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A satellite launcher has a unit feedback system, whose TF global open loop is given by:

$$G_c(s)G(s) = \frac{K(s^2-4s+18)(s+2)}{(s^2-2)(s+12)} $$

a) Draw the root locus for this function

b) Determine the range values of $K$ that make this system stable.

I don't know where start to evaluate a), because the TF has the same number of zeros and poles so, in this case, there's no branches and asymptotes?

In the item b), I achieve the following expression to evaluate the Routh–Hurwitz criterion:

$$s^3(1+K)+s^2(12-2K)+s(-2+10K) -24 + 36K = 0$$

However, when I finished the Routh–Hurwitz table and evaluated the inequalities, don't seems correct with the root locus provided by MATLAB.

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  • $\begingroup$ You mean you want the closed loop system to be stable for b) ? $\endgroup$ – Teo Protoulis Apr 25 at 23:48
  • $\begingroup$ Yes. Just take the denominator of G/(1+G), and obtain the posted equation. $\endgroup$ – Alessandro Melo Apr 26 at 2:59
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Suppose we have a third order polynomial in the form :

$$ s^3+a_2s^2+a_1s+a_0 = 0$$

There is nice caveat for third order systems which is derived from teh Routh-Hurwitz stability criterion. In order for this polynomial to be stable the following three conditions have to be met (trying to derive the Routh-Hurwitz table will be a total mess for this particular system):

  • $a_2 > 0$
  • $a_0 > 0$
  • $a_2a_1 > a_0$

The characteristic polynomial of the third order system is:

$$ (K+1)s^3+(12-2K)s^2+(10K-2)s+36K-24=0 $$

which by considering the fact that $K>0$ can be rewritten:

$$ s^3+\frac{12-2K}{K+1}s^2+\frac{10K-2}{K+1}s+\frac{36K-24}{K+1}=0 $$

The above requirements for this particular polynomial are:

  • $\frac{36K-24}{K+1} > 0 \ \Rightarrow \ K > 0.6667 $
  • $\frac{12-2K}{K+1} > 0 \ \Rightarrow \ K < 6$
  • $\frac{(12-2K)(10K-2)}{(K+1)^2} > \frac{36K-24}{K+1} \ \Rightarrow \ K\epsilon[0 \ 2]$

Taking these into consideration we conclude that the gain $K$ should lie somewhere in between the interval:

$$ 0.6667 \ \le \ K \ \le \ 2 $$

If you indeed try the values $0.6666$ or $2.01$ for $K$ you will see that your system goes unstable. For your information, there is a same caveat for the second order polynomials of the form:

$$ s^2+a_1s+a_0 = 0$$

This polynomial is stable if only and only if $a_1,a_0 > 0$.

Now, regarding you root locus of your open loop function, it is somewhat challenging to derive it since there is some complexity going on. You can always use some software to obtain it. This is the root locus from MATLAB:

enter image description here

Below is the root locus of the closed loop system for a specific value of $K=2$. Notice that the marks for the closed loop poles are indeed located on the imaginary axis which means that the system is critically stable (not strictly stable).

enter image description here

The system becomes strictly stable for values of the gain $K$ which lie in the interval: $(0.6667 \ 2)$. For $K=1$ the root locus of the closed loop system becomes:

enter image description here

And as a last test the root locus for $K=0.6667$ of the closed loop system also includes one pole of the closed loop system on the imaginary axis which implies again that the system is critically stable and not strictly stable:

enter image description here

As you see the stability of the system is very well stated throught the Routh-Hurwitz criterion. There are some rules of thumb in order to obtain the root locus of a system, such as that the poles of the system "go" towards the zeros of the system. However, I encourage you to try and obtain some on your own and have some software package to check them. Check also these series on how to draw them by hand. They are really good.

https://www.youtube.com/playlist?list=PLUMWjy5jgHK3-ca6GP6PL0AgcNGHqn33f

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  • $\begingroup$ In this case, to achieve the natural frequencies correspondents to maximum and minimum values of K, I should equate $(s+\alpha)(s^2+2 w_n \zeta s + w_n^2)$ with the characteristic polynomial replaced by K, right? $\endgroup$ – Alessandro Melo Apr 28 at 16:12
  • $\begingroup$ I'm not sure if I understand correctly what you mean. If you mean to get the values of $z$ & $ω_n$ for the closed loop system you have to obtain the transfer function of the closed loop system (with specific value for $K$) and then, as you say, match the coefficients with the general case. Don't forget to accept the answer if it solved your problem so the answer is characterised as a correct one ! $\endgroup$ – Teo Protoulis Apr 28 at 16:43
  • $\begingroup$ Sorry to come back here, but I can't reproduce your graphs. You achieved the characteristic polynomial assuming a negative feedback ( 1 + G = 0) but your graphs are ploted with positive feedback ( 1 - G = 0). So, when I try to visualise the poles and zeros from K = 2, for example, the rlocus function shows that the system are not stable. $\endgroup$ – Alessandro Melo May 2 at 19:54
  • $\begingroup$ The root locus graphs are the ones for the closed loop system (except the very first one). They are obtained by matlab using the feedback(G,1) command which assumes negative feedback. Is that what you mean or did I miss something ? $\endgroup$ – Teo Protoulis May 2 at 20:02
  • $\begingroup$ Theoretically, the Routh criteria changes if the system is negative or positive feedback? Because the criteria of [2/3 , 2] are obtained from a negative feedback. So, make sense use feedback(G,1) to make your plots and check the answers? $\endgroup$ – Alessandro Melo May 2 at 20:06

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