CS 367: Model-Based ReasoningLecture 11 (02/19/2002)

Gautam Biswas

Today’s Lecture

Today’s Lecture: Finish up Supervisory Control Onto Modeling of Continuous Systems: The Bond

Graph Approach

Supervisory Controller: Examples

Admissible strings: a1 precedes a2 iff b1 precedes b2

Build trim automata Ha such that Lm(Ha) contains only those strings that contain the above ordering constraints

Is Ha blocking?

In general, how do we build supervisors? If all events controllable and observable:

)()/()()/( 11 amma HLGSLandHLGSL

Realizing Supervisors

How to build an automaton that realizes S?

Build an automaton that marks K, i.e.,

)/()()()(

)/()()()()(

)()(

),,,,,,( 0

GSLGLRLGRL

GSLKGLKGLRLGRL

KRLRL

trimisRwhereYygEYR

mmmm

m

R

Note that R has the same event set as G, therefore,

Control action S(s) is encoded into transition structure of R

GRGR

))),,(((

)),((

}:{))],(([)(

00

0

0

sxyfg

syg

KsEsxfEsS

GR

R

cuc

Standard Realization of S

Start with G in state x, R in state y, following the execution of

G generates that is currently enabled, i.e., this event set is present in R’s active event set at y

R executes the event as a passive observer of G and the system now moves into states x’ and y’

Set of enabled events of G given by active event set of R at y’

)/( GSLs

Induced Supervisor

Reverse Question: Given C, can the product CG imply that C is controlling G

Depends on the controllability of L(C)

The supervisor for G induced by C is

ucCi EandGLwrtlecontrollabisCLiffGCLGSL )()()()/(

CiS

Reduced State Realization

L(S/G) = K may not be the most economical way to represent SS in terms of an automata (memory requirements)Relax requirements L(R) = K, andCome up with

Collapse 2,5,6,7, and 8 into one state

KRL rs )()/()( GSLGRL rs

Controllable sub languages and super languages of an uncontrollable language

K is not controllable wrt M and Euc

Two languages derived from K: The supremal controllable sub language K:

(Inside K) The infimal prefix-closed and controllable super

language of K: (Outside K)

MK

KMEK uc

CK

CK

MKKKK CC

K

Example: Supremally Controllable Language

2

121211222

1

12211121211221

21

22

2211212112121122

},{

Re

},,{

Re

},{

},,,{

)()(

KK

lecontrollabisthisbbaababaK

prefixasacontainthatstringsallmove

lecontrollabnotKbababbaababaK

prefixasaacontainthatstringsallKfrommove

ableuncontrollKmakesbaE

bababbaabbaababaK

HLkGLM

C

uc

am

Infimal Prefix-closed controllable language

}{

1

&

221

21

baaKK

lengthof

eventsableuncontrollofstringwithaastringExtend

beforeasKM

C

Supervisory Control Problems

BSCP: Basic Supervisory Control Problem Given G with event set E, and Euc E, and an admissible

language La = La L(G) find supervisor such that

Look up standard realization presented couple of lectures ago (sec. 3.4.2)

DuSCP: Dual Version of SCP:minimum required language Lr L(G)

)/()/()/(

.,.,)/(

)/(

GSLGSLLGSL

eibecanitlargesttheisGSL

LGSL

otheraother

a

)/()/()/(

.,.,)/(

)/(

GSLGSLLGSL

eibecanitmalleststheisGSL

LGSL

otherrother

r

Supervisory Controller Problems

SCPT: Supervisory Controller with Tolerance Ldes: desired language, try and achieve as much of it as

possible Ltot: tolerated language, do not exceed tolerated

langauge Solution:

Cdes

Ctol LLGSL )()/(

Non Blocking Supervisors

Controllable:

Non blocking: Lm(G) closure:typically holds by construction of KSupervisory Controller with BlockingTypically use two measures:

Blocking Measure: Satisficing Measure:

BM(S) and SM(S) conflicting, i.e., reducing one may increase the other

K G L E Kuc ) (

)(GLKK m

)/()(

)/()(

)/(\)/()(

GSLSSM

LGSLSSM

GSLGSLSBM

m

amm

m

Modular Control

Supervisor S combines the actions

of two or more supervisors, e.g.,

S1 and S2

We can always build R = R1 R2, but the point is to use R1 and R2 and take the active event sets of both at their respective states after execution of s

)/()/()/(

)/()/()/(

)()()(

2112mod

2112mod

2112mod

GSLGSLGSL

GSLGSLGSL

sSsSsS

mmm

Modular Control Example: Dining Philosophers

Philosopher i picks up for j is controllable

Philosopher putting down fork is uncontrollable

Remember there is only one marked state

Design two supervisors: one for each fork

2f

(1T,

Modular Control Example: Dining Philosophers

Modular supervisor Smod12 = R1 R2 G

Did not cover

Unobservability

Decentralized Control

Modeling of Continuous

Dynamic Systems

The Bond Graph

Bond Graph Methodology

From Systems Dynamics

•formal and systematic method for modeling physical systems

•forces one to make explicit: issues about system functionality and behavior assumptions

•unlike other modeling schemes…

directly grounded in physical reality…

1-1 correspondence with components and mechanisms of the physical system modeled…

(as opposed to formal languages, such as logic)

Bond Graphs… Modeling Language

(Ref: physical systems dynamics – Rosenberg and Karnopp, 1983)

NOTE: The Modeling Language is domain independent…

Bond Connection to enable Energy Transfer among components

(directed bond from A to B).

each bond: two associated variables effort, e flow, f

A Bef

Bond Graphs

•modeling language (based on small number of primitives)•dissipative elements: R•energy storage elements: C, I•source elements: Se, Sf

•Junctions: 0, 1

physical system mechanisms

R C, I Se, Sf 0,1forces you to make assumptions

explicituniform network – like representation:domain indep.

Generic Variables:

Signals effort, e elec. mechanicalflow, f voltage force

current velocityNOTE: power = effort × flow.

energy = (power) dt.

state/behavior of system: energy transfer between components…

rate of energy transfer = power flowEnergy Varibles

momentum, p= e dt : flux, momentumdisplacement, q = f dt : charge,

displacement

Effort Flow PowerEnergyMechanics Force, F Velocity, V FxV F. V.

Electricity Voltage, V Current, I VxI VI

Hydraulic Pressure, P Volume PxQ PQ(Acoustic) flow rate

(Q)

Thermo- Temperature, Entropy Q Q dynamics T flow rate

(thermal flow rate) Pseudo

Examples:

S

Q