k -Step Ahead Prediction Error Model 1. k -Step Ahead Prediction - - PowerPoint PPT Presentation

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k-Step Ahead Prediction Error Model

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k-Step Ahead Prediction Error Model ARMAX model is ARMA plus eXogeneous signal model: A(z)y(n) = B(z)u(n − k) + C(z)ξ(n)

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k-Step Ahead Prediction Error Model 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

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k-Step Ahead Prediction Error Model 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

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k-Step Ahead Prediction Error Model 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

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k-Step Ahead Prediction Error Model 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)

Digital Control

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Kannan M. Moudgalya, Autumn 2007

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k-Step Ahead Prediction Error Model

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k-Step Ahead Prediction Error Model Recall A(z)y(n) = B(z)u(n − k) + C(z)ξ(n)

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k-Step Ahead Prediction Error Model 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

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k-Step Ahead Prediction Error Model 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

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k-Step Ahead Prediction Error Model 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,

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k-Step Ahead Prediction Error Model 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

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k-Step Ahead Prediction Error Model 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)

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k-Step Ahead Prediction Error Model 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

Digital Control

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Kannan M. Moudgalya, Autumn 2007

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k-Step Ahead Prediction Error Model

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k-Step Ahead Prediction Error Model A(z)y(n) = B(z)u(n − k) + C(z)ξ(n)

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k-Step Ahead Prediction Error Model 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)

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k-Step Ahead Prediction Error Model 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,

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k-Step Ahead Prediction Error Model 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, ˆ y(n + k|n) =

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k-Step Ahead Prediction Error Model 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, ˆ y(n + k|n) = B(z) A(z)u(n)

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k-Step Ahead Prediction Error Model 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, ˆ y(n + k|n) = B(z) A(z)u(n) If C = A, divide C by A as follows, with j to be specified: C(z) A(z) = Ej(z) + z−jFj(z) A(z)

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k-Step Ahead Prediction Error Model 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, ˆ y(n + k|n) = B(z) A(z)u(n) If C = A, divide C by A as follows, with j to be specified: C(z) A(z) = Ej(z) + z−jFj(z) A(z) Ej(z) = ej,0 + ej,1z−1 + · · · + ej,j−1z−(j−1)

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k-Step Ahead Prediction Error Model 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, ˆ y(n + k|n) = B(z) A(z)u(n) If C = A, divide C by A as follows, with j to be specified: C(z) A(z) = Ej(z) + z−jFj(z) A(z) Ej(z) = ej,0 + ej,1z−1 + · · · + ej,j−1z−(j−1) Fj(z) = fj,0 + fj,1z−1 + · · · + fj,dFjz−dFj

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k-Step Ahead Prediction Error Model 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, ˆ y(n + k|n) = B(z) A(z)u(n) If C = A, divide C by A as follows, with j to be specified: C(z) A(z) = Ej(z) + z−jFj(z) A(z) Ej(z) = ej,0 + ej,1z−1 + · · · + ej,j−1z−(j−1) Fj(z) = fj,0 + fj,1z−1 + · · · + fj,dFjz−dFj Noise has past and future terms, to be split

Digital Control

3

Kannan M. Moudgalya, Autumn 2007

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Splitting Noise into Past and Future

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Splitting Noise into Past and Future y(n + j) = B(z) A(z)u(n + j − k) + C(z) A(z)ξ(n + j)

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Splitting Noise into Past and Future 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)

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Splitting Noise into Past and Future 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) +

• (ej,0 + ej,1z−1 + · · · + ej,j−1z−(j−1))

+ z−jfj,0 + fj,1z−1 + · · · + fj,dFjz−dFj A(z)

• ξ(n + j)

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Splitting Noise into Past and Future 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) +

• (ej,0 + ej,1z−1 + · · · + ej,j−1z−(j−1))

+ z−jfj,0 + fj,1z−1 + · · · + fj,dFjz−dFj A(z)

• ξ(n + j)

II = ej,0ξ(n + j)

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Splitting Noise into Past and Future 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) +

• (ej,0 + ej,1z−1 + · · · + ej,j−1z−(j−1))

+ z−jfj,0 + fj,1z−1 + · · · + fj,dFjz−dFj A(z)

• ξ(n + j)

II = ej,0ξ(n + j) + ej,1ξ(n + j − 1)

