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Chapter 5: Root Locus
Two Conditions for Plotting Root Locus
1
1
( )
1
( )
m
g i
ik n
j
j
K s z
G s
s p
Given open-loop transfer function Gk(s)
Characteristic equation
m
i
i
n
j
j
g
zs
ps
K
1
1
|)(|
|)(|
..2,1,0,)12()()(11
kkpszsn
j
j
m
i
i
Magnitude Condition and Argument Condition
Content Rules
1 Continuity and Symmetry Symmetry Rule
2
Starting and end points
Number of segments
n segments start from n open-loop
poles, and end at m open-loop zeros
and (n-m) zeros at infinity.
3 Segments on real axis On the left of an odd number of poles
or zeros
4 Asymptote n-m segments：
5 Asymptote
3
Rules for Plotting Root Locus
mn
zpn
j
m
i
ij
1 1
)()(
,2,1,0,)12(
k
mn
k＝
4
7
Angle of
emergence and
entry
Angle of emergence
Angle of entry
8 Cross on the
imaginary axis
Substitute s = j to characteristic equation
and solve
Routh’s formula
n
j
m
zii
ijz k1 1
)12(
m
i
n
pjj
jip k1 1
)12(
6
Breakaway
and break-in
points
0
][
ds
sFd 0 sZKsPsF g
0 sZsPsZsP
m
i
n
j ii pz1 1
11
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jBreak-in point
jBreakaway point
Rule 6: Breakaway and Break-in Points on the Real Axis
Use the following necessary condition
0 sZsPsZsP
0d
dor 0
1
d
dor 0
d
d
s
K
sHsGss
sHsG g
ji pszs
11
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-1-2-3
Example 5.3.1: Given the open-loop transfer function,
please draw the root locus.
)2)(1(
)3()()(
ss
sksHsG
2
* )1()(
s
sKsGk
Example 5.3.2:
j
0
s1
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.
please prove that the root locus in the
complex plane is a circle.
Given the open-loop transfer function
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8
Example 5.3.3:
)2)(1()()(
sss
KsHsG
-1-2 -0.42
j1.414
K=6
K=6
-j1.414
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Example 5.3.4:
-1
Breakaway
)181)(181()1()()(
2 jsjss
ksHsG
=121
-j3.16
k
j3.16
=121k
Example 5.3.5
s
K
15.0
5.0
ss
sR sC
2
Please sketch the root locus with respect to K=[0,+∞).
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Extension of Root Locus
Parameter Root Locus
Zero-degree Root Locus
Canonical form
1
)(
)(
)(
1
1
n
j
j
m
i
i
gk
ps
zs
KsG
Root locus gain
1. How to sketch the root locus
for other parameters?
2. How to sketch if Gk(s)=1
ConventionalRoot Locus
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Example 5.3.6: Ks is a ramp feedback
gain, please sketch the root locus with
respect to Ks=[0,+∞).
010)102(2 sKs s
01102
102
ss
sK s
0)(')('1 sHsG
102
10)(')('
2
ss
sKsHsG s
198)10890(180 90o
108o
198o
0101022 sKss s
5.3.2 Parameter Root Locus
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Example: Sketch the root loci of the system with the open loop transfer function:
0 :
)1(
)1()()( 1
1 Kss
sKsHsG
5.3.2 Zero-degree Root Locus
)1(
)1(
)1(
)1()()( 11
ss
sK
ss
sKsHsGAnalysis:
For this kind of systems, the characteristic equations are like as:
1)()( 0)()(1 111111 sHsGKsHsGK
0 : 1
)( g
g
k KK
sG
,2 ,1 ,0 ,2)( kksGk
Magnitude equation
Argument equation
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1 1
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
1
1
K
js
)16.0(
3.0
1
K
s
)17.6(
3.3
1
K
s
For Kg varying from －∞→0 together with Kg =[0,+∞)
simultaneously, the root loci are named the “complete
root loci”.
m
i
n
pjj
jip k1 1
2
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5-4 Application of Root Locus
Insert a zero )4)(1(1
)()(
)4)(1(1)(
sss
asKsG
sss
KsG
g
k
g
k
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Add a pole to the open-loop transfer function
))(2(
)3()(
)2(
)3()( 1
21
1asss
sKsGH
ss
sKsGH
Im
Re
Im
5
Re
Im
23 23
23
Re
Re
23
Im
1
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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.
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.
5.4.1 The effects of Zeros and Poles
Attracting effect
Repelling effect
)5.0)(2.0)(1.0(
)4.0()(
sss
sKsGk
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Example 5.4.1:
How to enlarge the stability region? )12)(15)(110(
6.3)(
sss
KsG c
k
)5.0)(2.0)(1.0(
036.0)(
sss
KsG c
k)5.0)(2.0)(1.0(
sss
K
60o
j0.2
ω1=0.17
Kc=0.778
-j0.2
ω2=-0.17
Kc=0.778
ω3=0
Kc=0.278
ω1=0
Kc=0.278
-0.06-0.027
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5.4.2 Performance Analysis Based on Root Locus
Example 5.4.2: Given the open-loop transfer function
)15.0()(
ss
KsGk
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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j
k→
∞
0
k＝ 0
k＝ 2
k＝ 0
k＝ 3
k＝ 10
k＝ 3
k＝ 2
3j
k→
∞
[s]
k＝ 1－2 －1
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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
311 2
12 jjs nn
316.016.3
1,16.310 n
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The criteria for transient performance can be given by
%35%100%100 05.11/ 2
eep
Peak time st
n
p 05.11 2
Settling time %)5(33
stn
s
Overshoot
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1. We can get the information of the system’s stability in terms of
that whether the root loci always are in the left-hand 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 s-
plane.
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. The root loci in the left-hand s-plane move to far from the
imaginary axis with the system’s parameter varying, the system’s
response is more rapid and the system is more stable, vice versa.
Complement
Use Matlab to sketch root locus
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2
1
2
1
( )+7 +12+3 +4
= =+1 +2 +3 +2
( )
m
g iggi
k n
j
j
K s zK s sK s s
G ss s s s
s p
In Matlab:
num=[1 7 12]
den=[1 3 2]
Rlocus(num,den)
sys=tf[num,den]
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