Systems for bond graph

Many systems can be expressed in terms used in bond graph. These terms are expressed in the table below.

Conventions for the table below:

$P$ is the active power;
${\hat {X))$ is a matrix object;
${\vec {x))$ is a vector object;
${\displaystyle x^{\dagger ))$ is the Hermitian conjugate of $x$ ; it is the complex conjugate of the transpose of $x$ . If $x$ is a scalar, then the Hermitian conjugate is the same as the complex conjugate;
${\displaystyle D_{t}^{n))$ is the Euler notation for differentiation, where: $D_{t}^{n}f(t)={\begin{cases}\displaystyle \int _{-\infty }^{t}f(s)\,ds,&n=-1\\[2pt]f(t),&n=0\\[2pt]{\dfrac {\partial ^{n}f(t)}{\partial t^{n))},&n>0\end{cases))$
${\begin{cases}\langle x\rangle ^{\alpha }:=|x|^{\alpha }\operatorname {sgn}(x)\\\langle {a}\rangle =k\langle b\rangle ^{\beta }\implies \langle b\rangle =\left({\frac {1}{k))\langle a\rangle \right)^{1/\beta }\end{cases))$
Vergent-factor: $\phi _{L}={\begin{cases}{\textrm {Prismatic)):\ {\dfrac {\textrm {length))((\textrm {cross-sectional))\ {\textrm {area))))\\{\textrm {Cylinder)):\ {\dfrac {\ln \left({\frac {\mathrm {radius_{out)) }{\mathrm {radius_{in)) ))\right)}{2\pi \cdot {\textrm {length))))\\{\textrm {Sphere)):\ {\dfrac {1}{4\pi \left(\mathrm {radius_{in)) \parallel \mathrm {-radius_{out)) \right)))\end{cases))$

	Generalized flow	Generalized displacement	Generalized effort	Generalized momentum	Generalized power (in watts for power systems)	Generalized energy (in joules for power systems)
Name	${\vec {f))(t)$	${\vec {q))(t)$	${\vec {e))(t)$	${\vec {p))(t)$	$P={\vec {f))(t)^{\dagger }{\vec {e))(t)$	$E={\vec {q))(t)^{\dagger }{\vec {e))(t)$
Description	Time derivative of displacement	A quality related to static behaviour.	The energy per unit of displacement	Time integral of effort	Transformation of energy from one to another form	Conserved quantity in closed systems
Elements
Name	Hyperance $H$ , hyperrigitance ${\displaystyle P=H^{-1))$	Compliance $C$ , rigitance ${\displaystyle K=C^{-1))$	Resistance $R$	Inertance $I$ (or $L$ )	Abrahance $A$	Magnance $M$
Properties	Power dissipative element	Charge storage element (State variable: displacement) (Costate variable: effort)	Power dissipative element	Momentum storage element (State variable: momentum) (Costate variable: flow)	Power dissipative element	Power dissipative element
Quantitative behaviour	For 1-dimension systems (linear): ${\displaystyle P=H\cdot \left(D_{t}^{0}q(t)\right)^{2))$ For 1-dimension systems: $e(t)=P\gamma \left[D_{t}^{-1}q(t)\right]$ Impedance: $Z(s)={\frac {1}{s^{2}H))={\frac {1}{s^{2))}P$	Potential energy for N-dimension systems: ${\begin{array}{lcl}V&=&{\frac {1}{2)){\vec {q))(t)^{\dagger }{\vec {e))(t)\\{\vec {q))(t)&=&{\hat {C)){\vec {e))(t)\end{array))$ Potential energy: $V=\int _{0}^{q}e(q)\,dq$ Potential coenergy: ${\overline {V))=\int _{0}^{e}q(e)de$ For 1-dimension systems: $f(t)=C\cdot {\frac {de}{dt))+e{\frac {dC}{dt))$ Impedance: $Z(s)={\frac {1}{sC))={\frac {1}{s))k$	For 1-dimension systems (linear): ${\displaystyle P=R\cdot \left(D_{t}q(t)\right)^{2))$ Power for 1-dimension non-linear resistances ( $e(f)$ is the effort developed by the element): $P=e(f)f$ Rayleigh power: ${\displaystyle {\mathfrak {R))={\frac {1}{2))R\cdot f(t)^{2))$ Rayleigh oower for non-linear resistances: ${\mathfrak {R))=\int _{0}^{f}e(f)\,df$ Rayleigh effort: $e_{\mathfrak {R))={\frac {d{\mathfrak {R))}{df))=e(f)$ For N-dimension systems: ${\begin{array}{lcr}P&=&{\vec {f))(t)^{\dagger }{\vec {e))(t)\\{\vec {e))(t)&=&{\hat {R)){\vec {f))(t)\end{array))$ For 1-dimension systems: $e(t)=R\cdot \gamma \left[D_{t}^{1}q(t)\right]$ Impedance: $Z(s)=R$	Kinetic energy for N-dimension systems: ${\begin{array}{lcl}T={\frac {1}{2)){\vec {\rho ))(t)^{\dagger }{\vec {f))(t)\\{\vec {\rho ))(t)={\hat {L)){\vec {f))(t)\end{array))$ Kinetic energy: $T=\int _{0}^{\rho }f(\rho )\,d\rho$ Kinetic coenergy: ${\overline {T))=\int _{0}^{f}\rho (f)\,df$ For 1-dimension systems: $e(t)=L\cdot {\frac {df}{dt))+f\cdot {\frac {dL}{dt))$ Impedance: $Z(s)=sL$	For 1-dimension systems (linear): ${\displaystyle P=A\cdot \left(D_{t}^{2}q(t)\right)^{2))$ For 1-dimension systems: $e(t)=A\cdot \gamma \left[D_{t}^{3}q(t)\right]$ Impedance $Z(s)=s^{2}A$	For 1-dimension systems (linear) ${\displaystyle P=A\cdot \left(D_{t}^{3}q(t)\right)^{2))$ For 1-dimension systems $e(t)=M\cdot \gamma \left[D_{t}^{5}q(t)\right]$ Impedance $Z(s)=s^{4}M$
Generalized behaviour	Energy from active effort sources: $W=\int _{0}^{q}{e_{\text{source))}\,dq$ Lagrangian: ${\mathfrak {L))=T-(V-W)$ Hamiltonian: ${\mathfrak {H))=T+(V-W)$ Hamiltonian effort: $e_{\mathfrak {H))={\frac {d{\mathfrak {H))}{dq))$ Lagrangian effort: $e_{\mathfrak {L))={\frac {d{\mathfrak {L))}{dq))$ Passive effort: $e_{\mathfrak {LH))={\frac {d}{dt)){\frac {d{\mathfrak {L))}{df))$ Power equation: ${\frac {d{\mathfrak {L))}{dt))+{\frac {d{\mathfrak {H))}{dt))=0$ Effort equation: $e_{\mathfrak {L))+e_{\mathfrak {H))=0$ Lagrangian equation: $e_{\mathfrak {R))+e_{\mathfrak {LH))=e_{\mathfrak {L))$ Hamiltonian equation: $e_{\mathfrak {R))+e_{\mathfrak {LH))=-e_{\mathfrak {H))$ If ${\overline {W))$ is the coenergy, $W$ is the energy, $S$ is the state variable and ${\overline {S))$ is the costate variable, ${\begin{aligned}{\overline {W))+W=S\cdot {\overline {S))\\{\overline {W))=\int _{0}^{\overline {S))S({\overline {S)))\,d{\overline {S))\\W=\int _{0}^{S}{\overline {S))(S)dS\end{aligned))$ For linear elements: ${\overline {W))=W={\frac {1}{2))S\cdot {\overline {S))$

