CDS Solutions to Final Exam


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1 CDS 22  Solutions to Final Exam Instructor: Danielle C Tarraf Fall 27 Problem (a) We will compute the H 2 norm of G using statespace methods (see Section 26 in DFT) We begin by finding a minimal statespace representation of the system with transfer function G(s) Assuming u( ) = y( ) = as usual, we have: s 3 y(s) + 2s 2 y(s) + sy(s) + y(s) = u(s) y (t) + 2ÿ(t) + ẏ(t) + y(t) = u(t) Defining x = y, x 2 = ẏ, x 3 = ÿ, the corresponding statespace representation is given by: ẋ x ẋ 2 = x 2 + u ẋ 3 2 x 3 y = [ Since A is Hurwitz (one way of verifying this is using the RouthHurwitz criterion), the Lyapunov equation x x 2 x 3 AL + LA + BB = has a unique solution L = L, and G 2 = CLC What is left is to solve the Lyapunov equation Letting: a a a 2 L = a a 3 a 4 a 2 a 4 a 5 and substituting, we get the following system of equations: 2a = a 2 + a 3 = 2a 4 = a 4 a a 2a 2 = a 5 a a 3 2a 4 = 2[a 2 + a 4 + 2a 5 =
2 The solution to this system of equations is: a =, a = a 4 =, a 2 = 2, a 3 = a 5 = 2 By direct substitution, the H 2 norm of G can now be computed to be equal to (b) We begin by noting that the relation between inputs u and u 2 and outputs y and y 2 given by: [ [ [ y (s) /s /s u (s) = y 2 (s) /s u 2 (s) can be represented in terms of the block diagram in Figure u (s) /s x (s) + + y (s) u 2 (s) /s x 2 (s) y 2 (s) Figure : Block diagram for problem (b) Let x (s) and x 2 (s) be the signals at the output of the two integrators in Figure, we have: sx (s) = u (s) sx 2 (s) = u 2 (s) y (s) = x (s) + x 2 (s) y 2 (s) = x 2 (s) The corresponding statespace representation is given by: [ [ [ [ [ ẋ x u = + ẋ 2 x 2 u 2 [ [ [ y x = y 2 Since both B and C have full rank, (A,B) and (C,A) are reachable and observable, respectively, and the realization is indeed minimal What is left is to verify (as a sanity check) that: x 2 C(sI A) B = C(sI) I = H(s) which is readily done by direct substitution 2
3 (c) By definition, the poles of the transfer matrix are those of the individual entries Hence, we have a pole at s = 2 and two repeated poles at s = To verify that, note that the number of poles of the system is equal to the order of the minimal realization of the system, which in this case is 3 (One way of seeing that is by drawing a block diagram as in part (b) and noting that we need exactly 3 first order, proper transfer blocks of order to obtain the desired inputoutput map) The zeroes of the transfer matrix are those values of s (finite or infinite) at which H(s) loses rank By inspection, note that: [ H() = 2 2 has rank, hence s = is a zero of the system Also note that: [ lim H(s) = 2 s also has rank, hence s = is an (infinite) zero of the system Finally, note that for [ 2(s 2) u(s) = (s + )(s ) we have: and: [ lim u(s) = s 2 3 [ lim H(s)u(s) = s 2 Hence s = 2 is also a zero of the system Note that there cannot be any more zeroes since the number of zeroes cannot exceed the number of poles Remark: Alternatively, the poles and finite zeroes of H(s) could have been computed by finding a minimal statespace realization of the system, computing the eigenvalues of A to determine the poles of the system, and finding the values of s C at which the system matrix [ A si B C D loses rank to compute the finite zeroes of the system (d) In order to verify stability of the equilibrium point at (,), we begin by finding the set of equilibrium points of the system The solutions to the system of equations: { 2x x 2 2 = 2x 2 x 2 4x 3 2 = 3
4 are the set of points (α,), α R, ie the x axis Since every neighborhood of the origin contains an equilibrium point distinct from (, ), the origin cannot be asymptotically stable To verify stability isl, consider the candidate Lyapunov function V (x,x 2 ) = x 2 + x 2 2 Note that V (x), x R 2, and that V (x) = x = x 2 = Also note that V (x) = [ V x [ V ẋ x 2 ẋ 2 = 2x 2 2(3x 2 + 2x 2 2), x R 2 V(x) is thus indeed a Lyapunov function for the system, and the origin is stable isl Problem 2 (a) Omitted (b) FALSE The system has a pole at s = 2, and is hence unstable To see that, consider the input { t [, u(t) = otherwise Note that u 2 2 = u 2 (t)dt = Also note that y() = and that y(t) = e 2(t ) y(), for t Hence y 2 2 = e 2( T) u(t)dt = e 2 e 2T dt = 2 (e2 ) y 2 (t)dt y 2 (t)dt = ( e 2 y() ) 2 ( e 2t ) 2 dt = Thus, the system can infinitely amplify an input signal with bounded L 2 norm (c) TRUE Let M(s) = s s + 5 and let (s) = e st We begin by s rewriting H(s), by simple algebraic manipulation, as: H(s) = s + 5 M(s) (s) 4
