Trajectory planning for the control of distributed parameter systems - PDF document

Trajectory planning for the control of distributed parameter systems with lumped controls Joachim Rudolph Institut für Regelungs- und Steuerungstheorie TU Dresden En l’honneur du 60 e anniversaire de Michel Fliess Joachim Rudolph (TU Dresden) 1 / 50 Aim: Transfer of a distributed parameter system with lumped controls ◮ between two regimes (most often stationary) ◮ following a reference trajectory. Tasks: Calculating possible trajectoires, including the (open-loop) control, (close the loop by feedback). Approach followed: extension of the flatness based control design for nonlinear finite dimensional systems (ode’s) and the module theoretic approach to linear systems. Work initiated together with H. Mounier, M. Fliess, P . Rouchon (1995), and done at Dresden with A. Lynch, F. Woittennek, J. Winkler, . . . Joachim Rudolph (TU Dresden) 2 / 50

Contents Wave Eq. and its Relatives (Hyperbolic Systems) Torsion of an Elastic Rod with a Tip Load Heavy Ropes in Horizontal Motion Heat Eq. and its Relatives (Parabolic Systems) Fundamentals on a Heat Conduction Example Examples of Beams and Plates Formal Framework for Linear Systems with Lumped Controls VGF Crystal Growth: A Quasi-linear Problem with Free Boundary Outline of a Flatness Based Approach and Simulation Flatness Based Parameterization for the Crystallization Stage Joachim Rudolph (TU Dresden) 3 / 50 Wave eq./hyperbolic syst. Ex.: Rod under torsion Elastic rod with tip load q ( x , t ) x 0 Math. model (wave eq.): ρ ∂ 2 q ∂ t 2 ( x , t ) = G ∂ 2 q 0 ≤ x ≤ l , t ≥ 0 ∂ x 2 ( x , t ) , ∂ q q ( x , 0 ) = 0 , ∂ t ( x , 0 ) = 0 Initial cond.: Boundary cond. (Torque µ ( t ) at x = 0 ): ∂ x ( l , t ) = − Θ ∂ 2 q GI ∂ q GI ∂ q ∂ x ( 0 , t ) = − µ ( t ) , ∂ t 2 ( l , t ) Aim: Turn the load in finite time following a trajectory for the angle while avoiding final oscillations. Joachim Rudolph (TU Dresden) 5 / 50

Wave eq./hyperbolic syst. Ex.: Rod under torsion With the derivation operator s one gets the boundary value pb. in x : G d 2 ˆ q GI d ˆ q GI d ˆ q dx 2 ( x ) − ρ s 2 ˆ dx ( l ) = − Θ s 2 ˆ q ( x ) = 0 , dx ( 0 ) = − ˆ µ ( s ) , q ( l ) � ρ/ G ) : Solution (with σ = q ( x ) = A exp ( σ s x ) + B exp ( − σ s x ) ˆ q ( x ) = C 1 cosh ( σ s ( x − l )) + C 2 sinh ( σ s ( x − l )) or, better, ˆ One wants to follow the angular trajectory t �→ y ( t ) = q ( l , t ) : y = ˆ q ( l ) = C 1 ˆ Boundary cond.s at x = l give GI d ˆ q dx ( l ) = GI C 2 σ s = − Θ s 2 ˆ q ( l ) = − Θ s 2 ˆ y whence the parameterization cosh ( σ s ( x − l )) − Θ s � � GI σ sinh ( σ s ( x − l )) q ( x ) = y ( s ) ˆ ˆ Joachim Rudolph (TU Dresden) 6 / 50 Wave eq./hyperbolic syst. Ex.: Rod under torsion With the shift (or delay) operator exp ( − σ ( x − l ) s ) : q ( x ) = 1 � Θ � e σ ( x − l ) s + e − σ ( x − l ) s − � e σ ( x − l ) s − e − σ ( x − l ) s � s y ˆ ˆ 2 GI σ Solution in t q ( x , t ) = 1 � y ( t + σ ( x − l )) + y ( t − σ ( x − l )) 2 Θ �� y ( t + σ ( x − l )) − ˙ y ( t − σ ( x − l )) � − ˙ GI σ µ = − GI d ˆ q dx ( 0 ) at x = 0 : The control follows from the gradient ˆ � GI σ − Θ �� � � � µ ( t ) = − y ( t − σ l ) − ˙ y ( t + σ l ) y ( t − σ l ) + ¨ y ( t + σ l ) ˙ ¨ 2 2 Choice of the trajectory for load position y by T = σ l ahead! ⇒ Lumped variable y is the free parameter: “basic (or flat) output”. Joachim Rudolph (TU Dresden) 7 / 50

