SINGULAR CONTROL IN MINIMUM TIME SPACECRAFT REORIENTATION Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 Hans Seywald* Analytical Mechanics Associates Inc., Spacecraft Control Branch - NASA Langley Renjith R. Kurnart Analytical Mechanics Associates Inc., Space Station Freedom Office - NASA Langley Abstract Time optimal solutions for the reorientation of an inertially symmetric rigid spacecraft with independent 3axes controls are investigated. All possible optimal control strategies are identified. These include bang-bang solutions, finite order singular arcs and infinite order singular arcs. Necessary conditions for optimality of finite order singular arcs are presented. Non-optimality of principal-axis rotation for rest-to-rest minimum-time reorientation is proved. Numerical examples of time-optimal solutions with finite and infinite order singular arcs are presented. A simple example of “a car in a parking lot” is discussed. It clearly illustrates that infinite order singular control may be the “unique” solution to an optimal control problem and cannot be ignored when choosing the optimal control strategies. 1 Introduction Optimal rigid body rotational maneuvers have been studied in detail by many researchers in the context of spacecraft attitude control systems [ 2 ] , [4], [8], [9], [14]. Minimum time and/or minimum control effort rotational maneuvers have been the main objectives in these studies. Recently, the minimum time reorientation problem has been studied by Li and Bainum [14]. The approach taken in that paper is to solve a sequence of minimum control effort problems with fixed final times. While subsequently reducing the final time, t f , it is assumed that the smallest value of t j for which a solution could be obtained, is close to the actual minimum final time and that the obtained control logic is “close” to the *Member AIAA t Member AIAA optimal minimum time solution. With this method Li and Bainum arrived at the conclusion that the minimum time solution involves singular control and that principal axis rotation is time optimal for rest-to-rest problems. It is clear that this result is based on some continuity assumption on the solution with respect to the prescribed final time, a dangerous assumption especially as the minimum time trajectory refers to an abnormal solution point of the minimum control effort problem. But even if this continuity assumption is satisfied, the minimum final time obtained by the sequential numerical solution of minimum control effort problems may be just a local minimum final time depending upon the specific choice of cost function. In fact, in a later study by Bilimoria and Wie [2] a variety of minimum time rest-to-rest reorientation problems has been solved and the non-optimality of principal axis rotation has been shown via numerical solutions of associated two-point boundary value problems (TPBVPs). However, the possibility of singular control has not been analyzed, except for the rather trivial case where all controls are singular at the same time. In this paper the authors investigate the problem of reorienting an inertially symmetric spacecraft in minimum time from given initial angular position and velocity to free or prescribed final angular position and velocity. Three bounded independent control torques are assumed with the control axes aligned along the principal axes. Emphasis is laid on identifying all possible coptimal control logics for the problem stated above. This includes bang-bang control arcs, finite order singular arcs and infinite order singular arcs. Higher order necessary conditions for optimality of finite order singular arcs are examined using Goh transformations of the associated Accessory Minimum Problem. Numerical examples of optimal solutions with fi- 432 nite and infinite order singular arcs are presented. Examples with bang-bang structure can be found in [a]. For rest-to-rest maneuvers, principal-axis rotations are identified as finite order singular arcs, the nonoptimality of which is shown in section 6.2. It is shown that principal axis rotations can appear in the synthesis of time optimal solutions only as realizations of infinite order singular control arcs. In this case the control history that furnishes minimum final time is not determined uniquely. Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 2 Dynamical System Consider an inertially symmetric rigid spacecraft with the control axes alligned with the principal axes. The J Euler rotational equations of motion can easily be derived as Figure 1: Definition of coordinate axes and Euler padw -= u (1) rameters dt where where (€1, € 2 , € 3 ) are the angles between the Euler axis is the vector of angular velocities of the rigid body about and an inertially fixed axis orthogonal coordinate systhe three body axes (shown in Figure 1) represented in tem (without loss of generality coincident with the body the body coordinate frame and t is the clock-time. The axis a t initial time) as shown in Figure 1. A principal axis rotation is defined as a rotation durthree independent components of control vector ing which the Euler axis is always aligned with a body axis (also same as a control axis). In the course of this paper we will find it very helpful are confined to the fixed limits to introduce the following matrices ui,max I ui I +Ui,rnaxc ro F, In physical terms ui = where E , i = 1 , 2 , 3 , are the three independent control torques represented in the body coordinate frame and I denotes the moments of inertia of the spacecraft about the principal axes. The kinematic equations of motion can be written as d!l dt -= 1 -Rq, 2 (5) (3) 0 where (4) QT 0 0 1 0 0 0 0 .]I .=[1 Here the quaternion vector o 01 -1 D1 := [ 0 0 1 0 0 0 1 0 0 1 0 - 1 1 0 1 0 0 0 0 ' (7) The benefit of these definitions lies in the fact that fl = w2D2 w3D3 along with the simple relations. WID1 = h o , Q1,42,431 + + is used in favor over the Euler angles to prevent singularities and to simplify the kinematic differential equations. However, a drawback in the use of quaternions is the increase in the state vector dimension by 1. The physical meaning of the quaternions is as follows: @ qo = cos T . @ qs = cos ~i sin 2 ; i = 1,2,3, where i, j E {1,2,3}. 433 i q= - - @21 A q . Let US now consider the following opimal control problem (Minimum-Time Spacecraft Reorientation MTSR): min tf (9) As all three controls u1, uz, '113 appear only linearly in the Hamiltonian (15) the Pontryagin Minimum Principle u=arg min H (see [3], [5], [13], [15]) yields for i = 1,2,3 U€(PWC[tO,t11)3 Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 (17) if A,,,, < 0 subject to the dynamical system ( l ) ,(3), the initial conui = ditions w ( 0 ) = w o , w o E R3 given singular if A,,,, E 0 q ( 0 ) = q o , qo E R4 given At times t,, where any of the control variables ui the final conditions switches between the possible control logics (18) a necessary condition for optimality is given by the continuity of the Hamiltonian and the control constraints (2) with ui,maT > 0, i = HI,: - HI,- = 0. 1,2,3. (19) f In (9) PWC[to,tf] denotes the set of all piecewise continuous functions on the interval [to, tf]. The analysis In practice this condition is used to determine the locain the remainder of this paper remains exactly the same tion of switching times. The transversality conditions , Hilbert space associated with boundary conditions (11) are given by if we replace PWC[to,tf] by L2[t0,t f ] the of all quadratically integrable functions on the interval dQ X,,,(tf) = vT[to,,tf] as the set of admissible control functions. (20) dW(tf 1 From (3) and the anti-symmetry of s2 given in (4) it follows readily that $ (qTq) = 0. Hence, the dynamia@ X , ( t f ) = vT(21) cal system (l),(3) of problem MTSR is not completely Wf) controllable. For this reason the function where !V is given by equation (12) and u E Rk is a constant multiplier vector. The optimal final time t f is determined by HI tf = - 1 . representing the final conditions (11) has to satisfy k 5 6 5 (13) Transformation of the and This represents a necessary condition for the existence of a non-degenerate solution of problem MTSR. It can be seen from the costate equations stated later in (17), that the lack of controllability in the q-equations translates into the same lack of controllability in the Xq-equations. This is an important point to notice when it comes to setting up a consistent boundary value problem for generating nu'merical solutions. ptimality Conditions Let the Hamiltonian H be defined by H = A,u Then the costate equations take the explicit form 1 + -XqRq. 2 (15) For analyzing the singular control cases XUi EZ 0 and for a better understanding of the costate dynamics we will find the following transformation of the costate dynamics helpful. Let S = [SI, S Z ,S3]' be the vector of switching functions, i.e. for i = 1 , 2 , 3 sj := A., (23) Then, by equation (16), the first time derivative of is given by Si = di Si dj := --q' T DiT A,. (24) 2 Using equations ( l ) ,(3), and (17) along with the simple relations given in (8) it is easy to verify that dT := [ d l , dz, d3] satisfies the differential equation is = - (E)' d=hd with (16) 434 h= [ 0 -wg w2 w3 0 -w1 -w2 W l ] 0 If conditions (32) are satisfied at a single point along the trajectory, then they are satisfied identically throughout the trajectory, irrespective of control u1. In fact, a t each instant of time the numerical value of control u1 is arbitrary as long as it satisfies the control constraint (2) and A, = d. (27) as long as it steers the state vector to a final position This implies that the state/costate system (l),(3), (16), that satisfies the boundary condition (11). (17) can be substituted equivalently by the simpler sysIf w2dz w3d3 # 0 , then we obtain singular control tem (l), (3), (27), (25). In analyzing the singular control of second order for control u1. Explicitly, we get cases we will make extensive use of this situation. This means, the dynamics of d can be expressed completely in terms of w and d itself. Vector d also carries all the information necessary to determine the optimal control. Equation (16) can be rewritten as + (33) Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 ar Contra1 Control ui is called singular on some non-zero time interval [ q ,TZ] if the associated switching function Si is identically zero on [TI, 7-21. As Si is independent of controls, successive differentiation of this identity with re(34) spect to time is well-defined and can be used to obtain in the interior of the singular arc. Goh’s Necessary additional conditions. We have Condition for singular control is satisfied if and only if wzdz w3d3 > 0 . In the remainder of this section we will give a short proof of the claims made above. From (as), (29), (30) we find that S 1 = 0, S I = 0, and SI = 0, imply + ,A, with fi given by (26). Conditions on singular optimal control are obtained from subsequently setting S = 0, S = 0, ... and so on. It is clear that all higher derivatives of S (if they exist) are linear in d . In particular, this implies that all derivatives of S are automatically zero if only d = 0. Before we proceed with analyzing the singular control cases for problem MTSR, it should be noted that not all possible combinations of singular and non-singular controls need to be considered. Physically it is clear that problem MTSR remains essentially unchanged if the coordinate axes are interchanged. This implies, for example, that in the case where only one control is singular, it can be assumed without loss of generality, that control 211 is singular, while 212, u3 are non-singular. In the following three sections a complete analysis of the singular control cases with one, two, and all three controls being singular is presented. 6.1 =0 =0 ~ 3 d 2 ~ 2 d 3 Using (36) in (31), Si3)= 0 yields Note that controls 212, u3 are assumed non-singular and hence are identically constant, except possibly at a finite number of points. Further differentiation of (38) yields If w2dz + w g d g # 0 then (39) can be solved for control u1 and we obtain the result stated in (33)) (34). Kelley’s Necessary Condition for optimality of singular control (see P O I , [111, [121, [161), One Control Singular with q E N denoting the order of the singular control, Without loss of generality we can assume that control yields wzdz w3d3 2 0. The same result can also be u1 is singular and controls 212, 213 are non-singular. Two obtained from Goh’s necessary condition (see [6], [17]). The assumption that w2d2 w3d3 = 0 together with possible cases have to be distinguished, namely w2d2 + (37) leads to w3d3 = 0 , and w2d2 w3d3 # 0 , If w2d2 wad3 = 0 , then we obtain singular control of infinite order for 211. Explicitly, we get + + + + ,A, dl =o d2 =0 =0 d3 =0 Here, det 435 Izz -ww”, I = 0 can be excluded as controls = 0 can appear at most at isolated points). Hence (40) implies d, = 0, d3 = 0. w3 But with (36), d l = 0, this implies immediately that all derivatives of S1 are automatically zero, without ever providing a condition on control u1 (singular control of infinite order). This is the result stated in (32). Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 6.2 Two Controls Singular Without loss of generality we can assume that controls u1 and u, are singular and control u3 is non-singular. Again, we have to distinguish two possible cases, namely d3 = 0, and d3 # 0. If d3 = 0, then we obtain singular control of infinite order for controls 211 and 212. Explicitly, we get ,A, =0 , A, =0 dl = 0 d2 = 0 d3 = 0 In complete analogy to the situation discussed in the previous section these conditions are satisfied identically throughout the trajectory, if they are only satisfied at a single point. Both controls, u1 and 212, can be chosen arbitrarily within their allowed bounds (2) as long as the boundary conditions (11) are satisfied. The case d3 # 0 can be excluded. Formally, conditions for a finite order singular control can be obtained in this case, but from Goh’s Necessary Condition it follows that this control logic is always suboptimal. In the remainder of this section we will give a short proof of the claims made above. We have If d3 # 0, then the switching function. Goh’s Necessary Condition for singular control states that along the singular arc some matrix Q2B2 has to be identically symmetric and another matrix, R3, has to be positive semi definite (for nomenclature, definitions and theorems see [6], [17]). These results are obtained by applying Goh’s transformation to the singular Accessory Minimum Problem and then appliying the usual Legendre-Clebsch condition to the transformed problem. If R3 E 0 is obtained, then the same transformation can be applied again to yield an additional set of conditions of the same form stated above but in terms of some new matrices Q;, B:, R$. For the present problem we obtain In the second cycle the symmetry condition on QhB; is satisfied if and only if d3 0. But this contradicts the assumption. Hence this control logic can be dismissed as non-optimal. For the case d3 = 0, conditions (41) follow immediately from (28), (29). 6.3 This case can be excluded for problem MTSR. We have (47) S. = O (29) d = O (43) It follows immediately that all higher order derivatives of S are automatically zero. Hence, all three controls are singular of infinite order. From (47), (48) we find immediately that in this case the Hamiltonian as defined in (15) satisfies H = 0. But for a minimum time problem this contradicts the transversality condition (22), HI tf =- 1. (44) 7 (44) implies w, = 0 All Three Controls Singular ernark on Finite ingular Controls In section 6.2 it is observed that in the finite order singular control case with controls u1, u2 singular, u3 nonboth u1 and u2 appear explicitly for and further differentiation yields explicit expressions for the first time in [Sy), S p ) ] (see equation (46)). In Bell controls u l , 212, namely and Jacobson [l]and in Kelley, Kopp, Moyer [ll]it is shown that in the case of a single scalar control being si3)= 0 ( 1 ) 212 = 0 (46) singular, the associated switching function has to be difu1 = 0 =0 ferentiated an even number of times before the control In this case Kelleys Necessary Condition for singular can appear explicitly for the first time. The result oboptimal control can not be applied as the controls ap- tained in (46) actually proves that this theorem can not pear for the first time explicitly in an odd derivative of be generalized to vector valued singular control. w1 sp =0 (45) I* 436 ri To obtain a consistent boundary value problem, only Q O ( t f ) = Q d t f ) = ! 7 3 ( t j ) = 0 can be prescribed and A rotation is called a principal axis rotation if, at ev- q l ( t f )formally has to be considered free which leads to ery instant of time, the rotation is performed about a the transversality condition A, ( t f )= 0. fixed principal axis of the spacecraft. By inspection it is If all four Euler parameters are prescribed explicitly observed that the results obtained in section 6 exclude at final time, then, depending on the software used for the possibility of principal axis rotations as minimum solving boundary value problems, it may be impossible time reorientation strategies, except possibly as realiza- to obtain a solution, or at least the convergence betions of the infinite order singular control case (41). For haviour would be very bad. The initial costate vector example, one such realization is given by associated with the solution to a rest-to-rest 180' turn is given by 21l(t) E Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 21g(t) U3(t) 0 vt ovt 1Vt A,,(O) A,(,O) A,,(O) = - 0.230330 = 0.076005 = - 0.693665 = 0. = - 0.356038 = 0.356038 = - 0.817612 Obviously, these controls represent a principal axis roA,O(O) tation. A,,(O) In general it can be stated that a principal axis roA,,(O) tation can furnish the minimum time solution only if A,,(O) the minimum time that it takes to satisfy all boundary conditions is equal to the minimum time that it takes to 9.2 Infinite Order Singular Control perform only the prescribed change in angular velocity. This follows immediately from the nature of the infi- Consider problem (l), (2), (3), (9) with initial condinite order singular control case (41). Specifically, this tions implies that a principal axis rotation can never furnish the solution t o a minimum time rest-to-rest reorientation problem. This contradicts a result stated earlier by and the final condition Li & Bainum [14]. W 3 ( t f ) = 1. 