Theory of Linear Systems Exercises. Luigi Palopoli and Daniele Fontanelli

Transcription

1 Theory of Linear Systems Exercises Luigi Palopoli and Daniele Fontanelli Dipartimento di Ingegneria e Scienza dell Informazione Università di Trento

2

3 Contents Chapter. Exercises on the Laplace Transform Chapter. Exercises on the Z-Transform 9 Chapter 3. Exercises on the State Space Representation 9 v

4

5 CHAPTER Exercises on the Laplace Transform Exercise Consider the following system expressed with the differential equation determine if it is BIBO stable. ÿ(t) = ẏ(t) 3y(t) u(t)+4 u(t), In order to determine the BIBO stability, we first compute the Laplace transform of the impulsive response imposing null initial conditions, i.e., that yields to s Y(s) = sy(s) 3Y(s) U(s)+4sU(s), G(s) = 4s s +s+3. Since the roots of the denominator are p = ±j, it follows that the system is BIBO stable. This fact can be proved by assuming u(t) = (t) and computing the forced response, i.e., 4s Y(s) = s(s++j )(s+ j ) = A s + A s++j + A 3 s+ j, where A = 3, A = +j3, A 3 = j3, where A 3 = A (the complex and conjugated). It then follows ( ) L (Y(s)) = 3 + +j3 e ( j )t + j3 e ( +j )t (t). By rewriting +j3 one has L (Y(s)) = ( 4+ 3 = Me jφ, with M =, φ = arctan 3 ), ( ) ( 3 +Me t e j( t φ) +Me t e j( t φ) (t) = 3 +Me t cos( ) t φ) (t), which is of course bounded.

6 . EXERCISES ON THE LAPLACE TRANSFORM Exercise Consider the following system expressed with the differential equation ÿ(t) = ẏ(t) 3y(t) u(t)+4 u(t), determine its unforced response given ẏ(0) = and y(0) =. To compute the unforced we need first to compute the Laplace transform of the system, that is s Y(s) sy(0) ẏ(0) = sy(s)+y(0) 3Y(s) U(s)+4sU(s) 4u(0), that yields to Y(s) = 4s s +s+3 U(s) 4 (s+)y(0)+ẏ(0) s u(0)+ +s+3 s. +s+3 Since we are interested in the unforced response, we have Y(s) = (s+)y(0)+ẏ(0) s +s+3 where = (s+)+ s +s+3 = A s++j + A s+ j, A =, A =, where again A = A. It then follows that L (Y(s)) = ( ) e ( j )t +e ( +j )t (t). Hence L (Y(s)) = e t( e j t +e j t ) (t) = e t cos( t)(t).

7 . EXERCISES ON THE LAPLACE TRANSFORM 3 Exercise 3 Consider the following system expressed with the differential equation ÿ(t) = 5 3ẏ(t) 4 y(t) 3 u(t)+ 4 3 u(t), determine its response to the input u(t) = k(t) given ẏ(0) = 0, y(0) = and u(0) =. To compute the response we need first to compute the Laplace transform of the system, that is 3s Y(s) 3sy(0) 3ẏ(0) = 5sY(s)+5y(0) 3 4 Y(s) U(s)+4sU(s) 4u(0), that yields to 4s 4 Y(s) = 3s +5s+ 3 U(s) 4 3s +5s+ 3 4 that turns to Y(s) = k(4s ) s(3s +5s+ 3 4 ) 4 3s +5s+ 3 4 u(0)+ (3s+5)y(0)+3ẏ(0) 3s +5s s+5 + 3s +5s+ 3, 4 i.e., the superposition of three different components. Therefore: For the first, we have where Y (s) = It then follows k(4s ) s(3s +5s+ 3 4 ) = A s + A s+ 6 A = k 4 3, A = k 5, A 3 = k 7 6. L (Y (s)) = k + A 3 s+ 3, ( e 6 t 76 e 3 t ) (t). For the second term, related to the initial condition of the input, we have 4 Y (s) = 3s +5s+ 3 = A 4 4 s+ + A 5 6 s+ 3, where It then follows L (Y (s)) = A 4 =, A 5 =. ( e 6 t +e 3 t) (t).