5 Since transfer function M(s) (s) is itself stable is stable, H(s) is stable iff the transfer function s + 5 Note that M = sup M(jw) = sup w w w = By the small gain w theorem, the feedback interconnection of M and is guaranteed to be internally stable if < In particular, the transfer function from u to y in Figure is then guaranteed to be stable u + + (s) M(s) y Figure 2: Block diagram for problem 2(c) Noting that the transfer function from u to y as shown in Figure is T u y = M(s) (s) it is clear that < is a sufficient condition for H(s) to be stable What is thus left to show is that < for all t [,) We have: and = sup w e jwt w e jwt = e j w w = tw = t w where the inequality uses the hint Hence for all t [,), we have: e jwt w t < and which concludes the proof sup w e jwt w < 5
6 (d) TRUE Problem 3 It is possible to derive a weak lower bound for T from the given bound on S(jw) in the interval w [ 4,4 Indeed, the algebraic constraint on S and T: S + T = and the triangle inequality allow us to conclude: = S + T S + T T S S(jw) w 4 = 9 A slightly stronger lower bound on T can be obtained by considering the effect of right half plane poles and zeroes of P (which are also right half plane poles and zeroes of the open loop transfer function L = PC, since there can be no unstable pole/zero cancellations between P and C) In particular, note that the pole at s = 3 and the zero at s = 2 impose the following interpolation constraints on S and T: S(2) = T(2) = S(3) = T(3) = We can immediately conclude, noting that T is stable hence analytic on the closed right half plane, and using the Maximum Modulus Principle that: T = sup T(jw) T(3) = w The lower bound can be improved further Since S is a stable nonminimum phase (S(3) = ) transfer function, it can be written as the product of a stable minimum phase system S MP and a stable allpass filter S AP of the form: S AP = s 3 s + 3 S AP where S AP is a stable allpass filter to account for other (potential) nonminimum phase zeroes of S: S = s 3 s + 3 S APS MP We have: S(jw) = S AP (jw)s MP (jw) = S AP (jw) S MP (jw) = S MP (jw), w We also have: S(2) = 5 S AP(2)S MP (2) 6
7 Hence: = S(2) = 5 S AP(2) S MP (2) S MP (2) = 5 S AP (2) 5 Finally, noting that for any stable, rational transfer function S, log S is analytic and bounded in the closed right half plane, we can apply Poisson s Integral Formula We have: log S MP (2) = 2 log S MP (jw) π 2 2 dw log(5) + w2 2 2 log S(jw) π 2 2 dw log(5) + w2 4 π 4 log S(jw) w 2 dw + 4 π 4 log S(jw) 2 2 dw log(5) + w2 Letting α = sup S(jw), and recalling that S(jw) for w < 4, we have: w >4 Noting that: and: We get: 4 4 π log() w 2 dw + 4 π log(α) dw log(5) + w w 2 dw = 2 arctan w 4 = arctan (2) w 2 dw = 2 arctan w 2 4 = 2 arctan (2) + π 4 log(α) log(5) 2 π log()arctan(2) 2 π arctan(2) 39 Hence S α Once again using the constraint S + T = and the triangle inequality, we have: S T S + T = T S 229 which is a much stronger lower bound Problem 4 Since the open loop system is reachable, for any choice of terminal state x o, there exists an input sequence, say u, u,, u n taking the system from the origin to x o in n steps Let, x, x 2,,x n be the state trajectory of the open 7
8 loop system corresponding to this input sequence, and note that for the closed loop system under the given nonlinear feedback, we have: By setting the exogenous input to be: x(k + ) = Ax(k) + B (w(k) + f(x(k))) w (k) = u k f(x k) the state evolution of the closed loop system (starting from the origin) under w is identical to that of the open loop system under u ; in particular, x(n) = x o Since the choice of x o was arbitrary, the system remains reachable under arbitrary nonlinear state feedback Problem 5 We begin by noting that the relation between the initial state and the inputs and outputs of the system up until time step n is given by: Y = y() y() y(n ) = C CA CA n = θx() + MU x() + D CB D CA n 2 B CB D The initial state can be uniquely determined from U and Y iff the observability matrix θ has full column rank (ie is left invertible), in which case we have: where θ L is the left inverse of θ x() = θ L (Y MU) (a) When the system is observable, the initial state can be uniquely determined from the output measurements alone iff: θ L M = () The given conditions are sufficient to ensure that () holds since M = under these conditions (b) When the output y is a scalar, the observability matrix is a square matrix of size n In this case, for an observable system, the observability matrix has a left and a right inverse, θ L = θ R = θ, and condition () holds iff M =, from which it follows that the conditions in part (a) are both necessary and sufficient u() u(n ) 8
