Exercise 8: Frequency Response of MIMO Systems

Transcription

1 Exercise 8: Frequency Response of MIMO Systems 8 Singular Value Decomposition (SVD The Singular Value Decomposition plays a central role in MIMO frequency response analysis Let s recall some concepts from the course Lineare Algebra I/II : 8 Preliminary Definitions The induced norm A of a matrix that describes a linear function like is defined as y = A u (8 A = max u 0 y u = max y u = (8 Remark In the course Lineare Algebra I/II you learned that there are many different norms and the expression A could look too generic However, in this course we will always use the euclidean norm A = A This norm is defined as A = µ max = maximal eigenvalue of A A (83 where A is the conjugate transpose or Hermitian transpose of the matrix A This is defined as A = (conj(a T (84 Example Let Then it holds A = i + i i A = (conj(a T T + i = i i i = + i i Remark If we deal with A R n m holds of course One can list a few useful properties of the euclidean norm: (i Remember: A T A is always a quadratic matrix! A = A T (85 (ii If A is orthogonal: A = (86

2 (iii If A is symmetric: A = max( e i (87 (iv If A is invertible: A = µmin = minimal eigenvalue of A A (88 (v If A is invertible and symmetric: A = min( e i (89 where e i are the eigenvalues of A In order to define the SVD we have to go a step further Let s consider a Matrix A and the linear function 8 It holds A = max u = y y = max u = (A u (A u = max u = u A A u = max i µ(a A = max i σ i (80 where σ i are the singular values of matrix A They are defined as where µ i are the eigenvalues of A A Combining 8 and 8 one gets σ i = µ i (8 σ min (A y u σ max(a (8 8 Singular Value Decomposition Our goal is to write a general matrix A C p m as product of three matrices: U, Σ and V It holds A = U Σ V with U C p p, Σ R p m, V C m m (83 Remark U and V are orthogonal, Σ is a diagonal matrix

3 y = A u u Figure : Illustration of the singular values Kochrezept: Let A C p m be given: (I Compute all the eigenvalues and eigenvectors of the matrix A A C m m and sort them as µ µ µ r > µ r+ = = µ m = 0 (84 (II Compute an orthogonal basis from the eigenvectors v i and write it in a matrix as V = (v v m C m m (85 (III We have already found the singular values: they are defined as σ i = µ i for i =,, min{p, m} (86 We can then write Σ as σ 0 0 Σ = R p m, p < m (87 σ m 0 0 (IV One finds u,, u r from σ Σ = σ n 0 0 R p m, p > m ( u i = σ i A v i for all i =,, r (for σ i 0 (89 3

4 (V If r < p one has to complete the basis u,, u r (with ONB to obtain an orthogonal basis, with U orthogonal (VI If you followed the previous steps, you can write A = U Σ V Motivation for the computation of Σ, U und V A A = (U Σ V (U Σ V = V Σ U U Σ V = V Σ Σ V = V Σ V (80 This is nothing else than the diagonalization of the matrix A A The columns of V are the eigenvectors of A A and the σi the eigenvalues For U: A A = (U Σ V (U Σ V = U Σ V V Σ U = U Σ Σ U = U Σ U (8 This is nothing else than the diagonalization of the matrix A A The columns of U are the eigenvectors of A A and the σi the eigenvalues Remark In order to derive the previous two equations I used that: The matrix A A is symmetric, ie U = U (because U is othogonal V = V (because V is othogonal (A A = A (A = A A Remark Since the matrix A A is always symmetric and positive semidefinite, the singular values are always real numbers Remark The MATLAB command for the singular value decomposition is [U,S,V]=svd One can write A T as A =transpose(a and A as A =conj(transpose(a Those two are equivalent for real numbers Remark Although I ve reported detailed informations about the calculation of U and V, this won t be relevant for the exam It is however good and useful to know the reasons that are behind this topic 4

5 Example Let u be with u = The matrix M is given as u = M = cos(x sin(x 0 0 We know that the product of M and u defines a linear function y = M u 0 cos(x = 0 sin(x cos(x = sin(x We need the maximum of y In order to avoid square roots, one can use that the x that maximizes y should also maximize y y = 4 cos (x + 4 sin (x has maximum d y dx = 8 cos(x sin(x + sin(x cos(x =! 0 { x max = 0, π, π, 3π } Inserting back for the maximal y one gets: y max =, y max = The singular values can be calculated with M M: M M = M M 4 0 = 0 4 λ i = { 4, } 4 σ i = {, } As stated before, one can see that y [σ min, σ max ] The matrix U has eigenvectors of M M as coulmns and the matrix V has eigenvectors of M M as columns In this case M M = M M, hence the two matrices are equal Since their product is a diagonal matrix one should recall from the theory that the eigenvectors are easy to determine: they are nothing else than the standard basis vectors This means Interpretation: U = [ 0 0 ] [ 0, Σ = 0 ], V = [ ] 0 0 Referring to Figure, let s interprete these calculations One can see that the maximal amplification occurs at v = V (:, and has direction u = U(:,, ie the vector u is doubled (σ max The minimal amplification occurs at v = V (:, and has direction u = U(:,, ie the vector u is halved (σ min 5

