---
title: "Stability of 56/[s(s⁴+7s³+6s²+42s+8)] feedback system"
subject: "Control Systems"
chapter: 4
key: "semester_3/control_systems:stability-56-s5"
source: https://www.bctengineer.com/q/semester_3/control_systems/stability-56-s5
site: BCT Engineering
---

# Stability of 56/[s(s⁴+7s³+6s²+42s+8)] feedback system

Control Systems · Chapter 4: Stability

- Asked in: 2080 Baisakh, 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/stability-56-s5
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4

## Question

Check stability of the system as given below in block diagram. [Figure: unity negative feedback system with forward path G(s) = 56/[s(s⁴ + 7s³ + 6s² + 42s + 8)]]

## Short answer

Characteristic equation s^5+7s^4+6s^3+42s^2+8s+56=0. Routh array yields a zero row; auxiliary equation gives roots ±j√2 and ±j2. No sign changes, so system is marginally stable with sustained oscillations at 1.414 and 2 rad/s.

## Answer

**Result:** the system is **marginally stable**. It has roots at $\pm j\sqrt2$ and $\pm j2$, one root at $s=-7$, and none in the RHP.

For unity negative feedback, the characteristic equation is $1+G(s)=0$. Its Routh array is formed and the first column is examined.

#### Characteristic equation

$$
\begin{aligned}
s(s^4+7s^3+6s^2+42s+8) + 56 &= 0 \\
s^5 + 7s^4 + 6s^3 + 42s^2 + 8s + 56 &= 0
\end{aligned}
$$

#### Routh array

| Row | Col 1 | Col 2 | Col 3 |
|---|---|---|---|
| $s^5$ | 1 | 6 | 8 |
| $s^4$ | 7 | 42 | 56 |
| $s^3$ | $\frac{7(6)-42}{7}=0$ | $\frac{7(8)-56}{7}=0$ | |

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

$$
A(s) = 7s^4 + 42s^2 + 56, \qquad \frac{dA}{ds} = 28s^3 + 84s
$$

| Row | Col 1 | Col 2 |
|---|---|---|
| $s^3$ | 28 | 84 |
| $s^2$ | $\frac{28(42)-7(84)}{28}=21$ | 56 |
| $s^1$ | $\frac{21(84)-28(56)}{21}=9.33$ | |
| $s^0$ | 56 | |

First column: $1,\ 7,\ 28,\ 21,\ 9.33,\ 56$. **No sign change**, so no root is in the RHP.

#### Roots on the imaginary axis

$$
7s^4+42s^2+56 = 7(s^2+2)(s^2+4) = 0 \Rightarrow s = \pm j1.414,\ \pm j2
$$

The fifth root: $\frac{q(s)}{(s^2+2)(s^2+4)} = s+7$, so $s=-7$.

| Region | Roots |
|---|---|
| RHP | none |
| $j\omega$ axis | $\pm j1.414$, $\pm j2$ (non-repeated) |
| LHP | $-7$ |

#### Conclusion

The poles on the $j\omega$ axis are simple (non-repeated), and none are in the RHP. The system is **marginally stable**: its response contains sustained oscillations at 1.414 rad/s and 2 rad/s.

## Check yourself

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

1. The characteristic equation is
   A. s^5+7s^4+6s^3+42s^2+8s+56=0
   B. s^4+7s^3+6s^2+42s+8=0
   C. s^5+7s^4+6s^3+42s^2+8s=0
   D. s^5+7s^4+6s^3+42s^2+8s-56=0
   Answer: A. Multiply by s and add 56. (Key point: Characteristic equation)

2. The s^3 row is zero. The auxiliary equation comes from
   A. The s^5 row
   B. The s^4 row: 7s^4+42s^2+56
   C. The s^2 row
   D. The s^3 row itself
   Answer: B. The auxiliary equation is formed from the row immediately above the zero row. (Key point: Zero row)

3. The auxiliary equation 7(s^2+2)(s^2+4)=0 gives
   A. +-1.414 and +-2
   B. -2 and -4
   C. +-j1.414 and +-j2
   D. +-j2 repeated
   Answer: C. s^2=-2 and s^2=-4 give +-j1.414 and +-j2. (Key point: Imaginary roots)

4. Why is the system marginally stable?
   A. All roots are in the LHP
   B. There is one sign change
   C. The jw roots are repeated
   D. No RHP roots and the jw-axis roots are simple
   Answer: D. No sign change, and non-repeated imaginary roots give sustained oscillation, which is marginal stability. (Key point: Verdict)

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