Given the transfer function \$G(s)=\frac{(1-\frac{s}{3})}{(\frac{s}{0.1}+1)(\frac{s}{100}+1)^2}\cdot 10\$, I would like to calculate the settling time.

From what I've learned, this is a third-order system, and the settling time can be calculated as \$T_{a}=\frac{3}{\zeta\omega_{n}}\$ (the same as a second-order system correct me if I'm wrong), where \$\omega_n\$ is the natural frequency, \$\zeta\$ is the damping factor.

However, there are no imaginary poles and I'm struggling to really understand the fundamentals. What's the "right" intuition/approach?

The relative bode diagram:

enter image description here

Update: I am also looking for solutions that involve simpler calculation than finding the Laplace inverse.

  • \$\begingroup\$ What's the "right" intuition/approach? - use a simulator (micro-cap works with s formulas). \$\endgroup\$
    – Andy aka
    Commented Apr 8, 2020 at 10:32
  • \$\begingroup\$ Settling time on this site is \$\dfrac{4}{\zeta \omega_n}\$ - this is based on reaching 98% of steady state. \$\endgroup\$
    – Andy aka
    Commented Apr 8, 2020 at 10:41

1 Answer 1


Well, first of all, we must find the steady state value. We can find that using the final value theorem of the Laplace transform:


Now, we can solve for \$\text{n}\text{%}\$ of the steady state value by solving:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{10}{\text{s}}\cdot\frac{1-\frac{\text{s}}{3}}{\left(1+10\text{s}\right)\left(1+\frac{\text{s}}{100}\right)^2}\right]_{\left(t\right)}=\frac{10\text{n}}{100}\space\Longleftrightarrow\space t=\dots\space\left[\text{sec}\right]\tag2$$

Using \$\text{n}=98\text{%}\$, we get:

        s)*((1 - (s/3))/((10*s + 1)*(1 + (s/100))^2))*(10), s, 
     t] == (10*98)/100, t > 0}, t]]

Out[1]={{t -> 39.4681}}
  • \$\begingroup\$ Thanks! Where does equation (2) come? Can you please elaborate a little bit more? \$\endgroup\$
    – Kevin
    Commented Apr 8, 2020 at 11:47
  • 1
    \$\begingroup\$ @Kevin Well the transfer function is given in your question and the right-hand side gives n% of the steady state value. And the 1/s is the step response function (Heaviside step function). \$\endgroup\$ Commented Apr 8, 2020 at 11:48
  • \$\begingroup\$ Thank you! Now it's clearer :) \$\endgroup\$
    – Kevin
    Commented Apr 8, 2020 at 11:54
  • \$\begingroup\$ @Kevin you're welcome \$\endgroup\$ Commented Apr 8, 2020 at 11:56
  • \$\begingroup\$ I was just wondering if there were also another method without calculating the Laplace Inverse, e.g. by looking for poles... \$\endgroup\$
    – Kevin
    Commented Apr 8, 2020 at 11:57

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