Question 10.1: Consider the system defined in Figure 10.1, with input x(t) ......

This problem is a constrained optimization problem, since the unknown parameter C must be found subject to both a constraint (1 ≤ C ≤ 2) and a minimization (or maximization) principle.

(a) It follows from Figure 10.1 that\ y(t)= \int{C[x(t) —y(t)]  dt} . Reformulating this as an equivalent initial-value problem and noting that x(t) = 1,

\ \dot{y}+Cy=C,\\y(0)=0.                         (10.2)

Equation (10.2) is a first-order linear differential equation, and its solution is straightforward:\ y(t) = 1 – e^{-Ct} . We wish to force y(t) close to unity. Using the MSE criteria of Equation (10.1) with\ z_{des}(t) = 1  and  z_{mod}(t, C) = 1 -e^{-Ct},

\ f(C)=\int_{0}^{1}{[1-y(t)]^2  dt}\\=\int_{0}^{1}{e^{-2Ct}  dt}\\= \frac{1-e^{-2C}}{2C}.                  (10.3)

The graph of f(C) is shown in Figure 10.2. Clearly, this is a monotonically decreasing function over the interval [1, 2], so the minimum point lies at the right-hand side of the constraint interval [1, 2] at C = 2. The corresponding objective functional value is\ f(2) = \frac{1}{4}(1 – e^{-4}) \approx 0.245.

(b) In this case, the output y(t) is unchanged, but the objective function is different. Since we must now attempt to track\ z_{des}(t) = \frac{1}{2},

\ f(C)=\int_{0}^{1}{[\frac{1}{2}-y(t)]^2  dt}\\=\int_{0}^{1}{(\frac{1}{2}-e^{-Ct})^2  dt}\\= \frac{1}{2C}(1-e^{-2C})+\frac{1}{C}(1-e^{-C})+\frac{1}{4}.                  (10.4)

In principle, a local minimum can be found by differentiating Equation (10.4), setting it equal to zero, and solving for C. However, even though f(C) can be differentiated, an exact solution for\ C_{min} cannot be found. Even so, a quick check of the graph in Figure 10.3 reveals a minimum on [1, 2] at about\ C_{min} = 1.125 . We need a systematic procedure to locate such minima.