Bond graph

Analogies between variables

The bond graph use energy flow, or power, as the basis for abstracting analogies between different physical domains^[4]. Power conjugate variables are a pair of variables whose product is power and a list of these appears in the table for various physical domains. As indicated in the table, the bond graph uses the effort/flow analogy to categorise each of the two conjugate variables; the across/through analogy is also possible but not commonly used.^[7][8] The conventional bond graph symbol for effort is e, and that for flow is f.^[4]

As well as the two conjugate power variables e and f, the bond graph uses two integrated variables p and q where:

${\displaystyle p=\int ^{t}e(\tau )d\tau ~~~{\text{and}}~~~q=\int ^{t}f(\tau )d\tau }$

equivalently:

${\displaystyle {\dot {p}}={\frac {dp}{dt}}=e~~~{\text{and}}~~~{\dot {q}}={\frac {dq}{dt}}=f}$

The harpoon symbol shown in the figure represents a power bond, or bond,  transferring energy; the harpoon direction corresponds to positive energy flow and corresponds to a sign convention. The conjugate variables are displayed on each bond for the purposes of illustration.

Analogies between components

The definitions relating p to e and q to f are indicated diagrammatically. Physical properties are encapsulated in constitutive equations relating the energy and power variables.  In the diagram, C represents the constitutive equation relating q and e,  I represents the constitutive equation relating p and f and  R represents the constitutive equation relating e and f. Instead of placing e, f, p and q on the four corners of a square as in the diagram, they can alternatively be placed at the four vertices of a tetrahedron; such a diagram is called the tetrahedron of state.^[4]

These three constitutive equations ${\displaystyle \Phi _{C},\Phi _{I}~{\text{and}}~\Phi _{R}}$ correspond to three components each relating e and f: two dynamic components C and I incorporating an integrator and the R component.^[4] The C and I components store, but do not dissipate energy, R dissipates, but does not store energy. It is also possible to define a memristor component linking p and q, but this component is not commonly used.^[9]

${\displaystyle \Phi _{C},\Phi _{I}~{\text{and}}~\Phi _{R}}$ may be linear or nonlinear functions relating the variables as:

${\displaystyle q=\Phi _{C}(e),~~p=\Phi _{I}(f)~~{\text{and}}~~e=\Phi _{R}(f)}$

in the linear case:

${\displaystyle q=Ce,~~p=If,~~e=Rf}$

where ${\displaystyle C,~I,~{\text{and}}~R}$ are scalar constants representing generalised capacitance, inertia and resistance respectively.

Because e and f are conjugate variables, they are carried on a single bond and therefore the three components C, I and R are connected to a single bond and thus have a single energy port though which energy flows. The components, and impinging bond are shown in the figure. By convention, bonds point into these three components.

Analogies between connections

Electrical circuit diagrams have two sorts of connection: parallel and series or common voltage and common current; they distribute, but do not store or dissipate energy. The bond graph analogy of the common voltage  and common current connections are the 0-junction (common effort) and 1-junction (common flow) respectively; they both distribute, but do not store or dissipate energy.^[4]

All bonds impinging on a 0-junctions have the same effort. As energy is distributed, not dissipated, it follows that the sum of energy inflows (indicated by bonds pointing in) must equal the sum of energy outflows (indicated by bonds pointing out). Hence, if the common effort is e, the power flow constraint implies that the sum of the   ${\displaystyle m}$ inflows ${\displaystyle f_{i}^{in}}$ must equal the sum of the ${\displaystyle n}$ outflows ${\displaystyle f_{j}^{out}}$:

${\displaystyle \sum _{i=1}^{m}f_{i}^{in}=\sum _{j=1}^{n}f_{j}^{out}}$

Using the same argument, the efforts impinging on a 1-junction are constrained by:

${\displaystyle \sum _{i=1}^{m}e_{i}^{in}=\sum _{j=1}^{n}e_{j}^{out}}$

Analogies between external connections

Systems such as that the simple electrical and mechanical systems in the figure have no connection to the environment. Such connections may include external voltages and external forces (that is efforts) and external currents and external velocities (that is flows). Similarly, external measurements of efforts and flows are also important. For the purposes of building hierarchical system, it is convenient to define energy ports though which energy can flow between systems. These five possibilities correspond to five bond graph components:^[3][4]

