---
title: "Transfer function from state model [0 1; −5 −6]"
subject: "Control Systems"
chapter: 8
key: "semester_3/control_systems:ss-tf-0-1-5-6"
source: https://www.bctengineer.com/q/semester_3/control_systems/ss-tf-0-1-5-6
site: BCT Engineering
---

# Transfer function from state model [0 1; −5 −6]

Control Systems · Chapter 8: State Space Analysis

- Asked in: 2080 Baisakh, 8 marks; 2073 Shrawan, 8 marks; 2081 Bhadra, 5+1 marks (asked 3 times)
- Video: [Transfer Function from State Model/Unit_5/Control Systems](https://www.youtube.com/watch?v=-3i7-tg8Kzo) — Dr.Jayaudhaya ,Simple and Easy Way
- 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-0-1-5-6
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8

## Question

The system equations are given by ẋ(t) = [0 1; −5 −6]x(t) + [0; 1]u(t) and y(t) = [1 0]x(t). Find transfer function of the system and also check stability.

## Short answer

The transfer function is Y(s)/U(s)=1/(s^2+6s+5). The characteristic equation s^2+6s+5=0 gives poles at s=-1 and s=-5, both in the left-half s-plane, so the system is stable.

## Answer

For a state model $\dot{x} = Ax + Bu$, $y = Cx + Du$, the transfer function is

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

Here

$$
A = \begin{bmatrix} 0 & 1 \\ -5 & -6 \end{bmatrix},\quad B = \begin{bmatrix} 0 \\ 1 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 0 \end{bmatrix},\quad D = 0
$$

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

$$
sI - A = \begin{bmatrix} s & -1 \\ 5 & s+6 \end{bmatrix}, \qquad |sI - A| = s(s+6) + 5 = s^2 + 6s + 5
$$

#### Step 2: Inverse

$$
(sI - A)^{-1} = \frac{\operatorname{adj}(sI-A)}{|sI-A|} = \frac{1}{s^2+6s+5}\begin{bmatrix} s+6 & 1 \\ -5 & s \end{bmatrix}
$$

#### Step 3: Transfer function

$$
\begin{aligned}
(sI-A)^{-1}B &= \frac{1}{s^2+6s+5}\begin{bmatrix} 1 \\ s \end{bmatrix} \\
T(s) &= \begin{bmatrix} 1 & 0 \end{bmatrix}\frac{1}{s^2+6s+5}\begin{bmatrix} 1 \\ s \end{bmatrix} = \frac{1}{s^2+6s+5}
\end{aligned}
$$

$$
\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 6s + 5} = \frac{1}{(s+1)(s+5)}
$$

#### Step 4: Stability
Characteristic equation: $|sI - A| = s^2 + 6s + 5 = 0$, so the eigenvalues of $A$ (poles) are

$$
s = -1, \qquad s = -5
$$

Both are real and negative (in the left half of the s-plane), so the system is **stable**. It is overdamped: comparing with $s^2 + 2\zeta\omega_n s + \omega_n^2$, $\omega_n = \sqrt5 = 2.24$ rad/s and $\zeta = 6/(2\sqrt5) = 1.34$. Its unit step response settles to $1/5 = 0.2$ without oscillation.

**Answer:** $\dfrac{Y(s)}{U(s)} = \dfrac{1}{s^2+6s+5}$; poles at $-1$ and $-5$, so the system is stable.

## Check yourself

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

1. Which formula gives the transfer function of x' = Ax + Bu, y = Cx + Du?
   A. (sI - A)^-1 B C + D
   B. C(sI - A)^-1 B + D
   C. C(A - sI)B + D
   D. B(sI - A)^-1 C + D
   Answer: B. T(s) = C(sI - A)^-1 B + D, with the dimensions matching so the result is scalar for a SISO system. (Key point: Transfer function from a state model)

2. For A = [0 1; -5 -6], what is |sI - A|?
   A. s^2 + 6s - 5
   B. s^2 - 6s + 5
   C. s^2 + 6s + 5
   D. s^2 + 5s + 6
   Answer: C. sI - A = [s -1; 5 s+6], so |sI - A| = s(s+6) + 5 = s^2 + 6s + 5. (Key point: Determinant of sI - A)

3. With B = [0;1] and C = [1 0], what is Y(s)/U(s)?
   A. s/(s^2 + 6s + 5)
   B. (s+6)/(s^2 + 6s + 5)
   C. 1/(s^2 + 6s - 5)
   D. 1/(s^2 + 6s + 5)
   Answer: D. (sI - A)^-1 B = [1; s]/(s^2+6s+5); C picks the first element, giving 1/(s^2+6s+5). (Key point: Resulting transfer function)

4. The poles are s = -1 and s = -5. What does this say about stability?
   A. Stable, because both poles are real and in the left half-plane
   B. Unstable, because both poles are real
   C. Marginally stable, because the poles are distinct
   D. Stable only if a zero exists in the left half-plane
   Answer: A. A system is stable when all poles (eigenvalues of A) lie in the left half of the s-plane. (Key point: Stability from poles)

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