Title

Root Locus Rules

We will assume that we are dealing with a closed loop system with this block diagram structure.

This closed loop system will have a transfer function. The transfer function is the ratio of the output to the input - as given below.

Y(s)/U(s) = KG(s)/[1 + KG(s)]

The closed loop poles are solutions of the characteristic equation:

1 + KG(s) = 0

Now, let's see how we can used information about the closed loop transfer function to figure out some facets of root locus behavior. Root locus behavior can be summarized in several rules. We'll present the rules and then discuss why each rule is true. And, we'll try to give an amusing paraphrase of each of those rules. Possibly you will remember them better with that.

Rule 0: The Root Locus has as many branches (closed loop poles migrating in the s-plane) as there are open loop poles."It was the best of times, it was the worst of times, it was the age of wisdom, it was the age of foolishness, it was an open loop pole, it was a closed loop pole", (falsely attributed to Charles Dickens)

The characteristic equation for the system is shown below. The transfer function, G(s) is assumed to be written as a ratio of polynomials. We will denote the numerator polynomial as N(s), and the denominator polynomial as D(s), and we use that convention below.

1 + KG(s) = 0

1 + KN(s)/D(s) = 0

D(s) + KN(s) = 0

Almost always the polynomial D(s) is of higher degree than N(s). Now consider this.

If D(s) is of higher degree than N(s), the degree of the closed loop system is equal to the degree of the open loop system. Call that degree N.
In other words, the closed loop system has as many closed loop poles (N) as it has open loop poles (N).
If there are N closed loop poles, and the root locus plot is a plot of where the closed loop poles go as the gain is increased, then there are N "branches" to the root locus. A branch is just the trace one pole makes as the gain, K, is adjusted starting at zero, and increasing indefinitely.

Rule 1:The locus of 1 + KG(s) = 0 begins at the open loop poles (of G(s)H(s)) for K=0, and ends at the open loop zeroes (of G(s)H(s)) as K goes becomes infinite. Yea, and each pole shall leave its open loop pole parent and cling to the open loop zero unless there be not enough zeroes, and then the closed loop pole will wander in the wasteland forever.

To see how this rule comes about, consider the characteristic equation again.

1 + KG(s) = 0

The transfer function, G(s) is assumed to be written as a ratio of polynomials. We will denote the numerator polynomial as N(s), and the denominator polynomial as D(s). Then, we will have:

1 + KG(s) = 0

1 + KN(s)/D(s) = 0

D(s) + KN(s) = 0

All of this is just rewriting the characteristic equation. However, in the last form above, you can see that when the gain, K, is really small, that the only thing left would be D(s). That allows us to conclude that the root locus starts on the open loop poles. Here is the train of logic.

Closed loop poles are solutions of the characteristic equation, D(s) + KN(s) = 0.
For very small gain values (K very small), closed loop poles will then be solutions of D(s) = 0.
Roots of D(s) (i.e. places where D(s) = 0) are poles of G(s) because D(s) is the denominator of G(s).
So, for very small values of K, the closed loop poles are the open loop poles, i.e., the poles of G(s).
As we increase the gain, K, the poles are continuous functions of K, so they move away from the open loop poles as the gain, K, is increases from zero.

And, that gives us the first part of Rule #1.

To get the second part of Rule #1, we need to consider what happens when we make the gain, K, very large. Let's go back to the characteristic equation in the form we were using.

D(s) + KN(s) = 0

Now, imagine what happens when the gain, K, becomes large. In that case, the only term in the equation is KN(s), and once again, we can trace out the logic.

To satisfy the characteristic equation we have to make KN(s) equal to zero.
To make KN(s) equal to zero, we have to make N(s) equal to zero.
In other words, we have to choose a value for s that is equal to one of the zeroes of N(s).
Zeroes of N(s) are zeroes of G(s) since N(s) is the numerator of G(s).

That almost sums up Rule #1, but there is a problem. To define the problem, consider an open loop system that has five poles and three zeroe. Here is what we know about the root locus for this system.

The root locus has five branches because the open loop system has five poles. (Rule 0)
The five branches start on the five open loop poles. (Rule 1)
Branches end on the open loop zeroes. (Rule 1)

The problem is that we know that branches end on the open loop zeroes, but we haven't discussed how many branches end on open loop zeroes. In the example above, what is really knowl is summarized in the following (corrected) analysis.