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Splitting Noise into Past and Future 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) +

• (ej,0 + ej,1z−1 + · · · + ej,j−1z−(j−1))

+ z−jfj,0 + fj,1z−1 + · · · + fj,dFjz−dFj A(z)

• ξ(n + j)

II = ej,0ξ(n + j) + ej,1ξ(n + j − 1) + · · · + ej,j−1ξ(n + 1)

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Splitting Noise into Past and Future 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) +

• (ej,0 + ej,1z−1 + · · · + ej,j−1z−(j−1))

+ z−jfj,0 + fj,1z−1 + · · · + fj,dFjz−dFj A(z)

• ξ(n + j)

II = ej,0ξ(n + j) + ej,1ξ(n + j − 1) + · · · + ej,j−1ξ(n + 1) All future terms.

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Splitting Noise into Past and Future 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) +

• (ej,0 + ej,1z−1 + · · · + ej,j−1z−(j−1))

+ z−jfj,0 + fj,1z−1 + · · · + fj,dFjz−dFj A(z)

• ξ(n + j)

II = ej,0ξ(n + j) + ej,1ξ(n + j − 1) + · · · + ej,j−1ξ(n + 1) All future terms. III =

• fj,0 + fj,1z−1 + · · · + fj,dFjz−dFj
• ξ(n)/A(z)

III term is known from previous measurements

Digital Control

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Kannan M. Moudgalya, Autumn 2007

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Example: Splitting Noise into Past and Future

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Example: Splitting Noise into Past and Future

y(n + j) = u(n + j − 2) 1 − 0.6z−1 − 0.16z−2 + 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2ξ(n + j)

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Example: Splitting Noise into Past and Future

y(n + j) = u(n + j − 2) 1 − 0.6z−1 − 0.16z−2 + 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2ξ(n + j)

Split C into Ej and Fj, for j = 2:

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Example: Splitting Noise into Past and Future

y(n + j) = u(n + j − 2) 1 − 0.6z−1 − 0.16z−2 + 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2ξ(n + j)

Split C into Ej and Fj, for j = 2:

1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = (1 + 1.1z−1) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2

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Example: Splitting Noise into Past and Future

y(n + j) = u(n + j − 2) 1 − 0.6z−1 − 0.16z−2 + 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2ξ(n + j)

Split C into Ej and Fj, for j = 2:

1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = (1 + 1.1z−1) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2

Substitute it in the expression for y(n + j),

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Example: Splitting Noise into Past and Future

y(n + j) = u(n + j − 2) 1 − 0.6z−1 − 0.16z−2 + 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2ξ(n + j)

Split C into Ej and Fj, for j = 2:

1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = (1 + 1.1z−1) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2

Substitute it in the expression for y(n + j), with j = 2:

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Example: Splitting Noise into Past and Future

y(n + j) = u(n + j − 2) 1 − 0.6z−1 − 0.16z−2 + 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2ξ(n + j)

Split C into Ej and Fj, for j = 2:

1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = (1 + 1.1z−1) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2

Substitute it in the expression for y(n + j), with j = 2: y(n + 2) = 1 1 − 0.6z−1 − 0.16z−2u(n)

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Example: Splitting Noise into Past and Future

y(n + j) = u(n + j − 2) 1 − 0.6z−1 − 0.16z−2 + 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2ξ(n + j)

Split C into Ej and Fj, for j = 2:

1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = (1 + 1.1z−1) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2

Substitute it in the expression for y(n + j), with j = 2: y(n + 2) = 1 1 − 0.6z−1 − 0.16z−2u(n) + (1 + 1.1z−1)ξ(n + 2)

5.

Example: Splitting Noise into Past and Future

y(n + j) = u(n + j − 2) 1 − 0.6z−1 − 0.16z−2 + 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2ξ(n + j)

Split C into Ej and Fj, for j = 2:

1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = (1 + 1.1z−1) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2

Substitute it in the expression for y(n + j), with j = 2: y(n + 2) = 1 1 − 0.6z−1 − 0.16z−2u(n) + (1 + 1.1z−1)ξ(n + 2) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2ξ(n + 2)

5.

