Control 20 0 10 20 5 0 2 10 1 10 0 10 1 10 2 10 pm

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Unformatted text preview: ing effect 2008/09 3 10 -160 Select z&p around ωg MECH466 : Automatic Control PM: 47 deg at ωg=60 rad/s 15 2008/09 PM -180 1 10 MECH466 : Automatic Control 2 10 3 10 16 4 Example of a lead-lag C(s) design Lead-lag compensator 100 20 50 15 10 0 5 -50 0 -2 10 10 -1 10 0 10 1 10 2 10 PM: 28 deg at ωg=47 rad/s 3 -100 -2 10 -1 0 10 10 1 10 2 10 3 10 40 -100 20 0 -120 -20 -140 -40 -160 -60 10 -2 10 -1 2008/09 10 0 10 1 10 2 10 PM: 47 deg at ωg=60 rad/s 3 MECH466 : Automatic Control 17 -180 -2 10 2008/09 PM -1 0 10 10 1 10 2 10 MECH466 : Automatic Control Step responses 3 10 18 Ramp responses 0.5 1.5 Uncompensated (C(s)=1) (C(s)=1) Ramp reference Lead-lag compensated Lead- 0.495 Uncompensated (C(s)=1) (C(s)=1) Kv=100 Kv=100 1 0.49 0.5 Less overshoot is due to larger PM. Faster response is due to larger wg. 0 2008/09 0.48 0.48 0 0.1 0.2 0.3 MECH466 : Automatic Control 0.4 Lead-lag compensated LeadKv=1000 Kv=1000 0.485 0.5 0.485 0.49 0.495 0.5 Smaller steady-state error is due to larger Kv. steadyKv. 19 2008/09 MECH466 : Automatic Control 20 5 An example Consider a system Consider Gain compensation C(s) C(s) Controller PM is specified to be 50 deg. PM In this example, to increase PM by gain In compensation, we need to lower the gain curve. G(s) G(s) Plant 50 Analysis for C(s)=1 Analysis C(s)=1 Stable Stable PM at least 12 deg PM GM at least 3.5 dB GM 0 -50 -100 10 -2 10 -1 10 0 10 1 10 2 -100 These values are too small for good transient response!...
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