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I am stuck on this exercise. I don't know how to deal with this squared y in the denominator. What am I supposed to do to obtain state space equations?

ex.1 Simplified dynamic model of the steel ball kept in the magnetic field has the following form: $$my'' = mg - C{u^2 \over y^2}.$$

u – voltage of the magnet, y – vertical ball position, g – gravitational constant, m – mass, c – constant. We consider normalization m = g = c = 1.

  1. Write down state space model equations.
  2. What is the operational point for y = 1?
  3. Write down the linear model equations around that point.
  4. Write down the respective transfer function.
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  • $\begingroup$ Hi and welcome to the site. $\endgroup$ – mike65535 Nov 26 '18 at 21:57
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Two principles here:

  1. When dealing with a differential equation, you define intermediate state variables so everything is in terms of first derivatives.
  2. This system is nonlinear, so the state-space equations won't be in terms of matrices.

Applying these principles, we define a state vector: $$ \mathbf x = [x_1, x_2]^T, $$ where: $$ x_1 = y \\ x_2 = \dot y $$ Note that: $$ \dot x_1 = x_2 $$ Substituting these into your original equation yields: $$ \dot x_1 = x_2 = f_1(x_2) \\ \dot x_2 = g - \frac{C}{m} \frac{u^2}{x_1^2} = f_2(x_1, u) $$ which is of the form: $$ \dot{\mathbf{x}} = \mathbf f(\mathbf x, u, t) $$ Since the model coefficients (i.e. $C, g, m$) don't depend on time, we can drop the $t$: $$ \dot{\mathbf{x}} = \mathbf f(\mathbf x, u) $$ Now that you have a put the ODE in explicit form, you can take derivatives to find the operating point.

The basic idea is to approximate $\mathbf f(\mathbf x, u)$ with a first-order Taylor approximation about some operating point $(\mathbf x_0, u_0)$: $$ \mathbf f(\mathbf x, u) \approx \mathbf f(\mathbf x_0, u_0) + (\mathbf x - \mathbf x_0) \frac{\partial \mathbf f}{\partial \mathbf x}\Bigg|_{(\mathbf x_0, u_0)} + (u - u_0) \frac{\partial \mathbf f}{\partial u} \Bigg|_{(\mathbf x_0, u_0)} $$

EDIT:

From the referenced lecture slides, note that: $$ \mathbf f(\mathbf x, u) = \mathbf f(\mathbf x_0 + \Delta \mathbf x, u_0 + \Delta u) = \dot{\mathbf{x}}|_{(\mathbf x_0, u_0)} + \dot{\Delta \mathbf{x}}, $$ where $\Delta \mathbf x = \mathbf x - \mathbf x_0$, $\Delta u = u - u_0$, and $$ \dot{\mathbf{x}}|_{(\mathbf x_0, u_0)} = f(\mathbf x_0, u_0). $$ This leads to a linear, time-invariant system in terms of the state variable $\Delta \mathbf x$: $$ \dot{\Delta \mathbf{x}} = \mathrm A \Delta \mathbf x + \mathrm B \Delta u, $$ where: $$ \mathrm A = \frac{\partial \mathbf f}{\partial \mathbf x}\Bigg|_{(\mathbf x_0, u_0)} \\ \mathrm B = \frac{\partial \mathbf f}{\partial u}\Bigg|_{(\mathbf x_0, u_0)} $$ The solution for this coupled set of equations yields the displacements $\Delta \mathbf x$ from the operating point $(\mathbf x_0, u_0)$. This is true for any operating point $(\mathbf x_0, u_0)$, regardless of whether or not $\mathbf f(\mathbf x_0, u_0) = 0$. Also note that this system is LTI mathematically, not physically. In reality, the matrices $\mathrm A$ and $\mathrm B$ can also change with time.

Note that a good control algorithm based on a linearized model will have to update the operating point at a fast enough rate that the nonlinearities don't cause any problems.

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  • 2
    $\begingroup$ i would call them Two "principles". $\endgroup$ – robert bristow-johnson Nov 27 '18 at 1:58
  • $\begingroup$ It might be useful to note that an operating point has to satisfy $f(x_0,u_0)=0$. $\endgroup$ – fibonatic Nov 27 '18 at 12:15
  • $\begingroup$ @fibonatic you can cheat it into an LTI system by solving for the perturbations $\Delta \mathbf x$ for any operating point $\endgroup$ – Robert L. Nov 27 '18 at 14:05
  • $\begingroup$ @CarlosDanger If $\dot{x}_0=f(x_0,u_0)\neq0$ then $A$ and $B$ might be time varying matrices, so in general you would have a linear time varying system. $\endgroup$ – fibonatic Nov 27 '18 at 14:26
  • $\begingroup$ @fibonatic The linearized system is LTI mathematically, not physically $\endgroup$ – Robert L. Nov 27 '18 at 14:27

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