# State will have dimension 5. One possible choice is given by y and its derivatives up to y (4)

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1 A Exercise State will have dimension 5. One possible choice is given by y and its derivatives up to y (4 x T (t [ y(t y ( (t y (2 (t y (3 (t y (4 (t ] T With this choice we obtain A B C [ ] D To find the transfer function we could use the formula involving (A, B, C, D but this would require the inversion of the 5 5 matrix (si A. More directly we can recognize in the structure of the obtained A, B and C the controller canonical form and therefore we can directly state that F (s s 5 + 4s 4 + 3s 3 2s 2 + s + Otherwise, since the transfer function relates the input to the output zero-state response (i.e. with x(, applying the derivative theorem to the differential equation leads to s 5 Y (s + 4s 4 Y (s + 3s 3 Y (s 2s 2 Y (s + sy (s + Y (s U(s and to the transfer function F (s Y (s/u(s previously found. System stability can be inferred from the eigenvalues, but since the denominator of the transfer function has degree equal to n 5, the poles coincide with the eigenvalues. The Routh necessary condition is not satisfied and therefore the system is not asymptotically stable. B Exercise 2 We can find the impulse response from the inverse Laplace transform of the transfer function which can be found as the ratio of the zero-state response transform with the input transform that is, being Y (s s s 2 + s 3 s + U(s 2 s + 3 we have F (s /(s3 (s + (s + 3 (s + /(s + 3 s 3 (s + 2 C Exercise 3 This exercise requires the correct use of the Laplace transform time shifting property and rewriting u(t as a linear combination of functions which have simple Laplace transform. The input u(t, as 4

2 u(t t Figure 3: Ex. 3, input u(t shown in Fig. 3, can also be written as u(t tδ (t 2(t δ (t + 2(t 2δ (t 2 2(t 3δ (t 3 + (t 4δ (t 4 4 a k (t kδ (t k k and therefore U(s ( 2e s + 2e 2s 2e 3s + e 4s s 2 and Y (s F (su(s. Recall that if, in general, Y (s Y (se st then and therefore defining we have, once the residues have been computed, y(t y (t T δ (t T Y (s F (s s 2 R s + R 2 s 2 + R 2 s R 3 s + y(t 4 a k y (t kδ (t k k with R 3/, R 2 F (, R 2 2/5 and R 3 /. The input and final output are plotted in Fig u y t Figure 4: Ex. 3, input u(t and corresponding response y(t D Exercise 4 When possible we try to exploit some particular matrix structure to simplify calculation as much as possible. 5

3 Note that matrix A is block diagonal ( A 3 Aα A 3 β and therefore We have aut A α } and ( Aα eig A β with A α ( A β } eig A α } eig A β } p Aβ (λ det (λi A β λ 2 eig A β } ( 3 3 +, } Since one eigenvalue is positive (and theefore has positive real part the system is unstable. As an alternative we could have applied the Routh criterion to the characteristic polynomial p A (λ λ 3 + λ 2 λ The necessary condition is not satisfied and therefore we can only assess that not all the eigenvalues have negative real part. Note that the Routh table has a row (with only one element equal to zero since the first two rows are linearly dependent. - - (N.B. There are rules to overcome this situation. Being the matrix A 2 upper triangular, the eigenvalues coincide with the elements on the diagonal λ, λ 2 3 and λ 3 3. The eigenvalue λ 3 3 makes the system unstable. Similarly, being A 3 lower triangular again the eigenvalues are the elements on the main diagonal and therefore, having λ 3 3, the system is unstable. Matrix A 4 is block triangular A 4 3 ( Aα A β (with matrix having the right dimensions and therefore we have ( } Aα eig eig A A α } eig A β } β Being p Aβ (λ λ 2 3λ + both roots (eigenvalues of p Aβ (λ have positive real part since the elements of the first column of the Routh table (which coincide for a second order equation with the polynomial coefficients change sign twice. The corresponding system is unstable. Note that A 5 is in the controller canonical form and therefore its characteristic polynomial is with Routh table p A5 (λ λ 3 + 2λ 2 + λ + 6

