DTIC ADA438593: Formation Dynamics under a Class of Control Laws

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T R R
ECHNICAL ESEARCH EPORT
Formation Dynamics under a Class of Control Laws
by Fumin Zhang, P.S. Krishnaprasad
CDCSS TR 2002-4
(ISR TR 2002-25)
+ CENTER FOR DYNAMICS
C D
- AND CONTROL OF
S SMART STRUCTURES
The Center for Dynamics and Control of Smart Structures (CDCSS) is a joint Harvard University, Boston University, University of Maryland center,
supported by the Army Research Office under the ODDR&E MURI97 Program Grant No. DAAG55-97-1-0114 (through Harvard University). This
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Formation Dynamics Under A Class of Control Laws1
Fumin Zhang and P.S. Krishnaprasad
Institute for Systems Research
University of Maryland at College Park
College Park, MD 20742
Abstract is kept unchangedoversufficiently long period is not a
trivial problem. Because the amount of fuel on board
A system of two earth satellites is analyzed as a con- is limited for each satellite, one needs to find a set of
trolled mechanical system. The orbit of an earth satel- orbits that demand the minimum control effort. This
lite can be represented by a point in the vector space of problem becomes more challenging when the effect of
ordered pairs of angular momentum and Laplace vec- J2 and other disturbances are considered. In [3] the
tors. Control laws are obtained by introducing a Lya- authors proposeda ring of evenly distributed satellites
punov function on this space. Formations of two satel- on the same circular orbit for communication purpose.
lites are achieved asymptotically by the controlled dy- The stabilityofsuchringisprovedin[4]. In[2]and[5]
namics. the investigations of Clohessy-Wiltshire equations re-
vealed possible formations with constantapparentdis-
tributions. The effects of perturbations are calculated
and possible station keeping strategy are proposed. In
1 Introduction
[7], the authors proposed that in the presence of J2, a
setofconstraintsontheorbitalelementsshallbesatis-
Thereisincreasinginterestinthebehaviorsofacluster fiedtopreventtheorbitsfromdriftingapart. However,
ofrelativelysmallsatellitesinspace. Suchsatellitesare extra station keeping is still necessarydue to the com-
usually inter-connected by wireless radio or laser links plicated nature of the disturbances. The adjustment
for communication. By keeping a cluster of such satel- can be performed periodically when the drifting error
litesinacertaingeometricalform,onecanacquireben- exceeds certain threshold.
efits for scientific observations. The information shar-
ing across the cluster will allow the satellites to work The initialization of a formation is another important
cooperatively to perform tasks impossible or difficult problem. The whole cluster can be launched together
for a single satellite. Compared to a single satellite by a space shuttle or rocket. Satellites will be first
providing the functionality of a cluster, a member of placedinaparkingorbitbeforetransferringtothefinal
a cluster can be smaller/lighter. Building and launch orbits. The final orbits must be such that the forma-
costs will then be reduced. In addition, A cluster can tion will be achieved. The orbital transfer can happen
alsobereconfiguredaccordingtodifferentmissiongoals individually. One idea is to develop a Lyapunov func-
or in the case that a member of the cluster fails. tionwhichwillachieveitsminimumwhencorrectorbit
isreached. In[6],aLyapunovfunctionisexpressedasa
Thesizeandshapeofaclusterorformationareusually quadraticfunctionofthedifferencesoforbitalelements
determined by the requiredfunctionality of the forma- betweencurrentorbitandthe destinationorbit. In[1],
tion. With J2 and higher order terms of the earth’s the authors proved that an elliptic orbit can be repre-
gravitational field ignored, the solution of the Kepler sented uniquely as a point on the linear space formed
problemtellsusthatasatellitewilltrackanellipticalor bytheangularmomentumvectorsandLaplacevectors.
circular orbit. To determine the orbit of each satellite A Lyapunovfunction was built naturally from the Eu-
in a cluster such that the size and shape of the cluster clideanmetriconthisspace. Ithasbeensuggestedthat
comparedtoHohmanntransfer,thisapproachwillcon-
1This research was supported in part by the National Aero-
sume less fuel.
nauticsandSpaceAdministrationunderNASA-GSFCGrantNo.