8 4. EXERCISES ON THE LAPLACE TRANSFORM For the third term, related to the initial condition of the output, we have 3s+5 Y 3 (s) = 3s +5s+ 3 = A 6 4 s+ + A 7 6 s+ 3, where It then follows L (Y 3 (s)) = A 6 = 9 8, A 7 = 8. ( 9 8 e 6 t ) 8 e 3 t (t). Therefore, we finally have by the superposition principle: [ y(t) = 4k 0k + + e 6 t 8k e 3 ](t). t 3 8 4

9 . EXERCISES ON THE LAPLACE TRANSFORM 5 Exercise 4 Consider the following system G(s) = 3s+7 s s 3 and determine if it is BIBO unstable. Since the coefficients of the polynomial at the denominator changes in sign, the system is BIBO unstable since it has one root with positive real part.

10 6. EXERCISES ON THE LAPLACE TRANSFORM Exercise 5 Consider the following Laplace transform of two transfer functions G (s) = 3s 7 s +s+3 s+ G (s) = s(s +s+3) determine if they are BIBO stable. Both of them are not unstable. However, the first one has two roots with negative real part (persistency of the sings of the polynomial), while the second has the same two roots plus a pole in the origin. Therefore, only the first one is BIBO stable. To verify this fact, it is sufficient to apply a unitary step as input and then compute the inverse Laplace transform of the output.

11 . EXERCISES ON THE LAPLACE TRANSFORM 7 Exercise 6 Given a system whose transfer function is s 5 G(s) = (s+) (s+) determine the time evolution of the output when the input is u(t) = e 3t + (t ). One way to takle this problem is to first compute the Laplace transform of the input, which is given by U(s) = s+3 + e s s, and then apply the inverse Laplace transform to the output Y(s) = s 5 (s+) (s+)(s+3) + (s 5)e s s(s+) (s+). Another, quite similar approach, would be to compute the inverse Laplace transform of the two output functions as they are applied at time t = 0, i.e., ( ) y (t) = L s 5 (s+), (s+)(s+3) ( ) y (t) = L s 5 s(s+), (s+) and then define y(t) = y (t)+y (t ).

12

13 CHAPTER Exercises on the Z-Transform Exercise Given the following Z-Transform of an input sequence U(z) = +3z z 5 +z 7, find the corresponding signal representation Straightforwardly, we apply the inverse Z-Transform of the Kronecker delta function to find: u(t) = δ(t)+3δ(t ) δ(t 5)+δ(t 7). 9

14 0. EXERCISES ON THE Z-TRANSFORM Exercise Given two sequences find y(t) = h(t) u(t). u(t) = (t) (t 4) and h(t) = (t) (t 5), We first recall that Then, we noticed that that yields Similarly y(t) = h(t) u(t) = Z (H(z)U(z)). u(t) = δ(t)+δ(t )+δ(t )+δ(t 3) U(z) = +z +z +z 3. h(t) = δ(t)+δ(t )+δ(t )+δ(t 3)+δ(t 4) H(z) = +z +z +z 3 +z 4. As a consequence: and then Y(z) = +z +3z +4z 3 +4z 4 +3z 5 +z 6 +z 7, y(t) = δ(t)+δ(t )+3δ(t )+4δ(t 3)+4δ(t 4)+3δ(t 5)+δ(t 6)+δ(t 7).

15 . EXERCISES ON THE Z-TRANSFORM Exercise 3 FindtheinverseZ-TransformwithROC z > forthefollowingtransfer function G(z) = +z +z 3 z + z. Furthermore, determine if the system is BIBO stable. We first rewrite G(z) in the more suitable form G(z) = z +z + z 3 z +. Being the roots of the denominator equal to z = and z =, we have where It then follows that and finally G(z) z = A z + A z + A 3 z, A =, A = 9, A 3 = 8. Z (G(z)) = Z () 9Z ( Z (G(z)) = δ(t)+ [ z z 8 9 ) ( ) z +8Z z ( ) ] t (t). Obviously, the system is BIBO unstable since z =. To further prove it, we compute the output of the system when the input u(t) is a unitary step, i.e., which yields to with Y(z) = G(z)U(z) = z +z + z 3 z + Y(z) z Z (Y(z)) = 9Z ( and then = A z z z y(t) = [ ) z z, + A 3 z + A 4 (z ), ( ( ) z 8Z )+8Z z z (z ) 8t 8+9 ( ) ] t (t).

