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NUAA-Control System Engineering

NUAA-Control System Engineering. Chapter 4 Root-locus Technique. 4-1 Introduction 4-2 Root-locus equation 4-3 Rules to draw regular root loci 4-4 Generalized root loci 4-5 Analysis of control system by RL method 4-6 Control system design by the root-locus method. Contents in Chapter 4.

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NUAA-Control System Engineering

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  1. NUAA-Control System Engineering Chapter 4 Root-locus Technique

  2. 4-1 Introduction 4-2 Root-locus equation 4-3 Rules to draw regular root loci 4-4 Generalized root loci 4-5 Analysis of control system by RL method 4-6 Control system design by the root-locus method Contents in Chapter 4 2

  3. Introduction Stability and the characteristics of transient response of closed-loop systems Locations of the closed-loop poles Problems to solve characteristic equation : 1. Difficult for a system of third or higher order. 2. Tedious for varying parameters. 3

  4. Varying the loop gain K In many systems, simple gain adjustment may move the closed-loop poles to desired locations. Then the design problem may become the selection of an appropriate gain value. It is important to know how the closed-loop poles move in the s plane as the loop gain K is varied. Y(s) R(s) K G(s) － H(s) The open loop gain K is an important parameter that can affect the performance of a system 4

  5. A simple method for finding the roots of the characteristic equation has been developed by W.R.Evans. This method is called root locus method. Graphical Analysis of Control System, AIEE Trans. Part II,67(1948),pp.547-551. Control System Synthesis by Root Locus Method, AIEE Trans. Part II,69(1950),pp.66-69 Root locus method 5

  6. Root Locus: the locus of roots of the characteristic equation of the closed-loop system as a specific parameter (usually, gain K) is varied form 0 to ∞. The advantages of RL approach: 1. Avoiding tedious and complex roots-solving calculation 2. Clearly showing the contributions of each loop poles or zeros to the location of the closed-loop poles. 3. Indicating the manner in which the loop poles and zeros should be modified so that the response meets system performance specifications. 6

  7. 4-2 Root-locus equation 7

  8. Consider a second-order system shown as follows: Start from an example R(s) Y(s) - Closed-loop TF: Characteristic equation (CE): Roots of CE: The roots of CE change as the value of k changes. When k changes from 0 to ∞, how will the locus of the roots of CE move? 8

  9. k＝0 0 －1/2 －1 As the value of k increases, the two negative real roots move closer to each other. A pair of complex-conjugate roots leave the negative real-axis and move upwards and downwards following the line s=-1/2. On the s plane, using arrows to denote the direction of characteristic roots move when k increases, by numerical value to denote the gain at the poles. 9

  10. By Root loci, we can analyze the system behaviors (1)Stability: when Root loci are on the left half plane, then the system is definitely stable for all k>0. (2)Steady-state performance: there ’s an open-loop pole at s=0, so the system is a type I system. The steady-state error is 0 under step input signal v0/Kv under ramp signal v0t ∞ under parabolic signal. 10

  11. (3)Transient performance: there’s a close relationship between root loci and system behavior on the real-axis: k<0.25underdamped; k=0.25critically damped k>0.25 underdamped. However, it’s difficult to draw root loci directly by closed-loop characteristic roots-solving method. The idea of root loci : by loop transfer function, draw closed-loop root loci directly.

  12. Relationship between zeros and poles of G(s)H(s) and closed-loop ones R(S) Y(s) G(s) - H(s) Forward path TF: Closed-loop TF: Feedback path TF: 12

  13. Write G(s) and H(s) into zero-pole-gain (zpk) form: RL gain of forward path RL gain of feedback path Poles of feedback path TF Zeros of forward path TF Zeros of feedback path TF Poles of forward path TF

  14. 1.The closed-loop zeros = feed forward path zeros + feedback path poles. 2. The closed-loop poles are related to poles and zeros of G(s)H(s) and loop gain K. 3.The closed-loop root loci gain = the feed forward path root loci gain .

