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Digital Control Systems (DCS)

Digital Control Systems (DCS). Lecture-1-2 Lead Compensation. Dr. Imtiaz Hussain Associate Professor Mehran University of Engineering & Technology Jamshoro, Pakistan email: imtiaz.hussain@faculty.muet.edu.pk URL : http://imtiazhussainkalwar.weebly.com/. 7 th Semester 14ES.

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Digital Control Systems (DCS)

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  1. Digital Control Systems (DCS) Lecture-1-2 Lead Compensation Dr. Imtiaz Hussain Associate Professor Mehran University of Engineering & Technology Jamshoro, Pakistan email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/ 7th Semester 14ES Note: I do not claim any originality in these lectures. The contents of this presentation are mostly taken from the book of Ogatta, Norman Nise, Bishop and B C. Kuo and various other internet sources.

  2. Introduction • A feedback control system that provides an optimum performance without any necessary adjustment is rare. • Compensator: A compensator is an additional component or circuit that is inserted into a control system to equalize or compensate for a deficient performance. • Necessities of compensation • A system may be unsatisfactory in: • Stability. • Speed of response. • Steady-state error.

  3. Compensation via Root Locus • Performance measures in the time domain: • Peak time; • Overshoot; • Settling time for a step input; • Steady-state error for test inputs

  4. Commonly Used Compensators • Among the many kinds of compensators, widely employed compensators are the • Lead compensators • Lag compensators • Lag–Lead compensators

  5. Lead Compensation • Generally Lead compensators are represented by following transfer function • or , () , ()

  6. Lead Compensation , ()

  7. Electronic Lead Compensator • Following figure shows an electronic lead compensator using operational amplifiers.

  8. Electronic Lead Compensator • This can be represented as • Where, • Then, , ()

  9. Lead Compensation Techniques Based on the Root-Locus Approach. • The procedures for designing a lead compensator by the root-locus method may be stated as follows: • Step-1: Analyze the given system via root locus.

  10. Step-2 • From the performance specifications, determine the desired location for the dominant closed-loop poles.

  11. Step-3 • From the root-locus plot of the uncompensated system (original system), ascertain whether or not the gain adjustment alone can yield the desired closed loop poles. • If not, calculate the angle deficiency. • This angle must be contributed by the lead compensator if the new root locus is to pass through the desired locations for the dominant closed-loop poles.

  12. Step-4 • Assume the Lead Compensator to be: • Where α and T are determined from the angle deficiency. • Kcis determined from the requirement of the open-loop gain.

  13. Step-5 • If static error constants are not specified, determine the location of the pole and zero of the lead compensator so that the lead compensator will contribute the necessary angle. • If no other requirements are imposed on the system, try to make the value of α as large as possible. • A larger value of α generally results in a larger value of Kv, which is desirable. • Larger value of αwill produce a larger value of Kvand in most cases, the larger the Kv is, the better the system performance.

  14. Step-6 • Determine the value of Kc of the lead compensator from the magnitude condition.

  15. Example-1 • Consider the position control system shown in following figure. • It is desired to design an Electronic lead compensator Gc(s) so that the dominant closed poles have the damping ratio 0.5 and undamped natural frequency 3 rad/sec.

  16. Step-1 (Example-1) • Draw the root Locus plot of the given system. • The closed loop transfer function of the given system is: • The closed loop poles are

  17. Step-1 (Example-1) • Determine the characteristics of given system using root loci. • The damping ratio of the closed-loop poles is 0.158. • The undamped natural frequency of the closed-loop poles is 3.1623 rad/sec. • Because the damping ratio is small, this system will have a large overshoot in the step response and is not desirable.

  18. Step-2 (Example-1) • From the performance specifications, determine the desired location for the dominant closed-loop poles. • Desired performance Specifications are: • It is desired to have damping ratio 0.5 and undamped natural frequency 3 rad/sec.

