-6...Justio cation for H2 Loop Shaping 3)3 Substituting these formulas into /)6)0)<we have 2)shP犯》Pe百P )-1n<-)3Ps) )-V<-P) 2 Theresults in Theorem )6u)0and)6))confirm that y/alternatively e)indicates the com patibility between the specified loop shape and clos ed.loop stability requirem entsw Theorem-6.-2 Let P he the nominal lant and let K e W,K2 W<he the associated controller obtained from loo sha ing desygn rocedure in the last section,Then if we have (K直kPK)1）≤☒i,下w,w) 1月6) (IPK)1）≤mm☒M)Ww,)K区N)-w} )6)x) (KIPK)1,P）≤mm)w,),)s/M)W,)} 1)6)3) MRPK):P)≤ y3/A) </w,)</W 月64) (IKP)1)≤mm)k区)w,M)-w,)} 月6)6) (GIKP)1,K)≤mm)k区/M)-W,区N)-W} /)6)6) where <N)EN.)e 3WPw.)' 、)sWPW,)/ 月67) M)e 3/M.)e )x sWw,) 月6)-) and N,M)Eres ectivelye M)Eis a normalized left co rime factorizationeres ec tively∈ight oo rime factorization∈of P.e WPW, / Proofw Note that M立se1PP)1, and MM E I-NN-
Justication for H Loop Shaping Substituting these formulas into we have K Ps Ps Ps p Ps p Ps The results in Theorem and con rm that alternatively indicates the compatibility between the speci ed loop shape and closedloop stability requirements Theorem Let P be the nominal plant and let K WKW be the associated control ler obtained from loop shaping design procedure in the last section Then if I K I PsKM s we have KI P K M s W W I P K min n M sW NsW o KI P KP min n N sW MsW o I P KP N s W W I KP min n N sW MsW o GI KP K min n M sW NsW o where N s Ns WPW WPW M s Ms WPW and N s M s respectively Ns Ms is a normalized left coprime factorization respec tively right coprime factorization of Ps WPW Proof Note that M s M s I PsP s and M sM s I N sN s
314 H LOOP SHAPING Then )=:m2,)=1+:n中P阿 1 1 1+P) 8,)=1(≤，) ☒P) 1+P) The proof for the normalized right coprime factorization i.imilaroAll other inequalitie. follow from noting A 王+PsK)Mgi≤y and I 王+PK✉)- K +PK W-'K W交PW, WP ◇ Thi.theorem shows that all closed-loop objective.are guaranteed to have bounded magnitude and the bound.depend only on y2W.,W 2 and P oo 16.4 Further Guidelines for Loop Shaping Let P =N M-,be a normalized right coprime factorizationodt was shown in Georgiou and Smith-1990)that s慧) Hence a small P)would nece.arily imply a small b P)o We hall now di.cu..the performance limitation.implied by thi.relation,hip or calar ystemoetz,-,m and p,,p---,pe be the open right half plane zero.and pole.of the plant PoDefine 名，+s2<+sm+N)=B(s8w(s N:s)=名，（s2<(s2m(s √D,+sp<+sP+s Then P can be written a. P-s)=Po-s)N:-)(N) where Po-s)has no open right half plane pole.or zero.odet No-s)and Mo-s)be table and minimum phase pectral factor. N-Ng)=(（+可）厂，MM)=1+P-P-》
H LOOP SHAPING Then M s maxM s M s minPsP s Ps N s M s Ps Ps The proof for the normalized right coprime factorization is similar All other inequalities follow from noting I K I PsKM s and I K I PsKM s W W K I P K h W PW i W WP I KP h W PW i This theorem shows that all closedloop ob jectives are guaranteed to have bounded magnitude and the bounds depend only on W W and P Further Guidelines for Loop Shaping Let P NM be a normalized right coprime factorization It was shown in Georgiou and Smith that boptP P inf Res Ms Ns Hence a small P would necessarily imply a small boptP We shall now discuss the performance limitations implied by this relationship for a scalar system Let z zzm and p ppk be the open right half plane zeros and poles of the plant P De ne Nz s z s z s z s z s zm s zm s Nps p s p s p s p s pk s pk s Then P can be written as P s PsNz sNps where Ps has no open right half plane poles or zeros Let Ns and Ms be stable and minimum phase spectral factors NsN s P sP s MsM s P sP s
