Lecture 31. Controllability

1. Lecture 31. Controllability The general idea of control The algebraic controllability theorem SISO systems to be controlled to zero

2. We know that dynamics can be modeled by We want to make some subset or combination of the states to do something we want. We call this the output (You will see some cases in which the output depends directly on the input, but we will by and large not do this, setting D equal to zero)

3. We are generally going to restrict ourselves to linear single input-single output (SISO) systems scalars Denote the dimension of the state vector by N Then A is a N x N matrix b a Nx 1 (column) vector c a I xN (row) vector

4. We want to make y do what we want. This has some implications for what x does For now, let me focus on x and not worry about y — If I can make x do what I want, y is easy Denote the desired behavior of x by xd, the error then I can write what we want

5. Write the differential equations Split the equations The idea is to find ud — we’ll work on this another time open loop control closed loop correction

6. We’re going to start with xd = 0 We will eventually (given “worlds enough and time”) look at tracking I don’t need the subscript, so I’ll stop writing it for now The control problem for now is to find u such that x —> 0 Is it possible to solve the problem?! closed loop correction BTW. If the problem is stable, then we don’t need control.

7. We know how to find xgiven u We simply find the state transition matrix and apply the formulas But that may be easier said than done; there are other things we can do. It also doesn’t tell us how to find u to make x go to zero

8. We saw last time that there are situations where the input cannot reach all the state The sense of this is not always clear If any of the elements of z is not connected to u, that element cannot be controlled This may or may not be an actual problem — more on that anon We need a criterion whereby we can tell if a system is controllable It should be based on A and B, so that we can figure this out before wasting too much time But first

9. What is stability and why is it important?

10. Stability for any problem there will be a homogeneous solution which we can write Suppose s ≠ 0: If Re(si) < 0 for alli the homogeneous solution will decay in time We call this (asymptotically) stable If Re(si) > 0 for anyi the homogeneous solution will grow in time We call this (linearly) unstable If Re(si) = 0 for alli the homogeneous solution will oscillate in time We call this marginally stable

11. If s = 0 we have algebraic eigenfunctions. We’ll be dealing with these as special cases. Stability is important because in real life there will always be some tiny part of the homogeneous solution in the answer If the solution contains an unstable part, it will grow without bound, and the uncontrolled system will “blow up” We need controls to prevent this Let’s suppose that we need or want control and return to the question IS CONTROL POSSIBLE?

12. The algebraic controllability theorem For a general linear system The controllability matrix is The system is controllable if and only if the rank of Q = N BTW: The theorem is true for multi-input systems

13. The matrix Q is square, NxN, for a single input system If there are M inputs then Q has N rows and MN columns An NxN matrix has rank Niff its determinant is nonzero, so the rank criterion works for square and nonsquare matrices Let’s look at a simple example to see how this goes

14. u and we can draw a block diagram of this x3 -3 x2 x1 + u - -2 -1

15. Now let’s carry on and find Q

16. You can find the rank by row reduction or find the determinant Either way, it is controllable

17. So, what happens next? Not only does this give us a controllability test, but it starts us on the road to putting the basic problem in companion form We want to modify A and b, and we do this using another TAT-1 transformation But, what is T? It is NOT the same T that we used to diagonalize

18. We start with Q (and restrict ourselves to SI systems) Take the last row of Q This is the first row of T, and we build the whole T on the next slide

19. and then the z equations follow the same way as before This is not the same z as that we found diagonalizing

20. What about doing this for the case we just looked at? Q-1 =

21. This can be done by hand, but is a little messy and lot tedious, so . . .

22. We can finally apply the differential equation in transformed form Of course, the eigenvalues of A1 are the same as those of A

23. This is not the same z as the z that we used in the diagonalization exercises and we have a nice block diagram for this + z1 u - -6 -11 -6

24. The old block diagram u x3 -3 x2 x1 + u - -2 -1

25. It is interesting note that the transfer function approach would give us the same dynamics I think I will leave that for you to establish. Let’s revisit the diagonalizing transformation for this case

26. What do we need to do? We need the matrix of the eigenvectors, V = T-1 We need its inverse, V-1 = T The new matrix is diagonal with its nonzero elements equal to the eigenvalues of A Summarize on the next slide

27. We can find Tb and draw the block diagram for this system

28. u -3 u -2 u -1

29. So we have three different pictures corresponding to the same dynamics

30. Original problem u x3 -3 x2 x1 + u - -2 -1

31. Companion (phase canonical) form + z1 u - -6 -11 -6

32. u Diagonalized form -3 u -2 u -1

33. Let’s get back to thinking about the companion form dynamics The eigenvalues are -3, -2, and -1 The characteristic polynomial must be

34. QUESTIONS?

35. Consider another three dimensional system, this one unstable, so we are more motivated to find a control Diagonalization The eigenvalues of A are -2, -1 and +1 The matrix of the eigenvectors is

36. Its inverse is T so we have Tb which suggest that this a controllable system

37. We can go through the controllability theorem for this one The determinant is +6, so it is invertible and the system is controllable Now we can move on to find the companion form of A and b

38. and the whole T matrix

39. So that companion form We can find A1 and b1 once we know the eigenvalues Why do we need to go through the ritual of finding T? Because we need to get x back!

40. Now let’s reflect on the control aspects of this problem Suppose we want x to go to zero. It will go to zero if z goes to zero so we need to address that issue. The system is unstable because we have a positive eigenvalue We have the transformed problem in z space

41. The bottom row of A1 contains the coefficients of the characteristic polynomial

42. We can move the eigenvalues by choosing u proportional to z and the z equation becomes

43. Now so our forced (inhomogeneous) differential equation becomes an unforced homogeneous differential equation

44. We know that the coefficients can be directly connected to the characteristic polynomial We have eigenvalues of -2, -1 and +1 The first two are fine — suppose we move the +1 to -1 The characteristic polynomial becomes so the last row coefficients must be: -2, -5, -4

45. So the closed loop A matrix is and the dynamics of the closed loop system are z1 - -4 -5 -2

46. The companion form matrix A1 was so the open loop z system is + z1 u - -2 1 2

47. Those are the gains in z space; we need gains in x space or, to put it another way and, putting in the numbers

48. So this makes the original problem into

49. I will leave it to you to draw the block diagram that goes with this matrix

50. QUESTIONS?