Designing Information Devices and Systems II Spring 2018 J. Roychowdhury and M. Maharbiz Discussion 6B

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1 EECS 16B Designing Information Devices and Systems II Spring 2018 J. Roychowdhury and M. Maharbiz Discussion 6B 1 Stability 1.1 Discrete time systems A discrete time system is of the form: xt + 1 A xt + B ut Let λ be any particular eigenvalue of A. This system is stable if λ < 1 for all λ. 1.2 Continuous time systems A continuous time system is of the form: d x (t) A x(t) + B u(t) Let λ be any particular eigenvalue of A. This system is stable if Re{λ} < 0 for all λ. 2 Controllability We are given a discrete time state space system, where x is our state vector, A is the state space model, B is the input matrix, and u is the control input. xt + 1 A xt + B ut We want to know if this system is controllable ; if given set of inputs, we can get the system from any initial state to any final state. This has an important physical meaning; if a physical system is controllable, that means that we can get anywhere in the state space. If a robot is controllable, it is able to travel anywhere in the system it is living in (given enough control inputs). 2.1 Controllability Matrix To figure out if a system is controllable, we can simplify the problem. If we want to reach any final state from any initial state, we can consider the initial state as the origin and the final state as any arbitrary point in the state space. A system is controllable if we start off at the initial state x0 0 at time t 0, and after some set of control inputs ut, we can reach an arbitrary final state x 0. Let s start the system off at x0 and see how the system evolves with each time step. x1 A x0 + B u0 A 0 + B u0 B u0 EECS 16B, Spring 2018, Discussion 6B 1

2 This shows us that we can go anywhere spanned by B in the first time step. Using our input vector u, we can push the system anywhere the matrix B lets us go. Now consider the next time step. x2 A x1 + B u1 AB u0 + B u1 Similarly, at this time step, we can go anywhere spanned by B of freedom to the system. If we go another time step, x3, we get the following: After k time steps, we get the following: xk A xk 1 + B uk 1 x3 A x2 + B u2 A 2 B u0 + AB u1 + B u2 AB. Every time step adds another degree A k 1 B u0 + A k 2 B u1 + A k 3 B u2 + + AB uk 2 + B uk 1 After 1 time step, we can go anywhere in the set of vectors spanned by B, after 2 time steps, we can go anywhere spanned by B AB, and after k time steps, we can go anywhere spanned by the columns of the matrix C defined below. This is called the controllability matrix. C B AB A 2 B A k 2 B A k 1 B If this matrix is of rank n (the dimension of our state space), then our system is controllable. It means that our control system is a surjection from the domain of control inputs to the state space. But what if these aren t enough steps and the system can be controlled only in k + 1{ steps? What is } the maximal number of steps we need to take to have a long sequence of control inputs that B,AB,A 2 B,... spans the state space? 2.2 Cayley-Hamilton Theorem These questions are answered by the Cayley-Hamilton theorem. The Cayley-Hamilton theorem says that higher order powers of A can be expressed as a linear combination of lower order matrix powers of A. Specifically if A is an n n, matrix, the highest order unique power of A is A n 1. Thus, if we keep applying control inputs past n time steps, our control inputs will be a linear combination of the previous control inputs and cannot increase the rank of the controllability matrix. 2.3 Controllability This also works for continuous time systems; instead of incrementing the time steps in our system by 1 every time, we increment by t, 2 t, etc. The math and our controllability test work out to be exactly the same! Putting all of this together, we get the following: x(t) A x(t) + B u(t) or xt + 1 A xt + B ut C B AB A 2 B A n 2 B A n 1 B EECS 16B, Spring 2018, Discussion 6B 2

3 Given a continuous or discrete time system x of dimension n, the system is controllable if its controllability matrix C is of rank n. If a system is controllable, then given a starting position x0 0, it takes a maximum of n control inputs over n time steps for the system to reach any final state x Discrete-Time Stability Determine which values of α and β will make the following discrete-time state space models stable: (a) xt + 1 αxt α < 1 (b) α xt + 1 β β xt α The eigenvalues of this system are: For this system to be stable, λ < 1, so λ α ± jβ λ α 2 + β 2 α 2 + β 2 < 1 (c) 1 α xt + 1 xt 0 1 The eigenvalues of this system are λ 1,1 This means that regardless of α, this system is always unstable. 2. Linearization and Controllability We have a system: dx 1 (t) x 1 (t)x 2 (t) 3 dx 2 (t) u(t)x 2 (t) + 8x 1 (t) x 2 (t)x 1 (t) 5 EECS 16B, Spring 2018, Discussion 6B 3

