--- title: "Root locus of K/[s(s+1)(s²+4s+5)] with ξ = 0.5" subject: "Control Systems" chapter: 5 key: "semester_3/control_systems:rl-k-s-s1-s2-4s" source: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s-s1-s2-4s site: BCT Engineering --- # Root locus of K/[s(s+1)(s²+4s+5)] with ξ = 0.5 Control Systems · Chapter 5: Root Locus Technique - Asked in: 2069 Chaitra, 12 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-s-s1-s2-4s - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5 ## Question Plot the root loci for closed loop system with G(s) = K/[s(s+1)(s²+4s+5)], H(s) = 1. Also determine the dominant closed loop pole with ξ = 0.5. ## Short answer For ξ = 0.5, the dominant closed-loop poles are s = −0.288 ± j0.500 (ωn = 0.577 rad/s) with gain K ≈ 2.04. The system is stable for 0 < K < 8. ## Answer $$ G(s) = \frac{K}{s(s+1)(s^2+4s+5)}, \qquad H(s) = 1 $$ Characteristic equation: $s^4 + 5s^3 + 9s^2 + 5s + K = 0$. #### 1. Poles, zeros, branches Poles: $0,\ -1,\ -2 \pm j1$ ($n = 4$). No zeros ($m = 0$). All four branches go to infinity. #### 2. Real-axis locus Only between $0$ and $-1$. #### 3. Asymptotes $$ \sigma_A = \frac{0 - 1 - 2 - 2}{4} = -1.25, \qquad \theta_A = \pm 45^\circ,\ \pm 135^\circ $$ #### 4. Breakaway point $$ \frac{dK}{ds} = -(4s^3 + 15s^2 + 18s + 5) = 0 $$ Real root: $s = -0.393$. Gain there: $K = 0.393 \times 0.607 \times 3.583 = 0.855$. The other roots, $-1.68 \pm j0.60$, are not on the locus. #### 5. Angle of departure from $-2 + j1$ From $0$: $153.43^\circ$; from $-1$: $135^\circ$; from $-2-j1$: $90^\circ$. $$ \phi_d = 180^\circ - 378.43^\circ = -198.43^\circ \equiv +161.57^\circ $$ #### 6. $j\omega$ crossing (Routh) | Row | Col 1 | Col 2 | Col 3 | |---|---|---|---| | $s^4$ | 1 | 9 | $K$ | | $s^3$ | 5 | 5 | | | $s^2$ | 8 | $K$ | | | $s^1$ | $(40-5K)/8$ | | | | $s^0$ | $K$ | | | $K_{mar} = 8$. Auxiliary equation: $8s^2 + 8 = 0 \Rightarrow s = \pm j1$. #### Sketch ```text | | ... | .. ..... | .... .... | .... .....x .X.. ...| .# | ---------------------------x....*...x------------- .# | ...| .....x .X.. .... | .... ..... | .... ... | .. | | -2 -1 0 x pole * breakaway (-0.393) X jw crossing (+-j1) # zeta = 0.5 poles (-0.288 +- j0.5, K = 2.04) ``` - Branches from $0$ and $-1$ break away at $-0.393$, curve toward the ±45° asymptotes, and cross the $j\omega$-axis at $\pm j1$. - Branches from $-2 \pm j1$ leave at $\pm 161.6^\circ$ and follow the ±135° asymptotes. #### 7. Dominant closed-loop poles for ξ = 0.5 Constant-ξ line: $\theta = \cos^{-1}0.5 = 60^\circ$ from the negative real axis, i.e. $s = \omega_n(-0.5 + j0.866)$. Search along it for $\angle G = -180^\circ$. The solution is at $\omega_n = 0.577$, giving $s = -0.288 + j0.500$. | From | Angle to $s$ | Distance | |---|---|---| | pole $0$ | $120.00^\circ$ | 0.577 | | pole $-1$ | $35.07^\circ$ | 0.870 | | pole $-2+j1$ | $-16.30^\circ$ | 1.783 | | pole $-2-j1$ | $41.22^\circ$ | 2.276 | $$ \angle G = -(120.00 + 35.07 - 16.30 + 41.22)^\circ = -180.0^\circ \;\checkmark $$ Gain from the magnitude criterion: $$ K = 0.577 \times 0.870 \times 1.783 \times 2.276 = 2.04 $$ With $K = 2.04$, the other two closed-loop poles are at $-2.21 \pm j1.11$. They are about 7.7 times farther from the $j\omega$-axis than the dominant pair, so the pair at $-0.288 \pm j0.5$ is truly dominant. **Answer:** The dominant closed-loop poles for ξ = 0.5 are **s = −0.288 ± j0.500** ($\omega_n = 0.577$ rad/s), at **K ≈ 2.04**. 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? A. Between 0 and -1 B. Left of -1 C. Right of 0 D. Between -1 and -2 Answer: A. Only between the real poles at 0 and -1 is there an odd count of poles to the right. (Key point: Real-axis locus) 2. Where is the breakaway point? A. s = -0.5 B. s = -0.393 (K = 0.855) C. s = -1.68 D. s = -1.25 Answer: B. The real root of dK/ds = 0 is -0.393; the others are complex and off the locus. (Key point: Breakaway) 3. The Routh array gives K_mar and crossing of: A. K = 8, s = ±j1 B. K = 2.04, s = ±j0.5 C. K = 5, s = ±j2 D. K = 0.855, s = ±j1 Answer: A. (40 - 5K)/8 = 0 gives K = 8; 8s^2 + 8 = 0 gives ±j1. (Key point: Marginal stability) 4. The dominant poles for zeta = 0.5 are at: A. -0.5 ± j0.866 B. -0.288 ± j0.500 with K = 2.04 C. -1 ± j1.73 D. -0.393 ± j0.68 Answer: B. On the 60° line, the angle -180° occurs at wn = 0.577, giving -0.288 + j0.5 and K = 2.04. 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