Chapter 5
Root Locus
1
In previous lecture

Chapter 4


Routh Hurwitz Criterion
Steady-state error analysis
2
Today’s plan

Construction of Root Locus
3
Learning Outcomes

At the end of this lecture, students should be
able to:


Sketch the root locus for a given system
Identify the stability of a system using root-locus
analysis
4
Introduction
 Root Locus illustrates how the poles of the
closed-loop system vary with the closed-loop
gain.
 Graphically, the locus is the set of paths in
the complex plane traced by the closed-loop
poles as the root locus gain is varied from
zero to infinity.
5
R(s)
+
-
K
C(s)
Gc
H
block diagram of the closed loop system
given a forward-loop transfer function;
KGc(s)H(s)
where K is the root locus gain, and the corresponding closed-loop
transfer function
KGc ( s)
G( s) 
1  KGc ( s) H ( s)
the root locus is the set of paths traced by the roots of
1 + KGc(s)H(s) =0
 as K varies from zero to infinity. As K changes, the solution to this
equation changes.
6
So basically, the root locus is sketch based on the characteristic
equation of a given transfer function.
Let say:
KGc ( s) H ( s )
G ( s) 
1  KGc ( s) H ( s )
Thus, the characteristic equation :
C.E  1  KGc (s) H ( s)
7
Root locus starts from the characteristic
equation.
1  KGc ( s) H ( s)  0
Split into 2 equations:
KGc (s) H (s)  1  j 0
Magnitude condition
| KGc ( s) H ( s) || 1 |
Angle condition
KGc ( s ) H ( s )  180 o (1  2m)
where
m  0,1,2, 
8
Remarks
If Si is a root of the characteristic equation, then 1+G(s)H(s) = 0 OR Both the
magnitude and Angle conditions must be satisfied
If not satisfied Not part of root locus
9
Example 1
K
G( s) 
s( s  2)
k
k
k
s ( s  2)

 2
k
s ( s  2)  k s  2 s  k
1
s ( s  2)
s1, 2 
 2  4  4k
 1  1  k
2
-2
-1

(ln P.O) 2
 2  (ln P.O) 2
  cos1 
0
10
S1  1  j1
Let say my first search point, S1
S1
1
2
0
-1
-2
(A) Satisfy the angle condition FIRST
G(s) |s 1 j1  1   2  135o  45o  180o
(B) Magnitude condition to find K
K
1
( 1  j1)( 1  j1  2)
K (
2 )(
2)  2
11
Let say the next search point
s2  2  j1
-2
-1
0
Check whether satisfy angle condition
G ( s ) |s  2 j1 
K
K

(2  j1)( 2  j1  2)
(2  j1)( j1)
G(s) |s 2 j1  [180o  tan 1 (1 2)  90o  270o  tan 1 (1 2)
G(s) |s 2 j1  180o OR  180o
Since angle condition was not satisfied  not part of root locus
12
CONSTRUCTION RULES
OF ROOT LOCUS
8 RULES TO FOLLOW
13
Construction Rules of Root Locus
KG ( s ) H ( s )  1
M
Assume
KG( s ) H ( s ) 
K  ( s  zi )
i 1
N
 (s  p j )
j 1
M
Then
KG( s ) H ( s ) 
K  ( s  zi )
i 1
N
 (s  p j )
 1
j 1
14
Thus
 (s  p j )
K 
 ( s  zi )
15
Rule 1 : When K = 0
s  pj
 Root at open loop poles
(a) The R-L starts from open loop poles
(b) The number of segments is equal to the number of open loop poles
16
Rule 2: When K = ∞
s   zi
(a) The R-L terminates (end) at the open loop zeros
17
Rule 3: Real-axis segments
On the real axis, for K > 0 the root locus exists to the left of an odd number
of real-axis, finite open-loop poles and/or finite open-loop zeros
Example 2
K ( s  3)( s  4)
G( s) 
( S  1)( S  2)
R-L doesn’t exist here
R-L exist here
R-L exist here
18
Rule 4: Angle of asymptote
180 (1  2m)
A 
, m  0,1,( N P  N Z  1)
NP  NZ
o
NP = number of poles
NZ = number of zeros
19
Example 3
K
G( s) 
s ( s  1)( s  2)
 A1
 A2
180o (1  2(0))