• ${\displaystyle {\text{S}}_{e}}$, an effort source analogous to applying external voltages and external forces
• ${\displaystyle {\text{S}}_{f}}$, a flow source analogous to applying external currents and external velocities
• ${\displaystyle {\text{D}}_{e}}$, an effort sensor (detector) analogous to measuring voltage and forces
• ${\displaystyle {\text{D}}_{f}}$, a flow sensor (detector) analogous to measuring currents and velocities
• ${\displaystyle {\text{SS}}}$, a source/sensor component which acts both as ${\displaystyle {\text{S}}_{e}}$-${\displaystyle {\text{D}}_{f}}$ and as ${\displaystyle {\text{S}}_{f}}$-${\displaystyle {\text{D}}_{e}}$ pairs as well as an energy port for external connections.^[3]

For example, the mass-spring-damper system can be augmented with an force ${\displaystyle F}$ applied to the mass and a measurement of the mass velocity ${\displaystyle v}$ using a single SS components. The colon notation has been included; the SS:io refers to the fact that the force and velocity pair could be considered as system input and output.

Analogies between energy transducers

The conjugate effort and flow variables have different units in each energy domain thus models in one domain cannot be directly be connected by bonds to a different energy domain. However, because power has the same units (J/s or W) in each domain, the two-port power transducing components TF and GY can be used to provide such connections.^[4] As the two components transmit, but do not store or dissipate power, it follows that the power associated with the conjugate variables of the left-hand and right-hand bonds must be the same:

${\displaystyle e_{2}f_{2}=e_{1}f_{1}}$

Each component has a modulus ${\displaystyle m}$ associated with it. In the case of the TF component:

${\displaystyle e_{2}=me_{1}~~{\text{and}}~~f_{1}=mf_{2}}$

In the case of the GY component:

${\displaystyle e_{2}=mf_{1}~~{\text{and}}~~e_{1}=mf_{2}}$

For example, a frictionless, massless piston of area ${\displaystyle A}$ converts hydraulic power to mechanical power so that the hydraulic pressure ${\displaystyle P}$ is related to mechanical force ${\displaystyle F}$ by:

${\displaystyle F=AP}$

and hydraulic flow ${\displaystyle V}$ to piston velocity ${\displaystyle v}$ by:

${\displaystyle V=Av}$

Thus the bond graph analogy is the TF component with modulus ${\displaystyle m=A}$ where ${\displaystyle e_{1}=P,~~f_{1}=V}$ and ${\displaystyle e_{2}=F,~~f_{2}=v}$.

For example,  an ideal DC motor converts electrical power into rotational mechanical power so that the mechanical torque ${\displaystyle T}$is related to the electrical current ${\displaystyle i}$ by:

${\displaystyle T=ki}$

and back EMF (voltage) ${\displaystyle E}$ to angular velocity ${\displaystyle \Omega }$ by

${\displaystyle E=k\Omega }$

Thus the bond graph analogy is the GY component with modulus ${\displaystyle m=k}$ where ${\displaystyle e_{1}=E,~~f_{1}=i}$ and ${\displaystyle e_{2}=T,~~f_{2}=\Omega }$.

In both cases, non-ideal transduction behaviour can be modelled by including C ,I and R bond graph components in the model.

Variables

The bond graph variables for biochemical systems are:

• Displacement: quantity of chemical species measured in moles, symbol ${\displaystyle x}$ (mol)
• Flow: rate of change of chemical species, symbol ${\displaystyle v}$ (mol/s)
• Effort: chemical potential, or Gibbs energy, per mole of a chemical species, symbol ${\displaystyle \mu }$ (J/mol)^[22]

Note that the product of effort and flow (μv) is, as always in the bond graph formulation, power (J/s).

Components

As detailed below, the main features of the components used to model biochemical systems are: the R and C components are nonlinear, there is no I component required and the R component is replaced by a two-port Re component.^[15]

Modelling simple reactions

Reaction ${\displaystyle {\ce {A <=> B}}}$

Sources:^[13][15]

The simple reaction ${\displaystyle {\ce {A <=> B}}}$ is represented by three components:

• C:A represents the species A with chemical potential ${\displaystyle \mu _{A}}$; the flow is ${\displaystyle -v}$.
• C:B represents the species B with chemical potential ${\displaystyle \mu _{B}}$; the flow is ${\displaystyle v}$.
• Re_r1 represents the reaction with flow ${\displaystyle v}$, forward affinity ${\displaystyle A^{f}=\mu _{A}}$and reverse affinity ${\displaystyle A^{r}=\mu _{B}}$.
• The bonds and junctions transfer chemical energy with effort and flow variables indicated.