6 05 u y = M u Figure : Illustration von der Singularwertzerlegung Example 3 Let 3 0 A = be given Question: Find the singular values of A and write down the matrix Σ Solution Let s compute A T A: A T A = ( One can see easily that the eigenvalues are = 3 λ = 6, λ = 9 ( The singular values are We write then σ = 4, σ = Σ =

7 Example 4 A transfer function G(s is given as ( s+3 s+ s+3 Find the singular values of G(s at ω = rad s s+ s+3 s+3 7

8 Solution The transfer function G(s evaluated at ω = rad s G(j = ( j+3 j+ j+3 j+ j+3 j+3 has the form In order to calculate the singular values, we have to compute the eigenvalues of H = G G: H = G G = = ( j+3 j+ ( 3 0 j+3 0 = j+ j+3 j+3 ( 3 3 ( j+3 j+ j+3 j+ j+3 j+3 For the eigenvalues it holds It follows det(h λ = det = ( 3 0 λ 0 0 ( 3 0 λ 3 λ 0 ( 0 = λ 6 0 λ + 5 ( 00 = λ 0 ( λ 5 0 λ = 0 and so λ = σ = σ =

9 8 Frequency Responses As we learned for SISO systems, if one excites a system with an harmonic signal u(t = h(t cos(ω t, (8 the answer after a big amount of time is still an harmonic function with equal frequency ω: y (t = P (j ω cos(ω t + (P (j ω (83 One can generalize this and apply it to MIMO systems With the assumption of p = m, ie equal number of inputs and outputs, one excite a system with µ cos(ω t + φ u(t = h(t (84 µ m cos(ω t + φ m and get Let s define two diagonal matrices and two vectors y (t = ν cos(ω t + ψ ν m cos(ω t + ψ m (85 Φ = diag(φ,, φ m R m m, Φ = diag(ψ,, ψ m R m m (86 µ = ( µ µ m T, ν = ( ν ν m T (87 With these one can compute the Laplace Transform of the two signals as: U(s = e Φ s ω µ s s + ω (88 and Y (s = e Ψ s s ω ν (89 s + ω With the general equation for a systems one gets Y (s = P (s U(s e Ψ s s Φ s s ω ν = P (s e s ω + ω µ s + ω e Ψ j ω ω ν = P (s e Φ j ω ω µ e Ψ j ν = P (s e Φ j µ (830 We then recall that the induced norm for the matrix of a linear transformation y = A u from 8 Here it holds P (j ω = e Ψ j ν max e Φ j µ 0 e Φ j µ = max e Ψ j ν e Φ j µ = (83 9

10 Since and One gets e Φ j µ = µ (83 e Ψ j ν = ν (833 P (j ω = max µ 0 ν µ = max µ = ν (834 Here one should get the feeling of why we introduced the singular value decomposition From the theory we ve learned, it is clear that and if µ σ min (P (j ω ν σ max (P (j ω (835 σ min (P (j ω ν µ σ max(p (j ω (836 with σ i singular values of P (j ω These two are worst case ranges and is important to notice that there is no exact formula for ν = f(µ 8 Maximal and minimal Gain You are given a singular value decomposition P (j ω = U Σ V (837 One can read out from this decomposition several informations: the maximal/minial gain will be reached with an excitation in the direction of the column vectors of U The response of the system will then be in the direction of the coulmn vectors of V Let s look at an example and try to understand how to use these informations: Example 5 We consider a system with m = inputs and p = 3 outputs We are given its singular value decomposition at ω = 5 rad: s Σ = 0 063, 0 0 ( V =, j j j j j U = j j j j j j For the singular value σ max = 0467 the eigenvectors are V (:, and U(:, : ( V =, V j =, (V =, j U = j, U = 0960, (U = j

11 The maximal gain is then reached with ( 0908 cos(5 t u(t = cos(5 t 068 The response of the system is then 075 cos(5 t cos(5 t 858 y(t = σ max 0960 cos(5 t 5548 = cos(5 t cos(5 t cos(5 t 474 Since the three signals y (t, y (t and y 3 (t are not in phase, the maximal gain will never be reached One can show that max y(t 0460 < 0467 = σ max t The reason for this difference stays in the phase deviation between y (t, y (t and y 3 (t The same analysis can be computed for σ min 8 Robustness and Disturbance Rejection Let s redefine the matrix norm as Remark It holds G(s = max (max (σ i (G(i ω (838 ω i G (s G (s G (s G (s (839 As we did for SISO systems, we can resume some important indicators for robustness and noise amplification: Robustness Noise Amplification SISO MIMO µ = min ( + L(j ω µ = min (σ min (I + L(j ω ω ω S = max ( S(j ω S = max (σ max(s(j ω ω ω

