ECE 486 Control Systems
Lecture 10

Routh test gives us a range of \(K\) to guarantee stability. On top of stability requirement, a further natural question would be, for what values of \(K\) do we best satisfy given design specs? Figure 2 shows the possible pole locations when time domain specifications are constrained as we discussed in Lecture 6.

We note time domain specifications are encoded in closed loop pole locations in Equation \eqref{d10_eq1}. Therefore we define the root locus for \(1+KL(s)\) as the set of all closed-loop poles, i.e., the roots of

\[ 1 + KL(s) = 0 \]

as \(K\) varies from \(0\) to \(\infty\).

Consider Figure 1 with

\begin{align*} L(s) &= \frac{1}{s^2+s} \text{, i.e.,}\\ b(s) &= 1, a(s) = s^2+s \end{align*}

as our first example. The characteristic equation \(a(s) + Kb(s) = 0\) becomes

\[ s^2 + s + K = 0. \]

By the quadratic formula, the roots are \[ s_{1,2} = - \dfrac{1\pm \sqrt{1-4K}}{2} = - \dfrac{1}{2} \pm \dfrac{\sqrt{1-4K}}{2}. \]

When the parameter \(K\) ranges from \(0\) to \(\infty\),

\[ \text{Root locus} = \left\{ - \dfrac{1}{2} \pm \dfrac{\sqrt{1-4K}}{2} : 0 < K < \infty \right\} \subset \CC. \]

Let’s sketch the root locus in the complex plane.

• Step 1: Start with \(K=0\), the two roots are \(s_{1,2} = -\frac{1}{2} \pm \frac{1}{2} = -1,0\). They are also open-loop poles since feedback \(K = 0\) brings no change to open loop \(L(s)\). As a start, we have two points in the graph as in Figure 3.

  

  Figure 3: Sketching root locus with \(L(s) = \frac{1}{s^2+s}\), step 1

• Step 2: As \(K\) increases from \(0\), the poles start to move.

  • (a) When \(1 - 4K \geq 0\), i.e., \(0 < K \leq \frac{1}4\), the poles are two real poles; they stay on the real axis as in Figure 4.

  

  Figure 4: Sketching root locus with \(L(s) = \frac{1}{s^2+s}\), step 2 (a)

  • (b) When \(1 - 4K < 0\), i.e., $ K > \frac{1}4$, the poles are two complex poles with constant real part \({\rm Re}(s) = -\frac{1}2\) as in Figure 5.

  

  Figure 5: Sketching root locus with \(L(s) = \frac{1}{s^2+s}\), step 2 (b)

  The point \(s=-\frac{1}2\) is the point of breakaway of root locus from the real axis.

Recall from Lecture 6 the relationship between time domain specifications and pole locations,

• \(t_s \approx \frac{3}{\sigma}\), a quickly settled response requires a large \(\sigma\). Here we have a fixed \({\rm Re} (s) = -\sigma = -\frac{1}{2}\), hence \(t_s \approx 6\).
• \(t_r \approx \frac{1.8}{\omega_n}\), a fast rising response requires a large \(\omega_n\), hence a large \(K\).

• \(M_p\) is associated with the boundary of the shaded cone as in Figure 6. A system response with small overshoot requires poles be inside the shaded region. From the figure, \(K\) shall remain small.

  

  Figure 6: Overlaid root locus and admissible pole locations with constrained time domain specifications, where \(L(s) = \frac{1}{s^2+s}\)

Thus, from this first example, we saw the root locus helps us visualize the trade-off between all the specs in terms of \(K\). However, for systems with order greater than \(2\), there will generally be no direct formula for the closed-loop poles as a function of \(K\). This is why we want to develop simple rules for sketching approximate root locus in the general case.

Recall our original definition, the root locus for \(1+KL(s)\) is the set of all closed-loop poles, i.e., the roots of \[ 1 + KL(s) = 0, \]

as \(K\) varies from \(0\) to \(\infty\).

A point \(s \in \CC\) on the complex plane is on the root locus if and only if

\[ \tag{2} \label{d10_eq2} L(s) = \underbrace{-\dfrac{1}{K}}_{\text{negative and real}} \text{ for some $K > 0$}. \]

Equation \eqref{d10_eq2} says its right hand side is real and negative since \(K\) is real and positive. This gives us an equivalent characterization of root locus

Let \(L(s) = \frac{b(s)}{a(s)}\). Write out the characteristic equation in \eqref{d10_eq1},

\begin{align*} &1 + K\frac{b(s)}{a(s)} = 1 + K \frac{s^m + b_1 s^{m-1} + \ldots + b_{m-1}s + b_m}{s^n + a_1 s^{n-1} + \ldots + a_{n-1}s + a_n} = 0 \\ \iff & (s^n + a_1 s^{n-1} + \ldots + a_{n-1}s + a_n) + K(s^m + b_1 s^{m-1} + \ldots + b_{m-1}s + b_m) = 0. \end{align*}

But \(\deg(a) = n \ge m = \deg(b)\) (\(L(s)\) is proper), the polynomial \(a(s) + Kb(s)\) has degree \(n\). Therefore, the characteristic equation has \(n\) not necessarily distinct roots, some of which may be repeated. As we vary \(K\), these \(n\) solutions also vary to form \(n\) branches.

