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 N0 Cauchy2s Theorem Suppose that n is a bounded open set with connected complement and a non.self.intersecting closed contour in n0 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 WiS0. 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
84 CHAPTER 6.DESIGN CONSTRAINTS Dez ne the ratior :=m,M ofp endulum mass to cart mass,The zero and pole of P in Res 0 are g 2= D=2P1+ Note that for r 2xedo a larger value of I means a smaller value of po and this in turn means that the sy stem is easier to stabilize (the time constant,is slower The foregoing two inequalities on WiS and WfiTllo apply,Since the cart-p endulum is a stabilization tasko let us focus on WT≥wapP+之 (6,5> p-z The robust stabilization problembecomes harder the larger the value of the right-hand side of (6,5 The scaling factor in this inequality is p+2_P1+F+1 =1+-1 (6,6> p-2 This quantity is always greater than 1o and it approaches 1 only whenr approaches ooo that iso only when the pendulum mass is much larger than the cart mass There is a tradecffo howevero in that a large value of r means a large value of po the unstable pole for a typical Wfi(highpass this in turn means a relatively large value ofWfi(p in (6,5 So at least for small uncertaintyo the worst case scenario is a short p endulum with a small mass m relative to the cart mass M, In contrasto the u-toy transfer function has no zeroso so the constraint there is simply WToo≥IWpH. If robust stabilization were the only objective we could achieve equality by careful selection of the controller,Ncte that for this case there is no apparent,tradeoff in making m,M large,The difference between the two caseso measuringx and measuring yo again highlights the imp ortant fact that sensor location can have a signiz cant effect on the difficulty of controlling a system or on the ultimate achiev able p erformance, Some simple exp eriments can be done to illustrate the points made in this erample Obtain several sticks of various lengths and try to balance them in the palm of your hand,You will notice that it is easier to balance longer sticso because the dy namics are slower and p above is smaller, It is also easier to balance the sticks if you look at the top of the stick (measuringy>rather than at the bottom(measuringx In facto even for a stick that is easily balanced when looking at the tcpo it will be impossible to balance it while lock ing only at the bcttom There is also feedbac fromthe forces that your hand feelso but this is similar to measuring x, The interested reader may repeat the analysis for the down position of the pendulum At this p oint it is useful to include the follow ing lemma which will be used subsequently, Lemma3 For every point s≤=（≤+j6≤with(）0, wlSwt-[g5s66g+二Zts
CHAPTER DESIGN CONSTRAINTS Dene the ratio r mM of pendulum mass to cart mass The zero and pole of P in Res are z r g l p zp r Note that for r xed a larger value of l means a smaller value of p and this in turn means that the system is easier to stabilize the time constant is slower The foregoing two inequalities on kWSk and kWT k apply Since the cartpendulum is a stabilization task let us focus on kWT k Wp p z p z The robust stabilization problem becomes harder the larger the value of the righthand side of The scaling factor in this inequality is p z p z p r p r This quantity is always greater than and it approaches only when r approaches that is only when the pendulum mass is much larger than the cart mass There is a tradeo however in that a large value of r means a large value of p the unstable pole for a typical W highpass this in turn means a relatively large value of jWp j in So at least for small uncertainty the worstcase scenario is a short pendulum with a small mass m relative to the cart mass M In contrast the utoy transfer function has no zeros so the constraint there is simply kWT k jWp j If robust stabilization were the only ob jective we could achieve equality by careful selection of the controller Note that for this case there is no apparent tradeo in making mM large The dierence between the two cases measuring x and measuring y again highlights the important fact that sensor location can have a signicant eect on the diculty of controlling a system or on the ultimate achievable performance Some simple experiments can be done to illustrate the points made in this example Obtain several sticks of various lengths and try to balance