Solution for Homework 5

Solution for Homework 5 ME243A/ECE23A Fall 27 Exercise 1 The computation of the reachable subspace in continuous time can be handled easily introducing the concepts of inner product, orthogonal complement of a subspace and adjoint operator Given a vector space V on a field K, an inner product between the vectors v and w in V, which we denote by the symbol v, w, is any operation from V V to K such that the following three properties are satisfied for every u, v, w V and for every α K: (positivity) v, v and v, v = if and only if v = ; (sesquilinearity) αv, w = α v, w and u + v, w = u, w + v, w ; (conjugate symmetry) v, w = w, v Here, and in the rest of this paper, the symbol denotes conjugation or, when applied to a matrix, conjugate transposition A vector space with a inner product is called an inner product space The most common example of inner product space is R n with the dot product: x, y = x 1y 1 + x 2y 2 + + x ny n = x y It is easy to check that the three properties above hold for the dot product If now our vector space is C 1 {[, t]}, that is the space of all the continuously differentiable, real and bounded functions defined from the interval [, ) to R, we may wonder what is a convenient way of defining an inner product for such a space It turns out that a good inner product for a space of functions is the convolution, so for f and g in C 1 {[, t]} we define: f, g = f g = f(τ)g(t τ)dτ The fact that the convolution satisfies the properties of inner products follows immediately from the properties of integrals 1

The main purpose of introducing inner products is to measure the angle between two vectors In particular, two vectors of an inner product space are orthogonal if and only if their inner product is zero If V is an inner product space, and W is a subspace of V, we define the orthogonal complement of W, which we denote by W, as the set of all the vectors of V that are orthogonal to every vector of W: W = {v V v, w = w W} If V is a finite-dimensional inner product space, then: V = W W The following two facts will be needed in the later proof Lemma 1 For every subspace W Lemma 2 For every matrix M (W ) = W ker M = (Im M ), where ker M is the null space of M (also called the kernel) and Im M is the range of M (also called the image) Let V and W be inner product spaces, and let T : V W be a linear operator We call the null space (or kernel, or nucleus) of T, and we write ker T, the set of all vectors of V that are mapped by T into the zero vector of W: ker T = {v V T (v) = } We call the range (or image) of T, and we write Im T, the set of all vectors of W that can be written as T (v) for some v V: Im T = {w W w = T (v) for some v V} We define the adjoint operator of T to be another linear operator T : W V such that the following property holds for every v V and w W: T (v), w = v, T (w), where the inner product in the right-hand side is the inner product of W and the inner product in the left-hand side is the inner product of V In the following, we will be using the key properties of adjoint operators stated below Lemma 3 For adjoint operators T and T Im T = Im T T 2

Lemma 4 For adjoint operators T and T Im T = ker(t ) With the algebraic tools we reviewed so far, we are now ready to compute the set of reachable states in continuous time In other words, we are asking what are the states to which we can drive our system under the action of an input u we are free to choose We know that the state response of the system is given by: x(t) = e A(t τ) Bu(τ)dτ (1) Equation (1) can be viewed as the definition of a linear operator x = T (u) that takes an input function from the space C 1 {[, t]} and maps it into the state x in R n, reached at time t Within this framework, computing the set of reachable states amounts to computing the image of the operator T In order to do so, let s first of all show that T (x) = B e A (t τ) x We have: ( T (u), x = e Bu(τ)dτ) A(t τ) x = = B e A (t τ) u(τ)dτx = u(τ)b e A (t τ) xdτ = u, T (x) Now, let s compute the null space of T (x), which is the set of all vectors x R n such that: B e A (t τ) x = Expanding the matrix exponential in Taylor series, we get: B x + B A(t τ)x + B 2 (t τ)2 A x + B 3 (t τ)3 A x + =, 2 3! which by the principle of polynomials identity can be satisfied for every (t τ) if and only if: B x = B Ax = B A 2 x = The Cayley-Hamilton theorem tells us that of these infinite equations only the first n are linearly independent, so we can reduce our system to: B B A B A 2 x = (2) B A n 1 3

