Robust Control - EPFL
ÉCOLE POLYTECHNIQUEFÉDÉRALE DE LAUSANNE Robust Control Alireza KarimiLaboratoire d’Automatique
Course Program1. Introduction2. Norms for Signals and Systems10. Linear Fractional Transformation3. Basic Concepts (stability and performance)11.4. Uncertainty and Robustness12. Model and Controller Reduction5. Stabilization (coprime factorization)13. Robust Control by Convex Optimization6. Design Constraints14. LMIs in Robust Control7. Loopshaping15. Robust Pole Placement8. Model Matching16. Parametric uncertaintyH Control9. Design for PerformanceReferences: Feedback Control Theory by Doyle, Francis and Tannenbaum (on the website of the course) Essentials of Robust Control by Kemin Zhou with Doyle, Prentice-Hall, 1998Robust Control2
IntroductionClassical ControlRobust Control-u(t)-P0y(t)-u(t) --P0? y(t)-P P0 P0 : nominal model : plant uncertaintyUncertainty sources:Structured: parametric uncertainty, multimodel uncertaintyUnstructured: frequency-domain uncertainty, unmodeled dynamics, nonlinearityRobust Control Objective: Design a controller satisfying stability andperformance for a set of modelsRobust Control3
Model Uncertainty and FeedbackFeedback control has been used for the first time to overcome the model uncertaintyr(t)-T S CP1 CP11 CPe(t)-6Cu(t)-Pv(t)- ? y(t)Transfer function between r and yTransfer function between v and yFor very large CP , T 1 (tracking) and S 0 (disturbance rejection) whatever the plant model is.For an open-loop stable system:C 0 robust stabilityLoopshaping:- C robust performance C(jω)P (jω) should be large in the frequencies where good performances aredesired and small where the stability is criticalIntroduction4
Norms for Signals and SystemsNorms for signals: Consider piecewise continuous signals mapping ( , ) to R. Anorm must have the following four properties:1. u 0 (positivity)2. au a u , a R (homogenity)3. u 0 u(t) 0 t (positive definiteness) u v u v (triangle inequality) 1-Norm: u 1 u(t) dt4. 2-Norm: u 2 1/2u2 (t)dt u 22 is the total signal energy -Norm: u sup u(t) p-Norm: u p Robust Controlt 1/p u(t) p dt1 p 5
Norms for SignalsAverage power of a signal is denoted by: pow(u) 1T 2T TlimRemark: One says u(t) 1/2u2 (t)dtpow is not a norm (it is not positive definite) T Lp if u p where Lp is an infinite-dimensional Banach space (L 2is an infinite-dimensional Hilbert space as well).Recall: A Banach space is a complete vector space with a norm. A Hilbert space is a complete vector space with an inner product x, y such that the norm defined by x x, x . A Hilbert space is always a Banach space, but the converse neednot hold.1(t) L but / L1 , / L2u(t) (1 e t )1(t) L but / L1 , / L2u(t) e t 1(t) L , L1 , L2 (Besides, u(t) is a power signal pow(u) 0)u(t) sin(t) L but / L1 , / L2 (u(t) is a power signal)Examples:Norms for Signals and Systems6
Norms for SystemsWe consider linear, time-invariant, causal and usually finite-dimensional systems. y(t) G(t) u(t),y(t) G(t τ )u(τ )dτ ,Ĝ(s) L[G]Some definitions: Ĝ(s) is stable if it is analytic in the closed RHP (Re s 0) Ĝ(s) is proper if Ĝ(j ) is finite (deg den deg num) Ĝ(s) is strictly proper if Ĝ(j ) 0 deg den deg num Ĝ(s) is biproper if (deg den deg num)Norms for SISO systems: 2-Norm: Ĝ 2 12π Ĝ(jω) 2 dωParsval’s theorem: (for stable systems) Ĝ 2 12π Ĝ(jω) 2 dωNorms for Signals and Systems 1/2 1/2 -Norm: Ĝ sup Ĝ(jω) ω 1/2 G(t) 2 dt7
Norms for SystemsRemarks: L2 is a Hilbert space of scalar-valued functions on jR. The inner product for this Hilbert space isdefined as:1 F̂ , Ĝ 2π F̂ (jω)Ĝ(jω)dω We say Ĝ(s) L2 if Ĝ 2 . It is the case iff Ĝ is strictly proper and has no poles on theimaginary axis. We say Ĝ(s) L if Ĝ . It is the case iff Ĝ is proper and has no poles on theimaginary axis. L is a Banach space of scalar-valued functions on jR. Ĝ is the peak value of the Bode magnitude plot of Ĝ. It is also the distance from the origin tothe farthest point on the Nyquist plot of Ĝ. H2 and H are respectively subspaces of L 2 and L with Ĝ(s) stable (Hp spaces areusually called Hardy spaces).1Examples: s 1 L2 , L but / H2 , H Norms for Signals and Systemss 1s 2 H but / H21s2 1 / L2 , L 8
