6.003: Signals and Systems Lecture 5 September 24, 200916.003: Signals and SystemsFeedback and ControlSeptember 24, 2009Last TimeUnderstanding the structure of a control problemautomatic control → feedbackAnalyzing feedback systemsfeedback → cyclic paths → persistent outputsDesigning control systemsconstructing well-behaved response propertiesExample: 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!Last TimeWe investigated a VERY simple model for the car.The car could move laterally in a lane without rotating!see previous (−1) → command 1see previous (0) → command 0see previous (+1) → command −1see previous (+1) → command −1see previous (0) → command 0Even that simple system could go unstable.Last time, we learned how to stablize it.TodayWe will investigate more realistic models of steering.We will analyze more realistic (and more complex) feedback systems.Steering ControllerDesign a system to automatically steer a car.pmoving car, velocity VAssume a sensor reports position p within the lane:p = 0: in centerp > 0: right of centerp < 0: left of center6.003: Signals and Systems Lecture 5 September 24, 20092Steering ControllerModel the system.Let X represent the desired position in the lane (normally 0).Turn the steering wheel (φ) in proportion to the difference betweenthe desired and current positions.+αsteering system−1XΦPThe relation between the angle of the steering wheel and the positionof the car in the lane is complicated:• turning the steering wheel causes the car to rotate.• forward motion of the car then alters its position in the lane.Check YourselfWhat is the (CT) relation between φ (angle of steeringwheel) and θ (angle of car)?θ1. θ ∝ φ2. θ ∝ sin φ3.˙θ ∝ φ4. θ ∝˙φ5. none of the aboveCheck YourselfWhat is the (CT) relation between θ (angle of car) and p(position of car in lane)?θp1. ˙p ∝ θ2. ˙p ∝ sin θ3. p ∝˙θ4. p ∝d sin θdt5. none of the aboveSteering ControllerCombining the relations for Φ, Θ, and P provides the following model.+αβR1 − RγR1 − R1−X YESPΦ Θcontrollersteerng systemsensorDetermine the system functional.Y = αβR1 − RγR1 − R(X − Y )Solving:YX=αβγR21 − 2R + (1 + αβγ)R2=KR21 − 2R + (1 + K)R2where K ≡ αβγ.Steering ControllerTo find the poles, replace R in the system functional by1zand solvefor the roots of the denominator.YX=KR21 − 2R + (1 + K)R2=Kz2− 2z + (1 + K)Poles are atz = 1 ± j√K .Steering ControllerIf K = 0, there is a double pole at z = 1.z = 1 ± j√KRe zIm zz-planeK = 0 2−1 0 1 2 34nmode 0−1 0 1 2 34nmode 1The output diverges: the system is “unstable.”6.003: Signals and Systems Lecture 5 September 24, 20093Steering ControllerIf K = 1, there are complex poles at z = 1 ± j.z = 1 ± j√K1Re zIm zz-planeK = 1−1 0 1 2 34nRe {mode 1}−1 0 1 2 34nIm {mode 1}The output oscillates: poles off the real axis.The output diverges: magnitude of poles greater than 1.Bigger K → even less stable.Steering ControllerNo value of K = αβγ gives acceptable performance.+αβR1 − RγR1 − R1−X YESPΦ Θcontrollersteerng systemsensorRe zIm z 2Need a better controller.Steering ControllerTry a controller based on first differences.+ α(1−R)βR1 − RγR1 − R1−X YESPΦ Θcontrollersteerng systemsensorY =αβγ(1 − R)R2(1 − R)(1 − R)(X − Y ) =KR21 − R(X − Y ) where K = αβγ.Solving,YX=KR21 − R + KR2Steering ControllerThe functional has the form of the “simple car” (last lecture).YX=KR21 − R + KR2The poles are at z =1 ±√1 − 4K2.Re zIm zz-planeThe system are stable for 0 < K < 1.Steering ControllerCan we really cancel the (1−R) terms in numerator and denominator?+ α(1−R)βR1 − RγR1 − R1−X YESPΦ Θcontrollersteerng systemsensorY =αβγ(1 − R)R2(1 − R)(1 − R)(X − Y ) =αβγR21 − R(X − Y )How could you test whether this is a valid operation?Steering ControllerSimulate the behavior of the system.Start by making a model with just adders, gains, and delays.+ α(1−R)βR1 − RγR1 − R1−X YESPΦ Θcontrollersteerng systemsensor+αβγ+ +R+RR− −X YΘ P6.003: Signals and Systems Lecture 5 September 24, 20094Steering ControllerUse step-by-step analysis.+αβγ+ +R+RR− −X YΘ Pα = 0.5 β = γ = 1nθ[0] = 0.1p[0] = 0.5With time, the value of θ goes to zero, but the value of p does not.Steering ControllerThe fact that the position does not decay to zero suggests that theclosed-loop response has a pole at z = 1.pPole at z = 1 → fundamental mode of 1n, n ≥ 0.Steering ControllerBut this is not consistent with our analysis after cancelling terms.+ α(1−R)βR1 − RγR1 − R1−X YESPΦ Θcontrollersteerng systemsensorY = α(1 − R)βR1 − RγR1 − RE =αβγR21 − R(X − Y )Solving:YX=KR21 − R + KR2where K = αβγSteering ControllerIf K =12, the poles are at z =1±√1−4K2=12± j12.Re zIm zz-planeThese poles correspond to decaying fundamental modes.Steering ControllerNow remove the parts that correspond to cancelled terms (red).+ α(1−R)βR1 − RγR1 − R1−X YESPΦ Θcontrollersteerng systemsensor+αβγ+ +R+RR− −X YΘ PSteering ControllerSimulate again, after removing cancelled parts (simulate blue part).+αβγ+ +R+RR− −X YΘ Pα = 0.5 β = γ = 1nθ[0] = 0.1p[0] = 0.5Now the value of p also goes to zero with time.6.003: Signals and Systems Lecture 5 September 24, 20095Cancelling Factors in the Numerator and DenominatorAnalysis of the block diagrams before and after cancelling factors inthe numerator and denominator gave different results.Why?Consider some simpler systems.Cancelling Factors in the Numerator and DenominatorAccumulate then take difference.+R R−1+X YWYX=1 − R1 − R= 1 ?Think through the steps carefully:W = X + RW(1 − R)W = XY = (1 − R)WY = XOutputs before and after cancelling (1 − R) are the same.Cancelling Factors in the Numerator and DenominatorTake difference then accumulate.R−1+ +RX YWYX=1 − R1 − R= 1 ?Again, think through the steps:W = (1 − R)XY = W + RY(1 − R)Y = W(1 − R)Y = (1 − R)XNot quite the same as Y = X.Cancelling Factors in the Numerator and DenominatorIf system is initially at rest, then Y = X.R−1+ +RX YW(1 − R)X = (1 − R)Yy[0] = x[0] = 0y[1] − y[0] = x[1] − x[0] → y[1] = x[1]y[2] − y[1] = x[2] −
View Full Document
Unlocking...