---
title: "K for sustained oscillation of k/[s(s+3)(s²+s+1)]"
subject: "Control Systems"
chapter: 4
key: "semester_3/control_systems:rh-k-s-s3-s2-s1"
source: https://www.bctengineer.com/q/semester_3/control_systems/rh-k-s-s3-s2-s1
site: BCT Engineering
---

# K for sustained oscillation of k/[s(s+3)(s²+s+1)]

Control Systems · Chapter 4: Stability

- Asked in: 2080 Bhadra, 6 marks
- Video: [Numerical 2 : Marginal value of K and Frequency of sustained oscillations](https://www.youtube.com/watch?v=OyDSqsS0eVc) — Learning Electronics
- 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-s-s3-s2-s1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-4

## Question

The open loop transfer function of a unity feedback system is given by G(s) = k/[s(s+3)(s²+s+1)]. Determine the value of k that will cause sustained oscillation in the closed loop system. Also determine the oscillation frequency.

## Short answer

Set the s¹ row of the Routh array to zero to find the marginal gain. Sustained oscillation occurs at k = 2.4375 with frequency ω = 0.866 rad/s.

## Answer

Sustained oscillation occurs at the marginal value of $k$, where the $s^1$ row of the Routh array is zero. The auxiliary equation from the $s^2$ row then gives the frequency.

#### Characteristic equation

$$
\begin{aligned}
s(s+3)(s^2+s+1) + k &= 0 \\
s^4 + 4s^3 + 4s^2 + 3s + k &= 0
\end{aligned}
$$

#### Routh array

| Row | Col 1 | Col 2 | Col 3 |
|---|---|---|---|
| $s^4$ | 1 | 4 | $k$ |
| $s^3$ | 4 | 3 | 0 |
| $s^2$ | $\frac{4(4)-1(3)}{4}=\frac{13}{4}$ | $k$ | |
| $s^1$ | $\frac{\frac{13}{4}(3)-4k}{\frac{13}{4}}=3-\frac{16k}{13}$ | 0 | |
| $s^0$ | $k$ | | |

For stability: $3-\frac{16k}{13}>0$ and $k>0$, so $0 < k < \frac{39}{16} = 2.4375$.

#### Value of k for sustained oscillation

$$
3 - \frac{16k}{13} = 0 \Rightarrow k = \frac{39}{16} = 2.4375
$$

#### Oscillation frequency

Auxiliary equation:

$$
\begin{aligned}
\frac{13}{4}s^2 + k &= 0 \\
\frac{13}{4}s^2 + \frac{39}{16} &= 0 \\
s^2 &= -\frac{3}{4} \Rightarrow s = \pm j0.866
\end{aligned}
$$

**Check:** at $k=2.4375$ the roots are $\pm j0.866$, $-2.866$ and $-1.134$.

**Answer:** $k = 2.4375$ gives sustained oscillation at $\omega = 0.866$ rad/s.

## Check yourself

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

1. What is the characteristic equation of k/[s(s+3)(s^2+s+1)] with unity feedback?
   A. s^4+3s^3+4s^2+4s+k=0
   B. s^4+4s^3+3s^2+k=0
   C. s^4+4s^3+4s^2+3s+k=0
   D. s^4+4s^3+4s^2+k s+3=0
   Answer: C. s(s+3)(s^2+s+1)=s^4+4s^3+4s^2+3s; add k. (Key point: Characteristic equation)

2. When does the closed loop oscillate with constant amplitude?
   A. When the s^0 element is zero
   B. When the s^4 element is zero
   C. When the s^1 row first element is zero
   D. When k = 0
   Answer: C. Marginal stability occurs when the s^1 first element is zero, i.e. 3-16k/13 = 0. (Key point: Condition for sustained oscillation)

3. What is k for sustained oscillation?
   A. k = 13/16
   B. k = 39/16 = 2.4375
   C. k = 3
   D. k = 4
   Answer: B. 3 = 16k/13 gives k = 39/16 = 2.4375. (Key point: Value of k)

4. How is the frequency obtained and what is it?
   A. From the s^3 row, ω = 2 rad/s
   B. From s^1 row, ω = 3 rad/s
   C. From the s^2 auxiliary equation (13/4)s^2+k=0, giving ω = 0.866 rad/s
   D. From k directly, ω = 2.4375 rad/s
   Answer: C. s^2 = -k/(13/4) = -0.75, so ω = 0.866 rad/s. (Key point: Frequency)

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