9 for the initial state to be uniquely determined from output measurements alone (c) The transfer function of the system is the ztransform of its impulse response, assuming zero initial conditions In this case, we have: y() y() D CB D = CAB CB D Hence G(z) = z n y(n) n= = D + z CB + CAB + z2 = D + z C[I + A z + A2 z 2 + B = D + z C(I A z ) B = D + C(zI A) B We wish to prove that in the SISO case, the conditions of part (a) are both necessary and sufficient conditions for G(z) to have no finite zeroes To prove necessity, note that if G(z) has no finite zeroes, then G(z) exists and we can write: G(z) y(z) = u(z) Thus the input corresponding to a given ouput signal is unique (for zero initial conditions), and the initial state can be uniquely determined from output measurements alone It follows from parts (a) and (b) that the conditions on the Markov parameters of the system stated in part (a) hold The proof of sufficiency is more involved We begin by making some basic observations First, consider the characteristic polynomial of matrix zi A: a (z) + a (z)λ + + λ n = Recall that by the CayleyHamilton Theorem, we have: Recalling the expansion: a (z)i + a (z)(zi A) + + (zi A) n = (2) (a b) i = i j= ( i j ) a i j ( b) j 9
10 we note that (zi A) i can be written as a sum of terms involving powers of A (and z) up to order i In particular, when i n 2, under the conditions in part (a), we have: C(zI A) i B = while: C(zI A) n B = ( ) n CA n B Now suppose that z o is a finite zero of G(z) and that the conditions of part (a) hold; that is G(z o ) = C(z o I A) B = (i) If a (z o ), by rearranging (2) and multiplying by C and B, then using the above observations, we have: G(z o ) = C(z o I A) B [ n 2 = a (z o ) C (z o I A) n a i (z o )(z o I A) i B = ( )n a (z o ) CAn B = Equivalently, CA n B =, θb =, and B = Hence, barring the trivial case where B = (in which case the system is unforced and its transfer function is identically ) the transfer function cannot have finite zeroes (ii) When a (z o ) =, it can be shown by a similar argument that z o is both a pole and zero of the transfer function and hence cancels out (hence it is not a zero of the transfer function, although it is one of the given nonminimal statespace realization) i= Problem 6 The plant [ can be represented as shown in Figure δ Let = The transfer function M(s) with inputs w δ, w 2 and outputs z, z 2 is given by: M(s) = [ CNδ N δ CD δ D δ Note that M is stable by design Moreover, note that M is a rank one matrix, and can be decomposed as follows: [ M(s) = Nδ D δ We are asked to compute: = ab [ C γ o = inf w γ(w)
11 z δ w N δ D w δ 2 δ z N o D o y K Figure 3: Block Diagram of problem 6 where We have: γ(w) = inf {max γ i : det(i M(jw) (jw)) = } i det(i M ) = det(i ab ) = det( b a) = b a = [ δ δ n Let c i = D δi CN δi We can rewrite γ(w) as: D δ CN δ D δn CN δ γ(w) = inf δ R n{max γ i : i n γ i c i = } i= This problem is substantially more difficult in this case than in the case where δ is complex To gain some intuition, consider the three dimensional case (n = 3, the axis is (δ,δ 2,δ 3 )); the generalization to higher n follows At each frequency w, define: P = {δ : P 2 = {δ : 3 δ i I(c i (jw)) = ) i= 3 δ i R(c i (jw)) = ) i=
12 L = P P 2 In general, L R 3 is a line This makes our problem equivalent to finding the smallest β such that the cube B β = {δ R n : max i δ i β} touches the line L That can be done by looking at the following projections: For every j such that I(c j ), do the following: Look for the smallest β j such that max i δ i = β j and δ i (R(c i ) α j I(c i )) = i where α j = R(cj) I(c For each such j, it can be shown that the minimum β j) j is Èi R(ci) αji(ci) and the optimal δ is δ i = β j sgn(r(c i ) α j I(c i )) By doing this, we have found the smallest β j such that the projection of B βj in δ j = touches the projection of L Among È all of the above candidate solutions, the only admissible is j i j such that δ j = I(ci)δi c j satisfies δ j β j Thus, at each frequency w, γ(w) is given by: γ(w) = i R(c i(jw)) α j I(c i (jw)) 2
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