Lecture 6 : Linear Fractional Transformations (LFTs) Dr.-Ing. Sudchai Boonto

Transcription

1 Lectre 6 : (LFTs) Dr-Ing Sdchai Boonto Department of Control System and Instrmentation Engineering King Mongkts Unniversity of Technology Thonbri Thailand

2 Feedback Strctre d i d r e y z K G g n The standard feedback configration is consisted of the interconnected plant P and controller K r is a reference signal n is a sensor noise d and d i are plant otpt distrbance and plant inpt distrbance g and y are plant inpt and otpt Lectre 6 : (LFTs) 2/19

3 Standard Problem: P K-Strctre w P z K v P cl [ z = v [ P11 P 12 P 21 P 22 }{{} P [ w external inpts: w external otpts: z controller inpt: v controller otpt: Lectre 6 : (LFTs) 3/19

4 Transformation into Standard Problem For any control strctre, perform the following steps: Collect all signals that are evalated for performance into the performance vector z Collect all signals from otside into generalized distrbance vector w Collect all signals that are fed to K into generalized measrement vector v Denote otpt of K by Ct ot K Determine transfer matrix [ [ z w = P v Lectre 6 : (LFTs) 4/19

5 Standard Problem Example For the classical control di d r e g y z K G z = d + G(d i + ) v = e = r (n + z) = r n d G(d i + ) r [ [ z 0 I G 0 G d = v I I G I G d i n [ [ P11 P P = 12 0 I G 0 G = P 21 P 22 I I G I G w = [ r T d T d T i n T T Lectre 6 : (LFTs) n 5/19

6 Standard Problem P K-Strctre Procedre leads to standard problem or the P K-Strctre: w P K v z [ [ [ z P 11 P 12 w = v P 21 P 22 }{{} P P cl Closed-loop interconnection described by Z(s) = P cl (s)w (s) or short z = P cl w with P cl = P 11 + P 12 (I KP 22 ) 1 KP 21 = P 11 + P 12 K(I P 22 K) 1 P 21 = F(P, K) Lectre 6 : (LFTs) 6/19

7 Linear Fractional Transformation (LFT) Consider a mapping F : C C of the form F (s) = a + bs c + ds with a, b, c, and d C is called a linear fractional transformation, if c 0 the F (s) can also be written as for some λ, β and γ C F (s) = α + βs(1 γs) 1 Lectre 6 : (LFTs) 7/19

8 Lower linear fractional transformation The lower LFT with respect to l is defined as [z w 1 z 1 1 P v 1 1 l v 1 = P [ 1 = l v 1 w 1 1 = [ A C [ B D w 1 1 F l (M, l ) = [ A C B l := A + B(I l D) 1 l C D = A + B l (I D l ) 1 C, provided that the inverse (I l D) 1 exists Lectre 6 : (LFTs) 8/19

9 Upper linear fractional transformation The pper LFT with respect to is defined as 2 v 2 [ v 2 z 2 = P [ 2 w 2 = [ A C [ B D 2 w 2 w 2 P z 2 2 = v 2 F (M, ) = [ A C B := D + C(I A) 1 B D = D + C (I A ) 1 B, provided that the inverse (I A) 1 exists Lectre 6 : (LFTs) 9/19

10 Example f W 1 d e y g K G W 2 f F n w P K v z [ z = P v w = [ d z = [ f [ w = n T f T W 2 G 0 W 2 G 0 0 W 1 F G F F G d n Lectre 6 : (LFTs) 10/19

11 Example Assming that the plant G is strictly proper and P, F, W 1, and W 2 have the following state-space realizations: [ [ Ag B G = g Af B, F = f, C g 0 C f D f [ [ Aw1 B W 1 = w1 Aw2 B, W 2 = w2 That is C w1 D w1 ẋ g = A g x g + B g (d + ), ẋ f = A f x f + B f (y g + n), C w2 D w2 y g = C g x g y = C f x f + D f (y g + n), ẋ w1 = A w1 x w1 + B w1, f = C w1 x + D w1, ẋ w2 = A w2 x w2 + B w2 y g, f = C w2 x w2 + D w2 y g Lectre 6 : (LFTs) 11/19

12 Example Define a new state vector x = [ x g x f x w1 x w2 T and elimainate the variable y g to get a realization of P as ẋ = Ax + B 1 w + B 2 z = C 1 x + D 11 w + D 12 v = C 2 x + D 21 w + D 22 with A g B g 0 B g A = B f C g A f A w1 0, B 1 = 0 B f 0 0, B 2 = 0 B w1 B w2 C g 0 0 A w [ [ Dw2 Cg 0 0 C C 1 = w2 0, D 0 0 C w = 0, D 12 = D w1 C 2 = [ D f C g C f 0 0, D 21 = [ 0 D f, D22 = 0 Lectre 6 : (LFTs) 12/19

13 A Mass/Spring/Damper System The dynamical eqation of the system motion can be described by ẍ + c mẋ + k m x = F m Sppose m, c, and k are not known exactly, bt are believed to lie in known intervals as m = m ± 10%, c = c ± 20%, k = k ± 30% Introdcing pertrbations δ m, δ c, δ k [ 1, 1 F 1 ẍ 1 ẋ 1 x m(1 + 01δ m ) s s c(1 + 02δ c ) k(1 + 03δ k ) Lectre 6 : (LFTs) 13/19

14 A Mass/Spring/Damper System It is easy to check that 1 m can be represented as an LFT in δ m: 1 m = 1 01 = m(1 + 01δ m ) 1 m m δ m(1 + 01δ m ) 1 = F l (M 1, δ m ), M 1 = 1 m 01 m 1 01 F ẍ = ẋ 2 1 ẋ = x 2 1 x = x 1 M s s m v m δ m c c δ c v c 02 k k δ k v k 03 Lectre 6 : (LFTs) 14/19

15 A Mass/Spring/Damper System ẋ x 1 ẋ 2 v k v = k m c m 1 01 x 2 m 1 m 1 m m F 03 k , c 0 02 c k v c m k c m [ẋ1 = F ẋ l (M, ) x 1 x 2 2 F k c = v k v c m v m where M = k m c m 1 01 m 1 m 1 m m 03 k c , = δ k δ c δ m k c Lectre 6 : (LFTs) 15/19

16 Basic Principle Consider an inpt/otpt relation z = a + bδ 2 + cδ 1 δ dδ 1 δ 2 + eδ1 2 w := Gw where a, b, c, d, and e are given constants or transfer fnctions we wold like to write G as an LFT in terms of δ 1 and δ 2 We can do this in three steps: 1 Draw a block diagram for the inpt/otpt relation with each δ separated as shown in the next Figre 2 Mark the inpts and otpts of the δ s as y s and s, respectively (This is essentially plling ot the s 3 Write z and v s in terms of w and s with all δ s taken ot Lectre 6 : (LFTs) 16/19

17 Basic Principle a b v 4 δ 2 4 z c w δ 1 v 3 1 δ 3 2 v 1 e 2 δ 1 v 2 d Lectre 6 : (LFTs) 17/19

18 Basic Principle where M = 0 e d be bd + c 0 b 0 ae ad 1 a v 1 1 v 2 2 v 3 = M 3 v 4 4 z w, then z = F (M, )w, = [ δ 1 I δ 2 I 2 Lectre 6 : (LFTs) 18/19

19 Reference 1 Kemin Zho and John Doyle Essentials of Robst Control, Prentice Hall, Carsten Scherer Lectre note on Theory of Robst Control, 2001 Lectre 6 : (LFTs) 19/19