Lecture 7 : Generalized Plant and LFT form Dr.-Ing. Sudchai Boonto Assistant Professor

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1 Dr.-Ing. Sudchai Boonto Assistant Professor Department of Control System and Instrumentation Engineering King Mongkuts Unniversity of Technology Thonburi Thailand

Linear Quadratic Gaussian The state space methods for optimal controller design developed in the 1960s Linear Quadratic Gaussian (LQG) control was recognized by the Apollo people, and the Kalman

2 Linear Quadratic Gaussian The state space methods for optimal controller design developed in the 1960s Linear Quadratic Gaussian (LQG) control was recognized by the Apollo people, and the Kalman filter became the first embedded system. In 1970s were found to suffer from being sensitive to modelling errors and parameters uncertainty. There were a lot of failures of the method in practical application: a Trident submarine caused the vessel to unexpectedly surface in a simulation of rough sea, the same year F-8c crusader aircraft led to disappointing results. J. Doyle, Guaranteed Margins for LQG Regulators, IEEE Transactions on Automatic Control, Vol. 23, No. 4, pp , /30

3 Robust Control George Zames posed the problem of robust control, also known as H -synthesis. In the 1980s research activities turned to a new approach, where design objectives are achieved by minimizing the H 2 norm or H norm of suitable closed-loop transfer functions. The new method is closely related to the familiar LQG methods the computation of both H 2 and H optimal controllers involves the solution of two algebraic Riccati equations. More efficient methods for such a design have been developed in the 1990s. Instead of solving Riccati equations, one can express H 2 and H constraints as linear matrix inequalities (LMI). The major problem with modern H 2 and H optimal control is the controllers have the same dynamic order as the plant. (This problem has been solved, they claimed, by Pierre Apkarian and Dominikus Noll since 2006.) If the plant to be controlled is of high dynamic order, the optimal design results in controllers that are difficult to implement. Moreover, the high order may cause numerical problems. 3/30

4 The Concept of a Generalized Plant LQG control In modern control, almost any design problem is represented in the form shown in the below Figure. w u P v z K We can show the problem of designing an LQG controller in the generalized plant format. Consider a state space realization of the plant with transfer function G(s) corrupted by process noise w x and measurement noise w y. ẋ = Ax + Bu + w x y = Cx + w y, where w x and w y are white noise processes. 4/30

5 The Concept of a Generalized Plant LQG control w x w y r K(s) u G(s) y A regulation problem with r = 0. The objective is to find a controller K(s) that minimizes the LQG performance index [ 1 T ( ) ] V = lim E x T Qx + u T Ru dt T T 0 The state space realization of the generalized P. It has two inputs w and u, and two output z and v: ẋ = A p x + B w w + B u u z = C zx + D zww + D zuu v = C v x + D vw w + D vu u 5/30

6 The Concept of a Generalized Plant LQG control The measured output v of the generalized plant to be the control error e = y in the LQG problem. Take the control input u of the generalized plant to be the control input of the LQG problem. Relate the plant model and the generalized plant: A p = A, B u = B, C v = C, D vu = 0 Select [ ] [ ] Q 1/2 0 C z =, D zu = 0 R 1/2 Assume w = 0, the square integral of the fictitious output z is ( ) z T zdt = x T Qx + u T Ru dt 0 0 6/30

7 The Concept of a Generalized Plant LQG control Assume that w is a white noise process satisfying E [ w(t)w T (t + τ) ] = δ(τ)i, and choose [ ] [ ] B w = Q 1/2 e 0, D vw = 0 Re 1/2 Then [ ] [ w x = B ww = Q 1/2 w 1 e 0 w 2 [ ] [ ] w y = D vw w = 0 Re 1/2 w 1 w 2 ] = Q 1/2 e w 1 = R 1/2 e w 2 It is easy to see that minimizing [ 1 T ] lim E z T (t)z(t)dt T T 0 is equivalent to minimizing the LQG performance index V. 7/30

