---
title: "Root locus of K(s+1)/[(s²+2s+2)(s²+2s+5)]"
subject: "Control Systems"
chapter: 5
key: "semester_3/control_systems:rl-k-s1-s2-2s-2"
source: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s1-s2-2s-2
site: BCT Engineering
---

# Root locus of K(s+1)/[(s²+2s+2)(s²+2s+5)]

Control Systems · Chapter 5: Root Locus Technique

- Asked in: 2068 Chaitra, 10 marks; 2080 Bhadra, 10 marks (asked 2 times)
- 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-s1-s2-2s-2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5

## Question

Sketch Root locus plot for the system having open loop transfer function G(s)H(s) = k(s+1)/[(s²+2s+2)(s²+2s+5)].

## Short answer

Poles at -1±j1 and -1±j2, zero at -1. Real-axis locus extends left of -1. Asymptotes are 60°,180°,300° from σ=-1. Departure angles are 180° from -1±j1 and 0° from -1±j2. Break-in at s=-1.77 (k=9.5). Imaginary-axis crossing at ±j2.74 (k=16.12), stable for 0<k<16.12.

## Answer

The root locus is drawn for $0 \le k < \infty$ with the standard construction rules.

$$
G(s)H(s) = \frac{k(s+1)}{(s^2+2s+2)(s^2+2s+5)}
$$

#### Step 1: Poles and zeros

- Poles: $s = -1\pm j1$ and $s = -1 \pm j2$, so $n = 4$.
- Zero: $s = -1$, so $m = 1$.
- Number of branches = 4. One branch ends at the zero and three go to infinity.

#### Step 2: Real-axis locus

There are no real poles. The only real-axis singularity is the zero at $-1$. Points to the **left of $-1$** have one zero to their right (odd), so the real axis from $-1$ to $-\infty$ is on the locus.

#### Step 3: Asymptotes

$$
\begin{aligned}
\theta_q &= \frac{(2q+1)180^\circ}{n-m} = 60^\circ,\ 180^\circ,\ 300^\circ \\
\sigma_A &= \frac{(-1-1-1-1)-(-1)}{3} = \frac{-3}{3} = -1
\end{aligned}
$$

#### Step 4: Angles of departure

**At $p_1 = -1+j1$**, angles from the other poles and the zero:

- from $-1-j1$: $90^\circ$; from $-1+j2$: $-90^\circ$; from $-1-j2$: $90^\circ$
- from zero $-1$: $90^\circ$

$$
\phi_{d1} = 180^\circ - (90^\circ - 90^\circ + 90^\circ) + 90^\circ = 180^\circ
$$

**At $p_2 = -1+j2$**:

- from $-1+j1$, $-1-j1$, $-1-j2$: $90^\circ$ each; from zero: $90^\circ$

$$
\phi_{d2} = 180^\circ - 270^\circ + 90^\circ = 0^\circ
$$

So the branch from $-1+j1$ leaves horizontally to the **left**, and the branch from $-1+j2$ leaves horizontally to the **right**. The lower half is the mirror image.

#### Step 5: Break-in point

$$
k = -\frac{(s^2+2s+2)(s^2+2s+5)}{s+1}
$$

Put $p = s+1$, so $k = -\frac{(p^2+1)(p^2+4)}{p}$:

$$
\frac{dk}{dp} = 0 \Rightarrow 3p^4 + 5p^2 - 4 = 0 \Rightarrow p^2 = 0.591 \ \text{(other root negative)}
$$

$p = -0.769$ gives $s = -1.77$ (on the locus) with $k = 9.50 > 0$. So $s = -1.77$ is a **break-in point**. The branches from $-1\pm j1$ meet here. After that, one goes right to the zero at $-1$ and the other goes left to $-\infty$. The roots $p=\pm j1.50$ give a complex $k$, so they are not on the locus.

#### Step 6: Imaginary-axis crossing

Characteristic equation: $s^4 + 4s^3 + 11s^2 + (14+k)s + (10+k) = 0$

| Row | Col 1 | Col 2 | Col 3 |
|---|---|---|---|
| $s^4$ | 1 | 11 | $10+k$ |
| $s^3$ | 4 | $14+k$ | |
| $s^2$ | $\frac{30-k}{4}$ | $10+k$ | |
| $s^1$ | $\frac{(30-k)(14+k)-16(10+k)}{30-k}$ | | |
| $s^0$ | $10+k$ | | |

$s^1$ row $=0$: $260 - k^2 = 0 \Rightarrow k = 16.12$.

Auxiliary equation: $\frac{30-16.12}{4}s^2 + 26.12 = 0 \Rightarrow s = \pm j2.74$.

#### Sketch

```text
 Upper half shown; lower half is its mirror image.

                         jw     /
                          |    /  60 deg asymptote
                          *   /   (from -1)
                         /| j2.74 (k=16.1)
              x-------- ' |
            -1+j2  (0 deg)|
                          |
         .--x (180 deg)   |
        /  -1+j1          |
       /                  |
 <====o=====o-------------+-----> sigma
   -1.77   -1             0
  (k=9.5)  zero
```

#### Results

| Item | Value |
|---|---|
| Asymptotes | $60^\circ, 180^\circ, 300^\circ$ from $\sigma_A=-1$ |
| Departure angles | $180^\circ$ at $-1\pm j1$; $0^\circ$ at $-1\pm j2$ |
| Break-in point | $s=-1.77$, $k=9.5$ |
| $j\omega$ crossing | $\pm j2.74$ at $k=16.12$ |
| Stability | Stable for $0<k<16.12$ |

## Check yourself

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

1. With 4 poles and 1 zero, how do the branches end?
   A. All 4 at infinity
   B. 1 at the zero and 3 at infinity
   C. 2 at the zero
   D. 3 at the zero and 1 at infinity
   Answer: B. Branches = n = 4; m = 1 end at zeros and n-m = 3 go to infinity. (Key point: Branch count)

2. Which part of the real axis is on the locus?
   A. From -1 to -infinity
   B. From -1 to 0
   C. The whole real axis
   D. None
   Answer: A. Points left of the zero at -1 have one zero to the right (odd count). (Key point: Real-axis locus)

3. What are the asymptote angles and centroid?
   A. 45°, 135° from -1
   B. 60°, 180°, 300° from -3
   C. 90°, 270° from -1
   D. 60°, 180°, 300° from -1
   Answer: D. n-m=3 gives angles 60°, 180°, 300°; centroid (-4+1)/3 = -1. (Key point: Asymptotes)

4. What are the departure angles at -1+j1 and -1+j2?
   A. 0° at -1+j1, 180° at -1+j2
   B. 180° at -1+j1, 0° at -1+j2
   C. 90° at both
   D. -90° at both
   Answer: B. The angle condition gives 180° and 0°, so the inner branch heads left and the outer one right. (Key point: Departure angles)

5. What is the break-in point, and what happens there?
   A. s = -1.77, branches from -1±j1 meet; one goes to the zero, one to -infinity
   B. s = -1, at the zero
   C. s = -2.74, at jω crossing
   D. There is none
   Answer: A. dk/dp=0 gives p=-0.769, s=-1.77, k=9.5, on the real-axis locus. (Key point: Break-in)

## Close variants set in other exams

- [Root locus of k(s−1)/[s(s−1)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-k-s-1-s-s) — Control Systems, chapter 5, asked 3×

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