Chapter 6 Design Constraints Before we see how to design control systems for the robust performance specification,it is useful to determine the basic limitations on achievable performance.In this chapter we study design constraints arising from two sources:from algebraic relationships that must hold among various transfer functions;from the fact that closed-loop transfer functions must be stable(ie.,analytic in the right half-plane).It is assumed throughout this chapter that the feedback system is internally st able. 6.1 Algebraic Constraints There are three items in this section. 1.The identity S+T=1 always holds.This is an immediate consequence of the definitions of S and T.So in particular,S(jw)and T(jw)cannot both be less than 1/2 at the same frequency w. 2.A necessary condit ion for robust performance is that the weighting functions satisfy min{Wi (jw)W2(jw)}<1,Vw. Proof Fix w and assume that W<W2 (the argument jw is suppressed).Then IW=IW(S+T)川 ≤IWSI+WT到 ≤IWSI+IW2T So robust performance (see Theorem 4.2),that is, WiS+W2Tl&<1, implies that W<1,and hence min{Wil,W2}<1. The same conclusion can be drawn when W2<Wil. So at every frequency either Wil or W2 must be lessthan 1.Typically,Wi(jw)is monoton- ically decreasing-for good track ing of low-frequency signals-and W2(jw)is monotonically increasing-uncertainty increases with increasing frequency. 79
Chapter Design Constraints Before we see how to design control systems for the robust performance specication it is useful to determine the basic limitations on achievable performance In this chapter we study design constraints arising from two sources from algebraic relationships that must hold among various transfer functions from the fact that closedloop transfer functions must be stable ie analytic in the right halfplane It is assumed throughout this chapter that the feedback system is internally stable Algebraic Constraints There are three items in this section The identity S T always holds This is an immediate consequence of the denitions of S and T So in particular jSj j and jT j j cannot both be less than at the same frequency A necessary condition for robust performance is that the weighting functions satisfy minfjWj j jW j jg Proof Fix and assume that jWjjWj the argument j is suppressed Then jWj jWS T j jWSj jWT j jWSj jWT j So robust performance see Theorem that is kjWSj jWT jk implies that jWj and hence minfjWj jWjg The same conclusion can be drawn when jWjjWj So at every frequency either jWj or jWj must be less than Typically jWj j is monoton ically decreasingfor good tracking of lowfrequency signalsand jWj j is monotonically increasinguncertainty increases with increasing frequency
80 CHAPTER五，DESIGN CONSTRAINTS 3.Ifp is apier L in Rezo ad is ae of L in tesanerar-paeth S(p)=0,S(z）=1, (6.1) T(p)=1,T(z）=0. (6.2) Treeintelpiction constants aeimmedicterm tedcGitions of s ad T.Foreanpe s0=1+阿=是=0 6.2 Analytic Constraints In this sefion wEdeliveomecons ticints concehing areblerelfo Iaceob taned f5m ady tic func tion treor.Tredstsubseetion peats somemcurenetica piminaie. Preliminaries web in with erollowing fundan conching comp functions eaimum mod- ulus treIGh,(aich's trelh,ad (ach's intelaforula Treeaestated reroroonve niAce Maximum Modulus Theorem Suppose that n is a region unonempty/open/connected set,in the complex plane and F is a function that is analytic in N<Suppose that F is not equal to a constant<Then F does not attain its marimum value at an interior point cf A simpealficaton of th Ih with uat te hka paesbws ttror Fin S Il Fllo=suplF(s)l. Res>0 Cauchy2s Theorem Suppose that n is a bounded open set with connected complement and a non.self.intersecting closed contour in n<lf F is analytic in then F(s)ds=0. dnhgmsgpiayp的cmanfmgaaomar F(o)= 1 F(s)ds. 2πjDs-So wera aso ncea eroisson inlatoInulawhhas tatteaueof abounded aay tic funcfion atarintin trerg rat-paeis competey deternined by trecoolainate of trepoint ogetrerwilntrevaue of trefunction on treimeinay &is. Lemma 1 Let F be analytic and of bounded magnitude in Res>0 and let So=00+j.o be a point in the complex plane with oo >0<Then s)-P6. 00 o+(.-·024
