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# RobustControlSystemsIII MEM635

Drexel

GPA 3.65

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This 5 page Class Notes was uploaded by Jada Daniel on Wednesday September 23, 2015. The Class Notes belongs to MEM635 at Drexel University taught by Bor-ChinChang in Fall. Since its upload, it has received 44 views. For similar materials see /class/212377/mem635-drexel-university in Mechanical Engineering at Drexel University.

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Date Created: 09/23/15

1 2 A Z 0 MEM635 Chapter 3b 2k 0k 0 2k diag 15 05k Model Reduction Problem Find Gk s of McMillan degree k lt 1 Examples of unitarily invariant norms Such that Gs Gk squot is minimized i the large Singl ar Value ii the sum of the singular values the trace norm Ch 39 th awe of e norm 111 the square root of the sum of the squared s1ngu1ar i Can the norm be minimized With reasonable values the Frobenius norm computational effort ii Is the chosen norm an appropriate measure of error 243 cannot be app11ed to our problem because the Hankel norm is appropriate in bOth counts39 resulting rank k approximation of H Will not generally be a Hankel matrix Discretetlme problem Find a Hankel matrix F of rank k such that Adamjan at al 1971 IH H ls mmlmlzed39 With quotH 3H the restriction that F Proposition 22 needs to be a Hankel matrix does not affect the achievable error ie there exists a Hankel matrix H X U 2V singular value decomposition A of rank k such that 3H H 039k1H U V unitary matrices 2 2 0 2diagoz1oz2oz 0 0 01120122201 Let be any unitarily invariant matrix norm ie 2 EU for anyb unitary U Then milnggk X Xl diaga1 052otkquot and the optimal If is J UikV 243 Where 3 Results on the Inertia 0f Matrices Def 31 The inertia of a complex square matriX A denoted by InA is the triple 7rAVA5A Where 7239A no of eigenvalues ofA in the open RHP VA no of eigenvalues ofA in the open LHP 5A no of eigenvalues ofA on the imaginary aXis Theorem 31 Ostrowski amp Schneider 1962 Given a complex square matriX A there is a unique Hermitian matriXX satisfying AquotX XA R Where R gt 0 ifand only if MA TjA 0 v 1 and then InA InX The condition R gt 0 is too strong for our application If R 2 0 the following corollary can be used Corollary 32 Ostrowski amp Schneider 1962 Given a compleX square matriXA such that A A 2 0 and X i X then 7239AX S 7239X VAX S 1X Theorem 33 Given Ae CW Be CMquot and Hermitian matriX compleX square matriX A there is a unique Hermitian matriX P 2 Pi satisfying APPAquotBBquot 0 34 Then 1 There lSil unique solution to 34 if and only if A jA 0 Vij 2 If 5P 0 then 7239A S 1P VA S 7239P 3 If 5A 0 then 7239P S 1A 1P S 7239A 4 If A B is controllable then 7239A 1P VA 7239P 5A 5P 0 5 rank P 703 VP 2 rankBABAquot lBand if MA 711w 0 v i j then the equality holds 7239A 1P VA 7239P 5A 5P 0 6 If 7239A n then 7239P 0 7 If VAn then 1P 0 Remark 31 Part 4 of the theorem gives the standard stability result If A B is controllable then A is stable iffP satisfying 34 is positive de nite Part 5 of the theorem gives the standard controllability result If A is stable then A B is controllable iffP is positive de nite Remark 32 If the condition A 1 tJA 0 V ij is not satis ed then the equality in part 5 may not hold 1 0 0 0 1 Let A 3 P 0 l 0 l 0 Then 7239P 1P l but A B is not controllable Lemma 35 Given Xe CW and Ye CW r 2 m such that XXi YY then there eXists Ue Cm such that UUi 1 and Y XU Balanced Realiz ations Given ABC minimal and stable LetP and Q be the Hermitian solutions to AP PAquot BBquot 0 AiQQACiC 0 Let Q have a Cholesky factorization Q RquotR Then RPRi gt 0 can be diagonalized as RPRquot UZZUquot with UquotU I 2 diag0391 0392 039quot 03912 0392 2 2 039quot Now a balancing transformation is given by T Z IZU R since TPTquot 2 Tquot 1QT1 2 That is after the balancing transformation the controllability and observability gramians are equal and diagonal Lemma 41 Given ABC minimal and stable It has a balanced realization that is unique up to an arbitrary transformation T such that T2 2 ET and TiT I In the case of distinct 01 and real Tand ABC then Tmust be a sign matriX ie the diagonal entries are i1 7 Theorem 42 truncations of a balanced realization A22Aquot BBquot 0 49 Aquot22ACquotC0 410 2 0 22 1 with 620 0 2 Partition ABC conformmaly with 2 as AA A12 BBl CC1 C2 A21 A22 B2 Then 1 If 121 0 then 7239A11 0 2 If 5A 0 and 13212 Zi Vij then InA11 In Zl and InA22 In 22 3 Any minimal realization of C1sI AU IB1 given by sI 2041 and 521 0 Remark For balanced realizations Z gt 0 Theorem 42 Part 2 implies that All is stable if that M212 M23 V 1395 Theorem 43 For any stable ABC there eXists T st TAT 1TBCT 1 has controllability and observability gramians given by P diag 212200 Q diag210230 Respectively with 212223 gt 0 and diagonal Allpass Transfer Function Theorem 51 Given ABC which is not necessarily stable then 1 If ABC is minimal the following two statements are equivalent a There eXists D st GsGs 2 0392 Vs where Gs CsI A IB D b There eXist P and Q st 0 PP 1QQquot ii AP PAquot BBquot 0 iii AiQ QA CC 0 iv PQ 2 0392 2 If part lb is satis ed then there eXists D satisfying DDquot 021 DquotC BquotQ 0 03quot CP 0 and any such D will satisfy part la Remark 51 This theorem does not assume ampA A A 0 V ij which would imply that the Lyapunov equations have unique solutions If A A A 0 for some i andj then there eXists an in nite number of solutions to part lb equations i ii and iii and when part la is satis ed only some of these solutions also satisfy part lb equation iv Theorem 52 Given Gs of McMillan degree n With a minimal realization ABCD and st GjwGja s I Va Then there eXist D D D12 E CPMXPm D21 D22 3 B BAG CMW39quot a C e Cplmw C2 Hs 2 651 12041 D is allpass H s That is Gja F 1 3 Gs is a submatriX of an all pass function With the same A matrix

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