Longitudinal mechanical power system^[2]
Passive elements
Flow-related variables	$D_{t}^{6}x$ : pounce / pop $\mathrm {[m/s^{6}]}$	$D_{t}^{5}x$ : flounce / crackle $\mathrm {[m/s^{5}]}$	$D_{t}^{4}x$ : jounce / snap $\mathrm {[m/s^{4}]}$	$D_{t}^{3}x$ : jerk $\mathrm {[m/s^{3}]}$
	$D_{t}^{2}x$ : acceleration $(x_{2t})\ \mathrm {[m/s^{2}]}$	$D_{t}^{1}x$ : velocity $(x_{t})\ \mathrm {[m/s]}$ (flow)	$D_{t}^{0}x$ : displacement $(x)\ \mathrm {[m]}$ (displacement)	$D_{t}^{-1}x$ : absement $\mathrm {[m\cdot s]}$
	$D_{t}^{-2}x$ : absity $\mathrm {[m\cdot s^{2}]}$	$D_{t}^{-3}x$ : abseleration $\mathrm {[m\cdot s^{3}]}$	$D_{t}^{-4}x$ : abserk $\mathrm {[m\cdot s^{4}]}$
Effort-related variables	$D_{t}^{1}F$ : yank $\mathrm {[N/s]}$	$D_{t}^{0}F$ : force $(P_{x}^{t})\ \mathrm {[N]}$ (effort)	$D_{t}^{-1}F$ : linear momentum $(P_{x}^{2t})\ \mathrm {[kg\cdot m/s]}$ (momentum)
Compliance (C)	Resistance (R)	Inertance (I)	Abrahance (A)	Magnance (M)
Spring $\langle P_{x}^{t}\rangle =k\langle x\rangle$ where $k$ is the spring stiffness	Damper $\langle P_{x}^{t}\rangle =b\langle x_{t}\rangle$ where $b$ is the damper parameter	Mass $\langle P_{x}^{t}\rangle =m\langle x_{2t}\rangle$ where $m$ is the mass	Abraham–Lorentz force $\langle P_{x}^{t}\rangle ={\frac {\mu _{0}q^{2)){6\pi c))\langle x_{3t}\rangle$ where ${\displaystyle \mu _{0))$ is the permeability $q$ is the electric charge $c$ is the speed of light	Magnetic radiation reaction force $\langle P_{q}^{t}\rangle ={\frac {\mu _{0}q^{2}R}{24\pi c^{3))}\langle x_{5t}\rangle$ where ${\displaystyle \mu _{0))$ is the permeability $q$ is the electric charge $c$ is the speed of light $R$ is the radius of the magnetic moment
Cantilever $\langle P_{x}^{t}\rangle ={\frac {3EI}{L^{3))}\langle x\rangle$ $L$ : cantilever length $E$ : Young's modulus $I$ : second moment of area	Cyclotron radiation resistance $\langle P_{x}^{t}\rangle ={\frac {\sigma _{t}B^{2)){c\mu _{0))}\langle x_{t}\rangle$ where ${\displaystyle \sigma _{t))$ : Thompson cross-section $c$ : speed of light ${\displaystyle \mu _{0))$ : permeability $B$ : magnetic field density
Prismatic floater in a wide waterbody $\langle P_{x}^{t}\rangle =\rho _{L}gA\langle x\rangle$ where ${\displaystyle \rho _{L))$ : liquid density $A$ : area $g$ : gravitational acceleration	Viscous friction $\langle P_{x}^{t}\rangle =b\langle x_{t}\rangle$ where $b$ is the viscous friction parameter
Elastic rod $\langle P_{x}^{t}\rangle ={\frac {EA}{L))\langle x\rangle$ where $E$ : Young's modulus $A$ : area $L$ : rod length	Inverse of kinetic mobility $\langle P_{x}^{t}\rangle ={\frac {1}{\mu ))\langle x_{t}\rangle$ where $\mu$ is the kinetic mobility
Newton's law of gravitation ${\displaystyle \langle P_{x}^{t}\rangle =GMm\langle x\rangle ^{-2))$ where $G$ : gravitational constant $M$ : mass of body 1 $m$ : mass of body 2	Geometry interacting with air (e.g. drag) ${\displaystyle \langle P_{x}^{t}\rangle ={\frac {1}{2))c\rho A\langle x_{t}\rangle ^{2))$ where $A$ : contact area $c$ : aerodynamic geometry coefficient $\rho$ : fluid density
Coulomb's law ${\displaystyle \langle P_{x}^{t}\rangle ={\frac {Qq}{4\pi \varepsilon _{0))}\langle x\rangle ^{-2))$ where ${\displaystyle \varepsilon _{0))$ : permittivity $Q$ : charge of body 1 $q$ : charge of body 2	Absquare^{[clarification needed]} damper ${\displaystyle \langle P_{x}^{t}\rangle =B\langle x_{t}\rangle ^{2))$ where $B$ is the Absquare damper parameter
Casimir force ${\displaystyle \langle P_{x}^{t}\rangle ={\frac {A\hbar c\pi ^{2)){240))\langle x\rangle ^{-4))$ where $A$ : area of plate $\hbar$ : Planck's reduced constant $c$ : speed of light	Dry friction ${\displaystyle \langle P_{x}^{t}\rangle =\mu F_{n}\langle x_{t}\rangle ^{0))$ where $\mu$ : kinetic friction coefficient ${\displaystyle F_{n))$ : normal force
Biot–Savart law ${\displaystyle \langle P_{x}^{t}\rangle ={\frac {\mu _{0}I_{1}I_{2}l}{2\pi ))\langle x\rangle ^{-1))$ where ${\displaystyle \mu _{0))$ : permeability ${\displaystyle I_{1))$ : current in wire 1 ${\displaystyle I_{2))$ : current in wire 2 $l$ : length of wires
Piston pressing fluid inside an adiabatic chamber ${\displaystyle P_{x}^{t}=AP_{0}\left(1+{\frac {x}{x_{0))}\right)^{-\gamma ))$ where $A$ : area of piston ${\displaystyle P_{0))$ : initial pressure inside ${\displaystyle x_{0))$ : initial position of piston $\gamma$ : poisson coefficient

