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MEM633 Lectures 4amp5 23 Linear Operators M function Let X and Y be arbitrary nonempty sets and let D be a nonempty subset of X A function f from D into Y is a rule that to each X in D assigns a unique element fX in Y M domain range The domain of f is the set D on which the function f is defined often written as Df The range of f is the set Rf fX X in D M linear operator X Y are linear spaces over F A function 139 is said to be a linear operator if and only if Tc1x1 of c1139x1 02139 for any vectors x1 3 in X and any scalars c in F C 1 2 Example Consider the operator that rotates a vector in a geometric plane counterclockwise 90 with respect to the origin y X2 X X3 X7 This operator is a linear operator Why Matrix Representation of a Linear Operator X and Y are n and m dimensional vector spaces over the same field P Let x1 x2 quot39 xr7 be a basis for X and y f quot39 ym a basis for Y Suppose the linear operator T X Y is defined by i 1 Tx any a2 anquot ym 1 12n where a a a are scalars in F 1i 2 ml Then the operator T can be represented by a matrix A T x l gt yAx with A ail i12m j12n Example Consider the rotation operator in the previous example again If we choose x1 3 as the basis of both X and Y then the matrix representation of 139 IS 0 1 1 0 If we choose x1 x2 as the basis of X and the basis f x3 of Y then the matrix representation of Tis 1 0 0 1 24 Linear Algebraic Eguations Consider the set of linear equations a X a X quot39a X y 111 122 1r7r7 1 a X a X a Xy 211 222 2r7r7 2 A xa xa xy m1 1 m2 2 mn n m where the given ail 39s and 39s are assumed to be elements of a field F the unknown s are also required to be in the same field This set of equations can be written in matrix form as Ax y where A is an mxn matrix aij x is an nX1 vector and y is an mX1 vector We can consider the matrix n m A as a linear operator which maps F into F Theorem The range of a linear operator A is a subspace of m P Let thej th column of A be denoted by a ie A a1 a2 ar7 then the matrix equation yAx can be written as yx a1x a2 x a 1 2 n It is easy to see that RA is the set of all the linear combinations of the columns of A and the dimension of RA is the max no of linearly independent columns in A Def rank The rank of a matrix A is the max no of linearly independent columns in A Le the dimension of the range space of A Example Consider the matrix 1 1 2 2 3 5 0 3 3 A has only two linearly independent columns Hence the rank of A is 2 and RA is a two dimensional space spanned by 0 1 3 and 1 2 0 Theorem Consider the linear operator n m AFFXI gtAx m a Given A and a vector y in F there exists a vector x such that Ax y if and only if the vector y is an element of RA m b Given A and for any y in F there exists a vector x such that Ax y if and only if RA Fm or equivalently rankA m Def null space The null space of a linear operator A is the set NA defined by NA x inFnAx0 ie N A is the set of all solutions of Ax0 Example Consider the matrix 0 1 1 2 3 A 1 2 3 5 7 3 0 3 3 3 3 5 which maps Fl into E Letx X1X2 X5 then 1 2 3 4 5 Axx a xa xa x a xa 1 2 3 4 5 1 2 1 2 xaxax aa 1 2 3 1 2 1 2 x4a2a x5a 3a xxx xa1xx2x3xa2 1 3 4 5 2 3 4 5 The vectors a1 and a2 are linearly independent hence Ax0 if and only if xxx x0 13 4 5 xx2x3x0 2 3 4 5 It is clear that the set of the three vectors 1 1 1 1 2 3 1 a 0 a and 0 0 1 0 0 0 1 form a basis of NA Theorem Let A be an mxn matrix then dim RA dim N A n 25 Eigenvalues and Eigenvectors M eigenvalue and eigenvector n Let A be a linear operator that maps lt3 into itself n and x a vector in Then x is an eigenvector of A corresponding to the eigenvalue xi if X750 and Axix To find an eigenvalue of A we write Axlx as A xi x 0 This equation has a nontrivial solution if and only if detAw 0 detA11 is a polynomial of degree n in xi and is called the characteristic polynomial of A Example 3 1 A 1 3 The eigenvalues of A are 11 2 and 2 4 and their