Lecture 09 EAILC Stochastic STR

7/30/2019 Lecture 09 EAILC Stochastic STR

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Lecture 9: Stochastic / Predictive Self-Tuning Regulators.

Minimum Variance Control.

Moving Average Controller.

c Leonid Freidovich. A ril 28, 2010. Elements of Iterative Learnin and Ada tive Control: Lecture 9 . 1/14


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The stochastic model

Assume the plant is represented by the special ARMAX model

A(q) y(t) = B(q) u(t) + C(q) e(t)

where {e(t)} is white noise,

A(q) = qn + a1 qn1 + + an, deg{A} = n

B(q) = b1 qnd0

+ + bn, deg{B} = n d0

C(q) = qn + c1 qn1 + + cn, deg{C} = n



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The stochastic model


A(q) y(t) = B(q) u(t) + C(q) e(t)

where {e(t)} is white noise,

A(q) = qn + a1 qn1 + + an, deg{A} = n

B(q) = b1 qnd0

+ + bn, deg{B} = n d0

C(q) = qn + c1 qn1 + + cn, deg{C} = n

Equivalently,

y(t) = a1 y(t 1) an y(t n)

+ bd0 u(t d0) + + bn u(t n)

+ e(t) + c1 e(t 1) + + cn e(t n)



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Remarks on modeling noise.

Assuming that C is stable is not restrictive:

Typically, such a model is obtained experimentally fromspectrum characteristics. It does not change if the unstableroots are substituted by stable, which are symmetrical to themwith respect to the unite circle.



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Assuming that C is stable is not restrictive:Typically, such a model is obtained experimentally fromspectrum characteristics. It does not change if the unstableroots are substituted by stable, which are symmetrical to themwith respect to the unite circle.

Apparently, one can also make deg{C} = deg{A} = n byrenaming signals.



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Apparently, one can also make deg{C} = deg{A} = n byrenaming signals. Let us see it on the example

y(t + 1) + a y(t) = b u(t) + c e(t)



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y(t + 1) + a y(t) = b u(t) + c e(t)

It can be rewritten as

(q + a) y(t) = b u(t) + c e(t)



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y(t + 1) + a y(t) = b u(t) + c e(t)


(q + a) y(t) = b u(t) + c q e(t 1)



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y(t + 1) + a y(t) = b u(t) + c e(t)


(q + a) y(t) = b u(t) + q enew(t)

introducing enew(t) = c e(t 1).



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y(t + 1) + a y(t) = b u(t) + c e(t)


y(t + 1) + a y(t) = b u(t) + enew(t + 1)

with enew(t) = c e(t 1) being white noise with

var{enew(t)} = c2 var{e(t)}.



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Minimum-variance control: Example

Consider the model

y(t) = a y(t 1) + b u(t 1) + c e(t 1) + e(t)

where |c| < 1 and {e(t)} is a sequence of random variableswith E{e(t)} = 0 and var{e(t)} = 2.

Our goal is to regulatey(t) to0 as close as possible.



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Consider the model

y(t) = a y(t 1) + b u(t 1) + c e(t 1) + e(t)


Our goal is to regulatey(t) to0 with minimal variance.



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Consider the model

y(t) = a y(t 1) + b u(t 1) + c e(t 1) + e(t)



In the absence of noise:

y(t) = a y(t 1) + b u(t 1)

the best strategy (dead beat design) is

u(t) =a

by(t) = y(t + 1) = 0

What should we do when noise is present?



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Consider the model

y(t) = a y(t 1) + b u(t 1) + c e(t 1) + e(t)



In the absence of noise or when c = 0:

y(t) = a y(t 1) + b u(t 1) + e(t)

the best strategy (dead beat design) is

u(t) =a

by(t) = y(t + 1) = e(t + 1)

What should we do when noise dynamics are present?