Wave eq./hyperbolic syst. Ex.: Heavy Ropes Heavy rope Math. model: − ∂ 2 w gx ∂ w ∂ � � ∂ t 2 ( x , t ) = 0 , x ∈ [ 0 , L ] , t ≥ 0 ∂ x ( x , t ) u ( t ) ∂ x x = L Initial cond.: ∂ w w ( x , 0 ) = 0 , ∂ t ( x , 0 ) = 0 , x ∈ [ 0 , L ] Boundary cond.: x ∂ w ∂ x ( 0 , t ) = 0 , t ≥ 0 w ( L , t ) = u ( t ) , x = 0 w ( x , t ) Aim: Horizontal transport without residual oscillations at the arrival Joachim Rudolph (TU Dresden) 8 / 50 Wave eq./hyperbolic syst. Ex.: Heavy Ropes P .d.e. (recall): − ∂ 2 w gx ∂ w ∂ � � ∂ t 2 ( x , t ) = 0 , x ∈ [ 0 , L ] , t ≥ 0 ∂ x ( x , t ) ∂ x Replacing derivation w.r.t. t by s yields o.d.e. d gx d ˆ w � � − s 2 ˆ w ( x ) = 0 , x ∈ [ 0 , L ] dx ( x ) dx with boundary cond. d ˆ w dx ( 0 ) = 0 . w ( L ) = ˆ u , ˆ Transformation of the independent variable z = 2 s x / g , x ∈ ( 0 , L ] , � (and limit for x = 0 ) yields modified Bessel equation z 2 d 2 ˆ dz 2 + z d ˆ w w dz − ( z 2 + n 2 ) ˆ w = 0 , with n = 0 Joachim Rudolph (TU Dresden) 9 / 50

Wave eq./hyperbolic syst. Ex.: Heavy Ropes Modif. Bessel eq. ( n = 0 , recall): z 2 d 2 ˆ dz 2 + z d ˆ w w dz − z 2 ˆ w = 0 Solution formula (with z = 2 s � x / g ) w ( x ) = C 1 I 0 ( 2 s x / g ) + C 2 K 0 ( 2 s � � x / g ) ˆ where I 0 and K 0 are Bessel fct.s of order 0 of the 1st and 2nd kind Boundary condition at x = 0 : 0 = C 1 I 1 ( 0 ) + C 2 K 1 ( 0 ) , I 1 ( 0 ) = 0 , K 1 ( x ) → ∞ for x ↓ 0 ⇒ C 2 = 0 w ( 0 ) ⇒ C 1 = ˆ y = ˆ y , thus: Position of free end ˆ w ( x ) = I 0 ( 2 s � x / g ) ˆ y ˆ Condition at x = L : u = I 0 ( 2 s � L / g )ˆ y ˆ w ( x ) and ˆ u : System equation, relating ˆ I 0 ( 2 s w ( x ) = I 0 ( 2 s � L / g ) ˆ � x / g ) ˆ u Joachim Rudolph (TU Dresden) 10 / 50 Wave eq./hyperbolic syst. Ex.: Heavy Ropes w ( x ) = I 0 ( 2 s x / g ) ˆ y � Recall: ˆ Representation of the function I 0 by Poisson integral I 0 ( x ) = 1 π � 2 exp ( x sin θ ) d θ π − π 2 yields y = 1 π � 2 I 0 ( 2 s exp ( 2 s x / g sin θ )ˆ � x / g )ˆ � y d θ π − π 2 With the exponential fct. as shift operator of amplitude 2 x / g sin θ : � w ( x , t ) = 1 π � 2 y ( t + 2 x / g sin θ ) d θ � π − π 2 where the basic output y ( t ) = w ( 0 , t ) is the position of the free end. We need values of y between t − 2 x / g and t + 2 x / g � � ⇒ distributed delay and advance (prediction). Joachim Rudolph (TU Dresden) 11 / 50

