GMV Control (Generalized Minimum Variance)

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GMV Control

(Generalized Minimum Variance)

MODEL IDENTIFICATION AND DATA ANALYSIS

Prof. S. Bittanti

POLITECNICO DI MILANO

(2)

GMV

Control

(Generalized Minimum Variance)

n 

Model reference (Q(z) = 0)

n 

Penalized control (P(z) = 1)

(3)

Summary

n 

Example

(minimum phase system)

n  System to be controlled

n  Model reference control design

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Example

n 

System

time domain description ARMAX (1,1,4)

n  y(t) = 0,8y(t – 1) +

+ u(t – 2) + 1,28u(t – 3) + 0,81u(t – 4) + e(t) + 0,6e(t – 1)

(5)

Example

System

n  A(z)y(t) = B(z)u(t-k) + C(z)e(t)

with

n  A(z) = 1 – 0,8z⁻¹

n  B(z) = 1 + 1,28z⁻¹ + 0,81z⁻² k = 2

(6)

System features

n 

Gain:

n  B(1) / A(1) = 15,45

n 

Zeros of A(z) (system poles)

n  z = 0,8

n 

Zeros of B(z):

n  z = 0,64 ± 0.63i

n 

Zeros of C(z):

(7)

Singularities

in the complex plane

n  A(z) x

n  B(z)

•

(8)

Open loop

step response

(9)

Model reference design

n 

System:

n  A(z)y(t) = B(z)u(t-k) + C(z)e(t)

n  e∼WN(0,s2)

n 

Model reference criterion

n  J = E[ (P(z)y(t+k) - y°(t))²]

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Overall control scheme

H(z) F(z) 1 / G(z) z⁻ B(z) 1 / A(z) C(z) yº(t) ₊ u(t) e(t) y(t) - + + C S k

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Model reference design

n 

Controller polynomials

n  F(z)

n  G(z) = PD(z)B(z)E(z)

n  H(z) = C(z)PD(z)

n 

E(z) and F(z) from Long Division

n  P_N(z)C(z) = P_D(z)A(z)E(z) + z-kF(z)

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Model reference design

n 

Main issue

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Model reference design

n 

Basic idea

P(z) = M(z)

-1

where M(z) is the

reference model

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Reference model

n 

M(z) =

(n poles in – f and gain 1)

n 

Step response of reference model

n 

Time response 90%

(the table shows the number of steps employed by the step response of M(z)to reach 90% of steady state value, as a

(1 + f)

ⁿ

(15)

Reference model choice

f n = 1 n = 2 n = 3 0.1 1 2 2 0.2 2 3 3 0.3 2 3 4 0.4 3 4 5 0.5 4 6 7 0.6 5 8 10 0.7 7 11 14 0.8 11 17 23 0.85 15 28 32 0.9 22 37 50 0.95 45 76 131

(16)

Model reference choice

n  choosing

n  n =2 and f = 0.4

n  ⇒

with such choice 4 steps are required to reach the 90% threshold

(17)

Determination of P(z)

n  Reference model: n  M(z) = n  P(z)= M(z)-1 n  P(z) = (1 – 0.8z-1 + 0.16z-2) / 0.36 n  P_N(z) = 1 – 0.8z-1 + 0.16z-2 (1 + 0.4)2 (1 + 0.4z-1₎2

(18)

Controller polynomials

n 

2 steps long division leads to

n  E(z) = 2,78 + 1,67z-1 n  F (z) = 0,16 + 0,096z-1 n 

thus:

n  F(z) = 0,16 + 0,096z-1 n  G(z) = PD(z)B(z)E(z) = 1 + 1,88z-1_{+ 1,58z}-2_{+ 0,49z}-3 H(z) = C(z)P (z)

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Overall control scheme

H(z) F(z) 1 / G(z) z⁻ B(z) 1 / A(z) C(z) yº(t) ₊ u(t) e(t) y(t) - + + C S k Characteristic Polynomial: B(z)C(z)PN(z)

