58 CHAPTER 5 STABILIZATION &tt品a.风知.Q, C Q:=1+P元5 Then,s aa (cCrse,supp to,s aade Q C:= 1-PQ5 (5.1) AceGail tOte den liticintimn2.te reedka systm i te rin rIster fuld 1 1+PC tftan Jainti 「1-PQ-P(1-PQ)-(1-PQ)1 Q 1-PQ -Q 5 PQ P(1-PQ) 1-PQ clay,tese rinertis batos. Nct ttalritater untaao a aingnaotrpaaetrt ecdhek The fen T5+T.Q fs Ghe T5 T.I.IIatcua the se IWI alac p emeIay se IWI fuldIsae S=1-PQ- T-PQ5 toaiu!apiaiotonac cttda ystm Iastt as arupkey taks apr (when=0).Pachecam the tecem.Thgn aympka tas atp It the belter fultGhr (le.,s)ha a3rCts=0,ti丛, t6a益Linzwr从aadii P(0)Q(0)=15 c=90Q,sQ0= P而5 oRrve tcte iatp atac.ALOy G caned ttacrcer tetr rGn naa c ts =0,aItmusthThe crem 3 (Cho.ter 3. Example Fte p P(S)=8+1)s+2 CHAPTER STABILIZATION Proof Suppose that C achieves internal stability Let Q denote the transfer function from r to u that is Q C P C Then Q S and C Q P Q Conversely suppose that Q S and dene C Q P Q According to the denition in Section the feedback system is internally stable i the nine transfer functions P C P C C P C P all are stable and proper After substitution from and clearing of fractions this matrix becomes P Q P P Q P Q Q P Q Q P Q P P Q P Q Clearly these nine entries belong to S Note that all nine transfer functions above are ane functions of the free parameter Q that is each is of the form T TQ for some T T in S In particular the sensitivity and complementary sensitivity functions are S P Q T P Q Let us look at a simple application Suppose that we want to nd a C so that the feedback system is internally stable and y asymptotically tracks a step r when d Parametrize C as in the theorem Then y asymptotically tracks a step i the transfer function from r to e ie S has a zero at s that is P Q This equation has a solution Q in S i P Conclusion The problem has a solution i P when this holds the set of all solutions is C Q P Q Q S Q P Observe that Q inverts P at dc Also you can check that a controller of the latter form has a pole at s as it must by Theorem of Chapter Example For the plant P s s s