Variable Structure Control ~ Disturbance

Rejection

Harry G. KwatnyDepartment of Mechanical Engineering & Mechanics

Drexel University

Outline

Linear Tracking & Disturbance Rejection

Variable Structure Servomechanism Nonlinear Tracking & Disturbance

Rejection

Disturbance Rejection Setup

state equationsdisturbanceperformancemeasurements

, , , ,n q m m r

x Ax Ew Buw Zwz Cx Fw Duy Cx Fw Du

x R w R u R z R y R

= + +== + += + +

∈ ∈ ∈ ∈ ∈

PlantController

performance variablesz

disturbancew

controlu

measurementsy

Eigenvalues of Zordinarily on Im axis

In effect, the error

Objectives

( )Regulator Problem: Find a controller to achieve the following

1) Regulation: 0 as 2) (Internal) Stability: Achieve specified transient response

Robust Regulator Problem: Find a solution to the Reg

z t t→ →∞

ulator Problem that satisfies

3) Robustness: 1) and 2) should be maintained under specified small perturbations of plant and/or control parameters

Solution: Part 1- Regulation( )

( )( )

0 0

Consider the possibility of a control that produces a trajectory

for some unspecified initial state and any initial disturbance vector ,

so that the corresponding 0. Then, , , must sati

u t

x t x w

z t x u w≡

( )

sfy

0Assume a solution of the form: ,

0Thus, the hypothesized control exists if there are , that satisfy

x Ax Ew Buw Zw

Cx Fw Dux Xw u Uw

XZw AXw Ew BUww

CXw Fw DUwu t X U

XZ AX BU ECX DU F

= + +== + +

= = ⇒= + +

∀= + +

− + + = −+ = −

Solutions typically are not unique

Solution: Part 2- Stability

( )( )

Define ,

Now, if , is controllable, it easy to choose

so that the closed loop has desired transientcharacteristics.With chosen, the control

u xu u u Uw u

x A x B ux x x Xw x

A B u K x

x A BK x

K

δ δδ δ

δ δ δδ δ

δ δ

δ δ

= + = +⇒ = +

= + = +

=

= +

( ) [ ]

can be written as a function of the systemstates ,

TOT

x wu u u Uw u Uw K x

x xu Uw K x Xw K U KX K

w w

δ δ δ= + = + = + ⇒

= + − = − =

Solution: Part 3- Observation

( )The control will be implemented using estimates of thecomposite state , . Consider the composite system

0 0

If the composite system is observable, we can cho

x w

x A E x Bd uw Z wdt

y Cx Fw

= +

= +

( )

ose a matrix, so that the following observer has the desired dynamics:

ˆ ˆˆ ˆ

ˆ ˆ0 0

Lx A E x Bd u L Cx Fw yw Z wdt

= + + + −

Properties of the Loop

U

Disturbance

PlantK

X

-

Observer

PlantObserver y

Disturbance

Compensator

u

wCompensator contains copy

of Z ~ internal model of disturbance

[ ]

[ ]

[ ]

ˆ ˆˆ0 0ˆ

ˆˆ

ˆ ˆˆ0 0ˆ

ˆˆ

x A E x BL C F u Ly

Z ww

xu K U KX

w

x A E B xL C F K U KX Ly

Z ww

xu K U KX

w

= + + −

= −

= + + − −

= −

Example

13s +

dc motor

,y motor speed

,y command

1,w load torque

controller

1s

1s

1s

Example

13s +

1w

x

K

3K− +2w

3wy

yu

1wcommandobserver

systemdisturbance

observer

−

13s +

yu

1w

3K− +2w

3wy command

observercompensator

−

( )19420.3 9.18897

49.83933 209csG

s s+

=++

VS Servo

U

Disturbance

Plant( )xψ ∆

X

-

Observer

( ) ( ) ( )( ) ( )

We make only one change

ˆ 0ˆ ˆ ˆ ˆ ˆ, ,

ˆ 0i i

ii i

u s xu K x u x x s x G x

u s xψ ψ

+

+

∆ ∆ >= ∆ ⇒ = ∆ ∆ = ∆ = ∆∆ ∆ <

Closed Loop Dynamics

( ) ( )

( ) ( )

( ) ( )

