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.
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,
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.
On Tue, 22 Jul 2008 01:58:06 +0100, ssylee wrote
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
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.)
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 :-).
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
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.
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.
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