--- title: "Root locus of K(s+1)/[s²(s+3.6)]" subject: "Control Systems" chapter: 5 key: "semester_3/control_systems:rl-k-s1-s2-s36" source: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s1-s2-s36 site: BCT Engineering --- # Root locus of K(s+1)/[s²(s+3.6)] Control Systems · Chapter 5: Root Locus Technique - Asked in: 2066 Jestha (old course), 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-s1-s2-s36 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5 ## Question Draw root locus for a unity feedback system with open loop transfer function G(s) = K(s+1)/[s²(s+3.6)]. Also determine the range of K for (i) overdamped response and (ii) unstable system. ## Short answer The root locus has a double pole at the origin, a pole at -3.6, and a zero at -1. Branches from the origin leave vertically, bend left toward asymptotes at -1.3 (±90°), and never cross the imaginary axis. Overdamped response is impossible for any K>0; the system is unstable for K≤0 and stable for all K>0. ## Answer $$ G(s) = \frac{K(s+1)}{s^2(s+3.6)} $$ Characteristic equation: $s^3 + 3.6s^2 + Ks + K = 0$. #### Root locus construction 1. **Poles:** $0, 0$ (double) and $-3.6$. **Zero:** $-1$. With $n = 3$ and $m = 1$, two branches go to infinity. 2. **Real-axis locus:** $[-3.6, -1]$ only. Left of a double pole the count is even, so $(-1, 0)$ is not on the locus. 3. **Asymptotes:** $$ \sigma_A = \frac{(0 + 0 - 3.6) - (-1)}{2} = -1.3, \qquad \theta_A = \pm 90^\circ $$ 4. **Departure from the double pole at the origin:** $2\phi = 180^\circ + \angle(\text{zero}) - \angle(\text{pole } {-3.6}) = 180^\circ + 0 - 0$, so $\phi = \pm 90^\circ$. The two branches leave the origin vertically. 5. **Breakaway points:** $$ K = -\frac{s^3 + 3.6s^2}{s + 1}, \qquad \frac{dK}{ds} = 0 \Rightarrow 2s^3 + 6.6s^2 + 7.2s = 0 $$ $$ s(2s^2 + 6.6s + 7.2) = 0 \Rightarrow s = 0 \text{ or } s = -1.65 \pm j0.94 $$ Apart from the double pole at the origin, the roots are complex. So there is **no breakaway or break-in point on the real axis**. A real break-in point exists only if the pole lies at more than 9 times the zero (here $3.6 < 9 \times 1$). 6. **Shape:** the two branches from the origin go up and down into the LHP, bend left, and approach the vertical asymptotes $\sigma = -1.3$ from the right. The branch from $-3.6$ moves right to the zero at $-1$. 7. **$j\omega$ crossing (Routh):** | Row | Col 1 | Col 2 | |---|---|---| | $s^3$ | 1 | $K$ | | $s^2$ | 3.6 | $K$ | | $s^1$ | $\dfrac{3.6K - K}{3.6} = 0.722K$ | | | $s^0$ | $K$ | | All entries are positive for every $K > 0$, so the locus never crosses into the RHP. ```text . | .. | . | .. | ... | .... | .... | ... ----x........................o---------x---------- ... .... | .... | ... | .. | . | .. | . | -3.6 -1 0 x pole (0, 0, -3.6) o zero (-1) asymptotes at sigma = -1.3; no real breakaway ``` The branches from 0 bend into the LHP toward the asymptotes σ = −1.3; the branch from −3.6 ends at −1. #### (i) Range of K for overdamped response Overdamping needs all closed-loop poles real. Because there is no real breakaway point, the two poles that leave the origin stay complex for every $K > 0$. (Numerical check: for K from 0.001 to 10 000, the cubic always has a complex pair.) Hence **there is no positive value of K for an overdamped response**; the response is always underdamped. #### (ii) Range of K for an unstable system For $K > 0$ all poles are in the LHP, so the system is stable for every positive $K$. At $K = 0$ there is a double pole at the origin. For $K < 0$ the constant term is negative and a root enters the RHP. **Answer:** Overdamped: **no value of K > 0**. Unstable: **K ≤ 0** (K < 0 unstable, K = 0 has a double pole at the origin). **Stable for all K > 0.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = K(s+1)/[s^2(s+3.6)], which part of the real axis lies on the root locus? A. From -3.6 to -1 only B. From -1 to 0 only C. From -3.6 to 0 D. Everything left of -3.6 Answer: A. A real-axis point is on the locus if an odd number of poles and zeros lie to its right. Between -3.6 and -1 there is one (the zero) to the right, so it qualifies; between -1 and 0 there are two (the double pole), an even count, so it does not. (Key point: Real-axis locus) 2. How are the asymptotes of this locus found, and what are they? A. Centroid -1.3, angles +-90 degrees B. Centroid -1.3, angles 0 and 180 degrees C. Centroid +1.3, angles +-90 degrees D. Centroid -4.6, angles +-90 degrees Answer: A. Centroid = (sum of poles - sum of zeros)/(n-m) = (-3.6 + 1)/2 = -1.3, and with n-m = 2 the angles are +-90 degrees. (Key point: Asymptotes) 3. Setting dK/ds = 0 gives s(2s^2 + 6.6s + 7.2) = 0. What does this show about breakaway points? A. Two real breakaway points exist at -1.65 and -0.94 B. A breakaway point exists at s = -3.6 C. There is no real breakaway or break-in point, since the non-zero roots are complex D. The only breakaway point is the origin, where K = 0 Answer: C. The quadratic has roots -1.65 +- j0.94, which are complex, so the only real root is s = 0 (the double pole itself); no real breakaway or break-in exists. (Key point: Breakaway points) 4. What is the conclusion about overdamped response and stability for K > 0? A. Overdamped for small K, unstable for large K B. Never overdamped and stable for every K > 0 C. Always overdamped and stable for every K > 0 D. Never overdamped and unstable for large K Answer: B. With no real breakaway the complex pair never becomes real, so the response is always underdamped, and the Routh array (3.6, 0.722K, K) shows no sign change, so it is stable for all K > 0. 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