R + C G c G p Σ - Figure 1: Closed Loop Control System Solution 6.8.4.2 (a) The answer is (b), with unityfeedbackyou need an integrator in the forward loop to havezerosteady state error to a step. (b) Fig. 2 shows the root locus. The crossing pointisats = ;6+j6so the gain is K c = js +2jjs +10j 10 s=;6+j6 = j;4+j6jj4+j6j 10 = 5:2 (c) Totheright. Atthepoint s = ;a + ja, the transportation lag con- tributes ;0:5 a 57:3  degrees of phase. So as shown in Fig. 3 the crossing pointmust be to the right. so wehavetohave ; ; ;0:5X  57:3  = ;180  : 1 Re(s) j Im(s) ζ = 1 2 -10 -2 Break-out at s = -6 Figure 2: Instersection of Root Locus with Line of ConstantDamping Re(s) j Im(s) ζ = 1 2 -10 -2 j x -x α β α β ++ (0.5 )(x)(57.3) = 180 o Figure 3: Root Locus with Transportation Lag Present 2