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 2 3 4 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
u(t + +2 + t 2 3 4-2 -2 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 + 5 + 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. 4..8.6 u y.4.2 2 3 4 5 6 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
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
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
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 2-2 2-2 shows that there is only one change of sign so one root with positive real part. numerically the roots are p.8297, p 2.959 and p 3.9256. As a check, Routh criterion shows asymptotic stability. The roots are p.7549, p 2/3.226±.7449j. Being the Routh table 2 /2 2 3 4-2 4-3 4 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
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
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 + 2 2 3 4 5 B 2 3 4 5 C 2 3 4 5 D 2 3 4 5 A Figure 5: Ex. 8, input u(t with R [(s + 2πjY (s] s 2πj π(4π2 3 ( + 4π 2 2 + 2π2 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
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
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