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Optimal and Robust Control

by: Candida Hammes DVM

Optimal and Robust Control MAE 7360

Candida Hammes DVM
Utah State University
GPA 3.54


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This 301 page Class Notes was uploaded by Candida Hammes DVM on Wednesday October 28, 2015. The Class Notes belongs to MAE 7360 at Utah State University taught by Staff in Fall. Since its upload, it has received 94 views. For similar materials see /class/230512/mae-7360-utah-state-university in Aerospace Engineering (AE) at Utah State University.

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Date Created: 10/28/15
Lecture Notes Weeks 8b 9a 9b Topic Hinfinity control ECEMAE 7360 Optimal and Robust Control Fall 2003 Offering Instructor Dr YangQuan Chen CSOIS ECE Dept Tel 4357970148 Email chen ieee0r 0rygcheneceusuedu 175 Chapter 1421 Understanding HCgt0 Control Objective Derivation of HCgt0 controller Methods lntuition ancl hanclvvaving Background State Feedback and Observer 0 Problem Formulation and Solutions 0 An intuitive Derivation 176 Problem Formulation and Solutions i 1431 is stabilizable and 0114 is detectable ii A Bg is stabilizable and 0214 is detectable iii nglol D12 0 I 31 0 iv D21 i Together with ii guarantees that the two ARES have nonnegative de nite stabilizing solutions ii Necessary and suf cient for G to be internally stabilizable iii The penalty on z 0136 Dlgu includes a nonsingular normalized penalty on the control u In the conventional H2 setting this means that there is no cross weighting between the state and control and that the control weight matrix is the identity iv in includes both plant disturbance and sensor noise these are orthog onal and the sensor noise weighting is normalized and nonsingular These assumptions can be relaxed 177 Output Feedback H00 Control EIK such that T2wcgt0 lt 7 if and only if 1 a X00 2 0 XOOA AXOO X04313 M 3233mm 001 0 n a YOO 2 0 AYOO YOOAquot YOOCfCl72 030mg 3le 0 iii pXOOYOO lt 72 A00 ZOOLOO Km 5 blt gt FOO O where A00 2 A V QBleXOO BQFOO 200L0002 FOO B XOO LCgt0 YOOC ZOO I 72Y00X00y1 178 Bounded Real Lemma z Glt5gtw Cs Csl A 1B E H00 00 2 z dt llzllg isu 0 Goo Sup w w 00 2 w2 0 w dt Goo lt 7 ll 000 M2 72 w2gt dt lt 0 Vw 0 ll El X X Z 0 such that XA AX XBBX72 00 0 and A BBquotX72 is stable ll El Y Y Z 0 such that YA AY YCquotCYM2 33 O and A YCquotC72 is stable 179 Let 135 72 G SGs Then GCgt0 lt y ltgt jw gt O Vw E R ltgt detlt1gtltjwgt y 0 since Moo 72 gt O and jw is continuous ltgt 135 has no imaginary axis zero ltgt 1345 has no imaginary axis pole A O B 138 qu A BB72 ltgt has no you aXis eigenvalues CC A Apply the following similarity transformation to 13 1 i I O i X I ABBXy2 BBWV2 By2 qu MX A XBB72 XB72 l BX72 3 l 7 21 MX XA AX XBBXyg 00 lf MX O we have 1371 72 lA BBX72 l Byg l A BBXyg XB72 l BX72 lIv2 372 l 172 jw gt 0 if A BBquotXy2 has no jw eigenvalue 180 System Equations in ABlwBgu z C1 D12U y 0233 D21 LU u F262 56 A BgF Blw z 01D12F By Bounded Real Lemma T2wCgt0 lt 7 ll El X X Z 0 such that XA B2F A B2FX XBleXyz 01 112qu D121 0 and A BQF BlBj Xv2 is stable complete 1 square El X X Z 0 such that XA AX XBleXyz 7 XBZng 001 F B XF ng 0 and A BQF BlBj Xv2 is stable lntuition gt F B Xl El X X Z 0 such that XA AX XBleXyz 7 XBngX 001 0 and A BlBj Xv2 Bng X is stable gt FFOO XXOO 181 Output Feedback Converting to State Estimation Suppose El a K such that lszwlloo lt 7 Then ltOOgt O by stability note also 230 0 1 W 22 w2gtdt 00 d 0 W 22 llwllg lt Xmsw dt A M 72 w2 XOOXOO dt Substituting 56 A36 Blw Em and z 0136 Dlgu 00 2 2 2 0 Call llull 72 llwll 2XOOA23 236XOOBlw 235XOOBgu dt OOOltCfC1 XOOA AXOOC Hun2 72 w2 236XOOBlw 2XOOBgugt dt using XCgt0 equation if 36 XoanBpgov2 XmBngxmn Hull 72 w2 236XOOBlw 2XOOBgugt dt if lt IIBiXooxW 72 w2 22X0031w B XOO2 u2 233X0032u dt completing the squares with respect to u and w 00 ltu 332902 72 Hw V QBonoasl dt 182 Summary 000 llzll2 72 w2gt dt OOO llvll2 72 T2gt dt v u BSXOOZC u F0035 7quot w V QBonosc Rewrite the system equation with w 7quot V QBonoa in ABleXOO72Blr32u OOU CQD21T w l l lszwlloo lt v ltgt llTwlloo lt v ltgt 0 ltu F0033 72 r2gt dt lt 0 lf state is available u F0036 worst disturbance 711 V QBfXOOa State is not available using estimated state u F0038 A standard Observer A Baum7 Bgu Lng y where L is the Observer gain to be determined 183 Let 6 z 6 55 Then A B1Bono72 LCQgt 31 LD21gtT v COe TmnCgt0 lt 7 gt El a Y Z O by bounded real lemma YABiBfXOO72LCQABiBfXOO72LCQYYFOFOOY72 031 LD2131 LD21 0 Complete square wrt L ABiBfXOO72ABiBfXOO72YYF OFOOY72Ble YC CQY L YO gtltL YC 0 Again intuition suggests that we can take L YC which gives YA BleXOOyg A BleXOOVQY YF OFOOYMQ Yogogi 3le 0 lt is easy to verify that Y YOOU V QXOOYOO 1 Since Y Z O we must have pltXooYoogt lt 72 Hence L ZOOLCgt0 and the controller is give by 55 A BleXOOvg Bgu ZooLooltC293 14 F003quot U 184 Chapter 14 H00 Control H00 background H00 1984 workshop approach Assumptions output feedback H00 control a matrix fact inequality characterization connection between ARE and ARI LMl proof for necessity proof for suf ciency comments optimality and dependence on 7 H00 controller structure example an optimal controller H00 control general case relaxing assumptions H2 and H00 integral control H00 ltering 185 H00 Background 0 Initial theory was SlSO Zames Helton Tannenbaum o Nevanlinna Pick interpolation o Operator theoretic methods Sarason Adamjan at al Ball Helton 0 Initial work handled restricted problems 1 block and 2 block 0 Solution to 2 X 2 block problem 1984 Honeywell ONE Workshop 186 H00 1984 HONR Workshop Approach Solution approach 0 Parameterize all stabilizing controllers Via Youla at all 0 Obtain realizations of the closed loop transfer matrix 0 Transform to 772 x 2 block77 general distance problem 0 Reduce to the Nehari problem and solve Via Glover Properties of the solution 0 Statespace using standard operations 0 Computationally intensive many Ric eqns 0 Potentially high order controllers c Find solution lt 7 iterate for optimal 187 Assumptions i 1431 is Controllable and 0114 is observable ii A Bg is stabilizable and 0214 is detectable iii nglol D12 0 l 31 0 iv D21 i Together with ii guarantees that the two AREs have positive de nite stabilizing solution ii Necessary and suf cient for G to be internally stabilizable iii The penalty on z 0136 Dlgu includes a nonsingular norrnalized penalty on the control u In the conventional H2 setting this means that there is no cross weighting between the state and control input and that the control weight matrix is the identity iv in includes both plant disturbance and sensor noise these are orthog onal and the sensor noise weighting is normalized and nonsingular These assumptions simplify the theorem statements and proofs and can be relaxed 188 Output Feedback H00 Control EIK such that T2wcgt0 lt 711 i 3 XCgt0 gt O XOOA AXOO X04313 M 3233mm 001 0 ii 3 YCgt0 gt O AYOO YOOAquot YOOCfCl72 030mg 3le 0 iii pXOOYOO lt 72 A00 ZOOLOO Km 5 blt gt FOO O where A00 2 A V QBleXOO BQFOO 200L0002 FOO B XOO LCgt0 YOOC ZOO I 72Y00X00y1 189 A Matrix Fact Packard 1994 Suppose X Y E RM and X X gt O Y Y gt 0 Let 7quot be a positive integer Then there exists matrices X12 E Rn X2 E RT such that X2 X5 and 71 X X12 X X12 Y gt O 3 X2 X1 X2 i if and only if X In In 2 O amp rank 3 n 7 1 Y n Proof lt2 By assumption there is a matrix X12 E Rn such that X Y 1 X12Xf2 De ning X2 I completes the construction gt Using Schur complements Y X4 X 1X12X2 ngxlxm xgrl lnverting using the matrix inversion lemma gives Y l X Xlgxfxg Hence X Y 1 X12X2 1Xf 2 Z O and indeed rankX Y l rankX12X2 1Xf2 S 7quot D 190 Inequality Characterization Lemma IC El r th order K such that Tzwcgt0 lt 7 only if i 3 Y1 gt 0 M1 mm moron1M 3131 72323 lt 0 11 3X1 gt O X1A AX1 XlBleXlvg 001 720302 lt 0 Xlfy In In Yiv Xlfy In 2 O and rank In Yiv nr iii Proof Suppose that there exists an r th order controller Ks such that T2wCgt0 lt 7 Let Ks have a state space realization A B Ks A A C D then A B ABQDCQ 320 131BQDD21 Tzw C C 302 A BDQl CC DC A A A l 01 13121302 D120 131213121 j Denote R 721 DZDC R 721 DUDE X X By Bounded Real Lernrna EIX 1 12 gt 0 such that ng X2 XltAC 36135119306 AC 363419306 191 9230135133 03112400 lt 0 This gives after much algebraic manipulation X124 AX1 XiBiBinV2 001 720302 XlBlDX121 VQC 721 DD 1X131DX121 7203 lt 0 which implies that Xmmxxammmhcmr galtn 31312 31336 Let 7 VQX l and partition 17 as l7 gt 0 then MHB mQW aamme Y03R4CCY BCPHB lt 0 This gives AM Mquot 3131 72323 Heroinv Agnewnxr Bgm nnn4oxXDMqCquivlt0 which implies that AY1 Y1A 3le 723ng moron1 m lt 0 By the matrix fact given X1 gt O and Y1 gt 0 there exists X12 and X2 such that Y 72X 1 or Yv Xvflz XiV X127 YiV Xl27 X2v X In X In 17 ZOanclrank 17 SHT D 192 Connection between ARE and ARI LMI Lemma ARE Ran and Vreugdenhil 1988 Suppose A B is con trollable and there is an X X such that QX XAAX XBBX Q lt 0 Then there exists a solution X gt X to the Riccati equation XAAXXBBXQO 07 such that A BBX is antistable Proof Let X be such that QX lt 0 Choose F0 such that A0 2 A BFO is antistable Let X0 X5 solve X0140 ASXO FSFFO Q 0 De ne F0 2 F0 BX Then X0 XA0 A3ltX0 X g o goo gt 0 and X0 gt X by anti stability of A0 De ne a non increasing sequence of hermitian matrices X2 XOZXi Z ZXnil gtX7 A A BFZ39 is antistable i O n 1 FZ39 BX 1 1771 X A AjX EE Qi01n 1 By Induction We show this sequence can indeed be de ned Introduce Fn BXn1 An A BF 193 We show that A is antistable Using 08 with 239 n 1 we get XnilAn AZXnil Q 1an Fm Fn71gtgtkltFn F7171 039 Let F 2 F BX then X7171 XgtAn AZltX7171 X QX 1533 F F1F EH gt 0 gt A is antistable by Lyapunov theorem since Xn1 X gt 0 Let X be the unique solution of XnAn AZXT Fan Q 09 Then Xn is hermitian Next we have X XA Aj X X QX gt 0 X7171 XngtAn AZltX7171 X71 Fm Fn71gtgtkltFn F7171 Z 0 Since A is antistable we have Xn1 Z X gt X We have a non increasing sequence X2 Since the sequence is bounded below by X gt X Hence the limit X 45210 Xn exists and is hermitian and we have X Z X Passing the limit at gt oo in 09 we get QltX 0 So X is a solution of 07 Note that X X Z O and ltX XgtA AltX X QltXgt X XBBX X gt O 010 hence X X gt O and A A BBquotX is antistable 194 Proof for Necessary There exists a controller such that Tzwcgt0 lt 7 only if the following three conditions hold i there exists a stabilizing solution XOO gt O to X AV XQVka iB W2 B X IX010 ii there exists a stabilizing solution YOO gt O to Ammmmmm 4mmmamo iii h I X71 gt0 or pXOOYOOlt72 n 7 oo Proof Applying Lemma ARE to part of Lemma lC we conclude that there exists a Y gt Y1 gt 0 such that muwwwymawam amo and A CfClYy2 is antistable Let X00 7214 we have x mm xaammtewmamao and A BiBiv2 BgB gtXoo Xo1ltA ClongoUXoo xgmmawmm is stable Similarly applying Lemma ARE to part ii of Lemma lC we conclude that there exists an X gt X1 gt 0 such that XANXXameM4mx Qo 195 and A BleXv2 is antistable Let YOO VQX l we have AYOO YOOAquot YOOCf01 M 030w 313 0 011 and A CfClvg 030mg is stable Finally note that the rank condition in part iii of Lemma IC is auto matically satis ed by 7quot Z n and 73021 In i XV In In VX0701 i In YV XlV In In Yiv or pXOOYOO lt 72 D 196 Proof for Suf ciency Show KW renders TzwCgt0 lt 7 The closed 100p transfer function with Km A BQFCgt0 31 A AC BC Tzw ZQOLOOCQ A00 ZOOLOOD21 1 CC DC i 01 D12Foo 0 De ne i 723021 72Y021Z0701 72ltZ ogt 1Yo1 7210le Then P gt O and PAC AZP PBCBjPvg 006 i 0 Moreover A B BY 1 B FOO B BY 1Z 1 ACBCBjP72 1 1 0 2 1 1 0 0 0 A BleXOOvg BQFOO has no eigenvalues on the imaginary axis since A BleYO1 is antistable and A BlBi XOOv2 BQFCgt0 is stable By Bounded Real Lemma TzwCgt0 lt 7 197 Comments The conditions in Lemma lC are in fact necessary and suf cient But the three conditions have to be checked simultaneously This is because if one nds an X1 gt O and a Y1 gt O satisfying conditions and but not condition this does not imply that there is no admissible H00 controller since there might be other X1 gt O and Y1 gt O that satisfy all three conditions For example consider 7 1 and lt is easy to check that X1 Y1 05 satisfy and but not Nevertheless we can show that yap 07321 and thus a suboptimal con troller exists for y 1 In fact we can check that 1 lt X1 lt 2 1 lt Y1 lt 2 also satisfy and For this reason Riccati equation approach is usually preferred over the Riccati inequality and MW approaches when ever possible 198 Example Consider the feedback system shown in Figure 04 with 50514 2 51 P W5 Wu s1s2 s02 s10 DesignaK to minimize the H00 norm fromw toz i3 239 U 6 i W5IPK 1 W5IPK 1P d 7T d a WuKIPK 1 WuKIPK 1P dz 2w dz LFT framework 7 02 2 2 0 2 0 0 0 1 0 0 0 20 20 We Wepi WEP 0 0 2 0 0 30 30 Gs0 oi Wu 0 00 1000 3 I P P 1 0 0 0 0 0 0 0 0 0 3 0 0 1 0 1 1 0 1 0 0 gt K TZW Vsubopt hinfsynG ny nu 7min vmax tol where my dimensions of y nu dimensions of u 7min a lower bound vmax an upper bound for 70m and tol is a tolerance to the optimal value Set my 1nu 1 7mm 0 vmax 10 tol 00001 we get vsubopt 07849 and a suboptimal controller 1282s10 1s727 1s14 1 53244944767 1s2219 1s14 1s02 1 199 lf we set tol 001 we would get ysubopt 07875 and a suboptimal controller 1278510 1s727 1s14 1 5233559 1s2197 1s14 1s02 139 N2 The only signi cant difference between K and K is the exact location of the far away stable controller pole Figure 025 shows the closed loop fre quency response of 6 Tm and Figure 026 shows the frequency responses of STKS and SP 10D 101 frequency radsec Figure 025 The closed loop frequency responses of 5Tm with K solid line and K dashed line 10392 10391 10B 101 102 10a 104 frequency radsec Figure 026 The frequency responses of STKS and SP with K 200 Consider again the two mass spring damper system shown in Figure 01 Assume that F1 is the control force F2 is the disturbance force and the measurements of 1 and 562 are corrupted by measurement noise 0015 10 0 yr 361 n1 W 7 Wm 8 100 9 y 2 2 O 10154710 5 100 Our objective is to design a control law so that the effect of the disturbance force F2 on the positions of the two masses 1 and 362 are reduced in a frequency range 0 g a g 2 The problem can be set up as shown in Figure 027 where We W 1 is the performance weight and Wu is the control weight In 2 order to limit the control force we shall choose 55 u is50 Figure 027 Rejecting the disturbance force F2 by a feedback control 201 F 2 LetuF1w n1 712 115131 0 W5P2 Cs 0 0 Wu P1 Wn P 2 6 where P1 and P2 denote the transfer matrices from F1 and F2 to 1 362 respectively 0 W1 SS W2 0 only reject the effect of the disturbance force F2 on the position 51 IIHltGK2gtH2 26584 feGK2OO 26079 feGKoooo 16101 This means that the effect of the disturbance force F2 in the desired frequency rang 0 s w s 2 will be effectively reduced with the H00 controller KOO by 516101 31054 times at 61 0 W1 0 W2 SS only reject the effect of the disturbance force F2 on the position 362 MG K2gt2 01659 fgG K2 OO 05202 fgG KOOCgt0 05189 This means that the effect of the disturbance force F2 in the desired frequency rang 0 s w s 2 will be effectively reduced with the H00 controller KOO by 505189 96358 times at 2 202 on The largest singular value o 101 o frequency Figure 028 The largest singular value plot of the closed loop system TAm with an H2 controller and an H00 controller 0W1W2 want to reject the effect of the disturbance force F2 on both 61 and 62 NEW Kglllg 4087 3a K2gtIOO 60921 llfelG Kallloo 43611 This means that the effect of the disturbance force F2 in the desired frequency rang 0 s w s 2 will only be effectively reduced with the H00 controller KOO by 543611 11465 times at both 61 and 2 This result shows clearly that it is very hard to reject the disturbance effect on both positions at the same time The largest singular value Bode plots of the closed loop system are shown in Figure 028 We note that the H00 controller typically gives a relatively flat frequency response since it tries to minimize the peak of the frequency response On the other hand the H2 controller would typically produce a freL quency response that