CDS 22 - Solutions to the Midterm Exam Instructor: Danielle C. Tarraf November 6, 27 Problem (a) Recall that the H norm of a transfer function is time-delay invariant. Hence: ( ) Ĝ(s) = s + a = sup /2 w R jw + a = sup w R a 2 + w 2 = a (b) The transfer function of the system is Ĝ(s) =, with corresponding s + H 2 norm: Ĝ 2 2 = Ĝ (jw)ĝ(jw)dw = = jw + w 2 = arctan w = 2 + jw dw hence Ĝ 2 = 2 (c) Recall that: y G u where G(t) is the impulse response of the system. Here, G(t) = te t when
û û 2 s s s + 2 ˆx ˆx 2 + + ŷ Figure : Figure for problem (d) t, with corresponding l norm: G = = G(t) dt te t dt = te t + = e t = and the amplitude of the output cannot exceed. e t dt (d) Consider the parallel interconnection of two first order systems as shown in the Figure above, and note by inspection that Ĝ is indeed the corresponding transfer matrix. Let x (t) = L (ˆx (s)) and x 2 (t) = L (ˆx 2 (s)). We have: ẋ 2 + 2x 2 = u 2 ẋ x = u y = x + x 2 Set x = x u, with ẋ = ẋ u = x = x + u. The state-space equations of the system are then: ] ] ] ] ] ẋ x u = + ẋ 2 2 x 2 u 2 y = ] ] x + ] ] u x 2 u 2 (e) For the autonomous system (i.e. input identically ), the state and output trajectories for t are given by: x(t) = e At x() y(t) = Ce At x() 2
where e At is defined by the infinite series: Note that here: Hence and e At = I + At + A2 t 2 A 2 = e At = x(t) = 2! + A3 t 3 3! +..., A k =, for k 3 t t 2 /2 t + 2t + t 2 /2 2 + t y(t) = + 2t + t 2 /2,, for t. (f) The transfer matrix is given by Ĝ(s) = C(sI A) B + D. Note that due to the structure of C and B, we only need to compute the upper right entry of (si A). Recall that We have: (si A) = si A = adj(si A) = adj(si A) det(si A) s s s, and Hence det(si A) = s s s + s = s3 Ĝ(s) = ] s 3 + = s3 + s 3 3
Problem 2 (a) FALSE. Consider a SISO system with A=B=, C=D=. A has an eigenvalue at while the transfer function is identically and hence has no poles - this is a very trivial example of an unobservable system. (b) TRUE. We have: Ĝ 2 2 = = = r Ĝ 2 2 Ĝ (rjw)ĝ(rjw)dw Ĝ (jv)ĝ(jv) r dv where the second equality follows by a change of the integration variable. (c) FALSE. Consider the case where M = ai for some scalar a <. Clearly, there exists a matrix D = I = D such that: DMD 2 = M 2 = a I 2 = a < However, the LMI: X a IXIa + X ] = X ( a 2 + )X ] < is not feasible, since it requires X to be both positive definite and negative definite. The correct statement will be derived later on in the class! (d) TRUE. One way of proving this is by using state-space methods to compute the H 2 norms of the relevant systems (see Section 2.6 in DFT). Possible state-space realizations of systems S and S 2 described by transfer functions Ĝ and Ĝ2 are given by: S = ( a ),S 2 = ( a2 System S, the cascade interconnection of S and S 2, with corresponding transfer function Ĝ(s) = Ĝ2(s)Ĝ(s), then has the following state space realization: ] ] a S = a 2 ] ] ) Recall that for a system ( A B S = C D ) 4
