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I could not find a Controls Stackexchange, I hope this is the right place, otherwise I would be glad to post it in the appropriate section. My question concerns the root locus method.

Given a plant with some transfer function that I want to control with gain, I would arrive to such an equation, $$1 + \frac{K}{s(s+0.5)(s^2 + a s + 4)} = 0,$$ where $a$ is a given constant (think $1$ for example), and $K$ is a gain I will be able to choose. I want to draw the root locus by hand to have a rough idea.

I place my poles, the root locus on the real axis, count the branches (four go to infinity, no zeros), find the centroid then asymptotes, find the breakouts on the real axis, find the intersection of the root locus with the imaginary axis and find the the angle of departure from the two complex poles.

Once I have done this, for $a$ close to $\frac 12$, I find myself with a departure angle which is close to $-90^\circ$ (for the upper pole), that is "downwards". This seems ambiguous regarding which asymptote the branch leaving from the upper complex pole will eventually reach. I attach two plots to the post with values of $a = 0.5 \pm 0.1$: $a = 0.49$ and $a = 0.51$.

Obviously now that I look at the plots, I can tell which branch goes where, and using some criterion like "if the branch leaving the complex pole goes to the right, it will go on the right asymptote, and vice versa", you would indeed get the right answer. My question is,

Is there a general way to tell which branch goes to which asymptote?

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I think you're over-analyzing. The root locus is an abstraction. While it usually makes it easier to think of the poles as "moving", you don't want to get too married to that thought.

It works better for me to just think of the poles as merging into one double pole (or triple, or n-tuple, for various contrived examples), and then breaking apart again into the asymptotic poles.

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  • $\begingroup$ I think I have understood that bit, it does not matter where it "comes from", all that matters is where the poles are at a given K. My concern was about the region around the bifurcation at 1.3919+0.25i, there's a slight difference depending on a. Sorry if that was not originally clear. I'm concerned about which branch goes where not to track a particular branch, but to get a more accurate drawing. $\endgroup$
    – Cryme
    Sep 9, 2019 at 18:55
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To echo the answer by TimWescott, consider the situation when $a=0.5$.

enter image description here

The complex pole on the top goes to the left (green) and that on the bottom goes to the right (orange). This is just one of many possibilities. Both could have gone to right, or both to the left, or one left and the other right. All are correct.

So, in the above case, there is no one correct way for a branch to go to an asymptote.

For cases $a=0.49$ and $0.51$, the poles of the former are more to the right and so they tend to go off to the right.

enter image description here

enter image description here

enter image description here

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    $\begingroup$ Thank you for the example! It was doing things like that which led me to think of the poles as merging and breaking apart, rather than going in one direction or another -- but it's been so long, I'd forgotten. $\endgroup$
    – TimWescott
    Sep 9, 2019 at 16:48
  • $\begingroup$ Yes I know that when branches cross, it does not make sense to track solutions (the implicit function theorem fails), but outside of the very particular value of a=1/2 (and outside of the real axis for the root locus), we do have separate branches. I just wanted to get a more accurate drawing. I think it can be worked out geometrically with the angles by drawing the segment with the complex poles as edges, which is the locus where they add no angle, then take the locus out of the real axis slightly and see where it goes. $\endgroup$
    – Cryme
    Sep 9, 2019 at 19:03
  • $\begingroup$ @OlivierMassicot, I would think if you are drawing it by hand to get a rough idea, as stated in your question, and the branches are passing by so closely, you already have a rough idea. The moment you start expending effort to get a more accurate plot, it is no longer a rough analysis. $\endgroup$
    – Suba Thomas
    Sep 9, 2019 at 19:22
  • $\begingroup$ Yes I agree it's good enough, I guess it's asking too much if you already see that they miss each other by a slight bit. I think my original problem was that I would not draw the branches stemming from the real axis vertical enough, so for a rough sketch it was quite off. $\endgroup$
    – Cryme
    Sep 9, 2019 at 19:25

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