þ
jxaÞ
¼ �1
2R0jðxaÞ�
2mj, j ¼ 1, 2 ð8Þ
with
R0jðxaÞ ¼dRjðxaÞ
dxa¼ aje
�bjxa þ cj, j ¼ 1, 2:
Clearly R01ðxaÞ < 0 and R02ðxaÞ > 0, so that Fm1 > 0(closing the valve) and Fm2 < 0 (opening the valve),accordingly to the convention on the force signs. It isclear that both electromagnets may be supplied at thesame time in order to impose some specific behaviouron the anchor. For the sake of the simplicity and
without loss of generality, we suppose that only theelectromagnet 1 (see figure 1) is supplied during theclosing phase of the valve, while during the openingphase that only the electromagnet 2 is supplied.
A torsional spring keeps the anchor in an intermediateposition. This force can be rendered equivalent (byequating the momenta) to a force Fel, a ¼ �kaðxa � x�aÞacting on the tip of the anchor, with x�a the intermediateposition, and ka the elastic coefficient for the torsion bar.Hence, considering the (equivalent) spring elasticforce Fel, a, the viscous friction force Ffr, a ¼ �ba _xa,the net magnetic force Fm ¼ Fm1 þ Fm2, and theconstraint force Fc,a due to the electromagnet surface,each considered with its sign, the anchor motionequation is given by
ma €xa ¼ Fel, a þ Ffr, a þ Fm þ Fc, a
¼ �kaðxa � x�aÞ � ba _xa þ Fm þ Fc, a ð9Þ
where ba is the viscous friction for anchor. Note that Fc,a
is always zero except for xa ¼ ��.Valve dynamics. The valve has two springs, one closingit and another opening it, which can be modelledequivalently by a single linear spring, preloaded tokeep the valve in the centre of its stroke x�v whenthe two electromagnets are not supplied, so that theelastic force Fel, v, due to the (equivalent) valve springis Fel, v ¼ �kvðxv � x�vÞ, with kv the elastic coefficientof the (equivalent) linear spring. Considering theother forces acting on the valve of mass mv,namely the viscous friction Ffr, v ¼ �bv _xv, thedisturbance force Fd due to the exhaust gases exitingthe cylinder, and the force Fc,v due to constraint givenby the valve seat, the valve dynamics are
mv €xv ¼ Fel, v þ Ffr, v þ Fd þ Fc, v
¼ �kvðxv � x�vÞ � bv _xv þ Fd þ Fc, v ð10Þ
with bv the viscous friction for the valve. The force Fd
is a disturbance acting on the valve. It depends onthe cam angle #c and parameters, such as thevalve equivalent area, the load, the gas turbulence, etc.A typical behaviour of Fd is given in figure 2 as afunction of #c, along with a valve lift reference.The model of Fd in a specific case will be given in thefollowing sections. Note that Fc,v is always zero exceptfor xv ¼ �.Mechanical dynamics. By combining (9), (10) one getsthe mechanical dynamics. As anticipated, we considerthe case where the anchor is connected with the valveduring its stroke, namely xa¼ xv, meaning that theconsidered position is that of the tip valve. Hence,adding (9) and (10) and considering (8), we work out
Figure 2. Reference trajectory xr and disturbance Fd.
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the mechanical dynamics
M €xv ¼ �kxv � b _xv þ Fm þ Fd þ Fc,
Fm ¼ �1
2
�R01ðxvÞ�
2m1 þR
02ðxvÞ�
2m2
�ð11Þ
where M ¼ ma þmv, b ¼ ba þ bv, k ¼ ka þ kv,Fc ¼ Fc, a þ Fc, v. Here we have set the centre of thevalve stroke x�v as the origin for xv. Equation (11)holds for xv 2 ½��, ��, where � is the maximal displace-ment with respect to x�v . Note that Fc is always zeroexcept for xv ¼ ��, where
Fcð�Þ ¼ k�þ1
2
�R01ð�Þ�
2m1þR
02ð�Þ�
2m2
��Fd� 0
Fcð��Þ ¼�k�þ1
2
�R01ð��Þ�
2m1þR
02ð��Þ�
2m2
��Fd� 0:
2.3 Mathematical model of the EMVS
From (11), (6), (7) one obtains the mathematical modelof an EMVS
_xv ¼ vv
_vv ¼1
M
��kxv� b _xvþFmþFdþFc
�_�mj ¼
Rj
N2
�Ipj�Mmjðxv,�mjÞ
�þ
1
Numj, j¼ 1, 2
_Ipj ¼�1
LpjRpjþ
Rj
N2
� �Ipjþ
Rj
LpjN2Mmjðxv,�mjÞ
�1
LpjNumj, j¼ 1,2
9>>>>>>>>>>>>>>=>>>>>>>>>>>>>>;
ð12Þ
with Fm as in (11). In the following we suppose that allthe state is available for measurement.
2.4 Hybrid model of the system and problem formulation
The mathematical model of the system is given byequation (12). The dynamics for the crankshaft angle# 2 ½0, 4�Þ, related with the cam’s angle#c ¼ #=2 2 ½0, 2�Þ, and velocity ! have not been consid-ered here for the sake of simplicity, even if the approachpresented in the following sections can be applied for thecomplete engine dynamics. Therefore, in what followswe consider that the engine has reached a steady-statewith ! ¼ !0 known.The control problem is to determine a controller
such that the valve is opened and closed following adesired valve lift reference trajectory (see figure 2),while the disturbance due to Fd is rejected. Due to thepeculiarity of the present application, it is important toensure exponential velocity of the convergence to the
valve lift reference, with velocity to be imposed by the
designer.A central aspect connected with the proper valve
motion is the so-called soft landing of the valve.
