Data mining for system identi cation

Transcription

1 LERTEKNIK REG AU T O MA RO T IC C O N T L applications to process André Carvalho Bittencourt Automatic Control, Linköping Sweden

2 Outline 1. Problem formulation. Theoretical guiding principles Modeling Data Estimation 3. Tests and outline of algorithm 4. Mining data from an entire plant 5. Concluding remarks / 15

3 Mining data Historical database continuous vars r, u, y discrete mode variable m 195 control loops 5 types of process variables 4 samples per minute data pts 3.1K/min, 4.5M/day Can we extract useful intervals of data for sysid? 3 / 15

4 Mining data Historical database continuous vars r, u, y discrete mode variable m 195 control loops 5 types of process variables 4 samples per minute data pts 3.1K/min, 4.5M/day Can we extract useful intervals of data for sysid? 3 / 15

5 Requisites and assumptions Assumptions SISO loops linear dynamics real-valued poles* Requisites minimal knowledge exible fast Approach take guidance from the theory use exible models and recursive solutions measure of quality 4 / 15

6 Modeling - de nitions e(k) m(k) r (k) System K (q) y (k) G0 (q) S M {M(θ) : θ Dθ } Model set y (k) = G0 (q) + H0 (q)e(k) Model structure H0 (q) M(θ) y (k) = G (q, θ) + H(q, θ)e(k) S M G0 (z) = G (z, θ0 ), H0 (z) = H(z, θ0 ) 0 0 for θ Dθ. S = M(θ ) Optimal one-step ahead predictor y (k θ) = H(q, θ) 1 G (q, θ) + 1 y (k) 1 H(q, θ) true set DT (S, M) {θ Dθ : S = M(θ)} 5 / 15

7 Modeling - de nitions e(k) m(k) r (k) System K (q) y (k) G0 (q) S M {M(θ) : θ Dθ } Model set y (k) = G0 (q) + H0 (q)e(k) Model structure H0 (q) M(θ) y (k) = G (q, θ) + H(q, θ)e(k) S M G0 (z) = G (z, θ0 ), H0 (z) = H(z, θ0 ) 0 0 for θ Dθ. S = M(θ ) Optimal one-step ahead predictor y (k θ) = H(q, θ) 1 G (q, θ) + 1 y (k) 1 H(q, θ) true set DT (S, M) {θ Dθ : S = M(θ)} 5 / 15

8 Modeling - de nitions e(k) m(k) r (k) System K (q) y (k) G0 (q) S M {M(θ) : θ Dθ } Model set y (k) = G0 (q) + H0 (q)e(k) Model structure H0 (q) M(θ) y (k) = G (q, θ) + H(q, θ)e(k) S M G0 (z) = G (z, θ0 ), H0 (z) = H(z, θ0 ) 0 0 for θ Dθ. S = M(θ ) Optimal one-step ahead predictor y (k θ) = H(q, θ) 1 G (q, θ) + 1 y (k) 1 H(q, θ) true set DT (S, M) {θ Dθ : S = M(θ)} 5 / 15

9 Modeling - de nitions e(k) m(k) r (k) System K (q) y (k) G0 (q) S M {M(θ) : θ Dθ } Model set y (k) = G0 (q) + H0 (q)e(k) Model structure H0 (q) M(θ) y (k) = G (q, θ) + H(q, θ)e(k) S M G0 (z) = G (z, θ0 ), H0 (z) = H(z, θ0 ) 0 0 for θ Dθ. S = M(θ ) Optimal one-step ahead predictor y (k θ) = H(q, θ) 1 G (q, θ) + 1 y (k) 1 H(q, θ) true set DT (S, M) {θ Dθ : S = M(θ)} 5 / 15

