Automatic Control
Instructor: Dr. Cailian Chen
Associate Professor
Department of Automation, SJTU
Contact: cailianchen@sjtu.edu.cn
Tel.: 021-3420 4279
Office: 2-515, Building of SEIEE
Office Time: 2:00pm-4:00pm, Tuesday
Slides and assignment can be downloaded from URL:
wicnc.sjtu.edu.cn/download.php
2015/4/13
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Norbert Wiener (1894 –1964)
Harvard PhD, 1912 (18 years old)
America Mathematician
Professor of Mathematics at MIT
控制科学的鼻祖
Cybernetics-2nd Edition : or the
Control and Communication in
the Animal and the Machine
Wiener is regarded as the originator of cybernetics, a
formalization of the notion of feedback, with many
implications for engineering, systems control,
computer science, biology, philosophy, and the
organization of society..
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Wiener:“Feedback is a method of controlling a
system by inserting into it the result of its past
performance”





Teaching and learning is a complex control system.
Input: Teaching (course, tutorial, office meeting)
Plant: Students
Output: Outcome of learning
Feedback—The key process for good output
Score of Quiz
6 students
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>80
<60
6 students
3
Course Outline
Week
Date
Content
9
06 Nov Ch5: Root locus
Assignment
08 Nov Ch5: Root locus
10
13 Nov Ch5: Root locus
15 Nov Ch5: Root locus
11
20 Nov Ch6: Frequency response
Assign 5 Due
22 Nov Ch6: Frequency response
12
27 Nov Ch6: Frequency response
29 Nov Ch6: Frequency response
13
04 Dec Break
06 Dec Break
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Course Outline
Week Date
Content
14
11 Dec
Ch6: Frequency response
13 Dec
Ch7: Design of control systems
18 Dec
Ch7: Design of control systems
20 Dec
Ch7: Design of control systems
25 Dec
Ch7: Design of control systems
27 Dec
Review
Assign 7 Due
27 Nov
Review
Test
15
16
17
Assignment
Assign 6 Due
Exam
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Chapter 5: Root Locus
Contents
5-1 Introduction of Root Locus
5-2 Properties of Root Locus
5-3 Root Locus Sketching Rules
5-4 Root Locus Based System Analysis and
Design
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5-1 Introduction of Root Locus
What we have known?
• We have had a quick look at modeling and analysis for
specifying system performance.
(Time domain method for first and second order systems)
• We will now look at a graphical approach, known
as the root locus method, for analyzing and designing
control systems
• As we will see, the root locations are important in
determining the nature of the system response
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Proportional controller
R(s)
-
G (s )
K
C (s )
H (s )
We have seen how this form of feedback is able to minimize the
effect of disturbances.
For investigating the performance of a system, we have to solve the
output response of the system.
Limitations:
(1) It is difficult to solve the output response of a system –
Root Locus
especially for a higher order system.
(2) We can’t easily investigate the changes of a system’s
performance from the time-domain
--especially when parameters of the system vary in a given
range, or, when devices are added to the system.
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History:
Root locus method was conceived by Evans in 1948.
Root locus method and Frequency Response method,
which was conceived by Nyquist in 1938 and Bode in 1945,
make up of the cores of the classical control theory for
designing and analyzing control systems.
By using root locus method，we can：
 analysis the performance of the system
determine the structure and parameter of the system
 design the compensator for control system
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[Definition]：The path traced by the roots of the
characteristic equation of the closed-loop system as
the gain K is varied from 0 to +∞ is called root locus.
K
R(s)
C (s )
Example：The second order system
with the open-loop transfer
- s (0.5s  1)
function
K
Gk ( s) 
s(0.5s  1)
2K
Closed-loop Transfer function：  ( s )  2
s  2s  2 K
Characteristic equation:
s 2  2s  2 K  0
Roots of characteristic equation： s1, 2  1  1  2K
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Roots： s1, 2  1  1  2K
[Discussion]：
① For K=0，s1=0 and s2=-2,
are the poles of closed-loop system.
② For K=0.32，s1=-0.4,s2=-1.6
③ For K=0.5，s1=-1,s2=-1
④ For K=1，s1=-1+j,s2=-1-j
⑤ For K=5，s1=-1+3j,s2=-1-3j
⑥ For K=∞，s1=-1+∞j,s2=-1-∞j



 

K 5
K 1
K 0
2
1
0
j1
K 0
 j1
[Terminology]：In the root locus， “  ”denotes the poles of the
open-loop transfer function， “ ” zeros of the open-loop
transfer function. The bode line represents the root locus，arrow
shows the root locus direction along some parameters.
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5-2 Properties of Root Locus
Control System：
R(s)
-
C (s )
G (s )
G( s)
Closed-loop transfer ( s ) 
function:
1  G( s) H (s)
Open-loop transfer
Gk (s)  G(s) H (s)
function:
H (s )
m
Canonical form of
Gk (s)
Gk ( s )  K g 
 (s  z )
i
i 1
n
 (s  p )
j
K g  root locus gain
j 1
zi and p j are zerosand polesof open- loop transferfunction
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5.2.1 Two canonical form of open-loop
transfer function
K  ( s  1)
nm （4-1）
m
Gk ( s )  G ( s ) H ( s ) 