Using the above equations, the flow ${\displaystyle v}$ is given by

${\displaystyle v=\kappa \left(\exp {\frac {\mu _{A}}{RT}}-\exp {\frac {\mu _{B}}{RT}}\right)=\kappa \left(K_{A}x_{A}-K_{B}x_{B}\right)}$

where the subscripts correspond to the species. This is the simple mass-action equation:

${\displaystyle v=k^{+}x_{A}-k^{-}x_{B}}$

where ${\displaystyle k^{+}=\kappa K_{A}{\text{ and }}k^{-}=\kappa K_{B}}$.

Enzyme-catalysed reaction

As discussed in section 1.4 of Keener & Sneyd,^[22] an enzyme-catalysed reaction reversibly transforming species ${\displaystyle {\ce {A}}}$ to species ${\displaystyle {\ce {B}}}$ via enzyme ${\displaystyle {\ce {E}}}$ and enzyme complex ${\displaystyle {\ce {C}}}$ can be written as the pair of reactions:

${\displaystyle {\ce {A + E <=> C <=> B + E}}}$

The enzyme complex ${\displaystyle {\ce {C}}}$ is formed from ${\displaystyle {\ce {A + E}}}$ and decomposes into the species ${\displaystyle {\ce {B}}}$ and releases enzyme ${\displaystyle {\ce {E}}}$. The bond graph shown in the figure shows how the enzyme is recycled.

The bond graph can be used to derive the properties of these reactions which are of generalised Michaelis-Menten form.^[13]

Energy transduction

The bond graph TF (transformer) component represents energy transduction either within or between energy domains.^[4] (Note that the TF component has been called the TD (transduction) component^[11][12] - TF is more widely used.^[4]) This section focuses on transduction between the chemical domain with effort ${\displaystyle \mu }$ (J/mol) and flow ${\displaystyle v}$ (mol/s) and a generic domain with effort ${\displaystyle e}$ and flow ${\displaystyle f}$.

The key feature of the TF component is that it transmits energy without dissipation;^[4] hence, with reference to the figure:

${\displaystyle ef=\mu v}$

The transformer has a modulus m (with appropriate units) so that:

${\displaystyle f=mv}$

the energy formula then implies that:

${\displaystyle \mu =me}$

Stoichiometry

The stoichiometry of a chemical reaction determines how many of each chemical species occurs. Thus, for example, the reaction ${\displaystyle {\ce {A <=> m B}}}$ converts one mol of species ${\displaystyle {\ce {A}}}$ to m mol of species ${\displaystyle {\ce {B}}}$.

The case where ${\displaystyle m=1}$ corresponds to the simple reaction of the first example above. Using the same approach for general ${\displaystyle m}$, the reaction flow is:^[13][15]

${\displaystyle v=\kappa \left(\exp {\frac {\mu _{A}}{RT}}-\exp {\frac {m\mu _{B}}{RT}}\right)=\kappa \left(K_{A}x_{A}-(K_{B}x_{B})^{m}\right)}$

Applications

A number of systems relevant to systems biology have been modelled using bond graphs. These include:

• Enterocyte homeostasis^[29][30]
• Glucose transport^[31]
• Cardiac Cellular Electrophysiological Modeling^[32]
• Cerebral Circulation^[33]
• Gene regulatory networks^[34]
• Actin filament polymerization^[27]
• Biochemical oscillators^[35]
• Photosynthesis^[15]
• Blood circulation^[36]
• Simplified E. coli^[37]
• Mitochondrial Electron Transport Chain^[38][26]
• Action potential^[23][24][25]

R, C and I components

The concept of causality  is visualised via the causal stroke notation.^[4] This notation is introduced in the three figures where the R, C and I components are connected to a bond augmented by the causal stoke: a short line perpendicular to the bond and located at either (but not both) end of the bond. (For clarity, the figures correspond to linear components; in the nonlinear case, ${\displaystyle Rf}$is replaced by ${\displaystyle \Phi _{R}(f)}$and ${\displaystyle e/R}$ by ${\displaystyle \Phi _{R}^{-1}(e)}$ and similarly for the C and I components. Whereas an acausal (without strokes) bond graph represents a set of equations (where the left and right sides of an equation can be swapped without change of meaning), a causal (with strokes on each bond) represents a set of assignment statements whereby the value of the left-hand side of the assignment statement (represented here by :=) becomes the value of the expression on the right-hand side of the assignment statement. Thus, for example, the constitutive equation of a linear resistor can be written as ${\displaystyle e=Rf}$and ${\displaystyle f=e/R}$ without changing the meaning; but in contrast, the two assignment statements e:=Rf and f := e/R are different. In particular, in the first case, f must be known to compute e and in the second case, e must be known to compute f.