Norms for SystemsNorms for matrices:1-Norm: The maximum absolute column sum norm is defined as A 12-Norm: The spectral norm or simply the norm of A is defined as: max A 2 j -Norm: The maximum absolute row sum norm is defined as A maxiF-Norm: Frobenius norm is defined as A F n aij .i 1λmax (A A).n aij .j 1trace(A A) Ax pInduced p-norm: The induced p-norm is defined from a vector p-norm: A p maxx 0 x pRemarks: The induced 2-norm and the norm of A are the same and called also the natural norm. This normis also equal to the maximum singular value of A. A σ̄(A). The spectral radius ρ(A) λmax (A) is not a norm.Norms for Signals and Systems9
Norms for SystemsNorms for MIMO systems: GivenĜ(s) a multi-input multi-output system2-Norm: This norm is defined as Ĝ 2 12π 1/2 trace Ĝ (jω)Ĝ(jω) dω -Norm: The H norm is defined as Ĝ sup Ĝ(jω) sup σ̄[Ĝ(jω)]ωωRemark: The infinity norm has an important property (submultiplicative) ĜĤ Ĝ Ĥ Norms for Signals and Systems10
Computing the NormsHow to compute the 2-norm: Suppose that Ĝ12 Ĝ 2 2π 12 Ĝ(jω) dω 2πj L2 , we have:j 1Ĝ( s)Ĝ(s)ds 2πj j Then by the residue theorem, Ĝ 22 equals the sum of the residues ofĜ( s)Ĝ(s)dsĜ( s)Ĝ(s) at its poles in theleft half-plane (LHP).How to compute the -norm:Choose a fine grid of frequency point {ω 1 , . . . , ωN }, then an estimate for Ĝ is:max Ĝ(jωk ) 1 k Nor for the MIMO systems:max σ̄[Ĝ(jωk )]1 k NMatlab commands:norm, bode, frsp, vsvdNorms for Signals and Systems11
Computing the NormsState-space methods for 2-norm: Consider a SISO state-space model L ẋ(t) Ax(t) Bu(t)y(t) Cx(t)Ĝ(s) C(sI A) 1 BFrom Parseval’s theorem: Ĝ 22 G 22 sx̂(s) Ax̂(s) B û(s)ŷ(s) C x̂(s)impulse responseCetA BH2T: G(t) CetA B B T etA C T dt CLC T0 where L TetA BB T etA dt0is the observability Gramian and can be obtained from following Lyapunov equation:AL LAT BB T 0 and the 2-norm is Ĝ 2 (CLC T )1/2 T 1/2For MIMO systems we have Ĝ 2 trace(CLC )Norms for Signals and Systems12
Computing the NormsState-space methods for -norm: Consider a SISO strictly proper state-space model Theorem:L . Ĝ γ iff the Hamiltonian matrix H has no eigenvalues on the imaginary axis: Aγ 2 BB T H C T C ATBisection algorithm:1. Select γu and γl such that γl Ĝ γu ;2. If (γu γl )/γl specified level., stop and Ĝ (γu γl )/2. Otherwise continue;3. Set γ (γu γl )/2 and test if Ĝ γ4. If H has no eigenvalue on jR, set γ uFor MIMO biproper (DH γ otherwise set γl γ ; go back to step 2. 0) systems:A BR 1 D T CBR 1 B T C T (I DR 1 D T )C (A BR 1 D T C)TNorms for Signals and Systems and R γ 2 I D T D13
Input-output relationshipsIf we know how big the input is, how big is the output going to be?Output Norms for Two InputsProofs: If u(t) δ(t) then y(t) u(t)δ(t)sin(ωt) y 2 Ĝ 2 y G Ĝ(jω) G(t τ )δ(τ )dτ G(t), so y 2 G 2 Ĝ 2 If u(t) δ(t) then y(t) G(t), so y G If u(t) sin(ωt) then y(t) Ĝ(jω) sin(ωt φ), so y 2 and y Ĝ(jω) Norms for Signals and Systems14
Input-output relationshipsSystem Gain:sup{ y : u 1}u(t) L2u(t) L y 2 Ĝ y Ĝ 2 G 1Proofs:Entry (1,1): We have122 y 2 ŷ 2 2π 22 Ĝ(jω) û(jω) dω2 1 Ĝ 2π û(jω) 2 dω Ĝ 2 û 22 Ĝ 2 u 22Entry(2,1): According to the Cauchy-Schwartz inequality y(t) G(t τ )u(τ )dτ G2 (t τ )dτ 1/2 u2 (τ )dτ 1/2 G 2 u 2 Ĝ 2 u 2 y Ĝ 2 u 2Norms for Signals and Systems15
