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MIT 6 003 - CT Feedback and Control

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QUIZ I

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6.003: Signals and SystemsCT Feedback and ControlOctober 29, 2009Feedback and ControlContinuous-time feedback has many applications.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 DistortionoriginaldistortedK = 2K = 4K = 8K = 16originalCrossover DistortionoriginaldistortedK = 2K = 4K = 8K = 16originalCrossover DistortionoriginaldistortedK = 2K = 4K = 8K = 16originalCrossover DistortionoriginaldistortedK = 2K = 4K = 8K = 16originalCrossover DistortionoriginaldistortedK = 2K = 4K = 8K = 16originalCrossover DistortionoriginaldistortedK = 2K = 4K = 8K = 16originalCrossover DistortionoriginaldistortedK = 2K = 4K = 8K = 16originalCrossover DistortionoriginaldistortedK = 2K = 4K = 8K = 16originalCrossover DistortionoriginaldistortedK = 2K = 4K = 8K = 16originalFeedback and ControlContinuous-time feedback has many applications.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 also useful for controlling unstable systems.Example: Magnetic levitation.i(t) = ioy(t)Control of Unstable SystemsMagnetic levitation is unstable.i(t) = ioy(t)f(t)MgEquilibrium (y = 0): magnetic force f(t) is equal to the weight Mg.Increase y → greater force → further increase y.Decrease y → reduced force → further decrease y.Positive feedback !Control of Unstable SystemsMagnetic levitation is unstable.i(t) = ioy(t)f(t)Mgf(t)y(t)Mgi(t) = i0Control of Unstable SystemsThe instability of magnetic levitation is similar to that of a ball poisedat the apex of perfectly smooth hill.“Levitation” with a SpringBy contrast, the mass and spring system is not unstable.F = Kx(t) − y(t)= M ¨y(t)x(t)y(t)“Levitation” with a SpringIf the body moves up (or down) the spring force decreases (or in-creases) and the body tends to fall back (or rise up).F = Kx(t) − y(t)= M ¨y(t)f(t)y(t)Mg−K“Levitation” with a SpringThe spring and mass system is a stable system, analogous to a ballin a valley.“Levitation” with a SpringThe block diagram for the spring and mass system has negativefeedback, because the slope of the force curve is negative.F = Kx(t) − y(t)= M ¨y(t)+KMA Ax(t) y(t)˙y(t)¨y(t)−Levitation with a SpringThis system is marginally stable (poles on imaginary axis), or stable(poles in left half plane) if effects of friction are included.s-planeω0≡qKMω0≡ −qKMMagnetic LevitationBy contrast, the system representing magnetic levitation has positivefeedback.f(t)y(t)Mgi(t) = i0+p20A Ax(t) y(t)˙y(t)¨y(t)+Magnetic LevitationThe poles are at s = ±p.s-planep0−p0Magnetic 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) = 0.9i0Magnetic LevitationMagnetic levitation can be “stabilized” using a controller with bothproportional plus derivative terms.+ α(1 + βs)A Ax(t) y(t)˙y(t)¨y(t)s-planep0−p0Magnetic LevitationWithout feedback, one pole is in the right half plane → unstable.With sufficient feedback, all poles are in left half plane → stable.s-planep0−p0Here, feedback is used to stabilize an otherwise unstable system.Try 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 PendulumAs a final example of stabilizing an unstable system, consider aninverted pendulum.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 stablized with


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