Usually the open loop gain (Let us say K) is varied. Based on the range of K. two types of root locus are defined.
K: 0 → ∞ Roots Locus Diagram
K: 0 → –∞ Complementary Root Locus Diagram
K: –∞ → ∞ Complete Root Locus
Roots locus uses the open loop transfer function to predict the closed loop stability.
We have transfer functions as
G(s)H(s) = k / (s + a) a > 0
Characteristic equation is q(s) = 1 + G(s)H(s) = 0
⇒ s + a + k = 0
S = (a + k)
Now, the root of closed loop equation is the function of a and system gain k. Now, if we vary the system gain, then
The variation of closed loop pole path with the variation of parameter k is given as
As seen, k is varied from 0 to ∞ the root tends to move left side of the splane which means gaining more stability. Thus the system is stable for k ϵ [0, ∞).
2. Root Locus for Second Order Open Loop Transfer FunctionSuppose G(s)H(s) = R / (s(s + a)a) > 0
The characteristic equation formed is
q(s) = 1 + G(s)H(s)
For q(s) = 0, 1 + (k / (s + a)) = 0
s^{2} + as + k = 0
The system of above equation is given as
Now k is varied from o to ∞ and the path of roots is traced.
For k = 0, s_{1, 2} = 0, a
For 0 < k < a^{2}/ 4,
For k = a^{2 }/ 4, S_{1, 2 }= a / 2, a / 2. The roots are real and equal.
For The roots are imaginary in this range
The variation of poles with variation in k for the given second order system is given as follows
For 0 < k < ∞, the poles are in negative splane and move away towards left side. Hence the system is stable.
In the root locus, lines start from poles and end at zeroes.
Now, let the number of zeroes be ‘z’ and number of poles be ‘p’
If z < p, then as ‘s’ tends to ∞, the value of transfer function becomes zero. Thus we can say that there are zeros at infinity and order of zeros at infinity is p–z. This means p–z lines will go towards infinity.
If p < z, then as ‘s’ tends to ∞, the value of transfer function becomes infinity. Hence we can say that there are poles at infinity and order of poles at infinity is z–p. It means z–p lines will come from infinity.
Now, if we have to construct root locus from a given transfer function, there are some rules that need to be followed
Example: We have a system with open loop gain G(s) = and feedback H(s) = 1 Sketch the root locus of the system and comment on the stability of the system.
Solution: We have G(s)H(s) =
Number of zeros =1 (–1)
Number of poles = 3 (0, 0, –4)
Number of asymptotes = p – z = 3 – 1 = 2.
∴ Angle of asymptotes = (2k + 1)180° / 2 for k = 0, 1
ϕA = 90°, 270°
Centroid,
For break  away point, put dk / ds = 0
Now, characteristic equation, 1 + G(s)H(s) = 0
⇒ 3s^{3} + 3s^{2} + 8s^{2} + 8s + s^{3} + 4s^{2} = 0
4s^{3} + 15s^{2} + 8s =0
s(4s^{2} + 15s + 8) =0
s = 0, (15 / 4) ± 2.462j (Calculate it yourself)
Since there are no complex poles in transfer function, hence there cannot be complex breakaway points. Hence s = 0 is the breakaway point Now, to find the point of intersection with jω axis, s^{3} + 4s^{2} + ks + k = 0 (From characteristic equation) Forming the Routh array, we get
For marginal stability, k = 0
Auxiliary equation is 4s^{2} + k = 0
s^{2 }= 0 s_{1, 2} = 0
Hence the root locus will not cut jω axis from the given information, it is constructed as follows
Since the root locus remains in left half, the system will be stable except for when it is marginally stable at origin point.
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