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I have equations of a dynamic system. I need to figure out what this physical system is.

The equations are:

\begin{align} \dot{x}_1&=bx_1+kx_2+x_3\\ \dot{x}_2&=x_1\\ \dot{x}_3&=\alpha (u-x_2)-\beta x_3 \end{align}

All I can figure out is that it is maybe a mass-spring-damper system, plus a feedback control, but I am not quite sure about the terms $x_3$ and $\dot{x}_3$. What do these two terms mean?

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You're right, it's a second order plant representing the mass-spring-damper system $P\left(s\right)=1/\left(s^2-bs -k\right)$ (stiffness $k$ and damping $b$ should be thus negative) under the closed-loop control action of the compensator $C\left(s\right)$, whose transfer function is:

$ C\left(s\right)=\frac{\alpha}{s+\beta}. $

The negative feedback is closed over the position of the mass, while the set-point is given by $u$.

We can easily derive these results by applying the Laplace transform to the system of equations and then substituting $x_1$:

$$ \begin{align} s^2x_2&=bsx_2+kx_2+x_3\\ sx_3&=\alpha (u-x_2)-\beta x_3 \end{align}, $$

from which we obtain the following transfer functions:

$$ \begin{align} P\left(s\right)&=\frac{x_2\left(s\right)}{x_3\left(s\right)}=\frac{1}{s^2-bs-k}\\ C\left(s\right)&=\frac{x_3\left(s\right)}{u\left(s\right)-x_2\left(s\right)}=\frac{\alpha}{s+\beta} \end{align}, $$

corresponding to the closed-loop system here below:

closed-loop system

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  • $\begingroup$ Hi, @Ugo thanks for the reply! It makes things clear. But I am not quite sure when doing the Laplace Transform of the equations, how do you deal with the x3 in the first equation and x2 in third equation. Thanks! $\endgroup$ – Zengrui Feb 24 '15 at 1:32
  • $\begingroup$ I've extended the answer. $\endgroup$ – Ugo Pattacini Feb 24 '15 at 9:30

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