---
title: "R-H stability of s⁴+8s³+18s²+16s+50"
subject: "Control Systems"
chapter: 4
key: "semester_3/control_systems:rh-s4-8-18-16-50"
source: https://www.bctengineer.com/q/semester_3/control_systems/rh-s4-8-18-16-50
site: BCT Engineering
---

# R-H stability of s⁴+8s³+18s²+16s+50

Control Systems · Chapter 4: Stability

- Asked in: 2070 Chaitra (old course), 8 marks
- Video: [Example on Routh Array Stable System](https://www.youtube.com/watch?v=BApzhX2WBe0) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/rh-s4-8-18-16-50
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4

## Question

Check the stability of the system represented by the following characteristic equation using R-H criteria: s⁴ + 8s³ + 18s² + 16s + 50 = 0

## Short answer

Routh array first column entries: 1, 8, 16, −9, 50. Two sign changes occur, so there are 2 roots in the right half-plane and 2 in the left half-plane. The system is unstable.

## Answer

**Result:** the system is **unstable**, with 2 roots in the right half plane.

The Routh–Hurwitz criterion says a system is stable only if all elements in the first column of the Routh array have the same sign. The number of sign changes equals the number of roots in the right half s-plane.

$$
s^4 + 8s^3 + 18s^2 + 16s + 50 = 0
$$

**Necessary condition:** all coefficients are present and positive, so it is satisfied. This alone does not prove stability, so the array is needed.

#### Routh array

| Row | Col 1 | Col 2 | Col 3 |
|---|---|---|---|
| $s^4$ | 1 | 18 | 50 |
| $s^3$ | 8 | 16 | 0 |
| $s^2$ | $\frac{8(18)-1(16)}{8}=16$ | $\frac{8(50)-1(0)}{8}=50$ | |
| $s^1$ | $\frac{16(16)-8(50)}{16}=-9$ | 0 | |
| $s^0$ | 50 | | |

#### Examination of the first column

First column: $1,\ 8,\ 16,\ -9,\ 50$

- $+16 \to -9$: first sign change
- $-9 \to +50$: second sign change

**2 sign changes**, so 2 roots lie in the right half of the s-plane.

| Region | Number of roots |
|---|---|
| Right half plane | 2 |
| Left half plane | 2 |
| $j\omega$ axis | 0 |

The actual roots are $s \approx 0.185 \pm j1.64$ and $s \approx -4.18 \pm j0.93$, which agrees with the Routh result.

**Answer:** the system is **unstable**. It has two roots in the RHP, so its response grows as an oscillation with an increasing envelope.

## Check yourself

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

1. In the Routh-Hurwitz criterion, what does the number of sign changes in the first column give?
   A. The number of roots on the jω axis
   B. The number of roots in the right half s-plane
   C. The number of roots in the left half s-plane
   D. The number of zeros of the open-loop system
   Answer: B. Each sign change in the first column of the Routh array corresponds to one root in the right half plane. (Key point: Stability rule)

2. All coefficients of s^4+8s^3+18s^2+16s+50 are positive. What does this tell us?
   A. The system is definitely stable
   B. The necessary condition holds, but the array is still needed to decide
   C. The system is definitely unstable
   D. The roots are all real
   Answer: B. Positive coefficients are only a necessary condition, so the Routh array must be built to confirm stability. (Key point: Positive coefficients are not enough)

3. In the Routh array, the s^1 row first element is (16(16)-8(50))/16. What is its value and effect?
   A. +9, so no sign change
   B. 0, giving an auxiliary equation
   C. -16, giving no sign change
   D. -9, which starts the sign changes
   Answer: D. (256-400)/16 = -9, a negative entry between positive entries in the first column. (Key point: s^1 row element)

4. The first column is 1, 8, 16, -9, 50. How many roots are in the RHP, and is the system stable?
   A. 0 roots; stable
   B. 1 root; unstable
   C. 2 roots; unstable
   D. 4 roots; unstable
   Answer: C. There are two sign changes (+16 to -9, -9 to +50), so 2 RHP roots and the system is unstable. (Key point: Root count)

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