--- 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-minus-1" source: https://www.bctengineer.com/q/semester_6/digital_control_system/liapunov-stability-0-5-minus-1 site: BCT Engineering --- # Liapunov stability test (G = [0 1; −0.5 −1]) Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2070 Magh, 6 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-minus-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 with G=[0 1; −0.5 −1]. The matrix P=[2.2 1.6; 1.6 4.8] has principal minors 2.2 and 8, both positive, confirming asymptotic stability. ## 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 $$ \mathbf{G} = \begin{bmatrix} 0 & 1 \\ -0.5 & -1 \end{bmatrix},\qquad \mathbf{G}^T\mathbf{P}\mathbf{G} = \begin{bmatrix} 0.25p_{22} & -0.5(p_{12}-p_{22}) \\ -0.5(p_{12}-p_{22}) & p_{11}-2p_{12}+p_{22} \end{bmatrix} $$ Setting $\mathbf{G}^T\mathbf{P}\mathbf{G} - \mathbf{P} = -\mathbf{I}$: $$ \begin{aligned} (1,1):&\quad 0.25p_{22} - p_{11} = -1\\ (1,2):&\quad -0.5p_{12} + 0.5p_{22} - p_{12} = 0 \;\Rightarrow\; p_{22} = 3p_{12}\\ (2,2):&\quad p_{11} - 2p_{12} = -1 \;\Rightarrow\; p_{11} = 2p_{12} - 1 \end{aligned} $$ #### Solving Substitute into (1,1): $0.75p_{12} - 2p_{12} + 1 = -1$, so $p_{12} = 1.6$. Then $$ p_{22} = 4.8,\qquad p_{11} = 2.2 $$ $$ \mathbf{P} = \begin{bmatrix} 2.2 & 1.6 \\ 1.6 & 4.8 \end{bmatrix} $$ #### Positive definiteness - $\Delta_1 = 2.2 > 0$ - $\Delta_2 = 2.2\times4.8 - 1.6^2 = 8 > 0$ So $\mathbf{P}$ is positive definite. #### Conclusion The equilibrium state $\mathbf{x} = \mathbf{0}$ is **asymptotically stable**, with Liapunov function $V(\mathbf{x}) = 2.2x_1^2 + 3.2x_1x_2 + 4.8x_2^2$ and $\Delta V(\mathbf{x}) = -(x_1^2 + x_2^2) < 0$. **Cross-check:** the characteristic equation $z^2 + z + 0.5 = 0$ gives $z = -0.5 \pm j0.5$, with $|z| = 0.707 < 1$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the test set up in this solution? A. Choose Q = I, then solve G^T P G - P = -Q for P B. Choose P = I, then solve for Q C. Choose Q = 0 and solve G^T P G = P D. Choose Q = I and solve G P G^T = Q Answer: A. Pick any positive-definite Q (here I) and solve G^T P G - P = -Q for P, then test whether P is positive definite. (Key point: Liapunov equation and choice of Q) 2. What is the resulting P? A. [2.2 -1.6; -1.6 4.8] B. [2.2 1.6; 1.6 4.8] C. [1.6 2.2; 2.2 4.8] D. [4.8 1.6; 1.6 2.2] Answer: B. Here p22 = 3p12, p11 = 2p12 - 1 and p12 = 1.6, so p22 = 4.8 and p11 = 2.2. The sign of p12 is positive, unlike the case with G22 = +1. (Key point: Solving for P) 3. With Delta1 = 2.2 and Delta2 = 8, what follows? A. P is positive definite, so the origin is asymptotically stable B. P is indefinite, so the system is unstable C. P is negative definite, so the system is unstable D. The test is inconclusive and the eigenvalues must be used Answer: A. Both leading minors are positive, so P is positive definite and V = x^T P x has delta V = -(x1^2 + x2^2) < 0. (Key point: Positive-definiteness test and conclusion) 4. Which characteristic equation and roots confirm the result? A. z^2 - z + 0.5 = 0, z = 0.5 +/- j0.5 B. z^2 + z + 0.5 = 0, z = -0.5 +/- j0.5, |z| = 0.707 < 1 C. z^2 + z - 0.5 = 0, z = 0.37 and -1.37 D. z^2 + 0.5z + 1 = 0, |z| = 1 Answer: B. For G = [0 1; -0.5 -1], |zI-G| = z^2 + z + 0.5, with roots -0.5 +/- j0.5 of magnitude 0.707, inside the unit circle. 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