---
title: "Poles and zeros from state model"
subject: "Control Systems"
chapter: 8
key: "semester_3/control_systems:ss-zeros-poles-5-6-1"
source: https://www.bctengineer.com/q/semester_3/control_systems/ss-zeros-poles-5-6-1
site: BCT Engineering
---

# Poles and zeros from state model

Control Systems · Chapter 8: State Space Analysis

- Asked in: 2071 Chaitra, 4 marks
- Video: [Finding Poles and Zeros using State-space Methods](https://www.youtube.com/watch?v=FPPWH797nFk) — richard pates
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/ss-zeros-poles-5-6-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8

## Question

For a system given by d/dt [x1; x2] = [−5 −6; 1 0][x1; x2] + [1; 0]u; y = [1 2][x1; x2], determine the zeros and the poles of the system.

## Short answer

The system poles are -2 and -3, zero is at -2. The pole-zero cancellation at -2 yields input-output dynamics 1/(s+3).

## Answer

Find the transfer function $C(sI-A)^{-1}B$; its denominator roots are the poles and numerator roots are the zeros.

$$
A = \begin{bmatrix} -5 & -6 \\ 1 & 0 \end{bmatrix},\quad B = \begin{bmatrix} 1 \\ 0 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 2 \end{bmatrix}
$$

#### Step 1

$$
sI - A = \begin{bmatrix} s+5 & 6 \\ -1 & s \end{bmatrix},\qquad |sI-A| = s(s+5) + 6 = s^2 + 5s + 6 = (s+2)(s+3)
$$

$$
\operatorname{adj}(sI-A) = \begin{bmatrix} s & -6 \\ 1 & s+5 \end{bmatrix},\qquad \operatorname{adj}(sI-A)B = \begin{bmatrix} s \\ 1 \end{bmatrix}
$$

#### Step 2

$$
C\,\operatorname{adj}(sI-A)B = s + 2
$$

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

#### Poles and zeros
- **Poles** (eigenvalues of $A$): $s = -2$ and $s = -3$
- **Zero:** $s = -2$

The zero at $-2$ cancels the pole at $-2$, so the input–output transfer function reduces to $\dfrac{1}{s+3}$. The mode $e^{-2t}$ exists inside the system but does not appear at the output: the system is controllable ($|[B\;\; AB]| = 1 \ne 0$) but **not observable** ($\left|\begin{smallmatrix} C \\ CA \end{smallmatrix}\right| = 0$). Both poles are in the left half-plane, so the system is stable.

## Check yourself

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

1. What are the poles?
   A. 2 and 3
   B. -2 and -3
   C. -1 and -6
   D. -5 and -6
   Answer: B. s^2 + 5s + 6 = (s+2)(s+3). (Key point: Poles)

2. What is C adj(sI-A) B?
   A. s + 1
   B. 2s
   C. s + 2
   D. s - 2
   Answer: C. adj B = [s; 1], and C = [1 2] gives s + 2. (Key point: Numerator)

3. What happens to the zero at -2?
   A. It makes the system unstable
   B. It doubles the pole
   C. It adds a pole at 2
   D. It cancels the pole at -2, reducing the TF to 1/(s+3)
   Answer: D. Same factor in numerator and denominator cancels. (Key point: Cancellation)

4. What does the cancellation imply?
   A. System is not observable (but controllable)
   B. System is not controllable
   C. System is unstable
   D. Both controllable and observable
   Answer: A. Observability matrix determinant is zero. (Key point: Hidden mode)

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