NPIC
Faculty of Electricity
Control Engineering
Properties of Root Locus
Lesson Objectives : After finish this lesson, students will be able to
 be aware of the control system problem
 be aware of the vector representation of complex
number
 define the root locus
 determine the properties of root locus
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Lesson Objectives 01
 Aware of Control
System Problem
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The Prerequisites To Root Locus Technique
1. Understanding the control system problems
2. Vector representation of complex numbers
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The Control System Problem
General Feedback Control System
Open Loop Transfer Function
Also called, Loop Gain
KG ( s ) H ( s )
Closed-loop Transfer Function
T (s) 
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KG ( s )
1  KG ( s ) H ( s )
The Control System Problem
What is the control system problem?
Difficult to obtain the poles
Poles location varied with the gain
T (s) 
KG ( s )
1  KG ( s ) H ( s )
Let’s find the closed-loop transfer function
K
T CL ( s ) 
KG ( s )
1  KG ( s ) H ( s )
N G (s)
DG (s)

1 K
N G (s) N H (s)
K

DG (s)
D G ( s ) D H ( s )  KN G ( s ) N H ( s )
DG (s) D H (s)
 K
N G (s)
DG (s)

DG (s) D H (s)
D G ( s ) D H ( s )  KN G ( s ) N H ( s )
TOL ( s )  KG ( s ) H ( s ) 
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KN G ( s ) N H ( s )
DG ( s ) D H ( s )
N G (s)
DG (s) D H (s)

KN G ( s ) D H ( s )
D G ( s ) D H ( s )  KN G ( s ) N H ( s )
The Control System Problem
Comparison Between Open Loop and Closed Loop Systems
T OL ( s ) 
KN
G
(s) N H (s)
DG (s) D H (s)
OL poles are roots of:
1. Forward transfer function
denominator
2. Feedback transfer
function denominator
T CL ( s ) 
KN
G
(s)D H (s)
D G ( s ) D H ( s )  KN
G
(s) N H (s)
CL poles are roots of:
1. Combinations of numerator and
denominator of forward and
feedback transfer functions
2. Poles depend on gain, K
Therefore, with CL system, the poles are not easily obtained
and change with the value of K
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The Control System Problem - Example
Given:
Open Loop System
G (s) 
H (s) 
s 1
s(s  2)
s3
s4
T OL ( s ) 
K ( s  1)( s  3 )
Poles: 0, -2 and -4
s ( s  2 )( s  4 )
Closed-loop System
T CL ( s ) 
T CL ( s ) 
KN
G
(s)D H (s)
D G ( s ) D H ( s )  KN
G
(s) N H (s)
K ( s  1)( s  4 )
s ( s  2 )( s  4 )  K ( s  1)( s  3 )
Poles: We have to factor and also depends on K. Root
locus technique help to find poles!
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Lesson Objectives 02
 Aware of
the Vector
Representation of Complex
Number
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Vector Representation of Complex Number
What is complex number?
s    j
Complex number is a vector
Vector has magnitude and direction
M 

  tan
2

1 
 
2
 
 
Therefore, we can also
represent complex number
s as:
s  M
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Vector Representation of Complex Number
But, if s is a variable in a
function, how to represent the
complex number. For example,
F (s)  s  a
Replacing s,
F ( s )  s  a    j    a
F ( s )    a  j
Another complex number
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Graphically,
Vector Representation of Complex Number
s    j
F ( s )  s  a    j    a
F ( s )    a  j
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Vector Representation of Complex Number
Same Vector
s  5  j2
Same Vector
F (s)  s  3
j
F (s)  5  3  j2  8  j2
2

3
5
8
is a complex number and can be represented by a
vector drawn from the zero of the function to the point s.
F (s)  s  a
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Vector Representation of Complex Number
General
F (s)  M
m
m
 (s  z )
i
F (s) 
i 1
n
 (s  p
M 
j
)
 zero length
 pole length

(s  zi )

(s  p j )

j 1
i 1
n
j 1
   zero angles   poles angle
m
 
  s  z
i 1
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Lecturer : IN SOKVAN
n
i
    s 
j 1
pj

