6.003: Signals and SystemsCT Feedback and ControlMarch 18, 2010Feedback and ControlFeedback: simple, elegant, and robust framework for control.+−X YESCcontrollerplantsensorWe started with robotic driving.di= desiredFrontdo= distanceFrontFeedback and ControlUsing feedback to enhance performance.Examples:• improve performance of an op amp circuit.• control position of a motor.• reduce sensitivity to unwanted parameter variation.• reduce distortions.• stabilize unstable systems− magnetic levitation− inverted pendulumFeedback and ControlReducing sensitivity to unwanted parameter variation.Example: power amplifierF0MP3 playerpoweramplifier8 < F0< 12speakerChanges in F0(due to changes in temperature, for example) lead toundesired changes in sound level.Feedback and ControlFeedback can be used to compensate for parameter variation.F0MP3 playerKβ+poweramplifier8 < F0< 12speakerX Y−H(s) =KF01 + βKF0If K is made large, so that βKF0 1, thenH(s) ≈1βindependent of K or F0!Feedback and ControlFeedback reduces the change in gain due to change in F0.F0MP3 player100110+8 < F0< 12X Y−0 10 20010208 < F0< 12F0Gain to SpeakerF0(no feedback)100F01 +100F010(feedback)Check YourselfF0MP3 playerKβ+poweramplifier8 < F0< 12speakerX Y−Feedback greatly reduces sensitivity to variations in K or F0.limK→∞H(s) =KF01 + βKF0→1βWhat about variations in β? Aren’t those important?Check YourselfWhat about variations in β? Aren’t those important?The value of β is typically determined with resistors, whose valuesare quite stable (compared to semiconductor devices).Crossover DistortionFeedback can compensate for parameter variation even when thevariation occurs rapidly.Example: using transistors to amplify power.MP3 playerspeaker+50V−50VCrossover DistortionThis circuit introduces “crossover distortion.”For the upper transistor to conduct, Vi− Vo> VT.For the lower transistor to conduct, Vi− Vo< −VT.+50V−50VViVoViVoVT−VTCrossover DistortionCrossover distortion can have dramatic effects.Example: crossover distortion when the input is Vi(t) = B sin(ω0t).+50V−50VViVotVo(t)Crossover DistortionFeedback can reduce the effects of crossover distortion.MP3 playerK+speaker+50V−50V−Crossover DistortionWhen K is small, feedback has little effect on crossover distortion.K++50V−50VViVo−tVo(t)K = 1Crossover DistortionAs K increases, feedback reduces crossover distortion.K++50V−50VViVo−tVo(t)K = 2Crossover DistortionAs K increases, feedback reduces crossover distortion.K++50V−50VViVo−tVo(t)K = 4Crossover DistortionAs K increases, feedback reduces crossover distortion.K++50V−50VViVo−tVo(t)K = 10Crossover DistortionK++50V−50VViVo−tVo(t)Demo• original• no feedback• K = 2• K = 4• K = 8• K = 16• originalJ.S. Bach, Sonata No. 1 in G minor Mvmt. IV. PrestoNathan Milstein, violinFeedback and ControlUsing feedback to enhance performance.Examples:• improve performance of an op amp circuit.• control position of a motor.• reduce sensitivity to unwanted parameter variation.• reduce distortions.