NAG5-10819, by the Air Force Office of Scientific Research un-
der AFOSR Grant No. F49620-01-0415, by the ArmyResearch Inthispaper,wewillshowthatcooperativeorbittrans-
OfficeunderODDR&EMURI97ProgramGrantNo. DAAG55- fer is possible in the setting of pairs of satellites. The
97-1-0114 to the Center for Dynamics and Control of Smart
satellites in a cluster can make the transfer together
Structures (through Harvard University), and under ODDR&E
and the cluster relationship can be established asymp-
MURI01ProgramGrant No. DAAD19-01-1-0465to theCenter
forCommunicatingNetworkedControlSystems(throughBoston totically. The same technique can be used to perform
University).
p. 1
station keeping task when the relative position errors Notice that
q
ofthe satellites exceeda certainthreshold. By extend- mq¨=−mµ (5)
(cid:2)q(cid:2)3
ingthemethodin[1]tomulti-satellitecase,wewillde-
signaLyapunovfunctionandshowthatcorrectcluster Thus
relationship between orbits can be established by the (q·p)q p
controlleddynamics which will minimize this function. mq¨×(q×p)=−mµ +mµ (6)
(cid:2)q(cid:2)3 (cid:2)q(cid:2)
Insection(2),wewillreviewtheKeplertwobodyprob-
lem and introduce the concept of the shape space of Comparing (6) with (4) we have A˙ =0
elliptic Keplerian orbits. In section (3), the Lyapunov
functionforapairofsatellitesisdevelopedandthecase Knowing l , W and A, we have seven integrals for the
of two satellites making a cooperative orbit transfer is two-body problem. They are not all independent be-
studied. Simulation results are given in section (4). cause there are two relations connecting them. They
are:
A·l = 0
2 The Kepler two body problem
(cid:2)A(cid:2)2 = m4µ2+2mW(cid:2)l(cid:2)2 (7)
For a system of two small satellites, one can make the
followingassumptions: (a)Thegravitationalattraction The space of ordered pairs (l,A) is R3×R3 on which
between the satellites can be omitted. (b) The total we define the metric:
mass of the satellites satisfies m1+m2 (cid:1)M where M
isthemassoftheEarth. Undertheseassumptions,the d((l1,A1),(l2,A2))=((cid:2)l1−l2(cid:2)2+(cid:2)A1−A2(cid:2)2)12
threebodysystemcanbeapproximatedbytwouncou- (8)
pled two body problems. Each of the two body prob-
Let P denote the phase space of the satellite, define a
lems can be further simplified to a one center problem
mapping π :P →R3×R3, (q,p)(cid:6)→(l,A). Let the set
with the center of Earth being the center of mass.
Σe be defined as
ThegravitationalpotentialfunctionV dependsonlyon
the distance (cid:2)q(cid:2) of the satellite from the center. Then Σe ={(q,p)∈P |W(q,p)<0,l(cid:4)=0} (9)
we have
and let the set D be defined as
mq¨=−∇V (1)
D ={(l,A)∈R3×R3|A·l =0,l(cid:4)=0,(cid:2)A(cid:2)<m2µ}
Let p = mq˙ be the momentum of the satellite. The (10)
angular momentum l = q × p and the energy W = In [1], the authors proved the following results:
1m(cid:2)q˙(cid:2)2+V are integrals of motion (i.e. they are con-
2
served). Since Theorem 2.1 (Chang-Chichka-Marsden)The follow-
ing hold:
l˙ = q˙×p+q×p˙
= 0+q×mq¨ 1. Σe is the union of all elliptic Keplerian orbits.
q ∂V
= −q×m =0 (2)
(cid:2)q(cid:2)∂(cid:2)q(cid:2) 2. π(Σe)=D and Σe =π−1(D).
and 3. The fiber π−1(l,A) is a unique (oriented) elliptic
Keplerian orbit for each (l,A)∈D.