This problem comes into play either when the valve is
closing or is reaching its maximum aperture, namely
when the anchor is approaching one of the
electromagnet surfaces (we have supposed the anchor
and the valve rigidly connected). Clearly, this ensures
also a soft landing for the valve in its seat during the
closing phase, and avoids the valve chattering in the
opening phase. Typical values of the valve velocity
approaching the mechanical constraints (seating
velocity) is 0.05 – 0.1m/s.One can translate the control problem with soft
landing into a regulation problem for the valve stem,
imposing that the valve position xv tracks a desired
reference xr, while the disturbance due to Fd is rejected.Various strategies can be followed to supply the elec-
tromagnets. For instance, some authors proposed that
the closed-loop control is unnecessary when the anchor
is in the intermediate position, since high currents
would be necessary to impose an effectual force
(Montanari et al. 2004). In order to reduce the control
effort and for the sake of simplicity we will suppose
that only one electromagnet is supplied when attracting
the anchor. Therefore, when the magnetomotive force is
positive (negative) only the electromagnet 1 (electro-
magnet 2) is supplied. This means that only one flux
dynamic equation is forced by the input, while the
other is in free evolution. Note that both the fluxes
appear in Fm, so that the flux in free evolution deter-
mines a term which can be considered a perturbation
to be compensated, see below.This control strategy yields to a description of system
(12) by means of a hybrid system with four discrete
states q 2 Q ¼ fq1, q2, q3, q4g of a finite state automaton
(see figure 3): one for the closing phase, one when the
valve is completely close, one for the opening phase,
and finally one when the valve is completely open. The
transitions among these states depend on the value of
the system state x ¼�Xv Vv �m Ip
�T2 R
4 and on
the value of the ‘‘perturbation’’ state
xw ¼�
�w Iw�T2 R
2. The resulting hybrid system has
continuous dynamics given by
_x ¼ fiðx,xw,w, uÞ ð13Þ
_xw ¼ fwiðx, xwÞ i 2 I ¼ f1, 2, . . . ,Ng ð14Þ
e ¼ hðx,wÞ ð15Þ
with N¼ 4, uðtÞ 2 Rm the input (m¼ 1), e 2 R
s thetracking error (s¼ 1), h ¼ Xv � xrðwÞ the output, xr(w)
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the output reference, and
fi¼
Vv
1
M
��kXv�bVvþFmðx,xwÞþFdðwÞþFc
�R
N2
Ip�MmðXv,�mÞþ
N
RU
!
�1
LpRpþ
R
N2
� �Ipþ
R
LpN2MmðXv,�mÞ�
1
LpNU
0BBBBBBBBBB@
1CCCCCCCCCCA
fwi¼
Rw
N2
�Iw�MmwðXv,�wÞ
��
1
LpwRpwþ
Rw
N2
� �Iwþ
Rw
LpwN2MmwðXv,�wÞ
0BB@
1CCA:
9>>>>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>>>>;
ð16Þ
In figure 2 a possible reference for the output isindicated. The vector w 2 R
r represents the (bounded)reference and disturbance signals. Finally, Fmðx, xwÞ isgiven by (11),
MmðXv,�mÞ ¼ N ð�mÞ þ RðXvÞ�m
MmwðXv,�wÞ ¼ N wð�wÞ þ RwðXvÞ�w
are the magnetomotive forces, while the state variablesXv, Vv, �m, Ip, �w, Iw, the controls U1, U2 and theconstraint force Fc are given in table 1, withFc2 ¼ k� þ ð1=2ÞR01ð�Þ�
2m1 � FdðwÞ � 0, Fc4 ¼ �k�þ
ð1=2Þ R02ð��Þ�2m2 � FdðwÞ � 0. In particular, when the
system is in the states q2, q4, the first twoequations in (13) are identically verified, due to theconstraint force Fc.
The transitions among these four models are forcedwhen the invariant conditions, given by
I1: x j xv 2 ½��, ��, Fm � 0 for q1
I2: x j xv � �, Fm � 0 for q2
I3: x j xv 2 ½��, ��, Fm � 0 for q3
I4:x j xv � ��, Fm � 0 for q4
9>>>=>>>;
ð17Þ
are violated (Lygeros et al. 1999). Note that xv � ��means that xv is equal to �� for more than a timeinstant. These are conditions on the values of theextended state
�xT xTw
�T. Moreover, the transitions
leaving a state qi 2 Q are regulated by the so-calledguard conditions, namely rules stating when a certaintransition can take place. Referring to figure 3, theguard conditions are
Gjk: x satisfies Ik, j, k ¼ 1, 2, 3, 4:
Finally, after a transition, x can possibly undergo a resetof its components. In the case under study the resetfunctions Rjk are the identities.
Remark 1: Clearly, the resulting hybrid system H
depends on the control strategy we have consideredhere. Indeed, it is easy to determine an analogoushybrid model in the case of different control strategies,possibly taking into account different ways in whichthe electromagnets can be supplied.
Moreover, it is easy to incorporate in the model alsovalve bounces due to collisions of the valve in its seat
Figure 3. Hybrid system H modelling the EMVS.
Table 1. Variables and parameters of functions (16).
q1 q2 q3 q4
Xv¼ xv Xv ¼ � Xv¼ xv Xv ¼ ��Vv¼ vv Vv ¼ 0 Vv¼ vv Vv¼ 0
�m ¼ �m1 �m ¼ �m1 �m¼�m2 �m ¼ �m2
�w ¼ �m2 �w ¼ �m2 �w ¼ �m1 �w ¼ �m1
Ip ¼ Ip1 Ip ¼ Ip1 Ip ¼ Ip2 Ip ¼ Ip2Iw ¼ Ip2 Iw ¼ Ip2 Iw ¼ Ip1 Iw ¼ Ip1U ¼ um1 U¼ 0 U¼ um2 U¼0R ¼ R1 R ¼ R1 R ¼ R2 R ¼ R2
N ¼ N 1 N ¼ N 1 N ¼ N 2 N ¼ N 2
Rw ¼ R2 Rw ¼ R2 Rw ¼ R1 Rw ¼ R1
N w ¼ N 2 N w ¼ N 2 N w ¼ N 1 N w ¼ N 1
R ¼ R1 R ¼ R1 R ¼ R2 R ¼ R2
Rp ¼ Rp1 Rp ¼ Rp1 Rp ¼ Rp2 Rp ¼ Rp2
Lp ¼ Lp1 Lp ¼ Lp1 Lp ¼ Lp2 Lp ¼ Lp2
Rw ¼ R2 Rw ¼ R2 Rw ¼ R1 Rw ¼ R1
Rpw ¼ Rp2 Rpw ¼ Rp2 Rpw ¼ Rp1 Rpw ¼ Rp1
Lpw ¼ Lp2 Lpw ¼ Lp2 Lpw ¼ Lp1 Lpw ¼ Lp1
Fc¼ 0 Fc ¼ Fc2 Fc¼ 0 Fc ¼ Fc4
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or of the anchor with the electromagnets. Indeed, in thecase of partially elastic collisions, for instance, one canconsider transitions with appropriate guard conditions(bounces are experienced for jvvj � vb, with vb a certainminimum velocity) and reset functions (v is set to �cvafter the bounce, with c � 1 a certain coefficient).
