Mechatronics 1: ME 392Q-6 & 348C 31-Aug-07 M.D. Bryant. Analogous Systems. e(t) Se: e. ef = p/i. q = p /I, p = " q C " R p I + e(t)

Similar documents
Modeling of Dynamic Systems: Notes on Bond Graphs Version 1.0 Copyright Diane L. Peters, Ph.D., P.E.

Modeling and Simulation Revision III D R. T A R E K A. T U T U N J I P H I L A D E L P H I A U N I V E R S I T Y, J O R D A N

ENGI9496 Modeling and Simulation of Dynamic Systems Bond Graphs

Modeling and Simulation Revision IV D R. T A R E K A. T U T U N J I P H I L A D E L P H I A U N I V E R S I T Y, J O R D A N

Louisiana State University Physics 2102, Exam 3 April 2nd, 2009.

Inductance, RL and RLC Circuits

Magnetic Energy Domain Magnetic Capacitance Magnetic Resistance Magnetic Domain Example: Inductor with Flux Return Magnetic Circuit

Scanned by CamScanner

Inductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits

Louisiana State University Physics 2102, Exam 2, March 5th, 2009.

AP Physics C Mechanics Objectives

) t 0(q+ t ) dt n t( t) dt ( rre i dq t 0 u = = t l C t) t) a i( ( q tric c le E

Physics 102 Spring 2006: Final Exam Multiple-Choice Questions

System variables. Basic Modeling Concepts. Basic elements single and. Power = effort x flow. Power = F x v. Power = V x i. Power = T x w.

Chapter In Fig , the magnetic flux through the loop increases according to the relation Φ B. =12.0t

EM Oscillations. David J. Starling Penn State Hazleton PHYS 212

NONLINEAR MECHANICAL SYSTEMS LAGRANGIAN AND HAMILTONIAN FORMULATIONS. Lagrangian formulation. Ek * (f,q) = 1 2 f t I(q) f

Physics 2020 Exam 2 Constants and Formulae

Part 4: Electromagnetism. 4.1: Induction. A. Faraday's Law. The magnetic flux through a loop of wire is

Chapter 32. Inductance

Physics 102 Spring 2007: Final Exam Multiple-Choice Questions

Basic Electronics. Introductory Lecture Course for. Technology and Instrumentation in Particle Physics Chicago, Illinois June 9-14, 2011

b) (4) How large is the current through the 2.00 Ω resistor, and in which direction?

Contents. Dynamics and control of mechanical systems. Focus on

Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the time-varying current.

Chapter 31 Electromagnetic Oscillations and Alternating Current LC Oscillations, Qualitatively

Index. Index. More information. in this web service Cambridge University Press

Chapter 30 Inductance

Lecture Fluid system elements

Chapter 30 Self Inductance, Inductors & DC Circuits Revisited

Alternating Current. Symbol for A.C. source. A.C.

Handout 10: Inductance. Self-Inductance and inductors

Dynamics and control of mechanical systems

Physics 208, Spring 2016 Exam #3

Exercise 5 - Hydraulic Turbines and Electromagnetic Systems

Modeling of Electromechanical Systems

NONLINEAR MECHANICAL SYSTEMS (MECHANISMS)

Name:... Section:... Physics 208 Quiz 8. April 11, 2008; due April 18, 2008

2.4 Models of Oscillation

Modeling and Computation

Chapter three. Mathematical Modeling of mechanical end electrical systems. Laith Batarseh

The POG Modeling Technique Applied to Electrical Systems

Modeling and Analysis of Dynamic Systems

Modeling of Electrical Elements

1 Phasors and Alternating Currents

Introduction to Controls

Electromagnetic Induction (Chapters 31-32)

Introduction to AC Circuits (Capacitors and Inductors)

Describe the forces and torques exerted on an electric dipole in a field.

2.4 Harmonic Oscillator Models

Lecture 1. Electrical Transport

The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d:

PROBLEMS - chapter 3 *

2.004 Dynamics and Control II Spring 2008

PES 1120 Spring 2014, Spendier Lecture 35/Page 1

Dynamic Modeling. For the mechanical translational system shown in Figure 1, determine a set of first order

2.004 Dynamics and Control II Spring 2008

8. (6) Consider the circuit here with resistors R A, R B and R C. Rank the

MATH 312 Section 3.1: Linear Models

Units (Different systems of units, SI units, fundamental and derived units)

Noise - irrelevant data; variability in a quantity that has no meaning or significance. In most cases this is modeled as a random variable.

