Lecture 25. Some Simple Control Examples

1. Lecture 25. Some Simple Control Examples in which we enlarge upon the simple intuitive control we’ve seen We generally want a system to be in some equilibrium state If the equilibrium is not stable, then we need a control to stabilize the state

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3. Last time we looked at an electric motor attached to a disk and set it up as a second order system or

4. Suppose we want the angle to be fixed at π/3 The desired state satisfies the differential equations with no input voltage

5. We can write the differential equations for the primed quantities We want the perturbations to go to zero Is the homogeneous solution stable? What are the eigenvalues of A?

6. There is one stable root and one marginally stable root. The homogeneous solution is

7. If the initial value of q’is not equal to its desired value then it will not decay. The problem as posed is satisfied for everything equal to zero but that’s not good enough We need control. If q’ is too big we want to make it smaller and vice versa Let’s look at this in block diagram mode

8. Here’s a block diagram representation of the differential equations open loop - + e q’ w’

9. We can close the loop by feeding the q’ signal back to the input closed loop - - w’ q’ feedback loop

10. So we have new equations governing the closed loop system

11. We’ve gone from an inhomogeneous set of equations to a homogeneous set This what closing the loop does; there’s no more undetermined external input. We want q and w to go to zero, and that will depend on the eigenvalues of the new system

12. This will converge to zero for any positive g Let’s put in some numbers: K = 0.429, R = 2.71, Ix = 0.061261 (10 cm steel disk) We are overdamped for small g and underdamped for large g We can get at the behavior by applying what we know about homogeneous problems

13. The eigenvectors I will select g = 0.2379 (to make some things come out nicely) This leads to s = -0.544 ± 0.544j

14. The homogeneous solution is With the numbers we have

15. At t = 0, we have A little algebra Now we have the complete solution in terms of the initial conditions

16. Let’s plot this and see what happens for q’0= π/3 and w’0= 0

17. Now let’s take a closer look at the cruise control we started with Tuesday evening This is a much harder problem than the one we just worked We designed a simple control to eliminate starting error — cruise control is meant to eliminate an ongoing error, an unknown external force Recall the abstract block diagram

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

19. We have some open loop control — a guess as to the fuel flow, the nominal fuel flow We have some closed loop control — correct the fuel flow if the speed is wrong We have disturbances that will make the speed wrong if we don’t do something about them We need to make a model, and then try to control the model Ultimately we’d want to run the model control on a simulation

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

21. 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

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

23. 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 I use K for gain here This is what we just did for the motor-disk system

24. The closed loop picture The open loop picture + + - f’ v’ 1/m - control feedback Note that this feedback really isn’t doing anything new — we already have negative feedback from the speed caused by air drag

25. But let’s carry on for a moment From the last lecture So we have Solution

26. 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)

27. 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 We won’t do D control today — it doesn’t buy us much for this problem

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

29. Look at the block diagram for this the red part is new f’ - y’ v’ 1/m - “natural” feedback + control feedback +

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

31. The homogeneous solution and we see that it will decay if k2 >0

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

34. Wait a minute here! What’s going on!? Have I pulled a fast one? No, let’s look at it more carefully using some of our mathematical development

35. Method 1 go “backwards” to get a second order ode — a familiar one The particular solution is constant, as we have seen Let’s look at the homogeneous solution

36. Method 1 so the system will be underdamped if and in that case we have

37. Method 1 Initial conditions: y’H+ y’P= 0, v’H+ v’P= 0 from which and its derivative clearly decays, leaving no permanent error in speed

38. Method 2 and we can use the state transition matrix We’ll need numbers to make that work: let k1 = 1 = k2 This satisfies the condition for an underdamped response And let g = 1, which we can do by choosing a time scale

39. Method 2 The eigenvalues and eigenvectors are We see that the system is stable, as we expected

40. Method 2 The state transition matrix is so that the response to any h(t) is given by

41. and the scaled response to a constant hill is

42. Suppose we have a more varied terrain? Let scaled response

43. The control still works very well, and tracks nicely once it is in place. Let’s look at a much more complicated roadway

44. The scaled response looks like the scaled forcing and the velocity error is minuscule — this really works

45. Overlay the two: forcing in black, response in yellow

46. What did we do here? We started with a one dimensional system — and tried to find a force to cancel and exterior force That didn’t work We added a variable to the mix found a new feedback got the velocity to be controlled at the expense of its integral about which we don’t care very much

47. Look at the block diagram for this the red part is new f’ - y’ v’ 1/m - “natural” feedback + control feedback +

48. QUESTIONS?

49. We need to say more about state transition matrices We will discover that the simple feedback games we’ve just played don’t always work We will learn how to control more complicated mechanisms and get the to do more complicated things