8 The Concept of a Generalized Plant LQG control The transfer function of a generalized plant that represents the LQG problem is P (s) = [ A p C p B p D p ] = A [ ] Q 1/2 0 C [ ] Q 1/2 e 0 0 [ ] 0 Re 1/2 B [ ] 0 R 1/2 0 where 0 stand for zero matrix blocks of appropriate dimensions. In MATLAB, use a command P = ss(ap, Bp, Cp, Dp) 8/30

9 The Concept of a Generalized Plant LQG control - reference tracking r w 1 Q 1/2 e w 2 R 1/2 e w x w y Q 1/2 z1 R 1/2 z 2 u B 1 s x C e A P (s) K(s) ] T the external input w = [r w 1 w 2 ] T the fictitious output z = [z 1 z 2 9/30

10 Aircraft control Assuming the state space model represents a linearized model of the vertical-plane dynamics of an aircraft is described below: A = B = , C = , D = /30

11 Aircraft control u 1 spoiler angle (in 0.1 deg) u 2 forward acceleration (in m s 2 ) u 3 elevator angle (in deg) x 1 relative altitude (in m) x 2 forward speed (in m s 1 ) x 3 pitch angle (in deg) x 4 pitch rate (in deg s 1 ) x 5 vertical speed (in m s 1 ) The design objectives are: fast tracking of step changes for all three reference inputs, with little or no overshoot control input must satisfy u 3 < 20. Hint: use H 2 control synthesis command, K = h2syn(gplant, nmeas, ncont), of MATLAB We will discuss how this function work later. 11/30

12 Aircraft control Design example 1.5 System outputs y(t) Control inputs u(t) This result has been done by selecting: R = I, Re = 0.1I, Q = C, and Qe = B. Noting that we did not use an integrator. 12/30

13 Feedback Structure d i d r e K u u g G y z n The standard feedback configuration 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 output disturbance and plant input disturbance u g and y are plant input and output. 13/30

14 Standard Problem: P K-Structure w P z u v K P cl [ ] z = v [ ] P11 P 12 P 21 P 22 }{{} P [ ] w u external inputs: w external outputs: z controller input: v controller output: u 14/30

15 Transformation into Standard Problem For any control structure, perform the following steps: Collect all signals that are evaluated for performance into the performance vector z Collect all signals from outside into generalized disturbance vector w Collect all signals that are fed to K into generalized measurement vector v Denote output of K by u Cut out K Determine transfer matrix [ ] [ ] z w = P v u 15/30

16 Standard Problem Example For the classical control d i d r e K u u g G y z n z = d + G(d i + u) v = e = r (n + z) = r n d G(d i + u) r [ [ ] d z 0 I G 0 G = d v] I I G I G i n u [ ] [ 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 ] 16/30

17 Standard Problem P K-Structure Procedure leads to standard problem or the P K-Structure: w u P K v z [ ] [ ] [ ] z P 11 P 12 w = v P 21 P 22 u }{{} 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) 17/30

18 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 F (s) = α + βs(1 γs) 1 for some λ, β and γ C. 18/30

19 Linear Fractional Transformations Lower linear fractional transformation The lower LFT with respect to l is defined as w 1 z 1 P [ ] [ ] z 1 w 1 = P = v 1 u 1 u 1 v 1 u 1 = l v 1 l [ A C ] [ ] B w 1 D u 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. 19/30

20 Linear Fractional Transformations Upper linear fractional transformation The upper LFT with respect to u is defined as u 2 u v 2 [ ] [ ] v 2 u 2 = P = z 2 w 2 [ A C ] [ ] B u 2 D w 2 w 2 P z 2 u 2 = uv 2. F u(m, u) = u [ A C ] B := D + C(I ua) 1 ub D = D + C u (I A u ) 1 B, provided that the inverse (I u A) 1 exists. 20/30

