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

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 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 ( 8 1 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. 2

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 3

Feedback and Control Reducing sensitivity to unwanted parameter variation. Example: power amplifier power amplifier F 0 MP3 player speaker 8 < F 0 < 12 Changes in F 0 (due to changes in temperature, for example) lead to undesired changes in sound level. 4

Feedback and Control Feedback can be used to compensate for parameter variation. power amplifier X Y MP3 player + F 0 K − 8 < F 0 < 12 speaker β KF 0 H ( s ) = 1 + βKF 0 If K is made large, so that βKF 0 » 1 , then 1 H ( s ) ≈ β independent of K or F 0 ! 5

Feedback and Control Feedback reduces the change in gain due to change in F 0 . X Y + F 0 MP3 player 100 − 8 < F 0 < 12 1 10 20 Gain to Speaker F 0 (no feedback) 100 F 0 (feedback) 10 1 + 100 F 0 10 8 < F 0 < 12 0 F 0 0 10 20 6

Check Yourself power amplifier X Y MP3 player + F 0 K − 8 < F 0 < 12 speaker β Feedback greatly reduces sensitivity to variations in K or F 0 . KF 0 → 1 K →∞ H ( s ) = lim 1 + βKF 0 β What about variations in β ? Aren’t those important? 7

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). 8

Crossover Distortion Feedback can compensate for parameter variation even when the variation occurs rapidly. Example: using transistors to amplify power. +50 V MP3 player speaker − 50 V 9

Crossover Distortion This circuit introduces “crossover distortion.” For the upper transistor to conduct, V i − V o > V T . For the lower transistor to conduct, V i − V o < − V T . V o +50 V − V T V i V o V i V T − 50 V 10

Crossover Distortion Crossover distortion changes the shapes of signals. Example: crossover distortion when the input is V i ( t ) = B sin( ω 0 t ) . V o ( t ) +50 V V i V o t − 50 V 11

Crossover Distortion Feedback can reduce the effects of crossover distortion. +50 V + MP3 player K − speaker − 50 V 12

Crossover Distortion When K is small, feedback has little effect on crossover distortion. +50 V V o ( t ) K = 1 V i + V o K − t − 50 V 13

Crossover Distortion Feedback reduces crossover distortion. +50 V V o ( t ) K = 2 V i + V o K − t − 50 V 14

Crossover Distortion Feedback reduces crossover distortion. +50 V V o ( t ) K = 4 V i + V o K − t − 50 V 15

Crossover Distortion Feedback reduces crossover distortion. +50 V V o ( t ) K = 10 V i + V o K − t − 50 V 16

Crossover Distortion +50 V Demo • original • no feedback V i + V o K − • K = 2 • K = 4 • K = 8 − 50 V • K = 16 • original V o ( t ) t J.S. Bach, Sonata No. 1 in G minor Mvmt. IV. Presto Nathan Milstein, violin 17

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 18

Control of Unstable Systems Feedback is useful for controlling unstable systems. Example: Magnetic levitation. i ( t ) = i o y ( t ) 19

Control of Unstable Systems Magnetic levitation is unstable. i ( t ) = i o f m ( t ) y ( t ) Mg Equilibrium ( y = 0 ): magnetic force f m ( t ) is equal to the weight Mg . Increase y → increased force → further increases y . Decrease y → decreased force → further decreases y . Positive feedback! 20

Modeling Magnetic Levitation The magnet generates a force that depends on the distance y ( t ) . i ( t ) = i o f m ( t ) y ( t ) Mg f m ( t ) i ( t ) = i 0 Mg y ( t ) 21

Modeling Magnetic Levitation The net force f ( t ) = f m ( t ) − Mg accelerates the mass. i ( t ) = i o f m ( t ) y ( t ) Mg f ( t ) = f m ( t ) − Mg = Ma = M ¨ y ( t ) i ( t ) = i 0 y ( t ) 22

Modeling Magnetic Levitation Represent the magnet as a system: input y ( t ) and output f ( t ) . i ( t ) = i o f m ( t ) y ( t ) Mg f ( t ) = f m ( t ) − Mg = Ma = M ¨ y ( t ) i ( t ) = i 0 y ( t ) y ( t ) magnet f ( t ) 23

Modeling Magnetic Levitation The magnet system is part of a feedback system. f ( t ) = f m ( t ) − Mg = Ma = M ¨ y ( t ) i ( t ) = i 0 y ( t ) y ( t ) magnet f ( t ) f ( t ) ¨ y ( t ) 1 y ( t ) magnet y ( t ) A A M 24

