I have previously discussed about why I kept the number of root-locus “rules” at a minimum in this post. One popular “rule” that I have omitted is the calculation of departure or arrival angles at complex poles or zeros.

The calculation of angles of departure or arrival are a direct consequence of the fact that, as discussed in Section 6.4, points in the root-locus must satisfy

$$\angle L(s) = \pi,$$

and that, in the case of a rational function

$$L = \frac{\beta \, (s – z_1)(s – z_2)\cdots(s – z_m)}{(s – p_1)(s – p_2)\cdots(s – p_n)},$$

the argument of $L(s)$ becomes simply

$$ \angle L(s) = \angle \beta + \sum_{i = 1}^{m} \angle (s – z_i) – \sum_{k = 1}^{n} \angle (s – p_k).$$

These quantities are illustrated in Fig. 6.9:

In this figure, if $\beta > 0$, then

$$\angle L(s_0)=\psi_1-\theta_1-\theta_2-\theta_3.$$

A rule for the departure and arrival angles follows directly from the above ideas.

For example, assume that $\beta > 0$ and let $s_0$ be a point on the root-locus that is near an open-loop simple pole $p_\ell$. It must be true that

$$\angle L(s_0) = -\pi$$

(the change from $\pi$ to $-\pi$ is to get a nicer formula). For a rational function that means that

$$ \angle (s_0 – p_\ell) = \pi + \sum_{1 \leq i \leq m}^{} \angle (s_0 – z_i) -\sum_{1 \leq k \leq n\\ \, \, \, k \neq \ell}^{} \angle (s_0 – p_k) = \pi – \bar{\phi}_{p_{k/\ell}}(s_0)$$

where

$$\bar{\phi}_{p_{k/\ell}}(s) = \sum_{1 \leq k \leq n\\ \, \, \, k \neq \ell}^{} \angle (s – p_k) -\sum_{1 \leq i \leq m}^{} \angle (s – z_i)$$

is the difference between the angles due to the open-loop poles

excluding the pole $p_\ell$ minus the angles due to the open-loop zeros evaluated at a point $s$.

As $s_0$ approaches $p_\ell$ via the root-locus, the quantity

$$\pi – \bar{\phi}_{p_{k/\ell}}(p_\ell)$$

must be the angle at which the root-locus branch emanating from $p_\ell$ *departs* from this pole. That is the pole’s *departure angle*.

For example, consider the pole-zero diagram in Fig. 6.9 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)},$$

where $a > 0$ and $b > 0$, is a compatible transfer-function. Because

$$

\begin{aligned}

\bar{\phi}_{p_{k/\ell}}(-a+jb) &= \angle (s_0 – a) + \angle(s_0 + a + j b) – \angle (s_0 + a) \\

&= \pi – \tan^{-1}(b/(2a)) +\pi/2 – \pi/2 = \pi – \tan^{-1}(b/(2a))

\end{aligned}

$$

the departure angle at the pole $-a+jb$ is

$$\pi – \bar{\phi}_{p_{k/\ell}}(-a+j b) =

\tan^{-1}(b/(2a))$$

which is positive and grows as $b/a$ grows.

By following the same reasoning one can conclude that the *arrival angle* at a complex zero $z_\ell$ must be equal to

$$\pi + \bar{\phi}_{z_{i/\ell}}(z_\ell)$$

where

$$\bar{\phi}_{z_{i/\ell}}(s) = \sum_{1 \leq k \leq n}^{} \angle (s – p_k) -\sum_{1 \leq i \leq m \\ \, \, \, i \neq \ell}^{} \angle (s – z_i)$$

is the exact same formula as for the departure angle, this time excluding the zero $z_\ell$ instead.

We close this post with a caveat: all of the above discussion assumes that the complex pole or zero at which you’re calculating the departure or arrival angle is simple. Can you think of what would be the required changes if the complex pole or zero had multiplicity greater than one? A hint is that the angles will be evenly split among the branches emanating/arriving at the pole/zero.