--- title: "Root locus of K/[s(s+1)(s²+4s+5)], critical damping" subject: "Control Systems" chapter: 5 key: "semester_3/control_systems:rl-s-s1-s2-4s-5" source: https://www.bctengineer.com/q/semester_3/control_systems/rl-s-s1-s2-4s-5 site: BCT Engineering --- # Root locus of K/[s(s+1)(s²+4s+5)], critical damping Control Systems · Chapter 5: Root Locus Technique - Asked in: 2082 Baisakh, 8+2 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-s-s1-s2-4s-5 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5 ## Question Draw root locus for unity feedback system with feedforward transfer function G(s) = K/[s(s+1)(s²+4s+5)]. Also, clearly identify the value of K for critical damping of the system. ## Answer The root locus is the path of the closed-loop poles in the s-plane as $K$ varies from 0 to $\infty$. The characteristic equation is $1 + G(s) = 0$, i.e. $$ s(s+1)(s^2+4s+5) + K = 0 \;\Rightarrow\; s^4 + 5s^3 + 9s^2 + 5s + K = 0 $$ #### Step 1: Poles, zeros and branches - Open-loop poles: $s = 0,\ -1,\ -2 \pm j1$ ($n = 4$); zeros: none ($m = 0$). - Number of branches = 4; all four go to infinity. #### Step 2: Real-axis locus A real-axis point is on the locus if the number of real poles + zeros to its right is odd. Only the segment **between 0 and −1** qualifies. #### Step 3: Asymptotes $$ \begin{aligned} \sigma_A &= \frac{\sum p - \sum z}{n-m} = \frac{0-1-2-2}{4} = -1.25 \\ \theta_A &= \frac{(2q+1)180^\circ}{4} = \pm 45^\circ,\ \pm 135^\circ \end{aligned} $$ #### Step 4: Breakaway point $K = -(s^4+5s^3+9s^2+5s)$, so $$ \frac{dK}{ds} = -(4s^3+15s^2+18s+5) = 0 $$ Roots: $s = -0.393$ and $s = -1.679 \pm j0.603$. Only $s = -0.393$ lies on the locus (between 0 and −1). Gain there: $$ K = 0.393 \times 0.607 \times |(-0.393)^2 + 4(-0.393) + 5| = 0.855 $$ #### Step 5: Angle of departure from $-2+j1$ Angles from the other poles to $-2+j1$: from $0$: $153.43^\circ$; from $-1$: $135^\circ$; from $-2-j1$: $90^\circ$. $$ \phi_d = 180^\circ - (153.43^\circ + 135^\circ + 90^\circ) = -198.43^\circ \equiv 161.57^\circ $$ By symmetry, the angle from $-2-j1$ is $-161.57^\circ$. #### Step 6: Imaginary-axis crossing (Routh array) | Row | Col 1 | Col 2 | Col 3 | |---|---|---|---| | $s^4$ | 1 | 9 | $K$ | | $s^3$ | 5 | 5 | | | $s^2$ | 8 | $K$ | | | $s^1$ | $(40-5K)/8$ | | | | $s^0$ | $K$ | | | Marginal stability: $40 - 5K = 0 \Rightarrow K = 8$. Auxiliary equation: $8s^2 + 8 = 0 \Rightarrow s = \pm j1$. The locus crosses the $j\omega$ axis at **$\pm j1$ rad/s when $K = 8$**. #### Sketch ```text | | ... | .. ..... | .... .... | .... .....x .X.. ...| .. | ---------------------------x....*...x------------- .. | ...| .....x .X.. .... | .... ..... | .... ... | .. | | -2 -1 0 x pole * breakaway (-0.393) X jw crossing (+-j1, K = 8) ``` Branches from 0 and −1 meet at −0.393, break away at ±90°, and bend toward the ±45° asymptotes, crossing at ±j1 (K = 8). Branches from $-2\pm j1$ leave at $\pm 161.6^\circ$ and approach the ±135° asymptotes. #### Gain for critical damping Critical damping occurs when the dominant closed-loop poles are real and equal, i.e. at the breakaway point $s = -0.393$. The other two poles are then at $-2.107 \pm j1.046$. **Answer:** Critical damping at **K ≈ 0.855** (double pole at s = −0.393). The system is stable for $0 < K < 8$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which real-axis segment lies on the locus for poles 0, -1, -2±j1? A. Between 0 and -1 B. Left of -1 C. Between -1 and -2 D. Whole negative axis Answer: A. Only between 0 and -1 is the number of real poles to the right odd. (Key point: Real-axis locus) 2. What is the asymptote centroid? A. -1.25 B. -2 C. -0.5 D. -1 Answer: A. (0-1-2-2)/4 = -1.25 with angles ±45°, ±135°. (Key point: Asymptote centroid) 3. Why does critical damping occur at the breakaway point? A. There the poles cross jω B. There K is maximum for stability C. There the dominant poles are real and equal D. There the poles are complex Answer: C. A double real root at the breakaway point means critical damping. (Key point: Critical damping) 4. What are the critical-damping gain and stability range? A. K ≈ 8; stable for 0