16 . EXERCISES ON THE Z-TRANSFORM Exercise 4 Consider the following difference equation representing the output y(t) of a system subjected to the input u(t) y(t+) = 3u(t+) u(t+)+u(t)+ 5 6 y(t+) 6 y(t) compute the impulse response of the system for null initial conditions. Since the initial conditions are null, we have that y(0) = y() = u(0) = u() = 0. Therefore the Z-Transform is given by which yields to z Y(z) = 3z U(z) zu(z)+u(z)+ 5 6 zy(z) 6 Y(z), Y(z) = 3z z + z 5 6 z + U(z) = 3z z + ( )( 6 z z )U(z). 3 The impulse response is then given by Hence, where It then follows that and finally H(z) = 3z z + ( )( z z ). 3 ( ) H(z) Z = A z z + A z A = 6, A = 9, A 3 =. Z (H(z)) = Z (6)+9Z ( Z (H(z)) = 6δ(t)+ [ 9 z z + A 3 z, 3 ) Z ( z z 3 ( ) t ( ) ] t (t). 3 Comparison Inordertoprovethecorrectnessoftheanalysis, wecancomputethetime evolution of the output y(t) when the input is a Kronecker delta u(t) = δ(t) using the difference equation and then compare it with the time evolution Z (H(z)) computed for every t, as reported in Table. As shown in the Table, the correctness of the analysis carried out is further verified. )

17 . EXERCISES ON THE Z-TRANSFORM 3 t y(t) Z (H(z)) t 0 3δ(t) = 3 6δ(t)+[9 ] = 3 δ(t)+ 5 6 y(0) = [ 9 ] 3 = δ(t)+ 5 6 y() 9 6y(0) = 4 9 = y() y() = = Table. Comparison of the two impulse responses.

18 4. EXERCISES ON THE Z-TRANSFORM Exercise 5 Consider the following difference equation representing the output y(t) of a system subjected to the input u(t) y(t) = 3u(t) u(t )+u(t )+ 5 6 y(t ) 6 y(t ) compute the forced response of the system for ( ) t u(t) = (t) 4 The forced response is computed for null initial conditions, i.e., y( ) = y( ) = 0. Therefore the Z-Transform is given by Y(z) = 3U(z) z U(z)+z U(z)+ 5 6 z Y(z) 6 z Y(z), which yields to Hence, Y(z) = 3z z + z 5 6 z + U(z) = 6 ( ) Y(z) Z = A z z ( z z(3z z +) )( z 3 + A z 3 + A 3 z, 4 )( z 4 where A = 8, A = 48, A 3 = 33. It then follows that ( ( ( ) Z (Y(z)) = 8Z z ) 48Z z )+33Z z z z 3 z 4 and finally Z (Y(z)) = [ 8 ( ) t 48 ( ) t ). ( ) ] t (t). 4 Comparison In order to prove the correctness of the analysis, we can compute the time evolution of the output y(t) when the input is u(t) using the difference equation and then compare it with the time evolution Z (Y(z)) computed for every t, as reported in Table. As shown in the Table, the correctness of the analysis carried out is further verified.

19 . EXERCISES ON THE Z-TRANSFORM 5 t y(t) Z (H(z)) t 0 3 [ ] = y(0) = 5 [ ] 4 = y() 6 y(0) = = Table. Comparison of the two vanishing step responses.

20 6. EXERCISES ON THE Z-TRANSFORM Exercise 6 Consider the following difference equation representing the output y(t) of a system subjected to the input u(t) y(t+) = 3u(t+) u(t+)+u(t)+ 5 6 y(t+) 6 y(t) compute the unforced response of the system for y() = 0 and y(0) =. The Z-Transform is given by z Y(z) z y(0) zy() = 3z U(z) z U(z)+U(z)+ 5 6 zy(z) 5 6 zy(0) 6 z Y(z), which yields to Y(z) = (3z z +) ( )( z z )U(z)+ z(z 5 ( 6 )y(0)+zy() )( ) 3 z z. 3 Hence (U(z) = 0), Y(z) = z(z 5 ( )( 6 ) ) z z 3 and, therefore where It then follows that and finally ( ) Y(z) Z = A z z Z (Y(z)) = Z ( Z (Y(z)) = A =, A = 3. [ z z + A z, 3 ) 3Z ( z z 3 ( ) t ( ) ] t 3 (t). 3 Comparison In order to prove the correctness of the analysis, we can compute the time evolution of the output y(t) when the input is u(t) using the difference equation and then compare it with the time evolution Z (Y(z)) computed for every t, as reported in Table 3. As shown in the Table, the correctness of the analysis carried out is further verified. )

21 . EXERCISES ON THE Z-TRANSFORM 7 t y(t) Z (H(z)) t 0 [ 3] = 0 [ ] = y() 6 y(0) = y() 6 y() = = = Table 3. Comparison of the responses for the selected initial conditions.