4 (a) Find the equilibrium point of this system when u(t) 0. To find the equilibrium point, we set both dx 1(t) and dx 2(t) equal to 0. 0 x 1 x 2 3 (1) 0 8x 1 x 2 x 1 5 (2) From Equation 1: Plugging into Equation 2: x 1 x x x 1 1 x 2 3 Therefore, our equilibrium point is x 1 (t) 1, x 2 (t) 3, and u(t) 0. (b) Linearize the system around its equilibrium point. To linearize the system, we take the Jacobian and plug in the equilibrium point values. d x(t) x 1 ( dx 1 ) x 2 ( dx 1 ) x(t) + u ( dx 1 ) u(t) d x(t) d x(t) x 1 ( dx 2 ) x 2 x 1 8 x 2 u x x(t) (c) Is the linearized system stable? x 2 ( dx 2 ) x1 1,x 2 3,u0 x1 1,x 2 3,u0 u(t) x(t) + 0 x 2 u ( dx 2 ) x1 1,x 2 3,u0 x1 1,x 2 3,u0 The system is stable if the real parts of both eigenvalues are negative. λ 1 > 0, so the system is unstable. (d) Is the linearized system controllable? det(λi A) (λ 3)(λ + 1) 5 0 λ 2 2λ 8 (λ 4)(λ + 2) 0 λ 1 4 λ 2 2 u(t) B 0 3 AB 3 3 The controllability matrix is full rank, so the system is controllable. EECS 16B, Spring 2018, Discussion 6B 4

5 3. Continuous-Time Stability Consider the linearized state space model for an inverted pendulum in the up position, which we found in lecture: where Is this system stable? No, it is not. The characteristic polynomial is From the quadratic formula: d x(t) x 0 1 x(t), g l k m θ dθ. λ 2 + k m λ g l 0 λ k m ± k g m 2 l 2 k 2 m g l > k m because 4 g l > 0. Therefore, one eigenvalue will have a positive real part, which In this case, means that the system is unstable. This fits our physical intuition since inverted pendula are expected to leave equilibrium. 4. Deadbeat Control Consider the system xt + 1 A xt + But xt + ut (a) Is this system controllable? We compute the controllability matrix C : C B 0 1 AB. 1 1 This matrix has a rank of 2, so the system is controllable. (b) For which initial states x0 is there a control that will bring the state to zero in a single time step? To find the initial states that can be brought to zero in a single step, we solve: x u x x 1 0 x 2 0 x 2 0 x u0 0 x 1 0 x 2 0. EECS 16B, Spring 2018, Discussion 6B 5

6 Therefore, there is a one-dimensional subspace {x 1 0 x 2 0 0} of initial states that can be brought to zero in one step. (c) For which initial states x0 is there a control that will bring the state to zero in two time steps? To find the initial states that can be brought to zero in two steps, we solve: x u x x 1 0 x u1 1 1 x 2 0 x u0 1 2x 1 0 2x 2 0 u0 2x 2 0 2x u0 + u1 Therefore, any initial state can be brought to zero in two steps using an appropriate choice of inputs u0 and u1. 5. Cayley and Hamilton Cayley is trying to control the system xt + 1 A xt + But xt + ut using feedback. (a) Is this system stable? We compute the characteristic equation: 0 det(a λi) det The eigenvalues of this system are 1 λ λ λ 2 ± ± 3. These are not both inside the unit circle, so the system is unstable. (b) Is this system controllable? We compute the controllability matrix C : C B 1 2 AB 1 2 This matrix has rank 1, so the system is not controllable. λ 2 2λ 8 EECS 16B, Spring 2018, Discussion 6B 6

7 (c) Cayley has been trying to find some k, such that the matrix C k B AB A 2 B A k B has rank 2 but still hasn t found one. Confirm that for k 3, this matrix still has rank 1. This matrix still has rank 1. C 3 B AB A 2 B A B (d) Cayley s friend Hamilton remembers hearing somewhere that for any n n matrix A, the matrix A n can always be written as a linear combination of A n 1,A n 2,...,A, and I. 1 Is this true for the A matrix of Cayley s system? We want to find some coefficients α and β, such that 10 6 A 2 β + α αa + βi α If we choose α 2 and β 8, we can make this equation hold. (e) Will Cayley ever find some k to make have rank 2? C k B AB A 2 B... A k B 3α. β + α No. Since A 2 is just a linear combination of A and I, repeatedly exponentiating A will never get him any more linearly independent matrices. Therefore, for any k, he will never be able to make C k have rank 2. Contributors: Siddharth Iyer. Saavan Patel. John Maidens. Kyle Tanghe. 1 Hamilton is right about this. It follows from the Cayley-Hamilton Theorem, which says that any n n matrix always satisfies its characteristic equation. Therefore, the characteristic equation λ 2 2λ 8 we derived above implies that A 2 2A 8 0. You ll learn more about this theorem if you take the advanced control course EE 221A. EECS 16B, Spring 2018, Discussion 6B 7