More root locus - System Analysis and Control

In this section, we cover some extensions and variations on the basic root locus plot we introduced in the previous section, and we finish with a discussion on how root locus can be used for control design.

Closed-loop calculations¶

Gain calculation¶

When sketching a root locus, we typically focus on the shape of the locus and not the specific locations of the poles for a given value of $k$ . However, it can be useful to calculate the gain $k$ for a specific point on the root locus. For example, if we want to know how much gain we need to add to move a pole from its open-loop location to a desired closed-loop location, we can use the magnitude condition to calculate $k$ .

Example: Consider the system $G(s) = \frac{1}{s(s+2)}$ . We want to calculate the proportional gain $k$ required to give the closed-loop poles a damping ratio of $\zeta = 1/\sqrt{2}$ .

Solution: Based on our intuition, the root locus has an initial real part, where the poles come together and meet at -1. Then, they break away and move vertically along the line $-1+j\omega$ . We reach the desired damping ratio when the poles are located at $-1\pm j$ . The magnitude condition states that for a point $s$ on the root locus, the gain $k$ is given by: $k = 1/|G(s)| = |s||s+2|$ . So the gain required to place a pole at $-1\pm j$ is

\begin{aligned} k &= |s|\cdot |s+2| \\ &=|-1+j|\cdot |-1+j+2| \\ &= \sqrt{2} \cdot \sqrt{2} = 2. \end{aligned}

(1)

We can verify this by plugging $k=2$ into the characteristic equation and solving for $s$ :

\begin{aligned} && 1+k G(s) &= 0 \\ \implies && 1 + 2\cdot \frac{1}{s(s+2)} &= 0 \\ \implies && s^2 + 2s + 2 &= 0 \\ \implies && s &= -1 \pm j. \end{aligned}

(2)

Sum of closed-loop poles¶

Using the gain condition tells us the gain $k$ for a specific location on the locus, but all closed-loop poles move simultaneously as $k$ varies! Here is a useful fact about how closed-loop poles move.

One useful consequence of this fact is that when closed-loop poles go off to infinity, they do so “at the same rate”. In other words, they all approach equally spaced points on an ever-growing circle centered at the centroid.

The negative locus¶

When we analyzed the root locus previously, we assumed that the plant had the form

G(s) = \frac{(s-z_1)\cdots (s-z_m)}{(s-p_1)\cdots (s-p_n)}.

(3)

If there is an additional positive constant in front of the transfer function, it does not affect the shape of the root locus (only our calculation of $k$ ). However, if there is a negative constant in front of the transfer function, then then this is equivalent to using a negative gain $k$ in the characteristic equation. So instead of $1+k G(s)=0$ , we have $1-k G(s)=0$ . This is called the negative locus.

Example. The root locus of $G(s) = \frac{1}{s}$ is the solution to $1+\frac{k}{s}=0$ , so $s=-k$ (negative real axis). If we instead consider $G(s) = -\frac{1}{s}$ , then the root locus is the solution to $1-\frac{k}{s}=0$ , so $s=k$ (positive real axis). The asymptote goes to the right instead of the left! This is a simple example, but the negative locus can be more complicated for higher-order systems.

In fact, each property for the positive locus has a corresponding property for the negative locus. Here is a summary:

Branches and limits. Unchanged. There are still $n$ branches that start at the open-loop poles. $m$ of them end up at open-loop zeros, and the remaining $n-m$ go to infinity along asymptotes.
Real axis. A point $s$ on the real axis is on the negative locus precisely when it is to the right of an odd number of poles and zeros. This is the complement of the condition for the positive locus. So together, the positive and negative loci cover the entire real axis.
Asymptotes and centroid. The centroid is the same as the positive locus. The asymptotes remain evenly spaced, but point in different directions. The formula for the angles of the negative locus asymptotes is
$\theta_\ell = \frac{2\pi \ell}{n-m}, \quad \ell = 0,1,\ldots,n-m-1$
(4)
Here is what they look like.

