Chapter 10 Design for Performance The performance criterion WiS0,or in ot her words,P-is stable.The weighting function Wi is assumed to be stable and strictly proper.The latter condition is not too serious a loss of generality.We will see that under these conditions it is always possible,indeed quite easy,to design a proper C which is internally stabilizing and makes WiSo0 and w>0.By the argument above regarding the Bode plot of J,if r is sufficiently small,then the Nyquist plot of J lies in the disk with center 1,radius e for ww1.Now G(1-J)oo equals the maximum of max G(jw)[1-J(jw)] 1 and G(j)[1-J(jwl. 153
Chapter Design for Performance The performance criterion kWSk was introduced in Section The associated design problem is to nd a proper C for which the feedback system is internally stable and kWSk When does such a C exist and how can it be computed These questions are easy when the inverse of the plant transfer function is stable When the inverse is unstable the questions are more interesting The solutions presented in this chapter use model matching theory P Stable We assume in this section that P has no zeros in Res or in other words P is stable The weighting function W is assumed to be stable and strictly proper The latter condition is not too serious a loss of generality We will see that under these conditions it is always possible indeed quite easy to design a proper C which is internally stabilizing and makes kWSk Let k be a positive integer and a positive real number and consider the transfer function J s s k Sketch the Bode plot of J The magnitude starts out at is relatively at out to the corner frequency and then rolls o to with slope k the phase starts out at is relatively at up to say and then rolls o to k radians So for low frequency J j This function has the useful property that it approximates beside a strictly proper function Lemma If G is stable and strictly proper then lim kG J k Proof Let and By the argument above regarding the Bode plot of J if is suciently small then the Nyquist plot of J lies in the disk with center radius for and in the disk with center radius for Now kG J k equals the maximum of max jG j J j j and max jG j J j j
154 CHAPTER 10.DESIGN FOR PERFORMANCE The first of these is bounded above by eGloo,and the second by l1-J‖xmax|G(jw)小. idil Since 1-J川∞≤1l+JI川o=2, we have IG1-∞≤mx{Gle,2nx1G(joj川. This holds for T sufficiently small.But the right-hand side can be made arbitrarily small by suitable choice of e and w because 1 im max|Gw)川=lGGo)川=0. 1→00Ww1 We conclude that for every 6>0,if r is small enough,then G(1-J)川≤d. This is the desired conclusion. We'll develop the design procedure first with the additional assumption that P is stable.By Theorem 5.1 the set of all internally stabilizing Cs is parametrized by the formula Q C=1-PQ° Q∈S. Then WiS is given in terms of Q by W1S=W1(1-PQ). To make WiSlloo<1 we are prompted to set Q=P-1.This is indeed stable,by assumption,but not proper,hence not in S.So let's try Q=P-1J with J as above and the integer k just large enough to make p-J proper (ie.,k equals the relative degree of P).Then W1S=W(1-J), whose oo-norm is 1 for sufficiently small T,by Lemma 1. In summary,the design procedure is as follows. Procedure:P and P-!Stable Input:P,Wi Step 1 Set k =the relative degree of P. Step 2 Choose r so small that ‖W(1-J)川o<1, where 1 J(s)=8+1
CHAPTER DESIGN FOR PERFORMANCE The rst of these is bounded above by kGk and the second by k J k max jG j j Since k J k kk kJ k we have kG J k max kGk max jG j j This holds for suciently small But the right hand side can be made arbitrarily small by suitable choice of and because lim max jG j j jG j j We conclude that for every if is small enough then kG J k This is the desired conclusion Well develop the design procedure rst with the additional assumption that P is stable By Theorem the set of all internally stabilizing Cs is parametrized by the formula C Q P Q Q S Then WS is given in terms of Q by WS W P Q To make kWSk we are prompted to set Q P This is indeed stable by assumption but not proper hence not in S So lets try Q P J with J as above and the integer k just large enough to make P J proper ie k equals the relative degree of P Then WS W J whose norm is for suciently small by Lemma In summary the design procedure is as follows Procedure P and P Stable Input P W Step Set k the relative degree of P Step Choose so small that kW J k where J s s k
