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Published byBethanie Robertson Modified over 9 years ago
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A Freq. Resp. Example Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design A Radar Tracking System Design a unity DC gain phase lead compensator Specs: o PM = 50 o T s < 4 sec Slide 1 of 8
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A Freq. Resp. Example Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design A proportional gain controller: The Bode diagram of the unity gain OL transfer fn GM = 3.5 dB (@1.4 rad/s) PM = 11 (@1.1 rad/s) Looks like we need to remove about 11 dB of gain 11 dB = 3.55 (gain) K = 1/3.55 = 0.282 11 dB Slide 2 of 8
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A Freq. Resp. Example Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design A proportional gain controller: The Bode diagram of the unity gain OL transfer fn GM = 14.5 dB (@1.4 rad/s) PM = 50 (@0.49 rad/s) Slide 3 of 8
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A Freq. Resp. Example Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Closed Loop Step Response The T s < 4 sec is NOT met PO 19% Increasing the PM by reducing K less PO, less CL BW, & slower T r K=0.455 PM of 35 K=0.158 PM of 65 Closed-Loop FR (magnitude) Slide 4 of 8
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A Freq. Resp. Example Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Phase Lead Compensator Design a unity gain (a 0 =1) phase lead comp From the T s < 4 requirement (and PM ( m =50 ) Slide 5 of 8
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A Freq. Resp. Example Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Phase Lead Compensator Slide 6 of 8
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A Freq. Resp. Example Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Phase Lead Compensator The OL Freq Resp GM = 23.8 dB (@ 8.8 rad/s) PM = 50 (@1.7 rad/s) Slide 7 of 8
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A Freq. Resp. Example Wednesday 25 Oct 2013 EE 401: Control Systems Analysis and Design Phase Lead Compensator The CL Unit Step Response: Comparing Lead & Lag The Lead compensator design meets both the T s and PM requirements Slide 8 of 8
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