# Root-Locus Plotting

For a system with Gc(s) = K(s+4)/s and G(s) = (s^2+2*s+2)^-1, I have the following question:
1. Why does this system not have a breakaway point? Is it b/c there is
no solution to d/ds(Gc(s)*G(s))=0? 2. To find the breakaway angle of a particular pole or zero (call it point p), do I draw the lines from the other poles/zeros to point p, and find out the angle between positive x-axis and the lines itself, and use the equation:
180 degrees = sum of all zero's angles - sum of all pole's angles?
Again, muchly to hear comments on this.
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ssylee wrote:

1) Why would you expect it to have a breakaway point? What is your definition of a breakaway point? With only one point and one zero on real axis, the pole goes to the zero and stops.
2) See Ogata, Modern Control Engineering, 1970, p336, item #6, and the test point example in Figure 8-19, p 337.
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I am not sure what you mean by "breakaway", but the plant has complex poles which will scatter and the real pole will migrate to the s+4 zero.
RayR
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Thank you all very much for your responses. To clarify my definition of "breakaway/entry point", it's defined as the value of s at which the polynomial d(1 + Gc(s)G(s))/ds = 0.
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ssylee wrote:

Perhaps your book defines it that way. I would define it as the point where the root locus breaks away from the real line. Then it becomes clear that the _symptom_ of a breakaway point is that the gain on the real line reaches a maximum (or an inflection point -- that's less clear, but a system that more than two poles all coming together at one spot on the real line is pretty contrived). Once you understand the symptom, then the _test_ for the symptom is to take the derivative of the gain, and look for zero.
So the basic answer to your question as stated is "because no matter how you vary the gain you always have one real pole and two complex ones".
That may seem smart-ass, but if you view the root locus as nothing more than a tool to visualize how roots move around as you vary some parameter of the system then it's exactly the only right answer.
You probably want more detail, so a better answer for you is that the farther away 'extraneous' poles are, the less they affect your little root locus plot. This is why root locus plots work at all -- real systems always have more poles than you can model, so if you had to know where _all_ the poles were you'd never be able to use a root locus plot. In this case, the imaginary poles are far enough away, or the nature of the plot is such that they are driven away from the real line instead of toward it.
Try this again with complex poles that are quite close to your zero; you'll probably see them meeting the real line, or at least dipping toward it before going off to infinity. Then try it again with your controller inverted (so it's a PD instead of a PI) and see how it goes.
--

Tim Wescott
Wescott Design Services
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On Tue, 22 Jul 2008 01:58:06 +0100, ssylee wrote (in article

I don't know whether this throws any more light on your question but if you move the complex poles from to (-1 +/- j1) to (-3 +/- j1), i.e. if you change the quadratic to (s^2 + 4*s + 5) then you do get a breakpoint at -2 on the real axis where all three locus branches touch
AAR
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The only way a root locus will have a breakaway (or entry) point is if there is a portion of the root locus on the real axis between two open loop (OL) poles (or zeros) where multiple closed loop (CL) poles exist. That is not the case in your example. You have a portion between an open loop pole and an open loop zero, and two other portions that leave complex poles and move away from the real axis.
Assume there is a portion of the root locus on the real axis between two OL poles. Since the root locus is the path that the CL poles follow from the open loop poles to the open loop zeros (or infinity) as a parameter is varied, when there are two OL poles with no zeros between them, the locus plots leave each OL pole and essentially "collide" at the breakaway point. At that point, the path the CL poles follow is forced to "break away" from the real axis. I encourage you to draw direction arrows on each segment of the locus leaving a pole so that you can see what is happening.
Think more about the concepts and less about the mathematics and you'll understand better. Concepts define the mathematics, not the other way around. In other words, a breakaway point is a necessary and sufficient condition to ensure there is no solution to d/ ds(Gc(s)*G(s))=0. (Forgive any mistakes in logical rigor I may have made by using the terms necessary and sufficient improperly - it's been a while since I've done any formal proofs.)
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rytko wrote:

(snip)
I _think_ you're saying that you have to have two OL poles on the real axis for breakaway points to exist -- it ain't so.
Just for fun, do a root-locus of this:
s^3 + 10s^2 + 30 H = --------------------- s^3 + 8s^2 + 10s + 80
That's zeros at s = 5 +/- j sqrt(5) and s = 0, and poles at s = +/- sqrt(10) and s = 8. It's just the sort of nice, practical system you run into every day, which is why you need to know how to do the root locus of it :-).
--

Tim Wescott
Wescott Design Services
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wrote:

Just for fun I put your tf into Matlab and did a root locus plot. I didn't see a breakaway there--what are we supposed to see?
It seems logical to me you need two poles or zeros on the real axis for a breakaway. Since a breakaway is a transition between real and complex, there must be two of them and they must be coming from real poles or heading towards real zeros. Is there some other combination that works?
dave y.
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dave y. wrote:

Whups. If you use my pole and zero locations it works. The numerator should be s^3 + 10s^2 + 80s, placing a zero on s = 0.
It also works with a single pole at s=0, a couple of resonant zeros on the imaginary axis, then a couple of well-damped but complex poles a good ways into the left-half plane.
It may be _logical_ that you need to start with two poles or zeros on the real axis, but you don't need that -- you just need to have a case at some gain where there are two poles on the real axis.
--

Tim Wescott
Wescott Design Services
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wrote:

Now I see it. Very interesting. Incidently, for those who don't use Matlab, the sisotool is neat, lets you drag a pole with the mouse and watch all the poles move in unison. Better than a static root locus plot to see how the poles are moving.