Root-locus breakaway/breakin points

As I mention in Chapter 6, one can provided additional root-locus “rules” that can help refine plots done by hand. My approach when writing the book was to keep the number of rules to a minimum, reflecting the fact that one will rarely draw a root-locus diagram by hand these days. In my opinion, the main goal here should be to learn how adding, removing, or moving poles and zeros can impact the overall root-locus. Not how to accurately sketch the root-locus. That goal can be accomplished comfortably with the limited set of rules provided in Section 6.4. Yet, every now and then I will get a question such as the one below:

I looked through the book, and I probably missed it, but if we have a root locus that diverges from the real axis and goes complex how do we determine the point at which it diverges? I get that for some plots it will diverge onto the imaginary axis at our center of asymptotes, but for some more complicated plots they diverge from a different point.


Yes, there’s a rule for that!

But before delving into the rule, note that the existence of such points, called breakaway or breakin points depending on whether the root-locus is leaving the real axis or joining the real axis, can many times be inferred without the need of the rule. Indeed, as I emphasize in Section 6.4, adding arrows to show how the roots flow from from poles to zeros will reveal the existence of such mandatory breakaway or breakin points on the real axis: if two arrows point toward each other then a breakaway point must exist somewhere between those two poles; if two arrows point away from each other then a breakin point must exist between those two zeros. The fact that one knows for sure when such point exists combined with the need for them to be located between pair of poles or zeros whose location is know mean that there is usually enough information for sketching the root-locus, even without calculating the exact breakaway/in point in such simple cases.

There are more complicated cases though, as illustrated in Fig. 6.13, which is reproduced below:

Fig 6.13 (a), (b) and (c): possible sketches of the root-locus for a transfer-function with three poles and one zero.

In this case, roots can break in as in (b) or (c) and then break away of the real root-locus. Each case will depend on the exact location of the roots and you should use a numeric calculator to help sort out which case you’re dealing with. We will get into more detail at the end of the post.

So if you are interested in calculating such breakaway/breakin points “by hand,” what should you do? Can they be easily calculated?

The answer is that you need to calculate the roots of a certain polynomial related to the loop transfer-function. If the polynomial is of low enough order you might be able to do it by hand. Otherwise you might need help from a numeric calculator, perhaps the same one that can plot you the entire root-locus… Do you see my point?

What polynomial roots need to be calculated? The key observation is that each breakaway or breakin point corresponds to a point in the root-locus for which the rational function $f(s)=1+α L(s)$ has at least a double root. That means $f(s)=0$, which will be true if $s$ is in the root-locus, and $f′(s)=0$. Of importance here is the fact that the second condition is the same as $L′(s)=0$, which is independent of $\alpha$, so breakaway/breakin points can be located without explicitly calculating the corresponding gain $\alpha$. Of course the value of the gain at a breakaway/breakin point, say $s_0$, can be calculated later if needed since $\alpha = -1/L(s_0)$.

If $L(s)=N(s)/D(s)$ is rational then 


and $L′(s)=0$ whenever 


Solve the above polynomial equation and determine all its real solutions. The ones which are on the root-locus will be your breakaway/breakin points.

For example, consider the root-locus plots in Fig. 6.13 reproduced above, for which

$$L(s) = \frac{(s + a)}{(s -a)(s+a+jb)(s+a-jb)}=
\frac{(s + a)}{(s -a)(s^2+2 a s+a^2-b^2)} $$


$$p(s) = D′(s)N(s) – N′(s)D(s) = 2 a b^2 + 2 a^2 s + 4 a s^2 + 2 s^3.$$

This is a third-order polynomial and a formula for the roots is already too complicated to be useful. As mentioned in the book, the roots of $p(s)$ are governed by the ratio $\rho = a/b$. Indeed, try replacing $s$ by $b z$ and divide by $b^3$ to reveal

$$p(b z)/b^3 = 2 \rho + 2 \rho^2 z + 4 \rho z^2 + 2 z^3.$$

For example, when $\rho = 1$ there are no real roots, and the root-locus is as in Fig. 6.13(a).

When $\rho = 4$ and $a = 1$ there are three real roots at

$$s \approx -1.2, \quad s \approx-0.7, \quad s \approx -0.07.$$

The last two of which are in the interval $[-a,a]=[-1,1]$, indicating two breakaway/breakin points as in Fig. 6.1.3(b). The leftmost root must be a breakin point because there is a zero to the left of it toward which one of the roots breaking in must flow to. Likewise, the rightmost root must be a breakaway point because there is a pole to the right of it from which one of the roots breaking away must flow from.

Finally, when $\rho=9/(2\sqrt{3})$ and $a = 1$, three are only two real roots

$$s \approx -1.33, \quad s \approx -0.33$$

but the root at $-0.33$ is a repeated root, and the root-locus is as in Fig. 6.13(c). At this special breakaway point the root-locus sports a triple root. That point is, at the same time, a breakin and a breakaway point. One can fathom that such a special point must exist as a transition between the diagrams from Fig. 6.13(a) and 6.13(b) implied by the continuity of the root-locus as a function of the coefficients of the loop transfer-function $L(s)$.

3 thoughts on “Root-locus breakaway/breakin points”

  1. Hi,
    Why can we assume that alpha is independent of s?
    And why did we need to go to the derivative to state that alpha is independent of s?
    I am tasked with proving that the gradient of alpha is zero with respect to s at the breakaway points.
    I like your reasoning based on the multiplicity, but it seems like you then assume what I’m trying to prove.

    1. You might be confused by the term “independent” here. For any given constant $\alpha > 0$, a point $s$ is in the root-locus if $f_{\alpha}(s) = 1 + \alpha L(s) = 0$. The calculation of the roots or of the derivatives of the univariate polynomial $f_{\alpha}(s)$ can be done assuming that $\alpha$ is a constant. There is no need to work with the object $f(\alpha, s) = 1 + \alpha L(s)$ for that purpose.

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