Variable Structure Control ~ Basics

Transcription

1 Varable Structure Control ~ Bac Harry G. Kwatny Department of Mechancal Engneerng & Mechanc Drexel Unverty

2 Outlne A prelmnary example VS ytem, ldng mode, reachng Bac of dcontnuou ytem Example: underea vehcle Degn baed on normal form

3 Example 1 2 x = ax bu, a, b > 0 u = 2 1 α x x> 0 = cx + x =, α x1 x 1 < 0 α, c > ldng mode: = cx + x 0 + cx = condton for extence: < 0

4 Setup Sytem: = f( xu, ) n m x R, u R, f mooth Control: u dcontnuou acro wtchng urface ( x) = 0 + u ( x), ( x) > 0 u ( x) =, = 1,, m u ( x), ( x) < 0 + u ( x), u ( x), mooth

5 Bac of Dcontnuou Sytem ~ 1 = f(,) xt f mooth n x( C k, k> 0) for each t except on m codmenon-one urface defned by ( x) = 0, = 1,, m What a oluton?

6 Bac of Dcontnuou Sytem ~ 2 If there a moton n the wtchng urface, t called a ldng moton. How t defned?

7 Bac: The Flppov Soluton dx() t dt F( xt, t) : = conv f( S( δ, xt ) Λ( δ, xt ), t) δ > 0 { y R y }} n S( δ, x) : = x < δ Λ( δ, x) : ubet of meaure zero on whch f not defned

8 Lyapunov Ung mooth Lyapunov functon wth dcontnuou vector feld. V V x C V V ( xt ) ξ ξ F ( xt, t) x Reult 1 ρ < 0 > 0 on P Λ V ρ < 1 ( V ρ ) 0 0 on ( V ρ > ) Reult 2 V ρ ( x), ρ > 0on P Λ V ρ ( x), ρ > 0 on P P

9 Sldng Doman { n } D M = x R x = 0 a ldng doman f: 1. trajectore begnnng n a δ -vcnty of D reman n an ε -vcnty untl reachng D 2. D doe not contan entre trajectore of the m 2 aocated contnuou ytem.

10 Propoton Lyapunov verfcaton of a ldng doman. 0 f ( x) 0 1 = = V( x) C, V( x): = > 0 otherwe D D an open, connected ubet of R V ρ x < 0 on D M D a ldng doman n

11 Propoton Fnte tme reachng of ldng doman. 2 V ( x) = σ x, σ > 0on a δ -vcnty of D D an open, connected ubet of R V ρ x < 0 on D M trajectore begnnng n a δ -vcnty of D n fnte tme D n D reach

12 VSC Degn va Sldng Mode: Setup Sytem: Control: = f( xu, ) n m x R, u R, f mooth u dcontnuou acro wtchng urface ( x) = 0 + u ( x), ( x) > 0 u ( x) =, = 1,, m u ( x), ( x) < 0 + u ( x), u ( x), mooth

13 VSC Degn va Sldng Mode: Strategy ( x) ( x) 1. Chooe wtchng urface,, o that ldng mode ha dered dynamc. ( x) ± 2. Chooe control functon, u, o that ldng mode reached n fnte tme. Approach: 1. How doe choce of equvalent control method ( x) affect ldng behavor? ± 2. How can we chooe u to nure fnte tme reachng? Lyapunov method

14 Reachng Chooe T V x = x x = x 2 T V ( x) = 2 ( x) f ( x, u) x T Suppoe 2 x x f xu, < ρ x on D M where D an open et contanng D M D a ldng doman and t reached n fnte tme from any ntal pont n D.

15 Example ~ Lnear SISO f x, u = Ax + bu, x = cx 2 V = x V = 2() x cax + 2 x cbu We can only affect the econd term. Chooe, u = k x gn x cb, k x > 0 x V = 2 ( x) cax cbk ( x) gn ( x) cb x provded cbk x cax ρ, x k x = ρ + cax / cb Th nure that x ρ a global ldng doman.

16 Equvalent Control ~ 1 + u tx x = f txu,,, u = u ( tx) ( x) Suppoe a ldng doman, > 0 { n } M = x R x =, < 0 0 contan

17 Equvalent Control ~ 2 ( x) = f( txu,, ) eq 0 x f a oluton ext for u, then on = 0, 0 and ( (, )) eq eq = f txu,, tx dynamc n ldng

18 Equvalent Control Specal Cae Sytem 'lnear n control' = + f x G x u x = S x f x + S x G x u 1 det SG 0 ueq = S x G x S x f x 1 = I G x S x G x S x f x Lnear Sytem 1 = Ax + Bu, x = Cx u = CB CAx (aumng det CB 0) eq 1 = I B CB C Ax eq 0

19 Geometry A ubet S of the lnear pace (over feld ) X a lnear ubpace of X f: x1, x2 S and c, c, cx + cx S { r r, } { } If x X ( = 1,, k), then pan x,, x a ubpace of X. R,S X then R + S = + R S { x S } R S = xx R& x Two ubpace R,S are ndependent f R S = 0 1 For our purpoe thnk Eucldean pace k n R

20 Geometry 2 If R, = 1,, k are ndependent ubpace, then the um R = R + + R 1 called an ndrect um and may be wrtten R = R R 1 k k The ymbol preupoe ndependence. Let X = R S. For each x X, there are unque r R, S o that x= r+. Th mple a unque functon r called the projecton on R along S.

