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Systems Theory

Discrete-time systems and exact discretisation

Solution of difference equations by induction, discrete modes and the unit circle, deadbeat response, exact discretisation proved, continuous and discrete eigenvalues, the z-transform.

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Many systems evolve in steps: a computer updating its state at every cycle, a digital controller, a yearly economic model, a population that breeds by seasons. Discrete-time systems have the same theory as continuous ones, with matrix powers in place of exponentials and the unit circle in place of the left half-plane. In this lesson we look at the solution of difference equations with proof, discrete modes, the exact discretisation of a continuous system with its proof and the link between eigenvalues in the two worlds.

Discrete-time state equations

The discrete-time linear time-invariant system is

    \[ x(k+1) = Ax(k) + Bu(k),\qquad y(k) = Cx(k) + Du(k) \]

where k is an integer, the step index.

Theorem. The solution with x(0) = x_0 is

    \[ x(k) = A^k x_0 + \sum_{j=0}^{k-1} A^{\,k-1-j}B\,u(j) \]

Proof. By induction. For k = 0 the sum is empty and x_0 remains. If the formula holds for k, then

    \[ x(k+1) = A\,x(k) + Bu(k) = A^{k+1}x_0 + \sum_{j=0}^{k-1}A^{\,k-j}B\,u(j) + Bu(k) \]

and the last term is exactly the term j = k of the sum, with A^0 = I. The formula therefore holds for k + 1. ∎

It is the discrete analogue of Lagrange’s formula: free response A^k x_0 plus a discrete convolution of the input with the impulse response A^{k-1}B.

Discrete-time modes

If A = V\Lambda V^{-1} is diagonalisable, A^k = V\Lambda^k V^{-1}, and the modes are the sequences \lambda_i^k. Their behaviour depends on the position of the eigenvalue in the complex plane relative to the circle of radius 1:

  • |\lambda| \lt 1: the mode tends to zero; |\lambda| \gt 1: it diverges;
  • \lambda real positive: monotonic behaviour; \lambda real negative: the sign alternates at every step, a behaviour with no equivalent in first-order continuous systems;
  • \lambda = \rho\,e^{\pm j\vartheta} complex: oscillation \rho^k\cos(k\vartheta), with \vartheta radians per step;
  • \lambda = 0: the mode vanishes after one step. If all the eigenvalues are zero, A^n = 0 and any initial state is cleared in at most n steps. This is the deadbeat response, impossible in continuous time.
Unit circle with the position of some discrete-time eigenvalues and the corresponding sequences
Inside the circle of radius 1 the modes decay: a positive real eigenvalue gives a monotonic sequence, a negative one a sequence with alternating signs, a complex pair a damped oscillation. An eigenvalue outside the circle gives a growing sequence.

Exact discretisation

A continuous system driven by a computer receives a piecewise-constant input: the computer updates it every T seconds and holds it until the next update (Process Control lesson on digital control). Seen only at the sampling instants, the continuous system becomes exactly a discrete system.

Theorem. If \dot{x} = Ax + Bu and u(t) = u(k) is constant for kT \le t \lt (k+1)T, then x_k = x(kT) satisfies

    \[ x_{k+1} = A_d\,x_k + B_d\,u(k),\qquad A_d = e^{AT},\quad B_d = \int_0^T e^{A\sigma}\,d\sigma\;B \]

Proof. Apply Lagrange’s formula starting at instant kT for a duration T:

    \[ x\big((k+1)T\big) = e^{AT}x(kT) + \int_{kT}^{(k+1)T} e^{A((k+1)T - \tau)}B\,u(k)\,d\tau \]

With the change of variable \sigma = (k+1)T - \tau the integral becomes \int_0^T e^{A\sigma}d\sigma\,B\,u(k), and u(k) comes out of the integral because it is constant. ∎

It is not an approximation: at the sampling instants the two models coincide exactly.

Example. The first-order system \dot{x} = (-x + u)/\tau has A = -1/\tau, B = 1/\tau. Then A_d = e^{-T/\tau} and B_d = \frac{1}{\tau}\int_0^T e^{-\sigma/\tau}d\sigma = 1 - e^{-T/\tau}. The static gain B_d/(1 - A_d) = 1 is the same as that of the continuous system, as it must be.

Eigenvalues in the two worlds

Theorem. If \lambda is an eigenvalue of A, then e^{\lambda T} is an eigenvalue of A_d = e^{AT}, with the same eigenvector.

Proof. If Av = \lambda v, then A^k v = \lambda^k v for every k, and summing the exponential series e^{AT}v = \sum_k \frac{T^k}{k!}A^k v = \sum_k \frac{(\lambda T)^k}{k!}v = e^{\lambda T}v. ∎

Since |e^{\lambda T}| = e^{T\,\mathrm{Re}\,\lambda}, an eigenvalue in the left half-plane (\mathrm{Re}\,\lambda \lt 0) goes inside the unit circle, the imaginary axis goes onto the circle, the right half-plane outside. Stability is preserved by sampling. Moreover, continuous eigenvalues differing by j\,2\pi/T end up at the same point: this is aliasing (lesson on sampling) seen from the side of the modes.

The z-transform

Like the Laplace transform for continuous systems, the z-transform turns difference equations into algebraic ones: x(k+1) becomes zX(z) - zx_0. The discrete transfer function is

    \[ G(z) = C(zI - A)^{-1}B + D \]

and its poles are the eigenvalues of A (unless there are cancellations). Stability requires them all to lie inside the unit circle.

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