Pulse transfer function matrix of a general companion form

The pulse transfer function is $F(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + D$, obtained by taking the z-transform of the state equations with zero initial state.

Given

$$\mathbf{G} = \begin{bmatrix} -a_1 & 1 & 0 \\ -a_2 & 0 & 1 \\ -a_3 & 0 & 0 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} h_1 \\ h_2 \\ h_3 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix},\quad D = b_0$$

Step 1: $z\mathbf{I} - \mathbf{G}$ and its determinant

$$z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z+a_1 & -1 & 0 \\ a_2 & z & -1 \\ a_3 & 0 & z \end{bmatrix}$$

Expanding along the first row:

$$\begin{aligned} |z\mathbf{I}-\mathbf{G}| &= (z+a_1)z^2 + 1\,(a_2z + a_3)\\ &= z^3 + a_1z^2 + a_2z + a_3 \end{aligned}$$

Step 2: First row of the adjoint

Only the first row of $(z\mathbf{I}-\mathbf{G})^{-1}$ is needed, because $\mathbf{C} = [1\ 0\ 0]$. The first row of the adjoint (the cofactors of the first column, transposed) is

$$\begin{bmatrix} z^2 & z & 1 \end{bmatrix}$$

Check: cofactor $C_{11} = z\cdot z - 0 = z^2$, $C_{21} = -(-1\cdot z - 0) = z$, $C_{31} = (-1)(-1) - 0 = 1$.

Step 3: $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}$

$$\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{\begin{bmatrix} z^2 & z & 1 \end{bmatrix}\begin{bmatrix} h_1 \\ h_2 \\ h_3 \end{bmatrix}}{z^3+a_1z^2+a_2z+a_3} = \frac{h_1z^2 + h_2z + h_3}{z^3+a_1z^2+a_2z+a_3}$$

Step 4: Add D

$$\begin{aligned} F(z) &= b_0 + \frac{h_1z^2 + h_2z + h_3}{z^3+a_1z^2+a_2z+a_3}\\ &= \frac{b_0z^3 + (h_1 + a_1b_0)z^2 + (h_2 + a_2b_0)z + (h_3 + a_3b_0)}{z^3 + a_1z^2 + a_2z + a_3} \end{aligned}$$

Remark

This is the observable canonical form written with the state order reversed. Comparing with $\dfrac{b_0z^3+b_1z^2+b_2z+b_3}{z^3+a_1z^2+a_2z+a_3}$ gives $h_i = b_i - a_ib_0$, which is the standard result for the observable form. Since there is one input and one output, the "matrix" is the 1×1 function above.

Answer: $F(z) = \dfrac{b_0z^3 + (h_1+a_1b_0)z^2 + (h_2+a_2b_0)z + (h_3+a_3b_0)}{z^3+a_1z^2+a_2z+a_3}$