rolls off fast in the high frequency range but with a large peak in the low frequency range 203 Optimality and dependence on y There exists an admissible controller such that lszwlloo lt 7 iff the follow ing three conditions hold i El a stabilizing XCgt0 gt 0 ii El a stabilizing YOO gt 0 iii MXOOYOO lt 72 o Denote by 70 the in mum over all 7 such that conditions i iii are satis ed 0 Descriptor formulae can be obtained for y yo 0 As 7 gt oo HCgt0 gt H2 XOO gt X2 etc and KM gt K2 391f72 Z 71gt 70 then Xmm Z X0472 and Yoolt71gt2 Yoolt72gt 0 Thus X00 and YOO are decreasing functions of y as is pXOOYOO 0 At 7 70 any one of the 3 conditions can fail o It is most likely that condition iii will fail rst 0 To understand this consider and let 71 be the largest y for which H 00 fails to be in d0mltRic because it fails to have either the stability property or the complementarity property The same remarks will apply to ii by duality o If the stability property fails at y 71 then HOO g d0mltRicgt but Ric can be extended to obtain XCgt0 and the controller u B XOO is stabilizing and makes Tzwcgt0 71 The stability property will also not hold for any 7 g 71 and no controller whatsoever exists which makes Tzwcgt0 lt 71 204 0 In other words if stability breaks down rst then the in rnurn over sta bilizing controllers equals the in rnurn over all controllers stabilizing or otherwise 0 In view of this we would expect that typically complementarity would fail rst o Cornplernentarity failing at y 71 means pXOO gt 00 as y gt 71 so condition iii would fail at even larger values of 7 unless the eigenvectors associated with pXOO as y gt 71 are in the null space of YOO 0 Thus condition iii is the most likely of all to fail rst 205 H00 Controller Structure A A00 ZooLoo Km 5 blt FOO O 2100 2 A 2313me BQFOO 200L0002 FOO B XOO LOO YOOC ZOO I V QYOOXOO 1 EC A3 Blwmt Bgu ZOOLOOCQ y A A 72 A u F0033 wmm 7 BlXoosc 1 10mm is the estimate of wworst 2 ZOOLOO is the lter gain for the OE problem of estimating F003 in the presence of the worst case w wwmt 3 The H00 controller has a separation interpretation Optimal Controller 1 appcomm A533 Looy 012 u F0038 013 A5 A Bchgt0 0002 H YOOAWOO vgpiBleXoo y YmeCH See the example below 206 Example 72 a 12 Y 1 a 1 a The eigenvalues of H00 and J00 are given respectively by 212 1 a212 1 0ltHOOgt iM 7 UUOO i 7 7 1 If 7 gt m then 26411100 and ALUOO exist and 021ngliav 26411100 1m 7 1 021ngliav PLUG 1m 7 1 Note that if 7 gt 1 then HOO E d0mltRic J00 E d0mltRic and X00 gt 0 a2 172 1 W Y i A gt O 00 xa2172 1 a7 207 lt can be shown that 72 2 Ma 1gtv2 1 cw lt 7 pltXooYoogt is satis ed if and only if 7 gt V a2 2 a So condition iii will fail before either or ii fails The optimal 7 for the output feedback is given by vopta22a and the optimal controller given by the descriptor formula in equations 012 and 013 is a constant ln fact U t 70m 1 0p lta2 973 1 aflOpt For instance let a 1 then you 1 07321 and um O7321 y Further 17321 1 O7321 Tzw O7321 O O7321 lt is easy to check that TzwCgt0 07321 208 An Optimal Controller There exists an admissible controller such that T2wOO g 7 iff the following three conditions hold X00 i there exists a full column rank matrix 1 E R271 such that 002 X001 X001 H00 TX Re ltTXgt S 0 V2 X002 X002 and X01Xoo2 X02X001 Yoo ii there exists a full colurnn rank matrix 1 E R271 such that 002 Yool Yool J00 Ty Re ltTygt S 0 V2 Yoo2 Yoo2 and Yoilyoo2 YoZQYOOl X350ng1 Vileogyoo2 iii 1 7 Yo 2Xoo2 Y2YW1 Moreover when these conditions hold one such controller is Koptlt8gt I CKlt8EK AKgtBK gt0 where EK Yglxool Wyggxoog BK 10220 CK B Xoog AK 2 EKTX BKCQXool YoilBQCK39 209 H00 Control General Case Assumptions A1 A Bg is stabilizable and 0214 is detectable A2 D12 and D21 0 I J A I B A3 jw 2 has full column rank for all w 1 D12 14 I B A4 jw 1 has full row rank for allw 02 D21 721m 0 R I Diph 1 O where D1ZD11 D12 2 O D R D1Df1 7 1 1 where D1 11 D21 A O B H00 Fi 150135 4101 A 41th A O 0 71 J00 R 1913 a B1Bf A B1Df1 X00 RicltHOO YOO RicUOO 210 F100 F200 L 2 L100 L200 BlDf1YOOCR 1 F 2 Pr1 D1101 BXOO D F100 and L100 are Partitioned as follows l lF11oo F1200 F220 F i LEGO D1111 D1112 0 L D Likgoo D1121 D1122 I Lgoooro There exists a stabilizing controller Ks such that HEW Kgtcgt0 lt 7 if and only if 1 7 gt malt5lD11117 D11127l75lDl1117 Dl121lgt ii HOO E d0mltRicgt with XCgt0 RicltHOO Z 0 iii J00 E d0mltRicgt with YOO RicUOO Z 0 iv 9000le lt 72 K fZltM007Qgt7 Q E RHOO lt 7 where D11 D1121Dl111721 D1111Dl111gt71D1112 D1122 211 D12 E Rmxm and D21 E Rmng are any matrices satisfying 13121515 I D1121lt721 Df111D1111gt71Df1217 D 1D21 I DTMQWQI D1111DT111gt71D11127 and B2 ZooB2 L1200D12 02 D21ltCQ F12oogt7 B1 ZooL2cgtoBQDf21D117 C1 F200D11D 11027 121 A BF Blbil g where ZOO I 72Y00X00y1 Some Special Cases 0 D12 I Then becomes 7 gt D1121 and D11 D1122 1512ng I 7 2D1121D1 121 13311521 I 0 D21 I Then becomes 7 gt 6D1112 and D11 D1122 1512ng I 13311321 I 77213312131112 0 D12 I 85 D21 I Then drops out and D11 D1122 1312ng I 13311321 I 212 Relaxing Assumptions Dpll Dp12 Dp21 Dp22 Assume Dplg has full column rank and Dim has full row rank Normalize D12 and D21 Perform SVD Dp12Up RP Dp2112p010p such that Up and 7 are square and unitary NOW let 2p Upz wp Ugw yp 15 up Rpu Klt5gt Rprlt5gtRp I 0 71 0 RP I 0 P5 ltgt 0 R171 AP 3110 3123 USDPHU UgDplgRgl RglppglU R7 P USCpl 31710192 1 71 1912ng 213 Then D12 D21l01l7 7eltPprgt II7eltGKgtlloo lloo Remove the Assumption D22 0 Suppose Ks is a controller for G with D22 set to zero Then the controller 101quot D22 7g 0 18 KltI D22Kgt71 Relaxing A3 and A4 Complicated Suppose that O O 1 G O O 1 1 1 O which violates both A3 and A4 and corresponds to the robust stabilization of an integrator If the controller u 6 where 6 gt O is used then Tm 65 with Tzwoo e s 6 Hence the norm can be made arbitrarily small as 6 gt 0 but 6 O is not stabilizing Relaxing A1 Complicated Relaxing A2 Singular Problem reduced ARE or LMl 214 H2 and H00 Integral Control H2 and H00 design frameworks do not in general produce integral con trol 21 Ways to achieve the integral control 1 introduce an integral in the performance weight We 21 We1 PK 1de Now if the norm 2 norm or oo norm between in and 21 is nite then K must have a pole at s O which is the zero of the sensitivity function The standard H2 or H00 control theory can not be applied to this problem formulation directly because the pole s O of We becomes an uncontrollable pole of the feedback system A1 is violated Suppose We can be factorized as follows We WesMs where M s is proper containing all the imaginary axis poles of We and M 45 E R7100 Wes is stable and minimum phase Now suppose there exists a controller Ks which contains the same imaginary axis poles that achieves the performance Then without loss of generality K can be factorized as Kltsgt K5Ms Now the problem can be reformulated as 215 A simple numerical example 2 O 10 8 P 3 21 W1 515 3 d 210 is10i 100 90 1 uis100i 1 1 579 Then we can choose without loss of generality that 5a 1 Wg Ozgt0 s sa 216 This gives the following generalized system a0 1 2110 0 1000000 90 0 00 2aaa ms 0 0 0 010 0 0 0 0 3 201 1 0 0 0 00 0 0 10 0 001 0 01 2110 suboptimal H00 controller Koo R 20603814s 1s as 100s 01557 00 5 1209 32175 262343s 1989 which gives the closed loop oo norm 7854 A i 206038145 1s 100s 01557 KOO KO lt5gtMlt5gt 55 32175 262343s 1989 N 78509 15 1005 01557 N 55 3217s 1989 An optimal H2 controller A i 434875 1s as 1005 0069 2 i s 12092 30945 411815 7964 and i 4348709 15 1005 0069 5 552 30945 411815 7964 195 K25Ms 2 An approximate integral control 1 WW 6 6 86 Ms1 217 for a suf ciently small 6 gt 0 For example a controller for 6 0001 is given by i 316880lt5 1s 100s 01545 00 i 5 00015 325 403705 20 N 785s 1s 100s 01545 N 55 325 20 which gives the closed loop H00 norm of 785 i 4347s 1s 100s 00679 2 i 5 000152 30935 41175 79718 218 H00 Filtering 3 l Ar Blind 330 O 02 D21wt Z 0133 311331 0 13211331 1 H00 Filtering Given a 7 gt 0 nd a causal lter Fs E RHOO if it exists such that A 2 J sup w lt 72 w 2l0 gt0gt lelQ with 2 Fsy Q l 2 M Z A This can be regarded as a H00 problem without internal stability There exists a causal lter Fs E R7100 such that J lt 72 if and only if J00 E d0mltRicgt and YOO RicUOO Z O A 1400302 1000 2Fs gt11 01 O where YOO is the stabilizing solution to YooA AYOO YooW QCi Cli C C2gtYoo 313i 039 Lecture Notes Week 5a Topic Uncertainty modeling and robustness stability vs performance ECElVIAE 7360 Optimal and Robust Control Fall 2003 Offering Instructor Dr YangQuan Chen CSOIS ECE Dept Tel 4357970148 Email chen ieee0r 0rygcheneceusuedu A Real Control Problem disturbances response sensor noise I measurements control mp ut Controller commands Controller Objective To provide desired responses in face of disturbances Uncertain plant dynamics External inputs sensor noise control input 78 Prepared by Ben M Chen Representation of Uncertain Plant Dynamics Perturbation disturbance response sensor noise Nominal Plant measurements control mp uts r Nominal Plant is a FDLTI System Perturbation is Member of Set of Possible Perturbations 79 Prepared by Ben M Chen Analysis Objectives Nominal Performance Question H2 Optimal Control Are closed loop responses acceptable for disturbances sensor noise commands Robust Stability Question Ha Optimal Control ls closed loop system stable for nominal plant for all possible perturbations Robust Performance Question Mixed H2H Optimal Control Are closed loop responses acceptable for all possible perturbations and all external inputs Simultaneously 80 Prepared by Ben M Chen Complete Picture of Robust Control Problem A EgtPEgt 81 Prepared by Ben M Chen Standard Feedback Loops in Terms of General Interconnection Structure 82 Prepared by Ben M Chen 108 Chapter 8 Uncertainty and Robustness model uncertainty small gain theorem additive uncertainty multiplicative uncertainty coprime factor uncertainty other tests robust performance skewed speci cations example siso vs mimo 109 Model Uncertainty PA5 135 wsAs Ajw lt1 Vw PMS IwsAsPs nominal model 39 log 0 actual model Suppose P E H is the nominal model and K is a controller Nominal Stability NS if K stabilizes the nominal P Robust Stability RS if K stabilizes every plant in H Nominal Performance NP if the performance objectives are satis ed for the nominal plant P Robust Performance RP if the performance objectives are satis ed for every plant in H 110 Examples iUMQO2QBQO a0g y 02 olumix 25o 3c a g E 17 Psoz E 130 WA g 1 with P0 Ps O O and 135 oz 6 10m2 cjsna P 11 P IV gt amm cbsnex es The frequency response P0 WA is shown in Figure 014 as circles 15 1 05 D 05 1 15 2 25 Figure 014 Nyquist diagram of uncertain system and disk covering Another way to bound the frequency response is to treat oz and 6 as norrn bounded uncertainties that is Psoz E P0 W1A1 WQAQI S 1 with P0 Ps O O and 10012524135 1V 7 1 cbsnex es 111 1004er 02 52 055 1s2 25 352 35 6 lt is in fact easy to show that P0 W1A1 W2A2 llAz39lloo 1P0 WAl llAlloo S 1 with W1 The frequency response P0 WA is shown in Figure 015 This bounding is clearly more conservative W2 5 Figure 015 A conservative covering Consider a process control model keiTS 4lt lt 2ltTlt 1lt lt2 Cs T81 k9 3 739 Take the nominal model as 65 G0lt8gt 255 1155 1 Then for each frequency all possible frequency responses are in a box as shown in Figure 016 Aan 00w 90le gtgt mf ginput50 pick 50 points the rst column of mf is the frequency points and the second column of mf is the corresponding magnitude responses 112 Figure 016 Uncertain delay system and G0 gt maggvpckmf2mf1 pack them as a varying ma trix gtgt Wa tmagmagg choose the order of Wu online A third order Wa is suf cient for this example gt ABCDuankWa converting into statespace gt Z P Kss2zpABCD converting into zeropolegain form We get W lt8 003765 1164808s 745145 02674 a 5 12436s O55755 49508 and the frequency response of Wu is also plotted in Figure 017 Similarly de ne the multiplicative uncertainty Amlt8gt Cs G0s G0lt8gt and a Wm can be found such that lAmjw g Wmjw as shown in Figure 018 A Wm is given by i 281695 0212s2 261285 1732 52 224255 26319 Wm Figure 018 Am dashed line and a bound Wm solid line 113 114 Small Gain Theorem W161 E ew 22 Small Gain Theorem Suppose M E RHOOY XQ Then the system is well posed and internally stable for all As E RHOO with a IIAHOO S 17 if and only if M8gtoo lt 7 b IIAHOO lt1Vifa d 0H1y1fM8gtlloo S 1 Proof Assume y 1 System is stable iff clet M A has no zero in the closed right half plane for all A E R7100 and ACgt0 g 1 ltclet1 MA y O for all A E R7100 and ACgt0 g 1 since I MA 21 maxMA Z 1 gt O gt Suppose Z 1 There exists a A E RHOO with ACgt0 g 1 such that det M5As has a zero on the imaginary axis so the system is unstable Suppose wo E R U 00 is such that 6Mjw0 Z 1 Let MUM Ujw2jw0Vjw0 be a singular value decomposition with Ujw0u1 U2 up Vjw0v1 v2 vql EM 02 115 We shall construct a A E R7100 such that Altjw0 0171mm ancl ACgt0 g 1 Indeed for such As detI Mjw0Ajw0 cletI UEVU1u f01 1 u 1 UEVv101 O and thus the closed loop system is either not well posed if wo 00 or unstable if on E n There are two different cases 1 wo O or 00 then U and V are real matrices Chose 1 gtllt X A ivlul E nq 1 01 2 O lt wo lt 00 write ul and 111 in the following form U11 j 1 U15 ullej l ulgejgg ulpejep 7 U1 v12e 2 Ulqejasq where uh Ulj E R are chosen so that 625 gbj E 7r 0 Z lt jw0gt 62 4 047 1910 Q 5239 1010 04739 Jron Choose gt O and 047 gt 0 so that Let 75 i 1 Bills 515 510 5 N8 7 U11 m5 mm H e R7100 01 04475 p Ulq aqs Then AOO 10 1 3 land mm 017W 116 The theorem still holds even if A and M are in nite dimensional This is summarized as the following corollary The following statements are equivalent i The system is well posed and internally stable for all A E H00 with IIAIIOO lt 17 ii The system is well posed and internally stable for all A E R7100 with IIAIIOO lt 17 iii The system is well posed and internally stable for all A E rcqxp with IIAII lt 17 1V IIMIIOO S m It can be shown that the small gain condition is suf cient to guaran tee internal stability even if A is a nonlinear and time varying stable operator with an appropriately de ned stability notion see Desoer and Vidyasagar 1975 117 Additive Uncertainty SO I PKW To PKU PK 1 Si I KP 1 n KPH KPH Let H 13 WlAWQ A E R7100 and let K stabilize P Then the closed loop system is well posed and internally stable for all ACgt0 lt 1 if and only if W2KSOW1Cgt0 g 1 E 1 WQKSOW1 118 Multiplicative Uncertainty H T H Let H 1 W1AW2gtP A E R7100 and let K stabilize P Then the closed loop system is well posed and internally stable for all A E R7100 with ACgt0 lt 1 if and only if W2TOW1Cgt0 g 1 119 Coprime Factor Uncertainty Let P MAN be stable left coprirne factorization and K stabilize P Suppose H M AMgt71ltN AN A 1 AN AMl with AM AN E R7100 Then the closed loop system is well posed and internally stable for all ACgt0 lt 1 if and only if 31 00 l 1 PK 1M 1 Other Tests W16 R7100 W2 6 RHOO A e R7100 ACgt0 lt 1 Perturbed Model Sets Representative Types of Robust Stability Tests H Uncertainty Characterized output sensor errors I WlAW2P neglected HF dynamics VV2TOVV1OO g 1 uncertain rhp zeros input actuators errors PI WlAlVg neglected HF dynarnics IIW2TiW1HOO S 1 uncertain rhp zeros LF pararneter errors I WlAW2 1P uncertain rhp poles VV230VV1OO g 1 LF pararneter errors PI WlAW2 1 uncertain rhp poles VV23iVV1OO g 1 additive plant errors P WlAlVg neglected HF dynarnics uncertain rhp zeros W2KSOW1OO g 1 LF pararneter errors PI WlAW2P 1 uncertain rhp poles HWQSOPWlHoo S 1 AM 1Z7 AN LF pararneter errors P M lN neglected HF dynarnics I J 501914 g 1 A AN A M l uncertain rhp poles amp zeros 00 N ANM AM 1 LF pararneter errors P NM 1 neglected HF dynarnics HM lsAK moo g1 A AN uncertain rhp poles amp zeros 121 Robust Performance sup 62 S 1 Hdugsl Ted W41 PAK 1 PA 6 H Suppose PA E 1 AW2P A E R7100 ACgt0 lt 1 and K inter nally stabilizes P Then robust performance is guaranteed if 504580 6ltW2T0 g 1 504580 and 6ltWesoaltr AWQTOW m 604580 6ltW6 1 awn2T0 1 6W2T0gt39 3 3 122 Skewed