Chapter 5: Root Locus
Nov. 13-15, 2012
Two Conditions for Plotting Root Locus
Given open-loop transfer function Gk(s)
m
Characteristic equation
Gk  s  
K g  ( s  zi )
i 1
n
 (s  p )
j 1
 1
j
Magnitude Condition and Argument Condition
n
 | (s  p ) |
j
Kg 
j 1
m
 | ( s  zi ) |
i 1
m
n
i 1
j 1
 (s  zi )   (s  p j )  (2k  1) , k  0,1,2..
Rules for Plotting Root Locus
Content
Rules
1
Continuity and Symmetry
Symmetry Rule
2
Starting and end points
Number of segments
3
Segments on real axis
4
Asymptote
n segments start from n open-loop
poles, and end at m open-loop zeros
and (n-m) zeros at infinity.
On the left of an odd number of poles
or zeros
n-m segments：
＝
5
(2k  1)
 , k  0,1,2,
nm
n
Asymptote

m
 ( p )   ( z )
j 1
j
i 1
i
nm
3
6
Breakaway
and break-in
points
d [ F s ]
 0
ds
F s   Ps   K g Z s   0
Ps Z s   P s Z s   0
m
n
1
1


i 1 zi  
j 1 pi  
Angle of emergence
7
Angle of
emergence and
entry
m
n
i 1
j 1
j p
m
 p   (2k  1)   i   j
Angle of entry
n
 z   (2k  1)    j   i
j 1
8
Cross on the
imaginary axis
i 1
i z
Substitute s = j to characteristic equation
and solve
Routh’s formula
4
Rule 6: Breakaway and Break-in Points on the
Real Axis
Breakaway point
j
Break-in point

j

Use the following necessary condition
dK g

dGs H s 
d 
1
 0 or
 0 or
0


ds
ds  Gs H s 
ds
1
1


Ps Z s   P s Z s   0
s  zi
s  pj
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Example 5.3.1: Given the open-loop transfer function,
please draw the root locus.
k ( s  3)
G( s) H ( s) 
zero pole
k =5.818
pole k =0.172
( s  1)(s  2)
6.Breakaway
and break-in
points
1.Draw
2.
Two
the
segments
open-loop
poles
and
4.
Segments
on
real
axis
3.Symmetry
5.Asymptote
zeros

 180
(2k  11)
1

 1801

2 1
3 a
Break--3
in
-2 breaka
-1
way

1

a
n
m
 p
 a
3zi 
j  
j 1
i 1
n
 s  pn  m
a 1 a  2
j
k
j 1
m
 s  zi

2a
2
(1  2)  3

0
2 1
a3
i 1
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Example 5.3.2: Given the open-loop transfer function
K * ( s  1)
Gk ( s ) 
s2
j
s1
please prove that the root locus in the
complex plane is a circle.

0
s2
Conclusion: For the open-loop transfer function with one
zero and two poles, the root locus of characteristic equation
is probably a circle in the complex plane.
Example 5.3.3:
G(s) H (s) 
K
s ( s  1)( s  2)
5.
Points
across
thebreak-in
imaginary
axis
4.
Breakaway
points
3.
2.
1.Asymptotes
Segments
Open-loop
onand
poles
real
axis
and zeros
j1.414
K=6
-2
-1
-0.42
K=6
-j1.414
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 s Z s   0
Pss3Z
s180
(22
k s1)K  0 
3s22P


  [3s  6s  2] 
60 ,180
yields

0
3
31 0 2
s
solution s1  0.42 s2  1.58
0 1 2
2
s  3
K1
0
63K
1
s
3s 2  6  0
3
s0
K K=6
8
Example 5.3.4:
G( s) H ( s) 
k
( s  1) 2 ( s  1  j18)(s  1  j18)
1.7.The
Find
point
poleswhere
and
zeros
the locus points
3.
Symmetry
6.
5.
Breakaway
Asymptote
and
break-in
2.No
4 segments
4.
segments
on realaxis
axis
across
the imaginary
j3.16
k =121
Breakaway
-1
k =121
-j3.16
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4
2

23180
sd [(

4
s

24
s
 k  0

(
2
k
) 1sj19
 )]
s 1) (s  1  j18)(1s40
18
 45 或0135
4
dsm2 24 19  k
s4
1
n 3
yields
sp j3
s z i12s  10  0
3

s
 1 
101

1 j3
j 4
1
i 11 40 1
Solutions
s

s
 
  2,3
 1
1
s2
14n  m 19  k 4
484  4k
1
s
02  140  0
14
s
14
s0
19  k
s   j3.16
484-4k=0 getk =121
9
Example 5.3.5
Rs  

K
s


0 .5
s 0.5s  1
C s 
2
Please sketch the root locus with respect to K=[0,+∞).
Extension of Root Locus
Canonical form
Gk ( s)  K g 
m
 (s  z )
i
i 1
n
 (s  p )
 1
Conventional
Root Locus
j
Root locus gain
j 1
1. How to sketch the root locus
for other parameters?
2. How to sketch if Gk(s)=1
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Parameter Root Locus
 Zero-degree Root Locus
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5.3.2 Parameter Root Locus
Example 5.3.6: Ks is a ramp feedback
gain, please sketch the root locus with
respect to Ks=[0,+∞).
s 2  (2  10K s )s  10  0 dFs2s  02s  10  10K s s  0
ds
10K s s
s  2 10  3.12  1  0
198o
s  n2 s  10
1  G
' ( s)sH 'p( sj )  0
108o
10K s 
j 1
m
 10 K s s
G ' ( s ) H 's(s )z  2 s  z
 i
i 1
90o
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s  p1 s  p2
s  2 s  10
1.2.Break-in
and
breakaway point
3.65of
3.65
Angle
emergence

