1. Feedback Control System THE ROOT-LOCUS DESIGN METHOD Dr.-Ing. Erwin Sitompul Chapter 5 http://zitompul.wordpress.com

2. 6/2 Erwin SitompulFeedback Control System Root Locus: Illustrative Example Examine the following closed-loop system, with unity negative feedback. The closed-loop transfer function is given as: The roots of the characteristic equation are: The characteristic equation The denominator of the closed-loop transfer function

3. 6/3 Erwin SitompulFeedback Control System Root Locus: Illustrative Example K=0 K=1/4 K=∞ : the poles of open-loop transfer function

4. 6/4 Erwin SitompulFeedback Control System Root Locus: Illustrative Example Where are the location of the closed-loop roots when K=1? K=1 –0.5 –0.866 0.866  There is a relation between gain K and the position of closed-loop poles, which also affects the dynamic properties of the system (ζ and ω d )

5. 6/5 Erwin SitompulFeedback Control System Root Locus of a Basic Feedback System The closed-loop transfer function of the basic feedback system above is: The characteristic equation, whose roots are the poles of this transfer function, is:

6. 6/6 Erwin SitompulFeedback Control System Root Locus of a Basic Feedback System To put the characteristic equation in a form suitable for study of the roots as a parameter changes, it is rewritten as: where  K is the gain of controller-plant-sensor combination.  K is selected as the parameter of interest.  W. R. Evans (in 1948, at the age of 28) suggested to plot the locus (location) of all possible roots of the characteristic equation as K varies from zero to infinity  root locus plot.  The resulting plot is to be used as an aid in selecting the best value of K.

7. 6/7 Erwin SitompulFeedback Control System Root Locus of a Basic Feedback System The root locus problem shall now be expressed in several equivalent but useful ways.  The equations above are sometimes referred to as “the root locus form of a characteristic equation.”  The root locus is the set of values of s for which the above equations hold for some positive real value of K.

8. 6/8 Erwin SitompulFeedback Control System Root Locus of a Basic Feedback System  Explicit solutions are difficult to obtain for higher-order system  General rules for the construction of a root locus were developed by Evans.  With the availability of MATLAB, plotting a root locus becomes very easy, using the command “rlocus(num,den)”.  However, in control design we are also interested in how to modify the dynamic response so that a system can meet the specifications for good control performance.  For this purpose, it is very useful to be able to roughly sketch a root locus which will be used to examine a system and to evaluate the consequences of possible compensation alternatives.  Also, it is important to be able to quickly evaluate the correctness of a MATLAB-generated locus to verify that what is plotted is in fact what was meant to be plotted.

9. 6/9 Erwin SitompulFeedback Control System Guideines for Sketching a Root Locus Deriving using the root locus form of characteristic equation, Taking the polynomial a(s) and b(s) to be monic, i.e., the coefficient of the highest power of s equals1, they can be factorized as: If any s = s 0 fulfills the equation above, then s 0 is said to be on the root locus.

10. 6/10 Erwin SitompulFeedback Control System Phase Condition Magnitude Condition Guideines for Sketching a Root Locus The magnitude condition implies: The phase condition implies: Defining and, the phase condition can be rewritten as:

11. 6/11 Erwin SitompulFeedback Control System Guideines for Sketching a Root Locus The root locus is the set of values of s for which 1 + KL(s) = 0 is satisfied as the real parameter K varies from 0 to ∞. Typically, 1 + KL(s) = 0 is the characteristic equation of the system, and in this case the roots on the locus are the closed-loop poles of that system. “ ” The root locus of L(s) is the set of points in the s-plane where the phase of L(s) is 180°. If the angle to a test point from a zero is defined as ψ i and the angle to a test point from a pole as Φ i, then the root locus of L(s) is expressed as those points in the s-plane where, for integer l, Σψ i – ΣΦ i = 180° + 360°(l–1). “ ”

12. 6/12 Erwin SitompulFeedback Control System Guideines for Sketching a Root Locus Consider the following example. : the poles of L(s) : the zero of L(s) Test point  s 0 is not on the root locus

13. 6/13 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus RULE 1: The n branches of the locus start at the poles of L(s) and m of these branches end on the zeros of L(s), while n–m branches terminate at infinity along asymptotes. Recollecting  The root locus starts at K = 0 at the poles of L(s) and ends at K = ∞ on the zeros of L(s)

14. 6/14 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus RULE 2: On the real axis, the loci (plural of locus) are to the left of an odd number of poles and zeros. 12345 : The root locus 1 1234  Angles from real poles or zeros are 0° if the test point is to the right and 180° if the test point is to the left of a given pole or zero.