The assignment statement representation can be graphically visualised as a block diagram where each assignment statement is represented as a block with input representing the right-hand of the assignment statement and output representing the left-hand side of the assignment statement.^[3][4] The block diagrams for each causality are shown in the figures for each component. Note that each causality of a component leads to a different block diagram.

R component

The figure shows the causality of the R component with linear constitutive equation. (a) Flow is imposed on R and R imposes effort; this corresponds to the assignment statement e := Rf and the block diagram. (b) Effort is imposed on R and R imposes flow; this corresponds to the assignment statement f := e/R and the corresponding block diagram.

Source-sensor components

The SS (source sensor) component acts  as an effort source (${\displaystyle S_{e}}$), flow detector (${\displaystyle D_{f}}$) combination when the causal stroke is distant from the SS component and vice versa. The figure shows the causality of the SS (source/sensor) component. (a) The SS acts as an effort source (${\displaystyle S_{e}}$) flow detector (${\displaystyle D_{f}}$) combination. (b) The SS acts as a flow source (${\displaystyle S_{f}}$), effort detector (${\displaystyle D_{e}}$) combination.

Junctions

As, by definition, all efforts associated with bonds impinging on a 0-junction are the same, it follows that exactly one bond can impose effort causality. Similarly, all flows associated with bonds impinging on a 1-junction are the same, it follows that exactly one bond can impose flow causality. Thus if a bond imposes effort causality on a 0-junction, the junction imposes effort on the other bonds and if  a bond imposes flow causality on a 1-junction, the junction imposes flow on the other bonds.

Causal propagation

When one-port components (sources, C, I and R) are connected by a junction structure consisting of 0-junctions, 1-junctions, TF and GY, the causality assigned to each one port component propagates though the junction structure because of the causal constraints imposed by the junction structure components. This propagation can be applied systematically using the sequential causal assignment procedure (SCAP):^[4]

1. Choose any source (SS, Se or Sf)  and assign the required causality. Immediately extend the causal implications though the bond graph as far as possible using the constraints on the junction elements (0 and 1) and the TF and GY elements.
2. Repeat step 1 for all of the source components.
3. Choose any storage element  (C or I)  and assign the integral causality. Immediately extend the causal implications though the bond graph as far as possible using the constraints on the junction elements (0 and 1) and the TF and GY elements.
4. Repeat step 3 until all C and I elements have been assigned causality.

The figure shows the result of this procedure on the simple example. In particular, the I component imposes flow causality onto the 1-junction thus the one junction imposes flow causality on to the other three components: SS, C and R. This means that the C, R and SS components impose effort causality onto the one junction; thus the C component is in integral causality and the R component corresponds to the assignment ${\displaystyle e_{R}:=Rf}$.

Although this procedure can be accomplished manually for small systems,  a computer-based approach is more generally useful.

There are three possible results of this procedure.

1. It completes with all C and I components in integral causality and all bond causalities assigned; the resultant system is a causal bond graph and can be converted into a state-space system.
2. All bonds have causality assigned but one or more C or I component has derivative causality.
3. All C and I components have integral causality, but some bond do not have causality assigned.

Whereas case 1 leads to ordinary differential equations (ode) in state-space form, case 2 gives rise to differential-algebraic equations (dae). Depending on the context, it may be better to reconsider the underlying physical system to avoid daes^[4][5][6], or it may be possible to reduce the dae to an ode^[6][39] or the daes can be solved using an appropriate dae solver.

State-space equations

A bond graph system representation contains the constitutive equations of each component, embedded in the structure of bond and junctions.

The question of how to manipulate a set of equations into a form suitable for analogue computation was posed, and partially answered by Lord Kelvin.^[47] This approach underlies the conversion of a bond graph model to a state-space representation suitable for digital computation.^[48]

The bond graph uses the notion of (bond graph) causality to provide a systematic and constructive way to investigate whether a state-space representation exists, and, if so, what is that representation; this causality approach is well suited to computational implementation and has an intuitive representation on the bond graph itself using the causal stroke notation.

A causal bond graph can be put into state-space form if^[4]:

• every bond has a causal stroke and
• every component has allowed causality
• all C and I components are in integral causality.

