Given the following transfer function:

$$H(s) = \frac{20ks + 1200k}{s^3 + 98s^2 + (20k - 191)s +900 + 1200k}$$

I'm asked to reduce this transfer function to a regular second order transfer function as \$k \to \infty\$:

$$H(s) = \frac{\omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2}$$

Here is the root locus of the original transfer function:

enter image description here

What do I have to take into account in generating the approximated transfer function?

Just use the dominant poles I've obtained from root-locus graph or is there more to it?

Another idea:

Neglect the term with \$s\$ in the denominator, omit \$s^3\$ in the denominator, leave \$s^2\$ as it is and ignore numbers in the lower degree monomial coefficients which are independent of \$k\$.

  • \$\begingroup\$ Thanks for your attention, I meant k, and fixed the other mistake too. \$\endgroup\$ Commented Dec 8, 2021 at 12:09
  • 2
    \$\begingroup\$ If k goes to infinity, the TF reduces to a single order equation. Just inspect the formula. In fact the TF reduces to unity (correction). \$\endgroup\$
    – Andy aka
    Commented Dec 8, 2021 at 12:14
  • \$\begingroup\$ Why is the root locus of the original transfer function drawn? Matlab's inbuilt rlocus command assumes that the varying gain is present only in the numerator as a multiplier to the whole numerator. The varying gain k in your example is not in that format. \$\endgroup\$
    – AJN
    Commented Dec 8, 2021 at 12:17
  • \$\begingroup\$ Maybe I should ask my teacher because I'm required to refer to it as a second order system and analyze its all typical parameters. The original instruction is to assume that the transfer function can be addressed as second order and I'm trying to find the logic behind it. Mathematically I can see what you said. \$\endgroup\$ Commented Dec 8, 2021 at 12:18
  • \$\begingroup\$ Won't do you much good at infinity, but here's a Pade Approximant with order 0,2 \$\endgroup\$
    – Pete W
    Commented Dec 9, 2021 at 0:05

4 Answers 4


For this type of exercise, you first need to rewrite the starting equation in a low-entropy format where the leading term is of similar dimension than the transfer function itself. The rest, the numerator and the denominator are unitless. It means that the coefficient \$b_1\$ of \$s\$ is in seconds, \$b_2\$ of \$s^2\$ is in time squared and \$b_3\$ of \$s^3\$ is in cubic seconds. It is important to check the unit of \$k\$ to respect the homogeneity of the transfer function. Once this is done, then you can try to approximate the denominator as a cascaded pole and a second-order polynomial form.

enter image description here

To succeed in doing so, you must consider the time constants affecting each of the coefficients and see which ones dominate at low and high frequency (well separated, or one dominates at low frequency while a resonance occurs in high frequency and so on). You will have to explain it in your answer to justify your choice. I recommend that you read this document, especially the section on the low-\$Q\$ approximation and after. It is very likely that the first pole and the zero are located in low frequency and do not affect the high-frequency response then dominated by the second-order polynomial.


The last poles when k=> infinity are confirmed.

For helping, this is the picture you get (unless I made an error ...) for value of k = 100 and k=1000.

It seems to be a "second-order" function.

enter image description here

For k= ~ 26, it is a CW wave. Diverging when k < 26 ... Converging when k > 26.

F(t) is ... picture for k = ~ 26

enter image description here


$$\text{expr}=\frac{20 k s+1200 k}{(20 k-191) s+1200 k+s^3+98 s^2+900}$$

EDIT Fixing wrong calc shown earlier after reminder from @aconcernedcitizen: Thanks.

Limit[expr, k -> \[Infinity]]


As k increases, the form is still different from what is requried:

enter image description here

  • \$\begingroup\$ When k -> inf it converges to 1. Both Wolfram and wxMaxima agree. How did you get that answer? It looks like your result comes from substituting k=6. \$\endgroup\$ Commented Dec 8, 2021 at 13:51
  • 1
    \$\begingroup\$ @aconcernedcitizen Actually now that I see it, I used exprInfinity instead of expr in my notebook. Thanks for the correction. I will edit it soon.. \$\endgroup\$
    – Syed
    Commented Dec 8, 2021 at 16:57

This was meant to be a comment, but I'm not really sure how you're meant to reduce that to a 2nd order, since the denominator's roots are steady even with \$k=10^6\$. The single, real root appears to converge to \$-6\$, and the real part of the other two converges to \$-1.9\$, but the imaginary parts keep growing to hundreds, thousands, or more. So if your teacher means to ignore the real \$-6\$ compared to the thousands and consider the complex conjugate pair as the only one advancing, then that's your solution. In which case you'll have two complex conjugate roots of the form \$2\pm jk\$ and the polynomimal will be \$s^2+4s+4+k^2\$. But that doesn't look right when you plot them. If you follow Verbal Kint's answer, you'll need to tinker with the coefficients but, while possible, I'm not sure that is what your teacher was aiming for. You could brutishly nullify the \$s^3\$ term and you'll have matching corner frequencies, but lower and lower quality factor for the 2nd order as \$k\rightarrow\infty\$.


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