Download SE302: Linear Control Systems

EE3511: Automatic Control Systems
Root Locus
Dr. Ahmed Nassef
EE3511_L11
Salman Bin Abdulaziz University
1
Learning Objectives
Understand the concept of root locus
and its role in control system design
 Recognize the role of root locus in
parameter design and sensitivity analysis

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Root locus
Location of the roots of the characteristic
equation (Poles) determines systems
response.
 Modifying one or more of the system’s
parameter cause the roots of the
characteristic equation to change.

 Root
locus is a graphical method
that describes the location of the
poles as one parameter change.
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Question
R(s)
E(s)
_
K
G (s )
Y(s)
H (s )
Y (s)
KG( s )
If T ( s ) 

,
R ( s ) 1  KG( s ) H ( s )
how do the roots of the characteri stic equation
1  KG( s ) H ( s )  0
change as K varies from 0 to  ?
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Root Locus (Introduction)
 1  KG( s) H ( s)  0
 KG( s) H ( s)  1,
however, G ( s ) H ( s ) is a complex vector
 KG( s) H ( s)  1   (2 N  1)180, N  0, 1, 2, ...
So, KG( s) H ( s )  1
( Magnitude Condition)
and KG( s) H ( s)  (2 N  1)180
( Angle Condition )
Let the open loop TF , GH , is in the form :
( s  z1 )( s  z 2 )( s  z3 )...( s  z m )
GH 
( s  p1 )( s  p2 )( s  p3 )...( s  pn )
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Root Locus (Introduction)
So from magnitude condition,
( s  z1 )( s  z 2 )( s  z3 )...( s  zm )
K
1
( s  p1 )( s  p2 )( s  p3 )...( s  pn )
Z1Z 2 Z 3 ...Z m
K
1
P1 P2 P3 ...Pn
P1 P2 P3 ...Pn
Magnitude of OL poles
or , K 

Z1Z 2 Z 3 ...Z m Magnitude of OL zeros
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Root Locus (Introduction)
So from angle condition ,
( s  z1 )( s  z 2 )( s  z3 )...( s  z m )
K
 (2 N  1)180
( s  p1 )( s  p2 )( s  p3 )...( s  pn )
Z11Z 2  2 Z 3 3 ...Z m  m
 K
P11 P2  2 P3 3 ...Pn  n
or , (1   2   3 ...   m )  ( 1   2   3 ...   n )   (2 N  1)180
m
n
i 1
i 1
  Zeros angles   Poles angles   (2 N  1)180
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Root Locus (Introduction)
From system Ch. Eq.,
( s  z1 )( s  z 2 )( s  z3 )...( s  z m )
1 K
0
( s  p1 )( s  p2 )( s  p3 )...( s  pn )
 (s  p1 )( s  p2 )( s  p3 )...( s  pn )  K[( s  z1 )( s  z2 )( s  z3 )...( s  zm )]  0
Note that S here are the closed loop poles
At K = 0, C.L. poles = O.L. poles
At K  ∞, C.L. poles = O.L. zeros
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Root Locus (Introduction)
• As K changes from 0  ∞ the root locus
starts from O.L. poles and ends at O.L. zeros.
•For a system with only one O.L. pole and no
zeros, the loci approaches from that pole (as
K=0) and reaches a zero at infinity along an
asymptote.
•If there are two poles, they will reach two
zeros at infinity along two asymptotes.
•So, in general the pole is looking for its zero
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Root Locus (Asymptotes)
• Number of asymptotes = n – m
= (# O.L. poles) – (# O.L. zeros)
 (2 N  1)180
• Angle of asymptotes,  
nm
If n – m = 1
if n – m = 2
If n – m = 3
If n – m = 4
α = 180
α = +90, -90
α = +60, -60, 180
α = +45, -45, +135, -135
• Position of intersection of asymptotes
O.L. poles   O.L. zeros

