I have a transfer function of the form:

$$H(s) = \frac{b\omega_n s^2 + a\omega_n^2 s + \omega_n^3}{s^3 + b\omega_n s^2 + a\omega_n^2 s + \omega_n^3}$$

If it matters, $a=b=2$. Is anyone aware of a simple explicit form of the step response for this transfer function (i.e. something directly in terms of $\omega_n$, $a$, and $b$)?

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    $\begingroup$ For those particular values of a and b, the denominator is quite easy to factor (poles at (s/w)= -1, and +/- 120 degrees). You can do partial fractions and solve H(s).s. $\endgroup$ – Juancho Sep 8 '19 at 16:28

It does matter that $a=b=2$, because this gives a relatively nice looking solution. As mentioned in a comment, you should split up the given transfer function $H(s)$. But first, let's introduce a normalized variable $p$:


Now we can write the given transfer function as


The denominator can be factored as


Partial fraction expansion of $\hat{H}(p)$ gives


The Laplace transform of the step response equals the transfer function divided by $p$:


The inverse Laplace transform of $(5)$ is (table)


Finally, denormalization according to $(1)$ taking into account that $\hat{A}(p)=\omega_nA(s)$ (because of division by $p$) gives the desired step response:

$$\begin{align}a(t)&=\hat{a}(\omega_nt)\\&= \left[1-e^{-\omega_nt}+\frac{2}{\sqrt{3}}e^{-\omega_nt/2}\sin(\sqrt{3}\omega_nt/2)\right]u(t)\tag{7}\end{align}$$

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  • $\begingroup$ This appears correct aside from an extra $\omega_n$ scale factor in front of the final expression. $\endgroup$ – rhz Sep 10 '19 at 1:20
  • $\begingroup$ I think I understand why you wrote (7) with the scaling (since it seems to follow from the properties of Laplace Transforms). However, clearly the system should settle at 1, not $\omega_n$. $\endgroup$ – rhz Sep 10 '19 at 1:35
  • $\begingroup$ @rhz: You're right, I've corrected it. The factor $\omega_n$ is already taken care of by division by $p$ in $(5)$. $\endgroup$ – Matt L. Sep 10 '19 at 7:33
  • $\begingroup$ Why is the extra factor accounted for by the 1/p associated with the step? $\endgroup$ – rhz Sep 10 '19 at 15:49
  • $\begingroup$ @rhz: Because $1/p=\omega_n/s$, so the necessary multiplication by $\omega_n$ happens by multiplying by $1/p$. What is left is the scaling of $t$ to obtain the denormalized result. $\endgroup$ – Matt L. Sep 10 '19 at 16:08

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