kb/data/en.wikipedia.org/wiki/Bond_graph-9.md

---
title: "Bond graph"
chunk: 10/11
source: "https://en.wikipedia.org/wiki/Bond_graph"
category: "reference"
tags: "science, encyclopedia"
date_saved: "2026-05-05T14:13:40.582713+00:00"
instance: "kb-cron"
---

These equations can be manipulated to yield the state equations. For this example, one is trying to find equations that relate 
  
    
      
        
          
            
              
                p
                ˙
              
            
          
          
            3
          
        
        (
        t
        )
      
    
    {\textstyle {\dot {p}}_{3}(t)}
  
 and 
  
    
      
        
          
            
              
                q
                ˙
              
            
          
          
            6
          
        
        (
        t
        )
      
    
    {\textstyle {\dot {q}}_{6}(t)}
  
 in terms of 
  
    
      
        
          p
          
            3
          
        
        (
        t
        )
      
    
    {\textstyle p_{3}(t)}
  
, 
  
    
      
        
          q
          
            6
          
        
        (
        t
        )
      
    
    {\textstyle q_{6}(t)}
  
, and 
  
    
      
        
          e
          
            1
          
        
        (
        t
        )
      
    
    {\textstyle e_{1}(t)}
  
.
To start, one should recall from the tetrahedron of state that 
  
    
      
        
          
            
              
                p
                ˙
              
            
          
          
            3
          
        
        (
        t
        )
        =
        
          e
          
            3
          
        
        (
        t
        )
      
    
    {\textstyle {\dot {p}}_{3}(t)=e_{3}(t)}
  
starting with equation 2, one can rearrange it so that 
  
    
      
        
          e
          
            3
          
        
        =
        
          e
          
            1
          
        
        −
        
          e
          
            2
          
        
        −
        
          e
          
            4
          
        
      
    
    {\displaystyle e_{3}=e_{1}-e_{2}-e_{4}}
  
. 
  
    
      
        
          e
          
            2
          
        
      
    
    {\displaystyle e_{2}}
  
 can be substituted for equation 4, while in equation 4, 
  
    
      
        
          f
          
            2
          
        
      
    
    {\displaystyle f_{2}}
  
 can be replaced by 
  
    
      
        
          f
          
            3
          
        
      
    
    {\displaystyle f_{3}}
  
due to equation 3, which can then be replaced by equation 5. 
  
    
      
        
          e
          
            4
          
        
      
    
    {\displaystyle e_{4}}
  
 can likewise be replaced using equation 7, in which 
  
    
      
        
          e
          
            5
          
        
      
    
    {\displaystyle e_{5}}
  
can be replaced with 
  
    
      
        
          e
          
            6
          
        
      
    
    {\displaystyle e_{6}}
  
 which can then be replaced with equation 10. Following these substituted yields the first state equation which is shown below.

  
    
      
        
          
            
              
                p
                ˙
              
            
          
          
            3
          
        
        (
        t
        )
        =
        
          e
          
            3
          
        
        (
        t
        )
        =
        
          e
          
            1
          
        
        (
        t
        )
        −
        
          
            
              R
              
                2
              
            
            
              I
              
                3
              
            
          
        
        
          p
          
            3
          
        
        (
        t
        )
        −
        
          
            r
            
              C
              
                6
              
            
          
        
        
          q
          
            6
          
        
        (
        t
        )
      
    
    {\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 
  
    
      
        
          
            
              
                q
                ˙
              
            
          
          
            6
          
        
        (
        t
        )
        =
        
          f
          
            6
          
        
        (
        t
        )
      
    
    {\textstyle {\dot {q}}_{6}(t)=f_{6}(t)}
  
. The second state equation is shown below.

  
    
      
        
          
            
              
                q
                ˙
              
            
          
          
            6
          
        
        (
        t
        )
        =
        
          f
          
            6
          
        
        (
        t
        )
        =
        
          
            r
            
              I
              
                3
              
            
          
        
        
          p
          
            3
          
        
        (
        t
        )
        −
        
          
            1
            
              
                R
                
                  7
                
              
              ⋅
              
                C
                
                  6
                
              
            
          
        
        
          q
          
            6
          
        
        (
        t
        )
      
    
    {\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.

  
    
      
        
          
            [
            
              
                
                  
                    
                      
                        
                          p
                          ˙
                        
                      
                    
                    
                      3
                    
                  
                  (
                  t
                  )
                
              
              
                
                  
                    
                      
                        
                          q
                          ˙
                        
                      
                    
                    
                      6
                    
                  
                  (
                  t
                  )
                
              
            
            ]
          
        
        =
        
          
            [
            
              
                
                  −
                  
                    
                      
                        R
                        
                          2
                        
                      
                      
                        I
                        
                          3
                        
                      
                    
                  
                
                
                  −
                  
                    
                      r
                      
                        C
                        
                          6
                        
                      
                    
                  
                
              
              
                
                  
                    
                      r
                      
                        I
                        
                          3
                        
                      
                    
                  
                
                
                  −
                  
                    
                      1
                      
                        
                          R
                          
                            7
                          
                        
                        ⋅
                        
                          C
                          
                            6
                          
                        
                      
                    
                  
                
              
            
            ]
          
        
        
          
            [
            
              
                
                  
                    p
                    
                      3
                    
                  
                  (
                  t
                  )
                
              
              
                
                  
                    q
                    
                      6
                    
                  
                  (
                  t
                  )
                
              
            
            ]
          
        
        +
        
          
            [
            
              
                
                  1
                
              
              
                
                  0
                
              
            
            ]
          
        
        
          
            [
            
              
                
                  
                    e
                    
                      1
                    
                  
                  (
                  t
                  )
                
              
            
            ]
          
        
      
    
    {\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.

