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LINEAR TIME-OPTIMAL CONTROL-EXAMPLES
المؤلف:
Lawrence C. Evans
المصدر:
An Introduction to Mathematical Optimal Control Theory
الجزء والصفحة:
35-40
8-10-2016
920
EXAMPLE 1: ROCKET RAILROAD CAR. We recall this example, introduced in §1.2. We have
For
According to the Pontryagin Maximum Principle, there exists h≠0 such that
We will extract the interesting fact that an optimal control α∗ switches at most one time.
We must compute etM. To do so, we observe
and therefore Mk = 0 for all k ≥ 2. Consequently,
The Maximum Principle asserts
and this implies that
α∗ (t) = sgn(−th1 + h2)
for the sign function
Therefore the optimal control α∗ switches at most once; and if h1 = 0, then α∗ is constant.
Since the optimal control switches at most once, then the control we constructed by a geometric method in §1.3 must have been optimal.
EXAMPLE 2: CONTROL OF A VIBRATING SPRING. Consider next the simple dynamics
where we interpret the control as an exterior force acting on an oscillating weight (of unit mass) hanging from a spring. Our goal is to design an optimal exterior forcing α∗(.) that brings the motion to a stop in minimum time.
We have n = 2, m = 1. The individual dynamical equations read:
which in vector notation become
for |α(t)| ≤ 1. That is, A = [−1, 1].
Using the maximum principle. We employ the Pontryagin Maximum Principle, which asserts that there exists h ≠ 0 such that
To extract useful information from (M) we must compute X(.). To do so, we observe that the matrix M is skew symmetric, and thus
According to condition (M), for each time t we have
Therefore
α∗(t) = sgn(−h1 sin t + h2 cos t).
Finding the optimal control. To simplify further, we may assume h21+h22 =1. Recall the trig identity sin(x + y) = sin x cos y + cos x sin y, and choose δ such that
−h1 = cos δ, h2 = sin δ. Then
α∗ (t) = sgn(cos δ sin t + sin δ cos t) = sgn(sin(t + δ)).
We deduce therefore that α∗ switches from +1 to −1, and vice versa, every π units of time.
Geometric interpretation. Next, we figure out the geometric consequences.
When α ≡ 1, our (ODE) becomes
In this case, we can calculate that
Consequently, the motion satisfies (x1(t) − 1)2 + (x2)2(t) ≡ r21, for some radius r1, and therefore the trajectory lies on a circle with center (1, 0), as illustrated.
If α ≡ −1, then (ODE) instead becomes
in which case
Thus (x1(t)+1)2 +(x2)2(t) = r22 for some radius r2, and the motion lies on a circle with center (−1, 0).
In summary, to get to the origin we must switch our control α(.) back and forth between the values ±1, causing the trajectory to switch between lying on circles centered at (±1, 0). The switches occur each π units of time.
References
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