Asymptotes of the negative locus for different values of n-m. Compare to the asymptotes of the positive locus in . — Figure 1:Asymptotes of the **negative locus** for different values of $n-m$ . Compare to the asymptotes of the positive locus in Figure 4.

Break-in and breakaway points Break-in and breakaway points occur in similar situations and behave similarly. In fact, the necessary condition (11) is the same for both the positive and negative loci. So when we apply the condition, we find all break-in and breakaway points. We can determine which ones belong to the positive locus and which ones belong to the negative locus by checking the real locus condition.
Imaginary axis crossings. The imaginary axis crossings are determined the same way for both the positive and negative loci.
Angles of departure and arrival. The angles of departure and arrival are rotated by $\pi$ for the negative locus compared to the positive locus. So the formulas for departure from a pole and arrival at a zero are, respectively,
$\begin{aligned} \phi_\textsf{dep} &= \sum_{i=1}^m \psi_i - \sum_{i=1}^{n-1} \phi_i \\ \psi_\textsf{arr} &= \sum_{i=1}^{n} \phi_i - \sum_{i=1}^{m-1} \psi_i \end{aligned}$
(5)
where $\phi_i$ and $\psi_i$ are the angles from the other poles and zeros, respectively, as before. An important consequence of this is that the root locus is smooth as it crosses $k=0$ at a pole. So the positive and negative loci connect together at the open-loop poles, and do so without any sharp corners.

Sketching a negative locus is similar to sketching a positive locus, and we can follow the same steps as before, keeping in mind the differences in the properties above.

More zeros than poles¶

Although physical systems always have $n\geq m$ (more poles than zeros), one can still think about what would happen if we had more zeros than poles. In this case, there is a simple trick we can use; just rewrite the characteristic equation as

1+k\cdot G(s)=0 \quad\implies\quad 1+\frac{1}{k}\cdot \frac{1}{G(s)}=0.

(6)

So instead of analyzing the root locus of $G(s)$ with gain $k$ , we can analyze the root locus of $\frac{1}{G(s)}$ with gain $\frac{1}{k}$ . Now $\frac{1}{G(s)}$ has more poles than zeros, so we can apply all the same properties as before. Practically speaking, we sketch as follows:

Sketch the root locus exactly as you would if the poles and zeros were reversed.
Flip the directions of the arrows on the branches, so they still start at the open-loop poles and end at the open-loop zeros.
For the asymptotes, imagine there are “poles at infinity”. So instead of the infinite branches starting at poles and heading to infinity along asymptotes, they arrive from infinity along asymptotes and end at zeros.

Exotic breakaway/break-in points¶

We originally discussed what happens if two branches meet at a point on the real axis and then break away or break-in. There are a couple additional cases that can occur and are worth knowing about, but they are quite rare.

Multiple branches colliding¶

It can happen that three or more branches collide at the same point on the real axis. In such cases, they always collide and equal angles and break away at equal angles. This occurs in a manner analogous to how the positive locus and negative locus have asymptotes that are offset from each other. Here are two examples of such cases.

A couple comments about such loci:

Fragility. These sorts of loci are fragile in the sense that small perturbations to the pole or zeros will cause the break-in/breakaway points to split apart. This is why they do not tend to occur in practice.
Nomenclature. When multiple branches collide, it doesn’t make sense to call them “break-in” or “breakaway” points, since some of the branches start on the real axis and some end on the real axis. There is no standard name for such points, so I will just call them “multiple collision points”.

Collisions outside the real axis¶

It can also happen that two branches collide at a point that is not on the real axis. In such cases, they break away at equal angles as well. Here are two examples of such cases.

Figure 3:Root loci with branches colliding outside the real axis.