1 1

2 2

1 2

Com

ˆ

bine th

ˆ

ˆ ˆ ˆ ˆ

ˆˆ ˆ ˆ ˆ

e estimator equation:

ˆ ˆ

with the estimator

ˆˆ

ˆ ˆ ˆ ˆappl

equ

y

ˆ

ation

L C x x L F w w

d w Zw L C x x L F w wdt

x xd x Xw Ax Ew XZw L XL C F

d x A

BUw B uw wdt

XZw AXw Ew BUw

d xdt

x Ew Budt

δ

+ − + −

= + − + −

− − = + − + − + + −

+

= + +

= +

−( ) ( ) ( ) ( )1 2

ˆˆ ˆ ˆ ˆ

ˆx x

Xw A x Xw E w w L XL C F B uw w

δ−

= − + − + − + −

Closed Loop Dynamics, 2

( ) ( )1 2 1 2

1 1

2 2

ˆ ˆˆˆ ˆ0 0ˆ ˆ0 0

x x A L XL C L XL F x x Bd x x A L C E L F x x udt

w w L C Z L F w wδ

− − − − − = − − − + − − − −

Sliding Behavior

( )

( )

( ) ( )

1 2 1 2

111 2

2

2

1 1 2 1 2

1 1

ˆˆ ˆLet and define a new coordinatesˆ , , ,

ˆ ˆ

Note: 0, 0, sliding 0

ˆ 0ˆ 0

n m m

x x xx R R

Mx N B x

KB K

MB KN

MAN M L XL C M L XL Fd x x A L C E L Fdt

w w

δ

δ δχ δχ δχ δχ

δχδ δχ δχ δ

δχ

δχ

δχ

−

−

= −

∈ ∈

= + ⇔ =

= = ⇔ ≡

− − − = − − − −

1

2 2

ˆˆ

x xL C Z L F w w

δχ − − −

Reaching Behavior

( )

Assume the switching control is designed to stabilizes theswitching manifold 0 for the perturbation system

,Two important results follow

trajectories are steered in finite time to

K xx A x B s s K x

ψδ

δ δ ψ δ=

= + =

•

a domain that ˆ contains the subspace 0

ˆ shrinks exponentially to the subspace 0K x

K xδ

δ=

• =

D

D

Reaching, 2( )

( )( )

ˆ For a fixed >0, there exists a finite time such that is confined to the domain

ˆ , 1, , , .

Moreover, 0, . this means that sliding does not occur on 0 but this mani

Ti

T x

k x i m t T

Ts

δ

δ

∆ ∆

≤ ∆ = ∀ ≥ ∆

∆→ ∆ →∞

• =

Theorem :

( ) ( ) ( ) ( )( ) ( )*

fold is reached ˆ ˆ asymptotically as ,

ˆ ˆ let denote an ideal sliding solution. can be viewed as non-ideal

sliding in that it can be shown that there exists a constant such th

x t x t w t w t

x t x t

c

δ δ

→ →

•

( )( ) ( )

( ) ( )( ) ( )

*

at for all

ˆ ˆ

The performance variables can be expressed asˆ ˆ

ˆ ˆ as , we have 0, 0, 0 0eq

t T

x t x t c

z C x x F w w D u

t x x w w u u z

δ δ

δ

δ δ

≥ ∆

− ≤ ∆

•

= − + − +

→∞ − → − → → → ⇒ →

Example: Drum Level Control

Drum

Downcomer

Riser

Feedwater Valves

Throttle Valve

Drum Level, 2N number of riser sectionsLdo,L downcomer length and riser section length (total riser length/N)Ado,A downcomer, riser cross section areaswi mass flow rate at ith nodePi pressure at ith nodeTi temperature at ith nodesi aggregate entropy at ith nodevi specific volume at ith nodewr,wdc,ws mass flow rates, riser, downcomer and turbine, respectivelyvdf,vdg drum specific volume, liquid and gas, respectivelyPd drum pressureTd drum temperatureV total drum volumeVw volume of water in drumxd net drum quality, xd=Vw/Vws0 throttle flow at rated conditionsPd0 drum pressure at rated conditionsAt normalized throttle valve position, at rated conditions At=1

