Example of Root Locus
The following is the example of root locus-
Draw the root locus diagram for a closed loop system whose loop transfer function is given by
G(s)H(s) = K/s(s + 3)(s + 6)
Solution:
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, -3, and -6
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: 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 = 3×180/3 = 180 degrees
For q = 2,
Angle = 5×180/3 = 300 degrees
Step 3: Centroid
The centroid is given by:
σ = Σ Real part of poles of G(S)H(S) – Σ Real part of Zeros of G(S)H(S)/P-Z
= 0 – 3 – 6 – 0/3
= -9/3
= -3
Thus, the centroid of the root locus is at -3 on the real axis.
Step 4: Breakaway point
We know that the breakaway point will lie between 0 and -3. 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 + 3)(s + 6) = 0
s(s + 3)(s + 6) + K = 0
s(s^2 + 15^s + 50) + K = 0
s^3 + 9s^2 + 18^s + K = 0
K = – s^3 – 9s^2 – 18^s
Differentiating both sides,
Dk/ds = – (3s^2 + 18^s + 18) = 0
3s^2 + 18^s + 18 = 0
Dividing the equation by 3, we get:
s^2 + 6^s + 6 = 0
Now, we will find the roots of the given equation by using the formula:
-b+-√(b^2-4ac)/2ac
Using the value, a = 1, b = 6, and c = 6
The roots of the equation will be -1.268 and -4.732.
Among the two roots, only -1.268 lie between 0 and -3. Hence, it will be the breakaway point.
Let’s verify by putting the value of the root in the equation K = – s^3 – 9s^2 – 18^s.
K = – (-1.268)^3 – 9(-1.268)^2 – 18(-1.268)
K = 9.374.
The value of K is found to be positive. Thus, it is a valid breakaway point.
Step 5: 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 + 9s^2 + 18^s + K = 0
The Roth table is shown below:
|
s^3 |
1 |
18 |
|---|---|---|
|
s^2 |
9 |
k |
|
s |
9*18 – 1k/k = 162 – k/k |
0 |
|
s^0 |
k |
From the third row s, 750 – K/K = 0
162 – K = 0
K = 162
From the second row s2,
9 s^2 + K = 0
Putting the value of Kin the above equation, we get:
9 s^2 = -162
s2 = -162/9
s2 = -18
s = j4.242 and -j4.242
Both the point lies on the positive and negative imaginary axis.
Step 6: There are no complex poles present in the given transfer function. Hence, the angle of departure is not required.
Step 7: Combining all the above steps.
The root locus thus formed after combining all the above steps is shown below:
construction of root locus
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 Routh Hurwitz criteria discussed above.
The system is stable if the value of K lies between 0 and 162. The root locus at such a value of K is in the left half of the s-plane. For a value greater than 162, the system becomes unstable, and it is because the roots start moving towards the right half of the s-plane. But, at K = 162, 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 s-plane.
Control Systems – Root Locus
The root locus is a procedure utilized in charge framework examination and plan. It centers around figuring out how the roots (or posts) of the trademark condition of a control framework change as a particular boundary, frequently the control gain, is changed. This graphical technique is especially useful in deciding the soundness and transient reaction of the framework.