12 Example 6 Given the MIMO system P (s = ( s+3 s+ s+ 3 s+ Starting at t = 0, the system is excited with the following input signal: ( cos(t u(t = µ cos(t + ϕ Find the parameters ϕ and µ such that for steady-state conditions the output signal y (t y (t has y (t equal to zero

13 Solution For a system excited using a harmonic input signal µ cos(ωt + ϕ u(t = µ cos(ωt + ϕ the output signal y(t, after a transient phase, will also be harmonic and hence have the form ν cos(ωt + ψ u(t = ν cos(ωt + ψ As we have learned, it holds For the first component one gets e Ψ j ν = P (jω e Φ j µ e ψ j ν = P (jω e ϕ j µ + P (jω e ϕ j µ For y (t = 0 to hold we must have ν = 0 In the given case, some parameters can be easily copied from the signals: With the given transfer functions, one gets 0 = j µ j + eϕ j 0 = 3 j 0 + µ j 0 = 3 j 0 + µ j e ϕ j µ = ϕ = 0 ω = (cos(ϕ + j sin(ϕ 0 = µ (cos(ϕ + sin(ϕ + j Splitting the real to the imaginary part, one gets two equations: µ (cos(ϕ + sin(ϕ = 0 µ (sin(ϕ cos(ϕ 0 = 0 ( µ (sin(ϕ cos(ϕ 0 Adding and subtracting the two equations one can reach two better equations: One of the solutions (periodicity reads µ sin(ϕ + 5 = 0 µ cos(ϕ + 5 = 0 µ = 5 ϕ = arctan + π 3

14 Example 7 A linear time invariant MIMO system with transfer function is excited with the signal u(t = P (s = ( s+ s + s+0 s+ s + µ cos(ω t + ϕ µ cos(ω t + ϕ Because we bought a cheap signal generator, we cannot know exactly the constants µ, and ϕ, A friend of you just found out with some measurements, that the excitation frequency is ω = rad The cheap generator, cannot produce signals with magnitude of µ bigger than s 0, ie µ + µ 0 This works always at maximal power, ie at 0 Choose all possible responses of the system after infinite time 5 sin(t + 04 y (t = cos(t 5 sin(t + 04 y (t = cos( t sin(t y (t = sin(t cos(t + 04 y (t = cos(t cos(t + 04 y (t = 5 cos(t 0 sin(t + 4 y (t = sin(t

15 Solution 5 sin(t + 04 y (t = cos(t 5 sin(t + 04 y (t = cos( t sin(t y (t = sin(t cos(t + 04 y (t = cos(t cos(t + 04 y (t = 5 cos(t 0 sin(t + 4 y (t = sin(t + 34 Explanation We have to compute the singular values of the matrix P (j These are With what we have learned it follows σ max = 8305 σ min = σ min = 3863 ν 8305 = 0 σ max The first response has ν = 6 that is in this range The second response also has ν = 6 but the frequency in its second element changes and that isn t possible for linear systems The third response has ν = that is too small to be in the range The fourth response has ν = 36 that is too big to be in the range The fifth response has ν = 50 that is in the range The sixth response has ν = that is in the range 5

16 Example 8 A 3 linear time invariant MIMO system is excited with the input 3 sin(30 t u(t = 4 cos(30 t You have forgot your PC and you don t know the transfer function of the system Before coming to school, however, you have saved the Matlab plot of the singular values of the system on your phone (see Figure 3 Choose all the possible responses of the system Figure 3: Singular values behaviour 05 sin(30 t y (t = 05 cos(30 t 05 cos(30 t + 4 sin(30 t y (t = 3 cos(30 t cos(30 t + 0 sin(30 t y (t = 0 cos(30 t 0 cos(30 t + y (t = 0 4 cos(30 t cos(30 t + cos(30 t y (t = cos(30 t + 04 cos(30 t

17 Solution 05 sin(30 t y (t = 05 cos(30 t 05 cos(30 t + 4 sin(30 t y (t = 3 cos(30 t cos(30 t + 0 sin(30 t y (t = 0 cos(30 t 0 cos(30 t + 0 y (t = 4 cos(30 t cos(30 t + cos(30 t y (t = cos(30 t + 04 cos(30 t + 05 Explanation From the given input one can read µ = = 5 From the plot one can read at ω = 30 rad s σ min = 0 and σ max = It follows 5 σ min = 05 ν 5 = 5 σ max The first response has ν = 075 that is in the range The second response has ν = 9 that is too big to be in the range The third response has ν = 003 that is to small to be in the range The fourth response has ν = 0 that is in the range The fifth response has ν = that is in the range 7