Solution 7.9.9.9 We rst close up the inner feedbacklooptoobtain G p (s) = 1 s 2 1+ 120s +100 s 2 (s 2 +12s +72) = (s 2 +12s +72) s 4 +12s 3 +72s 2 +120s +100 = (s +6;j6)(s+6+j6) (s +1;j)(s+1+j)(s+5;j5)(s+5+j5) Figure 1 shows a sketchoftherootlocus for the compensator G c (s)= K c (s +1) 2 s(s + p) For dominantpoles we pick s = ;4  j7. Weexpect the complex poles at s = ;5j5tomigrate to the zeros at s = ;6j6, and in the process become the dominantpoles. The loop transfer function poles at s = ;1j will migrate to the left and downward, ultimately breaking into the real axis. Wehave picked a low damping ratio because wewillhavetwopoles near s = ;1 with large time constants. To nd the gain and the location of the second pole of the compensator wewrite a short Matlab program. The program then yields K c =159:4 and p =24:71: The root locus is shown in Figure 2 The step response is shown in Figure 3. 1 XX X X X Re(s) Im(s) O O O X -p 2 Dominant Poles Figure 1: Preliminary Root Locus 2 -25 -20 -15 -10 -5 0 -8 -6 -4 -2 0 2 4 6 8 O O X X X X X X O 2 Figure 2: Accurate Root Locus 0 0.2 0.4 0.6 0.8 1 1.2 Step Response 01234 Time in Seconds Figure 3: Step Response 3