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lec06_print - MAE143b Linear Control - Theory and...

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Unformatted text preview: MAE143b Linear Control - Theory and Applications Lecture 6, Thursday Aug. 20, 2:00-4:50pm Prof. R.A. de Callafon [email protected] CONTENTS OF THIS LECTURE • Example of design specification ( sec. 3.4 ) • Effect of zero locations ( sec. 3.5 ) • Frequency response analysis ( sec. 6.1 ) – frequency response – Bode plots – example of RC-network – Bode plots and asymptotes of 1st order systems – Lead compensator and example MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 6, Page 1 Dynamic Response- time-domain characteristics Time domain characteristics of step response: rise time t r ≈ 1 . 8 ω n , settling time t s ≈ 4 . 6 βω n , peak time t p = π ω d , overshoot M p = e − πβ/ √ 1 − β 2 all hold for the standard 2nd order system G ( s ) = ω 2 n s 2 + 2 βω n s + ω 2 n with ω n > 0, 0 < β < 1 (underdamped) and corresponding step response y ( t ) = 1 − e − βω n t (cos ω d t + β ω n ω d sin ω d t ) MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 6, Page 2 Dynamic Response- transformation to s-domain Requirements on rise time, settling time, overshoot can be re- versely engineered to specification on pole locations in the s- domain – insight in pole locations is useful for (control) design purposes. Example Design mass/damper/spring system with a rise time t r ≤ . 6 sec, an overshoot m p ≤ 10% and a settling time t s ≤ 3 sec. t r ≈ 1 . 8 ω n ≤ . 6 → ω n = q k m ≥ 1 . 8 . 6 = 3 rad/s ≈ . 5 Hz M p = e − πβ/ √ 1 − β 2 < . 1 → β = 1 2 d √ mk ≥ . 6 t s ≈ 4 . 6 βω n ≤ 3 → βω n = 1 2 d m ≥ 4 . 6 3 ≈ 1 . 5 sec Smallest values for k and d for a given mass m : k = 3 2 · m , d = 2 · . 6 · √ mk , making βω n = 1 2 d m = 1 . 8 ≥ 1 . 5. MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 6, Page 3 Dynamic Response- effects of zeros and additional poles Guidelines for rise time t r ≈ 1 . 8 ω n , settling time t s ≈ 4 . 6 βω n , peak time t p = π ω d , overshoot M p = e − πβ/ √ 1 − β 2 only hold for the standard 2nd order system and should only be used as a guide line for more complicated systems : • rise time t r is too small – increase natural frequency • transient has too much overshoot – increase damping • transient too long to decay – move poles more left into LHP Effect of zeros: in general modify shape of response but not transient behavior! Example: G 1 ( s ) = 2 ( s +2)( s +3) = 2 s +2 − 2 s +3 G 2 ( s ) = 2( s +1) ( s +2)( s +3) = − 2 s +2 + 4 s +3 impulse response of G 1 ( s ) (solid) G 2 ( s ) (dashed) 0.5 1 1.5 2 2.5 3-0.5 0.5 1 1.5 2 MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 6, Page 4 Dynamic Response- effects of zeros and additional poles Difference in dynamic response can also be seen in step response....
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This note was uploaded on 05/27/2010 for the course MAE 143B taught by Professor Paoc.chau during the Spring '06 term at UCSD.

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lec06_print - MAE143b Linear Control - Theory and...

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