Basic Conceptsr--x1 -6du - ? x2 -CvBasic Feedback Loop:F x1 r F x3x1r10F x2 d Cx1 C10 x2 d x3 n P x2n0 P 1x3 1 x110Fr1 P F 1 x2 C C10 d 1 1 P CF 0 P 1nPCPx3 yPx3-? n r F CF d n1Well-posedness: The system is well-posed iff the above matrix is nonsingular. A stronger notion ofwell-posedness is that all the nine transfer functions be proper. A necessary and sufficient conditionfor this is that 1 P CF not be strictly proper (P CF ( )Robust Control 1)16
Internal StabilityIf the following nine transfer functions are stable then the feedback system is internally stable. 1 1 C1 P CF PC P F1P F CF 1P NPNCNF, C , F MPMCMFTheorem: The feedback system is internally stable iff there are no zeros in Re s 0 in the characteristic polynomialNP NC NF MP MC MF 0or the following two conditions hold:(a) The transfer function 1 P CF has no zeros in Re s(b) There is no pole-zero cancellation in Re sBasic Concepts 0. 0 when the product P CF is formed.17
Asymptotic TrackingInternal Model Principle: For perfect asymptotic tracking of r(t), the loop transfer function(with F̂L̂ P̂ Ĉ 1) must contain the unstable poles of r̂(s).Theorem: Assume that the feedback system is internally stable and n d 0.(a) If r(t) is a step, thenlim e(t) r(t) y(t) 0 iff Ŝ (1 L̂) 1 has at least one zero att the origin.(b) If r(t) is a ramp, then(c) If r(t)lim e(t) 0 iff Ŝ has at least two zeros at the origin.t sin(ωt), then lim e(t) 0 iff Ŝ has at least one zero at s jω .t Final-Value Theorem: If ŷ(s) has no poles in Re s 0 except possibly one pole at s 0 then:lim y(t) lim sŷ(s)t s 0cccProof (a): We have r̂(s) and ê(s) Ŝ(s) then lim e(t) lim sŜ(s) . The limit is zerot s 0sssiff Ŝ has at least one zero at origin. For this, P̂ or Ĉ should have a pole at origin.Basic Concepts18
PerformanceTracking performance can be quantified in terms of a weighted norm of the sensitivityfunction1Sensitivity Function: Transfer function from r to tracking error e : S 1 PCPCComplementary Sensitivity Function: Transfer function from r to y : T 1 PCS is the relative sensitivity of T with respect to relative perturbations in P:dT PC(1 P C) P C 2 P (1 P C)1 T /T S lim P 0 P/PdP T(1 P C)2PC1 PCPerformance Specification:1.r(t) is any sinusoid of amplitude 1 filtered by W 1 , then the max. amp. of e is W1 S .2. Suppose that {r W1 rpf , rpf 2 1}, then supr e 2 W1 S .3. In some applications good performance is achieved if S(jω) W 1 (jω) 1 , ω or W1 S 1 W1 (jω) 1 L(jω) , ωBasic Concepts19
Uncertainty and RobustnessPlant Uncertainty: We cannot exactly model the physical systems so there is always the modelingerrors. The best technic is to define a model set which can be structured or unstructured. Structured model set such as parametric uncertainty or multiple model setP {1: amin a amax } or P {P0 , P1 , P2 , P3 }2s as 1 Unstructured model set such as unmodeled dynamics or disk uncertaintyP {P0 : γ}Conservatism: Controller design for a model set greater than the real model set leads to aconservative design.Uncertainty Models (unstructured):Additive uncertaintyMultiplicative uncertaintyP̃ P W2P̃ P (1 W2 )P̃ : true modelRobust ControlP : nominal modelFeedback uncertaintyPPor P̃ 1 W2 P1 W2 : norm-bounded uncertainty W 2 : weighting filterP̃ 20
ExamplesExample 1:k frequency-response models are identified. Find the multiplicative uncertainty model andthe weighting filter.P̃P̃ P (1 W2 ) 1 W2P P̃ (jω) if 1 1 W2 (jω) P (jω) Let (Mik , φik ) be the magnitude-phase at ω i in k -th experiment and (M i , φi ) that of the nominalmodel (e.g. the mean value).Example 2: Suppose that Mik eφik W2 (jωi ) max 1 φkMi e i ikP̃ (s) { s 2: 0.1 k 10}. Represent this model by themultiplicative uncertainty. k0k P̃ (jω) 1 W2 (jω) max 1 W2 (jω) P (s) P (jω) 0.1 k 10 k0s 2The best value for k0 is 5.05 which gives W2 (s)Uncertainty and Robustness 4.95/5.0521
