QFT NO LINEAL. Alfonso Baños. Jornadas Red Temática Ingeniería de Control sobre Control No Lineal. Sevilla, 3 de Marzo de 2004

Transcription

1 QFT NO LINEAL Alfonso Baños Sevilla, 3 de Marzo de 2004 Jornadas Red Temática Ingeniería de Control sobre Control No Lineal 1

2 INDICE DE LA EXPOSICIÓN PARTE II: Nonlinear QFT 1 The robust control problem for SISO nonlinear plants Nonlinear Plant Modelling Problem 2 Robust control design: stability Robust stability Example Asymptotic tracking Nonlinear control proble QFT solution of the nonlinear control problem QFT solution of the linear equivalent problem 4 Robust control design 5 Conclusions and open problems 2

3 1 The robust control problem Nonlinear plant model A parameterized set of nonlinear systems Each element must be invertible P { Θ} = θ P θ Problem TDF feedback structure F - G Pσ θ Problem (robust stability and tracking): Given a set of references r, a set of acceptable outputs A r, designing a controller (F,G) for the closed loop system to be stable its output is in the set A r, for each P θ, θ Θ 3

4 2 Robust control design:stability Robust stability Assumption: The control system may be separated in a linear block K(s) and a nonlinear block N Σ K N Σ Conicity Analysis (Circle Criterion) Description of N: Sector [a,b] Result: Restrictions on K(jω) for L p -stability Passivity Analysis (Popov Criterion) Description of N: Sector [0,b] Result: Restrictions on K(jω) for L 2 -stability 4

5 2 Robust control design:stability Example: nonlinear system (a 1 [0.1,0.4], a 2 [0.3,0.5], m 1 = m 3 = 0.1 y m 2 [6,10]) P(s) m 1 m 2 m 3 a 1 a 2. Robust stability Boundaries (Circle Criterion) Magnitude (db) Phase (degrees) -40 5

6 Asymptotic properties (tracking) If stability via the Circle Criterion is assumed, the closed loop output has a unique value in its steady-state (Sandberg 92). More general condition: fading memory (Boyd and Chua 85): If a system has FM with respect to a set of inputs K, and X is the set os reachable states for any input in K, then there exist a unique steady-state for anu input in K, and for each initial condition in X Under what conditions a system has FM? 6

7 Asymptotic tracking Example: RC circuit with nonlinear resistence, a [0,2) 2 y& ( t) + (1 + ay ( t)) y( t) = u( t), y(0) = 0 uˆ( s) = G( s)( F( s)ˆ( r s) yˆ( s)) Asymptotic behaviour: 3 ay ( ) + ( 1+ G( 0)) y( ) G( 0) F( 0) r( ) = 0 2 (1 + ay ) y u = G0 ( F r 0 = u y ) Using root locus (G 0 = 12, F 0 = 1.18, r = 1) 1 11 ( + Gy 1 0) GFr 0 0 +, [ 05., ] 3 a y a Im ag Ax is a = 2 a = Real Axis 7

8 Nonlinear Control Problem Given the specification y [y l,,y u, ] for the reference r, F 0 y G 0 must satisfy the inequality GF 1 + G T : = u, 0 y u, r and also the real solution of must satisfy y r Tl, 0: = y + ( 1+ G) y GFr = 0 y l, r 8

9 QFT Solution to the nonlinear control problem Definition of the set of acceptable outputs: Acceptable values for r = 1 are y A 1 = [.,.] 0911 Computation of the equivalent linear family (ELF 0 ): y 2 ELF0 = u = ( 1+ ay ) y, y A1, a = u The close loop mapping ϕ a : [ 0,2 ] [0.29,1 ] y y PG 0 0F0 = PG r P ELF 1 + L,, 0 0 L, 0 0 y ay y GF ( 1+ ) = y 1+ ( 1+ ay ) y G 2 0 GF 0 0 = 2 = : ϕa( y ) 1 + ay + G 0 9