9 Numerical In this section, numerical examples are presented involving all the control logics analyzed in the previous sections. The control bounds ui,maz,i = 1 , 2 , 3 used in equation (2) are set equal 1. This choice is purely for convenience. 9.1 Non-Singular Control The problem of a minimum time 180-degree turn has been treated successfully in [2] and the resulting numerical solution has a switching structure of bang-bang type. The present study confirms these results and the convergence behaviour was excellent for all cases considered. Explicitly, the problem under consideration is given by equations (l), (2), (3), (9), subject to the initial conditions w ( 0 ) = [O,O,O] q(0) = [1,0,0,0], the final condition w ( t f ) = [O,O, 01 and a final condition associated with the prescribed final angular position. Even though the angular position is completely prescribed at final time (180-degree rotation w.r.t. initial position) not all Euler parameters can be prescribed a t t ! , This is due to the lack of controllability in the dynamics of q discussed earlier in section 3. In the case of a 180-degree rotation about the n1 axis (see Figure l), the desired final value of state q is q ( t f ) = [0, l , O , 01. Here, the costate equations (17) along with the transversality conditions A , ( t f ) = 0 imply A,(t) 0 V t . Hence, with definition (24), we get d 3 0 on [to,tf]which implies that singular control of infinite order is the only possible solution to the problem (see the analysis in section 6). The transversality conditions A,,(tj) = 0, A,,(t,) = 0 in conjunction with (16), (24) A, = d = 0 imply A,, ( t )E 0 V t , A w 2 ( t ) E 0 Vt. Finally, from condition (22) HI,, = -1, together with switching condition (18) and final condition ~ ( t = f )1 we get A W g ( t f )= -1. With A,, = 0 this implies A,,(t) = -1 Vt and all Lagrange multipliers are completely determined. Note that only control 213 is determined explicitly by the Minimum Principle. Controls 211, u2 are arbitrary as long as the boundary conditions are satisfied. Explicitly, we get u l ( t ) arbitrary, u ~ ( tarbitrary, ) u3(t) 1 vi!. Loosely speaking, this case of infinite order singular control is a consequence of the decoupling of controls ui and states w j for i # j. Note that controls u1, u g do not have any effect on angular velocity w3. The minimum maneuver time is solely determined by the initial / boundary conditions w3(O) = 0, w g ( t f ) = 1, and, in return, only control u3 is determined by the Minimum Principle. Obviously, this situation is preserved if any o f t nal states w l ( t f ) , w Z ( t f ) are prescribed such t can be reached within the minimum time that it takes control 213 to drive wg from 0 to 1, i.e. as long as Iw3(tf)l 2 I w i ( t f ) l for i = 1 , 2 . Then the optimal control histories are given by t=O with J' u1 ( t )dt = w1 ( t f ) , O1 uz(t) arbitrary withJ, u z ( t ) dt = ~ ~ ( t j ) , ug(t) E 1Vt. u1 ( t ) arbitrary In case of equality, i.e. w l ( t f ) = w g ( t f ) , or w z ( t f ) = problem is also solved by the infinite order singular control logic discussed in section 6.1, where two controls are bang-bang, one control is singular. Another, less trivial example is obtained by prescribing the final conditions Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 ~ g ( t f ) the , Q O ( t f ) free, 0.0126, q z ( t f ) = 0.0126, q 3 ( t f ) = 0.247383, U ( t f ) T = [O, 0,1]. ql(tf) Physically, these boundary conditions refer to a reorientation of the satellite that is obtained by a single 28.721°-rotation of the satellite about an axis whose inclination with respect to the inertial axes 721, 722, 723 is given by €1 = 87.088°, €2 = 87.088°, €3 = 4.120°, respectively (see Figures 1, 2). At the same time the satellite is prescribed to be brought to an angualr rate whose coordinates in body coordinates are ~ ( t f= )[0,~0,1], i.e. at final time the satellite is prescribed to spin about its body-z-axis at an angular rate w g ( t f ) = 1. For this problem the minimum maneuver time is again determined solely by the time that it takes control '113 to drive angular velocity wg from 0 to 1, which is 1 (second). Within this time it is possible to satisfy the remaining boundary conditions in