3. The regulation problem for hybrid systems
In this section we will solve the regulation problem fora hybrid system H, characterized by the continuousnon-linear dynamics (13), (15), along with the autono-mous non-linear dynamics
_w ¼ sie ðwÞ, ie 2 I e ¼ f1, 2, . . . ,Neg ð18Þ
I e a finite index set, wðtÞ 2Wi � Rr the state of the ieth
generator of bounded reference and disturbance signals,described by (18) and called exosystem (Isidori 1995).As already mentioned, the control problem is to
determine a controller such that the valve is openedand closed tracking (globally and exponentially) adesired valve lift reference trajectory xr(w), rejectingthe disturbance due to Fd(w). Since the state variables,references, disturbances, etc., in general may undergodiscontinuities, we require the uniform exponential con-vergence to a �-ball of the origin, namely we require
��xv � xrðwÞ�� � ke��ðt�t0Þ
��xvðt0Þ � xrðwðt0ÞÞ��þ �,
8xvðt0Þ, 8wðt0Þ ð19Þ
for appropriate k>0, �>0. When �¼ 0 this impliesthat the the so-called regulator equations for
system (13) have to be solved (Isidori and Byrnes
1990, Isidori 1995).In what follows we first consider the expression
of the reference trajectory to be tracked and of
the disturbance to be rejected, and then we solvethe regulator equations for system (13) in the state
qi, i 2 I .
3.1 The reference trajectory and disturbancesfor a EMVS
The exosystem models the reference trajectory andthe disturbances acting on the system. For the sake
of simplicity and without loss of generality, somesimplifications will be introduced in order to better
illustrate the proposed approach. For instance, weconsider that all the moving parts of the valve are
rigid bodies, and we consider as disturbances only thepressure of the intake/exhaust gases. Flexibility in
the valve mechanism, usually modeled as disturbance,
and/or other perturbations could easily be takeninto account at the expense of a more complicated
presentation.As far as the reference trajectory is concerned,
we consider a trajectory ensuring the soft landing
of the valve in its seat and of the anchor on the surfacesof the electromagnets. Such a trajectory is, for instance,
the one imposed by a mechanical cam. In our casethe EMVS allows more general references satisfying
some condition of obvious physical interpretation.In fact, it is possible to impose a trajectory composed
of four parts (see figure 2), depending on thecam’s angle #c ¼ #=2 2 ½0, 2�Þ, with # 2 ½0, 4�Þthe crankshaft’s angle, solution of _# ¼ !0 (since !0 ishere considered constant, this dynamic equation has
not been considered in the system model). We requirethat the reference be at least a C4 function of #c (and
hence of time). As it will be shown, this condition
implies that the input ensuring tracking is a continuousfunction.
The first part of the reference (corresponding toje¼ 1 in figure 2), valid for #c 2 ½#c0,#c1Þ, is parameter-
ized as
xr1ð#cÞ ¼X7k¼0
cr, 1k#kck!
ð20Þ
with the coefficients in cr, 1k, k ¼ 0, 1, . . . , 7, determinedso that at #c ¼ #c0 ¼ 0 the closed valve starts to
open with velocity, acceleration and jerk (namely up
to the third derivative) equal to zero, and isclosed (with zero velocity, acceleration and jerk) at
Figure 4. Reference hybrid system.
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#c ¼ #c1 (8 conditions, see appendix 2)
xr1ð#c0Þ ¼ � ¼X7k¼0
cr, 1k#kc0k!
,
xr1ð#c1Þ ¼ �� ¼X7k¼0
cr, 1k#kc1k!;
d‘xr1d#‘c
���#c0¼ 0 ¼
X7k¼‘
cr, 1k#k�‘c0
ðk� ‘Þ!,
d‘xr1d#‘c
���#c1¼ 0 ¼
X7k¼‘
cr, 1k#k�‘c1
ðk� ‘Þ!, ‘ ¼ 1, 2, 3:
9>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>;
ð21Þ
In this way one obtains the expressions of thecoefficients cr, 1k, k ¼ 0, 1, . . . , 7. In a similar way, it iseasy to check that the second part of the reference(corresponding to je¼ 2 in figure 2), is simplyxr2ð#cÞ ¼ ��, for #c 2 ½#c3,#c0 þ 2�Þ. Here cr, 20 ¼ ��,cr, 21 ¼ ¼ cr, 25 ¼ 0, cr, 40 ¼ �, cr, 41 ¼ ¼ cr, 45 ¼ 0,while cr, 3k ¼ �cr, 1k, k ¼ 1, . . . , 7.It is clear that these four references can be generated
by (recall that #c ¼ ð!0=2Þt)
_w0 ¼ 0
_wk ¼!0
2wk�1,
w0ð0Þ ¼ 1
wkð0Þ ¼ 0, k ¼ 1, . . . , 7
namely by the linear system
_wr ¼ srðwrÞ, wr ¼
w0
..
.
w7
0B@
1CA,
srðwrÞ ¼!0
2
0 0 0 0
1 0 0 0
0 1 0 0
..
. ... . .
. ... ..
.
0 0 1 0
0BBBBBB@
1CCCCCCAwr ð22Þ
with the switching among the following different outputs
xr, je ðwÞ ¼X7k¼0
cr, jekwk, je ¼ 1, . . . , 4: ð23Þ
The corresponding hybrid system generating thereferences is depicted in figure 4, where the invariantconditions are
Ir1:#c j #c 2 ½#c0,#c1Þ for the state je ¼ 1
Ir2:#c j #c 2 ½#c1,#c2Þ for the state je ¼ 2
Ir3:#c j #c 2 ½#c2,#c3Þ for the state je ¼ 3
Ir4:#c j #c 2 ½#c3,#c0 þ 2�Þ for the state je ¼ 4
while the guard and reset conditions are
Gr12:#c � #c1, Gr23:#c � #c2,
Gr34:#c � #c3, Gr41:#c � #c0 þ 2�
Rr12,Rr23,Rr34,Rr41:w0 ¼ 1, w1 ¼ ¼ w7 ¼ 0:
As far as the disturbance is concerned, it can begenerated by
_w8 ¼ 0
_w9 ¼!0
2w8
w8ð0Þ ¼ 1
w9ð0Þ ¼ 0
i.e., by the linear system
_wd ¼ sdðwdÞ, wd ¼w8
w9
� �,
sdðwdÞ ¼!0
2
0 0
1 0
� �wd ð24Þ
with the switching among the following disturbancesignals
Fd, he ðwÞ ¼ Fd0w8 þ cd, hepaw8
1þ ðw9 � pcÞ2=pb
, he ¼ 1, 2
ð25Þ
with
cd, he ¼0 if #c 2 ½#c4 þ Dd � 2�,#c4Þ
1 if #c 2 ½#c4,#c4 þ DdÞ,
�
where we have used the so-called Agnesi’s versiera (orAgnesi’s witch) to approximate the gas pressure profileduring the combustion phase in the cylinder. Thehybrid system generating the disturbance is depicted infigure 5, where the invariant conditions are
Id1:#c j #c 2 ½#c4 þ Dd � 2�,#c4Þ for the state he ¼ 1
Id2:#c j #c 2 ½#c4,#c4 þ DdÞ for the state he ¼ 2
Figure 5. Disturbance hybrid system.
