Control Systems/Root Locus

The Problem
Consider a system like a radio. The radio has a "volume" knob, that controls the amount of gain of the system. High volume means more power going to the speakers, low volume means less power to the speakers. As the volume value increases, the poles of the transfer function of the radio change, and they might potentially become unstable. We would like to find out if the radio becomes unstable, and if so, we would like to find out what values of the volume cause it to become unstable. Our current methods would require us to plug in each new value for the volume (gain, "K"), and solve the open-loop transfer function for the roots. This process can be a long one. Luckily, there is a method called the root-locus method, that allows us to graph the locations of all the poles of the system for all values of gain, K

Root-Locus
As we change gain, we notice that the system poles and zeros actually move around in the S-plane. This fact can make life particularly difficult, when we need to solve higher-order equations repeatedly, for each new gain value. The solution to this problem is a technique known as Root-Locus graphs. Root-Locus allows you to graph the locations of the poles and zeros for every value of gain, by following several simple rules. As we know that a fan switch also can control the speed of the fan.

Let's say we have a closed-loop transfer function for a particular system:


 * $$\frac{N(s)}{D(s)} = \frac{KG(s)}{1 + KG(s)H(s)}$$

Where N is the numerator polynomial and D is the denominator polynomial of the transfer functions, respectively. Now, we know that to find the poles of the equation, we must set the denominator to 0, and solve the characteristic equation. In other words, the locations of the poles of a specific equation must satisfy the following relationship:


 * $$D(s) = 1 + KG(s)H(s) = 0$$

from this same equation, we can manipulate the equation as such:


 * $$1 + KG(s)H(s) = 0$$


 * $$KG(s)H(s) = -1$$

And finally by converting to polar coordinates:


 * $$\angle KG(s)H(s) = 180^\circ$$

Now we have 2 equations that govern the locations of the poles of a system for all gain values:


 * $$1 + KG(s)H(s) = 0$$


 * $$\angle KG(s)H(s) = 180^\circ$$

Digital Systems
The same basic method can be used for considering digital systems in the Z-domain:


 * $$\frac{N(z)}{D(z)} = \frac{KG(z)}{1 + K\overline{GH}(z)}$$

Where N is the numerator polynomial in z, D is the denominator polynomial in z, and $$\overline{GH}(z)$$ is the open-loop transfer function of the system, in the Z domain.

The denominator D(z), by the definition of the characteristic equation is equal to:


 * $$D(z) = 1 + K\overline{GH}(z) = 0$$

We can manipulate this as follows:


 * $$1 + K\overline{GH}(z) = 0$$
 * $$K\overline{GH}(z) = -1$$

We can now convert this to polar coordinates, and take the angle of the polynomial:


 * $$\angle K\overline{GH}(z) = 180^\circ$$

We are now left with two important equations:


 * $$1 + K\overline{GH}(z) = 0$$


 * $$\angle K\overline{GH}(z) = 180^\circ$$

If you will compare the two, the Z-domain equations are nearly identical to the S-domain equations, and act exactly the same. For the remainder of the chapter, we will only consider the S-domain equations, with the understanding that digital systems operate in nearly the same manner.

The Root-Locus Procedure
In the transform domain (see note at right), when the gain is small, the poles start at the poles of the open-loop transfer function. When gain becomes infinity, the poles move to overlap the zeros of the system. This means that on a root-locus graph, all the poles move towards a zero. Only one pole may move towards one zero, and this means that there must be the same number of poles as zeros.

If there are fewer zeros than poles in the transfer function, there are a number of implicit zeros located at infinity, that the poles will approach.

First thing, we need to convert the magnitude equation into a slightly more convenient form:


 * $$KG(s)H(s) + 1 = 0 \to G(s)H(s) = \frac{-1}{K}$$

Now, we can assume that G(s)H(s) is a fraction of some sort, with a numerator and a denominator that are both polynomials. We can express this equation using arbitrary functions a(s) and b(s), as such:


 * $$\frac{a(s)}{b(s)} = \frac{-1}{K}$$

We will refer to these functions a(s) and b(s) later in the procedure.

We can start drawing the root-locus by first placing the roots of b(s) on the graph with an 'X'. Next, we place the roots of a(s) on the graph, and mark them with an 'O'.

Next, we examine the real-axis. starting from the right-hand side of the graph and traveling to the left, we draw a root-locus line on the real-axis at every point to the left of an odd number of poles or zeros on the real-axis. This may sound tricky at first, but it becomes easier with practice.

Now, a root-locus line starts at every pole. Therefore, any place that two poles appear to be connected by a root locus line on the real-axis, the two poles actually move towards each other, and then they "break away", and move off the axis. The point where the poles break off the axis is called the breakaway point. From here, the root locus lines travel towards the nearest zero.

