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Chapter 5

Chapter 5. Root Locus. In previous lecture. Chapter 4 Routh Hurwitz Criterion Steady-state error analysis. Today’s plan. Construction of Root Locus. Learning Outcomes. At the end of this lecture, students should be able to: Sketch the root locus for a given system

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Chapter 5

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  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. R(s) C(s) K Gc + - 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. 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 0 -1 -2

  11. Let say my first search point, S1 S1 0 -1 -2 (A) Satisfy the angle condition FIRST (B) Magnitude condition to find K

  12. Let say the next search point 0 -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 0 0 -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: 0 -1 -2

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

  28. 0 -1 -2

  29. Break away point Invalid, why ???

  30. How to determine these values ? 0 -1 -2 -0.42

  31. From characteristic equation Construct the Routh array

  32. Since Force S1 row to zero or K=6 Replace K=6 into S2 row

  33. 0 -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. Replace in this equation To make it easier to remember: 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. Replace in this equation To make it easier to remember: 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 0 -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 0 -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 departure 4 due to s=0 due to its complex conjugate -3 0 -4 How to determine θ?

  45. To determine θ 4 -3 0 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 0 -4 36.87o

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

  48. Construct the Routh array

  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 0 10 points -4 36.87o -5

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