W˙ = q˙·mq¨+V˙
= −q˙·∇V +V˙ =0 (3)
The mapping π is a continuousmapping becausel and
A are continuous with respect to (q,p). Furthermore,
If we make further assumptions that the shape of the following corollary hold.
the earth is a perfect homogeneous ball, then V =
−mµ/(cid:2)q(cid:2) where µ = kM and k is the gravity con- Corollary 2.2 π−1(K)iscompactforanycompactset
stant, the Laplace vector A = p×l−m2µ(cid:1)qq(cid:1) is also K ⊂D
conserved given q(t)(cid:4)=0 for all t>0. To see this,
d q Proof: SinceDisametricspace,Kiscompactimplies
A˙ = p˙×l−m2µ ( )
dt (cid:2)q(cid:2) that K is closed and bounded. By the continuity of π,
q˙ (q·q˙)q π−1(K) is closed. The only thing left to prove is the
= mq¨×(q×p)−m2µ((cid:2)q(cid:2) − (cid:2)q(cid:2)3 ) (4) boundedness of π−1(K).
p. 2
Kisclosedandboundedimpliesthatthereexistr0 >0, Proof: Use the definitions of ((cid:1)l,A(cid:1),Σ(cid:1)e,D(cid:1),π(cid:1),d(cid:1))
r1 >0 and 0≤r2 <m2µ s.t. and apply Theorem (2.1) and corollary (2.2) to
(li,Ai,Σei,Di,πi,di) for i = 1,2. The statements are
r0 ≤(cid:2)l(cid:2)≤r1 (cid:2)A(cid:2)≤r2 (11) immediately proved.
This is true since we already know (cid:2)A(cid:2) < m2µ and
(cid:2)l(cid:2)>0. Because(cid:2)·(cid:2)iscontinuousonthecompactset
K,wedeclarethatr0 andr2 existandcanbeachieved. 3 Control Strategies to Achieve Formations
The existence of r1 is based on the fact that (cid:2)l(cid:2)2 +
(cid:2)A(cid:2)2 is bounded. Based on the distance function on the shape space of
elliptic Keplerian orbits, a Lyapunov function can be
Withoutlossofgenerality,assumethesatellitehasunit
introducedonthephasespacefortheHamiltoniansys-
mass. From
tem of two satellites. By applying LaSalle’s invariance
(cid:2)A(cid:2)2 =µ2+2W(cid:2)l(cid:2)2 (12) principle,weprovethatthesystemcanbedriventoan
invariant manifold of the phase space where formation
We get is achieved.
µ2−(cid:2)A(cid:2)2
|W|= 2(cid:2)l(cid:2)2 (13) Thesystemoftwo(cid:3)sa2tellitesisaHamiltonian(Control)
system with H = i Hi where
Hence,
0< µ2−r22 ≤|W|≤ µ2 (14) Hi = 1 (cid:2)pi(cid:2)2−V((cid:2)qi(cid:2)) (17)
2r12 2r02 2mi
Let a denote the semi-major axis of an elliptic Kep- For each satellite the dynamics is:
lerian orbit and e denote the eccentricity. It is well
known that a=−2µ e= (cid:2)A(cid:2) (15) p˙i = −∂∂Hqii +ui =−miµ(cid:2)qqii(cid:2)3 +ui
W µ ∂Hi pi
q˙i = = (18)
Thus a,e are bounded. For any (q(t),p(t)) ∈ π−1(K), ∂pi mi
on an elliptic orbit q(t) is bounded below by a(1−e)
where i=1,2. Here ui are controls.
and bounded above by a(1+e). Furthermore, because
(cid:2)l(cid:2)=(cid:2)p(t)(cid:2)(cid:2)q(t)(cid:2)sinθ,(cid:2)p(t)(cid:2)isboundedgiventhe Using the notations at the end of the last section, let
fact that (cid:2)l(cid:2)(cid:4)=0. Hence (q(t),p(t)) is bounded which J1(q(cid:1),p(cid:1)) = (l1,A1), J2(q(cid:1),p(cid:1)) = (l2,A2) and J(q(cid:1),p(cid:1)) =
implies that π−1(K) is bounded. (J1,J2). Define a Lyapunov function as
1
Inthecaseoftwosatellites,let(li,Ai,Pi,Σei,Di,πi,di) V(J(q(cid:1),p(cid:1))) = [(cid:2)l1−l2−δl(cid:2)2+(cid:2)A1−A2−δA(cid:2)2
2
dsaetneolltietet.hLeetcoP(cid:1)rr=espPo1n×dPin2g,oq(cid:1)b(tj)ec=ts(qd1e(fit)n,eqd2(tfo))r,tp(cid:1)h(te))it=h +(cid:2)l1−ld(cid:2)2+(cid:2)A1−Ad(cid:2)2] (19)
(p1(t),p2(t)). Let Σ(cid:1)e = Σe1×Σe2. Let D(cid:1) = D1×D2 Here, δl , δA are constant vectors which specify the
(cid:1) (cid:1)
and l =(l1,l2), A=(A1,A2). Let desiredfinaldifferencebetween(l1,A1)and(l2,A2). ld
(cid:2) and Ad arealso constantvectorswhichspecify the ref-
d(cid:1)(((cid:1)l1,A(cid:1)1),((cid:1)l2,A(cid:1)2))= d21+d22 (16) erence orbitofthe formation. It canbe verifiedthat V
is continuously differentiable and bounded from below
and π(cid:1) = π1 ×π2. We define the shape space of ellip- by 0.