Some additional comments about the loci above:

For the locus on the left, there is nothing special about the imaginary axis. If we had translated all poles to the left by the same amount, the collision point would also have translated to the left by the same amount and would no longer be on the imaginary axis.
For the locus on the right, the complex collision point involves branches breaking in/out at equal angles, even though the angles themselves are not the usual multiples of $\frac{\pi}{2}$ we come to expect for real-axis collisions.
We can solve for any such collision point similarly to how we find real-axis break-in/breakaway points. We look for $s\in\C$ such that $G'(s)=0$ and $s$ belongs to the root locus. Points on the locus satisfy $1+kG(s)=0$ , i.e., $G(s) = -1/k$ . So $s\in\C$ belongs to the locus if $\Im(G(s))=0$ ( $k\in\R$ ) and $\Re(G(s))<0$ ( $k>0$ ).

Single-parameter locus¶

Root locus plots show us how closed-loop poles vary as we change the proportional gain $k$ . However, we can use the same idea to plot how poles vary as we change any single parameter. All we have to do is rewrite the characteristic equation in the form $1+k G(s)=0$ , where $k$ is the parameter we want to vary.

Example: variable damper¶

Consider a spring-mass-damper system

m \ddot x + b \dot x + k x = u

(7)

Suppose $m=1$ and $k=1$ are fixed, and we want to understand how the poles of the system vary as we change the damping coefficient $b$ . The characteristic equation is

s^2 + b s + 1 = 0.

(8)

Since $b$ is our parameter of interest, we can rewrite this as

1 + b\cdot \left( \frac{s}{s^2 + 1} \right) = 0.

(9)

So we can plot the root locus of $\frac{s}{s^2 + 1}$ to understand how the poles vary as we change $b>0$ . The root locus is shown below.

Parameter-locus of poles for a spring-mass-damper system as we vary the damping coefficient b. — Figure 4:Parameter-locus of poles for a spring-mass-damper system as we vary the damping coefficient $b$ .

Some observations:

When $b=0$ , the poles are at $\pm j$ (undamped oscillation). This makes sense since the system is just a spring-mass system with no damper.
As $b$ increases, the poles move along the root locus. Since $s^2+bs+1=0$ , changing $b$ is just changing $\zeta$ while keeping $\omega_n$ fixed. So the poles move along a circle of radius $\omega_n=1$ , as shown in the figure.
The break-in point satisfies $G'(s)=0$ , which yields $s=-1$ . Using the magnitude condition, this corresponds to $k = \frac{1}{|G(-1)|} = \frac{1}{|-1/2|} = 2$ . So the break-in point occurs at $b=2$ . We can confirm by plugging $b=2$ into the characteristic equation: $s^2 + 2s + 1 = (s+1)^2$ . As anticipated, we have a double root at $s=-1$ .

Integral gain design¶

Consider the system $G(s) = \frac{1}{s(s+2)}$ again. We design a PI controller for this system, so $C(s) = k_p + k_i/s$ . How do the closed-loop poles vary as we change $k_i$ while keeping $k_p$ fixed? The closed-loop poles satisfy the characteristic equation

1 + C(s)G(s) = 0 \quad\implies\quad 1 + \left(k_p + \frac{k_i}{s}\right)\cdot \frac{1}{s(s+2)} = 0.

(10)

We can rearrange this as:

\begin{aligned} && \left(1+\frac{k_p}{s(s+2)}\right) + k_i \cdot \frac{1}{s^2(s+2)} &= 0 \\ \implies && 1 + k_i \cdot \frac{\frac{1}{s^2(s+2)}}{1+\frac{k_p}{s(s+2)}} &= 0 \\ \implies && 1 + k_i \cdot \frac{1}{s^2(s+2)+k_ps} &= 0 \\ \implies && 1 + k_i \cdot \frac{1}{s(s^2+2s+k_p)} &= 0. \end{aligned}

(11)

We can explore what happens as we vary $k_i>0$ by plotting the root locus of $\frac{1}{s(s^2+2s+k_p)}$ . The root locus is shown below for $k_p=2$ .