4 2 /6 The corresponding system is asymptotically stable. First note that A 6 is lower triangular and therefore the system has the double eigenvalue λ which prevents the system from being asymptotically stable and λ 2 3. Moreover A 6 is also block diagonal with ( A α In order to understand if this double eigenvalue in leads to instability or not, recall that the eigenvalues of a matrix coincide with those of its transpose eig A α } eig A T } α This can be shown with the similarity transformation T such that ( ( ( A T α T A α T T T with T T and therefore A α and A T α are similar and share the same eigenvalues. We can then note that A T α is a Jordan block of dimension 2 (index 2 for the eigenvalue λ which makes the system unstable. E Exercise 5 For the considered systems we have the following results. The system has a double pole is s and therefore it is not asymptotically stable. To see it is unstable we can either recall that an eigenvalue will appear as a pole with at most multiplicity equal to its index (dimension of the largest Jordan block. Therefore here we would have an index equal to 2 and this leads to instability. This is evident from the realization A (, B (, C (, D which shows the presence of the Jordan block. Alternatively, since the transfer function is the Laplace transform of the impulse response p (t, we have P (s s s 2 R s + R 2 s 2, with R, R 2 and the impulse response p (t L [P (s] (R + R 2 t δ (t shows the presence of the diverging natural mode tδ (t. Being the poles of P 2 (s equal to p and p 2, the system is marginally stable (or more properly Lyapunov stable. 7

5 The denominator of P 3 (s can be factored as s 3 + 2s 2 + 3s s(s 2 + 2s + 3 thus we have a pole in s and two poles with negative real part. marginally stable (or more properly Lyapunov stable. The system is therefore The Routh criterion for P 4 (s is satisfied therefore the system is asymptotically stable. The roots are (found numerically p.9862, p 2/3.69 ±.934j. 2 /6 The system is not asymptotically stable since the necessary condition is not satisfied. Building the Routh table shows that there is only one change of sign so one root with positive real part. numerically the roots are p.8297, p and p As a check, Routh criterion shows asymptotic stability. The roots are p.7549, p 2/3.226±.7449j. Being the Routh table 2 / the system with transfer function P 7 (s is unstable due to the presence of two poles with positive real part (2 sign variations in the first column. The poles are p, p 2/3.777 ±.365j and p 4/5.777 ±.8j. F Exercise 6 Being the matrix triangular p A (λ (λ kλ eigenvalues: λ, λ 2 k, if k ; λ λ 2, if k. 8

6 If k the natural modes are e λt e kt and e λ2t ; if k > the system is unstable while for k < we have marginal stability (or preferably Lyapunov stability. For the case k we have several equivalent options. With k matrix A ( A has an obvious Jordan block of dimension 2 (index 2 and therefore the zero eigenvalue leads to instability. The natural modes are e t and te t t. Equivalently, with k, the dynamics equations are ẋ Ax, x R 2, ẋ x 2 ẋ 2 The second equation has the constant solution x 2 (t x 2 ( and therefore x (t x 2 (t + x (. These are the components of the free (unforced, zero-input state evolution x zi (t e At x2 (t + x x( ( x 2 ( which clearly shows the diverging behavior for generic initial conditions. In the Laplace domain, note that L [ e At] (si A s s 2 s which can be expanded as (Heaviside (si A s ( + s 2 ( Being the matrices (residues independent from s, the inverse Laplace transform is ( ( e At δ (t + tδ (t Post-multiplication (right multiplication by the initial condition x( leads to the same unforced response previously found. G Exercise 7 The forced response transform is given by and therefore, being Y (s P (su(s U(s s 2 s 2 s2 2s + 2 s 2 (s we get Y (s s s 2 2s + 2 s + s 2 s2 2s + 2 (s s 2 R (s + s + R 2 s 2 + R 2 s + with R 4, R 2 2 and R 2 5. An interesting aspect of this example is that the diverging exponential component of the input e t is not present at the output (while the polynomial component is since the system has filtered this diverging forcing term through the presence of the zero in s in the transfer function P (s. 9