The sequential causal assignment procedure (SCAP) provides a graphical approach to determining whether  a system represented by a bond graph  has a state-space representation.^[4][6]

The bond graph is analogous to an number of different systems; for clarity, the mechanical analogue is used as an example, but exactly the same method applied to electrical and other physical domains.

It is convenient to choose the system states as the integrated variables corresponding to the components in integral causality; in the case of the simple example, there are two states: ${\displaystyle p}$ corresponding to the mass M and ${\displaystyle q}$ corresponding to the spring. The system inputs and outputs correspond to the Se, Sf and SS components. In this case, there is one input and one output corresponding to the SS component: the effort (applied force ${\displaystyle F}$) and the measured flow (velocity ${\displaystyle v}$). Guided by the causal strokes, the state derivatives (effort applied to I components, flow applied to C components) can be written in terms of the states. Thus:

${\displaystyle {\begin{aligned}{\dot {q}}&=f={\frac {p}{m}}\\{\dot {p}}&=F-e_{c}-e_{r}=F-kq-df=F-kq-d{\frac {p}{m}}\end{aligned}}}$

This can be written in standard linear state space for as:

${\displaystyle {\begin{aligned}{\dot {x}}&=Ax+Bu\\y&=Cx\end{aligned}}}$

where the state ${\displaystyle x}$ input ${\displaystyle u}$ and output ${\displaystyle y}$ are given by:

${\displaystyle x={\begin{pmatrix}q\\p\end{pmatrix}},\;u=F{\text{ and }}y=v}$

and the matrices ${\displaystyle A}$, ${\displaystyle B}$ and ${\displaystyle C}$ are given by:

${\displaystyle A={\begin{pmatrix}0&{\frac {1}{m}}\\-k&-{\frac {d}{m}}\end{pmatrix}},\;B={\begin{pmatrix}0\\1\end{pmatrix}},\;C={\begin{pmatrix}0&{\frac {1}{m}}\end{pmatrix}}}$

The state equation has three parameters: the mass ${\displaystyle m}$, the spring constant ${\displaystyle k}$ and the damping constant ${\displaystyle d}$.

Using the usual Laplace transform notation of control systems theory, the system transfer function relating the input ${\displaystyle u}$ to output ${\displaystyle y}$ is:

${\displaystyle {s \over {\left(k+ds+ms^{2}\right)}}}$

Although it may be instructive to derive these formulae by hand, it is often convenient to make use of computer algebra to perform the same function.

System inversion

The behavior of a linear system is not only determined by the demominator of the system transfer function but also by the numerator. In the nonlinear case, the concept of the transfer function numerator is replaced by the concept of zero dynamics. Having a physical interpretation of zero dynamics is helpful in designing, and redesigning a dynamical system. For this reason, a bond graph approach to system inversion has been developed.^[49][50][51]

This procedure is illustrated using the simple mass-spring-damper system bond graph which, although linear, exemplifies the key ideas. In particular, system inversion is performed by reversing the causality on the source-sensor component; thus the role of input and output is reversed. In this particular case, reversing the causality of the SS component implies that the causality of the I component must also be reversed as only one component can impose flow onto the 1-junction. Thus only one component, the C component, remains in integral causality. Thus the zero dynamics are first order which, in this simple case, corresponds to the fact that the transfer function of the original system has a first order numerator.

Following the causal strokes, the remaining state ${\displaystyle q}$ can be written as in terms of the input ${\displaystyle v}$of the inverse system

${\displaystyle {\dot {q}}=v}$

${\displaystyle p}$ is no longer a state, but can be written as:

${\displaystyle p=mv}$

the output ${\displaystyle F}$ of the inverse system is:

${\displaystyle F=kp+dv+{\dot {p}}=kp+dv+m{\dot {v}}}$

This approach relies on the system input and output existing on the same SS component - the input and output are said to be colocated. If this is not so, the concept of bicausality must be used.

Source sensor (SS) components

The SS (source-sensor) component has two possible causal configurations corresponding to Se/Df (effort source, flow sensor) and De/Sf (effort sensor, flow source). As illustrated in the figure, the SS component has two bicausal configurations corresponding to Se/Sf (effort source, flow source) and De/Df (effort sensor, flow sensor). This is represented graphically using causal half-strokes where the half-stroke on the harpoon side of the bond corresponds to flow and the half-stroke on the other side of the bond corresponds to effort.