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while the guard and reset conditions are
Gd12:#c � #c4, Gd21:#c � #c4 þ Dd,
Rd12:w8 ¼ 1, w9 ¼ 0, Rd21: Id
with Id the identity map.The exosystem (18) is finally given by (22), (24), with
w ¼wr
wd
� �, sie ðwÞ ¼
srðwrÞ
sdðwdÞ
� �,
ie 2 I e ¼ f1, . . . , 6g ð26Þ
with outputs (23), (25) switching according to the hybridsystem depicted in figure 6, where the invariantconditions are
Ie1:#c j #c 2 ½#c0,#c1Þ for qe1,
Ie2:#c j #c 2 ½#c1,#c4Þ for qe2,
Ie3:#c j #c 2 ½#c4,#c2Þ for qe3,
Ie4:#c j #c 2 ½#c2,#c4 þ DdÞ for qe4;
Ie5:#c j #c 2 ½#c4 þ Dd,#c3Þ for qe5;
Ie6:#c j #c 2 ½#c3,#c0 þ 2�Þ for qe6;
while the guard and reset conditions are
Ge12:#c � #c1, Ge23:#c � #c4, Ge34:#c � #c2,
Ge45:#c � #c4þDd; Ge56:#c � #c3, Ge61:#c � #c0þ 2�
Re12ðwÞ ¼Rr12ðwrÞ
wd
� �, Re23ðwÞ ¼
wr
Rd12ðwdÞ
� �,
Re34ðwÞ ¼Rr23ðwrÞ
wd
� �; Re45ðwÞ ¼
wr
wd
� �,
Re56ðwÞ ¼Rr34ðwrÞ
wd
� �, Re61ðwÞ ¼
Rr41ðwrÞ
wd
� �:
3.2 Solution of the regulator equations for the EMVS
In this section we solve the regulator equations for thehybrid system. In the determination of the solution ofthe regulator equations it should be stressed that eachcomponent of the centre manifold depends on the ithsystem (13), with output (15), and on the ieth exosystem(26), with outputs (23), (25). For notational simplicity,we avoid indicating this dependence, which will beclear from the context.
Let us first consider the case in which system (13) is inthe state q1, where xv 2 ½��, ��, Fm � 0. In this case theregulator equations are given by
Ls�xv ¼ �vv
Ls�vv ¼1
M
��k�xv � b�vv þ �Fm
þ Fd, he ðwÞ�
Ls��m1¼
R1
N2
�Ip1 �Mm1ð�xv ,��m1
Þ þN
R1cm1
!
Ls�Ip1 ¼ �Rp1
Lp1�Ip1 �
1
Lp1Ls��m1
0 ¼ �xv � xr, je ðwÞ;
9>>>>>>>>>>>>>=>>>>>>>>>>>>>;
ð27Þ
where L stands for the classical Lie derivative, ands ¼ sie ðwÞ is the ieth exosystem function (26). Moreover,
�FmðwÞ ¼ML2s�xv þ bLs�xv þ k�xv � Fd, he ðwÞ
¼ �1
2R01ð�xv Þ�
2�m1: ð28Þ
From the first and the last equations of (27) it is easyto work out
�xv ðwÞ ¼ xr, je ¼X7k¼0
cr, jekwk
�vv ðwÞ ¼ Ls�xv ¼!0
2
X7k¼1
cr, jekwk�1
while from the second equation
0 � �1
2R01ð�xv Þ�
2�m1¼ �Fm
¼ k�xv þ bLs�xv þML2s�xv � Fd, he
¼X7k¼0
Cjekwk � Fd, he
Cjek ¼ kcr, jek þ bcr, jekþ1!0
2þMcr, jekþ2
!20
4,
k ¼ 0, . . . , 7;
Figure 6. Exosystem hybrid system.
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where cr, je8 ¼ cr, je9 ¼ 0. Hence, one obtains
��m1ðwÞ ¼
ffiffiffiffiffi�1p
if �1 � 0
0 otherwise
�
�1ðwÞ ¼ 2�FmðwÞ
�R01ð�xv Þ¼ 2
�FmðwÞ
jR01ð�xv Þj; ð29Þ
where R0jð�xv Þ 6¼ 0, j¼ 1, 2. The fourth equation of(27) admits a solution �Ip1 since the corresponding
differential equations (4) have eigenvalues �ðRp1=Lp1Þ
with negative real parts. This solution could
possibly be calculated in an approximated
way. Alternatively, one can consider a numeric
resolution; the problem is that the initial
conditions are unknown. Therefore, using the following
observer
Ls�Ip1 ¼ �Rp1
Lp1�Ip1 �
1
Lp1Ls��m1
where ��m1is given in (29), and �Ip1ð0Þ is known, one
obtains an estimate �Ip1 ðwÞ which exponentially tends
to �Ip1 ðwÞ.
Finally, from the third equation of (27) one works outthe steady-state input
cm1ðwÞ ¼ �R1
N
��Ip1 �Mm1ð�xv ,��m1
Þ
�þNLs��m1
:
This control exponentially tends to
cm1ðwÞ ¼ �R1
N
��Ip1 �Mm1ð�xv ,��m1
Þ
�þNLs��m1
satisfying the third equation of (27).When the system (13) is in the state q3, where
xv 2 ½��, ��, Fm � 0, in a similar way one determines
�xv ðwÞ ¼ xr, je ¼X7k¼0
cr, jekwk
�vv ðwÞ ¼ Ls�xv ¼!0
2
X7k¼1
cr, jekwk�1
��m2ðwÞ ¼
ffiffiffiffiffi�2p
if �2 � 0
0 otherwise
�
�2 ¼ 2��Fm
R02ð�xv Þ¼ 2
��Fm
jR02ð�xv Þj
cm2ðwÞ ¼ �R2
N
��Ip2 �Mm2ð�xv ,��m2
Þ
�þNLs��m2
0 0.05 0.1
−4
−3
−2
−1
0
1
2
3
4
0 0.01 0.02
−4
−3
−2
−1
0
1
2
3
4
× 10−3 × 10−3 × 10−3(a) (b) (c)
0 0.05 0.1−4
−3.5
−3
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
Figure 7. (a) Valve position xv and reference xr ¼ �xv ðwÞ [m]; (b) detail of xv, xr [m]; (c) error xv � xr [m].