Example: Splitting Noise into Past and Future

y(n + j) = u(n + j − 2) 1 − 0.6z−1 − 0.16z−2 + 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2ξ(n + j)

Split C into Ej and Fj, for j = 2:

1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = (1 + 1.1z−1) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2

Substitute it in the expression for y(n + j), with j = 2: y(n + 2) = 1 1 − 0.6z−1 − 0.16z−2u(n) + (1 + 1.1z−1)ξ(n + 2) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2ξ(n + 2) Second term is unknown;

5.

Example: Splitting Noise into Past and Future

y(n + j) = u(n + j − 2) 1 − 0.6z−1 − 0.16z−2 + 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2ξ(n + j)

Split C into Ej and Fj, for j = 2:

1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = (1 + 1.1z−1) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2

Substitute it in the expression for y(n + j), with j = 2: y(n + 2) = 1 1 − 0.6z−1 − 0.16z−2u(n) + (1 + 1.1z−1)ξ(n + 2) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2ξ(n + 2) Second term is unknown; Last term is known.

Digital Control

5

Kannan M. Moudgalya, Autumn 2007

6.

Splitting Noise into Past and Future

6.

Splitting Noise into Past and Future Ay(n) = Bu(n − k) + Cξ(n)

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Splitting Noise into Past and Future Ay(n) = Bu(n − k) + Cξ(n) y(n + j) = B Au(n + j − k) + C Aξ(n + j)

6.

Splitting Noise into Past and Future Ay(n) = Bu(n − k) + Cξ(n) y(n + j) = B Au(n + j − k) + C Aξ(n + j) = B Au(n + j − k) +

• Ej + z−jFj

A

• ξ(n + j)

6.

Splitting Noise into Past and Future Ay(n) = Bu(n − k) + Cξ(n) y(n + j) = B Au(n + j − k) + C Aξ(n + j) = B Au(n + j − k) +

• Ej + z−jFj

A

• ξ(n + j)

= B Au(n + j − k) + Fj A ξ(n) + Ejξ(n + j)

6.

Splitting Noise into Past and Future Ay(n) = Bu(n − k) + Cξ(n) y(n + j) = B Au(n + j − k) + C Aξ(n + j) = B Au(n + j − k) +

• Ej + z−jFj

A

• ξ(n + j)

= B Au(n + j − k) + Fj A ξ(n) + Ejξ(n + j) = B Au(n + j − k) + Fj A Ay(n) − Bu(n − k) C + Ejξ(n + j)

6.

Splitting Noise into Past and Future Ay(n) = Bu(n − k) + Cξ(n) y(n + j) = B Au(n + j − k) + C Aξ(n + j) = B Au(n + j − k) +

• Ej + z−jFj

A

• ξ(n + j)

= B Au(n + j − k) + Fj A ξ(n) + Ejξ(n + j) = B Au(n + j − k) + Fj A Ay(n) − Bu(n − k) C + Ejξ(n + j) = B Au(n + j − k) − FjB AC u(n − k) + Fj C y(n) + Ejξ(n + j)

6.

Splitting Noise into Past and Future Ay(n) = Bu(n − k) + Cξ(n) y(n + j) = B Au(n + j − k) + C Aξ(n + j) = B Au(n + j − k) +

• Ej + z−jFj

A

• ξ(n + j)

= B Au(n + j − k) + Fj A ξ(n) + Ejξ(n + j) = B Au(n + j − k) + Fj A Ay(n) − Bu(n − k) C + Ejξ(n + j) = B Au(n + j − k) − FjB AC u(n − k) + Fj C y(n) + Ejξ(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j)

Digital Control

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Kannan M. Moudgalya, Autumn 2007

7.

Splitting Noise into Past and Future

7.

Splitting Noise into Past and Future

From the previous slide, y(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j)

7.

Splitting Noise into Past and Future

From the previous slide, y(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j) C A = Ej + z−jFj A

7.

Splitting Noise into Past and Future

From the previous slide, y(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j) C A = Ej + z−jFj A ⇒ C A − z−jFj A = Ej

7.

Splitting Noise into Past and Future

From the previous slide, y(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j) C A = Ej + z−jFj A ⇒ C A − z−jFj A = Ej ⇒ C A

• 1 − z−jFj

C

• = Ej

7.

Splitting Noise into Past and Future

From the previous slide, y(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j) C A = Ej + z−jFj A ⇒ C A − z−jFj A = Ej ⇒ C A

• 1 − z−jFj

C

• = Ej

y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j)

7.