Angular mechanical power system
Passive elements
Flow-related variables	$D_{t}^{3}\theta$ : angular jerk $\mathrm {[rad/s^{3}]}$	$D_{t}^{2}\theta$ : angular acceleration $(\theta _{2t})\ \mathrm {[rad/s^{2}]}$
Flow-related variables	$D_{t}^{1}\theta$ : angular velocity $(\theta _{t})\ \mathrm {[rad/s]}$ (flow)	$D_{t}^{0}\theta$ : angular displacement $(\theta )\ \mathrm {[rad]}$ (displacement)
Effort-related variables	$D_{t}^{1}\tau$ : rotatum $\mathrm {[W/rad]}$	$D_{t}^{0}\tau$ : rorque $(P_{\theta }^{t})\ \mathrm {[J/rad]}$ (effort)
Effort-related variables	$D_{t}^{-1}\tau$ : angular momentum $(P_{\theta }^{2t})\ \mathrm {[J\cdot s/rad]}$ (momentum)
Compliance (C)	Resistance (R)	Inertance (I)
Inverse of the angular spring constant $\langle P_{\theta }^{t}\rangle =k_{r}\langle \theta \rangle$ where ${\displaystyle k_{r))$ is the angular spring constant	Angular damping $\langle P_{\theta }^{t}\rangle =R\langle \theta _{t}\rangle$ where $R$ is the damping constant	Mass moment of inertia Type $\langle P_{\theta }^{t}\rangle =I\langle \theta _{2t}\rangle$ where $I$ is the mass moment of inertia
Rod torsion $\langle P_{\theta }^{t}\rangle ={\frac {GJ}{L))\langle \theta \rangle$ where $G$ : shear modulus $J$ : polar moment of area $L$ : rod length	Governor (e.g. used in music boxes) ${\displaystyle \langle P_{\theta }^{t}\rangle =R\langle \theta _{t}\rangle ^{2))$ where $R$ is the governor constant
Bending moment (cantilever) $\langle P_{\theta }^{t}\rangle ={\frac {EI}{L))\langle \theta \rangle$ where $E$ : Young modulus $I$ : second moment of area $L$ : beam length
Parallel force field $\langle P_{\theta }^{t}\rangle =\langle P_{x}^{t}\rangle \sin \left(\alpha -\theta \right)$ where $\alpha$ : angle (counter-clockwise positive) between the polar axis and the force field $\theta$ : current angle of the object (e.g. pendulum)

Electric power system
Elements
Flow-related variables	$D_{t}^{2}q$ : electric inertia $\mathrm {[A/s]}$	$D_{t}^{1}q$ : electric current $(q_{t})\ \mathrm {[A]}$ (flow)	$D_{t}^{0}q$ : electric charge $(q)\ \mathrm {[C]}$ (displacement)
Effort-related variables	$D_{t}^{1}V$ : distension $\mathrm {[V/s]}$	$D_{t}^{0}V$ : voltage $(P_{q}^{t})\ \mathrm {[V]}$ (effort)	$D_{t}^{-1}V$ : flux linkage $(P_{q}^{2t})\ \mathrm {[V\cdot s]} {\text{ or ))\mathrm {[Wb\cdot turn]}$ (momentum)
Hyperance (H)	Compliance (C)	Resistance (R)	Inertance (I)	Abrahance (A)
Frequency-dependent negative resistor (FDNR) $\langle P_{q}^{t}\rangle ={\frac {1}{H))\langle q^{t}\rangle$	Linear capacitor $\langle P_{q}^{t}\rangle ={\frac {1}{\varepsilon ))\phi _{L}\langle q\rangle$ where $\varepsilon$ : Permittivity ${\displaystyle \phi _{L))$ : vergent factor	Linear resistor $\langle P_{q}^{t}\rangle =\rho \phi _{L}(1+\alpha (T-T_{0}))\langle q_{t}\rangle$ where $\rho$ : resistivity ${\displaystyle \phi _{L))$ : vergent factor $\alpha$ : thermal coefficient $T$ : current temperature ${\displaystyle T_{0))$ : reference temperature	Linear inductor (solenoid) $\langle P_{q}^{t}\rangle ={\frac {\mu _{0}N^{2}A}{L))\langle q_{2t}\rangle$ where ${\displaystyle \mu _{0))$ : permeability $N$ : number of turns $A$ : area $L$ : length	Frequency-dependent negative conductance (FDNC) Type $\gamma ^{1}\iff V=RD_{t}^{2}i\quad [H\cdot s]$
		Diode $P_{q}^{t}=nV_{T}\ln \left(1+{\frac {q_{t)){I_{s))}\right)$ where $n$ : ideality factor ${\displaystyle V_{T))$ : thermal voltage ${\displaystyle I_{s))$ : leakage current	Toroid $\langle P_{q}^{t}\rangle ={\frac {\mu N^{2}A}{2\pi r))\langle q_{2t}\rangle$ where $\mu$ : permeability $N$ : number of turns $A$ : cross-sectional area $r$ : toroid radius to centerline
Intra-gyrator			Inter-gyrator
Compliant gyrator	Resistive gyrator	Inertant gyrator	Inter-gyrator
	Hall effect device ${\begin{aligned}e_{x}=R_{H}{\frac {B_{z)){t_{z))}i_{y}\\e_{y}=R_{H}{\frac {B_{z)){t_{z))}i_{x}\end{aligned))$ where ${\displaystyle R_{H))$ : Hall coefficient ${\displaystyle B_{z))$ : vertical magnetic induction field ${\displaystyle t_{z))$ : vertical thickness	Induction motor ${\begin{aligned}e_{r}^{1}=M\cos(\theta )D_{t}f_{s}^{1}+f_{s}^{1}D_{t}\left[M\cos(\theta )\right]\\e_{s}^{1}=M\cos(\theta )D_{t}f_{r}^{1}+f_{r}^{1}D_{t}\left[M\cos(\theta )\right]\end{aligned))$ where $\theta$ : electrical angle ${\displaystyle e_{r}^{1))$ : voltage at rotor bar ${\displaystyle e_{s}^{1))$ : voltage at stator bar ${\displaystyle f_{s}^{1))$ : flow through stator bar ${\displaystyle f_{r}^{1))$ : flow through rotor bar	DC motor ${\begin{aligned}\tau _{e}=k_{a}\phi _{a}(i_{e})i_{a}\\\omega _{e}={\frac {1}{k_{a}\phi _{a}(i_{e})))e_{a}\end{aligned))$ where ${\displaystyle \tau _{e))$ : electromagnetic torque ${\displaystyle \omega _{e))$ : axis angular velocity ${\displaystyle i_{e))$ : field current ${\displaystyle i_{a))$ : armature current	Faraday gyrator ${\begin{aligned}F&=Bli\\V&={\frac {1}{Bl))v\end{aligned))$ where $F$ : force $v$ : rod velocity $V$ : rod voltage $B$ : magnetic induction field $l$ : rod length
			Faraday disk ${\begin{aligned}V&={\frac {1}{2))Br^{2}\omega \\\tau &={\frac {1}{2))Br^{2}i\end{aligned))$ where $B$ : magnetic induction field $r$ : disk radius
Intra-transformer			Inter-transformer
Electrical transformer (only for AC signals) ${\begin{aligned}V_{2}&={\frac {N_{2)){N_{1))}V_{1}\\f_{2}&={\frac {N_{1)){N_{2))}f_{1}\end{aligned))$