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where an estimate �Ip2 ðwÞ exponentially converging to�Ip2 ðwÞ is given by the solution of
Ls�Ip2 ¼ �Rp2
Lp2�Ip2 �
1
Lp2Ls��m2
with �Ip2 ð0Þ known.Let us finally consider the case in which system (13)
is in the state q2 (xv � �, Fm � 0, i.e., j¼ 1) or q4(xv � ��, Fm � 0, i.e., j¼ 2). In this case the regulatorequations are
Ls�xv ¼ 0
Ls�vv ¼ 0
Ls��mj¼
Rj
N2
��Ipj �Mmjð��,��mj
Þ þN
Rjcmj
�, j ¼ 1, 2
Ls�Ipj ¼ �Rpj
Lpj�Ipj �
1
LpjLs��mj
0 ¼ ��� xr ð30Þ
and it results
�xv ¼ ��, �vv ¼ 0
while ��mj, j¼ 1, 2, remain undetermined. It is hence
possible to choose
q2:��m1¼
ffiffiffiffiffi�1p
if �1 � 0
0 otherwise
�
q4:��m2¼
ffiffiffiffiffi�2p
if �2 � 0
0 otherwise
�
�1 ¼ 2k�� Fd, he
�R01ð�Þ¼ 2
�Fm
jR01ð�Þj
�2 ¼ 2k�þ Fd, he
R02ð��Þ¼ 2
��Fm
jR02ð��Þj:
Moreover, for �Ipj the same reasoning holds, and henceit is possibile to consider estimates �Ipj , solutions of
Ls�Ipj ¼ �Rpj
Lpj�Ipj �
1
LpjLs��mj
, j ¼ 1, 2
while the steady-state controls are given by
cmj ¼ NLs��mj�Rj
N�Ipj þ
Rj
NMmjð��,��mj
Þ, j ¼ 1, 2:
4. Full information output regulation for a EMVS
In this section we will determine a control law ensuringthat the reference is followed, rejecting the disturbance.This control will need the knowledge of the whole state.
0 0.05 0.1−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
0 0.01 0.02 0 0.05 0.1−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6(a) (b)(c)
Figure 8. (a) Valve velocity vv and reference vr ¼ �vv ðwÞ [m/s]; (b) detail of vv, vr [m/s]; (c) error vv � vr [m/s].
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The importance of such a control relies on the fact thatit represents the first step in the design of a dynamiccontroller from the measured variables (typically notincluding the valve position).First we consider the classical regulation problem for
the single state qi, i ¼ 1, . . . , 4 of H (system (13)), andthen we prove that the whole hybrid system fulfills therequirements of the regulation problem expressed by (19).
4.1 Output regulation for single states of H
The design of the desired control law is based on theLyapunov approach. Let us determine the control lawwhen system (13) is in the state q1. For the mechanicalsubsystem (first two equations of the EMVS model)
_xv
_vv
� �¼ A
xv
vv
� �þ BðFm þ Fd, he Þ,
A ¼0 1
�k
M�
b
M
!, B ¼
01
M
!ð31Þ
with Fm given by (11), let us consider thefollowing Lyapunov function candidate for the‘‘mechanical’’ dynamics
Vsm ¼1
2
xv � �xvvv � �vv
2Ps
, Ps ¼ PTs > 0
where kzk2P stands for zTPz. Moreover, note that
Ls�xvLs�vv
� �¼ A
�xv�vv
� �þ Bð�Fm
þ Fd, he Þ
with �Fmas in (28). Considering the derivative of Vsm
along the trajectories of (31), one obtains
_Vsm ¼ �xv � �xvvv � �vv
2Qs
þxv � �xvvv � �vv
� �T
PsBðFm � �FmÞ
with Ps solution of ðPsAþ ATPsÞ=2 ¼ �Qs,Qs ¼ QT
s > 0, and
Fm � �Fm¼ �
1
2R01ðxvÞ
��2m1 � �s1�
2�m1
��1
2R02ðxvÞ�
2m2,
�s1ðxv,�xv Þ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR01ð�xv Þ=R
01ðxvÞ
q:
Recall that R0jðxvÞ 6¼ 0, j¼ 1, 2, 8xv. In the following itwill be clear that xw ¼
��m2 Ip2
�T, acting as a distur-
bance, goes exponentially to zero, so that nocompensation is necessary.Therefore, consider the subsystem given by the
mechanical dynamics plus the dynamics of �m1, Ip1 and�Ip1 , and the following Lyapunov function candidate
Vs ¼ Vsm þ Vse ð32Þ
where the ‘‘electromagnetic’’ part is given by
Vse ¼N2
2R1
��m1 � �s1��m1
�2þ
1
2Rp1
��m1 þ Lp1Ip1 � ð��m1
þ Lp1�Ip1 Þ�2
þg1
4 ��
�TwPw
�2þ g2
Lp1
2Rp1
��Ip1 � �Ip1
�2
with g1, g2 > 0, and
w ¼�m2
�m2 þ Lp2Ip2
� �¼ Txw,
P ¼
1
Lp2
N2
R20
01
Lp2Rp2
0BBB@
1CCCA > 0, T ¼
1 0
1 Lp2
� �:
Notice that (see the appendix 1)
L2p1
���Ip1 � �Ip1 ���2 � ��m1 þ Lp1Ip1 � ð��m1þ Lp1�Ip1 Þ
�2� 3
��m1 � �s1��m1
�2þ 3L2
p1
���Ip1 � �Ip1 ���2 þ 3�2sjxv � �xvvv � �vv
2
�sj ¼ �sj maxw j��m1ðwÞj. The derivative of the Lyapunov
function candidate is given by
_Vs ¼ �xv � �xvvv � �vv
2
Qs
þ
��m1 � �s1��m1
�"�1
2R01ðxvÞ
��m1
þ �s1��m1
�BTPs
xv � �xvvv � �vv
� �
þ Ip1 � �s1�Ip1 ��N 1ð�m1Þ � �s1N 1ð��m1
Þ
þ R1ðxvÞ�m1 � �s1R1ð�xv Þ��m1
�
þN
R1
�um1 � �s1cm1
��N2
R1_�s1��m1
#
�1
2R02ðxvÞ�
2m2B
TPs
xv � �xvvv � �vv
