---
title: "Pulse transfer function from state space form"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:derive-pulse-tf-state-space"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/derive-pulse-tf-state-space
site: BCT Engineering
---

# Pulse transfer function from state space form

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2074 Bhadra, 6 marks; 2071 Bhadra, 8 marks; 2071 Magh, 8 marks (asked 3 times)
- Video: [Transfer Function Derivation from State Model Statement Problem No  4 - Control Systems](https://www.youtube.com/watch?v=n_spUERHzP8) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/derive-pulse-tf-state-space
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Derive the pulse transfer function (matrix) of the given state space representation form x(k+1)=Gx(k)+Hu(k) and y(k)=Cx(k)+Du(k).

## Short answer

Taking z-transforms of x(k+1)=Gx(k)+Hu(k) and y(k)=Cx(k)+Du(k) with zero initial conditions gives (zI-G)X(z)=HU(z) and Y(z)=CX(z)+DU(z). Substituting yields the pulse transfer function matrix F(z)=C(zI-G)⁻¹H+D = C·adj(zI-G)H/|zI-G|+D. This m×r matrix is invariant under state transformations.

## Answer

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 $\mathbf{G} = \begin{bmatrix} 0 & 1 \\ -0.16 & -1 \end{bmatrix}$, $\mathbf{H} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$, $\mathbf{C} = [1\ \ 0]$, $D = 0$:
$$
F(z) = \frac{[1\ \ 0]\begin{bmatrix} z+1 & 1 \\ -0.16 & z \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}}{z^2 + z + 0.16} = \frac{1}{z^2 + z + 0.16}
$$

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is x(0) = 0 set when deriving the pulse transfer function?
   A. Because G is always zero
   B. Because a transfer function is defined for zero initial conditions
   C. To make the system stable
   D. To remove D
   Answer: B. A transfer function relates output to input with all initial conditions zero. (Key point: Zero initial state)

2. From (zI - G)X(z) = HU(z), X(z) is:
   A. (zI - G) H U(z)
   B. H (zI - G)^-1 U(z)
   C. (zI - G)^-1 H U(z)
   D. (zI + G)^-1 H U(z)
   Answer: C. Pre-multiplying by the inverse gives X = (zI - G)^-1 H U. (Key point: Derivation step)

3. The poles of F(z) are:
   A. Eigenvalues of H
   B. Roots of C
   C. Eigenvalues of D
   D. Eigenvalues of G (roots of |zI - G| = 0), unless cancelled by zeros
   Answer: D. The denominator of F(z) is the characteristic polynomial of G. (Key point: Poles of F(z))

4. Under a change of state variables x = P x^, the pulse transfer function:
   A. Is unchanged, so F(z) is unique though the state-space form is not
   B. Is multiplied by P
   C. Gets divided by P
   D. Changes its poles
   Answer: A. C P (zI - P^-1 G P)^-1 P^-1 H + D simplifies back to F(z). (Key point: Invariance)

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