A Primer on State-Space Models

A transfer function describes how one input affects one output. A state-space model keeps track of the internal variables as well, which makes it a natural fit for systems with several inputs and outputs and for computation. This primer builds a state-space model from a familiar example and shows what its main matrix tells you.

From a differential equation to states

A mass $m$ on a spring with stiffness $k$ and a damper with coefficient $c$, driven by a force $u$, obeys

$$m\ddot{q} + c\dot{q} + kq = u,$$

where $q$ is the displacement. This is a second-order equation. To get a first-order vector equation, choose the state variables $x_1 = q$ and $x_2 = \dot{q}$. Then $\dot{x}_1 = x_2$ and $\dot{x}_2 = (u - c x_2 - k x_1)/m$, which in matrix form is

$$\dot{x} = \begin{bmatrix} 0 & 1 \\ -k/m & -c/m \end{bmatrix} x + \begin{bmatrix} 0 \\ 1/m \end{bmatrix} u, \qquad y = \begin{bmatrix} 1 & 0 \end{bmatrix} x .$$

The general linear time-invariant form is

$$\dot{x} = Ax + Bu, \qquad y = Cx + Du.$$

Symbol	Meaning	Size for $n$ states, $m$ inputs, $p$ outputs
$x$	State vector	$n \times 1$
$u$	Input vector	$m \times 1$
$y$	Output vector	$p \times 1$
$A$	How the state evolves on its own	$n \times n$
$B$	How inputs enter the state	$n \times m$
$C$	How the state appears in the output	$p \times n$
$D$	Direct path from input to output	$p \times m$

In the example $D = 0$, since the force does not affect the displacement instantaneously.

Stability from the eigenvalues of A

With $u = 0$ the solution is $x(t) = e^{At}x(0)$. It decays to zero for every initial condition if and only if every eigenvalue of $A$ has a negative real part. For the example the characteristic polynomial is

$$\det(sI - A) = s^2 + \frac{c}{m}\,s + \frac{k}{m},$$

with roots

$$s = \frac{-c \pm \sqrt{c^2 - 4mk}}{2m}.$$

For positive $m$, $c$ and $k$ both roots have negative real part, so the damped oscillator is stable. When $c^2 < 4mk$ the roots form a complex pair, and the system oscillates while it decays.

import numpy as np

m, c, k = 1.0, 0.5, 4.0
A = np.array([[0.0, 1.0],
              [-k / m, -c / m]])

print(np.linalg.eigvals(A))   # approximately -0.25 + 1.984j and -0.25 - 1.984j

The real part, $-c/(2m) = -0.25$, sets the decay rate. The imaginary part, $\sqrt{4mk - c^2}/(2m) \approx 1.984$, sets the oscillation frequency in radians per second.

Why it is worth the extra notation

The same $(A, B, C, D)$ form describes a single mass or a robot arm with many joints, and standard linear algebra routines operate on it directly. The limits matter too. The model above is linear, so a real spring that stiffens or a damper that saturates needs linearization around an operating point, and the conclusions then hold only near that point.