Splitting Noise into Past and Future

From the previous slide, y(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j) C A = Ej + z−jFj A ⇒ C A − z−jFj A = Ej ⇒ C A

• 1 − z−jFj

C

• = Ej

y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Last term has only future terms.

7.

Splitting Noise into Past and Future

From the previous slide, y(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j) C A = Ej + z−jFj A ⇒ C A − z−jFj A = Ej ⇒ C A

• 1 − z−jFj

C

• = Ej

y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Last term has only future terms. Hence, best prediction model:

7.

Splitting Noise into Past and Future

From the previous slide, y(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j) C A = Ej + z−jFj A ⇒ C A − z−jFj A = Ej ⇒ C A

• 1 − z−jFj

C

• = Ej

y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Last term has only future terms. Hence, best prediction model: ˆ y(n + j|n) = EjB C u(n + j − k) + Fj C y(n)

7.

Splitting Noise into Past and Future

From the previous slide, y(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j) C A = Ej + z−jFj A ⇒ C A − z−jFj A = Ej ⇒ C A

• 1 − z−jFj

C

• = Ej

y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Last term has only future terms. Hence, best prediction model: ˆ y(n + j|n) = EjB C u(n + j − k) + Fj C y(n) ˆmeans estimate.

7.

Splitting Noise into Past and Future

From the previous slide, y(n + j) = B A

• 1 − Fj

C z−j

• u(n + j − k) + Fj

C y(n) + Ejξ(n + j) C A = Ej + z−jFj A ⇒ C A − z−jFj A = Ej ⇒ C A

• 1 − z−jFj

C

• = Ej

y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Last term has only future terms. Hence, best prediction model: ˆ y(n + j|n) = EjB C u(n + j − k) + Fj C y(n) ˆmeans estimate.|n means “using measurements, available up to and including n”.

Digital Control

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Kannan M. Moudgalya, Autumn 2007

8.

Example: Splitting C/A into Ej and Fj

8.

Example: Splitting C/A into Ej and Fj 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = C A

8.

Example: Splitting C/A into Ej and Fj 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = C A = Ej + z−jFj A

8.

Example: Splitting C/A into Ej and Fj 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = C A = Ej + z−jFj A 1 + 1.1z−1 1 − 0.6z−1 − 0.16z−2 | 1 +0.5z−1 1 −0.6z−1 −0.16z−2 +1.1z−1 +0.16z−2 +1.1z−1 −0.66z−2 −0.176z−3 +0.82z−2 +0.176z−3

8.

Example: Splitting C/A into Ej and Fj 1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = C A = Ej + z−jFj A 1 + 1.1z−1 1 − 0.6z−1 − 0.16z−2 | 1 +0.5z−1 1 −0.6z−1 −0.16z−2 +1.1z−1 +0.16z−2 +1.1z−1 −0.66z−2 −0.176z−3 +0.82z−2 +0.176z−3

1 + 0.5z−1 1 − 0.6z−1 − 0.16z−2 = (1 + 1.1z−1) + z−2 0.82 + 0.176z−1 1 − 0.6z−1 − 0.16z−2

Digital Control

8

Kannan M. Moudgalya, Autumn 2007

9.

Another Method to Split C/A into Ej and Fj

9.

Another Method to Split C/A into Ej and Fj An easier method exists to solve C A = Ej + z−jFj A

9.

Another Method to Split C/A into Ej and Fj An easier method exists to solve C A = Ej + z−jFj A Cross multiply by A:

9.

Another Method to Split C/A into Ej and Fj An easier method exists to solve C A = Ej + z−jFj A Cross multiply by A: C = AEj + z−jFj

9.

Another Method to Split C/A into Ej and Fj An easier method exists to solve C A = Ej + z−jFj A Cross multiply by A: C = AEj + z−jFj

• C, A, z−j are known

9.

Another Method to Split C/A into Ej and Fj An easier method exists to solve C A = Ej + z−jFj A Cross multiply by A: C = AEj + z−jFj

• C, A, z−j are known
• Ej, Fj are to be calculated.

9.

Another Method to Split C/A into Ej and Fj An easier method exists to solve C A = Ej + z−jFj A Cross multiply by A: C = AEj + z−jFj

• C, A, z−j are known
• Ej, Fj are to be calculated.
• Think: How would you solve it?

Digital Control

9

Kannan M. Moudgalya, Autumn 2007

10.

