1. CS 367: Model-Based Reasoning Lecture 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 A B efef

4. Bond Graphs modeling language (based on small number of primitives) dissipative elements: R energy storage elements: C, I source elements: S e, S f Junctions: 0, 1 physical systemmechanisms R C, I S e, S f 0,1 forces you to make assumptions explicit uniform network – like representation: domain indep.

5. Generic Variables: Signalseffort, eelec.mechanical flow, fvoltageforce 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. EffortFlowPowerEnergy MechanicsForce, FVelocity, VF x V   F. V. ElectricityVoltage, VCurrent, IV x I  V. I HydraulicPressure, P VolumeP x Q  P.Q (Acoustic)flow rate, Q Thermo-Temperature, EntropyQ  Q dynamics Tflow rate (thermal flow rate) Pseudo bond graph Examples:

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

8. F:e V:f linear C: Capacitor non linear F P Q PQPQ C efef C g F C:k One Port Elements (continued ….) I: Inertia m x P1QP1Q P2QP2Q P = P 1 - P 2 PQPQ I I

9. Tetrahedron of State e  dt R  dt f P X I C x =  f d t E, f : Power Variables ; P = e.f. P, x : Energy Variables e = R.fDissipator (Instantaneous) x = c. eCapacitorPotential e = 1/c. x = 1/c  f d tEnergy p = I. f InductorKinetic f = 1/I. P = 1/I  e d tEnergy Integrating p =  e d t

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

11. How To Connect Elements: Ideal 3 – Ports v 1 = v 2 = v 3 = v There is no loss of energy at Junction; net power in = net power out therefore, F 3 v 3 = F 1 v 1 + F 2 v 2 i.e., F 3 = F 1 + F 2 In general, 1 F1v1F1v1 F3v3F3v3 F 2 v 2 1-junction: Common Flow junction equivalent of series junction 0 F1v1F1v1 F3v3F3v3 F 2 v 2 0-junction: Common Effort junction equivalent of parallel junction F 1 = F 2 = F 3 = F There is no loss of energy at Junction; net power in = net power out therefore, F 3 v 3 = F 1 v 1 + F 2 v 2 i.e., v 3 = v 1 + v 2 In general,

12. Ideal 3 – Port Junctions v F1 F1 F2F2 F3F3 V 1 = V 2 = V 3 = V :single flow var. F 3 – F 1 = F 2 or, F 1 + F 2 = F 3 algebraic sum of effort vars = 0 P1P1 P2P2 P3P3 b m F(t) k I:m 1 b:R SeSe C:k F(t) No power loss F1v1F1v1 F 2 v 2 F3v3F3v3 1

13. 0 – junction: dual of the 1 – junction Common Force Junction V1V1 V2V2 F1F1 F2F2 V 3 = V 1 – V 2 = rate of compression Q3Q3.P3.P3.P1.P1 P2P2 Q1Q1 Q2Q2 F1v1F1v1 F 3 v 3 F2v2F2v2 1 R F 1 = F 2 = F 3 = Fcommon effort V 3 = V 1 – V 2 sums of flow = 0

14. Others: 2 – Port Elements Transformers&Gyrators e2e2 e1e1 f1f1 TF e2e2 f2f2 b a e1e1 f1f1 GY f2f2 r e 2 = (b/a). e 1 e 1 = r. f 2 f 1 = (b/a). f 2 r. f 1 = e 2 Again: e 1. f 1 = (a/b). e 2 (b/a). f 2 = e 2. f 2 e 1. f 1 = r. f 2 (1/r). e 2 = e 2. f 2 No power loss

15. Example 1: Lever ab F1F1 F2F2 F 1 = (b/a). F 2 V 2 = (b/a). V 1 F P Q V e1e1 e2e2 i1i1 i2i2 Examples: Two ports Example 2: Electrical Transformer Example 3: Piston

16. Let’s model: V1V1 V2V2 m1m1 m2m2 k2k2 b k1k1 F(t) V 3 = V 1 – V 2 SeSe

17. Example: V1V1 V2V2 m1m1 m2m2 k2k2 r k1k1 F(t) (no friction) 1 0 1 1 C:k 1 I:m 1 I:m 2 SeSe F(t) V2V2 C:k 2 R:b V 3 = V 1 – V 2 V1V1

18. 3 Components : How To Connect 101 1 C I:m 1 I:m 2 SeSe V2V2 C:k 2 R V1V1 V3V3 Enforces the desire velocity relation.

19. V1V1 V2V2 m1m1 m2m2 k2k2 Rk1k1 F(t) 101C I:m 1 I:m 2 SeSe V2V2 V1V1 R C:k 1 F(t) m 1. a = - k 1 x 1 – k 2 x 2 m 2. a = F(t) – R. V 3 Another example:

20. Switch Domains i2i2 i3i3 i4i4 E C3C3 I5I5 R2R2 R4R4 101 R2R2 R4R4 I5I5 SeSe C3C3 e2e2 E i2i2 i2i2 i2i2 e3e3 i3i3 i4i4 i4i4 i4i4 ebeb eaea e5e5 i 1 = i 3 = i 4 E – i 2. R 2 = e 3 = e b