---
title: "Discretization and state transition matrix"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:discretization-state-transition"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/discretization-state-transition
site: BCT Engineering
---

# Discretization and state transition matrix

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2082 Chaitra (new course), 4 marks
- Video: [Discretization of State-Space Models](https://www.youtube.com/watch?v=ME8h5zEQE20) — NPTEL-NOC IITM
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/discretization-state-transition
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Describe discretization of continuous time state space representation. Obtain the state transition matrix for the discrete time system given below: x(k+1)=Gx(k)+Hu(k), y(k)=Cx(k), where G=[0 2; -0.2 -1.5], H=[1; 1] and C=[1 0].

## Short answer

Discretization uses G = e^(AT), H = ∫e^(Aλ)B dλ with ZOH. State transition matrix ψ(k) = G^k = Z^{-1}[(zI-G)^{-1}z]. For G=[[0,2],[-0.2,-1.5]], eigenvalues -0.3469, -1.1531; ψ(k) = M1(-0.3469)^k + M2(-1.1531)^k. System unstable (|λ2|>1).

## Answer

#### Discretization of a continuous-time state model
Consider $\dot{\mathbf{x}}(t) = \mathbf{A}\mathbf{x}(t) + \mathbf{B}\mathbf{u}(t)$, $\mathbf{y}(t) = \mathbf{C}\mathbf{x}(t) + \mathbf{D}\mathbf{u}(t)$, where the input comes from a zero-order hold, so $\mathbf{u}(t) = \mathbf{u}(kT)$ for $kT \le t < (k+1)T$.

The solution from $t = kT$ to $(k+1)T$ is
$$
\mathbf{x}((k+1)T) = e^{\mathbf{A}T}\mathbf{x}(kT) + \left(\int_0^T e^{\mathbf{A}\lambda}\,d\lambda\right)\mathbf{B}\,\mathbf{u}(kT)
$$
Hence
$$
\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) + \mathbf{H}\mathbf{u}(k),\qquad \mathbf{G} = e^{\mathbf{A}T},\quad \mathbf{H} = \left(\int_0^T e^{\mathbf{A}\lambda}d\lambda\right)\mathbf{B}
$$
$\mathbf{C}$ and $\mathbf{D}$ are unchanged. $e^{\mathbf{A}t}$ is found from $\mathcal{L}^{-1}[(s\mathbf{I}-\mathbf{A})^{-1}]$. If $\mathbf{A}$ is invertible, $\mathbf{H} = \mathbf{A}^{-1}(e^{\mathbf{A}T}-\mathbf{I})\mathbf{B}$.

#### State transition matrix of the given system
$$
\boldsymbol{\psi}(k) = \mathbf{G}^k = \mathcal{Z}^{-1}\left[(z\mathbf{I}-\mathbf{G})^{-1}z\right],\qquad \mathbf{G} = \begin{bmatrix} 0 & 2 \\ -0.2 & -1.5 \end{bmatrix}
$$
$$
|z\mathbf{I}-\mathbf{G}| = z^2 + 1.5z + 0.4 = 0 \;\Rightarrow\; \lambda_1 = -0.3469,\ \lambda_2 = -1.1531
$$
For distinct eigenvalues, $\mathbf{G}^k = \lambda_1^k\mathbf{M}_1 + \lambda_2^k\mathbf{M}_2$ with $\mathbf{M}_1 = \dfrac{\mathbf{G}-\lambda_2\mathbf{I}}{\lambda_1-\lambda_2}$ and $\mathbf{M}_2 = \dfrac{\mathbf{G}-\lambda_1\mathbf{I}}{\lambda_2-\lambda_1}$, where $\lambda_1-\lambda_2 = 0.8062$:
$$
\boldsymbol{\psi}(k) = \begin{bmatrix} 1.4303 & 2.4807 \\ -0.2481 & -0.4303 \end{bmatrix}(-0.3469)^k + \begin{bmatrix} -0.4303 & -2.4807 \\ 0.2481 & 1.4303 \end{bmatrix}(-1.1531)^k
$$
Check: $k = 0$ gives $\mathbf{I}$, and $k = 1$ gives $\mathbf{G}$. Since $|\lambda_2| > 1$, the system is unstable.

## Check yourself

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

1. What input assumption underlies discretising x' = Ax + Bu?
   A. u is zero
   B. u varies linearly
   C. The input is an impulse train
   D. Zero-order hold: u is constant over each sampling period
   Answer: D. The ZOH makes u(t) = u(kT) over kT ≤ t < (k+1)T. (Key point: Assumption in discretisation)

2. If A is invertible, how can H be written?
   A. H = A^-1 (e^(AT) - I) B
   B. H = A (e^(AT) - I) B
   C. H = e^(AT) B
   D. H = A^-1 B
   Answer: A. Integrating e^(A lambda) gives A^-1 (e^(AT) - I). (Key point: H when A invertible)

3. How is the state transition matrix of the discrete system obtained?
   A. psi(k) = e^(Gk)
   B. psi(k) = G^k = Z^-1[(zI-G)^-1 z]
   C. psi(k) = kG
   D. psi(k) = Z^-1[(zI-G)^-1]
   Answer: B. G^k is the inverse z-transform of z(zI-G)^-1. (Key point: State transition matrix)

4. Why is the given system unstable?
   A. Eigenvalue -0.3469 is negative
   B. H = [1;1]
   C. Eigenvalue -1.1531 has magnitude greater than 1
   D. C = [1 0]
   Answer: C. One eigenvalue outside the unit circle makes G^k grow. (Key point: Stability)

## Close variants set in other exams

- [State transition matrix of a 2×2 discrete system](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Astate-transition-matrix-g-0-2) — Digital Control System, chapter 5, asked 2×

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