Speci cations H mum A e 737100 HAHOO lt1 robust stability llez39lloo S 17 nominal performance IIWeSolloo S 1 Ted W554 PAwKSO 1 W580 I PAP lej l robust performance is guaranteed if 604580 KltPgt5ltIUT gt g 1 504580 HP wT0 g 1 123 Why Condition Number H12 H22 Hum A e R7100 AOO lt1 IthP A e R7100 ACgt0 lt1 H2 2 H1 if u1t Z lwtHP Vw since PM th I thAP l 1 15 5 s 1s 007 005 075 15 013 where 15 s 1s O1707s 002929 103 candman number 8 124 Example SISO vs MIMO c211 i O a wl ul i 1 a wi tag i a 0 tag U2 yi a 1 tag 1 5 612 as1 p lt8gt 52 a2 as 1 5 a2 SIP 1 1 5 a TPIP 1 1 1 a 51 a s 51 a1 39 1 Each loop has the open loop transfer function as i so each loop has phase margin Cbmax ltbmm 900 and gain margin kmin 0 kmax 00 Suppose one loop transfer function is perturbed wig rz yr 1 U1 P U2 112 Denote lt gt z 5 7 i T11 57 51 125 Then the maximum allowable perturbation is given by 1 i 1 TM 00 which is independent of a However if both loops are perturbed at the same time then the maximum allowable perturbation is much smaller as shown below Mi w Kb T 6oo lt 1umg AVHampQERHW 521 522 ACgt0 lt y The system is robustly stable for every such A iff 1 1 ygi ltlt11ragtgt1 TCgt0 x1 a2 In particular consider 511 522 2gtlt2 AAd ER Then the closed loop system is stable for every such A iff 52 2 611 6mg 1 611 622 1 a2gt611622gt detI TAd 8 1 126 has no zero in the Closed right half plane Hence the stability region is given by 2 611 622 gt 0 1 611 622 1 a2611622 gt 0 The system is unstable with 1 1a2 39 611 622 Lecture Notes Week 5b Topic LFT linear actional transformation ECEMAE 7360 Optimal and Robust Control Fall 2003 Offering Instructor Dr YangQuan Chen CSOIS ECE Dept Tel 4357970148 Email chen ieeeor orygcheneceusuedu 127 Chapter 9 Linear Fractional Transformation M11 M12 over A is de ned as M21 M22 A lower LFT of M fgM A M11 M12AI MQQA 1M21 Similarly an upper LFT fuM Au M22 M21Ault1 MiiAquMlQ 21 111 U1 A111 1111 U1 1111 U1 M11 M12 M21 M22 128 Properties o J quotAM A is well posed if I MQQA is invertible MM Aw am A with N given by M11 M12M231M21 MiQME21 N MilMgl M231 0 Suppose C is invertible Then A BQC DQ 1 0 QDWA QB with AC4 B AC lD 0 1 C 1D C lA 0 1 B DC lA DC 1 Ml l if M12 is invertible then fgM Q C QD 1A QB Wlth A MQ1M11B M21 M22M 1M1170 M131 and D M22Mfgl N if M21 is invertible then MM Q A mm DQ 1 Wlth A MllM l B M12 MllM lMgg C M2711 and D M2711M22 129 Example The following diagram can be rearranged as an LFT z fgG Ow with A B A B Au Bu Av Bu Pic opinidDiiW1iouDqui That is 55p Apgvhp Bpltd 7 112 01551 93f A f Bfltyp n y Cf33f Dfyp 71 Stu Auscu Bun Uf Cuasu Duu div A0561 vap v 071337 Dvyp Now de ne a new state vector 55p 33f u 1 and eliminate the variable yp to get a realization of G as 56 ABlwBgu z 0136 Duw Dlgu y 0236 D2111 Dggu 130 with BPOBUO OAOO P P pCOC AB B A O Parametric Uncertainty A MassSpring amper System k1036k Then 1 33 1 HMA 32 362 F where 0 1 0 0 0 ii Q 6k00 M 031 0 030 0 0 A 0660 00250300 0 006m k c 13 1 1 01 131 IE pulling ouf fho As where Then y1 U1 142 U2 143 M U3 1J4 U4 Z U 0 e d 0 1 1 0 0 0 0 M 1 0 0 0 0 0 be bdc oib quotOwl c i 77777 d i 777 zfuMAw A 6112 O 0 6212 133 134 HIMAT Example Wdel 505 100 055 3 O s 10000 W s 003 505 100 7 P O 0 55 3 s 10000 5 003 25 128 8 320 O W 25 128 7 8 320 00226 366 189 321 0 O 0 19 0983 0 0414 0 7 00123 117 263 0 778 224 of 0 0 1 0 0 0 0 573 0 0 0 0 0 0 0 573 0 0 4 1 121 1 Tr m 2 Figure 019 HIMAT Closed loop interconnection 135 The open loop interconnection is P1 21 P2 Z2 d1 61 d2 62 m 31 712 32 U1 U2 The SIMULINK block diagram aircraftm Eile Edit gations gimulation Ser I x AxBu 1 u w 1 Demux p State Space Wdel Mux1 n2 2 x RxBu pe x AxBu y CxDu y CXDU 3931 Siate Space Wp sumz State SPaCEI himat dist1 Demugtlt1 Mux n J WUXZ distz noise1 X AxBu J y CxDu quot39 52 S1ate Space wn W3 nonsez Mux 7 M Mux u2 f H J I Figure 020 SIMULINK block diagram for HIMAT aircraftm A A B The Cs C D can be computed by gtgt A B C D linmod aircraft 136 which gives 77035624 00000 70018 0 0 0 0 704140 7778 0 0 0 0 709439 0 252476 0 71000012 0 0 0 0 0 700226 7366 7189 7321 0 0 719 0983 0 A 7 0 00123 7117 7263 0 0 0 0 1 0 0 0 754087 0 0 0 0 0 0 754087 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 B 704140 0 0 0 7778 224 0 0 0 0 0 0 0 0 70943912 0 0 0 0 725247612 703562412 0 0 0 0 0 0 2865 0 0 709439 C 0 0 0 2865 0 0 0 573 0 0 0 0 0 0 573 0 0 0 0 0 0 0 50 0 0 0 0 0 0 0 D 0 0 05 0 0 0 0 0 0 0 05 0 0 0 0 0 1 0 2 0 0 0 0 0 1 0 2 0 00000 000000 0 70018 0 0 732012 0 77035624 0 0 224 0 0 0 0 0 0 0 0 0 0 252476 137 Redheffer Star Products P P11 P12 K11 K12 P21 P22 7 K21 K22 PK Fl P K11 13121 K11P2271K12 K21ltI 1322K11gt71 P21 Fu K P22 z 1 w z w P z PK w y u 2 K w kl has a representation PK 138 where hm b3 Q D x 7 A 19212711 ng 13215ch1 7 BK1PFlCQ AK BK1PL 1D220K1 7 31 BQR71DK11D21 BQR71DK12 7 BK1RAD21 BK2 BK1RAD22DK12 7 01 D12DK11PL 102 D123710K1 7 DK21PL 102 0K2 DK21PL 1D220K1 7 D11 D12DK11PL 1D21 D12371DK12 DK21PL 1D21 DK22 DK21PL 1D22DK12 R I DQQDKH R I DK11D22 A 7A 32 DK11CK1 7 702 D22 BKl AK B B1 32 DK11DK12 7 7D21 D22 BK1 BK2 O 01 D12 DK11 0K1 7 702 D22 DK21 0K2 D D11 D12 DKll DK12 7 7D21 D22 DK21 DK22 gt P 9lt K starpP K dimy dimu gtgt HP K starpP K Unstructured Uncertainties An unstructured perturbation can be connected to the plant in a number of ways each generating a unique set of possible plant model In this presentation we present 5 basic connections of the perturbation to the nominal plant Additive Perturbation Input Multiplicative Perturbation Output Multiplicative Perturbation Input Feedback Perturbation Output Feedback Perturbation Additive Perturbation Model GS Gos Ads Input Multiplicative Perturbation W Gosgt1 Aim Output Multiplicative Perturbation GS1 AosGaS gt Input Feedback Perturbation GS Gos1 A 31 1 gt Output Feedback Perturbation 05 1 Af0s lGoS Linear Fractional Transformations LFT What are they M M Let M 12 171p2 Xq1q2 M21 M22 6 C and let Al 6 C qzxp2and A u e C 3971 be two other complex matrices and let s have two Systems connected as follows u2 y2 w2 I M z2 I Let s take Case 1 21M11 M12W1 yl M21 M22 1 u1Aly1 y1 M21w1M22A1y1 71 3 y1IM22Al Mziwi Substituting this value of yl in the equation corresponding to zl we have Z1 M11w1M12Aly1 1 3Z1 M11 M12A11M22A1 M21W1 or we can represent 21 as 21 F1M A1w1 Where F1M A1 2 M11 MizAia 39M 22A11 M21 We see that F1 M Al is a transform from w1 to z1 hence we can represent F1MAlas TZW1 gt 21 Tzwlw1 Similar is with Case 2 w2 M z2 we haVe Z2 M22M21Au1 39 M22 AID 1M1 W2 or we can also write z2 FuM Auw2 Where FuM7 Au M22M21Au1 39 M22 AID 1M1 Also Note that FuM Au is a transformation from w2 to z2 so we can represent FuM Au as TZW2 So z2 TZW2 w2 A Few De nitions and Lemma s An LFT F MA is said to be well defined or well posed if IM22 A is invertible Lemma 1 If ABCD and Q are appropriately sized matrices or transfer functions and FQ ABQ CDQ and if C is invertible Then ABQCDQ1 F M Q and CQD391 AQB F 1N Q with AC 1 B AC ID C IA C4 M 1 1 N 1 1 C C D B DC A DC Lemma 2 Let FMQ be a given LFT with M M1 M2 If M 12 is invertible then FMMQ CQD391AQB With A M121M11B M21 M22M121M112C M121aandD 4224121 that is 0 A CWF M M 71 B D l 21 21 K i lllllllll I 11 0 0 J E 21 0 M22 E 1 F or any n0nsingular matrix E NOW if M21 is invertible then EltMQ 1 1BQCDQ1 With A MHM21 IDB 2M M11M21 1M22C M21 1and D Z21 1M22 that is 5 0 I E1 quotquotquotquot for any nonsingular matrix E Lemma 3 LelMM11 M12 M21 M 22 and M 22 is non Singular Then FuMA 1 FuNA With N is given by M11M12M22 M21 M12M221 1 1 M22 M21 M 22 1 N Big Question Why are we doing all this LFT is a very convenient tool to formulate many mathematical objects Let s have a feedback system with the following block diagram Now we want the above block diagram to be rearranged as an LFT that looks like Step 1 Identify the Inputs and Outputs of the System We see that the Inputs to the system are d n and u and the Outputs are v uf and y We see that W2Pd u v gt v WZPd WZPu 1 Also ufW1u 2 Fan y but a Pdu gt F Pdu n y gty FPd Fn FPu So we have 3 equations representing the input output relations of the system Now representing these equations in matrix form we have v sz O sz d uf O 0 W1 n y FP F FP u v d Letz andw uf n We have Zi Giwi U K G11 G12 WhereG G21 G22 wP 0 WP and Gll2 OG12 G21 FP FG22 FP 1 Another Simple Example A Mass Spring Damper System with Parametric Uncertainty Consider the Mass Spring Damper System as shown F The dynamical equation of the system motion can be described by The System Description We have 3 physical parameters In c k Their values are not known exactly but are believed to lie in known intervals m is within 10 of a nominal mass mg C is within 20 of a nominal value c0 k is within 30 of a nominal value k0 We now introduce perturbations 8m 50 8k and they are assumed to lie between the interval ll If we knew the exact values of m k and c we could have simulated the system as follows lm 1m01015m 1 01 3 6m1016m 11 mo mo mll m12A11m22A171m21 where mu m12 39 m21land m222 01 m0 m0 1 gt FZM5m m 1 01 Where M 2 m0 m0 1 01 Similarly c c01 0256 and k k01 035k k01036k The First step is to de ne the inputs and outputs of the system The input signals of the dynamical system are selected as x1xx2 x andF The output signals are selected as x1 and x2 The next step is to isolate the parametric uncertainties and denote the inputs and outputs of 8m Sc 8k as yk yc ym and uk uc um respectively With these parametric uncertainties isolated our system looks like X1 9 7 C0 7 i A 02 ye uc r lk JD 0 Now with these uncertainty parameters isolated we literally have 6 Inputs which are x1x2F uk uc um and 5 Outputs which are x1 9xz 9ykycym X1 X1 ls I 3x2 m0 But a Fbc with b cox2uc andc kox uk soak0x1ukFc0x2uc k andxg Ox Cox2i 1uk 1uC O391um m0 711 1 0 m0 m0 m0 m0 We also seeyk 0310 x1 yc 02c0x2 andym a 01 um gtymk0x1c0x2Fukuc 01um J 1 000 0 1 00 0 1 00 0 001 0 0 C 0 02 C n 0 1 2 X XF 0 0k 00 co 30k 00 A 7 0 k N 00 00 l F N A With Basic Principle for Writing a Matrix LFT The basic principle in writing a LFT Matrix is called PULLING OUT OF A S Consider an Input Output relation a 95 c5 5 2 z Z IZZW Gw 1 61 5152 651 a b c d e are given constants or transfer functions We would like to write G as an LFT in terms of 81 and 82 To do this we follow a 3 step Procedure Step 1 Draw a block diagram for the IO relation with each 5 separated a 1362 c51522 o H 45152 e512 Step 2 Mark the inputs and outputs of the 8 s as y s and u s resp This is essentially called PULLING OF A S Step 3 Write z s and y s in terms of w s and u s with all 8 s taken out This is basically called CALCULATING MATRIX M M W eu2 60 yZZul y3u1 y4zcu3by1 gt y4cu3bw eu2 du3 gt y4 beu2 bdcu3bw zaylu4 zz aeuz adu3u4aw yl 1 yz 2 ys 2M 3 y4 4 Z W Where OOHHO O 0 H0000 WOOH Lecture Notes Week 3b Topic Performance Speci cations and Limitations ECEMAE 7360 Optimal and Robust Control Fall 2003 Offering Instructor Dr YangQuan Chen CSOIS ECE Dept Tel 4357970148 Email chen ieeeor or ygcheneceusuedu 77 Chapter 6 Performance Speci cations and Limitations 0 Feedback Properties 0 Weighted H2 and H00 Performance 0 Selection of Weighting Performance 0 Bode7s Gain and Phase Relation o Bode7s Sensitivity Integral o Analyticity Constraints 78 Feedback Properties L K u ldap ld If TL i S 1 KP 1 so 1 PK T 1 S KP1 KP 1 T0 1 so PK1 PK1 3 T07quot n SOPer Sod up K800quot n KSOd Sid Disturbance rejection at the plant output low frequency 680 6 1 P104 ltlt 1 l Q PK 68013 6 1 PK 1P 5PSZ39 ltlt 1 Disturbance rejection at the plant input low frequency 68 6 1 KP 1 ltlt 1 1 1 Q KP air 5K1 P104 was lt 1 Sensor noise rejection and robust stability high frequency 53 6PK1PK 1 ltlt 1 79 Note that 9 SOltlt1ltgt qPKgtgt1 S ltlt1ltgt KPgtgt1 T0ltlt1 ltgt PKltlt1 Now suppose P and K are invertible then Qiq Q 5 QPKgtgt1orQKPgtgt1 1 PK 1P K 1 dig KU PKW W134 1 P39 Desired Loop Shape 80 Weighted H2 and H00 Performance Figure 04 Standard feedback con guration with weights H2 Performance Assume Rt 776t and E7777 I Minimize the expected energy of the error e E EU llellgdt IIWeSoWdllg Include the control signal u in the cost function WESOWd quKSOWd 2 2 2 E Mtp wm 2 Robustness problem H00 Performance under worst possible case 111 2 llWESOWdHOO lldllg 1 restrictions on the control energy or control bandwidth 111 ll llg IIWuKSoWdlloo lldllg 1 Combined cost 2 WESOW HF md6p wLMKamp l I 2 S 00 81 Selection of Weighting Functions SISO Let L PK be a standard second order system 2 w L ss2 wn 775 06 216 4 trzz 0335308 tag Mpe V1752 0lt5lt1 sensitivity function 101 Figure 05 Sensitivity function S for g 00501020508 and 1 with normalized frequency wWn i 1 i ss2 wn i1Li522 wnswg SjWn gt 1 closed loop bandwidth cub tun since Sjw Z 1 Vw 2 tab 82 A good control design M5 SCgt0 not too large peak sensmvny rallo Figure 06 Peak sensitivity MS versus damping ratio 1iWei MS Figure 07 Performance weight We and desired S Werequire s S lt 39 V SgtI 8M8wb 8 you w Ms ltgt WCS 1 WesSW M Practical cons1deration We 51 83 lSG Cl Figure 08 Practical performance weight We and desired S Control weighting function Wu i s cubeMu 515 wbc 1qu l Mu l a lKSGmI Figure 09 Control weight Wu and desired KS 84 Bode s Gain and Phase Relation L stable and minimum phase 1 00 l L 4Ljw0 gioo d3yln coth lgldu V lnww0 1n coth lv 1 Z w 3 w Figure 010 The function In 00th vs V dl L 39 4Ljw0 depends mostly on the behavior of W near woz 1 M 11406 rad oz ln3 6530 oz ln3 7 ln coth idu 13146 rad oz ln5 7530 oz ln5 W 1443 rad a ln 10 8270 a ln 10 4Ljw0 large if L attenuates slowly near wo and small if it attenuates rapidly near wo For example it is reasonable to expect E X 6530 if the slope of L E for i g 50 g 3 Ljw0 lt E X 7530 if the slope of L E for g 50 g 5 E X 8270 if the slope of L E for 1170 g 50 g 10 The behavior of 4Ljw is particularly important near the crossover fre quency we where Ljwc 1 since 7r Ljwc is the phase margin of 85 the feedback system Further the return difference is given by L C 1Ljwc 1L 1jwc 2 sinw 7 which must not be too small for good stability robustness It is important to keep the slope of L near we not much smaller than 1 for a reasonably wide range of frequencies in order to guarantee some reasonable performance L stable and nonminimum phase with RHP zeros 21 22 2k 521 522 szk Lltsgt meltsgt s zl s 22 s 2 where me is stable and minimum phase and Ljw mejw Hence k jw0 2239 LL w sz w 4 j 0 plt 0 2131 ij Zi 1 00 all Lm k 39 Mmmh d 2 4M 7T 00 d 2 21 jwo 2239 which gives 1 oo dlnL y k jw0 z 4L i l th 7d jwol 7r 00 dy HCO 2 V 23921 jwo Z mg 239 Since 4 g 0 for each 2 a nonminimum phase zero contributes 0 239 an additional pllase lag and imposes limitations on the rolloff rate of the open loop gain For example suppose L has a zero at z gt 0 then mg z mom 2 N 9OO 53130 280 jwo z w0zz2z4 Since the slope of L near the crossover frequency is in general no greater than 1 which means that the phase due to the minimum phase part me of L will in general be no greater than 900 the crossover frequency or the closed loop bandwidth must satisfy wc lt z2 86 phase 1m0 z in degree Figure 011 Phase 1w0z due to a real zero 2 gt 0 coo 2 in degree phase b2 Figure 012 Phase mayo 2 due to a pair of complex zeros z z i jy and z gt 0 87 for closed loop stability and some reasonable closed loop performance Next suppose L has a pair of complex right half zeros at z 5 j jy with 6 gt 0 then WOIZD L jwo z jwoz jwo 2 jwo 2 woiziizi2izi3izi4 1800 1O626O 7370 560 gtgt z 1800 8670 5590 4130 z 3600 00 00 00 ltlt ln this case we conclude that the crossover frequency must satisfy IZI4 392 gtgt 92 we lt lzlS z lzl ltlt in order to guarantee the closed loop stability and some reasonable closed loop performance 88 Bode s Sensitivity Integral Let p1 p2 pm be the open right half plane poles of L 000 ln Sjwdw 7r Wm 03 In the case where L is stable the integral simpli es to Oooln Sjwdw O 04 water bed effect lSG CON Figure 013 Water bed e ect of sensitivity function Suppose wll Bandwidth constraints and stability robustness M Lltjwgt s 6 lt1 w e m 00gt a 1 win Ari max SltJwgt26 lt1 egt ltwrwzgt 6 we lwWhl where m a 221 WM Uh w The above