1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Dan Boschen Apr 26 '17 at 11:41
  • $\begingroup$ 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... $\endgroup$ – user28187 Apr 26 '17 at 11:48
  • $\begingroup$ 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? $\endgroup$ – Dan Boschen Apr 26 '17 at 11:50
  • $\begingroup$ Very good with your updated question in showing your work; please check my answer thoroughly and let me know if you agree $\endgroup$ – Dan Boschen Apr 26 '17 at 12:18
0
$\begingroup$

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.

root locus

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);
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.