--- title: "Root locus: poles −2±j3, zero at origin" subject: "Control Systems" chapter: 5 key: "semester_3/control_systems:rl-poles-2j3-zero-0-critical" source: https://www.bctengineer.com/q/semester_3/control_systems/rl-poles-2j3-zero-0-critical site: BCT Engineering --- # Root locus: poles −2±j3, zero at origin Control Systems · Chapter 5: Root Locus Technique - Asked in: 2070 Chaitra, 8 marks - Video: [Intro to Control - 12.3 Root Locus Basics Part 1](https://www.youtube.com/watch?v=-eaJDo9Bvy8) — Engineering with Prof. Kim - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/rl-poles-2j3-zero-0-critical - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5 ## Question Draw Root Locus for the system that has open-loop pole/zero plot in s-plane as below in figure. Also estimate the system gain at the point where the system exhibits critical damping. [Figure: open-loop poles at s = −2 ± j3 and a zero at the origin] ## Short answer The break-in point is at s = −3.606, corresponding to critical damping. At that point the gain is K ≈ 3.21 and both closed-loop poles are real and equal at −3.606. ## Answer Open-loop poles at $-2 \pm j3$ and a zero at the origin give $$ G(s)H(s) = \frac{Ks}{(s+2)^2 + 9} = \frac{Ks}{s^2 + 4s + 13} $$ Characteristic equation: $s^2 + (4+K)s + 13 = 0$. #### Root locus construction 1. **Branches:** $n = 2$, $m = 1$. One branch ends at the zero ($s = 0$); the other goes to $-\infty$. 2. **Real-axis locus:** $(-\infty, 0]$, the whole negative real axis. 3. **Asymptote:** $n - m = 1$, angle $180^\circ$ (the negative real axis). 4. **Angle of departure from $-2 + j3$:** angle from zero $0$: $123.69^\circ$; from pole $-2-j3$: $90^\circ$. $$ \phi_d = 180^\circ + 123.69^\circ - 90^\circ = 213.69^\circ \;(\equiv -146.31^\circ) $$ The branch leaves moving left and downward. 5. **Break-in point:** $$ K = -\frac{s^2 + 4s + 13}{s}, \qquad \frac{dK}{ds} = 0 \;\Rightarrow\; s^2 - 13 = 0 \;\Rightarrow\; s = -\sqrt{13} = -3.606 $$ ($s = +3.606$ is not on the locus.) 6. **Shape:** a pole-pair with one zero gives a circle centred at the zero, with radius $\sqrt{13} = 3.606$ (the distance from the zero to the poles). The complex poles move along this circle and meet the real axis at $-3.606$. One branch then moves right to the zero at $0$ and the other goes to $-\infty$. 7. **$j\omega$ crossing:** none for $K > 0$, since all coefficients of $s^2 + (4+K)s + 13$ are positive. The system is stable for all $K > 0$. ```text | | .x | .... | .. | .. | .. | . | ......................*................o---------- . | .. | .. | .. | .... | .x | | | -6 -3.6 0 x pole (-2+-j3) o zero (0) * break-in (-3.606) ``` #### Gain at critical damping Critical damping corresponds to the break-in point $s = -3.606$, where the two closed-loop poles are real and equal. $$ K = \frac{|s^2 + 4s + 13|}{|s|} = \frac{13 - 14.422 + 13}{3.606} = \frac{11.578}{3.606} = 3.21 $$ Check: critical damping means $(4+K)^2 = 4 \times 13 = 52$, so $K = \sqrt{52} - 4 = 3.21$. ✓ **Answer:** **K ≈ 3.21**. Both closed-loop poles are then at **s = −3.606**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For two complex poles and one zero, the complex part of the locus is: A. A vertical line B. A circle centred at the zero C. A straight line at 45° D. A parabola Answer: B. It is a circle centred on the zero with radius equal to the zero-pole distance, sqrt(13) = 3.606. (Key point: Shape of locus) 2. The break-in point comes from dK/ds = 0 giving: A. s = -sqrt(13) = -3.606 B. s = -2 C. s = -13 D. s = -4 Answer: A. s^2 - 13 = 0, and only s = -3.606 is on the locus. (Key point: Break-in point) 3. Why is the system stable for all K > 0? A. The characteristic equation s^2 + (4+K)s + 13 has all positive coefficients B. There are no zeros C. The poles are real D. K is bounded Answer: A. No jw crossing exists, so all roots remain in the LHP. (Key point: Stability) 4. Critical damping is reached when (4+K)^2 = 52, so K is: A. 13 B. 3.21 C. 7.2 D. 4 Answer: B. K = sqrt(52) - 4 = 3.21, with both poles at -3.606. 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