---
title: "Root locus of k(s−4)/(s²+6s+18)"
subject: "Control Systems"
chapter: 5
key: "semester_3/control_systems:rl-k-s-4-s2-6s"
source: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s-4-s2-6s
site: BCT Engineering
---

# Root locus of k(s−4)/(s²+6s+18)

Control Systems · Chapter 5: Root Locus Technique

- Asked in: 2079 Bhadra, 8 marks
- Video: [root locus in control system](https://www.youtube.com/watch?v=MIVkzOiFvhI) — Smart Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s-4-s2-6s
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5

## Question

Sketch the root locus of unity feedback system with G(s) = k(s−4)/(s²+6s+18); hence find the range of parameter k for stability of the closed loop system. Does the system exhibit sustained oscillation for any value of parameter k?

## Short answer

The root locus has complex poles at -3±j3 and a non-minimum-phase zero at +4. A break-in occurs at -3.616 (k=1.23). The system is stable for 0<k<4.5. No sustained oscillation occurs for any positive k; oscillation would require k=-6, which is outside the positive range.

## Answer

$$
G(s) = \frac{k(s-4)}{s^2 + 6s + 18}
$$

Poles: $-3 \pm j3$. Zero: $+4$ (a right-half-plane, non-minimum-phase zero).
Characteristic equation: $s^2 + (6+k)s + (18 - 4k) = 0$.

#### Root locus construction
1. **Branches:** $n = 2$, $m = 1$. One branch ends at the zero $+4$; the other goes to $-\infty$ (asymptote at $180^\circ$).
2. **Real-axis locus:** $(-\infty, +4]$, i.e. everything left of the zero.
3. **Angle of departure from $-3 + j3$:** angle from zero $+4$: $180^\circ - \tan^{-1}(3/7) = 156.80^\circ$; from pole $-3-j3$: $90^\circ$.
$$
\phi_d = 180^\circ + 156.80^\circ - 90^\circ = 246.80^\circ \;(\equiv -113.2^\circ)
$$
4. **Break-in point:**
$$
k = -\frac{s^2 + 6s + 18}{s - 4}, \qquad \frac{dk}{ds} = 0 \Rightarrow s^2 - 8s - 42 = 0 \Rightarrow s = -3.616 \text{ or } 11.616
$$
$s = 11.616$ gives $k < 0$. **Break-in at s = −3.616**, where $k = 1.23$.
5. **Shape:** the complex poles move on a circle centred at the zero $+4$ with radius $\sqrt{7^2 + 3^2} = 7.616$ and meet at $-3.616$. One branch then goes left to $-\infty$. The other moves right along the real axis, passes through the origin, and ends at $+4$.

```text
                                 |
                                 |
                                 |
                        x        |
                        .        |
                       ..        |
                       .         |
                       .         |
.......................*....................o-----
                       .         |
                       .         |
                       ..        |
                        .        |
                        x        |
                                 |
                                 |
                                 |
    -10              -3.6        0          4
x pole (-3+-j3)  o zero (+4)  * break-in (-3.616)
```

#### Range of k for stability
For a quadratic, all coefficients must be positive:
- $6 + k > 0 \Rightarrow k > -6$
- $18 - 4k > 0 \Rightarrow k < 4.5$

At $k = 4.5$ a closed-loop pole is exactly at $s = 0$. For $k > 4.5$ that pole moves into the RHP toward $+4$.

**Stable for 0 < k < 4.5** (for the root locus with k ≥ 0; in general −6 < k < 4.5).

#### Sustained oscillation?
Sustained oscillation needs a pair of poles on the $j\omega$-axis at $\omega \ne 0$. That requires $6 + k = 0$ with $18 - 4k > 0$, i.e. $k = -6$, which lies outside the $k \ge 0$ range. For $k > 0$ the locus crosses the imaginary axis only at the origin ($k = 4.5$), on the real axis. The system then becomes unstable through a real pole (non-oscillatory, exponential growth), not through oscillation.

**Answer:** The system is stable for **0 < k < 4.5**. **No**, there is no sustained oscillation for any positive k. (Only k = −6 would give oscillation at $\omega = \sqrt{42} = 6.48$ rad/s.)

## Check yourself

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

1. For G(s) = k(s-4)/(s^2+6s+18) with k >= 0, which part of the real axis is on the locus?
   A. Everything to the right of +4
   B. From 0 to +4 only
   C. Everything to the left of the zero at +4
   D. Only the segment between -3 and +4
   Answer: C. There is one real zero at +4 and no real poles, so every real point left of +4 has exactly one singularity to its right (odd). (Key point: Real-axis locus)

2. Where do the complex branches meet the real axis?
   A. At s = 11.616
   B. At s = -3.616
   C. At s = -3
   D. At s = -7.6
   Answer: B. dk/ds = 0 gives s^2 - 8s - 42 = 0, so s = -3.616 or 11.616; the root 11.616 gives k < 0, so the break-in is at -3.616 (k = 1.23). (Key point: Break-in point)

3. Using the characteristic equation s^2 + (6+k)s + (18-4k) = 0, what is the stable range for k >= 0?
   A. 0 < k < 4.5
   B. 0 < k < 6
   C. k > 4.5
   D. 0 < k < 18
   Answer: A. Positive coefficients require 6 + k > 0 and 18 - 4k > 0, i.e. k < 4.5. (Key point: Stability range)

4. Why is there no sustained oscillation for any k > 0?
   A. The locus never leaves the left half plane
   B. A jw pair needs 6 + k = 0, i.e. k = -6, so for k > 0 the locus only crosses the imaginary axis at the origin
   C. The poles are real for all k
   D. The zero at +4 cancels the poles
   Answer: B. Sustained oscillation needs a conjugate pair on the jw axis, which requires the s-coefficient 6 + k to be zero; for k > 0 the crossing is only at s = 0 (k = 4.5), a real pole. (Key point: Sustained oscillation)

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