---
title: "R-H range of K for s⁴+20s³+15s²+2s+K"
subject: "Control Systems"
chapter: 4
key: "semester_3/control_systems:rh-s4-20-15-2-k"
source: https://www.bctengineer.com/q/semester_3/control_systems/rh-s4-20-15-2-k
site: BCT Engineering
---

# R-H range of K for s⁴+20s³+15s²+2s+K

Control Systems · Chapter 4: Stability

- Asked in: 2078 Kartik, 4 marks; 2078 Kartik, 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-s4-20-15-2-k
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4

## Question

The characteristic equation of a feedback control system is s⁴ + 20s³ + 15s² + 2s + K = 0. (i) Determine the range of K for the system to be stable. (ii) Can the system be marginally stable? If so, find the required value of K and the frequency of sustained oscillation.

## Short answer

Construct the Routh array: stability requires all first-column entries positive, so 0 < K < 1.49. The system becomes marginally stable at K = 1.49 where the s¹ row vanishes. The auxiliary equation 14.9s² + 1.49 = 0 gives sustained oscillation at ω = 0.316 rad/s.

## Answer

Apply the Routh array with $K$ in it. For stability, every element of the first column must be positive. Marginal stability occurs when the $s^1$ row becomes zero.

$$
s^4+20s^3+15s^2+2s+K = 0
$$

#### Routh array

| Row | Col 1 | Col 2 | Col 3 |
|---|---|---|---|
| $s^4$ | 1 | 15 | $K$ |
| $s^3$ | 20 | 2 | 0 |
| $s^2$ | $\frac{20(15)-1(2)}{20}=14.9$ | $K$ | |
| $s^1$ | $\frac{14.9(2)-20K}{14.9}=\frac{29.8-20K}{14.9}$ | 0 | |
| $s^0$ | $K$ | | |

#### (i) Range of K for stability

- From $s^1$ row: $29.8 - 20K > 0 \Rightarrow K < 1.49$
- From $s^0$ row: $K > 0$

$$
0 < K < 1.49
$$

#### (ii) Marginal stability

Yes. At $K = 1.49$ the $s^1$ row becomes all zero, and the system has a pair of roots on the $j\omega$ axis (marginally stable, sustained oscillation).

Auxiliary equation from the $s^2$ row:

$$
\begin{aligned}
14.9s^2 + K &= 0 \\
14.9s^2 + 1.49 &= 0 \\
s^2 &= -0.1 \Rightarrow s = \pm j0.316
\end{aligned}
$$

So the frequency of sustained oscillation is $\omega = \sqrt{0.1} = 0.316$ rad/s.

**Answer:** stable for $0 < K < 1.49$; marginally stable at $K_{mar} = 1.49$, oscillating at $\omega = 0.316$ rad/s.

## Check yourself

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

1. For stability, the Routh first column must have
   A. All elements positive (no sign change)
   B. At least one zero
   C. Alternating signs
   D. Only the first element positive
   Answer: A. Any sign change indicates RHP roots, so all first-column elements must have the same sign. (Key point: Stability condition)

2. What is the range of K for stability?
   A. K > 1.49
   B. 0 < K < 1.49
   C. 0 < K < 14.9
   D. K < 0
   Answer: B. The s^1 element (29.8-20K)/14.9 > 0 gives K < 1.49, and the s^0 element gives K > 0. (Key point: Stability range)

3. At what condition is the system marginally stable?
   A. The s^4 row becomes zero
   B. K = 0
   C. The s^1 row becomes zero (K = 1.49)
   D. The s^0 element is negative
   Answer: C. At K = 1.49 the s^1 element is zero, giving a pair of imaginary roots. (Key point: Marginal condition)

4. The frequency of sustained oscillation is found from 14.9s^2 + 1.49 = 0, so w is
   A. 1.49 rad/s
   B. 0.1 rad/s
   C. 3.16 rad/s
   D. 0.316 rad/s
   Answer: D. s^2 = -0.1, so w = sqrt(0.1) = 0.316 rad/s. (Key point: Oscillation frequency)

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