6 003 Signals and Systems Lecture 5 September 24 2009 6 003 Signals and Systems Last Time Feedback and Control Understanding the structure of a control problem automatic control feedback Analyzing feedback systems feedback cyclic paths persistent outputs Designing control systems constructing well behaved response properties September 24 2009 Example Steering a Car Last Time Algorithm steer left when car is right of center and vice versa We investigated a VERY simple model for the car The car could move laterally in a lane without rotating straight ahead steer right steer right see previous 0 command 0 steer right see previous 1 command 1 straight ahead see previous 1 command 1 steer left see previous 0 command 0 steer left see previous 1 command 1 Bad algorithm poor performance Here we get persistent oscillations Even that simple system could go unstable Last time we learned how to stablize it Today Steering Controller We will investigate more realistic models of steering Design a system to automatically steer a car We will analyze more realistic and more complex feedback systems moving car velocity V p Assume a sensor reports position p within the lane p 0 in center p 0 right of center p 0 left of center 1 6 003 Signals and Systems Lecture 5 Steering Controller September 24 2009 Check Yourself Model the system What is the CT relation between angle of steering wheel and angle of car Let X represent the desired position in the lane normally 0 Turn the steering wheel in proportion to the difference between the desired and current positions X steering system P 1 1 2 3 4 5 The relation between the angle of the steering wheel and the position of 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 Yourself sin none of the above Steering Controller Combining the relations for and P provides the following model What is the CT relation between angle of car and p position of car in lane X E controller S P Y 1 sensor Determine the system functional p 1 2 3 4 5 R R 1 R 1 R steerng system p p sin p p d sin dt none of the above Y R R X Y 1 R 1 R Solving KR2 Y R2 X 1 2R 1 R2 1 2R 1 K R2 where K Steering Controller Steering Controller To find the poles replace R in the system functional by z1 and solve for the roots of the denominator If K 0 there is a double pole at z 1 z 1 j K K Y KR2 2 X 1 2R 1 K R2 z 2z 1 K K 0 Poles are at z 1 j K Im z mode 0 z plane 1 0 1 2 3 4 2 Re z mode 1 1 0 1 2 3 4 The output diverges the system is unstable 2 n n 6 003 Signals and Systems Lecture 5 September 24 2009 Steering Controller Steering Controller If K 1 there are complex poles at z 1 j z 1 j K No value of K gives acceptable performance Im z K 1 X Re mode 1 z plane 1 0 1 2 3 4 E controller n S P Y 1 sensor Re z 1 R R 1 R 1 R steerng system Im z Im mode 1 1 0 1 2 3 4 n 2 The output oscillates poles off the real axis The output diverges magnitude of poles greater than 1 Bigger K even less stable Re z Need a better controller Steering Controller Steering Controller Try a controller based on first differences The functional has the form of the simple car last lecture X E 1 R R R 1 R 1 R steerng system controller P Y KR2 X 1 R KR2 Y The poles are at z 1 1 4K 2 Im z S z plane 1 sensor Y 1 R R2 1 R 1 R KR2 X Y 1 R X Y Re z where K Solving Y KR2 X 1 R KR2 The system are stable for 0 K 1 Steering Controller Steering Controller Can we really cancel the 1 R terms in numerator and denominator Simulate the behavior of the system Start by making a model with just adders gains and delays X E 1 R R R 1 R 1 R steerng system controller S P Y X E controller 1 S sensor Y 1 R R2 1 R 1 R R R 1 R 1 R steerng system 1 R P Y 1 sensor X Y R2 X Y 1 R X R How could you test whether this is a valid operation 3 R R P Y 6 003 Signals and Systems Lecture 5 September 24 2009 Steering Controller Steering Controller Use step by step analysis The fact that the position does not decay to zero suggests that the closed loop response has a pole at z 1 X R R P Y R p 0 0 5 0 5 1 0 0 1 p n Pole at z 1 fundamental mode of 1n n 0 With time the value of goes to zero but the value of p does not Steering Controller Steering Controller But this is not consistent with our analysis after cancelling terms If K 12 the poles are at z 1 1 4K 12 j 12 2 X E R R 1 R 1 R steerng system 1 R controller S P Im z z plane Y Re z 1 sensor Y 1 R R R 1 R1 R E R2 X Y 1 R These poles correspond to decaying fundamental modes Solving Y KR2 X 1 R KR2 where K Steering Controller Steering Controller Now remove the parts that correspond to cancelled terms red Simulate again after removing cancelled parts simulate blue part X E R R 1 R 1 R steerng system 1 R controller S P X Y R R R p 0 0 5 1 0 5 1 sensor X R R P 0 0 1 Y R Now the value of p also goes to zero with time 4 n P Y 6 003 Signals and Systems Lecture 5 September 24 2009 Cancelling Factors in the Numerator and Denominator Cancelling Factors in the Numerator and Denominator Analysis of the block diagrams before and after cancelling factors in the numerator and denominator gave different results Accumulate then take difference X W Why R R Y 1 1 R Y 1 X 1 R Consider some simpler systems Think through the steps carefully W X RW 1 R W X Y 1 R W Y X Outputs before and after cancelling 1 R are the same Cancelling Factors in the Numerator and Denominator Cancelling Factors in the Numerator and Denominator Take difference then accumulate If system is initially at rest then Y X X R W Y 1 X R 1 R 1 R Y 1 X 1 R Again think through the steps W Y …
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