Lifted approach to ILC/Repetitive Control Okko H. Bosgra Maarten Steinbuch TUD Delft Centre for Systems and Control TU/e Control System Technology Dutch Institute of Systems and Control DISC winter semester 2003/2004
Contents Formulation lifted approach Internal model principle Design of ILC without Q-filter using LQ theory Analysis of existing design results Analysis of Repetitive Control Properties, convergence, performance
classical ILC structure 1 L(z) Q(z) f kn C(z) z N f k P (z) y ref ε k y k d Update in trial space: f kn = Q(z)(f k L(z)ε k )
classical ILC structure 2 y ref, d assumed to be periodic inputs, period length N ε k f k d y ref C(z) P (z) y k control input f k, disturbance input y ref d, output ε k input/output relationship: ε k = (I P C) 1 (y ref d) (I P C) 1 P fk
System formulation 1 y ref d S(z) f k P (z) ε k Define P (z) : = (I P C) 1 P 1 S(z) : = (I P C)
System formulation 2 Better to represent these relations in the time-domain: (A p, B p, C p ) minimal realization for P (z), (A c, B c, C c, D c ) minimal realization for C(z): P (z) = C p (zi A p ) 1 B p C(z) = C c (zi A c ) 1 B c D c i.e. P (z) strictly proper, C(z) proper. State space representation for joint feedback system: x k1 = Ax k Bf k N r k where r k = y ref d ε k = Cx k r k
System formulation 3 ( ) ( ) ( ) A = C = ( A p B p D c C p B p C c B c C p A c ) C p 0 N = B p D c B c B = B p 0 r k f k A, B, C, N ε k
System formulation 4 Split up of system: r k periodic, r k asymptotically periodic r k A, N, C f k A, B, C r k ε k Simplification: consider r k as being periodic
System formulation 5 r k f k A, B, C ε k Servo problem ILC: x 0 = 0, x N = 0, x 2N = 0, x 3N = 0,... typical pattern of r k : pick and place r k 0 N 2N 3N time
System formulation 6 Periodic disturbance suppression Repetitive Control typical pattern of r k : r k 0 N 2N 3N time e.g. drive control with nonuniform torque pattern x 0 = 0, x N, x 2N, x 3N,... follow from RC algorithm Thus ILC and RC require slightly different formulations
System formulation in trial domain 1 Causal LTI system (A, B, C) yields impulse response h k R m l h k = CA k 1 B k = 1, 2, 3,... = 0 k 0 The system has m outputs, l inputs Let j = 0, 1, 2,... denote the trial number and define y j = y Nj y Nj1 y Nj2. f j = f Nj f Nj1 f Nj2. r = r 0 r 1 r 2. ε j = ε Nj ε Nj1 ε Nj2. y NjN 1 f NjN 1 r N 1 ε NjN 1
System formulation in trial domain 2 The lifted system representation in trial domain is now: x NjN = F x Nj Gf j y j = Hx Nj Jf j ε j = y j r where h 0 0 0...... 0 h 1 h 0 0...... 0 h 2 h 1 h 0... 0 J =........................ 0 h N 1...... h 2 h 1 h 0 ( ) G = A N 1 B A N 2 B... AB B C CA H = CA 2. F = A N CA N 1
System formulation in trial domain 3 time invariance J is Toeplitz causality J is lower triangular Toeplitz matrix J represents convolution (A, B, C) minimal (F, G, H) minimal In trial domain, r is constant disturbance vector For ILC, x k = 0 for k = Nj, j = 0, 1, 2,... system model reduces to ε j = Jf j r
Internal model principle 1 Asymptotic rejection of constant disturbance provided that 1. Error-driven dynamic disturbance model is added to controller 2. Each controller disturbance mode can propagate throught plant, i.e. is not cancelled against transmission zero 3. Dynamics of plant and controller are asymptotically stabilised by feedback Vector-valued constant disturbance r generated in trial domain by integrator having initial value ic = r z 1 r ic