  19. Step-2 (Example-1) • Alternatively desired location of closed loop poles can also be determined graphically • Desired ωn= 3 rad/sec • Desired damping ratio= 0.5 Desired Closed Loop Pole

  20. Step-3 (Exampl-1) • From the root-locus plot of the uncompensated system ascertain whether or not the gain adjustment alone can yield the desired closed loop poles. Desired Closed Loop Pole

  21. Step-3 (Exampl-1) • If not, calculate the angle deficiency. • To calculate the angle of deficiency apply Angle Condition at desired closed loop pole. Desired Closed Loop Pole -2 -1 -2 -1 100.8o 120o

  22. Step-3 (Exampl-1) • Alternatively angle of deficiency can be calculated as. Where are desired closed loop poles

  23. Step-4 (Exampl-1) • This angle must be contributed by the lead compensator if the new root locus is to pass through the desired locations for the dominant closed-loop poles. • Note that the solution to such a problem is not unique. There are infinitely many solutions.

  24. Step-5 (Exampl-1) Solution-1 • Solution-1 • If we choose the zero of the lead compensator at s = -1 so that it will cancel the plant pole at s =-1, then the compensator pole must be located at s =-3.

  25. Solution-1 Step-5 (Example-1) = • The pole and zero of compensator are determined as • The Value of can be determined as

  26. Solution-1 Step-6 (Example-1) • The Value of Kc can be determined using magnitude condition.

  27. Solution-1 Example-1 • Determine the circuit values • We must choose standard values of components • Let us choose

  28. Solution-1 Example-1 • Determine the circuit values • We must choose standard values of components • Let us choose

  29. Solution-1 Example-1 • Determine the circuit values • We must choose standard values of components • If we choose (not a standard value)

  30. Final Design Check Solution-1 • The open loop transfer function of the designed system then becomes • The closed loop transfer function of compensated system becomes.

  31. Final Design Check Solution-1

  32. Final Design Check Solution-1

  33. Final Design Check Solution-1 • The static velocity error constant for original system is obtained as follows. • The steady state error is then calculated as

  34. Final Design Check Solution-1 • The static velocity error constant for the compensated system can be calculated as • The steady state error is then calculated as

  35. Final Design Check Solution-1

  36. Solution-2 Step-5 (Exampl-1) • Solution-2 -2 -1 90o 49.2o -3 -2 -1

  37. Solution-2 Step-5 (Exampl-1) • Solution-2 -2 -1 90o 49.2o -3 -2 -1

  38. Solution-2 Step-5 (Exampl-1) • Solution-2 Uncompensated System Compensated System

  39. Root Locus

  40. Step Response

  41. Steady State Error

  42. Solution-3 Step-5 (Example-1) • If no other requirements are imposed on the system, try to make the value of α as large as possible. A larger value of α generally results in a larger value of Kv, which is desirable. • Procedure to obtain a largest possible value for α. • First, draw a horizontal line passing through point P, the desired location for one of the dominant closed-loop poles. This is shown as line PA in following figure. • Draw also a line connecting point P and the origin O. A P -2 O -1 -3 -2 -1

  43. Solution-3 Step-5 (Example-1) • Bisect the angle between the lines PA and PO, as shown in following figure. -2 -1 P A -3 -2 -1 O

  44. Solution-3 Step-5 (Example-1) • Draw two lines PC and PD that make angles with the the bisector PB. • The intersections of PC and PD with the negative real axis give the necessary locations for the pole and zero of the lead network. P A -2 -1 O -3 -2 -1 C B D

  45. Solution-3 Step-5 (Example-1) • The lead compensator has zero at s=–1.9432 and pole at s=–4.6458. • Thus, Gc(s) can be given as P -2 A -1 = -3 -2 -1 O B C D

  46. Solution-3 Step-5 (Example-1) = • For this compensator value of is • Also

  47. Solution-3 Step-6 (Example-1) • Determine the value of Kc of the lead compensator from the magnitude condition.

  48. Solution-3 Step-6 (Example-1) • The Kc is calculated as • Hence, the lead compensator Gc(s) just designed is given by

  49. Solution-3 Final Design Check Desired Closed Loop Pole Desired Closed Loop Pole Compensated System Uncompensated System

  50. Solution-3 Final Design Check

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