---
title: "Root locus of K(s+9)/[s(s²+4s+11)]"
subject: "Control Systems"
chapter: 5
key: "semester_3/control_systems:rl-k-s9-s-s2-4s"
source: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s9-s-s2-4s
site: BCT Engineering
---

# Root locus of K(s+9)/[s(s²+4s+11)]

Control Systems · Chapter 5: Root Locus Technique

- Asked in: 2073 Shrawan, 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-s9-s-s2-4s
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5

## Question

A unity feedback control system has an open loop transfer function G(s) = K(s+9)/[s(s²+4s+11)]. Sketch the root locus and determine: i) The range of 'K' for system to be stable ii) Undamped natural frequency of oscillation.

## Short answer

Real-axis locus: [-9,0]. Asymptotes: σ=+2.5, ±90°. No breakaway point. Departure angle from -2+j2.646 is -16.38°. Stable range: 0<K<8.8. At the marginal limit K=8.8, the undamped natural frequency is 4.45 rad/s.

## Answer

$$
G(s) = \frac{K(s+9)}{s(s^2+4s+11)}
$$

Characteristic equation: $s^3 + 4s^2 + (11+K)s + 9K = 0$.

#### Root locus data
1. **Poles:** $0,\ -2 \pm j2.646$. **Zero:** $-9$. With $n = 3$ and $m = 1$, two branches go to infinity.
2. **Real-axis locus:** $[-9, 0]$.
3. **Asymptotes:**
$$
\sigma_A = \frac{(0 - 2 - 2) - (-9)}{2} = +2.5, \qquad \theta_A = \pm 90^\circ
$$
The asymptotes lie in the right half-plane, so the complex branches must cross the $j\omega$-axis.
4. **Breakaway:** $dK/ds = 0 \Rightarrow 2s^3 + 31s^2 + 72s + 99 = 0$. Roots: $s = -13.03$ (gives $K < 0$; not on the locus) and $-1.24 \pm j1.51$ (complex). So there is **no breakaway point**. The pole at $0$ moves along the real axis to the zero at $-9$.
5. **Angle of departure from $-2 + j2.646$:** from zero $-9$: $20.71^\circ$; from pole $0$: $127.09^\circ$; from the conjugate pole: $90^\circ$.
$$
\phi_d = 180^\circ + 20.71^\circ - (127.09^\circ + 90^\circ) = -16.38^\circ
$$
6. **$j\omega$ crossing (Routh):**

| Row | Col 1 | Col 2 |
|---|---|---|
| $s^3$ | 1 | $11+K$ |
| $s^2$ | 4 | $9K$ |
| $s^1$ | $\dfrac{4(11+K) - 9K}{4} = \dfrac{44 - 5K}{4}$ | |
| $s^0$ | $9K$ | |

#### Sketch
```text
                                   | ..
                                   |..
                                   ..
                                  .X
                                ...|
                            x....  |
                                   |
                                   |
----o----..........................x--------------
                                   |
                                   |
                            x....  |
                                ...|
                                  .X
                                   ..
                                   |..
                                   | ..
   -9               -4             0       2.5
x pole   o zero (-9)
X jw crossing (+-j4.45, K = 8.8)
```

#### (i) Range of K for stability
All first-column entries must be positive:
- $9K > 0 \Rightarrow K > 0$
- $44 - 5K > 0 \Rightarrow K < 8.8$

**Answer (i): 0 < K < 8.8**

#### (ii) Undamped natural frequency of oscillation
At $K = 8.8$ the auxiliary equation is

$$
4s^2 + 9(8.8) = 0 \;\Rightarrow\; s^2 = -19.8 \;\Rightarrow\; s = \pm j4.45
$$

**Answer (ii): ω = 4.45 rad/s** (sustained oscillation when K = 8.8).

## Check yourself

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

1. What does the asymptote centroid +2.5 imply?
   A. The system is stable for all K
   B. The complex branches must cross the jw axis
   C. There are breakaway points
   D. The locus has no asymptotes
   Answer: B. Asymptotes in the right half-plane force the branches to cross into the RHP. (Key point: Asymptotes in RHP)

2. Why is there no breakaway point?
   A. The real root -13.03 gives K < 0 and the others are complex
   B. There are no real axis segments
   C. All roots are positive
   D. The zero is at the origin
   Answer: A. dK/ds roots are either on a part with negative K or complex, so none is valid. (Key point: Breakaway)

3. The Routh s^1 entry (44 - 5K)/4 and s^0 entry 9K give stability for:
   A. 0 < K < 4
   B. 0 < K < 8.8
   C. K > 8.8
   D. 0 < K < 44
   Answer: B. K > 0 and 44 - 5K > 0 give K < 8.8. (Key point: Stability range)

4. At K = 8.8 the frequency of oscillation is:
   A. 2.5 rad/s
   B. 8.8 rad/s
   C. 4.45 rad/s
   D. 9 rad/s
   Answer: C. 4s^2 + 9(8.8) = 0 gives s = ±j4.45. (Key point: Oscillation frequency)

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