I am looking at an old exam question and I am wondering giving a certain G(s) how can I choose the points of the zeros in a PID Controller to have a second order response for small values of K and a first order response for large values of K. So far I have found the transfer function but I am unsure how to determine the zeros. If anyone could even provide an explanation I would be very grateful!

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For a) finding the transfer function I get enter image description here

From here I can see that if K is small I will get a second order response, and if K is large I can see where I would get the first order response.

What I do not understand is how do I find the Zeroes (a and b) for the PID to give the desired response?

Edit: I apologize, both K and k are the same I will fix that. In the question for small K the system should have a second order response to a step input and a first order response for large values of K.

  • 1
    \$\begingroup\$ What do you want for non-small values of K? Is "K" the same as "k"? \$\endgroup\$
    – Andy aka
    Dec 10, 2015 at 15:12
  • \$\begingroup\$ I have updated my question in answer your questions. \$\endgroup\$ Dec 10, 2015 at 15:32
  • \$\begingroup\$ Try again - hint - 3rd line from the top! \$\endgroup\$
    – Andy aka
    Dec 10, 2015 at 15:35
  • \$\begingroup\$ In the problem statement? \$\endgroup\$ Dec 10, 2015 at 15:38
  • \$\begingroup\$ Also if k is small don't you actually get a third order response? \$\endgroup\$
    – Andy aka
    Dec 10, 2015 at 15:38

1 Answer 1


CLTF is: $$\small \frac{Y(s)}{R(s)}=\frac{K(s+a)(s+b)}{s^3+(2+K)s^2+(2+K(a+b))s+abK}$$

To produce a 1st order TF, we would like \$\small (s+a)(s+b)\$ to cancel two of the roots in the denominator. Exact cancellation is not normally attainable, so let's go for approximate.

Thus, write the 'ideal' denominator as \$\small(s+a)(s+b)(s+c)\$, and find the value of \$\small c\$ that approximates this denominator when \$\small K\$ is large.

Expanding \$\small(s+a)(s+b)(s+c)\$ and comparing denominators:

\$\small s^3+(a+b+c)s^2+(ab+bc+ca)s+abc\: \approx\: s^3+(2+K)s^2+(2+K(a+b))s+abK\$

Comparing constant terms, \$\small abc=abK\$, hence \$\small c\rightarrow K\$, and substituting:

\$\small s^3+(a+b+c)s^2+(ab+bc+ca)s+abc\: \approx\: s^3+(a+b+K)s^2+(ab+K(a+b))s+abK\$

Now, let \$\small K\$ become large, so that:

\$\small [(a+b)+K]\approx [2+K]\$ and \$\small [ab+K(a+b)]\approx [2+K(a+b)]\$

Hence we have the denominator:\$\small\ (s^2+(a+b)s+ab)(s+K)\$, and the required 1st order CLTF is: $$\small \frac{Y(s)}{R(s)}\approx \frac{K}{s+K}$$

To illustrate, consider \$\small a=1\$; \$\small b=1\$; \$\small K=100\$.

The CLTF is: $$\small \frac{Y(s)}{R(s)}= \frac{100(s^2+2s+1)}{s^3+102s^2+202s+100}$$

It can be seen that \$\small (s^2+2s+1)\$ is an approximate factor of \$\small (s^3+102s^2+202s+100)\$ [it's an exact factor of \$\small (s^3+102s^2+201s+100)\$], giving the 1st order CLTF: $$\small\frac{Y(s)}{R(s)}\approx\frac{100}{s+100}$$

Analysis for small \$\small K\$ can proceed similarly.


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