Yell if you have any questions

ENGG4420 LECTURE 7. CHAPTER 1 BY RADU MURESAN Page 1. September :29 PM

Solution: Based on the slope of q(t): 20 A for 0 t 1 s dt = 0 for 3 t 4 s. 20 A for 4 t 5 s 0 for t 5 s 20 C. t (s) 20 C. i (A) Fig. P1.

ELECTRONICS E # 1 FUNDAMENTALS 2/2/2011

Induction_P1. 1. [1 mark]

Here are some internet links to instructional and necessary background materials:

P441 Analytical Mechanics - I. RLC Circuits. c Alex R. Dzierba. In this note we discuss electrical oscillating circuits: undamped, damped and driven.

Chapter 33. Alternating Current Circuits

/20 /20 /20 /60. Dr. Galeazzi PHY207 Test #3 November 20, I.D. number:

Solved Problems. Electric Circuits & Components. 1-1 Write the KVL equation for the circuit shown.

TELLEGEN S THEOREM APPLIED TO MECHANICAL, FLUID AND THERMAL SYSTEMS

Version 001 CIRCUITS holland (1290) 1

Mathematical Modeling and response analysis of mechanical systems are the subjects of this chapter.

First-order transient

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2003 Experiment 17: RLC Circuit (modified 4/15/2003) OBJECTIVES

RLC Circuit (3) We can then write the differential equation for charge on the capacitor. The solution of this differential equation is

Physics 2B Winter 2012 Final Exam Practice

ET3-7: Modelling II(V) Electrical, Mechanical and Thermal Systems

SUBJECT & PEDAGOGICAL CONTENT STANDARDS FOR PHYSICS TEACHERS (GRADES 9-10)

Electrical polarization. Figure 19-5 [1]

Circuits. David J. Starling Penn State Hazleton PHYS 212

General Physics (PHY 2140)

Physics Jonathan Dowling. Final Exam Review

Mod. Sim. Dyn. Sys. Amplifiers page 1

Exam 3 Topics. Displacement Current Poynting Vector. Faraday s Law Self Inductance. Circuits. Energy Stored in Inductor/Magnetic Field

SCHOOL OF COMPUTING, ENGINEERING AND MATHEMATICS SEMESTER 1 EXAMINATIONS 2012/2013 XE121. ENGINEERING CONCEPTS (Test)

Name (Last, First): You may use only scientific or graphing calculators. In particular you may not use the calculator app on your phone or tablet!

Exam 2 Fall 2014

INDUCTANCE Self Inductance

Mod. Sim. Dyn. Sys. Amplifiers page 1

Yell if you have any questions

Review. Spring Semester /21/14. Physics for Scientists & Engineers 2 1

Name: Class: Date: Multiple Choice Identify the letter of the choice that best completes the statement or answers the question.

Chapter 3: Fundamentals of Mechanics and Heat. 1/11/00 Electromechanical Dynamics 1

Harold s AP Physics Cheat Sheet 23 February Electricity / Magnetism

Energy Storage Elements: Capacitors and Inductors

Chapter 31: AC Circuits

Modeling, Identification and Control of Aerospace Systems. Luiz Carlos S. Góes and Euler G. Barbosa Instituto Tecnológico de Aeronáutica DCTA

Gen. Phys. II Exam 2 - Chs. 21,22,23 - Circuits, Magnetism, EM Induction Mar. 5, 2018

Transcription:

V + - K R + - - k b V R V L L J + V C M B Analogous Systems i = q. + ω = θ. C -. λ/l = q v = x F T. Se: e e(t) e = p/i R: R 1 I: I e C = q/c C = dq/dt e I = dp/dt Identical dierential equations & bond graph or Series electrical elements Parallel mechanical elements Each system has Power input, eort source S e :e = V, F, T Potential energy (o position) storage o Capacitance C:C = C, k, K o Displacement q = q, x, θ Kinetic energy (o motion) storage, o Inertance I:I= L, M, J o Momentum p = λ, p, h Power loss, resistance R:R = R, b, B Common (same) low, all elements: = i, v, ω C: C q = p /I, p = " q C " R p I + e(t) I q + Rq + 1 C q = e(t) 1

R: R i L R C S : (t) 0 ec = q/c C = dq/dt C: C I = p/i e I = dp/dt M k v I: I b p = q C, q = " q RC " p I + (t) p + 1 RC p + 1 IC p = 1 C (t) Identical dierential equations & bond graph or Parallel electrical elements Series mechanical elements Each system has Power input, low source S : = i, v Capacitance C:C = C, k Inertance I:I= L, M Power loss, resistance R:R = R, B Common (same) eort, all elements: e = V, F 2

