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 )
0
