6.003: Signals and Systems CT Feedback and Control October 25, 2011 - - PowerPoint PPT Presentation

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6.003: Signals and Systems CT Feedback and Control October 25, 2011 - - PowerPoint PPT Presentation

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6.003: Signals and Systems CT Feedback and Control October 25, 2011 1 Mid-term Examination #2 Tomorrow, October 26, 7:30-9:30pm, No recitations on the day of the exam. Coverage: Lectures 112 Recitations 112 Homeworks 17 Homework 7 will not

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6.003: Signals and Systems

CT Feedback and Control

October 25, 2011

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Mid-term Examination #2

Tomorrow, October 26, 7:30-9:30pm, No recitations on the day of the exam. Coverage: Lectures 1–12 Recitations 1–12 Homeworks 1–7 Homework 7 will not be collected or graded. Solutions are posted. Closed book: 2 pages of notes (81

2 × 11 inches; front and back).

No calculators, computers, cell phones, music players, or other aids. Designed as 1-hour exam; two hours to complete. Old exams and solutions are posted on the 6.003 website.

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Feedback and Control

Using 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 pendulum

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Feedback and Control

Reducing sensitivity to unwanted parameter variation. Example: power amplifier Changes in F0 (due to changes in temperature, for example) lead to undesired changes in sound level. F0 MP3 player power amplifier 8 < F0 < 12 speaker

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Feedback and Control

Feedback can be used to compensate for parameter variation. F0 MP3 player K β + power amplifier 8 < F0 < 12 speaker X Y − H(s) = KF0 1 + βKF0 If K is made large, so that βKF0 » 1, then 1 H(s) ≈ β independent of K or F0!

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Feedback and Control

Feedback reduces the change in gain due to change in F0. F0 MP3 player 100

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+ 8 < F0 < 12 X Y − 10 20 10 20 8 < F0 < 12 F0 Gain to Speaker F0 (no feedback) 100F0 1 + 100F0

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(feedback)

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Check Yourself

F0 MP3 player K β + power amplifier 8 < F0 < 12 speaker X Y − Feedback greatly reduces sensitivity to variations in K or F0. lim

K→∞ H(s) =

KF0 1 + βKF0 → 1 β What about variations in β? Aren’t those important?

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Check Yourself

What about variations in β? Aren’t those important? The value of β is typically determined with resistors, whose values are quite stable (compared to semiconductor devices).

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Crossover Distortion

Feedback can compensate for parameter variation even when the variation occurs rapidly. Example: using transistors to amplify power. MP3 player speaker +50V −50V

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Crossover Distortion

This circuit introduces “crossover distortion.” For the upper transistor to conduct, Vi − Vo > VT . For the lower transistor to conduct, Vi − Vo < −VT . +50V −50V Vi Vo Vi Vo VT −VT

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Crossover Distortion

Crossover distortion changes the shapes of signals. Example: crossover distortion when the input is Vi(t) = B sin(ω0t). +50V −50V Vi Vo t Vo(t)

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Crossover Distortion

Feedback can reduce the effects of crossover distortion. MP3 player K + speaker +50V −50V −

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Crossover Distortion

When K is small, feedback has little effect on crossover distortion. K + +50V −50V Vi Vo − t Vo(t) K = 1

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Crossover Distortion

K + +50V −50V Vi Vo − Feedback reduces crossover distortion. t Vo(t) K = 2

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Crossover Distortion

K + +50V −50V Vi Vo − Feedback reduces crossover distortion. t Vo(t) K = 4

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SLIDE 16

Crossover Distortion

K + +50V −50V Vi Vo − Feedback reduces crossover distortion. t Vo(t) K = 10

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SLIDE 17

K + +50V −50V Vi Vo − t Vo(t)

Demo

  • original
  • no feedback
  • K = 2
  • K = 4
  • K = 8
  • K = 16
  • original

Crossover Distortion

J.S. Bach, Sonata No. 1 in G minor Mvmt. IV. Presto Nathan Milstein, violin

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Feedback and Control

Using 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 pendulum

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Control of Unstable Systems

Feedback is useful for controlling unstable systems. Example: Magnetic levitation. i(t) = io y(t)

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Control of Unstable Systems

Magnetic levitation is unstable. i(t) = io y(t) fm(t) Mg Equilibrium (y = 0): magnetic force fm(t) is equal to the weight Mg. Increase y → increased force → further increases y. Decrease y → decreased force → further decreases y. Positive feedback!

