---
title: "R-H range of K for K(s+1)/(s³+bs²+3s+1)"
subject: "Control Systems"
chapter: 4
key: "semester_3/control_systems:rh-range-k-ks1-b"
source: https://www.bctengineer.com/q/semester_3/control_systems/rh-range-k-ks1-b
site: BCT Engineering
---

# R-H range of K for K(s+1)/(s³+bs²+3s+1)

Control Systems · Chapter 4: Stability

- Asked in: 2082 Baisakh, 6 marks
- Video: [Applications of R-H Criteria (Part 1)](https://www.youtube.com/watch?v=S0_Hd5M7aJY) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/rh-range-k-ks1-b
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4

## Question

A unity feedback system has the feedforward transfer function G(s) = K(s+1)/(s³+bs²+3s+1). Using R-H Criterion, find the range of K for the system to be stable. When does the system just oscillate and what would be that frequency during sustained oscillation?

## Short answer

The characteristic equation is s³ + bs² + (3+K)s + (1+K)=0. Routh conditions: K > −1 and b(3+K) > 1+K. For K>0, stable range is 0 < K < (3b−1)/(1−b) when 1/3 < b < 1. Marginal stability at K = (3b−1)/(1−b) with oscillation frequency ω = √(3+K) = √(2/(1−b)) rad/s.

## Answer

**Result:** stable for $K>-1$ and $b(3+K)>1+K$. For $K>0$ this means any $K$ when $b\ge1$, and $0<K<\frac{3b-1}{1-b}$ when $\frac13<b<1$. It just oscillates at $K=\frac{3b-1}{1-b}$, with $\omega=\sqrt{3+K}=\sqrt{\frac{2}{1-b}}$ rad/s.

The characteristic equation is $1+G(s)=0$. The Routh array gives the stability conditions in terms of $K$ and $b$. Sustained oscillation occurs when the $s^1$ row becomes zero.

#### 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 stability

1. $b > 0$
2. $1+K>0 \Rightarrow K>-1$
3. $b(3+K) - (1+K) > 0 \Rightarrow K(b-1) > 1-3b$

Interpreting condition 3 (for $K>0$, $b>0$):

- If $b \ge 1$: condition 3 holds for every $K>0$, so the system is stable for all $K>0$.
- If $b < 1$: $K < \dfrac{3b-1}{1-b}$. A positive range exists only if $b > \frac{1}{3}$.

$$
\text{Range: } 0 < K < \frac{3b-1}{1-b}\quad\left(\tfrac13<b<1\right)
$$

#### Sustained oscillation

The system just oscillates when the $s^1$ row is zero:

$$
b(3+K) = 1+K \quad\Rightarrow\quad K_{mar} = \frac{3b-1}{1-b}
$$

Auxiliary equation from the $s^2$ row:

$$
\begin{aligned}
bs^2 + (1+K) &= 0 \\
\omega^2 &= \frac{1+K}{b} = 3+K = \frac{2}{1-b} \\
\omega &= \sqrt{\frac{2}{1-b}}\ \text{rad/s}
\end{aligned}
$$

**Example:** for $b = 0.5$: $K_{mar} = \frac{0.5}{0.5}=1$ and $\omega = \sqrt{4} = 2$ rad/s. Check: $s^3+0.5s^2+4s+2 = (s^2+4)(s+0.5)$, which has roots $\pm j2$.

## Check yourself

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

1. The characteristic equation is
   A. s^3+bs^2+(3+K)s+(1+K)=0
   B. s^3+bs^2+3s+K=0
   C. s^3+bs^2+(3-K)s+(1-K)=0
   D. K(s+1)=0
   Answer: A. Adding K(s+1) to the denominator gives s^3 + bs^2 + (3+K)s + (1+K). (Key point: Characteristic equation)

2. Stability requires
   A. b<0, K<-1
   B. b>0, K>-1, b(3+K)>1+K
   C. b(3+K)<1+K only
   D. K>1 only
   Answer: B. The s^1 element is [b(3+K)-(1+K)]/b and s^0 is 1+K; all must be positive. (Key point: Stability conditions)

3. The system just oscillates when the s^1 row is zero, giving K =
   A. (1-b)/(3b-1)
   B. 3b-1
   C. (3b-1)/(1-b)
   D. b/(1-b)
   Answer: C. b(3+K) = 1+K gives K(b-1) = 1-3b, so K = (3b-1)/(1-b). (Key point: Marginal gain)

4. The oscillation frequency is
   A. sqrt(1+K)
   B. sqrt(b)
   C. 2/(1-b)
   D. sqrt(3+K) = sqrt(2/(1-b))
   Answer: D. From bs^2+(1+K)=0, w^2 = (1+K)/b, which equals 3+K = 2/(1-b). (Key point: Oscillation frequency)

## Close variants set in other exams

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