# PID to generate different order Systems

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!

For a) finding the transfer function I get

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.

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

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.