o 
nm
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First-Order Example

The closed loop transfer function is
C ( s)
K

R( s ) s  1  K



which has a single, real pole at s = −(K + 1).
At K = 0,
s = −1
At K= ∞,
s = −∞
So, as K increases, the locus starts from -1 and
moves to the left towards ∞ along the real axis .
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Root Locus
(First-Order System Example 1)
There is only one O.L. pole
at -1
# of asymp. =n-m=1-0=1
Angle of asymp.=180o
The pole is looking for a
zero at infinity along one
asymptote with angle 180o
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Root Locus
(First-Order System Example 2)
2K
Ch. Eq. s 
0
K 1
2K
s
K 1
At K  0 s  0
At
K   s  2
E(s)
R(s)
K
_
( s  2)
s
Y(s)
Root Locus

OL pole at 0
 OL zero at -2
 # of asymp. =n-m=1-1=0
 So, as K increases, the loci
starts from 0 and moves
towards -2 along the real axis.
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X
–2
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0
13
Root Locus (Second-Order System Example 1)
The characteri stic equation
s  2s  K  0
2
E(s)
R(s)
 2  4  4K
roots are : r1, 2 
2
r1  1  1  K , r2  1  1  K
_
K
1
s ( s  2)
Y(s)
At K = 0, s1=0 and s2=−2 (CL poles = OL poles).
 At K= 1 s1=-1 and s2=−1
 At K= ∞, s1=-1+j∞ and s2 = -1−j∞ (CL poles = zeros at ∞)

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Root Locus (Second-Order System Example 1)

for 0<K ≤1, we have two real poles located at
r1  1  1  K , r2  1  1  K

For K > 1, we have a pair of complex
conjugate poles at:
r1  1  j K  1, r2  1  j K  1
Root Locus
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X
-2
X
0
15
Root Locus Example:
Step Responses
Step Responses
1.6
Amplitude
1.4
K = 50.0
K = 15.0
1.2
K = 2.0
1
0.8
K = 1.0
0.6
K = 0.5
K=1
X
–2
0.4
s
X
K=0
0.2
0
0
jw
K
2
4
6
8
10
Time (sec.)



OL poles at 0 and -2
# of asymp. =n-m=2-0=2 their angles are +90 and -90
So, as K increases, the loci starts simultaneously from 0 and -2
and moves along the real axis until it breaks away at -1 to ±j∞.
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Root Locus (Second-Order System Example 2)





OL poles at 0 and -3
OL zero at -2
# of asymp. =n-m=2-1=1
Asymp. angle is 180o.
E(s)
R(s)
_
K
( s  2)
s ( s  3)
As K increases, one part of the locus starts
from the pole at 0 and ends at the a zero at -2
(along the real axis)
The other part starts at the pole at -3 and ends at a
zero at -∞ along the asymptote.
X
–3
–2
Y(s)
X
0
Rule: A segment of real axis is a part of root locus if and
only if it lies to the left of an odd number of poles and
zeros.
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Example (odd number rule)
Double poles
6
No
5
Yes
X
X
X
X
3
Yes
2
1
0
No
Yes
No
Complex poles or
zeros do not affect
the count
X
X
5
Yes
EE3511_L11
4
No
X
3
Yes
X
2
No
X1
Yes
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0
No
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Example 1

Draw Root Locus of the following system
K
s 3  3s 2  2 s
1  K G( s) H ( s)  0 _
1
1 K 3
0
2
s  3s  2s
1
1 K
0
s( s  1)( s  2)
n = 3, m = 0
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Segments of real axis
X
X
X
3
2
1
Yes
No
Yes
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0
No
19
Example 1

# of asymptotes = n – m = 3
centroid   o
poles   zeros {0  1  2}  {0}



 1
nm
30
2N 1
Angles   A 
180, N  0,1,..., n  m  1  2
nm
  A  60, 180, 300
60
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X
X
-2
-1
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X
20
Root Locus Procedure
Breakaway points
Determine the breakaway points on
the real axis (if any):
 Breakaway points are points at which the
root loci breakaway from real axis or the
root loci return to real axis.
 The locus breakaway from the real axis
occurs where there is a multiplicity of
roots.