10 Modeling - de nitions e(k) m(k) r (k) System K (q) y (k) G0 (q) S M {M(θ) : θ Dθ } Model set y (k) = G0 (q) + H0 (q)e(k) Model structure H0 (q) M(θ) y (k) = G (q, θ) + H(q, θ)e(k) S M G0 (z) = G (z, θ0 ), H0 (z) = H(z, θ0 ) 0 0 for θ Dθ. S = M(θ ) Optimal one-step ahead predictor y (k θ) = H(q, θ) 1 G (q, θ) + 1 y (k) 1 H(q, θ) true set DT (S, M) {θ Dθ : S = M(θ)} 5 / 15

11 Properties of models One-to-one relation T (q, θ) W (q, θ) = T (q, θ)x(k) e(k) 1 1 y (k θ) = [H(q, θ) G (q, θ) 1 H(q, θ) ] = W (q, θ)z(k) y (k) y (k) = [G (q, θ) H(q, θ)] Identi ability at θ0 θ gives the same freq W (z, θ) = W (z, θ0 ), z = θ = θ0 Whether no other resp: Remarks choose True parameter If θ0 S M what about the data? S M and M is globally identi able, DT (S, M) = θ0. then 6 / 15

12 Properties of models One-to-one relation T (q, θ) W (q, θ) = T (q, θ)x(k) e(k) 1 1 y (k θ) = [H(q, θ) G (q, θ) 1 H(q, θ) ] = W (q, θ)z(k) y (k) y (k) = [G (q, θ) H(q, θ)] Identi ability at θ0 θ gives the same freq W (z, θ) = W (z, θ0 ), z = θ = θ0 Whether no other resp: Remarks choose True parameter If θ0 S M what about the data? S M and M is globally identi able, DT (S, M) = θ0. then 6 / 15

13 Properties of models One-to-one relation T (q, θ) W (q, θ) = T (q, θ)x(k) e(k) 1 1 y (k θ) = [H(q, θ) G (q, θ) 1 H(q, θ) ] = W (q, θ)z(k) y (k) y (k) = [G (q, θ) H(q, θ)] Identi ability at θ0 θ gives the same freq W (z, θ) = W (z, θ0 ), z = θ = θ0 Whether no other resp: Remarks choose True parameter If θ0 S M what about the data? S M and M is globally identi able, DT (S, M) = θ0. then 6 / 15

14 Properties of models One-to-one relation T (q, θ) W (q, θ) = T (q, θ)x(k) e(k) 1 1 y (k θ) = [H(q, θ) G (q, θ) 1 H(q, θ) ] = W (q, θ)z(k) y (k) y (k) = [G (q, θ) H(q, θ)] Identi ability at θ0 θ gives the same freq W (z, θ) = W (z, θ0 ), z = θ = θ0 Whether no other resp: Remarks choose True parameter If θ0 S M what about the data? S M and M is globally identi able, DT (S, M) = θ0. then 6 / 15

15 Flexible model structures - S M ARX: delay expansion L-ARX: Laguerre expansion P Wu (z, θ) Pni=1 bi z i, Bn (z) Wy (z, θ) ni=1 ai z i, An (z) P Wu (z, θ) Pni=1 bi Li (z, α), B n (z, α) Wy (z, θ) ni=1 ai Li (z, α), A n (z, α) exact if n exact if n convergence depends on Ts convergence depends on α more e cient with delays unknown delays real poles Remarks describe any linear Li (q, α) = (1 α )Ts q α αq q α 1 i 1 dynamics linear regressions Wu (z, θb ) = B nb (z, α) Wy (z, θa ) = Ana (z) Laguerre better for delays integrate globally identi able choice of n and α if integrating plant guess of largest delay and time cte 7 / 15

16 Flexible model structures - S M ARX: delay expansion L-ARX: Laguerre expansion P Wu (z, θ) Pni=1 bi z i, Bn (z) Wy (z, θ) ni=1 ai z i, An (z) P Wu (z, θ) Pni=1 bi Li (z, α), B n (z, α) Wy (z, θ) ni=1 ai Li (z, α), A n (z, α) exact if n exact if n convergence depends on Ts convergence depends on α more e cient with delays unknown delays real poles Remarks describe any linear Li (q, α) = (1 α )Ts q α αq q α 1 i 1 dynamics linear regressions Wu (z, θb ) = B nb (z, α) Wy (z, θa ) = Ana (z) Laguerre better for delays integrate globally identi able choice of n and α if integrating plant guess of largest delay and time cte 7 / 15

