Lecture 13: Stability and Control I We are learning how to analyze mechanisms but what we’d like to do is make them do our bidding We want to be able to control them — to make a robot trace a particular path, say A full study of this is well beyond the scope of this course we’ll just do enough to give you the flavor of the thing and to let you do some things 1

2 You might think that if we knew the forces/torques to hold a robot somewhere that we could just set them to those values and we’d be home free Not true, and the key lies in stability We want to have a system that is close to where it should be when the forces are close to where they should be Eventually we can design transient forces that will control our systems

3 Restrict the discussion (at least for now) to holonomic systems Industrial robots are holonomic I will use Hamilton’s equations in the form The system is holonomic, so there is no Lagrange multiplier term

4 Equilibrium There are equilibria for which the system moves (and we’ll look at those) but all of our equilibria have constant generalized (conjugate) momenta In many cases the equilibrium conjugate momentum is equal to zero This discussion of stability is limited to those cases, making the left hand side of the first equation equal to zero

5 It is generally possible to solve the equilibrium equations for the equilibrium (generalized) forces I’ll do this eventually using a robot example Let’s suppose we’ve done that The big question is: What happens if we don’t get the forces exactly right? Will the position of the robot, say, be close to what we want or will it be completely different?

6 In other words Is the robot stable or unstable? I need some vocabulary (all for linear stability) I’m talking about small errors, which I can address using linear stability Nonlinear stability is beyond the scope of the course stable: if we move the system away from equilibrium, the system will go back unstable: if we move the system away from equilibrium, the error will grow (initially) exponentially marginally stable: if we move the system away from equilibrium, the error will oscillate about its reference position

7 No system without damping (dissipation) can be stable without control The best you can hope for is marginal stability. How do we assess (linear) stability? We have the equations and we suppose that we have found the equilibrium forces

8 Note that  is a placeholder. It has no physical meaning; it merely marks what is “small” Let Substitute

9 The two functions need to be expanded to get their O(1) and O(  ) terms separated Plug all of this in and collect like powers of , and we can drop the O(  2 ) terms

10 The  is no longer necessary, and we have the perturbation equation for q’ The last term will be zero for the current argument

11 Repeat for the momentum equation

12 If we restrict the discussion to cases where p 0 = 0, then we have a pair of coupled homogeneous linear equations with constant coefficients These admit exponential solutions, and the real part of the exponents determine stability

13 Denote the exponents by s If Re(s) < 0 for all s, the system is asymptotically stable If Re(s) > 0 for any s, the system is unstable If Re(s) = 0 for all s, the system is marginally stable

14 ??

15 Let’s see how this plays out a simple setting 

16 If we apply the various holonomic constraints we arrive at a one DOF system

17 I like state space for this sort of problem We have two equilibria: q = 0, q = π We can cover both at the same time by using  0 to denote the equilibrium We linearize

18 We can expand the sine function, noting that sinq 0 is zero, to find

19 We seek exponential solutions

20 if q 0 = 0, the cosine is positive and the system is marginally stable if q 0 = π, the cosine is negative and the system is unstable There’s no damping, so the unforced system cannot be stable

21 We can generalize a bit... by adding a torque we can place the equilibrium anywhere This doesn’t change the perturbation analysis Q 0 does not enter that

22 The system is unstable if the equilibrium position has  in the second or third quadrants More colloquially: if the pendulum is above the horizontal We can illustrate this more physically by an artificial example

23 Add a wheel and a counterweight  m1 m1 m2m2 r z constraint

24 equilibrium is at It is stable down and unstable up Can we envision how this goes?

25  z up increases restoring torque down decreases restoring torque STABLE  z up decreases restoring torque down increases restoring torque UNSTABLE

26 ??

Inverted pendulum on a cart M m l  y 27 (y 1, z 1 )

28 Equilibrium requires sin  = 0 constraint (nonsimple holonomic) Euler-Lagrange equations (I won’t use these for stability here) New Lagrangian

29 Hamilton’s equations. Start with the Lagrangian again. As before, equilibrium requires sinq 2 = 0 assign the generalized coordinates; find the conjugate momenta The evolution equations (velocity and momentum)

30 Horizontal momentum is conserved, we can use this to reduce the problem If we start from rest If the bob starts to fall forward, the cart starts to move backwards

31 To understand stability we need to go a little further on the path to solution We can throw out the p 1 equation, reducing the problem to three equations— two velocity and one momentum We can build a three dimensional state for this problem

32 To understand stability we need to linearize this system The nonlinear state equation

33 Let me introduce a nice new trick for linearizing Expand

34 Seek exponential solutions to the perturbation problem The determinant is a polynomial in s from which we can find the possible values

35 where the bullets stand for fairly complicated terms The characteristic polynomial is a cubic, and it has only cubic and linear terms (the cubic term is unity — the leading term in these is always unity We can review this in Mathematica shortly In the present case

36 For l = 1, m = 1, M = 10 we can plot the linear term as a function of  0 and p 1 Unstable if  0 π/2

37 ??

38 What about something a little more real: the three link robot. We’ll look at this one in Mathematica as well, but we can say some things Recall that we have three generalized coordinates after applying all the holonomic constraints

39 The components of the momentum are too big to transcribe here the form is as below The linearizing matrix has the form

40 The characteristic polynomial is a quintic equation with no constant term The zero eigenvalue means that we are at best marginally stable All but the quintic and cubic terms drop out when the robot is stationary so I have three zero eigenvalues and two nonzero eigenvalues I can plot their square

41 unstable marginally stable

42 ?? On to Mathematica: the inverted pendulum and the three-link robot