(cid:1)
tic Keplerian orbits to be the set D with the distance
function d(cid:1). Toinvestigatethederivationofthecontrollaw,wecan
rewrite (18) as:
Proposition 2.3 The following hold: ∂H
p(cid:1)˙ = − +u=B(q(cid:1),p(cid:1))+u
∂q(cid:1)
(cid:1)
1. Σe is the union of all pairs of elliptic Keplerian ∂H
orbits. q(cid:1)˙ = (20)
∂p(cid:1)
2. π(cid:1)(Σ(cid:1)e)=D(cid:1) and Σ(cid:1)e =π(cid:1)−1(D(cid:1)). and J(q(cid:1),p(cid:1)) is conserved when u=0. Then we have
3. The fiber π(cid:1)−1((cid:1)l,A(cid:1)) is a unique pair of (oriented) ∂V ∂J ∂V ∂J
elliptic Keplerian orbit for each ((cid:1)l,A(cid:1))∈D(cid:1). V˙(J(q(cid:1),p(cid:1))) = ∂J · ∂q(cid:1)q(cid:1)˙ + ∂J · ∂p(cid:1)p(cid:1)˙
∂V ∂J ∂V ∂J ∂V ∂J
4. π(cid:1)−1(K(cid:1)) is compact for any compact set K(cid:1) ⊂D(cid:1) = ∂J · ∂q(cid:1)q(cid:1)˙ + ∂J · ∂p(cid:1)B(q(cid:1),p(cid:1))+ ∂J · ∂p(cid:1)u
p. 3
= ∂∂VJ ·J˙|u=0+ ∂∂VJ · ∂∂Jp(cid:1)u c2 = 21(cid:2)ld(cid:2)2
= ∂V · ∂Ju=(∂J)T∂V ·u (21) c3 = 1(m21µ−(cid:2)Ad(cid:2))2
∂J ∂p(cid:1) ∂p(cid:1) ∂J 2
1
By letting c4 = (m22µ−(cid:2)Ad−δA(cid:2))2
4
u=−λ((∂J)T∂V ) (22) c < min{c1,c2,c3,c4} (25)
∂p(cid:1) ∂J
i.e Let the set
(cid:4)2
ui =−λi (∂V ∂Jj) (23) G={((cid:1)l,A(cid:1))∈R6×R6|V((cid:1)l,A(cid:1))≤c} (26)
∂Jj ∂pi
j=1
where V is defined in equation (19).
where λ = diag(λi) and λi > 0 for i = 1,2, we have
V˙ ≤0alongtheintegralcurvesofthesystem. In[1]the Then the set Ω = π(cid:1)−1(G∩D(cid:1)) is a compact subset of
authors used this observation for the setting of a sin-
(cid:1)
gle satellite to derive a Lyapunov based orbit transfer. Σe.
The following theorem based on LaSalle’s invariance
principle holds. Proof: According to proposition(2.3), allwe needto
show is that the set G∩D(cid:1) is a compact subset of the
(cid:1)
set D. According to our definitions,
Theorem 3.1 Given the system as defined in (20),
Let V : P(cid:1) → R be a continuously differentiable func- D(cid:1) = {((cid:1)l,A(cid:1))∈R6×R6|li·Ai =0, (cid:1)li (cid:4)=0,
taiocnomwphaiccht sisetboΩun⊂deΣ(cid:1)defroonmwbheilcohwV.˙S≤up0p.osLeetthEere⊂eΩxisbtes (cid:2)Ai(cid:2)<m2iµ, i=1,2} (27)
the set on which V˙ = 0. Let M be the largest invari-
ant set in E. Then by applying the control u as defined Obviously,thissetD(cid:1) isnotaclosedsubsetofR6×R6.
