Some solutions of the written exam of January 27th, 2014

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1 TEORIA DEI SISTEMI Systems Theory) Prof. C. Manes, Prof. A. Germani Some solutions of the written exam of January 7th, 0 Problem. Consider a feedback control system with unit feedback gain, with the following transfer function in open-loop Ws) = K s+) s +).. Draw the amplitude and phase Bode diagrams, and the polar diagram for K = ;. Compute the denominator of the closed-loop transfer function; 3. Compute the number of poles with negative real part of the closed loop transfer function as a function of the gain K, ), using both the Nyquist criterion and the Routh criterion. Solution of problem. Let Ws) denote Ws) for K = : Ws) = s+) s +) We consider the Bode plots and the polar plot of Ws). In Bode form we have Ws) = K Ws) = K ) ) +s) + s The low frequency gain is K 0 = lim s 0 Ws) = K 0 db = 0log 0 K 0 = 0log 0 = 0log 0 = 6 db. Thus Wjω) = +jω +jω/) = +jω ω / Wjω) = +jω ω /. Note that the term ω / is real, and therefore its phase is 0 when ω < because ω / > 0) and is π when ω > because ω / < 0). the Bode plots and the Nyquist plot of the open loop transfer function are in the enclosed file). The closed-loop transfer function is s+) s +) W CL s) = K Ws) +K Ws) = K +K s+) s +) = K s+) s +) K and the denominator of W CL s) is: d CL s) = s+) s +) K = s +s+)s +) K = s +s 3 +5s +8s+ K NYQUIST ANALYSIS Let N be the number of times that the Nyquist plot of Wjω) encircles the point in the counterclockwise direction. From the plot it is clear that For K 0,) we have N = 0: the Nyquist plot does not encircle the point ; For K > we have N = : the Nyquist plot encircles one time the point in the clockwise negative) direction; For K < 0 we have N = : the Nyquist plot encircles two times the point in the clockwise negative) direction;

2 Recall the Nyquist formula in the form p CL = p OL N where, p CL is the number of poles with positive real part of the Closed Loop CL) system, and p OL is the number of poles with positive real part of the Open Loop OL) system. Since for the given Ws) we have p OL = 0 there isn t any unstable pole in the open-loop system) we have p CL = N, and therefore For K 0,) we have N = 0, and thus p CL = 0 stable closed loop system); For K > we have N =, and thus p CL = unstable closed loop system); For K < 0 we have N =, and thus p CL = unstable closed loop system). Note that for K = the denominator of W CL s) is 0 for s = 0, and thus s = 0 is a pole of W CL s). ROUTH ANALYSIS The case K = 0 will be not analyzed because it corresponds to the trivial case where the open-loop transfer function is zero. The characteristic polynomial of the closed-loop system is the denominator of W CL s): d CL s) = s +s 3 +5s +8s+ K The first two rows rows and 3) of the Routh table are: 5 K 3 8 The computation of the elements in the third row row number ) gives a, = 5 8 = 8 0 =, a, = -K 0 = K) = K Thus we have 5 K 3 8 K The computation of the element in the fourth row row number ) gives a, = 8 K = K) 8 = K, and the last element, a 0, is K 5 K 3 8 K K 0 K Analyzing the signs of the first column we have: For K 0,) we have no sign variation, so that p CL = 0; For K > we have one sign variation 0), so that p CL = ; For K < we have two sign variation and 0) so that p CL =. For the particular case of K =, the characteristic polynomial of the closed-loop system is d CL s) == s +s 3 +5s +8s = ss 3 +s +5s+8), thus, we have a pole in the origin. The sign of the remaining poles can be studied again by means of the Routh criterion:

3 There is no sign variation in the first column, and therefore the remaining poles have all negative real part simple stability of the origin of the state space). These results coincide with those obtained using the Nyquist analysis.

