# Find the point where locus crosses the damping ratio line?

Given a transfer function, find the point where locus crosses the damping ratio of 0.5?

$$G(s)= \dfrac{K(s-2)(s-4)}{s^2+6s+25}$$

The textbook only shows this solved by Matlab or program, I would like to know how to do it by hand calculations?

Solve the closed-loop equation: $$T(s) = \frac{G(s)}{1+G(s)} = \frac{K(s^2-6s+8)}{s^2+6s+25+K(s^2-6s+8)}$$
structure the denominator to the following: $$s^2 + as+b$$ and relate to the denominator of the standard second order mass model: $$H(s) = \frac{Q\omega^2}{s^2+2\zeta\omega s+ \omega^2}$$ where $$Q$$ is any arbitrary gain. Relate $$K$$ to $$\zeta$$ such that $$\zeta = 0.5$$. $$T(s) = \frac{K(s^2-6s+8)}{s^2(1+K)+s(6-6K)+25+8K}$$ $$T(s) = \frac{K(s^2-6s+8)/(1+K)}{s^2+s(6-6K)/(1+K) + (25+8K)/(1+K)}$$ $$2\zeta = \frac{6-6K}{1+K}\cdot\frac{\sqrt{1+K}}{\sqrt{25+8K}}$$ $$2\zeta = \frac{(6-6K)\sqrt{1+K}}{\sqrt{1+K}^2\sqrt{25+8K}} = \frac{(6-6K)}{\sqrt{(1+K)(25+8K)}} = 1$$ $$\rightarrow 8K^2+33K+25 = (6-6K)^2$$ Solve this and mind the sign. due to the squared structure, on $$K$$ value will yield a damping value of $$-0.5$$. so plug the result back into the equation to verify the correct one.