  15. Suppose that G(s)H(s) has m zeros (Zi) and n poles（Pi), the above equation can be re-written as Root-locus equation To draw Closed-loop root locus is to solve the CE That is RL equation K* varies from 0 to ∞ 15

  16. RL equation: Magnitude equation (ME) Since G(s)H(s) is function of a complex variable s, the root locus equation can be described by the following two equations: Angle equation (AE) Magnitude equation is related not only to zeros and poles of G(s)H(s), but also to RL gain ； Angel equation is only related to zero and poles of G(s)H(s). Use AE to draw root loci and use ME to determine the value of K on root loci. 16

  17. Example 1 Poles of G(s)H(s)（×） Zeros of G(s)H(s)（〇） For a point s1on the root loci, use AE Use ME Angel is in the direction of anti-clockwise 17

  18. Example 2 Unity-feedback transfer function: One pole of G(s)H(s): No zero. Test a points1 on the negative real-axis All the points on the negative real-axis are on RL. Test a point outside the negative real-axis s2=-1-j All the points outside the negative real-axis are not on RL. 18

  19. Probe by each point: 1. Find all the points that satisfy the Angle Equation on the s-plane, and then link all these points into a smooth curve, thus we have the system root locus when k* changes from 0 to ∞; 2. As for the given k, find the points that satisfy the Magnitude Equation on the root locus, then these points are required closed-loop poles. However, it’s unrealistic to apply such “probe by each point” method. W.R. Evans (1948) proposed a set of root loci drawing rules which simplify the our drawing work.

  20. 4-3 Rules to draw regular root loci (suppose the varying parameter is open-loop gain K ) 20

  21. Properties of Root Loci 1 and points of Root Loci 2 Number of Branches on the RL 3 Symmetry of the RL 4 Root Loci on the real-axis 5 Asymptotes of the RL 6 Breakaway points on the RL 7 Departure angle and arrival angle of RL 8 Intersection of the RL with the imaginary axis 9 The sum of the roots and the product of the roots of the closed-loop characteristic equation 21

  22. 1 and points Root loci originate on the poles of G(s)H(s) (for K=0) and terminates on the zeros of G(s)H(s) (as K=∞). RL Equation: Magnitude Equation: Root loci start from poles of G(s)H(s) Root loci end at zeros of G(s)H(s). 22

  23. 2 Number of branches on the RL nth-order system, RL have n starting points and RL have n branches RL Equation: The order of the characteristic equation is n as K varies from 0 to ∞,n roots changen root loci. For a real physical system, the number of poles of G(s)H(s) are more than zeros，i.e. n > m. n root loci end at open-loop zeros（finite zeros)； m root loci end at open-loop zeros（finite zeros)； （n－m）root loci end at (n－m) infinite zeros. 23

  24. 3 Symmetry of the RL The RL are symmetrical with respect to the real axis of the s-plane. The roots of characteristic equation are real or complex-conjugate. Therefore, we only need to draw the RL on the up half s-plane and on the real-axis, the rest can be obtained by plotting its mirror image. 24

  25. ？ 4 RL on the Real Axis On a given section of the real axis, RL for k>0 are found in the section only if the total number of poles and zeros of G(s)H(s) to the right of the section is odd. zero：z1 poles：p1、p2、p3、p4、p5 Pick a test point s1 on [p2，p3] The sum of angles provided by every pair ofcomplex conjugate poles are 360°; The angle provided by all the poles and zeros on the right of s1 is 180°; The angle provided by all the poles and zeros on the left of s1 is 0°. 25

  26. Example Consider the following loop transfer function Determine its root loci on the real axis. jw Poles: Zeros: S-plane Repeated poles: -6 -5 -4 -3 -2 -1 0 On the right of [-2，-1] the number of real zeros and poles=3. On the right of [-6，-4], the number of real zeros and poles=7. 26

  27. 5 Asymptotes of RL When n ≠ m, there will be 2|n-m| asymptotes that describe the behavior of the RL at |s|=∞. RL Equation: The angles between the asymptotes and the real axis are（i= 0，1，2，… ,n-m-1） ： The asymptotes intersect the real axis at： 27

  28. Example Consider the following loop transfer function Determine the asymptotes of its root loci. 3poles：0、-1、-5 n-m = 3 -1 = 2 1zero：-4 The asymptotes intersect the real axis at The angles between the asymptotes and the real axis are 28

  29. Example Consider the following loop transfer function Determine the asymptotes of its root loci. 4poles：0、-1+j、-1-j、-4 n-m=4-1=3 1zero：-1 The asymptotes intersect the real axis at The angles between the asymptotes and the real axis are 29

  30. 6 Breakaway points on the RL Breakaway points on the RL correspond to multi-order roots of the RL equation. The breakaway points on the RL are determined by finding the roots of dK/ds=0 or dG(s)H(s)=0. 30