21 Linear Fractional Transformations Example uf W1 d e K u G yg W2 f F n w u P K v z w = [ d z = [ f [ [ z w = P = v] u] n ] T u f ] T W 2 G 0 W 2 G 0 0 W 1 F G F F G d n u 21/30

22 Linear Fractional Transformations Example Assuming that the plant G is strictly proper and P, F, W 1, and W 2 have the following state-space realizations: That is [ Ag B g G = C g 0 [ Aw1 B w1 W 1 = C w1 D w1 ẋ g = A gx g + B g(d + u), ẋ f = A f x f + B f (y g + n), ] [ Af B, F = f C f D f ], W 2 = y g = C gx g ], [ Aw2 B w2 C w2 D w2 y = C f x f + D f (y g + n), ẋ w1 = A w1 x w1 + B w1 u, u f = C w1 x u + D w1 u, ẋ w2 = A w2 x w2 + B w2 y g, f = C w2 x w2 + D w2 y g. ] 22/30

23 Linear Fractional Transformations Example Define a new state vector x = [ ] T x g x f x w1 x w2 and elimainate the variable y g to get a realization of P as ẋ = Ax + B 1 w + B 2 u z = C 1 x + D 11 w + D 12 u v = C 2 x + D 21 w + D 22 u 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. 23/30

24 Linear Fractional Transformations A Mass/Spring/Damper System The dynamical equation of the system motion can be described by ẍ + c m ẋ + k m x = F m. Suppose m, c, and k are not known exactly, but are believed to lie in known intervals as m = m ± 10%, c = c ± 20%, k = k ± 30% Introducing perturbations δ m, δ c, δ k [ 1, 1]. F 1 m( δ m) ẍ 1 s ẋ 1 s x c( δ c) k( δ k ) 24/30

25 Linear Fractional Transformations A Mass/Spring/Damper System It is easy to check that 1 m can be represented as an LFT in δ m: 1 m = = m( δ m) 1 m m δm( δm) 1 = F l (M 1, δ m), M 1 = 1 m 0.1 m F M ẍ = ẋ 2 1 s ẋ = x 2 1 s x = x 1 u m v m δ m c u c δ c v c 0.2 k u k δ k v k /30

26 Linear Fractional Transformations A Mass/Spring/Damper System x ẋ 1 1 ẋ 2 k m c m x 2 m 1 m 1 m m v k v = F 0.3 k , u k c c u v c m k c u m ] [ẋ1 = F ẋ l (M, ) x 1 x 2 2 F u k u c = v k v c u m v m where M = k m c m 1 m 1 m 1 m 0.1 m 0.3 k c k c , = δ k δ c δ m 26/30

27 Linear Fractional Transformations Basic Principle Consider an input/output 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 functions. we would 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 input/output relation with each δ separated as shown in the next Figure. 2. Mark the inputs and outputs of the δ s as y s and u s, respectively. (This is essentially pulling out the s 3. Write z and v s in terms of w and u s with all δ s taken out. 27/30

28 Linear Fractional Transformations Basic Principle a b v 4 δ 2 u 4 z c w v 1 δ 1 u 1 v 3 δ 2 u 3 e u 2 δ 1 v 2 d 28/30

29 Linear Fractional Transformations Basic Principle v 1 u 1 v 2 u 2 v 3 = M u 3 v 4 u 4 z w where M = 0 e d be bd + c 0 b 0 ae ad 1 a, then z = F u (M, )w, = [ ] δ 1 I δ 2 I 2 29/30

30 Reference 1 Herbert Werner Lecture Notes on Control Systems Theory and Design, Mathwork Control System Toolbox: User s Guide, Kemin Zhou and John Doyle Essentials of Robust Control, Prentice Hall, /30

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

Lecture 6 : Linear Fractional Transformations (LFTs) Dr.-Ing. Sudchai Boonto Lectre 6 : (LFTs) Dr-Ing Sdchai Boonto Department of Control System and Instrmentation Engineering King Mongkts Unniversity of Technology Thonbri Thailand Feedback Strctre d i d r e y z K G g n The standard