1. Salman Bin Abdulaziz University
EE3511: Automatic Control Systems
Root Locus
Dr. Ahmed Nassef
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2. Salman Bin Abdulaziz University
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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3. Salman Bin Abdulaziz University
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
E(s)
Y(s) _
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5. Root Locus (Introduction)
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6. Root Locus (Introduction)
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7. Root Locus (Introduction)
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8. Root Locus (Introduction)
At K = 0, C.L. poles = O.L. poles
At K  ∞, C.L. poles = O.L. zeros
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9. 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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10. Root Locus (Asymptotes)
Number of asymptotes = n – m
= (# O.L. poles) – (# O.L. zeros)
Angle of asymptotes,
If n – m = 1 α = 180
if n – m = 2 α = +90, -90
If n – m = 3 α = +60, -60, 180
If n – m = 4 α = +45, -45, +135, -135
Position of intersection of asymptotes
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First-Order Example
The closed loop transfer function is
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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12. 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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13. Root Locus (First-Order System Example 2)
E(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. X
– 2
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14. Root Locus (Second-Order System Example 1)
E(s)
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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15. Root Locus (Second-Order System Example 1)
for 0<K ≤1, we have two real poles located at
For K > 1, we have a pair of complex conjugate poles at: X X
Root Locus -2
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16. Root Locus Example: Step Responses
Time (sec.)
Amplitude
Step Responses 2 4 6 8 10
0.2
0.4
0.6
0.8 1
1.2
1.4
1.6
K = 0.5
K = 1.0
K = 2.0
K = 15.0
K = 50.0 jw s X
– 2
K = 0
K = 1
K  
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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17. 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)
Y(s) _
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 X
– 3
– 2
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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18. Example (odd number rule)
No Yes Yes No Yes No
Double poles X
Yes No Yes No Yes No
Complex poles or zeros do not affect the count
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19. Salman Bin Abdulaziz University
Example 1
Draw Root Locus of the following system _ X
Yes No Yes No
Segments of real axis
n = 3, m = 0
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20. Salman Bin Abdulaziz University
Example 1
# of asymptotes = n – m = 3 60 X X X -2 -1
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21. 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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22. Salman Bin Abdulaziz University
Breakaway points
At breakaway points
Solve the above equation to determine the breakaway point. Select solution that are in segments of real axis that is part of root locus
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23. Salman Bin Abdulaziz University
Breakaway points
To find breakaway points X X X -2 -1
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24. Salman Bin Abdulaziz University
Root locus of Example 1 60 X X X -2 -1
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25. 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) = k
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26. Salman Bin Abdulaziz University
Example 2
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Departure Angles
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28. Salman Bin Abdulaziz University
Departure Angles
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29. Salman Bin Abdulaziz University
Root Locus Rules
Plot O.L. poles and zeros of the O.L. gain function;
Identify # asymptotes n-m
Where, n = # O.L. poles and m = # O.L. zeros.
Determine the asymptotes angles from:
Determine the position of intersection along the real axis, if n – m > 1 from:
Use odd rule to sketch the loci along the real axis
Determine the break-away or break-in points along the real axis using:
Calculate the angle of departure or angle of arrival using angle condition.
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30. Salman Bin Abdulaziz University
Example 3
Draw root locus of the following system _
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31. Salman Bin Abdulaziz University
Example 3 X X X
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33. Salman Bin Abdulaziz University
Example 4
How do we draw root locus for this system? _
Initial Step: Express the characteristics as
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34. Example 4 Initial Step: Express the characteristics as
Continue the Root Locus Procedure
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35. Salman Bin Abdulaziz University
Poles = [ i i]
Centroid = -3
Angles = 60, -60, 180
Intersection points =+-sqrt(2)
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36. Salman Bin Abdulaziz University
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37. Salman Bin Abdulaziz University
Example 5
How do we draw root locus for this system? _
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38. Salman Bin Abdulaziz University
Example 5
Initial Step: Express the characteristics as
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39. Salman Bin Abdulaziz University
Example 5 β X +j X
26.6o X -j X +j X X -j
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40. Salman Bin Abdulaziz University
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41. Examples of Root Locus sketches
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