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Chapter 5

Feedback Control Systems. Chapter 5. The Root-Locus Design Method. Chapter 5. The Root-Locus Design Method. Root Locus: Illustrative Example. Examine the following closed-loop system, with unity negative feedback. The closed-loop transfer function is given as:.

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Chapter 5

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  1. Feedback Control Systems Chapter 5 The Root-Locus Design Method

  2. Chapter 5 The Root-Locus Design Method 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. Chapter 5 The Root-Locus Design Method Root Locus: Illustrative Example K=∞ : the poles of open-loop transfer function K=1/4 K=0 K=0 K=∞

  4. Chapter 5 The Root-Locus Design Method Root Locus: Illustrative Example • Where are the location of the closed-loop roots when K=1? K=1 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 ωd). –0.866 K=1 –0.5

  5. Chapter 5 The Root-Locus Design Method 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. Chapter 5 The Root-Locus Design Method 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. Chapter 5 The Root-Locus Design Method 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. Chapter 5 The Root-Locus Design Method 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. It can be done simply by 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. Chapter 5 The Root-Locus Design Method Guidelines for Sketching a Human Face

  10. Chapter 5 The Root-Locus Design Method 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 sequals 1, they can be factorized as: • If any s = s0 fulfills the equation above, then s0 is said to be on the root locus.

  11. Chapter 5 The Root-Locus Design Method Guideines for Sketching a Root Locus • The magnitude condition implies: Magnitude Condition • The phase condition implies: Phase Condition • Defining and , the phase condition can be rewritten as:

  12. Chapter 5 The Root-Locus Design Method 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 ψi and the angle to a test point from a pole as Φi, then the root locus of L(s) is expressed as those points in the s-plane where, for integer l, Σψi – ΣΦi =180°+360°(l–1). ”

  13. Chapter 5 The Root-Locus Design Method Guideines for Sketching a Root Locus • Consider the following example. : the poles of L(s) : the zero of L(s) Test point  s0 is not on the root locus

  14. Chapter 5 The Root-Locus Design Method Rules for Plotting a Root Locus RULE 1: The n branches of the locus start at the poles of L(s) and mof these branches end on the zeros of L(s), while n–mbranches 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)

  15. Chapter 5 The Root-Locus Design Method 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. 5 4 3 2 1 : The root locus 1 4 3 2 1 • 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.

  16. Chapter 5 The Root-Locus Design Method Rules for Plotting a Root Locus • The rule is now applied to obtain the root locus of:

  17. Chapter 5 The Root-Locus Design Method Rules for Plotting a Root Locus • For any test point s0 on the real axis, the angles φ1 andφ2 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 s1  s0 is not on the root locus Now, check the phase condition of s1!

  18. Chapter 5 The Root-Locus Design Method Rules for Plotting a Root Locus RULE 3: For large s and K, n–m of the loci are asymptotic to lines at angles Φl radiating out from the point s= αon the real axis, where: Angles of Asymptotes Center of Asymptotes

  19. Chapter 5 The Root-Locus Design Method Rules for Plotting a Root Locus • For , we obtain 60° 180° –2.67 300°

  20. Chapter 5 The Root-Locus Design Method Rules for Plotting a Root Locus RULE 4: The angle of departure of a branch of a locus from a pole is given by: and the angle of arrival of a branch of a locus to a zero is given by:

  21. Chapter 5 The Root-Locus Design Method 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°.

  22. Chapter 5 The Root-Locus Design Method 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°.

  23. Chapter 5 The Root-Locus Design Method Rules for Plotting a Root Locus • So, the root loci will start their journey from –4±j4 towards ∞ with the direction of ±45°.

  24. Chapter 5 The Root-Locus Design Method 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 = ± jω0 are the solution of the characteristic equation in root locus form, 1 + KL(s) = 0. • The points ± jω0are the points of cross-over on the imaginary axis.