Modeling Magnetic Levitation For small distances, force grows approximately linearly with distance. f ( t ) = f m ( t ) − Mg = Ma = M ¨ y ( t ) i ( t ) = i 0 K y ( t ) y ( t ) f ( t ) K f ( t ) y ( t ) ¨ 1 y ( t ) y ( t ) K A A M 25

“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 ) Mg − K y ( t ) 26

Block Diagrams Block diagrams for magnetic levitation and spring/mass are similar. Spring and mass � � � � F = K x ( t ) − y ( t ) = M ¨ y ( t ) y ( t ) ¨ y ( t ) ˙ K x ( t ) + y ( t ) A A M − Magnetic levitation F = Ky ( t ) = M ¨ y ( t ) y ( t ) ¨ y ( t ) ˙ K x ( t ) = 0 + y ( t ) A A M + 27

Check Yourself How do the poles of these two systems differ? Spring and mass � � � � F = K x ( t ) − y ( t ) = M ¨ y ( t ) y ( t ) ¨ y ( t ) ˙ K x ( t ) + y ( t ) A A M − Magnetic levitation F = Ky ( t ) = M ¨ y ( t ) y ( t ) ¨ y ( t ) ˙ K x ( t ) = 0 + y ( t ) A A M + 28

Check Yourself How do the poles of the two systems differ? Spring and mass � � s -plane F = K x ( t ) − y ( t ) = M ¨ y ( t ) � K Y K M X = → s = ± j s 2 + K M M Magnetic levitation s -plane F = Ky ( t ) = M ¨ y ( t ) � s 2 = K K → s = ± M M 29

Magnetic Levitation is Unstable i ( t ) = i o f m ( t ) y ( t ) Mg f ( t ) ¨ y ( t ) 1 y ( t ) magnet y ( t ) A A M 30

Magnetic Levitation We can stabilize this system by adding an additional feedback loop to control i ( t ) . f ( t ) i ( t ) = 1 . 1 i 0 i ( t ) = i 0 i ( t ) = 0 . 9 i 0 Mg y ( t ) 31

Stabilizing Magnetic Levitation Stabilize magnetic levitation by controlling the magnet current. i ( t ) = i o f m ( t ) y ( t ) Mg i ( t ) α f ( t ) y ( t ) ¨ 1 y ( t ) magnet y ( t ) A A M 32

Stabilizing Magnetic Levitation Stabilize magnetic levitation by controlling the magnet current. i ( t ) = i o f m ( t ) y ( t ) Mg f i ( t ) − K 2 1 + y ( t ) A A M K f o ( t ) 33

Magnetic Levitation Increasing K 2 moves poles toward the origin and then onto jω axis. ¨ y ( t ) y ( t ) ˙ K − K 2 x ( t ) + y ( t ) A A M s -plane But the poles are still marginally stable. 34

Magnetic Levitation Adding a zero makes the poles stable for sufficiently large K 2 . y ( t ) ¨ y ( t ) ˙ K − K 2 x ( t ) + y ( t ) ( s + z 0 ) A A M s -plane Try it: Demo [designed by Prof. James Roberge]. 35

Inverted Pendulum As a final example of stabilizing an unstable system, consider an inverted pendulum. md 2 x ( t ) θ ( t ) dt 2 θ ( t ) mg l mg l x ( t ) lab frame cart frame (inertial) (non-inertial) d 2 θ ( t ) d 2 x ( t ) ml 2 = mg l sin θ ( t ) − m l cos θ ( t ) m-l2 dt 2 dt 2 2 m -l m-l2 m -l 2 2 m -l I force distance distance force 36

Check Yourself: Inverted Pendulum Where are the poles of this system? md 2 x ( t ) θ ( t ) dt 2 mg θ ( t ) l mg x ( t ) l ml 2 d 2 θ ( t ) = mgl sin θ ( t ) − m d 2 x ( t ) dt 2 l cos θ ( t ) dt 2 37

Check Yourself: Inverted Pendulum Where are the poles of this system? md 2 x ( t ) θ ( t ) dt 2 mg θ ( t ) l mg x ( t ) l ml 2 d 2 θ ( t ) = mgl sin θ ( t ) − m d 2 x ( t ) dt 2 l cos θ ( t ) dt 2 2 d 2 θ ( t ) d 2 x ( t ) ml − mglθ ( t ) = − ml dt 2 dt 2 � − mls 2 − s 2 /l Θ g H ( s ) = = = poles at s = ± X ml 2 s 2 − mgl s 2 − g/l l 38

Inverted Pendulum This unstable system can be stablized with feedback. md 2 x ( t ) θ ( t ) dt 2 mg θ ( t ) l mg x ( t ) l Try it. Demo. [originally designed by Marcel Gaudreau] 39

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 40

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