 60o
30
180o (1  2(1))

 180o
30
NP=3
NZ=0
m  0,1,2
 A2

 A1
 A3
Centroid
 A3
180o (1  2(2))

 300o  60o
30
20
Rule 5: Centroid
A
Re( p )   Re( z )


j
i
NP  NZ
From example 3
K
G( s) 
s ( s  1)( s  2)
Recall s  Re  j Im
Thus, s  0,1,2
1 2
A 
 1
3
-1
21
Rule 6: Break away & break in points
-2
-1
Break away point
0
-2
-1
0
Break in point
How to find these points ?
22
KG ( s ) H ( s )  1
1
K 
G( s) H ( s)
Differentiate K with respect to S & equate to zero
dK d 
1  d
 
 G ( s) H ( s)  0

ds ds  G ( s) H ( s)  ds
How ?
Solve for S  this value will either be the break away or break in point
23
Let say
N1 N 2
G( s) H (s) 
D1 D2
d
D1 D2 ( N1 N 2 )' N1 N 2 ( D1 D2 )'
G ( s) H ( s) 
0
2
ds
( D1 D2 )
Thus,
D1D2 ( N1N2 )'N1N2 (D1D2 )'  0
24
Now, find
dK d   1  d   D1D2 
 
 


ds ds  G(s) H (s)  ds  N1 N 2 
dK N1 N 2 ( D1 D2 )' D1 D2 ( N1 N 2 )'

0
ds
N1 N 2
D1D2 ( N1 N2 )' N1 N2 ( D1D2 )'  0
Thus,
So, notice that
dK d 
1  d
 
 G ( s) H ( s)

ds ds  G ( s) H ( s)  ds
25
Example 4
1
G( s) 
s ( s  2)
Solution:
dK d
d  1 
 G ( s)  
0

ds ds
ds  s( s  2) 
 (2s  2)

0
2
( s ( s  2))
s  1
-2
0
-1
26
Example 5
1
G( s) 
s ( s  1)( s  2)
-2
-1
0
NP = 3
NZ = 0
27
180 (1  2m)
A 
, m  0,1,( N P  N Z  1)
NP  NZ
o
 A  60o ,180o ,60o
A 
 Re( p
j
)   Re( zi )
NP  NZ
1 2

 1
3
-2
-1
0
28
dK
d 
1


ds
ds  s 3  2 s 2  2 s 
 1(3s 2  6s  2)
 3
0
2
2
( s  3s  2s)
(3s 2  6s  2)  0
 6  36  4(3)( 2)
s1 , s2 
2(3)
Break away point
s1 , s2  1.6,0.42
Invalid, why ???
29
1
G( s) 
s ( s  1)( s  2)
-2
-1
-0.42
0
How to determine
these values ?
30
From characteristic equation
1  KG( s )  0


1
1 K 
0

 s ( s  1)( s  2) 
s  3s  2 s  K  0
3
2
Construct the Routh array
31
S3
1
2
S2
3
K
S1
6-K/3
S0
K
Force S1 row to zero or K=6
Replace K=6 into S2 row
3s  6  0
2
s  2
s  j 2
2
Since
s  j
j  j  2
 2
32

-2
-1
-0.42
2
0

2
33
Rule 7 : Angle of departure (arrival)
Finding angles of departure and arrival from complex poles & zero
Root locus start from open loop poles
 Angle of departure
p
Root locus terminates at open loop zeros  Angle of departure
z
34
Angle of Departure
1   2   3   4   5   6  (2k  1)180o
35
To make it easier to remember:
Let’s change:
 p  1
Angles due to poles, keep using symbol θ
4  4
5  5
Angles due to zeros, change to symbol β
2
3
6
2
 3
 6

