---
title: "State transition matrix (G = [0 1; −0.21 −1])"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:state-transition-matrix-0-21"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/state-transition-matrix-0-21
site: BCT Engineering
---

# State transition matrix (G = [0 1; −0.21 −1])

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2067 Mangsir, 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/state-transition-matrix-0-21
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Determine the state transition matrix for the system given by: x(k+1)=Gx(k)+Hu(k), y(k)=Cx(k)+Du(k), where G=[0 1; -0.21 -1], H=[1; 1], C=[1 0], D=[1].

## Short answer

Compute G^k by inverse Z-transform of (zI-G)^(-1)z. With eigenvalues -0.3 and -0.7, the state transition matrix ψ(k) = [[1.75(-0.3)^k - 0.75(-0.7)^k, 2.5(-0.3)^k - 2.5(-0.7)^k], [-0.525(-0.3)^k + 0.525(-0.7)^k, -0.75(-0.3)^k + 1.75(-0.7)^k]].

## Answer

The state transition matrix of $\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) + \mathbf{H}u(k)$ is
$$
\boldsymbol{\psi}(k) = \mathbf{G}^k = \mathcal{Z}^{-1}\left[(z\mathbf{I}-\mathbf{G})^{-1}z\right]
$$
It depends only on $\mathbf{G}$. $\mathbf{H}$, $\mathbf{C}$ and $\mathbf{D}$ are needed only for the response.

#### Step 1: $(z\mathbf{I}-\mathbf{G})^{-1}$
$$
z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z & -1 \\ 0.21 & z+1 \end{bmatrix},\qquad |z\mathbf{I}-\mathbf{G}| = z^2 + z + 0.21 = (z+0.3)(z+0.7)
$$
$$
(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z+0.3)(z+0.7)}\begin{bmatrix} z+1 & 1 \\ -0.21 & z \end{bmatrix}
$$

#### Step 2: Partial fractions of $(z\mathbf{I}-\mathbf{G})^{-1}z$
$$
\begin{aligned}
\frac{(z+1)z}{(z+0.3)(z+0.7)} &= \frac{1.75z}{z+0.3} - \frac{0.75z}{z+0.7}\\
\frac{z}{(z+0.3)(z+0.7)} &= \frac{2.5z}{z+0.3} - \frac{2.5z}{z+0.7}\\
\frac{-0.21z}{(z+0.3)(z+0.7)} &= \frac{-0.525z}{z+0.3} + \frac{0.525z}{z+0.7}\\
\frac{z\cdot z}{(z+0.3)(z+0.7)} &= \frac{-0.75z}{z+0.3} + \frac{1.75z}{z+0.7}
\end{aligned}
$$
(For example, for (1,1): at $z = -0.3$, $\frac{0.7}{0.4} = 1.75$; at $z = -0.7$, $\frac{0.3}{-0.4} = -0.75$.)

#### Step 3: Inverse z-transform
Using $\mathcal{Z}^{-1}\left[\dfrac{z}{z+a}\right] = (-a)^k$:
$$
\boldsymbol{\psi}(k) = \begin{bmatrix} 1.75(-0.3)^k - 0.75(-0.7)^k & 2.5(-0.3)^k - 2.5(-0.7)^k \\ -0.525(-0.3)^k + 0.525(-0.7)^k & -0.75(-0.3)^k + 1.75(-0.7)^k \end{bmatrix}
$$

#### Check
- $k = 0$: $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \mathbf{I}$
- $k = 1$: $\begin{bmatrix} -0.525+0.525 & -0.75+1.75 \\ 0.1575-0.3675 & 0.225-1.225 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.21 & -1 \end{bmatrix} = \mathbf{G}$
- $k = 2$: $\begin{bmatrix} -0.21 & -1 \\ 0.21 & 0.79 \end{bmatrix} = \mathbf{G}^2$

Both eigenvalues ($-0.3$, $-0.7$) are inside the unit circle, so $\boldsymbol{\psi}(k) \to \mathbf{0}$ and the system is asymptotically stable.

## Check yourself

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

1. How is the state transition matrix psi(k) found, and what does it depend on?
   A. psi(k) = Z^-1[(zI-G)^-1 z]; it depends only on G
   B. psi(k) = Z^-1[(zI-G)^-1 H]; it depends on G and H
   C. psi(k) = Z^-1[C(zI-G)^-1]; it depends on G and C
   D. psi(k) = Z^-1[(zI-G)^-1]; it depends on G, H, C, D
   Answer: A. psi(k) = G^k = Z^-1[(zI-G)^-1 z]. H, C and D matter only for the response, not for psi. (Key point: Definition of state transition matrix)

2. What is |zI - G| and its factorisation for G = [0 1; -0.21 -1]?
   A. z^2 + z + 0.21 = (z+0.3)(z+0.7)
   B. z^2 - z + 0.21 = (z-0.3)(z-0.7)
   C. z^2 + z - 0.21 = (z+0.3)(z-0.7)
   D. z^2 + 0.21z + 1 = (z+0.3)(z+0.7)
   Answer: A. |zI-G| = z(z+1) + 0.21 = z^2 + z + 0.21, whose roots are -0.3 and -0.7. (Key point: Characteristic polynomial)

3. Which transform pair converts the partial fractions to functions of k?
   A. Z^-1[z/(z+a)] = a^k
   B. Z^-1[z/(z+a)] = (-a)^k
   C. Z^-1[z/(z+a)] = -a^k
   D. Z^-1[z/(z+a)] = a^-k
   Answer: B. z/(z+a) has pole at -a, so its inverse is (-a)^k. With a = 0.3 and 0.7 this gives (-0.3)^k and (-0.7)^k. (Key point: Inverse z-transform of each term)

4. What does psi(1) equal, and what does the eigenvalue location say?
   A. psi(1) = G, and the system is asymptotically stable since -0.3 and -0.7 are inside the unit circle
   B. psi(1) = I, and the system is unstable
   C. psi(1) = H, and the system is marginally stable
   D. psi(1) = G, and the system is unstable since the eigenvalues are negative
   Answer: A. psi(0) = I and psi(1) = G. Both eigenvalues have magnitude below 1, so psi(k) tends to 0 and the system is asymptotically stable. (Key point: Checking and stability)

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