--- title: "Root locus of k(s+4)/[s(s²+2s+2)] and stable k" subject: "Control Systems" chapter: 5 key: "semester_3/control_systems:rl-k-s4-s-s2-2s" source: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s4-s-s2-2s site: BCT Engineering --- # Root locus of k(s+4)/[s(s²+2s+2)] and stable k Control Systems · Chapter 5: Root Locus Technique - Asked in: 2066 Bhadra (old course), 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-s4-s-s2-2s - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5 ## Question Sketch the root locus of a unity feedback control system with open loop transfer function G(s) = k(s+4)/[s(s²+2s+2)] and find the range of k for which the system will be stable. ## Short answer The system is stable for 0 < k < 2. At k = 2 it is marginally stable and oscillates at ω = 2 rad/s; for k > 2 it becomes unstable. ## Answer $$ G(s) = \frac{k(s+4)}{s(s^2+2s+2)} $$ Characteristic equation: $s^3 + 2s^2 + (2+k)s + 4k = 0$. #### Root locus construction 1. **Poles:** $0,\ -1 \pm j1$. **Zero:** $-4$. With $n = 3$ and $m = 1$, two branches go to infinity. 2. **Real-axis locus:** $[-4, 0]$. The pole at $0$ travels to the zero at $-4$. 3. **Asymptotes:** $$ \sigma_A = \frac{(0 - 1 - 1) - (-4)}{2} = +1, \qquad \theta_A = \pm 90^\circ $$ The asymptotes lie in the RHP at $\sigma = +1$, so the complex branches will cross the $j\omega$-axis. 4. **Breakaway:** $dk/ds = 0 \Rightarrow 2s^3 + 14s^2 + 16s + 8 = 0$. Roots: $-5.72$ (gives $k < 0$) and $-0.64 \pm j0.54$ (complex). Hence **no breakaway point** for $k > 0$. 5. **Angle of departure from $-1 + j1$:** from zero $-4$: $18.43^\circ$; from pole $0$: $135^\circ$; from pole $-1-j1$: $90^\circ$. $$ \phi_d = 180^\circ + 18.43^\circ - (135^\circ + 90^\circ) = -26.57^\circ $$ 6. **$j\omega$ crossing (Routh):** | Row | Col 1 | Col 2 | |---|---|---| | $s^3$ | 1 | $2+k$ | | $s^2$ | 2 | $4k$ | | $s^1$ | $\dfrac{2(2+k) - 4k}{2} = 2 - k$ | | | $s^0$ | $4k$ | | $k_{mar} = 2$. Auxiliary equation: $2s^2 + 8 = 0 \Rightarrow s = \pm j2$. ```text | .. | . |.. .. .X ..| x..... | | ------o........................x------------------ | x..... | ..| .X .. |.. | . | .. -4 -1 0 1 x pole o zero (-4) X jw crossing (+-j2, k = 2) ``` The complex poles leave at $\mp 26.6^\circ$, curve right, cross the $j\omega$-axis at $\pm j2$, and approach the vertical asymptotes at $\sigma = +1$. #### Range of k for stability First column positive: $4k > 0$ and $2 - k > 0$. **Answer:** The system is stable for **0 < k < 2**. At k = 2 it oscillates at **ω = 2 rad/s**; for k > 2 it is unstable. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With centroid +1 and angles ±90°, what happens to the complex branches? A. They stay in the LHP B. They cross the jw axis into the RHP C. They break away on the real axis D. They end at the zero Answer: B. The asymptotes are in the RHP so the branches cross the jw axis. (Key point: Asymptotes) 2. The angle of departure from -1 + j1 is: A. +26.57° B. -26.57° C. +135° D. -90° Answer: B. 180° + 18.43° - (135° + 90°) = -26.57°. (Key point: Departure angle) 3. The Routh s^1 entry (2 - k) gives stability for: A. k > 2 B. 0 < k < 2 C. 0 < k < 4 D. k < 0 Answer: B. 4k > 0 and 2 - k > 0 give 0 < k < 2, with oscillation at w = 2 rad/s at k = 2. 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