8)y Further Guidelines for Loop Shaping 315 Then Po=NoMois a normalized coprime factorization and NoNz>and MoNp>form a pair of normalized coprime factorization of P.Thus bgtP≥×√N3Xz3+MsXp31 ∀Res≥>0 -16.192 Now InNoos>>and In-Moos>>are both analytic in Re-s>>0 and so can be determined from their boundary values on Re-s>=0 via Poisson integrals Freudenberg and Looze< 1911page37]: Z口 In Noore斗= lnN.斗K.-(r≥d-(r2 ,701+P-(2-(2 三 2 Za In Mocrel.生= lnM.K.-(r≥d-(r2 70+P-(2d(2 ,□ 2 where r 0<- (201 2<and 00S0> K.(r≥=- 1-2-(r2n0+-(r The function K.(r>is shown in Figure 16.6 for various values of 0.Note that the function is symmetric to 0 frequency when 00<i.e.<when Noos or Moosis evaluated on the real axis.The function converges to an impulse function at.=r when 0 approaches to 90i.e.<when Noos or Moosis evaluated close to the imaginary axls. De ne Zb at=。K.2d a Then it is easy to see that I(,60>=1.The/intogral I ab 0>for different values of a,b and 0 are shown in Figure 16.7 and Figure 16.1.Figure 16.7 shows that 50%of weighting is in the frequency range-r<.<ri =1210%or more of weighting is in the frequency range-3r <3r fi =32and 95%or more of weighting is in the frequency range-l0r≤.≤10rf=l0≥Thusit is dear that N-rel:丬will be small if P.is small in the range of frequency 0<.<r -note thatP.=P-j. Similarly<M-rel will be small if P.islarge in the range of frequency 0<.<r. It is also important to note from Figure 16.1 that a very large percentage of weighting is concentrated in a very narrow frequency range<05r <1-5r<for a large 0<i.e.< when s=rel.has a much larger imaginary part than the real part. If we now return to the bound -16.19>and note that Nzzi>=0 and Np->=0 we see that there are several ways in which the bound may be small<i.e.<guaralteeing that boptP≥is poor
Further Guidelines for Loop Shaping Then P NM is a normalized coprime factorization and NNz and MNp form a pair of normalized coprime factorization of P Thus boptP q jNsNz sj jMsNpsj Res Now lnNs and lnMs are both analytic in Res and so can be determined from their boundary values on Res via Poisson integrals Freudenberg and Looze page ln jNrej j Z ln jNjj Kr dr Z ln jP jj Kr dr ln jMrej j Z ln jMjj Kr dr Z ln jP jj Kr dr where r and Kr cos r sin r The function Kr is shown in Figure for various values of Note that the function is symmetric to frequency when ie when jNsj or jMsj is evaluated on the real axis The function converges to an impulse function at r when approaches to ie when jNsj or jMsj is evaluated close to the imaginary axis De ne I ab Z b a K d Then it is easy to see that I The integral I ab for di erent values of a b and are shown in Figure and Figure Figure shows that ! of weighting is in the frequency range r r ! or more of weighting is in the frequency range r r and ! or more of weighting is in the frequency range r r Thus it is clear that jNrej j will be small if jP jj is small in the range of frequency r note that jP jj jP jj Similarly jMrej j will be small if jP jj is large in the range of frequency r It is also important to note from Figure that a very large percentage of weighting is concentrated in a very narrow frequency range r r for a large ie when s rej has a much larger imaginary part than the real part If we now return to the bound and note that Nz zi and Nppj we see that there are several ways in which the bound may be small ie guaranteeing that boptP is poor
316 H2 LOOP SHAPING 180■-80 0■80 160=-70 9■70 1.4 1.20=-60 ! 6=-45 0.6 8=-30 9■30 0.4 ■0 -3 用市 -1 /r0 1 2 Figure 16.6:K3--r6vs Normalized Frequency.-r <N:s6j and 6j both small for some s:This corresponds to P-s6 having a right half planevero near a right half plane pole.See Exanple 16.1 below. <N:-s6j and jMoos6j both small for some s:This corresponds to either jp-j.6j being large in an entire frequency range of [01r]or iP-j.6;being large aroumnd =r where r is the modulus of a right half plane zero of P.See Example 16. below. <Ns6j and Noos6j both small for some s:This corresponds to either ip-j.6j beivg small in an entire frequency range of [01r]or iP-j.6j being large around =r where r is the modulus of a right half plane pole of P.See Example 16.3 below. <iNos6;and iMos6;both small for some s:The only way in which these can both be small is if jP-j.6 is both small and large at frequencies near.isi~ i.e.~jP-j.6j is approximately equal to 1 and InjP-j.6j has a large slope.See Exanple 16.4 below. Eram le Consider a ustable and non2minimumphase system P36=-K3(r6 s+16s(16 The frequency responses of P_s6 with r =0.9 and K=0.111 and 10 are shown in Figure 16.9.The following table shows that bot P6 will be very small for all K whenever r is close to 1~ie.~whenever there is a unstable pole close to a unstable zero
