MIT OpenCourseWare http://ocw.it.edu 16.323 Principles of Optial Control Spring 2008 For inforation about citing these aterials or our Ters of Use, visit: http://ocw.it.edu/ters.
16.323 Lecture 10 Singular Arcs Bryson Chapter 8 Kirk Section 5.6
Spr 2008 16.323 10 1 Singular Probles There are occasions when the PMP { } u (t) = arg in H(x, u, p, t) u(t) U fails to define u (t) can an extreal control still exist? Typically occurs when the Hailtonian is linear in the control, and the coefficient of the control ter equals zero. Exaple: on page 9-10 we wrote the control law: u b < p 2 (t) u(t) = 0 b < p 2 (t) < b u p 2 (t) < b but we do not know what happens if p 2 = b for an interval of tie. Called a singular arc. Botto line is that the straightforward solution approach does not work, and we need to investigate the PMP conditions in ore detail. Key point: depending on the syste and the cost, singular arcs ight exist, and we ust deterine their existence to fully characterize the set of possible control solutions. Note: control on the singular arc is deterined by the requireents that the coefficient of the linear control ters in H u reain zero on the singular arc and so ust the tie derivatives of H u. Necessary condition for scalar u can be stated as [( ) ] d 2k ( 1) k H u dt 2k u 0 k = 0, 1, 2...
Spr 2008 Singular Arc Exaple 1 16.323 10 2 With ẋ = u, x(0) = 1 and 0 u(t) 4, consider objective in 2 0 (x(t) t 2 ) 2 dt First for standard Hailtonian H = (x(t) t 2 ) 2 + p(t)u(t) which gives H u = p(t) and ṗ(t) = H x = 2(x t 2 ), with p(2) = 0 (10.15) Note that if p(t) > 0, then PMP indicates that we should take the iniu possible value of u(t) = 0. Siilarly, if p(t) < 0, we should take u(t) = 4. Question: can we get that H u 0 for soe interval of tie? Note: H u 0 iplies p(t) 0, which eans ṗ(t) 0, and thus ṗ(t) 0 x(t) = t 2, u(t) = ẋ = 2t Thus we get the control law that 0 p(t) > 0 u(t) = 2t when p(t) = 0 4 p(t) < 0
Spr 2008 16.323 10 3 Can show by contradiction that optial solution has x(t) t 2 for t [0, 2]. And thus we know that ṗ(t) 0 for t [0, 2] But p(2) = 0 and ṗ(t) 0 iply that p(t) 0 for t [0, 2] So there ust be a point in tie k [0, 2] after which p(t) = 0 (soe steps skipped here...) Check options: k = 0? contradiction Check options: k = 2? contradiction So ust have 0 < k < 2. How find it? Control law will be { 0 when 0 t < k u(t) = 2t k t < 2 apply this control to the state equations and get: { 1 when 0 t k x(t) = 2 t + (1 k 2 ) k t 2 To find k, note that ust have p(t) 0 for t [k, 2], so in this tie range ṗ(t) 0 = 2(1 k 2 ) k = 1 So now both u(t) and x(t) are known, and the optial solution is to bang off and then follow a singular arc.
Spr 2008 Singular Arc Exaple 2 16.323 10 4 LTI syste, x 1 (0), x 2 (0), t f given; x 1 (t f ) = x 2 (t f ) = 0 [ ] [ ] 0 1 1 A = B = 0 0 1 t f and J = 1 2 x 2 1 dt (see Bryson and Ho, p. 248) 0 So H = 1 x 1 (t) 2 2 + p 1 (t)x 2 (t) + p 1 (t)u(t) p 2 (t)u(t) ṗ 1 (t) = x 1 (t), ṗ 2 (t) = p 1 (t) For a singular arc, we ust have H u = 0 for a finite tie interval Thus, during that interval H u = p 1 (t) p 2 (t) = 0? d Hu = ṗ 1 (t) ṗ 2 (t) dt = x 1 (t) + p 1 (t) = 0 Note that H is not an explicit function of tie, so H is a constant for all tie H = 1 x 1 (t) 2 + p 1 (t)x 2 (t) + [p 1 (t) p 2 (t)] u(t) = C 2 but can now substitute fro above along the singular arc 1 x1 (t) 2 + x 1 (t)x 2 (t) = C 2 which gives a faily of singular arcs in the state x 1, x 2 To find the appropriate control to stay on the arc, use d 2 dt 2 (H u) = ẋ 1 + ṗ 1 = (x 2 (t) + u(t)) x 1 (t) = 0 or that u(t) = (x 1 (t) + x 2 (t)) which is a linear feedback law to use along the singular arc.
