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CS 367: Model-Based Reasoning Lecture 13 (02/26/2002)

CS 367: Model-Based Reasoning Lecture 13 (02/26/2002). Gautam Biswas. Today’s Lecture. Today’s Lecture: Modeling of Continuous Systems: The Bond Graph Approach (Basics) Next Lecture: Causality and Bond Graphs State Space Equations More Complex Examples 20-sim.

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CS 367: Model-Based Reasoning Lecture 13 (02/26/2002)

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  1. CS 367: Model-Based ReasoningLecture 13 (02/26/2002) Gautam Biswas

  2. Today’s Lecture • Today’s Lecture: • Modeling of Continuous Systems: The Bond Graph Approach (Basics) • Next Lecture: • Causality and Bond Graphs • State Space Equations • More Complex Examples • 20-sim

  3. 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 e f B A

  4. 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 systemmechanisms • R C, I Se, Sf 0,1 • forces you to make assumptions • explicit uniform network – like representation: domain indep.

  5. Generic Variables: Signals effort, eelec. mechanical flow, f voltageforce currentvelocity NOTE: power = effort×flow. energy = (power) dt. 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

  6. Examples: Effort Flow Power Energy Mechanics Force, F Velocity, V F x V  F. V. Electricity Voltage, V Current, I V x I  V. I Hydraulic Pressure, P Volume P x Q  P.Q (Acoustic) flow rate, Q Thermo- Temperature, Entropy Q  Q dynamics T flow rate (thermal flow rate) Pseudo bond graph

  7. Constituent Relations R, C, I : passive 1-ports -- one port through which they exchange energy. R R: resistor In mechanical systems = DASHPOT F = b•V R Sys • F V R: b e f R R v = i•R (Electricity) e = R•f F v (e) F Linear (R constant) non linear e (t) = R (f)• f (t) V (f)

  8. e f C P Q P C Q P Q I One Port Elements (continued ….) I: Inertia C: Capacitor g F m x F C:k P =P1 -P2 F:e P1 Q P2 Q linear non linear I V:f

  9. Tetrahedron of State e dt R E, f : Power Variables ; P = e.f. P, x : Energy Variables C p =  e d t P X I dt f x =  f d t e = R.f Dissipator (Instantaneous) x = c . e Capacitor Potential e = 1/c . x = 1/c  f d t Energy p = I . fInductor Kinetic f = 1/I . P = 1/I  e d t Energy Integrating

  10. e(t) f Se I:m Se Two Other 1 – Ports Flow Source f(t) e Effort Source Sf e(t) independent of flow f If e(t) = constant Constant Effort Source f(t) independent of effort e If f(t) = constant Constant Flow Source F m F Drive System F v F v(t) v(t) C:k Sf

  11. 1-junction: Common Flow junction F3 v3 F1 v1 1 F2 v2 equivalent of series junction How To Connect Elements: Ideal 3 – Ports 0-junction: Common Effort junction F3 v3 F1 v1 0 F2 v2 equivalent of parallel junction F1 = F2 = F3 = F There isno loss of energy at Junction; net power in = net power out therefore, F3v3 = F1v1 + F2v2 i.e.,v3 = v1 + v2 In general, v1 = v2 = v3 = v There isno loss of energy at Junction; net power in = net power out therefore, F3v3 = F1v1 + F2v2 i.e.,F3 = F1 + F2 In general,

  12. F2 v2 F1 v1 F3 v3 1 k F(t) b m P2 C:k 1 P3 I:m b:R P1 F(t) Se Ideal 3 – Port Junctions v V1 = V2 = V3 = V :single flow var. F3 – F1 = F2 or, F1 + F2 = F3 algebraic sum of effort vars = 0 F1 F3 F2 No power loss

  13. V1 V2 F3 v3 F1 v1 F2 v2 F1 F2 1 Q3 .P3 Q1 Q2 .P1 P2 0 – junction: dual of the 1 – junction Common Force Junction V3 = V1 – V2 = rate of compression R F1 = F2 = F3 = F common effort V3 = V1 – V2sums of flow = 0

  14. Others: 2 – Port Elements Transformers & Gyrators e1 e1 e2 GY TF f2 f1 f2 f1 r b a e2 e2 = (b/a) . e1 e1 = r . f2 f1 = (b/a) . f2 r . f1 = e2 Again: e1. f1 = (a/b) . e2 (b/a) . f2 = e2. f2 e1. f1 = r . f2 (1/r) . e2 = e2. f2 No power loss

  15. F Examples: Two ports i2 i1 Example 1: Lever • • a b e1 e2 • • Example 2: Electrical Transformer F1 F2 V F1 = (b/a) . F2 V2 = (b/a) . V1 Q P Example 3: Piston

  16. Se V1 V2 k2 k1 F(t) m1 m2 b V3 = V1 – V2 Let’s model:

  17. Example: V1 V2 k2 k1 F(t) m1 m2 r (no friction) I:m1 I:m2 F(t) Se C:k1 1 1 V1 V2 0 C:k2 R:b 1 V3 = V1 – V2

  18. I:m1 I:m2 Se C 1 0 1 V1 V2 R 1 C:k2 V3 Enforces the desire velocity relation. 3 Components : How To Connect

  19. Another example: V1 V2 k2 k1 R F(t) m1 m2 I:m1 C:k1 I:m2 F(t) Se C 1 0 1 V1 V2 m1. a = - k1x1 – k2x2 m2. a = F(t) – R . V3 R

  20. R2 R4 i3 i2 i4 I5 C3 E Switch Domains R4 R2 i4 ea e2 i2 eb e5 E Se 1 0 1 I5 i4 i4 i2 i2 e3 i3 i1 = i3 = i4 E – i2 . R2 = e3 = eb C3

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