Linear Systems Notes Lecture Proposition. A M n (R) is positive definite iff all nested minors are greater than or equal to zero. n Proof. ( ): Positive definite iff λ i >. Let det(a) = λj and H = {x D = } and take Q H, so if Q is positive on all vectors then it s clearly positive on any subspace of V, so det(a k ) >. ( ): We know that det(a k ) > including det(a) >. By interlacing property, the matrix is positive definite.. Lyapunov functions Theorem. ẋ = Ax is globally exponentially stable (GES) iff A is Hurwitz (i.e., max(re(σ(a))) < ). Idea is to transform basis in order to take a trajectory which doesn t admit the application of Lyapunov function to basis wherein the system decomposes in Jordan form, with each Jordan block admitting a quadratic form (Lyapunov function) for each direct summand; take Lyapunov function to be the direct sum of these pieces. E.g. If U = x 2 = x x, then U = ẋ x + x ẋ = x J x + x Jx, i.e. 2λ 2λ... 2λ ẋ = (λi k + N k )x () The determinant is 4λ 2, but this probably won t work, we should change the basis so that instead of having ones on the diagonal we have ɛ, which can be made arbitrarily small. Write x = y, x 2 = ay 2,..., x d a d y d, and also write, x k = λx k + x k so a k y k = λa k y k + a k 2 y k and y k = λy k + a y k. If λ / R, if A : R d R d, then P A (λ) = z d + tr(a)z d +... + det(a); then P A (λ) = implies P A (λ) = too, which implies ( that nonreal ) complex roots always come in roots. If z λz, then a b the matrix of this operator is b a ( ) ( ) 2 5 + 2i Let A =, char A (z) = z 2 + 2z + 5 = (λ + + 2i)(λ + 2i), v = and ( ) ( ) ( ) ( ) 2i 2i 2 v 2 = ; A f =, let e + 2i = and e 2 = be the new basis. Then to calculate: ( ) ( ) ( ) ( ) 2 5 2 2 = (3) /2 /2 2 (2)
Linear Systems Notes 2 This matrix is simply multiplication by 2i, as expected. If y = ay by 2 and ẏ 2 = by + ay 2, then (y 2 + y2 2 ). = 2ẏ y + 2ẏ 2 y 2. x Qx positive definite quadratic form such that d/dx(x Qx) = ẋ Qx + x Qẋ = x A Qx + x QAx = x (A Q + QA)x, negative definite, i.e. the derivative of a quadratic form is still quadratic. x Qx a x 2, for a > and d/dx(x Qx) a x 2. Then let take U(x(t)) = f(t), so f a x 2, then f a x 2 a 2 f, from which it follows that f(t) f()e a2t.
Linear Systems Notes 3 2 Lecture Theorem 2. Let ẋ = Ax. Then A is Hurwitz iff the origin is globally exponentially stable. Proof. ( ): Done. ( ): Assume that Re(λ), with λ an eigenvalue. Then we have ẋ = λx +..., ẋ 2 = λx 2 +..., etc. On the other hand, if λ C \ R, then ẋ = ax bx +... and ẋ 2 = bx + ax 2. Corollary. Let ẋ = Ax+g(x) where g(x) c x 2, for small x, ẋ = f(x) = Ax+g(x) (Taylor), f() =. If A is Hurwitz, then any quadratic strict Lyapunov function for ẋ = Ax will be also strict Lyapunov for a neighborhood of x. Definition. Domain (region) of stability: If U is positive definite in Ω and U f(x) is negative definite in Ω; the region of attraction is the set of {ω Ω : x() = ω lim t x(t) }. Let ẋ = Ax, x R d. We want a quadratic form; want U = x P x, positive definit, and U Ax negative definite. Plugging in, we have d dt U(x(t)) = ẋ P x + x ẋ = x (A P + P A)x. We want A P +P A = Q with Q negative definite symmetric matrix. To solve there are d(d+) 2 equations to solve. Having solved for P, you then have to verify that P is positive definite. Proposition 2. If A is Hurwitz, then such P exists and is positive definite. Proof. Take x P x, and look at x(t) at the solution (trajectory) of ẋ = Ax, and x (t)qx(t)dt which exists. This integral is simply x e A t Qe At xdt = x e A t Qe At dtx. Claim: this integral (-) solves the Lyapunov equation. Positive definiteness of P follows immediately from Q being negative definite and its integral will still be negative definite. The only thing to check is that it solves the equation. d δ Calculate: dtp x(t) = lim e A t Qe At dt = Q, hence d/dtx P x = Q. δ δ 2. Discrete Time Systems Given ẋ = Ax, the analogous discrete time system is x[k + ] = Ax[k], and stability is given not by max Re(σ(A)) <, but that max σ(a) < (global exponential stability, or just exponential stability for nonlinear perturbations x[k + ] = Ax[k] + g(x[k])).