10 QFT Solution to the nonlinear control problem Design is based on application of Schauder fixed point theorem: Acontinuous map of a convex and compact subset of a Banach space on itself has a fixed point. Example: A 1 = [0.9,1.1] a convex and compact subset of R ϕ is continuous over A 1 If the linear equivalent problem is solved for G 0 and F 0, then ϕ(a 1 ) A 1 Result: There exists a fixed point y *, = ϕ a (y *, ), for each value of a. y *, is the output of the nonlinear control system y *, A 1 G 0 y F 0 son are a valid design for the nonlinear problem!! Nonlinear robust control problem linear equivalent robust control problem 10

11 QFT solution to the equivalent linear control problem Equivalent problem: to compute G 0 y F 0 such that for each P 0 [0.29,1], Using logs, PG 0 0F0 1+ PG 0 0 [ 0911.,.] G0 0.29G0 log( ) log( ) < G G G0 1 + G0 0.9 < F0 < G0 G0 Solution (more economic ): G 0 = 10.1, F 0 = 1.2 Another solution wiht a SDF Problema equivalente : calcular G 0 tal que para cualquier P 0 [0.29,1], PG PG 0 0 [ 0911.,.] Solución más económica : G 0 = 31 11

12 Solution of the nonlinear control problem The QFT solution of the equivalent linear problem is a solution of the original nonlinear control problem (Schauder fixed point theorem) a ELF 0 y L, y *, 0 {1} {1.09} [0.62,0.71] [1.04,1.06] [0.45,0.55] [0.99,1.02] [0.35,0.45] [0.93,0.98] [0.29,0.38] [0.90,0.95] The solution is not overly conservative (in this case)! 12

13 Solution to the nonlinear control problem: more general references Specification (linear): Ar = [ 0911.,.] r Linear equivalent Problem: y ELF u = (1 + ay u Given α r, to compute G 0 and F 0 such that [ 0,2] = [ α,1] 2 0 = ) y, y A r, a r G0 α rg0 log( ) log( ) < 0.087, α r 1 + G0 1 + α rg0 0.9(1 + α rg0) 1.1(1 + G0) r < F0 < r α rg0 G0 Solution: r ELF 0 G 0 F 0 [0,1] [0.29,1] [0,3] [0.17,1] [0,5] [0.016,1]

14 4 Robust control design: transitory QFT solution to the nonlinear control probleml Definition of the set of acceptable outputs: A R := {A r, r R} Computation of the linear equivalent family (ELF): y yˆ( s) 1 ELF= Pθ ( s) = u = s ( y), y YR,θ Θ uˆ( s) The closed loop operator ϕ a : r F Controller K - Σ G u + y P y L ϕ : A A, y = ϕ ( y) r, θ r r L r, θ Transformation to a linear equivalent problem : ϕ θ If r, is continuous over A r, a compact and convex subset of a Banach space Y S (representing stable signals), then a valid controller for ELF is also valid for the original nonlinear plant P 14

15 4 Robust control design: transitory Solución QFT al problema de control lineal equivalente (Ejemplo: Circuito RC no lineal) Especificación: A r = 8t 12t { y() t 0.9( 1 e ) u () t y() t 1.1( 1 e ) u () t } Problema lineal equivalente : s 1+ G5() s = 12 7 s s ( 1+ )( 1+ ) 3 70 s s ( 1+ )( 1+ F5 ( s) ) = s s s ( 1+ )( 1+ )( 1+ ) Time-sec 15

16 5 Conclusions and open problems The structure of the uncertainty is directly related with the control effort and the complexity of the control problem: A bigger structure means lesser control effort A bigger structure means more mathematical complexity QFT gives a quantitative relationship between the quantity of feedback and the system uncertainty QFT gives un good balance between the structure level and the complexity of the resulting problem QFT solves the scalar LTI problem very efficiently QFT solves the nonlinear control problem, solving an equivalent linear problem, using as basis the Schauder fixed point theorem 16

17 5 Conclusiones y problemas abiertos Conjetura: En el caso no lineal, el controlador será menos conservativo dependiendo de la incertidumbre Líneas básicas de desarrollo futuro: Derive weakest conditions for a (uncertain) nonlinear plant to be properly controlled by a linear controller (high frequency behaviour appears to be crucial) Computation of ELF (discretization of infinite sets) Computation of ELF using a Volterra Series description of the nonlinear plant Use other definitions of ELF Use other Fixed Point Results to relaxed acceptable sets Uso de técnicas de linealización: pseudolinealización, etc. 17