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Breakaway points

At breakaway points
dK
0
ds

Solve the above equation to determine the
breakaway point. Select solution that are in
segments of real axis that is part of root locus
1  K G( s) H ( s)  0
1
K 
G( s) H ( s)
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 K  s  3s  2s
Salman Bin Abdulaziz University
3
2
22
Breakaway points

To find breakaway points


d 3
2
s  3s  2 s  0
ds
2
3s  6 s  2  0
X
-2
X
-1
X
break away points  1.5774 and  0.4226
Select  0.4226
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Root locus of Example 1
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X
X
-2
-1
60
X
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Angle of departure and arrival
Determine the the angle of locus
departure from complex poles and the
angle of locus arrival at complex
zeros, using the phase criterion
 Sum of angle contributions of poles and
zeros (measured with standard
reference) = 180+360k

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Example 2
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Departure Angles
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Departure Angles
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Root Locus Rules
1. Plot O.L. poles and zeros of the O.L. gain function;
2. Identify # asymptotes n-m
Where, n = # O.L. poles and m = # O.L. zeros.
3. Determine the asymptotes angles from:   (2 N  1)180
nm
4. Determine the position of intersection along the real
O.L. poles   O.L. zeros
axis, if n – m > 1 from:

o 
nm
5. Use odd rule to sketch the loci along the real axis
6. Determine the break-away or break-in points along the
real axis using: dk  0
ds
7. Calculate the angle of departure or angle of arrival using
angle condition.
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Example 3

Draw root locus of the following system
_
K ( s  5)
( s  1)( s  2)( s  3)
Characteristic Equation :
s5
1 K
0
( s  1)( s  2)( s  3)
s5
G(s) 
, n  3, m  1
( s  1)( s  2)( s  3)
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Example 3
s5
G (s) 
, n  3, m  1
( s  1)( s  2)( s  3)
Poles :  1,  2,  3 , Zero : 5
# of symptotes  3  1  2
Angles 90,  90
 1  2  3  (5)  1
Centroid 

3 1
2
dk
 0  2 s 3  21s 2  60 s  49
-5
ds
roots are  6.42,  2.62 and  1.45
X
-3
X
-2
X
-1
select  1.45
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root locus of
s5
1 K
0
( s  1)( s  2)( s  3)
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Example 4

How do we draw root locus for this system?
_
1
3
2
s  3s  2s  K
Initial Step: Express the characteristics as
1  K G (s)  0
What is G ( s ) ?
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Example 4
Initial Step: Express the characteristics as
1
Characteri stic Equation : 1  3
0
2
s  3s  2 s  K
s 3  3s 2  2 s  K  1  0  s 3  3s 2  2 s  1  K


1
1 K 3
0
2
s  3s  2 s  1
1
 G (s)  3
s  3s 2  2 s  1
Continue the Root Locus Procedure
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
Poles = [-2.3247
-0.3376 +0.5623i
-0.3376 - 0.5623i]
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Centroid = -3
 Angles = 60, -60, 180


Intersection points
=+-sqrt(2)
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Example 5

How do we draw root locus for this system?
R(s)
_
EE3511_L11
s2
s( s 2  2s  K )
Salman Bin Abdulaziz University
C (s )
37
Example 5
Initial Step: Express the characteristics as
s2
Char acteristic Equation : 1 
0
2
s( s  2s  K )
s ( s 2  2 s  K )  s  2  0  s 3  2 s 2  Ks  s  2
s
3

 2 s 2  s  2  Ks
1 K
s
0
s  2s  s  2
s
 G (s)  3
s  2s 2  s  2
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3
2
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Example 5
We have a zero at 0 and poles at  2,  j
β
X +j
# Asymptotes  2, angles  90,90
int er sec tion at    1
-3
X
-2
26.6o
-1
X -j
from angle condition :
 zeros   poles  180
90  (26.6  90   )  180
  180  26.6  153.4
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o
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X +j
-3
X
-2
-1
X -j
39
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Examples of Root Locus sketches
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