17 Flexible model structures - S M ARX: delay expansion L-ARX: Laguerre expansion P Wu (z, θ) Pni=1 bi z i, Bn (z) Wy (z, θ) ni=1 ai z i, An (z) P Wu (z, θ) Pni=1 bi Li (z, α), B n (z, α) Wy (z, θ) ni=1 ai Li (z, α), A n (z, α) exact if n exact if n convergence depends on Ts convergence depends on α more e cient with delays unknown delays real poles Remarks describe any linear Li (q, α) = (1 α )Ts q α αq q α 1 i 1 dynamics linear regressions Wu (z, θb ) = B nb (z, α) Wy (z, θa ) = Ana (z) Laguerre better for delays integrate globally identi able choice of n and α if integrating plant guess of largest delay and time cte 7 / 15

18 Flexible model structures - S M ARX: delay expansion L-ARX: Laguerre expansion P Wu (z, θ) Pni=1 bi z i, Bn (z) Wy (z, θ) ni=1 ai z i, An (z) P Wu (z, θ) Pni=1 bi Li (z, α), B n (z, α) Wy (z, θ) ni=1 ai Li (z, α), A n (z, α) exact if n exact if n convergence depends on Ts convergence depends on α more e cient with delays unknown delays real poles Remarks describe any linear Li (q, α) = (1 α )Ts q α αq q α 1 i 1 dynamics linear regressions Wu (z, θb ) = B nb (z, α) Wy (z, θa ) = Ana (z) Laguerre better for delays integrate globally identi able choice of n and α if integrating plant guess of largest delay and time cte 7 / 15

19 Properties of the data Informative enough {z(k)}n 1 ARX informative enough data Distinguishes non-equiv. models E [(W (z, θ1 ) W (z, θ )) z(k)] = 0 W (e ıω, θ1 ) W (e ıω, θ ) open: i is SRnb closed: let K (q) = YX (q) (q) r (k) 0, dist. rejection (nx na, ny nb ) 0 i Persistent excitation (PE) of φ(k) Full rank information matrix E φ(k)φ(k)t > 0 n-su. if rich signal r (k) 6 0, servo i r (k) is SRnr nr min(na nx, nb ny ) (SRn) φ(k) = [u(k 1),, u(k n)] is PE. 8 / 15

20 Properties of the data Informative enough {z(k)}n 1 ARX informative enough data Distinguishes non-equiv. models E [(W (z, θ1 ) W (z, θ )) z(k)] = 0 W (e ıω, θ1 ) W (e ıω, θ ) open: i is SRnb closed: let K (q) = YX (q) (q) r (k) 0, dist. rejection (nx na, ny nb ) 0 i Persistent excitation (PE) of φ(k) Full rank information matrix E φ(k)φ(k)t > 0 n-su. if rich signal r (k) 6 0, servo i r (k) is SRnr nr min(na nx, nb ny ) (SRn) φ(k) = [u(k 1),, u(k n)] is PE. 8 / 15

21 Properties of the data Informative enough {z(k)}n 1 ARX informative enough data Distinguishes non-equiv. models E [(W (z, θ1 ) W (z, θ )) z(k)] = 0 W (e ıω, θ1 ) W (e ıω, θ ) open: i is SRnb closed: let K (q) = YX (q) (q) r (k) 0, dist. rejection (nx na, ny nb ) 0 i Persistent excitation (PE) of φ(k) Full rank information matrix E φ(k)φ(k)t > 0 n-su. if rich signal r (k) 6 0, servo i r (k) is SRnr nr min(na nx, nb ny ) (SRn) φ(k) = [u(k 1),, u(k n)] is PE. 8 / 15

