Control Systems. System response. L. Lanari

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1 Control Systems m i l e r p r a in r e v y n is o System response L. Lanari

2 Outline What we are going to see: how to compute in the s-domain the forced response (zero-state response) using the transfer function how to inverse transform the resulting responses so to obtain the time response analyze the relationship between eigenvalues and poles in deeper details Lanari: CS - System response 2

3 Zero State Response (forced response) Y ZS (s) =W (s)u(s) see table of common Laplace transforms (rational functions) (rational function).(rational function) = rational function Y ZS (s) y ZS (t) Partial fraction expansion (Heaviside) for rational functions Basic idea: write the function, we want to find the inverse Laplace transform of, as a linear combination of easy to transform terms and then use the linearity property of the inverse transformation For rational functions: represent a complicated fraction as the sum of simpler fractions for which we know the inverse Laplace transform (partial fraction expansion or decomposition) Lanari: CS - System response 3

4 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 with distinct roots of D(s) i.e. D(s) =a n n i= (s p i ) F (s) = N(s) a n n i= (s p i) then F(s) can be expanded as F (s) = n i= R i s p i with the residues Ri computed as R i =[(s p i )F (s)] s=pi Lanari: CS - System response 4

5 Partial fraction expansion example: given the transfer function W(s) of a system S, find its impulse response w(t) W (s) = s +2 s(s + )(s + 0) Lanari: CS - System response 5

6 Partial fraction expansion (general case) Let F (s) = N(s) D(s) be a strictly proper rational function with coprime N(s) and D(s). Let D(s) have m roots each with multiplicity ni that is then F(s) can be expanded as m i= n i = n F (s) = m i= n i j= R ij (s p i ) j with the residues Rij computed as R ij = (n i j)! d n i j ds n i j {(s p i) n i F (s)} s=p i Lanari: CS - System response 6

7 Partial fraction expansion example: find the zero-state output response (or output forced response) of the system characterized by the transfer function W(s) to the input u(t) = t (remember the function u(t) is assumed to be zero for t < 0) W (s) = s (s + )(s + 0) U(s) =L[u(t)] = s 2 Y (s) =W (s)u(s) = Y (s) = R s + R 2 s 2 + R 2 s + + R 3 s + 0 d 2 R = s 2 (2 )! ds 2 Y (s) d s = s=0 ds (s + )(s + 0) d 2 2 R 2 = s 2 (2 2)! ds 2 2 Y (s) s = s=0 (s + )(s + 0) s=0 s R 2 = [(s + )Y (s)] s= = s 2 = 2 (s + 0) s= 9 s R 3 = [(s + 0)Y (s)] s= 0 = s 2 = (s + ) 900 s (s + )(s + 0) y(t) =L (Y (s)) = R + R 2 t + R 2 e t + R 3 e 0t δ (t) Lanari: CS - System response 7 s= 0 s=0 = 0 = 2 00 s 2

8 Partial fraction expansion special case: F(s) has k poles in s = 0 (generic rational function) then F (s) = N(s) D(s) = N(s) s k D (s) with R k = F (s) = R s + R 2 s 2 leading coefficient (k k)! d k k s k ds k k F (s) + + R k s k + R ij (s p i ) j s=0 = s k F (s) s=0 = N(0) D (0) (this result will be useful for the steady-state response to an order k input ) t k k! Lanari: CS - System response 8

9 Partial fraction expansion for complex poles p = α + jβ residue R = a + jb p = α jβ R R (s p) k + R (s p ) k = As + B ((s α) 2 + β 2 ) k Lanari: CS - System response 9

10 Partial fraction expansion for k = R s p + R s p = R(s p )+R (s p) (s α) 2 + β 2 = s(r + R ) (Rp + R p) (s α) 2 + β 2 = s(r + R ) α(r + R ) jβ(r R) (s α) 2 + β 2 R + R =2a, R R = 2jb A =2a B = 2(aα + bβ) R s p + R 2as 2(aα + bβ) = s p (s α) 2 + β 2 = As + B (s α) 2 + β 2 with As + B = A(s α)+β(aα + B)/β R s p + R A(s α) β(aα + B)/β = s p (s α) β2 (s α) 2 + β 2 in t e αt [A cos βt +(Aα + B)/β sin βt] Lanari: CS - System response 0

