Why state representation; controllable canonical form

State representation

The state of a system is the smallest set of variables $x_1(k), \ldots, x_n(k)$ such that knowing them at $k = k_0$, together with the input for $k \ge k_0$, completely fixes the future behaviour. A discrete-time system is then described by first-order vector difference equations:

$$\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) + \mathbf{H}\mathbf{u}(k),\qquad \mathbf{y}(k) = \mathbf{C}\mathbf{x}(k) + \mathbf{D}\mathbf{u}(k)$$

Why it is required

• It handles multi-input multi-output systems as easily as single-input ones; transfer functions become clumsy for these.
• It includes initial conditions, whereas a transfer function assumes zero initial state.
• It applies to time-varying and nonlinear systems, not only LTI ones.
• It shows internal behaviour: hidden modes, controllability and observability. A transfer function can hide pole–zero cancellations.
• It is the basis of modern design: pole placement, state observers, optimal (LQR) control and Liapunov stability.
• The first-order form is ideal for computer simulation and implementation.

Coefficients of the given system

$$\frac{Y(z)}{U(z)} = \frac{z+1}{z^2+1.3z+0.4} = \frac{0\cdot z^2 + 1\cdot z + 1}{z^2 + 1.3z + 0.4}$$

$a_1 = 1.3$, $a_2 = 0.4$, $b_0 = 0$, $b_1 = 1$, $b_2 = 1$.

Derivation

Introduce $Q(z)$ such that

$$\frac{Q(z)}{U(z)} = \frac{1}{z^2+1.3z+0.4},\qquad Y(z) = (z+1)Q(z)$$

In the time domain, $q(k+2) + 1.3q(k+1) + 0.4q(k) = u(k)$ and $y(k) = q(k+1) + q(k)$. Choose $x_1(k) = q(k)$ and $x_2(k) = q(k+1)$:

$$\begin{aligned} x_1(k+1) &= x_2(k)\\ x_2(k+1) &= -0.4x_1(k) - 1.3x_2(k) + u(k)\\ y(k) &= x_1(k) + x_2(k) \end{aligned}$$

Controllable canonical form

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.4 & -1.3 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 1 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}$$

Check

$$\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{\begin{bmatrix} 1 & 1 \end{bmatrix}\begin{bmatrix} z+1.3 & 1 \\ -0.4 & z \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}}{z^2+1.3z+0.4} = \frac{1 + z}{z^2+1.3z+0.4}$$