(22), every solution of thesystem starting within Ω ap- If we let
proaches M as t→∞. Furthermore, Let Γ denote the
subset of Ω where u(t)≡0 holds for all t, then M =Γ K ={((cid:1)l,A(cid:1))∈R6×R6|li·Ai =0, i=1,2} (28)
This set K is a closed subset of R6×R6. The set D(cid:1)
Proof: It is easy to verify that the conditions for
is a subset of K. It is not difficult to see that the set
LaSalle’s invariance principle are all satisfied. Hence (K −D(cid:1)) is a closed subset of R6×R6.
(q(t),p(t))→M as t→∞.
In the next step we want to show that if the value of c
The set Γ where u(t)≡0 is an invariant set contained
is chosen as proposed, we must have
in E. This is because we must have V˙ = u·u ≡ 0 on
Γ. Hence Γ ⊂ M. On the other hand, M is invariant G∩(K−D(cid:1))=∅ (29)
implies V˙ ≡ 0 on M. Furthermore, u(t) ≡ 0 on M.
Hence M ⊂Γ. We conclude that M =Γ. Supposethisisnottrue. NoticethatsinceG∩(K−D(cid:1))
is a compact subset of R6×R6, the function V has a
From this theorem we know that the largest invariant minimumandamaximumvalueonthisset. Themaxi-
set is M = Ω ∩ C where C = {(q(cid:1),p(cid:1)) ∈ Σ(cid:1)e|u(t) ≡ mumvalueshallnotexceedc. Theminimumvaluecan
0}. Then two questions should be answered. The first be found by solving a constrained minimization prob-
question is whether a suitable compact set Ω exists in lem.One should compare the unconstrained minimum
(cid:1)
Σe. The second question is to determine C. value with the minimum value achieved on the bound-
ary of (K −D(cid:1)). In this case, the unconstrained mini-
The following proposition gives the answer to the first mum value of V is 0 which can not be achieved since
question. (ld,Ad) ∈ D(cid:1).On the other hand, c1 can be achieved
on the boundary of (K − D(cid:1)) where l1 = 0. c2 can
Proposition 3.2 Given be achieved on the boundary where l2 = 0. c3 can be
achieved on the boundary where A1 =m21µ. c4 can be
(cid:2)ld(cid:2) (cid:4)= 0 achieved on the boundary where A2 =m22µ. However,
(cid:2)Ad(cid:2) < m21µ ifc<min{c1,c2,c3,c4},thenthemaximumislessthan
Ad·ld = 0 (24) the minimum. We have a contradiction. Thus the set
G must have no intersection with the set (K −D(cid:1)).
Let c,c1,c2,c3,c4 be positive numbers which satisfy
On the other hand, we know that G∩K (cid:4)= ∅. This
c1 = 1(cid:2)ld−δl(cid:2)2 means G∩K ⊂ D(cid:1). The compactness of G∩K comes
4
p. 4
from the fact that G is compact and K is closed in Problem 3.4 Suppose two satellites are controlled by
R6×R6 (30) and (31) respectively. If we select ld (cid:4)= 0 and
(cid:2)Ad(cid:2) < m21µ, we want to calculate the maximal set
C ⊂Σ(cid:1)e where u1(t)≡0 and u2(t)≡0 are satisfied.
Next we want to characterize the invariant set. By Solution: From u1(t)≡0, we have
equation (23), without loss of generality, we let λ1 = (l1−l2−δl+l1−ld)×q1+l1×(A1−A2−δA+A1
λ2 =1, the controls can be calculated as −Ad)+((A1−A2−δA+A1−Ad)×p1)×q1 ≡0(38)
u1 = −[∂∂Vl1 ∂∂pl11 + ∂∂AV1∂∂Ap11] From u2(t)≡0, we have
= −[(l1−l2−δl+l1−ld)×q1 (l1−l2−δl)×q2+l2×(A1−A2−δA)
+l1×(A1−A2−δA+A1−Ad) +((A1−A2−δA)×p2)×q2 ≡0 (39)
+((A1−A2−δA+A1−Ad)×p1)×q1](30)
We want to solve these two equations for l1,l2,A1 and
u2 = −[∂V ∂l2 + ∂V ∂A2] A2. Notice that these unknowns are time invariant.
