---
title: "Pulse TF matrix of a continuous state model"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:pulse-tf-matrix-continuous-model"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/pulse-tf-matrix-continuous-model
site: BCT Engineering
---

# Pulse TF matrix of a continuous state model

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2072 Asoj, 6 marks
- Video: [27   State Space Representation To Transfer Function](https://www.youtube.com/watch?v=5aYchzrEK0M) — SOUMYA RANJAN SAHOO
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/pulse-tf-matrix-continuous-model
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

State space representation of a system is given by [ẋ₁; ẋ₂]=[0 1; 0 -2][x₁; x₂]+[0; 1]u. Obtain the pulse transfer function matrix.

## Short answer

Discretize with a zero-order hold: G=e^{AT}, H=∫e^{Aλ}dλ B. For T=1 s, the transfer functions from u to the state are X1(z)/U(z)=(0.2838z+0.1485)/((z−1)(z−0.1353)) and X2(z)/U(z)=0.4323/(z−0.1353).

## Answer

The continuous model is first discretized with a zero-order hold ($\mathbf{G} = e^{\mathbf{A}T}$, $\mathbf{H} = \int_0^T e^{\mathbf{A}\lambda}d\lambda\,\mathbf{B}$). The pulse transfer function matrix is then $(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}$. No output equation is given, so the matrix is found **from u to the state vector** $\mathbf{x}$, for general T and for T = 1 s.

#### Step 1: $e^{\mathbf{A}t}$
$$
\mathbf{A} = \begin{bmatrix} 0 & 1 \\ 0 & -2 \end{bmatrix},\qquad (s\mathbf{I}-\mathbf{A})^{-1} = \begin{bmatrix} \dfrac{1}{s} & \dfrac{1}{s(s+2)} \\[2mm] 0 & \dfrac{1}{s+2} \end{bmatrix}
$$
$$
e^{\mathbf{A}t} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2t}) \\ 0 & e^{-2t} \end{bmatrix}
$$

#### Step 2: G and H
$$
\mathbf{G} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & e^{-2T} \end{bmatrix},\qquad
\mathbf{H} = \int_0^T\begin{bmatrix} \tfrac{1}{2}(1-e^{-2\lambda}) \\ e^{-2\lambda} \end{bmatrix}d\lambda = \begin{bmatrix} \tfrac{1}{4}(2T-1+e^{-2T}) \\ \tfrac{1}{2}(1-e^{-2T}) \end{bmatrix}
$$
For T = 1 s: $\mathbf{G} = \begin{bmatrix} 1 & 0.4323 \\ 0 & 0.1353 \end{bmatrix}$, $\mathbf{H} = \begin{bmatrix} 0.2838 \\ 0.4323 \end{bmatrix}$.

#### Step 3: $(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}$
$$
(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z-1)(z-e^{-2T})}\begin{bmatrix} z-e^{-2T} & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & z-1 \end{bmatrix}
$$
$$
\frac{\mathbf{X}(z)}{U(z)} = \begin{bmatrix} \dfrac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} \\[3mm] \dfrac{\tfrac{1}{2}(1-e^{-2T})}{z-e^{-2T}} \end{bmatrix}
$$

#### For T = 1 s
$$
\frac{\mathbf{X}(z)}{U(z)} = \begin{bmatrix} \dfrac{0.2838z + 0.1485}{(z-1)(z-0.1353)} \\[3mm] \dfrac{0.4323}{z-0.1353} \end{bmatrix}
$$

#### Interpretation
- If the output is $y = x_1$ (position), the pulse transfer function is the first element. This is the ZOH equivalent of $1/[s(s+2)]$.
- If the output is $y = x_2$ (velocity), it is the second element, the ZOH equivalent of $1/(s+2)$.
- For any output $y = \mathbf{C}\mathbf{x}$, the pulse transfer function is $\mathbf{C}$ times this vector.

## Check yourself

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

1. What must be done first for a continuous model?
   A. Take the Laplace transform of C
   B. Set T = 0
   C. Ignore A
   D. Discretise it with a zero-order hold to get G and H
   Answer: D. The pulse transfer function applies to the discrete model. (Key point: First step)

2. Since no C is given, the matrix found is from:
   A. u to the state vector x
   B. y to u
   C. x to u
   D. u to y only
   Answer: A. (zI-G)^-1 H maps the input to all states. (Key point: No output equation)

3. The second element 0.4323/(z-0.1353) corresponds to:
   A. Output y = x1, ZOH of 1/(s(s+2))
   B. Output y = x2 (velocity), ZOH of 1/(s+2)
   C. The input u
   D. The poles
   Answer: B. x2 is velocity with pole -2. (Key point: Second element)

4. If y = x1, the pulse transfer function is:
   A. 0.4323/(z-0.1353)
   B. 1/(z-1)
   C. (0.2838z+0.1485)/((z-1)(z-0.1353))
   D. 0.2838/(z-0.1353)
   Answer: C. This is the ZOH equivalent of 1/(s(s+2)). (Key point: Position output)

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