Hydraulic / pneumatic power system
Elements
Flow-related variables	$D_{t}^{1}V$ : volumetric flow rate $(V_{t})\ \mathrm {[m^{3}/s]}$ (flow)	$D_{t}^{0}V$ : volume $(V)\ \mathrm {[m^{3}]}$ (displacement)
Effort-related variables	$D_{t}^{0}P$ : pressure $(P_{V}^{t})\ \mathrm {[Pa]}$ (effort)	$D_{t}^{-1}P$ : fluid momentum $(P_{V}^{2t})\ \mathrm {[Pa\cdot s]}$ (momentum)
Compliance (C)	Resistance (R)	Inertance (I)
Pipe elasticity $\langle P_{V}^{t}\rangle =\left({\frac {t_{W}E}{2r_{0}V_{0))}\right)\langle V\rangle$ where ${\displaystyle r_{0))$ : nominal pipe radius ${\displaystyle V_{0))$ : pipe volume (without stress) ${\displaystyle t_{W))$ : wall thickness $E$ : Young's modulus	Darcy sponge $\langle P_{V}^{t}\rangle =\left({\frac {\mu }{k))\phi _{L}\right)\langle V_{t}\rangle$ where $\mu$ : dynamic viscosity $k$ : permeability ${\displaystyle \phi _{L))$ : vergent factor	Fluid inertia in pipes $\langle P_{V}^{t}\rangle =\left(\rho \phi _{L}\right)\langle V_{2t}\rangle$ where $\rho$ : fluid density ${\displaystyle \phi _{L))$ : vergent factor
Compressible fluid (approximation) $\langle P_{V}^{t}\rangle =\left({\frac {B}{V_{0))}\right)\langle V\rangle =\underbrace {\left({\frac {\rho _{0}c^{2)){V_{0))}\right)} _{\begin{array}{c}{\text{Acoustic))\\{\text{Approximation))\end{array))\langle V\rangle$ where ${\displaystyle V_{0))$ : pipe volume (without stress) $B$ : bulk modulus ${\displaystyle \rho _{0))$ : gas density at reference pressure $c$ : speed of sound	Valve ${\displaystyle \langle P_{V}^{t}\rangle =\left({\frac {\rho }{2C_{d}^{2}A_{0}^{2))}\right)\langle V_{t}\rangle ^{2))$ ${\displaystyle C_{d))$ : discharge coefficient $\rho$ : fluid density ${\displaystyle A_{0))$ : smallest area for fluid passage
Tank with area ${\textstyle A{\frac {1}{[m]^{n-1))}\left(h+h_{0}\right)^{n-1))$ : $P_{V}^{t}=\rho gh_{0}\left[\left(1+{\frac {nV}{A{\frac {1}{[m]^{n-1))}h_{0}^{n))}\right)^{1/n}-1\right]$ where $h$ : height with respect to the ground ${\displaystyle h_{0))$ : inlet height $\rho$ : fluid density $A$ : curve scaling (dimension of area) ${\displaystyle [m]^{n-1))$ : correction of dimension $n$ : curve parameter $g$ : acceleration of gravity Case $n=1$ (prismatic case): $P_{V}^{t}={\frac {\rho g}{A))V$ Case $n=0$ (current diode case): $P_{V}^{t}=\rho gh_{0}\left(e^{\frac {V}{A[m]))-1\right)$	Poiseuille resistance for cylinders $\langle P_{V}^{t}\rangle =\left({\frac {8\mu L}{\pi R^{4))}\right)\langle V_{t}\rangle$ where $\mu$ : dynamic viscosity $L$ : pipe length $R$ : pipe radius
Isothermal chamber ${\displaystyle \langle P_{V}^{t}\rangle =k\langle V_{t}\rangle ^{-1))$ where $k$ is chamber's constant	Turbulence resistance $\langle P_{V}^{t}\rangle =a_{t}\langle V_{t}\rangle ^{\frac {7}{4))$ where ${\displaystyle a_{t))$ is an empirical parameter
Compressible fluid $P_{V}^{t}=-B\ln V$ where $B$ is the bulk modulus	Nozzle ${\displaystyle \langle P_{V}^{t}\rangle ={\frac {\rho }{2\left(A_{\text{out))^{2}\parallel -A_{\text{in))^{2}\right)))\langle V_{t}\rangle ^{2))$ where $\rho$ : fluid density $A_{\text{out))$ : outlet area $A_{\text{in))$ : inlet area
Adiabatic bladder ${\displaystyle P_{V}^{t}=P_{V,0}^{t}\left(1-{\frac {V}{V_{0))}\right)^{-\gamma ))$ where ${\displaystyle P_{V,0}^{t))$ : reference pressure ${\displaystyle V_{0))$ : reference volume $\gamma$ : Poisson coefficient	Check valve $P_{V}^{t}=k\ln \left(1+{\frac {V_{t)){V_{t,0))}\right)$ where $k$ : empirical constant ${\displaystyle V_{t,0))$ : reverse bias absolute flow

Gyrator–capacitor power system
Elements
Flow-related variables	$D_{t}^{0}\varphi$ : magnetic flux $(\varphi )\ \mathrm {[Wb]}$ (displacement)
Effort-related variables	$D_{t}^{0}{\mathcal {F))$ : magnetomotive force $({\mathcal {F)))\ \mathrm {[A\cdot turn]}$ (momentum)
Compliance (C)	Resistance (R)	Inertance (I)
Permeance ( ${\mathcal {P))$ ) $\langle {\mathcal {F))\rangle ={\frac {1}{\mu ))\phi _{L}\langle \varphi \rangle \quad \mathrm {[H/turn^{2}]}$ where $\mu$ : permeability ${\displaystyle \phi _{L))$ : vergent factor	Magnetic complex impedance ( ${\displaystyle Z_{M))$ ) ${\mathcal {F))=Z_{M}{\varphi _{t))\quad \mathrm {[{turn}^{2}/\Omega ]}$	Magnetic complex inductance ( ${\displaystyle L_{M))$ ) ${\mathcal {F))=L_{M}{\varphi _{2t))\quad \mathrm {[{turn}^{2}F]}$

Gravitational power system
Elements
Flow-related variables	${\displaystyle D_{t}^{0}i_{g))$ : gravitational current Loop $\left(i_{g}=2\varepsilon _{g}v_{\text{orbit))^{3}\right)\ \mathrm {[kg/s]}$ (flow)	${\displaystyle D_{t}^{-1}i_{g))$ : gravitational charge $(M)\ \mathrm {[kg]}$ (displacement)
Effort-related variables	${\displaystyle D_{t}^{0}V_{g))$ : gravitational voltage $\left(V_{g}={\frac {1}{2)){\frac {v_{\text{orbit))^{4)){c^{2))}\right)\ \mathrm {[m^{2}/s^{2}]}$ (effort)	${\displaystyle D_{t}^{-1}V_{g))$ : gravitational momentum $\left(\phi _{g}={\frac {\pi v_{\text{orbit))^{3}r}{c^{2))}\right)\ \mathrm {[m^{2}/s]}$ (momentum)
Compliance (C)	Resistance (R)	Inertance (I)
Gravitational capacitance Type $\gamma ^{1}\iff M_{g}=CV_{g}\quad \mathrm {[kg/m^{2}]}$ $C_{g}={\frac {2c^{2}r^{2)){GM))$	Gravitational orbital resistance Type $\gamma ^{1}\iff V_{g}=R_{g}i_{g}\quad \mathrm {[m^{2}/kg\cdot s]}$ $R_{g}={\frac {\mu _{g)){4))v_{\text{orbit))$	Gravitational inductance Type $\gamma ^{1}\iff \phi _{g}=Ii_{g}\quad \mathrm {[m^{2}/kg\cdot s^{2}]}$ $L_{g}={\frac {2\pi ^{2}G}{c^{2))}r$