• stabilize unstable systems− magnetic levitation− inverted pendulumControl of Unstable SystemsFeedback is useful for controlling unstable systems.Example: Magnetic levitation.i(t) = ioy(t)Control of Unstable SystemsMagnetic levitation is unstable.i(t) = ioy(t)fm(t)MgEquilibrium (y = 0): magnetic force fm(t) is equal to the weight M g.Increase y → increased force → further increases y.Decrease y → decreased force → further decreases y.Positive feedback!Modeling Magnetic LevitationThe magnet generates a force that depends on the distance y(t).i(t) = ioy(t)fm(t)Mgfm(t)y(t)Mgi(t) = i0Modeling Magnetic LevitationThe net force accelerates the mass.i(t) = ioy(t)fm(t)Mgfm(t) − Mg = f(t) = M a = M ¨y(t)magnet1MA Ay(t)y(t)f(t)Modeling Magnetic LevitationOver small distances, magnetic force grows ≈ linearly with distance.f(t)y(t)Mgi(t) = i0magnet1MA Ay(t)y(t)f(t)“Levitation” with a SpringRelation between force and distance for a spring is opposite in sign.F = Kx(t) − y(t)= M ¨y(t)x(t)y(t)f(t)y(t)Mg−KModeling Magnetic LevitationOver small distances, magnetic force nearly proportional to distance.f(t)y(t)Mgi(t) = i0Kf(t) ≈ Ky(t)K1MA Ay(t)y(t)f(t)Block DiagramsBlock diagrams for magnetic levitation and spring/mass are similar.Spring and massF = Kx(t) − y(t)= M ¨y(t)+KMA Ax(t) y(t)˙y(t)¨y(t)−Magnetic levitationF = Ky(t) = M ¨y(t)+KMA Ax(t) = 0 y(t)˙y(t)¨y(t)+Check YourselfHow do the poles of these two systems differ?Spring and massF = Kx(t) − y(t)= M ¨y(t)+KMA Ax(t) y(t)˙y(t)¨y(t)−Magnetic levitationF = Ky(t) = M ¨y(t)+KMA Ax(t) = 0 y(t)˙y(t)¨y(t)+Check YourselfHow do the poles of the two systems differ?s-planeSpring and massF = Kx(t) − y(t)= M ¨y(t)YX=KMs2+KM→ s = ±jrKMs-planeMagnetic levitationF = Ky(t) = M¨y(t)s2=KM→ s = ±rKMMagnetic Levitation is Unstablei(t) = ioy(t)fm(t)Mgmagnet1MA Ay(t)y(t)f(t)Magnetic LevitationWe can stabilize this system by adding an additional feedback loopto control i(t).f(t)y(t)Mgi(t) = 1.1i0i(t) = i0i(t) = 0.9i0Stabilizing Magnetic LevitationStabilize magnetic levitation by controlling the magnet current.i(t) = ioy(t)fm(t)Mgmagnet1MA Aαy(t)y(t)f(t)i(t)Stabilizing Magnetic LevitationStabilize magnetic levitation by controlling the magnet current.i(t) = ioy(t)fm(t)Mg+1MA Ay(t)fi(t)fo(t)−K2KMagnetic LevitationIncreasing K2moves poles toward the origin and then onto jω axis.+K−K2MA Ax(t) y(t)˙y(t)¨y(t)s-planeBut the poles are still marginally stable.Magnetic LevitationAdding a zero makes the poles stable for sufficiently large K2.+K−K2M(s + z0)A Ax(t) y(t)˙y(t)¨y(t)s-planeTry it: Demo [designed by Prof. James Roberge].Inverted PendulumAs a final example of stabilizing an unstable system, consider aninverted pendulum.x(t)θ(t)mglmd2x(t)dt2θ(t)mgllab frame(inertial)cart frame(non-inertial)ml2|{z}Id2θ(t)dt2= mg|{z}forcel sin θ(t)| {z }distance− md2x(t)dt2| {z }forcel cos θ(t)| {z }distanceCheck Yourself: Inverted PendulumWhere are the poles of this system?x(t)θ(t)mglmd2x(t)dt2θ(t)mglml2d2θ(t)dt2= mgl sin θ(t) − md2x(t)dt2l cos θ(t)Check Yourself: Inverted PendulumWhere are the poles of this system?x(t)θ(t)mglmd2x(t)dt2θ(t)mglml2d2θ(t)dt2= mgl sin θ(t) − md2x(t)dt2l cos θ(t)ml2d2θ(t)dt2− mglθ(t) = −mld2x(t)dt2H(s) =ΘX=−mls2ml2s2− mgl=−s2/ls2− g/lpoles at s = ±rglInverted PendulumThis unstable system can be
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