4.2.ROBUST STABILITY 45 Proof (Assume that W2Tlloo 1.Construct the Nyquist plot of L,indenting D to the left around poles on the imaginary axis.Since the nominal feedback system is internally stable, we know this from the Nyquist criterion:The Nyquist plot of L does not pass through-1 and its number of counterclockwise encirclements equals the number of poles of P in Res >0 plus the number of poles of C in Res >0. Fix an allowable A.Construct the Nyquist plot of PC =(1+AW2)L.No additional inden tations are required since AW2 introduces no additional imaginary axis poles.We have to show that the Nyquist plot of (1+AW2)L does not pass through-1 and its number of counterclockwise encirclements equals the number of poles of(1+AW2)P in Re s >0 plus the number of poles of C in Re s >0;equivalently,the Nyquist plot of(1+AW2)L does not pass through-1 and encircles it exactly as many times as does the Nyquist plot of L.We must show,in other words,that the perturbation does not change the number of encirclements. The key equation is 1+(1+△W2)L=(1+)(1+△W2T) (4.1) Since ‖△W2TI‖o≤IW2Tlo<1, the point 1+AW2T always lies in some closed disk with center 1,radius 1,for all points s on D. Thus from (4.1),as s goes once around D,the net change in the angle of 1+(1+AW2)L equals the net change in the angle of 1+L.This gives the desired result. (→)Suppose t hat‖W2T‖o≥l.We will construct an allowable△t hat destabilizes the feedback system.Since T is strictly proper,at some frequency w, W2(0w)T(jw)1=1. (4.2) Suppose that w=0.Then W2(0)T(0)is a real number,either +1 or-1.If A =-W2(0)T(0),then △is allowable and 1+△W2(0)T(0)=0. From (4.1)the Nyquist plot of (1+AW2)L passes through the critical point,so the perturbed feedback sy stem is not internally stable. If w >0,constructing an admissible A takes a little more work;the details are omitted. The theorem can be used effectively to find the stability margin Bsup defined previously.The simple scaling technique {P=(1+△W2)P:‖△I‖l≤3}={P=(1+B-1△6W2)P:IB-1△‖≤1} ={P=(1+△13W2)P:‖△l≤1} toget her with the theorem shows that Fsup=sup{B:IBW2Tl‖lo<1}=1/川W2Tlo The condition W2T<1 also has a nice graphical interpretat ion.Note that lW2Tlo<1÷ W2(jw)L(jw) <1,w 1+L(w) 台IW2(w)L(w)川<|1+L(w),w. ROBUST STABILITY  Proof  Assume that kW￾T k￾   Construct the Nyquist plot of L indenting D to the left around poles on the imaginary axis Since the nominal feedback system is internally stable we know this from the Nyquist criterion The Nyquist plot of L does not pass through  and its number of counterclockwise encirclements equals the number of poles of P in Res   plus the number of poles of C in Res   Fix an allowable  Construct the Nyquist plot of P C    W￾L No additional inden tations are required since W￾ introduces no additional imaginary axis poles We have to show that the Nyquist plot of   W￾L does not pass through  and its number of counterclockwise encirclements equals the number of poles of   W￾P in Re s   plus the number of poles of C in Re s   equivalently the Nyquist plot of   W￾L does not pass through  and encircles it exactly as many times as does the Nyquist plot of L We must show in other words that the perturbation does not change the number of encirclements The key equation is     W￾L    L  W￾T   Since kW￾T k￾ kW￾T k￾   the point W￾T always lies in some closed disk with center  radius   for all points s on D Thus from  as s goes once around D the net change in the angle of     W￾L equals the net change in the angle of   L This gives the desired result   Suppose that kW￾T k￾   We will construct an allowable  that destabilizes the feedback system Since T is strictly proper at some frequency  jW￾jT jj     Suppose that    Then W￾T  is a real number either  or  If   W￾T  then  is allowable and W￾T    From  the Nyquist plot of   W￾L passes through the critical point so the perturbed feedback system is not internally stable If   constructing an admissible  takes a little more work the details are omitted ￾ The theorem can be used eectively to nd the stability margin sup dened previously The simple scaling technique fP    W￾P  kk￾ g  fP     W￾P  k k￾ g  fP     W￾P  kk￾ g together with the theorem shows that sup  supf  k W￾T k￾  g  kW￾T k￾ The condition kW￾T k￾   also has a nice graphical interpretation Note that kW￾T k￾   W￾jLj   Lj    jW￾jLjj  j  Ljj