Electric field volumetric power density system
Elements
Flow-related variables	$D_{t}^{0}J$ : current density $(J)\ \mathrm {[A/m^{2}]}$ (flow)	$D_{t}^{-1}J$ : electric displacement field $(D)\ \mathrm {[C/m^{2}]}$ (displacement)
Effort-related variables	$D_{t}^{0}E$ : electric field $(E)\ \mathrm {[V/m]}$ (effort)	$D_{t}^{-1}E$ : magnetic potential vector $(A)\ \mathrm {[V\cdot s/m]}$ (momentum)
Compliance (C)	Resistance (R)	Volumetric power density inertance ( ${\displaystyle I_{V))$ )
Electrical permittivity Type $\gamma ^{1}\iff D=\epsilon _{0}E\quad \mathrm {[F/m]}$	Electrical resistivity Type $\gamma ^{1}\iff E=\rho J\quad \mathrm {[\Omega m]}$	Magnetic permeability $\mu _{0}\ \mathrm {[H/m]}$

Magnetic field volumetric power density system
Elements
Flow-related variables	$D_{t}^{0}B$ : magnetic flux density $(B)\ \mathrm {[T]} {\text{ or ))\mathrm {[Wb/m^{2}]}$ (displacement)
Effort-related variables	$D_{t}^{0}H$ : magnetic field strength $(H)\ \mathrm {[A\cdot {turn}/m]}$ (effort)
Compliance (C)	Resistance (R)	Volumetric power density inertance ( ${\displaystyle I_{V))$ )
Magnetic permeability for magnetic circuits Type $\gamma ^{1}\iff B=\mu H\quad \mathrm {[H/(m\cdot {turn}^{2})]}$

Gravitoelectric volumetric power density system
Elements
Flow-related variables	${\displaystyle D_{t}^{0}J_{g))$ : flux of mass $(J_{g})\ \mathrm {[kg/m^{2}s]}$ (flow)	${\displaystyle D_{t}^{-1}J_{g))$ : accumulated flux of mass $(D_{g})\ \mathrm {[kg/m^{2}]}$ (displacement)
Effort-related variables	$D_{t}^{0}g$ : acceleration of gravity $(g)\ \mathrm {[m/s^{2}]}$ (effort)
Compliance (C)	Resistance (R)	Volumetric power density inertance ( ${\displaystyle I_{V))$ )
Gravitational permittivity Type $\gamma ^{1}\iff D_{g}=\epsilon _{g}g\quad \mathrm {[kg\cdot s^{2}/m^{3}]}$ $\varepsilon _{g}={\frac {1}{4\pi G))$		Gravitational permeability $\mathrm {[m/kg]}$ ${\displaystyle I_{V}=\mu _{g}={\frac {4\pi G}{c^{2))))$

Gravitomagnetic volumetric power density system
Elements
Flow-related variables	${\displaystyle D_{t}^{0}B_{g))$ : gravitomagnetic field $\left(B_{g}=\omega _{\text{orbit))\left({\frac {v_{\text{orbit))}{c))\right)^{2}\right)\ \mathrm {[Hz]}$ (displacement)
Effort-related variables	${\displaystyle D_{t}^{0}H_{g))$ : gravitomagnetic field strength $(H_{g})\ \mathrm {[Pa\cdot s]}$ (effort)
Compliance (C)	Resistance (R)	Volumetric power density inertance ( ${\displaystyle I_{V))$ )
Gravitational permeability $-{\frac {4\pi G}{c^{2))}\ \mathrm {[m/kg]}$

Thermal power-temperature system
Elements
Flow-related variables	$D_{t}^{1}Q$ : heat rate $(\psi _{t})\ \mathrm {[W]}$ (flow)	$D_{t}^{0}Q$ : total heat $(\psi )\ \mathrm {[J]}$ (displacement)
Effort-related variables	$D_{t}^{0}T$ : temperature $(T)\ \mathrm {[K]}$ (Effort)
Compliance (C)	Resistance (R)	Inertance (I)
Isobaric heat $T={\frac {1}{\rho Vc_{p))}\psi ={\frac {1}{C_{P))}\psi$ where $\rho$ : object mass density $V$ : volume of object ${\displaystyle c_{p))$ : pressure specific heat capacity	Conduction resistance ${\displaystyle T={\frac {1}{k))\phi _{L}\psi _{t))$ where $k$ : thermal conductivity ${\displaystyle \phi _{L))$ : vergent factor
Isocoric heat $T={\frac {1}{C_{V))}\psi$ where ${\displaystyle C_{V))$ is the constant-volume heat capacitance	Convection resistance ${\displaystyle T={\frac {1}{hA))\psi _{t))$ where $h$ : convection coefficient $A$ : interface area
Isothermal heat $T={\frac {1}{nR\ln \left({\frac {V_{f)){V_{i))}\right)))\psi$ where $R$ : universal gas constant $n$ : number of moles ${\displaystyle V_{f))$ : final volume ${\displaystyle V_{i))$ : initial volume	Stefan–Boltzmann law ${\displaystyle \langle T\rangle =\left({\frac {1}{e\sigma A))\right)^{0.25}\langle \psi _{t}\rangle ^{0.25))$ where $e$ : emissivity $\sigma$ : Stefan–Boltzmann constant $A$ : interface area

Continuum mechanics volumetric power density system
Elements
Flow-related variables	$D_{t}^{1}\varepsilon$ : strain rate $({\dot {\varepsilon )))\ \mathrm {[Hz]}$ (flow)	$D_{t}^{0}\varepsilon$ : strain $(\varepsilon )\ \mathrm {[1]}$ (displacement)
Effort-related variables	$D_{t}^{0}\sigma$ : stress $(\sigma )\ \mathrm {[Pa]}$ (effort)
Compliance (C)	Resistance (R)	Volumetric power density inertance ( ${\displaystyle I_{V))$ )
Inverse of rigidity Type $\gamma ^{1}\iff \varepsilon =C\sigma \quad \mathrm {[Pa^{-1}]}$ $C={\frac {1}{K))$	Viscosity Type $\gamma ^{1}\iff \sigma =R{\dot {\varepsilon ))\quad \mathrm {[Pa\cdot s]}$	Power density inertance: density of material $\rho \ \mathrm {[kg/m^{3}]}$

Other systems:

Thermodynamic power system (flow is entropy-rate and effort is temperature)
Electrochemical power system (flow is chemical activity and effort is chemical potential)
Thermochemical power system (flow is mass-rate and effort is mass specific enthalpy)
Macroeconomics currency-rate system (displacement is commodity and effort is price per commodity)
Microeconomics currency-rate system (displacement is population and effort is GDP per capita)

Tetrahedron of state

The tetrahedron of state is a tetrahedron that graphically shows the conversion between effort and flow. The adjacent image shows the tetrahedron in its generalized form. The tetrahedron can be modified depending on the energy domain.

