Consider the system of Example 8.7.
(a) Design a phase-lag compensator such that with \(K=0.5\), the system is critically damped with roots having a time constant of approximately \(2.03 \mathrm{~s}\). This time constant is the same as that of the uncompensated system with \(K=0.244\).
(b) By computer, verify the characteristic-equation roots in part (a).

Example 8.7

Consider the system of Fig. 8-29, with T = 1 s and 1 s(s + 1) Gp(s) = 1. Ku = 2. K = 0.8. The root locus for the uncompensated system is illustrated in Fig. 8-28(a), with Z1 = -0.718, 22 = 0.368, and z3 = 1. To find za and K, use MATLAB to generate the root locus of the uncompensated system. For our design, choose za to be the point of critical damp- ing. The following MATLAB code generates the needed root-locus plot displayed in Fig. 30(a): 0.196 and Za = 0.648. R(s) + >> T = 1; Gp=tf ( [1], [1 1 0]), Gz=c2d (Gp, T) rlocus (Gz), axis ([0 1 -0.5 0.5]), grid This system is shown to be critically damped with two poles at z = 0.648 for K = 0.196. We will design a phase-lag compensator that results in critically damping with two poles at z 0.65 with K = 0.8. Hence the steady-state errors are reduced by a factor of approximately 4. From the design procedure: G(z) T = Filter D(z) FIGURE 8-29 System for Examples 8.7 and 8.8. 0.368 (z+0.718) (z 1) (z - 0.368) 1--Ts S K Plant Gp(s) C(s)