What (2) shows is that x is a vector of the null space of T if and only if x is a vector of the null space of the matrix U Now, using Lemma 1, Lemma 2 and Lemma 4, we can write: Im T = (ker T ) = (ker U ) = ((Im U) ) = Im U This computation proves the equivalence R = Im U, that is second equivalence in part (c) All the others are consequences of this main fact In particular, note that the proof is completely independent of the time t we pick as final integration time in (1) This proves part (a), and means that the set of reachable states in continuous time is independent of time Once we observe that T T = W tf x, the first equivalence in part (c) is a direct consequence of Lemma 3 The chain of implications in part (b) can be stated as the set of equivalences: all of which have been proved already Exercise 2 Im W tf = (ker T ) = (ker U ) = R, The proof needs to be carried out only for values of λ that are eigenvalues of A, because for these values we have the lowest possible rank for the matrix λi A The proof goes by contraddiction Suppose that for the eigenvalue λ we have: rank [ λi A B ] < n Then, there exists a non-zero vector v such that: This condition can be split into: v T [ λi A B ] = v T A = λv T (3) v T B = (4) Equation (3) shows that v is an eigenvector of A, so we can write: v T A k = λ k v T for k = 1, 2,, n 1 (5) Post-multiplying equations (5) by B, and considering (4), we get: v T A k B = λ k v T B =, 4

and, finally: v T [ B AB A 2 B A n 1 B ] = v T U = So, since v, the controllability matrix U cannot have full rank, and the system is not controllable Let s now suppose that U is not full rank Then, there exists a non-zero vector such that: v T [ B AB A 2 B A n 1 B ] = Let p(s) be the characteristic polynomial of A and A T, then p(a T )v = v T p(a) = If λ is a zero of p(s), meaning that it is an eigenvalue of A or A T, then we can factor p(s) in the following way: and we have: p(s) = g(s)(s λ), v T p(a) = v T g(a) }{{} w T (λi A) = The vector w T is non-zero and satisfies: w T (λi A) = (6) Moreover, setting g(s) = α +α 1 s+ +α n 1 s n 1, the vector w T also satisfies: w T B = v T g(a)b = v T (α B + α 1 AB + + α n 1 A n 1 B) = (7) Equations (6) and (7) show that w T [ λi A B ] =, for a non-zero vector w, so the matrix [ λi A B ] is not full rank Exercise 3 For the given system, we have 1 1 U = 1 2 2 2 and 1 1 V = 3 1 2 4 Since rank U = rank V = 3, the system is both controllable and observable 5

Exercise 4 Using long division, we can rewrite G(s) as: [ 1 s+2 1 ] [ ] G(s) = s+1 s 2 1 s 1 2 1 + s 2 s+1 s 1 1 1 }{{}}{{} G( ) Ĝ(s) Let ψ(s) be the least common multiple of all the denominators of the elements of Ĝ(s), then: ψ(s) = s 2 (s + 1)(s 1) = s 4 s 2 = a + a 1 s + a 2 s 2 + a 3 s 3 + s 4 The matrix ψ(s)ĝ(s) can be written as: [ ] s 3 s 2 s 3 + 2s 2 s 3 s s 2 1 2s 3 + 2s 2 s 3 + s 2 = [ ] [ ] [ ] [ ] 1 1 2 = + s + s 2 1 1 1 + s 3 1 1 2 1 2 1 }{{}}{{}}{{}}{{} C C 1 C 2 C 3 A controllable state space realization of G(s) is the following: I A = I I, a I a 1 I a 2 I a 3 I B =, I C = [ C C 1 C 2 C 3 ], D = G( ) Note, however, that this is not the controllable canonical form of the system This realization solves parts (a) and (b) To get an observable one, let s first of all check the rank of the observability matrix We get rank V = 7, meaning that we can get rid of 5 states out of the total 12 In order to do that, we pick 7 linearly independent rows of V, which we call v 1,, v 7 and we choose 5 more vectors e 8,, e 12 such that matrix: v 1 v 7 e 8 e 12 6

has full rank Calling this matrix T 1, the similarity transformation: Ā = T 1 AT, B = T 1 B, C = CT, puts the system into the following form: ] [Ā11 Ā =, Ā 21 Ā 22 [ ] B1 B =, B 2 C = [ C1 ] The system { Ā 11, B 1, C 1, D } is the controllable realization we are looking for, which solves part (c) This procedure, called Kalman decomposition with respect to observability, can be used in general to spot the observable subsystem in a larger system that is not observable Exercise 5 Inserting a dummy state in the minimal realization in canonical controllable form, together with suitable zeros in the B and C matrices, we get: 1 A = 1 3 B = 1 C = [ 1 1 ], 1 which is not controllable, nor observable Note the selection of a negative number for a 33 This is to avoid introducing a zero (or, even worse, positive) eigenvalue in A, which would make A no longer Hurwitz 7