---
title: "R-H root distribution for 200/[s(s³+6s²+11s+6)]"
subject: "Control Systems"
chapter: 4
key: "semester_3/control_systems:rh-200-s-s3-6-11"
source: https://www.bctengineer.com/q/semester_3/control_systems/rh-200-s-s3-6-11
site: BCT Engineering
---

# R-H root distribution for 200/[s(s³+6s²+11s+6)]

Control Systems · Chapter 4: Stability

- Asked in: 2078 Bhadra, 6 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-200-s-s3-6-11
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4

## Question

Use RH criterion to determine number of roots on right side, left side and on the imaginary axis itself of s-plane for the system below. [Figure: unity negative feedback system with forward path G(s) = 200/[s(s³ + 6s² + 11s + 6)]]

## Short answer

Characteristic equation s^4+6s^3+11s^2+6s+200=0. Routh array shows two sign changes, so 2 roots lie in RHP, 2 in LHP, and none on the jω axis. The system is unstable.

## Answer

**Result:** RHP = 2, LHP = 2, $j\omega$ axis = 0. The system is **unstable**.

For unity negative feedback the characteristic equation is $1+G(s)=0$. The number of sign changes in the first column of the Routh array gives the number of roots in the right half plane.

#### Characteristic equation

$$
\begin{aligned}
s(s^3+6s^2+11s+6) + 200 &= 0 \\
s^4 + 6s^3 + 11s^2 + 6s + 200 &= 0
\end{aligned}
$$

#### Routh array

| Row | Col 1 | Col 2 | Col 3 |
|---|---|---|---|
| $s^4$ | 1 | 11 | 200 |
| $s^3$ | 6 | 6 | 0 |
| $s^2$ | $\frac{6(11)-1(6)}{6}=10$ | $\frac{6(200)-0}{6}=200$ | |
| $s^1$ | $\frac{10(6)-6(200)}{10}=-114$ | 0 | |
| $s^0$ | 200 | | |

First column: $1,\ 6,\ 10,\ -114,\ 200$.

Sign changes: $+10 \to -114$ (one) and $-114 \to +200$ (two), so there are **2 sign changes**.

#### Root location

| Region | Number of roots |
|---|---|
| Right half s-plane | 2 |
| Left half s-plane | $4-2 = 2$ |
| On $j\omega$ axis | 0 (no zero row) |

The numerical roots ($s \approx 1.28 \pm j2.54$ and $s \approx -4.28 \pm j2.54$) agree with this.

**Answer:** 2 roots in the RHP, 2 in the LHP, none on the imaginary axis. The closed-loop system is **unstable**; the gain of 200 is too high.

## Check yourself

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

1. The characteristic equation is
   A. s^4+6s^3+11s^2+6s+200=0
   B. s^3+6s^2+11s+200=0
   C. s^4+6s^3+11s^2+6s-200=0
   D. s^4+6s^3+11s^2+206s=0
   Answer: A. s(s^3+6s^2+11s+6)+200. (Key point: Characteristic equation)

2. Number of RHP roots equals
   A. The number of zero rows
   B. The number of sign changes in the first column
   C. The number of positive coefficients
   D. The order of the system
   Answer: B. Each sign change in the first column corresponds to one RHP root. (Key point: Counting RHP roots)

3. The first column 1, 6, 10, -114, 200 has how many sign changes?
   A. 1
   B. 0
   C. 2
   D. 3
   Answer: C. 10 to -114 and -114 to 200. (Key point: First column)

4. The root distribution (RHP, LHP, jw) is
   A. 4, 0, 0
   B. 0, 4, 0
   C. 1, 3, 0
   D. 2, 2, 0
   Answer: D. 2 RHP roots, the remaining 4-2 = 2 in the LHP, and no zero row so none on the axis. (Key point: Root distribution)

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