Dynamic Systems Elements Sources: supply power, prescribe eort or low Resistance: direct relation between eort & low Energy Storage Devices (single or multiports) Inertance: kinetic energy Capacitance: potential energy IC: kinetic & potential energies Lossless (conserve power, " P j = 0) multiports bonds Junctions o 0 junction: common eort, balance lows o 1 junction: common low, balance eorts Transormers & Gyrators For Bond graphs Bond: indicates power transer between elements Hal arrow: indicates direction o postive power transer between elements 3

Sources Eort source o Prescribes eort e = e(t), eort labeled on hal arrow side o bond o Flow can be anything (rest o system determines low) o Se prescribes eort (onto A), causal stroke (short vertical bar) away (ram against A) Flow source S : (t) A o Prescribes low = (t), non-hal arrow side o Eort can be anything o S prescribes low (to A), causal stroke toward (hose squirts away) 4

S e : e e(t) A V(t) + F(t) T(t) - P(t) Figure 3.6: a) A bond graph o an eort source S e :e(t), with maniestations in the various energy domains: b) electrical voltage source V (t); c) mechanical translational applied orce F(t); d) mechanical rotational applied torque T(t); and e) luidic applied pressure P(t). 5

Flow and Flow Sources S : (t) A I(t) v(t) Ω(t) Q(t) Figure 3.7: a) A bond graph o a low source S : (t), with maniestations in energy domains: b) electrical current source I(t); c) mechanical translational prescribed linear velocity v(t); d) mechanical rotational prescribed angular velocity Ω(t); and e) hydraulic prescribed volumetric low Q(t). 6

Resistance Direct relation between eort & low: e = e() or = (e) I e vs. plots in quadrants 1 & 3, dissipates power P = e 0 2 Causality choices: Eort Controlled: i = V/R orm o Ohm s law Action: A rams R with eort e Reaction: R accepts eort e rom A, then hoses A with low = (e) 7

Flow Controlled: A e = e( ) R V = i R orm o Ohm s law Action: A hoses R with low Reaction: R accepts low rom R, then R rams A with eort e = e() I e vs. plots in quadrants 1 & 3, dissipates power P = e 0 e < 0 < 0 e e > 0 > 0 8

e R e R = e R ( R ) R R R = R (e R ) R T b Ω b F b b β p 1 i R R v b Ω b Q c + - V R F b T b p o Figure 3.10: Bond graphs o a resistance in its causal orms a) and b), with maniestations in energy domains: c) electrical resistor; d) mechanical translational linear dashpot; e) mechanical rotational rotary dashpot; and ) hydraulic low constriction or turbulence. 9

Capacitance Kinematic constraint, low C & displacement q: C = q Stores "potential" energy, energy o position or coniguration U(q) = P dt = e dt = e dq dt dt = e(q) dq => Eort-displacement dependence e = e(q) Linear capacitance: e(q) = q/c U(q) = q 2 /2C Energy variable: displacement q Note: e = e(q) = U/ q Relates eort & displacement 10

Capacitances in various power domains system kinematics displacement eort physics general = q. q e = e(q) e = e(q) electrical i = q. charge voltage Gauss law q V = V(q) C: capacitance mech. translation mech. rotation magnetic luidic v = ẋ ω = θ. displacement x angular displacement θ " = " Magnetic lux Q = v. φ luid volume v V=q/C orce F = F(x) F = kx torque T = T(θ) T = K θ Magnetomotive orce M = M(φ) M = R φ pressure P = P(v) P t = v t /(A/ρg) Spring law k: elastic stiness Torsion K: torsion stiness Magnetic R: reluctance pressure P = P(v) P t = v t /(A/ρg) 11

Capacitance e c = e c (q ) q. c = C e c = e c (q ) q. c = T c C ic C + - Vc F c F c v 2 Ω 2 v 1 T c Ω 1 A Q 2 h P = ρgh Q 1 Figure 3.9: Bond graph o capacitances in a) integral causality (preerred), b) derivative causality with maniestations in energy domains: c) electrical capacitor; d) mechanical translational linear stiness; e) mechanical rotational torsional stiness; and ) hydraulic luid 12 tank.