∂l2 ∂p2 ∂A2 ∂p2
By applying the results inthe lemma to equation(38),
= (l1−l2−δl)×q2+l2×(A1−A2−δA) we get
+((A1−A2−δA)×p2)×q2 (31)
l1−l2−δl+l1−ld = αA1
We need the following lemma to solve the equations A1−A2−δA+A1−Ad = αl1 (40)
obtained by letting u1(t)≡0 and u2(t)≡0. From equation (38) we get
Lemma 3.3 Suppose a single satellite has external A1−A2−δA = βl2
control u(t) ≡ 0. Let x, y be time invariant unknown l1−l2−δl = βA2 (41)
vectors. Suppose (q(t),p(t))∈Σe, the solution of equa- Hence,
tion
x×q+l×y+(y×p)×q ≡0 (32) l1−ld = αA1−βA2
A1−Ad = αl1−βl2 (42)
is
x=αA y =αl (33) (cid:1)
The maximal invariantset C within Σe is where equa-
For some α∈R tion (41) and (42) are satisfied.
Noticethatalthoughwehaveintroducedtwounknown
Proof: Taketheinnerproductwithq(t)onbothsides
variables α and β, we can still solve for (l1,l2,A1,A2)
of equation (32) we get:
intermsof(ld,Ad,δl,δA)becausewehavetwo“extra”
(l×y)·q(t)≡0 (34) equations
l1·A1 =0 l2·A2 =0 (43)
Since the orbitofthe satellite is elliptic, q(t)(cid:4)=0. Fur-
thermore,q(t)wouldstayintheorbitalplanel⊥ which
is perpendicular to l. Notice because l×y is time in- Let C1 denote the set where α=0 and β =0. Let C2
variantandq(t)isnot0,weconcludethatl×y mustbe denote the set where α = 0 and β (cid:4)= 0. Let C3 denote
the set where α(cid:4)=0 and β =0. Let C4 denote the set
either parallel to l or vanish. However, since the rela-
where α(cid:4)=0 and β (cid:4)=0. Then
tion (l×y)·l=0 must be satisfied, the only possibility
is to havel×y =0. Thus,there exists a number α∈R C =C1∪C2∪C3∪C4 (44)
such that y =αl. Equation (32) can be simplified to
(x+αl×p)×q ≡0 (35) Among all the possible invariant sets we have calcu-
lated, only C1 is the one we want the system to ap-
Since
q proach. Thus, we shall pick suitable values for δl, δA,
p×l =A+m2µ (36)
(cid:2)q(cid:2) ld,Ad andinitialconditionssothatC1 is the onlypos-
sible invariant set within Ω. The following proposition
we now have,
gives a set of sufficient conditions to achieve this goal.
(x−αA)×q(t)≡0 (37)
Again, since q(t) is tracking an elliptic orbit, from the Proposition 3.5 Let δl, δA, ld,Ad satisfies all the as-
fact that x and A are time invariant, we must have sumptions in proposition (3.2) and
x=αA.
δl·δA = 0
p. 5
δl·(ld−δl) = 0 Because β (cid:4)= 0, this implies that l2 and A2 vanish.
δA·(Ad−δA) = 0 Hence C3 vanishes.
(ld−δl)·(Ad−δA) = 0 (45)
On C4, the following equations are satisfied
and one of the following inequalities
l1−l2 = δl+βA2
ld (cid:4)= δl−σδA A1−A2 = δA+βl2 (56)
Ad (cid:4)= δA−σδl (46)
√
where σ =± 2. l1−αA1+βA2 = ld
A1−αl1+βl2 = Ad (57)
Select the initial condition (q(cid:1)0,p(cid:1)0) s.t
Replace βA2 and βl2 of equation (57), we have
V(q(cid:1)0,p(cid:1)0)<c (47)
2l1−l2−αA1 = ld+δl
where cis defined in proposition (3.2). The system will 2A1−A2−αl1 = Ad+δA (58)
be controlled to the set C1 where
From (56) we can solve for l1 and A1 and put them in
l1−l2−δl = 0 (58). We get
A1−A2−δA = 0
l1 = ld (1−αβ)l2+(2β−α)A2 = ld−δl+αδA
A1 = Ad (48) (2β−α)l2+(1−αβ)A2 = Ad−δA+αδl(59)
are satisfied. According to (45), we have
Proof: Wewillshowthatthemaximalinvariantsub- (ld−δl+αδA)·(Ad−δA+αδl)=0
set of the compact set determined by the given initial
so
conditions contains only the set C1. Other invariant (1−αβ)(2β−α)((cid:2)l2(cid:2)2+(cid:2)A2(cid:2)2)=0
sets will vanish or will be impossible to reach.
