# Second Order Transfer Function Question regarding Overshoot Formula

If I have a closed loop second order transfer function such as:

$$\frac{10-s}{0.3s^2+3.1s+(1+24K_{C})}$$

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?

• I guess the formula you are referring to is derived for 2nd order transfer functions with no s in the numerator. – Karlo Apr 18 '16 at 11:17

What is the step input in the s-domain, $\frac{B}{s}$?