Pulse transfer function matrix (G = [−4 1; −2 −1], D = [1; 0])

The pulse transfer function matrix is $\mathbf{F}(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + \mathbf{D}$. There is one input and two outputs, so $\mathbf{F}(z)$ is a 2×1 matrix.

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

$$z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z+4 & -1 \\ 2 & z+1 \end{bmatrix}$$ $$|z\mathbf{I}-\mathbf{G}| = (z+4)(z+1) + 2 = z^2 + 5z + 6 = (z+2)(z+3)$$ $$(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z+2)(z+3)}\begin{bmatrix} z+1 & 1 \\ -2 & z+4 \end{bmatrix}$$

Step 2: Multiply by H

$$(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{1}{(z+2)(z+3)}\begin{bmatrix} z+1+1 \\ -2+z+4 \end{bmatrix} = \frac{1}{(z+2)(z+3)}\begin{bmatrix} z+2 \\ z+2 \end{bmatrix} = \begin{bmatrix} \dfrac{1}{z+3} \\[2mm] \dfrac{1}{z+3} \end{bmatrix}$$

Step 3: Multiply by C

$$\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} \dfrac{1}{z+3} \\[2mm] \dfrac{1}{z+3} \end{bmatrix} = \begin{bmatrix} \dfrac{2}{z+3} \\[2mm] \dfrac{3}{z+3} \end{bmatrix}$$

Step 4: Add D

$$\mathbf{F}(z) = \begin{bmatrix} \dfrac{2}{z+3} + 1 \\[2mm] \dfrac{3}{z+3} + 0 \end{bmatrix} = \begin{bmatrix} \dfrac{z+5}{z+3} \\[2mm] \dfrac{3}{z+3} \end{bmatrix}$$

Remarks

• The factor $(z+2)$ cancels: the mode at $z = -2$ is not excited by the input, so it does not appear in $\mathbf{F}(z)$.
• The non-zero $D_1 = 1$ makes $Y_1/U$ have equal numerator and denominator degree (direct feed-through).

Answer: $\dfrac{Y_1(z)}{U(z)} = \dfrac{z+5}{z+3}$, $\dfrac{Y_2(z)}{U(z)} = \dfrac{3}{z+3}$