A root locus plot is simply a plot of the s zero values and the s poles on a graph with real and imaginary coordinates. This method is very powerful graphical technique for investigating the effects of the variation of a system parameter on the locations of the closed loop poles.

1. ROOT LOCUS
KHALID SAEED AL BADRI

2. A system to automatically
track a subject in a visual
image can be modeled as
follows

3. A root locus plot is simply a
plot of the s zero values and
the s poles on a graph with real
and imaginary coordinates.

4. • This method is very powerful
graphical technique for
investigating the effects of the
variation of a system parameter
on the locations of the closed
loop poles.

5. The root locus of an (open-loop)
transfer function G(s) is a plot of the
locations of all possible closed loop
poles and zeros.

6. Poles and zeros
)())((
)())((
)(
21
21
n
m
pspsps
zszszsk
sF





n
m
ppp
zzz


,,
,,
21
21 zeros
poles
axisRe
axisIm
pole
zero

7. Root Locus Sketching Rules
Rule 1:Find the number of
poles Np and zeros Nz
Rule 2:Root locus is
symmetric about the
real axis.
Rule 3: Along the real
axis, the root locus
includes all
segments that are to
the left of an odd
number of poles and
zeros.

8. Root Locus Sketching Rules
Rule4: find the asymptotic location at 0
and angles k.

9. Example 1
Sketch the root locus of the closed-loop
system.
Ei
Plant G(s)Controller
DV
16
0 0174 1s s( . )KP0.03
V
D
0.03
   
 
 
 
1 0
0.48
1 0
0.0174 1
p
N s
p
D s
K G s H s
K
s s
 
 


10. Example 1
Step 1:Transform the closed-loop characteristic
equation into the standard forms:
Step 2:Find the open-loop zeros, zi , and the open-
loop poles, pi :
 
 
 
1
1 27.58 0
57.47
N s
p
D s
K
s s
 

No open-loop zeros
open-loop poles 47.57,0 21  pp
K

11. Example 1
Step 3:Determine the real axis segments that are to
be included in the root locus.
1 0p 2 57.47p  

12. Example 1
Step 4: Determine the asymptotes 0 and angles k .
(2 1) [rad]k
P Z
k
N N


 

2
2


 


0
i i
P Z
p z
N N




  57.47
28.74
2

  

13. Example 1
Step 5: Determine the break-away and break-in points .
( ) ( )
0 or 0,
( ) ( )
d N s d D s
ds D s ds N s
   
    
   
 0.0174 1
0,0.0348 1 0, 28.74
0.48
s sd
s s
ds
 
     
 

14. Example 1
-60 -50 -40 -30 -20 -10 0
Real Axis
-30
-20
-10
0
10
20
30
Img. Axis
-57.47
-28.74

15. Example 2
A positioning feedback control system is proposed.
The corresponding block diagram is:
Sketch the root locus.
Y(s)U(s)
Plant G(s)
Controller
R(s) 16
0 0174 1s s( . )
K(s + 80)
+


16. Example 2
122
-140 -120 -100 -80 -60 -40 -20 0
-40
-30
-20
-10
0
10
20
30
40
Real Axis
Imag Axis
1 0p 2 57.47p   37.6

17. Example 3
A feedback control system is proposed. The corresponding
block diagram is:
Sketch the root locus.
Y(s)U(s)
Plant G(s)Controller
R(s) 1
4 202
s s s( ) 
+

K
s( ) 4
     
 2
1 0
1
1 0
4 4 20
cG s G s H s
K
s s s s
 
 
  
Find closed-loop characteristic equation:

18. Example 3
Step 1:
Step 2:
Step 3:
 
  
 
2
1
1 0
4 20 4
N s
D s
K
s s s s
 
  
1 2 3,40, 4, 2 4p p p j     
open-loop zeros
open-loop poles
No open-loop zeros
1 0p 2 4p  

19. Example 3
Step 4:
Step 5:
(2 1) [rad]k
P Z
k
N N


 

4
3
4
5
4
7
4









 





0
i i
P Z
p z
N N




       0 4 2 4 2 4
2
4 0
j j       
  

( ) ( )
0 or 0,
( ) ( )
d N s d D s
ds D s ds N s
   
    
   
 
 
  
 
2
4 3 2
3 2
8 36 80
1
4 24 72 80
4 20 4
0
D sd d d
s s s s
ds N s ds d
s s s s
s
s s s
  
      
 
 
   
    

1 2,32, 2 2.45s s j    

20. Example 3
Step 6:Determine the imaginary axis crossings
  2
1
1 0
4 20 4
K
s s s s
  
  
  2
4 3 2
4 20 4 0
8 36 80 0
s s s s K
s s s s K
    
     
s j
   4 2 3
36 8 80 0K j        
4 2
21
3
1 2
260036 0
,
08 80 0 10 3.16
KKK 
  
     
   
     

21. Example 3
-6 -5 -4 -3 -2 -1 0
-4
-3
-2
-1
0
1
2
3
4
Real Axis
Imag Axis

22. Example 1/ MATLAB
 
 
 
1
1 27.58 0
57.47
N s
p
D s
K
s s
 

num=[0 1];
den=[1 57.47 0];
rlocus(num,den)

23. Example 2/ MATLAB
num=[1 80];
den=[1 57.47 0];
rlocus(num,den)

24. Example 3/ MATLAB
num=[0 1];
den=[1 8 36 80 0];
rlocus(num,den)
  2
1
1 0
4 20 4
K
s s s s
  
  

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