1. **Problem Statement:** Given the unity feedback system with open-loop transfer function $$G(s) = \frac{K(s-1)(s-2)}{s(s+1)}$$ we need to find: a. Breakaway and break-in points b. j\omega-axis crossing c. Range of gain $K$ for stability d. Value of $K$ for second-order complex poles with damping ratio $\zeta=0.5$ 2. **Root Locus Basics:** The root locus shows the locations of closed-loop poles as $K$ varies. Closed-loop characteristic equation: $$1 + G(s) = 0 \implies 1 + \frac{K(s-1)(s-2)}{s(s+1)} = 0$$ which simplifies to $$s(s+1) + K(s-1)(s-2) = 0$$ 3. **Breakaway and Break-in Points:** These occur where multiple roots coincide on the real axis. Find $K$ as a function of $s$: $$K = -\frac{s(s+1)}{(s-1)(s-2)}$$ Break points satisfy: $$\frac{dK}{ds} = 0$$ Calculate derivative: $$\frac{dK}{ds} = -\frac{(2s+1)(s-1)(s-2) - s(s+1)(2s-3)}{(s-1)^2 (s-2)^2} = 0$$ Simplify numerator: $$N = (2s+1)(s-1)(s-2) - s(s+1)(2s-3)$$ Expand: $(2s+1)(s-1)(s-2) = (2s+1)(s^2 - 3s + 2) = 2s^3 - 6s^2 + 4s + s^2 - 3s + 2 = 2s^3 - 5s^2 + s + 2$ $s(s+1)(2s-3) = s(s+1)(2s-3) = s(2s^2 - 3s + 2s - 3) = s(2s^2 - s - 3) = 2s^3 - s^2 - 3s$ So, $$N = (2s^3 - 5s^2 + s + 2) - (2s^3 - s^2 - 3s) = -4s^2 + 4s + 2 = 0$$ Divide by -2: $$2s^2 - 2s - 1 = 0$$ Solve quadratic: $$s = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2}$$ Numerical values: $$s_1 = \frac{1 + \sqrt{3}}{2} \approx 1.366, \quad s_2 = \frac{1 - \sqrt{3}}{2} \approx -0.366$$ Check if these points lie on root locus segments (between poles and zeros on real axis). 4. **j\omega-axis Crossing:** Substitute $s = j\omega$ into characteristic equation: $$1 + G(j\omega) = 0$$ $$1 + \frac{K(j\omega -1)(j\omega -2)}{j\omega (j\omega +1)} = 0$$ Separate real and imaginary parts and solve for $\omega$ and $K$. Calculate numerator: $$(j\omega -1)(j\omega -2) = (j\omega)(j\omega) - 2j\omega - j\omega + 2 = -\omega^2 - 3j\omega + 2$$ Denominator: $$j\omega (j\omega +1) = j\omega (j\omega) + j\omega = -\omega^2 + j\omega$$ So, $$G(j\omega) = K \frac{-\omega^2 - 3j\omega + 2}{-\omega^2 + j\omega}$$ Set $1 + G(j\omega) = 0$: $$1 + K \frac{-\omega^2 - 3j\omega + 2}{-\omega^2 + j\omega} = 0$$ Multiply both sides by denominator: $$-\omega^2 + j\omega + K(-\omega^2 - 3j\omega + 2) = 0$$ Separate real and imaginary parts: Real: $$-\omega^2 + 2K = 0 \implies 2K = \omega^2 \implies K = \frac{\omega^2}{2}$$ Imaginary: $$\omega - 3K \omega = 0 \implies \omega(1 - 3K) = 0$$ For $\omega \neq 0$: $$1 - 3K = 0 \implies K = \frac{1}{3}$$ From real part: $$K = \frac{\omega^2}{2} = \frac{1}{3} \implies \omega^2 = \frac{2}{3} \implies \omega = \sqrt{\frac{2}{3}} \approx 0.816$$ 5. **Range of Gain for Stability:** Poles start at $s=0$ and $s=-1$ (poles of $G(s)$) and zeros at $s=1$ and $s=2$. System is stable if all closed-loop poles have negative real parts. From j\omega crossing, stability limit is at $K = \frac{1}{3}$. For $0 < K < \frac{1}{3}$, system is stable. 6. **Value of $K$ for Damping Ratio $\zeta=0.5$:** Damping ratio $\zeta = \frac{-\text{Re}(s)}{|s|}$ for complex poles. Use characteristic equation: $$s^2 + s + K(s-1)(s-2) = 0$$ Rewrite: $$s^2 + s + K(s^2 - 3s + 2) = 0$$ $$s^2 + s + Ks^2 - 3Ks + 2K = 0$$ $$s^2(1+K) + s(1 - 3K) + 2K = 0$$ Assuming complex conjugate poles: $$s = -\sigma \pm j\omega_d$$ Damping ratio: $$\zeta = \frac{\sigma}{\sqrt{\sigma^2 + \omega_d^2}} = 0.5$$ Use quadratic formula for $s$: $$s = \frac{-(1 - 3K) \pm \sqrt{(1 - 3K)^2 - 8K(1+K)}}{2(1+K)}$$ Set real part $\sigma = \frac{1 - 3K}{2(1+K)}$ and solve for $K$ such that $\zeta=0.5$. This requires numerical or algebraic solving; approximate solution: Try $K=0.2$: $$\sigma = \frac{1 - 0.6}{2(1.2)} = \frac{0.4}{2.4} = 0.1667$$ Calculate $\omega_d$ from discriminant: $$D = (1 - 3K)^2 - 8K(1+K)$$ $$= (1 - 0.6)^2 - 8(0.2)(1.2) = 0.16 - 1.92 = -1.76 < 0$$ Imaginary part: $$\omega_d = \frac{\sqrt{1.76}}{2(1.2)} = \frac{1.326}{2.4} = 0.5525$$ Damping ratio: $$\zeta = \frac{0.1667}{\sqrt{0.1667^2 + 0.5525^2}} = \frac{0.1667}{0.576} = 0.289$$ Try $K=0.5$: $$\sigma = \frac{1 - 1.5}{2(1.5)} = \frac{-0.5}{3} = -0.1667$$ Negative real part means unstable, so $K$ must be less than about 0.4. Try $K=0.3$: $$\sigma = \frac{1 - 0.9}{2(1.3)} = \frac{0.1}{2.6} = 0.0385$$ $$D = (1 - 0.9)^2 - 8(0.3)(1.3) = 0.01 - 3.12 = -3.11$$ $$\omega_d = \frac{\sqrt{3.11}}{2(1.3)} = \frac{1.764}{2.6} = 0.678$$ $$\zeta = \frac{0.0385}{\sqrt{0.0385^2 + 0.678^2}} = 0.0567$$ Try $K=0.1$: $$\sigma = \frac{1 - 0.3}{2(1.1)} = \frac{0.7}{2.2} = 0.318$$ $$D = (1 - 0.3)^2 - 8(0.1)(1.1) = 0.49 - 0.88 = -0.39$$ $$\omega_d = \frac{\sqrt{0.39}}{2(1.1)} = \frac{0.624}{2.2} = 0.283$$ $$\zeta = \frac{0.318}{\sqrt{0.318^2 + 0.283^2}} = 0.67$$ By interpolation, $K \approx 0.15$ yields $\zeta = 0.5$. **Final answers:** - Break points at $s \approx 1.366$ and $s \approx -0.366$ - j\omega-axis crossing at $\omega \approx 0.816$, $K = \frac{1}{3}$ - Stability range: $0 < K < \frac{1}{3}$ - $K \approx 0.15$ for $\zeta = 0.5$ complex poles