10.1.P11 STABLE 155 Step 3 Set Q=Pl1J> Step 4 Set C=Q When P is unstable,the parametrization in Theorem 52 is used> Procedure P11 Stable Input:P,Wi Step-Do a coprime factorization of P:Find four functions in S satisfying the equations N P二M NX +MY 100 Step 2 Set=the relative degree of P> Step 3 Choose-so small that WiMY(1-J)loo Step 5 Set C=(X +MQ) Example Consider the unstable plant and weighting function P)=a-W(回=10x 8+7o This weight has bandwidth 1 rad/s,so it might be used to get good tracking (iez approximately 1%tracking error,up to this frequency)>The prev ious procedure for these data goes as follows: Step-First,do a coprime factorization of P over S: 1 N(s)= (8+1)7 M(s)= (s-2)7 (s+1)7 X(s)= 2781 8+1: Y(s)= 8+7 8+700 Step2‖=2
P STABLE Step Set Q P J Step Set C Q P Q When P is unstable the parametrization in Theorem is used Procedure P Stable Input P W Step Do a coprime factorization of P Find four functions in S satisfying the equations P N M NX M Y Step Set k the relative degree of P Step Choose so small that kWM Y J k where J s s k Step Set Q Y NJ Step Set C X MQ Y NQ Example Consider the unstable plant and weighting function P s s W s s This weight has bandwidth rads so it might be used to get good tracking ie approximately tracking error up to this frequency The previous procedure for these data goes as follows Step First do a coprime factorization of P over S N s s M s s s X s s s Y s s s Step k
156 CHAPTER 10.DESIGN FOR PERFORMANCE Step 3 Choose r so that the oo-norm of 100(s-2)2(s+7) (s+1)A (Ts+1)2 is 1.The norm is computed for decreasing values of r: oo-Norm 10- 199.0 10-2 19.97 1.997 10 0.1997 So take T 10-4. Step 4 Q(s)= (s+1)(s+7) (10-4s+1)2 Step 5 C(s)=104 (s+1)3 s(s+7)(10-4s+2) This section concludes with a result stated but not proved in Section 6.2.It concerns the performance problem where the weight Wi sat isfies w1≤w≤w2 else Thus the criterion WiSo0 and 6>1,there erists a proper C such that the feedback system is internally stable and (10.1)holds.I Proof The idea is to approximately invert P over the frequency range [0,w2]while rolling off fast enough at higher frequencies.From Theorem 5.2 again,the formula for all internally stabilizing proper controllers is C X+MQ Q∈S. Y-NQ' For such C S=M(Y-NQ). (10.2) Now fix e>0 and 6>1.We may as well suppose that e0 so small that clMY‖o<e, (10.3) The assumption on P in Lemma 2 is slightly stronger than necessary;see the statement in Section 6.2
CHAPTER DESIGN FOR PERFORMANCE Step Choose so that the norm of s s s s is The norm is computed for decreasing values of Norm So take Step Q s s s s Step C s s s s s This section concludes with a result stated but not proved in Section It concerns the performance problem where the weight W satises jW j j else Thus the criterion kWSk is equivalent to the conditions jS j j jS j j else Lemma If P is stable then for every and there exists a proper C such that the feedback system is internal ly stable and holds Proof The idea is to approximately invert P over the frequency range while rolling o fast enough at higher frequencies From Theorem again the formula for all internally stabilizing proper controllers is C X MQ Y NQ Q S For such C S M Y NQ Now x and We may as well suppose that Choose c so small that ckM Y k The assumption on P in Lemma is slightly stronger than necessary see the statement in Section