lower bound shows that the sensitivity can be very signi cant in the transition band 89 Poisson integral relation Suppose L has at least one more poles than zeros and suppose z 0 jyo with 0 gt O is a right half plane zero of L Then oo 330 m 2 Pz39 1 S d 1 7 05 00 HI ltjwgtlw yo2 w W H E13p lt gt De ne a i W 0 6z 7 La 35 w y0gt2dw Then m Zp SC 239 oo 1 0 win 7 I LOO n Sjw w y0gt2dw S W 0Zgtgt1HSltjwgtlloo 92 We Isltsgtooz n z391 which gives 2 13239 2 H 90 Analyticity Constraints Let phpg pm and 21 22 2 be the open right half plane poles and zeros of L respectively Sltp gt0 TltP gt1i12m and 1 0 j 12k Suppose S IL 1 and T LIL 1 are stable Then 191192 pm are the right half plane zeros of S and 21 22 2 are the right half plane zeros of T Let m r k 325 H M B s pl 18p j182j Then Bpjw 1 and Bzjw 1 for all frequencies and moreover B 15Ss E H00 B 1STs E H00 p 2 Hence by the maximum modulus theorem we have Slt8gtlloo H3171lt8gtSlt8gtH00 gt Bglltzgt8ltzgtl IBQ1ltZgt for any z with gt 0 Let z be a right half plane zero of L then 7 ZP S gt B 1 7 ltsgtool p ltzgt gm Similarly one can obtain k pzr Tlt8gtlloo Z lellpll H 7 7 J1 p Zj where p is a right half plane pole of L The weighted problem can be considered in the same fashion Let We be a weight such that WES is stable Then m H I Z 13239 21 W5lt8gtSlt8gtlloo 2 WW Zp 91 i sMS wb Now suppose W45 7 H w VVESCgt0 g 1 and z is areal right half be plane zero Then ZMswb m 2 H alt17 zwbe 21 zp which gives lt Z lt gt lt 1 gt w oz 7 z 2 oz 7 b 1 046 M5 M5 bandwidth must be much smaller than the right half plane zero Lecture Notes Week 6aamp6b Topic SSV mu and musynthesis ECEMAE 7360 Optimal and Robust Control Fall 2003 Offering Instructor Dr YangQuan Chen CSOIS ECE Dept Tel 4357970148 Email chen ieee0r 0rygcheneceusuedu 139 Chapter 10 a and a Synthesis general framework analysis and synthesis methods for unstructured uncertainty stability with structured uncertainties structured singular value structured robust stability robust performance extension to nonlinear time varying uncertainties skewed problem overview on a synthesis 140 General Framework General Framework Analysis Framework M11lt8gt M12lt8gt Mltsgt ltPlt5gt7Klt5gtgt Mm M209 7 Z M22 M21AltI M11Agt71M12 U E zw M Analysis and Synthesis Methods for Unstructured 141 Uncertainty Input Performance Perturbation Analysis Synthesis Assumptions Speci cations Assumptions Tests Methods Ewtw7 Eztzt g 1 LQG 6t 7 7 A 0 HMZZHZ S 1 Wiener Hopf U 6 t w 0 W EltHzH gt1 EU0U3 I H2 HwHZ g 1 g 1 A 0 HMZZHOO g 1 Singular Value Loop Shaping HwHZ g 1 Internal Stability HAHOO lt1 HMHHOO g 1 H00 142 Stability with Structured Uncertainties Assume Alt8gt 611717 7 65119 A1 7 E with 6 OO lt 1 and AjOO lt1 Robust Stability ltgt The following interconnection is stable ANS 61 M11lt8gt Stability Conditions 1 suf cient conditions 111Cgt0 g 1 Conservative ignoring structure of the uncertainties 2 necessary conditions Test for each 6239 ltAjgt individually assuming no uncertainty in other channels 111mr00 g 1 Optimistic because it ignores interaction between the 6239 ltAjgt 143 Structured Singular Value 1111 61 E ew 22 Unstructured A Problem Given M E cpxq nd a smallest A E qup in the sense of 6A such that detI MA 0 It is easy to see that 04min inf 6A det MA O A E qup inf dz detI dMA O 6A g 1 A E qup and Egg AMA f1 5W with a smallest destabilizing A Ades vlu f detI MAdes O WM where M Mu1vf 021121 SO 6M can be de ned as altMgt mam detI MA 07 A E W 144 Structured A A diag 6llrl6slr A1AFI6 ECCAjECijgtltmj S7 amm inf 6A detltl MA O A E A infa detltl onA O A S 1 A E A and rnax1 pMA ofl g M WA rnin De nition of SSV For M 6 CW MAM is de ned as 1 MW 2 H1me A e A det lt1 Mm 0 0396 unless no A E A makes 1 MA singular in WlllCll case MAM O o If A 6126 E C 81FOr1n then MAM pM the spectral radius of M If A CW 50 F1 m1n then MM aw MM SMM SWM 145 O 1 M 03 for any 6 gt 0 Then pM O and WM 6 But MM 0 since detU MA 1 for all admissible A 12 12 Then M Oand M 1 Since 1212 plt gt lt gt 2M detI MA16162 O if61 62 1 so MM 1 Thus neither p nor 6 provide useful bounds even in these simple cases 146 2 UEAz UUIn diagD1Dgd1m1dF1ImF71ImF I meal 513F133 gt0dj endj gt0 39 Note that for any A E A U E 21 and D E D D WEIl UAEA AUEA 6UA6AU6A DAAD ForaIIUEZl andDED i 7 71 Igl agMUM gArgngpAMi1AM I1I 10ltDMD gt i 71 I613 pUM 3 MM 3 Igg JaltDMD gt Doer 1982 pMU MAM Not Convex MAMI391E16DMD 1H2S F g 3 F01234 0 yes yes yes no yes yes no no no gtgt b0undsr0wd mu1Iblk 6112 O O O O O O 62 O O O O A O 0 A3 0 O O O O 0 A4 0 O O O O O 6513 O O O O O 0 A6 61 62 65 E C Ag G ngg A4 E ngg A6 E C2X1 can be speci ed by blk DOOOOkDr N HOOOwr CD gtgt D5 Dr unwrapdltr0wd blk 148 Structured Robust Stability How large A in the sense of can be Without destabilizing the feedback system Since the closed loop poles are given by detl MA O the feedback system becomes unstable if detl M5As O for some 5 E 6 NOW let or gt 0 be a su iciently small number such that the closed loop system is stable for all stable ACgt0 lt oz Next increase a until ozmax so that the closed loop system becomes unstable So 0mm is the robust stability margin De ne A z E RHOO Aso E A for all 50 E 6 Let 6 gt O The system is well posed and internally stable for all E A With ACgt0 lt if and only if 31g MAltGltjwgtgt S 6 149 Robust Performance G lt8 G11 012 p 021 022 A 0 Apz ZAEAAfECCQ2Xp2 0 Af N5 Z Gplt5gt w Let 6 gt O For all A5 6 A With ACgt0 lt 1 the system is B well posed internally stable and GP A g 6 if and only if oo 31g MAP Gpltjwgtgt S 6 E 150 Extension to Nonlinear Time Varying Uncertainty Suppose A E AN is a structured Nonlinear Time varying Uncer tainty and suppose D is constant scaling matrix such that DAD 1 E AN Then a su icient condition for stability is by small gain theorern Hmm 1 151 HIMAT Example gtgt A B C D linm0d aircraft gt G pckA B C D gt K Gpm hinfsynG 2 2 0 10 0001 2 Which gives 7 18612 Gpoo a stabilizing controller K and a Closed loop transfer matrix GP P1 Z1 quotPquot 39 Gpltsgt d1 Gpltsgt GP G 61 d2 Gp21 Gp22 62 m 712 maximum singular value X 1 0 3 1 0 2 10 1 10D 10 1o2 103 39equency radsec Figure 022 Singular values of Gpjw Now generate the singular value frequency responses of GP 152 gtgt wlogspace33300 gtgt pr frspGp W pr is the frequency response of GP gt u sv vsvdpr gtgt vplot liv m s The singular value frequency responses of GP are shown in Figure 022 To test the robust stability we need to compute Gp1100 gtgt Gp11 S 1ltGp1212 gtgt norm0fGp11 hinfnormltGp11I 0001 which gives G1211Cgt0 0933 lt 1 So the system is robustly stable To check the robust performance we shall compute the uApGpjw for each frequency with A E CCQXQ Af E C4X2 Ar Maximum Singular Value and mu I I I maximum singular value mu bounds 10 1 10D 10 39equency radsec Figure 023 MAP Gpjw and 5Gpjw gtgt blk2242 153 gtgt bndsdvecsenspvecmuprblk gtgt vplot liv m vnormltpr buds gtgt title MaXimum Singular Value and mu gtgt Xlabel frequencyradsec gtgt text00117 maximum singular value gt text05 08 mu bounds The structured singular value uApGpjw and 6Gpjw are shown in Figure 023 It is clear that the robust performance is not satis ed Note that Gpll 0pm Gp217 Gp227 ag llfultGp7 AMIGO S i ltgt 8319 MAP Using a bisection algorithm we can also nd the worst performance 221 mop mum 127824 154 Skewed Problem W2TZ39W1 W2K80Wd WESOPIVl WCSOWd 39 robust performance condition W2T W1 de2KSOWd iWESOPwl Wesowd MltGltjwgtgt dinf a J g 1 WER for all a Z 0 An upper bound MMGUW S iHltWJ1PW1gtW2T W1 IIWeSoWdH u is proportional to the square root of the plant condition number Assumptions We wsl Wd 1 W1 1 W2 wtl and P is stable and has a stable inverse ie minimum phase and Klt8gt P 18gtllt8gt such that Ks is proper and the closed loop is stable Then so 52 1 as To n 5 I Tltsgt1 thI thP71 1115613 11155 Then Ut7391 thltdPgt71 63 711de 11155 New dim a 155 Let the SVD of Pjw be Pjw UEVquot Z3 diag01 02 am with 01 6 and 0m Q where m is the dimension of P Then thI wt7d2 1 wsedE ruse PldiagM1 M2 7 dETR New inf 0 wt739I wt7d2 1 wsedE ruse where P1 and P2 are permutation matrices and where d Z 1 Mi th wt7 a wsedazr wse Hence d6 Z wsedazr wse 1 1004 inf maxW1 da 2wsed0 2 mm dETR Z MltGltjwgtgt inf maxa wt7 wt7da 1 wt739 inf max d6 Z wsedazr wt7392 inf max wsel2 Iw l2 steda 2 dER Z dog The maximum is achieved at 2 lthI lwselg and 1 MAGjw lesel2 thrl2 lwsellwt7cP MAltGltjwgtgt lws llw ll ltPgt 156 Overview on u Synthesis fgltG K G11 G12KltI G22Kgt71G2p We Ion The u synthesis is not yet fully solved But a reasonable approach is to solve rriin 13f Home KD 1H DD EH00 00 by iteratively solving for K and D ie rst minimizing over K with D xed then minimizing pointvvise over D with K xed then again over K and again over D etc This is the so called D K Iteration 0 Fix D Inin Home KD 1HOO is a standard H00 optimization problem 0 Fix K ppngwllD ltGvKgtD lm is a standard convex optimization problem and it can be solved point wise in the frequency domain sup DinEfD DwJquotAG KjwD1l Note that when S 0 no scalar blocks Dw diagdf 11111 1 e D 157 D K Iterations i Fix an initial estimate of the scaling matrix DW E D pointwise across frequency ii Find scalar transfer functions Ms d1s E R7100 for 239 1 F 1 such that d jw df iii Let Ds diagd1s1dF1sII Construct a state space model for system iv Solve an HOG optimization problem to minimize lWG ml over all stabilizing Ks Denote the minimizing controller by K v Minimize DwfgG K D 1 over Dw pointwise across frequency The minimization itself produces a new scaling function vi Compare Dw with the previous estimate Dw Stop if they are close otherwise replace DW with DW and return to step ii The joint optimization of D and K is not convex and the global con vergence is not guaranteed many designs have shown that this approach works very well Lecture Notes Week 7b Topic LQRKalman FilterLQG ECEMAE 7360 Optimal and Robust Control Fall 2003 Offering Instructor Dr YangQuan Chen CSOIS ECE Dept Tel 4357970148 Email chen ieee0r 0rygcheneceusuedu Properties of LQR Control 45 Linear Quadratic Regulator L 0R Consider a linear system characterized by szxBu where A B is stabilizable We de ne the cost index JxuQR IxTQx uTRudt Q 2 0 R gt 0 0 and A Q is detectable The linear quadratic regulation problem is to find a control law u Fx such that A B F is stable and J is minimized ltwas shown in the rst Part of this course that the solution is given by F R IBTP T 1 T IMAP RM BPQO 46 Prepared by Ben M Chen If we arrange the LQR control in the following block diagram ape a we can find its gain margin and phase margin as we have done in classical control It is clear that the openloop transfer function Open loop transfer function 2 FsI A IB R 1BTPsI A lB The block diagram can be redrawn as follows 0 R lBTPsI A IB gt 47 Prepared by Ben M Chen Return Difference Equality and Inequality Consider the LQR control law The following socalled return difference equality hold R BT ja1 AT 1QjaI AB 1 BT ja1 AT 1FTRI Fja1 A 1B The following is called the return difference inequality 1BT jaI AT 1FTRI Fja1 A IB 2 R Proof Recall that F R lBTP PA ATP PBR 1BTP Q 0 Then we have ijI PA ijI ATP PBR 1RR IBTP Q 0 PjaI A jaI ATPFTRF Q 48 Prepared by Ben M Chen uaqo 39w uaa q paJedaJd 617 EIJV MM IHid4iV loaf Lg I 9V Imf54iV loaf Lg 21 f EIJV 09051409 ImI Lg 51141 IMMHidIiiV loaf Lg 511407 Whig Highly 10M ig am am Hid ld 75gt dydig c d l d I new 1019 9H1 5U10N EIJV 1090511 iV loaf Lg 5114097 Kraftmidlly loaf Lg 51140 Mylar sally loaf Lg 11407 10905149197 loaf Lg 51141 mng lidlliV loaf Lg 511407 ImdiV Imf 49V loaf Lg 511407 0900 1mfd 49V loaf Lg ugelqo am g4V 101 Aq 1u5u 9L1 uo pue 4iV loaf ig Aq 149 9L1 uo 1 BugMpan Single Input Case In the single input case the transfer function Open loop transfer function 2 f s1 A 1b is a scalar function Let Q h M Then the return difference equation is reduced to r bT jaI AT 1hhTja1 Ab r 1bT ja1 AT 1fT1fjaI A 1b iv w hw m Ayfrhfrmy Ayf ii rhfrmy A bfzr ii 1fjaI A 1b 2 21 Return Difference Inequality 50 Prepared by Ben M Chen Graphically i1fjw1 A 1bi221 3 i fij A 1b 1j0 21 impliesthat Clearly the phase margin resulting from the LQR design is at least 60 degrees 51 Prepared by Ben M Chen Example Consider a given plant characterized by 1 liclfl Solving the LQR problem which minimizes the following cost function 1 0 JxuQRTxTQxuTRudt with Q0 0 R01 we obtain 06872 02317 02317 01373 and F23166 13734 which results the closedloop eigenvalues at 11867 4 j13814 Clearly the closedloop system is asymptotically stable 52 Prepared by Ben M Chen Phase deg Magnitude dB T0Y1 Bode Diagram s From U1 1 0 F 0 1 0 gt20 Frequency PM 84 100 radsec 53 Prepared by Ben M Chen Kalman Filter 54 Prepared by Ben M Chen Review Random Process A random variableX is a mapping between the sample space and the real numbers A random process aka stochastic process is a mapping from the sample space into an ensemble of time functions known as sample functions To every member in the sample space there corresponds a function of time a sample function Xz AXU I y w 39 m 5 39 z a E x mquot quotwe 5 5 Prepared by Ben M Chen Mean Moment Variance Covariance of Stationary Random Process Let f x r be the probability density function pdf associated with a random process Xr If the pdf is independent of time t ie f x r f x then the corresponding random process is said to be stationary We will focus our attention only on this class of random processes in this course For this type of random processes we de ne 1 mean or expectation 2 moment jth order moment m EX 2 Tx fxdx EXf 2 Tx fxdx 3 variance 7w 4 covariance oitowo random processes 02 Ex m2 To m2fxdx convw Ev Evw Ew Two random processes v and w are said to be independent if I vafv wdvdw 0 fvw is the joint pdf of v and w 700700 5 Prepared by Ben M Chen Autocorrelation Function and Power Spectrum Autocorrelation function is used to describe the time domain property of a random process Given a random process v its autocorrelation function is defined as follows Rxc r2 E v01 gtvltz2gti If v is a stationary process RAIDIZ RxI2 I1 RAT Rxtrr E vrvZ T Note that Rx0 is the time average of the power or energy of the random process Power spectrum of a random process is the Fourier transform of its autocorrelation function It is a frequency domain property of the random process To be more speci c it is de ned as Sxa ijrerr 57 Prepared by Ben M Chen VWiite Noise Color Noise and Gaussian Random Process White Noise is a random process with a constant power spectrum and an autocorrelation function Rx T Z q which implies that a white noise has an in nite power and thus it is non existent in our real life However many noises or the socalled color noises or noises with nite energy and finite frequency components can be modeled as the outputs of linear systems with an injection of white noise into their inputs ie any color noise can be generated by a white noise while noise Linear System gt color noise Gaussian Process v is also known as normal process has a pdf 1 viuf 2 v e 2quot f a 2 2 J I mean 0 Z varlance 5 8 Prepared by Ben M Chen K alman Filter for a Linear Time In variant L TI System Consider a LT system characterized by x Ax Bu vl v is the input noise y Cx wl w is the measurement noise Assume 1 AC is observable 2 vl and wl are independent white noises with the following properties Evr 0 Ewr 0 EvzvT 1 Q60 t Q QT 2 0 EwlwTT R60 t R RT gt 0 3 A Q is stabilizable to guarantee closedloop stability The problem of Kalman Filter is to design a state estimator to estimate the state xZ by 5ct such that the estimation error covariance is minimized ie the following index is minimized Je Exl lTXlf1 59 Prepared by Ben M Chen Construction of Steady State K aIman Filter Kalman lter is a state observer with a specially selected observer gain or Kalman lter gain It has the dynamic equation 2 A Bu Key 0 is given C with the Kalman lter gain Ke being given as Kg 2 PQCTR I where P8 is the positive definite solution of the following Riccati equation PgAT APe PeCTR 1CPe Q 0 Let e x fc We can show see next that such a Kalman filter has the following