--- title: "Root locus of K/[s(s+6)(s²+4s+13)]" subject: "Control Systems" chapter: 5 key: "semester_3/control_systems:rl-k-s-s6-s2-4s" source: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s-s6-s2-4s site: BCT Engineering --- # Root locus of K/[s(s+6)(s²+4s+13)] Control Systems · Chapter 5: Root Locus Technique - Asked in: 2071 Shrawan, 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-s-s6-s2-4s - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5 ## Question The open loop transfer function of a control system is given by G(s)H(s) = K/[s(s+6)(s²+4s+13)]. Sketch the root locus for 0 ≤ K ≤ ∞ and determine the breakaway point, the angle of departure from complex poles and the stability conditions. ## Short answer Breakaway point is at s = −4.20. Angles of departure from the complex poles −2 ± j3 are ∓70.56°. The system is stable for 0 < K < 227.8; at K = 227.8 it oscillates at 2.79 rad/s, and for larger K it is unstable. ## Answer $$ G(s)H(s) = \frac{K}{s(s+6)(s^2+4s+13)} $$ Characteristic equation: $s^4 + 10s^3 + 37s^2 + 78s + K = 0$. #### 1. Poles and zeros Poles: $0,\ -6,\ -2 \pm j3$. No zeros. Hence $n = 4$, $m = 0$, and 4 branches go to infinity. #### 2. Real-axis locus The segment $[-6, 0]$. #### 3. Asymptotes $$ \sigma_A = \frac{0 - 6 - 2 - 2}{4} = -2.5, \qquad \theta_A = \pm 45^\circ,\ \pm 135^\circ $$ #### 4. Breakaway point $$ K = -(s^4 + 10s^3 + 37s^2 + 78s), \qquad \frac{dK}{ds} = -(4s^3 + 30s^2 + 74s + 78) = 0 $$ Roots: $s = -4.20$, $s = -1.65 \pm j1.39$. The complex roots give complex $K$ and are rejected. **Breakaway point: s = −4.20**, with $$ K = |{-4.2}|\,|{1.8}|\,|17.64 - 16.8 + 13| = 4.2 \times 1.8 \times 13.84 = 104.6 $$ #### 5. Angle of departure from $-2 + j3$ | From | Angle to $-2+j3$ | |---|---| | pole $0$ | $123.69^\circ$ | | pole $-6$ | $36.87^\circ$ | | pole $-2-j3$ | $90^\circ$ | $$ \phi_d = 180^\circ - (123.69^\circ + 36.87^\circ + 90^\circ) = -70.56^\circ $$ From $-2 - j3$, the departure angle is $+70.56^\circ$. #### 6. $j\omega$-axis crossing | Row | Col 1 | Col 2 | Col 3 | |---|---|---|---| | $s^4$ | 1 | 37 | $K$ | | $s^3$ | 10 | 78 | | | $s^2$ | $(370-78)/10 = 29.2$ | $K$ | | | $s^1$ | $(2277.6 - 10K)/29.2$ | | | | $s^0$ | $K$ | | | $K_{mar} = 227.76$. Auxiliary equation: $29.2s^2 + 227.76 = 0 \Rightarrow s = \pm j2.79$. #### Sketch ```text | ... | .... .... | .... .... x. ..... ... .........X ... | ... | .. | ----------x........*...................x---------- .. | ... | ... | ... .........X .... x. ..... .... | .... ... | .... | -6 -4.2 -2 0 x pole * breakaway (-4.2) X jw crossing (+-j2.79, K = 227.8) ``` - Branches from $0$ and $-6$ meet at $-4.2$, break away vertically, and follow the ±135° asymptotes. - Branches from $-2 \pm j3$ leave at $\mp 70.6^\circ$, cross the $j\omega$-axis at $\pm j2.79$, and follow the ±45° asymptotes. #### Stability conditions | K | System | |---|---| | $0 < K < 227.76$ | Stable | | $K = 227.76$ | Marginally stable, oscillates at 2.79 rad/s | | $K > 227.76$ | Unstable | **Answer:** Breakaway at **s = −4.20**. Departure angles are **∓70.56°**. The system is stable for **0 < K < 227.8**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The asymptote centroid is: A. -2.5 B. -3.0 C. -2.0 D. -6.0 Answer: A. (0 - 6 - 2 - 2)/4 = -2.5, with angles ±45°, ±135°. (Key point: Asymptote centroid) 2. Why are -1.65 ± j1.39 rejected as breakaway points? A. They give complex K B. They lie on the real axis C. They are in the RHP D. They give K = 0 Answer: A. Complex roots give complex K, so they are not on the locus; breakaway is at -4.20. (Key point: Rejecting complex roots) 3. The angle of departure from -2 + j3 is: A. +70.56° B. -70.56° C. +109.4° D. -123.69° Answer: B. 180° - (123.69° + 36.87° + 90°) = -70.56°. (Key point: Departure angle) 4. The Routh s^1 entry gives the system stable for: A. 0 < K < 104.6 B. 0 < K < 78 C. 0 < K < 227.76 D. K > 227.76 Answer: C. 2277.6 - 10K > 0 gives K < 227.76; crossing at ±j2.79. (Key point: Stable range) ## Close variants set in other exams - [Root locus of k/[s(s+4)(s²+4s+20)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-k-s-s4-s2-4s) — Control Systems, chapter 5, asked 6× - [Root locus of K(s²−2s+5)/(s²+1.5s−1)](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-k-s2-2s5-s2-15s) — Control Systems, chapter 5, asked 6× ## Related in Control Systems - [Root locus of K(s+2)/(s²−4s+13)](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-k-s2-s2-4s-13) — Control Systems, chapter 5 - [Root locus of K(s+9)/[s(s²+4s+11)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-k-s9-s-s2-4s) — Control Systems, chapter 5 - [Root locus criteria; poles 0, −1±j2](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-criteria-poles-0-1j2) — Control Systems, chapter 5 - [Root locus with poles 0, −1 and zero −3](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-poles-0-1-zero-3) — Control Systems, chapter 5 - [Root locus: poles −2±j3, zero at origin](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-poles-2j3-zero-0-critical) — Control Systems, chapter 5 - [Root locus of K/[s(s+1)(s²+4s+5)], critical damping](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arl-s-s1-s2-4s-5) — Control Systems, chapter 5 ## Related in other subjects - [Root locus of a sampled system (T = 0.5 s)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adraw-root-locus-t-0-5) — Digital Control System, chapter 4 - [Root-locus design: 8 samples per cycle, T = 0.2 s](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aroot-locus-8-samples-t02) — Digital Control System, chapter 4 - [Open loop versus closed loop control](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Aopen-vs-closed-loop) — Embedded System, chapter 7, asked 6×