Find trim point of dynamic system

`[x,u,y,dx] = trim('`

* sys*')

[x,u,y,dx] = trim('

`sys`

[x,u,y,dx] = trim('

`sys`

[x,u,y,dx] = trim('

`sys`

[x,u,y,dx,options] = trim('

`sys`

[x,u,y,dx,options] = trim('

`sys`

A trim point, also known as an equilibrium
point, is a point in the parameter space of a dynamic system at which
the system is in a steady state. For example, a trim point of an aircraft
is a setting of its controls that causes the aircraft to fly straight
and level. Mathematically, a trim point is a point where the system's
state derivatives equal zero. `trim`

starts from
an initial point and searches, using a sequential quadratic programming
algorithm, until it finds the nearest trim point. You must supply
the initial point implicitly or explicitly. If `trim`

cannot
find a trim point, it returns the point encountered in its search
where the state derivatives are closest to zero in a min-max sense;
that is, it returns the point that minimizes the maximum deviation
from zero of the derivatives. `trim`

can find trim
points that meet specific input, output, or state conditions, and
it can find points where a system is changing in a specified manner,
that is, points where the system's state derivatives equal specific
nonzero values.

`[x,u,y,dx] = trim('`

finds
the equilibrium point of the model '* sys*')

`sys`

`x0`

. Specifically, `trim`

finds
the equilibrium point that minimizes the maximum absolute value of `[x-x0,u,y]`

.
If `trim`

cannot find an equilibrium point near the
system's initial state, it returns the point at which the system is
nearest to equilibrium. Specifically, it returns the point that minimizes
`abs(dx)`

where `dx`

represents
the derivative of the system. You can obtain `x0`

using
this command.[sizes,x0,xstr] = sys([],[],[],0)

`[x,u,y,dx] = trim('`

finds
the trim point nearest to * sys*',x0,u0,y0)

`x0`

, `u0`

, `y0`

,
that is, the point that minimizes the maximum value of abs([x-x0; u-u0; y-y0])

`[x,u,y,dx] = trim('`

finds
the trim point closest to * sys*',x0,u0,y0,ix,iu,iy)

`x0`

, `u0`

, `y0`

that
satisfies a specified set of state, input, and/or output conditions.
The integer vectors `ix`

, `iu`

,
and `iy`

select the values in `x0`

, `u0`

,
and `y0`

that must be satisfied. If `trim`

cannot
find an equilibrium point that satisfies the specified set of conditions
exactly, it returns the nearest point that satisfies the conditions,
namely,abs([x(ix)-x0(ix); u(iu)-u0(iu); y(iy)-y0(iy)])

`[x,u,y,dx] = trim('`

finds
specific nonequilibrium points, that is, points at which the system's
state derivatives have some specified nonzero value. Here, * sys*',x0,u0,y0,ix,iu,iy,dx0,idx)

`dx0`

specifies
the state derivative values at the search's starting point and `idx`

selects
the values in `dx0`

that the search must satisfy
exactly.`[x,u,y,dx,options] = trim('`

specifies
an array of optimization parameters that * sys*',x0,u0,y0,ix,iu,iy,dx0,idx,options)

`trim`

passes
to the optimization function that it uses to find trim points. The
optimization function, in turn, uses this array to control the optimization
process and to return information about the process. `trim`

returns
the `options`

array at the end of the search process.
By exposing the underlying optimization process in this way, `trim`

allows
you to monitor and fine-tune the search for trim points.The following table describes how each element affects the search for a trim point. Array elements 1, 2, 3, 4, and 10 are particularly useful for finding trim points.

No. | Default | Description |
---|---|---|

1 | 0 | Specifies display options. 0 specifies no display; 1 specifies tabular output; -1 suppresses warning messages. |

2 | 10^{–4} | Precision the computed trim point must attain to terminate the search. |

3 | 10^{–4} | Precision the trim search goal function must attain to terminate the search. |

4 | 10^{–6} | Precision the state derivatives must attain to terminate the search. |

5 | N/A | Not used. |

6 | N/A | Not used. |

7 | N/A | Used internally. |

8 | N/A | Returns the value of the trim search goal function (λ in goal attainment). |

9 | N/A | Not used. |

10 | N/A | Returns the number of iterations used to find a trim point. |

11 | N/A | Returns the number of function gradient evaluations. |

12 | 0 | Not used. |

13 | 0 | Number of equality constraints. |

14 | 100*(Number of variables) | Maximum number of function evaluations to use to find a trim point. |

15 | N/A | Not used. |

16 | 10^{–8} | Used internally. |

17 | 0.1 | Used internally. |

18 | N/A | Returns the step length. |

`[x,u,y,dx,options] = trim('`

sets
the time to * sys*',x0,u0,y0,ix,iu,iy,dx0,idx,options,t)

`t`

if the system is dependent on time.If you fix any of the state, input or output values, `trim`

uses
the unspecified free variables to derive the solution that satisfies
these constraints.

Consider a linear state-space system modeled using a State-Space block

$$\begin{array}{l}\dot{x}=Ax+Bu\\ y=Cx+Du\end{array}$$

The *A*, *B*, *C*,
and *D* matrices to enter at the command line or
in the block parameters dialog are:.

A = [-0.09 -0.01; 1 0]; B = [ 0 -7; 0 -2]; C = [ 0 2; 1 -5]; D = [-3 0; 1 0];

To find an equilibrium point in this model called `sys`

,
use:

[x,u,y,dx,options] = trim('sys') x = 0 0 u = 0 0 y = 0 0 dx = 0 0

The number of iterations taken is:

options(10) ans = 7

To find an equilibrium point near `x = [1;1], u = [1;1]`

,
enter

x0 = [1;1]; u0 = [1;1]; [x,u,y,dx,options] = trim('sys', x0, u0); x = 1.0e-13 * -0.5160 -0.5169 u = 0.3333 0.0000 y = -1.0000 0.3333 dx = 1.0e-12 * 0.1979 0.0035

The number of iterations taken is

options(10) ans = 25

To find an equilibrium point with the outputs fixed to `1`

,
use:

y = [1;1]; iy = [1;2]; [x,u,y,dx] = trim('sys', [], [], y, [], [], iy) x = 0.0009 -0.3075 u = -0.5383 0.0004 y = 1.0000 1.0000 dx = 1.0e-15 * -0.0170 0.1483

To find an equilibrium point with the outputs fixed to 1 and the derivatives set to 0 and 1, use

y = [1;1]; iy = [1;2]; dx = [0;1]; idx = [1;2]; [x,u,y,dx,options] = trim('sys',[],[],y,[],[],iy,dx,idx) x = 0.9752 -0.0827 u = -0.3884 -0.0124 y = 1.0000 1.0000 dx = 0.0000 1.0000

The number of iterations taken is

options(10) ans = 13

The trim point found by `trim`

starting from
any given initial point is only a local value. Other, more suitable
trim points may exist. Thus, if you want to find the most suitable
trim point for a particular application, it is important to try a
number of initial guesses for `x`

, `u`

,
and `y`

.

`trim`

uses a sequential quadratic programming
algorithm to find trim points. See Sequential Quadratic Programming (SQP) (Optimization Toolbox) for a description
of this algorithm.