CS 367: Model-Based Reasoning Lecture 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: L ( S / G ) L ( H ) and L ( S / G ) L 1 a m 1 m (H a ) Realizing Supervisors How to build an automaton that realizes S? Build an automaton that marks K, i.e., R (Y , E , g , R , y 0 , Y ), where R is trim Lm ( R ) L ( R ) K L ( R G ) L ( R ) L (G ) K L (G ) K L ( S / G ) L m ( R G ) Lm ( R ) Lm (G ) Lm ( S / G ) Note that R has the same event set as G, therefore, R G RG Control action S(s) is encoded into transition structure of R S ( s ) [ E uc ( f ( x 0 , s ))] { E c : s K } R ( g ( y 0 , s )) R G ( g f (( y 0 , x 0 ), s )) Standard Realization of S Start with G in state x, R in state y, following the execution of s L ( S / G ) 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’ 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 S C i L ( S i / G ) L ( C G ) iff L ( C ) is controllab le wrt L ( G ) and E uc C Reduced State Realization L(S/G) = K may not be the most economical way to represent S in terms of an automata (memory requirements) Relax requirements L(R) = K, and Come up with L ( R rs ) K L ( R rs G ) L ( S / G ) Collapse 2,5,6,7, and 8 into one state Controllable sub languages and super languages of an uncontrollable language K is not controllable wrt M and Euc K E uc M K K M Two languages derived from K: The supremal controllable sub language K: K C K (Inside K) The infimal prefix-closed and controllable super language of K: K C (Outside K) K C K K K C M Example: Supremally Controllable Language M L (G ) k L m ( H a ) K { a 2 b 2 a 1b1 , a 2 a 1b 2 b1 , a 1 a 2 b1b 2 , a 1b1 a 2 b 2 } E uc { a 2 , b 2 } makes K uncontroll able Re move from K all strings that contain a 1 a 2 as prefix K 1 { a 2 b 2 a 1b1 , a 2 a 1b 2 b1 , a 1b1 a 2 b 2 } K 1 not controllab le Re move all strings that contain a 1 as prefix K 2 { a 2 b 2 a 1b1 , a 2 a 1b 2 b1 } this is controllab le K C K2 Infimal Prefix-closed controllable language M & K as before Extend string a1 a 2 with string of uncontroll able events of length 1 K C K { a1 a 2 b 2 } 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 L ( S / G ) La L ( S / G ) is the largest L ( S other / G ) L a it can be , i .e ., L ( S other / G ) L ( S / G ) Look up standard realization presented couple of lectures ago (sec. 3.4.2) DuSCP: Dual Version of SCP:minimum required language Lr L(G) L ( S / G ) Lr L ( S / G ) is the s mallest L ( S other / G ) L r it can be , i .e ., L ( S other / G ) L ( S / G ) 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 C C Solution: L ( S / G ) ( L tol L des ) Non Blocking Supervisors K E uc L ( G ) K Controllable: Non blocking: Lm(G) closure: K K L m (G ) typically holds by construction of K Supervisory Controller with Blocking Typically use two measures: Blocking Measure: Satisficing Measure: BM ( S ) L ( S / G ) \ L m ( S / G ) SM ( S ) L m ( S / G ) L am SM ( S ) L m ( S / G ) BM(S) and SM(S) conflicting, i.e., reducing one may increase the other Modular Control Supervisor S combines the actions of two or more supervisors, e.g., S1 and S2 S mod 12 ( s ) S 1 ( s ) S 2 ( s ) L ( S mod 12 / G ) L ( S 1 / G ) L ( S 2 / G ) L m ( S mod 12 / G ) L m ( S 1 / G ) L m ( S 2 / G ) 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 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 (1T, 2f 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 e B A f (directed bond from A to B). each bond: two associated variables effort, e flow, f 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 forces you to make assumptions explicit C, I Se, Sf 0,1 uniform network – like representation: domain indep. Generic Variables: Signals effort, e flow, f NOTE: power = effort × flow. energy = (power) dt. elec. voltage current mechanical force velocity state/behavior of system: energy transfer between components… rate of energy transfer = power flow Energy Varibles momentum, p= e dt : flux, momentum displacement, q = f dt : charge, displacement Examples: Mechanics Effort Force, F Flow Velocity, V Power FxV Electricity Voltage, V Current, I VxI VI Hydraulic (Acoustic) Pressure, P Volume flow rate (Q) PxQ PQ ThermoTemperature, dynamics T Entropy Q flow rate S (thermal flow rate) Pseudo Energy F. V. Q Q

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