# Second Order Transfer Function Question regarding Overshoot

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1. Apr 17, 2016

### Tom Hardy

1. The problem statement, all variables and given/known data
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?

2. Apr 17, 2016

### Staff: Mentor

3. Apr 19, 2016

### donpacino

https://en.wikipedia.org/wiki/Overshoot_(signal)

4. Apr 22, 2016

### rude man

Split your xfr function into two parts: one is a low-pass 2nd order, the other a bandpass 2nd order. For both, there are expressions for overshoot in textbooks etc. Your answer depends on the numerical value of KC.

You then just add the time responses of the two separated xfr functions.
@nascent, thanks for the link! wow, even does nyquist plot!