Using the tetrahedron of state, one can find a mathematical relationship between any variables on the tetrahedron. This is done by following the arrows around the diagram and multiplying any constants along the way. For example, if you wanted to find the relationship between generalized flow and generalized displacement, you would start at the $f (t)$ and then integrate it to get $q (t)$ . More examples of equations can be seen below.

Relationship between generalized displacement and generalized flow.

q(t)=\int f(t)\,dt

Relationship between generalized flow and generalized effort.

f(t)={\frac {1}{R))\cdot e(t)

Relationship between generalized flow and generalized momentum.

f(t)={\frac {1}{I))\cdot p(t)

Relationship between generalized momentum and generalized effort.

p(t)=\int e(t)\,dt

Relationship between generalized flow and generalized effort, involving the constant C.

e(t)={\frac {1}{C))\int f(t)\,dt

All of the mathematical relationships remain the same when switching energy domains, only the symbols change. This can be seen with the following examples.

Relationship between displacement and velocity.

x(t)=\int v(t)\,dt

Relationship between current and voltage, this is also known as Ohm's law.

i(t)={\frac {1}{R))V(t)

Relationship between force and displacement, also known as Hooke's law. The negative sign is dropped in this equation because the sign is factored into the way the arrow is pointing in the bond graph.

F(t)=kx(t)

For power systems, the formula for the frequency of resonance is as follows:

{\displaystyle \omega ={\sqrt {\frac {1}{LC))))

For power density systems, the formula for the velocity of the resonance wave is as follows:

{\displaystyle c={\sqrt {\frac {1}{LC))))

Components

If an engine is connected to a wheel through a shaft, the power is being transmitted in the rotational mechanical domain, meaning the effort and the flow are torque (τ) and angular velocity (ω) respectively. A word bond graph is a first step towards a bond graph, in which words define the components. As a word bond graph, this system would look like:

{\text{engine))\;{\overset {\textstyle \tau }{\underset {\textstyle \omega }{-\!\!\!-\!\!\!-\!\!\!-\!\!\!-))}\;{\text{wheel))

A half-arrow is used to provide a sign convention, so if the engine is doing work when τ and ω are positive, then the diagram would be drawn:

{\text{engine))\;{\overset {\textstyle \tau }{\underset {\textstyle \omega }{-\!\!\!-\!\!\!-\!\!\!\rightharpoondown ))}\;{\text{wheel))

This system can also be represented in a more general method. This involves changing from using the words, to symbols representing the same items. These symbols are based on the generalized form, as explained above. As the engine is applying a torque to the wheel, it will be represented as a source of effort for the system. The wheel can be presented by an impedance on the system. Further, the torque and angular velocity symbols are dropped and replaced with the generalized symbols for effort and flow. While not necessary in the example, it is common to number the bonds, to keep track of in equations. The simplified diagram can be seen below.

{S_{e))\;{\overset {\textstyle e_{1)){\underset {\textstyle f_{1)){-\!\!\!-\!\!\!-\!\!\!\rightharpoondown ))}\;{\text{I))

Given that effort is always above the flow on the bond, it is also possible to drop the effort and flow symbols altogether, without losing any relevant information. However, the bond number should not be dropped. The example can be seen below.

{S_{e))\;{\overset {\textstyle _{1)){\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoondown ))}\;{\text{I))

The bond number will be important later when converting from the bond graph to state-space equations.

Association of elements

Series association

Suppose that an element has the following behavior:

e(t)=\alpha g(q(t))

where

g(x)

is a generic function (it can even differentiate/integrate its input) and

\alpha

is the element's constant. Then, suppose that in a 1-junction you have many of this type of element. Then the total voltage across the junction is:

e(t)=\left(\sum _{i}\alpha _{i}\right)g(q(t))\implies {\begin{array}{||c||}\hline \displaystyle \alpha _{\text{eq))=\sum _{i=1}^{N}\alpha _{i}\\\hline \end{array))

Parallel association

Suppose that an element has the following behavior:

e(t)=g(\alpha q(t))

where

g(x)

is a generic function (it can even differentiate/integrate its input) and

\alpha

is the element's constant. Then, suppose that in a 0-junction you have many of this type of element. Then it is valid:

g^{-1}\left(e(t)\right)=\alpha _{i}q_{i}(t)\implies {\frac {1}{\alpha _{i))}g^{-1}(e(t))=q_{i}(t)\implies \left(\sum _{i}{\frac {1}{\alpha _{i))}\right)g^{-1}(e(t))=q(t)\implies g(g^{-1}(e(t)))=g\left({\frac {1}{\sum _{i}{\frac {1}{\alpha _{i))))}q(t)\right)\implies {\begin{array}{|c|}\hline \alpha _{\text{eq))=\parallel _{i=1}^{N}\alpha _{i}\\\hline \end{array))

Single-port elements

Single-port elements are elements in a bond graph that can have only one port.

Sources and sinks

Sources are elements that represent the input for a system. They will either input effort or flow into a system. They are denoted by a capital "S" with either a lower case "e" or "f" for effort or flow respectively. Sources will always have the arrow pointing away from the element. Examples of sources include: motors (source of effort, torque), voltage sources (source of effort), and current sources (source of flow).

S_{e}\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!))}\;\ J\qquad {\text{and))\qquad S_{f}\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!))}\;\ J

where J indicates a junction.

Sinks are elements that represent the output for a system. They are represented the same way as sources, but have the arrow pointing into the element instead of away from it.

{\displaystyle J\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!))}\;\ S_{e}\qquad {\text{and))\qquad J\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!))}\;\ S_{f))

Inertia

Inertia elements are denoted by a capital "I", and always have power flowing into them. Inertia elements are elements that store energy. Most commonly these are a mass for mechanical systems, and inductors for electrical systems.

J\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!))}\;\ I

Resistance

Resistance elements are denoted by a capital "R", and always have power flowing into them. Resistance elements are elements that dissipate energy. Most commonly these are a damper, for mechanical systems, and resistors for electrical systems.

J\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!))}\;\ R

Compliance

Compliance elements are denoted by a capital "C", and always have power flowing into them. Compliance elements are elements that store potential energy. Most commonly these are springs for mechanical systems, and capacitors for electrical systems.

J\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!))}\;\ C

Two-port elements

These elements have two ports. They are used to change the power between or within a system. When converting from one to the other, no power is lost during the transfer. The elements have a constant that will be given with it. The constant is called a transformer constant or gyrator constant depending on which element is being used. These constants will commonly be displayed as a ratio below the element.

Transformer

A transformer applies a relationship between flow in flow out, and effort in effort out. Examples include an ideal electrical transformer or a lever.

Denoted

{\begin{matrix}{\overset {\textstyle _{1)){\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}\ \ \ TR\ \ {\overset {\textstyle _{2)){\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}\ \\^{r:1}\end{matrix))

where the r denotes the modulus of the transformer. This means

{\displaystyle f_{1}r=f_{2))

and

{\displaystyle e_{2}r=e_{1))

Gyrator

A gyrator applies a relationship between flow in effort out, and effort in flow out. An example of a gyrator is a DC motor, which converts voltage (electrical effort) into angular velocity (angular mechanical flow).

{\begin{matrix}{\overset {\textstyle _{1)){\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}\ \ \ GY\ \ {\overset {\textstyle _{2)){\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}\ \\^{g:1}\end{matrix))

meaning that

{\displaystyle e_{2}=gf_{1))

and

e_{1}=gf_{2}.