Because(cid:2)l2(cid:2)and(cid:2)A2(cid:2)cannotvanish,therearethree
On C2, the following equations are satisfied possibilities:
l1−l2 = δl (1) α=2β but αβ (cid:4)=1
A1−A2 = δA (49)
From equation (59) we have
l1−αA1 = ld 1
A1−αl1 = Ad (50) l2 = 1−2β2(ld−δl+2βδA)
1
from (50), since ld·Ad =0 and l1·A1 =0 we have A2 = 1−2β2(Ad−δA+2βδl) (60)
α((cid:2)l1(cid:2)2+(cid:2)A1(cid:2)2)=0 (51) Solve for l1,A1 from equation (56), we have
Because α (cid:4)= 0, this implies that l1 and A1 vanish. 1
Thus equation (50) will not be satisfied. C2 vanishes. l1 = 1−2β2(ld+β(Ad+δA))
1
On C3, the following equations are satisfied A1 = 1−2β2(Ad+β(ld+δl)) (61)
l1−l2 = δl+βA2 Now, since l1·A1 =0 we have
A1−A2 = δA+βl2 (52)
(ld+β(Ad+δA))·(Ad+β(ld+δl))=0
l1+βA2 = ld which implies
A1+βl2 = Ad (53)
β((cid:2)ld(cid:2)2+(cid:2)Ad(cid:2)2+(cid:2)δl(cid:2)2+(cid:2)δA(cid:2)2)=0
From (52) we can solve for l1 and A1 and put them in
(53). We get This result is impossible.
l2+2βA2 = ld−δl (2) αβ =1 but α(cid:4)=2β
A2+2βl2 = Ad−δA (54)
From equation (59) we have
Since (ld−δl)·(Ad−δA)=0 and l2·A2 =0 we have
1
2β((cid:2)l2(cid:2)2+(cid:2)A2(cid:2)2)=0 (55) l2 = 2β−α(Ad−δA+αδl)
p. 6
1
A2 = 2β−α(ld−δl+αδA) (62) htieorne,swoethpautttfhaectcoornstbr1olainsdwbe2igihntteodt.hOeuLryacponutnroovl ifsunc-
Solve for l1,A1 from equation (56), we have u1 = −ξ[b1(l1−l2−δl+l1−ld)×q1
+l1×b2(A1−A2−δA+A1−Ad)
l1 = 2β1−α(Ad+β(ld+δl)) u2 = η+[(bb12((l1A−1−l2A−2δ−l)×δAq2)×+pl12)××bq21(]A1−A2−δA)
A1 = 2β1−α(ld+β(Ad+δA)) (63) +(b2(A1−A2−δA)×p2)×q2] (66)
whereξ ,η arepositivenumberswhichcanbeadjusted
Now, since l1·A1 =0 we have for numerical performance.
(Ad+β(ld+δl))·(ld+β(Ad+δA))=0 In practical applications this control law needs to be
discretized. Here we use a simple technique to ob-
which implies
tain a discrete control law from (66). Assume that
β((cid:2)ld(cid:2)2+(cid:2)Ad(cid:2)2+(cid:2)δl(cid:2)2+(cid:2)δA(cid:2)2)=0 trhecetitohnr.usLteetrTofbaestahteeltliitmeecianntebrevafilrbeedtwtoewenartdwsoafinryindgis-
of the thruster pulses. Let t0 denote the starting time
This result is also impossible.
of the controller. Denote u¯i the discretizedcontrol. At
(3) αβ =1 and α=2β. time t0+nT, we have:
This implies that α=±√2. Equation (59) will be u¯i(n)=ui(t0+nT)·T (67)
Thatis,we assumethatduring the time intervalT the
ld−δl+αδA = 0 control ui is constant. Of course, if T is too large the
Ad−δA+αδl = 0 (64) algorithm may fail to converge.
This is impossible becauseit violatedthe conditionsin
(46).
Hence C4 vanishes.
A special case of this proposition is when we choose
δl=0andδA=0. Thisisthecasewheretwosatellites
are driven to the same orbit.
