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MIT16_30F10_lec04

# MIT16_30F10_lec04 - Topic#4 16.30/31 Feedback Control...

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Topic #4 16.30/31 Feedback Control Systems Control Design using Bode Plots Performance Issues Synthesis Lead/Lag examples

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Fall 2010 16.30/31 4–2 Bode’s Gain Phase Relationship Control synthesis by classical means would be very hard if we had to consider both the magnitude and phase plots of the loop, but that is not the case. Theorem : For any stable, minimum phase system with transfer func- tion G ( s ) , G ( j ω ) is uniquely related to the slope of | G ( j ω ) | . Relationship is that, on a log-log plot, if slope of the magnitude plot is constant over a decade in frequency, with slope n , then G ( j ω ) 90 n So in the crossover region, where L ( j ω ) 1 if the magnitude plot is (locally): s 0 slope of 0, so no crossover possible s 1 slope of -1, so about 90 PM s 2 slope of -2, so PM very small Basic rule of classical control design: Select G c ( s ) so that LTF crosses over with a slope of -1. September 19, 2010
Fall 2010 16.30/31 4–3 Performance Issues Step error response 1 e ss = 1 + G c (0) G p (0) and we can determine G c (0) G p (0) from the low frequency Bode plot for a type 0 system. For a type 1 system, the DC gain is infinite, but define K v = lim sG c ( s ) G p ( s ) e ss = 1 /K v s 0 So can easily determine this from the low frequency slope of the Bode plot. September 19, 2010

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Fall 2010 16.30/31 4–4 Performance Issues II Classic question: how much phase margin do we need? Time response of a second order system gives: 1. Closed-loop pole damping ratio ζ PM/ 100 , PM < 70 2. Closed-loop resonant peak M r = 2 ζ 1 1 ζ 2 1 , near 2 sin( PM/ 2) ω r = ω n 1 2 ζ 2 3. Closed-loop bandwidth ω BW = ω n 1 2 ζ 2 + 2 4 ζ 2 + 4 ζ 4 and ω c = ω n 1 + 4 ζ 4 2 ζ 2 Fig. 1: Frequency domain performance specifications. So typically specify ω c , PM, and error constant as design goals September 19, 2010 Band width ω BW 10 1 0.7 Amplitude ratio |y|/|r| 0.1 20 0 -3 -20 Decibels ω, ω c M r rad/s resonant peak Image by MIT OpenCourseWare.
Fall 2010 16.30/31 4–5 Fig. 2: Crossover frequency for second order system 1 figure(1) % 2 z=[0:.01:1]'; zs=[0:.01:1/sqrt(2) eps]'; 3 wbw=sqrt(1 2 * z.ˆ2+sqrt(2 4 * z.ˆ2+4 * z.ˆ4)); % 4 wc=sqrt(sqrt(1+4 * z.ˆ4) 2 * z.ˆ2); % 5 wr=sqrt(1 2 * zs.ˆ2); % 6 set(gcf, 'DefaultLineLineWidth' ,2) % 7 plot(z,wbw,z,wc,zs,wr) % 8 legend( ' \ omega { BW } / \ omega n' , ' \ omega c/ \ omega n' , ' \ omega r/ \ omega n' ) % 9 xlabel( ' \ zeta' );ylabel( ' \ omega' ) % 10 print dpng r300 fresp.png Other rules of thumb come from approximating the system as having a 2nd order dominant response: 1 + 1 . 1 ζ + 1 . 4 ζ 2 10-90% rise time t r = ω n Settling time (5%) t s = 3 ζω n Time to peak amplitude t p = π ω n

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