Question 10.11: Repeat Example 10.10 for the case where tracking is desired ......
Repeat Example 10.10 for the case where tracking is desired for\ y_{ss} = 6 and v_{ss} = 3 . Again assume that input to be x(t) = 10 and the time horizon to be [0, 20].
Using the criteria for steady-state linear systems,\ C_1 = x_{ss}/y_{ss} = \frac{10}{6} \approx 1.67 and C_2 = v_{ss}/y_{ss} = \frac{3}{6} = 0.5 . Unfortunately, the sequence does not appear to converge to these values. The reason for this can be seen from observing the response shown in Figure 10.23, where the error is evident. In this case, the system response has a significant amount of overshoot and there is considerable error accumulated before the steady-state phase actually begins. This is true even though the time horizon in this example is [0, 2] instead of [0, 10] as it was in Example 10.10.
There are two ways in which this problem can be remedied. First, as was done in this example, we can extend the time horizon. An even better approach is to trim the objective function limits to the range [10, 20] so that the error is not measured so long over the transient phase. Application of the cyclic coordinates technique over the horizon [10, 20] shows how effective this is. The cyclic coordinates sequence shown in Figure 10.24 indicates that convergence seems swift and true.