with A Hurwitz, the H 2 norm of the corresponding transfer function Ĝ can be computed as: Ĝ 2 2 = CLC where L is the solution to the Lyapunov equation: AL + LA = BB For first order systems S and S 2, we thus have: Ĝi 2 2 = c2 i b2 i 2a i For the second order system S, the relevant Lyapunov equation is: ] ] ] ] ] a l l l l + a ] = a 2 l l 2 l l 2 a 2 Solving the corresponding system of three equations in three unknowns: 2a l + = (a + a 2 )l + l = l a 2 l 2 = we get: l = 2a, l = Hence Ĝ 2 2 = l 2, and 2a (a + a 2 ), l 2 = Ĝ 2 Ĝ 2 Ĝ2 2 Ĝ 2 2 Ĝ 2 2 Ĝ2 2 2 2a a 2 (a + a 2 ) 2a 2a 2 a a 2 2. 2a a 2 (a + a 2 ) (e) TRUE. Note that I A being singular is equivalent to the existence of a v such that: (I A)v = Iv = Av v 2 = Av 2 2 A 2 v 2 2 = A 2 σ max (A) To show that this lower bound can be achieved, let UΣV be a singular value decomposition of A and consider the perturbation matrix = V ΣU 5
where Note that 2 = Σ = σ max (A) σ max (A)............... and that: I A = I V ΣU UΣV = I V ΣΣV... = V I............... V with (I A)v = ; hence I A is singular. 6
Problem 3 (a) Let x = y and x 2 = ẏ u. We then have: and ẋ 2 = ÿ u ẋ = ẏ = x 2 + u = aẏ by cy 2 + u = a(x 2 + u) bx cx 2 + u = bx ax 2 cx 2 + ( a)u The corresponding second order state space realization is: ẋ = x 2 + u ẋ 2 = bx ax 2 cx 2 + ( a)u y = x Note that this realization is not unique; it is simply one of many possible realizations. (b) When a = 3, b = 2 and c = 2, the state equations of the autonomous system reduce to: { ẋ = x 2 ẋ 2 = 2x 3x 2 2x 2 and the corresponding linearized (about the origin) dynamics are given by: ] ] ] δ δ = δ 2 2 3 The eigenvalues of the linearized dynamics are λ = and λ 2 = 2. Since the linearized dynamics are asymptotically stable, and for f(x) = 2x 2, we have: f(δ) f() f (x) δ 2 2 δ 2 lim = lim = δ 2 δ 2 δ δ we conclude that the equilibrium point at the origin is locally asymptotically stable. Clearly, it cannot be globally stable as there exists another equilibrium point at (,). δ 2 7
(c) The dynamics of the autonomous nonlinear system are given by: where A = 2 3 ẋ = Ax + f(x) ] and f(x) = ] 2x 2 Let V (x) = x Px be a Lyapunov function for the linearized system; then P is a solution of the Lyapunov equation A P +PA = Q for some Q >, and along the trajectories of the nonlinear system, we have: V (x) = ẋ Px + x Pẋ = x A + f (x)]px + x PAx + f(x)] = x Qx + 2x Pf(x) λ min (Q) x 2 + 2λ max (P) f(x) 2 x 2 λ min (Q) + 2λ max (P)ɛ] x 2 whenever x ɛ/2 Thus, the function V : B ɛ R is a Lyapunov function for the nonlinear system in the neighborhood B ɛ = {x R 2 x 2 < ɛ/2} for ɛ < λ min(q), for any Q > and corresponding solution P to the 2λ max (P) Lyapunov equation; in particular, trajectories starting in this neighborhood will converge asymptotically to the origin. For instance, for Q = I with λ min (Q) =, we have P = ] 5 4 with λ max (P) = 3+2 5 and the corresponding provable region of attraction: { } B = x R 2 x 2 < 2 3 + 2 5 Ideally, we would like to choose Q > to maximize the ratio λ min(q) 2λ max (P), which is not an easy problem. What is straightforward though is computing an upper bound for this ratio, which establishes the limitations of this approach in finding a region of attraction. We have: 2λ max (P) A = 2 P A 2 PA A P + PA = Q λ min (Q) which implies: λ min (Q) 2λ max (P) A 2 8