H LOOP SHAPING −3 −2 −1 0 1 2 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 θ = 80 o θ = 70 o θ = 80 o θ = 70 o θ = 60 o θ = 80 o θ = 70 o θ = 60 o θ = 45 o θ = 80 o θ = 70 o θ = 60 o θ = 45 o θ = 30 o θ = 80 o θ = 70 o θ = 60 o θ = 45 o θ = 30 o θ = 0 o θ = 80 o θ = 70 o θ = 60 o θ = 45 o θ = 30 o θ = 0 o θ = −80 o θ = 80 o θ = 70 o θ = 60 o θ = 45 o θ = 30 o θ = 0 o θ = −80 o θ = −70 o θ = 80 o θ = 70 o θ = 60 o θ = 45 o θ = 30 o θ = 0 o θ = −80 o θ = −70 o θ = −60 o θ = 80 o θ = 70 o θ = 60 o θ = 45 o θ = 30 o θ = 0 o θ = −80 o θ = −70 o θ = −60 o θ = −45 o θ = 80 o θ = 70 o θ = 60 o θ = 45 o θ = 30 o θ = 0 o θ = −80 o θ = −70 o θ = −60 o θ = −45 o θ = −30 o ω/ r Kθ ( ω/ r) Figure Kr vs Normalized Frequency r jNz sj and jNpsj both small for some s This corresponds to P s having a right half plane zero near a right half plane pole See Example below jNz sj and jMsj both small for some s This corresponds to either jP jj being large in an entire frequency range of r or jP jj being large around r where r is the modulus of a right half plane zero of P See Example below jNpsj and jNsj both small for some s This corresponds to either jP jj being small in an entire frequency range of r or jP jj being large around r where r is the modulus of a right half plane pole of P See Example below jNsj and jMsj both small for some s The only way in which these can both be small is if jP jj is both small and large at frequencies near jsj ie jP jj is approximately equal to and ln jP jj has a large slope See Example below Example Consider a unstable and nonminimum phase system Ps Ks r s s The frequency responses of Ps with r and K and are shown in Figure The following table shows that boptP will be very small for all K whenever r is close to ie whenever there is a unstable pole close to a unstable zero
16.4.Further Guidelines for Loop Shaping 317 年■0 09 a■3 0.8 a■2 0.7 aa15 a■1 a=0.8 0.2 -80 -60-40 f沿egree 204060 80 Figure 16.7:I a.a() 0.5 0.7 J0.9 1.1 1.3 1.5 K=0.1 bopt(P) 0.01250.0075 0.00250.00250.00740.0124 0.5 0.7 0.9 1.1 1.3 1.5 K=1 bopt(P) 0.10360.05790.01790.01650.0457 0.0706 0.5 0.7 0.9 1.1 1.3 1.5 K=10 bopt(h)） 0.06510.03120.0011 0.00770.02010.0311 Example 16.2 Consider a non-minimum phase plant n(同= K(s-1) s(s+1) The frequency responses of P2(s)with K=0.1,1 and 10 are shown in Figure 16.10. The following table shows clearly that bopt(P2)will be small if P2(jw)is large around w=12the modulus of the right half plane zero. K 0.01 0.11 10 100 bpt(P)0.70010.64510.31270.01410.0091 Note that bopt (L/s)=0.707 for any L and bopt(P2)-0.707 as K-0.This is because P(jw)around the frequency of the right half plane zero is very small. Next consider a plant with a pair of complex right half plane zeros: 乃3(s)= K[(s-cose)2+sin20] s[(s+cose)2+sin20
Further Guidelines for Loop Shaping −80 −60 −40 −20 0 20 40 60 80 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 α = 10 α = 3 α = 2 α = 10 α = 3 α = 2 α = 1.5 α = 10 α = 3 α = 2 α = 1.5 α = 1 α = 10 α = 3 α = 2 α = 1.5 α = 1 α =0.8 α = 10 α = 3 α = 2 α = 1.5 α = 1 α =0.8 α =0.5 θ (degree) I[−α, α] (θ) Figure I r K boptP r K boptP r K boptP Example Consider a nonminimum phase plant Ps Ks ss The frequency responses of Ps with K and are shown in Figure The following table shows clearly that boptP will be small if jPjj is large around the modulus of the right half plane zero K boptP Note that boptLs for any L and boptP as K This is because jPjj around the frequency of the right half plane zero is very small Next consider a plant with a pair of complex right half plane zeros Ps Ks cos sin ss cos sin