Internal model principle 2 Adding disturbance model in feedback loop with gain L: r f j1 f j x j1 x j y j ε j z 1 G z 1 H F J L Asymptotic rejection: ε j 0 for j while r = constant 0 Then y j must compensate r y j r for j
Internal model principle 3 Only possible if rank of steady-state gain of (F, G, H, J) is Nm and if f j has at least dimension Nm Closed-loop state-space model in trial domain: ( ) ( ) ( ) ( ) f j1 I Nl LJ LH f j L = r G F 0 x j1 ε j = ( ) ( ) f j J H r Closed-loop system matrix: ( ) ( ) ( I Nl LJ LH I Nl 0 = G F G F x j x j I Nl 0 ) ( ) L J H
Internal model principle 4 System is controllable if controllability matrix has rank N l n: [ ] I Nl I Nl I Nl I Nl... rank = 0 G G F G G F G F 2 G... [ ] I Nl 0 0 0... rank = n Nl 0 G F G F 2 G... System is observable if observability matrix has rank N l n: J H I Nm 0 [ ] J HG HF rank J HG HF G HF 2 = rank I Nm H J H I Nm H F H G F I n.... This requires both matrices to have rank at least Nl n:
Internal model principle 5 I Nm 0 I Nm 0 0 H I rank Nm H I Nm H F H = rank 0 F H = Nm n 0 F 2 H.... This requires m l, i.e. number of outputs number of inputs [ ] [ ] J H J H(F I n ) 1 rank = rank = G F I n G I n [ ] J H(I n F ) 1 G 0 rank = nrank (J H(zI n F ) 1 G) 0 I z=1 n Nm Nl plant steady-state gain matrix must have rank Nl
Internal model principle 6 If (F, G, H) is of order zero (ILC case) then the Nm Nl matrix J must have rank Nl, i.e. its columns must be linearly independent If (F, G, H) is of order greater than zero (RC case) then we may assume F = A N 0 Then the requirement is rank(j HG) = Nl where C h N h 3 h 2 h 1 CA. ( )...... h3 h 2 HG = CA 2 A N 1 B... AB B =.......... h3............ CA N 1 h 2N 1 Contribution to rank from upper right influence of previous trial h N
Internal model principle 7 Result: Suppose L asymptotically stabilizes system. Then disturbance r is asymptotically rejected if m l. Proof: L asymptotically stabilizes all dynamics ( ) λ I Nl LJ LH i G F < 1 i = 1, 2,..., Nl n which implies ( ) LJ LH rank = Nl n G I n F i.e. this (Nl n) (Nl n) matrix is invertible For asymptotically stable transfer function matrix P (z), the steady state gain matrix is P (z) z=1
Internal model principle 8 Steady state gain matrix between r and ε j : ( ) [ ( )] 1 ( ) I Nl LJ LH L I Nm J H I Nln = G F 0 ( ) ( ) 1 ( ) LJ LH L = I Nm J H =: Ω G I n F 0 The rank of this steady state gain matrix Ω follows from I Nm J H I Nm Γ 1 Γ 2 rank L LJ LH = rank L I Nl 0 0 G I n F 0 0 I n where ( ) ( ) 1 LJ LH Γ = J H G I n F
The rank of this expression now equals Ω 0 0 rank 0 I Nl 0 = rank Ω Nl n 0 0 I n Internal model principle 9 From its structure, the rank of the first matrix is I Nm J H rank L LJ LH Nm n 0 G I n F so that rank Ω Nm Nl There is asymptotic rejection of r if this rank is zero, i.e. for m l, or number of inputs number of outputs
Internal model principle: conclusion 10 J has dimension Nm Nl. Feedback L requires: system controllable: always system observable if m l and rank(j HG) = Nl i.e. HG can contribute in making J full rank disturbance rejection if m l Thus requirements: m = l and (J HG) square and full rank If not satisfied: not all modes asymptotically stable r not fully compensated
Iterative learning control 1 For ILC, the initial state in each trial is zero, which applies in machine operations like pick-and-place tasks. Thus the system is r f j1 f j y j ε j z 1 J L
Iterative learning control 2 Controller contains all dynamics in system no Q-filter thus uncompromised convergence J square, not necessarily of full rank convergence if eigenvalues of I LJ smaller than 1 faster convergence for smaller eigenvalues of I LJ L can be time-varying, non-causal, i.e. no reason to restrict L to be Toeplitz additive noise disturbances may act on J If J full rank, feedback system is first order with state feedback good robustness properties attainable