Linear Systems Notes 4 2.2 Hamiltonian Systems Let ẋ = Ax and x R 2d, (q, ( p). ) Hamiltonian matrices are never Hurwitz. Example: q = ap + bq a b and ṗ + cp + dq, the matrix which is Hamiltonian iff tr(a) = ; can t have Hurwitz because c d R(λ + λ 2 ) < contradicts λ + λ 2 =. Nevertheless, such systems can still be Lyapunov stable (only Lyapunov stable), if det(a) >. If det(a) <, then the system is unstable.
Linear Systems Notes 5 3 Lecture 2 3. BIBO Stability Consider the LTV system ẋ = A(t)x + B(t)u (4) C(t)D(t)u (5) Question: if input is bounded, will output also be? Take first an example: say ẋ = v, and v = x + u, y = x; if ɛ sin(t) we have ẍ + x = ɛ sin t which has solution x(t) = A sin(t) + B cos(t) ± ɛt cos(t); hence with bounded u sup t [,T ] So to formalize this discussion, we consider bounded inputs in the L sense, namely u = u(t), if u < C R, will y < M C R as well? There is not much to say in this case, just take y(t) = t Cϕ(t, s)bu(s)ds + Du(t). For one thing, D should be bounded in the matrix norm, i.e. D(t) M D for all t; where M is as follows: For a matrix transformation M L(V, W ), where V comes equipped with a norm V and W as well with a norm W, then we define M M = sup Mv W. What remains is to demand that the integrand is bounded in v V = the operator norm. Precisely: Theorem 3. If there exists constants E, D R such that for all t, s the above bounded conditions hold, then the system is BIBO stable. Proposition 3. For a linear time invariant system (with ẋ = Ax + Bu), if A is Hurwitz then the system is BIBO stable. 3.2 Controllability Take ẋ = Ax + Bu and a finite time interval [, t ]. Define the reachable subspace R(, t ) = {x V : {u(t)} t [t,t ] s.t. x t = x t = x}. Note that R is a linear subspace in V, for if x, x 2 R(, t ), then λx + x 2 R as well because if you have controls u x and u 2 x 2, then λu + u 2 λx + x 2. ( ) Take ẋ = x. This system is asymptotically stable. If we add controls, which states 2 3 ( ) ( ) can we reach? Well if our control is say u, then if you start at the origin, since is an ( ) eigenvector of A, the trajectory x(t) will remain on that eigenvalue, i.e. traj(x(t)) span( ). ( ) If on the other hand, the control is given by u then you have control over one coordinate
Linear Systems Notes 6 direction of travel. This control does allow one to reach the whole space, for if you apply a control you can transverse trajectories then turn off the control and follow the natural dynamics for a period of time then turn on control again, etc. (Intuition for why R = R 2 here). On the other hand, we have controllable subspaces, defined similarly: C(, t ) = {x V : x( ) = x u t [t,t ] : x x(t ) = }. For ϕ(t, ), V t V t where C sits in the first space and R in the second. Recall ϕ(t, s) : V s V t satisfying ϕ t Define W r (, t ) = ϕ(t, s)b(s)b(s) ϕ (t, s)ds. Theorem 4. R(, t ) = Im(W R ). = Aϕ, and ϕ(t, t) = I. Proof. Define u(t) = B (t)ϕ (t, t)z, where assuming x W R z. Then simply compute t ϕ(t, s)b(s)b (s)ϕ(t, s x(t ) = W r z = x. In the other direction, if x / Im(W R ), then we want to show that X / R(, t ). But x Im(W R ) iff (c, x) = for all CW R = ; x = W R z cx = cw R z =. x / Im c row vector such that cx, cw R =, so = c t ϕ(t, s)b(s)b (s)ϕ(t, s)dsc = t (cϕ(t, s)b(s))(cϕb) ds = t cϕ(t, s)b ds which implies that all cϕ(t, s)b(s) is uniformly zero, so that x(t ) = ϕ(t, s)b(s)u(s)ds
