---
title: "R-H root distribution of s⁵+4s⁴+2s³+8s²+s+4"
subject: "Control Systems"
chapter: 4
key: "semester_3/control_systems:rh-s5-4-2-8-1"
source: https://www.bctengineer.com/q/semester_3/control_systems/rh-s5-4-2-8-1
site: BCT Engineering
---

# R-H root distribution of s⁵+4s⁴+2s³+8s²+s+4

Control Systems · Chapter 4: Stability

- Asked in: 2075 Asoj, 4 marks; 2079 Bhadra, 8 marks (asked 2 times)
- 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-s5-4-2-8-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4

## Question

Using R-H criteria, tell how many roots of polynomial are in right half s-plane, in left half s-plane and on jω axis. Comment on stability. s⁵ + 4s⁴ + 2s³ + 8s² + s + 4 = 0

## Short answer

A zero row appears in the Routh array; the auxiliary polynomial yields repeated roots at s = ±j1 twice. The first column has no sign changes, so 0 RHP roots, 1 LHP root at s = −4, and 4 on the jω axis. Repeated imaginary-axis poles make the system unstable.

## Answer

**Result:** RHP = 0, LHP = 1 (at $s=-4$), $j\omega$ axis = 4 ($\pm j1$, each twice). The system is **unstable**.

The number of sign changes in the first column gives the RHP roots. A zero row gives the roots that are symmetric about the origin; these are found from the auxiliary polynomial.

$$
s^5+4s^4+2s^3+8s^2+s+4=0
$$

#### Routh array

| Row | Col 1 | Col 2 | Col 3 |
|---|---|---|---|
| $s^5$ | 1 | 2 | 1 |
| $s^4$ | 4 | 8 | 4 |
| $s^3$ | $\frac{4(2)-1(8)}{4}=0$ | $\frac{4(1)-1(4)}{4}=0$ | |

The $s^3$ row is all zeros. Auxiliary polynomial from the $s^4$ row:

$$
A(s) = 4s^4+8s^2+4, \qquad \frac{dA}{ds} = 16s^3+16s
$$

| Row | Col 1 | Col 2 |
|---|---|---|
| $s^3$ | 16 | 16 |
| $s^2$ | $\frac{16(8)-4(16)}{16}=4$ | $\frac{16(4)-0}{16}=4$ |
| $s^1$ | 0 (zero row again) | |

New auxiliary polynomial $A_1(s)=4s^2+4$, $\frac{dA_1}{ds}=8s$:

| Row | Col 1 |
|---|---|
| $s^1$ | 8 |
| $s^0$ | 4 |

First column: $1,\ 4,\ 16,\ 4,\ 8,\ 4$. **No sign change**, so there is no root in the RHP.

#### Roots from the auxiliary polynomial

$$
4s^4+8s^2+4 = 4(s^2+1)^2 = 0 \Rightarrow s=\pm j1,\ \pm j1
$$

These 4 roots lie on the $j\omega$ axis (repeated pair). The remaining root: $\frac{s^5+4s^4+2s^3+8s^2+s+4}{(s^2+1)^2} = s+4$, so $s=-4$.

| Region | Number of roots |
|---|---|
| Right half s-plane | 0 |
| Left half s-plane | 1 ($s=-4$) |
| On $j\omega$ axis | 4 ($\pm j1$ repeated) |

#### Stability

No root is in the RHP, but the roots on the $j\omega$ axis are **repeated**. They produce a response term like $t\sin t$, which grows with time. So the system is **unstable**.

## Check yourself

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

1. The s^3 row of the array is all zeros. What does this indicate?
   A. The polynomial has roots symmetric about the origin, found from the auxiliary polynomial
   B. All roots are in the LHP
   C. The system is stable
   D. The array must be restarted with a different polynomial
   Answer: A. A zero row means roots placed symmetrically about the origin, found from the auxiliary polynomial of the row above. (Key point: Zero row in s^3)

2. The auxiliary polynomial 4s^4+8s^2+4 = 4(s^2+1)^2 gives
   A. +-j1 and +-j2
   B. +-j1 repeated
   C. -1 repeated
   D. +-1
   Answer: B. (s^2+1)^2 = 0 gives s = +-j1, each twice. (Key point: Auxiliary roots)

3. Dividing the polynomial by (s^2+1)^2 leaves
   A. s - 4
   B. s + 1
   C. s + 4, so s = -4
   D. s^2 + 4
   Answer: C. The quotient is s+4, so the fifth root is -4 in the LHP. (Key point: Remaining root)

4. Why is the system called unstable?
   A. One sign change in the first column
   B. The repeated jw roots give t sin t growth
   C. The root -4 is too far left
   D. The polynomial has odd order
   Answer: B. No RHP root exists, but the jw-axis roots are repeated and so the response grows. (Key point: Stability verdict)

## Close variants set in other exams

- [R-H root distribution of s⁶+2s⁵+8s⁴+12s³+20s²+16s+16](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arh-s6-2-8-12-20) — Control Systems, chapter 4, asked 4×
- [R-H root distribution of s⁵+3s⁴+5s³+4s²+s+3](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arh-s5-3-5-4-1) — Control Systems, chapter 4, asked 4×

## Related in Control Systems

- [R-H root distribution of s⁶+3s⁵+4s⁴+6s³+5s²+3s+2](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arh-s6-3-4-6-5) — Control Systems, chapter 4, asked 5×
- [R-H stability of s⁵+2s⁴+2s³+4s²+s+2](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arh-s5-2-2-4-1) — Control Systems, chapter 4
- [Relative and absolute stability; R-H of s⁶+2s⁵+8s⁴+12s³+20s²+16s+16](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arh-relative-absolute-s6) — Control Systems, chapter 4, asked 2×
- [Nyquist stability of (4s+1)/[s²(s+1)(2s+1)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Anyquist-4s1-s2-s1-2s1) — Control Systems, chapter 6, asked 3×
- [Root locus of k(s+2)/[s(s+1)(s²+8s+64)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-k-s2-s-s1-s2) — Control Systems, chapter 5
- [K and T for roots left of s=−a](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arh-k-t-left-of-a) — Control Systems, chapter 4