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ece580 lecture 04 - Lecture 4 ECE 580 Feedback Control...

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Sept. 15, 2011 Feedback Control Systems (I) © Douglas Looze 1 Lecture 4 ECE 580 Feedback Control Systems (I) MIE 444 Automatic Controls Doug Looze
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Sept. 15, 2011 Feedback Control Systems (I) © Douglas Looze 2 Announce Ø Problem set 1 solutions will be posted on the course website this afternoon Ø Problem set 2 available Due next Thursday Ø Matlab diary on course website http://ece580.ecs.umass.edu/ Ø TA: Arun Saranathan [email protected]
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Sept. 15, 2011 Feedback Control Systems (I) © Douglas Looze 3 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Step Response Time (sec) Amplitude Ø Step response (2nd order system) ss y p y p t r y r t d y d t s t ±4% p ss p ss y y M y - =
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Sept. 15, 2011 Feedback Control Systems (I) © Douglas Looze 4 0.3 0.7 ζ < < Ø Rise time for 2nd order system Damping ratio 0 1 ζ < < r d t π β ϖ - = 100 % Damping ratio 1.8 r n t ϖ = Empirical 10 - 90% Ø Peak time and value p d t π ϖ = 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 0.2 0.4 0.6 0.8 1 Damping Ratio Peak Overshoot 2 1 p M e πζ ζ - - @
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Sept. 15, 2011 Feedback Control Systems (I) © Douglas Looze 5 Ø Settling time 2% settling time ( 29 ln 0.02 4 s n n t ζϖ ζϖ = - 1% settling time ( 29 ln 0.01 4.6 s n n t ζϖ ζϖ = -
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Sept. 15, 2011 Feedback Control Systems (I) © Douglas Looze 6 Today Ø Additional poles/zeros Ø Stability Routh-Hurwitz criterion Ø Reading FPE 3.5 3.6
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Sept. 15, 2011 Feedback Control Systems (I) © Douglas Looze 7 Additional Poles/Zeros Ø Have considered the transfer function ( 29 2 2 2 2 n n n G s s s ϖ ζϖ ϖ = + + 2 poles 0 zeros Normalized (gain 1) Ø Note: can temporally normalize also ( 29 2 1 2 1 n n G s s s ζ ϖ ϖ = + + ÷ ÷ 2 1 2 1 n n s s ζ = + + ( 29 u t ( 29 y t G
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Sept. 15, 2011 Feedback Control Systems (I) © Douglas Looze 8 Ø Step response (2nd order system) 0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Step Response Time (sec) Amplitude ss y p y p t r y r t d y d t s t ±4% 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 0.2 0.4 0.6 0.8 1
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