Control engineering sample exam paper - Model answers

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1 Question Control engineering sample exam paper - Model answers a) By a direct computation we obtain x() =, x(2) =, x(3) =, x(4) = = x(). This trajectory is sketched in Figure (left). Note that A 2 = I and A 4 = I. b) By a direct computation we obtain x() =,x(2) =, x(3) = This trajectory is sketched in Figure (right)., x(4) = = x(). x 2 x 2 x(3) x()=x(4) x(2) x x(3) x() x x()=x(4) x() x(2) Figure : Sketch of the trajectories of the considered system. c) Using the results in a) and b) yields β α x() =, x(2) = α β This trajectory is sketched in Figure 2., x(3) = β α, x(4) = α β = x(). d) We have to show that for any ɛ>thereexistsaδ(ɛ) such that x() <δ(ɛ) implies x(k) <ɛfor all k. Note that for any initial condition the distance of x(k) from the origin is constant, i.e. x(k) is constant for all k. Therefore, the selection δ(ɛ) = ɛ makes the above implication true. The state does not converge to the origin, hence the equilibrium is not attractive. Finally, the system is time-invariant, hence stability is uniform. e) Given a continuous-time system ẋ = A c x, its Euler approximate model, with sampling time T,is x(k +)=(I + TA c )x(k) =Ax(k).

2 x 2 x(3) β x()=x(4) x(2) α x x() Figure 2: Sketch of a generic trajectory of the considered system. Hence, A c = A I T = T. The characteristic polynomial of the matrix A c is s 2 +2s+2, hence the continuous-time system is asymptotically stable. (Note that, in this case, this is true regardless of the value of T.) 2

3 Question 2 a) The equilibrium points are the solutions of the equations = L x 2u + E L From the first equation we have = RC x 2 + C x u. x 2 = E u and from the second equation x = Ru x 2 = E Ru 2. b) The linearized model is given by δ x = Aδ x + Bδ u = ẋ x ẋ x 2 ẋ 2 x ẋ 2 x 2 δ x + ẋ u ẋ 2 u δ u = u L u C RC δ x + E Lu E CRu 2 δ u. c) The characteristic polynomial of the matrix A is p(s) =s 2 + RC s + u2 LC, hence, by Routh test, its roots have negative real part. This implies that the linearized system is asymptotically stable for any positive R, L, C and u. d) By the principle of stability in the first approximation, the equilibrium of the nonlinear system is locally asymptotically stable. e) If u =wehave ẋ = E L, ẋ 2 = RC x 2. The system does not have any equilibrium because the equation = E does not have any solution. Moreover, x (t) =x () + Et and x 2 (t) =e t RC x 2 () which shows that x (t) and x 2 (t). 3

4 Question 3 a) The reachability matrix is R = B AB A 2 B = The matrix R has rank 2. In fact, its determinant is zero, and the first two columns are linearly independent. This implies that the system is not reachable. A basis for the reachable space is given by the image of R, namely ImR =Im. 2 2 b) The set of reachable states in one step is given by x() = Bu() = u(), 2 with u() IR. The set of reachable states in two steps is given by u() x(2) = ABu() + Bu() =, u() 2 2 with u() IR and u() IR. c) The condition for controllability in two steps is ImA 2 ImB,AB, which is equivalent to rank B AB A 2 =rank B AB =2. Note that B AB A 2 = and this has rank two. Hence the system is controllable in two steps. d) To write the system is the canonical form for unreachable systems we define a matrix L from the first two columns of R and adding a third colum which makes it invertible, namely L = The transformed system is described by ˆx(k +)=L ALˆx(k)+L Bu(k) = 2 2. /2 /2. ˆx(k)+ u(k), from which we see that the unreachable system is described by the equation ˆx 2 (k +) =. This is consistent with the fact that the unreachable system has dimension one, because rankr = 2, and the system is controllable, i.e. the unreachable modes are at zero. 4

5 Question 4 a) Note that, if w(t) = sin t, cos t, then ẇ(t) =cost, sin t, and cos t sin t sin t = 2 cos t. This implies that w(t) belongs to the class of signals for which asymptotic tracking is achievable. b) Let e = x w and note that Let now ė =ẋ ẇ = Ax + Bu ẇ = Ax + Bu Aw = Ae + Bu. u = Ke = K(x w) and select K =k,k 2,k 3 such that the eigenvalues of A + BK are all equal to. Note that k +k 2 +k 3 A + BK = 2. The characteristic polynomial of A + BK is p(s) =(s +)(s 2 k s ++k 2 ) and this should be equal to (s +) 3. This is achieved by selecting k = 2 andk 2 =, while k 3 can be arbitrarily assigned. This is due to the fact that the pair (A, B) isnot reachable, however it is stabilizable and the unreachable mode is s =. This can be seen considering the reachability pencil si A B = s s 2 s + and noting that it has rank 2 for s = and rank 3 for any other s. 5

6 Question 5 a) The reachability matrix is R = 2 α + and the system is reachable if α. Ifα = the reachability pencil is s si A B = s and this has rank for s = and rank 2 for any other s. Therefore, the system is controllable and stabilizable. b) The observability matrix is O = α and the system is observable if α. Ifα = the observability pencil is s si A = s C and this has rank for s = 2 and rank 2 for any other s. Therefore the system is not detectable. c) Let K =k k 2 and note that A + BK = k + +k 2 α + k k 2 + and that the characteristic polynomial of this matrix is s 2 +( 2 k k 2 )s +(k 2 α + k 2 α). Hence the selection k = k 2 = is such that the eigenvalues of A + BK are equal to. Let L =l l 2 T and note that l + l A + LC =, α + l 2 l 2 and that the characteristic polynomial of this matrix is Hence the selection s 2 +( 2 l + l 2 )s +( 2l 2 + l ++l α α) l = 3+α α l 2 = +3α α is such that the eigenvalues of A + LC are equal to. Finally, the controller is ξ = (A + BK + LC)ξ Ly, u = Kξ. d) The limit for α of K = k 2 + k2 2 is 2, whereas the limit for α of L is equal to +. This is in agreement with the fact that, for α =, the system is not reachable but stabilizable, with a non-reachable mode equal to, but not detectable., 6

7 Question 6 a) Let i and i 2 be the currents through L and C, respectively. Then y = i = i + i 2. Moreover, Hence, A = RC i = x 2 = u Lẋ 2 R R L b) The reachability matrix is B = RC L C = RC R 2 C 2 R L L 2 i 2 = Cẋ = u x R. C = R D = R. and it is full rank if R 2 C L. If R 2 C = L the system is not reachable. However, as both the eigenvalues of A have negative real part, the system is stabilizable. c) The observability matrix is O = R R 2 C and it is full rank if R 2 C L. If R 2 C = L the system is not observable. However, as both the eigenvalues of A have negative real part, the system is detectable. d) Consider new coordinates ˆx such that x = Lˆx and L is constructed from the first column of the reachability matrix, multiplied by R 2 C, and a second column which renders L invertible: R L =. Then R L ˆx = RC RC y = ˆx +. u These equations show that the unreachable subsystem, namely ˆx 2 = RC ˆx 2 is observable, and that the unobservable subsystem, namely ˆx = RC ˆx + u is reachable. Hence, there is no subsystem which is both reachable and observable. 7

8 e) The input-output behaviour of the first electrical network is described by Ce At B + D = e R L t L e RC t R 2 C +/R. This, for L = R 2 C,isequalto/R, which describes the input-output behaviour of the second circuit. 8