Control Systems. Laplace domain analysis

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1 Control Systems Laplace domain analysis L. Lanari

2 outline introduce the Laplace unilateral transform define its properties show its advantages in turning ODEs to algebraic equations define an Input/Output representation of the system through the transfer function exploring the structure of the transfer function solve a realization problem Lanari: CS - Laplace domain analysis

3 Laplace transform real valued function in the real variable t f(t) L F (s) complex valued function in the complex variable s t R s C F (s) = Z f(t)e st dt Laplace (unilateral) transform for f(t) with no impulse and no discontinuities at t = F (s) =L[f(t)] Im Region of existence: for all s with real part greater equal to an abscissa of convergence ¾ example f(t) =e at, a >, = a -a region of convergence Re Lanari: CS - Laplace domain analysis

4 inverse Laplace transform f(t) L F (s) L f(t) =L [F (s)] = j Z j j F (s)e st ds unique for one-sided functions f(t) defined for t or f(t) = for t < (one-to-one) all give the same F(s) with this assumption the inverse Laplace transform is unique F(s)? f(t) t Lanari: CS - Laplace domain analysis 4

5 Laplace transform Laplace transform of the impulse L[ (t)] = Z F (s) = Z (t)e st dt = f(t)e st dt Z (t)e st dt = e s = correct definition of the Laplace transform Linearity L[ f(t)+ g(t)] = L[f(t)] + L[g(t)] Derivative property L apple df (t) dt = sl[f(t)] f() can be applied iteratively L h f(t) i = s L[f(t)] sf() f() very useful in model derivation Lanari: CS - Laplace domain analysis 5

6 LTI systems also useful here ẋ(t) =Ax(t)+Bu(t) x() = x (proof): linearity + derivative property algebraic solution X(s) =(si A) x +(si A) BU(s) and therefore Y (s) =C(sI A) x + C(sI A) B + D U(s) Lanari: CS - Laplace domain analysis 6

7 LTI systems state ZIR transform state ZSR transform X(s) =(si A) x +(si A) BU(s) {z } {z } x(t) =e At x + Z t e A(t ) Bu( )d and therefore, comparing L e At =(si A) Heaviside step function L e at = s a L [ (t)] = s (t) for t for t< Lanari: CS - Laplace domain analysis 7 t

8 LTI systems y(t) =Ce At x + Z t w(t )u( )d with w(t) =Ce At B + D (t) transform W (s) =C(sI A) B + D Convolution theorem Y (s) = C(sI A) x + C(sI A) B + D U(s) = C(sI A) x + W (s) U(s) L applez t w(t )u( )d = W (s) U(s) being L[ (t)] = the output response transform corresponding to u(t) = ±(t) is indeed W(s) same for H(s) Lanari: CS - Laplace domain analysis 8

9 transfer function Input/Output behavior x = (ZSR) y ZS (t) = Z t w(t )u( )d Y ZS (s) =W (s) U(s) Transfer function W (s) = C(sI A) B + D = Y ZS(s) U(s) = L [w(t)] Input/Output behavior independent from state choice? independent from state space representation? Lanari: CS - Laplace domain analysis 9

10 transfer function (A, B, C, D) ẋ = Ax + Bu y = Cx + Du z = Tx det(t ) 6= ( e A, e B, e C, e D) ż = e Az + e Bu y = e Cz + e Du ea = TAT e B = TB e C = CT e D = D W (s) =C(sI A) B + D = e C(sI e A) e B + e D for a given system, the transfer function is unique Lanari: CS - Laplace domain analysis

11 shape of the transfer function W (s) =C(sI A) B + D inverse through the adjoint (transpose of the cofactor matrix) C (adjoint of (si A)) B + D det(si A) cofactor(i, j) =( ) i+j minor(i, j) polynomial of order n - polynomial of order n - det(si A) C (cofactor of (si A))T B + D polynomial of order n rational function (strictly proper) strictly proper rational function: proper rational function: degree of numerator < degree of denominator degree of numerator = degree of denominator Lanari: CS - Laplace domain analysis