21 Geometry 3 The projecton a lnear map Q : X X, uch that Im Q= R and ker Q= S, and Q ( I Q) ( I Q) 0 2 X = X X =R S Note that the projecton on S along R. Thu, Q I Q = Q = Q Converly, for any map : X X uch that = X = Im Q ker Q 2 Q Q Q.e., Q the projecton on Im Q along ker Q.

22 Geometry 4 Recall 1 = I B CB C Ax 1 Q = B CB C 1 ( 1 ) Q = B CB CB CB C = B CB C ImQ = Im B ker Q = ker C X = Im B ker C B CB C the projecton on Im B along ker C I B CB C the projecton on ker C along Im B

23 Lnear Example ~ 1 x = cx, 0 0 [ ] 1 n 1 c = cax + cbu : = 0 u = cb cax eq 1 = I b cb c Ax Sldng dynamc Defne a matrx V whoe column pan ker c,.e., V = v v, cv = 0 Notce that b ker c and X = Imb ker c. Defne a tate tranformaton x w, z x = Vw + bz, eq 1 n 1 w R, z R 1 Vw + bz = I b cb c A Vw + bz

24 Lnear Example ~ 2 w w z z 1 [ V b] = I b( cb) c A[ V b] 1 U UV Ub In [ V b] = d ( cb) c d dv db = 0 1 = w U 1 w U w I b( cb) c AV [ b] AV [ b] z = = d z 0 z note: d I 1 c 0 b cb = w UAV UAb w z = 0 0 z ldng: w = [ ] UAV w

25 Degnng the Sldng Surface Conder the ytem = f x + G x u rank G x = m atfe controllablty rank condton Tranform to regular form: = f x x (, ) (, ) (, ) = f x x + G x x u x R, x R,det G 0 around x n m m around x Strategy: 0 1) chooe x = x o that (, ) = f x x ha dered behavor, 2) chooe u to enforce ldng on (, ) x x = x + x

26 Lnear Cae = Ax + Bu, rank B = m, A, B controllable reorder tate to obtan B1 m m B=, wth B2 R,det B2 0 B 2 tranform to regular form: 1 In m BB 1 2 z = Tx, T = 1 0 B2 A A z = z u A11 A12 A21 A + 22 I the par, controllable

27 Lnear Cae 2 To hape ldng dynamc, Chooe Now, by any mean, pole placement, LQG, etc. 1 2 To degn the control u take T T V ( z) = ( z) = ( z) Q ( z), Q = Q > 0 Q 2 2 T T T V z = Q= z K I Q K I Az+ Qu u z = Kz z = A + A K z K z = Kz + z T [- m] [- m] * * = κ ( z) gn, ( z) = Q ( z), κ T T > [- ] [- ] z z K I Q K I Az m m

28 Example: Underwater Vehcle [ ] mx + cxx = u, mc, uncertan, u UU, = v c 1 v = vv+ u m m Chooe a ldng urface: = v+ λx Why? Becaue 0 = λx ( v = λx tablze frt eq) V xv = = T 2 Chooe control baed on,, v 1 U V 2[ λ 1] c > 0 = + 2 u u = vv m U < 0 m

29 Control Baed on Normal Form ξ = F( ξ, zu, ) z = Az + E[ αξ (, z) + ρξ (, z) u] y = Cz Chooe x uch that x = 0 Kz x = 0 + u ( x) ( x) > 0 u ( x) = u ( x) ( x) < 0 Recall Brunovky tructure of A,E

30 Sldng Dynamc Note : ame a feedback lnearzng control ( x) = 0 Kz( x) = 0 Kz = KAz + KE α( x) + ρ( x) ueq = 0 KE = I 1 1 ueq = ρ x KAz x ρ x α x z = I EK Az, Kz t = 0 [ ] 0 Sldng Dynamc

31 Choong K One choce: r1 k1 0 0 K =, k = a,1 ar, 1 1 r m 0 0 k m Sldng egenvalue Egenvalue of (A+EK) are: mare 0 and r mare 0 1 0, 1,, m λ = a,1 a,2 a r, 1

32 Reachng Conder the potve defnte quadratc form n V ( x) = T Q Upon dfferentaton we obtan d dt V T T T [ KAz + α] QKz 2u ρ QKz T = 2 Q = 2 + If the control are bounded, u U > 0 ( 0 > U mn, u U max, > 0 ) then chooe u = R S * U x T mn, max, * U x a f a f > 0 < 0, = 1,, m, * T x = ρ ( x ) QKz ( x )

33 VSC Summary