--- title: "State transition matrix of a 2×2 discrete system" subject: "Digital Control System" chapter: 5 key: "semester_6/digital_control_system:state-transition-matrix-g-0-2" source: https://www.bctengineer.com/q/semester_6/digital_control_system/state-transition-matrix-g-0-2 site: BCT Engineering --- # State transition matrix of a 2×2 discrete system Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2079 Chaitra, 8 marks - Video: [Solutions of Sampled-Data State-Space Equations (Dr. Jake Abbott, University of Utah)](https://www.youtube.com/watch?v=O1gnwcJEW7w) — JJAbbottatUtah - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/state-transition-matrix-g-0-2 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5 ## Question 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 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. Verified ψ(0)=I, ψ(1)=G. System unstable (|λ2|>1). ## Answer The state transition matrix of $\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) + \mathbf{H}u(k)$ is $\boldsymbol{\psi}(k) = \mathbf{G}^k$. It is found as $$ \boldsymbol{\psi}(k) = \mathcal{Z}^{-1}\left[(z\mathbf{I}-\mathbf{G})^{-1}z\right] $$ $\mathbf{H}$ and $\mathbf{C}$ are not needed for $\boldsymbol{\psi}(k)$. #### Step 1: $(z\mathbf{I}-\mathbf{G})^{-1}$ $$ z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z & -2 \\ 0.2 & z+1.5 \end{bmatrix},\qquad |z\mathbf{I}-\mathbf{G}| = z^2 + 1.5z + 0.4 $$ $$ (z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{z^2+1.5z+0.4}\begin{bmatrix} z+1.5 & 2 \\ -0.2 & z \end{bmatrix} $$ #### Step 2: Eigenvalues $$ z = \frac{-1.5 \pm \sqrt{2.25 - 1.6}}{2} = -0.75 \pm 0.4031 \;\Rightarrow\; \lambda_1 = -0.3469,\quad \lambda_2 = -1.1531 $$ $\lambda_1 - \lambda_2 = 0.8062$. #### Step 3: Partial fractions of each element of $(z\mathbf{I}-\mathbf{G})^{-1}z$ Each element has the form $\dfrac{N(z)\,z}{(z-\lambda_1)(z-\lambda_2)} = \dfrac{N(\lambda_1)}{\lambda_1-\lambda_2}\dfrac{z}{z-\lambda_1} + \dfrac{N(\lambda_2)}{\lambda_2-\lambda_1}\dfrac{z}{z-\lambda_2}$. | Element | $N(z)$ | coeff. of $\lambda_1^k$ | coeff. of $\lambda_2^k$ | |---|---|---|---| | (1,1) | $z+1.5$ | $1.1531/0.8062 = 1.4303$ | $0.3469/(-0.8062) = -0.4303$ | | (1,2) | $2$ | $2.4807$ | $-2.4807$ | | (2,1) | $-0.2$ | $-0.2481$ | $0.2481$ | | (2,2) | $z$ | $-0.3469/0.8062 = -0.4303$ | $-1.1531/(-0.8062) = 1.4303$ | Using $\mathcal{Z}^{-1}\left[\dfrac{z}{z-\lambda}\right] = \lambda^k$: #### Result $$ \boldsymbol{\psi}(k) = \begin{bmatrix} 1.4303(-0.3469)^k - 0.4303(-1.1531)^k & 2.4807(-0.3469)^k - 2.4807(-1.1531)^k \\ -0.2481(-0.3469)^k + 0.2481(-1.1531)^k & -0.4303(-0.3469)^k + 1.4303(-1.1531)^k \end{bmatrix} $$ #### Check - $k = 0$: $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ - $k = 1$: $\begin{bmatrix} 0 & 2 \\ -0.2 & -1.5 \end{bmatrix} = \mathbf{G}$ - $k = 2$: $\begin{bmatrix} -0.4 & -3 \\ 0.3 & 1.85 \end{bmatrix} = \mathbf{G}^2$ Exact form: $\lambda_{1,2} = -0.75 \pm \sqrt{65}/20$ and $1.4303 = 0.5 + 7.5/\sqrt{65}$, $2.4807 = 20/\sqrt{65}$. **Note:** $|\lambda_2| = 1.153 > 1$, so $\boldsymbol{\psi}(k)$ grows without bound and the system is unstable. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which matrices are needed for psi(k)? A. G and H B. G, H and C C. C only D. Only G Answer: D. The state transition matrix depends only on the system matrix G. (Key point: Which matrices are needed) 2. What are the eigenvalues of G? A. -0.3469 and -1.1531 B. 0.3469 and 1.1531 C. -0.4 and -1.5 D. -0.75 and 0.4031 Answer: A. Roots of z^2+1.5z+0.4: -0.75 ± 0.4031. (Key point: Eigenvalues) 3. How is the coefficient of lambda1^k found for an element N(z)z/((z-l1)(z-l2))? A. N(l2)/(l1-l2) B. N(l1)/(l1-l2) C. N(l1)(l1-l2) D. N(0)/l1 Answer: B. This is the residue at z = l1. (Key point: Partial fraction coefficient) 4. Which check confirms psi(k) is right? A. k = 0 gives G B. k = 1 gives I C. k = 0 gives I and k = 1 gives G D. psi(k) tends to zero Answer: C. G^0 = I and G^1 = G. 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