Transcript Document

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
e
B
A
f
(directed bond from A to B).
each bond: two associated variables effort, e
flow, 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
forces you to make assumptions
explicit
C, I
Se, Sf 0,1
uniform network – like representation:
domain indep.
Generic Variables:
Signals
effort, e
flow, f
NOTE: power = effort × flow.
elec.
voltage
current
mechanical
force
velocity
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
Examples:
Effort
Flow
Power
Mechanics
Force, F
Velocity, V
FxV

F. V.
Electricity
Voltage, V
Current, I
VxI

V. I
Hydraulic
(Acoustic)
Pressure, P
VolumeP x Q
flow rate, Q
Thermodynamics
Temperature,
T
Entropy
flow rate

S
(thermal flow

rate)
Q
Pseudo
bond graph
Energy

Q
P.Q

Q
Constituent Relations
R, C, I : passive 1-ports -- one port through which
they exchange energy.
R
Sys
R: resistor
e = R•f
In mechanical systems = DASHPOT
F = b•V
F
R: b
(e)
V
Linear
F
e
(R constant)
f
non linear
e (t) = R (f) • f (t)
V (f)
R
v = i•R
(Electricity)
•
F
v
R
One Port Elements (continued ….)
C: Capacitor
F
F
V  x
C:k
V  x
e
f
C
P
g
I: Inertia
P
Q
Q
F:e
F  dt
v  x
p  F
P = P1 - P2
P2
Q
non linear
q  Ce
V:f
m
x
m om e n tu mp  m .v   F  dt
P1
linear
F  mv
v1
1
F  k  x  k  v  dt (C  )
k
m F  p
C
 i  dt C  e
e  1  i  dt
C
P
Q
Q
I
I
V  L dtdi
i(t) 1
I
 v  dt
Tetrahedron of State
e
dt
R
p=edt
P
C
X
I
f
Integrating
E, f : Power Variables ; P = e.f.
P, x : Energy Variables
dt
x=fdt
e = R.f
Dissipator
x=c.e
Capacitor
e = 1/c . x = 1/c  f d t
p=I.f
Inductor
f = 1/I . P = 1/I  e d t
(Instantaneous)
Potential
Energy
Kinetic
Energy
Two Other 1 – Ports
Effort Source
Se
e(t)
f
e(t) independent of flow f
If e(t) = constant
Constant Effort Source
F
Flow Source
f(t)
e
f(t) independent of effort e
If f(t) = constant
Constant Flow Source
m
F
Drive System
Se
Sf
F
v
I:m
v(t)
Sf F C:k
v(t)
How To Connect Elements:
Ideal 3 – Ports
1-junction: Common Flow junction
F1
v1 1
F2 v2
F3
v3
equivalent of series
junction
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, all flows equal&
e
i
0
0-junction: Common Effort junction
F1
v1 0
F2 v2
F3
v3
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, all efforts equal&
f
i
0
Ideal 3 – Port Junctions
v
F1
F2 v2
F3
F2
V1 = V2 = V3 = V :single flow var.
F3 – F1 = F2 or, F1 + F2 = F3
algebraic sum of effort vars = 0
F1
v1 1
F3
k
v3
F(t)
b
m
P2
C:k
P1
P3
No power loss
b:R
F(t)  Fs  Fm  Fd  0
F(t).v  Fs .v  Fd .v  Fm .v  0
1
F(t)
Se
I:m
0 – junction: dual of the 1 – junction
Common Force Junction
V1
V2
V3 = V1 – V2 = rate of compression
R
F1
F3 v3
F2
F1
v1 1
Q3
F1 = F2 = F3 = F
V3 = V1 – V2
.P3
Q1
.P1
P2
P1  P2  P3
F2
v2
common effort
sums of flow = 0
Q2 P Q  P Q  P Q  0
1 1
2 2
3 3
Others: 2 – Port Elements
Transformers
e1
e2
TF
f1
f2
b
a
e2 = (b/a) . e1
f1 = (b/a) . f2
&
Gyrators
e1
GY
f1
Again:
e1 . f1 = (a/b) . e2 (b/a) . f2 = e2 . f2
e1 . f1 = r . f2 (1/r) . e2 = e2 . f2
r
e2
f2
e1 = r . f2
r . f1 = e2
No power loss
Examples: Two ports
Example 1: Lever
a
i2
i1
•
b
•
e1
e2
•
•
Example 2: Electrical Transformer
F1
F1 = (b/a) . F2
V2 = (b/a) . V1
F2
V
F
P
Example 3: Piston
Q
Let’s model:
V1
k1
m1
V2
k2
b
Se
F(t)
m2
V3 = V1 – V2
Example:
V1
k1
m1
V2
k2
r
(no friction)
I:m1
C:k1
1
V1
R:b
F(t)
m2
I:m2
0
1
1
V2
F(t)
C:k2
V3 = V1 – V2
Se
3 Components : How To Connect
I:m1
C
1
V1
R
I:m2
0
1
V3
Enforces the desire velocity relation.
S
1
V2
e
C:k2
Another example:
V1
k1
k2
R
m1
m1 . a = - k1x1 – k2x2
m2 . a = F(t) – R . V3
F(t)
m2
I:m1
C
V2
1
V1
C:k1
0
R
I:m2
1
V2
F(t)
Se
Switch Domains
R2
i2
E
R4
i3
C3
i4
R2
e2 i2
I5
S
e
i1 = i3 = i4
E – i2 . R2 = e3 = eb
E
i2
1
eb
0 i
4
i2
e3
i3
C3
R4
ea i4
e5
1 i
4
I5