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## Directed Triangular Formations

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**Supelec**EECI Graduate School in Control Directed Triangular Formations A. S. Morse Yale University Gif – sur - Yvette May 24, 2012 TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAAAAAAAAAAAA**FORMATION CONTROL**with Leader – Follower Constraints**2**d12 3 d13 1 FORMATION CONTROL with Leader – Follower Constraints 1 follows 2 and 3 Agent i maintains distance dij This plus rigidity maintains the formation Closed cycles can present problems because of miss-registration of agents’ positions**d12**d13 FORMATION CONTROL with Leader – Follower Constraints 1 follows 2 and 3 Agent i maintains distance dij This plus rigidity maintains the formation Closed cycles can present problems because of miss-registration of agents’ positions**[ i ] = i + 1 mod 3**Can this formation be maintained? 2 d1 > d2 + d3 d1 d2 > d1 + d3 yij = position of agent j in local coordinate system of agent i 1 d3 > d1 + d2 d2 d3 3 xi = position of agent i in global coordinates**[ i ] = i + 1 mod 3**2 d1 1 d2 d3 3**d1**d1 d2 d2 d3 d3 [ i ] = i + 1 mod 3 2 2 1 1 Target set: 2 disconnected subsets 3 3 Can’t have just one ei = 0 Can’t have all ei = 0 with two agents at same position Equilibrium set = Equilibrium set ¾ So E is disjoint from Two agents at same point; third correctly positioned Target All agents at same point All trajectories starting outside of equilibrium set converge to target set E**d1**d2 d3 2 d1 > d2 + d3 d2 > d1 + d3 1 d3 > d1 + d2 Target set: 3 Equilibrium set = x1 , x2 , x3 are co- linear iff rank [z1z2z3]2£3 < 2 verify Co-linear set: If all three agents are properly positioned, they cannot be in a line E and N are disjoint**d1**d2 d3 2 1 Target set: 3 Equilibrium set = verify = {x: det [z1z2] = 0} Co-linear set: N is an invariant set E and N are disjoint**d1**d2 d3 self-contained 2 1 3 All trajectories of the z system exist and are bounded on [0, 1) All trajectories of the x system exist on [0, 1) but not necessarily bounded!**d1**d2 d3 self-contained 2 transpose of rigidity matrix of {G , x} 1 3**d1**d2 d3 2 transpose of rigidity matrix of {G , x} 1 3 compact closed Pick ½* small so that for T = closure of S(½*) as fast as**d1**d2 d3 2 1 3 V decreasing if z is not in M If z is bounded away from M, even in the limit, then because V is decreasing, z must enter S (½*) in some finite time. _ If z starts outside of N it cannot enter N and thus M in finite time _ as fast as**d1**d2 d3 Claim: for any z2M, e1 + e2 + e3 < 0 Implication of claim: 2 Suppose: 1 Beyond some time T, z must be close enough to M that for t¸T, e1 + e2 +e3 < 0 Thus 3 Therefore if If z is bounded away from M, even in the limit, then because V is decreasing, z must enter S (½*) in some finite time. _ If z starts outside of N it cannot enter N and thus M in finite time _ as fast as**d1**d2 d3 Claim: for any z2M, e1 + e2 + e3 < 0 2 1 verify 3 Suppose z2M1 Four cases to consider: 1. ||z2|| = ||z3|| = 0 ||z1|| = 0 ei = -di2 , i = 1,2,3 e1 = e3 = 0 and e2 = -d22 ||z1|| 0 2. ||z2|| = 0 and ||z3|| 0 3. ||z2|| 0 and ||z3|| = 0 Same type reasoning as case 2. 4. ||z2|| 0 and ||z3|| 0**d1**d2 d3 Claim: for any z2M, e1 + e2 + e3 < 0 2 1 3 Suppose z2M1 4. ||z2|| 0 and ||z3|| 0**d1**d2 d3 Claim: for any z2M, e1 + e2 + e3 < 0 2 1 Suppose z2M1 3 4. ||z2|| 0 and ||z3|| 0 ||z1|| > ||z2|| so all ||zi|| 0 If any ei = 0then all ei = 0 because of def. M So all ei 0 |e3| > |e1| |e2| > |e1| If e1 < 0 then e2 > 0 and e3 > 0 d1 > ||z1|| d2 < ||z2|| d3 < ||z3|| e1 > 0 e2 < 0 e3< 0 e1 + e2 + e3 < 0**What about the xi?**We already know that the xi and zi exist globally and that the zi are bounded . If the ei tend to 0 exponentially fast, the xi then to finite constants If the ei do not tend to 0, the xi drift off to 1.**d1**d2 d3 Summary [ i ] = i + 1 mod 3 2 1 Target set: 3 Co-linear set: All xtrajectories start outside of the co-linear set converge to the target set exponentially fast. All trajectories start inside of the co-linear set which are not in the equilibrium set drift off to infinity as t!1. So there is a pretty good understanding of directed triangular formations**Remark**Results like the preceding also exist for more complicated graphs which are cycle free. For example: We will {probably not} talk about such formations next directed However there is no general theory for directed graphs which have cycles. For undirected formations however, more is known. We will talk about this later.