1 / 43

Lecture 25 Assorted Control Preparation

Lecture 25 Assorted Control Preparation. Block diagrams. Cruise control. Stability (the simple inverted pendulum). BLOCK DIAGRAMS. Friedland uses a lot of these. Sometimes they are helpful sometimes they are not. He gives symbols for multiply, integrate and add.

nita
Télécharger la présentation

Lecture 25 Assorted Control Preparation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 25 Assorted Control Preparation Block diagrams Cruise control Stability (the simple inverted pendulum)

  2. BLOCK DIAGRAMS Friedland uses a lot of these. Sometimes they are helpful sometimes they are not. He gives symbols for multiply, integrate and add We can see how this works by looking at the usual first order system block diagram

  3. second order ode representation state space representation or simple block diagram representation - + a v y -

  4. integrate v to get y “feedback” loops integrate the acceleration to get v - + a u y -

  5. Here are some examples from Friedland

  6. translated Friedland’s equations two inputs u two outputs x

  7. All the lines in this figure represent vectors

  8. The equations for that diagram are dynamics output We won’t be concerned much, if at all, with vector inputs We will not look at systems for which the output depends directly on the input — for us D = 0

  9. Inverted pendulum on a cart which problem we will work much later m q l u M y

  10. Energies Constraints The Lagrangian Euler-Lagrange equations

  11. Linearize Solve for the second derivatives Convert to state space

  12. y output B input q output internal “feedback”

  13. CRUISE CONTROL desired speed INVERSE PLANT GOAL: SPEED nominal fuel flow Actual speed + + PLANT: DRIVE TRAIN Input: fuel flow - - error disturbance Feedback: fuel flow CONTROL

  14. disturbance simple first order model divide the force open loop part linearize and the goal is to make v’ go to zero

  15. Let’s say a little about possible disturbances hills are probably the easiest to deal with analytically mgsinf f I’ll say more as we go on

  16. The open loop picture + + f v 1/m -

  17. I’m not in a position to simply ask f to cancel h(t) (because I don’t know what it is!) I want some feedback mechanism to give me more fuel when I am going too slow and less fuel when I am going too fast This system is already stable in the sense that the air drag acts in the same way as the artificial gain I don’t really need the primes anymore, so I will drop them

  18. The closed loop picture The open loop picture + + - f v 1/m -

  19. From the last lecture So we have Solution

  20. We do not need this whole apparatus to get a sense of how this works Consider a hill, for which s(t) is constant, call it s0 We can find the particular solution by inspection The homogeneous solution decays, and we see that we have a permanent error in the speed The bigger K, the smaller the error, but we can’t make it go away (and K will be limited by physical considerations in any case)

  21. What we’ve done so far is called proportional (P) control We can fix this problem by adding integral (I) control. There is also derivative (D) control PID control incorporates all three types, and you’ll hear the term

  22. Add a variable and its ode Let the force depend on both variables Then

  23. Convert to state space We remember that x denotes the error so the initial condition for this problem is y = 0 = v

  24. The homogeneous solution and we see that it will decay

  25. What happens now when we go up a hill? We can now let the displacement take care of the particular solution

  26. Let’s look at the behavior of first and second order cruise controls in Mathematica

  27. Simple inverted pendulum

  28. Inverted pendulum on a cart which problem we will work much later for now, fix the cart: y = 0 = u m q Add a torque l M

  29. Energies Constraints The Lagrangian Euler-Lagrange equation

  30. How much of this would you like to see?

  31. Linearize Convert to state space

  32. Look at the homogeneous solution The system is unstable If there is no torque, the pendulum will fall down

  33. Suppose the torque depends on the angle We are introducing feedback now where it is essential

  34. I have converted the open loop problem to a closed loop problem and said problem is a new homogeneous problem and we want the solution to decay to zero to hold the pendulum up

  35. Inverted pendulum: position only feedback t - w q +

  36. If the eigenvalues are pure imaginary The inverted pendulum will oscillate

  37. But we would like to hold it erect

  38. In order to keep the square root part less than the outside piece so that the real part of both values of s remain negative We need both feedback elements to make hold the pendulum up

  39. Inverted pendulum: full state feedback t - q w + -

  40. WE’RE DONE

More Related