The rules for the bicausality of junctions are the same as those for causality: only one bond imposes an effort onto a 0-junction and only one bond imposes a flow onto a 1-junction. In the context of inversion, the bonds connected to I, C and components cannot be bicausal.

These ideas are illustrated using an example with non-collocated source and sensor.

Example: inversion of non-collocated source-sensor system

The example is extended by adding a zero junction and a further SS component. In the mechanical case, this corresponds to inserting a force sensor and velocity source between the spring and the ground; in the electrical case it corresponds to adding a current source and voltage sensor in parallel to the capacitor. The example is simplified by setting the flow of the added flow source to zero as indicated on the bond graph.

As in the collocated example, the mechanical system is considered and system input is taken to be the force ${\displaystyle F}$ acting on the mass M. However, the output is taken to be the force ${\displaystyle F_{s}}$ of the spring which is not collocated with the applied force ${\displaystyle F}$.

The corresponding state-space system is the same except that the output matrix C is given by:

${\displaystyle C={\begin{pmatrix}k&0\end{pmatrix}}}$

reflecting the change of the sensor. The transfer function relating input ${\displaystyle F}$ to output ${\displaystyle F_{s}}$becomes:

${\displaystyle {k \over {\left(k+ds+ms^{2}\right)}}}$

The denominator remains unchanged, but the numerator is different reflecting the change of sensor.

The system is inverted by making the output ${\displaystyle F_{s}}$an input and the input ${\displaystyle F}$ an output. Thus the left hand SS becomes both an effort and flow sensor and the right hand SS becomes both and effort and flow source with the corresponding causalities. As the I, C and R components must retain conventional causality, the bicausality propagates as shown. Both the I and C components are in derivative causality and thus the zero dynamics are zero order which, in this simple case, corresponds to the fact that the transfer function of the original system has a zero order numerator.

As in the collocated case, the transfer function of the inverse system is the reciprocal of the transfer function of the system; again, the bond graph causality method can also be used to examine the zero dynamics of nonlinear systems.^[65][54]

Electromagnetic

The steps for solving an Electromagnetic problem as a bond graph are as follows:

1. Place an 0-junction at each node
2. Insert Sources, R, I, C, TR, and GY bonds with 1 junctions
3. Ground (both sides if a transformer or gyrator is present)
4. Assign power flow direction
5. Simplify

These steps are shown more clearly in the examples below.

Linear mechanical

The steps for solving a Linear Mechanical problem as a bond graph are as follows:

1. Place 1-junctions for each distinct velocity (usually at a mass)
2. Insert R and C bonds at their own 0-junctions between the 1 junctions where they act
3. Insert Sources and I bonds on the 1 junctions where they act
4. Assign power flow direction
5. Simplify

These steps are shown more clearly in the examples below.

Simplifying

The simplifying step is the same regardless if the system was electromagnetic or linear mechanical. The steps are:

1. Remove Bond of zero power (due to ground or zero velocity)
2. Remove 0 and 1 junctions with less than three bonds
3. Simplify parallel power
4. Combine 0 junctions in series
5. Combine 1 junctions in series

These steps are shown more clearly in the examples below.

Parallel power

Parallel power is when power runs in parallel in a bond graph. An example of parallel power is shown below.

Parallel power can be simplified, by recalling the relationship between effort and flow for 0 and 1-junctions. To solve parallel power you will first want to write down all of the equations for the junctions. For the example provided, the equations can be seen below. (Please make note of the number bond the effort/flow variable represents). $${\displaystyle {\begin{matrix}f_{1}=f_{2}=f_{3}&&e_{2}=e_{4}=e_{7}\\e_{1}=e_{2}+e_{3}&&f_{2}=f_{4}+f_{7}\\&&\\e_{3}=e_{5}=e_{6}&&f_{7}=f_{6}=f_{8}\\f_{3}=f_{5}+f_{6}&&e_{7}+e_{6}=e_{8}\end{matrix}}}$$

By manipulating these equations you can arrange them such that you can find an equivalent set of 0 and 1-junctions to describe the parallel power.

For example, because ${\textstyle e_{3}=e_{6}}$ and ${\textstyle e_{2}=e_{7}}$ you can replace the variables in the equation ${\textstyle e_{1}=e_{2}+e_{3}}$ resulting in ${\textstyle e_{1}=e_{6}+e_{7}}$ and since ${\textstyle e_{6}+e_{7}=e_{8}}$, we now know that ${\displaystyle e_{1}=e_{8}}$. This relationship of two effort variables equaling can be explained by an 0-junction. Manipulating other equations you can find that ${\displaystyle f_{4}=f_{5}}$ which describes the relationship of a 1-junction. Once you have determined the relationships that you need you can redraw the parallel power section with the new junctions. The result for the example show is seen below.

