6.003: Signals and Systems Lecture 4 September 22, 200916.003: Signals and SystemsFeedback and ControlSeptember 22, 2009Feedback and ControlFeedback is pervasive in natural and artificial systems.pVTurn steering wheel to stay centered in the lane.cardriverdesiredpositionactualpositionFeedback and ControlFeedback is useful for regulating a system’s behavior, as when athermostat regulates the temperature of a house.thermostatdesiredtemperatureactualtemperatureheatingsystemFeedback and ControlConcentration of glucose in blood is highly regulated and remainsnearly constant despite episodic ingestion and use.fooddigestivesystemglucosecirculatorysystemglucoseinsulincells &tissuesglucoseinsulinpancreas(β cells)+−glucoseinglucoseconcentrationpancreas(β cells)insulincirculatorysystemcells &tissuesstoredglucoseFeedback and ControlMotor control relies on feedback from pressure sensors in the skinas well as proprioceptors in muscles, tendons, and joints.Try building a robotic hand to unscrew a lightbulb!Shadow Dexterous Robot Hand (Wikipedia)Today’s goalUse systems theory to gain insight into how to control a system.6.003: Signals and Systems Lecture 4 September 22, 20092Example: Steering a CarAlgorithm: steer left when car is right of center and vice versa.steer leftsteer leftstraight ahead?steer rightsteer rightsteer rightstraight ahead?Bad algorithm → poor performance.Here we get persistent oscillations!Outline of the lectureUnderstanding the structure of a control problemautomatic control → feedbackAnalyzing feedback systemsfeedback → cyclic paths → persistent outputsDesigning control systemsconstructing well-behaved response propertiesStructure of a Control Problem(Simple) Control systems have three parts.+−X YESCcontrollerplantsensorThe plant is the system to be controlled.Thesensor measures the output of the plant.The controller specifies a command C to the plant based on thedifference between the input X and sensor output S.Three Parts in the Car Steering Problemplant = car perfect sensor proportional controller+αcar1−X YESCcontrollerplantsensorX = desired left/right position in lane (= 0)C = steering wheel position (C = αE = α(X − S) = α(X − Y ))Y = actual left/right position in laneCar Steering ProblemA realistic model for the car is complicated.+αcar1−X YESCcontrollerplantsensorTurning the steering wheel rotates the car body so that forwardmotion of the vehicle changes the position in the lane.Simpler Car Steering ProblemStart with a simpler problem: Imagine a (very non-physical) car thatcan instantly change its lateral position from y[n] to y[n] + c[n].0 1yshift −12shift −14shift −18shift −116shift −132shift −164shift −1128+αcar1−X YESCcontrollerplantsensorWhat is the value of α in this example?What would be a better value of α?6.003: Signals and Systems Lecture 4 September 22, 20093Check YourselfOn each step, the car changes its position from y[n] to y[n]+c[n].0 1yshift −12shift −14shift −18shift −116shift −132+αcar1−X YESCcontrollerplantsensorFind the appropriate model for this (very unusual) car.1. R 2.1R3.11 + R4.R1 − R5. none of the aboveCheck YourselfWhat is the system functionalYXfor the entire controllersystem?0 1yshift −12shift −14shift −18shift −116shift −132+αcar1−X YESCcontrollerplantsensor1.R1 − R2. αR1 − R3.αR1 − (1 −α)R4.αR1 − αR5. none of the aboveDesigning a ControllerThe “closed-loop” system has a single pole at (1 − α).YX=αR1 − (1 −α)R0 1yshift −12shift −14shift −18shift −116shift −132+α =12car1−X YESCcontrollerplantsensorTherefore, the output can deviate from the input for long periodsof time, as we saw when α =12.Designing a ControllerRange of responses y[n] given y[0] = 1 and x[n] = 0.y[n]nα = 0.1y[n]nα =12y[n]nα = 1y[n]nα =32Designing a ControllerLong-term deviations of the output from the input represent a failureto control the system in the desired fashion.We would like the car to instantly move to the desired position.0 1yshift −12shift −14shift −18shift −116shift −132+αcar1−X YESCcontrollerplantsensorFortunately, we can control the shape of the outputYX=αR1 − (1 −α)Rby controlling the “closed-loop” pole at (1 − α).Designing a ControllerThe “open-loop” system (car) has a pole at z = 1 (an accumulator).The “closed-loop” system (with feedback) has a pole at z = 1 − α.ReImWe can adjust α to optimize performance.YX=αR1 − (1 −α)R6.003: Signals and Systems Lecture 4 September 22, 20094Designing a ControllerThe “open-loop” system (car) has a pole at z = 1 (an accumulator).The “closed-loop” system (with feedback) has a pole at z = 1 − α.ReImWe can adjust α to optimize performance.YX=αR1 − (1 −α)RThe value with the fastest response results when α = 1.Modeling Sensor DelayThe system that we just analyzed is idealized.Real systems may deviate from this analysis in important ways.+α = 1R1 − RR−X YESCcontrollerplantsensorFor example, what would happen if the sensor introduced a delay?Modeling Sensor DelayWhat would happen if the sensor introduced a delay?see previous (−1) → command 1see previous (0) → command 0see previous (+1) → command −1see previous (+1) → command −1see previous (0) → command 0Introducing delay can destabilize a control system.Key issue in biological and artificial control systems.Check Yourself+αR1 − RR−X YESCcontrollerplantsensorWhich is true?1.YX=αR1 − R2.YX=αR1 − R + αR23.YX=αR1 + R −αR24.YX=αR1 − R− αR5. none of the aboveModeling Sensor DelayFind the closed-loop poles.Substitute R →1zin the system functional:YX=αR1 − R + αR2=α1z1 −1z+ α1z2=αzz2− z + αThe poles are atz =1 ±√1 − 4α2.Modeling Sensor DelayIf α is small, the fundamental modes occur at z ≈ α and z ≈ 1 − α.z =1 ±√1 − 4α2≈1 ± (1 −2α)2= 1 − α , α1Re zIm zz-planeα ≈ 00 1 2 34nmode 10 1 2 34nmode 2Little feedback (i.e., α small) → slow decay of fundamental mode 2→ slow system response.6.003: Signals and Systems Lecture 4 September 22, 20095Modeling Sensor DelayAs α increases, the fundamental modes move toward each other andcollide at z =12when α =14.z =1 ±√1 − 4α2=1 ±√1 − 12=12,1221Re zIm zz-planeα =140 1 2 34nmode 10 1 2 34nmode 2Persistent responses decay. Car moves toward desired location y = 0.Modeling Sensor DelayIf α > 1/4, the fundamental modes become complex.z =1 ±√1 − 4α2=1 ∓
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