Linear Systems Notes 7 4 Lecture 3 4. Lagrange Multiplier Given f : V R, with unconstrained optimization, one approach would be to find all the critical points and test whether they are local extremum. More generally, ( f x,..., f x d ), at a mininum point (assuming differentiability of f), f =. With constrained optimization, we have a set of conditions which we want to be satisfied, namely g i : V R (g : V V = R I, with (g) i = g i ), i =,..., I, with each g i = (g = ). Difficult problem. So we introduce a new vector, λ R I. We now consider the function F (x, λ) = f(x) + λg(x). Extremum points of f subject to g = corresponds exactly to extremum points of F (x, λ) on V W. In this case, F = ( F x,..., F x d, F λ,..., F λ I ) (6) The result is that f x j + λ i g x j =, with j =,..., d, g i (x) =. Optimization with Lagrange multipliers will help us in reachability. As usual we have ẋ = A(t)x(t) + B(t)u(t), with x V = R d, Take x = t ϕ(t, s)b(s)u(s)ds (7) {u} = U, where U is the space of all control functions. Given g : U V, with u ϕbuds, want g x = while minimizing the energy /2 u 2 ds (note that here what we want most, namely a solution to the dynamic system, gives us the constraint in the optimization problem). As before, λ V, form λ(g(u) x ) + 2 u uds = F (u, λ). The result t t λ( ϕ(t, s)b(s)u(s)ds x ) + /2 u (s)u(s)ds (8) equivalent to t /2u u + λϕ(t, s)b9s)u(s)ds λx (9) Now we need to take the gradient; the tricky part is the differentiating w.r.t. u, because it is a function. To address this difficulty, we consider the function at u, at u(s), some arbitrary s, denoted F u(s). F u(s) = u (s) + λϕ(t, s)b(s) () We know that if we solve this problem, then u (s) = λϕ(t, s)b(s), this is the optimal control. Alternatively u(s) = B (s)ϕ (t, s)z, with z = λ. The result: x = t ϕ(t, s)b(s)b (s)ϕ (t, s)dsz = W r z ()
Linear Systems Notes 8 The basis of this argument works so long as a feasible point exists. In other words, assuming that a minimal control exists, then it has this form. Controllability is similar. C(, t ) = {x : u x + ϕ(t, s)bu(s)ds = }. The controllability grammian W C = t ϕ(, s)b(s)b (s)ϕ (, s)ds. So we want W C = x V t. We have almost the same setup as before, = ϕ(t, )x + t ϕ(t, s)b(s)u(s)ds = g with u 2 ds minimum. The solution in this case follows the same process of computations as in reachability. 4.2 LTI Case Let ẋ = Ax + Bu, x V d, u U k, define the controllability matrix C = [B AB... A d B] which has kd columns. Theorem 5. R(, t ) = C(, t ) = ImC. Proof. Let x R, iff x = d t e A(t s) Bu(s)ds. By Cayley Hamilton, A d = cj A j so we can d express e A = αj A j, so t α j A j Bu(s)ds = A j B t α j (s)u(s)ds, and hence x ImC. In the other direction, let x ImC, and suppose that CR =. Want to show that c x =. R = ImW R, so = cw R = c t e A(t s) BB e A (t s) dsc = ce A(t s) B 2 ds ce A(t s) B =. Taking derivatives, of any order, will also be zero. So e.g. cb =, cae (t s) B = (when t = s, equality still holds) or cab =, etc. Hence cw R = ca k B =, but x = A k Bu k. Conclusion: if R = C = ImC = V. Theorem 6. LTI system is controllable iff there are no left eigenvectors of A in the kernel of B. u Ker(B), means Bu =, where B : U V. Similarly, B : V U, then c ker(b ) if cb =. Proof. dim R < d rank(c) < d c cc =. So cb =, i.e. c kerb. Now, we want to prove that among these c s, one of them is an eigenvector. Consider L = ker(c ) = {c : C c = } V. Claim: LA L; cb = cab =... = ca d B = (ca)b = (ca)ab = = (ca)a d B = c( α j A j )B =. Hence it s a linear operator on this space, and therefore has an eigenvector. In the other direction, if cb =, ca = λc, then cab = λcb =. Iterating, gives the same. Therefore, the rank cannot be full.