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

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 e

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 C, I Se, Sf 0,1 forces you to make assumptions explicit

uniform network – like representation: domain indep.

Generic Variables: Signals effort, e elec. mechanical flow, f voltage force current velocity 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

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

Examples:

S

Q

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 R

v = i•R (Electricity) e = R•f

• F

v

Linear (R constant)

V (f)

F V

R: b (e) F

non linear

R e

f

e (t) = R (f) • f (t)

F:e

V:f

linear

C: Capacitor

non linear

F

xV

P

Q

P Q

C

e

f C

g

)k

1(Ct dvkxkF

F

xV C:k

dtiC

1e

eCdti

eCq

One Port Elements (continued ….)

I: Inertia

m x

pF

xv

P1

Q

P2

Q

P = P1 - P2

P

Q I

I

dtvI

1i(t)

LVdtdi

Fp

dtFm.vpmomentum

dtFm

1v

vmF

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.f Dissipator (Instantaneous) x = c . e Capacitor Potential e = 1/c . x = 1/c f d t Energy p = I . f Inductor Kinetic f = 1/I . P = 1/I e d t Energy

Integrating

p = e d t

Se

e(t)

f

e(t) independent of flow f If e(t) = constant Constant Effort Source

F m

F v

Se I:m

Effort Source

Two Other 1 – Ports

Flow Source

Sf

f(t)

e

f(t) independent of effort e If f(t) = constant Constant Flow Source

v(t)

F Drive System

Sf C:k F

v(t)

How To Connect Elements: Ideal 3 – Ports

v1 = v2 = v3 = v There is no loss of energy at

Junction;

net power in = net power out

therefore,

F3v3 = F1v1 + F2v2

i.e., F3 = F1 + F2

In general,

1 F1

v1

F3

v3 F2 v2

1-junction: Common Flow junction

0

&

ie

equalflowsall

equivalent of series

junction

0 F1

v1

F3

v3 F2 v2

0-junction: Common Effort junction

0

&

if

equaleffortsall

equivalent of parallel

junction

F1 = F2 = F3 = F There is no loss of energy at

Junction;

net power in = net power out

therefore,

F3v3 = F1v1 + F2v2

i.e., v3 = v1 + v2

In general,

Ideal 3 – Port Junctions

v

F1

F2

F3

V1 = V2 = V3 = V :single flow var. F3 – F1 = F2 or, F1 + F2 = F3

algebraic sum of effort vars = 0

P1

P2

P3

b m

F(t) k

I:m 1 b:R

Se

C:k

F(t)

0.vF.vF.vFF(t).v

0FFFF(t)

mds

dms

No power loss

F1

v1

F2 v2

F3

v3 1

0 – junction: dual of the 1 – junction Common Force Junction

V1 V2

F1 F2

V3 = V1 – V2 = rate of compression

Q3

.P3

.P1 P2 Q1 Q2

F1

v1

F3 v3

F2

v2 1

R

0QPQPQP

PPP

332211

321

F1 = F2 = F3 = F common effort V3 = V1 – V2 sums of flow = 0

Others: 2 – Port Elements Transformers & Gyrators

e2 e1

f1

TF e2

f2 b a

e1

f1

GY f2 r

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

Example 1: Lever

a b

F1 F2

F1 = (b/a) . F2

V2 = (b/a) . V1

•

• •

•

F

P Q

V

e1 e2

i1 i2

Examples: Two ports

Example 2: Electrical Transformer

Example 3: Piston

Let’s model:

V1 V2

m1 m2

k2

b

k1 F(t)

V3 = V1 – V2

Se

Example:

V1 V2

m1 m2

k2

r

k1

F(t)

(no friction)

1

0

1

1

C:k1

I:m1 I:m2

Se

F(t)

V2

C:k2 R:b V3 = V1 – V2

V1

3 Components : How To Connect

1 0 1

1

C

I:m1 I:m2

Se

V2

C:k2 R

V1

V3

Enforces the desire velocity relation.

V1 V2

m1 m2

k2 R k1 F(t)

1 0 1 C

I:m1 I:m2

Se

V2 V1

R

C:k1

F(t)

m1 . a = - k1x1 – k2x2

m2 . a = F(t) – R . V3

Another example:

Switch Domains

i2

i3 i4

E C3

I5

R2 R4

1 0 1

R2 R4

I5 Se

C3

e2

E

i2

i2 i2

e3 i3

i4 i4

i4

eb

ea

e5

i1 = i3 = i4

E – i2 . R2 = e3 = eb