properties 1imEet1imExt 541 0 lim J9 lim EeTt et trace Pg taco law law law 60 Prepared by Ben M Chen K alman Filter and Linear Quadratic Regulator They Are Dual Recall the optimal regulator problem 5c Ax Bu x0 xO given JzojiltxTQxuTRudl QQT 20 and RzRT gt0 0 The LQR problem is to nd a state feedback law u F x such thatJ is minimized It was shown that the solution to the above problem is given by FzR lBTP PAATP PBR 1BTPQ0 PPT gt0 and the optimal value of J is given by J ng x0 Note that x0 is arbitrary Let us consider a special case when x0 is a random vector with Ex0 0 Ex0x 1 Then we have ELI ExoTPxo E py39inxoj39 ZZpy39EUOixoy an39 tracep 1 11 11 H 61 11 Prepared by Ben M Chen The Duality 4 Linear Quadratic Regulator a Kalman Filter F R IBTP Ke PQCTR 1 PAATP PBR IBTPQ0 PQAT APe PeCTR 1CPe Q 0 JOptjmal trace P JOptimal trace Pe 39 These two problems are equivalent or dual if we let ATHA BTHC FTlt gtKe Plt gtP e 62 Prepared by Ben M Chen Proof of the Properties of K alman Filter Recall that the dynamics of the given plant and Kalman lter ie szxBuvz A BuKey f amp yCxwl J7CA We have 639 x p AxBuvl AE Bu KeCxwt CE A KQCx Evt Kewt A KCe1 K lvl Zedz W with V0 2 EVZ 0 Ewan Eli in 1 KeEme 1 KJQ 0 Next it is reasonable to assume that initial error e0 and dz are independent ie EreogtdTltrgto 3 Prepared by Ben M Chen Furthermore T T I EWWTHZU KeEvltrgtv oi Evlw WM KT EWIVTT EWIWTT Q6t T o 1 U Ke o Ran 1 16 QKQRKQT 6rz r van r We will next show Z is asymptotically stable and lim Eet eT t 2 Fe and PQZT ZPQ V Recall that the solution to the following state equation from your linear systems course notes 639 Ze dz el eZ e0 jeA dT dzquot 64 Prepared by Ben M Chen Also recall that Kg PQCTR 1 and PeAT APe PeCTR 1CPe Q 0 We have PQAT PQCTR 1CPQ APe PeCTR 1CPe PQCTR 1CPQ Q 0 gt 11AT CTR ICPQ A PQCTR ICPQ PQCTR ICPQ Q 0 gt 11AT CTKA KQCPQ PQCTR 1CPQ Q 0 gt PQZT 211 2 PQCTR 1CPQ Q V g 0 Since Q QT 2 0 and A Q is assumed to be stabilizable it follows from Lyapunov stability theory that matrix X A KQC is asymptotically stable 65 Prepared by Ben M Chen Noting that eg is deterministic we have Pt EezeT 1 EH6 e0 jeAHdT d1 em e0 jeAHdT dr 0 0 eri Ee0eT 0eZT jeA TEdTeT 013 d1quot 0 jeA EeOdT leZT H d1quot jeAltMgtdrjEdrdT 6eA7T0 dc 0 0 0 z eZIEe0eT 01 jeA Tder6T WET dc 0 0 eZ EeOeTOeZT 351 VeATO Udr eZ EeOeTOeA7T jeAnVeATquot d17 0 0 Since Z is stable we have eg gt 0 as z gt 00 Thus Poo WWW dn 0 66 Prepared by Ben M Chen We next show that Poo P8 ie the solution associated with the Kalman filter ARE Let 239 2 2 20 given 3 20 2 920 200 2 o In view of PQZT Zpe V we have zT PEZT 213812 zTVz gt zTPeZTz ZTZPEZ ZTVZ gt ZTPEZ Z39TPEZ ZTVZ gt ZTPEZ ZTVZ Next we have zTVzdt T ZT0eZTtVeth0dt o eATtVeAtdtz0 zT 0Pooz0 1 dt j zTPezdt zTtPezt ZToOPezoO ZT0PEZ0 o zT0Pez0 Thus we have for every given 20 zT0Pez0 zT0Pooz0 gt P8PoojeZT VeZ dn 67 0 Prepared by an M Chen It is now simple to see that 1imEeleTl Poo Pg gt 1imEeT Zet trace Pe Finally we have Eel eZ Ee0 jeA E d139 d1quot o Example Consider a given plant characterized by the following state space model c 3 i x u vt EvtvTT Q50 L39 061 350 L39 y 1 01x wt EwtwT 1 R6t 1 0250 r Solving the Kalman filter ARE we obtain 00792 00343 03962 A e Z 7 Kg 2 x AxBuKey y 00343 00314 01715 902 68 Prepared by Ben M Chen Linear Quadratic Gaussian LQG 69 Prepared by Ben M Chen Problem Statement It is very often in control system design for a real life problem that one cannot measure all the state variables of the given plant Thus the linear quadratic regulator although it has a very impressive gain and phase margins GM 00 and PM 60 degrees is impractical as it utilizes all state variables in the feedback ie u 7 F x In most of practical situations only partial information of the state of the given plant is accessible or can be measured for feedback The natural questions one would ask Can we recover or estimate the state variables of the plant through the partially measurable information The answer is yes The solution is Kalman filter Can we replace x the control law in LQR ie u i F x by the estimated state to carry out a meaningful control system design The answer is yes The solution is called LQG Do we still have impressive properties associated with LQG The answer is no Any solution Yes It is called loop transfer recovery LTR technique to be covered later 70 Prepared by Ben M Chen Linear Quadratic Gaussian Design Consider a given plant characterized by x Ax Bu vl v is the input noise y Cx wl w is the measurement noise where vl and wl are white with zero means vl wl and x0 are independent and EVIVTT Qe5Z T Q 2 0 EWIWTT R9504 Re gt 0 Ex0 x0 The performance index has to be modi ed as follows T J limEI xTQxuTRudt Q 2 0 R gt 0 Tgtoo 0 The Linear Quadratic Gaussian LQG control is to design a control law that only requires the measurable information such that when it is applied to the given plant the overall system is stable and the performance index is minimized 71 Prepared by Ben M Chen Solution to the LOG Problem Separation Principle Step 1 Design an LQR control law u 7F x which solves the following problem XAxBu JxuQRIxTQxuTRudt Q20 Rgt0 0 Le compute PA ATP PBR IBTP Q 0 P gt 0 F R lBTP Step 2 Design a Kalman lter for the given plant ie 2 A Bu Key f Cx lt where PeAT APe PeCTR1CPe Qe 0 Pg gt 0 Kg PQCTR I Step 3 The LQG control law is given by u F 5c ie A BuKey C5c A BF KQCEKey Liz F Liz F 72 Prepared by Ben M Chen Block Diagram Implementation of LQG Control Law Reference u y I gt Plant 2 J LQRControl Kalman lter A More Detailed Block Diagram Reference u y r gt Plant gt LQG Controller 73 Prepared by Ben M Chen ClosedLoop Dynamics of the Given Plant together with LOG Controller icAxBuvt A BF KQCEKey yCxwt u Recall the plant A F xr and the controller We define a new variable 6 x 2 and thus 639 x p Ax BchBr vz A BF KeC Ker Kewt Ax 2 KeCx 2 Br vt Kewt A KeCe Br vt Kewl and 5c AxBu vt Ax BFEBrvt Ax BFx eBrvt A BFxBFe Brvt Clearly the closedloop system is characterized by the following state space equation LP BF BF Jlxh rw V V j be 0 A KeC be B v Kew y c olxlw W The closedloop poles are given by MA BF uMA KeC which are stable 74 Prepared by Ben M Chen Lecture Notes Week 4b Topic Balanced model reduction ECEMAE 7360 Optimal and Robust Control Fall 2003 Offering Instructor Dr YangQuan Chen CSOIS ECE Dept Tel 4357970148 Email chen ieee0r 0rygcheneceusuedu 92 Chapter 7 Balanced Model Reduction o Balanced Realization o Balanced Model Reduction 0 Frequency Weighted Balanced Model Reduction 0 Relative Reduction 93 Balanced Realization Consider the following Lyapunov equation AX X A Q 0 Assume that A is stable then the following statements hold o X 5 0 eAthAtdt oXgtOifQgtOandXZOifQZO o if Q 2 0 then Q A is observable iff X gt 0 Suppose X is the solution of the Lyapunov equation then 0 Re A 0ifXgtOandQZO o AisstableifX gtOandQ gt0 0 A is stable if X Z O Q 2 O and Q A is detectable Let A be stable Then a pair C A is observable iff the obsemability Gramz an Q gt O AQ QA 00 0 Similarly A B is controllable iff the controllabz lz ty Gramz an P gt O APPABB0 A B 0 Let C D be a state space realization of a not necessarily stable transfer matrix Cs Suppose that there exists a symmetric matrix P1 0 l O O with P1 nonsingular such that AP PAquot BB 0 94 Now partition the realization A B C D compatibly with P as Then is also a realization of G Moreover A11 B1 is controllable if A11 is stable Proof Using 0 APPA BB to get B2 O and A21 0 Hence part of the realization is not controllable A11 A12 Bl A11 A12 Bl All 31 A21 A22 B2 0 A22 0 lo 12le lologln 1 A B 0 Let C D be a state space realization of a not necessarily stable transfer matrix Cs Suppose that there exists a symmetric matrix i gtk i Q1 0 with Q1 nonsingular such that QA AQ 00 0 Now partition the realization A B C D compatibly with Q as 95 Then A11 Bl 01 D is also a realization of G Moreover Cl A11 is observable if A11 is stable 0 Let P and Q be the controllability and observability Grarnians APPA BB O AQ QA 00 0 Suppose P Q E diag01ogon Then the state space realization is called internally balanced realiza tion and 01 Z 02 Z Z on 2 O are called the Hankel singular ualues of the system Two other closely related realizations are called input normal real ization with P I and Q 22 and output normal realization with P 22 and Q 1 Both realizations can be obtained easily from the balanced realization by a suitable scaling on the states 0 Let P and Q be two positive sernide nite rnatrices Then there exists a nonsingular matrix T such that 21 E2 TPT 0 E1 ltT1gtQT1 0 E3 0 respectively with 21 22 23 diagonal and positive de nite ln the special case where l l is a minimal realization a balanced C D realization can be obtained through the following simpli ed procedure 1 Compute P gt O and Q gt O 2 Find a matrix R such that P RR 3 Diagonalize RQR to get RQR UEQUquot 4 Let T 1 PJ UE lQ Then TPT T 1QT 1 2 and TAT 1 TB CT 1 D Suppose or gtgt 0H1 for some r then the balanced realization implies that those states corresponding to the singular values of 0TH on are less controllable and observable than those states corresponding to 01 or Therefore truncating those less controllable and ob servable states will not lose much information about the system J is balanced input normal realization P I and Q 22 output normal realization P 22 and Q I 97 Suppose E R7100 is a balanced realization that is there exists Z diaglt01151 02152 UNISN Z O with 01gt 02 gt gt UN 2 0 such that AEEABB0 AE2ACCO Then N 01 IIGIIOO if llglttgtll dt 3 2g at where gt CeAtB Proof A Bw Cm A B is controllable and C A is observable g l 621 E 1E 1 AE 1E 1A2ltw BE 1 d i 7 it E 156 llwllg llw B E 15th Integration from t oo to t O with oo O and 330 330 gives 936271360 llwllg llw 3E 1 Iw 2 i i gtk 71 wdg mp llwlb mltogt i x0 7 x02 x0 Given 330 0 and w O for t Z O the norm of zt CeAtmo can be found from 0 Man dt 00 mgeAtooeAtx0dt 32330 98 To ShOW 01 GOO note that 2 HQ w2 f3 Ztgt dt IIGII sup 00 wE 2O000 we c iooioo W 2 WE 2000l fgoo dt 9607 0 33021 60 We shall now ShOW the other inequalities Since Glsl I Ooogte 5tdt Res gt O by the de nition of H00 norrn we have Goo Resggo Oooglttgte dtll dt R230lo llglttgt OOO glttgt dt To prove the last inequality let 6239 be the 2th unit vector and de ne E1 1651l EN 6515N711 6515N N Then 2 E Ef I and 21 00 00 N 0 llglttgtlldt 0 CeAtQZE EfeAt23dt z391 3 000 HCeAtQEZrEfeAtQBH dt l 3 AGOHCeAtQEZr EjeAtQBHdt l 2 y Oo llCeAt2E ll2 W 1 HEg eAtQBHQ dt 3 N 2 Z 0239 21 99 where we have used Cauchy Schwarz inequality and the following relations 0C HCeAt2E 2 dt 0C Am EjemQorroer tQEQ dt 2AMX EfEEZr 20 EfeAtQBHQ dt 0C Am EjeAtQBB FeAWQEQ dt th D gt Ab Eb Cb sig TinvbalrealA B C sig is avector of Hankel singular values and Tinv T l gtgt GD7 Sig sysbalG gtgt Gr struncltGb 2 truncate to the second order 100 Balanced Model Reduction GGrAa gt inf G GTOO degGT r 0 Suppose A11 A12 Bl Cs A21 A22 32 01 C2 is a balanced realization with Gramian E diaglt217 22 AEEABBO AE2ACCO where 21 diaglt01151 02152 JTIST 22 diagltar115r170r215T27 39 39 39 70N15Ngt ancl 01gt02gtquot39gt0rgt0r1gt0r2gtquot39gt0N where Q has multiplicity 52 239 1 2 N and 5152 5N n Then the truncated system G45 A11 Bl 01 D is balanced and asymptotically stable Furthermore Gltsgt emu g 2am 0m 0N 39 Glt8gt Gltooloo g 201 0N Gn71lt5gtlloo 2039N 101 Proof We shall rst show the one step model reduction Hence we shall assume 22 UNISN De ne the approximation error 01 Apply a similarity transformation T to the preceding statespace realiza tion with 2 2 O I I O T 12 120 T4 10 O O I O O I to get 1411 0 14122 Bl E11 0 A11 A122 0 A21 A21 A22 32 l 0 201 02 lo Consider a dilation of E11lt8gt2 E11lt8gt E12lt8gt Elt8gt i E21lt8gt E22lt8gt A11 0 24122 31 O 0 A11 A122 O O39szlof A21 A21 A22 32 03 O 2C391 Cg O 20NI 20NBfEf1 0 B 2aNI 0 J ea 102 Then it is easy to verify that 15 O UIQVEfl O O O 2039NISN satis es 21151521quot 83 0 150 BIquot 0 Using these two equations we have 21 BB 315 E5EN5 O 21 7 DB DDquot 21 2115 1521 BB PCBD 0 A 3 l c C5 DB 915 j A O DIquot 401 where the second equality is obtained by applying a similarity transfor rnation T I 15 O I Hence E11Cgt0 g 201v which is the desired result The remainder of the proof is achieved by using the order reduction A11 Bl by onestep results and by noting that Gks obtained by the kth order partitioning is internally balanced with balanced Grarnian given by 21 diag0151 02152 O39klsk 103 Let E45 Gk1s Gks for k 12N 1 and let GN5 Cs Then aEkUMl S 201m since Gk s is a reduced order model obtained from the internally balanced realization of Gk1s and the bound for onestep order reduction holds Noting that N71 Gltsgt Grltsgt I Ekltsgt by the de nition of E45 we have N71 N71 EthYw Grlt1wgtl S I ElEkltJwgtl S 2 I am This is the desired upper bound D 104 o bound can be tight For example a1 V171 n a2 b72 95 Z s 315aZ an hm l 0 J witha gt0andb gt0ThenPQ and Hogan 00 i Q 2traceP 2 i 0 397 21 271 239 o bound can also be loose for systems with Hankel singular values Close to each other For example 199579 54682 96954 09160 63180 54682 0 0 02378 00020 Cs 96954 0 0 40051 00067 09160 02378 40051 00420 02893 63180 00020 00067 02893 0 7 with Hankel singular values given by 01 1 02 09977 03 09957 04 09952 7quot 0 1 2 3 G G cgt0 2 1996 1991 19904 Bounds 2Zf 1a 79772 59772 39818 19904 20711 2 19954 19914 19904 105 FrequencyWeighted Balanced Model Reduction General Case inf W0G Grgtmlloo d Gag eg AB A B i A Bo Glo ol m oi Dil Woloo Dal WOGM Let P and Q be the solutions to the following Lyapunov equations APPABB 0 6221 21 00 0 The input output weighted Gramians P and Q are de ned by In 0 P and Q satisfy the following lower order equations A BCl P P12 P P12 A BCl BDi BDi 70 0 Al P13 P22 sz P22 0 Al Bl Bl Q on A 0 A 0 Q on 0D 0D 0 Q12 Q22 B00 A0 B00 A0 Q12 Q22 0 C l m I gt P can be obtained from PIn 0PM QIn Ole PAquot APBB 0 W0 I gt Q can be obtained from QAAQCC 0 106 Now let T be a nonsingular matrix such that 21 TPT T 1QT 1 E2 ie balanced and partition the system accordingly as TAT 1 TB i CT 1 0 Then a reduced order model GT is Obtained as i A11 Bl Gr 39 Works well but with guarantee 107 Relative Reduction Gr Arel G71ltG Gr inf degGT r 00 and a related problem is G Ann Let Cs E R7100 be minimum phase and D be nonsingular A BD lC39 BD 1 D710 D 1 39 a Then the inputoutput weighted Gramians P and Q are given by PAAPBB 0 QltA BD 10 A BD 10Q 0D 1D 1C 0 b Suppose P and Q are balanced Then W0 6145 l P Q diaglt01151 O39TISTO39T1IST1 O39NISNgt diaglt2h 22 and let G be partitioned compatibly with El and 22 as A11 A12 Bl Then i A11 Bl Gr is stable and minimum phase Furthermore N Amnmg H1ltl20 10220 gt 1 z39r N Amuloog H1ltl20 10 20 gt 1 z39r Lecture Notes Week 8a Topic LTR H2 and ARE ECEMAE 7360 Optimal and Robust Control Fall 2003 Offering Instructor Dr YangQuan Chen CSOIS ECE Dept Tel 4357970148 Email chen ieee0r 0rygcheneceusuedu Loop Transfer Recovery Design 1 15 Prepared by Ben M Chen Is an LOG Controller Robust It is now wellknown that the linear quadratic regulator LQR has very impressive robustness properties including guaranteed in nite gain margins and 60 degrees phase margins in all channels The result is only valid however for the full state feedback case If observers or Kalman lters ie LQG regulators are used in implementation no guaranteed robustness properties hold Still worse the closedloop system may become unstable if you do not design the observer of Kalman filter properly The following example given in Doyle 1978 shows the unrobustness of the LQG regulators Example Consider the following system characterized by 39 11O 1 10 X O 1x lu 1v y xw where x u and y denote the usual states control input and measured output and where w and v are white noises with intensities 1 and c gt 0 respectively 116 Prepared by Ben M Chen The LQG controller consists of an LQR control law a Kalman lter LQR Design Suppose we wish to minimize the performance index 1 1 jxTQxuTRudr R 1 Q q H 1 1M gt 0 0 It is known that the state feedback law u i F x which minimize the performance index J is given by F R IBTP PA ATP