Pulse transfer function from state space form

The pulse transfer function (matrix) relates the z-transform of the output to that of the input, with all initial conditions zero. It is found by taking the z-transform of the state and output equations.

Given system

$$\begin{aligned} \mathbf{x}(k+1) &= \mathbf{G}\mathbf{x}(k) + \mathbf{H}\mathbf{u}(k)\\ \mathbf{y}(k) &= \mathbf{C}\mathbf{x}(k) + \mathbf{D}\mathbf{u}(k) \end{aligned}$$

Here $\mathbf{x}$ is an n-vector, $\mathbf{u}$ an r-vector, $\mathbf{y}$ an m-vector, and $\mathbf{G}$ ($n\times n$), $\mathbf{H}$ ($n\times r$), $\mathbf{C}$ ($m\times n$), $\mathbf{D}$ ($m\times r$) are constant matrices.

Derivation

1. Take the z-transform of the state equation, using $\mathcal{Z}[\mathbf{x}(k+1)] = z\mathbf{X}(z) - z\mathbf{x}(0)$:

$$z\mathbf{X}(z) - z\mathbf{x}(0) = \mathbf{G}\mathbf{X}(z) + \mathbf{H}\mathbf{U}(z)$$

2. Put $\mathbf{x}(0) = \mathbf{0}$ (a transfer function is defined for zero initial state):

$$(z\mathbf{I} - \mathbf{G})\mathbf{X}(z) = \mathbf{H}\mathbf{U}(z)$$

3. Pre-multiply by $(z\mathbf{I}-\mathbf{G})^{-1}$, which exists for all $z$ except the eigenvalues of $\mathbf{G}$:

$$\mathbf{X}(z) = (z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}\mathbf{U}(z)$$

4. Take the z-transform of the output equation:

$$\mathbf{Y}(z) = \mathbf{C}\mathbf{X}(z) + \mathbf{D}\mathbf{U}(z)$$

5. Substitute $\mathbf{X}(z)$:

$$\mathbf{Y}(z) = \left[\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + \mathbf{D}\right]\mathbf{U}(z)$$

Result

$$\mathbf{F}(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + \mathbf{D} = \frac{\mathbf{C}\,\operatorname{adj}(z\mathbf{I}-\mathbf{G})\,\mathbf{H}}{|z\mathbf{I}-\mathbf{G}|} + \mathbf{D}$$

$\mathbf{F}(z)$ is an $m\times r$ matrix. Element $F_{ij}(z)$ is the pulse transfer function from input $u_j$ to output $y_i$. For a single-input single-output system it is a scalar.

Properties

• Characteristic equation: $|z\mathbf{I}-\mathbf{G}| = 0$. The poles of $\mathbf{F}(z)$ are eigenvalues of $\mathbf{G}$, unless a pole is cancelled by a zero.
• Invariance: a change of state variables $\mathbf{x} = \mathbf{P}\hat{\mathbf{x}}$ gives $\hat{\mathbf{G}} = \mathbf{P}^{-1}\mathbf{G}\mathbf{P}$, $\hat{\mathbf{H}} = \mathbf{P}^{-1}\mathbf{H}$, $\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}$, and

$$\hat{\mathbf{C}}(z\mathbf{I}-\hat{\mathbf{G}})^{-1}\hat{\mathbf{H}} + \mathbf{D} = \mathbf{C}\mathbf{P}\,\mathbf{P}^{-1}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{P}\,\mathbf{P}^{-1}\mathbf{H} + \mathbf{D} = \mathbf{F}(z)$$

So the pulse transfer function is unique, even though the state-space form is not.

• $\mathbf{D} \neq 0$ means a direct feed-through, and the numerator degree equals the denominator degree.

Example

For $\mathbf{G} = \begin{bmatrix} 0 & 1 \\ -0.16 & -1 \end{bmatrix}$, $\mathbf{H} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$, $\mathbf{C} = [1\ \ 0]$, $D = 0$:

$$F(z) = \frac{[1\ \ 0]\begin{bmatrix} z+1 & 1 \\ -0.16 & z \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}}{z^2 + z + 0.16} = \frac{1}{z^2 + z + 0.16}$$