(20)

Closed loop

step response

(21)

Closed loop

step response

n 

input u(t) with

s

2

= 0

Transfer function from yo_{to u:}

(22)

Penalized control design

n 

System to be controlled:

n  A(z)y(t) = B(z)u(t-k) + C(z)e(t)

n  e∼WN(0,s2)

n 

Performance criterion

n  J = E[(y(t + k) + Q(z)u(t) - yº(t))²]

n  P(z) = 1

(23)

Penalized control design

n 

Controller polynomials

n  F(z) = F˜(z)QD(z)

n  G(z) = B(z)QD(z)E(z) + C(z)QN(z)

n  H(z) = C(z)QD(z)

n 

E(z) and F˜(z) from Long Division

n  C(z) = A(z)E(z) + z-kF˜(z)

(24)

Penalized control design

n 

Closed loop tr function from y° to y

n  S(z) = n 

Characteristic polynomial

n  B(z)QD(z) + A(z)QN(z) z⁻ 1 + Q(z) A(z) _B(z) k

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Choice of Q(z)

n 

Typical:

n  Q(z) = q constant n  Q(z) = q (1 - z⁻¹) n  Q(z) = q 1 - z⁻¹ 1 – gz⁻¹

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Choice of Q(z)

n 

Q(z) =

q

n 

Poles: B(z) + q A(z) = 0

n  q = 0 : zeros of B(z)

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Poles

n 

Root locus of B(z) + q A(z)

q = 0

q ≃ 8,57

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Closed loop

step response (q = 1)

n 

Output y(t) with

s

2

= 0

(29)

Closed loop

step response (

q

= 1)

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Closed loop

step response (q = 8,57)

n 

Output y(t) with

s

2

= 0

(31)

Closed loop

step response (q = 8,57)

n 

Input u(t) with

s

2

= 0

(32)

Gain of control system

n 

S(1) =

n 

q

≠

0 ⇒

gain S(1)

≠

1 non zero

steady state error

1

1 +

q

A(1) B(1)

(33)

Choice of Q(z)

n 

To avoid bias in steady state

Q(z) =

q

(1 - z

⁻

¹

)

n 

Poles: B(z) + q(1 - z

⁻

¹

)A(z) = 0

n  q = 0 : zeros of B(z)

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Closed loop poles

n 

Root locus of B(z) +

q

(1 - z

⁻

¹

)A(z)

q = 0

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Closed loop gain

n 

S(z) =

for z = 1 the gain is 1

n 

Unitary gain guaranteed for all q

1

1 + q(1 - z⁻¹) A(z)

(36)

Closed loop

step response (q = 1)

n 

Output y(t) with

s

2

= 0

(37)

Closed loop

step response (q = 1)

n 

Input u(t) with

s

2

= 0

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Closed loop

step response (

q

= 50)

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Closed loop

step response (q = 50)

n 

Input u(t) with

s

2

= 0

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Choice of Q(z)

n 

Q(z) = q(1 - z

⁻

¹

) / (1 – 0.9z

⁻

¹

)

n 

Poles:

n  q = 0 : zeros of (1 – 0.9z⁻¹)B(z)

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Closed loop poles

n  Root locus of (1 – 0.9z⁻¹)B(z) + q(1 - z⁻¹)A(z)

q → ∞

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Closed loop

step response (q = 1)

n 

Output y(t) with

s

2

= 0

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Closed loop

step response (q = 1)

n 

Input u(t) with

s

2

= 0

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Closed loop

step response (q = 7,3)

n 

Output y(t) with

s

2

= 0

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Closed loop

step response (q = 7,3)

n 

Input u(t) with

s

2

= 0

(46)

Choice of Q(z)

n 

Q(z) = q(1 - z

⁻

¹

) / (1 – 0.8z

⁻

¹

)

n 

Closed loop poles:

n  q = 0 : zeros of (1 – 0.8z⁻¹)B(z)

(47)

Closed loop poles

q = 0

(48)

