Mass-Spring Damper State Space Model

Suppose we have the following system:

where $y$ is the displacement of the spring from its natural rest length.

Newton’s 2nd Law states that

$$M\ddot{y}=u-ky-b\dot{y}$$

Choosing states:

$$\begin{array}{r}{x}_1:=y\\ x_2:=\dot{y}\end{array}$$

This then gives us:

$$\begin{array}{rl}{\dot{x}}_1& =\dot{y}=x_2\\ {\dot{x}}_2& =\ddot{y}=\frac{1}{M}(u-k{x}_1-b{x}_2)\\ ⟹y& =x_1\end{array}$$

Thus:

$$\begin{array}{rl}\dot{x}& =\underset{A}{\underbrace{\left[\begin{array}{cc}0& 1\\ -\frac{k}{M}& -\frac{b}{M}\end{array}\right]}}x+\underset{B}{\underbrace{\left[\begin{array}{c}0\\ \frac{1}{M}\end{array}\right]}}u\\ y& =\underset{C}{\underbrace{\left[\begin{array}{cc}1& 0\end{array}\right]}}x+\underset{D}{\underbrace{0}}u\end{array}$$

Sub-xample: Extra State

What if we were to add an unnecessary extra state $x_3$?

$${\dot{x}}_3=2x_2$$

• Note that $x_3(t)=e^{2t}{x}_3(0)$ has unstable dynamics!

$$\begin{array}{rl}\dot{x}& =\left[\begin{array}{ccc}0& 1& 0\\ -\frac{k}{M}& -\frac{b}{M}& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]+\left[\begin{array}{c}0\\ \frac{1}{M}\\ 0\end{array}\right]u\\ y& =\left[\begin{array}{ccc}1& 0& 0\end{array}\right]x+0u\end{array}$$

Thus, the dimension (number of states) of the state space model is not unique either!