ExamplesExample 3: Assume that P (s) 1s2 andP̃ (s) e τ s s12 where 0 τ 0.1. Find W2 (s) for themultiplicative uncertainty model. P̃ (jω) 1 W2 (jω) e τ jω 1 W2 (jω) ω, τ P (jω)Using the Bode diagram we can find W 2 (s) 1Example 4: Consider the model set { s2 as 10.21s0.1s 1 .: 0.4 a 0.8}. Find W2 (s) for Feedbackuncertainty model.Take:a 0.6 0.2 , 1 1So1P (s)P̃ (s) 2 s 0.6s 0.2 s 11 W2 (s)P (s)whereP (s) Uncertainty and Robustness1, W2 (s) 0.2s2s 0.6s 122
Robust StabilityRobustness: A controller is robust with respect to a closed-loop characteristic, if this characteristicholds for every plant in PRobust Stability: A controller is robust in stability if it provides internal stability for every plant in PStability margin: For a given model set with an associate size, it can be defined as the largest modelset stabilized by a controller. P (1 W2 ) with β , thestability margin for a controller C is the least upper bound of β .Stability margin for an uncertainty model: Given P̃Im LModulus margin: The distance from -1 to theopen-loop Nyquist curve.Mm inf 1 L(jω) inf 1 L(jω) ωω 1 1 S 1 sup ω 1 L(jω)Uncertainty and Robustness 11Re LMmL(jω)23
Robust StabilitySmall Gain Theorem: Suppose (s)H RH γ 0. The following feedback loop isinternally stable for all (s) RH withand let- 1/γ if and only if H γH(s)Remark: For a given with 1/γ the condition H γ is only sufficient and veryconservative. However for all RH , it is a necessary and sufficient condition.Robust stability condition for plants with additive uncertainty:P̃ P W2 H W2 C1 CPClosed-loop system is internally stable forall 1iff W2 CS 1Uncertainty and Robustness- - W2r(t)-6C-P- ? y(t)24
Robust StabilityRobust stability condition for plants with multiplicative uncertainty: CPP̃ P (1 W2 ) H W21 CPClosed-loop system is internally stable forall 1iff W2 T 1Proof: Assume that W2 T - - W2r(t)- - P- C- ? y(t)6 1. We show that the winding number of 1 CP around zero isequal to that of 1 C P̃ .1 C P̃ 1 CP (1 W2 ) 1 CP CP W2 1 CP (1 CP )T W21 C P̃ (1 CP )(1 W2 T )so Wno { (1 C P̃ )} Wno{(1 CP )} Wno{(1 W2 T )}.But Wno {(1 W2 T )} 0 because W2 T 1Uncertainty and Robustness25
Robust StabilityRobust stability condition for plants with feedback uncertainty (1):P 1P̃ H W21 W21 CPClosed-loop system is internally stable forall 1iffW2 ?r(t)- P- Cy(t)-6 W2 S 1Robust stability condition for plants with feedback uncertainty (2):P PP̃ H W21 W2 P1 CPClosed-loop system is internally stable forall 1iff W2 P S 1Uncertainty and RobustnessW2 r(t)-6C-?-Py(t)26
Robust PerformanceNominal performance condition: W 1 S 1Robust stability condition for multiplicative uncertainty:Robust performance for multiplicative uncertainty: W 2 T 1 W 2 T 1 and W1 S̃ 1 where:111S S̃ 1 CP(1 W)(1 CP)(1 WT)1 W2 T1 C P̃22 W1 S Robust performance conditions: W2 T 1 and 1 W2 T 1 Theorem: A necessary and sufficient condition for robust performance is W1 S W2 T 1 W2 CS 1 and W1 S̃ 1 where: 11SW1 S 1 S̃ 1 CP C W21 W2 CS1 W2 CS 1 C P̃Robust performance for additive uncertainty:Or equivalently in one inequality condition:Uncertainty and Robustness W 1 S W2 CS 127
StabilizationThe main objective is to parameterize all of the controllers which provide internal stability for agiven plantTheorem: Assume that PProof: (F 1) 1 1 C1 P CF PC RH (P is stable). The set of all stabilizing controllers is given by: Q Q RH C : 1 PQ P F1P 1 P Q P (1 P Q) 1(1 P Q) F RH 1 PQ Q CF QPQP (1 P Q)1 PQ1 On the other hand, suppose that C stabilizes P then defineCQ RH which leads to C Q : 1 CP1 PQIn this parameterization sensitivity and complementary sensitivity areRobust ControlS 1 PQT PQ28
Coprime FactorizationObjective: Given P , find M, N, X and Y RH such that:NP MNX MY 1Remarks: N and M are called coprime factors of G over RH N and M can have no common zeros in Re s 0 nor at s N (s0 )X(s0 ) M (s0 )Y (s0 ) 0 1 If P is stable we have : M 1, N P, X 0, Y 1 It is easy to obtain N and M , for example:P (s) if kN (s)1 s 1M (s) N (s) 1s 1,M(s) (s 1)k(s 1)k 1 then M and N have a common zero at s , so k 1How to compute X(s) and Y (s)?Stabilization29
