Table of contents 
Uses of Root Locus 
Advantages of Root Locus 
Angle and Magnitude Condition of Root Locus 
Graphical Method to determine the value of 'K' 
Rules of Root Locus 
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We have studied the stability of a system, it depends on the location of the roots of the characteristic equation also, the stability of the system depends on the location of closedloop poles also.
The root locus method was introduced by W.R Evans in 1948. Root locus is a graphical method in which the movement of poles in the splane can be located when a specific parameter is varied from 0 to infinity. The parameter assumed to be varied is generally the gain of the system.
Consider the below closed loop system.
The equation of a closed loop system is given by:
1 + G(s)H(s) = 0
Where,
In the case of root locus, the gain K is also assumed as part of the closedloop system. K is known as system gain or the gain in the forward path.
The characteristic equation after including the forward gain can be represented as:
1 + KG'(s)H(s) = 0
Where,
G(s) = KG'(s)
When the system includes the variable parameter K, the roots of the closed loop system are now dependent on the values of 'K.'
The value of 'K' variable can vary in two cases, as shown below:
In addition in determining the stability of the system, root locus also helps to determine:
The root locus's angle conditions help us determine whether the given point exists on the root locus branch or not. We can find the value of the system gain 'K' with the help of the magnitude condition on the root locus.
For a general closed loop system, the characteristic equation is given by:
1 + G(s)H(s) = 0
G(s)H(s) = 0  1
G(s)H(s) = 1
We know that the splane is complex. Hence, the above equation in terms of a complex variable can be written as:
G(s)H(s) = 1 + j0
The two conditions of the root locus are angle condition and magnitude condition. Let's start.
G(s)H(s) = 1 + j0
The above equation in terms of angle can be represented as:
Where,
q = 0, 1, 2 ...
The root locus plot of the point 1 + j0 is shown below:
Thus, we can define the angle condition as any value of the 's' for which the root of the equation (1 + G(s)H(s) = 0) is given by:
Uses of Angle Condition in Root Locus
The common use of the angle condition in root locus is to test any point in the splane. It is used to find if the given point exists or not.
The magnitude condition is calculated by equating both the sides of the characteristic equation, which is given by:
1 + G(s)H(s) = 0
G(s)H(s) = 1
G(s)H(s) = 1 + j0 = 1
The magnitude condition is given by:
G(s)H(s) = 1
Uses of Magnitude Condition
The value of K can be determined with the help of magnitude condition, if the point is known to be on the root locus verified by the angle condition.
Consider the below example.
Example: Test whether the point 2 + 5j is present on the root locus or not. Consider the system with G(s)H(s) = K/s(s + 4).
We will first verify using the angle condition.
Angle and Magnitude condition of root locus
G(s)H(s) = K/s(s + 4)
Put s = 2 + j5, we get:
= K +j0/ (2 + j5)(2 + j5 + 4)
= K +j0/ (2 + j5)(2 + j5)
Converting it into polar and considering angles, we get:
Angle G(s)H(s) = 0 /111.8 x 68.2 = 180 degrees
The resulted angle is the multiple of positive and negative value of 180 degrees. Thus, the angle condition is verified.
 The angle value of (2 + 5j) = 111.8 degrees
 The angle value of (2 + 5j) = 68.2 degrees
We can use any scientific calculator to convert from rectangular to polar.
Now, we know that the given point 2 + 5j exits on the root locus.
We can use the magnitude condition, which is given by:
G(s)H(s) = 1
Value of G(s)H(s) at s = 2 +5j will be:
K/5.3851 X 5.3851 = 1
After solving, K = 29
Thus, s=2 + 5j is one of the roots of the equation.
To determine the value of the system gain K, we need a prior knowledge of the points on the root locus, i.e., the location of the point known to be on the root locus.
The value of K is given by:
K = product of phasor lengths drawn from the open loop poles up to a point on root locus/ product of phasor lengths drawn from the open loop zeroes up to a point on root locus
Consider the below example.
Example: Find the value of system gain K for the system G(s)H(s) = K/s(s + 4)
Given that a point 2 + j5 is already present on the root locus plot.
We are given that a point 2 + j5 is confirmed on the root locus.
The system G(s)H(s) = K/s(s + 4) has two poles and no zeroes. It is because the numerator has no. values of s.
Equating the denominator equal to zero, we get:
s(s + 4) = 0
s = 0 and s = 4
Thus, the openloop poles are 0 and 4.
