Question

Let us look again at the time-optimal positioning problem (Section 1.4 of the textbook): with initial state ri (0) = X, X < 0, and x2(0) = 0, to be steered to the final state x1 (t) = 0, 2 (t1)0, and with cost given by J (t, r1, 2, u) t. Assume now that we want to use an nsymmetric bang bang control; thai i, ea only asse valu u=-a or u=b where a, b > 0. Show that the target can be reached with a single switch. Find t2, the time of the switch, and ti, the time taken to reach the target.

Answer 1

Since X<0; and at time t=t1, x1 comes at 0; we can say that u is positive before switching time t2 and then it goes negative after t2.

So, we have u=b for 0<t<t2 and u=-a for t2<t<t1.

We write expression for x1 in both regions and apply boundary conditions in both viz.

x1(0)=X, x1(t1)=0; x2(0)=x2(t1)=0

Then we equate x1 and x2 from both regions at t=t2 and hence find the expression for t1 and t2 in terms of a, b and X:

ん te Co, t.) 2 2. b (a t b) ao a+ b 12 a.

So,

                    

2aX 70.5 ; t1 ba ba+b)