k -Step Ahead Prediction Error Model 1. ARMAX model is ARMA plus eXogeneous signal model: A ( z ) y ( n ) = B ( z ) u ( n − k ) + C ( z ) ξ ( n ) u - input y - output ξ - white noise k - delay • A , B , C are polynomials in z − 1 • All delay is factored into k so the constant terms of A , B , C are not zero • Constant terms of A and C are one (that is, A , C are monic) 1 Digital Control Kannan M. Moudgalya, Autumn 2007
k -Step Ahead Prediction Error Model 2. Recall A ( z ) y ( n ) = B ( z ) u ( n − k ) + C ( z ) ξ ( n ) • Any change in u can affect y only after k samples • But white noise starts affecting the process right away • Want to get the best estimate of the output so as to take corrective action, starting now The above equation can be rewritten as, A ( z ) y ( n + j ) = B ( z ) u ( n + j − k ) + C ( z ) ξ ( n + j ) Want to predict output from n + k onwards or for n + j , j ≥ k 2 Digital Control Kannan M. Moudgalya, Autumn 2007
k -Step Ahead Prediction Error Model 3. A ( z ) y ( n ) = B ( z ) u ( n − k ) + C ( z ) ξ ( n ) y ( n + k ) = B ( z ) A ( z ) u ( n ) + C ( z ) A ( z ) ξ ( n + k ) If C = A , the best prediction model is, y ( n + k | n ) = B ( z ) ˆ A ( z ) u ( n ) If C � = A , divide C by A as follows, with j to be specified: C ( z ) A ( z ) = E j ( z ) + z − j F j ( z ) A ( z ) E j ( z ) = e j, 0 + e j, 1 z − 1 + · · · + e j,j − 1 z − ( j − 1) F j ( z ) = f j, 0 + f j, 1 z − 1 + · · · + f j, d F j z − d F j Noise has past and future terms, to be split 3 Digital Control Kannan M. Moudgalya, Autumn 2007
Splitting Noise into Past and Future 4. y ( n + j ) = B ( z ) A ( z ) u ( n + j − k ) + C ( z ) A ( z ) ξ ( n + j ) y ( n + j ) = B ( z ) A ( z ) u ( n + j − k ) � ( e j, 0 + e j, 1 z − 1 + · · · + e j,j − 1 z − ( j − 1) ) + + z − j f j, 0 + f j, 1 z − 1 + · · · + f j, d F j z − d F j � ξ ( n + j ) A ( z ) II = e j, 0 ξ ( n + j ) + e j, 1 ξ ( n + j − 1) + · · · + e j,j − 1 ξ ( n + 1) All future terms. � f j, 0 + f j, 1 z − 1 + · · · + f j, d F j z − d F j � III = ξ ( n ) /A ( z ) III term is known from previous measurements 4 Digital Control Kannan M. Moudgalya, Autumn 2007
Example: Splitting Noise into Past and Future 5. 1 + 0 . 5 z − 1 u ( n + j − 2) y ( n + j ) = 1 − 0 . 6 z − 1 − 0 . 16 z − 2 + 1 − 0 . 6 z − 1 − 0 . 16 z − 2 ξ ( n + j ) Split C into E j and F j , for j = 2 : 1 + 0 . 5 z − 1 0 . 82 + 0 . 176 z − 1 1 − 0 . 6 z − 1 − 0 . 16 z − 2 = (1 + 1 . 1 z − 1 ) + z − 2 1 − 0 . 6 z − 1 − 0 . 16 z − 2 Substitute it in the expression for y ( n + j ) , with j = 2 : 1 y ( n + 2) = 1 − 0 . 6 z − 1 − 0 . 16 z − 2 u ( n ) + (1 + 1 . 1 z − 1 ) ξ ( n + 2) 0 . 82 + 0 . 176 z − 1 + z − 2 1 − 0 . 6 z − 1 − 0 . 16 z − 2 ξ ( n + 2) Second term is unknown; Last term is known. 5 Digital Control Kannan M. Moudgalya, Autumn 2007
Splitting Noise into Past and Future 6. Ay ( n ) = Bu ( n − k ) + Cξ ( n ) y ( n + j ) = B Au ( n + j − k ) + C Aξ ( n + j ) = B E j + z − j F j � � Au ( n + j − k ) + ξ ( n + j ) A = B Au ( n + j − k ) + F j A ξ ( n ) + E j ξ ( n + j ) = B Au ( n + j − k ) + F j Ay ( n ) − Bu ( n − k ) + E j ξ ( n + j ) A C = B Au ( n + j − k ) − F j B AC u ( n − k ) + F j C y ( n ) + E j ξ ( n + j ) = B 1 − F j u ( n + j − k ) + F j � � C z − j C y ( n ) + E j ξ ( n + j ) A 6 Digital Control Kannan M. Moudgalya, Autumn 2007
Splitting Noise into Past and Future 7. From the previous slide, y ( n + j ) = B � 1 − F j � u ( n + j − k ) + F j C z − j C y ( n ) + E j ξ ( n + j ) A C A = E j + z − j F j A ⇒ C A − z − j F j A = E j ⇒ C 1 − z − j F j � � = E j A C y ( n + j ) = E j B C u ( n + j − k ) + F j C y ( n ) + E j ξ ( n + j ) Last term has only future terms. Hence, best prediction model: y ( n + j | n ) = E j B C u ( n + j − k ) + F j ˆ C y ( n ) ˆ means estimate. | n means “using measurements, available up to and including n ”. 7 Digital Control Kannan M. Moudgalya, Autumn 2007