CHAPTER DESIGN CONSTRAINTS If p is a pole of L in Res and z is a zero of L in the same halfplane then Sp Sz T p T z These interpolation constraints are immediate from the denitions of S and T For example Sp Lp Analytic Constraints In this section we derive some constraints concerning achievable performance obtained from analytic function theory The rst subsection presents some mathematical preliminaries Preliminaries We begin with the following fundamental facts concerning complex functions the maximum mod ulus theorem Cauchys theorem and Cauchys integral formula These are stated here for conve nience Maximum Modulus Theorem Suppose that is a region nonempty open connected set in the complex plane and F is a function that is analytic in Suppose that F is not equal to a constant Then jF j does not attain its maximum value at an interior point of A simple application of this theorem with equal to the open right halfplane shows that for F in S kF k sup Res jF s j Cauchys Theorem Suppose that is a bounded open set with connected complement and D is a nonselfintersecting closed contour in If F is analytic in then ID F s ds Cauchys Integral Formula Suppose that F is analytic on a nonselfintersecting closed contour D and in its interior Let s be a point in Then F s j ID F s s s ds We shall also need the Poisson integral formula which says that the value of a bounded analytic function at a point in the right halfplane is completely determined by the coordinates of the point together with the values of the function on the imaginary axis Lemma Let F be analytic and of bounded magnitude in Res and let s j beapoint in the complex plane with Then F s Z F j d
6.2.ANALYTIC CONSTRAINTS 81 Proof Construct the Nyquist contour D in the complex plane taking the radius,r,large enough so that the point so is encircled by D. Cauchy's integral formula gives F(s0)= 1 F(s)ds. 2TjJDs-80 Also,since-3o is not encircled by D,Cauchy's theorem gives 1 0= F(s)ds. 2πjyDs+s0 Subtract these two equations to get F(so)= 人P8g-8+ 1 30+80 ds Thus F(so)=I1+12, where = 00 F(us,-s+河4 00 F(0w) 哈+@-w0r, I2= 1 CT/2 00 AsT→0 F(u)). So it remains to show that I2→0asT→o. We have h≤9r2p-wE+咖 The integral 1 -2e0-s0r1k0+50r可0 converges as r→o.Thus 2≤constant×F, which gives the desired result
ANALYTIC CONSTRAINTS Proof Construct the Nyquist contour D in the complex plane taking the radius r large enough so that the point s is encircled by D Cauchys integral formula gives F s j ID F s s s ds Also since s is not encircled by D Cauchys theorem gives j ID F s s s ds Subtract these two equations to get F s j ID F s s s s s s s ds Thus F s I I where I Z rr F j s j s j d Z rr F j d I j Z F re j rej s rej s rje jd As r I Z F j d So it remains to show that I as r We have I kF k r Z jej sr jjej sr j d The integral Z jej sr jjej sr j d converges as r Thus I constant r which gives the desired result
82 CHAPTER 6.DESIGN CONSTRAINTS Bounds on the Weights Wi and W2 Suppose that the loop transfer function L has a zero z in Res,0.Then IW1S‖o,W1)i 6.3) This is a direct consequence of the maximum modulus theorem and 6.1)oo JW1)j=JW1)S)j≤sup j18)S8)j=WSIo· Res>0 So a necessary condition that the performance criterion WiS<1 be achievable is that the weight satisfy Wix)j<1.In words<the magnitude of the weight at a right half-plane zero of P or C must be less than 1. Similarly<suppose that L has a pole p in Res,0.Then IW2Tllo,JW2)方 6.4) so a necessary condition for the robust stability criterion W2To<1 is that the weight W2 satisfy W2)j<1. All2Pass and Minimum2Phase Transfer Functions Two types of transfer functions play a critical role in the rest of this boo oall-pass and minimum- phase.A function in S is all-pass if its magnitude equals 1 at all points on the imaginary axis. The terminology comes from the fact that a filter with an all-pass transfer funct ion passes without attenuation input sinusoids of all frequencies.It is not difficult to show that such a function has pole-zero symmetry about the imaginary axis in the sense that a point so is a zero iff its reflection< So<is a pole.Consequently<the function being stable all its zeros lie in the right half-plane.Thus an all-pass function is<up to sign<the product of factors of the form 8·80 Reso >0. 8+80 Examples of all-pass functions are 8.1 s2.8+2 1 8+1 s2+8+2 A function in S is minimum-phase if it has no zeros in Res >0.This terminology can be explained as follows.Let G be a minimum-phase transfer function.There are many ot her transfer functions having the same magnit ude as G<for example FG where F is all-pass.But all these other transfer funct ions have greater phase.Thus<of all the transfer functions hav ing Gs magnit ude<the one with the minimum phase is G.Examples of minimum-phase functions are 1 1, 8+2 8+18+1's2+8+1 It is a useful fact that every function in S can be written as the product of two such factorsoo for example 48.2) 8.2 4冷+2) 82+8+1 +2 2+8+1