Examples

State equations

Once a bond graph is complete, it can be utilized to generate the state-space representation equations of the system. State-space representation is especially powerful as it allows complex multi-order differential system to be solved as a system of first-order equations instead. The general form of the state equation is $${\displaystyle {\dot {\mathbf {x} }}(t)=\mathbf {A} \mathbf {x} (t)+\mathbf {B} \mathbf {u} (t)}$$ where ${\textstyle \mathbf {x} (t)}$ is a column matrix of the state variables, or the unknowns of the system. ${\textstyle {\dot {\mathbf {x} }}(t)}$ is the time derivative of the state variables. ${\textstyle \mathbf {u} (t)}$ is a column matrix of the inputs of the system. And ${\textstyle \mathbf {A} }$ and ${\textstyle \mathbf {B} }$ are matrices of constants based on the system. The state variables of a system are ${\textstyle q(t)}$ and ${\textstyle p(t)}$ values for each C and I bond without a causal conflict. Each I bond gets a ${\textstyle p(t)}$ while each C bond gets a ${\textstyle q(t)}$.

For example, if you have the following bond graph

you would have the following ${\textstyle {\dot {\mathbf {x} }}(t)}$, ${\textstyle \mathbf {x} (t)}$, and ${\textstyle \mathbf {u} (t)}$ matrices:

$${\displaystyle {\dot {\mathbf {x} }}(t)={\begin{bmatrix}{\dot {p}}_{3}(t)\\{\dot {q}}_{6}(t)\end{bmatrix}}\qquad {\text{and}}\qquad \mathbf {x} (t)={\begin{bmatrix}p_{3}(t)\\q_{6}(t)\end{bmatrix}}\qquad {\text{and}}\qquad \mathbf {u} (t)={\begin{bmatrix}e_{1}(t)\end{bmatrix}}}$$

The matrices of ${\textstyle \mathbf {A} }$ and ${\textstyle \mathbf {B} }$ are solved by determining the relationship of the state variables and their respective elements, as was described in the tetrahedron of state. The first step to solve the state equations is to list all of the governing equations for the bond graph. The table below shows the relationship between bonds and their governing equations.

More information , ...

Element type	Bond Name	Bond with causality	Governing equation(s)
Single-port elements	Source/ Sink, S	${\displaystyle S_{e}\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!|}}}\;}$	${\displaystyle {\text{input}}=e(t)}$
		${\displaystyle S_{f}\;{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}\;}$	${\displaystyle {\text{input}}=f(t)}$
	Resistance, R: Dissipated Energy	${\displaystyle \;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!|}}}\ R}$	${\displaystyle f(t)={\frac {1}{R}}e(t)}$
		${\displaystyle \;{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}\ R}$	${\displaystyle e(t)=Rf(t)}$
	Inertance, I: Kinetic Energy	♦${\displaystyle \;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!|}}}\ I}$	${\displaystyle f(t)={\frac {1}{I}}\int e(t)\,dt}$
		${\displaystyle \;{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}\ I}$	${\displaystyle e(t)=I{\dot {f}}(t)}$
	Compliance, C: Potential Energy	${\displaystyle \;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!|}}}\ C}$	${\displaystyle f(t)=C{\dot {e}}(t)}$
		♦${\displaystyle \;{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}\ C}$	${\displaystyle e(t)={\frac {1}{C}}\int f(t)\,dt}$
Double-port elements	Transformer, TR	${\displaystyle {\begin{matrix}{\overset {\textstyle }{{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}|}}\ TR\ \ {\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}|\ \\^{r:1}\end{matrix}}}$	${\displaystyle f_{1}={\frac {1}{r}}f_{2}}$ ${\displaystyle e_{2}={\frac {1}{r}}e_{1}}$
		${\displaystyle {\begin{matrix}|\ {\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}\ TR\ \ |\ {\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}\ \\^{r:1}\end{matrix}}}$	${\displaystyle e_{1}=re_{2}}$ ${\displaystyle f_{2}=rf_{1}}$
	Gyrator, GY	${\displaystyle {\begin{matrix}|\ {\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}\ GY{\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}|\ \\^{g:1}\end{matrix}}}$	${\displaystyle e_{1}=gf_{2}}$ ${\displaystyle e_{2}=gf_{1}}$
		${\displaystyle {\begin{matrix}\ {\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}|\ GY{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup }}}\ \\^{g:1}\end{matrix}}}$	${\displaystyle f_{1}={\frac {1}{g}}e_{2}}$ ${\displaystyle f_{2}={\frac {1}{g}}e_{1}}$
Multi-port elements	0 junction	One and only one causal bar at the junction	${\displaystyle {\text{all }}e(t)=}$
			${\displaystyle \sum f(t)_{\text{in}}=\sum f(t)_{\text{out}}}$
	1 junction	one and only one causal bar away from the junction	${\displaystyle \sum e(t)_{\text{in}}=\sum e(t)_{\text{out}}}$
			${\displaystyle {\text{all }}f(t)=}$