PBR IBTP Q 0 P gt 0 For this particular example we can obtain a closedform solution F214q1 1f1 1 It can be verified that the open loop of L0 regulator with any q gt 0 has an infinite gain margin and a phase margin over 105 degrees Thus it is very robust 1 17 Prepared by Ben M Chen It can also be shown that the Kalman lter gain for this problem can be expressed as Klt2Wgtlilklil which together with the LQR law result an LQG controller A or u FsI ABFKC 1Ky uz Fx Suppose that the resulting closedloop controller has a scalar gain 1 s nominally unity A BF KCEKy associated with the input matrix ie O the actual input matrix 1 8B J 1 8 Tedious manipulations show that the characteristic function of the closedloop system comprising the given system an the LQG controller is given by KSH39S4 S3QS228kfkf 4s1 gkf 118 Prepared by Ben M Chen A necessary condition for stability is that 28kfkf 4gtO and 1 slg gt0 It is easy to see that for sufficient large q and a the closedloop could be unstable for a small perturbation in B in either direction For instance let us choose q a 60 Then it is simple to verify the closedloop system remains stable only when 008 lt e lt 001 The above example shows that the LQG controller is not robust at all Mat is wrong The answer is that the openloop transfer function of the LQR design and the openloop transfer function of the LQG design are totally different and thus all the nice properties associated with the LQR design vanish in the LQG controller It can be seen more clearly from the precise mathematical expressions of these two openloop transfer functions and this leads to the birth of the socalled Loop Transfer Recovery technique 1 19 Prepared by Ben M Chen OpenLoop Transfer Function of LQR L F Openloop transfer function When the loop is broken at the input point of the plant ie the point marked X we have L FsI A lBu Thus the loop transfer matrix from u to L is given by Lts Fs1 AIB We have learnt from our previous lectures that the open loop transferLts have very impressive properties if the gain matrixF comes from LQR design 120 Prepared by Ben M Chen OpenLoop Transfer Function of LOG l FinABFKC1K lt Openloop transfer function When the loop is broken at the input point of the plant ie the point marked X we have n Fs1 A BF KC 1KC s1 A lBu Thus the loop transfer matrix from u to L is given by L0s Fs1 A BF KC 1KCs1 A 1B Clearly Lts and L0s are very different and that is why LQG in general does not have nice properties as LQR does 121 Prepared by Ben M Chen Loop Transfer Recovery The above problem can be fixed by choosing an appropriate Kalman filter gain matrixK such that Lts and L0s are exactly identical or almost matched over a certain range of frequencies Such a technique is called Loop Transfer Recovery The idea was first pointed out by Doyle and Stein in 1979 They had given a suf cient condition under which L0s Lts They had also develop a procedure to design the Kalman filter gain matrix K in terms of a tuning parameter q such that the resulting L0s gt Lts as q gt 00 for the invertible and minimum phase systems DoyleStein Conditions It can be shown that L0s and Lts are identical if the observer gain Ksatisfies K C13K 1 BClt1gtB 1 lt1 2 s1 A 1 which is equivalent to B 0 prove it Thus it is impractical 122 Prepared by Ben M Chen Classical L TR Design The following procedure was proposed by Doyle and Stein in 1979 for left invertible and minimum phase systems De ne Qq Q0 quVBT R R0 where Q0 and R0 are noise intensities appropriate for the nominal plant in fact QO can be chosen as a zero matrix and R0 1 and Vis any positive de nite symmetric matrix Vcan be chosen as an identity matrix Then the observer or Kalman lter gain is given by K PC TR 1 where P is the positive de nite solution of AP PAT Qq PCTR 1CP 0 It can be shown that the resulting openloop transfer function L0s from the above observer or Kalman lter has L0 Lt as q 00 123 Prepared by Ben M Chen Example Consider a given plant characterized by O 1 0 35 21xw X 3 4 x1 61v y with Evt Ewt o and Evlv1 Ewtw139 5z 139 This system is of minimum phase with one invariant zero at s 2 The LQR control law is given by uz sz SO 10x The resulting openloop transfer function Lts has an in nity gain margin and a phase margin over 85 We apply DoyleStein LTR procedure to design an observer based controller ie u FCID 1 BF KC 1K y where K is computed as on the previous page with Q 35 35 61 2 O O 1 1225 2135 q 61 q 1 2135 3721q2 124 Prepared by Ben M Chen Mngniludc an LOG with q2 100000 Frcqucncy masec Lac with q2 10000 and by m u an New Formulation for Loop Transfer Recovery Consider a general stabilizable and detectable plant X A x B u y C x D u The transfer function is given by Ps CCDB D CD s1 A Also letF be a state feedback gain matrix such that under the state feedback control law u 7 F x has the following properties the resulting closedloop system is asymptotically stable and the resulting target loop Lts FCDB meets design speci cations GM PM etc Such a state feedback can be obtained using LQR design or any other design methods so long as it meets your design specifications Usually a desired target loop would have the shape as given in the following figure 126 Prepared by Ben M Chen A 7 To avoid highfrequcncy unmodolled dynamics To have a good disturbance rejection min singular Values 127 mm by Ben 1 Gen The problem of loop transfer recovery LTR is to nd a stabilizing controller u Cs y such that the resulting openloop transfer function from u to ie L0s CSPs is either exactly or approximately equal to the target loop Lts Let us define the recovery error as the difference between the target loop and the achieved loop ie Es Lts L0 s 2 FCle CsPs Then we say exact LTR is achievable if Es can be made identically zero or almost LTR is achievable if Es can be made arbitrarily small 128 Prepared by Ben M Chen Observer Based Structure for Cs A Full Order Observer Based Controller A BuKy CE Du Dynamic equations of Cs A u F x Transfer function of Cs C0s F cIr1 BFKC KDF 1K Achieved openloop L0s C0SPSFCD1BFKC KDF1KCCDBD 129 Prepared by Ben M Chen Lemma Recovery error E0s ie the mismatch between the target loop and the resulting openloop of the observer based controller is given by E0s Ms I Msl11 Flt1gtB Ms Fc1gt KC 1B KD Proof L0 S C0sPs Flt1gt 1 BFKC KDF 1KClt1gtBD FIlt1gt 1KO 1B KDF 1lt1gt 1KC 1KClt1gtBD Ms 1Flt1gt 1KCYKClt1gtBFlt1gt 1KO 1KD IMs 1HI 14KO 1lt1gt 1lt1gtBFlt1gt 1KO 1KD 1Flt1gtB Flt1gt 1 KQ IBFlt1gt 1 KC 1KD 1Flt1gtB FCD 1 K04 B KD IMs 1Flt1gtB Ms Note that we have used CID 1KCT1KC I CID 1 KC 1CD 1 Thus E0 Lt L0 FCDB 1 M 1FCDB M MI M 1I FCDB 130 Prepared by Ben M Chen Loop Transfer Reco very Design It is simple to observe from the above lemma that the loop transfer recovery is achievable if and only if we can design a gain matrixK such that Ms can be made either identically zero or arbitrarily small where M s F lt1gt KC 1B KD Let us de ne an auxiliary system XATxCTuFTw 2W y x u KTx zzBTxDTu gt Closedloop transfer function from w to z is BT DTKTsI AT CTKT 1FT MTs Thus LTR design is equivalent to design a state feedback law for the above auxiliary system such that certain norm of the resulting closedloop transfer function is made either identically zero or arbitrarily small As such all the design techniques in HZ and H00 can be applied to design such a gain There is no need to repeat all over again once this is formulated 1 41 Prepared by Ben M Chen L TR Design via CSS Architecture Based Controller CSS Based Controller CSS Structure was proposed by Chen Saberi and Sannuti in 1992 It has the following VA KCVKy Dynamic equations of Cs u u F v Transfer function of Cs Ccs F 13 1 KC1K Achieved openloop LcsCcsPsFCID1 KC 1KCCDBD 132 Prepared by Ben M Chen Lemma Recovery error Ecs ie the mismatch between the target loop and the resulting openloop of the CSS architecture based controller is given by Ecs Ms Fc1gt KC 1B KD Proof Ecs L s Us FCDB FCIgt 1 KC 1KCCIgtB D FCIgt 1 KC 1CD 1 KCCIgtB KCCDB KD FCIgt 1 KC B KCCDB KCCDB KD MS It is clear that LTR via the CSS architecture based controller is achievable if and only if one can design a gain matrixK such that the resulting Ms can be made either identically zero or arbitrarily small This is identical to the LTR design via the observer based controller Thus one can again using the H2 and H00 techniques to carry out the design of such a gain matrix 133 Prepared by Ben M Chen What is the Advantage of CSS Structure Theorem Consider a stabilizable and detectable system 2 characterized by A B CD and target loop transfer function Lts FCDB Assume that 2 is left invertible and of minimum phase which implies that the target loop Lts is recoverable by both observer based and CSS architecture based controllers Also assume that the same gain K is used for both observer based controller and CSS architecture based controller and is such that for all a e Q where Q is some frequency region of interest 0 lt amaxiMjwltlt1 Gnunilxjw GnuJFle A IB gtgt1 Then for all a e Q GmaxlEc 110 ltlt GmaxlEo 110 Proof See Chen Saberi and Sannuti Automatica vol 27 pp 257280 1991 See also Saberi Chen and Sannuti Loop Transfer Recovery Analysis and Design Springer 1993 134 Prepared by Ben M Chen Remark In order to have good command following and disturbance rejection properties the target loop transfer function Liam has to be large and consequently the minimum singular value 6minLtjw should be relatively large in the appropriate frequency region Thus the assumption in the above theorem is very practical Example Consider a given plant characterized by 0 1 0 2 1 0 X 3 4x1uy x u Let the target loop Lts F QB be characterized by a state feedback gain F 50 10 Using MATLAB we know that the above system has an invariant zero at s 2 Hence it is of minimum phase Also it is invertible Thus the target loop Lts is recoverable by both the observer based and CSS architecture based controllers The following gain matrixK is 69 K 846 135 Prepared by Ben M Chen obtained by using the H2 optimization method Magnitude LU Frequency radsec MUN mm by Ben in men Magnitude Frequency masac EJJ39M A r r 1 37 Ram by Ben u an Chapter 13 H2 Optimal Control 169 0 H2 Optimal control 0 stability margins of H2 controllers 170 H2 Optimal Control 02 D21 0 Assumptions i A 32 is stabilizable and Cg A is detectable ii D12 has full column rank with D12 D L l unitary and D21 has full D21 row rank With unitary l A 39 I B iii jw 2 has full column rank for all w 01 D12 A 39 I B iv jw 1 has full row rank for all w 2 21 H2 Problem nd a stabilizing controller K that minimizes Tzw2 X2ltA BQDT201gtltA BQDf201gtX2 XQBQBX2CTDLD01 O Y2A Bngng A Blpglogm Ygogogig 31131113 0 171 De ne F2 3 ltB2FX2 Dl201gt7 L2 ltY203 l Blpgll A B2F2 I 01 D12F2 0 There exists a unique Optimal controller A LC B LD G48 7 G 8 2 2 1 2 21 f I 0 A BQFQ L202 L2 K0 5 M l F2 0 Moreover minlszwllg lch313 F2Gf3 lchL2 016 A B2F2 B2 E R7100 is inner and UNGC E RH 01 D12F2 D12 A L202 31 L2D21 02 D21 0 V E RHOO is CO inner and GfVN E RH c all stabilizing controllers Ks fgM2 Q Q E RHOO with l A B2F2 L202 L2 B2 Mm F2 0 1 CQ 0 Closed loop with K Tm 0031 UFQGf UQV 172 2 2 2 lszwllg lchB1 F2Gf QVllg lchB12 F2Gf2Q2 and Q 0 gives the unique Optimal control K 7quotgM2 O 173 Stability Margins of H2 Controllers o LQR margin 2 600 phase margin and Z 6dB gain margin 0 LQG or H2 Controller No guaranteed margin Then and where Glt gt 11 o 0 01 e 1 mm 0 0 0 0 1 10 01 0 X220z0z7Y22 0404 55 oz24 240 1 5 1 Kopt ltOZ Suppose the controller implemented in the system or plant G22 is actually K kKopt 174 with a nominal value k 1 Then the closed loop system A matrix becomes 11 O O A 01 ka ka 60 1 5 1 6 O oz 1 a The characteristic polynomial has the form det5 121 L454 1353 1252 als do with a1oz 42k 1oz a011 koz necessary for stability do gt O and a1 gt O ozgtgt1and gtgt1andk1gta01 koz anda1 2k Doz oz gtgt 1 and 6 gtgt 1 or q and a the system is unstable for arbitrarily small perturbations in k in either direction Thus by choice of q and a the gain margins may be made arbitrarily small lt is interesting to note that the margins deteriorate as control weight 1q gets small large q and or system driving noise gets large large 0 In modern control folklore these have often been considered ad hoc means of improving sensitivity H2 LQG controllers have no global system independent guaranteed robustness properties lmprove the robustness of a given design by relaxing the optimality of the lter or PC controller with respect to error properties LQG loop transfer recovery LQGLTR design technique The idea is to design a ltering gain or PC control law in such way so that the LQG or H2 control law will approximate the loop properties of the regular LQR control 161 Chapter 12 Algebraic Riccati Equations AXXAXHXQO HH QQ The associated Hamiltonian matrix Then 0 I J lHJ JHJ H J I 0 so H and H are similar Thus is an eigenvalue iff is eigltHgt 75 jw ltgt H has n eigenvalues in Re 5 lt O and n in Re 5 gt 0 Let XH be the n climensional spectral subspace corresponding to eigenvalues in Re 5 lt 0 X1 XH lm X2 where X1 X2 6 CW X1 and X2 can be chosen to be real matrices lf X1 is nonsingular de ne X HicltHgt Xng1 d0mltHicgt C RQWQ I gt Rm where d0mltHicgt consists of all H matrices such that o H has no eigenvalues on the imaginary axis 0 XH lm are complementary or X1 is nonsingular 162 Theorem Suppose H E d0mltRicgt and X RicH Then i X is real symmetric ii X satis es the algebraic Riccati equation AXXAXRXQO iii A RX is stable X1 Proof Let XXH lm We show XfX2 is symmetric Note 2 that there exists a stable matrix H in Rm such that X1 X1 H X2 X2 Premultiply this equation by X1 J X2 to get X X X X 1 JH 1 1 J 1 H X2 X2 X2 X2 Since J H is symmetric gt Xf X2 X X1H Hi XfX2 X X1 Hilt XfX2 XgXil This is a Lyapunov equation Since H is stable the unique solution is XfX2 X X1 O 163 ie XfXg is symmetric gt X Xf1XfX2Xf1 is symmetric ii Start with the equation X1 i X1 X2 X2 H i H and post multiply by X f 1 to get H NOW premultiply by X I This is precisely the Riccati equation I I 111 I OH X XlHXll A RX XJLXfl Thus A RX is stable because H is D gt X1X2 I iCSChI ltHgt X X2X1 164 Theorem Suppose eigH 75 jw and R is semi de nite Z O or g 0 Then H E d0mHz39c ltgt A H is stabilizable X1 X1 Proof lt2 Note that XH lm H X2 X2 We need to show that X1 is nonsingular ie Ker X1 0 X1 X2 H Claim Ker X1 is H invariant Let 6 E Ker X1 and note that X X1 is symmetric and AX1 RXQ X1H Premultiply by 33X5 post multiply by 6 to get X HX2 O gt RXQLC O gt XlHJc O ie H4 E KerX1 Suppose Ker X1 75 0 Then HKerX1 has an eigenvalue A and a corresponding eigenvector 2 Hcc ReltO 07 EKerX1 Note that QX1 AX2 XgH Post multiply the above equation by SE A MX2c 0 Recall that RXQLC O we have X A X R O 1 X1 X A R stabilizable gt X23 O gt 6 O gt 6 0 since 2 full column rank which is a contradiction gt H E d0mHz39c gt A RX stable gt A H stabilizable has 165 Bounded Real Lemma Let y gt 0 C5 E R7100 A B C D A BR 1DC BR lBquot CI DR 1DC A BR 1DC where R 72 DD Then the following conditions are equivalent 1 Goo lt 1 ii 6D lt y and H has no eigenvalues on the imaginary axis iii 6D lt y and H E d0mHz39c gt D lt y and H E d0mHz39c ancl HicH Z O RicH gt 0 if C A is observable D lt y and there exists an X Z 0 such that and q l iv A q l V XABR 1DCABR 1DCXXBR 1BXCDR 1DC 0 and ABR 1DCBPF1BX has no eigenvalues on the imaginary axis vi 6D lt y and there exists an X gt 0 such that XABR 1DCABR 1DCXXBR 1BXC1DR 1DC lt 0 vii there exists an X gt 0 such that X A AX X B 0 BX w 19 lt 0 C D y Proof 1 gt Assume D O for simplicity Then there is an X Z O XA AX XBBX72 0 0 166 and A BBquotXy2 has no jw axis eigenvalue Hence A B W 2 mm w has no zeros on the imaginary axis since A BBquotXy2 By W7 lt8 BX72 17 has no poles on the imaginary axis Next note that Xjw A ij AX XBBX72 00 0 Multiply Bjw A 1 on the left and jaw A 1B on the right of the above equation to get Bjw A 1XB BXjw A 1B Bjw A 1XBBXij A 1B72 Bjw A 1CCjw1 A 1B 0 Completing square we have GltjwgtGltjwgt 721 WltjwgtWltjwgt Since Ws has no jw axis