Drum Level, 3 u1 = q, u2 = ωe, u3 = At

dωav

dt = f1(ωav,s1,s2,s3,Pav,Pd)

ds1dt = f2(ωav,s1,Pav)+g21(Pav,s1)u1+g22(ωav,Pd)u2

ds2dt = f3(ωav,s1,s2,Pav)+g31(Pav,s2)u1

ds3dt = f4(ωav,s2,s3,Pav)+g41(Pav,s3)u1

ds4dt = f5(ωav,s3,s4,Pav)+g51(Pav,s4)u1

dPav

dt = f6(ωav,s1,s2,s3,s4,Pav,Pd)+g61(ωav,s1,s2,s3,s4,Pav)u1

dPddt = f7(ωav,s1,s2,s3,s4,Pav,Pd,Vw)+g71(ωav,s1,s2,s3,s4,Pav,Pd,Vw)u1+g72(Pd,Vw)u2-g73(Pd,Vw)u3

dVw

dt = f8(ωav,s1,s2,s3,s4,Pav,Pd,Vw)+g81(ωav,s1,s2,s3,s4,Pav,Pd,Vw)u1+g82(Pd,Vw)u2-g83(Pd,Vw)u3

y1 = Pd, y2 = , = h2(Vw), y3 = ωs = h3(Pd)+d3(Pd)u3

Linearized Dynamics, Poles

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

-1 -0.8 -0.6 -0.4 -0.2 0 0.2

real

imag

inar

y

Poles as a function of load level, 5%-100%

Linearized Dynamics, Zeros

Transmission Zeros

Transmission zeros

Drum Level, Normal Form

1s

At

ω e

q

,

Pd

ω s

y1

y2

y3

u1

u3

u2 Valve Actuator

Valve Actuator

FurnaceDrum

& Circulation

Loop

1s

,

Pd

ω s

y1

y2

y3

v 1

v 3

v 2

1s

1s

1s

1= z2z

3= z

5= z

1s

1s

4z

6z

Drum Level, Switching Controller

( )

( )

( ) ( ) ( )

*11 1 1

2 2 2 2

*3 3 3 3

* *

/0.50.50.0001 0.02 /

sgn , 1,2,3

0.0071 0.09436 0.9350 5 0 00.0031 0.06793 0.1461 , 0 0.1 0

0 0 0.9176 0 0 0.001

d d d

s s s

Ti i i

z P P Ps z zs z z zs z z z dt

u x U s i s Qs z

Q

ω ω ω

ρ

ρ

= −= += + == + = −

= − = =

− ≈ − =

∫

Drum Level - Conventional, PID with Steam/Water FF load change 80-75%

0.98

0.99

1

1.01

0 50 100 150 200time (s)

Drum Pressure

-0.2

0

0.2

0.4

0 50 100 150 200time (s)

Drum Level

0.74

0.75

0.76

0.77

0 50 100 150 200time (s)

Steam Flow

0.7

0.75

0.8

0.85

0 50 100 150 200time (s)

Feedwater Flow

Drum Level - Conventional, PID load change 15-20%

0.78

0.8

0.82

0.84

0 50 100 150 200time (s)

Drum Pressure

-1

0

1

2

0 50 100 150 200time (s)

Drum Level

0.19

0.195

0.2

0.205

0 50 100 150 200time (s)

Steam Flow

0

0.1

0.2

0.3

0 50 100 150 200time (s)

Feedwater Flow

Drum Level - VS Control, load change 80-75%

0.985

0.99

0.995

1

0 100 200 300 400time (s)

Dru

m P

ress

ure

-0.2

0

0.2

0 100 200 300 400time (s)

Dru

m L

evel

0.75

0.8

0.85

0 100 200 300 400time (s)

Stea

m F

low

0.65

0.7

0.75

0.8

0 100 200 300 400time (s)

Feed

wat

er F

low

, , , ,e t d sQ A Pω ω→

Drum Level – VS Control 15-20%

0.8

0.81

0.82

0.83

0 50 100 150 200time (s)

Dru

m P

ress

ure

-0.2

0

0.2

0.4

0 50 100 150 200time (s)

Dru

m L

evel

0.1

0.15

0.2

0.25

0 50 100 150 200time (s)

Stea

m F

low

0.1

0.2

0.3

0.4

0 50 100 150 200time (s)