corresponding eigenvectors are x1 1 1 39 and f 1 1 139 Example Suppose that an nxn matrix A has n linearly 1 2 n Independent eigenvectors x x quot39 x correspond ing to eigenvalues A1 12 xi Then any vector y can be written in the form yax1afma r7 1 2 n where a1 a2 ar7 is called the representation of y with respect to the basis x1 x2 xr7 From the fact that i i Ax 1 x I it follows that k k 1 k 2 k n Ayi1 a1x 012 a2x xin anx lf 11 gt1i 23n and a1 0 then Aky will tend to lie along the vector x7 when k is large Example Consider the differential equation y39t Ayn Suppose that Ax ix and set m yte X Then 39 m Yt19 x1yt m m AytAe xe xixiyt Theorem A and P are nxn matrices with P nonsingular Then xi is an eigenvalue of A with eigenvector x 1 if and only if xi is an eigenvalue of P AP with eigenvector P x 26 Norms and Inner Products M vector norm The norm of a vector x in a linear space X is a realvalued function of x denoted by x II which satisfies the following axioms 1 x 2 0 for all x in X x 0 if and only if X0 2 Max M c 39x H for all scalars in F and each x in X 3 xy H s x y for each x y in X triangular inequality Example n Consider the linear space C Let x x1 x2 quot39Xn 39 Then the norm of x can be chosen as n iim or n 12 2 nw mw 11 or 2 maxlx I be J J H is easy to verify that each of them satisfies all the axioms of a norm E operator norm The norm of a linear operator A is defined as llell A sup sup Ax quotxquot Wquot l where quotsupquot stands for supremum the least upper bound The operator norm is defined through the vector norm Therefore for different quotxquot we have different Theorem Let A be an mxn matrix Then IIAIL mjaxgi n HAIL mitXltZi at 0 1 Theorem Let A be in men and the maximum eigenvalue of AA is 02 Then llAllz 0 Def inner product The inner product of xand y in X is a function of x and y denoted by ltxy gt which satisfies the following axioms 1 ltxygtltyxgt 2 ltxyzgtltxzgtltyzgt 3 ltcxygtcltxygtfor all scalars c 4ltxxgt20 and ltxxgt 0 if x 0 Example n In the inner product is defined as n x ltxygtxy xzyz 21 Theorem CauchySchwarz Inequality ltxygtSltxxgt12 ltyygt12 The equality holds if and only if x and y are linearly dependent Theorem llt xx gt has the property of a norm 27 Normed Linear Spaces and Inner Product Spaces M normed linear spaces A linear space on which a norm is defined is called a normed linear space Example n The space C equipped with the norm 1p p p X Xquotquot39X p21 P 1 n is a normed linear space The space is denoted by PM Example 2P spaces Let 1 s p lt co The space F consists of all infinite sequences of scalarsX X such that 1 n i x lPlt so 11 The norm in p is defined by Be Up lell 2 x v 11 The space M is defined to consist of the bounded sequences X1 X withthe norm n quotxllw supl x1 l l Example Lp spaces For p 2 1 the space Lp ab consists of those real valued functions X on the interval ab for which p xt is Lebesgue integrable The norm on this space is defined as b 1 lell fixer air The space Loo ab is the space of all functions on ab which are bounded except possibly on a set of measure zero The norm of a function in this space is defined as lellw ess sup i xti M Cauchy sequence i A sequence X in a normed linear space is said to be a Cauchy sequence if n m x x H gt 0 as nm gtOo ie for each s gt 0 there exists an integer N such that n m x x lts forallnmgtN In a normed linear space every convergent sequence is a Cauchy sequence Why However a Cauchy sequence may not be convergent M Banach space A normed linear space X is complete if every Cauchy sequence of X has a limit in X A complete normed linear space is called a Banach space Example P 1 s p 3 co is a Banach space M Innerproduct space A linear space on which an innerproduct is defined is called an innerproduct space An innerproduct space X can be also considered as a normed linear space with the norm Jlt xxgt M Hilbert space A Hilbert space is an innerproduct space which is complete with respect to the