UNIVERSITY OF KWA-ZULU NATAL HOWARD COLLEGE CAMPUS CONTROL SYSTEMS II Tutorial: Frequency-Based Controller Design 1. Evaluate the error constant, and the steady state error of the following Systems. 32 (a) πΊ(π ) = π (π 2 +12π +32) 20(π +2)(π +3)(π +6)(π +8) (b) πΊ(π ) = π 2 (π +7)(π +9)(π +10)(π +15) 2 (c) πΊ(π ) = (π +1)((π +2) 2. For the three systems given in question 1, determine the new error constant value required to have a steady-state error of 0.0001. What is the proportional gain adjustment required in each case? 3. Make the Nyquist plot of the following systems. Determine the gain and phase margins. 32 π (π 2 +3π +10) (a) πΊ(π ) = (b) πΊ(π ) = (c) πΊ(π ) = (π +1)(π +2)(π +5) 20(π +50) (π +7)(π +9)(π +10) π +2.5 (πΊπ ≅ 21.6ππ΅, ππ ≅ .60 ) (πΊπ ≅ 20.6ππ΅, ππ ≅ 93.20 ) (πΊπ = ∞, ππ = ∞) 4. For the following three functions, draw the Bode plot, and determine the gain and phase margins: 15(π +3) (πΊπ ≅ ∞, ππ ≅ 67.50 ) 20(π +6)(π +8) (πΊπ ≅ −23.2ππ΅, ππ ≅ −46.10 ) (a) πΊ(π ) = π (π +2)(π +5) (b) πΊ(π ) = π (π +2)(π 2 +3π +9) (c) πΊ(π ) = π 2 (π +1)((π +2) 20 (πΊπ ≅ ∞, ππ ≅ −10.50 ) 5. Draw the Nichols plot, and determine the gain and phase margins of the following systems: (a) 15(π +3) πΊ(π ) = π (π +2)(π +5) 20(π +6)(π +8) π (π +2)(π 2 +3π +9) (b) πΊ(π ) = (c) πΊ(π ) = π 2 (π +1)((π +2) 20 πΎ 6. The system πΊ(π ) = π (π +50)(π +12) is used in unity feedback. Using frequency based methods: (i) Find the value of K required to yield a closed-loop response of 20%. (K=194,200) (ii) Design a lag controller to reduce the steady-state error by a factor of 10. ( πΊπΆ πππ (π ) = 0.0691(π +2.04) π +0.141 ) 100πΎ 7. It is required to compensate the system πΊ(π ) = π (π +100)(π +36) , so that the closed-loop response yields a peak time π‘π = 0.1 π ππ, a steady-state error constant πΎπ = 40 and a damping factor π = 0.456 (20% overshoot). Design the lead controller required. Add a phase π +25.3 adjustment of 100 . (ππ = 48.10 , ππππ₯ = 24.10 , π½ = 0.42, πΊπΆ ππππ (π ) = 2.38 π +60.2 ). 8. For the system πΊ(π ) = πΎ , π (π +50)(π +12) ππ = 0.2 π ππ. ( πΊπΆ ππππ (π ) = design a lead controller for πΎπ = 50, %ππ = 20, and 2.27(π +33.2) π +75.4 ) πΎ π 9. For πΎπ = 12/π ππ, design a lead controller for the system πΊ(π ) = π (π +1) such that the resulting closed-loop system has a phase margin of 400 . Adjust the resultant angle by +50 . (ππππ₯ = 300 , πΊπΆ ππππ (π ) = 0.337π π +1 0.125π +1 ) πΎ 10. Design a double lead controller for the system πΊ(π ) = π 2 (0.2π +1) for an acceleration error constant πΎπ = 10 and a phase margin of 350 . Why is a double lead control network required in this particular problem? Add a 140 phase adjustment. (K=10, ππππ₯ = 35 − (−33) + 14 = 820 . Divide the 82 degrees into 41 degrees each. For (0.358π +1)2 0.358π +1 ∗ each of the lead controllers, ππππ₯ = 410 , π½ = 0.2. πΊπΆ1 ππππ = 0.077π +1 ; πΊπΆ ππππ = (0.077π +1)2 ) πΎ 11. Design a lag controller for πΊ(π ) = π (π +1)(π +4) so that the closed-loops system has the following performance. The tolerance band is 5% (not the usual 2%). Damping factor π = 0.4 3 Settling time π‘π = 10 π ππ. (For a 5% tolerance band, π‘π = ππ π Velocity error constant πΎπ ≥ 5π ππ Use an angular adjustment of 120 . ). −1 ( ππ = 430 , ππ = 1.02 ππππ ππ −1 , ππππ₯ = 550 , ππ = 0.522 ππππ ππ.−1 )