 4.3
3.12



  180  (90  108 )  198
12
5.3.2 Zero-degree Root Locus
Example: Sketch the root loci of the system with the open
loop transfer function:
K1 (1  s)
G( s) H ( s) 
s( s  1)
Analysis:
K1 : 0  
K1 (1  s)  K1 ( s  1)
G(s) H (s) 

s( s  1)
s( s  1)
For this kind of systems, the characteristic equations are like as:
1  K1G1 (s) H1 (s)  0  K1G1 (s) H1 (s)  1
Magnitude equation
Argument equation
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Gk (s) 
1
Kg
K g : 0  
Gk (s)  2k , k  0,  1,  2,   
13
Root locus by using the
sketching rules with the
following modification:

Real-axis: Left of even
number of zeros or poles

Asymptote

Angles of emergence and
entry
m
n
i 1
j 1
j p
s  3.3
( K 1  6.17)
s j
K1  1
1
1
s  0.3
( K 1  0.16)
 p  2k  i   j
For Kg varying from －∞→0 together with Kg =[0,+∞)
simultaneously, the root loci are named the “complete
root loci”.
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5-4 Application of Root Locus
Insert a zero
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Gk (s) 
Kg
s  1(s  1)(s  4)
 Gk (s) 
K g ( s  a)
s  1(s  1)(s  4)
15
Add a pole to the open-loop transfer function
K1 ( s  3)
K1 ( s  3)
GH1 ( s) 
 GH 2 ( s) 
s( s  2)
s( s  2)(s  a)
Im
Im
Re
3
2
Re
5
3
2
Im
Im
Re
3
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2
1
Re
3
2
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5.4.1 The effects of Zeros and Poles
Attracting effect
Generally, adding an open zero in the left s-plane will lead the
root loci to be bended to the left.
The more closer to the imaginary axis the open zero is, the more
prominent the effect on the system’s performance is.
Repelling effect
Generally, adding an open pole in the left s-plane will lead the
root loci to be bended to the right.
The more closer to the imaginary axis the open pole is, the more
prominent the effect on the system’s performance is.
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3.6 K c
Gk ( s) 
(10s  1)(5s  1)(2s  1)
How to enlarge the stability region?
0.036K c
K
Gk ( s) 

( s  0.1)( s  0.2)( s  0.5) ( s  0.1)(s  0.2)(s  0.5)
Example 5.4.1:
j0.2
ω1=0.17
Kc=0.778
K ( s  0.4)
Gk ( s) 
( s  0.1)( s  0.2)( s  0.5)
ω1=0
Asymptote:Kc=0.278
  0.2
Asympt ot e:   0.1
12  60, 3  180
1260o 90
point: a  -0.06
0.027
Breakawaypoint-0.027
:  aω=0Breakaway
0.06
3
Intercept
Int erceptwit h imaginary
axis with imaginaryaxis :
Kc=0.278
1  0-j0.2
for K c  0.278 12  0.17 for K c  0.778
ω2=-0.17
3  0 for K c  0.278
Kc=0.778
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5.4.2 Performance Analysis Based on Root Locus
Example 5.4.2: Given the open-loop transfer function
K
Gk ( s) 
s(0.5s  1)
please analyze the effect of open-loop gain K on the
system performance.
Calculate the dynamic performance criteria for K＝5.
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k¡ú¡Þ
j
3j
k£ ½1 0
[s]
k£ ½3
k£ ½2
k£ ½0
£ -2

k£ ½1 £ -1
k£ ½0
0

k£ ½2
k¡ú¡Þ
k£ ½3
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It is observed from the root locus that the system is
stable for any K.
For 0＜K＜0.5(0＜k＜1), there are two different negative
real roots.
For K=0.5(k=1), there are two same negative real roots.
For K＞0.5(k＞1), there are a pair of conjugate complex
poles.
For K=5( k=10), the closed-loop poles are
s12   n  j n 1   2  1  j 3
1
n  10  3.16,  
 0.316
3.16
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The criteria for transient performance can be given by
Overshoot
p e
 / 1 2
Peak time
tp 
Settling time t s 
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 100%  e1.05  100%  35%

n 1   2
3
n
 1.05s
 3s (  5%)
22
1. We can get the information of the system’s stability by checking
whether or not the root loci are on the left half s-plane with the
system’s parameter varying.
2. We can get some information of the system’s steady-state error in
terms of the number of the open-loop poles at the origin of the splane.
3. We can get some information of the system’s transient
performance in terms of the tendencies of the root loci with the
system’s parameter varying.
4. If the root loci in the left half s-plane move to somewhere far away
from the imaginary axis with the system’s parameter varying, the
system’s response decays more rapidly and the system is more stable,
vice versa.
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Complement

Use Matlab to sketch root locus
m
Gk  s  
K g  ( s  zi )
i 1
n
 (s  p )
j 1
=
K g  s +3 s +4 
 s +1 s +2 
=
K g  s 2 +7 s +12 
s 2 +3s +2
j
In Matlab:
num=[1 7 12]
sys=tf[num,den]
den=[1 3 2]
rlocus(num,den)
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