norm Jlt xx gt Example n The space C is a Hilbert space with inner product ltxygtx y M orthogonal complement Two vectors x y in a Hilbert space X are orthogonal if lt x y gt 0 If M is a closed subspace of X then Mi the orthogonal complement of M is the set of all vectors in X which are orthogonal to every vector in M That is XML M 28 Timedomain and Frequencydomain Spaces Timedomain spaces L2oooo Consider a signal vector xt X t X t t 1 2 n where Xlt i 12quot39n is a complexvalued function I defined for all time 00 lt t lt co Assume that the signals under consideration satisfy IZxZt2 dt lt co mp1 The space of all such signals is denoted by n L2 0000 or simply by L20000 This space is a Hilbert space with innerproduct lt xy gt J xtytdt Then the norm of X is llxllz i glxmr on Frequencydomain space L2 Consider a function xjwx1jw xzow xjwT Where xlja i 12quot39n is a complexvalued function defined for all frequencies 00 lt 03 lt 0 31 Assume that the functions under consideration satisfy IZxljw2 dwlt co m 11 The space of all such functions is denoted by n L2 or simply by L2 This space is a Hilbert space with innerproduct 1 w x lt xy gt lx1w yJwdw 27A The norm on L2 is 1 w n 12 2 llxllz glwml cm RL2 space the set of rational functions in L2 with real coeffi cients These functions are strictly proper and have no poles on the imaginary axis n H2 H2 space a closed subspace of L2 with functions Xs analytic in Res gt 0 J H2 space the orthogonal complement of H2 in L2 nxm 0c Loo space L the space of nxm complexvalued matrix functions whose largest singular values are essentially bounded The Loonorm ofamatrix function s inL is 00 quotCDIL 2 ess supECDja RLOO space s E RLOO iff s is proper rational with real coefficients and has no poles on the imaginary axis nxm 0c H0 space H a closed subspace of Loo with functions s analytic in Res gt 0 RH0 space the set of proper stable rational matrices with real coefficients Notes on Linear Robust Control MEM 633 October 2 2002 Professor Harry G Kwatny Of ce 3151A hkwatnycoedrexeledu I a CS 6 e 6 ac MEM 633634 Notes Professor K warn y Contents 1 Introduction to the Robust Control Prnhlem 1 2 State Space Models 2 21 Solutions of Linear Systems 2 22 The MatriXF r quot 39 2 23 C quot 39 quotquot 2 More on Invariance 7 24 Observability 7 More on Invariance 9 25 Kalman D r quot39 9 26 ThorpMorse Form 9 27 Zeros 9 3 Nominal Controller Design State Space Perspective 12 31 State Feedback 12 Pole Placement 12 The Linear Regulator Problem 12 32 Observers amp the r quot Principle 12 33 Disturbance Rejection 12 4 Transfer Function Mm ek 13 41 State Space to Transfer Function 13 42 Frequency Response 13 43 Poles amp Zeros of Transfer Functions 13 44 Reali atinn 13 5 Closed Loop Transfer F quot 15 51 Well Posedness 15 52 Closed Loop Transfer F quot 15 Output 15 Error 15 Control 15 MEM 633634 Notes Professor K warn y 6 Performance in the Frequency nnmain 61 a R J F U A fundamental tradeoff 62 quot39 quotJ Peaks 63 B 1 39Ju 64 Limits on Performance 7 Nominal Controller Design Frequency Domain Perspective 71Eguati0n Section 1 Full State Feedback Controllers The Quadratic Regulator Problem MinMax Control Solving the Riccati Equation 72 Output Feedback Controllers The Classical Hz Problem 7 LQG The Modern Paradigm Solution Summary 8 Robust Stability amp Nyquist Analvsis 81 SISO Nyquist Analysis Guaranteed Gain and Phase Margin 82 MIMO Nyquist Analvltilt 83 M A Structure 84 Small Gain Theorem 85 Robust Stability of the M A Structure 9 Robust Performance 10 Some Complex Variable Concepts 101 Analytic Functions Residue Theorem Poisson Integrals Bode Waterbed Formula 102 Parseval s Theorem 11 Normed Linear Spaces iii MEM 633634 Notes Professor Kwatn y 111 Norms and Normed Linear Spaces 40 Examples of normed linear spares39 40 112 System Norms Induced Norms 42 Transfer matrix norms 42 MEM 633634 Notes Professor Kwatn y 1 Introduction to the Robust Control Problem MEM 633634 Notes