zeros we conclude that GCgt0 lt 7 vi gt follows from Schur complement vi gt by following the similar procedure as above gt vi let C applying part 11 to D Then there exists an 6 gt 0 such that 00 lt 7 Now follows by 167 Theorem Suppose H has the form A BB CC A Then H E d0mHic iff A B is stabilizable and C A has no unob servable modes on the imaginary axis Furthermore X RicH Z 0 And X gt 0 if and only if C A has no stable unobservable modes Proof Only need to show that assuming A B is stabilizable H has no imaginary eigenvalues iff C A has no unobservable modes on the a imaginary axis Suppose that jw is an eigenvalue and O 75 is a 2 corresponding eigenvector Then A36 BBz jam CC Az jwz Rearrange A jwc BB2 A jw1z 0033 Thus z A jwc21313Bz2 A ijz CC Cas2 so as A ijz is real and C2 ltltA jw1gtzgt lt27 A jw1gtgt BZ2 Therefore Bz O and Cm 0 So A jw1 O A ijz 0 Combine the last four equations to get A jw 2A a B 0 l J l C 0 168 The stabilizability of A B gives z 0 Now it is clear that jw is an eigenvalue of H iff jw is an unobservable mode of C A A BBXX XA BBX XBBX CC O X Z 0 since A BBX is stable D Corollary Suppose AB is stabilizable and CA is detectable Then AXXA XBBXCC 0 has a unique positive semide nite solution Moreover it is stabilizing Corollary Suppose D has full column rank and denote R DD gt 0 Let H have the form A 0 B 1 H R oo A CD A BR 1DC BR 1B or DR 1DC A BR 1DC A jw B Then H E d0mHzc iff A B is stabilizable and C D has full column rank for all to Furthermore X RiCH Z 0 if H E d0mRic and KerX 0 if and only if DjCA BRADY has no stable unobservable modes A jw B This is because has full column rank for all a ltgt I DR 1DC A BR 1DC has no unobservable modes on jw 8X18 Lecture Notes Week 3a Topic NormsH2Hin nity Spaces Internal Stability ECEMAE 7360 Optimal and Robust Control Fall 2003 Offering Instructor Dr YangQuan Chen CSOIS ECE Dept Tel 4357970148 Email chen ieee0r or ygcheneceusuedu Chapter 4 H2 and H00 Spaces 53 o Hilbert space 0 H2 and H00 Functions 0 State Space Computation of H2 and H00 norms Hilbert Spaces Inner product on C n 331 111 lt7ygt3y 352 V56 3 7y 3 EC 21 2 gm SUN 1 ltgt lt7ygt cos 43611 43611 E Om y orthogonal if 43511 De nition 01 Let V be a vector space over C An inner product on V is a complex valued function lt V X V i gt no such that for any 3611 z E V and o 6 E no 1 lt36 Ody W 04 ygt 6 2gt 11 571 W iii as 5 gt 0 if a y O A vector space V with an inner product is called an inner product space inner product induced norrn z lt36 6 distance between vectors 6 and y d y Two vectors 6 and y orthogonal if 3631 O denoted 6 J y 55 o ltygt g Cauchy Schwarz inequality Equality holds iff 6 ay for some constant or or y O o lel2 1 2 2 2 y2 Parallelograrn law 2 2 2 39 Writll H33 llyll 1 i y Hilbert space a complete inner product space Examples 0 C with the usual inner product ngtltm o r with the inner product A B 2 Trace AB i g mysz VA B 6 CW z391j1 o 2a b all square integrable and Lebesgue rneasurable functions de ned on an interval 1 b with the inner product lt12 ggt abflttgtglttgtdt Matrix form f g 2 5 Trace ftgt dt 39 2 Lil 007 00gt1 ltf79gt 1 f3 Tracelflttgt9lttgtl di 0 132 2O oo subspace of 2 oo oo 0 2 2 oo O subspace of 2 oo oo Analytic Functions Let S C C be an open set and let fs be a complex valued function de ned on S fs S gt G Then fs is analytic at a point zo in S if it is differentiable at 20 and also at each point in some neighborhood of 20 It is a fact that if fs is analytic at 20 then f has continuous derivatives of all orders at zo Hence a function analytic at 20 has a power series representation at zo A function fs is said to be analytic m S if it has a derivative or is analytic at each point of S Maximum Modulus Theorem lf fs is de ned and continuous on a closed bounded set S and analytic on the interior of S then gag flt8gt 2533 flt8gt where 8S denotes the boundary of S 57 2 and H2 Spaces 2ltjRgt Space all complex matrix functions F such that the integral below is bounded L Trace FjwFjw dw lt 00 with the inner product 1 00 F G gioo Trace F jwGjw dw and the inner product induced norm is given by F21ltFFgt R 2ltjRgt or simply 7332 all real rational strictly proper transfer matrices with no poles on the imaginary axis H2 Space a closed subspace of 20k with functions Fs analytic in Res gt O i Trace F0 jwFa jw dw m C U Ugt0 1 00 ioo Trace F jwFjw dw R712 real rational subspace of H2 all strictly proper and real ra tional stable transfer matrices Hgf Space the orthogonal complement of H2 in L2 ie the closed subspace of functions in L2 that are analytic in Res lt O 73712L the real rational subspace of H3 all strictly proper rational antistable transfer matrices Parseval s relations 2lt O000gt gigUR 2000gt r3712 2lt O00 G2 llgllg where Glt8gt l9lttgtl E 132 00 and H00 Spaces LOOUR Space 00ltjRgt or simply LOO is a Banach space of matrix valued or scalar Valued functions that are essentially bounded on jn with norm Foo I essigg lFle 731304 jn or simply 73300 all proper and real rational transfer ma trices with no poles on the imaginary axis H00 Space H00 is a closed subspace of 00 with functions that are analytic and bounded in the open right half plane The H00 norm is de ned as sup 6Fs sup Fjw Resgt0 we The second equality can be regarded as a generalization of the max imum modulus theorem for matrix functions See Boyd and Desoer 1985 for a proof R7100 all proper and real rational stable transfer matrices Hg Space Hgo is a closed subspace of 00 with functions that are analytic and bounded in the open left half plane The Hg norm is de ned as sup 6Fs sup6Fjw Reslt0 we 73ng all proper real rational antistable transfer matrices 59 H00 Norm as Induced H2 Norm Let Cs E 00 be a p x q transfer matrix Then a multiplication operator is de ned as MG 2 2 gt 2 Mgf Gf G Then IIMGII sup quot fquot HGHOO fez f2 new 21W ffltjwgtGltjwgtGltjwgtfltjwgt dw 1 oo 2 llGlliogmll jwll dw angling To show that GCgt0 is the least upper bound rst choose a frequency wo where 6 Gjw is maximum ie lGUwoll Goo and denote the singular value decomposition of Cjwo by Gltjw0 u1jwovfjwo Uz39uzUwolvawo where 7quot is the rank of Cjwo and uzr vi have unit length If wo lt 00 write 14ij as alejgl oz 792 v1ltjwo 2 0416qu 60 where 0 E n is such that 6239 E 7r 0 Now let 0 g 6239 3 00 be such that 6 Z 3239 7010 5239 jwo with 00 if 6 O and let f be given by 121 W A fltsgt 3 fltsgt B is O q with 1 replacing if 6239 0 where a scalar function f is chosen so that i c iflw wol lteor ww0 lt6 7 i 0 otherwise where 6 is a small positive number and c is chosen so that f has unit 2 norrn ie c Mr2e This in turn implies that f has unit 2 norrn Then llellg 22 l lG jwogtl2r 5lGltongtl2 Wl 5 G 30102 G o Similarly if too 00 the conclusion follows by letting wo gt oo in the above 61 Computing 2 and H2 Norms Let Cs E 2 and gt 1Gs Then Hang 21 f TraceGjwGjw dw 74 TraceG sGs d5 the residues of TraceG sGs at its poles in the left half plane L Trace9lttgtglttgt dt llgllg Consider Cs traceBLOB traceCLCC vvhere L0 and LC are observability and controllability Gramians ALC LCAquot BBquot 0 AL0 LOA 00 O E R712 Then we have CeAtB t Z O O t lt 0 L0 OOO ethWCeAt dt LC OOO eAtBBeAt dt Note that gt E 1G OOO Tracegtgt dt OOO TraceBeAtCCeAtB dt TraceB OOO eAtCCeAtdtB traceBLOB OOO Tracegtgt dt OOO TraceCeAtBBeAtC dt hypothetical input output experiments Apply the impulsive input 6te 6t is the unit impulse and 6239 is the ith standard basis vector and denote the output by z t gte Then zzr E 132 assuming D O and m IIGIIE Hang Can be used for nonlinear time varying systems 62 Example Consider a transfer matrix 3s3 2 i 5 152 5 1 i Gilt 511 gt 1 7G5Gu s2s3 5 4 with Then the command h2n0rmltGs gives G52 06055 and h2n0rmltcjtltGu gives Gu2 3132 Hence HG 2 Mugs Gun 32393 gt P gramA B Q gramA C or P lyapA B gtk B gt G5 GU sdecompG decompose into stable and antistable parts 63 Computing 00 and H00 Norms Let Cs E LOO Goo I 688 Sgp GUw o the farthest distance the Nyquist plot of G from the origin 0 the peak on the Bode magnitude plot o estimation set up a ne grid of frequency points cub wN llGlloo z lawman 7A B Letygt0andGsi 70 D 6731300 GCgt0 lt y ltgt 6D lt y 85 H has no jw eigenvalues h r H 7 A BR 1DC 3343 W e e o1 DR 1DC A BR 1DC and R 72 DD Let 135 72 G SGs Goo lt 7 ltgt jw gt O Vw E R ltgt det 130w 75 0 since 1300 R gt O and jw is continuous ltgt 135 has no imaginary axis zero ltgt 1345 has no imaginary axis pole BPK1 H Low 1345 HPHD C HARM PH 64 ltgt H has no jw axis eigenvalues if the above realization has neither uncontrollable modes nor unobservable modes on the imaginary axis Assume that jwo is an eigenvalue of H but not a pole of 1345 Then jwo must be either an unobservable mode of R 1DC R lBquot l H 113le1 CDPJ1 unobservable mode of R 1DC R lBquot l H Then there exists an or an uncontrollable mode of H Suppose jwo is an 330 75 0 such that 1 362 H330 jwoaso R 1DC 3713 30 0 1 ijI A1 O ijIA2 CC1 DC1B2 0 Since A has no imaginary axis eigenvalues we have 61 O and 2 O Contradiction Similarly a contradiction will also be arrived if jwo is assumed to be an BR 1 b uncontrollable mode of H CDR1 65 Bisection Algorithm a select an upper bound 7 and a lower bound 7 such that 7 g IIGIIOO S 71 b if 7 77 gspeci ecl level stop z 7 72 Otherwise go to next step 0 set 7 71 7W2 cl test if GCgt0 lt 7 by calculating the eigenvalues of H for the given 7 e if H has an eigenvalue on jn set 7 7 otherwise set 7 7 go back to step WLOG assume 7 1 since GCgt0 lt 7 iff H7 1GHOO lt 1 66 Estimating the H00 norm Estimating the H00 norm experimentally the maximum magnitude of the steady state response to all possible unit amplitude sinusoidal input signals z Gltjwgt sinwt lt Gltjwgtgt u sinwt Let the sinusoidal inputs be ul sinwot 1 U1 7 Mt U2 s1ncu0t qbg 7 a ug uq sinw0t gbq uq 7 Then the steady state response of the system can be written as yl sinw0t 61 141 7 wt yg sinwz0t 62 7 g yzg yp Sinltw0t 6p yp for some 11239 6239 239 1 2 p and furthermore Hana sup quot1quot U where is the Euclidean norm 67 Examples Consider a massspring damper system as shown in Figure 01 Thef largest Ingular ilaluef frequency radfsuec Figure 02 HGHOO is the peak of the largest singular value of Gjw The dynamical system can be described by the following differential equations 1 1 39quot A 2 walla 333 333 F2 334 334 68 with 0 0 1 0 0 0 0 0 0 1 0 0 A E E i E B i 0 m1 m1 m1 m1 Mr E m h M 0 1 m2 m2 m2 m2 m2 Suppose that Cs is the transfer matrix from F1 F2 to 1 332 that is 1000 00100 7 DO7 and suppose k1 1 k2 4171 02 b2 01 m1 1 and 7712 2 with appropriate units gtgt GkaABCD gtgt hinfnormG00001 or linfnormG00001 relative error S 00001 gtgt wlogspace11200 200 points between 1 10 1 ancl10 101 gtgt GffrspGw computing frequency response gt usvvsvdGf SVD at each frequency gtgt vplot liv lm s grid plot both singular values and grid Gscgt0 1147 the peak of the largest singular value Bode plot in Figure 02 Since the peak is achieved at wmax 08483 exciting the system using the following sinusoidal input F1 O9614sin08483t F2 02753 sin08483t 012 69 gives the steady state response of the system as 1147 x 09614 sin08483t 15483 1147 x 02753 sin08483t 14283 39 1 2 This shows that the system response will be ampli ed 1147 times for an input signal at the frequency wmax which could be undesirable if F1 and F2 are disturbance force and 1 and 562 are the positions to be kept steady Consider a two by two transfer matrix 105 1 1 n 52025100 51 Glsl s 2 55 1 52 015 10 s 25 3 A statespace realization of G can be obtained using the following MATLAB commands gtgt G11nd2sys1010102100 gtgt G12nd2sys111 gtgt G21nd2sys1210110 gtgt G22nd2sys55156 gtgt GsbsabvG11G21abvG12G22 Next we set up a frequency grid to compute the frequency response of G and the singular values of Cjw over a suitable range of frequency gtgt Wlogspace02200 200 points between 1 100 and 100 102 gtgt GffrspGW computing frequency response gt usvvsvdGf SVD at each frequency 70 gtgt vplot liv lm s grid plot both singular values and grid gtgt pkvnorms nd the norm from the frequency response of the singular values The singular values of Cjw are plotted in Figure 03 which gives an es timate of GCgt0 32861 The state space bisection algorithm described previously leads to GCgt0 5025 d 001 and the corresponding MATLAB command is gtgt hinfnormG00001 or linfnormG00001 relative error 3 00001 I H l 101 102 Figure 03 The largest and the smallest singular values of Gjw The preceding computational results show clearly that the graphical method can lead to a wrong answer for a lightly clamped system if the frequency grid is not suf ciently dense lncleed we would get GCgt0 43525 48286 and 49737 from the graphical method if 400 800 and 1600 frequency points are used respectively Chapter 5 Internal Stability 71 o internal stability 0 coprime factorization over RHOO 0 performance 72 Internal Stability Consider the following feedback system LU1 61 T 623 o well posed if I KltoogtPltoogt is invertible 0 Internal Stability if I K PI 1 KP 1 fro PK 1 PU KP 1 I PK 1 lERHOO 0 Need to check all Four transfer matrices For example i5 1 Ai 1 i5Y i5 I A 1 51L 1 K 52 5 152 p 1 5 1 51 52 52 0 Suppose X E H00 lnternal stability ltgt PU KP 1 E H00 0 Suppose P E H00 lnternal stability ltgt RU PK 1 E H00 0 Suppose P K E H00 lnternal stability ltgt I PK 1 E H00 0 Suppose no unstable pole zero cancellation in PK lnternal stability ltgt I P5IA5 1 E H00 73 Example Let P and K be two by two transfer matrices 1 0 1 8 1 P 8 1 1 K 51 O O 1 51 Then 1 1 51 512 An 515 1 A 1i 52 5225 1 PKi 1 1 PK i O lt Sig gt 51 52 So the closed loop system is not stable even though A 522 cltI PK elt gt 512 has no zero in the closed right half plane and the number of unstable poles of PK W W 1 Hence in general clet PK having no zeros in the closed right half plane does not necessarily imply I PK 1 E R7100 74 Coprime Factorization over R7100 0 two polynomials 7715 and 715 are coprime if the only common factors are constants 0 two transfer functions 7715 and 715 in R7100 are copm me over R7100 if the only common factors are stable and invertible transfer functions units h mh l nh l e 737100 gt h 1 e 737100 Equivalent there exists as y E R7100 such that mm yn 1 o Matrices M and N in R7100 are right copm me over R7100 if there exist matrices XT and K in R7100 such that M mi N XTMKNI o Matrices M and N in R7100 are left copm me over R7100 if there exist matrices X l and Y in R7100 such that X1 MN Y1 lMXlJerll 75 Let P NIW 1 MAN and K UV 1 V lU be rcfancl lcf respectively Then the following conditions are equivalent 1 The feedback system is internally stable 2 M U N V l is invertible in R7100 17 U 3 N M l is invertible in RHOO 4 MV NU is invertible in R7100 5 VM UN is invertible in R7100 AB LetP C D F and L be such that A BF and A LC are both stable De ne be a stabilizable ancl detectable realization and let F I O C D I Then l lM N Xlll39 76 Example L P 1 Th P et 5 8lt83gt and 07 5 s3 en 5 5 s 1s 3 tion To nd an 335 E H00 and a ys E H00 such that sns s 1 A A s 1039 Then K 1711 with u K and v 1 is a coprirne factorization and i 5 117085s 22145 0077 f mlt8gtvlt8gt nlt8gtult8gt 8 ms SW 10 7 68 5 With 715 and 7715 1 forms a coprirne factoriza s ysms 1 consider a stabilizing controller for P K Then we can take i i s ls1s3 355 lt5gt lt5gt 5 1170855 22145 0077 s 1s 3s 10 5 1170855 22145 0077 148 USgt lt8gt MATLAB programs can be used to nd the appropriate F and L rnatri ces in statespace so that the desired coprirne factorization can be obtained Let A E Rm B E nnxm and C E Rpm Then an F and an L can be obtained from gt FlqrA B eyen eyem Of gtgt FplaceA B Pf Pf poles of ABF gt L lqrA C eyen eyep OF gt L placeA C Pl P1poles of ALC Lecture Notes Week 7a Topic Youla Controller parameterization ECEMAE 7360 Optimal and Robust Control Fall 2003 Offering Instructor Dr YangQuan Chen CSOIS ECE Dept Tel 4357970148 Email chen ieee0r 0rygcheneceusuedu 111 Controller Parameterization Suppose 132 is stabilizable and C2 A is detectable Youla parameterization all controllers K that internally stabilize G 0 Suppose