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?
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1 Answer
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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.
