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Adaptive Pole Placement (or pole-assignment)

In pole placement we aim to place the poles of

the closed loop transfer function in reasonable

positions.

controller

command signal

calculations estimation

parameter

output y(t)Process

signal

control

u(t)

Block diagram of a self tuning regulator

1. Recursive least squares (RLS) for the iden-

tification.

2. Pole-placement for the controller design.

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Where Process dynamics can be circuits, ac-

tuator, sensors, whose behavior can be slow,or even unstable. e.g. robotics, automo-

bile, magnetic suspension systems.

Why It is possible to shape the closed system

response by assigning closed loop poles.The simple design principle is by assigning

closed loop poles as desired, making the

closed loop system (under control) faster

and stable when following command signal.

How Based on polynomials manipulation.

e.g. Process

B(q1)

A(q

1)

= 0.00415(1 + 0.65q1) ( 1 + 0.35q1)

(1 0.95q1)(1 0.9q1) ( 1 + 0.65q1)(1)

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Desired closed-loop denominator of

T(q1) = 1 0.21q1 + 0.386q2 (2)

Both open and closed loop systems have a

steady state gain of 1.

0 200 400 600 800 1000 12001.5

1

0.5

0

0.5

1

1.5

t

systemre

sponses

command signalclosed loopopen loop

Result of Matlab example polas.m

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Start with

Ay(t) =Bu(t 1) + e(t) (3)

and

u(t) = G

Fy(t) (4)

(dont know what F, G are yet !)

Closed loop equation

Ay(t) = q1BG

F y(t) + e(t) (5)

or closed loop equation

[AF+ q1

BG] closed loop denominator

y(t) =F e(t) (6)

The dynamics of the closed loop are governby the denominator, and we would like this tofollow something sensible, say

T = 1 + t1

q1 + ... + tnqn (7)

We choose t1, t2,..., etc, as we desire. F, Gare then found as the solution to

AF+ q1BG=T (8)

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Example: A= 1 + a1q1, B =b1. Substituting

this into

AF+ q1BG=T (9)

Set F = 1,

(1 + a1q1)(1)+ q1b1(g0 + g1q

1) = 1 + t1q1

(10)gives

g0 =t1 a1

b1, F = 1. (11)

We choose t1. Let t1= 0, 5.

The controller is

u(t) = g0y(t) (12)

Then for a1 = 0.5, b1= 2, g0 = 0.5.

For the real time self-tuning controller, use

RLS to find a1, b1, then enter to

g0 =t1 a1

b1(13)

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Adaptive Pole Placement (or pole-assignment)

The idea is based on servo following.

A(q )1

1C(q )

A(q )

B(q )1

1

y(t)

noise e(t)

+

+

G(q )1

H(q )1

1

F(q )

1

set point

r(t)

plant

u(t)

Controller

Adaptive pole-placement control structure

The feedback control structure ( with pre- com-pensator ) is the above, plus noise dynamics tobe controlled

Ay(t) =Bu(t 1) + Ce(t) (14)

which has at least a one-step delay in input,and the control is of the form

F u(t) =Hr(t) Gy(t) (15)

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where

F(q1) = 1 + f1q1 + ... + fnfq

nf, (16)

G(q1

) =g0+ g1q1

+ ...gngqng

, (17)

H(q1) =h0+ h1q1 + ...hnhq

nh. (18)

The closed loop description of the system is

[AF+ q1BG]

closed loop denominator

y(t) =q1BHr(t)+CF e(t)

(19)

The dynamics of the closed loop are govern

by the denominator, and we must specify by

assigning stable poles (inside unit circle).

T(q1) = 1 + t1q1 + ... + tnqn (20)

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Basic Idea of Adaptive Pole-Placement Control

1. The customer specifies the closed loop poles

in T.

2. The plant is given as InputOutput data

with underlying dynamics governed by B/A,

which we have to estimate. (system iden-

tification)

3. The control designer is to choose the con-

troller polynomials F,G,H, such that the

desired closed loop response is obtained.

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Pole assignment by

AF+ q1BG=T C, (21)

which may have many solutions, can cancel the

noise polynomial.

To get a unique solution, we should selectnf =nb, ng =na 1, nt na+ nb nc.

1. Often, C= 1.

2. A, B must be coprime (no common factors)

3. C(which has been cancelled) must be sta-

ble.

4. SolvingF, G from (21) is a problem of solv-

ing the Diophatine equation.

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Solving the Diophatine equation.

We will consider a reduced version (servo con-

trol), using a noise-free model

Ay(t) =Bu(t 1) (22)

with a controller

F u(t) =Hr(t) Gy(t) (23)

giving in a closed loop

y(t) = BH

F A + q1BGr(t 1) (24)

We need to find the polynomial F, G such that

T =AF+ q1

BG, (25)

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Example: We have a model with na= 3, nb=

2. We only want to place one pole, T = 1 +t1q

1.

nf = nb = 2, ng = na 1 = 2, nt = 1