12 shape of the transfer function W (s) = strictly proper rational function + D proper rational function W (s) = N(s) D(s) D = D 6= strictly proper rational function proper rational function W (s) = N(s) D(s) roots zeros poles (for coprime N(s) & D(s)) i.e. no common roots from previous analysis the poles are a subset of the eigenvalues of A {poles} {eigenvalues} more on this later Lanari: CS - Laplace domain analysis

13 partial fraction expansion (distinct roots case) Let F (s) = N(s) D(s) be a strictly proper rational function with coprime N(s) and D(s) and distinct roots of D(s) i.e. D(s) =a n n Y i= (s p i ) ) F (s) = N(s) a n Q n i= (s p i) then F(s) can be expanded as F (s) = nx i= s R i p i with the residues Ri computed as R i =[(s p i )F (s)] s=p i this result will be used for the transfer function W(s) output zero-state response Y(s) Lanari: CS - Laplace domain analysis

14 ( poles & eigenvalues W (s) = N(s) D(s)? from def W (s) = det(si A) N (s) characteristic polynomial if det(si A) and N (s) coprime {poles} = {eigenvalues} if det(si A) and N (s) not coprime {poles} subset of {eigenvalues} Input/Output representation Transfer function vs Input/State/Output representation State space here visible we need to understand when & why this happens (so to understand when we can consider the transfer function equivalent to a state space representation) Lanari: CS - Laplace domain analysis 4

15 poles & eigenvalues (distinct eigenvalues of A case) n = state space dimension = dimension of A = number of eigenvalues np = number of poles in W(s) W (s) = N(s) D(s) = N(s) Q np i= (s p i) = partial fraction expansion n p X R i p i s i= nx W (s) =L[w(t)] = L[Ce At e jt u j vj T j= A B] = done the same analysis in t for n = nx j= Cu j v T j B s j if v T j B = and/or Cu j = spectral form the eigenvalue j does not appear as a pole we have a hidden mode associated to the eigenvalue j (see structural properties) NB distinct eigenvalues is different from diagonalizable Lanari: CS - Laplace domain analysis 5

16 poles & eigenvalues (distinct eigenvalues of A case) fact: all the natural modes appear in the state ZIR for some generic initial condition e At If for an eigenvalue j we have that v T j B = implies the corresponding mode will not appear in the state impulsive response e At B = H(t) the corresponding mode (or eigenvalue) is said to be uncontrollable Cu j = implies the corresponding mode will not appear in the output transition matrix Ce At = (t) the corresponding mode (or eigenvalue) is said to be unobservable Lanari: CS - Laplace domain analysis 6

17 poles & eigenvalues (distinct eigenvalues of A case) Theorem Every pole is an eigenvalue. An eigenvalue i becomes a pole if and only if it is both controllable and observable vi T B 6= and Cu i 6= or equivalently the following two PBH rank tests are both verified rank A ii B = n and rank A C ii! = n Popov-Belevitch-Hautus controllability test Popov-Belevitch-Hautus observability test (the PBH test could be tested for a generic but matrix A - I loses rank only for = i ) Lanari: CS - Laplace domain analysis 7

18 poles & eigenvalues (distinct eigenvalues of A case) Where does the PBH test comes from? Observability (sketch): rank A C ii! <n means that the rectangular (n+) x n matrix has not full column rank and therefore it has a non-zero nullspace, that is there exists a n vector ui such that A C ii! u i = ( ( (A ii) u i = Au i = i u i Cu i = Cu i = that is there exists an eigenvector which belongs to the nullspace of C (or the corresponding mode is unobservable) Lanari: CS - Laplace domain analysis 8

19 example (si A) = with s+ A = (s+)(s ) s B = = M s + + M M = (s + )(si A) / = s= M = (s )(si A) / = s= C = s fractional decomposition works also for rational matrices a different way to compute the matrix exponential e At = M e t + M e t both natural modes appear (as it should be) in the state transition matrix Lanari: CS - Laplace domain analysis 9

