Solution: 7.9.9.1 R + C G c G p Σ - The system is actually critically damped twice. The following MATLAB dialogue draws the root locus shown in Figure 1. EDU>gcgp = zpk([-3.6],[0 -1],1) Zero/pole/gain: (s+3.6) ------- s(s+1) EDU>K = linspace(0,50,1000);; EDU>rlocus(gcgp,K) EDU>print -deps rl7991.eps EDU> From Figure 1 weseethat there are twopoints where the sytem has a double real pole. The rst is near s = ;0:6, and the second is near s = ;6:4. Figure 2 shows howwecalculate the gain to nd the break-out pointnear s = ;0:6 Then K = jsjjs +1j js +3:6j Table 1 summarizes the search. The composite gain to place twoclosed loop s -0.53 -0.535 -0.54 -0.545 -0.55 K 0.08113 0.081166 0.081176 0.081170 0.08115 Table 1: Searchforbreakout pointnearorigin poles at s = ;0:54 is K =0:081176: The following MATLAB dialogue veri es this search -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 -4 -3 -2 -1 0 1 2 3 4 Real Axis Imag Axis Figure 1: Accurate root locus XXO -1-3.6 s + 1 s s + 3 Re (s) Im(s) Figure 2: Calculating gain near s = ;0:1 s -6.5 -6.55 -6.6 -6.64 -6.65 -6.66 -6.67 K 12.3276 12.3229 12.32 12.31895 12.31885 12.31882 12.31886 Table 2: Searchforbreakout pointnearorigin EDU>gcgp = zpk([-3.6],[0 -1], 0.081176) Zero/pole/gain: 0.081176 (s+3.6) ---------------- s (s+1) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 0.081176 (s+3.6) --------------------- (s+0.5419) (s+0.5393) EDU>step(tc,15) EDU>print -deps sr7991a.eps EDU> The step response is shown if Figure 3 Asimilar searchnears = ;6:4issummarized in Table 2 Thus the second pointofcritical damping is for s = ;6:667 with the composite gain of K =12:31882The following MATLAB dialogue veri es this computation. EDU>gcgp = zpk([-3.6],[0 -1],12.31882) Zero/pole/gain: 12.3188 (s+3.6) --------------- s (s+1) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 12.3188 (s+3.6) ---------------------- Time (sec.) A mp li tu d e Step Response 0 5 10 15 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 3: Unit step response for double pole at s = ;0:54 (s^2 + 13.32s + 44.35) EDU>p = [1 13.32 44.35] p= 1.000000000000000e+00 1.332000000000000e+01 4.435000000000000e+01 EDU>roots(p) ans = -6.734833147735482e+00 -6.585166852264519e+00 EDU>step(tc,2) EDU> The step response is shown in Figure 4. Note that because of the zero we actually have some overshoot with tworealpoles. Time (sec.) A mp li tu d e Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Unit step response for double pole at s = ;6:66 Finally,from Figure 1 wesee that a tangenttotheroot locus has angle  =tan ;1 (1:75=1)  60  : Thus, the minimum damping ratio is around  =0:5,