4 Problem. Given the following discrete-time system [ ] xt+) = Axt)+But) dove A = yt) = Cxt), discuss the properties of the natural modes;. compute the state-transition matrix A t ; 3. compute the input-output impulse response and transfer function B = [ ] 0 C = [ 0 ] Solution of problem. Let s start computing the eigenvalues of A, which are the roots of the characteristic polynomial of A. The computation of the characteristic polynomial gives A = [ ] λi A = λ ) + ) The eigenvalues of A are computed solving λi A = 0. Instead of using the classical formula for second-degree equations, we can proceed as follows λ ) + ) = 0 From this we have the two complex-conjugates eigenvalues: λ ) = λ ) = ±. λ = +j, λ = j We denote r C and l C the right and left eigenvalues associated to λ. Their computation is made by solving the two systems of equation λ I A)r = 0, l λ I A) = 0, and taking, among the infinite solutions, one solution such that l r =. The right and left eigenvalues associated to λ are the conjugates of r and l so that λ = λ, r = r and l = l ). We choose the following set of eigenvalues λ = +j r = [ ], l j = [ ] j, λ = j r = [ ], l j = [ ] j. check that l r = l r = and l r = l r = 0). The properties of the natural modes to be discussed are the following: - Stability, Eccitability, Observability. Both modes are asymptotically stable, because they satisfy the stability condition λ k < note that the system is a discrete-time system). In particular ) λ = λ = + ) = 8 = < Moreover, both natural modes are eccitable because l B = j/ 0 and l B = j/ 0, and are observable because Cr = Cr = 0. The transition matrix can be computed using the spectral decomposition A t = λ t r l +λ t r l = R λ t r l ) The expression A t = R +j ) t [ ] [ ]) j = R j +j ) t [ ]) j j is not useful if we can not expand the power +j ) t. Thus, it is important to write λ using the magnitude and phase form Euler formula) λ = λ e j λ = λ cos λ )+j sin λ ))

5 so that λ t = λ t e j λ t = λ t cos λ t)+j sin λ t)) The magnitude and phase of λ = +j are λ = / 8 and λ = arctan) = π. Thus ) t ) t λ t = e j π t = 8 cos π ) 8 t)+j sinπ t) Thus the transition matrix A t becomes ) t A t = R cos π ) [ ] ) j 8 t)+j sinπ t) j ) t = R cos π )[ ]) j 8 t)+j sinπ t). j Taking the real part we have ) [ t cos π A t = t) sin π t) ] 8 sin π t) cos π t) The impulse response is w0) = D and wt) = CA t B for t > 0. Then we have w0) = 0 and fort > 0 wt) = [ 0 ] ) [ t cos π 8 t )) sin π t )) ] [0 ] ) t π ) sin π t )) cos π t )) = sin 8 t ). The transfer function Wz) can be computed in two ways: ) by taking the Z-transform of the impulse response wt); ) by using the formula Wz) = CzI A) B +D, with D = 0. Using the second method we easily get Wz) = z ) + = z ) + ) [ 0 ] [ z z ) ] [0 ] As an exercise you can compute Wz) by transforming wt). Of course, you should obtain the same result! I suggest to develop the Z-transform of a generic function ft) = ca t sin bt ) and then to apply the delay theorem: Consider a function ft), t 0, and its Z-transform Fz) = Zft)). Let gt) = ft ). Then, the Z-transform of gt) is such that Gz) = z Fz). When ft) = ca t sin bt ), then gt) = ca t sin bt ) ). Notice that the impulse response wt) has the same structure of gt) when c =, a = / 8 and b = π/. Thus, the computation of Wz) from Gz) and from Fz) is straightforward. The computation of the Z-transform Fz) is easy if we rewrite the sinus in the exponential form: ft) = ca t e jbt e jbt )/j, so that ft) = c ae jb ) t ae jb ) t) /j. The only transformation we need to remember is Zp t ) = z/z p). From this, Z ft) ) = c ) z j z ae z jb z ae jb Replacing c =, a = / 8 and b = π/, and noting that ae jb = / 8)e jπ/ = / 8)+j)/ = +j)/, after few computation we get the expression for Wz).