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Modeling Magnetic Levitation

The magnet generates a force that depends on the distance y(t). i(t) = io y(t) fm(t) Mg fm(t) y(t) Mg i(t) = i0

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Modeling Magnetic Levitation

The net force f(t) = fm(t) − Mg accelerates the mass. i(t) = io y(t) fm(t) Mg f(t) = fm(t) − Mg = Ma = M ¨ y(t) y(t) i(t) = i0

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SLIDE 23

Modeling Magnetic Levitation

Represent the magnet as a system: input y(t) and output f(t). i(t) = io y(t) fm(t) Mg f(t) = fm(t) − Mg = Ma = M ¨ y(t) y(t) i(t) = i0 magnet y(t) f(t)

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Modeling Magnetic Levitation

The magnet system is part of a feedback system. f(t) = fm(t) − Mg = Ma = M ¨ y(t) y(t) i(t) = i0 magnet y(t) f(t) magnet

1 M

A A y(t) y(t) ¨ y(t) f(t)

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Modeling Magnetic Levitation

For small distances, force grows approximately linearly with distance. f(t) = fm(t) − Mg = Ma = M ¨ y(t) y(t) i(t) = i0 K K y(t) f(t) K

1 M

A A y(t) y(t) ¨ y(t) f(t)

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“Levitation” with a Spring

Relation between force and distance for a spring is opposite in sign.

  • F = K x(t) − y(t)

= M ¨ y(t) x(t) y(t) f(t) y(t) Mg −K

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  • Block Diagrams

Block diagrams for magnetic levitation and spring/mass are similar. Spring and mass F = K x(t) − y(t) = M ¨ y(t) + K M A A x(t) y(t) ˙ y(t) ¨ y(t) − Magnetic levitation F = Ky(t) = M ¨ y(t) + K M A A x(t) = 0 y(t) ˙ y(t) ¨ y(t) +

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  • Check Yourself

+ K M A A x(t) = 0 y(t) ˙ y(t) ¨ y(t) + How do the poles of these two systems differ? Spring and mass F = K x(t) − y(t) = M ¨ y(t) + K M A A x(t) y(t) ˙ y(t) ¨ y(t) − Magnetic levitation F = Ky(t) = M ¨ y(t)

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Check Yourself

How do the poles of the two systems differ? s-plane Spring and mass F = K

  • x(t) − y(t)
  • = M ¨

y(t) Y X =

K M

s2 + K

M

→ s = ±j

  • K

M s-plane Magnetic levitation F = Ky(t) = M ¨ y(t) s2 = K M → s = ±

  • K

M

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Magnetic Levitation is Unstable

i(t) = io y(t) fm(t) Mg magnet

1 M

A A y(t) y(t) ¨ y(t) f(t)

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Magnetic Levitation

We can stabilize this system by adding an additional feedback loop to control i(t). f(t) y(t) Mg i(t) = 1.1i0 i(t) = i0 i(t) = 0.9i0

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Stabilizing Magnetic Levitation

Stabilize magnetic levitation by controlling the magnet current. i(t) = io y(t) fm(t) Mg magnet

1 M

A A α y(t) y(t) ¨ y(t) f(t) i(t)

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Stabilizing Magnetic Levitation

Stabilize magnetic levitation by controlling the magnet current. i(t) = io y(t) fm(t) Mg +

1 M

A A y(t) fi(t) fo(t) −K2 K

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Magnetic Levitation

Increasing K2 moves poles toward the origin and then onto jω axis. +

K−K2 M

A A x(t) y(t) ˙ y(t) ¨ y(t) s-plane But the poles are still marginally stable.

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Magnetic Levitation

Adding a zero makes the poles stable for sufficiently large K2. +

K−K2 M

(s + z0) A A x(t) y(t) ˙ y(t) ¨ y(t) s-plane Try it: Demo [designed by Prof. James Roberge].

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inverted pendulum. x(t) θ(t) mg l md2x(t) dt2 θ(t) mg l lab frame cart frame

Inverted Pendulum

As a final example of stabilizing an unstable system, consider an (inertial) (non-inertial) d2θ(t) d2x(t) ml2 = mg l sin θ(t) − m l cos θ(t) m-l2 dt2 m-l2 m

  • l

2 m -l dt2 2 m -l 2

I force distance distance force

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Check Yourself: Inverted Pendulum

Where are the poles of this system? x(t) θ(t) mg l md2x(t) dt2 θ(t) mg l ml2 d2θ(t) dt2 = mgl sin θ(t) − m d2x(t) dt2 l cos θ(t)

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Check Yourself: Inverted Pendulum

Where are the poles of this system? x(t) θ(t) mg l md2x(t) dt2 θ(t) mg l ml2 d2θ(t) dt2 = mgl sin θ(t) − m d2x(t) dt2 l cos θ(t)

2 d2θ(t)

d2x(t) ml − mglθ(t) = −ml dt2 dt2

  • Θ

−mls2 −s2/l g H(s) = = = poles at s = ± X ml2s2 − mgl s2 − g/l l

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Inverted Pendulum

This unstable system can be stablized with feedback. x(t) θ(t) mg l md2x(t) dt2 θ(t) mg l Try it. Demo. [originally designed by Marcel Gaudreau]

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Feedback and Control

Using 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 pendulum

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MIT OpenCourseWare http://ocw.mit.edu

6.003 Signals and Systems

Fall 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.