# Overshoot computations not matching

Take the following transfer function of a 3rd order system:

$$H(s)=\dfrac{2.302~s+0.3548}{s^3+0.739~s^2+3.223~s+0.3548}$$

with poles:

         Pole              Damping       Frequency      Time Constant

-1.13e-01                 1.00e+00       1.13e-01         8.89e+00
-3.13e-01 + 1.75e+00i     1.76e-01       1.78e+00         3.19e+00
-3.13e-01 - 1.75e+00i     1.76e-01       1.78e+00         3.19e+00


and with the following unitary step response:

If I compute the percentage overshoot (PO) based on the damping ratio $\zeta=0.176$, I get:

$$PO=100{\times}e^{\dfrac{-\zeta\pi}{\sqrt{1-\zeta^2}}}=100{\times}e^{\dfrac{-0.176\pi}{\sqrt{1-0.176^2}}}=\boxed{57.02\%}$$

However, if I compute the PO using the graphical method (comparing the peak value with the final value) I get a completely different result:

$$PO=\dfrac{v_{peak}-v_{final}}{v_{final}}{\times}100=\dfrac{1.2-1}{1}{\times}100=\boxed{20\%}$$

I don't understand why such discrepancy. Why are my PO computations not matching?

• remember that 3rd-order is not the same as 2nd-order. – robert bristow-johnson Oct 9 '17 at 22:17
• @robertbristow-johnson Thank you for the reminder. Do you imply that one of the methods is not suitable for 3rd-order systems? – codeaviator Oct 9 '17 at 22:31
• read the sources you cite regarding percentage overshoot. what is all of that math, from which you derive a mathematical expression apply to? – robert bristow-johnson Oct 9 '17 at 23:21

Computing the partial fraction expansion,

$$H (s) = \frac{2.302 s + 0.3548}{s^{3} + 0.739 s^{2} + 3.223 s + 0.3548} \approx \frac{2.2861 - 0.0309306 s}{s^2 + 0.626454 s + 3.1525} + \frac{0.0309306}{s + 0.112546}$$

For the time being, let us neglect the step response of the 1st order subsystem. Note that the 2nd order subsystem has the following form

$$\pm \gamma (s - z) \left(\frac{\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}\right)$$

where $\gamma \in \mathbb R$ is the gain (or attenuation) and $z \in \mathbb R$ is a (finite) zero. However, the formula you used to calculate the step response overshoot is only applicable for 2nd order systems of the form

$$G (s) := \frac{\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}$$

whose zeros are at infinity. To summarize, the overshoot is smaller than you expected because:

• you neglected the 1st order subsystem.
• you neglected the attenuation $\gamma \neq 1$.
• you did not subtract $\gamma$ times the derivative of the step response of $G (s)$ at the peak time.

In any case, not neglecting the above would only provide one estimate of the overshoot. To compute a satisfactory approximation of its exact value, you can take the inverse Laplace transform, differentiate and find where the derivative vanishes.

• You are absolutely right. Thank you for your enlightening answer! – codeaviator Oct 14 '17 at 13:19