LMIs in Control/Matrix and LMI Properties and Tools/Schur Complement Lemma-Based Properties

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LMI Condition

Consider P11π•Šn, P12ℝn×m, P22, Xπ•Šm, P13ℝn×p,P23ℝm×p, and P33π•Šp.

There exists X such thatTemplate:NumBlk if and only if Template:NumBlkAny matrix Xπ•Šm satisfying

X< P22 + [P22TP23][P11P13*P33]1[P12P23T] is a solution to (1)


Consider P11π•Šn, P12, Xℝn×m, P22π•Šm,P13ℝn×p, P23ℝm×p and P33π•Šp.

There exists X such thatTemplate:NumBlk

if and only if

Template:NumBlk

If two inequalities in (4) hold, then a solution to (3) is given by

X=P23P331P13TP12T

Consider P11, Xπ•Šn, P12ℝn×m, and P22π•Šm, where X>0.

There exists X such thatTemplate:NumBlk if and only ifTemplate:NumBlk


Consider Xπ•Šn, Hℝm×n,Gℝm×m , and Pπ•Šm, where P>0.Template:NumBlkimplies

X>HTG1PGTH


Consider P1π•Šn, P2, Xπ•Šq, Q1ℝn×m, Q2ℝq×p, R1π•Šm, and R2π•Šp

LMI givesTemplate:NumBlkif and only ifTemplate:NumBlk


Consider Pπ•Šn, Rπ•Šm, Sπ•Šp, Qℝn×m, Xℝn×p, Vℝm×p, and Eℝp×m.

LMI givesTemplate:NumBlkare satisfied if and only ifTemplate:NumBlk


Consider P1, Qπ•Šn, P2, Q2ℝn×m, and P3, Q3π•Šm, where P1>0, P3>0, Q1>0, and Q3>0.

There exists P2, P3, Q2, and Q3 such thatTemplate:NumBlkif and only ifTemplate:NumBlk

Proof

Proof for (3)

Necessity ((3) (4)) comes from the requirement that the submatrices corresponding to the principle minors of (3) are negative definite

Sufficiency ((4) (3)) is shown by rewriting the matrix inequalities of (4) in the equivalent form

P11P13TP331P13<0, and P22P23TP331P23<0

Concatenating the two matrices and choosing X=P23TP331P23P12T gives the equivalent matrix inequality Template:NumBlkorTemplate:NumBlkwhich is equivalent to (3) using the Schur complement lemma.

Proof for (6)

the LMI in (5) can be written using the Schur complement lemma asTemplate:NumBlk

Template:NumBlk

Template:NumBlk

Proof for (7)

Using the Schur complement lemma on (7) for X>HTG1PGTH

Using the property G+GTPGTP1G or equivalent (G+GTP)1G1PGT gives

X>HT(G+GTP)1HHTG1PGTH


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