  25. Chapter 5 The Root-Locus Design Method 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.

  26. Chapter 5 The Root-Locus Design Method Rules for Plotting a Root Locus • The characteristic equation is now solved using K = 256. Points of Cross-over • Another way to solve for ω0 is by simply replacing any s with jω0without finding the value of K first. ≡ 0 ≡ 0 Same results for K and ω0

  27. Chapter 5 The Root-Locus Design Method Rules for Plotting a Root Locus • The points of cross-over are now inserted to the plot. 5.66 –5.66

  28. Chapter 5 The Root-Locus Design Method Rules for Plotting a Root Locus • The complete root locus plot can be shown as:

  29. Chapter 5 The Root-Locus Design Method 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:

  30. Chapter 5 The Root-Locus Design Method 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. Imagaxis Imagaxis Realaxis Realaxis Break-awaypoint Break-inpoint

  31. Chapter 5 The Root-Locus Design Method Example 1: Plotting a Root Locus Draw the root locus plot of the system shown below. RULE 1 3 zeros at infinity

  32. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 1: Plotting a Root Locus RULE 2

  33. Chapter 5 The Root-Locus Design Method Example 1: Plotting a Root Locus Angles of Asymptotes Center of Asymptotes

  34. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 1: Plotting a Root Locus 60° 180° RULE 3 RULE 4 Not applicable. The angles of departure or the angles of arrival must be calculated only if there are any complex poles or zeros. 300°

  35. Chapter 5 The Root-Locus Design Method Example 1: Plotting a Root Locus Replacing s with jω0, ≡ 0 ≡ 0 Points of Cross-over

  36. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 1: Plotting a Root Locus RULE 5 1.414 –1.414

  37. Chapter 5 The Root-Locus Design Method Example 1: Plotting a Root Locus The root locus must have a break-away point on the real axis, between –1 and 0. It can be found by solving:

  38. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 1: Plotting a Root Locus Not on the root locus On the root locus The break-away point RULE 6 1.414 –0.423 –1.414

  39. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 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. 1.414 –0.423 –1.414

  40. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 1: Plotting a Root Locus The final sketch, with direction of root movements as K increases from 0 to ∞ can be shown as: Final Result 1.414 –0.423 –1.414 Determine the locus of all roots when K = 6!

  41. Chapter 5 The Root-Locus Design Method Example 2: Plotting a Root Locus • Draw the root locus plot of the system. • Define the value of K where the system is stable. • Find the value of K so that the system has a root at s = –2.

  42. Chapter 5 The Root-Locus Design Method Example 2: Plotting a Root Locus RULE 1 There is one branch, starts from the pole and approaches the zero

  43. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 2: Plotting a Root Locus RULE 2 RULE 3 Not applicable, since n=m. RULE 4 Not applicable. The angles of departure or the angles of arrival must be calculated only if there are any complex poles or zeros.

  44. Chapter 5 The Root-Locus Design Method Example 2: Plotting a Root Locus Replacing s with jω0, ≡ 0 ≡ 0 Points of Cross-over

  45. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 2: Plotting a Root Locus RULE 5 The point of cross-over, as can readily be guessed, is at s = 0. K = 2 RULE 6 Not applicable. There is no break-in or break-away point.

  46. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 2: Plotting a Root Locus • Draw the root locus plot of the system. The final sketch, with direction of root movements as K increases from 0 to ∞ can be shown as: Final Result

  47. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 2: Plotting a Root Locus b) Define the value of K where the system is stable. K = 0 K = 2 K = ∞ System is stable when the root of the characteristic equation is on the LHP, that is when 0 ≤ K < 2.

  48. Chapter 5 The Root-Locus Design Method Imagaxis 3 2 1 Realaxis –4 –3 –2 –1 0 1 2 –1 –2 –3 Example 2: Plotting a Root Locus c) Find the value of K so that the system has a root at s = –2. K = 0.5 K = 0 K = 2 K = ∞ Inserting the value of s = –2 in the characteristic equation,

  49. Chapter 5 The Root-Locus Design Method Homework 6 • No.1, FPE (6thEd.), 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.2, FPE (6thEd.), 5.7.(b) • Hint: After completing the hand sketch, verify your result using MATLAB. Try to play around with Data Cursor. • No.3 • Sketch the root locus diagram of the following closed-loop system as accurate as possible. • Due: 06.11.2013.

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