10.1.P-STABLE 157 (1+c)2w2 such that |M(0w)Y(jw)川≤1+c,w≥w3. (10.5) The assumpt ion on P implies that N-is stable (but not proper).Choose a function V in S with the following three properties: 1.VN-is proper. 2.|1-V(0w川≤C,w≤w3. 3.l1-Vo≤1+c The idea behind the choice of V can be explained in terms of its Nyquist plot:It should lie in the disk with center 1,radius c up to frequency w3(property 2)and in the disk with center 1,radius 1+c thereafter (property 3).In addition,V should roll off fast enough so that VN-is proper.It is left as an exercise to conv ince yourself that such a V exists-a function of the form (T8+1)(T2s+1)k will work. Finally,take Q to be Q:-VN-Y. Substit ution into (10.2)gives S=MY(1-V). Thus for w≤w3 IS(jw)l≤clMY‖o from proprty2 e from(10.3) and for w>w3 lS(jw)川≤(1+cM(jw)Y(jw)from property3 ≤(1+c)2from(10.5) <6from(10.4).■
P STABLE c Since P is strictly proper so is N This fact together with the equation NX M Y shows that M j Y j Since jM j Y j j is a continuous function of it is possible to choose such that jM j Y j j c The assumption on P implies that N is stable but not proper Choose a function V in S with the following three properties V N is proper j V j j c k V k c The idea behind the choice of V can be explained in terms of its Nyquist plot It should lie in the disk with center radius c up to frequency property and in the disk with center radius c thereafter property In addition V should roll o fast enough so that V N is proper It is left as an exercise to convince yourself that such a V exists a function of the form s s k will work Finally take Q to be Q V NY Substitution into gives S M Y V Thus for jS j j ckM Y k from proprty from and for jS j j c jM j Y j j from property c from from
158 CHAPTER 10.DESIGN FOR PERFORMANCE 10.2 P-1 Unstable We come now to the first time in this book that we need a nonclassical met hod,namely,interpolation theory.To simplify matters we will assume in this section that P has no poles or zeros on the imaginary axis,only distinct poles and zeros in the right half-plane,and at least one zero in the right half-plane (i.e.,P-is unstable). Wi is stable and strictly proper. It would be possible to relax these assumptions,but the development would be messier. To motivate the procedure to follow,let's see roughly how the design problem of finding an internally stabilizing C so that WiS0, S(p)=0 for p a pole of P in Res >0. The weighted sensitivity function G:=WiS must therefore satisfy G(z)=Wi(z)for z a zero of P in Res >0, G(p)=0 for p a pole of P in Res >0. So the requirement of internal stability imposes interpolation constraints on G.The performance spec WiSoo<1 translates into Go<1.Finally,the condition S E S requires that G be analytic in the right half-plane. One approach to the design problem might be to find a function G satisfying these conditions, then to get S,and finally to get C by back-substitution.This has a technical snag because the requirement that C be proper places an additional constraint on G not handled by our NP theory of the Chapter 9.For this reason we proceed via controller parametrizat ion. Bring in again a coprime factorization of P: P=N M NX +MY =1. The controller parametrization formula is C X+MQ Y-NQ Q∈S, and for such C the weighted sensitiv ity function is WIS=WM(Y-NQ). The parameter Q must be both stable and proper.Our approach is first to drop the properness requirement and find a suitable parameter,say,Qim,which is improper but stable,and then to get a suitable Q by rolling Qim off at high frequency.The reason this works is that Wi is strictly proper, so there is no performance requirement at high frequency.The method is out lined as follows:
CHAPTER DESIGN FOR PERFORMANCE P Unstable We come now to the rst time in this book that we need a nonclassical method namely interpolation theory To simplify matters we will assume in this section that P has no poles or zeros on the imaginary axis only distinct poles and zeros in the right half plane and at least one zero in the right half plane ie P is unstable W is stable and strictly proper It would be possible to relax these assumptions but the development would be messier To motivate the procedure to follow lets see roughly how the design problem of nding an internally stabilizing C so that kWSk can be translated into an NP problem The