Solved Example of Root Locus
Draw the root locus for the open loop transfer function of unity feedback control system given below
G (s) H(s) = K / s (s+1) (s+3)
SOLUTION
STEP 1 : Open loop Poles and Zeros
s (s+1) (s+3) = 0
s= 0 ,-1 , -3 (poles)
There is no zeros
STEP 2 : Number of branches of Root Locus , N
N = P = 3
STEP 3 : Existence of root locus on real axis , as per the rule the existence of root locus on real axis is (-∞ to -3 ) , (-1 to 0)
STEP 4 : Break away Point
CE can be expressed as 1+ G(s) H(s) = 0
1 + k / s (s+1)(s+3) = 0
k = -s3 – 4s2 – 3s
dk / ds = – (3s2 + 8s + 3)
dk / ds = 0
=> -( 3s2 + 8s + 3 ) =0
s = -0.45 , -2.21 (neglect)
STEP 5 : Centroid point , x = [ 0+ (-1) + (-3) ] – 0 / (3 – 0)
x = – 1. 33
STEP 6 : Angle of Asymptotes
θ = ( 2 m + 1 ) 1800 / ( P -Z ) ; where m = 0 , 1 , 2 {here P-Z-1 =2}
θ = 600 , 1800 , 3000
STEP 7 : Intersection with imaginary axis by applying Routh Array Criteria
s3 |
1 |
3 |
s2 |
4 |
k |
s1 |
(12 – k )/ 4 |
0 |
s0 = 1 |
k |
0 |
12 – k / 4 =0
=> k = 12
auxiliary equation is 4s2 + k =0
=> 4s2 + 12 =0
=> s = 1.73 j , – 1.73 j
Since there is no complex poles and zeros of transfer function therefore there is no angle of departure and arrival .
Root Locus of given transfer function is as below
Construction of Root Locus
The Root Locus is the Technique to identify the roots of Characteristics Equations Within a Transfer Function. It Follows the Process of Plotting the roots on a graph, Showcasing their Variations across Different Parametric Values. In this article, we will be going through What is Root Locus?, Angle and Magnitude Conditions, Construction Rules of Root Locus, and At last we will solve the Examples.
Table of Content
- What is Root Locus?
- Angle and Magnitude Condition
- Construction Rules of Root Locus
- Effects of adding Open Loop Poles and Zeros on Root Locus
- Solved Example
- Application
- FAQs on Root Locus in Control System
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