Wave eq./hyperbolic syst. Ex.: Heavy Ropes Simulation of the heavy rope Joachim Rudolph (TU Dresden) 12 / 50 Wave eq./hyperbolic syst. Ex.: Heavy Ropes Two ropes attached on one trolley Math. model, with lengths L 1 � = L 2 and deviations from vertical w 1 , w 2 I 0 ( 2 s w 1 ( x ) = I 0 ( 2 s � L 1 / g ) ˆ � x / g ) ˆ u I 0 ( 2 s w 2 ( x ) = I 0 ( 2 s � L 2 / g ) ˆ � x / g ) ˆ u w ( 0 ) , then w ( x ) for a ficticious rope of length L 1 + L 2 and ˆ y = ˆ Introduce ˆ w ( x ) = I 0 ( 2 s � x / g ) ˆ y ˆ Now w ( L 2 ) = I 0 ( 2 s w 1 ( 0 ) , w ( L 1 ) = I 0 ( 2 s w 2 ( 0 ) � L 2 / g ) ˆ y =: ˆ � L 1 / g ) ˆ y =: ˆ ˆ ˆ whence w 1 ( x ) = I 0 ( 2 s x / g ) I 0 ( 2 s w 2 ( x ) = I 0 ( 2 s x / g ) I 0 ( 2 s � � L 2 / g ) ˆ y , � � L 1 / g ) ˆ y ˆ ˆ u = I 0 ( 2 s L 1 / g ) I 0 ( 2 s u L = I 0 ( 2 s � � L 2 / g ) ˆ y , � ( L 1 + L 2 ) / g ) ˆ y ˆ and ˆ Joachim Rudolph (TU Dresden) 13 / 50

Wave eq./hyperbolic syst. Ex.: Heavy Ropes Simulation of two ropes Joachim Rudolph (TU Dresden) 14 / 50 Wave eq./hyperbolic syst. Ex.: Heavy Ropes Related problems ◮ Telegrapher’s equation ◮ Timoshenko beam equation ◮ Heat Exchangers ◮ Networks of strings and Timoshenko-beams ◮ etc. Joachim Rudolph (TU Dresden) 15 / 50

Parabolic syst./beams/plates Ex.: Heat Eq. Heat conduction problem u ( t ) T ( x , t ) x x = 0 x = − L Math. Modell: λ ∂ 2 T ∂ x 2 ( x , t ) = ( ρ c ) ∂ T − L ≤ x ≤ 0 , t ≥ 0 ∂ t ( x , t ) , T ( x , 0 ) = 0 Initial cond.: Boundary cond.: ∂ T λ ∂ T ∂ x ( 0 , t ) = 0 , ∂ x ( x , t ) = β T ( − L , t ) − u ( t ) � � Task: Finite time transition between stationary temperatures, using the boundary control u . Joachim Rudolph (TU Dresden) 17 / 50 Parabolic syst./beams/plates Ex.: Heat Eq. Using derivation operator s yields ordinary boundary value pb.: λ d 2 ˆ T d ˆ T − λ d ˆ T T ( x ) = 0 , dx ( 0 ) = 0 , dx 2 ( x ) − ( ρ c ) s ˆ dx ( − L ) + β ˆ T ( − L ) = β ˆ u Solution: �� � �� � ( ρ c ) s ( ρ c ) s T ( x ) = C 1 cosh + C 2 sinh ˆ x x λ λ Evaluating boundary cond.: �� � �� � d ˆ � � T ( ρ c ) s ( ρ c ) s ( ρ c ) s ( ρ c ) s sinh cosh dx ( x ) = C 1 x + C 2 x λ λ λ λ d ˆ T dx ( 0 ) = 0 ⇔ C 2 = 0 − λ d ˆ T dx ( − L ) + β ˆ T ( − L ) = β ˆ u � �� � �� �� � ( ρ c ) s ( ρ c ) s ( ρ c ) s sinh + β cosh C 1 ( − L ) ( − L ) u ⇔ − λ = β ˆ λ λ λ Joachim Rudolph (TU Dresden) 18 / 50