11 Partial fraction expansion H(s) = (s 2 + )(s 2) 2 H(s) = R s 2 + R 2 (s 2) 2 + R 2 s j + R 3 s + j d R = ds (s 2 = 4 + ) s=2 25 R 2 = (s 2 = + ) s=2 5 R 2 = [(s j)h(s)] s=j = (s + j)(s 2) 2 R 3 = [(s + j)h(s)] s= j = (s j)(s 2) 2 s=j s= j = 8+6j = 2 25 j 3 50 = 8+6j = j 3 50 = R 2 H(s) = R s 2 + R 2 (s 2) 2 + As + B s 2 +, with A = 4 25, B = 6 50 h(t) = R e 2t + R 2 te 2t + A cos t + B sin t δ (t) Lanari: CS - System response

12 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 - System response 2

13 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) np i= (s p i) = n p W (s) =L[w(t)] = L[Ce At B]=L[C i= R i s p i n e λjt u j vj T j= B] = done the same analysis in t for n = 2 n j= Cu j v T j B s λ j if v T j B =0 and/or Cu j =0 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 - System response 3

14 Poles & eigenvalues (distinct eigenvalues of A case) If for an eigenvalue j we have that v T j B =0 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 =0 implies the corresponding mode will not appear in the output transition matrix Ce At = Ψ(t) the corresponding mode (or eigenvalue) is said to be unobservable Lanari: CS - System response 4

15 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 v T i B = 0 Cu i =0 or equivalently the following two PBH rank tests are both verified rank A λ i I B = n and rank A λi I C = 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 - System response 5

16 Poles & eigenvalues (distinct eigenvalues of A case) Where does the PBH test comes from? Observability (sketch): rank A λi I C <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 λi I C u i =0 (A λ i I) u i =0 Au i = λ i u i Cu i =0 Cu i =0 that is there exists an eigenvector which belongs to the nullspace of C (or the corresponding mode is unobservable) Lanari: CS - System response 6

17 example (si A) = with s+ A = 0 (s+)(s ) 0 s B = 0 = M s + + M 2 M = (s + )(si A) /2 = s= 0 0 M 2 = (s )(si A) 0 /2 = s= 0 C = 0 s fractional decomposition works also for rational matrices a different way to compute the matrix exponential e At = M e t + M 2 e t both natural modes appear (as it should be) in the state transition matrix Lanari: CS - System response 7

18 example (si A) B = s+ 0 C(sI A) = 0 s mode corresponding to 2 does not appear mode corresponding to does not appear e At B Ce At W (s) =0 no poles equivalently w(t) =Ce At B =0 forced response will always be zero independently from the input applied (look at the 2 first order ODE) u / (A λ I)u =0 u 2 / (A λ 2 I)u 2 =0 0 u 0 2 =0 u = 0 2 u =0 u 2 = 2 v T = /2 v T 2 = 0 /2 v T B = 0 v T 2 B =0 Cu =0 Cu 2 =0 Lanari: CS - System response 8

19 example equivalently with the PBH rank test mode corresponding to is controllable rank controllability test =2=n mode corresponding to 2 is uncontrollable rank =<n observability test rank =<n rank =2=n 0 0 mode corresponding to is unobservable mode corresponding to 2 is observable Lanari: CS - System response 9

20 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 rank A λ i I B = n rank A λi I = n C controllability observability 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 0 (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). Lanari: CS - System response 20

21 example A 0 = λ 0 0 λ 0 0 λ (si A 0 ) = (s λ ) 3 PBH rank test verified for B = (s λ ) 2 (s λ ) 0 (s λ ) 2 (s λ ) 0 0 (s λ ) 2 C = =0 = 0 easily seen from A 0 λ I = B = B 2 = 0 0 C = 0 0 F (s) = (s λ ) 3 0 C 2 = C = 0 0 F 2 (s) = (s λ ) 2 0 (s λ ) 3 = s λ B 3 = B 2 = 0 C 3 = 0 0 F 3 (s) =0 0 Lanari: CS - System response 2

22 example A 4 = λ 0 0 λ λ since A 4 λ I = 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 = 2 (si A 4 ) = (s λ ) 3 (s λ ) 2 (s λ ) 0 0 (s λ ) 2 0 = 0 0 (s λ ) 2 (s λ ) 2 (s λ ) 0 0 (s λ ) (s λ ) B 4 = 0 C 4 = 0 0 F 4 (s) = 0 (s λ ) 2 B 5 = 0 C 5 = C = F 5 (s) = s λ (s λ ) 2 = s λ B 6 = B = 0 0 C 6 = C = 0 0 F 6 (s) =0 Lanari: CS - System response 22

23 example A 7 = λ λ 0 (si A 7 ) = s λ 0 0 λ 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 - System response 23

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