Solution: 7.9.4.4 R + C G c G p Σ - Figure 1: Unityfeedback For the system 0f Figure 1 wehave G p (s)= 1 (s +0:1)(s+ 1)(s +20) (a) Figure 2 shows the poles of the plantand the zero of the PD compensator and the desired closed loop poles. XXX O Re(s) Im(s) -20 -a -1 -0.1 -4 3 Figure 2: Angle calculation The zero is at ;a = ; K d K pd : 1 The gain is K pd .Thus, for the root locus to pass through s = ;4+j3we must satisfy the angle condition at the point, that is, wemust have ; 1 ; 2 ; 3 = ;180  ;; or =  1 +  2 +  3 ;180  = [180  ;tan ;1 (3=3:9)]+ [180  ;tan ;1 (3=3)]+ tan ;1 (3=16);180  = 142:43  +135  +10:62  ;180  = 108:05  : Thus, the zero lies to the rightofs = ;4, so by simple trigonometry we have a = 4; 3 tan(180  ;108:05  ) = 4; 3 tan(71:95  ) = 3:022: Then K pd = p 3 2 +3:9 2  p 3 2 +3 2  p 3 2 +16 2 p 3 2 +0:978 2 = 4:924:2416:28 3:155 = 107:7: Finally, K d = 3:022K pd = 3:022108 = 325:5: b The closed loop transfer function is of the form T c (s)= 108(s+3:022) (s +4;j3)(s+4+j3)(s+ ) 2 The simplest wayto nd is just to searchalong the real axis to the right of s = ;20 until we ndagain of 108. Try s = ;16. Then K = (4)(15:9)(15) 12:978 = 73:51: Too small try s = ;14 K = (6)(13:9)(13) 10:978 = 98:76 Close, try s = ;13 K = (7)(12:9)(12) 9:978 = 108:6 Just a little too large, try s = ;13:1 K = (6:9)(13)(12:1) 10:078 = 107:7 So wesettle for third pole at s = ;13:1and T c (s)= 108(s+3:022) (s +4;j3)(s+4+j3)(s+13:1) The following MATLAB dialogue locates the third pole exactly and then nds the unit step response shown in Figure EDU>g= zpk([-3.022],[-0.1 -1 -20],108) Zero/pole/gain: 108 (s+3.022) -------------------- (s+0.1) (s+1) (s+20) EDU>h = 1 h= 3 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 1 EDU>tc = feedback(g,h) Zero/pole/gain: 108 (s+3.022) -------------------------------- (s+13.07) (s^2 + 8.033s + 25.13) EDU>step(tc,4) EDU>step(tc,3) EDU>print -deps pd4step.eps EDU> 4