---
title: "Find K and b for K(s+1)/(s³+bs²+3s+1)"
subject: "Control Systems"
chapter: 4
key: "semester_3/control_systems:rh-k-b-oscillate"
source: https://www.bctengineer.com/q/semester_3/control_systems/rh-k-b-oscillate
site: BCT Engineering
---

# Find K and b for K(s+1)/(s³+bs²+3s+1)

Control Systems · Chapter 4: Stability

- Asked in: 2072 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/rh-k-b-oscillate
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4

## Question

Determine value of 'K' and 'b' so that the unity feedback system with open loop transfer function G(s) = K(s+1)/(s³ + bs² + 3s + 1) [rest of the question is not printed in the paper].

## Short answer

For sustained oscillation the s¹ row must be zero and the auxiliary equation gives ω² = 3+K. With ω = 2 rad/s, K = 1. Then b = (1+K)/ω² = 0.5. Required values: K = 1, b = 0.5.

## Answer

**Result** (assuming oscillation at 2 rad/s): $K = 1$, $b = 0.5$. In general, $K = \frac{3b-1}{1-b}$ and $\omega=\sqrt{\frac{2}{1-b}}$.

**Assumption:** the end of the question is missing. It is taken in its usual form: "...so that the system **oscillates at a frequency of 2 rad/s**". The general relations are also derived, so any other given frequency can be substituted.

#### Characteristic equation

$$
\begin{aligned}
s^3+bs^2+3s+1 + K(s+1) &= 0 \\
s^3 + bs^2 + (3+K)s + (1+K) &= 0
\end{aligned}
$$

#### Routh array

| Row | Col 1 | Col 2 |
|---|---|---|
| $s^3$ | 1 | $3+K$ |
| $s^2$ | $b$ | $1+K$ |
| $s^1$ | $\frac{b(3+K)-(1+K)}{b}$ | 0 |
| $s^0$ | $1+K$ | |

#### Conditions for sustained oscillation

The $s^1$ row must be zero:

$$
b(3+K) = 1+K \qquad \text{...(1)}
$$

Auxiliary equation (from the $s^2$ row) with $s=j\omega$:

$$
bs^2 + (1+K) = 0 \Rightarrow \omega^2 = \frac{1+K}{b} \qquad \text{...(2)}
$$

From (1) and (2): $\omega^2 = 3 + K$.

#### Substitute $\omega = 2$ rad/s

$$
\begin{aligned}
4 &= 3 + K \Rightarrow K = 1 \\
b &= \frac{1+K}{\omega^2} = \frac{2}{4} = 0.5
\end{aligned}
$$

**Check:** $s^3+0.5s^2+4s+2 = (s^2+4)(s+0.5)$. The roots are $\pm j2$ and $-0.5$, so the system oscillates at 2 rad/s.

**Answer:** $K = 1$, $b = 0.5$. In general: $K = \frac{3b-1}{1-b}$ and $\omega = \sqrt{\frac{2}{1-b}}$. The system is stable when $b(3+K) > 1+K$.

## Check yourself

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

1. The s^1 row zero gives
   A. b(3+K) = 1+K
   B. b(3+K) = K
   C. b = K
   D. 3+K = 0
   Answer: A. [b(3+K)-(1+K)]/b = 0. (Key point: Zero-row condition)

2. Combining conditions gives w^2 =
   A. 1 + K
   B. 3 + K
   C. b
   D. K/b
   Answer: B. w^2 = (1+K)/b = 3+K using the zero-row condition. (Key point: Frequency relation)

3. For w = 2 rad/s, K and b are
   A. K = 2, b = 0.75
   B. K = 4, b = 1
   C. K = 1, b = 0.5
   D. K = 1, b = 2
   Answer: C. 4 = 3+K gives K = 1; b = (1+K)/w^2 = 2/4 = 0.5. (Key point: Values)

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