1. **Stating the problem:** We are asked to analyze the root locus of the system with open-loop transfer function $$G(s) = \frac{3.92K}{s(s+2)(2s+10)}$$ and feedback $$H(s) = 1$$. 2. **Formula and rules:** The characteristic equation for the closed-loop system is $$1 + G(s)H(s) = 0$$, which gives: $$1 + \frac{3.92K}{s(s+2)(2s+10)} = 0$$ or equivalently, $$s(s+2)(2s+10) + 3.92K = 0$$ The root locus shows the locations of the closed-loop poles as the gain $$K$$ varies from 0 to $$\infty$$. 3. **Step 1: Identify poles and zeros** - Poles are roots of the denominator of $$G(s)$$: $$s=0, \quad s=-2, \quad 2s+10=0 \Rightarrow s=-5$$ - There are no finite zeros. 4. **Step 2: Number of branches** Number of poles $$n=3$$, zeros $$m=0$$, so there are 3 branches starting at poles and going to zeros at infinity. 5. **Step 3: Determine asymptotes** Since $$n>m$$, there are $$n-m=3$$ asymptotes. Their angles are given by: $$\theta_k = \frac{(2k+1)180^\circ}{n-m} = \frac{(2k+1)180^\circ}{3}$$ for $$k=0,1,2$$. So the asymptote angles are: $$60^\circ, 180^\circ, 300^\circ$$. The centroid of the asymptotes is: $$\sigma = \frac{\sum \text{poles} - \sum \text{zeros}}{n-m} = \frac{(0) + (-2) + (-5) - 0}{3} = \frac{-7}{3} = -2.33$$ 6. **Step 4: Real axis segments** Root locus exists on real axis segments to the left of an odd number of poles and zeros. Poles on real axis: -5, -2, 0 Segments: - $$(-\infty, -5)$$: to the left of 1 pole (odd) => locus exists - $$(-5, -2)$$: between 2 poles (even) => no locus - $$(-2, 0)$$: between 2 poles (even) => no locus - $$(0, \infty)$$: to the right of all poles (0 poles) => no locus 7. **Step 5: Breakaway and break-in points** These occur where multiple roots meet on the real axis. They satisfy: $$\frac{dK}{ds} = 0$$ where $$K = -\frac{s(s+2)(2s+10)}{3.92}$$. Expanding: $$K = -\frac{2s^3 + 14s^2 + 20s}{3.92}$$ Differentiating: $$\frac{dK}{ds} = -\frac{6s^2 + 28s + 20}{3.92} = 0$$ Multiply both sides by 3.92: $$6s^2 + 28s + 20 = 0$$ Divide by 2: $$3s^2 + 14s + 10 = 0$$ Solve quadratic: $$s = \frac{-14 \pm \sqrt{14^2 - 4 \times 3 \times 10}}{2 \times 3} = \frac{-14 \pm \sqrt{196 - 120}}{6} = \frac{-14 \pm \sqrt{76}}{6}$$ $$\sqrt{76} \approx 8.7178$$ So, $$s_1 = \frac{-14 + 8.7178}{6} = -0.88$$ $$s_2 = \frac{-14 - 8.7178}{6} = -3.12$$ Only $$s_2 = -3.12$$ lies on the real axis segment $$(-\infty, -5)$$? No, it lies between -5 and -2, where no locus exists. $$s_1 = -0.88$$ lies between -2 and 0, no locus there either. So no breakaway points on the locus segments. 8. **Step 6: Stability and final analysis** - Poles start at $$0, -2, -5$$. - As $$K$$ increases, poles move along root locus branches. - One branch goes to $$-\infty$$ along the real axis. - Two branches go to infinity along asymptotes at angles $$60^\circ$$ and $$300^\circ$$ from centroid $$-2.33$$. - The system is stable for small $$K$$ since poles are in left half-plane. - As $$K$$ increases, poles may cross imaginary axis, causing instability. **Final answer:** The root locus has 3 branches starting at poles $$0, -2, -5$$ with asymptotes at angles $$60^\circ, 180^\circ, 300^\circ$$ centered at $$-2.33$$. The locus exists on $$(-\infty, -5)$$ real axis segment. No breakaway points lie on locus segments. Stability depends on $$K$$ value.