Example: Splitting C/A into E j and F j 8. 1 + 0 . 5 z − 1 1 − 0 . 6 z − 1 − 0 . 16 z − 2 = C A = E j + z − j F j A 1 + 1 . 1 z − 1 1 − 0 . 6 z − 1 − 0 . 16 z − 2 | 1 +0 . 5 z − 1 1 − 0 . 6 z − 1 − 0 . 16 z − 2 +1 . 1 z − 1 +0 . 16 z − 2 +1 . 1 z − 1 − 0 . 66 z − 2 − 0 . 176 z − 3 +0 . 82 z − 2 +0 . 176 z − 3 1 + 0 . 5 z − 1 0 . 82 + 0 . 176 z − 1 1 − 0 . 6 z − 1 − 0 . 16 z − 2 = (1 + 1 . 1 z − 1 ) + z − 2 1 − 0 . 6 z − 1 − 0 . 16 z − 2 8 Digital Control Kannan M. Moudgalya, Autumn 2007
Another Method to Split C/A into E j and F j 9. An easier method exists to solve C A = E j + z − j F j A Cross multiply by A : C = AE j + z − j F j • C , A , z − j are known • E j , F j are to be calculated. • Think: How would you solve it? 9 Digital Control Kannan M. Moudgalya, Autumn 2007
Different Noise and Prediction Models: AR- 10. MAX ARMAX Model : Ay ( n ) = Bu ( n − k ) + Cξ ( n ) C = E j A + z − j F j y ( n + j | t ) = E j B C u ( n + j − k ) + F j ˆ C y ( n ) 10 Digital Control Kannan M. Moudgalya, Autumn 2007
Different Noise and Prediction Models: ARI- 11. MAX ARIMAX model with ∆ = 1 − z − 1 : Ay ( n ) = Bu ( n − k ) + C ∆ ξ ( n ) A ∆ y ( n ) = B ∆ u ( n − k ) + Cξ ( n ) Recall ARMAX model: Ay ( n ) = Bu ( n − k ) + Cξ ( n ) Is the solution for ARMAX model useful? A ← A ∆ , B ← B ∆ C = E j A ∆ + z − j F j y ( n + j | n ) = E j B ∆ u ( n + j − k ) + F j ˆ C y ( n ) C 11 Digital Control Kannan M. Moudgalya, Autumn 2007
Different Noise and Prediction Models: ARIX 12. Recall ARIMAX model from previous slide: A ∆ y ( n ) = B ∆ u ( n − k ) + Cξ ( n ) y ( n + j | n ) = E j B ∆ u ( n + j − k ) + F j ˆ C y ( n ) C ARIX model, obtained with C = 1 in ARIMAX: Ay ( n ) = Bu ( n − k ) + 1 ∆ ξ ( n ) 1 = E j A ∆ + z − j F j y ( n + j | t ) = E j B ∆ u ( n + j − k ) + F j y ( n ) ˆ 12 Digital Control Kannan M. Moudgalya, Autumn 2007
Minimum Variance Control: Regulation 13. ARMAX Model: Ay ( n ) = Bu ( n − k ) + Cξ ( n ) C = E j A + z − j F j y ( n + j ) = E j B C u ( n + j − k ) + F j C y ( n ) + E j ξ ( n + j ) Minimum variance control: Minimize the variations in y at k : y ( n + k ) = E k B C u ( n ) + F k C y ( n ) + E k ξ ( n + k ) � y 2 ( n + k ) � To minimize E . ξ ( n + k ) is ind. of u ( n ) , y ( n ) E k Bu ( n ) + F k y ( n ) = 0 u ( n ) = − F k E k By ( n ) 13 Digital Control Kannan M. Moudgalya, Autumn 2007
Example: Minimum Variance Control 14. 0 . 5 1 y ( n ) = 1 − 0 . 5 z − 1 u ( n − 1) + 1 − 0 . 9 z − 1 ξ ( n ) A = (1 − 0 . 5 z − 1 )(1 − 0 . 9 z − 1 ) = 1 − 1 . 4 z − 1 + 0 . 45 z − 2 B = 0 . 5(1 − 0 . 9 z − 1 ) C = (1 − 0 . 5 z − 1 ) k = 1 C = E k A + z − k F k 1 − 0 . 5 z − 1 = E 1 (1 − 1 . 4 z − 1 + 0 . 45 z − 2 ) + z − 1 F 1 Solving, E 1 = 1 F 1 = 0 . 9 − 0 . 45 z − 1 14 Digital Control Kannan M. Moudgalya, Autumn 2007
Example: Minimum Variance Control 15. B = 0 . 5(1 − 0 . 9 z − 1 ) E 1 = 1 F 1 = 0 . 9 − 0 . 45 z − 1 E k By ( n ) = − 0 . 9 − 0 . 45 z − 1 u ( n ) = − F k 0 . 5(1 − 0 . 9 z − 1 ) y ( n ) = − 0 . 9 2 − z − 1 1 − 0 . 9 z − 1 y ( n ) � ( E k ξ ( n + k )) 2 � � ( ξ ( n + 1)) 2 � y 2 ( n + k ) � � = E = E E = σ 2 15 Digital Control Kannan M. Moudgalya, Autumn 2007
Minimum Variance Control for ARIX Model 16. Recall Ay ( n ) = Bu ( n − k ) + 1 ∆ ξ ( n ) y ( n + j | n ) = E j B ∆ u ( n + j − k ) + F j y ( n ) ˆ 1 = E j A ∆ + z − j F j Minimum variance control law is obtained by forcing ˆ y ( n + j | n ) to be zero: E k B ∆ u ( n ) = − F k y ( n ) ∆ u ( n ) = − F k E k By ( n ) For nonminimum phase systems, use an alternate approach 16 Digital Control Kannan M. Moudgalya, Autumn 2007
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