22

23 CHAPTER 3 Exercises on the State Space Representation Exercise Given a continuous time linear system [ ] [ 0 ẋ(t) = x(t)+ u(t) = Ax(t)+Bu(t), 0 0 ] compute the unforced response for a generic x(0) and verify if the system is BIBO stable. The unforced response is given by x u (t) = e At x(0) = Te Jt T x(0), where T is the transformation matrix that transforms the dynamic matrix A in the Jordan form J. Therefore, T is a base of generalised eigenvectors of the matrix A. However, it is evident that the matrix A has two eigenvalues (i.e., it is upper triangular), that are λ = and λ = 0. It is then clear that [ ] 0 J =. 0 0 The transformation matrix T = [v,v ], where v i = [v i,,v i, ] T, is then obtained by solving the eigenvector problem, i.e., { v +v = v Av i = λ i v i For the first eigenvector v, we have while for the second v, we have v = v = 0 = v { v +v = 0 0 = 0 [ α 0], [ β β], 9

24 0 3. EXERCISES ON THE STATE SPACE REPRESENTATION where α R and β R, and α 0 and β 0. Therefore, a possible choice for T is [ ] [ ] T = T 0 =. 0 It then follows that for a generic x(0) = [x (0),x (0)] T, we have [ ][ ][ ][ ] [ x u (t) = Te Jt T e t 0 x (0) x (0)+e x(0) = = t ] (x (0) x (0)) x (0) x (0) For what concerns the BIBO stability, it is immediate to notice that the system has an eigenvalue in zero, so it is marginally stable. Therefore, it has a real pole in the origin, that corresponds to an integrator. It finally yields that the system is not BIBO stable.

25 3. EXERCISES ON THE STATE SPACE REPRESENTATION Exercise Given a continuous time linear system in Jordan form whose unforced response for the initial condition x(0) = [,,0,3] T is x u (t) = e Jt x(0) = cos(t)+sin(t) cos(t) + sin(t) 3te t 3e t determine the explicit expression of the matrix J., We first recall that a matrix in Jordan form is block diagonal. We then recall that in order to have sinusoidal functions in the expression of x u (t), at least two complex and conjugated eigenvalues are needed. Since the state has dimension 4 and the sinusoidal functions comes without the presence of the time t, we conclude that we have only two complex and conjugated eigenvalues that has geometric and algebraic multiplicity equals to one (i.e., eigenvalues λ = σ + jω and λ = σ jω). Hence, the submatrix associated to such eigenvalues is diagonalisable. On the contrary, the other two eigenvalues are coincident, with geometric multiplicity equals to one (since we have the explicit presence of the time t that multiplies the exponential function). To summarise, we have two Jordan blocks, one associated to a pair of complex and conjugated eigenvalues, the other with only one mini-block (hence associated to a chain of generalised eigenvalues whose length is two). It then follows that e Jt = [ ] e J t 0 0 e J t = e σt cos(ωt) e σt sin(ωt) 0 0 e σt sin(ωt) e σt cos(ωt) e λ 3t te λ 3t e λ 3t Therefore, by recalling the initial condition x(0) and the expression of x u (t), it follows that σ = 0, ω = and λ 3 = and, then [ ] J 0 J = = J

26 3. EXERCISES ON THE STATE SPACE REPRESENTATION Exercise 3 Given the following continuous time linear invariant system [ ] [ ẋ(t) = x(t)+ u(t) = Ax(t)+Bu(t), 4 3 ] compute the response of the system assuming an initial condition x(0) = [0,] T and an input u(t) = α(t t 0 ), where t 0 > 0. To solve this problem we have to recall the explicit form of the response of a linear continuous time system, i.e., x(t) = e At x(0)+ t 0 e A(t τ) Bu(τ)dτ. Due to the particular form of the input, we have x(t) = e At x(0)+ t t 0 e A(t τ) Bu(τ)dτ = e At x(0)+α t t 0 e A(t τ) dτb, where the second equation is obtained by noticing that both B and u(t) are constant for t t 0. We then recall that the exponential of the matrix can be readily obtained using the Jordan form, i.e., x(t) = Te Jt T x(0)+α = Te Jt T x(0)+αt t Te J(t τ) T dτb = t 0. t t 0 e J(t τ) dτt B It is then evident that it is necessary to compute the Jordan form J of the matrix A to solve the described problem. To this end, let us first compute the eigenvalues, which are det(λi A) = 0 λ +λ+ = 0 λ = λ =, so an eigenvalue with algebraic multiplicity equals to. Let us check the geometric multiplicity by tackling the eigenvector problem { vi v i = v i Av i = λv i 4v i 3v i = v i whichyieldstoonlyoneconstraintv i = v i, hencetheequationissolvedby eigenvectors linearly dependent with v = [,] T. Therefore, the geometric multiplicity is one, so the matrix is defective. It follows that the second generalised eigenvector will solve (A λi)v () = v () = v Av () = λv () +v () v () v () = v () + 4v () 3v () = v () +