denition of S is S P C For C to be internally stabilizing it is necessary and sucient that S S and P C have no right half plane pole zero cancellations Theorem Thus S must interpolate the value at the right half plane zeros of P and the value at the right half plane poles see also Section that is S must satisfy the conditions S z for z a zero of P in Res S p for p a pole of P in Res The weighted sensitivity function G WS must therefore satisfy G z W z for z a zero of P in Res G p for p a pole of P in Res So the requirement of internal stability imposes interpolation constraints on G The performance spec kWSk translates into kGk Finally the condition S S requires that G be analytic in the right half plane One approach to the design problem might be to nd a function G satisfying these conditions then to get S and nally to get C by back substitution This has a technical snag because the requirement that C be proper places an additional constraint on G not handled by our NP theory of the Chapter For this reason we proceed via controller parametrization Bring in again a coprime factorization of P P N M NX M Y The controller parametrization formula is C X MQ Y NQ Q S and for such C the weighted sensitivity function is WS WM Y NQ The parameter Q must be both stable and proper Our approach is rst to drop the properness requirement and nd a suitable parameter say Qim which is improper but stable and then to get a suitable Q by rolling Qim o at high frequency The reason this works is that W is strictly proper so there is no performance requirement at high frequency The method is outlined as follows
1-c3%DESIGN EXAMPLE<FLEXIBLE BEAM 159 Procedure Input:P,Wi Step 1 Do a coprime factorization of P:Find four functions in S satisfying the equations NX +MY =1. Step.Find a stable function Qim such that WiM(Y NQim)oo 1 1. Step 3 Set 1 J(6)=09+可 where k is just large enough that QimJ is proper and 0 is just small enough that IWiM(Yr NQimJ)川o11. Step fi Set Q=QimJ2 Step 5 Set C=(X+MQ)/(Y NQ)2 That Step 3 is feasible follows from the equation WiM(Y NQimJ)=WiM(Y NQim)J+WiMY(1 J). The 4rst term on the righthand side has -norm less than 1 from Step 2 and the fact that 1,whilethe. -norm of the second term goes to 0 as 0 goes to 0 by Lemma 12 Step 2 is the model-matching problem,4nd a stable function Qim to minimize ‖TrT,Qiml‖o, where Ti:=WIMY and T:=WiMN2 Step 2 is feasible iff Yopt,the minimim model-matching error,is 1 12 10.3 Design Example:Flexible Beam This section presents an example to illustrate the procedure of the preceding section2 The example is based on a real experimental setup at the University of Toronto2 The control system depicted in Figure 1021,has the following components:a flexible beam,a high-torque dc motor at one end of the beam a sonar position sensor at the other end,a digital computer as the controller with analog-to digital interface hardware,a power ampli4er to drive the mctor,and an antialiasing 4lter2 The cbjective is to control the position of the sensed end of the beam2 A plant model was obtained as follows2 The beam is pinned to the mctor shaft and is free at the sensed end2 First the beam itself was modeled as an ideal Euler-Bernoulli beam with no damping: this yielded a partial differential equation model,reflecting the fact that the physical model of the
DESIGN EXAMPLE FLEXIBLE BEAM Procedure Input P W Step Do a coprime factorization of P Find four functions in S satisfying the equations P N M NX M Y Step Find a stable function Qim such that kWM Y NQim k Step Set J s s k where k is just large enough that QimJ is proper and is just small enough that kWM Y NQimJ k Step Set Q QimJ Step Set C X MQ Y NQ That Step is feasible follows from the equation WM Y NQimJ WM Y NQim J WM Y J The rst term on the right hand side has norm less than from Step and the fact that kJ k while the norm of the second term goes to as goes to by Lemma Step is the model matching problem nd a stable function Qim to minimize kT TQimk where T WM Y and T WMN Step is feasible i opt the minimum model