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Minimum-variance control: Example (contd)

We have the process

y(t + 1) + a y(t) = b u(t) + e(t + 1) + c e(t)



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We have the process

y(t + 1) + a y(t) = b u(t) + e(t + 1) + c e(t)

Let us try to compute e(t) symbolically

(q + a) y(t) = b u(t) + (q + c) e(t)



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We have the process

y(t + 1) + a y(t) = b u(t) + e(t + 1) + c e(t)


e(t) =q + a

q + cy(t)

b

q + cu(t)



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We have the process

y(t + 1) + a y(t) = b u(t) + e(t + 1) + c e(t)


e(t) =q + a

q + cy(t)

b

q + cu(t)

Substituting back into the model

y(t+1) = a y(t)+b u(t)+e(t+1)+c q + a

q + cy(t)

b

q + cu(t)



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We have the process

y(t + 1) + a y(t) = b u(t) + e(t + 1) + c e(t)


e(t) =q + a

q + cy(t)

b

q + cu(t)


y(t + 1) =(c a) q

q + cy(t) +

b q

q + cu(t)

known and independent from e(t + 1)!

+e(t + 1)



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We have the process

y(t + 1) + a y(t) = b u(t) + e(t + 1) + c e(t)


e(t) =q + a

q + cy(t)

b

q + cu(t)


y(t + 1) =(c a) q

q + cy(t) +

b q

q + cu(t)


+e(t + 1)

Clearly, var{y(t + 1)} var{e(t + 1)} =

2

.



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We have the process

y(t + 1) + a y(t) = b u(t) + e(t + 1) + c e(t)


e(t) =q + a

q + cy(t)

b

q + cu(t)


y(t + 1) =(c a) q

q + cy(t) +

b q

q + cu(t)


+e(t + 1)

The minimal variance is achieved with: u(t) = c a

b y(t).



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Minimum-variance control: General case


A(q) y(t) = B(q) u(t) + C(q) e(t)

where {e(t)} is white noise with var{e(t)} = 2.



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A(q) y(t) = B(q) u(t) + C(q) e(t)

where {e(t)} is white noise with var{e(t)} = 2.

Additional assumptions:

The polynomials A and B are monic (right rescaling).



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A(q) y(t) = B(q) u(t) + C(q) e(t)

where {e(t)} is white noise with var{e(t)} = 2

.


The polynomials A and B are monic (right rescaling). deg{A} = deg{C} = n 1 (always can be done).



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A(q) y(t) = B(q) u(t) + C(q) e(t)


.



deg{A} deg{B} = d0 1 known relative degree.



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A(q) y(t) = B(q) u(t) + C(q) e(t)


.




The polynomial C is stable (always can be done).



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A(q) y(t) = B(q) u(t) + C(q) e(t)


.




The polynomial C is stable (always can be done).

The polynomial B is stable a serious restriction.



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Minimum-variance control: General case (contd)

We have the process

A(q) y(t) = B(q) u(t) + C(q) e(t)



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We have the process

A(q) y(t) = B(q) u(t) + C(q) e(t)

Let us solve it for y(t)

y(t) =B(q)

A(q)u(t) +

C(q)

A(q)e(t)



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We have the process

A(q) y(t) = B(q) u(t) + C(q) e(t)

Shifting by the relative degree

y(t + d0) =B(q)

A(q)u(t + d0) +

C(q)

A(q)e(t + d0)



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We have the process

A(q) y(t) = B(q) u(t) + C(q) e(t)

Shifting by the relative degree and rewriting

y(t + d0) =B(q)

A(q)u(t + d0) +

C(q) qd01

A(q)e(t + 1)



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We have the process

A(q) y(t) = B(q) u(t) + C(q) e(t)

Shifting by the relative degree and rewriting

y(t + d0) =B(q)

A(q)u(t + d0) +

C(q) qd01

A(q)e(t + 1)

Using polynomial long division

C(q) qd01

A(q) = F(q) +

G(q)

A(q)

whereG(q)A(q)

is strictly proper and

deg{F} = deg{C(q) qd01} deg{A} = d0 1.



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We have rewritten the process as

y(t + d0) =B(q)

A(q)u(t + d0) +

F(q) +

G(q)

A(q)

e(t + 1)



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Finally the process can be represented by

y(t + d0) =qd0 B(q)

A(q) proper fraction

u(t) +q G(q)


e(t) + F(q) e(t + 1)



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y(t + d0) =qd0 B(q)


u(t) +q G(q)


e(t) + F(q) e(t + 1)

Solving the model for e(t), we have

e(t) =A(q)

C(q)y(t)

B(q)

C(q)u(t)



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y(t + d0) =qd0 B(q)


u(t) +q G(q)


e(t) + F(q) e(t + 1)


e(t) =A(q)

C(q)y(t)

B(q)

C(q)u(t)