27 3. EXERCISES ON THE STATE SPACE REPRESENTATION 3 which yields to v () = v (), therefore a solution like v () = [,] T is linear independent to v () = [,] T. We finally have that T = [ ] T = [ with [ ] [ J = T AT = e 0 Jt e t te = t ] 0 e t, that is one Jordan block with an associated mini block. Finally, the unforced response is given by [ x u (t) = Te Jt T te x(0) = t ] e t te t, while the forced response is t [ t x f (t) = αt e J(t τ) dτt B = αt t 0 e τ t t ] dτ t 0 (t τ)e τ t dτ t t 0 0 t 0 e τ t T B = dτ [ e t 0 t (t = αt 0 )e t0 t t(e t0 t ] ) t+ 0 e t 0 t T B = [ α(t+e t 0 = t +t(e t0 t ) e t0 t ] (t 0 ) ) α(t+e t0 t +t(e t0 t ) e t0 t (t 0 ) 3) Finally, we have x(t) = x u (t)+x f (t). ],

28 4 3. EXERCISES ON THE STATE SPACE REPRESENTATION Exercise 4 Given a discrete time linear system [ ] [ x(k+) = 0 0 x(k)+ u(k) = Ax(k)+Bu(k), 3 ] compute the unforced response for a generic x(0) and the steady state value (if any). The unforced response is given by x u (k) = A k x(0) = TJ k T x(0), where T is the transformation matrix that transforms the dynamic matrix A in the Jordan form J. Therefore, T is a base of generalised eigenvectors of the matrix A. However, it is evident that the matrix A has two eigenvalues (i.e., it is lower triangular), that are λ = and λ =. It is then clear that [ J = 0 0 The transformation matrix T = [v,v ], where v i = [v i,,v i, ] T, is then obtained by solving the eigenvector problem, i.e., v = v Av i = λ i v i For the first eigenvector v, we have while for the second v, we have ]. 3v +v = v v = v 3v +v = v v = v = [ ] α, α [ 0 β], where α R and β R, and α 0 and β 0. Therefore, a possible choice for T is [ ] [ ] 0 T = T 0 =. It then follows that for a generic x(0) = [x (0),x (0)] T, we have [ ][ x u (k) = TJ k T 0 x(0) = k ][ ][ ] [ ] 0 0 x (0) = k x (0) 0 x (0) ( k. )x (0)+x (0)

29 3. EXERCISES ON THE STATE SPACE REPRESENTATION 5 For the steady state value, we have just to compute [ ] lim x 0 u(k) =. k + x (0)+x (0)

30 6 3. EXERCISES ON THE STATE SPACE REPRESENTATION Exercise 5 Given a discrete time linear system in Jordan form whose unforced response for the initial condition x(0) = [,,0,3] T is k x u (k) = J k x(0) = k cos(k π ) k sin(k π ), 3 k 4 determine the explicit expression of the matrix J. We first recall that a matrix in Jordan form is block diagonal. We then recall that in order to have sinusoidal functions in the expression of x u (k), at least two complex and conjugated eigenvalues are needed. Since the state has dimension 4 and the sinusoidal functions comes without a multiplication by the time k, we conclude that we have only two complex and conjugated eigenvalues that has geometric and algebraic multiplicity equals to one (i.e., eigenvalues λ = σ+jω and λ 3 = σ jω). Hence, the sub-matrix associated to such eigenvalues is diagonalisable. By the particular form of the other two entries, we conclude that there are two additional and distinct real eigenvalues (λ and λ 4 ). To summarise, we have three Jordan blocks, one associated to a pair of complex and conjugated eigenvalues, and the other two associated with two distinct mini-blocks (in other words, the matrix is diagonalisable). It then follows that J k J k 0 0 = 0 J k 0 = 0 0 J3 k λ k ρ k cos(kθ) ρ k sin(kθ) 0 0 ρ k sin(kθ) ρ k cos(kθ) λ k 4 where ρ = λ = σ +ω and θ = arctan( Imag(λ ) Real(λ ) ) = arctan(ω σ ). Therefore, by recalling the initial condition x(0) and the expression of x u (k), it follows that σ = 0, ω =, λ = and λ 4 = 4, then J = J J 0 = 0 cos( π ) sin(π ) sin( 0 0 J π ) cos(π ) 0 = ,