Root locus of the closed-loop poles for a system with a PI controller as we vary the integral gain k_i while keeping a fixed proportional gain k_p=2. — Figure 5:Root locus of the closed-loop poles for a system with a PI controller as we vary the integral gain $k_i$ while keeping a fixed proportional gain $k_p=2$ .

No matter what $k_p$ is, the root locus for $k_i>0$ will have three poles and no zeros. Therefore, there will be three asymptotes, two of which will point to the right and therefore eventually cross into the RHP (instability). This confirms what we know about integral gains; making them too large can lead to instability!

Root locus design¶

Root locus plots are not just a tool for analysis; they can also be used for control design. The usual design workflow is as follows:

Root locus design workflow

Identify design goals. Identify design goals in terms of desired closed-loop pole locations (or regions in the complex plane). For example, we might want a certain damping ratio or settling time for our dominant poles. Or we might want zero steady-state error to a certain test input such as a step.
Try P-control. Start by plotting the root locus of $G(s)$ . This tells us how the system will perform if we use P-control. If the root locus passes through our desired pole locations, then we can just pick a gain $k$ that gives us the desired closed-loop poles and we’re done.
Add poles/zeros if necessary. If the root locus does not pass through our desired pole locations or we need to change the system type to improve steady-state error, then we need to add poles and/or zeros to our compensator $C(s)$ . Use your root locus knowledge to decide what to add and where (approximately).
Plot new root locus. Plot the new root locus with the added poles/zeros; i.e., plot the root locus of $G(s)C(s)$ and check if it passes through the desired pole locations. Keep iterating until the locus is satisfactory.
Calculate gain. The root locus tells us that there is a gain $k$ such that the compensator $k C(s)$ gives us the desired closed-loop poles. Use the magnitude condition to calculate this gain.

The key to using root locus for design is to separate the problem into two parts: (1) shaping the root locus by adding poles and zeros, and (2) picking a gain $k$ to achieve the desired closed-loop poles.

Example: PI control design¶

For example, if we want to design a PI controller for a plant $G(s)$ , we shouldn’t think of this as “picking $k_p$ and $k_i$ ”. Instead, we should rewrite the PI compensator as:

\begin{aligned} C(s) &= k_p + \frac{k_i}{s} \\ &= k_p \cdot \left( 1 + \frac{k_i/k_p}{s} \right) \\ &= k_p \cdot \left( \frac{s + k_i/k_p}{s} \right) \\ &= k_p \cdot \left( \frac{s+a}{s} \right), \end{aligned}

(12)

where $a = k_i/k_p$ . Therefore, our characteristic equation is:

1 + \left( k_p + \frac{k_i}{s} \right) G(s) = 0 \quad\implies\quad 1 + k_p \cdot \left( \frac{s+a}{s} \right) G(s) = 0.

(13)

We proceed in two phases:

Shape the root locus. First, we pick $a$ to shape the root locus of $\frac{s+a}{s} G(s)$ so that it passes through our desired closed-loop pole locations. This is the “compensator design” phase and in this case, it amounts to placing a zero somewhere on the real axis. For example, we could use it to cancel a LHP pole.
Pick the gain. Once we have the desired shape of the root locus, we can pick $k_p$ to achieve the desired closed-loop poles. This is the “gain design” phase.
Calculate $k_i$ . Finally, we can calculate $k_i$ using the relationship $a = k_i/k_p$ , and we’re done!

Test your knowledge¶

Let’s return to the interactive root locus demo:

Use the demo to test your understanding of the properties of root locus plots we saw in this section. Specifically:

Negative locus. Reveal the negative locus and verify the real axis, asymptote, and continuity properties for the negative locus.
More zeros than poles. Test your skills at sketching root loci with some examples that contain more poles than zeros by hiding the locus, placing the poles and zeros, and then revealing the locus again to check your work.
Exotic break-in/breakaway points. Reproduce the examples from Figure 2 and Figure 3 and try moving the poles and zeros slightly to verify the fragility of such loci.