Professor Kwatn y 2 State Space Models 21 Solutions of Linear Systems 5c Atx Btu xt x0u ltIgttt0 x0 j t sBsusds ltIgtt 7 Atlt1gtt 7 ltIgtt t I 22 The Matrix Exponential Cayley Hamilton Theorem Every square matrix satis es its own characteristic equation ie if l M A xiquot aHl a0 Then A Aquot an1Aquot 1a01 0 From this we obtain a w0t1 051tA0 n1tA 1 23 Controllability We brie y review some basic concepts and results for linear autonomous systems c Ax Bu y Cx Where xE Rquotu E R39quot ye RF Recall that given an initial state x0 x0 and a control ut tgt 0 the corresponding trajectory is de ne by the variations of parameters formula x t39xo l u eA xo IeM SUBuwds 0 MEM 633634 Notes Professor Kwatn y De nition A state x1 E Rquot is reachable from x0 if there exists a nite time tgt 0 and a piecewise continuous control u such that xtx0u x1 RXU denotes the set of states reachable from x0 Let us make a few preliminary observations If x1 is reachable from x0 in some time II gt 0 it is reachable in every time I To see this simply rescale s 21 2 IeA quotSBusds IeA rBud 0 0 0 Thus we have the replacement us gt e quot u Notice that x1 is reachable from x0 if and only if x1 eA x0 is reachable from the origin for 0lttlt viz l l cl eA xO IeM SUBuMdS ltgt cl eA xO IeA SBusds 0 0 As a result we focus on characterizing the set of states reachable from the origin Let denote the linear vector space of control functions 117 TE 0J1 and 3272Rnthe space of states x01 The inner product on 26 is the usual 21 tau 0 77u7d7 The mapping A 26 gt 3 defined by x01 eA 1 Burdr 11 defines the state reached from the origin at time II when the control u is applied on the time interval 0 t1 A state x1 is reachable from the origin over the time interval 0 t1 if and only if the relation Au x1 has a solution ut Such a solution exists if and only if x1 E ImA It is more convenient to apply the equivalent condition x1 E ImAA because AAquot 32 gt3 So let s us prove this result MEM 633634 Notes Professor Kwatn y Lemma 1 There exists a solution u of Au x1 if and only if x1 E ImAAquot Proof Sufficiency xl E ImAA 3 cl E ImA is obvious To prove necessity assume cl Au and c1 Elm114 Since 3 hnA kerA x1 has a component in kerAi It follows that there exists an x2 such that 1414sz 0 and x1sz 0 Now XZTAAXZ 0 3 Aix2 0 3 Alx2 0 x1sz Au x2 9 Lgfix2 gt7 0 contradicting the condition x1sz 0 Qed To use this result we need to calculate the adjoint mapping Al 3 gt 9K It is defined by ltxAug ltAxugt Thus AxT mm x7 eA SBuSds 1 T A7014 T J1 B e x utdt From this we identify Aquotx BTeAquot 1quotx 12 so that AA x eAquot TBBTeAT d7 x Consequently in view of Lemma 1 we have the following result Proposition The set of states reachable from the origin over the time interval 0t1 is Im AAquot Im eAlt 1 BBTeA 1 517 De nition The matrix 2 7 T 7 Go t1 0 e BBTeA 1 an MEM 633634 Notes Professor Kwatn y is called the controllabz39lz39ty Gramian De nition The system or the matrix pair AB is said to be completely controllable if any state x1 E Rquot is reachable from any other state x0 E R in nite time Proposition The system is completely controllable if and only if rank GO t1 71 Choose a point 776 3 and use it to construct a control ut 14 77 BTeArlt T W Starting at x0 0 apply the control to obtain x01 AAquot 77 eA l BBTeAquot 1 d7 77 Gct177 13 Any x1 reachable from the origin satis es this relation for some 7 Hence a control that steers the system from the origin to x1 on the interval 0 II is ut An BTeAquot 1 n 14 for any 77 that satis es Gct177xl 15 If the system is completely controllable there is a unique 77 yielding ut BTeAquot 1quotGt1yg 16 It is interesting to note that if the system is unstable the control at any time t becomes small as II gt 0 The opposite occurs if the system is stable This is consistent with the well known fact that highly maneuverable vehicles tend to be unstable or marginally stable Let B ImB and Al BBABA 1B ImBABA lB Then we have the following