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Systems Theory

State-space representation, equilibria and linearisation

The concept of state, classification of systems, linear time-invariant systems, an RLC circuit example, equilibrium points, linearisation with proof, the pendulum, changes of coordinates.

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Systems theory studies in a unified way very different objects, circuits, machines, chemical reactions, populations, economies, looking only at how causes (inputs) produce effects (outputs) over time. Its central tool is the state-space representation. In this lesson we look at what the state is, how systems are classified, equilibrium points, linearisation with its proof and the example of the pendulum, and changes of coordinates.

The state

A dynamical system has memory: the output at an instant depends not only on the input at that instant, but on the whole past history. The state x(t) is the minimal information about the past that, together with the future input, determines the whole future evolution. In an electric circuit it is the voltages across capacitors and the currents in inductors; in a mechanical system the positions and velocities; in a tank the levels.

The continuous-time input-state-output representation is

    \[ \dot{x}(t) = f\big(x(t), u(t), t\big),\qquad y(t) = g\big(x(t), u(t), t\big) \]

with x \in \mathbb{R}^n (the number n is the order of the system), input u \in \mathbb{R}^m and output y \in \mathbb{R}^p. The first is the state equation, the second the output map. In discrete time the derivative is replaced by the next step, x(k+1) = f(x(k), u(k), k).

Classification

  • Linear if f and g are linear in x and u; superposition then holds.
  • Time-invariant if f and g do not depend explicitly on time.
  • Strictly proper if the output depends on the input only through the state, i.e. there is no instantaneous link.

The most studied case, and the basis of the whole course, is the linear time-invariant system:

    \[ \dot{x} = Ax + Bu,\qquad y = Cx + Du \]

with constant matrices A (n\times n), B (n\times m), C (p\times n) and D (p\times m).

Block diagram of the state-space representation of a linear time-invariant system
The input u goes through matrix B and is added to the term Ax; the integrator produces the state x, which through C, plus the direct link D, gives the output y. The internal feedback through A describes the system’s own dynamics.

Example: a series RLC circuit. With state x_1 = v_C (capacitor voltage) and x_2 = i_L (inductor current), input the source voltage u and output v_C:

    \[ \dot{x}_1 = \frac{1}{C}x_2,\qquad \dot{x}_2 = \frac{1}{L}\left(u - x_1 - R x_2\right) \]

i.e. A = \begin{pmatrix} 0 & 1/C \\ -1/L & -R/L \end{pmatrix}, B = \begin{pmatrix} 0 \\ 1/L \end{pmatrix}, C = (1\ \ 0), D = 0.

Equilibrium points

With a constant input u = \bar{u}, a state \bar{x} is an equilibrium if, starting there, the system stays there: f(\bar{x}, \bar{u}) = 0. A linear system with invertible A has a single equilibrium, \bar{x} = -A^{-1}B\bar{u}; a non-linear system can have many, of different nature.

Linearisation

Almost all real systems are non-linear, but around an equilibrium they behave like a linear system.

Theorem. Let \bar{x}, \bar{u} be an equilibrium of \dot{x} = f(x, u) with differentiable f. For small deviations \delta x = x - \bar{x}, \delta u = u - \bar{u}, to first order,

    \[ \dot{\delta x} = A\,\delta x + B\,\delta u,\qquad A = \left.\frac{\partial f}{\partial x}\right|_{\bar{x},\bar{u}},\quad B = \left.\frac{\partial f}{\partial u}\right|_{\bar{x},\bar{u}} \]

Proof. Since \bar{x} is constant, \dot{\delta x} = \dot{x} = f(\bar{x} + \delta x, \bar{u} + \delta u). The first-order Taylor expansion around (\bar{x}, \bar{u}) gives

    \[ f(\bar{x} + \delta x, \bar{u} + \delta u) = f(\bar{x}, \bar{u}) + \frac{\partial f}{\partial x}\,\delta x + \frac{\partial f}{\partial u}\,\delta u + o(\|\delta x\| + \|\delta u\|) \]

and the first term is zero because the point is an equilibrium. Neglecting the higher-order terms, the linear system remains. ∎

The matrices A and B are the Jacobians of f. The output is linearised in the same way, with C = \partial g/\partial x and D = \partial g/\partial u.

The example of the pendulum

A pendulum of length L, with viscous friction and an applied torque u, with state x_1 = \theta (angle from the downward vertical) and x_2 = \omega = \dot\theta:

    \[ \dot{x}_1 = x_2,\qquad \dot{x}_2 = -\frac{g}{L}\sin x_1 - \beta x_2 + \gamma u \]

With u = 0 the equilibria are x_2 = 0, \sin x_1 = 0: the pendulum hanging down (\theta = 0) and the inverted pendulum (\theta = \pi). The Jacobians are

    \[ A_0 = \begin{pmatrix} 0 & 1 \\ -g/L & -\beta \end{pmatrix},\qquad A_\pi = \begin{pmatrix} 0 & 1 \\ g/L & -\beta \end{pmatrix} \]

because the derivative of \sin x_1 is \cos x_1, which is +1 at 0 and -1 at \pi. The two matrices differ by a sign, and that sign decides everything: the first describes damped oscillations, the second an unstable system (lesson on stability). It is the same model as the inverted pendulum on a cart and the Segway, which stay upright only thanks to feedback.

Changes of coordinates

The state is not unique: if T is an invertible matrix, z = T^{-1}x is also a state. Substituting x = Tz:

    \[ \dot{z} = T^{-1}AT\,z + T^{-1}B\,u,\qquad y = CT\,z + Du \]

The two representations describe the same system: they have the same eigenvalues (the matrices A and T^{-1}AT are similar) and the same input-output relation. Choosing T well gives more convenient forms, such as the diagonal one, in which each coordinate evolves on its own, or the canonical forms (lesson on realisation).

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