6.003: Signals and Systems Lecture 12 March 16, 201016.003: Signals and SystemsCT Feedback and ControlMarch 16, 2010Feedback and ControlFeedback: simple, elegant, and robust framework for control.+−X YESCcontrollerplantsensorLast time: robotic driving.di= desiredFrontdo= distanceFrontFeedback and ControlThis week: using feedback to enhance performance.Examples:• increasing speed and bandwidth• controlling position instead of speed• reducing sensitivity to parameter variation• reducing distortion• stabilizing unstable systems− magnetic levitation− inverted pendulumOp-ampsAn “ideal” op-amp has many desireable characteristics.V+V−Vo= K (V+− V−)K• high speed• large bandwidth• high input impedance• low output impedance• ...It is difficult to build a circuit with all of these features.Op-AmpThe gain of an op-amp depends on frequency.1021031041051 10 102103104105106ω [log scale]|K(jω)| [log scale]0−π21 10 102103104105106ω [log scale]∠K(jω)|Frequency dependence of LM741 op-amp.Op-AmpLow-gain at high frequencies limits applications.1021031041051 10 102103104105106ω [log scale]|K(jω)| [log scale]audiofrequencies0−π21 10 102103104105106ω [log scale]∠K(jω)|Unacceptable frequency response for an audio amplifier.6.003: Signals and Systems Lecture 12 March 16, 20102Op-AmpAn ideal op-amp has fast time response.ViVoKStep response:1tVi(t) = u(t)AtVo(t) = s(t)Check YourselfDetermine τ for the unit-step response s(t) of an LM741.τAt[seconds]s(t)102103104105|K(jω)| [log scale]0−π21 10 102103104105ω [log scale]∠K(jω)|1. 40 s 2.402πs 3.140s 4.2π40s 5.12π×40s0. none of the aboveOp-AmpWe can use feedback to improve performance of op-amps.ViVoK(s)R1R2βV0V−= βVo=R2R1+ R2VocircuitβK(s)+−ViVoVoVi=K(s)1 + βK(s)6.003 modelDominant PoleOp-amps are designed to have a dominant pole at low frequencies:→ simplifies the application of feedback.s-plane−40α = 40 rad/s =40 rad/s2π rad/cycle≈ 6.4 HzImproving PerformanceUsing feedback to improve performance parameters.ViVoK(s)R1R2βV0V−= βVo=R2R1+ R2VocircuitβK(s)+−ViVoVoVi=K(s)1 + βK(s)=αK0s+α1 + βαK0s+α=αK0s + α + αβK06.003 modelCheck YourselfWhat is the most negative value of the closed-loop polethat can be achieved with feedback?s-plane−α?1. −α(1 + β) 2. −α(1 + βK0)3. −α(1 + K0) 4. −∞5. none of the above6.003: Signals and Systems Lecture 12 March 16, 20103Improving PerformanceFeedback extends frequency response by a factor of 1 + βK0(K0= 2 × 105).1 + βK00.0010.010.111 10 102103104105106ω [log scale]|H(jω)||H(j0)|[log scale]1 + βK00−π21 10 102103104105106ω [log scale]∠H(jω)|Improving PerformanceFeedback produces higher bandwidths by reducing the gain at lowfrequencies. It trades gain for bandwidth.ω [log scale]1 1021041061080.1110102103104105|H(jω)| [log scale]β = 0 (open loop)β = 10−4β = 10−2β = 1Improving PerformanceFeedback makes the time response faster by a factor of 1 + βK0(K0= 2 × 105).Step responses(t) =K01 + βK0(1 − e−α(1+βK0)t)u(t)t [seconds]s(t)K01 + βK01/40β = 0β = 1Improving PerformanceFeedback produces faster responses by reducing the final value ofthe step response. It trades gain for speed.Step responses(t) =K01 + βK0(1 − e−α(1+βK0)t)u(t)t [seconds]s(t)2 × 1051/40β00.5 × 10−51.5 × 10−5The maximum rate of voltage changeds(t)dtt=0+is not increased.Improving PerformanceFeedback improves performance parameters of op-amp circuits.• can extend frequency response• can increase speedPerformance enhancements are achieved through a reduction of gain.Motor ControllerWe wish to build a robot arm (actually its elbow). The input shouldbe voltage v(t), and the output should be the elbow angle θ(t).roboticarmv(t) θ(t) ∝ v(t)We wish to build the robot arm with a DC motor.DC motorv(t) θ(t)This problem is similar to the head-turning servo in 6.01 !6.003: Signals and Systems Lecture 12 March 16, 20104Check YourselfWhat is the relation between v(t) and θ(t) for a DC motor?DC motorv(t) θ(t)1. θ(t) ∝ v(t)2. cos θ(t) ∝ v(t)3. θ(t) ∝ ˙v(t)4. cos θ(t) ∝ ˙v(t)5. none of the aboveMotor ControllerUse proportional feedback to control the angle of the motor’s shaft.+αγAβv(t) θ(t)DC motoramplifierfeedback(potentiometer)−ΘV=αγA1 + αβγA=αγ1s1 + αβγ1s=αγs + αβγMotor ControllerThe closed loop system has a single pole at s = −αβγ.ΘV=αγs + αβγσωs-planeAs α increases, the closed-loop pole becomes increasingly negative.Motor ControllerFind the impulse and step response.The system function isΘV=αγs + αβγ.The impulse response ish(t) = αγe−αβγtu(t)and the step response is therefores(t) =1β1 − e−αβγtu(t) .θ(t)t1βα ↑The response is faster for larger values of α.Try it: Demo.Motor ControllerThe speed of a DC motor does not change instantly if the voltageis stepped. There is lag due to rotational inertia.γAVΘ1First-order modelintegratorγApA1 + pAVΘ2Second-order modelintegrator with lagStep response of the models:v(t)1ttθ(t)θ1(t) = γtu(t)θ2(t) =γt −γp(1 − e−pt)u(t)Motor ControllerAnalyze second-order model.+αγpA21 + pAβv(t) θ(t)DC motoramplifierfeedback(potentiometer)−ΘV=αγpA21+pA1 +αβγpA21+pA=αγpA21 + pA + αβγpA2=αγps2+ ps + αβγps = −p2±rp22− αβγp6.003: Signals and Systems Lecture 12 March 16, 20105Motor ControllerFor second-order model, increasing α causes the poles at 0 and −pto approach each other, collide at s = −p/2, then split into two poleswith imaginary parts.σωs-plane−pIncreasing the gain α does not increase speed of convergence.Motor ControllerStep response.ts(t)1βMotor ControllerStep response.ts(t)1βMotor ControllerStep response.ts(t)1βcos(ωdt + φ)e−pt/2Feedback and Control: SummaryCT feedback is useful for many reasons. Today we saw two:• increasing speed and bandwidth• controlling position instead of speedNext time we will look at several others:• reduce sensitivity to parameter variation• reduce distortion• stabilize unstable systems− magnetic levitation− inverted
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