basic result Theorem R0 ltAB Proof We need to show that ltABgtImGCttgt0 MEM 633634 Notes Professor Kwatn y Since GO t is symmetric 3 3 ImGC JD ker GO this is equivalent to l ltABgt kerGCt t gt 0 First show xe ker GO t 3 xe ltABgtL If xE ker GO t then xTGCx 0 so that L HxTemeHZ d7 0 0 lt 7 ltr Therefore Jae 30 0ltsltt Expanding e and comparing coefficients leads to xTA B 0 139 0n l Consequently xTBAB AHB 0 which implies that x is orthogonal to AIB ie xe AIRi Now show xe kerGCt lt xe ltABgti But xe AIBi implies xTBABAquot lB 0 so reversing the above steps leads to xTGCx 0 This is true only if xe ker GO t Qed Theorem The system or the matrix pair AB is completely controllable if and only if R0 Rquot or equivalently rankBABAquot lB n Proof Qed We wish to emphasize the geometric aspects of controllability and observability To do so fully requires the concept of an invariant subspace De nition A subspace V g Rquot is invariant with respect toA if AV g V Clearly every eigenvector de nes a onedimensional invariant subspace Furthermore the set of all vectors h satisfying An in is called the eigenspace ofA associated with the eigenvalue 1 Every eigenspace ofA is an invariant subspace as is every subspace that can be constructed as the sum of 6 MEM 633634 Notes Professor Kwatn y eigenspaces Perhaps less obvious is the fact that every invariant subspace is the direct sum of eigenspaces RO is Ainvariant ie AR0 g R0 In fact RO is the smallest Ainvariant subspace of Rquotcontaining B Moreover if dimR0 n1 there exists a system of coordinates in which the state equations take the form 9h A11 A12 x1 Bl u X16 Rnl X26 Rnn1 x2 0 A22 x2 0 such that the pair 141131 is completely controllable ie All B1gt Rquot and in fact x1 are coordinates in R0 Hence the restriction of the system to RO x20 results in a controllable system Thus we refer to RO as the controllable subspace More on Invariance Recall that application of the linear control u Kx v results in the closed loop system 5c ABKxBv This motivates the following de nition De nition A subspace VQR is 1437mvariant if there exists a state feedback matrix K such that ABKVg V Now the following theorem can be established Theorem V g Rquot is AB7invariant if and only if A V g V B 24 Observability We brie y review some basic concepts and results for linear autonomous systems X Ax Bu y Cx where xE Rquotu E R ye RF Recall that given an initial state x0 x0 and a control ut tgt 0 the corresponding trajectory is define by the variations of parameters formula MEM 633634 Notes Professor Kwatn y l xtx0u eA x0 IeM SlBuSds 0 De nition A state x16 Rn is indistinguishable from x0 if for every nite time t and piecewise continuous control ut ytx1u ytx2u 1x0 denotes the set of states indistinguishable from x0 De ne 972 ker CAH 11 Theorem 10 91 proof De nition The system or the matrix pair CA is said to be completely observable if knowledge of ut and yt on a nite time interval determines the state trajectory on that interval Theorem The system or the matrix pair CA is completely observable if and only if 10 Q proof Wonham 10 is Ainvariant In fact 10 is the largest Ainvariant subspace of Rn contained in kerC If dim10 M1 there exists a set of coordinates in which the system equations are in the form u x16 x2e x2 A21 A22 x2 B2 X1 y010 X2 MEM 633634 Notes Professor Kwatn y such that the pair Cl111 is completely observable and x2 are coordinates in 10 We call I 0 971 the unobservable subspace More on Invariance II39I Similar results to those 39 quot 39 J for quot 39 quotquoty can be A for observability De nition A subspace V g Rquot is CA7invariant if there exists a matrix F such that A F C V g V Note that AFC is the closed loop matrix resulting from output injection Theorem V g Rquot is CAiinvariant if and only if AVnkerc g V 25 Kalman Decomposition 26 TharpMorse Form 27 Zeros Recall for a linear system the output is related to the input and initial state by m C31 A 1x0 Cw