CHAPTER DESIGN CONSTRAINTS Bounds on the Weights W and W Suppose that the loop transfer function L has a zero z in Res Then kWSk jWz j This is a direct consequence of the maximum modulus theorem and jWz j jWz Sz j sup Res jWs Ss j kWSk So a necessary condition that the performance criterion kWSk be achievable is that the weight satisfy jWz j In words the magnitude of the weight at a right halfplane zero of P or C must be less than Similarly suppose that L has a pole p in Res Then kWT k jWp j so a necessary condition for the robust stability criterion kWT k is that the weight W satisfy jWp j AllPass and MinimumPhase Transfer Functions Two types of transfer functions play a critical role in the rest of this book allpass and minimum phase A function in S is al lpass if its magnitude equals at all points on the imaginary axis The terminology comes from the fact that a lter with an allpass transfer function passes without attenuation input sinusoids of all frequencies It is not dicult to show that such a function has polezero symmetry about the imaginary axis in the sense that a point s is a zero i its reection s is a pole Consequently the function being stable all its zeros lie in the right halfplane Thus an allpass function is up to sign the product of factors of the form s s s s Res Examples of allpass functions are s s s s s s A function in S is minimumphase if it has no zeros in Res This terminology can be explained as follows Let G be a minimumphase transfer function There are many other transfer functions having the same magnitude as G for example F G where F is allpass But all these other transfer functions have greater phase Thus of all the transfer functions having Gs magnitude the one with the minimum phase is G Examples of minimumphase functions are s s s s s s It is a useful fact that every function in S can be written as the product of two such factors for example s s s s s s s s
6,.,ANALYTIC CONSTRAINTS 83 Lemma.For each function G in S there erist an all-pass function Gap and a minimum-phase function Gmp such that G-GapGmp.The factors are unique up to sign. Proof Let Gap be the product of all factors of the form S.So S+S0' where so ranges over all zeros of G in Res >0,counting mult iplicities,and then de-ne Gmp一 Gap The proof of uniqueness is left as an exercise, For technical reasons we assume for the remainder of this section that L has no poles on the imaginary axis,Factor the sensitivity function as S=SapSmp Then Smp has no zeros on the imaginary axis(such zeros would be poles of L>and Smp is not strictly proper((since s is not子Thus Smp∈S, As a simple example of the use of all-pass functions,suppose that P has a zero at z with z>0,a pole at p withp >0.also,suppose that C has neither poles nor zeros in the closed right half-plane, Then Sop( S.2 s+p p(s>- S+z It follows from the preceding section that S(>1,and hence Smn(2>=5p(el-2+2 2.p Similarly, Tnpp>-(pl=p+之 D.z Then IlWi Slloo-IWiSmplloo (2Smp(zj- W and llWTllao wo Thus,if there are a pole and zero close to each other in the right half-plane,they can greatly amplify the effect that eit her would have alone, Example These inequalities are effectively illustrated with the cart-pendulum example of Sec- tion 5,7,Let P(s>be the uto-r transfer function for the up position of the pendulum,that is, 1s2.g P(s>-s2MIs2.(M+mg
ANALYTIC CONSTRAINTS Lemma For each function G in S there exist an al lpass function Gap and a minimumphase function Gmp such that G GapGmp The factors are unique up to sign Proof Let Gap be the product of all factors of the form s s s s where s ranges over all zeros of G in Res counting multiplicities and then dene Gmp G Gap The proof of uniqueness is left as an exercise For technical reasons we assume for the remainder of this section that L has no poles on the imaginary axis Factor the sensitivity function as S SapSmp Then Smp has no zeros on the imaginary axis such zeros would be poles of L and Smp is not strictly proper since S is not Thus S mp S As a simple example of the use of allpass functions suppose that P has a zero at z with z a pole at p with p also suppose that C has neither poles nor zeros in the closed right halfplane Then Saps s p s p Taps s z s z It follows from the preceding section that Sz and hence Smpz Sapz z p z p Similarly Tmpp Tapp p z p z Then kWSk kWSmpk jWz Smpz j Wz z p z p and kWT k Wp p z p z Thus if there are a pole and zero close to each other in the right halfplane they can greatly amplify the eect that either would have alone Example These inequalities are eectively illustrated with the cartpendulum example of Sec tion Let P s be the utox transfer function for the up position of the pendulum that is P s ls g s M ls M m g