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Modeling of Physical Systems: Lecture Summaries Page 1 8 Buildi ng Bond Gr ap h Models 8.1 The C- 1-R Exampl e A simple RC circuit system has a bond graph as follows: When assigning causality, the C element has preferred integral causality, dictating that the R element determines the current ow. Since there is one (1) independent energy-storage element (ESE) with integral causality, this is a 1st order system. The state of the system is the state variable on the independent ESE, or q C . We seek a state equation in the form, ˙ x = f (x,u,t). In this case, the state equation derivation begins with the rate law, ˙ q C = i C = i R , where it is emphasized that the capacitor current, i C , is causally determined by the resistor current, i R . Now, using the constitutive relation for the resistor, i R = v R R = v C R = 1 RC q, whe re the last two assignmen ts follo w fro m: 1) v R = v C (f rom bond gr aph) , and 2) v C = q/C (constitutive relation). This is the single state equation for this system. R.G. Longoria, F all 2006 UT-Austin

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Modeling of Physical Systems: Lecture Summaries Page 1

8 Building Bond Graph Models

8.1 The C-1-R Example

A simple RC circuit system has a bond graph as follows:

When assigning causality, the C element has preferred integral causality, dictating that theR element determines the current flow.

Since there is one (1) independent energy-storage element (ESE) with integral causality, thisis a 1storder system. The state of the system is the state variable on the independent ESE,or qC.

We seek a state equation in the form, x = f(x,u,t). In this case, the state equation derivationbegins with the rate law,

qC = iC = iR,

where it is emphasized that the capacitor current, iC, is causally determined by the resistorcurrent, iR.

Now, using the constitutive relation for the resistor,

iR =vRR

=vCR

= 1

RCq,

where the last two assignments follow from: 1) vR =vC (from bond graph), and 2)

vC = q/C (constitutive relation).

This is the single state equation for this system.

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Modeling of Physical Systems: Lecture Summaries Page 2

8.2 Review: Junction Elements and Physical Laws

The interconnection of bond graph elements is achieved using junction elements, whichrepresent commonly used physical laws/constraints. The common physical junctions are

summarized below.

It is helpful to begin associating with each of these junctions the key algebraic relationsbetween the physical variables involved. That is,

for 0 junctions:

f= 0

for 1 junctions:

e = 0

The sign convention on these relations reflects power bonds that all have power positivetoward the junctions.

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Modeling of Physical Systems: Lecture Summaries Page 3

With sign convention, follow logical structure, keeping arrows on power bonds directed fromone side to the other, except when there is some explicit reason to do otherwise. Future

examples will help to illustrate this further.

It is also possible to work through some basic examples to convince yourself of some patternsin assignments, as below.

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Modeling of Physical Systems: Lecture Summaries Page 4

8.4 Translational Mechanical Systems

In the system to the right, power variables, (F, xo), are indi-cated but it is not clear which one is a causal input. Lets

sketch the bond graph, leaving the causality unspecified.

First, identify that the damper and spring b and k2, respec-tively, are mechanically connected in series, so they are mod-eled by ideal R and C elements that have common force (ef-fort). This is indicated by a 0 junction. Note that the velocity,V2 = Vb + Vk2, is the total velocity of those cascaded elements.

Second, recognize that this velocity, V2, is the same as for spring k1, so the two subsystemsare joined by a 1 junction.

Finally, the F/xo bond is attached at the same velocity.

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Modeling of Physical Systems: Lecture Summaries Page 5

Now, consider the basic mass-spring-damper system:

How would you attach a compliance model at the base. Now, the mass and original spring-damper combination do not see the same velocity. The mass now has a velocity differentfrom the relative velocity seen by the spring-damper elements. The bond graph shownbelow can be drawn by inspection, if you can see that the base compliance will see the

same force as the original spring-damper combination.

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Modeling of Physical Systems: Lecture Summaries Page 6

How do we find the value of the transmitted force? Here is where causality helps. But first,re-draw the bond graph as below, and then reduce to the bond graph with causality on theright.

One of the key techniques introduced in this bond graph is the use of an explicit flow sourceto represent the ground velocity. For a fixed ground, just make Vg = 0.

If you can work through these examples, the following should make sense.

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Modeling of Physical Systems: Lecture Summaries Page 7

The same basic bond graph interconnections can be seen in the two examples shown below.However, it is necessary to recognize for each case how the mass, m, and the force, Fo(t),should be attached in each case.

A key step is to recognize that the mass and the force are each associated with a distinctvelocity in this system, and should thus be attached to a 1 junction that identifies thisvelocity. For the mass, the force due to gravity also acts at the same velocity, so an effortsource, mg, is attached to the 1 junction as well.

It can be helpful to define a 1 junction for every distinct velocity in a system, and thenattached elements known to have the same velocity. This is done on the far right of thefigure above. This can be condensed to the bond graph on the far left if no other elementsare attached to the 1 junction (see rules in Ch. 3 of BP).

On the gravitational force, why is the power bond directed positive in?

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Modeling of Physical Systems: Lecture Summaries Page 8

Lastly, the standard vehicle suspension and another vibration example.

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Modeling of Physical Systems: Lecture Summaries Page 9

Before going on to rotational examples, consider some simple examples with levers - needfor a transformer.

...and another

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Modeling of Physical Systems: Lecture Summaries Page 10

8.5 Rotational Mechanical Systems

Basic (fixed-axis) rotational systems should be examined using methods similar to thoseemployed for translational systems. The example shown below illustrates how you might

take apart such a system.

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Modeling of Physical Systems: Lecture Summaries Page 11

8.6 Electrical Systems

Consider now some examples of electrical systems. The succession of the modeling is num-bered in the first case:

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Modeling of Physical Systems: Lecture Summaries Page 13

8.7 Hydraulic Systems

Finally, we examine hydraulic system examples. First, a basic hydraulic system.

Then, the reservoir-tunnel-tank problem from homework 1, as worked in class:

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Modeling of Physical Systems: Lecture Summaries Page 14

When studying hydraulic systems with motors/pumps, one key element is the positive-displacement machine (PDM). Recall, that this is a transformer-based system, as indicatedbelow.

Below, this machine appears as both pump and motor.

References

 Beaman, J.J., and H.M. Paynter, Modeling of Physical Systems, Book in progress;notes for ME 383Q.

 D. Karnopp, D. Margolis and R. Rosenberg, System Dynamics, Wiley, 2000 (3rd ed.)

R.G. Longoria, Fall 2006 UT-Austin