AFB DUs G C I A IBD k k SZlquot39szm S l S 1 do s ps p Y0 CNsx0 ns l ds ds s X A partial fraction expansion of the second term shows that this equation can be written in the form CNsx0 c1 YS ds dsS1w1th c GM Moreover it can be shown that it is always possible to choose x0 so that the term in brackets vanishes If x0 is so chosen then YS 51 S MEM 633634 Notes Professor Kwatn y Moreover if X is a system zero c1 GM 0 Thus Ys 0 3 yt 0 In summary if X is a system zero there exists x0 such that xt x0 and ut e results in yt 0 This is the essential property of 8180 system zeros that we intend to generalize Consider the MIMO system 5c Ax Bu y Cx Du Suppose ut gel g E R39quot We ask if there exists a solution of the form xt x06 x0 E Rquot such that yt E 0 The assumed solution must satisfy 2x06 Aerh Bgeh 0 Cxoeh Dgeu This leads to lI Ax0 Bg 0 Cx0 Dg 0 or M A B x0 0 17 C D g This represents n p equations in nm unknowns Suppose M A B r rank C D Let us consider the following cases p lt m equation 17 always has nontriVial solutions If the system matrix has maximum rank r n p there are m p independent solutions m lt p and r 2 71 m there are no nontriVial solutions ofl7 r lt n minm p there are nontriVial solutions MEM 633634 Notes Professor Kwatn y MEM 633634 Notes Professor Kwatn y 3 Nominal Controller Design State Space Perspective 31 State Feedback Pole Placement The Linear Regulator Problem 32 Observers amp the Separation Principle 33 Disturbance Rejection MEM 633634 Notes Professor Kwatn y 4 Transfer Function Models 41 State Space to Transfer Function 42 Frequency Response 43 Poles amp Zeros of Transfer Functions 44 Realizations MEM 633634 Notes Professor Kwatn y MEM 633634 Notes Professor Kwatn y 5 Closed Loop Transfer Functions 51 Well Posedness 52 Closed Loop Transfer Functions d 1 Y GU D1 U KR Y D2 E R Y Output Y0 I GsKs lGsKsRs 1 GsKs lGsD1s 1 GsKs 1 GsKsD2 s Error Es1 GsKs 1Rs 1 GsKs lGsD1s 1 GsKs39lGsKsDz s Control Um Ks1 GsKs391Rs Ks1 GsKs lGsD1s Ks1 GsKs 1I 26sKsD2 s MEM 633634 Notes Professor Kwatn y 6 Performance in the Frequency Domain 61 Sensitivity Functions Es 1L 1Rs 1L IGD1s1L 1LD2s where L GK Sensitivityfunction S I LII Complementary sensitivity function T I L71L Consider a scalar system in which L GK is the open loop transfer function and T 1L 1L is the closed loop transfer function Then compute the relative variation of the closed loop with respect to relative variation of the open loop transfer function dTT dT dLL dL T 2 1 L 1L L1L 1L71L 1L 1L1 1LS This is Bode s original reason for the terminology sensitivity function for S A fundamental tradeoff Note that I L 11 L 1L I STI T 1 L 1L 1 I1 LI L 1 GI KG 1 1 GK 1 G GKI GK 1 GI KG 1K 1 GK 1 GK Es SSRs SSGD1S TSD2s Us Ks1 GsKs 1Rs Ks1 GsKs lGsD1s Ks1 GsKs l1 2GsKsD2 s MEM 633634 Notes Professor Kwatn y 62 Sensitivity Peaks MS maXSja MT maXTja Sensitivity peaks are related to gain and phase margin Sensitivity peaks are related to overshoot and damping ratio Im A L plane Eli Slt1 I 5 a Re L060 ISI S 1 1L 63 Bandwidth Bandwidth sensitivityaBS maXVISjw lt lJE Va E 0 v Bandwidth complementary sensitivity a M mvinivSja lt 1 J5 Va E v 00 Crossover frequency a mVaXvLja Z 1 Va E 0 v Bandwidth is related to rise time and settling time 64 Limits on Performance MEM 633634 Notes Professor Kwatn y 7 Nominal Controller Design Frequency Domain Perspective 71 Full State Feedback Controllers The Quadratic Regulator Problem Jx t xTTQfxT f x7rQxr uTrRu7d7 Principle of Optimality amp HJB Equation Stability of Sol ns to LQR Consider the linear system 5c Ax Suppose Vx xTPx with P gt 0 Compute the time rate of change of Valong trajectories V VxxTPAxxTATPx 11 x Suppose V xTQx Q gt 0 Then clearly the system is asymptotically stable xt gt 0 as t gt 0 Actually if detA 0 we require only Q 20 Hence if there exists a P gt 0 that satis es the Liapunov equation PAATP Q 12 for any Q 2 0 the system is asymptotically stable provided detA 0 As a matter of fact this