7 H Exercise 8 The forced response transform is given by Y (s P (su(s and therefore the only difficulty lies in finding U(s from known Laplace transform and properties. The input u(t is a truncated sinusoidal function with frequency Hz or 2π rad/s as shown in Fig. 5-A. As shown in Fig. 5-D, the input u(t can be seen as the result of a time shift of sec (Fig. 5-B to which an opposite and time shifted of 2 sec sinusoid (Fig. 5-C has been added and therefore u(t [sin(2π(t ] δ (t [sin(2π(t 2] δ (t 2 Defining y (t as the output corresponding to the input sin 2πt, the output y(t is given by y(t y (t δ (t y (t 2δ (t 2 We therefore just need to compute y (t as the inverse Laplace transform of 2π Y (s P (s s 2 + 4π 2 with P (s the transfer function of the given system (state space representation is in the controller canonical form P (s s + s 2 + 2s + s (s + 2 We have Y (s s (s + 2 2π s 2 + 4π 2 R s + 2πj + R s 2πj + R 2 s + + R 22 (s B C D A Figure 5: Ex. 8, input u(t with R [(s + 2πjY (s] s 2πj π(4π2 3 ( + 4π π2 2( + 4π 2 2 j [ d ( R 2 (s + 2 Y (s ] 2π((2π2 3 ds s ( + (2π 2 2 R 22 [ (s + 2 Y (s ] s 4π + (2π 2

8 and therefore, defining the residue R as R a + jb, we have R s + 2πj + which admits the inverse Laplace transform The overall y (t is then given by I Exercise 9 Ra st s 2πj 2a s s 2 + (2π 2 + 2b 2π s 2 + (2π 2 2a cos 2πt + 2b sin 2πt y (t 2a cos 2πt + 2b sin 2πt + R 2 e t + R 22 te t Being the matrix block diagonal, the eigenvalues are the union of the eigenvalues of ( A, A 2 2 ( that is λ, λ 2 i and λ 3 λ 2 i (the characteristic polynomial of A is p A(λ (λ (λ 2 +. The natural modes are therefore J Exercise e λ t e t, sin t (or equivalently cos t The characteristic polynomial is p A (λ λ 2 + 4λ + 3 (λ + (λ + 3 and therefore λ, λ 2 3. The eigenvector associated to λ is ( u or any parallel. the eigenvector associated to λ 2 3 is ( u 2 Choose U ( u u 2 so that U 2 ( ( v T v T 2 From the spectral form we obtain ( x zi (t e At x( e λt u v T + e λ2t u 2 v2 T x( ( e t ( /2 /2 + e 3t ( e t /2 /2 /2 /2 ( ( e t + e 3t ( + e 3t ( /2 /2 /2 /2 ( } ( 2 /2 /2 } ( 2

9 while y zi (t Ce At x( Cx zi (t e t + 3e 3t As an alternative, we can find the coefficients c v T x and c 2 v T 2 x which give the initial condition x in the (u, u 2 basis x( c u + c 2 u 2 u + u 2 and therefore, being vi T u j δ ij, c and c 2 scalars, ( e At x( e λt u v T + e λ2t u 2 v2 T x( ( e λt u v T + e λ2t u 2 v2 T (c u + c 2 u 2 c e λ t u + c 2 e λ 2t u 2 K Exercise The eigenvalues and associated eigenvectors are ( λ u 3, λ 2 u 2 The output zero-input response is therefore y zi (t Ce At x( C e λt u v T + e λt u 2 v2 T } e λt Cu v T + e λt Cu 2 v2 T x( } e λt Cu v T x( ( } x( being Cu 2 and therefore any initial condition solves the problem. This is understandable since Cu 2 implies that the diverging natural mode e λ 2t e t is unobservable. L Exercise 2 The eigenvalues and associated eigenvectors are ( λ u 2, λ 2 5 u 2 and therefore we have a constant natural mode (for λ and a diverging one (for λ 2 5. In order for the diverging natural mode not to compare in the output free response, the initial state needs to belong to the eigenspace relative to λ, that is ( α x( αu 2α Since λ the output response from x( not only will be non diverging but also constant. ( 3 2

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