If we let the Lyapunov function V(J(q,p)) =
12[(cid:2)l1−ld(cid:2)2 + (cid:2)A1−Ad(cid:2)2], we can control a single
satellite to transfer between elliptic orbits. This case
is first analyzed in [1]. In this case, the invariant set
is simply the set where l1 = ld and A1 = Ad are satis-
fied. Comparing to this case, the invariant set of two
satellites case is more complicated.
Tocompletelysetuptheformationcompletely,weneed
to know the final positions of the satellites on their Figure 1: Two Satellites Formation
orbits. This can be computed off-line by simulation.
Thenwecanchooseproperstartingseparationbetween Figure 1 shows when δl = 0 and δA =0. By applying
two satellites. the discretized control law we can drive two satellites
Sat1 and Sat2 onto the same orbit. We take the unit
lengthtobetheradiusoftheearth,theunittimetobe
one minute and the unit mass to be 1000kg. The ini-
4 Simulation Results
tialconditionsofthetwosatellitesaregivenbyspecify-
ing their six orbitalelements (a,e,i,ω,Ω,τ) . We have
In order to verify the control algorithm, we wrote a
(3,0.3,0,π/2,0,0)forSat1and(4,0.2,π/4,π/4,π/3,0)
simulation of controlling two satellites into formation
for Sat2. The destinationorbitis acircularorbitgiven
on MATLAB. We take the Lyapunov function as
as (3,0,π/6,π/2,0,0). Hence, ld and Ad are deter-
1 mined. During the simulationprocess,we noticed that
V(J(q,p))= 2(b1(cid:2)l1−l2−δl(cid:2)2+ by choosing b2 = 1000b1 and ξ = η = 0.01 we can get
b2(cid:2)A1−A2−δA(cid:2)2+b1(cid:2)l1−ld(cid:2)2+b2(cid:2)A1−Ad(cid:2)2)(65) a decent result.
p. 7
Figure2showsthechangeof(cid:2)l1−l2(cid:2),100(cid:2)A1−A2(cid:2) 5 Acknowledgement
andtheLyapunovfunctionV withrespecttotimedur-
ing the whole process. As we can see, the Lyapunov The secondauthorwouldliketo thankDavidChichka,
function is being reduced during the whole process. Dong Eui Chang and Jerry Marsden for conversations
and preprint of their work [1].
References
[1] D.E. Chang, D. Chichka, and J.E. Marsden.
Lyapunov-based transfer between elliptic keplerian or-
bits. accepted by Discrete and Continuous Dynamical
Systems, B,2:57–67,2002.
[2] D.F.Chichka.Satelliteclusterswithconstantap-
parent distribution. Journal of Guidance,Control,and
Dynamics, 24(1):117–122,2001.
[3] A.G.Y. Johnston and C.R. McInnes. Au-
tonomous control of a ring of satellites. Advances in
the Astronautical Sciences, 95 part I:93–105,1997.
[4] P.S. Krishnaprasad. Relative equilibria and sta-
bility ofringsofsatellites. InProc. 39th IEEE Confer-
ence on Decision and Control (CDROM), pages 1285–
Figure 2: Above: (cid:1)l1−l2(cid:1)and(cid:1)A1−A2(cid:1)asafunction 1288, New York, 2000. IEEE.
of time; Below: the Lyapunov function as a [5] R.J.Sedwick,D.W.Miller,andE.M.C.Kong.Mit-
function of time igationof differentialperturbations in formationflying
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2.5
Final [6] H. Schaub, S.R. Vadali, J.L. Junkins, and K.T.
Sepration
Alfriend. Spacecraft formation flying control using
2 mean orbit elements. to appear in Journal of the As-
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[7] Hanspeter Schaub and Kyle T. Alfriend.
1.5
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1
0.5
0
0 0.5 1 1.5 2 2.5
Initial Seperation
Figure 3: The final separation between two Satellites as
a function of starting separation
Thefinalrelativepositionofthe twosatellitesdepends
on the starting relative position of the two satellites.
In the following experiments, two LEO satellites are
first placed on the same circular orbit specified as
[1.1,0,0,π/2,0,0] with certain amount of initial sep-
aration. Then they are transfered to a final orbit of
[1.3,0,0,π/2,0,0]. Figure 3 shows the final separation
versusthe initialseparationbetweenthe twosatellites.
Itcanbe seenthatthe separationis increasedsince we
are transferring to a higher obit.
p. 8