Solution: 7.9.3.12 G pG c + - R C For the system shown above, let G p (s)= 1 (s + 3)(s +10) and G c (s)= K c (s + a) s + b Toachieve zero steady state error to a step input we need to increase the system type to one, so the pole of the compensator must be at s =0. Fig. 1 shows the angles involved in making the root locus pass through s = ;4j3. Then for the root locus to pass through s = ;4j3, wemust have ; 1 ; 2 ; 3 = ;180  ;; or =  1 +  2 +  3 ;180  = [180  ;tan ;1 (3=4)]+ [180  ;tan ;1 (3=1)]+ tan ;1 (3=6);180  = 143:1+108:4+26:6;180  = 98:13  Then a =3+ 3 tan(98:13  ) =2:571 Finally K c = p 3 2 +4 2  p 3 2 +6 2  p 3 2 +1 2 p 3 2 +0:4286 2 = 31:92 Then the compensator is G c = 31:92(s+2:571) s The following MATLAB dialogue nds the step response of the closed loop system. 1 -10 θ 1 α Re(s) Im(s) -a 3 -3 θ 2 θ 3 -4 Figure 1: Vector Diagram to Find Pole of Lead Compensator EDU>g = zpk([-2.571],[0 -3 -10],31.92) Zero/pole/gain: 31.92 (s+2.571) --------------- s(s+3) (s+10) EDU>h = 1 h= 1 EDU>tc = feedback(g,h) Zero/pole/gain: 31.92 (s+2.571) -------------------------------- (s+2.105) (s^2 + 10.89s + 38.99) EDU>step(tc,3) EDU>print -deps sr79312.eps EDU> 2 Time (sec.) Amplitude Step Response 0 0.5 1 1.5 2 2.5 3 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Step response of compensated system The step response is shown in Figure 2. 3