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

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

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

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

Step 2: $(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}$

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

Step 3: Multiply by C (D = 0)

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

Remarks

• $\dfrac{Y_1(z)}{U(z)} = \dfrac{2}{z+3}$ and $\dfrac{Y_2(z)}{U(z)} = \dfrac{3}{z+3}$.
• The eigenvalue $z = -2$ is cancelled by a zero: the mode at $-2$ is not excited by $u$ (it is uncontrollable). This is why the transfer functions are first order although the system is second order.
• Both eigenvalues ($-2$, $-3$) lie outside the unit circle, so the system is unstable.

Answer: $\mathbf{F}(z) = \begin{bmatrix} 2/(z+3) \\ 3/(z+3) \end{bmatrix}$