If I have a closed loop second order transfer function such as:
$$T(s) = \frac{10-s}{0.3s^2+3.1s+(1+24K_{c})}$$
Can I still use this formula for overshoot (when a step input is applied) ?: $$\frac{A}{B}=e^{\frac{-\pi \zeta}{\sqrt{1-\zeta^2}}}$$ Where $B$ is the step input size
I don't think you can but I'm not sure, can someone confirm?