15. 6/15 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus The rule is now applied to obtain the root locus of:

16. 6/16 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus For any test point s 0 on the real axis, the angles φ 1 and φ 2 of two complex conjugate poles cancel each other, as would the angles of two complex conjugate zeros (see figure below). The pair does not give contribution to the phase condition  s 0 is not on the root locus Now, check the phase condition of s 1 ! s1s1

17. 6/17 Erwin SitompulFeedback Control System RULE 3: For large K and s, n–m of the loci are asymptotic to lines at angles Φ l radiating out from the point s = α on the real axis, where: Rules for Plotting a Root Locus Center of Asymptotes Angles of Asymptotes

18. 6/18 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus For, we obtain –2.67 60° 180° 300°

19. 6/19 Erwin SitompulFeedback Control System RULE 4: The angle of departure of a branch of a locus from a pole is given by: Rules for Plotting a Root Locus and the angle of arrival of a branch of a locus to a zero is given by:

20. 6/20 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus For the example, the root loci must depart with certain angles from the complex conjugate poles at –4 ± j4, and go to the zero at ∞ with the angles of asymptotes 60° and 300°.

21. 6/21 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus From the figure, But Thus By the complex conjugate symmetry of the roots, the angle of departure of the locus from –4 – j4 will be +45°.

22. 6/22 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus So, the root loci will start their journey from –4 ± j4 towards ∞ with the direction of 45°. ±

23. 6/23 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus RULE 5: The locus crosses the jω axis (imaginary axis ) at points where:  The Routh criterion shows a transition from roots in the left half-plane to roots in the right half-plane.  This transition means that the closed-loop system is becoming unstable.  This fact can be tested by Routh’s stability criterion, with K as the parameter, where an incremental change of K will cause the sign change of an element in the first column of Routh’s array.  The values of s = ± jω 0 are the solution of the characteristic equation in root locus form, 1 + KL(s) = 0.  The points ± jω 0 are the points of cross-over on the imaginary axis.

24. 6/24 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus For the example, the characteristic equation can be written as:  The closed-loop system is stable for K > 0 and K < 256  for 0 < K < 256.  For K > 256 there are 2 roots in the RHP (two sign changes in the first column).  For K = 256 the roots must be on the imaginary axis.

25. 6/25 Erwin SitompulFeedback Control System The characteristic equation is now solved using K = 256. Points of Cross-over Another way to solve for ω 0 is by simply replacing any s with jω 0 without finding the value of K first. ≡ 0 Rules for Plotting a Root Locus Same results for K and ω 0

26. 6/26 Erwin SitompulFeedback Control System 5.66 –5.66 Rules for Plotting a Root Locus The points of cross-over are now inserted to the plot.

27. 6/27 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus The complete root locus plot can be shown as:

28. 6/28 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus RULE 6: The locus will have multiple roots at points on the locus where: The branches will approach and depart a point of q roots at angles separated by:

29. 6/29 Erwin SitompulFeedback Control System Rules for Plotting a Root Locus  A special case of point of multiple roots is the intersection point of 2 roots that lies on the real axis.  If the branches is leaving the real axis and entering the complex plane, the point is called the break-away point.  If the branches is leaving the complex plane and entering the real axis, the point is called the break-in point. Real axis Imag axis Break-away point Real axis Imag axis Break-in point

30. 6/30 Erwin SitompulFeedback Control System Example 1: Plotting a Root Locus Draw the root locus plot of the system shown below. 3 zeros at infinity RULE 1