Spr 2008 Details for LTI Systes 16.323 10 5 Consider the in tie-fuel proble for the general syste with M u i M + and ẋ = Ax + Bu tf J = (1 + c i u i )dt 0 i=1 t f is free and we want to drive the state to the origin We studied this case before, and showed that H = 1 + (c i u i + p T B i u i ) + p T Ax i=1 d On a singular arc, k dt k (H u ) = 0 coefficient of u in H is zero p T (t)b i = ±c i for non-zero period of tie and ( ) T d k (p T d k p(t) (t)b i ) = B i = 0 k 1 dt k dt k Recall the necessary conditions ṗ T = H x = p T A, which iply. ( ) T d k p(t) dt k p T = ṗ T A = p T A 2... p T = p TA = p T A 3 ( 1) k p T A k and cobining with the above gives ( ) T d k p(t) B i = ( 1) k p T A k B i = 0 dt k
Spr 2008 16.323 10 6 Rewriting these equations yields the conditions that p T AB i = 0, p T A 2 B i = 0, [ ] p T A B i AB i A n 1 B i = 0 There are three ways to get: [ ] p T A B i AB i A n 1 B i = 0 On a singular arc, we know that p(t) = 0 so this does not cause the condition to be zero. What if A singular, and p(t) T A = 0 on the arc? Then ṗ T = p T A = 0. In this case, p(t) is constant over [t 0, t f ] Indicates that if the proble is singular at any tie, it is singular for all tie. This also indicates that u is a constant. A possible case, but would be unusual since it is very restrictive set of control inputs. [ ] Third possibility is that B i AB i A n 1 B i is singular, eaning that the syste is not controllable by the individual control inputs. Very likely scenario ost coon cause of singularity conditions. Lack of controllability by a control input does not necessarily ean that a singular arc has to exist, but it is a possibility.
Spr 2008 16.323 10 7 For Min Tie probles, now c i = 0, so things are a bit different In this case the switchings are at p T B i = 0 and a siilar analysis as before gives the condition that [ ] p T B i AB i A n 1 B i = 0 Now there are only 2 possibilities p = 0 is one, but in that case, but we would expect that H = 0 H = 1 + p T (Ax + Bu) = 1 Second condition is obviously the lack of controllability again. Suary (Min tie): If the syste is copletely controllable by B i, then u i can have no singular intervals Not shown, but if the syste is not copletely controllable by B i, then u i ust have a singular interval. Suary (Min tie-fuel): If the syste is copletely controllable by B i and A is non-singular, then there can be no singular intervals
Spr 2008 Nonlinear Systes 16.323 10 8 Consider systes that are nonlinear in the state, but linear in the control ẋ(t) = a(x(t)) + b(x(t))u(t) with cost J = tf t 0 g(x(t))dt For a singular arc, in general you will find that d k dt k (H u i ) = 0 k = 0,..., r 1 but these conditions provide no indication of the control required to keep the syste on the singular arc i.e. the coefficient of the control ters is zero. d But then for soe r and i, r dt r (H ui ) = 0 does retain u i. So if u j (x, p) are the other control inputs, then d r (H ui ) = C(x, p, u j (x, p)) + D(x, p, u j (x, p))u i = 0 dt r with D = 0, so the condition does depend on u i. Then can define the appropriate control law to stay on the singular arc as C(x, p, u j (x, p, )) u i = D(x, p, uj (x, p, ))
Spr 2008 16.323 10 9 Properties of this solution are: r 2 is even Singular surface of diension 2n r in space of (x, p) in general, but 2n r 1 if t f is free (additional constraint that H(t) = 0) Additional necessary condition for the singular arc to be extreal is that: [ ] d r ( 1) r/2 H u 0 u i dt r Note that in the exaple above, [ ] d r H u D u i dt r
Spr 2008 Last Exaple 16.323 10 10 Goddard proble: thrust progra for axiu altitude of a sounding rocket [Bryson and Ho, p. 253]. Given the EOM: 1 v = [F (t) D(v, h)] g ḣ = v F (t) ṁ = c where g is a constant, and drag odel is D(v, h) = 1 2 ρv2 C d Se βh Proble: Find 0 F (t) F ax to axiize h(t f ) with v(0) = h(0) = 0 and (0), (t f ) are given The Hailtonian is ( ) 1 F (t) H = p 1 [F (t) D(v, h)] g + p 2 v p 3 c and since v(t f ) is not specified and we are axiizing h(t f ), p 2 (t f ) = 1 p 1 (t f ) = 0 Note that H(t) = 0 since the final tie is not specified. The costate EOM are: 1 D v 1 0 1 D ṗ = 0 0 p h F D 2 0 0 H is linear in the controls, and the iniu is found by iniizing ( p 1 p c 3 )F (t), which clearly has 3 possible solutions: F = F ax ( p 1 p c 3 ) < 0 0 < F < F ax if ( p 1 p c 3 ) = 0 1 F = 0 p c 3 ) > 0 ( p Middle expression corresponds to a singular arc.
Spr 2008 16.323 10 11 Note: on a singular arc, ust have H u = p 1 c p 3 = 0 for finite interval, so then Ḣ u = 0 and Ḧ u = 0, which eans ( ) D D + p 1 p 2 = 0 v c and F = D + g + (10.16) D + 2c D + c 2 2 D [ v v 2 ] D D 2 D g(d + c ) + c(c v) vc 2 v h v h which is a nonlinear feedback control law for thrust on a singular arc. For this particular drag odel, the feedback law siplifies to: [ g βc 2 ( v ) c ] F = D + g + 1 + 1 2 1 + 4(c/v) + 2(c/v) 2 g c v ( ) and the singular surface is: g = 1 + v c D Constraints H(t) = 0, H u = 0, and Ḣ u = 0 provide a condition that defines a surface for the singular arc in v, h, space: v D D + g D v = 0 (10.17) c v It can then be shown that the solution typically consists of 3 arcs: 1. F = F ax until 10.17 is satisfied. 2. Follow singular arc using 10.16 feedback law until (t) = (t f ). 3. F = 0 until v = 0. which is of the for bang-singular-bang
Spr 2008 16.323 10 12 Figure 10.1: Goddard Proble
Spr 2008 16.323 10 13 Figure 10.2: Goddard Proble
Spr 2008 16.323 10 14 Figure 10.3: Goddard Proble