3
$\begingroup$

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  
                                       (rad/seconds)      (seconds)    

 -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:

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?

$\endgroup$
  • 2
    $\begingroup$ remember that 3rd-order is not the same as 2nd-order. $\endgroup$ – robert bristow-johnson Oct 9 '17 at 22:17
  • $\begingroup$ @robertbristow-johnson Thank you for the reminder. Do you imply that one of the methods is not suitable for 3rd-order systems? $\endgroup$ – codeaviator Oct 9 '17 at 22:31
  • 1
    $\begingroup$ read the sources you cite regarding percentage overshoot. what is all of that math, from which you derive a mathematical expression apply to? $\endgroup$ – robert bristow-johnson Oct 9 '17 at 23:21
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.