requirement is necessary as well as sufficient for asymptotic stability Now consider the linear control system c Ax Bu 13 Suppose that u Kx so that the closed loop system is cABKx 14 Moreover choose K R IBTPx where PAATP PBR IBTP Q 15 Now 15 can be rewritten PABK ABKTP Q PBR IBTP MEM 633634 Notes Professor Kwatn y A Generalization of the Standard LQR Consider the more general performance index J T T T T T x t x TQfxT 2 x 7Qx7 2x 7Sut u 7Ru7d7 with R gt 0 and Q SRTIS 20 We can reduce this problem to the standard case by noting the identity complete the square T T fo ZxTSu uTRu u R 1Sx R u R 1Sx x7 Q SR 1S7x De ne a new control L7 u RTISTx so that the system equations become x AxBu 2 x A BR 1STxB and in terms of the new control the performance index is Jx t xTTQfxT I r Q SR 1S7x2 77R 7d7 Thus the problem has been recast in the from of the standard regulator problem The solution is u R 1BTP S x P A BR 1ST A BR 1STT P PBR IBTP Q SR IST MinMax Control Consider a system by disturbances wt with performance outputs yt 5c Ax Bu Ew 16 y Cx The disturbance is norm bounded but otherwise unknown Our goal is to find a state feedback control that produces minimum quadratic cost for the worstcase disturbance To make this statement precise we set up the performance index Ju w Imiympu7rur y2wTrwrdr 17 MEM 633634 Notes Professor Kwatn y where the weighting constants pygt 0 By explicitly including the disturbances in the cost in a negative way stationary points of J u w will be saddle points We seek ut that minimizes J while a perverse nature seeks wt that maximizes it The optimization problem is to nd min max Ju w 14 w Proposition Consider the system 16 with performance index 17 Suppose that l wt has bounded energy 2 A B and AE are stabilizable 3 A C is detectable Then if the optimal solution exists it is a unique saddle point of J u w where l The optimal minmax control is ut Kxt with K iBTS p 2 The worst case disturbance is wt LZETSxU V S is the unique symmetric nonnegative solution of the algebraic Riccati equation ATSSA SiBBT i2EET CTC 0 9 Recall that solution of the Riccati equation can be carried out by eigendecomposition of its associated Hamiltonian matrix As long as the Hamiltonian matrix has no eigenvalues on the imaginary axis the required decomposition can be performed The Hamiltonian matrix for the minmax optimization problem is A LZEET iBBT H V 0 C TC AT 20 MEM 633634 Notes Professor Kwatn y Given the detectabilitystabilizabilty assumptions it is possible to prove there exists a ymin so that there are no eigenvalues of H on th imaginary axis provided 9 gt ymin In fact when 9 gt Do we approach the standard LQR solution For 9 ymin the controller is the full state feedback H M controller All other values of ymin S ylt 0 produce valid min max controllers The condition that Re 1H 0 is equivalent to stability of the matrix A izEET 337 9 0 Notice that the closed loop system matrix is A iBBT 0 Since EET 92 is destabilizing the feedback system has some margin of stability Solving the Riccati Equation Constructing the Solution Computing Tools CARE Solve continuous time algebraic Riccati equations XLGRR CAREABQRSE computes the unique symmetric stabilizing solution X of the continuous time algebraic Riccati equation 1 A XE E XA E XB SR B XE S Q O or equivalently l l l F XE E XF E39XBR B XE Q SR 8 O with FABR 839 When omitted RS and E are set to the default values RI 80 and EI Additional optional outputs include the gain matrix 1 G R B XE S39 the vector L of closed loop eigenvalues ie EIGA BGE 21 MEM 633634 Notes Professor Kwatn y and the Frobenius norm RR of the relative residual matrix XLGREPORT CAREABQ report turns off error messages and returns a successfailure diagnosis REPORT instead The value of REPORT is 1 if Hamiltonian matrix has eigenvalues too close to jw axis 2 if XX2Xl with X1 singular the relative residual RR when CARE succeeds X1X2LREPORT CAREABQ implicit also turns off error messages