--- title: "Root locus of K(s+2)/(s²−4s+13)" subject: "Control Systems" chapter: 5 key: "semester_3/control_systems:rl-k-s2-s2-4s-13" source: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s2-s2-4s-13 site: BCT Engineering --- # Root locus of K(s+2)/(s²−4s+13) Control Systems · Chapter 5: Root Locus Technique - Asked in: 2067 Asar (old course), 8 marks - Video: [Root locus solved example](https://www.youtube.com/watch?v=ww9rHEAwTnE) — S&T Dude - 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-s2-4s-13 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5 ## Question For a unity feedback system that has the forward transfer function G(s) = K(s+2)/(s² − 4s + 13): i) Sketch the root locus. ii) Find the imaginary axis crossing. iii) Find the gain K at the jω axis crossing. iv) Find the break-in point. ## Short answer The jω-axis crossing occurs at ±j4.58 rad/s with K = 4. The break-in point is at s = −7 (K = 18). Angles of departure are ±126.87°. ## Answer $$ G(s) = \frac{K(s+2)}{s^2 - 4s + 13} $$ Poles: $s = 2 \pm j3$ (in the right half-plane). Zero: $s = -2$. Characteristic equation: $s^2 + (K-4)s + (13 + 2K) = 0$. #### (i) Sketch of the root locus 1. **Branches:** $n = 2$, $m = 1$. One branch ends at the zero $-2$; the other goes to $-\infty$. 2. **Real-axis locus:** $(-\infty, -2]$, i.e. to the left of the zero. 3. **Asymptote:** one, at $180^\circ$. 4. **Angle of departure from $2 + j3$:** angle from zero $-2$: $\tan^{-1}(3/4) = 36.87^\circ$; from pole $2-j3$: $90^\circ$. $$ \phi_d = 180^\circ + 36.87^\circ - 90^\circ = 126.87^\circ $$ 5. **Shape:** the complex part is a circle centred at the zero $-2$ with radius $\sqrt{4^2 + 3^2} = 5$. The poles move left along this circle, cross the $j\omega$-axis, and meet on the real axis at $-7$. From there, one branch goes right to $-2$ and the other goes left to $-\infty$. ```text | ......... | ....... ...X... .... | .... ... | .x .. | .. | . | ..........*.................o------+-------------- . | .. | .. | ... | .x .... | .... ....... ...X... ......... | | -7 -2 0 2 x pole (2 +- j3) o zero (-2) * break-in (-7) X jw crossing (+-j4.58, K = 4) ``` #### (ii) Imaginary-axis crossing At $s = j\omega$, the real and imaginary parts of the characteristic equation must both vanish: $$ \begin{aligned} \text{Imag: } & (K-4)\omega = 0 \Rightarrow K = 4 \\ \text{Real: } & -\omega^2 + 13 + 2K = 0 \Rightarrow \omega^2 = 21 \end{aligned} $$ The locus crosses at **$s = \pm j4.583$**. (Check: the circle $|s+2| = 5$ at $\sigma = 0$ gives $\omega = \sqrt{25 - 4} = \sqrt{21}$.) #### (iii) Gain at the crossing **K = 4**. The system is unstable for $K < 4$ and stable for $K > 4$. #### (iv) Break-in point $$ K = -\frac{s^2 - 4s + 13}{s + 2}, \qquad \frac{dK}{ds} = 0 \Rightarrow s^2 + 4s - 21 = 0 \Rightarrow s = -7 \text{ or } 3 $$ $s = 3$ is not on the locus. **Break-in at s = −7**, where $$ K = -\frac{49 + 28 + 13}{-5} = 18 $$ **Answer:** $j\omega$ crossing at **±j4.583 rad/s** with **K = 4**. The break-in point is **s = −7** (K = 18). The departure angle is **±126.87°**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The open-loop poles at 2 ± j3 imply the system is: A. Stable at K = 0 B. Unstable at low gain and stabilised for large K C. Always stable D. Always unstable Answer: B. Open-loop poles in the RHP mean instability for K < 4; it is stable for K > 4. (Key point: Poles in RHP) 2. At s = jw, setting imaginary part (K-4)w = 0 gives: A. K = 4 and w = sqrt(21) B. K = 2 and w = 3 C. K = 18 and w = 7 D. K = 13 and w = 4 Answer: A. K = 4, then -w^2 + 13 + 8 = 0, w^2 = 21, w = 4.583. (Key point: Jw crossing method) 3. The break-in point is: A. s = 3 B. s = -7 (K = 18) C. s = -2 D. s = -5 Answer: B. s^2 + 4s - 21 = 0 gives -7 or 3; only -7 is on the locus. (Key point: Break-in point) 4. The complex part of the locus is a circle with centre and radius: A. Origin, radius 5 B. Zero at -2, radius 5 C. Pole 2, radius 3 D. Zero at -2, radius 3 Answer: B. The circle is centred at the zero -2 with radius sqrt(16 + 9) = 5. 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