Different Noise and Prediction Models: AR- MAX

10.

Different Noise and Prediction Models: AR- MAX ARMAX Model

10.

Different Noise and Prediction Models: AR- MAX ARMAX Model : Ay(n) = Bu(n − k) + Cξ(n)

10.

Different Noise and Prediction Models: AR- MAX ARMAX Model : Ay(n) = Bu(n − k) + Cξ(n) C = EjA + z−jFj

10.

Different Noise and Prediction Models: AR- MAX ARMAX Model : Ay(n) = Bu(n − k) + Cξ(n) C = EjA + z−jFj ˆ y(n + j|t) = EjB C u(n + j − k) + Fj C y(n)

Digital Control

10

Kannan M. Moudgalya, Autumn 2007

11.

Different Noise and Prediction Models: ARI- MAX

11.

Different Noise and Prediction Models: ARI- MAX ARIMAX model

11.

Different Noise and Prediction Models: ARI- MAX ARIMAX model with ∆ = 1 − z−1:

11.

Different Noise and Prediction Models: ARI- MAX ARIMAX model with ∆ = 1 − z−1: Ay(n) = Bu(n − k) + C ∆ξ(n)

11.

Different Noise and Prediction Models: ARI- MAX ARIMAX model with ∆ = 1 − z−1: Ay(n) = Bu(n − k) + C ∆ξ(n) A∆y(n) = B∆u(n − k) + Cξ(n)

11.

Different Noise and Prediction Models: ARI- 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)

11.

Different Noise and Prediction Models: ARI- 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?

11.

Different Noise and Prediction Models: ARI- 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∆

11.

Different Noise and Prediction Models: ARI- 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∆

11.

Different Noise and Prediction Models: ARI- 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 = EjA∆ + z−jFj

11.

Different Noise and Prediction Models: ARI- 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 = EjA∆ + z−jFj ˆ y(n + j|n) = EjB∆ C u(n + j − k) + Fj C y(n)

Digital Control

11

Kannan M. Moudgalya, Autumn 2007

12.

Different Noise and Prediction Models: ARIX

12.

Different Noise and Prediction Models: ARIX Recall ARIMAX model from previous slide: A∆y(n) = B∆u(n − k) + Cξ(n) ˆ y(n + j|n) = EjB∆ C u(n + j − k) + Fj C y(n)

12.

Different Noise and Prediction Models: ARIX Recall ARIMAX model from previous slide: A∆y(n) = B∆u(n − k) + Cξ(n) ˆ y(n + j|n) = EjB∆ C u(n + j − k) + Fj C y(n) ARIX model,

12.

Different Noise and Prediction Models: ARIX Recall ARIMAX model from previous slide: A∆y(n) = B∆u(n − k) + Cξ(n) ˆ y(n + j|n) = EjB∆ C u(n + j − k) + Fj C y(n) ARIX model, obtained with C = 1 in ARIMAX:

12.

Different Noise and Prediction Models: ARIX Recall ARIMAX model from previous slide: A∆y(n) = B∆u(n − k) + Cξ(n) ˆ y(n + j|n) = EjB∆ C u(n + j − k) + Fj C y(n) ARIX model, obtained with C = 1 in ARIMAX: Ay(n) = Bu(n − k) + 1 ∆ξ(n)

12.

Different Noise and Prediction Models: ARIX Recall ARIMAX model from previous slide: A∆y(n) = B∆u(n − k) + Cξ(n) ˆ y(n + j|n) = EjB∆ C u(n + j − k) + Fj C y(n) ARIX model, obtained with C = 1 in ARIMAX: Ay(n) = Bu(n − k) + 1 ∆ξ(n) 1 = EjA∆ + z−jFj

12.

Different Noise and Prediction Models: ARIX Recall ARIMAX model from previous slide: A∆y(n) = B∆u(n − k) + Cξ(n) ˆ y(n + j|n) = EjB∆ C u(n + j − k) + Fj C y(n) ARIX model, obtained with C = 1 in ARIMAX: Ay(n) = Bu(n − k) + 1 ∆ξ(n) 1 = EjA∆ + z−jFj ˆ y(n + j|t) = EjB∆u(n + j − k) + Fjy(n)

Digital Control

12

Kannan M. Moudgalya, Autumn 2007

13.

Minimum Variance Control: Regulation

13.