22 Properties of the data Informative enough {z(k)}n 1 ARX informative enough data Distinguishes non-equiv. models E [(W (z, θ1 ) W (z, θ )) z(k)] = 0 W (e ıω, θ1 ) W (e ıω, θ ) open: i is SRnb closed: let K (q) = YX (q) (q) r (k) 0, dist. rejection (nx na, ny nb ) 0 i Persistent excitation (PE) of φ(k) Full rank information matrix E φ(k)φ(k)t > 0 n-su. if rich signal r (k) 6 0, servo i r (k) is SRnr nr min(na nx, nb ny ) (SRn) φ(k) = [u(k 1),, u(k n)] is PE. 8 / 15

23 Properties of the data Informative enough {z(k)}n 1 ARX informative enough data Distinguishes non-equiv. models E [(W (z, θ1 ) W (z, θ )) z(k)] = 0 W (e ıω, θ1 ) W (e ıω, θ ) open: i is SRnb closed: let K (q) = YX (q) (q) r (k) 0, dist. rejection (nx na, ny nb ) 0 i Persistent excitation (PE) of φ(k) r (k) 6 0, servo i r (k) is SRnr nr min(na nx, nb ny ) Full rank information matrix E φ(k)φ(k)t > 0 n-su. if rich signal (SRn) φ(k) = [u(k 1),, u(k n)] PE. Step signal example is = (k), φ(k) = [ (k 1),..., (k n)] Let E [φ(k)φ(k)t ] = , SR1! 8 / 15

24 Properties of the data Informative enough {z(k)}n 1 ARX informative enough data Distinguishes non-equiv. models E [(W (z, θ1 ) W (z, θ )) z(k)] = 0 W (e ıω, θ1 ) W (e ıω, θ ) open: i is SRnb closed: let K (q) = YX (q) (q) r (k) 0, dist. rejection (nx na, ny nb ) 0 i Persistent excitation (PE) of φ(k) Full rank information matrix E φ(k)φ(k)t > 0 n-su. if rich signal PE. Remarks (SRn) φ(k) = [u(k 1),, u(k n)] r (k) 6 0, servo i r (k) is SRnr nr min(na nx, nb ny ) is how to verify? nite sample? disturbance rejection? look at the estimate! 8 / 15

25 Properties of the data Informative enough {z(k)}n 1 ARX informative enough data Distinguishes non-equiv. models E [(W (z, θ1 ) W (z, θ )) z(k)] = 0 W (e ıω, θ1 ) W (e ıω, θ ) open: i is SRnb closed: let K (q) = YX (q) (q) r (k) 0, dist. rejection (nx na, ny nb ) 0 i Persistent excitation (PE) of φ(k) Full rank information matrix E φ(k)φ(k)t > 0 n-su. if rich signal PE. Remarks (SRn) φ(k) = [u(k 1),, u(k n)] r (k) 6 0, servo i r (k) is SRnr nr min(na nx, nb ny ) is how to verify? nite sample? disturbance rejection? look at the estimate! 8 / 15

26 Recursive least squares The RLS estimate θ k = arg min θ Dθ Vk (θ) = arg Consistent, i.e. Remarks k i λ ε (k, θ) i=1 θ DT (S, M)(= θ0 ), S M, M(θ) λ min θ Dθ k X if open, excitation in u servo, excitation in r through feedback is identi able (globally) N is informative enough 1, Z Frequency representation 9 / 15

27 Recursive least squares The RLS estimate θ k = arg min θ Dθ Vk (θ) = arg Consistent, i.e. Remarks k i λ ε (k, θ) i=1 θ DT (S, M)(= θ0 ), S M, M(θ) λ min θ Dθ k X if open, excitation in u servo, excitation in r through feedback is identi able (globally) N is informative enough 1, Z Frequency representation k i For constant λ = 1 N and N. V (θ) = 1 4π G0 + Bθ Gθ H0 Hθ Φ + u Hθ Hθ Φue Bθ = (H0 Hθ ) Φu Φε = Z π Φε (ω, θ)dω and π Φue γ0 + γ0 Φu 9 / 15