Feed

wat

er F

low

Regulation of Nonlinear Systems

The nonlinear regulator problem Exponential stabilizability and detectability Existence and construction of controllers

The (local) Nonlinear Regulator Problem

( )( )( )( )

( ) ( ) ( ) ( )

, , statedisturbanceerror,measurement,

Assume an equilibrium point at the origin, i.e.0 0,0,0 , 0 0 , 0 0,0 , 0 0,0

Assume disturbance dynamics are neutrally stable

x f x w u

w w

z h x w

y g x w

f h g

ω

ω

=

=

=

=

= = = =

The Nonlinear Regulator Problem, 2

( ) ( ) ( )( )

( ) ( ) ( )( )( )( )

( )( )

Consider either a state feedback controller:

, , : , , , ,

or an output feedback controller

, , , ,, , , , , :

, ,cl

ncl cl cl cl cl cl

rcl cl cl

xx

v

u k x w x f x w f x w k x w x x R

f x w v g x wu v y v v y v R x f x w

v g x w

ηη φ

φ →

= ⇒ = = = ∈

= = ∈ ⇒ = =

Local

( )( )( )

( )

Determine a feedback control law such that1) , for each 0 , a neighborhood of the origin, the closed

loop has an exponentially stable trajectory ,

2) , 0, as

w W

x w t

z t

∈

→

Regulator Problem :Stability

Regulation ( ) ( ) for each 0 and 0 , a neighborhood of the origin.

clt w W x X→∞ ∈ ∈

Exponential Stabilizability and Detectability

( ) ( )( ) ( )

( )( )

: The system, with 0,0 0 is exponentially stabilizable

if there exists a feedback control with 0 0

such that , is exponentially stable.

x f x u f

u k x k

x f x k x

= =

= =

=

Definition (Exponential Stabilizability)

Definiti

( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( )

: The system, , with 0,0 0 and 0 0 is

exponentially detectable around the origin if there exists

a system , , where 0,0 0, , ,0

and 0 is an expone

n

x f x u y g x f g

y R x g x f xξ γ ξ ξ γ γ

ξ

= = = =

= ∈ = =

=

on (Exponential Detectability)

( )ntially stable equilibrium point of ,0 .ξ γ ξ=

Remarks

Regulation of Feedback Linearizable Systems - 1

( ) ( )( )( )( )

( ) ( ) ( ) ( )

2, statedisturbanceerror,measurement,

Assume an equilibrium point at the origin, i.e.0 0,0 , 0 0 , 0 0,0 , 0 0,0

Assume disturbance dynamics are neutrally stable

x f x w G x u

w w

z h x w

y g x w

f h g

ω

ω

= +

=

=

=

= = = =

Regulation of Feedback Linearizable Systems - 2

( )( ) ( )

( ) ( ){ }( ) ( )( )

1 1 2

2 0 2 0

0 2

1

2 0 0

, ,

, ,

choose: , ,

take , so that Re 0

lim 0t

F u

A E x w x w u

z C

u x w x w v

v K x w A E K

z t

ξ ξ ξ

ξ ξ α ρ

ξ

ρ α

ξ λ

−

→∞

=

= + + =

= − +

= + <

⇒ =

Example: SISO Linear Systemsx Ax Ew buw Zwz cx fw

z cAx cEw fZw cb

= + +== +

= + + +

[ ]0A E x

u c f cbZ w

= +

[ ]2

0

u

A E xz c f cAb

Z w

= +

( ) [ ]

[ ]

1

1

0

,0

rr r

rr

u

A E xz c f cA bu

Z w

A E xcA b c f

Z wρ α

−

−

= +

⇒ = =

Example Continued( )

( )

[ ]

[ ]

( )

1 2 2

1

1

1

1

1

1 2 2 1

3

2

2 2 1

1

r r r r r

r

r

r r r r r

r r r r r

r

xcA cE A A Z AZ Z

w

x

cA b

ucA b

c f

cA cE fZ

cA cE A A Z Z

cA cE A A Z AZ Z vw

xw

xw

wfZ

x

ρ

ξ

ξ

ξ

α − − − −

− − −−

− − −

−

−

− −

= + + + +

+ + + + +

=

= −

=

=

+

+ + +

=