6 003 Signals and Systems Lecture 12 6 003 Signals and Systems March 16 2010 Feedback and Control Feedback simple elegant and robust framework for control CT Feedback and Control X E C Y controller plant S sensor Last time robotic driving di desiredFront do distanceFront March 16 2010 Feedback and Control Op amps This week using feedback to enhance performance An ideal op amp has many desireable characteristics Examples increasing speed and bandwidth controlling position instead of speed reducing sensitivity to parameter variation reducing distortion stabilizing unstable systems magnetic levitation inverted pendulum V Vo K V V K V high speed large bandwidth high input impedance low output impedance It is difficult to build a circuit with all of these features Op Amp The gain of an op amp depends on frequency Low gain at high frequencies limits applications K j log scale 105 104 103 102 log scale 1 10 102 103 104 105 106 K j K j K j log scale Op Amp 0 2 log scale 1 10 102 103 104 105 106 Frequency dependence of LM741 op amp audio frequencies 105 104 103 102 log scale 1 10 102 103 104 105 106 1 10 102 103 104 105 106 0 2 log scale Unacceptable frequency response for an audio amplifier 1 6 003 Signals and Systems Lecture 12 Op Amp March 16 2010 Check Yourself An ideal op amp has fast time response Determine for the unit step response s t of an LM741 Vi K j K j log scale Vo K s t Step response A Vo t s t Vi t u t 1 t seconds A 105 104 103 102 0 2 t t 1 40 s 2 2 1 40 s 1 s 3 40 10 102 103 log scale 4 2 40 s 104 105 1 5 2 40 s 0 none of the above Op Amp Dominant Pole We can use feedback to improve performance of op amps Op amps are designed to have a dominant pole at low frequencies simplifies the application of feedback circuit Vi 6 003 model s plane K s Vo Vi K s Vo R1 V0 40 R2 Vo K s Vi 1 K s V Vo R2 R1 R2 40 rad s Vo Improving Performance 40 rad s 6 4 Hz 2 rad cycle Check Yourself Using feedback to improve performance parameters circuit Vi What is the most negative value of the closed loop pole that can be achieved with feedback 6 003 model K s Vo R1 Vi K s V0 Vo K s Vi 1 K s Vo s plane R2 V Vo R2 R1 R2 K0 s 0 1 K s K0 s K0 Vo 1 1 2 1 K0 3 1 K0 4 5 none of the above 2 6 003 Signals and Systems Lecture 12 March 16 2010 Improving Performance Feedback extends frequency response by a factor of 1 K0 K0 2 105 Feedback produces higher bandwidths by reducing the gain at low frequencies It trades gain for bandwidth 0 open loop 1 105 0 1 H j log scale 1 K0 0 01 0 001 log scale 1 H j H j log scale H j0 Improving Performance 10 102 103 0 104 105 106 1 K0 log scale 10 102 103 104 105 103 102 10 4 10 2 10 1 2 1 104 1 0 1 106 log scale 102 1 104 106 108 Improving Performance Improving Performance Feedback makes the time response faster by a factor of 1 K0 K0 2 105 Feedback produces faster responses by reducing the final value of the step response It trades gain for speed Step response Step response s t K0 1 e 1 K0 t u t 1 K0 K0 1 K0 s t K0 1 e 1 K0 t u t 1 K0 s t s t 0 2 105 1 0 5 10 5 1 5 10 5 t seconds 0 t seconds 1 40 1 40 The maximum rate of voltage change ds t is not increased dt t 0 Improving Performance Motor Controller Feedback improves performance parameters of op amp circuits We wish to build a robot arm actually its elbow The input should be voltage v t and the output should be the elbow angle t can extend frequency response can increase speed Performance enhancements are achieved through a reduction of gain v t robotic arm t v t We wish to build the robot arm with a DC motor v t DC motor t This problem is similar to the head turning servo in 6 01 3 6 003 Signals and Systems Lecture 12 Check Yourself March 16 2010 Motor Controller Use proportional feedback to control the angle of the motor s shaft What is the relation between v t and t for a DC motor v t DC motor v t 1 2 3 4 5 amplifier DC motor A t feedback potentiometer t v t cos t v t t v t cos t v t none of the above 1s A V 1 A s 1 1s Motor Controller Motor Controller The closed loop system has a single pole at s Find the impulse and step response The system function is V s The impulse response is V s t s plane h t e t u t and the step response is therefore 1 s t 1 e t u t t 1 As increases the closed loop pole becomes increasingly negative t The response is faster for larger values of Try it Demo Motor Controller Motor Controller The speed of a DC motor does not change instantly if the voltage is stepped There is lag due to rotational inertia Analyze second order model First order model integrator V A 1 amplifier Second order model integrator with lag pA A 2 V 1 pA v t v t t feedback potentiometer t t Step response of the models 1 DC motor pA2 1 pA pA2 pA2 p 1 pA 2 pA2 V 1 pA pA2 s ps p 1 1 pA r p p 2 s p 2 2 1 t tu t 2 t t p 1 e pt u t t 4 6 003 Signals and Systems Lecture 12 March 16 2010 Motor Controller Motor Controller For second order model increasing causes the poles at 0 and p to approach each other collide at s p 2 then split into two poles with imaginary parts Step response s t 1 s plane t p Increasing the gain does not increase speed of convergence Motor Controller Motor Controller Step response Step response s t s t 1 1 e pt 2 cos d t t Feedback and Control Summary CT feedback is useful for many reasons Today we saw two increasing speed and bandwidth controlling position instead of speed Next time we will look at several others reduce sensitivity to parameter variation reduce distortion stabilize unstable systems magnetic levitation inverted pendulum 5 t
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