28 Recursive least squares The RLS estimate θ k = arg min θ Dθ Vk (θ) = arg Consistent, i.e. Remarks k i λ ε (k, θ) i=1 θ DT (S, M)(= θ0 ), S M, M(θ) λ min θ Dθ k X if open, excitation in u servo, excitation in r through feedback is identi able (globally) N is informative enough 1, Z Frequency representation Open-loop, Φue 0. Unbiased estimate. Bθ = 0 9 / 15

29 Recursive least squares The RLS estimate θ k = arg min θ Dθ Vk (θ) = arg Consistent, i.e. Remarks k i λ ε (k, θ) i=1 θ DT (S, M)(= θ0 ), S M, M(θ) λ min θ Dθ k X if open, excitation in u servo, excitation in r through feedback is identi able (globally) N is informative enough 1, Z Frequency representation Servo, Φu = Φru +Φeu and Φue = Φeu γ0. Bθ = H0 Hθ Excitation in r through feedback! γ0 Φeu (Φru + Φeu ) 9 / 15

30 Recursive least squares The RLS estimate θ k = arg min θ Dθ Vk (θ) = arg Consistent, i.e. min θ Dθ k X Remarks λk i ε (k, θ) i=1 θ DT (S, M)(= θ0 ), S M, M(θ) λ 1, Z N open, excitation in u servo, excitation in r through feedback if dist, excitation in is identi able (globally) e through feedback is informative enough Frequency representation Dist. rejection r, Φu 0. Noise must a ect the input! Bθ = H0 Hθ γ0 Φeu 9 / 15

31 Recursive least squares The RLS estimate θ k = arg min θ Dθ Vk (θ) = arg Consistent, i.e. min θ Dθ k X Remarks λk i ε (k, θ) i=1 θ DT (S, M)(= θ0 ), S M, M(θ) λ 1, Z N open, excitation in u servo, excitation in r through feedback if dist, excitation in is identi able (globally) e through feedback is informative enough Frequency representation Dist. rejection r, Φu 0. Noise must a ect the input! Bθ = H0 Hθ γ0 Φeu 9 / 15

32 RLS for linear regressions Linear regression (L-)ARX: y (k θ) = ϕ(k)t θ Asymptotics Let Recursive solution θ k θ k 1 = R (k) 1 ϕ(k)ε(k, θ k 1 ) R (k) = λr (k 1) + ϕ(k)ϕ(k)t 1 λ 1 and λ(θ k θ0 ) As N (0, Pθ ), h i 1 1 Pθ, γ0 E ϕ(k)ϕ(k)t Σθ = (1 λ) Vk (θ k ) = λvk 1 (θ k 1 ) + ε(k, θ k 1 )ε(k, θ k ) Remarks closed-form recursive sol invertibility of R k γ0 R 1 Finite sample estimates γ k = (1 λ)vk (θ k ) ˆ = (1 λ)r (k) R k (1 λ) Σ k = Vk (θ k )R (k) 1 QR-RLS solution 10 / 15

33 RLS for linear regressions Linear regression (L-)ARX: y (k θ) = ϕ(k)t θ Asymptotics Let Recursive solution θ k θ k 1 = R (k) 1 ϕ(k)ε(k, θ k 1 ) R (k) = λr (k 1) + ϕ(k)ϕ(k)t 1 λ 1 and λ(θ k θ0 ) As N (0, Pθ ), h i 1 1 Pθ, γ0 E ϕ(k)ϕ(k)t Σθ = (1 λ) Vk (θ k ) = λvk 1 (θ k 1 ) + ε(k, θ k 1 )ε(k, θ k ) Remarks closed-form recursive sol invertibility of R k γ0 R 1 Finite sample estimates γ k = (1 λ)vk (θ k ) ˆ = (1 λ)r (k) R k (1 λ) Σ k = Vk (θ k )R (k) 1 QR-RLS solution 10 / 15

