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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 2730, 2013, Ålesund, Norway ECMS-2008 22nd European Conference on Modelling and Simulation, June 36, 2008 Nicosia, Cyprus ICBGM-2007: 8th International Conference on Bond Graph Modeling And Simulation, January 1517, 2007, San Diego, California, U.S.A. ECMS-2006 20TH European Conference on Modelling and Simulation, May 2831, 2006, Bonn, Germany IMAACA-2005 International Mediterranean Modeling Multiconference ICBGM-2005 International Conference on Bond Graph Modeling and Simulation, January 2327, 2005, New Orleans, Louisiana, U.S.A. Papers ICBGM-2003 International Conference on Bond Graph Modeling and Simulation (ICBGM'2003) January 1923, 2003, Orlando, Florida, USA Papers 14TH European Simulation symposium October 2326, 2002 Dresden, Germany ESS'2001 13th European Simulation symposium, Marseilles, France October 1820, 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 2628, 1999 Castle, Friedrich-Alexander University, Erlangen-Nuremberg, Germany ICBGM-1999 International Conference on Bond Graph Modeling and Simulation January 1720, 1999 San Francisco, California ESS-97 9TH European Simulation Symposium and Exhibition Simulation in Industry, Passau, Germany, October 1922, 1997 ICBGM-1997 3rd International Conference on Bond Graph Modeling And Simulation, January 1215, 1997, Sheraton-Crescent Hotel, Phoenix, Arizona 11th European Simulation Multiconference Istanbul, Turkey, June 14, 1997 ESM-1996 10th annual European Simulation Multiconference Budapest, Hungary, June 26, 1996 ICBGM-1995 Int. Conf. on Bond Graph Modeling and Simulation (ICBGM'95), January 1518, 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}}}