1. Chapter 5
Root Locus

2. In previous lecture Chapter 4 Routh Hurwitz Criterion
Steady-state error analysis

3. Today’s plan
Construction of Root Locus

4. Learning Outcomes
At the end of this lecture, students should be able to:
Sketch the root locus for a given system
Identify the stability of a system using root-locus analysis

5. Introduction
Root Locus illustrates how the poles of the closed-loop system vary with the closed-loop gain.
Graphically, the locus is the set of paths in the complex plane traced by the closed-loop poles as the root locus gain is varied from zero to infinity.

6. KGc(s)H(s) 1 + KGc(s)H(s) =0 K Gc H-
C(s) K Gc
R(s) H
block diagram of the closed loop system
given a forward-loop transfer function;
KGc(s)H(s)
where K is the root locus gain, and the corresponding closed-loop transfer function
the root locus is the set of paths traced by the roots of
1 + KGc(s)H(s) =0
as K varies from zero to infinity. As K changes, the solution to this equation changes.

7. So basically, the root locus is sketch based on the characteristic equation of a given transfer function.
Let say:
Thus, the characteristic equation :

8. Root locus starts from the characteristic equation.
Split into 2 equations:
Magnitude condition
Angle condition
where

9. If not satisfied Not part of root locus
Remarks
If Si is a root of the characteristic equation, then 1+G(s)H(s) = 0 OR Both the magnitude and Angle conditions must be satisfied
If not satisfied Not part of root locus

10. Example 1 -1 -2

11. Let say my first search point, S1-1 -2
(A) Satisfy the angle condition FIRST
(B) Magnitude condition to find K

12. Let say the next search point-1 -2
Check whether satisfy angle condition
Since angle condition was not satisfied  not part of root locus

13. CONSTRUCTION RULES OF ROOT LOCUS
8 RULES TO FOLLOW

14. Construction Rules of Root Locus
Assume
Then

15. Thus

16. Rule 1 : When K = 0  Root at open loop poles
The R-L starts from open loop poles
The number of segments is equal to the number of open loop poles

17. Rule 2: When K = ∞
The R-L terminates (end) at the open loop zeros

18. Rule 3: Real-axis segments
On the real axis, for K > 0 the root locus exists to the left of an odd number of real-axis, finite open-loop poles and/or finite open-loop zeros
Example 2
R-L doesn’t exist here
R-L exist here
R-L exist here

19. Rule 4: Angle of asymptote
NP = number of poles
NZ = number of zeros

20. Example 3
NP=3
NZ=0
Centroid

21. Rule 5: Centroid
From example 3
Recall
Thus, -1

22. Rule 6: Break away & break in points-1 -1 -2 -2
Break away point
Break in point
How to find these points ?

23. Differentiate K with respect to S & equate to zero
How ?
Solve for S  this value will either be the break away or break in point

24. Let say
Thus,

25. Now, find
Thus,
So, notice that

26. Example 4
Solution: -1 -2

27. Example 5 -1 -2
NP = 3
NZ = 0

28. -1 -2

29. Break away point
Invalid, why ???

30. How to determine these values ?-1 -2
-0.42

31. From characteristic equation
Construct the Routh array

32. S3 1 2 S2 3 K S1 6-K/3 S0 Since Force S1 row to zero or K=6
Replace K=6 into S2 row

33. -1 -2
-0.42

34. Rule 7 : Angle of departure (arrival)
Finding angles of departure and arrival from complex poles & zero
Root locus start from open loop poles
 Angle of departure
Root locus terminates at open loop zeros
 Angle of departure

35. Angle of Departure

36. To make it easier to remember:
Replace in this equation
Let’s change:
and let k=0, we get
Angles due to poles, keep using symbol θ
Since Φp= Φp+360o
Re-arrange
Angles due to zeros, change to symbol β
Thus, the angle of departure:

37. Angle of Arrival

38. To make it easier to remember:
Replace in this equation
Let’s change:
and let k=0, we get
Angles due to poles, keep using symbol θ
Re-arrange
Thus, the angle of arrival:
Angles due to zeros, change to symbol β

39. Angle of Departure
Angle of Arrival

40. Example 6 Step 1: Determine poles & zeros Np=3 Nz=0 poles at :
Step 2: Maps poles & zeros on the s-plane 4 -3 -4

41. Step 3: Calculate angle of asymptote

42. Step 4: Determine centroid
poles:
Step 5: Sketch asymptotes line with dotted line 4 -3 -2 -4

43. Step 6: Determine the break-away or break in points
Complex numbers
NO break away or break in points
 only exist on real-axis of s-plane

44. Step 7: Since complex poles exist, calculate the angle of departure4
due to s=0
due to its complex conjugate -3 -4
How to determine θ?

45. To determine θ 4 -3
Thus: -4
Angle of departure :
-36.87o
Cross Hair

46. Step 8: Determine jω-crossing
see RULE 8 : After this
Step 9: Sketch the root locus
-36.87o 4
Rule 8 -3 -2 -4
36.87o

47. Rule 8 : R-L crosses jω-axis
From Characteristic equation:

48. Construct the Routh array1 25 s2 6 K s1
(6(25)-K)/6=
(150-K)/6 s0

49. Force row s1 to be zero, thus 150-K=0; K = 150
Substitute K = 150 into row s2
Since s = jω
Thus

50. Final Sketch 5
-36.87o 4 -3 -2
10 points -4
36.87o -5

51. Try it
Sketch the root locus for:

53. How to use Matlab to sketch root locus
In Matlab:
num = [1 7 12]
den = [1 3 2]
rlocus(num,den)

54. Figure 8:10 page 438

55. Any Q’s ???