Now, we will join the points 0, 4, and 2 + j5, as shown below:
Angle and Magnitude condition of root locus
Length from s = 0 to the point = p
Length from s = 4 to the point = q
P = (2^{2} + 5^{2})^{1/2}
P = (29)^{1/2}
Similarly, q = (2^{2} + 5^{2})^{1/2}
q = (29)^{1/2}
We know,
K = product of phasor lengths drawn from the openloop poles up to a point on root locus/ product of phasor lengths drawn from the openloop zeroes up to a point on root locus
Since, there is no openloop zero, the value of the denominator will be assumed as unity.
K = p x q = 29
Here, we will discuss the six basic rules required to plot a root locus. We will also discuss an example that will help us to implement the rules easily.
There are two conditions where the poles can be greater than the number of zeroes, or the number of zeroes can be greater than the number of poles in the given characteristic equation.
Let the number of branches in the root locus is N. Both the cases arise when we plot the root locus. The default conditions are given for each case that helps in determining the number of branches terminating or approaching infinity.
Case 1: P > Z
For example,
Let P = 3, and Z = 1
Then,
The number of root locus branches = 3 = no. of poles
P  Z = 3  1 = 2
It means that 3 branches will start from the location of open loop poles
No. of branches terminating at the open loop zero location = 1
No. of branches approaching to infinity = P  Z = 2
Case 2: Z > P
For example,
Let P = 1, and Z = 3
Then,
The number of root locus branches = 3 = no. of zeroes
Z  P = 3  1 = 2
It means that 1 branch will start from the location of open loop poles
No. of branches terminating at the finite open loop pole location = 3 = all the root locus branches
No. of branches originating from infinity = Z  P = 2
For example,
Let poles be 2 and 4, and zeroes are 1 and 3. We need to find that points 2.2 and 3.4 lies on the root locus or not.
We know that 2.2 lies between 2 and 3, while point 3.4 lies between 3 and 4.
We know that 2.2 lie between 2 and 3, while point 3.4 lies between 3 and 4.
The area of the root locus between the points is shown below:
Here, orange line represents the area where root locus lies.
Let's consider an example for better understanding.
Example: G(s)H(s) = K(s + 1)(s + 4)/ s(s + 3)(s + 5). Find on which sections of the real axis, the root locus exists.
We know that the denominator signifies the poles, and the numerator signifies the zeroes. Thus, 0, 3, and 5 are the poles, and 1 and 4 are the zeroes per the given transfer function. It means that there are 3 poles and 2 zeroes.
These poles and zeroes on the real axis will appear as:
According to rule number 3,
 The sections between 0 and 1 (for example, point 0.4) contain only one pole and no zeroes on the righthand side. It means that the sum is odd (i.e. 1). So, it exists on the root locus.
 The sections between 1 and 3 (for example, point 2.1) contain only one pole and one zero on the righthand side. It means that the sum of poles and zeroes is even (i.e. 2). So, it does not exist on the root locus.
 The sections between 3 and 4 (for example, point 3.5) contain two poles and one zero on the righthand side. It means that the sum is odd (i.e. 3). So, it exists on the root locus.
 The sections between 4 and 3 (for example, point 4.3) contain two poles and two zeroes on the righthand side. It means that the sum of poles and zeroes is even (i.e. 4). So, it does not exist on the root locus.
 The sections greater than 5 (for example, point 8.6) contain three poles and two zeroes on the righthand side. It means that the sum is odd (i.e. 5). So, it exists on the root locus.
Thus, the line marked with orange depicts the sections where the root locus exists. It is shown below:
We have already discussed that (P  Z) provides the number of branches approaching infinity for the given transfer function. The information about such branches approaching infinity is defined under rule number 4, known as asymptotes.
The angle of such asymptotes is given by:
= (2q + 1)180 / P  Z
Where,
q = 0, 1, 2, 3, 4 ... (P  Z  1)
These are always symmetric about the real axis.
Rule number 4 describes the guidelines or information about the branches approaching infinity, known as the asymptotes. But, the angles are insufficient to plot the root locus, and the location of such branches in the splane is equally important, defined by rule number 5.
Centroid is a point where the asymptotes intersect at a common point on the real axis. It can be calculated as:
The last rule is the breakaway point. It is also a point on the root locus where multiple roots of the given equation occurs. It is calculated for a specific value of system gain K.
Or
It can be defined as a point on the root locus where two or more roots occur for a particular value of K.
The root locus branches always leave breakaway points at an angle of 180/n.
Where,
N = number of branches approaching at the breakaway point.
The value of the angle can be positive or negative.
Let's discuss some predictions about the existence of the breakaway points:
The above transfer function has two poles at 0 and 3. According to rule number 3, the point on the section between 0 and 3 (for example, point 2.2) has one pole and no zero on the righthand side. It signifies that the sum of zeroes and poles is 1, i.e. odd. Thus, the section between 0 and 3 exists of the root locus.
Hence, there must exist a minimum of one breakaway point in between them.
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