Solution 7.9.9.8 This is an example of a problem where a rst order compensator will work just about as well as a second order. We use the the PI-compensator G c (s)= K c (s+2:5) s : Figure 1 shows how the gain calculation to nd the break-out pointTable XXX s + 3 s + 3.2 -3 -3.2 Im(s) Re(s) s Figure 1: Gain Computation to Find Break-out Point shows that the break-out pointisats = ;1:03. s -1.05 -1.04 -1.03 -1.02 -1.01 K c 4.4021 4.4029 4.4031 4.4027 4.4017 Table 1: Locating Break-out Point Thus the compensator is G c (s)= 0:4403(s+2:5) s : Figure 2 shows the unit step response. Given that a speed controller is initiated at the speed that is to be maintained, the factthat the step response takes six seconds to reachnearsteady state should not be a problem. 1 0 0.2 0.4 0.6 0.8 1 1.2 Step Response 02468 Time in Seconds Figure 2: Gain Computation to Find Break-out Point 2