1. Prof. Wahied Gharieb Ali Abdelaal
Faculty of Engineering
Computer and Systems Engineering Department
Master and Diploma Students
CSE 502: Control Systems (1)
Topic# 8
Root Locus Technique
Prof. Wahied Gharieb Ali Abdelaal

2. TOPICS
Introduction
Basics of the Root Locus
Summary of the Root Locus steps
Examples

3. Introduction
Importance of poles and zeros of the closed loop transfer function of a linear control system:
The behavior of the roots of the characteristic equation (poles of the closed loop transfer function) affect the system stability
The transient behavior of the system is governed by the poles and zeros of the closed loop transfer function

4. (usually gain, K) is varied from 0 to infinity.
Introduction
The closed-loop poles of the negative feedback control:
are the roots of the characteristic equation:
The root locus is the locus of the closed-loop poles when a specific parameter
(usually gain, K) is varied from 0 to infinity.

5. Introduction
The value of s in the s-plane that make the loop gain KG(s)H(s) equal to -1 are the closed-loop poles
(i.e )
KG(s)H(s) = -1 can be split into two equations by equating the magnitudes and angles of both sides of the equation.

6. Angle and Magnitude Conditions
Introduction
Angle and Magnitude Conditions
Independent of K

7. Learning by doing – Example 1
Basics of Root Locus
Learning by doing – Example 1
Sketch the root locus of the following system:
Determine the value of K such that the damping ratio ζ of a pair of dominant complex conjugate closed-loop is 0.5.

8. Basics of Root Locus
Step# 1: Draw the n poles and m zeros of the open loop G(s)H(s) using X and O respectively.
The n branches of the root locus start at the n poles.
m of these n branches end on the m zeros
The n-m other branches terminate at infinity along asymptotes.

9. Basics of Root Locus Applying Step #1
Draw the n poles and m zeros of G(s)H(s) using X and O respectively.
Three open loop poles:
p1 = 0; p2 = -1; p3 = -2
No finite zeros

10. Basics of Root Locus Step# 2:
Determine the loci on the real axis. Choose an arbitrary test point. If the TOTAL number of both real poles and zeros is to the RIGHT of this point is ODD, then this point is on the root locus
The loci on the real axis are to the left of an ODD number of REAL poles and REAL zeros of G(s)H(s)

11. Basics of Root Locus Applying Step #2
Determine the loci on the real axis:
Choose an arbitrary test point.
If the TOTAL number of both real poles and zeros is to the RIGHT of this point is ODD, then this point is on the root locus

12. Basics of Root Locus Step# 3
Determine the n - m asymptotes of the root loci. Locate s = α on the real axis. Compute and draw angles. Draw the asymptotes using dashed lines.
The root loci for very large values of K must be asymptotic to straight lines originate on the real axis at point:
radiating out from this point at angles:

13. Basics of Root Locus Applying Step #3 Determine the n - m asymptotes:
Locate s = α on the real axis:
Compute and draw angles:
Draw the asymptotes using dash lines.

14. Basics of Root Locus Step# 4 Find the breakpoints. Express K such as:
Set dK/ds = 0 and solve for the poles.
The breakpoints are the closed-loop poles that satisfy:

15. Basics of Root Locus Applying Step #4 Find the breakpoints.
Express K such as:
Set dK/ds = 0 and solve for the poles.

16. Basics of Root Locus Last step:
Draw the n-m branches that terminate at infinity along asymptotes
The n branches of the root locus start at the n poles.
m of these n branches end on the m zeros
The n-m other branches terminate at infinity along asymptotes.

17. Basics of Root Locus Applying Last Step
Draw the n-m branches that terminate at infinity along asymptotes

18. Basics of Root Locus Using MATLAB >>num=[1]
>>den=[ ]
>>sys=tf(num,den)
>>rlocus(sys)

19. Points on both root locus & imaginary axis?
Basics of Root Locus
Points on both root locus & imaginary axis?
Points on imaginary axis satisfy:
Points on root locus satisfy:
Substitute s=jω into the characteristic equation and solve the phase equation for ω.
Then calculate K from the gain equation at this value of
jω?
- jω

20. Learning by doing – Example 1
Basics of Root Locus
Learning by doing – Example 1
Sketch the root locus of the following system:
Determine the value of K such that the damping ratio ζ of a pair of dominant complex conjugate closed-loop is 0.5.

21. Summary of Root Locus Root Locus Basic Steps:
The Loci starts (K=0) at the poles of G(s) and terminates (K=) at the zeros of G(s) at infinity
2. The Loci exits on the real axis only to the left of an odd number of real poles and zeros of G(s)
3. If the loci leaves the real axis, it must be symmetrical about this axis because complex poles appear in complex conjugate pairs
4. Loci which terminates at  must follow asymptotes
which make an angle  with the real axis where :
P=number of poles , Z: number of finite zeros l=0,1,2,…

22. Summary of Root Locus Steps
5. The asymptotes intersect at point  on the real axis given by:
P: Sum of the poles of G(s)
Z: Sum of the zeros of G(s)
P, Z: The number of poles and zeros of G(s)
6. The point of Breakaway from the real axis can be determined by solving the equation (where 1+G(s)=0 must be satisfied).
7. The point at which the loci cross the imaginary axis can be determined by letting s=j in the characteristic equation.

23. Examples
Example 2
Let s=j  s+2=(+2)j
K=0
K=4
K=6 -2 -6
-10 ….

24. Examples Example 3 One zero at: s=-3 One pole at: s= 1-3K/K+1
Consider
There is a pole at s=1
The system is unstable. Suppose we want to make it stable and we want to find the poles of the closed-loop as a function of K using the unity feed back.
One zero at: s=-3
One pole at: s= 1-3K/K+1
K=1
K=3
K= -1 -2 -3
End
K=1/ 3
K=0 +1

25. Examples Example 4 Consider:
The are no finite zeros. Therefore, all loci must terminate at infinity.
2. The loci must exits on the real axis in the regions: -4<s<0 and s<-10
3. The loci must be symmetrical when the have to leave the real axis.
4. The angle of the asymptotes must be:

26. Examples Example 4 5. The asymptote will intersect the real axis at:
6. The point of breakaway from the real axis can be determined from:
The C.E. is given by 1+G(s)=0
K+s(s+4)(s+10)=0

27. Examples
Example 4
s=-7.57 (not valid because it does not belong to the loci) or s=-1.76 which is valid
7. The point at which the loci cross the imaginary axis is given by replacing s=j in the C.E.
=0 and =40
K=0 K= 560

28. Examples

29. Examples

30. Examples