By Cornelius T Leondes

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**Additional resources for Control and dynamic systems : advances in theory and applications ; 10**

**Example text**

Extrapolate (or truncate) the terminal portion of the second subarc to Tk as follows. Yk(Tk ) = Uk (Tk-1)+ Fkail)(Tk-1)(Tk - Tk-1). ~ 154) 7. Compute a measure of the difference between Yk and One such measure is as follows. Uk-l' p(k,k - 1) = max 1Uk(t) - Yk-1(t) tE [t0, T] (155) where p is an N-dimensional vector. If p(k,k- 1) < ~, where 7 is an N-dimensional vector of small positive numbers, the solution has converged. If p(k,k - 1) > T, let Yk be the new nominal solution and return to step 2.

GRAHAM AND C. T. LEONDES equation (107) can be solved in closed form for any of the components of u, it is desirable to do so; however, equations (105) and (107) represent the more general case. The system variables, Y, are discontinuous at the corner times. The magnitude of the discontinuity is defined as: 11m Y(t + e) e-0 x where a (108) t- = 11m [Y(t - e) + x[g( t a - e), a e]} a e- 0 is an N-dimensional vector function of Y and t which possesses first derivatives with respect to both Y and t.

T. LEONDES Figure 1 shows the minimum time solutions for various values of the parameter b. The following values are used for the remaining parameters. 3. 5 (67) Minimum Time Problem with Free Corners This example is similar to the preceding one except that the corner time is free and the magnitude of the discontinuity is define in a different manner. The equations of motion, i00 I I e• io 1 ~ 0. 5 1 1 . 0 20 60 40 Q0 loi U FIG. 1. , they are given by equations (47) through (52). The magnitude of the discontinuity is defined as follows: 0 2 z- cl)+ c2(y - c 3) (68) • Since the corner time is free, it can be selected to minimize the final time.