matching error is Design Example Flexible Beam This section presents an example to illustrate the procedure of the preceding section The example is based on a real experimental setup at the University of Toronto The control system depicted in Figure has the following components a exible beam a high torque dc motor at one end of the beam a sonar position sensor at the other end a digital computer as the controller with analog to digital interface hardware a power amplier to drive the motor and an antialiasing lter The ob jective is to control the position of the sensed end of the beam A plant model was obtained as follows The beam is pinned to the motor shaft and is free at the sensed end First the beam itself was modeled as an ideal Euler Bernoulli beam with no damping this yielded a partial dierential equation model reecting the fact that the physical model of the
160 CHAPTER Q2 DESIGN FOR PERFORMANCE PC D/A amp motor beam A/D filter Figure 10.1:Flexible beam setup. beam has an infinite number of modes.The model is therefore linear but infinite-dimensional.The corresponding transfer function from torque input at the motor end to tip deflection at the sensed end has the form 00 + i=0 Then damping was introduced,yielding the form 00 s2 +2Giwis +w? =0 The first term is co/s2 and corresponds to the rigid-body slewing motion about the pinned end. The second term, C- 82+2(w-8+w2 corresponds to the first flexible mode.And so on.The motion was found to be adequately modeled by the first four flexible modes.Then the damping rat ios and natural frequencies were determined experimentally.Finally,the amplifier,motor,and sensor were introduced into the model.The antialiasing filter was ignored for the purpose of design. For simplicity we shall take the plant transfer function to be -6.4750s2+4.0002s+175.7700 P(8)=35s+Q5682s2+109.50218+0.0929 The poles are 0,-0.0007,-0.0565±5.2700j. The first two poles correspond to the rigid-body motion;the one at s=-0.0007 has been perturbed away from the origin by the back EMF in the motor.The two complex poles correspond to the first flexible mode,the damping ratio being 0.0675.The zeros are -4.9081,5.5008. Because of the zero at s =5.5008 the plant is non-minimum phase,reflecting the fact that the actuator (the motor)and the sensor are not located at the same point on the beam.The procedure of the preceding section requires no poles on the imaginary axis,so the model is (harmlessly) perturbed to -6.4750s2+4.0002s+175.7700 P(8)=53+0.5682s9+109.502152+0.0929s+10
CHAPTER DESIGN FOR PERFORMANCE PC DA amp motor beam AD lter sensor Figure Flexible beam setup beam has an innite number of modes The model is therefore linear but innite dimensional The corresponding transfer function from torque input at the motor end to tip deection at the sensed end has the form X i ci s i Then damping was introduced yielding the form X i ci s iis i The rst term is cs and corresponds to the rigid body slewing motion about the pinned end The second term c s s corresponds to the rst exible mode And so on The motion was found to be adequately modeled by the rst four exible modes Then the damping ratios and natural frequencies were determined experimentally Finally the amplier motor and sensor were introduced into the model The antialiasing lter was ignored for the purpose of design For simplicity we shall take the plant transfer function to be P s s s s s s s The poles are j The rst two poles correspond to the rigid body motion the one at s has been perturbed away from the origin by the back EMF in the motor The two complex poles correspond to the rst exible mode the damping ratio being The zeros are Because of the zero at s the plant is non minimum phase reecting the fact that the actuator the motor and the sensor are not located at the same point on the beam The procedure of the preceding section requires no poles on the imaginary axis so the model is harmlessly perturbed to P s s s s s s s