Minimum Variance Control: Regulation ARMAX Model: Ay(n) = Bu(n − k) + Cξ(n)

13.

Minimum Variance Control: Regulation ARMAX Model: Ay(n) = Bu(n − k) + Cξ(n) C = EjA + z−jFj

13.

Minimum Variance Control: Regulation ARMAX Model: Ay(n) = Bu(n − k) + Cξ(n) C = EjA + z−jFj y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j)

13.

Minimum Variance Control: Regulation ARMAX Model: Ay(n) = Bu(n − k) + Cξ(n) C = EjA + z−jFj y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Minimum variance control: Minimize the variations in y at k:

13.

Minimum Variance Control: Regulation ARMAX Model: Ay(n) = Bu(n − k) + Cξ(n) C = EjA + z−jFj y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Minimum variance control: Minimize the variations in y at k: y(n + k) = EkB C u(n) + Fk C y(n) + Ekξ(n + k)

13.

Minimum Variance Control: Regulation ARMAX Model: Ay(n) = Bu(n − k) + Cξ(n) C = EjA + z−jFj y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Minimum variance control: Minimize the variations in y at k: y(n + k) = EkB C u(n) + Fk C y(n) + Ekξ(n + k) To minimize E

• y2(n + k)
• .

13.

Minimum Variance Control: Regulation ARMAX Model: Ay(n) = Bu(n − k) + Cξ(n) C = EjA + z−jFj y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Minimum variance control: Minimize the variations in y at k: y(n + k) = EkB C u(n) + Fk C y(n) + Ekξ(n + k) To minimize E

• y2(n + k)
• . ξ(n + k) is ind. of u(n), y(n)

13.

Minimum Variance Control: Regulation ARMAX Model: Ay(n) = Bu(n − k) + Cξ(n) C = EjA + z−jFj y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Minimum variance control: Minimize the variations in y at k: y(n + k) = EkB C u(n) + Fk C y(n) + Ekξ(n + k) To minimize E

• y2(n + k)
• . ξ(n + k) is ind. of u(n), y(n)

EkBu(n) + Fky(n) = 0

13.

Minimum Variance Control: Regulation ARMAX Model: Ay(n) = Bu(n − k) + Cξ(n) C = EjA + z−jFj y(n + j) = EjB C u(n + j − k) + Fj C y(n) + Ejξ(n + j) Minimum variance control: Minimize the variations in y at k: y(n + k) = EkB C u(n) + Fk C y(n) + Ekξ(n + k) To minimize E

• y2(n + k)
• . ξ(n + k) is ind. of u(n), y(n)

EkBu(n) + Fky(n) = 0 u(n) = − Fk EkBy(n)

Digital Control

13

Kannan M. Moudgalya, Autumn 2007

14.

Example: Minimum Variance Control

14.

Example: Minimum Variance Control y(n) = 0.5 1 − 0.5z−1u(n − 1) + 1 1 − 0.9z−1ξ(n)

14.

Example: Minimum Variance Control y(n) = 0.5 1 − 0.5z−1u(n − 1) + 1 1 − 0.9z−1ξ(n) A = (1 − 0.5z−1)(1 − 0.9z−1) = 1 − 1.4z−1 + 0.45z−2 B = 0.5(1 − 0.9z−1) C = (1 − 0.5z−1) k = 1

14.

Example: Minimum Variance Control y(n) = 0.5 1 − 0.5z−1u(n − 1) + 1 1 − 0.9z−1ξ(n) A = (1 − 0.5z−1)(1 − 0.9z−1) = 1 − 1.4z−1 + 0.45z−2 B = 0.5(1 − 0.9z−1) C = (1 − 0.5z−1) k = 1 C = EkA + z−kFk 1 − 0.5z−1 = E1(1 − 1.4z−1 + 0.45z−2) + z−1F1

14.

Example: Minimum Variance Control y(n) = 0.5 1 − 0.5z−1u(n − 1) + 1 1 − 0.9z−1ξ(n) A = (1 − 0.5z−1)(1 − 0.9z−1) = 1 − 1.4z−1 + 0.45z−2 B = 0.5(1 − 0.9z−1) C = (1 − 0.5z−1) k = 1 C = EkA + z−kFk 1 − 0.5z−1 = E1(1 − 1.4z−1 + 0.45z−2) + z−1F1 Solving, E1 = 1 F1 = 0.9 − 0.45z−1

Digital Control

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Kannan M. Moudgalya, Autumn 2007

15.

