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

# Transfer function and stability from [1 0; 1 1]

Control Systems · Chapter 8: State Space Analysis

- Asked in: 2081 Bhadra, 6+2 marks
- 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-1-0-1-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8

## Question

Determine transfer function from the given state space model and check stability of the system. [ẋ1; ẋ2] = [1 0; 1 1][x1; x2] + [1; 1]u, y = [1 1][x1; x2]

## Short answer

The transfer function is (2s-1)/(s-1)^2, with a repeated pole at s = 1 in the right half-plane; the system is unstable.

## Answer

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

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

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

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

$$
sI - A = \begin{bmatrix} s-1 & 0 \\ -1 & s-1 \end{bmatrix},\qquad |sI-A| = (s-1)^2 = s^2 - 2s + 1
$$

#### Step 2: Inverse

$$
(sI-A)^{-1} = \frac{1}{(s-1)^2}\begin{bmatrix} s-1 & 0 \\ 1 & s-1 \end{bmatrix}
$$

#### Step 3: Transfer function

$$
\begin{aligned}
(sI-A)^{-1}B &= \frac{1}{(s-1)^2}\begin{bmatrix} s-1 \\ 1 + s - 1 \end{bmatrix} = \frac{1}{(s-1)^2}\begin{bmatrix} s-1 \\ s \end{bmatrix} \\
\frac{Y(s)}{U(s)} &= \begin{bmatrix} 1 & 1 \end{bmatrix}\frac{1}{(s-1)^2}\begin{bmatrix} s-1 \\ s \end{bmatrix} = \frac{2s-1}{(s-1)^2} = \frac{2s-1}{s^2-2s+1}
\end{aligned}
$$

#### Stability
Characteristic equation: $(s-1)^2 = 0$, so there is a **repeated pole at $s = +1$** in the right half of the s-plane (also visible because $A$ is lower triangular with diagonal entries 1, 1).

The impulse response contains terms $e^{t}$ and $te^{t}$, which grow without bound. The system is **unstable**.

**Answer:** $\dfrac{Y(s)}{U(s)} = \dfrac{2s-1}{(s-1)^2}$; poles at $s = 1, 1$, zero at $s = 0.5$; unstable.

## Check yourself

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

1. A = [1 0; 1 1]. What are the eigenvalues?
   A. 0 and 1
   B. 1 and 1
   C. -1 and -1
   D. 1 and 2
   Answer: B. For a triangular matrix the eigenvalues are the diagonal entries. (Key point: Poles of a triangular A)

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

3. With B = [1;1], C = [1 1], what is Y/U?
   A. (2s + 1)/(s+1)^2
   B. 2/(s-1)^2
   C. (s - 1)/(s-1)^2
   D. (2s - 1)/(s-1)^2
   Answer: D. (sI-A)^-1 B = [s-1; s]/(s-1)^2; C adds the two elements to give 2s - 1. (Key point: Transfer function)

4. Why is the system unstable?
   A. Repeated pole at s = +1 gives terms e^t and t e^t that grow
   B. Because the zero is positive
   C. Because the poles are repeated, which is always unstable
   D. Because D = 0
   Answer: A. Poles in the right half-plane give growing exponentials. (Key point: Stability reasoning)

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