K g  ( s  zi )
 (s  p )
j
j 1
 (T s  1)
j
m
i 1
n
i 1
n
j 1
m
i
K  Kg
z
i 1
n

j 1
i
pj
Time constant
canonical form: K is
open-loop gain
（4-2）
Zero-pole canonical form: Kg is
root locus gain
In root locus method, we use (4-2）.
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5.2.2 Magnitude and Phase Equations
R(s) +
-
E(S)
G(s)
C(s)
H(s)
Characteristic equation of closed-loop transfer function
1  G(s)H  s   0
G(s) H  s   1
（4-3）
The objective is to determine values of s that
satisfy the equation. Such s are the on the root
locus.
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Definition
m
Gk ( s)  1 or k g 
 (s  z )
i
i 1
n
 (s  p )
 1 is the root-locus equation.
j
j 1
Remembering that s may be complex, it root-locus
equation can be equivalently written into
m
Kg 
 | (s  z ) |
i
i 1
n
 | (s  p ) |
Magnitude equation
1
j
j 1
m
n
i 1
j 1
Phase equation
Argument equation
 ( s  zi )   ( s  p j )  (2k  1) , k  0,1,2...
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Root locus method is performed in two stages:
1. Finding all values of s satisfying the argument equation
2. Finding particular values of s that satisfy the magnitude
equation
Phase equation is the sufficient and necessary condition
for root locus.
—— All the values of s satisfying the argument equation
constitute the root locus.
——Magnitude equation is used to determine the gain Kg
for a particular root.
——Argument equation is independent on Kg.
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Trial guess method：
Example 5.1
Gk ( s) 
k
s(2s  1)
Solution：
1）Find zeros and poles of open-loop transfer function on
s plane
K
0.5k
Gk (s) 
s(s  0.5)

g
s(s  0.5)
,
K g  0.5k
jω
Poles:
-p2
×
-0.5
-p1
×
0
σ
 p1  0,
 p2  0.5
There are no zeros.
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2）Inspecting the real axis：
s1>0

(s1  p1 )  (s1  p2 )  0
-< s1 <-p2 (s  p )  (s  p )  180
1
1
1
2
-p2
×
-0.5
Argument equation
is not satisfied.
jω
S1
-p1
×
0
σ
(s1  p1 ) (s1  p2 )  180  0  180
The all values between  p1 ~  p2 of the real axis are
-p2< s1 <-p1
the points of root locus
For s1=-0.1
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K g  s1  p1  s1  p2  0.1 0.4  0.04
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3）Examine locations off the real axis: arbitrarily
choose an s1
S1
•
In order to satisfy
 (s1  p1 )  (s1  p2 )  180
s1 should be on the
perpendicular bisector of the
line between the two poles.
-p2
×
-0.5
jω
-p1
-0.25
σ
×
0
Let s1=-0.25 + j0.25
 (s1  p1 )  (s1  p2 )  1350  450  1800
2
K g  s1  p1  s1  p2  0.25 2  0.125
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Example: For the second order system，draw the root
locus with respect to Kg from 0→∞
kg
(s)  2
Solution:
kg
s  s  k g R(s)
C (s )
Characteristic equation of closed-loop transfer s( s  1)
function：
1 1
2
s  s  k g  0，s1, 2   
1  4k g
2 2
[Discussion]：
①
For kg  0，we haves1,2  0 and -1.
②
k g  ，s1 goes to the leftalong the negative real axis，
and s2 goes to theright along negativereal axis from 1.
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kg
R(s)
C (s )
s( s  1)
-
1
1
③ For k g  ，s1, 2   .
4
1 2
For any 0  k g  ，s1, 2 are on thenegativereal axis.
4
1
④ For k g  ，s1, 2 are complex numbers .
4
As Kg increases, location of roots branch out along the
vertical line.
1
2
⑤ k g  时， s1, 2    j
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kg  
[Summary]
When Kg changes from zero to infinity,
there are two segments starting from the
open-loop poles to infinity.
For higher order system, it is very
difficult to sketch all the root loci. But
graphical method shows its advantage.
s 1  s 
A

 A2
B 1

s( s  1)


A'
 A1
kg  0 

0 .5 0 k g  0
kg  
Are Points A and A’ on the root locus？(Trial guess
method)
kg
j1
 j1

 s  ( s  1)   OA  BA   A1   A2  (2k  1)
Obviously，A is on the root locus, but A’ is not on the root locus.
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Example 5-2: The open-loop transfer function is given by
k ( s  4)
G ( s) H ( s) 
s ( s  2)(s  6.6)
Check if s1= -1.5+j2.5 is on the root locus. If yes, please
determine the parameter gain k
1)Via argument equation
( s1  z1 )  ( s1  p1 ) 
( s1  p2 )  ( s1  p3 )
S1
k =12.15
 45 - 120 - 79 - 26  180
2)Via magnitude equation
k
26O 45O 79O 120O
-P2 -P1
-P3 -Z1
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s1  p1  s1  p2  s1  p3
s1  z1
2 . 9  2 . 6  5 .8

 12.15
3 .6
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We can rapidly sketch the root-loci of a control system.
Sketch the root-loci for the following open-loop transfer functions:
K ( s  2)(s  4)
s( s  6)
K ( s  2)(s  6)
(1) G ( s) 
s( s  4)
(2) G ( s) 
K ( s  2)(s  3)
(4) G( s) 
s( s  1)
K ( s 2  6s  13)
(5) G( s) 
s( s  2)
Im
Im
Re
Im
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(6) G ( s) 
K ( s  3)
s( s  2)
K ( s  1)(s  2)
( s 2  6s  13)
Im
Re
Im
Re
(3) G( s) 
Re
Im
Re
Re
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Summary



The preceding angle and magnitude criteria can be used to
verify which points in the s-plane form part of the root
locus
It is not practical to evaluate all points in the s-plane to
find the root locus
We can formulate a number of rules that allow us to sketch
the root locus
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