--- title: "Root locus of s³+9s²+Ks+K = 0" subject: "Control Systems" chapter: 5 key: "semester_3/control_systems:rl-s3-9s2-sk-k" source: https://www.bctengineer.com/q/semester_3/control_systems/rl-s3-9s2-sk-k site: BCT Engineering --- # Root locus of s³+9s²+Ks+K = 0 Control Systems · Chapter 5: Root Locus Technique - Asked in: 2078 Bhadra, 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-s3-9s2-sk-k - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5 ## Question The characteristic equation of the system is s³ + 9s² + sK + K = 0. Sketch the complete root locus and comment on stability. ## Short answer The root locus has a triple meeting point at s=-3 (K=27), with asymptotes at -4 (±90°). Branches from the double pole at the origin leave vertically, meet the real branch from -9 at -3, then one goes to the zero at -1 and two go to infinity along the asymptotes. The system is stable for all K>0. ## Answer #### Converting to root-locus form $$ s^3 + 9s^2 + K(s + 1) = 0 \;\Rightarrow\; 1 + \frac{K(s+1)}{s^2(s+9)} = 0 $$ Equivalent open-loop transfer function: $G(s)H(s) = \dfrac{K(s+1)}{s^2(s+9)}$. #### Root locus construction 1. **Poles:** $0, 0, -9$. **Zero:** $-1$. With $n = 3$ and $m = 1$, two branches go to infinity. 2. **Real-axis locus:** $[-9, -1]$. 3. **Asymptotes:** $$ \sigma_A = \frac{(0 + 0 - 9) - (-1)}{2} = -4, \qquad \theta_A = \pm 90^\circ $$ 4. **Departure from the double pole at 0:** $2\phi = 180^\circ$, so the branches leave vertically at $\pm 90^\circ$ and curve into the LHP. 5. **Breakaway point:** $$ K = -\frac{s^3 + 9s^2}{s + 1}, \qquad \frac{dK}{ds} = 0 \Rightarrow 2s^3 + 12s^2 + 18s = 0 $$ $$ 2s(s + 3)^2 = 0 \Rightarrow s = 0,\ -3,\ -3 $$ $s = -3$ is a **triple root point**: $$ K = -\frac{-27 + 81}{-2} = 27 $$ Check: $(s + 3)^3 = s^3 + 9s^2 + 27s + 27$, which matches the characteristic equation with $K = 27$. ✓ 6. **Branch angles at $s = -3$:** three branches meet, so arriving and leaving branches are spaced $60^\circ$ apart. The two complex branches from the origin and the real branch from $-9$ arrive (from directions $\pm 60^\circ$ and $180^\circ$). They leave at $0^\circ$ (to the zero at $-1$) and $\pm 120^\circ$ (toward the asymptotes $\sigma = -4$). 7. **$j\omega$ crossing (Routh):** | Row | Col 1 | Col 2 | |---|---|---| | $s^3$ | 1 | $K$ | | $s^2$ | 9 | $K$ | | $s^1$ | $8K/9$ | | | $s^0$ | $K$ | | No sign change for $K > 0$, so there is no $j\omega$ crossing. ```text . | . | . | .. | . | . | .. ..... | .. ...... ....| ----x..........................*........o---.x---- .. ...... ....| .. ..... | . | . | .. | . | . | . | -9 -4 -3 0 x pole (0, 0, -9) o zero (-1) * triple point (-3, K=27) ``` #### Comment on stability - For $0 < K < 27$: one real pole lies between $-9$ and $-3$, plus a complex pair moving from the origin toward $-3$. The system is stable and underdamped. - At $K = 27$: there is a triple pole at $s = -3$. The system is stable with a critically damped type response. - For $K > 27$: one real pole lies between $-3$ and $-1$, and a complex pair moves toward $\sigma = -4$. The system is stable and underdamped. - The first column of the Routh array is positive for all $K > 0$. **Answer:** The locus has a triple meeting point at **s = −3 (K = 27)**, asymptotes at **σ = −4 (±90°)**, and stays entirely in the LHP. The system is **stable for all K > 0**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is s^3 + 9s^2 + Ks + K = 0 put into root locus form? A. 1 + K/[s^2(s+9)] = 0 B. 1 + K(s+9)/s^2 = 0 C. 1 + K(s+1)/[s^2(s+9)] = 0 D. 1 + Ks/[(s+1)(s+9)] = 0 Answer: C. Grouping the K terms gives s^3 + 9s^2 + K(s+1) = 0; dividing by s^2(s+9) gives 1 + K(s+1)/[s^2(s+9)] = 0, so poles 0, 0, -9 and a zero at -1. (Key point: Root locus form) 2. What are the asymptote centroid and angles? A. Centroid -4, angles +-90 degrees B. Centroid -5, angles +-90 degrees C. Centroid -4, angles 0 and 180 degrees D. Centroid -10, angles +-60 degrees Answer: A. Centroid = (0 + 0 - 9 + 1)/2 = -4, and with two branches going to infinity the angles are +-90 degrees. (Key point: Asymptotes) 3. What special feature occurs at s = -3? A. A single breakaway point where K = 9 B. A jw-axis crossing at K = 27 C. A zero of the system D. A triple root point where K = 27, since (s+3)^3 = s^3 + 9s^2 + 27s + 27 Answer: D. dK/ds = 0 gives 2s(s+3)^2 = 0, so s = -3 is a double root of dK/ds; at K = 27 the characteristic polynomial equals (s+3)^3, three roots coinciding. (Key point: Triple breakaway point) 4. What does the Routh array conclude about stability? A. Unstable for K > 27 B. Stable only for 0 < K < 27 C. Stable for all K > 0, since there is no sign change in the first column D. Marginally stable at K = 9 Answer: C. The first column is 1, 9, 8K/9, K, all positive for K > 0, so no roots reach the right half plane and the locus never crosses the jw axis. 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