MECH 466 - Lecture21-LeadLagDesigninFrequencyDomain-2009W

50 0 50 1 to satisfy low frequency requirement adjust

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Unformatted text preview: is for C(s)=1 Analysis C(s)=1 G(s) G(s) Plant -1 10 0 10 1 10 2 -150 Stable Stable PM at least 12 deg PM GM at least 3.5 dB GM 2008/09 10 -100 Controller -2 MECH466 : Automatic Control -200 -250 10 13 -2 10 2008/09 -1 10 0 10 1 10 2 MECH466 : Automatic Control Phase-lag C(s) design 14 After Step 2 2. Find the frequency ωg (which will become gain crossover frequency after compensation) where 50 0 -50 -100 10 In this example, -2 10 -1 10 0 10 1 10 2 -100 PM=55 -150 -200 -250 Note: The reason of +5 deg is explained later. Note: 10 2008/09 MECH466 : Automatic Control 15 2008/09 -2 10 -1 10 0 MECH466 : Automatic Control 10 1 10 2 16 4 Phase-lag C(s) design After Step 3 50 3. Set z and p as deg 0 -50 For small phase lag at ωg -100 -2 -45 -1 10 Actual phase 10 10 0 10 1 2 10 -100 dB PM=50 -150 For setting new gain crossover at ωg -20 -200 -250 -2 -1 10 2008/09 MECH466 : Automatic Control 17 2008/09 C(s)=1 C(s)=1 10 1 2 10 18 Ramp responses (PM=12deg, ωg=1.1) C(s)=0.286 C(s)=0.286 (PM=50deg, ωg=0.5) 70 C(s)=CLag(s) (PM=52.3deg, ωg=0.4) C(s)=C (s) 1.5 10 MECH466 : Automatic Control Step responses 2 0 10 69 68 67 C(s)=0.286 C(s)=0.286 66 1 Ramp reference C(s)=1 (Kv=2) C(s)...
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