20 example (si A) B = s+ mode corresponding to does not appear e At B W (s) = C(sI A) = s mode corresponding to does not appear Ce At no poles equivalently w(t) =Ce At B = forced response will always be zero independently from the input applied (look at the first order ODE) u / (A I)u = u / (A I)u = apple apple u = u = apple apple u = u = v T = / v T = / v T B 6= v T B = Cu = Cu 6= Lanari: CS - Laplace domain analysis

21 example equivalently with the PBH rank test controllability test mode corresponding to is controllable rank ==n mode corresponding to is uncontrollable rank =<n observability test rank C A =<n rank C A ==n mode corresponding to is unobservable mode corresponding to is observable Lanari: CS - Laplace domain analysis

22 poles & eigenvalues (general case) Theorem Every pole is an eigenvalue. An eigenvalue i becomes a pole with the multiplicity ma (algebraic multiplicity) if and only if both PBH rank tests are verified A ii rank A ii B = n rank C controllability observability! = n NB - If one of the two conditions is not verified then the eigenvalue i will appear as a pole with multiplicity strictly less than the algebraic multiplicity, possibly even (in this case we will have a hidden eigenvalue). In particular the eigenvalue will appear at most as a pole with multiplicity equal to its index (dimension of the largest Jordan block). NB - If for an eigenvalue i the geometric mg( i) > then there exists a hidden dynamics. Lanari: CS - Laplace domain analysis

23 example A = 4 5 (si A ) = (s ) PBH rank test verified for B = (s ) (s ) 6 4 (s ) 7 (s ) 5 (s ) C = 6= 8 8 6= easily seen from A I = 4 5 B = B = 4 5 C = F (s) = (s ) 4 5 C = C = F (s) = (s ) (s ) = index of s B = B = 4 5 C = F (s) = Lanari: CS - Laplace domain analysis

24 example A 4 = 4 5 since A 4 I = 4 5 the PBH rank test will never be satisfied independently from B and C. At most the eigenvalue will appear as a pole with multiplicity = index of = (si A 4 ) = (s ) (s ) (s ) 4 (s ) 5 = (s ) (s ) (s ) 4 (s ) 5 (s ) B 4 = 4 5 C 4 = F 4 (s) = (s ) B 5 = 4 5 C 5 = C = F 5 (s) = s (s ) = s B 6 = B = 4 5 C 6 = C = F 6 (s) = Lanari: CS - Laplace domain analysis 4

25 example A 7 = 4 5 (si A 7 ) = s 4 5 etc... the PBH rank test will never be satisfied independently from B and C. At most the eigenvalue will appear as a pole with multiplicity = index of = Lanari: CS - Laplace domain analysis 5

26 Laplace transform tables (t) sin t s + = /j s j /j s + j (t) s cos t s s + = / s j + / s + j e at s a sin( t + ) s sin + cos s + t k k! s k+ e at sin t (s a) + t k k! eat (s a) k+ e at cos t (s a) (s a) + Lanari: CS - Laplace domain analysis 6

27 realizations ok (A, B, C, D) W (s) =C(sI A) B + D how? W (s) =C(sI A) B + D realization infinite solutions (A, B, C, D) state dimension? how can we easily find one state space representation (A, B, C, D)? may be complicated for MIMO systems (here SISO) we see only one, obtainable directly from the coefficients of the transfer function (others are obtainable by simple similarity transformations) Lanari: CS - Laplace domain analysis 7

28 realizations W (s) = N(s) D(s) coprime N(s) & D(s) - find D W (s) = b n s n + b n s n + + b s + b s n + a n s n + a n s n + + a s + a + D - state with dimension n - one possible realization for A, B, C A c = 6 4 a a a a n C c = b b b b n 7 5 B c = controller canonical form (useful for eigenvalue assignment) Lanari: CS - Laplace domain analysis 8

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