31 3. EXERCISES ON THE STATE SPACE REPRESENTATION 7 Exercise 6 Given the following discrete time linear invariant system 3 x(k+) = 4 4 x(k)+ 0 0 u(k) = Ax(k)+Bu(k), 0 0 compute the response of the system assuming an initial condition x(0) = [,, ] T and an input u(k) = α(k k 0 ), with k 0 Z and k 0 > 0. To solve this problem we have to recall the explicit form of the response of a linear discrete time system, i.e., k x(k) = A k x(0)+ A k i Bu(i). i=0 Due to the particular form of the input, we have k x(k) = A k x(0)+ A k i Bu(i) = A k x(0)+α for k > k 0 and i=0 k k x(k) = A k x(0)+ A k i Bu(i) = A k x(0), i=0 i=k 0 A k i B, for k k 0. We then recall that the power of the matrix can be readily obtained using the Jordan form, i.e., k x(k) = TJ k T x(0)+α TJ k i T B = i=k 0. k = Te Jt T x(0)+αt T B i=k 0 J k i It is then evident that it is necessary to compute the Jordan form J of the matrix A to solve the described problem. To this end, let us first compute the eigenvalues, which are det(λi A) = 0 λ 3 3λ +3λ = 0 λ = λ = λ 3 = λ =, so an eigenvalue with algebraic multiplicity equal to 3. In other words, the Jordan form will have only one Jordan block. Let us check the geometric multiplicity by tackling the eigenvector problem 3v i v i v i3 = v i Av i = λv i 4v i v i 4v i3 = v i v i3 = v i3

32 8 3. EXERCISES ON THE STATE SPACE REPRESENTATION which yields to only one constraint v i = v i v i3, hence the equation is solved by linearly independent eigenvectors, for example (3.) v = η and v = β 0, 0 β or the more straightforward (3.) v = η η and v = 0 β. 0 β η In both cases, η,β R and η 0 and β 0. It is then obvious that the geometric multiplicity is, which is less than 3, hence we have two Jordan mini-blocks of dimensions and, respectively, associated to the eigenvalue λ. We can then readily define the Jordan form J of the matrix A as J = T AT = In order to fully characterise the Jordan form we need also to define the transformation matrix T, which is formed by a basis of generalised eigenvectors. To this end, let us first define v () = v and v () = v, making the choice reported in (3.). Then, (A λi)v () = v () v () v() v() 3 = η 4v () v() 4v() 3 = η 0 = 0 which yields to v () = v() v() 3 η. Following the same steps of the eigenvector, in this particular case we have two straightforward solutions v = γ γ η and v = 0 δ, 0 δ or v = γ 0 γ and v = δ η, 0 δ where γ,δ R and γ 0 and δ 0. For the first choice, v is linearly dependent to v (), hence v() = v. For the second choice, v is linearly dependent to v (), hence v() = v. Notice that if one tries to continue the chain by computing v (3), it will obtain from the first choice η = 0, which violates the hypothesis, and from the second choice δ = 0, which again violates the hypothesis. So the chain comprises only v () and v () (se the mini-block of dimension ).

33 3. EXERCISES ON THE STATE SPACE REPRESENTATION 9 Finally, by computing v () starting from v (), one immediately has that β = 0, which violates the hypothesis. It then follows that the matrix T = [v (),v(),v() ] has two possible values T = η γ 0 η γ η β and T = η 0 0 η δ η β, 0 0 β 0 δ β that, by simply choosing η = β = γ = δ =, yields to T = 0 and T =, and T = and T = Whatever is the choice made (i.e., T = T or T = T ), we have J = T AT = J k = k 0 0 0, that is one Jordan block with two mini blocks. Notice that the same results would be obtained if the choice of (3.) would be selected, instead (prove it on your own). Finally, the unforced response is given by x u (k) = TJ k T x(0) = +k +4k, while the forced response is k x f (k) = αt T B = = αt = αk αk. α J k i i=k 0 k k 0 Finally, we have λ k i k k 0 (k i )λ k i 0 0 k k 0 λ k i x(t) = x u (t)+x f (t). k k 0 λ k i T B =

34