== International conferences on bond graph modeling (ECMS and ICBGM) ==
A bibliography on bond graph modeling may be extracted from the following conferences :

ECMS-2013 27th European Conference on Modelling and Simulation, May 27–30, 2013, Ålesund, Norway
ECMS-2008 22nd European Conference on Modelling and Simulation, June 3–6, 2008 Nicosia, Cyprus
ICBGM-2007: 8th International Conference on Bond Graph Modeling And Simulation, January 15–17, 2007, San Diego, California, U.S.A.
ECMS-2006 20TH European Conference on Modelling and Simulation, May 28–31, 2006, Bonn, Germany
IMAACA-2005 International Mediterranean Modeling Multiconference
ICBGM-2005 International Conference on Bond Graph Modeling and Simulation, January 23–27, 2005, New Orleans, Louisiana, U.S.A. – Papers
ICBGM-2003 International Conference on Bond Graph Modeling and Simulation (ICBGM'2003) January 19–23, 2003, Orlando, Florida, USA – Papers
14TH European Simulation symposium October 23–26, 2002 Dresden, Germany
ESS'2001 13th European Simulation symposium, Marseilles, France October 18–20, 2001
ICBGM-2001 International Conference on Bond Graph Modeling and Simulation (ICBGM 2001), Phoenix, Arizona U.S.A.
European Simulation Multi-conference 23-26 May, 2000, Gent, Belgium
11th European Simulation symposium, October 26–28, 1999 Castle, Friedrich-Alexander University, Erlangen-Nuremberg, Germany
ICBGM-1999 International Conference on Bond Graph Modeling and Simulation January 17–20, 1999 San Francisco, California
ESS-97 9TH European Simulation Symposium and Exhibition Simulation in Industry, Passau, Germany, October 19–22, 1997
ICBGM-1997 3rd International Conference on Bond Graph Modeling And Simulation, January 12–15, 1997, Sheraton-Crescent Hotel, Phoenix, Arizona
11th European Simulation Multiconference Istanbul, Turkey, June 1–4, 1997
ESM-1996 10th annual European Simulation Multiconference Budapest, Hungary, June 2–6, 1996
ICBGM-1995 Int. Conf. on Bond Graph Modeling and Simulation (ICBGM'95), January 15–18, 1995, Las Vegas, Nevada.

== See also ==
20-sim simulation software based on the bond graph theory
AMESim simulation software based on the bond graph theory
Hybrid bond graph
Coenergy

== 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
      
    
    {\displaystyle P}
  
 is the active power;

  
    
      
        
          
            
              X
              ^
            
          
        
      
    
    {\displaystyle {\hat {X}}}
  
 is a matrix object;

  
    
      
        
          
            
              x
              →
            
          
        
      
    
    {\displaystyle {\vec {x}}}
  
 is a vector object;

  
    
      
        
          x
          
            †
          
        
      
    
    {\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;

  
    
      
        
          D
          
            t
          
          
            n
          
        
      
    
    {\displaystyle D_{t}^{n}}
  
 is the Euler notation for differentiation, where: 
  
    
      
        
          D
          
            t
          
          
            n
          
        
        f
        (
        t
        )
        =
        
          
            {
            
              
                
                  
                    
                      ∫
                      
                        −
                        ∞
                      
                      
                        t
                      
                    
                    f
                    (
                    s
                    )
                    
                    d
                    s
                    ,
                  
                
                
                  n
                  =
                  −
                  1
                
              
              
                
                  f
                  (
                  t
                  )
                  ,
                
                
                  n
                  =
                  0
                
              
              
                
                  
                    
                      
                        
                          
                            ∂
                            
                              n
                            
                          
                          f
                          (
                          t
                          )
                        
                        
                          ∂
                          
                            t
                            
                              n
                            
                          
                        
                      
                    
                  
                  ,
                
                
                  n
                  >
                  0
                
              
            
            
          
        
      
    
    {\displaystyle 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}}}
  

  
    
      
        
          
            {
            
              
                
                  ⟨
                  x
                  
                    ⟩
                    
                      α
                    
                  
                  :=
                  
                    |
                  
                  x
                  
                    
                      |
                    
                    
                      α
                    
                  
                  sgn
                  ⁡
                  (
                  x
                  )
                
              
              
                
                  ⟨
                  
                    a
                  
                  ⟩
                  =
                  k
                  ⟨
                  b
                  
                    ⟩
                    
                      β
                    
                  
                  
                  ⟹
                  
                  ⟨
                  b
                  ⟩
                  =
                  
                    
                      (
                      
                        
                          
                            1
                            k
                          
                        
                        ⟨
                        a
                        ⟩
                      
                      )
                    
                    
                      1
                      
                        /
                      
                      β
                    
                  
                
              
            
            
          
        
      
    
    {\displaystyle {\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}}}