infinitely many ways, hence yielding infinite order singular control. A family of possible controls u l ( t ) ,ug(t) that satisfy the boundary conditions can be obtained by setting Figure 2: A non-trivial reorientation problem that leads to infinite order singular control Control u3 is identically 1. In Table 1 the numerical values for the parameters t l , ta, c l , c2, c3, c4 are given for four selected cases, namely: t l = 0.0, tl = 0.1, tl = 0.2, and tl = 0.3. In Figures 3 - 11 the associated time histories for controls u1, u2, states w , q and costates A,, A, are displayed. inite Order Singular Control ible control logic involving finite order sin- Table 1: Parameters associated with 4 possible solutions in an infinite order singular control case is given in section 6.1. Possibly after a 438 0.20 , 0.02 -J 0.01 - n J 3 -0.44 : Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 -0.80 $0.01 -aolution 1 -solution 2 3 _---solution solution 4 I Cr I I I I I---- 0.00 + 0.00 0 t 0 Figure 6: Time histories of state q1 for 4 possible solutions in an infinite order singular control case 0.20 2 0.02 0.01 aolution solution solution solution - zo.01 1 2 I 4 I-___ % I I I 3 0.00 --0.800. 0 o t8 . 0 ~ o 0.00 0 t Figure 4: Time histories of controls uz for 4 possible solutions in an infinite order singular control case 0 Figure 7: Time histories of state q 2 for 4 possible solutions in an infinite order singular control case 1 .00 0.99 n 50.98 U 0.97 0.96 0.95 0 t 0 ! Figure 5: Time histories of state qo for 4 possible solutions in an infinite order singular control case tions in an infinite order singular control case 439 solution 1 -solution 2 Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 solution - - - - solution ,1 3 4 .I! Time (sec) Figure 12: Time history of controls u in a second order singular control case ~i~~~~9: ~i~~ histories of state w1 for 4 possible solutions in an infinite order singular control case permutation of the axis, it is of the form: u1 singular of second order, u2, u3 bang-bang. At the beginning of a singular arc of this type conditions (33) have to be satisfied and in the interior of the singular arc control u1 is given by equation (34). For a numerical example, consider problem (l), (2), (3), (9) with initial conditions 0.08 0.06 +d ;;;OD4 3 ~ ( 0 =) [-0.045, ~ -1.1, -1.01, dolT 0.02 = [I,o,o, 01, and final conditions ~ ( t j =) [-0.472, ~ -1.1, -1.8991, 0.00 0 t 0 q ( t f ) T = [-0.274,0.076, -0.356, -0.8901. Figure 10: Time histories of state w2 for 4 possible so- These boundary conditions are chosen such that the relutions in an infinite order singular control case sulting trajectory is singular throughout the time interval. The optimal controls as well as state and costate functions of time are given in Figures 12 - 16. 10 Conclusions The minimum time reorientation problem for an inertially symmetric rigid spacecraft with three bounded independent control torques aligned with the principal axes is investigated. All possible control logics are clearly identified. These include bang-bang type control and singular control of finite and infinite order. Numerical examples are obtained for all cases. Principal axis rotations for rest-to-rest maneuvers are proved to be suboptimal. It is also shown in this paper that vector valued sint gular control can appear explicitly for the first time in an odd derivative of the switching function. (It is wellFigure 11: Time histories of state w3 for 4 possible soknown that this can be excluded for scalar singular conlutions in an infinite order singular control case trol). 440 Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 -0.21 0 0.4 0.2 0.6 0.8 I Time (sec) 'I'ime (sec) Figure 13: Time history of angular velocities w in a Figure 16: Time history of costates A, in a second order singular control case second order singular control case Furthermore] the concept of singular control of infinite order is addressed in this paper and it is shown that for certain problems all controls that furnish minimum cost are of this type. ppendix In the following we present a simple optimal control problem for which, in general, singular control of infinite order furnishes the only solution. Consider the problem min t j u,u€PWC[lo,t.fl (49) subject to the dynamical system 'I'irnc (sec) Figure 14: Time history of Euler parameters q in a second order singular control case 2 = u, Y = 211 the boundary conditions z(0) = 0, 2 ( t f ) = 0, Y(0) = 01 Y ( t j ) = 1 1 and the control constraints u E [-1, +1]] 21 E [-1,+1]. 