Pulse transfer function from a 3×3 state model

The pulse transfer function is $F(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + D$. Because $\mathbf{C} = [1\ 0\ 0]$, only the first row of the inverse is needed.

Step 1: Characteristic polynomial

$$z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z+0.3679 & -1 & 0 \\ 0.1353 & z & -1 \\ 0.0497 & 0 & z \end{bmatrix}$$ $$\begin{aligned} |z\mathbf{I}-\mathbf{G}| &= (z+0.3679)(z^2) + 1\,(0.1353z + 0.0497)\\ &= z^3 + 0.3679z^2 + 0.1353z + 0.0497 \end{aligned}$$

Step 2: First row of adj$(z\mathbf{I}-\mathbf{G})$

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

(Cofactors of the first column: $z\cdot z - 0$, $-(-z - 0)$, $(+1) - 0$.)

Step 3: $\mathbf{C}\,\text{adj}\,\mathbf{H}$

$$\begin{bmatrix} z^2 & z & 1 \end{bmatrix}\begin{bmatrix} 0.5 \\ 0.25 \\ 0.125 \end{bmatrix} = 0.5z^2 + 0.25z + 0.125$$

Step 4: Add D

$$\begin{aligned} F(z) &= 1.25 + \frac{0.5z^2 + 0.25z + 0.125}{z^3 + 0.3679z^2 + 0.1353z + 0.0497}\\ &= \frac{1.25z^3 + (0.5 + 1.25\times0.3679)z^2 + (0.25 + 1.25\times0.1353)z + (0.125 + 1.25\times0.0497)}{z^3 + 0.3679z^2 + 0.1353z + 0.0497} \end{aligned}$$

Result

$$F(z) = \frac{Y(z)}{U(z)} = \frac{1.25z^3 + 0.9599z^2 + 0.4191z + 0.1871}{z^3 + 0.3679z^2 + 0.1353z + 0.0497}$$

In powers of $z^{-1}$:

$$F(z) = \frac{1.25 + 0.9599z^{-1} + 0.4191z^{-2} + 0.1871z^{-3}}{1 + 0.3679z^{-1} + 0.1353z^{-2} + 0.0497z^{-3}}$$

The given model is an observable-type companion form, so in general $b_i = h_i + a_ib_0$. This matches the numerator found above.