After substituting it back

y(t+d0) =qd0 B

Au(t)+

q G

A

A

Cy(t)

B

Cu(t)

+F e(t+1)



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y(t + d0) =qd0 B(q)


u(t) +q G(q)


e(t) + F(q) e(t + 1)


e(t) =A(q)

C(q)y(t)

B(q)

C(q)u(t)

After substituting it back and collecting terms

y(t + d0) =q G

Cy(t) +

q B

A C

qd0 C G

u(t) + F e(t + 1)



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y(t + d0) =qd0 B(q)


u(t) +q G(q)


e(t) + F(q) e(t + 1)


e(t) =A(q)

C(q)y(t)

B(q)

C(q)u(t)

After substituting it back and collecting terms

y(t + d0) =q G

Cy(t) +

q B

C

(qd0 C G)

Au(t) + F e(t + 1)



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y(t + d0) =qd0 B(q)


u(t) +q G(q)


e(t) + F(q) e(t + 1)


e(t) =A(q)

C(q)y(t)

B(q)

C(q)u(t)

Substituting F(q) fromC(q) qd01

A(q) = F(q) +G(q)A(q)

y(t + d0) =q G(q)

C(q)y(t) +

q B(q) F(q)

C(q)u(t) + F(q) e(t + 1)



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y(t + d0) =q G(q)

C(q)y(t) +

q B(q) F(q)

C(q)u(t)

known and independent from F(q) e(t + 1)

+F(q) e(t + 1)


Mi i i l G l ( d)


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y(t + d0) =q G(q)

C(q)y(t) +

q B(q) F(q)

C(q)u(t)


+F(q) e(t + 1)

Clearly, var{y(t + d0)} var{F(q) e(t + 1)}


Mi i i t l G l ( td)


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y(t + d0) =q G(q)

C(q)y(t) +

q B(q) F(q)

C(q)u(t)


+F(q) e(t + 1)

var{y(t+d0)} var{F(q) e(t+1)} = 1 + f21 + + f

2d01

2

where F(q) = qd01 + f1 qd02 + + fd01.


Mi i i t l G l ( td)


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y(t + d0) =q G(q)

C(q)y(t) +

q B(q) F(q)

C(q)u(t)


+F(q) e(t + 1)


2d01

2

where F(q) = qd01 + f1 qd02 + + fd01.

The minimal variance is achieved with: u(t) =

G(q)

B(q) F(q) y(t) .


Minimum variance control: General case (contd)


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y(t + d0) =q G(q)

C(q)y(t) +

q B(q) F(q)

C(q)u(t)


+F(q) e(t + 1)


2d01

2

where F(q) = qd01 + f1 qd02 + + fd01.


G(q)

B(q) F(q) y(t) .The closed-loop system is

A(q) y(t) = B(q) u(t) + C(q) e(t)




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Minimum-variance control: General case (cont d)


y(t + d0) =q G(q)

C(q)y(t) +

q B(q) F(q)

C(q)u(t)


+F(q) e(t + 1)


2d01

2

where F(q) = qd01 + f1 qd02 + + fd01.


G(q)


A(q) y(t) = B(q) G(q)

B(q) F(q) y(t) + C(q) e(t)




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y(t + d0) =q G(q)

C(q)y(t) +

q B(q) F(q)

C(q)u(t)


+F(q) e(t + 1)


2d01

2

where F(q) = qd01 + f1 qd02 + + fd01.


G(q)


B(q) (A(q) F(q) + G(q)) y(t) = B(q) F(q) C(q) e(t)




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y(t + d0) =q G(q)

C(q)y(t) +

q B(q) F(q)

C(q)u(t)


+F(q) e(t + 1)


2d01

2

where F(q) = qd01 + f1 qd02 + + fd01.


G(q)

B(q) F(q) y(t) .

The closed-loop system (using: qd01 C = A F + G) is

qd0

1 B(q) C(q) y(t) = B(q) F(q) C(q) e(t)


Minimum-variance control: Remarks


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(1) The closed-loop system

qd01 B(q) C(q) y(t) = B(q) F(q) C(q) e(t)

defines the noise-to-output relation

y(t) =B(q) F(q) C(q)

qd01 B(q) C(q)e(t)




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Minimum variance control: Remarks




y(t) =F(q)

qd01e(t) =

qd01 + f1 qd02 + + fd01

qd01e(t)




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y(t) = (1 + f1 q1 + + f

d0+1) e(t)




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y(t) = e(t) + f1 e(t 1) + + fd0+1 e(t d0 + 1)

which is a moving average of order d0

1.




