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

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

Control Systems · Chapter 5: Root Locus Technique

- Asked in: 2075 Asoj, 10 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-s2-2s-2-s2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5

## Question

Sketch the root locus for the unity feedback system having the forward path transfer function G(s) = K/[(s²+2s+2)(s²+2s+5)].

## Short answer

No real-axis locus. Asymptotes intersect at σ=-1 with angles ±45°,±135°. Departure angles: +90° from -1+j1, -90° from -1+j2. Complex breakaway at -1±j1.581 (K=2.25). Imaginary-axis crossing occurs at ω=1.871 rad/s (K=16.25). System stable for 0<K<16.25.

## Answer

$$
G(s) = \frac{K}{(s^2+2s+2)(s^2+2s+5)}
$$

Characteristic equation: $s^4 + 4s^3 + 11s^2 + 14s + 10 + K = 0$.

#### 1. Poles and zeros
- Poles: $-1 \pm j1$ and $-1 \pm j2$ ($n = 4$). No zeros.
- All 4 branches go to infinity.

#### 2. Real-axis locus
There are no real poles or zeros, so **no part of the real axis** is on the locus.

#### 3. Asymptotes
$$
\sigma_A = \frac{-1-1-1-1}{4} = -1, \qquad \theta_A = \pm 45^\circ,\ \pm 135^\circ
$$

#### 4. Angles of departure
- At $-1 + j1$: angles from $-1-j1$: $90^\circ$; from $-1+j2$: $-90^\circ$; from $-1-j2$: $90^\circ$.
$$
\phi_d = 180^\circ - (90^\circ - 90^\circ + 90^\circ) = 90^\circ
$$
The branch leaves **straight up**.
- At $-1 + j2$: the three angles are all $90^\circ$.
$$
\phi_d = 180^\circ - 270^\circ = -90^\circ
$$
The branch leaves **straight down**.

So the two branches move toward each other along the line $\sigma = -1$ and meet.

#### 5. Breakaway points
$K = -(s^4 + 4s^3 + 11s^2 + 14s + 10)$

$$
\frac{dK}{ds} = -(4s^3 + 12s^2 + 22s + 14) = -2(s+1)(2s^2+4s+7) = 0
$$

- $s = -1$: $K = -(1)(4) = -4 < 0$, so it is not on the locus.
- $s = -1 \pm j1.581$: at $s = -1 + jb$, $(s^2+2s+2) = 1 - b^2 = -1.5$ and $(s^2+2s+5) = 4 - b^2 = 1.5$. Hence $K = 2.25 > 0$, a **valid complex breakaway point**.

At $K = 2.25$ the poles meet at $-1 \pm j1.581$ and then split horizontally (±90° from the vertical) toward the 45° and 135° asymptotes.

#### 6. Imaginary-axis crossing (Routh)

| Row | Col 1 | Col 2 | Col 3 |
|---|---|---|---|
| $s^4$ | 1 | 11 | $10+K$ |
| $s^3$ | 4 | 14 | |
| $s^2$ | 7.5 | $10+K$ | |
| $s^1$ | $(65 - 4K)/7.5$ | | |
| $s^0$ | $10+K$ | | |

$K_{mar} = 16.25$. Auxiliary equation: $7.5s^2 + 26.25 = 0 \Rightarrow \omega = \sqrt{3.5} = 1.871$ rad/s.

#### Sketch
```text
                                 |
                                 |
     ....                        |       ....
        ......                   |  ......
             ......     .x     ..X...
                  ......*....... |
                        x.       |
                                 |
                                 |
---------------------------------+----------------
                                 |
                                 |
                        x.       |
                  ......*....... |
             ......     .x     ..X...
        ......                   |  ......
     ....                        |       ....
                                 |
                                 |
       -3              -1        0
x pole   * breakaway (-1 +- j1.58, K = 2.25)
X jw crossing (+-j1.87, K = 16.25)
```
Each pair moves along $\sigma = -1$, meets at $-1 + j1.581$, and turns. One branch heads to the 135° asymptote; the other heads right, crossing the $j\omega$-axis at $j1.871$ and following the 45° asymptote.

**Answer:** Complex breakaway at **−1 ± j1.581 (K = 2.25)**. Departure angles are **+90° from −1+j1** and **−90° from −1+j2**. The system is stable for **0 < K < 16.25** and oscillates at **1.871 rad/s** when K = 16.25.

## Check yourself

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

1. Why is no part of the real axis on this root locus?
   A. There are no real poles or zeros
   B. The poles are repeated
   C. There are four branches
   D. The asymptotes are at 45°
   Answer: A. All poles are complex, so no real-axis point has an odd count of poles/zeros to its right. (Key point: Real-axis locus)

2. Why is s = -1 rejected as breakaway but -1 ± j1.581 accepted?
   A. s = -1 gives K = -4 < 0, while -1 ± j1.581 gives K = 2.25 > 0
   B. Both give negative K
   C. s = -1 is complex
   D. -1 ± j1.581 gives K = 0
   Answer: A. A breakaway point must have positive real K. (Key point: Valid breakaway)

3. The departure angles from -1+j1 and -1+j2 are:
   A. -90° and +90°
   B. +90° and -90°
   C. 0° and 180°
   D. +45° and -45°
   Answer: B. Computed: 180° - (90-90+90) = +90° (up) and 180° - 270° = -90° (down), so the branches approach each other. (Key point: Departure angles)

4. The Routh s^1 row (65 - 4K)/7.5 = 0 gives:
   A. K = 2.25 and w = 1
   B. K = 10 and w = 3.5
   C. K = 16.25 and w = 1.871 rad/s
   D. K = 65 and w = 7.5
   Answer: C. K = 65/4 = 16.25; auxiliary equation 7.5s^2 + 26.25 = 0 gives w = 1.871. (Key point: Marginal gain)

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