It is important to note that the s-plane is symmetrical about the real axis, so whatever is drawn on the top-half of the S-plane, must be drawn in mirror-image on the bottom-half plane.

Once a pole breaks away from the real axis, they can either travel out towards infinity (to meet an implicit zero), or they can travel to meet an explicit zero, or they can re-join the real-axis to meet a zero that is located on the real-axis. If a pole is traveling towards infinity, it always follows an asymptote. The number of asymptotes is equal to the number of implicit zeros at infinity.

Root Locus Rules
Here is the complete set of rules for drawing the root-locus graph. We will use p and z to denote the number of poles and the number of zeros of the open-loop transfer function, respectively. We will use Pi and Zi to denote the location of the ith pole and the ith zero, respectively. Likewise, we will use &psi;i and &rho;i to denote the angle from a given point to the ith pole and zero, respectively. All angles are given in radians (&pi; denotes &pi; radians).

There are 11 rules that, if followed correctly, will allow you to create a correct root-locus graph.

We will explain these rules in the rest of the chapter.

Root Locus Equations
Here are the two major equations:


 * {| class="wikitable"

! S-Domain Equations !! Z-Domain Equations
 * $$1 + KG(s)H(s) = 0$$
 * $$1 + K\overline{GH}(z) = 0$$
 * $$\angle KG(s)H(s) = 180^o$$
 * $$\angle K\overline{GH}(z) = 180^o$$
 * }
 * $$\angle K\overline{GH}(z) = 180^o$$
 * }

Note that the sum of the angles of all the poles and zeros must equal to 180.

Number of Asymptotes
If the number of explicit zeros of the system is denoted by Z (uppercase z), and the number of poles of the system is given by P, then the number of asymptotes (Na) is given by:


 * $$N_a = P - Z$$

The angles of the asymptotes are given by:


 * $$\phi_k = (2k + 1)\frac{\pi}{P - Z}$$

for values of $$k = [0, 1, ... N_a - 1]$$.

Asymptote Intersection Point
The asymptotes intersect the real axis at the point:


 * $$\sigma_0 = \frac{\sum_P - \sum_Z}{P - Z}$$

Where $$\sum_P$$ is the sum of all the locations of the poles, and $$\sum_Z$$ is the sum of all the locations of the explicit zeros.

Breakaway Points
The breakaway points are located at the roots of the following equation:


 * $$\frac{dG(s)H(s)}{ds} = 0$$ or $$\frac{d\overline{GH}(z)}{dz} = 0$$

Once you solve for z, the real roots give you the breakaway/reentry points. Complex roots correspond to a lack of breakaway/reentry.

The breakaway point equation can be difficult to solve, so many times the actual location is approximated.

Root Locus and Stability
The root locus procedure should produce a graph of where the poles of the system are for all values of gain K. When any or all of the roots of D are in the unstable region, the system is unstable. When any of the roots are in the marginally stable region, the system is marginally stable (oscillatory). When all of the roots of D are in the stable region, then the system is stable.

It is important to note that a system that is stable for gain K1 may become unstable for a different gain K2. Some systems may have poles that cross over from stable to unstable multiple times, giving multiple gain values for which the system is unstable.

Here is a quick refresher:


 * {|class="wikitable"

! Region ! colspan=2 | S-Domain ! colspan=2 | Z-Domain ! Stable Region ! Marginally Stable Region ! Unstable Region
 * Left-Hand S Plane || $$ \sigma < 0$$|| Inside the Unit Circle || $$|z| < 1$$
 * The vertical axis || $$ \sigma = 0$$ || The Unit Circle || $$|z| = 1$$
 * Right-Hand S Plane || $$ \sigma > 0$$ || Outside the Unit Circle, || $$|z| > 1$$
 * }

Example: Root-Locus Using MATLAB/Octave
{{TextBox|1=Use MATLAB, Octave, or another piece of mathematical simulation software to produce the root-locus graph for the following system:


 * $$T(s) = K\frac{s^2+7s+12}{(s^2 + 3s + 6)}$$

First, we must multiply through in the denominator:


 * $$N(s) = S^2+7S+12$$
 * $$D(s) = S^2+3S+2$$

Now, we can generate the coefficient vectors from the numerator and denominator: num = [0 1 7 12]; den = [0 1 3 2]; Next, we can feed these vectors into the rlocus command: rlocus(num, den); Note:In Octave, we need to create a system structure first, by typing: sys = tf(num, den); rlocus(sys); Either way, we generate the following graph:



{{Control Systems/Nav|Jurys Test|Nyquist Stability Criteria}}