Multiport Capacitance e 1 1... C e k k... e m m Energy stored in ield m-ports into Capacitance, m-power lows o Flows & displacement via kinematics: k = qk. o Displacements: q k m o Power: P = Pk k = 1 o Total potential energy: m = ek k k = 1 E = Pdt = m ek k dt k = 1 = m. ek qk k = 1 dt = m ek dqk k = 1 via integral, E = E(q 1, q 2,..., q m ) 13

depends on all displacements q k Energy & Power: de m dt = P = ek k k = 1 Derivative o E = E(q 1, q 2,..., q m ), chain rule: de dt m E = qk k = 1 dqk dt m E = qk k = 1 Equate coeicients o k, in blue terms: ek = ek(q1, q2,..., qm) = E qk Eort on k th bond rom partial o energy w.r.t. displacement q k on k th bond. k 14

Inertance Physics constraint, eort e I & momentum p: e I = p inertial orce Stores "kinetic" energy T(p) = P dt = e dt = dp dt dt = (p) dp => Flow-momentum dependence = (p) Linear inertance: (p) = p/m T(p) = p 2 /2I Energy variable: momentum p Note: = (p) = T/ p Relates low & momentum e = ṗ = (p) I 15

Inertance A. e I = p I = I (p) I A. e I = p p = p( I ) I i L = i L (λ) + Ω = Ω(h) L V L = λ. v = v(p) J Q(p) - M F I = ṗ T I = ḣ P = ṗ Figure 3.8: Bond graphs o an inertance in a) integral causality, and b) derivative causality, with maniestations in energy domains: c) electrical inductance; d) mechanical translational mass inertia; e) mechanical rotational rotational inertia; and ) hydraulic low inertia. There is no magnetic or thermal inertia. 16

Table 3.3: Inertances or the various power domains used in this book. Magnetic systems, which lack inertial eects, were omitted. mechanical mechanical general electrical translation rotation luidic dynamics e I = ṗ V L = λ F I = ṗ T I = ḣ P = ṗ momentum p λ [V s] p [N s] h [N m s] p [N m 2 s] low = (p) i = i(λ) v = v(p) Ω = Ω(h) Q = Q(p) (linear I) = p/i = λ/l = p/m = h/j = p/i physics law Faraday D Alembert: Newton D Alembert:Euler Newton 17

Inertances in various power domains system type physics momentum low dependence Physics law general e = ṗ p = (p) electrical Inductor voltage V = λ. lux linkage λ current i = i(λ) mech. inertial orce linear mom. velocity v Faraday Newton translation mech. FI = ṗ inertial torque p ang. mom. v = v(p) ang.vel. ω F = ma = ṗ Euler rotation luidic TI = ḣ inertial pressure PI = ṗ h luidic momentum p ω = ω(h) luid volume Q = Q(p) T = Iα = ḣ unsteady low terms in momentum equations No magnetic or thermal inertance! (no kinetic energy in those domains)! 18

Multiport Inertance e 1 1... I e k k... e m m via integral, E = E(p 1, p 2,..., p m ) k = k (p 1, p 2,..., p m ) = E/ p k Eort on k th bond rom partial o energy w.r.t. displacement q k on k th bond. 19

IC Device Stores kinetic & potential energies in same ield E = E(p 1, p 2,..., p m, q 1, q 2,..., q n ) Ports with momenta p k & displacements q l Flows on I bonds: k = E/ p k = k (p 1, p 2,..., p m, q 1, q 2,..., q n ) Eorts on C bonds: e l = E/ q l = e l (p 1, p 2,..., p m, q 1, q 2,..., q n ) IC 20

0 & 1 Junctions No power loss or storage Power balance: n P = k=1 P in k m - i=1 P out i n = k=1 e in k in k m - i=1 e out i out i = 0 e 2 2 e 1 1 0 0 junction: common (same) eort, all bonds: e1 = e2 =... = en = em = e n in m k - out n+m i = k = 0 k=1 i=1 k=1 e 3 Flow balance 0 junction incorporates: Electrical Kircho's Current Law ( currents into = 0 ) node Mechanical kinematics (balance o velocities & rate o displacements) 3 21

Kircho s current law, n k=1 i k = 0, or electrical power domains, wherein the sum o the currents i k lowing into a circuit node must equal zero. Translation kinematics n k=1 v k = 0, or mechanical translational domains, which equates translational velocities v k along some direction across a body to zero. Rotational kinematics n k=1 Ω k = 0, or mechanical rotational domains, wherein the rotational velocities Ω k along some axis through a body must equate to zero. Continuity equation n k=1 Q k = 0,orincompressibleluidicpowerdomains,wherein the sum o the volumetric lows Q k into and out o a control volume must equate to zero. Flux rate continuity equation n k=1 φ k = 0,ormagneticpowerdomains,wherein the sum o the lux lows φ k over a node in a magnetic circuit must equal zero. 22

e 2 2 e 1 1 1 1 junction: common (same) low, all bonds: 1 = 2 =... = n = m = n e in m k - e out n+m i = ek = 0 k=1 i=1 k=1 1 junction incorporates: Electrical Kircho's Voltage Law (over loop) Mechanical D'Alembert's dynamic equilibrium e 3 3 23

