1. OBJECTIVE  Determination of root from the characteristic equation by using graphical solution.  Rules on sketching the root locus.  Analysis of closed-loop using root locus 4.0 ROOT LOCUS

2. Root locus concept  Consider  Open-loop transfer function And Where If And KG(s) H(s)H(s) R(s)R(s) + E(s)E(s)Y(s)Y(s) B(s)B(s)

3. Closed-loop transfer function ~ Number and position of zeros for open-loop and closed-loop are the same ~ Position of poles for the closed-loop depend on the position of poles, zeros and K. Characteristic equation is. If Let and Its open-loop transfer function and its closed-loop as ~ Position of poles for the closed-loop depend on the position of poles, zeros and K of the open-loop. transfer function. Root locus concept Where H(s)=1

4. As K varies the closed-loop poles follow similarly and form a locus. For that we define Root locus is a locus of characteristic equation as K varies from 0 to . Referring to a characteristic equation Let say Re-arrange -1 is a complex number where Root locus concept

5. Magnitude condition Re-arrange where is the magnitude from a test point to open-loop poles and the magnitude from a test point to open-loop zeros From the magnitude condition, we can determine K. Let s be the test point, the magnitude are and from open-loop zeros and poles respectively. Hence the magnitude condition is

6. s-plane n1n1 n m m1m1 s Magnitude condition

7. Angle condition Revisiting the complex number of -1 Its angle Expand where is the angle from a test point to open-loop zeros and whereis the angle from a test point to open-loop poles

8. From the angle condition, we can determine the angle contribution by s the test point, the angle are and from open-loop zeros and poles respectively. Hence the angle condition is Angle condition

9. s s-plane Angle condition

10. Example Give the unity feedback system that has the forward transfer function. a)Calculate the angle of G(s) at the point (-2+j0.707) by finding the algebraic sum of angles of the vectors drawn from the zeros and poles of G(s) to the given point. b)Determine if the point specified in a) is on the root locus. c)If the point specified in a) is on the root locus, find the gain, K, using the lengths of the vectors.

11. Example a)  1 +  2 -  3 -  4 =19.471 o +35.264 o -90 o -144.736 o =-180 o b)Point is on the root locus. c) -4 -3 -2 j0.707 11 22 33 44 s

12. Procedure for drawing a Root Locus (2) The locus is symmetrical at the real axis. Consider Characteristic equation is If » K=3;a=1;roots([1 2*a 4+K]) ans = -1.0000 + 2.4495i -1.0000 - 2.4495i (1) Start with the characteristic equation.

13. That shows the locus is symmetrical at-axis. s-plane Procedure for drawing a Root Locus

14. 3)The number of locus depends on the number of open loop poles  Open-loop transfer function  Closed-loop transfer function  Its characteristic equation  Number of locus is reflected by the order of characteristic equation which is the same as the number of open-loop poles of the system. - Procedure for drawing a Root Locus +

15. 4)Root locus begin from open loop poles and end at open loop zeros or symmetrical lines  Consider closed-loop transfer function,  If K=0, then the characteristic equation becomes P 1 (s)P 2 (s)=0 which are open-loop poles.  If K  0, characteristic equation is  If K→∞, the equation becomes Z 1 (s)Z 2 (s)≈0, which are open-loop zeros.  Locus will begin from the open-loop poles, K=0, and end at the open- loop zeros, K→∞.  If the number of open-loop zeros are less than the open-loop poles, then the locus will end at asymptote lines given by angle. Procedure for drawing a Root Locus

16.  where m and n are the order of numerator and denominator of the open- loop transfer function respectively.  And r=  1,  3,  5,  7,…..  And the line intersect the real-axis at  where and are the sum of all poles and zeros of open- loop transfer function respectively. Procedure for drawing a Root Locus

17. - Example Consider a P-compensator Find the open-loop zero and poles. Obtain the closed-loop poles at K p =0 and K p →∞. Consequently, shows where the locus end.

18. Solution  which give an open-loop zero at –2  open-loop poles at 0, –5 and –8  Its characteristic equation  As we know that the locus start at K p =0 substitute this to the characteristic equation  s 3 +13s+s(40)=s(s+5)(s+8)=0, give closed-loop poles at 0, –5 and –8, which is the same  as open-loop poles. Example

19. To find the closed-loop poles at We rearrange the characteristic equation becomes If, the root of the characteristic equation is –2, which is the same as the open-loop zero. As there is 3 open-loop poles, there will be 3 loci, one will end at open-loop zero of -2 and the other two will end at the asymptote lines, with angles at where with intersection of the real-axis at Example

