Table of contents  
Rule Number 1  
Rule Number 2  
Rule Number 3  
Rule Number 4  
Rule Number 5  
Rule Number 6  
Key Points of Sketching Root Locus  
Let's Discuss an Example and use these Rules to Solve it 
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The root locus is a graphical representation in sdomain and it is symmetrical about the real axis. Because the open loop poles and zeros exist in the sdomain having the values either as real or as complex conjugate pairs. In this document, let us discuss Rules to construct (draw) the root locus.
Root Locus
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:
Thus, the line marked with orange depicts the sections where the root locus exists. It is shown below:
Centroid is a point where the asymptotes intersect at a common point on the real axis. It can be calculated as:
Note: The value of the centroid is always real, which can be positive or negative. It can be a part of root locus or sometimes not.
Or
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.
Example: Draw the root locus diagram for a closed loop system whose loop transfer function is given by:
G(s)H(s) = K/s(s + 5)(s + 10)
Also find if the system is stable or not.
Solution: We will follow the procedure according to the steps discussed above.
Step 1: Finding the poles, zeroes, and branches.
The denominator of the given transfer function signifies the poles and the numerator signifies the zeroes. Hence, there are 3 poles and no zeroes.
Poles = 0, 5, and 10
Zeroes = No zero
P  Z = 3  0 = 3
There are three branches (P  Z) approaching to infinity and there are no open loop zeroes. Hence infinity will be the terminating point of the root locus.
Step 2: Section of the real axis where the root locus lies.
There are three poles, which are shown below:
Step 3: Angle of asymptotes.
Angle of such asymptotes is given by:
= (2q + 1)180 / P  Z
q = 0, 1, and 2
For q = 0,
Angle = 180/3 = 60 degrees
For q = 1,
Angle = 3x180/3 = 180 degrees
For q = 2,
Angle = 5x180/3 = 300 degrees
Step 4: Centroid
The centroid is given by:
= 0  5  10  0/3
= 15/3
= 5
Thus, the centroid of the root locus is at 5 on the real axis.
The plot showing the centroid and the angle of asymptotes is given below:
Step 5: Breakaway Point
We know that the breakaway point will lie between 0 and 5. Let's find the valid breakaway point.
1 + G(s)H(s) = 0
Putting the value of the given transfer function in the above equation, we get:
1 + K/s(s + 5)(s + 10) = 0
s(s + 5)(s + 10) + K = 0
s(s^{2} + 15s + 50) + K = 0
s^{3} + 15s^{2} + 50s + K = 0
K =  s^{3}  15s^{2}  50s
Differentiating both sides,
Dk/ds =  (3s^{2} + 30s + 50) = 0
3s^{2} + 30s + 50 = 0
Dividing the equation by 3, we get:
s^{2} + 10s + 16.667 = 0
Now, we will find the roots of the given equation by using the formula:
Using the value, a = 1, b = 10, and c = 16.667
The roots of the equation will be 2.113 and 7.88.
Among the two roots, only 2.113 lie between 0 and 5. Hence, it will be the breakaway point.
Let's verify by putting the value of the root in the equation K =  s^{3}  15s^{2}  50s.
K =  2.113 ^{3}  15(2.113)^{2}  50(2.113)
K = 48.112
The value of K is found to be positive. Thus, it is a valid breakaway point.
Step 6: Intersection with the negative real axis.
Here, we will found the intersection points of the root locus on the imaginary axis using the Routh Hurwitz criteria using the equation s^{3} + 15s^{2} + 50s + K = 0
The Routh table is shown below:
From the third rows, 750  K/K = 0
750  K = 0
K = 750
From the second row s^{2},
15 s^{2} + K = 0
Putting the value of Kin the above equation, we get:
15 s^{2} = 750
s^{2} = 750/15
s^{2} = 50
s = j7.071 and j7.071
Both the point lies on the positive and negative imaginary axis.
Step 7: There are no complex poles present in the given transfer function. Hence, the angle of departure is not required.
Step 8: Combining all the above steps.
The root locus thus formed after combining all the above steps is shown below:
Step 9: Stability of the system
The system can be stable, marginally stable, or unstable. Here, we will determine the system's stability for different values of K based on the Roth Hurwitz criteria discussed above.
The system is stable if the value of K lies between 0 and 750. The root locus at such a value of K is in the left half of the splane. For a value greater than 750, the system becomes unstable, and it is because the roots start moving towards the right half of the splane. But, at K = 750, the system is marginally stable.
We can conclude that stability is based on the location of roots in the left half or right half of the splane.
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