Example: Minimum Variance Control

15.

Example: Minimum Variance Control B = 0.5(1 − 0.9z−1) E1 = 1 F1 = 0.9 − 0.45z−1

15.

Example: Minimum Variance Control B = 0.5(1 − 0.9z−1) E1 = 1 F1 = 0.9 − 0.45z−1 u(n) = − Fk EkBy(n)

15.

Example: Minimum Variance Control B = 0.5(1 − 0.9z−1) E1 = 1 F1 = 0.9 − 0.45z−1 u(n) = − Fk EkBy(n) = − 0.9 − 0.45z−1 0.5(1 − 0.9z−1)y(n)

15.

Example: Minimum Variance Control B = 0.5(1 − 0.9z−1) E1 = 1 F1 = 0.9 − 0.45z−1 u(n) = − Fk EkBy(n) = − 0.9 − 0.45z−1 0.5(1 − 0.9z−1)y(n) = −0.9 2 − z−1 1 − 0.9z−1y(n) E

• y2(n + k)
• = E
• (Ekξ(n + k))2

15.

Example: Minimum Variance Control B = 0.5(1 − 0.9z−1) E1 = 1 F1 = 0.9 − 0.45z−1 u(n) = − Fk EkBy(n) = − 0.9 − 0.45z−1 0.5(1 − 0.9z−1)y(n) = −0.9 2 − z−1 1 − 0.9z−1y(n) E

• y2(n + k)
• = E
• (Ekξ(n + k))2

= E

• (ξ(n + 1))2

15.

Example: Minimum Variance Control B = 0.5(1 − 0.9z−1) E1 = 1 F1 = 0.9 − 0.45z−1 u(n) = − Fk EkBy(n) = − 0.9 − 0.45z−1 0.5(1 − 0.9z−1)y(n) = −0.9 2 − z−1 1 − 0.9z−1y(n) E

• y2(n + k)
• = E
• (Ekξ(n + k))2

= E

• (ξ(n + 1))2

= σ2

Digital Control

15

Kannan M. Moudgalya, Autumn 2007

16.

Minimum Variance Control for ARIX Model

16.

Minimum Variance Control for ARIX Model Recall Ay(n) = Bu(n − k) + 1 ∆ξ(n)

16.

Minimum Variance Control for ARIX Model Recall Ay(n) = Bu(n − k) + 1 ∆ξ(n) ˆ y(n + j|n) = EjB∆u(n + j − k) + Fjy(n) 1 = EjA∆ + z−jFj

16.

Minimum Variance Control for ARIX Model Recall Ay(n) = Bu(n − k) + 1 ∆ξ(n) ˆ y(n + j|n) = EjB∆u(n + j − k) + Fjy(n) 1 = EjA∆ + z−jFj Minimum variance control law is obtained by forcing ˆ y(n+j|n) to be zero:

16.

Minimum Variance Control for ARIX Model Recall Ay(n) = Bu(n − k) + 1 ∆ξ(n) ˆ y(n + j|n) = EjB∆u(n + j − k) + Fjy(n) 1 = EjA∆ + z−jFj Minimum variance control law is obtained by forcing ˆ y(n+j|n) to be zero: EkB∆u(n) = −Fky(n)

16.

Minimum Variance Control for ARIX Model Recall Ay(n) = Bu(n − k) + 1 ∆ξ(n) ˆ y(n + j|n) = EjB∆u(n + j − k) + Fjy(n) 1 = EjA∆ + z−jFj Minimum variance control law is obtained by forcing ˆ y(n+j|n) to be zero: EkB∆u(n) = −Fky(n) ∆u(n) = − Fk EkBy(n)

16.

Minimum Variance Control for ARIX Model Recall Ay(n) = Bu(n − k) + 1 ∆ξ(n) ˆ y(n + j|n) = EjB∆u(n + j − k) + Fjy(n) 1 = EjA∆ + z−jFj Minimum variance control law is obtained by forcing ˆ y(n+j|n) to be zero: EkB∆u(n) = −Fky(n) ∆u(n) = − Fk EkBy(n) For nonminimum phase systems, use an alternate approach

Digital Control

16

Kannan M. Moudgalya, Autumn 2007