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Tutorial Frequency based design

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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) 𝐺(𝑠) = 𝑠(𝑠2 +12𝑠+32)
20(𝑠+2)(𝑠+3)(𝑠+6)(𝑠+8)
(b) 𝐺(𝑠) = 𝑠2 (𝑠+7)(𝑠+9)(𝑠+10)(𝑠+15)
2
(c) 𝐺(𝑠) = (𝑠+1)((𝑠+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
𝑠(𝑠2 +3𝑠+10)
(a)
𝐺(𝑠) =
(b)
𝐺(𝑠) =
(c)
𝐺(𝑠) = (𝑠+1)(𝑠+2)(𝑠+5)
20(𝑠+50)
(𝑠+7)(𝑠+9)(𝑠+10)
𝑠+2.5
(𝐺𝑀 ≅ 21.6𝑑𝐡, πœ‘π‘€ ≅ .60 )
(𝐺𝑀 ≅ 20.6𝑑𝐡, πœ‘π‘€ ≅ 93.20 )
(𝐺𝑀 = ∞, πœ‘π‘€ = ∞)
4. For the following three functions, draw the Bode plot, and determine the gain and phase
margins:
15(𝑠+3)
(𝐺𝑀 ≅ ∞, πœ‘π‘€ ≅ 67.50 )
20(𝑠+6)(𝑠+8)
(𝐺𝑀 ≅ −23.2𝑑𝐡, πœ‘π‘€ ≅ −46.10 )
(a)
𝐺(𝑠) = 𝑠(𝑠+2)(𝑠+5)
(b)
𝐺(𝑠) = 𝑠(𝑠+2)(𝑠2 +3𝑠+9)
(c)
𝐺(𝑠) = 𝑠2 (𝑠+1)((𝑠+2)
20
(𝐺𝑀 ≅ ∞, πœ‘π‘€ ≅ −10.50 )
5. Draw the Nichols plot, and determine the gain and phase margins of the following systems:
(a)
15(𝑠+3)
𝐺(𝑠) = 𝑠(𝑠+2)(𝑠+5)
20(𝑠+6)(𝑠+8)
𝑠(𝑠+2)(𝑠2 +3𝑠+9)
(b)
𝐺(𝑠) =
(c)
𝐺(𝑠) = 𝑠2 (𝑠+1)((𝑠+2)
20
𝐾
6. The system 𝐺(𝑠) = 𝑠(𝑠+50)(𝑠+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.
( 𝐺𝐢 π‘™π‘Žπ‘” (𝑠) =
0.0691(𝑠+2.04)
𝑠+0.141
)
100𝐾
7. It is required to compensate the system 𝐺(𝑠) = 𝑠(𝑠+100)(𝑠+36) , so that the closed-loop
response yields a peak time 𝑑𝑝 = 0.1 𝑠𝑒𝑐, a steady-state error constant 𝐾𝑉 = 40 and a
damping factor πœ‰ = 0.456 (20% overshoot). Design the lead controller required. Add a phase
𝑠+25.3
adjustment of 100 . (πœ‘π‘‘ = 48.10 , πœ‘π‘šπ‘Žπ‘₯ = 24.10 , 𝛽 = 0.42, 𝐺𝐢 π‘™π‘’π‘Žπ‘‘ (𝑠) = 2.38 𝑠+60.2 ).
8. For the system 𝐺(𝑠) =
𝐾
,
𝑠(𝑠+50)(𝑠+12)
𝑇𝑠 = 0.2 𝑠𝑒𝑐. ( 𝐺𝐢 π‘™π‘’π‘Žπ‘‘ (𝑠) =
design a lead controller for 𝐾𝑉 = 50, %𝑂𝑆 = 20, and
2.27(𝑠+33.2)
𝑠+75.4
)
𝐾
𝑉
9. For 𝐾𝑉 = 12/𝑠𝑒𝑐, design a lead controller for the system 𝐺(𝑠) = 𝑠(𝑠+1)
such that the
resulting closed-loop system has a phase margin of 400 . Adjust the resultant angle by +50 .
(πœ‘π‘šπ‘Žπ‘₯ = 300 , 𝐺𝐢 π‘™π‘’π‘Žπ‘‘ (𝑠) =
0.337𝑠𝑠+1
0.125𝑠+1
)
𝐾
10. Design a double lead controller for the system 𝐺(𝑠) = 𝑠2 (0.2𝑠+1) for an acceleration error
constant πΎπ‘Ž = 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, πœ‘π‘šπ‘Žπ‘₯ = 35 − (−33) + 14 = 820 . Divide the 82 degrees into 41 degrees each. For
(0.358𝑠+1)2
0.358𝑠+1
∗
each of the lead controllers, πœ‘π‘šπ‘Žπ‘₯
= 410 , 𝛽 = 0.2. 𝐺𝐢1 π‘™π‘’π‘Žπ‘‘ = 0.077𝑠+1 ; 𝐺𝐢 π‘™π‘’π‘Žπ‘‘ = (0.077𝑠+1)2 )
𝐾
11. Design a lag controller for 𝐺(𝑠) = 𝑠(𝑠+1)(𝑠+4) so that the closed-loops system has the
following performance. The tolerance band is 5% (not the usual 2%).
Damping factor πœ‰ = 0.4
3
Settling time 𝑑𝑠 = 10 𝑠𝑒𝑐. (For a 5% tolerance band, 𝑑𝑠 = πœ‰πœ”
𝑛
Velocity error constant 𝐾𝑉 ≥ 5𝑠𝑒𝑐
Use an angular adjustment of 120 .
).
−1
( πœ‘π‘‘ = 430 , πœ”π‘ = 1.02 π‘Ÿπ‘Žπ‘‘π‘ π‘’π‘ −1 , πœ‘π‘šπ‘Žπ‘₯ = 550 , πœ”π‘ = 0.522 π‘Ÿπ‘Žπ‘‘π‘ π‘’π‘.−1 )
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