Coprime FactorizationEuclid’s algorithm: Given polynomials m(λ) and n(λ) (deg nand y(λ) such that nx my deg m ) find polynomials x(λ) 1.Step 1: Divide m into n to get quotient q1 and remainder r1 :n mq1 r1 , deg r1 deg mStep 2: Divide r1 into m to get quotient q2 and remainder r2 :m r1 q2 r2 , deg r2 deg r1Step 3: Divide r2 into r1 to get quotient q3 and remainder r3 : r1 r2 q3 r3 , deg r3 deg r2Continue Stop at step k when rk is a nonzero constant.Find r3 as a function of m, n and q i :r2r3 (n mq1 ) (m (n mq1 ) q2 ) q3 n(1 q2 q3 ) m( q3 q1 q1 q2 q3 )r1which gives:Stabilizationr111x (1 q2 q3 ) and y ( q3 q1 q1 q2 q3 )r3r330
Coprime FactorizationProcedure to find M, N, X and Y for an unstable plant G:n(λ)Step 1: Transform G(s) to G̃(λ) under the mapping s (1 λ)/λ. Write G̃ m(λ)Step 2: Using Euclid’s algorithm, find x(λ) and y(λ) such that:nx my 1 1/(s 1)Step 3: Find M, N, X and Y from m, n, x and y under the mapping λState-Space Method:Step 1: Transform G(s) to A, B, C and D (state space realization)Step 2: Compute F and H so that A BF and A HC are stable (F -place(A,B,Pf))Step 3: Compute M, N, X and Y as follows: M (s) : X(s) : StabilizationA BFBF1A HCHF0 N (s) : Y (s) : A BFBC DFDA HC B HDF1 31
Controller ParametrizationTheorem: The set of all C s for which the feedback system is internally stable equal: C X MQ:Y NQQ RH NcProof: For C , the stability condition is: (N Nc M Mc ) 1 RH , but we have:McN (X M Q) M (Y N Q) N X M Y 1 (N Nc M Mc ) 1 RH Conversely, if C stabilizes the closed-loop system we should show that it belongs to the above set.C is stabilizing V : (N Nc M Mc ) 1 RH N Nc V M Mc V 1 Y N Q. From the above equation and N X M Y 1 we findNc Vthat Nc V X M Q so the controller C the set of all stabilizing controller. It is easyMc Vto verify that Q RH Let Q be the solution of Mc VRemark: The sensitivity functions are:1 M (Y N Q)S 1 CPStabilizationCPT N (X M Q)1 CP32
ExampleLetP (s) 1(s 1)(s 2)Compute a proper controller C so that:1. The feedback system is internally stable.2. Perfect asymptotic tracking of step reference (d 0).3. Perfect asymptotic disturbance rejection when d sin 10t (r 0).Procedure: Parameterize all stabilizing controllers. Reduce the asymptotic specs to interpolation constraints on the parameters. Find (if possible) a parameter to satisfy these constraints. Back-substitute to get the controller.Stabilization33
Design ConstraintsAlgebraic Constraints: S T 1 so S(jω) and T (jω) cannot both be less than 1/2 at the same frequency. A necessary condition for robust performance is that:min{ W1 (jω) , W2 (jω) } 1, ωSo at every frequency either W 1 or W2 must be less than 1. Typically W 1 is monotonicallydecreasing and W 2 is monotonically increasing. If p is a pole and z a zero of L both in Re s 0 then:S(p) 0 S(z) 1 T (p) 1 T (z) 0Analytic Constraints: Bounds on the weights W 1 and W2 : W1 S W1 (z) Proof from the Maximum Modulus Theorem: W2 T W2 (p) F sup F (s) Re s 0Robust Control34
Analytic ConstraintsAll-Pass and Minimum-Phase Transfer Functions: F (s) RH is all-pass if F (jω) 1 ω G(s) RH is minimum-phase if it has no zeros in Re s 0. It has the minimum phaseamong all transfer functions with the same magnitude (F G where F is all-pass). Every function G in RH can be presented as G G ap Gmp Suppose that L CP has no poles on the imaginary axis, so S (1 L) 1 Sap Smp and 1Smp has no zeros on the imaginary axis. Thus S mp RH . Suppose that z and p are the only zero and pole of P in the closed RHP and C has neither polesnor zeros there. Then:s p s pz pS(z) 1 Smp (z) Sap z p z p Then: W1 S W1 Smp W1 (z)Smp (z) W1 (z)z p s zp z Similarly: Tap and T (p) 1 W2 T W2 (p)s zp z Design Constraints 1Sap(z)35