but now returns matrices X1X2 such that XX2Xl REPORTO indicates success 72 Output Feedback Controllers The Classical H2 Problem LQG The classical output feedback optimal control problem for SISO systems was solved during the 2quotd World War using a frequency domain formulation It is referred to as the WienerHopfKolmogorov problem Attempts to extend this result to the MIMO case using frequency domain techniques were not fruitful The MIMO problem was formulated and solved in the state space by Kalman and coworkers around 1960 We summarize the result here The Linear Quadratic Guassian LQG Problem Setup The standard problem formulation is as follows The plant is described by 5c Ax Bu w y Cx v The disturbances w v are independent zeromean white noise processes have covariances Ewtw7 7 W50 7 Evtv7 7 V50 7 and EwtvT7 0 22 MEM 633634 Notes Professor Kwatn y We seek ut that minimizes the performance index 7 J Eii jo xTQxuTRudt Q QT 201 RT gt0 Solution Summary u Kfct K R IBTP PA ATP PBR IBTP Q P 2 0 fc AfcBu Ly CcL SCTV 1 SAT AS SCTV 1CS W S 20 The Modern Paradigm We View the control design problem in terms of the diagram shown in Figure 1 W Z gt P gt gt u y K Figure 1 The socalled modern paradigm Views the plant in terms of two input sets disturbance and control inputs and two out put sets performance and measured variables In the frequency domain the plant is characterized in terms of a transfer matrix H i 213 uKv Closing the loop with Produces the closed loop transfer function ZFwa FP11P12IPZZK71P21 23 MEM 633634 Notes Professor Kwatn y In the state space the plant is de ned as follows in terms of differential equations 5c AxBlwBzu z C1xD11wDuu y C2xD21wD22u and for convenience we define the data structure A B1 32 P C1 D11 D12 C2 D21 D22 In the following we will make the following standing assumptions about the plant 1 D11 0 2 A Bl is stabilizable 3 A C2 is detectable V B D271 V zo witthygt0 9 JV m u 5 R1TC D R Rmgt0withRu gt0 39 1 12 39 RT Ru 11 Our goal is to find an output y feedback controller that optimizes a performance measure defined in terms of the performance variables 2 for some specified class of disturbance inputs w 24 MEM 633634 Notes Professor Kwatn y PLANT I W 21 Figure 2 The standard LQG problem recast in the modern paradigm Frequency Domain Formulation of the H2 Problem The energy norm or 2norm of any scalar function f t is de ned by 2 w 2 L w 2 ft2 L f tdt 2 L F 1al da The latter obtained from Parseval s theorem This is easily generalized to vector a function fr I fTrFTdr jFTjwFjwdw Now let us consider the control system shown in the gure Suppose we consider the class of disturbances to be a zero mean white noise with EwT twt 7 157 We seek to choose K so that the expected value of zt2 is a minimum Thus we seek to Z mgnEllzltrgtll Recall 150 1 Now compute 25 MEM 633634 Notes Professor Kwatn y I zTtztdt ij ZT jaZjadcu em 27 w tr i cZUw K jwww 2 l T Ez2 trELGEZjwZ jwda 1 T tr j FjaF jwda 27239 A 2 art The H Problem We consider the same control design problem as above except with the class of disturbances de ned by quotW0L1 We seek to choose K such that the maximum of quot20quot2 over all disturbance inputs wt is a minimum Thus we seek to min max quot20 2 K W21 Now compute I zTtztdt ir ZT jwZ 1mm 7 4a 1 M E MWT JwFT JWFJCUWJCUdw Now we seek the maximum performance energy over all disturbances with unit norm It occurs when W060 is aligned with the maximum eigenvalue of F F 1 m W7 to F 3960 today Z La o o WW1 13quot 52 Fm IIFII max 27 tztdt Mfl Solution Summary 26 MEM 633634 Notes Professor Kwatn y Proposition H2 Output Feedback Suppose wt is a unit intensity white noise signal E wtwT 7 I 5 t 7 Then the unique stabilizing optimal controller that minimizes we e m 0 A2 43ze L2C2 L2D22F2 Fz R R B Xz lszwSllz is were L2 Y2CZT VW Vy1 X 2 Y2 satisfy the two Riccati equations XZAr ArTX2 XszR7uleTX2 Rgtoc RmRIZR AZYZ Y2A Y2CTV 1C2YZ Vu V V IV 2 JV Wyy A7 ABZR74R 0 W 2 A2 A V VIC 27