--- title: "Root locus of k/[s(s+4)(s²+4s+20)]" subject: "Control Systems" chapter: 5 key: "semester_3/control_systems:rl-k-s-s4-s2-4s" source: https://www.bctengineer.com/q/semester_3/control_systems/rl-k-s-s4-s2-4s site: BCT Engineering --- # Root locus of k/[s(s+4)(s²+4s+20)] Control Systems · Chapter 5: Root Locus Technique - Asked in: 2074 Chaitra, 10 marks; 2080 Baisakh, 10 marks; 2078 Kartik, 10 marks; 2076 Chaitra, 10 marks (asked 4 times) - 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-s4-s2-4s - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-5 ## Question For a unity feedback system the open loop transfer function of a control system is given by G(s) = k/[s(s+4)(s²+4s+20)]. Sketch the root locus for 0 ≤ K ≤ ∞ and determine the breakaway point, the angle of departure from complex poles and the stability conditions. ## Short answer Breakaway points are at s=-2 (K=64) and s=-2±j2.45 (K=100). The departure angle from -2+j4 is -90° (and +90° from -2-j4). The system is stable for 0260. ## Answer The root locus is the path traced by the closed-loop poles (roots of $1+G(s)H(s)=0$) as $K$ varies from 0 to $\infty$. It is sketched using the standard construction rules. $$ G(s)H(s) = \frac{K}{s(s+4)(s^2+4s+20)} $$ #### Step 1: Poles and zeros - Poles: $s=0$, $s=-4$, $s=-2\pm j4$ (from $s^2+4s+20=0$). So $n=4$. - Zeros: none, $m=0$. All 4 branches go to infinity. #### Step 2: Real-axis locus A point on the real axis is on the locus if the number of poles + zeros to its right is odd. So the locus lies **between $0$ and $-4$**. #### Step 3: Asymptotes $$ \begin{aligned} \theta_q &= \frac{(2q+1)180^\circ}{n-m} = 45^\circ,\ 135^\circ,\ 225^\circ,\ 315^\circ \\ \sigma_A &= \frac{\sum \text{poles} - \sum\text{zeros}}{n-m} = \frac{0-4-2-2}{4} = -2 \end{aligned} $$ #### Step 4: Breakaway points $$ \begin{aligned} K &= -s(s+4)(s^2+4s+20) = -(s^4+8s^3+36s^2+80s) \\ \frac{dK}{ds} &= -(4s^3+24s^2+72s+80) = 0 \\ &\Rightarrow s^3 + 6s^2 + 18s + 20 = 0 \\ &\Rightarrow (s+2)(s^2+4s+10) = 0 \\ s &= -2,\quad s = -2\pm j2.45 \end{aligned} $$ - At $s=-2$: $K = -(16-64+144-160) = 64 > 0$, a valid breakaway point (on the real-axis segment). - At $s=-2\pm j2.45$: $K = 100$ (real and positive), so these are also valid break points. So the branches from $0$ and $-4$ meet at $-2$ ($K=64$) and leave vertically along $s=-2$. They meet the branches coming down from $-2\pm j4$ at $-2\pm j2.45$ ($K=100$). Then they break away again towards the $45^\circ$ and $135^\circ$ asymptotes. #### Step 5: Angle of departure from complex pole $-2+j4$ Angles from the other poles to $-2+j4$: - From $s=0$: $180^\circ - \tan^{-1}(4/2) = 116.57^\circ$ - From $s=-4$: $\tan^{-1}(4/2) = 63.43^\circ$ - From $s=-2-j4$: $90^\circ$ $$ \phi_d = 180^\circ - (116.57^\circ + 63.43^\circ + 90^\circ) = -90^\circ $$ So the locus leaves $-2+j4$ straight downwards ($-90^\circ$), and leaves $-2-j4$ at $+90^\circ$. #### Step 6: Crossing of the imaginary axis (Routh) Characteristic equation: $s^4+8s^3+36s^2+80s+K=0$ | Row | Col 1 | Col 2 | Col 3 | |---|---|---|---| | $s^4$ | 1 | 36 | $K$ | | $s^3$ | 8 | 80 | | | $s^2$ | $\frac{8(36)-80}{8}=26$ | $K$ | | | $s^1$ | $\frac{26(80)-8K}{26}$ | | | | $s^0$ | $K$ | | | $s^1$ row $=0$: $2080 - 8K = 0 \Rightarrow K = 260$. Auxiliary equation: $26s^2 + 260 = 0 \Rightarrow s = \pm j3.16$. #### Step 7: Sketch ```text Upper half shown; lower half is its mirror image. \ jw / \ 135 deg asymptote | / 45 deg \ | / asymptote \__ * j3.16 (K=260) \ x -2+j4 _/| \__ | _/ | \ v _/ | `-o------' | -2+j2.45 (K=100) | ^ | | | ----x=======o=============x---- sigma -4 -2 (K=64) 0 ``` #### Results | Item | Value | |---|---| | Breakaway points | $s=-2$ ($K=64$); $s=-2\pm j2.45$ ($K=100$) | | Angle of departure | $-90^\circ$ at $-2+j4$, $+90^\circ$ at $-2-j4$ | | $j\omega$ crossing | $s=\pm j3.16$ at $K=260$ | | Stability | Stable for $0260$ | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which real-axis segment is on the locus for poles 0, -4, -2±j4? A. From -4 to -infinity B. Between 0 and -4 C. From 0 to +infinity D. No real-axis segment Answer: B. Only the segment between 0 and -4 has an odd number of poles to its right. (Key point: Real-axis locus) 2. What are the asymptote angles and centroid? A. 60°, 180°, 300° with centroid -2 B. 45°, 135°, 225°, 315° with centroid -2 C. 45°, 135°, 225°, 315° with centroid -4 D. 90° and 270° with centroid -2 Answer: B. n-m=4 gives 180(2q+1)/4; centroid = (0-4-2-2)/4 = -2. (Key point: Asymptotes) 3. Where do the branches from 0 and -4 break away? A. s = -1.57, K = 17 B. s = -4, K = 0 C. s = -2, K = 260 D. s = -2, with K = 64 Answer: D. dK/ds=0 gives s=-2 and K = 2 x 2 x 16 = 64 (|s^2+4s+20| = 16 at s = -2); a valid breakaway on the real segment. (Key point: Breakaway) 4. What is the angle of departure from -2+j4? A. +90° B. -63.4° C. 180° D. -90° Answer: D. 180° - (116.57°+63.43°+90°) = -90°. (Key point: Angle of departure) 5. What gives the stability limit and what is it? A. Routh: K = 260 where the locus crosses jω at ±j3.16 B. Magnitude condition at breakaway: K = 64 C. Angle of departure: K = 90 D. Asymptote centroid: K = 2 Answer: A. The s^1 row 2080-8K=0 gives K=260; stable for 0