Vector Representation of Complex Number
Example
F (s) 
s 1
s(s  2)
20  116 . 6
s  3  j4
What is F(s)?
F (s) 
20

17  104 . 0


 116 . 6  126 . 9  104 . 0
5 17
F ( s )  0 . 217   114 . 3
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
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



5126 . 9

Lesson Objectives 03
 define the root
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locus
What is Root Locus?
The root locus is the path of the roots of the
characteristic equation shown out in the
s-plane as a system parameter is changed.
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Defining Root Locus
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Defining Root Locus
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Defining Root Locus
Root Locus
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Defining Root Locus
Root Locus
Representation of the paths of closedloop poles as the gain is varied
What can we learn from this graphic?
0<K<25
The system is over-damped
K=25
The system is critically damped
K>25
The system is under-damped
The system is stable
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Lesson Objectives 04
 Determine the properties
of root locus
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Properties of Root Locus
T (s) 
j
KG ( s )
1  KG ( s ) H ( s )
How do we get poles?
180


1  KG ( s ) H ( s )  0
KG ( s ) H ( s )   1
The value “-1” is a
complex number
1
KG ( s ) H ( s )   1  1 180
M  1    ( 2 k  1)180
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
Properties of Root Locus
Magnitude Condition
KG ( s ) H ( s )   1  1 180
j

KG ( s ) H ( s )  1
K 
180
1


G (s)H (s)
Angle condition
1
KG ( s ) H ( s )   1  1180

 KG ( s ) H ( s )  ( 2 k  1)180
k  0 ,1, 2 , 
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
Properties of Root Locus - Example
Given a unity feedback system
forward transfer function
G (s) 
K (s  2)
2
( s  4 s  13 )
1. Is point -3+0j is on a root
locus?
2. If the point is on the root
locus, find the value of K?
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Steps:
1. Determine zeros and poles of the
forward transfer function
2. Determine angles from zeros and
poles to the interested point
3. Determine the length of vector
from zeros and poles to the
interested point
4. Add all angles. If it is equal to
multiple of 180, then the point is
on the root locus.
5. Determine K using length of zero
and pole vectors
Properties of Root Locus - Example
1. Determine zeros and poles of the forward transfer
function
Using quadratic equation,
G (s) 
G (s) 
25
b
2
b  4 ac
2a

K (s  2)
( s  4 s  13 )
j
3
2
K (s  2)
( s  2  j 3 )( s  2  j 3 )
Lecturer : IN SOKVAN
3


2

3
Properties of Root Locus - Example
2. Determine angles from zeros and poles to the
interested point
1

j
3
2
3


2
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 2  180
3
Lecturer : IN SOKVAN
1
   251 . 57
3

 3  90  tan
3

 1  270  tan
1 
1 
1

   108 . 43
3

Properties of Root Locus - Example
3. Determine the length of vector from zeros and poles
to the interested point

j
3
L1 
L1
L2
3
L3
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

2

3
Lecturer : IN SOKVAN
2
2

10
2
2

10
3 1
L2  1
L3 
3 1
Properties of Root Locus - Example
4. Add all angles. If it is equal to multiple of 180, then
the point is on the root locus.
 1  251 . 57

 2  180

 3  108 . 43

 2  ( 1   3 )  180  251 . 57  108 . 43   180




Therefore, the point -3+0j is a point on the root
locus
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Properties of Root Locus - Example
5. Determine K using length of zero and pole vectors
L1 
K 
2
3 1
2

L2  1
10
1
G (s)H (s)

1
M 
M
K 
1
M
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L3 
2
3 1
 zero length
 pole length
 10
2

10
1

10

10
1
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Home work
Given a unity feedback system forward transfer function
a / . G (s) 
K ( s  2 )( s  6 )
( s  8 s  25 )
2
K ( s  1)
2
,
b / . G (s) 
2
c /. G ( s ) 
s
2
K (s  4)
,
d /. G ( s ) 
( s  1)
2
K
( s  1) ( s  4 )
3
1. Is point -3+0j is on a root locus?
2. If the point is on the root locus, find the value of K?
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