Analytic ConstraintsExample: Consider the inverse pendulum problem.(M m)ẍ ml(θ̈ cos θ θ̇ 2 sin θ) u m(ẍ cos θ lθ̈ g sin θ) dLinearized model: xls2 g1 22s [M ls (M m)g] s2θ ls2u M m 2dm s2Tuxls g 2s [M ls2 (M m)g]RHP poles and zeros:z ydq- mx -θ luM%g/l p 0, 0,(M m)gMl 1 g no RHP zeroTuy222M ls (M m)gs [M ls (M m)g]For Tux if m M W2 T 1 ( W2 (p) is an increasing function) the system is difficult tocontrol. The best case is m/M and l large.Tuθ For Tuθ and Tuy a larger l gives a smaller p so the system is easier to stabilize.Design Constraints36
Analytic ConstraintsThe Waterbed Effect σ0 jω0 with σ0 0, 1σ0log Smp (s0 ) log S(jω) 2dω2π σ0 (ω ω0 )Lemma: For every point s 0Theorem: Suppose that P has a zero at z with Re zM1 : maxω1 ω ω2 0 and: S(jω) M2 : S Then there exist positive constants c 1 and c2 , depending only on ω 1 , ω2 and z , such that : 1c1 log M1 c2 log M2 log Sap(z) 0Theorem (The Area Formula): Assume that the relative degree of L is at least 2. Then 0log S(jω) dω π(log e)where {pi } denotes the set of poles of L in Re sDesign Constraints Re pii 0.37
LoopshapingObjective: Given P, W1 and W2 find controller C providing internal stability and robust performance: W1 S W2 T 1 or W1 (jω) W2 (jω)L(jω) 1Γ(jω) : 1 L(jω)1 L(jω) Idea: Find graphically L(jω) satisfying the above condition and then compute C ω L/PNote that we assume P is minimum phase and stable.We have:Γ 1 L W1 W2 L and 1 L 1 L 1 L W1 W2 L W1 W2 L Γ 1 L 1 L So if W1 W2 L 1 L Γ 1:In low frequencies L 1 L W1 W1 1 1 W2 1 W2 W1 1 W2 In high frequencies L 1 L 1 W1 1 W1 1 W2 W2 W2 1 W1 Robust Control38
Procedurestep 1: Plot two curves on log-log scale:at LF ( W1 1 W2 ) W1 and at HF1 W2 ( W2 1 W1 )1 W1 W2 step 2: Fit the graph of L on the same plot such that: at low frequency it lies above the first curve and also 1 at high frequency it lies below the second curve and 1 at very high frequency let it roll off at least as fast as does P (so C is proper) near crossover frequency do a smooth transition, keeping the slope as gentle as possible.Because the slope of L determines the phase of L (Bode’s integral): ν 1d ln L ln cothdν L(jω0 ) where ν ln(ω/ω0 )π d ν2The steeper the graph of L near the crossover frequency, the smaller the value of L andlarger the phase marginstep 3: Get a stable, minimum-phase TF for L such that L(0)Loopshaping 0 and compute C L/P39
ExampleAssume that the relative degree of P equals 1. Find L for robust performance if the objective is totrack sinusoidal signals over the frequency range from 0 to 1 rad/s and the weighting function W 2 is:s 1W2 (s) 20(0.01s 1)We can define W 1 as follows (in loopshaping design it is not necessary to have a rational TF for W 1 ): a W1 (jω) 00 ω 1else LF ( W1 1): ω 1 PlotThe larger the value of a, the smaller the tracking errorHF ( W2 1): ω 20 W1 1 W1 in LF (ω 1) andin HF (ω 20)1 W2 W2 Choose L 1b1 W1 and find b such that in HF L ( b 20)s 1 W2 W2 a W1 a 13.15 Find the maximum value of a such that in LF L 1 W2 1 W2 Loopshaping40
Model MatchingT1 (s) and T2 (s), stableproper transfer functions, find a stable Q(s) tominimize T1 T2 Q Trivial case: If T1 /T2 is stable then the uniqueoptimal Q is T1 /T2 and-Objective: GivenQr(t)- T2- T1--? ε(t)γopt min T1 T2 Q 0Simplest nontrivial case:T2 has only one RHP zero at s s 0 . Then by the maximum modulustheorem: T1 T2 Q T1 (s0 ) T2 (s0 )Q(s0 ) T1 (s0 ) γopt T1 (s0 ) T1 T1 (s0 )Note that Q is stable and leads to γopt T1 (s0 ) .T2Example:T1 (s) Robust Control4s 3 ,T2 (s) s 2(s 1)3 Q T1 T1 (2)T2 4(s 1)3 5(s 3)41
