Feedback and feedforward control system ppt

comparison between feedback and feedforward control system ppt and feedback and feedforward control ppt
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Prof.EvanBaros,United Kingdom,Teacher
Published Date:26-07-2017
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Feedback and Control Feedback: simple, elegant, and robust framework for control. E C + X Y − controller plant S sensor Last time: robotic driving. d = desiredFront i d = distanceFront o 3Feedback and Control 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 pendulum 4Op-amps An “ideal” op-amp has many desireable characteristics. V + V =K (V −V ) K o + − V − • high speed • large bandwidth • high input impedance • low output impedance • ... It is difficult to build a circuit with all of these features. 5Op-Amp The gain of an op-amp depends on frequency. 5 10 4 10 3 10 2 10 ω log scale 2 3 4 5 6 1 10 10 10 10 10 10 0 π − 2 ω log scale 2 3 4 5 6 1 10 10 10 10 10 10 Frequency dependence of LM741 op-amp. 6 \K(jω) K(jω) log scaleOp-Amp Low-gain at high frequencies limits applications. audio 5 10 frequencies 4 10 3 10 2 10 ω log scale 2 3 4 5 6 1 10 10 10 10 10 10 0 π − 2 ω log scale 2 3 4 5 6 1 10 10 10 10 10 10 Unacceptable frequency response for an audio amplifier. 7 \K(jω) K(jω) log scaleOp-Amp An ideal op-amp has fast time response. V i V K o Step response: V (t) =u(t) V (t) =s(t) o i 1 A t t 8Check Yourself Determine τ for the unit-step response s(t) of an LM741. 5 10 4 10 s(t) 3 A 10 2 t 10 τ seconds 0 π − 2 2 3 4 5 1 10 10 10 10 10 ω log scale 40 1 2π 1 1. 40 s 2. s 3. s 4. s 5. s 2π 40 40 2π×40 0. none of the above 9 \K(jω) K(jω) log scaleCheck Yourself Determine the step response of an LM741. System function: αK 0 K(s) = s + α Impulse response: −αt h(t) = αK e u(t) 0 Step response: t t t −ατ αK e 0 −ατ −αt s(t) = h(τ )dτ = αK e dτ = = K (1 − e )u(t) 0 0 −α −∞ 0 0 Parameters: 5 A = K = 2 × 10 0 1 1 τ = = s α 40 10Check Yourself Determine τ for the unit-step response s(t) of an LM741. 5 10 4 10 s(t) 3 A 10 2 t 10 τ seconds 0 π − 2 2 3 4 5 1 10 10 10 10 10 ω log scale 40 1 2π 1 1. 40 s 2. s 3. s 4. s 5. s 2π 40 40 2π×40 0. none of the above 11 \K(jω) K(jω) log scaleOp-Amp Performance parameters for real op-amps fall short of the ideal. Frequency Response: high gain but only at low frequencies. 5 10 audio frequencies 4 10 3 10 2 10 ω log scale 2 3 4 5 6 1 10 10 10 10 10 10 Step Response: slow by electronic standards. 5 2× 10 t seconds 1/40 12 K(jω) log scaleOp-Amp We can use feedback to improve performance of op-amps. circuit 6.003 model V i K(s) V V + K(s) V o o i − R 1 βV β 0 R 2 V K(s) o = V 1 +βK(s) i   R 2 V =βV = V − o o R +R 1 2 13Dominant Pole Op-amps are designed to have a dominant pole at low frequencies: → simplifies the application of feedback. s-plane −40 40 rad/s α = 40 rad/s = ≈ 6.4 Hz 2π rad/cycle 14Improving Performance Using feedback to improve performance parameters. circuit 6.003 model V i K(s) V o V + K(s) V i o − R 1 βV 0 β R 2 V K(s) o = V 1 +βK(s)   i R 2 αK 0 V =βV = V − o o s+α R +R 1 2 = αK 0 1 +β s+α αK 0 = s +α +αβK 0 15Check Yourself What is the most negative value of the closed-loop pole that can be achieved with feedback? s-plane −α ? 1. −α(1 + β) 2. −α(1 + βK ) 0 3. −α(1 + K ) 4. −∞ 0 5. none of the above 16Improving Performance Using feedback to improve performance parameters. circuit 6.003 model V i K(s) V o V + K(s) V i o − R 1 βV 0 β R 2 V K(s) o = V 1 +βK(s)   i R 2 αK 0 V =βV = V − o o s+α R +R 1 2 = αK 0 1 +β s+α αK 0 = s +α +αβK 0 17Check Yourself What is the most negative value of the closed-loop pole that can be achieved with feedback? αK 0 Open loop system function: s + α → pole: s = −α. αK 0 Closed-loop system function: s + α + αβK 0 → pole: s = −α(1 + βK ). 0 The feedback constant is 0 ≤ β ≤ 1. → most negative value of the closed-loop pole is s = −α(1 + K ). 0 18Check Yourself What is the most negative value of the closed-loop pole that can be achieved with feedback? 3 s-plane −α −α(1 +K ) 0 1. −α(1 + β) 2. −α(1 + βK ) 0 3. −α(1 + K ) 4. −∞ 0 5. none of the above 19Improving Performance Feedback extends frequency response by a factor of 1 + βK 0 5 (K = 2 × 10 ). 0 1 0.1 1 +βK 0 0.01 0.001 ω log scale 2 3 4 5 6 1 10 10 10 10 10 10 0 1 +βK 0 π − 2 ω log scale 2 3 4 5 6 1 10 10 10 10 10 10 20 H(jω) \H(jω) log scale H(j0)Improving Performance Feedback produces higher bandwidths by reducing the gain at low frequencies. It trades gain for bandwidth. β = 0 (open loop) 5 10 −4 β = 10 4 10 3 10 −2 β = 10 2 10 10 β = 1 1 0.1 ω log scale 2 4 6 8 1 10 10 10 10 21 H(jω) log scaleImproving Performance Feedback makes the time response faster by a factor of 1 + βK 0 5 (K = 2 × 10 ). 0 Step response K 0 −α(1+βK )t 0 s(t) = (1 − e )u(t) 1 + βK 0 s(t) K 0 1 +βK 0 β = 1 β = 0 t seconds 1/40 22

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