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lec09_print - MAE143b Linear Control Theory and...

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MAE143b Linear Control - Theory and Applications Lecture 9, Tuesday Sep. 1, 2:00-4:50pm Prof. R.A. de Callafon [email protected] CONTENTS OF THIS LECTURE Recap PID Controller and Design ( sec. 4.1 - 4.3 ) – properties – design for 2nd order system via pole placement State Space Design Methodologies – design by pole placement ( sec. 7.5 ) – Ackermanns’ formula ( sec. 7.5 + 7.6.1 ) – state feedback + observer ( sec. 7.7.1, 7.7.3 & 7.8 ) Frequency Response Design Methodologies ( Chap 6 ) – recap on frequency response ( sec. 6.1 ) – Nyquist plot & stability criterion ( sec. 6.2 & 6.3 ) – stability margins & loop shaping ( sec. 6.4 ) MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 9, Page 1 PID Control - properties Recap: ideal PID controller C ( s ) = K p + K i s + K d s = K d s 2 + K p s + K i s behaves like an integrator in low frequencies range , proportional controller in middle frequency range and differentiator at high frequency range . Distinction between low, high and middle frequency range is determined by the location of the zeros of the PID controller EXAMPLE: PID controller with K d = 1, K p = 11, K i = 10 having zeros at z 1 = 1 and z 2 = 10. 10 -2 10 0 10 2 10 0 10 2 10 4 10 -2 10 0 10 2 -100 -50 0 50 100 MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 9, Page 2
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PID Control - properties For actual implementation we use the stable and proper transfer function of PID controller ¯ C ( s ) = K p + K i s + K d s ( s + τ 1 )( τ 2 s + 1) = ( s + z 1 ) ( s + τ 1 ) ( s + z 2 ) ( τ 2 s + 1) and is a series connection of a lag compensator (at low frequen- cies) and a lead compensator (at high frequencies) . EXAMPLE: PID controller with K d = 1, K p = 11, K i = 10 creating zeros at z 1 = 1 and z 2 = 10 and τ 1 = 0 . 01 and τ 2 = 0 . 01 approximating the integrator with a pole at 0 . 01 and limiting the differ- entiating with a pole at 100 10 -2 10 0 10 2 10 0 10 2 10 4 10 -2 10 0 10 2 -100 -50 0 50 100 MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 9, Page 3 PID Control - design for 2nd order system via pole placement Application of the PID controller C ( s ) = K p + K i s + K d s to the standard 2nd order system G ( s ) will yield S ( s ) = s ( s 2 + 2 βω n s + ω 2 n ) s 3 + ω n (2 β + ω n K d ) s 2 + ω 2 n (1 + K p ) s + ω 2 n K i for which design rules based on closed-loop pole placement can be applied . Given a desired location of closed-loop poles p 1 , p 2 and p 3 , then: Compute desired closed-loop polynomial ( s p 1 )( s p 2 )( s p 3 ) = s 3 + a 2 s 2 + a 1 s + a 0 Solve K p , K i and K d from a 2 = ω n (2 β + ω n K d ) a 1 = ω 2 n (1 + K p ) a 0 = ω 2 n K i (1) The idea of closed-loop pole assignment is crucial in the analytic design of feedback controllers ! MAE143b, UCSD, Summer II 2009, R.A. de Callafon – Lecture 9, Page 4
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State Space Control - design by pole placement Feedback control design by pole placement gets much easier (in terms of formulae) if one switches to a state space model representation of the plant G ( s )!
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