Iterative learning control 3 System relations: f j1 = (I Nl LJ)f j Lr (1) ε j = Jf j r (2) Asymptotically stable if λ i (I Nl LJ) < 1 i = 1, 2,..., Nl This implies that LJ is non-singular, so that non-singularity of LJ is necessary in order that the system is asymptotically stable. Transfer function matrix betreen r and ε j (sensitivity function) ε j = [I Nl J(zI Nl I Nl LJ) 1 L]r For z = 1 we have steady state gain I Nl J( LJ) 1 L
Iterative learning control 4 Thus LJ non-singular and steady-state gain zero requires J and L to be square and non-singular If J singular: stabilization and full rejection of r not possible Two reasons for loss of rank for J: strictly proper system, or additional delays non-minimum phase zeroes in i/o behaviour of system If Markov parameters h 0, h 1, h 2,..., h d 1 are zero, then rank J is at most Nm d In that case: no full stabilization, no full rejection
ILC example 1 0 1 r = 1 1 4 0 0 0 0 im J = im 1 0 = im 1 0 1 1 0 1 2 J = J is output matrix in system with unit system matrix third mode unobservable feedback has no effect on pole location of third mode 0 0 0 1 0 0 1 1 0
ILC example 2.1 H(z) = z 2 z(z 1) = 2 z 3 z 1 [ ] zi A B = C zero at z = 2 0 1 impulse response h = 3 3 3 z 0 2 0 z 1 3 1 1 1 1 0 1 observability matrix H = 0 1 0 1 0 1
ILC example 2.2 0 0 0 0 0 1 0 0 0 0 J = 3 1 0 0 0 3 3 1 0 0 3 3 3 1 0 By definition of zeros there exists x 0 and exponential input signal containing zero as exponential factor, such that resulting output is zero: 0 0 0 0 0 1 1 1 1 0 0 0 0 2 0 1 ( ) 3 1 0 0 0 4 0 1 1 = 0 1 3 3 1 0 0 8 0 1 3 3 3 1 0 16 0 1
ILC example 2.3 or J f j H x 0 = 0 Then upper bound on smallest non-zero singular value of J is given by σ(j) < H x 0 f j For nonminimum-phase zero f j will grow for larger dimensions N forcing σ(j) to be small Thus delays and non-minimum phase zeroes create loss of rank for J
ILC singular values 1 If J is square but rank-deficient, the singular value decomposition provides the rank as the dimension of Σ 1 : ( ) ( ) ( ) Σ 1 0 V1 T J = U 1 U 2 0 Σ 2 V2 T where Σ 2 is zero. Then decompose f, the input to J, as where Jf = Jf 1 Jf 2 f 1 imv 1, f 2 imv 2 Thus Jf = U 1 Σ 1 V T 1 f 1 as V T 1 f 2 = 0, V T 2 f 1 = 0
and replace f 1 by V 1 f, i.e dimension reduction of state ILC singular values 2 r u j f j1 f j y j ε j z 1 JV 1 L All poles can now be stabilised or assigned
ILC pole assignment f j1 = (I LJV 1 )f k Lr f 0 = 0 ε j = JV 1 f j r If or then LJV 1 = α I LU 1 Σ 1 V T 1 V 1 = α I L = ασ 1 1 U T 1 If α = 1 then dead-beat. The output y follows r only in subspace im J =im U 1
Alternative for pole assignment: LQ optimal control f j1 = f j u j ILC LQ optimal control 1 y j = J V 1 f j LQ criterion: Cr = = (yk T Q y k u T k R u k ) k=1 k=1 (f T k V T 1 J T QJV 1 f k u T k R u k ) Useful choice: Q = I, R = βi
ILC LQ optimal control 2 Solution to LQ-optimal control problem with unit system and input matrices: u j = (βi P ) 1 P f j where P is stabilizing solution of algebraic Riccati equation or P = P V T 1 J T JV 1 P (βi P ) 1 P 0 = Σ 2 1 P (βi P ) 1 P Thus P is diagonal, with entries p i on diagonal, and Σ 1 = diag (σ i ): ( ) p i = 1 2 σ2 i 1 1 4β σ 2 σi 2 i β
ILC LQ optimal control 3 u j = (βi Σ 2 1 βi) 1 (Σ 2 1 βi) f j = LJV 1 f j Solving for L gives the result L = (2βI Σ 2 1) 1 (Σ 1 βσ 1 )U T 1