31. 6/31 Erwin SitompulFeedback Control System Example 1: Plotting a Root Locus Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 RULE 2

32. 6/32 Erwin SitompulFeedback Control System Example 1: Plotting a Root Locus Center of Asymptotes Angles of Asymptotes

33. 6/33 Erwin SitompulFeedback Control System Example 1: Plotting a Root Locus Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 RULE 3 60° 180° 300° Not applicable. The angles of departure or the angles of arrival must be calculated only if there are any complex poles or zeros. RULE 4

34. 6/34 Erwin SitompulFeedback Control System Example 1: Plotting a Root Locus Replacing s with jω 0, ≡ 0 Points of Cross-over

35. 6/35 Erwin SitompulFeedback Control System Example 1: Plotting a Root Locus Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 RULE 5 1.414 –1.414

36. 6/36 Erwin SitompulFeedback Control System Example 1: Plotting a Root Locus The root locus must have a break-away point, which can be found by solving:

37. 6/37 Erwin SitompulFeedback Control System Example 1: Plotting a Root Locus Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 RULE 6 1.414 –1.414 On the root locus  The break-away point Not on the root locus –0.423

38. 6/38 Erwin SitompulFeedback Control System Example 1: Plotting a Root Locus After examining RULE 1 up to RULE 6, now there is enough information to draw the root locus plot. Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 1.414 –1.414 –0.423

39. 6/39 Erwin SitompulFeedback Control System Example 1: Plotting a Root Locus Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 1.414 –1.414 –0.423 The final sketch, with direction of root movements as K increases from 0 to ∞ can be shown as: Final Result Determine the locus of all roots when K = 6!

40. 6/40 Erwin SitompulFeedback Control System Example 2: Plotting a Root Locus a)Draw the root locus plot of the system. b)Define the value of K where the system is stable. c)Find the value of K so that the system has a root at s = –2.

41. 6/41 Erwin SitompulFeedback Control System Example 2: Plotting a Root Locus RULE 1 There is one branch, starts from the pole and approaches the zero

42. 6/42 Erwin SitompulFeedback Control System Example 2: Plotting a Root Locus Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 RULE 2 Not applicable, since n = m. RULE 3 Not applicable. The angles of departure or the angles of arrival must be calculated only if there are any complex poles or zeros. RULE 4

43. 6/43 Erwin SitompulFeedback Control System Example 2: Plotting a Root Locus Replacing s with jω 0, ≡ 0 Points of Cross-over

44. 6/44 Erwin SitompulFeedback Control System Example 2: Plotting a Root Locus Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 The point of cross-over, as can readily be guessed, is at s = 0. RULE 5 Not applicable. There is no break-in or break-away point. RULE 6 K = 2

45. 6/45 Erwin SitompulFeedback Control System Example 2: Plotting a Root Locus Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 The final sketch, with direction of root movements as K increases from 0 to ∞ can be shown as: Final Result a)Draw the root locus plot of the system.

46. 6/46 Erwin SitompulFeedback Control System Example 2: Plotting a Root Locus Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 b)Define the value of K where the system is stable. System is stable when the root of the characteristic equation is on the LHP, that is when 0 ≤ K < 2. K = 2 K = 0 K = ∞

47. 6/47 Erwin SitompulFeedback Control System Example 2: Plotting a Root Locus Real axis Imag axis 120–1–2–3–4 1 2 –1 –2 3 –3 K = 2 K = 0 K = ∞ c)Find the value of K so that the system has a root at s = –2. Inserting the value of s = –2 in the characteristic equation, K = 0.5

48. 6/48 Erwin SitompulFeedback Control System Homework 6  No.1, FPE (5 th Ed.), 5.2. Hint: Easier way is to assign reasonable values for the zeros and poles in each figure. Later, use MATLAB to draw the root locus.  No.3 Sketch the root locus diagram of the following closed- loop system as accurate as possible.  No.2, FPE (5 th Ed.), 5.7.(b) Hint: After completing the hand sketch, verify your result using MATLAB. Try to play around with Data Cursor.