Transcript Document
CS 367: Model-Based Reasoning Lecture 15 (03/12/2002) Gautam Biswas 7/17/2015 1 Today’s Lecture Last Lectures: Modeling with Bond Graphs Today’s Lecture: 7/17/2015 Review Bond Graphs and Causality State Space Equations from Bond Graphs More Complex Examples 20-SIM 2 Review: Modeling with Bond Graphs • Based on concept of reticulation • Properties of system lumped into processes with distinct parameter values Lumped Parameter Modeling • Dynamic System Behavior: function of energy exchange between components • State of physical system – defined by distribution of energy at any particular time Dynamic Behavior: Current State + Energy exchange mechanisms 7/17/2015 3 Review: Modeling with Bond Graphs Exchange of energy in system through ports 1 ports: C, I: energy storage elements; R: dissipator 2 ports: TF, GY Exchange with environment: through sources and sinks: Se & Sf Behavior Generation: two primary principles Continuity of power Conservation of energy enforced at junctions: 3 ports 0- (parallel) junction 1- (series) junction 7/17/2015 4 Review: Junctions Electrical Domain: 0- enforces Kirchoff’s current law, 1enforces Kirchoff’s voltage law Mechanical Domain: 0- enforces geometric compatibility of single force + set of velocities that must sum to 0; 1- enforces dynamic equilibrium of forces associated with a single velocity Hydraulic Domain: 0- conservation of volume flow rate, when a set of pipes join 1- sum of pressure drops across a circuit (loop) involving a single flow must sum to 0. Sometimes junction structures are not obvious. 7/17/2015 5 Component Behaviors Mechanics Electricity Hydraulic Thermal T, temperature Effort e(t) F, force V, voltage P, pressure Flow f(t) v, velocity i, current Q, volume flow rate Q , heat flow rate Momentum p =e.dt P, momentum , flux p =P.dt P.dt = Pp Displacement q =f.dt x, distance q, charge q =Q.dt volume Q, heat energy Power P(t)=e(t).f(t) F(t).v(t) V(t).i(t) P(t).Q(t) Energy E(p)=f.dp E(q)=e.dq v.dP (kinetic) F.dx (potential) i.d v.dq Q.dp P.dq 7/17/2015 Q Q 6 Building Electrical Models For each node in circuit with a distinct potential create a 0-junction Insert each 1 port circuit element by adjoining it to a 1junction and inserting the 1-junction between the appropriate of 0-junctions. Assign power directions to bonds If explicit ground potential, delete corresponding 0junction and its adjacent bonds Simplify bond graph (remove extraneous junctions) Hydraulic, thermal systems similar, but mechanical different 7/17/2015 7 Electrical Circuit: Example 7/17/2015 8 Electrical Circuits: Example 2 Try this one: 7/17/2015 9 Building Mechanical Models For each distinct velocity, establish a 1-junction (consider both absolute and relative velocities) Insert the 1-port force-generating elements between appropriate pairs of 1-junctions; using 0-junctions; also add inertias to respective 1-junctions (be sure they are properly defined wrt inertial frame) Assign power directions Eliminate 0 velocity 1-junctions and their bonds Simplify bond graph 7/17/2015 10 Example: Mechanical Model 7/17/2015 11 Mechanical Model: Example 2 Try this one: 7/17/2015 12 Behavior of System: State Space Equations Linear System x A. x B.u ; x : n 1 state vector u : m 1 input vector y C . x D.u y : p 1 output vector Nonlinear System x ( x, u) ; x : n 1 state vector u : m 1 input vector y ( x, u) y : p 1 output vector 7/17/2015 13 State Equations Linear x A. x B.u x1 a11 . x1 .... a1n x n b11u1 .... b1m um x 2 a 21. x1 .... a 2 n x n b21u1 .... b2 m um .. . x n a n1. x1 .... a nn x n bn1u1 .... bnm um Nonlinear x ( x, u) x 1 1 ( x1 ,...., x n , u1 ,....,um ) x 2 2 ( x1 ,...., x n , u1 ,....,um ) 7/17/2015 .. . x n n ( x1 ,...., x n , u1 ,....,um ) 14 State Space: Standard form m. x k. x b. x m. x b. x k. x 0 Single nth order form Or Write in terms of state variablesx and v m.v b.v k.x x v n first-order coupled equations In general, can have any combination in between 7/17/2015 15 More complex example g Fourth order form with x 2 as system variable k2 d4 1 1 d2 k .k 1 1 ( x2 ) k1 ( ) ( x 2 ) 1 2 . x 2 k1 ( )g dt m1 m 2 dt m1.m 2 m1 m 2 m2 7/17/2015 16 More complex example (2): Second order form with two displacements x3 and x4 m1. x3 m1. x4 k1. x3 m2 .g m1. x3 (m1 m2 ) x4 k2 . x4 (m1 m2 ) g First order form with x3 and x 4 , and v1 and v 2 m1.v1 k1. x3 m1. g m 2 .v2 k1. x3 k 2 x 4 m 2 .g x 3 v1 v 2 x 4 v 2 7/17/2015 17 Causality in Bond Graphs To aid equation generation, use causality relations among variables Bond graph looks upon system variables as interacting variable pairs Cause effect relation: effort pushes, response is a flow Indicated by causal stroke on a bond A 7/17/2015 e f B 18 Causality for basic multiports Note that a lot of the causal considerations are based on 7/17/2015 algebraic relations 19 Causality Assignment Procedure 7/17/2015 20 Causality Assignment: Example 7/17/2015 21 Causality Assignment: Double Oscillator 7/17/2015 22 Causality Assignment: Example 3 Try this one: 7/17/2015 23 Generate equations from Bond Graphs Step 1: Augment bond graph by adding 1) 2) 3) 7/17/2015 Numbers to bonds Reference power direction to each bond A causal sense to each e,f variable of bond 24 Equation generation procedure 7/17/2015 25 Equation generation example p 2 E (t ) e3 e 4 E (t ) R3 . f 3 e5 E (t ) R3I.2P2 q5 C5 q5 f 4 f 6 f 2 7/17/2015 q p2 5 I 2 C 5 .R6 e6 e p 2 5 R6 I 2 R6 26 Equation Generation: Example 2 7/17/2015 27 H. W. Problem 1 : k1 k2 m1 m2 b1 b2 F0(t) Two springs, masses, & damper friction all linear. F0(t) = f1 = constant. Build bond graph; state equations. Simulate for various parameter values. 7/17/2015 28 H. W. Problem 2 : V (a) Bond graph. g M (b) Derive state equations in terms of energy variables. B K V m g (c) Simulate in 20-Sim with diff. Parameter values. Comment on results. k V0(t) 7/17/2015 Input : Velocity at bottom of tire 29 Extending Modeling to other domains Fluid Systems e(t) – Pressure, P(t) f(t) – Volume flow rate, Q(t) Momentum, p = e.dt = Pp, integral of pressure Displacement, q = Q.dt = V, volume of flow Power, P(t).Q(t) Energy (kinetic): Q(t).dPp Energy (potential): P(t).dV Fluid Port: a place where we can define an average pressure, P and a volume flow rate, Q Examples of ports: (i) end of a pipe or tube (ii) threaded hole in a hydraulic pump 7/17/2015 30 Fluid Ports Flow through ports transfers energy P – force/unit area Q – volume flow rate P.Q = power = force . displacement / time Moving fluid also has kinetic energy But it can be ignored if 1 Q 2 P ( ) 2 A Next time: fluid capacitors (tanks), resistances (pipes), and sources (pumps) 7/17/2015 31