them in the palm of your hand You will notice that it is easier to balance longer sticks because the dynamics are slower and p above is smaller It is also easier to balance the sticks if you look at the top of the stick measuring y rather than at the bottom measuring x In fact even for a stick that is easily balanced when looking at the top it will be impossible to balance it while looking only at the bottom There is also feedback from the forces that your hand feels but this is similar to measuring x The interested reader may repeat the analysis for the down position of the pendulum At this point it is useful to include the following lemma which will be used subsequently Lemma For every point s j with log jSmps j Z log jSj j d
6,.,ANALYTIC CONSTRAINTS 85 Proof Set F(s>:=InSmp(s Then F is analytic and of bounded magnitude in Res.0,(This follows from the prop erties Sup)S.the idea is that since Smp has no poles or zeros in the right halfplaneo InSmp is well-behaved there>Apply Lemma 1 to get 00 o听+@-o≥- Now take real parts of bcth sides: Ref(so-- 0 (6,7> But so ISmpl eReF that iso In Snp ReF- Thus from (6,7> nl5mmso斗=mSmp+hy> 00 or since S=Snpl on the imaginary ariso 血sos0ugT- Finallyo since logx =logelnzo the result follows up on muiltiply ing the last equation by loge, The Waterbed Effect Consider a trackingproblemwhere the reference signals have their energy sp ectra concentrated in aknown frequency range say w2],This is the idealized situationwhere Wi is a bandp ass2 Iter, Let M denote the marimmm magnitude of S on this frequency band M:=mar|S(w≥ ■】9■年。2 and let M2 denote the marimummmagnitude over all frequencieso that isoS,Then good track ing capability is characterized by the inequality M 1 1,On the other hando we cannot permit M2 to be too large:Remember (Section 4,2>that 1>M2 equals the distance from the critical p oint to the Nyquist plct of Lo so large M2 means small stability margin (a typical upper bound for M2 is 2 Notice that M must be at least 1 because this is the value of S at in nite frequency,So the question arises:Can we have M very small and M2 not too large?Or does it happ en that very small Mi necessarily means very large M2?The latter situation might be compared to a waterbed:AsS ispushed down on one frequency rangeo it pops up somew here else It turns out that non minimmp hasep lants exhibit the waterbed effect, Theorem.Suppose that P has a zero at z with Rez >0.Then there erist positive constants c and c2,depending only on w1,w2,and z,such that ci logMi +c2 logM2.log Sap(.0-
ANALYTIC CONSTRAINTS Proof Set F s ln Smps Then F is analytic and of bounded magnitude in Res This follows from the properties Smp S mp S the idea is that since Smp has no poles or zeros in the right halfplane ln Smp is wellbehaved there Apply Lemma to get F s Z F j d Now take real parts of both sides ReF s Z ReF j d But Smp eF eReF e jImF so jSmpj eReF that is ln jSmpj ReF Thus from ln jSmps j Z ln jSmpj j d or since jSj jSmpj on the imaginary axis ln jSmps j Z ln jSj j d Finally since log x log e ln x the result follows upon multiplying the last equation by log e The Waterbed Eect Consider a tracking problem where the reference signals have their energy spectra concentrated in a known frequency range say This is the idealized situation where W is a bandpass lter Let M denote the maximum magnitude of S on this frequency band M max jSj j and let M denote the maximum magnitude over all frequencies that is kSk Then good tracking capability is characterized by the inequality M On the other hand we cannot permit M to be too large Remember Section that M equals the distance from the critical point to the Nyquist plot of L so large M means small stability margin a typical upper bound for M is Notice that M must be at least because this is the value of S at innite frequency So the question arises Can we have M very small and M not too large Or does it happen that very small M necessarily means very large M The latter situation might be compared to a waterbed As jSj is pushed down on one frequency range it pops up somewhere else It turns out that nonminimumphase plants exhibit the waterbed eect Theorem Suppose that P has a zero at z with Rez Then there exist positive constants c and c depending only on and z such that c log M c log M log jSapz j
