Unformatted text preview: -150
-200
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10 2008/09 -2 10 -1 10 0 10 1 MECH466 : Automatic Control 10 2 21 2008/09 MECH466 : Automatic Control Bode plot for C(s)=0.286 22 Step responses 50 1.4 Low freq. gain 0
decreases. 1.2 -50 1 -100
10 -2 10 -1 10 0 10 1 10 0.8 2 Uncompensated (C(s)=1)
Uncompensated (C(s)=1) 0.4 K=0.286 (PM=50deg) 0.2 -150 0.6 Gain compensated -100 K=0.158 (PM=65deg) -200
-250
10 2008/09 -2 10 -1 10 0 MECH466 : Automatic Control 10 1 10 0
0 2 23 2008/09 K=0.455 (PM=35deg) 5 10 MECH466 : Automatic Control 15
24 6 Phase-lag compensator (review) Phase-lag C(s) design
We try to design phase-lag C(s) which gives
phaseC(s)
• PM 50deg
• Low frequency gain same as the original plant. dB
0 -20 -5 1. To satisfy low frequency requirement, adjust
DC gain of OL system by a constant gain K. -10
-15
-20 -2
10 deg -1 0 10 10 1 2 10 C(s)
C(s) 3 10 10 Analysis for C(s)=1
Analysis
C(s)=1 0 -40 -45 2008/09 -1 0 10 10 1 2 10 3 10 MECH466 : Automatic Control Plant 10 25 2008/09 After Step 1
OK Controller Stable
Stable
PM at least 12 deg
PM
GM at least 3.5 dB
GM -20 -60 -2
10 G(s)
G(s) MECH466 : Automatic Control 26 Phase-lag C(s) design
2. Find the frequency ωg (which will become gain
crossover frequency after compensation)
where 50
0
-50
-100
10 -2 10 -1 10 0 10 1 10 2 In this example, -100
-150
-200
-250
10
2008/09 -2 10 -1 10 0 MECH466 : Automatic Control 10 1 10 Note: The reason of +5 deg is explained later.
Note: 2 27 2008/09 MECH466 : Automatic Control 28 7 A...
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- Fall '08
- RyozoNagamune
- Laplace, automatic control
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