Replace in this equation
1   2   3   4   5   6  (2k  1)180o
and let k=0, we get
 p  2  3  4  5  6 180o
Since Φp= Φp+360o
 p  2  3  4  5  6  180o
Re-arrange
 p  180o  2  3  6  4  5
Thus, the angle of departure:
 p  180o   zi   p j
36
Angle of Arrival
 2  1   3   4   5   6  (2k  1)180o
37
To make it easier to remember:
Let’s change:
z   2
Angles due to poles, keep using symbol θ
1  1
4  4
5  5
Replace in this equation
 2  1   3   4   5   6  (2k  1)180o
and let k=0, we get
z  1   3   4  5   6  180o
Re-arrange
 z  180o  1   4   5   3   6
Thus, the angle of arrival:
Angles due to zeros, change to symbol β
3
6
3
 6

z  180o   p j   zi
38
Angle of Departure
Angle of Arrival
 p  180o   zi   p j
z  180o   p j   zi
39
Example 6
K
G( s) 
s( s 2  6s  25)
Step 1: Determine poles & zeros
poles at :
0,
6
36  100
 0,3  j 4
2
Np=3
Nz=0
Step 2: Maps poles & zeros on the s-plane
4
-3
0
-4
40
Step 3: Calculate angle of asymptote
180o (1  2m)
A 
, m  0,1,  (3  0  1)  0,1,2
NP  NZ
 A1
180o (1  2(0))

 60o
30
 A2
180o (1  2(1))

 180o
30
 A3
180o (1  2(2))

 300o  60o
30
41
Step 4: Determine centroid
poles:
 0,3  j 4
A
Re( p


j
)   Re( zi )
NP  NZ
33
A 
 2
30
Step 5: Sketch asymptotes line with dotted line
4
-3
-2
0
-4
42
Step 6: Determine the break-away or break in points


dK d  1  d
2
 


s
(
s
 6s  25)  0

ds ds  G( s)  ds
 [3s 2  12s  25]  0
 12  144  300  12   156
s1 , s2 

6
6
Complex
numbers
NO break away or break in points
 only exist on real-axis of s-plane
43
Step 7: Since complex poles exist, calculate the angle of departure
 p |s 3 j 4  180o   zi  p j
4

-3
0
 p  180    90
o
o
due to s=0 due to its
complex
conjugate
-4
How to determine θ?
44
  180  
o
To determine θ
4
4
  tan ( )
3
1


-3
0
Thus:
-4
4
  180  tan    126.87 o
3
1
o
 p 126.87  90  180
Angle of departure :
 p  180 126.87  90  36.87
o
o
o
-36.87o
Cross Hair
45
Step 8: Determine jω-crossing
see RULE 8 : After this
Step 9: Sketch the root locus
-36.87o
4
Rule 8
-3
-2
36.87o
0
-4
46
Rule 8 : R-L crosses jω-axis
K
G( s) 
s( s 2  6s  25)
From Characteristic equation:
1  G( s)  0
1
s( s 2
K
0
 6 s  25)
s 3  6 s 2  25s  K  0
47
Construct the Routh array
s3
1
25
s2
6
K
s1
(6(25)-K)/6=
(150-K)/6
s0
K
48
Force row s1 to be zero, thus 150-K=0; K = 150
Substitute K = 150 into row s2
6 s 2  150  0
s  25
2
s

 25   j 5
Since s = jω
j   j 5
Thus
  5
49
Final Sketch
-36.87o
-3
-2
36.87o
5
4
0
10 points
-4
-5
50
Try it
Sketch the root locus for:
K ( s  2s  2)
G( s) 
s( s  1)( s  2)
2
51
52
How to use Matlab to sketch root locus
K ( s  3)( s  4)
K [ s 2  7 s  12]
KG( s) 

( s  1)( s  2)
s 2  3s  2
In Matlab:
num = [1 7 12]
den = [1 3 2]
rlocus(num,den)
53
Figure 8:10 page 438
54
Any Q’s ???
55