---
title: "Transfer function from state model [0 −1; −2 −3]"
subject: "Control Systems"
chapter: 8
key: "semester_3/control_systems:ss-tf-0-1-2-2"
source: https://www.bctengineer.com/q/semester_3/control_systems/ss-tf-0-1-2-2
site: BCT Engineering
---

# Transfer function from state model [0 −1; −2 −3]

Control Systems · Chapter 8: State Space Analysis

- Asked in: 2068 Chaitra, 8 marks
- Video: [Derivation of Transfer Function from State Model](https://www.youtube.com/watch?v=vY2Byo2SJ48) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/ss-tf-0-1-2-2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8

## Question

Determine TF for the system whose state space representation is given by: [ẋ1; ẋ2] = [0 −1; −2 −3][x1; x2] + [1; 2]u, y = [1 0][x1; x2].

## Short answer

Compute G(s)=C(sI-A)⁻¹B with A=[0 -1;-2 -3], B=[1;2], C=[1 0]. Result: Y(s)/U(s) = (s+1)/(s²+3s-2). The denominator has a root at +0.562, so the system is unstable.

## Answer

For $\dot{x} = Ax + Bu$, $y = Cx + Du$:

$$
\frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D = \frac{C\,\operatorname{adj}(sI-A)\,B}{|sI-A|} + D
$$

$$
A = \begin{bmatrix} 0 & -1 \\ -2 & -3 \end{bmatrix},\quad B = \begin{bmatrix} 1 \\ 2 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 0 \end{bmatrix},\quad D = 0
$$

#### Step 1: $(sI - A)$

$$
sI - A = \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix} 0 & -1 \\ -2 & -3 \end{bmatrix} = \begin{bmatrix} s & 1 \\ 2 & s+3 \end{bmatrix}
$$

#### Step 2: Determinant and adjoint

$$
|sI-A| = s(s+3) - (1)(2) = s^2 + 3s - 2
$$

$$
\operatorname{adj}(sI-A) = \begin{bmatrix} s+3 & -1 \\ -2 & s \end{bmatrix}
$$

#### Step 3: Multiply by $B$ and $C$

$$
\operatorname{adj}(sI-A)B = \begin{bmatrix} (s+3)(1) + (-1)(2) \\ (-2)(1) + s(2) \end{bmatrix} = \begin{bmatrix} s+1 \\ 2s-2 \end{bmatrix}
$$

$$
C\,\operatorname{adj}(sI-A)B = \begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} s+1 \\ 2s-2 \end{bmatrix} = s + 1
$$

#### Step 4: Transfer function

$$
\frac{Y(s)}{U(s)} = \frac{s+1}{s^2+3s-2}
$$

- Zero: $s = -1$
- Poles: $s = \dfrac{-3 \pm \sqrt{17}}{2} = 0.562$ and $-3.562$

The pole at $+0.562$ lies in the right half-plane, so this system is unstable.

**Answer:** $\dfrac{Y(s)}{U(s)} = \dfrac{s+1}{s^2+3s-2}$.

## Check yourself

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

1. For A = [0 -1; -2 -3], what is |sI - A|?
   A. s^2 + 3s + 2
   B. s^2 + 3s - 2
   C. s^2 - 3s - 2
   D. s^2 + 3s + 1
   Answer: B. sI - A = [s 1; 2 s+3], so determinant = s(s+3) - 2 = s^2 + 3s - 2. (Key point: Determinant)

2. With B = [1;2], C = [1 0], what is Y/U?
   A. (2s-2)/(s^2 + 3s - 2)
   B. 1/(s^2 + 3s - 2)
   C. (s+1)/(s^2 + 3s - 2)
   D. (s+1)/(s^2 + 3s + 2)
   Answer: C. adj(sI-A)B = [s+1; 2s-2], and C picks the first element, s+1. (Key point: Transfer function)

3. What is the stability of this system?
   A. Stable, since the zero is at -1
   B. Stable, since the sum of poles is negative
   C. Marginally stable, since one pole is near zero
   D. Unstable, since a pole lies at s = +0.562
   Answer: D. The roots of s^2 + 3s - 2 are 0.562 and -3.562; a right half-plane pole means instability. (Key point: Stability decision)

4. What shows at once that s^2 + 3s - 2 cannot be stable?
   A. Its coefficients do not all have the same sign
   B. Its degree is two
   C. It has a constant term
   D. Its leading coefficient is 1
   Answer: A. For stability all coefficients of the characteristic polynomial must be positive. (Key point: Quick necessary condition)

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