Multi-port elements

Junctions, unlike the other elements can have any number of ports either in or out. Junctions split power across their ports. There are two distinct junctions, the 0-junction and the 1-junction which differ only in how effort and flow are carried across. The same junction in series can be combined, but different junctions in series cannot.

0-junctions

0-junctions behave such that all effort values (and its time integral/derivative) are equal across the bonds, but the sum of the flow values in equals the sum of the flow values out, or equivalently, all flows sum to zero. In an electrical circuit, the 0-junction is a node and represents a voltage shared by all components at that node. In a mechanical circuit, the 0-junction is a joint among components, and represents a force shared by all components connected to it.

{\text{all ))e{\text{'s are equal))

\sum f_{\text{in))=\sum f_{\text{out))

An example is shown below.

{\displaystyle {\overset {\textstyle _{1)){\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoondown ))}{\stackrel {\textstyle {\stackrel {\textstyle _{2)){\upharpoonright ))}{0)){\overset {\textstyle _{3)){\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoondown ))))

Resulting equations:

{\displaystyle e_{1}=e_{2}=e_{3))

{\displaystyle f_{1}=f_{2}+f_{3))

1-junctions

1-junctions behave opposite of 0-junctions. 1-junctions behave such that all flow values (and its time integral/derivative) are equal across the bonds, but the sum of the effort values in equals the sum the effort values out, or equivalently, all efforts sum to zero. In an electrical circuit, the 1 junction represents a series connection among components. In a mechanical circuit, the 1-junction represents a velocity shared by all components connected to it.

{\text{all ))f{\text{'s are equal))

\sum e_{\text{in))=\sum e_{\text{out))

An example is shown below.

{\displaystyle {\overset {\textstyle _{1)){\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoondown ))}{\stackrel {\textstyle {\stackrel {\textstyle _{2)){\upharpoonright ))}{1)){\overset {\textstyle _{3)){\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoondown ))))

Resulting equations:

{\displaystyle f_{1}=f_{2}=f_{3))

{\displaystyle e_{1}=e_{2}+e_{3))

Causality

Bond graphs have a notion of causality, indicating which side of a bond determines the instantaneous effort and which determines the instantaneous flow. In formulating the dynamic equations that describe the system, causality defines, for each modeling element, which variable is dependent and which is independent. By propagating the causation graphically from one modeling element to the other, analysis of large-scale models becomes easier. Completing causal assignment in a bond graph model will allow the detection of modeling situation where an algebraic loop exists; that is the situation when a variable is defined recursively as a function of itself.

As an example of causality, consider a capacitor in series with a battery. It is not physically possible to charge a capacitor instantly, so anything connected in parallel with a capacitor will necessarily have the same voltage (effort variable) as that across the capacitor. Similarly, an inductor cannot change flux instantly and so any component in series with an inductor will necessarily have the same flow as the inductor. Because capacitors and inductors are passive devices, they cannot maintain their respective voltage and flow indefinitely—the components to which they are attached will affect their respective voltage and flow, but only indirectly by affecting their current and voltage respectively.

Note: Causality is a symmetric relationship. When one side "causes" effort, the other side "causes" flow.

In bond graph notation, a causal stroke may be added to one end of the power bond to indicate that this side is defining the flow. Consequently, the side opposite from the casual stroke controls the effort.

Sources of flow ( ${\displaystyle S_{f))$ ) define flow, so they host the causal stroke:

S_{f}\;|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!

Sources of effort (

{\displaystyle S_{e))

) define effort, so the other end hosts the causal stroke:

S_{e}\;-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!|

Consider a constant-torque motor driving a wheel, i.e. a source of effort ( ${\displaystyle S_{e))$ ). That would be drawn as follows:

{\begin{array}{r}{\text{motor))\\S_{e}\end{array))\;{\overset {\textstyle \tau }{\underset {\textstyle \omega }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!|))}\;{\text{wheel))

Symmetrically, the side with the causal stroke (in this case the wheel) defines the flow for the bond.

Causality results in compatibility constraints. Clearly only one end of a power bond can define the effort and so only one end of a bond can (the other end) have a causal stroke. In addition, the two passive components with time-dependent behavior, $I$ and $C$ , can only have one sort of causation: an $I$ component determines flow; a $C$ component defines effort. So from a junction, $J$ , the preferred causal orientation is as follows:

J\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!|))}\;I\qquad {\text{and))\qquad J\;{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}\;C

The reason that this is the preferred method for these elements can be further analyzed if you consider the equations they would give shown by the tetrahedron of state.

f(t)={\frac {1}{I))\int e(t)\,dt\qquad {\text{and))\qquad e(t)={\frac {1}{C))\int f(t)\,dt

The resulting equations involve the integral of the independent power variable. This is preferred over the result of having the causality the other way, which results in derivative. The equations can be seen below.

e(t)=I{\dot {f))(t)\qquad {\text{and))\qquad f(t)=C{\dot {e))(t)

It is possible for a bond graph to have a causal bar on one of these elements in the non-preferred manner. In such a case a "causal conflict" is said to have occurred at that bond. The results of a causal conflict are only seen when writing the state-space equations for the graph. It is explained in more details in that section.

A resistor has no time-dependent behavior: apply a voltage and get a flow instantly, or apply a flow and get a voltage instantly, thus a resistor can be at either end of a causal bond:

J\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!\rightharpoonup \!\!\!|))}\;R\qquad {\text{and))\qquad J\;{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}\;R

Transformers are passive, neither dissipating nor storing energy, so causality passes through them:

\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!|))}\;TF\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!|))}\;\qquad {\text{or))\qquad \;{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-))}\;TF\;{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-))}\;

A gyrator transforms flow to effort and effort to flow, so if flow is caused on one side, effort is caused on the other side and vice versa:

\;{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-))}\;GY\;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!|))}\;\qquad {\text{or))\qquad \;{\overset {\textstyle }{\underset {\textstyle }{-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!|))}\;GY\;{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-))}\;

Junctions

In a 0-junction, efforts are equal; in a 1-junction, flows are equal. Thus, with causal bonds, only one bond can cause the effort in a 0-junction and only one can cause the flow in a 1-junction. Thus, if the causality of one bond of a junction is known, the causality of the others is also known. That one bond is called the 'strong bond'

{\text{strong bond))\rightarrow \;\dashv \!{\overset {\textstyle \top }{\underset {\textstyle \bot }{0))}\!\dashv \qquad {\text{and))\qquad {\text{strong bond))\rightarrow \;\vdash \!{\overset {\textstyle \bot }{\underset {\textstyle \top }{1))}\!\vdash

In a nutshell, 0-junctions must have a single causal bar, 1-junctions must have all but one causal bars.

Determining causality

In order to determine the causality of a bond graph certain steps must be followed. Those steps are:

Draw Source Causal Bars
Draw Preferred causality for C and I bonds
Draw causal bars for 0 and 1 junctions, transformers and gyrators
Draw R bond causal bars
If a causal conflict occurs, change C or I bond to differentiation

A walk-through of the steps is shown below.

