Lecture 22: Introduction to state space formulation

1. Lecture 22: Introduction to state space formulation The idea of state space The automobile suspension as an example Simple dc motors and their uses

2. We know that we need two initial conditions to solve a second order ode That means that if we know these two conditions at one moment in time we can use the differential equation to find the evolution of those conditions In the simple 1DOF case the two conditions are the values of y and its derivative We can call that the state of the system and the behavior of the state can be described by the evolution of the state

3. We can use the idea of state space is to transform our second order equations to pairs of first order equations Each Euler-Lagrange equation becomes two first order equations This is visualizable for one degree of freedom problems for which the state and we can plot in what is called the phase plane

4. I have notation conventions An upper case bold Roman letter denotes a matrix, A A lower case bold Roman letter denotes a (column) vector, b Upper and lower case italic denote scalars Let me illustrate the phase plane using the pendulum

5. The state space equations for the pendulum are I can integrate these numerically and plot the phase plane for different starting points If the initial energy is below a critical value the pendulum will swing back and forth (the critical value for a scaled pendulum is √2) Let’s see some examples

6. I’m going to use this as a reference state

10. We can’t draw the motion of the state for systems with more dimensions Are there questions before we look at this in general?

11. We have been working with second order equations These generally take the form

12. In most cases it is relatively easy to convert them to the form (often they come out that way to begin with) Then we can write

13. I have twice as many equations in twice as many unknowns, but the equations are all first order equations, and we’ll see how good that is I can define a state vector and a forcing vector The differential equations in vector form become these are functions ofpand q

14. The state again contains the terms that I would need to solve an initial value problem The state thus describes the system and its evolution

15. Under circumstances for which the system is linear or for which it makes sense to linearize the system we get the following very nice-looking matrix equation This is not the same A that we had before. We get it by inspection, and we’ll learn how

16. View the automobile suspension system, which is linear, from this perspective Two degrees of freedom four initial conditions a four dimensional state

17. From Gillispie, TD Fundamentals of Vehicle Dynamics (1992) We can fit this to our vertical model

18. Rotate our picture to the vertical c2 k2 z2 f2 m2 c3 k3 f1 z1 m1 c1 k1

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

20. Free Body Diagrams m2 m1

21. The differential equations, supposing z positive up We want to know both the (damped) natural frequencies and the response to rough roads This means a homogeneous and a particular solution We can approach this from our new viewpoint

22. Define the derivatives and substitute Choose the obvious state vector

23. We need to do some manipulation (which you ought to be able to reconstruct)

24. This is of the form This is a single-input (SI) system, on which we will focus for the second half of the course

25. What do we do with this? The same ritual as before: homogeneous solution particular solution Start with the homogeneous solution Suppose solutions that depend on exp(st) These are still a set of homogeneous odes with constant coefficients

27. and we need the determinant of this to vanish Expand by minors

28. first minor expanded by minors second minor expanded by minors See this in Mathematica

29. At this point I have reduced the determinant to three second order determinants. I leave it to you to finish the chore. Here’s the result in Mathematica, using the determinant operator and cleaning the result up to look nice.

30. If we put the numbers into the latter form, we get the same result we had before Our old analysis is on the next two slides

31. OLD 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

32. OLD The determinant with the parameters substituted in is expanding

33. We should think about a more methodical method to get at these things. We are now in a set up where it is possible to use the words eigenvalues and eigenvectors in a correct mathematical sense The values of s that make the determinant of A* equal to zero are the eigenvalues of A, and the eigenvectors that go with them are the eigenvectors of A. We can write our homogeneous state vector in terms of these eigenvalues and eigenvectors

34. One big advantage of this is that Mathematica (and MatLab) can find the eigenvalues and eigenvectors for you

35. Note that the eigenvalues are the same as the s values we found using the previous approach “It has four roots, which come in complex conjugate pairs: -39.3025 ± j70.4975, -2.5494 ± j6.9412 rad/sec” (from a previous lecture)

36. (in Mathematica) We can write the homogeneous state vector in terms of eigenvectors and eigenvalues

37. Now let’s make a start on the particular solution in this world Suppose we have harmonic forcing and let’s take a complex notation approach to that (I’m using the cosine for a change of pace)

38. cross out the exp(jwt) and rearrange

39. Let me write this as from which

40. and the particular solution is The homogeneous solution is

41. What happens to the initial conditions here? We don’t need any differentiations, so we have

42. We can play with this by noting that we can rewrite the homogeneous part

43. You all can solve that problem and you know that once you’ve done that you have to take the real part And then you have to apply initial conditions Since everything is first order, you don’t need derivatives the initial condition is simply

44. This is not how we are going to solve these problems later on but I want to keep things simple and familiar for today. I want to take a brief segue into the subject of dc motors We are going to want to drive and control mechanisms and one way to do that is through motors There’s a reference posted at today’s lecture, but I’m going to tell you all you need to know for the class I cover motors in §3.5 of the text

45. voltage in torque and rotation out

47. Simple dc motor equations torque rotation rate current back emf We will use, K1 = K = K2 input voltage - back emf = current x resistance

48. We can do a little physics torque = rate of change of angular momentum plug in the motor torque The maximum torque occurs at zero speed The maximum speed occurs at zero torque starting torque no-load speed

50. We don’t always get the motor constants, but we can find them from the no load speed and the stall torque