The locus starts from \(K = 0\). What happens near \(K = 0\)?

By \(a(s) + Kb(s) = 0\), if \(K \to 0\), then \(a(s) = - Kb(s) \to 0\). (Why? Hint: \(b(s)\) is bounded except for \(s = \infty\) since \(b(s)\) is a polynomial.)

Therefore either

• \(s\) is close to a root of \(a(s) = 0\), or equivalently
• \(s\) is close to a pole of \(L(s)\).

We considered one end point of the range of \(K \in (0, \infty)\) in Rule B. Now let’s consider the other end point at infinity \(K \to \infty\). Since \(K\) is not zero, we can divide out \(K\) on both sides,

\begin{align*} 0 &= a(s) + Kb(s), \\ b(s) &= -\frac{1}{K}a(s). \end{align*}

As \(K \to \infty\), \(b(s) \to 0\) (since \(a(s)\) is finite), i.e., \(s\) is around some root of \(b(s)\).

• branches end at the roots of \(b(s) = 0\), or equivalently
• branches end at zeros of \(L(s)\).

Note if \(n > m\) when \(L(s)\) is strictly proper, we have \(n\) branches, but only \(m\) zeros. This is not a problem since the remaining \(n-m\) branches go off to infinity; they end at “zeros at infinity”.

The branches of the RL start at the open-loop poles. We ask the question which way do they go, left or right?

Recall the phase condition \[ 1 + KL(s) = 0 \iff \angle L(s) = 180^\circ \] for a positive \(K\). But the angle of \(L(s) = \frac{b(s)}{a(s)}\) is

\begin{align*} \angle L(s) &= \angle \frac{b(s)}{a(s)} \\ &= \angle \frac{(s-z_1) (s-z_2) \cdots (s-z_m)}{(s-p_1)(s-p_2)\cdots(s-p_n)} \\ &= \sum^m_{i=1}\angle (s-z_i) - \sum^n_{j=1}\angle (s-p_j). \hspace{3cm} \text{(by De Moivre's formula)} \end{align*}

This angles sum must be \(\pm 180^\circ\) for any \(s\) that lies on the root locus. Note again, the form \(L(s)\) assumes here is zero-pole form, which relies on the fact that both \(a(s)\) and \(b(s)\) are monic. One associated matlab command is zero-pole-gain zpk().

Now we try some test points as shown in Figure 10.

\begin{align*} \begin{array}{ll} \angle (s_1 - z_1) = 0^\circ & \text{(by $s_1 > z_1$)} \\ \angle (s_1 - p_1) = 180^\circ & \text{(by $s_1 < p_1$)} \\ \angle (s_1 - p_2) = 0^\circ & \text{(by $s_1 > p_2$)} \\ \angle (s_1 - p_3) = -\angle(s_1 - p_4). & \text{(by conjugacy of $p_3, p_4$)} \end{array} \end{align*}

Check angles sum,

\begin{align*} &\angle (s_1 - z_1) - \left[ \angle(s_1-p_1) + \angle (s_1 - p_2) + \angle(s_1 - p_3) + \angle(s_1 - p_4)\right] \\ & = 0^\circ - \left[180^\circ + 0^\circ + 0^\circ\right] \\ & = - 180^\circ . \end{align*}

We conclude \(s_1\) in Figure 10 is on the root locus.

Similarly if we continue to test more points

\begin{align*} \begin{array}{ll} \angle (s_2 - z_1) = 180^\circ & \text{(by $s_2 < z_1$)} \\ \angle (s_2 - p_1) = 180^\circ & \text{(by $s_2 < p_1$)} \\ \angle (s_2 - p_2) = 0^\circ & \text{(by $s_2 > p_2$)} \\ \angle (s_2 - p_3) = -\angle(s_2 - p_4). & \text{(by conjugacy of $p_3, p_4$)} \end{array} \end{align*}

Check angles sum,

\begin{align*} &\angle (s_2 - z_1) - \left[ \angle(s_2-p_1) + \angle (s_2 - p_2) + \angle(s_2 - p_3) + \angle(s_2 - p_4)\right] \\ & = 180^\circ - \left[180^\circ + 0^\circ + 0^\circ\right] \\ & = 0^\circ \neq \pm 180^\circ. \end{align*}

We conclude \(s_2\) in Figure 11 is not on the root locus.

Rules E and F will be discussed next time, and we shall complete Example 2 then.