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u a a ce co t o e a s




y(t) = e(t) + f1 e(t 1) + + fd0+1 e(t d0 + 1)


1.

(2) The closed-loop poles consist of (a) zeros of B(q), (b) zerosof C(q), and (c) d0 1 zeros.




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y(t) = e(t) + f1 e(t 1) + + fd0+1 e(t d0 + 1)


1.

(2) The closed-loop poles consist of (a) zeros of B(q), (b) zerosof C(q), and (c) d0 1 zeros.

(3) The minimum variance controller

u(t) = G(q)

B(q) F(q)y(t)

can be interpreted as a pole-placement controller.


Minimum-variance control as pole-placement


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p p

Sinceqd01 C(q) = A(q) F(q) + G(q)

the closed-loop characteristic polynomial

Ac(q) = qd01 B(q) C(q) = A(q) B(q) F(q) R(q)

+B(q) G(q) S(q)




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p p




+B(q) G(q) S(q)

Note that for the Diophantine equation Ac = A R+ B S

deg{S(q)} = deg{G(q)} = n 1

andS/R

is proper since

deg{R} = deg{B} +deg{F} = (n d0) + (d0 1) = n 1.




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+B(q) G(q) S(q)

Note that for the Diophantine equation Ac = A R+ B S

deg{S(q)} = deg{G(q)} = n 1

andS/R

is proper since

deg{R} = deg{B} +deg{F} = (n d0) + (d0 1) = n 1.

Can we use an analogous pole-placement technique for thecase whenB(q) has unstable zeros?


Moving Average Controller


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Design the controller

u(t) = S(q)

R(q)y(t) for A(q) y(t) = B(q) u(t)+C(q) e(t)

as follows.




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u(t) = S(q)


as follows.

Factor B as B = B+ B, where B+ is monic and stable.




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u(t) = S(q)


as follows.


Set d = deg{A} deg{B+

}.




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u(t) = S(q)


as follows.



}. Find Rp and S solving the Diophantine equation

qd1 C = A Rp + B S, d d0




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u(t) = S(q)


as follows.



}. Find Rp and S solving the Diophantine equation

qd1 C = A Rp + B S, d d0

Let R = Rp B+, where deg{Rp} = d 1.

The obtained transfer function for the closed-loop system is

y(t) = q1d Rp(q) e(t) = e(t)+r1 e(t1)+ +rd1 e(td+1)


Example 4.3


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Consider the plant described by

(q2 + a1 q + a2) y(t) = (b0 q + b1) u(t) + (q2 + c1 q + c2) e(t)


Example 4.3


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1. If |b1/b0| < 1 minimum variance (MV) controller can be

designed noticing that

q11 (q2 + c1 q + c2)

q2 + a1 q + a2= 1 +

(c1 a1) q + (c2 a2)

q2 + a1 q + a2

as follows

u(t) = G(q)

B(q) F(q)y(t) =

(c1 a1) q + (c2 a2)

(b0 q + b1) 1y(t)


Example 4.3


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1. If |b1/b0| < 1 minimum variance (MV) controller can be

designed noticing that

q11 (q2 + c1 q + c2)

q2 + a1 q + a2= 1 +

(c1 a1) q + (c2 a2)

q2 + a1 q + a2

as follows

u(t) = G(q)

B(q) F(q)y(t) =

(c1 a1) q + (c2 a2)

(b0 q + b1) 1y(t)

2. If |b1/b0| > 1 minimum variance (MV) controller cannot bedesigned but we can apply moving average controller (MA) withd = 2 using the solution of the Diophantine equation

q (q2+c1 q+c2) = (q2+a1 q+a2)(q+r1)+(b0 q+b1)(s0 q+s1)


Next Lecture / Assignments:


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Next meeting (May 10, 13:00-15:00, in A208Tekn): Recitations.

Homework problems: Consider the process in Example 4.3

with a1 = 1.5, a2 = 0.7, b0 = 1, c1 = 1, and c2 = 0.2.

Determine the variance of the output in the closed-loop systemas a function of b1 when the Moving Average controller is used.

Compare with the lowest achievable variance.


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