Lecture 28. Putting what we have together

1. Lecture 28. Putting what we have together State space Eigenvalues and eigenvectors State transition matrices Transfer functions

2. We are going to look at how these things connect starting with the automobile suspension moving on to some degenerate problems Rederive the equations of motion and convert to state space

3. Gillespie Us z2 m2 k3 c3 z1 m1 k1 zG

4. Free Body Diagrams m2 m1 Be careful with the arrows and the signs I suppose z to be positive up

5. Collect the forces on each mass and write Newton’s law which we can rearrange into differential equations with left and right hand sides

6. The differential equations, supposing z positive up Define a physically reasonable state

7. and with zG as the input

8. It would be nice to make a block diagram of this, but it’s pretty messy The two tracks are somewhat different, and there’s a lot of cross feedback x3 x1 u = zG x4 x2 All four variables contribute to both inputs. I can draw this symbolically.

9. red denotes the upper pair x3 x1 u = zG x4 x2 blue denotes the lower pair

10. Some numbers from Gillispie, TD Fundamentals of Vehicle Dynamics (1992) 5.0 L Mustang front suspension (year unspecified) Tire spring constant: 1198 lb/in Spring spring constant: 143 lb/in Weight: 957 lb “The 40% damping ratio is reasonably representative of most cars” It’s not clear to me exactly what this represents, but I’m guessing it refers to a damping ratio with respect to the body mass-spring spring system

11. If we are to do this correctly, we need to guess the effective weight of the wheel-axle-tire segment: 75 lb, say Standard US customary gravity acceleration is 386 in/s2; we take mass to be weight in pounds divided by 386 We take c = 2zwm ≈ 2z√(km), where we use k from the spring spring and m of the vehicle. In our case this will be 2 x 0.4 x √(143 x 2.479) = 15.06 lb-sec/in

12. Put these numbers into our picture z2 m2 = 2.479 lb-sec2/in m2 k3 c3 k3 = 143 lb/in c3 = 15.06 lb-sec/in z1 m1 = 0.194 lb-sec2/in m1 k1 = 1198 lb/in k1 zG

13. Let me define an output. I’ll choose body position: z2 = x2 Whether we are going to use a transfer function approach or the state transition matrix we will need the resolvent — it’s the Laplace transform of the state transition matrix and it can be used to find the transfer function This is all very messy, and so I am going to use the numbers rather than doing it all symbolically

14. The inverse of this is pretty awful, even with the numbers I can display the columns one at a time

17. Now that I have b, C and the resolvent, I can form the transfer function The numerator is not constant, so we will want to use the method we discussed last time

18. This defines a state, but it is not the state we started with!

19. The new state block diagram — much simpler than the old one + u - 83.7 6970 37515 356217

20. and we can add the output y 37515 356217 + u - 83.7 6970 37515 356217

21. red denotes the upper pair x3 x1 u = zG y x4 x2 blue denotes the lower pair

22. Let’s compare the A matrices for the two systems The original matrix The new matrix

23. Let’s talk about the new matrix It has the same eigenvalues as the old matrix Its last row corresponds to the denominator of the transfer function The matrix is in phase canonical form, or what Friedland calls companion form

24. We’ll talk a lot more about these forms next week We’ll also be able to answer the questions that are in the air: How does the new state relate to the old state? What is the physical interpretation of the new state? For now, let’s also note that the eigenvalues of either form of A are also the roots of the denominator of the transfer function This was a nice problem. The new state had the same dimensions as the old state Let’s look at a case where this fails.

25. QUESTIONS?

26. Let’s work in the context of a single problem: the inverted pendulum on a cart m q Now it won’t hurt to start from square one Derive the equations of motion l u M y

27. Energies Constraints The Lagrangian Euler-Lagrange equations

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

29. Friedland’s picture

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

31. Now I want to put in the motor that drives the force force on the cart the motor torque wheel radius

32. This will modify the matrix A and make the input u the voltage e so that we have a modified A and b

33. Put in numbers: K = 49/20, R = 4, r = ¼, M = 50, m = 1, l = 1, g = 98/10

34. The eigenvalues of A: -3.1672, -0.4706, 0, 3.2576 The system is unstable, but we know that from intuition As before, we want the resolvent, with a view to exploring both the state transition matrix and the transfer function (once we’ve picked an output) And, of course, the details are going to have to take place in Mathematica

35. Even with the numbers, the inverse of this requires considerable simplification and we wind up with a resolvent best displayed by column

38. Almost none of the entries have fourth order denominators! Let’s the take pendulum angle as our output So we have a third order state instead of the fourth order state we started with

39. The new state block diagram — much simpler than the old one + u - 0.48 -10 -4.706

40. and we can add the output y 0.049 + u - 0.48 -10 -4.706

41. What does all this mean? Can we figure out what is going on? Does this make any sort of sense? The A matrix for the new reduced state is The eigenvalues of this matrix are the same as the nonzero eigenvalues of the original matrix

42. I’ll defer thinking about the physical meaning of the new state I will say that the reduction in order is associated with the zero eigenvalue Recall that the eigenvector associated with the original zero eigenvalue had only a z1 (= y, the position) component The physics doesn’t care about the position of the cart The text has an example of order reduction — 5A and I’d like to take a look at how that goes — what happens It’s not a real physical system, but nonetheless instructive

43. The eigenvalues of A are -4, -3, -2, -1

44. The eigenvectors are They are independent and span the space. We could use them to find the state transition matrix if we wanted but I don’t right now Let me continue with the resolvent

46. I can find the state transition matrix from the resolvent as usual One can see all four eigenvalues appear as we expect. This look like a perfectly well-behaved problem, but wait . . .

47. The first thing we note is that the resolvent has no fourth order denominators (after simplification — there’re plenty of them before) We can write the transfer function + u - 1

48. This is very much a cobbled together system to make the reduction dramatic But even the product is severely reduced in order

49. How can we reconcile this weird behavior with the perfectly well-behaved state transition matrix?? The answer lies in B Only two of the eigenvalues show up here

50. QUESTIONS? We’ll make more sense of this on Tuesday, when we push forward a bit