34 RLS for linear regressions Linear regression (L-)ARX: y (k θ) = ϕ(k)t θ Asymptotics Let Recursive solution θ k θ k 1 = R (k) 1 ϕ(k)ε(k, θ k 1 ) R (k) = λr (k 1) + ϕ(k)ϕ(k)t 1 λ 1 and λ(θ k θ0 ) As N (0, Pθ ), h i 1 1 Pθ, γ0 E ϕ(k)ϕ(k)t Σθ = (1 λ) Vk (θ k ) = λvk 1 (θ k 1 ) + ε(k, θ k 1 )ε(k, θ k ) Remarks closed-form recursive sol invertibility of R k γ0 R 1 Finite sample estimates γ k = (1 λ)vk (θ k ) ˆ = (1 λ)r (k) R k (1 λ) Σ k = Vk (θ k )R (k) 1 QR-RLS solution 10 / 15

35 RLS for linear regressions Linear regression (L-)ARX: y (k θ) = ϕ(k)t θ Asymptotics Let Recursive solution θ k θ k 1 = R (k) 1 ϕ(k)ε(k, θ k 1 ) R (k) = λr (k 1) + ϕ(k)ϕ(k)t 1 λ 1 and λ(θ k θ0 ) As N (0, Pθ ), h i 1 1 Pθ, γ0 E ϕ(k)ϕ(k)t Σθ = (1 λ) Vk (θ k ) = λvk 1 (θ k 1 ) + ε(k, θ k 1 )ε(k, θ k ) Remarks closed-form recursive sol invertibility of R k γ0 R 1 Finite sample estimates γ k = (1 λ)vk (θ k ) ˆ = (1 λ)r (k) R k (1 λ) Σ k = Vk (θ k )R (k) 1 QR-RLS solution 10 / 15

36 RLS for linear regressions Linear regression (L-)ARX: y (k θ) = ϕ(k)t θ Asymptotics Let Recursive solution θ k θ k 1 = R (k) 1 ϕ(k)ε(k, θ k 1 ) R (k) = λr (k 1) + ϕ(k)ϕ(k)t 1 λ 1 and λ(θ k θ0 ) As N (0, Pθ ), h i 1 1 Pθ, γ0 E ϕ(k)ϕ(k)t Σθ = (1 λ) Vk (θ k ) = λvk 1 (θ k 1 ) + ε(k, θ k 1 )ε(k, θ k ) Remarks closed-form recursive sol invertibility of R k γ0 R 1 Finite sample estimates γ k = (1 λ)vk (θ k ) ˆ = (1 λ)r (k) R k (1 λ) Σ k = Vk (θ k )R (k) 1 QR-RLS solution 10 / 15

37 Tests T4 : 0: Normalize the data Granger causality test Can T1 : step changes in or r (k) That is how the plant is operated! > η1 or For help predict y (k θ) = ϕu (k)t θkb + ϕy (k)t θka r (k) > η1 under (θ kb )T T : y (k) variability in y (k) s(k) = s(k) is used y γy (k) > η as a quality measure! 5: Logical conditions Ti+1 conditioning of info matrix Check whether 1 H0 : θ b = 0 (Σbθ ) 1 θ kb As Xnb should vary after step Monitor variance of T3 : y (k)? κ (R (k)) = R (k) is invertible σmin (R (k)) σmax (R (k)) > η3 only computed if Ti passed Exit if any test fails Accept interval if T4 Interval goes from passed T1 to exit 11 / 15

38 Tests T4 : 0: Normalize the data Granger causality test Can T1 : step changes in or r (k) That is how the plant is operated! > η1 or For help predict y (k θ) = ϕu (k)t θkb + ϕy (k)t θka r (k) > η1 under (θ kb )T T : y (k) variability in y (k) s(k) = s(k) is used y γy (k) > η as a quality measure! 5: Logical conditions Ti+1 conditioning of info matrix Check whether 1 H0 : θ b = 0 (Σbθ ) 1 θ kb As Xnb should vary after step Monitor variance of T3 : y (k)? κ (R (k)) = R (k) is invertible σmin (R (k)) σmax (R (k)) > η3 only computed if Ti passed Exit if any test fails Accept interval if T4 Interval goes from passed T1 to exit 11 / 15

