Root Locus Steps and Procedure

1. Gyanmanjari Institute of Technology
Department of Electrical Engineering
Subject:- Controll System Engineering (2150909)
Topic: Root - Locus Plot steps and procedure
Name Enrollment No.
Jay Makwana 151290109027
Dhruv Pandya 151290109032
Root Locus StepsAnd Procedure1
Department of Electrical Engineering
Subject:- Controll System Engineering (2150909)
Topic: Root - Locus Plot steps and procedure
Name Enrollment No.
Jay Makwana 151290109027
Dhruv Pandya 151290109032

2. Root – Locus Plot
Determination of The Stability of a System
Root Locus StepsAnd Procedure2

3. Index
1. Introduction
2. Steps to find Root-locus
3. Example
a. Pole zero
b. centroid
c.Asymptotes
d. Break away point
e. Intersection point
f.Angle of Departure
4. Root Locus On Matlab
Root Locus StepsAnd Procedure3
1. Introduction
2. Steps to find Root-locus
3. Example
a. Pole zero
b. centroid
c.Asymptotes
d. Break away point
e. Intersection point
f.Angle of Departure
4. Root Locus On Matlab

4. 1. Introduction to Root Locus:
 The Stability of a given closed loop system depends upon the location of
the roots of the characteristics equation, which is the location of the closed
loop poles. If we change some parameter of a system, then the location of
closed loop pole changes in 's' plane.
 This movement of poles in 's' plane is called as 'Root Locus'.
 Root Locus is a simple graphical method for determining the roots of the
characteristic equation which was invented by W.R. Evans in 1948. It can
be drawn by varying the parameter (usually gain of the system) from zero
to infinity.
Root Locus StepsAnd Procedure4
 The Stability of a given closed loop system depends upon the location of
the roots of the characteristics equation, which is the location of the closed
loop poles. If we change some parameter of a system, then the location of
closed loop pole changes in 's' plane.
 This movement of poles in 's' plane is called as 'Root Locus'.
 Root Locus is a simple graphical method for determining the roots of the
characteristic equation which was invented by W.R. Evans in 1948. It can
be drawn by varying the parameter (usually gain of the system) from zero
to infinity.

5. 2. General steps for drawing the Root
Locus of the given system:
1. Determine the open loop poles, zeros and a number of branches from
given G(s)H(s).
2. Draw the pole-zero plot and determine the region of real axis for which
the root locus exists.Also, determine the number of breakaway points
(This will be explained while solving the problems).
3. Calculate the angle of asymptote.
4. Determine the centroid.
5. Calculate the breakaway points (if any).
6. Calculate the intersection point of root locus with the imaginary axis.
7. Calculate the angle of departure or angle of arrivals if any.
8. From above steps draw the overall sketch of the root locus.
9. Predict the stability and performance of the given system by the root
locus.
Root Locus StepsAnd Procedure5
1. Determine the open loop poles, zeros and a number of branches from
given G(s)H(s).
2. Draw the pole-zero plot and determine the region of real axis for which
the root locus exists.Also, determine the number of breakaway points
(This will be explained while solving the problems).
3. Calculate the angle of asymptote.
4. Determine the centroid.
5. Calculate the breakaway points (if any).
6. Calculate the intersection point of root locus with the imaginary axis.
7. Calculate the angle of departure or angle of arrivals if any.
8. From above steps draw the overall sketch of the root locus.
9. Predict the stability and performance of the given system by the root
locus.

6. 3. Let us learn the Root Locus method by
solving a problem as given below:
A feedback control system has an open loop transfer
function,
Find the root locus as K varies from zero to infinity
   
)22)(1( 2


ssss
K
sHsG
Root Locus StepsAnd Procedure6
A feedback control system has an open loop transfer
function,
Find the root locus as K varies from zero to infinity
   
)22)(1( 2


ssss
K
sHsG

7. From the numerator, there is no 0’s term present, So, number of zeros
(z) = 0
From the denominator, equating it to zero we get,
:. s = 0, -3,
   
)22)(1( 2


ssss
K
sHsG
Poles And Zeroes:
Root Locus StepsAnd Procedure7
From the numerator, there is no 0’s term present, So, number of zeros
(z) = 0
From the denominator, equating it to zero we get,
:. s = 0, -3, j1- 

8. centroid
Centroid:
25.1
4
5
0-4
0-(-1)+(-1)+(-3)+0

Root Locus StepsAnd Procedure8

9. Asymptotes:
Angle of asymptotes(Өp)=
Where , q = 0,1,2 … (p-z-1)
So here, q = 0,1,2,3
z-p
1)180(2q 0

Root Locus StepsAnd Procedure9
Angle of asymptotes(Өp)=
Where , q = 0,1,2 … (p-z-1)
So here, q = 0,1,2,3

10. zp
q



0
180)12(

0
3
0
2
0
1
0
0
45
4
7180
135
4
5180
135
4
3180
45
4
180














0
0
45
0
135
j1
0
3
0
2
0
1
0
0
45
4
7180
135
4
5180
135
4
3180
45
4
180














0
-1-2-3
Root Locus StepsAnd Procedure10
j1

11. Breakaway I Break in point:
The characteristic equation of the transfer
function is given by,
0616s15s4s
0
ds
dK
6)16s15s-(4s
ds
dK
6s)8s5s(s-K
0K6s8s5ss
0K2)2s3)(ss(s
23
23
234
234
2






Root Locus StepsAnd Procedure11
0616s15s4s
0
ds
dK
6)16s15s-(4s
ds
dK
6s)8s5s(s-K
0K6s8s5ss
0K2)2s3)(ss(s
23
23
234
234
2






-2.28s 

12. Intersection Point:
From the characteristic equation of the transfer function,
0K6s8s5ss 234

Ks
Ks
Ks
s
Ks
0
1
2
3
4
073.06
08.6
065
81

js
s
s
Ks
09.1
8.6
2.8
02.88.6
08.6
2
2
2





Routh Array,
Root Locus StepsAnd Procedure12
Ks
Ks
Ks
s
Ks
0
1
2
3
4
073.06
08.6
065
81

js
s
s
Ks
09.1
8.6
2.8
02.88.6
08.6
2
2
2






13. Angle of Departure/Arrival:
Angle of departure, w.r.t p2
431
0
0
180
pppp
z
zpd






0
2p
1p4p
Root Locus StepsAnd Procedure13
431
0
0
180
pppp
z
zpd






0
-1-2-3
3p

14. From the figure,
000
0000
01
4
0
3
010
1
56.7156.251180
56.25156.2690135
56.26
2
1
tan
90
1351tan180













d
p
p
p
p





0
2p
1p4p
Root Locus StepsAnd Procedure14
000
0000
01
4
0
3
010
1
56.7156.251180
56.25156.2690135
56.26
2
1
tan
90
1351tan180













d
p
p
p
p





0
-1-2-3
3p

15. 0
1.09j
-2.28
j1
-3 -2 -1 0
-1.28
-1.09j
-2.28
Root Locus StepsAnd Procedure15
j1

16. 4. Matlab Program:
n=input('enter numerator ')
d=input('enter denominator')
g=tf(n,d)
rlocus(g)
Root Locus StepsAnd Procedure16
n=input('enter numerator ')
d=input('enter denominator')
g=tf(n,d)
rlocus(g)

17. Root Locus StepsAnd Procedure17

18. Thank YouThank You
Root Locus StepsAnd Procedure18