� �� ð�m1 � ��m1
ÞðIp1 � �Ip1 Þ � Lp1ðIp1 � �Ip1 Þ2
�g1
��TwPw
�1
Lp2Mm2ðxv,�m2Þ�m2 þ I2p2
�
� g2
��Ip1 � �Ip1
�2;
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where
_�s1 ¼1
2�s1
R001ð�xv ÞR01ðxvÞ�vv �R
01ð�xv ÞR
001ðxvÞvv
R01ðxvÞ2
:
Since R2ðxvÞ � 0
�1
Lp2Mm2ðxv,�m2Þ�m2
¼ �1
Lp2
�N 02e
n2�m2�2m2 þR2ðxvÞ�2m2
�
� �N 02
Lp2�2m2 � ��
2m2
�TwPw ¼ ��m2
Ip2
� �T
�P�m2
Ip2
� �� � ��
��2m2 þ I2p2
�,
�P ¼ TTPT;
where �� > 0 is the minimum eigenvalue of �P, andN 02=Lp2 > 1, as shown in the simulation section. Hence,
_Vs � �xv � �xvvv � �vv
2
Qs
þ
��m1 � �s1��m1
�
"�1
2R01ðxvÞ
��m1 þ �s1��m1
�BTPs
xv � �xvvv � �vv
� �
þ Ip1 � �s1�Ip1 ��N 1ð�m1Þ � �s1N 1ð��m1
Þ
þ R1ðxvÞ�m1 � �s1R1ð�xv Þ��m1
�
þN
R1
�um1 � �s1cm1
��N2
R1_�s1��m1
#
�1
2R02ðxvÞ�
2m2B
TPs
xv � �xvvv � �vv
� �� ð�m1 � ��m1
ÞðIp1 � �Ip1Þ � Lp1ðIp1 � �Ip1Þ2
� g1
��2m2 þ I2p2
�2� g2
��Ip1 � �Ip1
�2:
Therefore, when system (13) is in the state q1, the control
um1 ¼ �s1cm1 þR1
N
"1
2R01ðxvÞ
��m1 þ �s1��m1
�
BTPs
xv � �xvvv � �vv
� �� k3
��m1 � �s1��m1
��
�Ip1 � �s1�Ip1
�þN 1ð�m1Þ � �s1N 1ð��m1
Þ
þ R1ðxvÞ�m1 � �s1R1ð�xv Þ��m1þN2
R1_�s1��m1
#
ð33Þ
k3>0 (recall that um2 ¼ 0), is such that
_Vs � �xv � �xv
vv � �vv
2
Qs
� k3ð�m1 � �s1��m1Þ2
�1
2R02ðxvÞ�
2m2B
TPs
xv � �xv
vv � �vv
!
þ j�m1 � �s1��m1jjIp1 � �Ip1 j
þ �sjjIp1 � �Ip1 jxv � �xv
vv � �vv
� Lp1ðIp1 � �Ip1 Þ
2
� g1
��2m2 þ I2p2
�2� g2ð�Ip1 � �Ip1Þ
2;
where we have considered that (see the appendix 1)
� ð�m1� �s1��m1þ ð�s1� 1Þ��m1
ÞðIp1��Ip1 Þ
� j�m1� �s1��m1jjIp1� �Ip1 j þ �sjjIp1� �Ip1 j
xv� �xvvv� �vv
:
Since R02ðxvÞ is bounded, k3, g1, g2 can be made
large enough to make the right hand termnegative. This shows that x ¼ ð xv vv �m1 Ip1 Þ
T expo-
nentially tends to � ¼ ð�xv �vv ��m1�Ip1 Þ
T since �s1tends to 1. Finally, it is easy to check thatxw ¼ ð�m2 Ip2 Þ
T exponentially tends to
�w ¼ ð��m2�Ip2 Þ
T¼ 0.
Analogously, when (13) is in q3 (um1 ¼ 0) thecontrol
um2 ¼ �s2cm2 þR2
N
"1
2R02ðxvÞ
��m2 þ �s2��m2
�
BTPs
xv � �xvvv � �vv
� �� k3ð�m2 � �s2��m2
Þ
� ðIp2 � �s2�Ip2 Þ þ N 2ð�m2Þ � �s2N 2ð��m2Þ
þ R2ðxvÞ�m2 � �s2R2ð�xv Þ��m2þN2
R2_�s2��m2
#ð34Þ
is such that x ¼�xv vv �m2 Ip2
�Texponentially
tends to � ¼��xv �vv ��m2
�Ip2�T, while
xw ¼��m1 Ip1
�Texponentially tends to
�w ¼���m1
�Ip1�T¼ 0.
The exponential stability when (13) is in q2 or q4 can
be shown analogously.
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4.2 Solution of the regulation problem for the hybridsystem H
The solvability of the regulation problem for H,expressed by (19), can be proven on the basis of simpleconsiderations about the so called dwelltime (Liberzon 2003). In particular, we will show thatthe ith system remains in the state qi at least for acertain time �d, i, namely for a time intervalTi ¼ ½ti, t
0i ¼ ti þ �d, iÞ. In fact, H leaves qi when the
invariant condition Ii in (17) is not fulfilled. Theseconditions depend on the values of xv and Fm. Theabsence of Zeno behaviour like chattering is physicallyobvious in the present application, and can be provedmathematically showing that xv and Fm are uniformlycontinuous. Note that the control inputs umiðx,wÞ,i ¼ 1, . . . , 4, designed in x 4.1 are continuous functionsof their arguments x, w. Then the system (13) in closedloop, rewritten as
_x ¼ fiðx, xw,w, uiðx,wÞÞ ¼ �fiðt, xÞ
is such that �fi is continuous in its arguments. Moreover,since only polynomial and exponential terms appear in�fi, @ �fi=@x is continuous. Therefore, �fi is Lipschitzand hence its solution x(t) is locally uniformlycontinuous in a compact set. Hence, xv(t) and
FmðxvðtÞ,�m1ðtÞ,�m2ðtÞÞ in (11) are uniformly continuous.This implies that no Zeno behaviours can occur, evenfor t!1. In fact, if a function �(t) is uniformly contin-uous, for a generic k1>0 let �ti be the time instant suchthat j�ð�tiÞj ¼ k1. For the uniform continuity, thereexists a positive constant �d, i such thatj�ðtþ Þ ��ðtÞj < k1=2 for all t � �ti and for all 2 ½0, �d, i�. Hence,
j�ðtÞj ¼ j�ðtÞ ��ð�tiÞ þ�ð�tiÞj � j�ð �tiÞj
� j�ðtÞ ��ð�tiÞj > k1 �k12¼
k12¼ �k1
for all t 2 ½ �ti, �ti þ �d, i�. Since �(t) maintains its sign in aneighbourhood of �ti, which does not depend on t, chat-tering can not occur.