{\begin{matrix}S_{f}&{\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&0&{\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&TR&{\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&0&{\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&C_{5}\\&&\downharpoonleft &&^{r:1}&&\downharpoonleft &&\\&&C_{2}&&&&R_{6}&&\end{matrix))

The first step is to draw causality for the sources, over which there is only one. This results in the graph below.

{\begin{matrix}S_{f}&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&0&{\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&TR&{\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&0&{\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&C_{5}\\&&\downharpoonleft &&^{r:1}&&\downharpoonleft &&\\&&C_{2}&&&&R_{6}&&\end{matrix))

The next step is to draw the preferred causality for the C bonds.

{\begin{matrix}S_{f}&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&0&{\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&TR&{\overset {\textstyle }{\underset {\textstyle }{\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&0&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&C_{5}\\&&{\bar {\downharpoonleft ))&&^{r:1}&&\downharpoonleft &&\\&&C_{2}&&&&R_{6}&&\end{matrix))

Next apply the causality for the 0 and 1 junctions, transformers, and gyrators.

{\begin{matrix}S_{f}&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&0&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&TR&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&0&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&C_{5}\\&&{\bar {\downharpoonleft ))&&^{r:1}&&{\underline {\downharpoonleft ))&&\\&&C_{2}&&&&R_{6}&&\end{matrix))

However, there is an issue with 0-junction on the left. The 0-junction has two causal bars at the junction, but the 0-junction wants one and only one at the junction. This was caused by having ${\textstyle C_{2))$ be in the preferred causality. The only way to fix this is to flip that causal bar. This results in a causal conflict, the corrected version of the graph is below, with the ${\textstyle \star }$ representing the causal conflict.

{\begin{matrix}S_{f}&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&0&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&TR&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&0&{\overset {\textstyle }{\underset {\textstyle }{|\!\!\!-\!\!\!-\!\!\!-\!\!\!\rightharpoonup ))}&C_{5}\\&&{\underline {\downharpoonleft ))\star &&^{r:1}&&{\underline {\downharpoonleft ))&&\\&&C_{2}&&&&R_{6}&&\end{matrix))

Converting from other systems

One of the main advantages of using bond graphs is that once you have a bond graph it doesn't matter the original energy domain. Below are some of the steps to apply when converting from the energy domain to a bond graph.

Electromagnetic

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

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

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

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

{\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

Simple electrical system

A simple electrical circuit consisting of a voltage source, resistor, and capacitor in series.

The first step is to draw 0-junctions at all of the nodes:

{\begin{matrix}&0&&0&\\&&&&\\&&&&\\&0&&0&\end{matrix))

The next step is to add all of the elements acting at their own 1-junction:

{\begin{matrix}&&&&R&&&&\\&&&&|&&&&\\&&0&-&1&-&0&&\\&&|&&&&|&&\\S_{e}&-&1&&&&1&-&C\\&&|&&&&|&&\\&&{\underline {0))&-&-&-&0&&\end{matrix))

The next step is to pick a ground. The ground is simply an 0-junction that is going to be assumed to have no voltage. For this case, the ground will be chosen to be the lower left 0-junction, that is underlined above. The next step is to draw all of the arrows for the bond graph. The arrows on junctions should point towards ground (following a similar path to current). For resistance, inertance, and compliance elements, the arrows always point towards the elements. The result of drawing the arrows can be seen below, with the 0-junction marked with a star as the ground.

Now that we have the Bond graph, we can start the process of simplifying it. The first step is to remove all the ground nodes. Both of the bottom 0-junctions can be removed, because they are both grounded. The result is shown below.

Next, the junctions with less than three bonds can be removed. This is because flow and effort pass through these junctions without being modified, so they can be removed to allow us to draw less. The result can be seen below.

The final step is to apply causality to the bond graph. Applying causality was explained above. The final bond graph is shown below.

Advanced electrical system

A more advanced electrical system with a current source, resistors, capacitors, and a transformer

Following the steps with this circuit will result in the bond graph below, before it is simplified. The nodes marked with the star denote the ground.

Simplifying the bond graph will result in the image below.

Lastly, applying causality will result in the bond graph below. The bond with star denotes a causal conflict.

Simple linear mechanical

A simple linear mechanical system, consisting of a mass on a spring that is attached to a wall. The mass has some force being applied to it. An image of the system is shown below.

For a mechanical system, the first step is to place a 1-junction at each distinct velocity, in this case there are two distinct velocities, the mass and the wall. It is usually helpful to label the 1-junctions for reference. The result is below.

{\begin{matrix}&&\\&&\\1_{\text{mass))&&\\&&\\&&\\&&\\1_{\text{wall))&&\end{matrix))

The next step is to draw the R and C bonds at their own 0-junctions between the 1-junctions where they act. For this example there is only one of these bonds, the C bond for the spring. It acts between the 1-junction representing the mass and the 1-junction representing the wall. The result is below.

{\begin{matrix}&&\\&&\\1_{\text{mass))&&\\|&&\\0&-&C:{\frac {1}{k))\\|&&\\1_{\text{wall))&&\end{matrix))

Next you want to add the sources and I bonds on the 1-junction where they act. There is one source, the source of effort (force) and one I bond, the mass of the mass both of which act on the 1-junction of the mass. The result is shown below.

{\begin{matrix}S_{e}:F(t)&&\\|&&\\1_{\text{mass))&-&I:m\\|&&\\0&-&C:{\frac {1}{k))\\|&&\\1_{\text{wall))&&\end{matrix))

Next power flow is to be assigned. Like the electrical examples, power should flow towards ground, in this case the 1-junction of the wall. Exceptions to this are R, C, or I bond, which always point towards the element. The resulting bond graph is below.

Now that the bond graph has been generated, it can be simplified. Because the wall is grounded (has zero velocity), you can remove that junction. As such the 0-junction the C bond is on, can also be removed because it will then have less than three bonds. The simplified bond graph can be seen below.

The last step is to apply causality, the final bond graph can be seen below.

Advanced linear mechanical

A more advanced linear mechanical system can be seen below.

Just like the above example, the first step is to make 1-junctions at each of the distant velocities. In this example there are three distant velocity, Mass 1, Mass 2, and the wall. Then you connect all of the bonds and assign power flow. The bond can be seen below.

Next you start the process of simplifying the bond graph, by removing the 1-junction of the wall, and removing junctions with less than three bonds. The bond graph can be seen below.

There is parallel power in the bond graph. Solving parallel power was explained above. The result of solving it can be seen below.

Lastly, apply causality, the final bond graph can be seen below.

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

{\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:

{\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.

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

"♦" denotes preferred causality.

For the example provided,

the governing equations are the following.

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

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.

{\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.

{\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.

{\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.

Systems for bond graph

Tetrahedron of state

Components

Association of elements

Series association

Parallel association

Single-port elements

Sources and sinks

Inertia

Resistance

Compliance

Two-port elements

Transformer

Gyrator

Multi-port elements

0-junctions

1-junctions

Causality

Junctions

Determining causality

Converting from other systems

Electromagnetic

Linear mechanical

Simplifying

Parallel power

Examples

Simple electrical system

Advanced electrical system

Simple linear mechanical

Advanced linear mechanical

State equations

International conferences on bond graph modeling (ECMS and ICBGM)

See also

References

Further reading

External links