6 Problem 3. Consider a continuous-time system characterized by the following impulse response: wt) = 3e t. Compute the forced response and the harmonic response to the input ut) = 5 sint). Solution of problem 3. Both responses are easily computed in the Laplace domain. First of all we need to compute the transfer function of the system by transforming the impulse response: Ws) = L wt) ). Recalling that L e at) = /s a) we get Ws) = 3 s+. Forced response. The forced response in the Laplace domain is easily computed as Ys) = Ws)Us), where Us) is the Laplace transform of the input function. Recalling that L c sinωt) ) = cω/s +ω ), we have and then Ys) = Ws)Us) = Us) = 0 s + 30 s+)s +) = 30 s+)s j)s+j) Now must compute the partial fraction decomposition of Ys), and we can choose one of the two forms: or Ys) = R s+) + R s j) + R 3 s+j) Ys) = R s+) + as+b s +) recall that R 3 = R )...to be continued... the back-transformation of Ys) gives the forced response yt) =......to be computed... Harmonic response. The harmonic response exists if an only if the system is asymptotically stable or at least the controllable and observable subsystem should be asymptotically stable). This means that all poles of the transfer function that are also the eigenvalues of the controllable and observable subsystem), must have negative part. Since this condition is verified in this problem the pole is -), then the harmonic response exists and we can proceed in its computation. The general formula for the harmonic response of continuous-time systems to an input of the form ut) = c sinωt) is y h t) = c Wjω) sinωt+ Wjω) ). In our problem Ws) = 3/s+), and ω =. Thus we need to compute Wj) and Wj) ): Wjω) = 3 Wjω) = jω + = 3 ω + 3 ω = 3 jω + = jω + = arctan. jω + ) From these, when ω = we have 3 Wj) = = 3 ) +, Wj) = arctan = π. Thus, the harmonic response to the input ut) = 5 sint) is y h t) = 5 Wj) sint+ Wj) ) = 5 sint π/)

7 Problem. Given the system xt+) = Axt)+But), yt) = Cxt) with matrices 0 0 A = B = 0 C = [ 0 0 ]. Find a basis for the space of reachable states, and a basis for the space of unobservable states;. Find the subspaces X, X, X 3 and X of the Kalman structural decomposition; 3. Find an input sequence that brings the state from x0) = [0 0 0] T to x a = [ 0] T.. Find an input sequence that brings the state from x0) = [0 0 0] T to x b = [ 0 ] T. Solution of problem. The computation of the reachability matrix gives P 3 = [ B AB A B ] 0 0 = 0 0 P = RP 3 ) the range of P 3 ) is the space of reachable states. Note that only the first two columns of P 3 are independent, so the rank of P 3 is, and is the dimension of P. The first two columns of P 3 are two admissible basis vectors: 0 P = R[B AB A B]) = R[B AB]) = span 0, 0, v = 0, v = The computation of the observability matrix gives C 0 0 Q 3 = CA = CA 0 0 Q = NQ 3 ) the null-space of Q 3 ) is the space of unobservable states. Note that only the first two rows of Q 3 are independent, so the rank of Q 3 is, and is the dimension of Q is the rank drop of Q 3 ). Thus, we need to find one basis vector for Q, that is a solution of the homogeneous system Q 3 b = 0 3. Observing that the first two columns of Q 3 are equal, we have the following solution b = [ 0] T, which is a basis for Q. Thus C Q = N CA = N CA [ ]) C CA = span, b = 0 Now we are ready to find the four subspaces X, X, X 3 and X. Let s start with X = Q P. Being Q a one-dimensional space and P a two-dimensional space, the intersections is either 0 or Q itself. Thus, we must only check if b spanv,v ) or not. This can be made by checking whether rank[v v ]) = rank[v v b]) or not. It is easy to check that 0 0 rank 0 = rank 0 0 In other words, b is linearly dependent on v,v it is clear that b = v +v ). Thus, b is a basis for X. The space X is any space such that X X = P. We can choose either X = spanv ) or X = spanv ), because we have P = span[b v )], and also P = span[b v )]. Let s choose X = spanv ), so that a basis for X is v = [0 0 ] T. The space X 3 is any space such that X X 3 = Q. However, since we have X = Q, then X 3 =. The space X is any space such that X X X 3 X = C 3. Then, we only need to find a vector linearly independent of b,v. Let s choose X = spand), with d = [ 0 0] T, such that X =spanb), rank [ b v d ]) 0 X = rank 0 0 =spanv ), = 3, and then 0 0 X 3 =, X =spand).. 0.

