---
title: "Liapunov stability test (G = [0 1; −0.5 1])"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:liapunov-stability-0-5-plus-1"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/liapunov-stability-0-5-plus-1
site: BCT Engineering
---

# Liapunov stability test (G = [0 1; −0.5 1])

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2073 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/liapunov-stability-0-5-plus-1
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Check the stability of given system by Liapunov stability test: [x₁(k+1); x₂(k+1)]=[0 1; -0.5 1][x₁(k); x₂(k)].

## Short answer

Solve GᵀPG − P = −I using G=[0 1; −0.5 1]. The solution P=[2.2 −1.6; −1.6 4.8] has leading minors 2.2>0 and determinant 8>0, so it is positive definite and the system is asymptotically stable.

## Answer

**Liapunov stability theorem (discrete time):** the equilibrium $\mathbf{x} = \mathbf{0}$ of $\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k)$ is asymptotically stable if and only if, for any positive-definite Hermitian matrix $\mathbf{Q}$, there is a positive-definite matrix $\mathbf{P}$ satisfying
$$
\mathbf{G}^T\mathbf{P}\mathbf{G} - \mathbf{P} = -\mathbf{Q}
$$
Then $V(\mathbf{x}) = \mathbf{x}^T\mathbf{P}\mathbf{x}$ is a Liapunov function, and $\Delta V = \mathbf{x}^T(\mathbf{G}^T\mathbf{P}\mathbf{G}-\mathbf{P})\mathbf{x} = -\mathbf{x}^T\mathbf{Q}\mathbf{x} < 0$.

Choose $\mathbf{Q} = \mathbf{I}$ and $\mathbf{P} = \begin{bmatrix} p_{11} & p_{12} \\ p_{12} & p_{22} \end{bmatrix}$.

#### Forming the equations
$$
\mathbf{G} = \begin{bmatrix} 0 & 1 \\ -0.5 & 1 \end{bmatrix},\qquad \mathbf{G}^T = \begin{bmatrix} 0 & -0.5 \\ 1 & 1 \end{bmatrix}
$$
$$
\mathbf{P}\mathbf{G} = \begin{bmatrix} -0.5p_{12} & p_{11}+p_{12} \\ -0.5p_{22} & p_{12}+p_{22} \end{bmatrix}
$$
$$
\mathbf{G}^T\mathbf{P}\mathbf{G} = \begin{bmatrix} 0.25p_{22} & -0.5(p_{12}+p_{22}) \\ -0.5(p_{12}+p_{22}) & p_{11}+2p_{12}+p_{22} \end{bmatrix}
$$
Setting $\mathbf{G}^T\mathbf{P}\mathbf{G} - \mathbf{P} = -\mathbf{I}$ element by element:
$$
\begin{aligned}
(1,1):&\quad 0.25p_{22} - p_{11} = -1\\
(1,2):&\quad -0.5p_{12} - 0.5p_{22} - p_{12} = 0 \;\Rightarrow\; p_{22} = -3p_{12}\\
(2,2):&\quad p_{11} + 2p_{12} + p_{22} - p_{22} = -1 \;\Rightarrow\; p_{11} = -1 - 2p_{12}
\end{aligned}
$$

#### Solving
Substitute into (1,1): $0.25(-3p_{12}) - (-1-2p_{12}) = -1$, so $1.25p_{12} = -2$ and $p_{12} = -1.6$. Then
$$
p_{22} = 4.8,\qquad p_{11} = -1 + 3.2 = 2.2
$$
$$
\mathbf{P} = \begin{bmatrix} 2.2 & -1.6 \\ -1.6 & 4.8 \end{bmatrix}
$$

#### Test for positive definiteness (Sylvester's criterion)
- $\Delta_1 = p_{11} = 2.2 > 0$
- $\Delta_2 = |\mathbf{P}| = 2.2\times4.8 - (1.6)^2 = 10.56 - 2.56 = 8 > 0$

$\mathbf{P}$ is positive definite (its eigenvalues are 1.44 and 5.56).

#### Conclusion
The origin is **asymptotically stable in the large**, and $V(\mathbf{x}) = 2.2x_1^2 - 3.2x_1x_2 + 4.8x_2^2$ is a Liapunov function.

**Cross-check:** $|z\mathbf{I}-\mathbf{G}| = z^2 - z + 0.5 = 0$ gives $z = 0.5 \pm j0.5$, with $|z| = 0.707 < 1$. Both eigenvalues are inside the unit circle, which agrees.

## Check yourself

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

1. Which equation must be solved for P in the discrete-time Liapunov test of x(k+1) = Gx(k)?
   A. G P + P G^T = -Q
   B. G^T P G - P = -Q
   C. G^T P + P G = -Q
   D. G P G^T + P = Q
   Answer: B. The discrete-time equation is G^T P G - P = -Q, which makes the change in V = x^T P x equal to -x^T Q x. (Key point: Discrete Liapunov equation)

2. From the (1,2) element equation, what relation links p22 and p12 for this G?
   A. p22 = 3 p12
   B. p22 = -3 p12
   C. p22 = -0.5 p12
   D. p22 = p12
   Answer: B. The (1,2) equation -0.5p12 - 0.5p22 - p12 = 0 gives p22 = -3 p12. Together with the other equations this leads to p12 = -1.6. (Key point: Solving for the elements of P)

3. For P = [2.2 -1.6; -1.6 4.8], how is positive definiteness confirmed?
   A. Both off-diagonal terms are non-zero
   B. The leading minors are 2.2 > 0 and det P = 8 > 0
   C. The trace of P is positive and det P is zero
   D. P is symmetric, which is enough
   Answer: B. Sylvester's criterion needs all leading principal minors positive: 2.2 and 2.2 x 4.8 - 1.6^2 = 8. (Key point: Sylvester's criterion)

4. What does the answer conclude, and how does the eigenvalue cross-check agree?
   A. Unstable, since p12 is negative
   B. Asymptotically stable, since the eigenvalues are 0.5 +/- j0.5 with |z| = 0.707 < 1
   C. Marginally stable, since the eigenvalues lie on the unit circle
   D. Asymptotically stable, since the eigenvalues are 0.5 and 1
   Answer: B. P is positive definite, so the origin is asymptotically stable. The characteristic equation z^2 - z + 0.5 = 0 gives z = 0.5 +/- j0.5, inside the unit circle. (Key point: Conclusion and cross-check)

## Close variants set in other exams

- [Liapunov stability test (G = [0 1; −0.5 −1])](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aliapunov-stability-0-5-minus-1) — Digital Control System, chapter 5, asked 2×

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