13.3.DESIGN EXAMPLERFLEXIBLE BEAM C60 A commonway to specify desired closed-lop peformarceis by astep responsetest2 For this flexiblebeam thespe isthat astep refererceirput (r)should produceaplart output (y)satisfy ing settlirg time≈8s0 overshoot≤00%. We will accomplish this by shapirg T(s),the trarsfer furctionfrom r to y,so that it approximates astandad second-orde system:Theideal T(s)is 4n a(s):-8-+2:4ns+4n A settlirg timeof 8 srequires 4.6 :4n ≈8 and anovershoot of 0%requires =0.0 Thesolutiors are:=0.5902 and 4n=0.958.2 Lets rourd to =0.6 and 4n=C2 So theideal T(s)is 0 Tid(s)= s-+02s+0 Then the ideal sensitiv ity furtionis Sa(s)=0-a(6)=s6+02) s-+02s+0° Now take the weightirg furctionwi(s)to be sia(s),that is W(s)=s+02s+0 s(s+02), Theratiorale for this choice is arough agumert that goes as follows2 Corsider Step 2 of the procedurein the precedirg section from it thefurtion F:=WiM(Y-NQim) equals a constart times an al-pass furction2 The procedure then rolls off Qim to result in the weighted sersitivity furction WiS :=WiM(Y-NQimJ). So WiSF except at high frequerey,that is S≈FSid. Now F behaves approximately like a time delay except at high frequercy (this is a propety of all passfurctions)2So wearrive at therough approximation S≈(time delay）×Sid
DESIGN EXAMPLE FLEXIBLE BEAM A common way to specify desired closed loop performance is by a step response test For this exible beam the spec is that a step reference input r should produce a plant output y satisfying settling time s overshoot We will accomplish this by shaping T s the transfer function from r to y so that it approximates a standard second order system The ideal T s is Tid s n s ns n A settling time of s requires n and an overshoot of requires exp p The solutions are and n Lets round to and n So the ideal T s is Tid s s s Then the ideal sensitivity function is Sid s Tid s s s s s Now take the weighting function W s to be Sid s that is W s s s s s The rationale for this choice is a rough argument that goes as follows Consider Step of the procedure in the preceding section from it the function F WM Y NQim equals a constant times an all pass function The procedure then rolls o Qim to result in the weighted sensitivity function WS WM Y NQimJ So WS F except at high frequency that is S F Sid Now F behaves approximately like a time delay except at high frequency this is a property of all pass functions So we arrive at the rough approximation S time delay Sid
62 CHAPTER 10.DESIGN FOR PERFORMANCE Hence our design should produce actual step response 4 delayed ideal step response4 One furt her adjustment is required in the problem setup:W must be stable and strictly proper, so the function above is modi<ed to 83+-28+- Ws)- 8+000-）8+-2)0400-8+可1 The procedure can now be applied. Step<Since P∈S we take N-PM--X=0、Y-- Step.The model-matching problem is to minimize kW M.NQim)k:=kW PQim)k14 Since P has only one right half-plane zero,at s=545-08,we have from Section 9.- minkW×.PQimk=|W545-08川=-402-04 Thus the spec kW Ski o -is not achievable for this P and W.Let us scale W as w) 049 -W4 -402- Then W 545-08)=049 and the optimal Qim is W:09 Qim- W P that is Qim0)-s00008s+002-s+04-768s3+0s707s3+-89-0s+00026） s+64-08-33+6488978+980- Step 3 Set 8):= 28+4 Compute kW PQim)k for decreasing values of 2 until the norm is o 2 2-Norm 0.- 一-2 0.05 =0- 0.0≈ 0.988 Take2=040≈ Step 4 Q=Qim/ Step5C=Q=×.PQ)
CHAPTER DESIGN FOR PERFORMANCE Hence our design should produce actual step response delayed ideal step response One further adjustment is required in the problem setup W must be stable and strictly proper so the function above is modied to W s s s s s s The procedure can now be applied Step Since P S we take N P M X Y Step The model matching problem is to minimize kWM Y NQim k kW P Qim k Since P has only one right half plane zero at s we have from Section min kW P Qim k jW j Thus the spec kWSk is not achievable for this P and W Let us scale W as W W Then jW j and the optimal Qim is Qim W WP that is Qim s s s s s s s s s s Step Set J s s Compute kW P QimJ k for decreasing values of until the norm is Norm Take Step Q QimJ Step C Q P Q
