6.003: Signals and SystemsCT Feedback and ControlOctober 27, 2009Mid-term Examination #2Tomorrow, October 28, 7:30-9:30pm, Walker Memorial.No recitations tomorrow.Coverage: cumulative with more emphasis on recent materiallectures 1–12homeworks 1–7Homework 7 includes practice problems for mid-term 2.It will not collected or graded. Solutions are posted.Closed book: 2 pages of notes (812× 11 inches; front and back).Designed as 1-hour exam; two hours to complete.Feedback and ControlFeedback is pervasive in natural and artificial systems.We have previously investigated feedback in DT systems.Example: robotic steering+plantcontrollersensorx[n] y[n]−ny[n]Feedback and ControlToday: investigate continuous-time feedback and control.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 pendulumOp-AmpMuch of the utility of op-amps depends on their having high gain.Ideal op-amp: lim K → ∞V+V−Vo= K (V+− V−)KOp-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.Check YourselfWhat system function has this frequency response?102103104105|K(jω)| [log scale]0−π21 10 102103104105106ω [log scale]∠K(jω)|1.αK0s + α2.K0ss + α3.K0(s + α)s4.K0(s + α)α5. noneCheck YourselfWhat system function has this frequency response? 1102103104105|K(jω)| [log scale]0−π21 10 102103104105106ω [log scale]∠K(jω)|1.αK0s + α2.K0ss + α3.K0(s + α)s4.K0(s + α)α5. noneOp-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.Op-AmpAn ideal op-amp has fast time response.ViVoKStep response:1tVi(t) = u(t)AtVo(t) = s(t)Check YourselfDetermine the step response of an LM741.τAt [seconds]s(t)1. A = 2 × 105; τ = 40 s2. A = 2 × 105; τ =140s3. A = 80 × 105; τ = 40 s4. A = 80 × 105; τ =140s5. none of the aboveCheck YourselfDetermine the step response of an LM741.System function:K(s) =αK0s + αImpulse response:h(t) = αK0e−αtu(t)Step response:s(t) =Zt−∞h(τ)dτ =Zt0αK0e−ατdτ =αK0e−ατ−αt0= K0(1 − e−αt)u(t)Parameters:A = K0= 2 × 105τ =1α=140sCheck YourselfDetermine the step response of an LM741.t [seconds]s(t)1/402 × 1051. A = 2 × 105; τ = 40 s2. A = 2 × 105; τ =140s3. A = 80 × 105; τ = 40 s4. A = 80 × 105; τ =140s5. none of the aboveOp-AmpPerformance parameters for real op-amps fall short of the ideal.Frequency Response: high gain but only at low frequencies.1021031041051 10 102103104105106ω [log scale]|K(jω)| [log scale]audiofrequenciesStep Response: slow by electronic standards.t [seconds]1/402 × 105Op-AmpThe op-amp’s poor performance parameters result from a dominantpole at a relatively low frequency.s-plane−40α = 40 rad/s =40 rad/s2π rad/cycle≈ 6.4 HzOp-amps are designed to have a dominant pole:→ easy to customize their behavior using feedback !Improving 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 + αβK0LTI 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 aboveImproving 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 + αβK0LTI modelCheck YourselfWhat is the most negative value of the closed-loop pole that can beachieved with feedback?Open loop system function:αK0s + α→ pole: s = −α.Closed-loop system function:αK0s + α + αβK0→ pole: s = −α(1 + βK0).The feedback constant is 0 ≤ β ≤ 1.→ most negative value of the closed-loop pole is s = −α(1 + K0).Check YourselfWhat is the most negative value of the closed-loop polethat can be achieved with feedback? 3s-plane−α−α(1 + K0)1. −α(1 + β) 2. −α(1 + βK0)3. −α(1 + K0) 4. −∞5. none of the aboveImproving PerformanceFeedback extends frequency response by a factor of 1 + βK0(K0= 2 × 105).1 + βK00.0010.010.111 10 102103104105106ω [log scale]|K(jω)||K(j0)|[log scale]1 + βK00−π21 10 102103104105106ω [log scale]∠K(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 !Check 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 aboveCheck YourselfWhat is the relation between v(t) and θ(t) for a DC motor?To first order, the rotational speed˙θ(t) of a DC motor is proportionalto the input voltage v(t).DC motorv(t) θ(t)v(t)tθ(t)tFirst-order model: integratorγAVΘCheck 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.
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