The root locus has five branches because the open loop system has five poles. (Rule 0)
The five branches start on the five open loop poles. (Rule 1)
Three branches end on the open loop zeroes. (Rule 1).
Two branches go to infinity.

That will be further explained in Rule 3, but first, there is another rule we need to examine.

Rule 2:The Root Locus lies on the real axis to the left of an odd number of singularities - poles and zeroes. Sinister survival! - Actually, sinister means "left", and left handed people were once considered "sinister". Be careful how you use the word "sinister" from now on.

To see why this rule is true, consider the characteristic equation again, and rewrite it. We can rewrite the characteristic equation as:

KG(s)H(s) = -1 = -1 + j0 = e^j^p

There are really two equations here since -1 can be viewed as a complex number. That means the real part and the imaginary part of the equation both have to be satisfied. (That's the reason for the exponential on the RHS of the last equation above. Instead of a real part and an imaginary part, we can think in terms of a magnitude and an angle (And, that's the real reason for the exponential!). We will focus on the idea of the angle. We have to have:

Angle(KG(s)) = -180^o

Actually, any odd multiple of -180^o will work, so the angle can also be -540^o, +180^o, etc. But, the angle has to be an odd multiple of 180 degrees because that's the angle of -1 when -1 is expressed in angular form.

Here is an example of a root locus. In G(s) there are poles at s = 0, -1, -2 and -5. There is a zero at s = -3. On the plot, we have shown an arbitrary point, s, between 0 and -1. That point is shown with a purple dot.

Now, examine the transfer function, KG(s).

KG(s) = K(s + 3)/[(s(s + 1)(s + 2)(s + 5)]

Every factor in the transfer function has an angle, and the total angle is the sum of the angles of all the factors. (The factors in the denominator contribute negative angles, but that won't be an issue here.) If you have a term like (s + 2), then you can think of that as (s - (-2)). In turn, you can think of that as a vector quantity that starts at -2 and goes to s. Here's that plot, expanded with that vector shown.

For this point, s, the angle is zero degrees! And, you can see that the angle will be zero degrees for every factor to the left of the point, s, that you choose (on the real axis). So, if you choose a point, s, on the real axis anywhere, points to the left will contribute zero angle to KG(s). Only points to the right of the point s contribute angle.

Now, let's look at the angle contributed by the pole at s = 0.

In this case, the arrow from s = 0 (i.e. the origin in the s-plane!) to the point, s, (indicated by the purple dot) is -180^o. So, for the point we have chosen, the total angle is -180^o and because of that we know that the point we have chosen (the purple dot) is going to be on the root locus. In other words, there is going to be some gain value that will produce a closed looop pole at the point we have chosen. Now, we don't know what gain will put a closed loop pole at that point, nor do we have any idea where the other closed loop poles are going to be, but we know we wil have some gain that will produce a pole there.

Anyhow, looking at the big picture, the plot below shows where the poles

Rule 3: The Root Locus goes to infinity at angles of (2k+1)180^o/(#P-#Z) for a sequence of integer k's. Gentlemen, please go to your corners.

Note that #P is the number of open loop poles and #Z is the number of open loop zeroes.

The explanation for this rule takes a different route - although is also starts with the characteristic equation. In Rule #1, we encountered the characteristic equation in this form.

D(s) + KN(s) = 0

And, we have to remember that we know that D(s) is almost always of a higher degree than N(s).

Each factor in the transfer function is of the form (s - p_n) or (s - z_n). Each such factor can be considered to be a vector (complex number) that starts at p_n or z_n and ends at the point s. The drawing shows an s between two real poles.

The total angle for G(s) is the sum of the angles from all poles and zeroes to the point s. Remember to take into account the fact that poles - being in the denominator - contribute negatively to the total angle.

v For the s shown as a blue dot, the total angle is -180^o.

v For the s shown as a green dot, the total angle is 0^o.

We can generalize this result. You should be able to see that it does not matter whether the singularity is a pole or a zero. If the singularity is to the right of the point you are examining, it will contribute 180^o to the angle. If the singularity is to the left, it will contribute nothing. Thus, we can conclude the rule:
This section explains where one of the rules comes from, but it is not yet complete. For the rest of the rules derivations consult one of the many excellent texts available in this area.