Nevanlinna-Pick ProblemProblem: Let {a1 , . . . , an } be a set of points in the open RHP and {b 1 , . . . , bn } a set of distinctpoints in complex plane. Find a stable, proper, complex-rational function G satisfying: G 1 and G(ai ) bi , i 1, . . . , n1 bi bjSolvability: The NP problem is solvable iff the n n Pick matrix Q, whose ij th element isai ajis positive semidefinite (Q 0). Note that Q is Hermitian (Q Q where Q is the complexconjugate transpose of Q). Q 0 iff all its eigenvalues are 0.Mobius Function: A Mobius function has the form:z bMb (z) 1 zbwhere b 1 Mb has a zero at z b and a pole at z 1/b so M b is analytic in open unit disk. Mb maps the unit disk onto the unit disk and the unit circle onto the unit circle. The inverse map Mb 1 Model Matchingz b M b is a Mobius function too.1 zb42
Nevanlinna-Pick Problem 1: Find a stable, proper G(s) such that G 1 and G(a1 ) b1 where b1 1 and Re a1 0.NP problem for nCase 1 b1 1: The unique solution is G(s) b1 .Case 2 b1 1: The set of all solutions is:{G : G(s) M b1 [G1 (s)Aa1 (s)], G1 CRH , G1 1]}where the all-pass function A a (s)Example: For a1: s as a 2 and b1 0.6 we have: G(s) G1 (s) s 2s 2 0.61 0.6G1 (s) s 2s 2s 0.5G1 (s) 1 results in G(s) s 0.5Remark 1: If G1 is an all-pass function, so is GRemark 2: When ai are the complex-conjugate pairs, if G G R jGI is the solution of the NPproblem then GR is also a solution to the NP problem.Model Matching43
Nevanlinna-Pick ProblemConsider the NP problem with n points:Case 1 b1 1: G(s) b1 is the unique solution (and hence b 1 b2 · · · bn ). 1: Pose the NP’ problem with n 1 data points: {a 2 , . . . an } and {b 2 , . . . , b n }where b i : Mb1 (bi )/Aa1 (ai ) i 2, . . . , nCase 2 b1 Lemma: The set of all solutions to the NP problem is G(s) M b1 [G1 (s)Aa1 (s)] where G1 (s)ranges over the solutions to the NP’ problem.Example: Consider the NP problem with a {1, 2} and b {1/2, 1/3}.Solvability:solvable, The problem is becauseQ 1 b21 2a11 b2 b1a2 a11 b1 b2a1 a2 1 b222a2 3/85/18 5/18 eig(Q) [0.5867 0.0105] Q 02/9b2 b1 0.21 b2 b1 0.6NP’ problem: a2 2, b2 a a 211/3a2 a1s 8 s 11 s2 3s 8s 8 s 12NP problem: G(s) 1 s 8 s 1 s2 3s 81 2 s 8 s 1Model Matching G1 (s) s 2s 2 0.61 0.6 s 2s 2s 8 s 844
Model Matching ProblemFind Q such thatγopt min{ T1 T2 Q γ}γDefine:G 1(T1 T2 Q)γT1 γG. However, to ensure theT2stability of Q, T1 γG should contain the RHP zeros of T 2 (i.e. zi ), that is:We find first G such that G 1 then we compute Q 1γG(zi ) T1 (zi ) G(zi ) T1 (zi )γThis is a NP problem and γopt is the smallest γ for which the problem has a solution. That is, theassociated Pick matrix is positive semidefinite.Aij 1zi zjA γ 2 B 0 where :Bij T1 (zi )T1 (zj )zi zjLemma: γopt equals the square root of the largest eigenvalue of the matrix A 1/2 B A 1/2 .Model Matching45
Model Matching ProblemProcedure: Given T1 and T2 find a stable Q to minimize T 1 T2 Q (T1 tf(num,den)) 0.zz zero(T2);z zz(find(real(zz) 0))Step 1: Determine zi the zeros of T2 in ResStep 2: Form the matrices A and B :Aij 1zi zjBij T1 (zi )T1 (zj )zi zjStep 3: Compute γopt as the square root of the largest eigenvalue of the matrix A 1/2 B A 1/2 .gamma sqrt(max(eig(inv(sqrtm(A)) *B*inv(sqrtm(A)))))Step 4: Find G, the solution of the NP problem with data:z1.zn 1 1γoptT1 (z1 ) . . . γoptT1 (zn )Step 5: SetQ Model MatchingT1 γopt GT2Q minreal((T1-gamma*G)/T2,0.01)46
Model Matching ProblemState-Space Procedure:Step 1: Factor T2 as the product of an all-pass T 2ap and a minimum phase factor T 2mpStep 2: Define R: T1and factor R as R R1 R2 with R1 strictly proper with all poles inT2ap RHP and R2 H and find a minimum realization of R 1 (s) ABC0 Step 3: Solve the Lyapunove equations:ALc Lc A BB A Lo Lo A C CStep 4: Find the maximum eigenvalue λ 2 of Lc Lo and a corresponding eigenvector w . Step 5: Define:f (s) Step 6: Then γoptModel MatchingAwC0 g(s) A λ 1 Lo wB 0 (s) λ and Q (R λ fg(s))/T2mp47