On the basis of definitions (17) and of the previousdiscussion, instantaneous switchings can not occur,and H remains in a state qi at least for a finite time inter-val Ti ¼ ½ti, t
0i ¼ ti þ �d, i�, with �d, i that does not go to
zero. This shows the existence of a minimum dwell time.The solvability of the regulation problem forH is now
proved as follows. Let us suppose thatH is in the genericith discrete state qi, and consider a generic switchingsequence I ¼ i, i2, . . . , j, . . . , i��1, i driving H again in qi(cycle). Since the discrete structure of H in our case isa finite state machine, this sequence always exists and
0 0.02 0.04 0.06 0.08 0.1−500
−400
−300
−200
−100
0
100
200
300
400
500
0 0.02 0.04 0.06 0.08 0.1−500
−400
−300
−200
−100
0
100
200
300
400
500(a) (b)
Figure 9. (a) Input um1 [V]; (b) input um2 [V].
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can not be infinite. Moreover, when H is in qi we have
determined in the previous section a Lyapunov function
Vs, i ¼ Vsm, i þ Vse, i such that
k1, i
zm
ze, i
zw, i
2
� Vsm, i þ Vse, i � k2, i
zm
ze, i
zw, i
2
_Vsm, i þ _Vse, i � ��s, i
zm
ze, i
zw, i
2
zm ¼xv � �xvvv � �vv
; ze, i ¼
�mj � �sj��mj
Ipj � �Ipj
zw, i ¼�m‘
Ip‘
2
; ‘ ¼ ð jþ 1Þmod 2
for appropriate values k1, i, k2, i. Note that �s, i can bechosen arbitrarily large. Therefore,
_Vs, i � ��iVs, i
�i ¼ �s, i=k2, i, so that Vs, iðtÞ � e��iðt�tiÞVs, iðtiÞ,t 2 Ti ¼ ½ti, t
0i�, and setting �d ¼ min
i2I�d;i
Vs, iðt0iÞ ¼ Vsm, iðt
0iÞ þ Vse, iðt
0iÞ � Vs, iðti þ �dÞ
� e��i�dVs, iðtiÞ ¼ e��i�d�Vsm, iðtiÞ þ Vse, iðtiÞ
�:
When H switches in qiþ1 at tiþ1 ¼ t0i, from the definitionsof Vsm, �xv , �vv , xr(w), one checks that
Vsm, iðtiþ1Þ ¼ Vsm, iðt0iÞ, so that
Vsm, iðtiþ1Þ � e��i�dVsm, iðtiÞ þ De, i,
De, i ¼ e��i�dVse, iðtiÞ � Vse, iðt0iÞ:
On the contrary, at the switching instantVse, iðtiþ1Þ 6¼ Vse, iðt
0iÞ. In fact, when in qi, �mj! ��mj
,
Ipj ! �Ipj , �Ipj ! �Ipj , �m‘ ! 0, Ip‘ ! 0,
‘ ¼ ðjþ 1Þmod 2, while when in qiþ1, �m‘ ! ��m‘ ,Ip‘ ! �Ip‘ , �Ip‘ ! �Ip‘ , �mj! 0, Ipj ! 0. Moreover, at
the switching ��m‘ ðtiþ1Þ 6¼ ��mjðt0iÞ. This will determine
the presence of a term � in (19).If �
Ps, i
min, �Ps, imax are the minimum and maximum eigen-
values of Ps,i, �Ps, i
min=2kzmk2 � Vsm, i � �
Ps, imax=2kzmk
2, and
Vsm, iþ1ðtiþ1Þ � kiVsm, iðtiþ1Þ � kie��i�dVsm, iðtiÞ þ kiDe, i
with ki ¼ �Ps, iþ1max =�
Ps, i
min. Hence, it is easy to check that
Vsm, iþ1ðtiþ1Þ � kie��i�dVsm, iðtiÞþki�e, i
Vsm, iþ2ðtiþ2Þ � kikiþ1e�ð�iþ�iþ1Þ�dVsm, iðtiÞ
þkiþ1�e, iþ1
Vsm, iþ3ðtiþ3Þ � kikiþ1kiþ2e�ð�iþ�iþ1þ�iþ2Þ�dVsm, iðtiÞ
þkiþ2�e, iþ2
..
.
Vsm, iðtiþ�Þ ¼Vsm, iþ�ðtiþ�Þ
� kikiþ1 kiþ��1e�ð�iþ�iþ1þþ�iþ��1Þ�dVsm, iðtiÞ
þkiþ��1�e, iþ��1
9>>>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>>>;
ð35Þ
with jI j ¼ �þ 1 the cardinality of I , and
�e, i ¼ De, i
�e, iþ1 ¼ kie��iþ1�dDe, i þ De, iþ1
�e, iþ2 ¼ kikiþ1e�ð�iþ1þ�iþ2Þ�dDe, i
þ kiþ1e��iþ2�dDe, iþ1 þ De, iþ2
..
.
De, iþh ¼ e��iþh�dVse, iþhðtiþhÞ � Vse, iþhðt0iþhÞ,
h ¼ 0, . . . , �� 1:
It is now sufficient to show that for every pair of timeintervals Ti, Tiþ�, in which H is in qi, one has
Vs, iðtiþ�Þ � Vs, iðtiÞ � �WiðtiÞ
for a (family) of positive definite continuous functionsWi, i 2 I . For instance, if one takes WiðtiÞ ¼ �Vsm, iðtiÞ,for an arbitrarily small �>0, the inequality
Vsm, iðtiþ�Þ � Vsm, iðtiÞ
�
�kikiþ1 kiþ��1e
�ð�iþ�iþ1þþ�iþ��1Þ�d � 1�
Vsm, iðtiÞ þ kiþ��1�e, iþ��1
� ��Vsm, iðtiÞ þ kiþ��1�e, iþ��1
is verified if
kikiþ1 kiþ��1e�ð�iþ�iþ1þþ�iþ��1Þ�d � 1 � ��:
Since �>0 is arbitrarily small, one ensures that theprevious inequality is verified when
kikiþ1 kiþ��1e�ð�iþ�iþ1þþ�iþ��1Þ�d � 1 < 0;
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i.e.,
�i þ �iþ1 þ þ �iþ��1 >1
�dln�kikiþ1 kiþ��1
�:
If we choose �s, i ¼ ¼ �s, iþ��1 ¼ �s, under thecondition
�s >1
�dDiln�kikiþ1 kiþ��1
�,
Di ¼1
k2, iþ
1
k2, iþ1þ þ
1
k2, iþ��1
we ensure the uniform exponential convergence of zm tozero in a practical sense. Finally, to show that the
regulation problem is solved, note that for
t 2 Tiþ� ¼ ½tiþ�, t0iþ��,
_Vs, iþ� � ��iþ�Vs, iþ�, so that, with
the same arguments
k1, iþ�kzmðtÞk2 � Vsm, iþ�ðtÞ � Vs, iþ�ðtÞ � e��iþ�ðt�tiþ�Þ
�Vsm, iþ�ðtiþ�Þ þ Vse, iþ�ðtiþ�Þ
�� e��iþ�ðt�tiþ�Þ
��kVsm, iðtiÞ þ �e, iþ�
�� e��iþ�ðt�tiþ�Þ
��kk2, ikzmðtiÞk
2 þ �e, iþ�
�;
where the last of (35) has been used, �k ¼ kikiþ1 kiþ��1e
�ð�iþ�iþ1þþ�iþ��1Þ�d , �e, iþ� ¼ kiþ��1�e, iþ��1þVse, iþ�ðtiþ�Þ. Hence, using the Bernoulli’s inequality
(see appendix 1), for t 2 Tiþ�
kzmðtÞk�
ffiffiffiffiffiffiffiffiffiffiffi�e, iþ�k1, iþ�
seð��iþ�=2Þðt�tiþ�Þ
1þ
�kk2, i�e, iþ�
kzmðtiÞk2
!1=2
� e��ðt�tiþ�Þ�kkzmðtiÞkþ �
�� ke��ðt�tiþ�ÞkzmðtiÞkþ �
k ¼ ð1=2Þð �kk2, iÞ=ð�e, iþ�Þ�kzmðtiÞk, � ¼ ð�iþ�Þ=2, � ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið�e, iþ�Þ=ðk1, iþ�Þ
p, where ti is the initial time instant.