Equilibrium o orces, n k=1 F k = 0, or mechanical translational domains, wherein the sum o orces F k on a body along some direction must equal zero. Equilibrium o moments, n k=1 T k = 0,ormechanicalrotationaldomains,wherein the sum o moments T k over a body along some axis must equal zero. Momentum equation, n k=1 P k = 0, or luidic power domains, wherein the sum o the pressure drops P k along a low path must equal zero. Magnetomotive orce equilibrium, n k=1 M k = 0, or magnetic power domains, wherein the sum o the magnetomotive orce drops M k along a lux path must equal zero. Note that inertial eects such as F I = ṗ arise as separate terms in these balances. 24

Causality: 0 & 1 Junctions e 2 2 0 junction common eort e 1 = e 2 =... = e n = e e 1 1 0 e 3 3 only ONE bond can set common eort e ONE ram against 0 (otherwise contradiction) Note: 0 junction has only 1 ram, but MUST have e 1 1 e 2 1 2 e 3 3 1 junction: common low 1 = 2 =... = n = only ONE bond can set common low ONE hose squirts 1 (otherwise contradiction) Note: 1 junction has only 1 hose, but MUST have 25

Transormers & Gyrators Converts power, spans domains Lossless 2 port : P1 = P2 Transormer e 1 1 TF: n e 2 2 o relates eort to eort: e 1 = n e 2 o & low to low: 2 = n 1 o conserves power: (ne 2 ) 1 = e 1 1 = P 1 = P 2 = e 2 2 = e 2 (n 1 ) Gyrator e 1 1 GY: r e 2 2 o relates eort to low: e 1 = r 2 & e 2 = r 1 o conserves power: (r 2 ) 1 = e 1 1 = P 1 = P 2 = e 2 2 = (r 1 ) 2 26

e 1 1 TF: n e 2 2 e 1 1 TF: n e 2 2 + + i 1 i 2 V 1 - F V 2 2 F 1 B n 1 n 2 l 1 l 2 - v 1 v 2 R 1 R 2 T 1 T 2 Ω 1 Ω 2 Q T, Ω A F = pa p v v = R Ω Ω R F = T/R P, Q Figure 3.11: Bond graphs o transormers in its allowed causal orms a) and b), with maniestations in energy domains: c) electrical transormer with turns ratio n = n 1 /n 2 ; d) mechanical translational lever mechanism with leverage n = l 2 /l 1 ; e) mechanical rotational gears and rollers with gear ratio n = R 1 /R 2. Transormers can also span power domains. Examples include ) translational to hydraulic piston with n = A; g) rotational to translational roller on lat, or rack and pinion with n = 1/R; and h) rotational to hydraulic positive displacement 27 pump.

e 1 1 GY: r e 2 2 e 1 1 GY: r e 2 2 + i 1 i 2 + - V 1 i + V - lines o induction B total lux φ V 2 magnetomotive orce M = n i n turn coil servo motor -V mim T, Ω B Figure 3.12: Bond graphs o gyrators in allowed causal orms a) and b), with examples: c) electrical: electrical gyrator ormed by matched pairs o ield eect transistors, or transconductance ampliiers; d) electrical-mechanical rotational: DC servo motor; e) mechanicaltranslational-mechanical rotational: gyroscope; and ) electrical-magnetic: solenoid. Gyrators oten span power domains. 28

TF & GY Causality Transormer: e1 = n e2 eort evokes eort 1 = n-1 2 low evokes low 2 choices: Gyrator: e1 = r 2 low evokes eort 1 = r-1 e2 eort evokes low 2 choices e 1 e 2 GY 1 2 29

Table 3.2: A list o bond graph elements or various energy domains, with corresponding element constants and SI units in square brackets. mechanical mechanical element general electrical translation rotation luidic eort source S e : e(t) S e : V (t) S e : F(t) S e : T(t) S e : P(t) low source S : (t) S : I(t) S : v(t) S : Ω(t) S : Q(t) capacitance C C [F] k [N m 1 ] κ [N m rad 1 ] C t [m 5 N 1 ] inertance I L [H] m [kg] J [kg m 2 ] I [N m 2 s] resistance R R [Ω= V A 1 ] b [N m 1 s] β [N m s] R c [N m 5 ] transormer TF transormer levers gears & rollers gyrator GY transconductance ampliiers 0 junction k k = 0 Kircho s kinematics: kinematics: Continuity e k = e in Current Law velocities angular velocities equation 1 junction k e k = 0 Kircho s Equilibrium: Equilibrium: Momentum k = in Voltage Law orces moments equation 30