Basic State-Space Model Examples

Basic Example

Find the state-space model for:

$$\stackrel{...}{x}+a_1\ddot{x}+a_2\dot{x}+a_{3}x=u$$

Since the highest order is $\stackrel{...}{x}$, we use $x,\dot{x},\ddot{x}$ as state variables in a vector:

$$\bar{x}=\left[\begin{array}{c}x\\ \dot{x}\\ \ddot{x}\end{array}\right]$$

We can then write the derivative of $\bar{x}$ in terms of the state variables, which gives us a three equation system including the original equation we wanted to model:

$$\dot{\bar{x}}=\left[\begin{array}{c}\dot{x}\\ \ddot{x}\\ \stackrel{...}{x}\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -a_3& -a_2& -a_1\end{array}\right]\left[\begin{array}{c}x\\ \dot{x}\\ \ddot{x}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u$$

The first row gives:

$$\dot{x}=\left[\begin{array}{ccc}0& 1& 0\end{array}\right]\left[\begin{array}{c}x\\ \dot{x}\\ \ddot{x}\end{array}\right]+0u=\dot{x}$$

The second row gives:

$$\ddot{x}=\left[\begin{array}{ccc}0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ \dot{x}\\ \ddot{x}\end{array}\right]+0u=\ddot{x}$$

The third row gives:

$$\stackrel{...}{x}=\left[\begin{array}{ccc}-a_3& -a_2& -a_1\end{array}\right]\left[\begin{array}{c}x\\ \dot{x}\\ \ddot{x}\end{array}\right]+1u=-a_{3}x-a_2\dot{x}-a_1\ddot{x}+u$$

Mass-Spring-Damper

An example is the typical mass-spring-damper system:

$$\begin{array}{rl}& \\ m\ddot{x}+c\dot{x}+kx& =f\\ \ddot{x}+\frac{c}{m}\dot{x}+\frac{k}{m}x& =\frac{f}{m}\end{array}$$

We define:

$$\begin{array}{rl}{x}_1& =x\\ x_2& ={\dot{x}}_1=\dot{x}\\ x_3& =\dot{x_2}=\ddot{x}\end{array}$$

Then, we can re-write the original equation as:

$$\begin{array}{rl}{\dot{x}}_2+\frac{c}{m}{x}_2+\frac{k}{m}{x}_1& =\frac{f}{m}\\ {\dot{x}}_2& =\frac{f}{m}-\frac{c}{m}{x}_2-\frac{k}{m}{x}_1\end{array}$$

Thus, we’ve reduced our original 2nd-order equation to two first-order equations:

$$\{\begin{array}{l}{\dot{x}}_2=\frac{f}{m}-\frac{c}{m}{x}_2-\frac{k}{m}{x}_1\\ {\dot{x}}_1=x_2\end{array}$$

This is our state-variable model. The variables $x_1$ and $x_2$ are the state variables.

3rd Order System

A third order system:

$$a_3\stackrel{...}{y}(t)+a_2\ddot{y}(t)+a_1\dot{y}(t)+a_{0}y(t)=f(t)$$

Re-arrange so that highest order variable is isolated:

$$\stackrel{...}{y}(t)=\frac{1}{a_3}[f(t)-a_2\ddot{y}(t)-a_1\dot{y}(t)-a_{0}y(t)]$$

Since this is 3rd order, we expect three 1st-order ODEs to describe the system.

We define

$$\begin{array}{rl}{x}_1(t)& =y(t)\\ x_2(t)& ={\dot{x}}_1(t)=\dot{y}(t)\\ x_3(t)& ={\dot{x}}_2(t)=\ddot{y}(t)\\ {\dot{x}}_3(t)& =\stackrel{...}{y}(t)\end{array}$$

Now, our state-variable model is::

$$\begin{array}{rl}{\dot{x}}_1(t)& =x_2(t)\\ {\dot{x}}_2(t)& =x_3(t)\\ {\dot{x}}_3(t)& =\frac{f(t)}{a_3}-\frac{a_2}{a_3}{x}_3(t)-\frac{a_1}{a_3}{x}_2(t)-\frac{a_0}{a_3}{x}_1(t)\end{array}$$

In vector-matrix form, we can write:

$$\left[\begin{array}{c}{\dot{x}}_1\\ {\dot{x}}_2\\ {\dot{x}}_3\end{array}\right]=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -\frac{a_0}{a_3}& -\frac{a_1}{a_3}& -\frac{a_2}{a_3}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ \frac{1}{a_3}\end{array}\right]f(t)$$

The output $y(t)=x_1(t)$ can also be described in matrix form:

$$y(t)=\left[\begin{array}{ccc}1& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]+\left[\begin{array}{c}0\end{array}\right]f(t)$$