Solution: 7.9.9.2 R + C G c G p Σ - Referring to Figure 1 Wehave the angle condition  3 = +180  ; 1 ;  2 = 108:43  +180  ; 135  ;123:69  = 29:74  : Then p 1 =3+ 3 tan27:74  =8:25 The gain is K c = jsjjs +1jjs +8;;25j 10js +2j j s=;3+j3 = 2:95 Then K v = 10K c 2 1 8:25 = 7:09;; XX XO Re(s) Im(s) -1-p -2 θ α 1 θ 2θ 3 ?3 3 Figure 1: Accurate root locus 1 and e ss (ramp)= 1 K v =0:141: If werepeat this process for a zero at s = ;3:3, weget  3 = +180  ; 1 ; 2 = 84:29  +180  ;135  ; 123:69  = 5:6  : Then p 1 =3+ 3 tan5:6  =33:6 The gain is K c = jsjjs +1jjs +33:6j 10js +3:3j j s=;3+j3 = 15:6 Then K v = 10 K c 3:3 1 33:6 = 15:32;; and e ss (ramp)= 1 K v =0:065: For the compensator determined in Example 7.3.1 wehave K v = 107:8 3 118 =13;; and e ss =0:0769 Thus, as wemovethezero to the left wegetasmaller steady error, but the price wepayishigher gain. 2