39 Tests T4 : 0: Normalize the data Granger causality test Can T1 : step changes in or r (k) That is how the plant is operated! > η1 or For help predict y (k θ) = ϕu (k)t θkb + ϕy (k)t θka r (k) > η1 under (θ kb )T T : y (k) variability in y (k) s(k) = s(k) is used y γy (k) > η as a quality measure! 5: Logical conditions Ti+1 conditioning of info matrix Check whether 1 H0 : θ b = 0 (Σbθ ) 1 θ kb As Xnb should vary after step Monitor variance of T3 : y (k)? κ (R (k)) = R (k) is invertible σmin (R (k)) σmax (R (k)) > η3 only computed if Ti passed Exit if any test fails Accept interval if T4 Interval goes from passed T1 to exit 11 / 15

40 Tests T4 : 0: Normalize the data Granger causality test Can T1 : step changes in or r (k) That is how the plant is operated! > η1 or For help predict y (k θ) = ϕu (k)t θkb + ϕy (k)t θka r (k) > η1 under (θ kb )T T : y (k) variability in y (k) s(k) = s(k) is used y γy (k) > η as a quality measure! 5: Logical conditions Ti+1 conditioning of info matrix Check whether 1 H0 : θ b = 0 (Σbθ ) 1 θ kb As Xnb should vary after step Monitor variance of T3 : y (k)? κ (R (k)) = R (k) is invertible σmin (R (k)) σmax (R (k)) > η3 only computed if Ti passed Exit if any test fails Accept interval if T4 Interval goes from passed T1 to exit 11 / 15

41 Tests T4 : 0: Normalize the data Granger causality test Can T1 : step changes in or r (k) That is how the plant is operated! > η1 or For help predict y (k θ) = ϕu (k)t θkb + ϕy (k)t θka r (k) > η1 under (θ kb )T T : y (k) variability in y (k) s(k) = s(k) is used y γy (k) > η as a quality measure! 5: Logical conditions Ti+1 conditioning of info matrix Check whether 1 H0 : θ b = 0 (Σbθ ) 1 θ kb As Xnb should vary after step Monitor variance of T3 : y (k)? κ (R (k)) = R (k) is invertible σmin (R (k)) σmax (R (k)) > η3 only computed if Ti passed Exit if any test fails Accept interval if T4 Interval goes from passed T1 to exit 11 / 15

42 FIR Example, Granger causality test y (k) = B10 (q) ( + d(k)) + e(k) Remarks fails as a causality test θ 6= 0,k > 1000 ( 500 < k < 1000 d(k) 6= 0, 1500 < k < 000 but works well to select data! statistical signi cance of (any) parameter SNRu =10, SNRd =30 1 / 15

43 Outline of the algorithm k k 1 k0 k0 +η0 k1 k k3 k4 ke N E5 m(k) c(k) η0 E0 ζ(k) > η1 µy (k) I γy (k) I γy (k) > η E ϕ(k) I3 R (k) I3 1 κ (k) I3 1 E3 κ (k) > η3 s(k) I4 s(k) > η4 E4 13 / 15

44 Mining data from an entire plant Evaluation 170 minutes to run Plant 195 control loops selects 1.46% of all samples 37 months of data nds all bump tests 1.15G samples every test is important quality measure supports further analysis Mode of operation type count (%) Average length Loop type open open Density (i) closed closed Flow Level (i) Pressure (i) Temperature / 15

45 Mining data from an entire plant Evaluation 170 minutes to run Plant 195 control loops selects 1.46% of all samples 37 months of data nds all bump tests 1.15G samples every test is important quality measure supports further analysis Level, open Density, closed y u y u r k k 14 / 15

46 Summary Requisites range of variables (normalization) mode operation type (change in u or r) integrating plant ( nite gain models) guess of largest delay (tuning) guess of largest time cte (tuning) 7 tuning vars, 5 thresholds for entire plant Extensions Kautz polynominals (complex poles) nding the topology MIMO case 15 / 15

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