20. -8 -5.5 -5 -2 s-plane Example

21. Consider the following closed-loop transfer function If, find the closed-loop poles at and 100. Finally, trace the loci for Solution: As the characteristic equation is. By substituting the above values to the equation will give the following table: Example

22. Gain, KClosed-loop pole 0 -9.5826 -0.4174 3-9.2426 -0.7574 10 -8.3166 -1.6834 21 40 100 -5 -5.0000 + 4.3589i -5.0000 - 4.3589i -5.0000 + 8.8882i -5.0000 - 8.8882i Example

23. -5.0 s­-plane - j8.8882 -j4.3589 j4.3589 j8.8882 Example

24. For we can trace the locus as it move from 0 to ∞ as shown below s-plane -j4.3589 - j8.8882 j4.3589 j8.8882 Example

25. (5) Locus on the real-axis is It should fulfill the angle condition. Consider HG E DC B A s-plane Procedure for drawing a Root Locus F

26.  Take a test point at A-H and use the angle condition Angle condition A0No B-180Yes C-180-180 = -360No D-180-180+180 = -180Yes E-180-180+180-180 = -360No F-180-180+180-180+180 = -180Yes G-180-180+180-180+180-180 = -360No H-180-180+180-180+180-180-180 = -540Yes Test point Procedure for drawing a Root Locus

27. (6) Breakaway point of locus from the real-axis  It is the point where the root locus originating from the open poles meet at the real-axis and breakaway  Consider back the closed-loop transfer function of  If we plot a graph of the closed-loop position against the gain K. We notice that at the point of -5,  before the closed-loop poles become complex, the gain is maximum for any real poles. After this the  locus will breakaway from the real-axis  We can determine this by taking the differential of the characteristic equation, Procedure for drawing a Root Locus

28. A unity feedback system with an open-loop system as given below Determine the maximum gain K before the system starts to oscillate. Solution: The characteristic equation is Rearrange Differentiate to obtain the maximum gain Example

29. which gives Out of the two values, we choose -0.93 as –5.74 is not on the locus. Use the magnitude condition Using s=-0.93, maximum K before the starts to oscillate Example

30. (7) Stability boundary of the locus  We use the Routh-Hurwitz criteria determine the maximum gain before instability and its frequency. This can be obtained through Procedure for drawing a Root Locus

31. (8) Angle of departure and arrival  For complex poles, the locus will leave the poles from an angle called angle of departure.  By selecting a test point very near to the open-loop pole, -p y the angle of departure,  is given by  or  = (Sum of angle from all open-loop zeros to test point - (Sum of angle from all open-loop poles to test point ) -r  Procedure for drawing a Root Locus

32. Example: Find the angle of departure of the complex poles s-plane Example

33. As for the above diagram, the angle of departure from the complex pole is while the conjugate pair, its angle of departure is For complex open-loop zero, the locus will end here and the angle of arrival, , given by an angle condition as or  = r  - (Sum of angle from all open-loop zeros to ) + (Sum of angle from all open-loop poles to) Example

34. Example Given a unity feedback system as given below. Find the angle of departure from the complex pole - + R(s) C(s)

35. Example s-plane Solution: -3 -2 j1 -j1

36. A plant of transfer function is feedback with a unity feedback. Sketch a root locus for 0<K<∞. Solution  Its characteristic equation is  As the number of open-loop poles is 3, there will be 3 loci and the loci start at 0,-1 dan –2.  The locus on the real-axis Example

37. -2  Satah-s As there is no zeros, the locus will end at asymptotes lines, where This gives asymptote angles at, with real-axis crossing at Example

38.   -2 -1 s-plane The loci will meet and depart at breakaway point Rearrange Differentiate for the maximum value of K Example

39. Its root The probable root is –0.4 as it is on the locus The crossing at the boundary of stability 12 3K K and Range of gain for stability. The natural frequency is determined by taking, replacing to gives. Example

40. » num=[1];den=[1 3 2 0];%Takrifkan rangkap pindah gelung buka » rlocus(num,den);%Kerahan We can determine the gain at particular points through » [k,p]=rlocfind(num,den) Example

41. Determination of gain Can be determined using magnitude condition. O  ss s-plane m1 n1 m2 m3 Zero and poles magnitude from test point Take a test point, s the gain K is

42. Pole Determination If the order of the characteristic equation is n, there will be n poles. If n-1 poles are known we know the last pole by comparing the coefficient. Consider Which can be factored towhere its poles are. Expanding the factor form Equating the coefficient will give the n pole as

43. Example For a third order system with, two of the poles are –2 and –3, determine its third pole. + - Solution: Characteristic equation Comparing and which give hence the third pole is Example