'lime (sec) Figure 15: Time history of costates A, in a second order singular control case (52) It is easy t o see that the dynamics of states 2 and y are completely decoupled and the minimum final time for problem (49) - (52) is exactly the minimum time that it takes to drive state y to its prescribed final value y ( t j ) = 1. For control u we can choose u ( t ) 0, but any other choice is also possible as long as z ( t j ) = 0 remains satisfied. Hence we see that there is a welldefined minimum cost associated with problem (49) (52). Even though the controls that furnish this mini- 441 be of engineering interest to determine one set of control functions that furnishes this minimum cost. It is also clear that any solution to (51), (52) that minimizes the cost criterion has to satisfy the first-order necessary conditions of optimal control. Applying these conditions we find the costate equations x, x, = 0, = 0, (53) Downloaded by UNIVERSITY OF NEW SOUTH WALES (UNSW) on October 26, 2017 | http://arc.aiaa.org | DOI: 10.2514/6.1991-2645 that means, both costates are constant throughout the time interval [to,tf]. In the non-singular case A, # 0, A, # 0, the contros u , u are determined by u = { $1 -1 if if A, < 0 A,>O ’ u = { +1 if if A, < 0 A, > O ’ -1 [6] B.S. Goh, “Necessary Conditions for Singular Extremals Involving Multiple Control Variables”, SIAM Journal of Control, Vol. 4, No. 4, 1966. [7] H. Goldstein, “Classical Mechanics”, AddisonWesley, second edition, 1980. [8] J.L. Junkins, J.P. Turner, “Optimal Spacecraft Rotational Maneuvers”, Elsevier Scientific Publishing, New York, 1985. [9] J.L. Junkins, J.P. Turner, “Optimal Continuous Torque Attitude Maneuvers”, Journal of Guidance and Control, vo1.3, No.3, pp 210-217, MayJune 1980. (54) [lo] H.J. Kelley,“A Second Variation Test for Singular Extremals”, AIAA J., 2 (1964), pp. 1380-1382. (55) It is clear that problem (49) - (52) can not be solved without using singular control, as the boundary value problem (50), (53), (51) with control logic (54), (55) does not have a solution for A, # 0, A, # 0. Let us now consider the case where control u is singular. Control u is singular if and only if the associated switching function S, := A, is identically zero. From (53) it is clear that all time derivatives of S, are automatically zero. In particular,l this implies that the condition S, E 0 can never provide any condition on control u. If S, is zero at a single point along the trajectory, then S, will be identically zero throughout the trajectory. The stationary nature of the cost criterion is not influenced by the choice of control u , in this case. Any control function of time u ( t ) can be chosen as long as it steers the state to the prescribed final position. References [ll] H.J. Kelley, R.E. Kopp, H.G. Moyer, “Singular Extremals in Optimal Control”, Optimization Techniques, vol 2, G . Leitmann, ed., Academic Press, New York, 1967. [12] R.E. Kopp, H.G. Moyer, (‘Necessary Conditions for Singular Extremals”, AIAA Journal, Vo1.3, No.8, August 1965. [13] E.B. Lee, L. Markus,“Foundations of Optimal Control Theory”, Robert E. Krieger Publishing Company, Malabar, Florida, 1986. [14] F. Li, P.M. Bainum, “Numerical Approach for Solving Rigid Spacecraft Minimum-Time Attitude Maneuvers”, Journal of Guidance, Control and Dynamics, Vo1.13, No.1, 1990, pp 38-45. [15] L.W. Neustadt,“A Theory of Necessary Conditions”, Princeton University Press, Princeton, New Jersey, 1976. [l] D.J. Bell, D.H. Jacobson, “Singular Optimal Control Problems”, Mathematics in Science and Engineering, Academic Press, 1975, pp 61-100. [16] H.M. Robbins, “A Generalized Legendre-Clebsch Condition for the Singular Cases of Optimal Control”, IBM Journal, July 1967. [a] [17] H. Seywald, “Optimal Control Problems with Switching Points” Ph.D. Thesis, Virginia Polytechnic Institute and State University, July 1990. K.D. Bilimoria, B. Wie, “Minimum-Time LargeAngle Reorientation of a Rigid Spacecraft”, Paper No. 90-3486, AIAA Guidance, Navigation and Control Conference, Portland, Oregon, August 1990. [3] A.E. Bryson, Y.C. Ho,“Applied Optimal Control”, Hemisphere Publishing Corporation, New York, 1975. [4] R.S. Chowdhry, E.M. Cliff, “Optimal Rigid Body Reorientation Problem”, Paper No: 90-3485, AIAA Guidance, Navigation and Control Conference, Portland, Oregon, August 1990. [5] G.M. Ewing,“Calculus of Variations”, Dover Publications, Inc., New York, 1985. 442 f18] J . Stoer, R. Bulirsch, “Introduction to Numerical Analysis”, Springer Verlag, New York, Heidelberg, Berlin, 1980.

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