530_343lecture20

# 530_343lecture20 - ME 530.343 Design and Analysis of...

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ME 530.343: Design and Analysis of Dynamic Systems Spring 2009 Lecture 20 - Cruise Control Example April 10, 2009 Today’s Objectives Cruise control example Root locus design Cruise Control m bv Damping from air resistance resistance. Velocity, Force from Engine f m ˙ v + bv = f b = 50 Ns/m f = 500 N Design criteria: If engine gives a 500 N force, the velocity at steady-state will be v = f b = 500 50 = 10 m/s (22 mph) We desire to accelerate up to that speed in less than 5 seconds ( t r 5 sec ) Since this is only a cruise control system, a 10% overshoot on the velocity will not do much damage (overshoot < 10%) A 2% steady-state error is acceptable (ss error < 2%) Summary of design goals: rise time < 5 sec overshoot < 10% steady-state error < 2% Transfer function for original system msV ( s ) + bV ( s ) = F ( s ) V ( s ) F ( s ) = 1 ms + b To plot step response in Matlab: m = 1000; b = 50 f = 500; 1

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num = [1]; den = [m b]; step(f*num,den); % assumes zero initial conditions and unit step input * f
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## This note was uploaded on 03/02/2010 for the course MECH 530.343 taught by Professor Sun during the Spring '08 term at Johns Hopkins.

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530_343lecture20 - ME 530.343 Design and Analysis of...

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