8 Note that other choices for X and X are possible. Question. and.: - Find an input sequence that brings the state from x0) = [0 0 0] T to x a = [ 0] T ; - Find an input sequence that brings the state from x0) = [0 0 0] T to x b = [ 0 ] T. Note that x a P, and therefore we know that there exists an input sequence that brings x0) = 0 3 to x a. From the general explicit form of a discrete-time system: ut ) t xt) = A t x0)+ A t τ Buτ) = A t x0)+ [ B AB A t B ]. τ=0 u). u0) We know that the matrix P 3 transforms an input sequence [u0) u) u)] T to x3), starting with x0): x3) = [ B AB A B ] u) u). u0) Thus, if a vector x P 3, then there exists a triple u0),u),u) ) that transfers x0) = 0 to x. However, since only the first two columns of P 3 are independent, when x P, then we can also look for a pair u0),u) ) that transfers x0) = 0 to x, by solving x = [ B AB ][ ] u). u0) We see that x a P x a is reachable) while x b P x b is not reachable). Thus, for x a = [ 0] T we can find a pair u0),u) ) that transfers x0) = 0 to x a. We only have to solve: x a = [ B AB ][ ] [ ] 0 u) u) = 0. u0) 0 u0) Solving this vector equation it is easy to see that u0) = and u) = bring the state x0) = 0 to the state x) = x a.

9 Problem 5. Given the system {ẋ t) = α)x t)+αx t)x t) ẋ t) = α x t) x t). study the stability of the equilibrium point x e = 0,0) for all the values of the parameter α, ), using the method of linear approximation at the equilibrium point, and the Lyapunov function method;. study for what values of α the system admits other equilibrium points, and study their stability properties. Suggestion for the Lyapunov function: Vx) = x x e, ) +βx x e, ), with suitable β > 0.) Solution of problem 5. The system considered is of the form ẋt) = f xt);α ), where xt) is the state and α is a constant parameter. The vector function fx;α) is as follows { f x;α) = α)x +αx x The Jacobian is The Jacobian computed at x e = 0,0) is f x;α) = α x x [ ] [ ] x f x f α+αx αx Jx) = = x f x f x α Jx e ) = [ α ] 0 0 α Since Jx e ) is diagonal, the two eigenvalues are the diagonal terms: Thus: λ = α, λ = α. 0,0) is asymptotically stable when both eigenvalues are strictly negative, that is when α < 0) and α < 0). That means that 0,0) is asymptotically stable for α > ; 0,0) is unstable when at least one eigenvalue is strictly positive i.e. when α > 0) or α > 0). That means that 0,0) is unstable for α <. The only case where we can not conclude anything about the stability of 0,0) is when α = in this case the origin is a simply stable equilibrium point of the linear approximation of the nonlinear system, but nothing can be concluded about the nonlinear system). In this case we must study the stability of the point x e = 0,0) by using a suitable Lyapunov function. For α = the function fx;α) is { f x;) = x x f x;) = x x. Following the suggestion we consider the Lyapunov function Vx) = x +βx, which is positive definite for any β > 0. According to the Lyapunov theorem, if the derivative Vx) = dv/dx)fx;) is semidefinite negative, then the equilibrium x e is simply) stable, while if Vx) is definite negative, then the equilibrium is asymptotically stable. The computation of Vx) gives Vx) = dv dx fx;) = x V)f x;)+ x V)f x;) = x f x;)+βx f x;) = x x βx βx x.

10 By choosing β = we can eliminate the term x x which is sign-indefinite, so that we get Vx) = x. which is semidefinite negative it is never positive, but it is zero for any pair x,0), and not only at the equilibrium point x e = 0,0)). As a consequence, when α = the equilibrium 0,0) is simply stable stable, but not asymptotically stable). All other equilibrium points can be found by solving the system of equations fx;α) = 0, i.e. { α)x +αx x = 0 From the first equation we have α x x = 0. x α)+αx ) = 0 α+αx = 0 x = α α, α 0 the other solution is x = 0, and leads to the equilibrium point 0,0) already studied in the previous question). From the second equation we have α x x = 0 α α α x = 0 x = α α) Thus, there exists other equilibrium points in R only if α 0 and α α) > 0, that is for α 0,). In this case we have two more equilibrium points: ) α α), α x e, =, x e, = α α), α ). α α Note that when α = all the equilibrium points collapse to 0,0). The Jacobian computed in the two equilibrium points is [ 0 ±α ] α α) Jx e,, ) = x e,, = ± α α), α ). α α) α α The characteristic polynomial is λi Jx e,, ) [ λ α ] α α) = ± α α) λ+α = λ +α λ+α α). Note that the two Jacobians have the same characteristic polynomial. Recall that the roots of a second order polynomial all have negative real part if and only if the three coefficients have the same sign check this property using the Routh table). Thus, both equilibrium points are asymptotically stable, because they exists if and only if α 0 and α 0,), which imply that α > 0 and α α) > 0.