Note that if tiþ� ¼ 1, then one can prove the uniform
exponential stability.
5. Simulation results
The control law has been implemented on a digital
computer to test the performance. A maximal current
of 30A, and a maximal dissipated power of 1 kW have
been considered. In the following we give the values of
the system’s parameters.
k ¼ 1:17 105 N=m b ¼ 6Ns=m
M ¼ 0:1054Kg N ¼ 50
Lp, 1 ¼ 4:8000 10�6 Lp, 2 ¼ 8:0745 10�6 H
Rp, 1 ¼ 0:0451� Rp, 2 ¼ 0:0234�
R1 ¼ 0:2040� R2 ¼ 0:2440�
rl, 1 ¼ 0:57813 rl, 2 ¼ 0:452
N j0 ¼ 31:66 nj ¼ 8:64 103
aj ¼ bjpj1e�pj2rl, j� bj ¼ ð�1Þ
jrl, jpj2
cj ¼ ð�1Þjrl, jpj3 dj ¼ pj1 þ rl, jpj3
� ¼ 0:004m
j¼ 1, 2, and (in Nm/Wb2)
p11 ¼ 1:5167 106, p21 ¼ 1:5330 106
p12 ¼ 1:1473 103, p22 ¼ 996:5755
p13 ¼ 3:8869 108, p23 ¼ 1:8226 108:
The control law has been determined settingQ ¼ 2 102I, k3 ¼ 1010. Figures 7–9 summarize thesimulation results. After a short transient, the trackingerror for xv is of the order 10
�6 m, and the seating velo-city results to be about 0.01 m/s. This implies the respectof the control requirements.
6. Conclusions
In this paper a controller has been designed for acamless engine, in which the main control problem isrepresented by the so-called soft landing. The approachfollows the regulation theory. The main limitation of thederived controller is the use of the whole state. In fact,a technological request is the elimination of the valveposition sensor. We are currently working on anextension of the proposed controller to a dynamic one,which only uses output measurements.
Appendix 1
In this appendix we prove some useful relations about�sj, j¼ 1, 2. Let us consider first
0 <R0ðy1Þ
R0ðy2Þ¼
ae�by1 þ c
ae�by2 þ c¼ 1þ z,
z ¼e�by1 � e�by2
e�by2 þ c=a> �1
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y1, y2 2 ½��, ��. Note that
e�by1 � e�by2 ¼X1k¼1
ð�1Þkbk
k!ðyk1 � yk2Þ
¼ �bðy1 � y2ÞX1k¼1
ð�1Þk�1bk�1
ðk� 1Þ!
Sk�1
k
� jbj jy1 � y2jX1i¼0
jbji
i!�i
¼ jbj jy1 � y2j ejbj�
with Sk�1 ¼Pk�1
h¼0 yk�1�h1 yh2, jSk�1j � k�k�1. Hence, the
Bernoulli’s inequality
ð1þ zÞr � 1þ rz, 8r 2 ½0, 1� � R,
8z > �1, z 2 R
can be applied with r ¼ 1=2, obtaining
ð1þ zÞ1=2 � 1þ1
2z � 1þ �sjy1 � y2j � 1þ �s
y1 � y2
_y1 � _y2
,
�s ¼1
2
jbj ejbj�
e�jbj� þ c=a:
Using this relation, if �sj � 1 (this happens when�xv � xv if j¼ 1, and �xv � xv if j¼ 2)
�sj ¼
"R0jð�xv Þ
R0jðxvÞ
#1=2
¼
"1þ
e�bj�xv � e�bjxv
e�bjxv þ cj=aj
#1=2
� 1þ �sjxv � �xvvv � �vv
namely
�sj � 1 � �sjxv � �xvvv � �vv
while if �sj < 1 (this happens when �xv < xv if j¼ 1, and�xv > xv if j¼ 2)
�sj ¼1"
R0jðxvÞ
R0jð�xv Þ
#1=2¼
1"1þ e�bjxv�e�bj�xv
e�bj�xvþcj=aj
#1=2
�1
1þ �sjxv � �xvvv � �vv
so that
1� �sj �
�sj
xv � �xvvv � �vv
1þ �sj
xv � �xvvv � �vv
� �sj
xv � �xvvv � �vv
:
Appendix 2
The coefficients in (20), satisfying (21), are given by
cr, 10 ¼ �� 1� 70#40D4� 168
#50D5� 140
#60D6� 40
#70D7
� �
cr, 11 ¼ �280
D�#30D3
Dþ #0ð Þ3
D3
cr, 12 ¼840
D2�#20D2
1þ 4#0Dþ 5
#20D2þ 2
#30D3
� �
cr, 13 ¼ �1680
D3�#0Dþ 6
#20D2þ 10
#30D3þ 5
#40D4
� �
cr, 14 ¼1680
D4� 1þ 12
#0Dþ 30
#20D2þ 20
#30D3
� �
cr, 15 ¼ �20160
D5� 1þ 5
#0Dþ 5
#20D2
� �
cr, 16 ¼100800
D6� 1þ 2
#0D
� �
cr, 17 ¼ �201600
D7�
with D ¼ #c1 � #c0.
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