86 CHAPTER 6.DESIGN CONSTRAINTS Proof Since z is a zero of P,it follows from the preceding section that S(2)=1,and hence Smp(2)=Sap(z)1.Apply Lemma 3 with 80=2=00+jw0 to get 00 -00 Thus log Sap(z)0,p1.As observed in the preceding section,S must interpolate zero at the unstable poles of P,so S(p)=0.Thus the all-pass factor of S must contain the factor 8-p 8+p that is, Sap(s)= 8一卫G(s) s+p for some all-pass function G.Since G(1)0 nor on the imaginary axis
CHAPTER DESIGN CONSTRAINTS Proof Since z is a zero of P it follows from the preceding section that Sz and hence Smpz Sapz Apply Lemma with s z j to get log jSapz j Z log jSj j d Thus log jSapz j c log M c log M where c is dened to be the integral of over the set and c equals the same integral but over the complementary set It remains to observe that jSapz j by the maximum modulus theorem so log jSapz j Example As an illustration of the theorem consider the plant transfer function P s s s s p where p p As observed in the preceding section S must interpolate zero at the unstable poles of P so Sp Thus the allpass factor of S must contain the factor s p s p that is Saps s p s p Gs for some allpass function G Since jG j maximum modulus theorem there follows jSap j p p So the theorem gives c log M c log M log p p Note that the righthand side is very large if p is close to This example illustrates again a general fact The waterbed eect is amplied if the plant has a pole and a zero close together in the right halfplane We would expect such a plant to be very dicult to control It is emphasized that the waterbed eect applies to nonminimumphase plants only In fact the following can be proved Section If P has no zeros in Res nor on the imaginary axis
-.2.ANALYTIC CONSTRAINTS 87 in the frequency range [61,62],then for every e>0 and 6>1 there exists a controller C so that the feedback system is internally stable,Mi0,and then S()=8+1 s+1+k So Sloo =1 and,for every e>0 and 62,if k is large enough,then 1S(06)川0. Theorem 2 Assume that the relative degree of L is at least 2.Then glsG6)a6=xs9∑Rnl Proof In Lemma 3 take 6o=0 to get g,wl-gso2oo or equivalent ly, ssG6Na平o6-bzlSwpiao). Multiply by oo: kg 6 o 5ngdoo) 06+62 It can be shown that the left-hand side converges to og \s (6)ld6 as oo-oo.[The idea is that for very large oo the function 6+6 equals nearly 1 up to large values of 6.On the other hand,log S(j6)tends to zero as 6 tends to oo.So it remains to show that g|Smla)=-(loge)(∑Rwl. (6.8)
ANALYTIC CONSTRAINTS in the frequency range then for every and there exists a controller C so that the feedback system is internally stable M and M As a very easy example take P s s The controller Cs k is internally stabilizing for all k and then Ss s s k So kSk and for every and if k is large enough then jSj j The Area Formula Herein is derived a formula for the area bounded by the graph of jSj j log scale plotted as a function of linear scale The formula is valid when the relative degree of L is large enough Relative degree equals degree of denominator minus degree of numerator Let fpig denote the set of poles of L in Res Theorem Assume that the relative degree of L is at least Then Z log jSj jd log e X Repi Proof In Lemma take to get log jSmp j Z log jSj j d or equivalently Z log jSj j d log jSmp j Multiply by Z log jSj j d log jSmp j It can be shown that the lefthand side converges to Z log jSj jd as The idea is that for very large the function equals nearly up to large values of On the other hand log jSj j tends to zero as tends to So it remains to show that lim log jSmp j log e X Repi
88 CHAPTER 6.DESIGN CONSTRAINTS We can write S=SapSnp where SEp)= Y s-pi is+厉 It is claimed that 思（血sX)=0. To see this6.8)it remains to show that 思hX)y3-∑脚 6.10) Now InxSapx )t0o9 In so to prove >6.10)it suffices to prove 血-2 =Repi. 6.11) Let pi=r+jy and use 6.9)again as followsoo 纵+西多 专1n气-p3 +(u9 2 方血2-p3 2 n1+x(00+为(0y I 4 1-x(w9+为(W90 年1+2(“y+(1-11-(u”+(9) 2+ 0 2 =4( 十… =x+. Repi+.... Letting（→3 gives6.11).■
CHAPTER DESIGN CONSTRAINTS We can write S SapSmp where Saps Y i s pi s pi It is claimed that lim ln S To see this note that since L has relative degree at least we can write L c k as for some constant c and some integer k Thus as ln S ln L ln c k Now use the Maclaurins series ln x x x to get ln S c k The righthand side converges to zero as tends to This proves the claim In view of the claim to prove it remains to show that lim ln Sap X Repi Now lnSap lnY i pi pi X i ln pi pi so to prove it suces to prove lim ln pi pi Repi Let pi x jy and use again as follows ln pi pi ln pi pi ln x y x y ln x y ln x y n x x o x Repi Letting gives