G E R HOC Then K I GggQ1 Q E and I D22Qoo nonsingular K stabilizes a stable plant G22 i K I ngK1 is stable So let Q I GggK1 0 General Case Let F and L be such that A LC2 and A 32F are stable Then K IVA lAB2F L02 LD22Fl L 32 LD22 J 2 F 0 I I D22 Cz D2217 with any Q E quotR Hoc and I D22Qoo nonsingular 112 Figure 01 Structure of Stabilizing Controllers 113 o Closedloop Matrix EH K EKG EU Q 7471 Q T11 T12QT21 3 Q E RHom I D22QOO illVOI39tibh Where T is given by A 32F iB2F B1 32 T11 T12 0 A LC2 Bl I LD2L 0 i T21 T22 7 Cr D12F D12F D11 D12 0 C2 D21 0 o Coprirne factorization approach Let G22 Nil1 1 Ajfil he ref and ief of G22 over RHOC respectively And let Uo VI Uo VI 6 RHOC satisfy the Bezout identity 71 700 M U0 1 o 7N M N v1 0 I 39 Then K lt30 mgrm New m QyN 1 a QM f5 397Q7 Q E RHOC Where r 271 771 L0H h 271 271 7 i0 to A and I V 117Qyoo is invertible J Lecture 16 Controller Structures Introduction K J Astr m Many common issues in design of machines electronics computer software mechatronics 39 IntrOdUCtlon o How to deal with complexity Feedback and Feedforward o Modularization 1 2 339 LmearSChemes o Standardization 4 Nonlinear Schemes 5 0 Structures Gain Scheduling and Adaptation o Paradigms Design prinCIpIes 639 summary 0 Top Down and Bottom Up Theme Building complex control systems Bottom Up Design Bottom Up Design of Control Systems A way to VieW SYStemS Components System principles 0 A number of building blocks o Controllers 0 Feedback 0 Ideas to combine them 0 Observers o Feedforward o What are the building blocks of control o Estimators 0 Model followmg o What prinCIpIes can be used to select and combine them o Filters 0 Cascade o The danger Can it be done better C C I b o Limiters 0 Split range 0 ommissioning ose oops one y one Dead zones Gain scheduling o Selectors o Adaptation K J Astro39m August 2001 Top Down Design of Control Systems 0 Model complete system 0 Design an integrated system 0 System concepts State feedback Observers Model predictive control 0 Commissioning Needs careful consideration Feed back A very powerful idea with dramatic impact Controller Process Reduce effect of disturbances Reduce effect of process variations Linearize nonlinear systems Does not require accurate process model Measurement noise is injected into the system Risk for instability Feedforward Disturbance Feedforward Control signal Reduce effects of disturbances that can be measured Improve response to reference signals No risk for instability Feedback and Feedforward Feedback 0 Feedforward Closed loop 0 Open loop Acts only when there 0 Acts before deviations are deviations show up Market Driven 0 Planning Robust to model errors 0 Not robust to model S lt 1 for some some errors S 1 for all frequencies frequencies Risk for instability o No risk for instability Requires good models K J Astro39m August 2001 Combination of Feedback and Feedforward Linear Schemes Feedforward Model following Systems with two degrees of freedom 2DOF Filters Cascade control State feedback Observers o Attenuation of disturbances with specific character The Smith Predictor Model Predictive Control Model Following 2DOF Applications of Model Following yo ys 3 P Coordination in multiaxis motion control Robotics Path following Mixing in chemical processes Coordinated production changes 99 K J Astro39m August 2001 Filters Cascade Control How to use several sensors State feedback is the ultimate Typical filters case 0 Low pass 0 High pass 0 Band pass 0 Notch 0 Body bending filters Typical applications 0 Reduce disturbances 0 Improve robustness high frequency rolloff 0 Smooth reference signals When is Cascade Control Useful Key idea make tight feedback around essential places where A V D U there are essential perturbations disturbances or uncertain ties R lt m lt Guidelines 3 0 Well defined relation between primary and secondary u variables 0 Essential disturbances and process variations in inner loop 0 Inner loop faster 0 Tight feedback high gain and high bandwidth in inner loop K J Astro39m August 2001 Attenuation of Disturbances with Specific Character Idea Exploit model of distur bances internal model principle 0 Constant disturbances Inte gral Action o Sinusoidal disturbances o M C 2 as J 32 2 as a2 0 Periodic disturbances A disturbance observer is an Systems with Time Delays o The derivative of the output gives poor prediction for systems with time delay 0 Better predictions are possible by using past control signals ut T O lt T lt Td Replace the regular PID controller 1 t d u 2 139 esds 301 by the PPI Predictive PI controller alternative 6 2 D 1 t t u ke esds kp usds eT ki t L A simple form of the Smith predictor The Smith Predictor Model Predictive Control Controller ys I Process all 59 Z 2 CO v P u Po quot e 139 Design controller C as if there were no time delays in the process I I gt gt I gt e I I I ml I I e I I I o Beautifully simple 0 Are there some snags Cancellations may degrade performance Does not work if process is unstable Less general than state feedback 0 Widely used in process industries C K J Astrbm August 2001 State Feedback and Observers 0 Use model to estimate variables that are not directly measurable 0 States are the variables required to account for storage of mass momentum and energy 0 Estimate the state 0 Feedback from full state deviation 0 Feedforward to generate um and ym Nonlinear Schemes o Limiters 0 Split range 0 Ratio control 0 Selectors 0 Fuzzy control 0 Gain scheduling o Neural networks 0 Adaptation Limiters Limiters are often used 0 To avoid saturation 0 An element in circuits for windup protection 0 To protect equipment to rapid changes A simple amplitude limiter u uh y H Rate Limiter A rate limiter causes delays JAS K J Astro39m August 2001 Jump and Rate Limiter yum 2 alan n 5 i Commonly used in the power industry for load changes to save boilers Split Range A simple way to use one controller to control two actuators Commonly used for heating and cooling Heating valve Open Cooling valve Closed Ratio Control Arrangement to obtain two flows that are proportional to each other eg oil and air in boilers The scheme B is preferable Why Selector Control Scheme used to achieve several control objectives eg control temperature unless pressure is too high A way to constrain process variables during operation y 2 Process K J Astro39m August 2001 Control of Fuel and Air in a Boiler An elegant solution Power demand Fuzzy Control 0 Rule based control 0 Linguistic variables high low medium 0 Membership functions o If temperature high then increase flow a little 5 099 0 0 0000 o 90 OQOOOO o v oooo Wkg o o39o vo o o o o Linguistic variables 2 n 2L dt u Fuzzy Control 0 A nonlinear state feedback How do we get the states What does the nonlinearity look like Rules and interpolation Why so few rules When is it useful 0 Excellent to automate successful manual operations 0 Intuitive o A lot of controversy The No Model Myth 0 Fuzzy control is more useful than its detractors claim but less useful than the propagandists claim 0 Neurofuzzy Neural Networks Representation of functions of many variables yt HZ WAD Feedforward neural network Real and artificial neurons A nonlinear function with a learning mechanism K J Astro39m August 2001 Gain Scheduling Gain sch edule Controller parameters operating condition Command 39 al Controller 5 5 Example of scheduling variables 0 Production rate 0 Machine speed Mach number and dynamic pressure 0 Room occupancy Uses of Gain Scheduling Many uses Linearization of actuators Surge tank control Control over wide operating regions Important issues Choice of scheduling variables Granularity of scheduling table Interpolation schemes Bumpless parameter changes Man machine interfaces Importance of autotuning Adaptation 39Eelftnnmg regulator Speci cation Pr oeess parameters I Certainty Equivalence Many control and estimation schemes Dual control Control should be directing as well as investigating Uses of Adaptation Process dynamics Tuning Tools Var l Cons an Automatic Tuning W L Gain scheduling Useacontrollerwrtn Useacontrollerwrtn varylrg parameters constant parameters Adaptive feedback Adaptive feedforward Unpredictable Predictable variations varla lo s Use an adaptrve Use gam halth Integrated systems K J Astro39m August 2001 Lecture Notes for ECEMAE 7360 Robust and Optimal Control Fall 2003 Jinsong Liang July 17 2003 What will we study 0 Optimal Control Theory The calculus of variations a little bit Solution of general optimization problems Optimal closed loop control LQR problem Pontryagin s minimum principle What will we study cont 0 RIOTS Introduction to numerical optimal control Introduction to RIOTS Background Usage and demo Warm up An Static Optimization Problem min LIu 1 E R u E Rm subject to fxu O f E R Define Hamiltonian function Hwu A Mm ATfxu A e Rquot Necessary conditions 8H 0 8A f 8H 2 L96 jg 0 81 811 Lufg Ao 8U What is an optimal control problem System model fzut 5105 E R ut E Rm Performance index cost function JuT gbzTT ALxtuttdt Final state constraint wTT 0 w 6 RP Some COSt fUI ICtiOI I examples 0 Minimum fuel problem J 2 AT u2tdt O o Minimum time problem T J 1dt to o Minimum energy problem J xTTgtRxT f Momma uTtRutdt O 5 A little bit of calculus of variations o Why calculus of variation We are dealing with a function of functions LutT called functional rather than a function of scalar variables LIu We need a new mathematical tool 0 What is 6106 a Relationship between dxT 61T and dT Relationship between dacT 623T and dT xt xh quot39 quot 39 dacT 633T abTdT dxT llAa 1quot L ide d xt xt dT T TdT Leibniz s rule T Jzt 2 Lo hmttdt T CL hxTTdT t h xt t6zdt 0 where hm Solution of the general optimization problem JuT IIITT ALwtuttdt 1 92315 f90ut 1305 E R HO 6 Rm 2 wTT 0 3 Using Lagrange multipliers Mt and 1 to join the constraints 2 and 3 to the performance index 1 T J wTTvaxTTt Man U tATtfx u t irdt 0 Note 1 constant Mt function Define the Hamiltonian function Hxut Lxut ATfxut then T J xTgtT uTwxTT t Hay m ATdrldt O Ugng Lennnzs nnetheincnyhentin J aszafuncUon ofincn rnentsin xpxu1h and tis df m wgv dw cm winWT wTTdv H ATirdtT 4 ft Hgax Hgdu ATM HA i2T6gtdt TO eHthate the vaHatkanin integrate by parts T T ATdirdtz gtT6zT A ATdacdt to 0 10 Remember dxT 61T iTdT so M m My MdelT cit thv HdtlT Jim T t Hx AT6x Hfau HA 22T6Aidt 0 According to the Lagrange theory the constrained minimum of J is attained at the unconstrained minimum of J This is achieved when dJ O for all independent increments in its arguments Setting to zero the coefficients of the independent increments du 612 6a and 6A yields following necessary conditions for a minimum 11 System model it fxut t Z to to fixed Cost function T mm xTgtT t Lltxutdt O Final state constraint wmm 0 State equation 8H m a f Costate adjoint equation 85 aT aL A fgt 81 5 12 Stationarity condition aH aL afT 8U 8U 8U Boundary transversality condition six wgv MTITdMT Cbt My HITdT 0 Note that in the boundary condition since dxT and dT are not independent we cannot simply set the coefficients of dxT and dT equal to zero If dxT 0 fixed final state or dT 0 fixed final time the boundary condition is simplified What if neither is equal to zero 13 An optimal control problem example temperature control in a room It is desired to heat a room using the least possible energy If 6t is the temperature in the room 6a the ambient air temperature outside a constant and ut the rate of heat supply to the room then the dynamics are 9 a0 0a bu for some constants a and b which depend on the room insulation and so on By defining the state as wet 606 6a we can write the state equation it az bu 14 In order to control the temperature on the fixed time interval 0T with the least supplied energy define the cost function as M gum g OT u2rtgtdt for some weighting 3 The Hamiltonian is H 1 gt ax bu The optimal control ut is determined by solving irzHAZ ax I bu 5 A Hx 2 LA 6 o Hu 2 u bk 7 15 From the stationarity condition 7 the optimal control is given by Mt bgtt 8 so to determine ut we need to only find the optimal costate At Substitute 8 into 5 yields the state costate equations d ax b2gt 9 A 2 ca 10 Sounds trivial Think about the boundary condition 16 From the boundary condition m My MTITdMT git thv HTdT 0 dT is zero dxT is free and there is no final state constraint So MT 2 3 371 2 s1T 10 So the boundary condition of 1 and A are specified at to and T respectively This is called two point boundary value TPBV problem Let s assume MT is known From 10 we have Mt 6 TtgtT 17 So i az b2gtTeaTt Solving the above ODE we have 5106 2 1Oeat ATeaTsmhat Now we have the second equation about xT and MT 1T xOeaT Exam e M 2a Assuming 120 2 00 MT can now be solved 20a5 ACT 2 2a b251 e 2aT 18 Now the costate equation becomes 1Oase t X t aeaT 5b2sinhaT Finally we obtain the optimal control 100LlseaIt aeaT 5b2sinhaT ut 2 Conclusion TPBV makes it hard to solve even for simple OCP problems In most cases we have to rely on numerical methods and dedicated OCP software package such as RIOTS 19 Closed loop optimal control LQR problem Problems with the optimal controller obtained so far 0 solutions are hard to compute o open loop For Linear Quadratic Regulation LQR problems a closed loop controller exists 20 System model i Atx Btu Objective function Jul 2 xTTSTxT AfaToam uTRtudt where ST and Q06 are symmetric and positive semidefinite weighting matrices Rt is symmetric and positive definite for all t e t0T We are assuming T is fixed and the final state MT is free State and costate equations d Act BR lBTA A QxATA 21 Control input ut R1BTgt Terminal condition MT STxT Considering the terminal condition let s assume that 1205 and Mt satisfy a linear relation for all t e t0T for some unknown matrix St Mt Stxt To find St differentiate the costate to get A 2 Sq Si 2 Sq 3mm BR1BTSac 22 Taking into account the costate equation we have 3x 2 ATS SA SBR lBTS on Since the above equation holds for all 106 we have the Riccati equation 3 2 ATS SA SBR lBTS Q t g T Now the optimal controller is given by w R1BTS06J06 and K06 R lBTS06x06 is called Kaman gain Note that solution of 806 does not require 1206 so K06 can be computed off line and stored 23 Pontryagin s Minimum Principle a bang bang control case study So far the solution to an optimal control problem depends on the stationarity condition 290 0 What if the control ut is constrained to lie in an admissible region which is usually defined by a requirement that its magnitude be less than a given value Pontryagin s Minimum Principle the Hamiltonian must be minimized over all admissible u for optimal values of the state and costate HIut g HIugtt for all admissible u 24 Example bang bang control of systems obeying Newton s laws System model CL1 1 1 2 CL1 2 U Objective function time optimal JuTOTldt Input constraints ut s 1 25 End point constraint womm lt 323 gt o The Hamiltonian is H 1 A1122 AQU where A A1A2T is the costate Costate equation A1 O gtgt1 constant A2 gt1 gt A205 is a linear function of t remember it Boundary condition A2TuT 1 26 Pontryagin s minimum principle requires that A3tut s A3tut How to make sure A tut is less or equal than A tut for any admissible ut Answer ut sgnAEt 8 What if A306 2 0 Then u is undetermined Since A306 is linear it changes sign at most once 50 does ut 27 Since rigorous derivation of ut is still a little bit complicated an intuitive method using phase plane will be shown below Going backward x1 O and x2 O in time from T with ut 1 or ut 1 we obtain a trajectories or switching curve because switching of control if any must occur on this curve Why 28 10 u1 Switching Curve 50 40 30 20 10 O 10 20 30 40 50 References F L Lewis and V L Syrmos Optimal Control John Wiley amp Sons Inc 1997 A E Bryson and Y C Ho Applied Optimal Control New York Hemisphere 1975 J T Betts Practical Methods for Optimal Control Using Non linear Programming SIAM 2001 I M Gelfand and S V Fomin Calculus of Variations New York Dover Publications 1991 30


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