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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?

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    $\begingroup$ I guess the formula you are referring to is derived for 2nd order transfer functions with no s in the numerator. $\endgroup$
    – Karlo
    Apr 18, 2016 at 11:17

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Formula is derived for pure 2nd order TF without any zeros. The validity of the formula for cases with zeros depends on relative position of the zero with regard to poles. The poles of $T(s)$ are \begin{equation} p_{12}=\pm\frac{\sqrt{841-2880\,K_c }}{6}-\frac{31}{6}, \end{equation} therefore $K_c>-1/24$ is required for a system to be stable. Zero is in RHP.

Transfer function $T(s)$ can be rewritten as \begin{equation} T(s) = G\frac{s/(\alpha\zeta\omega_n) + 1}{(s/\omega_n)^2 + 2\zeta(s/\omega_n) + 1} \end{equation}

where $G=(1+24K_c)/10$ and \begin{equation} \omega_n = \sqrt{\frac{1+24K_c}{0.3}}, \quad \zeta=\frac{3.1\omega_n}{2(1+24K_c)} = \frac{3.1}{2 \sqrt{1+24K_c}\sqrt{0.3}} \end{equation}

\begin{equation} \alpha=-\frac{10}{\zeta\omega_n} = -10\cdot\frac{2\cdot0.3}{3.1} = -1.9355 \end{equation}

If $|\alpha|> 3$ the zero has very little effect on overshoot, but as $\alpha$ decreases below 3, it has an increasing effect, especially when $\alpha=1$ or less. So with $\alpha=-1.93$ you are somewhere in the middle.

Here $\alpha$ is always negative for any choice of $K_c$. This implies that zero is in RHP and the time response is thus somewhat different (derivative part is in opposite direction). In your particular case the overshoot is almost the same as it would be with pure 2nd order TF (see plot on left).

Left plot shows pure 2nd order TF ($T_0(s)$) and a 'derivative' part $T_d(s)$, so that $T(s)=T_0(s)+T_d(s)$. In contrast, zero in LHP (lets say at $s=-10$) would produce plot $H(s)$ on the right.

enter image description here

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I think you should still be able to find the inverse laplace transforms of the closed loop TF, at set values of Kc. Such that the overshoot can be determined in the time domain.

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

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Yes you can, but only for underdamped response. The formula is derived from a second order system response, the one in your question looks to be a first order system but under a PI controller. That usually isn't relevant for the formula, given that the parameters are so arranged that the response is underdamped.

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