(How to ask a Homework Question): Define poles by using proportional controller

Given is a process with the transfer function

$$G(s) = \frac{s - 1}{s^2 + 3s + 2}$$

I want to create a controller so that the poles of the controlled system are

$$p_{1,2} = -4 \pm i$$

Is it possible to do this using a proportional controller?

I already calculated the transfer function of the closed loop system, which is:

$$G_1(s) = \frac{k \cdot (s-1)}{(s+1) \cdot (s+2) + k \cdot (s-1)}$$

Then I tried to figure out whether there exists a value for k which leads to the wanted poles and saw that there exists no such value, so my answer would be that it is not possible to use a proportional controller. Please correct me if my solution is wrong.

• The answer is quite simple but I want to make sure you understand why, especially if this is a homework question. Please indicate why you think you want to use a proportional controller (or if this is a HW question), and more importantly what you already know about what a proportional controller is. It is good we add a little thought before giving an answer that is simply copied- but this won't be too painful if you are willing to play along. – Dan Boschen Apr 26 '17 at 11:41
• It is a homework question. I know what a proportional controller is. If we use a fixed gain, we can modify the position of the poles, since the polynome in the denominator would change. I just don't understand if we can set the poles exactly to the desired value... – user28187 Apr 26 '17 at 11:48
• ok great, have you come across a root locus yet? Also do you know what the denominator will look like in the closed loop system? – Dan Boschen Apr 26 '17 at 11:50
• Very good with your updated question in showing your work; please check my answer thoroughly and let me know if you agree – Dan Boschen Apr 26 '17 at 12:18

Nice work and your conclusion is correct.

Further confirmation is the poles in the closed loop system are the roots of $$1+K G(s)=0$$

Which results in the polynomial with a numerator

$$s^2+(3+K)s+(2-K)$$

In order to have complex roots, using the quadratic formula $\sqrt{b^2-4ac}$ must be negative.

So we want to test if $4(2-K)>(3+K)^2$ for any K

Since $4(2-K)=(3+K)^2$ has roots at approximately -.101, -9.898:

$$K^2+10K+1=0$$

The range of K where the poles can be complex is therefore from -.101 to -9.898.

Showing all poles for the closed loop system as a function of K (root locus) over the range of K=-10.5 to 0 as plotted below confirms that $-4 \pm i$ is not a closed loop pole.

We could mathematically prove this by using the relationship of "Break-Outs" on the root locus:

(See Dr. Cheever's excellent pages at http://lpsa.swarthmore.edu/Root_Locus/Example2/Example2.html)

As Dr. Cheever summarized, break-out or break-in points occur where N(s)D'(s)-N'(s)D(s)=0 where N(s) and D(s) are the numerator and denominator polynomial of the open loop transfer function, and N'(s) and D'(s) are the differentiation.

$N(s)= s-1$

$N'(s) = 1$

$D(s) = s^2+s3+2$

$D'(s) = 2s+3$

Solving this results in roots at approximately -1.4495 and 3.4495 indicating the break out and break in locations. The root locus will break out at -1.4495 when K=-0.101 and follow a trajectory as shown in the diagram above to the break-in at 3.4495 when K=-9.898. Therefore $-4 \pm i$ is not a closed loop pole.

Note: Root locus was plotted using Octave with the rlocus command as part of the control toolbox

pkg load control %if not already loaded
sys=tf([1 -1],[1 3 2]);
rlocus(sys,.1,-10.5,0);