  31. Method－1 The breakaway point bis the solution of Proof： Suppose the multi-order root is s1，then we have 31

  32. (2) divided by (1) yields Thus Or Solutions1 is the breakaway point b 32

  33. Example The poles and zeros of G(s)H(s) are shown in the following figure, determine its root loci. jω Rule 1、2、3 σ 0 RL have three branches，starting from poles 0、－2、－3，ending at on finite zero －1and two infinite zeros. The RL are symmetrical with respect to the real axis. -3 -2 -1 Rule 5The RL have two asymptotes(n－m＝2) Rule 6The RL have breakaway points on the real axis (within[-3,-2]) Ruel 4The intersections [-1,0] and[-3,-2] on the real axis are RL. 33

  34. The breakaway point bis the solution of Method－2 Proof： Suppose smoves from p2to p1, if we increase K from zero，K reaches its maximum when s reaches b. Then K decreases and reaches zero at p1. K corresponding to the breakaway point has the maximum value.Thus 34

  35. Example jω j4 σ -2 0 -4 -j4 Rule 1、2、3、4 n=4,m=0 the RL are symmetrical with respect to the real axis; the RL have four branches which start from poles 0,-4 and -2±j4; the RL end at infinite zeros; the intersection [-4,0] on the real-axis is RL Rule 5 The RL have four asymptotes. 35

  36. Example jω j4 σ -2 0 -4 -j4 Rule 6the breakaway point of the RL 36

  37. 7 Angles of departure and angles of arrival of the RL The angle of departure or arrival of a root locus at a pole or zero, respectively, of G(s)H(s) denotes the angle of the tangent to the locus near the point. 37

  38. Angle of Departure: Pick up a point s1 that is close to p1 Applying Angle Equation (AE) s1p1 angle of departure θp1 38

  39. Angle of Arrival: 39

  40. j -0.5 -1 1 Example Consider the following loop transfer function Determine its RL when K varies from 0 to ∞. 3 poles P1,2=-1(repeated poles) P3=1；1zero Z1=0，n-m=2。 3 branches ，2 asymptotes Angle of departure: 40

  41. Method 2 8 Intersection of the RL with the Imaginary Axis The characteristic equation have roots on the Im-axis and the system is marginally stable. Intersection of the RL with Im-axis? Method 1 Use Routh’s criterion to obtain the value of K when the system is marginally stable, the get ω from K. 41

  42. Consider the following loop transfer function Example Determine the intersection of the RL with Im-axis. Method 1 Closed-loop CE: Routh’s Tabulation: Marginally stable: K=6 Auxiliary equation: Intersection point: 42

  43. Consider the following loop transfer function Example →Closed-loop CE Determine the intersection of the RL with Im-axis. Method 2 Closed-loop CE: 43

  44. Summary of General Rules for Constructing Root Loci R(S) Y(s) G(s) - H(s) First, obtain the characteristic equation (CE) Then, rearrange the equation so that the parameters of interest appear as the multiplying factor in the form We sketch the Root Loci when K>0 varies from 0 to ∞.

  45. The angles between the asymptotes and the real axis are（i= 0，1，2，… ,n-m-1） ： The asymptotes intersect the real-axis at： Step 1: Locate the poles and zeros of G(s)H(s) on the s-plane. The RL start from poles of G(s)H(s) and terminate at zeros (finite zeros or zeros at infinity) Step 2: Determine the RL on the real-axis. Choose a test point on the real-axis, if the total number of real poles and real zeros to the right of the test is odd, then the test point is on the RL. Step 3: Determine the asymptotes of RL.

  46. Step 4: Determine the breakaway and break-in points. The breakaway and break-in points can be determined by finding the roots of Step 5: Determine the angle of departure (angle of arrival) of the RL from a complex pole (at a complex zero) Angle of departure from a complex pole pj= Angle of arrival at a complex zero zk=

  47. Method 2 Step 6: Find the points where the RL may cross the imaginary-axis. Method 1 Use Routh’s criterion to obtain the value of K when the system is marginally stable, the get ω from K.

  48. Example Consider a unity feedback system with the open-loop transfer function as: loop TF: Sketch RL of the closed-loop system when K varies from 0 to ∞. Solution. Step 1: Locate the poles and zeros of G(s)H(s) on the s-plane. loop poles: p1=0, p2=-3, p3,4=-1±j, n=4; no loop zeros: m=0; 48

  49. 1.5 1 0.5 0 -3 -2.5 -2 -1.5 -1 -0.5 Step 2: Determine the RL on the real-axis. The section [0,-3] of the real axis are RL. Step 3: Determine the asymptotes of RL The angles between the asymptotes and the real axis are（i= 0，1，2，… ,n-m-1） ： The asymptotes intersect the real-axis at： 49

  50. 1.5 1 0.5 0 -3 -2.5 -2 -1.5 -1 -0.5 Step 4: Determine the breakaway and break-in points. 50

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