Design for PerformanceObjective: Find a proper C for which the feedback system is internally stable and W 1 S Lemma: If G is stable and strictly proper, thenlim G(1 J) τ 0 11 0 where J(s) (τ s 1)kP and P 1 stable: In this case the set of all stabilizing controller is:QC 1 PQQ H andW1 S W1 (1 P Q) P 1 is stable but not proper, so let’s try Q P 1 J to make it proper. ThenW1 S W1 (1 J) whose -norm is less than 1 for sufficiently small τ .Clearly, QP 1 stable: Do a coprime factorization of P N/M, N X M Y 1 Set J (τ s 1) k with k the relative degree of P Choose τ so small that W1 M Y (1 J) 1 Set Q Y N 1 J and C (X M Q)/(Y N Q)Robust Control48
P 1 Unstable (General Case)P has no poles or zeros on the imaginary axis, only distinct poles and zeros in theRHP and at least one zero in the RHP. W 1 is stable and strictly proper.Assumptions:Proce
H Control 12. Model and Controller Reduction 13. Robust Control by Convex Optimization 14. LMIs in Robust Control 15. Robust Pole Placement 16. Parametric uncertainty References: Feedback Control Theory by Doyle, Francis and Tannenbaum (on the website of the course) Essentials of Robust Control by Kemin Zhou with Doyle, Prentice-Hall .File Size: 1MB
Everything you always wanted to know about multicore graph processing but were afraid to ask Jasmina Malicevic EPFL Baptiste Lepers EPFL Willy Zwaenepoel EPFL Abstract Graph processing systems are used in a wide variety of fields, ranging from biology to social networks, and a large num
robust loop-shaping POD controller design in large power systems. By applying the model reduction and modern robust loop-shaping control technique, the FACTS robust loop-shaping POD controller is realized. This controller exploits the advantages of both conventional loop-shaping and modern . H robust control technique.
Modern robust control theories have been developed significantly in the past years. The key idea in a robust control paradigm is to check whether the design specifications are satisfied even for the “worst-case” uncertainty. Many efforts have been taken to investigate the application of robust control techniques to power systems.
BIENVENUE À L’EPFL Informations pour les nouveaux étudiants Pour tout secours . elle vous rend la vie plus simple et plus fun : consulter l’horaire et le lieu des cours (IS-Academia) . accessible sur place et en ligne des films et des romans de science-fiction,
Élodie Lamothe, Gaëtan Lévêque, and Olivier J. F. Martin Nanophotonics and Metrology Laboratory Swiss Federal Institute of Technology EPFL-STI-NAM, ELG, Station 11 CH-1015 Lausanne, Switzerland elodie.lamothe@epfl.ch . Raether, Plasmons on Smooth and Rough
Compact Modeling of Ultra Deep Submicron CMOS Devices Wladyslaw Grabinski*, Matthias Bucher, Jean-Michel Sallese, François Krummenacher Swiss Federal Institute of Technology (EPFL), Electronics Laboratory, CH-1015 Lausanne, Switzerland *Motorola, Modeling and Characterization Lab, CH-1218 Le Grand Saconnex, Switzerland
The \Robust" Approach: Cluster-Robust Standard Errors The cluster-robust approach is a generalization of the Eicker-Huber-White-\robust" to the case of observations that are correlated within but not across groups. Instead of just summing across observations, we take the crossproducts of x and for each group m to get what looks like (but S .
measured by ASTM test method C 173 or C 231. Dimensions – Unless otherwise specified, the minimum length of each barrier section will be 10 feet. It is common for DOTs to ask for lengths of 20 feet or even 30 feet. ASTM C 825 Design Steel Reinforcement – Unless designated by the purchaser, reinforcement shall be designed by the producer and be sufficient to permit handling, delivery .
