Root Locus

Root Locus Steps and Procedure

1. GyanmanjariInstitute of Technology
Department of Electrical Engineering
Subject:-ControllSystem Engineering (2150909)
Topic: Root-Locus Plot steps and procedure
Name Enrollment No.
JayMakwana 151290109027
DhruvPandya 151290109032
Root Locus Steps And Procedure
1
Department of Electrical Engineering
Subject:-ControllSystem Engineering (2150909)
Topic: Root-Locus Plot steps and procedure
Name Enrollment No.
JayMakwana 151290109027
DhruvPandya 151290109032

2. Root–LocusPlot
Determination ofTheStability of a System
Root Locus Steps And Procedure
2

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 OnMatlab
Root Locus Steps And Procedure
3
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 OnMatlab

4. 1.Introduction to Root Locus:
TheStabilityofagivenclosedloopsystemdependsuponthelocationof
therootsofthecharacteristicsequation,whichisthelocationoftheclosed
looppoles.Ifwechangesomeparameterofasystem,thenthelocationof
closedlooppolechangesin's'plane.
Thismovementofpolesin's'planeiscalledas'RootLocus'.
RootLocusisasimplegraphicalmethodfordeterminingtherootsofthe
characteristicequationwhichwasinventedbyW.R.Evansin1948.Itcan
bedrawnbyvaryingtheparameter(usuallygainofthesystem)fromzero
toinfinity.
Root Locus Steps And Procedure
4
TheStabilityofagivenclosedloopsystemdependsuponthelocationof
therootsofthecharacteristicsequation,whichisthelocationoftheclosed
looppoles.Ifwechangesomeparameterofasystem,thenthelocationof
closedlooppolechangesin's'plane.
Thismovementofpolesin's'planeiscalledas'RootLocus'.
RootLocusisasimplegraphicalmethodfordeterminingtherootsofthe
characteristicequationwhichwasinventedbyW.R.Evansin1948.Itcan
bedrawnbyvaryingtheparameter(usuallygainofthesystem)fromzero
toinfinity.

5. 2. Generalsteps for drawing the Root
Locusofthe given system:
1.Determine the open loop poles, zeros and a number of branches from
given G(s)H(s).
2.Draw thepole-zero plotand 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 ofasymptote.
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 Steps And Procedure
5
1.Determine the open loop poles, zeros and a number of branches from
given G(s)H(s).
2.Draw thepole-zero plotand 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 ofasymptote.
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. Letus 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 Steps And Procedure
6
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 Steps And Procedure
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,j1-

8. centroid
Centroid:
25.1
4
5
0-4
0-(-1)+(-1)+(-3)+0

Root Locus Steps And Procedure
8

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 Steps And Procedure
9
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 Steps And Procedure
10
j1

11. BreakawayIBreak 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 Steps And Procedure
11
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





RouthArray,
Root Locus Steps And Procedure
12
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.tp2
431
0
0
180
pppp
z
zpd






0
2p
1p4p
Root Locus Steps And Procedure
13
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 Steps And Procedure
14
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 Steps And Procedure
15
j1

16. 4.MatlabProgram:
n=input('enter numerator ')
d=input('enter denominator')
g=tf(n,d)
rlocus(g)
Root Locus Steps And Procedure
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n=input('enter numerator ')
d=input('enter denominator')
g=tf(n,d)
rlocus(g)

17. Root Locus Steps And Procedure
17

18. Thank YouThank You
Root Locus Steps And Procedure
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