14 KiB
| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| Bond graph | 9/11 | https://en.wikipedia.org/wiki/Bond_graph | reference | science, encyclopedia | 2026-05-05T14:13:40.582713+00:00 | kb-cron |
1
mass
|
0
−
C
:
1
k
|
1
wall
{\displaystyle {\begin{matrix}&&\\&&\\1_{\text{mass}}&&\\|&&\\0&-&C:{\frac {1}{k}}\\|&&\\1_{\text{wall}}&&\end{matrix}}}
Next one wants 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.
S
e
:
F
(
t
)
|
1
mass
−
I
:
m
|
0
−
C
:
1
k
|
1
wall
{\displaystyle {\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), one 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 one connects all of the bonds and assign power flow. The bond can be seen below.
Next one starts 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
x
˙
(
t
)
=
A
x
(
t
)
+
B
u
(
t
)
{\displaystyle {\dot {\mathbf {x} }}(t)=\mathbf {A} \mathbf {x} (t)+\mathbf {B} \mathbf {u} (t)}
where
x
(
t
)
{\textstyle \mathbf {x} (t)}
is a column matrix of the state variables, or the unknowns of the system.
x
˙
(
t
)
{\textstyle {\dot {\mathbf {x} }}(t)}
is the time derivative of the state variables.
u
(
t
)
{\textstyle \mathbf {u} (t)}
is a column matrix of the inputs of the system. And
A
{\textstyle \mathbf {A} }
and
B
{\textstyle \mathbf {B} }
are matrices of constants based on the system. The state variables of a system are
q
(
t
)
{\textstyle q(t)}
and
p
(
t
)
{\textstyle p(t)}
values for each C and I bond without a causal conflict. Each I bond gets a
p
(
t
)
{\textstyle p(t)}
while each C bond gets a
q
(
t
)
{\textstyle q(t)}
. For example, if one has the following bond graph
one would have the following
x
˙
(
t
)
{\textstyle {\dot {\mathbf {x} }}(t)}
,
x
(
t
)
{\textstyle \mathbf {x} (t)}
, and
u
(
t
)
{\textstyle \mathbf {u} (t)}
matrices:
x
˙
(
t
)
=
[
p
˙
3
(
t
)
q
˙
6
(
t
)
]
and
x
(
t
)
=
[
p
3
(
t
)
q
6
(
t
)
]
and
u
(
t
)
=
[
e
1
(
t
)
]
{\displaystyle {\dot {\mathbf {x} }}(t)={\begin{bmatrix}{\dot {p}}_{3}(t)\\{\dot {q}}_{6}(t)\end{bmatrix}}\qquad {\text{and}}\qquad \mathbf {x} (t)={\begin{bmatrix}p_{3}(t)\\q_{6}(t)\end{bmatrix}}\qquad {\text{and}}\qquad \mathbf {u} (t)={\begin{bmatrix}e_{1}(t)\end{bmatrix}}}
The matrices of
A
{\textstyle \mathbf {A} }
and
B
{\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.
"♦" denotes preferred causality. For the example provided,
the governing equations are the following.
e
1
=
input
{\textstyle e_{1}={\text{input}}}
e
3
=
e
1
−
e
2
−
e
4
{\textstyle e_{3}=e_{1}-e_{2}-e_{4}}
f
1
=
f
2
=
f
4
=
f
3
{\textstyle f_{1}=f_{2}=f_{4}=f_{3}}
e
2
=
R
2
f
2
{\textstyle e_{2}=R_{2}f_{2}}
f
3
=
1
I
3
∫
e
3
d
t
=
1
I
3
p
3
{\textstyle f_{3}={\frac {1}{I_{3}}}\int e_{3}\,dt={\frac {1}{I_{3}}}p_{3}}
f
5
=
f
4
⋅
r
{\textstyle f_{5}=f_{4}\cdot r}
e
4
=
e
5
⋅
r
{\textstyle e_{4}=e_{5}\cdot r}
e
5
=
e
7
=
e
6
{\textstyle e_{5}=e_{7}=e_{6}}
f
6
=
f
5
−
f
7
{\textstyle f_{6}=f_{5}-f_{7}}
e
6
=
1
C
6
∫
f
6
d
t
=
1
C
6
q
6
{\textstyle e_{6}={\frac {1}{C_{6}}}\int f_{6}\,dt={\frac {1}{C_{6}}}q_{6}}
f
7
=
1
R
7
e
7
{\textstyle f_{7}={\frac {1}{R_{7}}}e_{7}}