Close

"♦" denotes preferred causality.

For the example provided,

the governing equations are the following.

1. ${\textstyle e_{1}={\text{input}}}$
2. ${\textstyle e_{3}=e_{1}-e_{2}-e_{4}}$
3. ${\textstyle f_{1}=f_{2}=f_{4}=f_{3}}$
4. ${\textstyle e_{2}=R_{2}f_{2}}$
5. ${\textstyle f_{3}={\frac {1}{I_{3}}}\int e_{3}\,dt={\frac {1}{I_{3}}}p_{3}}$
6. ${\textstyle f_{5}=f_{4}\cdot r}$
7. ${\textstyle e_{4}=e_{5}\cdot r}$
8. ${\textstyle e_{5}=e_{7}=e_{6}}$
9. ${\textstyle f_{6}=f_{5}-f_{7}}$
10. ${\textstyle e_{6}={\frac {1}{C_{6}}}\int f_{6}\,dt={\frac {1}{C_{6}}}q_{6}}$
11. ${\textstyle f_{7}={\frac {1}{R_{7}}}e_{7}}$

These equations can be manipulated to yield the state equations. For this example, you are trying to find equations that relate ${\textstyle {\dot {p}}_{3}(t)}$ and ${\textstyle {\dot {q}}_{6}(t)}$ in terms of ${\textstyle p_{3}(t)}$, ${\textstyle q_{6}(t)}$, and ${\textstyle e_{1}(t)}$.

To start you should recall from the tetrahedron of state that ${\textstyle {\dot {p}}_{3}(t)=e_{3}(t)}$starting with equation 2, you can rearrange it so that ${\displaystyle e_{3}=e_{1}-e_{2}-e_{4}}$. ${\displaystyle e_{2}}$ can be substituted for equation 4, while in equation 4, ${\displaystyle f_{2}}$ can be replaced by ${\displaystyle f_{3}}$due to equation 3, which can then be replaced by equation 5. ${\displaystyle e_{4}}$ can likewise be replaced using equation 7, in which ${\displaystyle e_{5}}$can be replaced with ${\displaystyle e_{6}}$ which can then be replaced with equation 10. Following these substituted yields the first state equation which is shown below.

$${\displaystyle {\dot {p}}_{3}(t)=e_{3}(t)=e_{1}(t)-{\frac {R_{2}}{I_{3}}}p_{3}(t)-{\frac {r}{C_{6}}}q_{6}(t)}$$

The second state equation can likewise be solved, by recalling that ${\textstyle {\dot {q}}_{6}(t)=f_{6}(t)}$. The second state equation is shown below.

$${\displaystyle {\dot {q}}_{6}(t)=f_{6}(t)={\frac {r}{I_{3}}}p_{3}(t)-{\frac {1}{R_{7}\cdot C_{6}}}q_{6}(t)}$$

Both equations can further be rearranged into matrix form. The result of which is below.

$${\displaystyle {\begin{bmatrix}{\dot {p}}_{3}(t)\\{\dot {q}}_{6}(t)\end{bmatrix}}={\begin{bmatrix}-{\frac {R_{2}}{I_{3}}}&-{\frac {r}{C_{6}}}\\{\frac {r}{I_{3}}}&-{\frac {1}{R_{7}\cdot C_{6}}}\end{bmatrix}}{\begin{bmatrix}p_{3}(t)\\q_{6}(t)\end{bmatrix}}+{\begin{bmatrix}1\\0\end{bmatrix}}{\begin{bmatrix}e_{1}(t)\end{bmatrix}}}$$

At this point the equations can be treated as any other state-space representation problem.