Closed loop

step response (q = 6,1)

n 

Output y(t) with

s

2

= 0

(49)

Closed loop

step response (q = 6,1)

n 

Input u(t) with

s

2

= 0

(50)

Closed loop

step response (q = 6,1)

n 

Output y(t) with

s

2

= 10

-4

(51)

Closed loop

step response (q = 6,1)

n 

Input u(t) with

s

2

= 10

-4

(52)

GMV Control

(Generalized Minimum Variance)

Prof. S. Bittanti

POLITECNICO DI MILANO

(53)

Ezample 2:

non-minimum phase system

n 

System ARMAX (1,2,3)

n  A(z) = 1 - 0,5z⁻²

n  B(z) = 1 – 2z⁻¹ + 2z⁻² + z⁻³ k = 1

n  C(z) = 1 – 1,4z⁻¹ + 0,7z⁻²

(54)

System features

n  Gain

n  B(1) / A(1) = 4

n  Zeros of A(z) (poles)

n  z = 0,71 n  z = -0,71 n  Zeros of B(z): n  z = -0,35 n  z = 1,18 ± 1,20i n  Zeros of C(z):

(55)

Singularities

in the complex plane

n  A(z) x

n  B(z)

•

(56)

Open loop

step response

(57)

Choosing Q(z)

n 

With: Q(z) = q(1 - z

⁻

¹

) / (1 – 0,5z

⁻

¹

)

n 

Poles:

n  q = 0 : zeros of (1 – 0.5z⁻¹)B(z)

(58)

Closed loop poles

q = 0

(59)

Closed loop

step response (

q

= 1,5)

(60)

Closed loop

step response (

q

= 1,5)

(61)

Closed loop

step response (

q

= 2,1)

(62)

Closed loop

step response (

q

= 2,1)

(63)

Closed loop

step response (

q

= 19)

(64)

Closed loop

step response (

q

= 19)

(65)

Closed loop

step response (

q

= 19)

(66)

Closed loop

step response (

q

= 19)

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GMV Control

(Generalized Minimum Variance)

Prof. S. Bittanti

POLITECNICO DI MILANO

(68)

Example 3

n 

Non-minimum phase system

(zeros outside the unit disk)

and

oscillatory open loop behaviour

(poles located near the border of

stability region)

(69)

Singularities

n  Zeros of A(z) (open loop poles)

n  z = - 0,95 ± 0,1i

n  z = -0,5 ± 0,6i

n  Zeros of B(z) (open loop seros)

n  z = 1 ± i

n  z = 0,2 ± 0,6i

n  Zeros of C(z):

(70)

ARMAX model

n  (4,1,4)

n  A(z) = 1 + 2,9z⁻¹ + 3,422z-2 + 2,072z-3 + 0,557z-4

n  B(z) = 1 – 2,4z-1 + 3,2z-2 – 1,6z-3 + 0,8z-4

n  C(z) = 1 + 0,7z⁻¹

n  Input output delay: k = 1

n  Time domain equation

n  y(t) = -2,9y(t-1) - 3,422y(t-2) - 2,072y(t-3) -

0,557y(t-4) + + u(t-1) - 2,4u(t-2) + 3,2u(t-3) - 1,6u(t-4) + 0,8u(t-5) +

(71)

Singularities

in the complex plane

n  A(z) x

n  B(z)

•

(72)

Open loop

step response

(73)

Choosing Q(z)

n 

Q(z) =

q

(1 - z

⁻

¹

)/(1 + 0,3z

-1

)

n 

Closed loop poles:

n  q = 0 : zeros of (1 + 0,3z-1)B(z)

(74)

Closed loop poles

n  Root locus of (1 + 0,3z-1)B(z) + l(1 - z⁻¹)A(z)

l → ∞

(75)

Closed loop

step response (q = 30)

n 

Output y(t) with

s

2

= 0

(76)

Closed loop

step response (q = 30)

n 

Input u(t) with

s

2